Algebra II (Math 201)

Robert Boltje Department of Mathematics

University of California Santa Cruz, CA 95064

Email: boltje@math.ucsc.edu

Winter 1999

I. Modules over Principal Ideal Domains

1 Modules, Module Homomorphisms, and Basic Concepts

1.1 Remark Let M be an Abelian group (written additively).

(a) The set End(M) := Hom(M, M) of group homomorphisms is a ring under

(f + g)(m) := f(m) + g(m) , (f · g)(m) := f(g(m)) ,

for f, g ∈ End(M) and m ∈ M. Indeed, it is very easy to verify that f +g, f ·g ∈ End(M) and that the ring axioms are satisfied. The 0-element and 1-element are the constant function m ↦ 0 and the identity map m ↦ m, respectively. End(M) is called the endomorphism ring of M. Any subring of End(M) is called a ring of endomorphisms of M.

(b) If M is generated by elements m_{i}, i ∈ I, and if f, g ∈ End(M) with f(m_{i}) = g(m_{i}) for all i ∈ I, then f = g.

(c) Recall that f(a · m) = a · f(m) for all a ∈ ℤ, m ∈ M, and f ∈ End(M).

1.2 Example Let M = ℤ. Since ℤ is generated by 1, the map

α: End(ℤ) → ℤ , f ↦ f(1) ,

is injective. It is also surjective, since for given a ∈ ℤ there is the endomorphism ℤ → ℤ, z ↦ nz. Moreover α is a ring homomorphism: For f, g ∈ End(ℤ) we have

α(f + g) = (f + g)(1) = f(1) + g(1) = α(f) + α(g)

and

α(f · g) = (f ∘ g)(1) = f(g(1)) = f(g(1) · 1) =
g(1) · f(1) = f(1) · g(1) = α(f) · α(g) ,

and finally α(id_{ℤ}) = id_{ℤ}(1) = 1. There fore α is a ring isomorphism.

1.3 Remark Each ring R is isomorphic to a ring of endomorphisms on R itself. For a ∈ R let a_{l} denote the left multiplication by a, so

a_{l}: R → R: r ↦ ar .

Then a_{l} is a group homomorphism, since

a_{l}(r + s) = a(r + s) = ar + as = a_{l}(r) + a_{l}(s) .

Therefore, the map

α: R → End(R) , a ↦ a_{l} ,

is well-defined. We will show that α is an injective ring homomorphism so that R is isomorphic to the ring α(R) of endomorphisms of R. Let a, b ∈ R. If a_{l} = b_{l}, then a = a_{l}(1) = b_{l}(1) = b. Moreover,

(α(a + b))(r) = (a + b)_{l}(r) = (a + b)r = ar + br =
a_{l}(r) + b_{l}(r) = (a_{l} + b_{l})(r) = (α(a) + α(b))(r)

which implies that α is additive, and

(α(ab))(r) = (ab)_{l}(r) = (ab)r = a(br) = a_{l}(b_{l}(r)) = (α(a) ∘ α(b))(r)

which implies that α is multiplicative. Finally, α(1) = 1_{l} = id_{R}.

1.4 Remark Let M be an Abelian group. Then the evaluation map

φ: End(M) × M → M , (f, m) ↦ f(m) =: f · m ,

satisfies

f · (m + n) = f · m + f · n  (f is a homomorphism)
(f + g) · m = f · m + g · m  (definition of f + g)
(f · g)(m) = f · (g · m))  (definition of f · g)
1 · m = m  (1 = id_{M})

for all f, g ∈ End(M) and m, n ∈ M.

More generally, if R is any ring and ρ: R → End(M) is a ring homomorphism, then the map

φ: R × M , (r, m) = (ρ(r))(m) =: r · m ,

satisfies

(i) r · (m + n) = r · m + r · n  (ρ is a homomorphism)
(ii) (r + s) · m = r · m + s · m  (ρ is additive)
(iii) (r · s) · m = r · (s · m)  (ρ is multiplicative)
(iv)  1_{R} · m = m  (ρ(1_{R}) = id_{M})

1.5 Definition Let R be a ring. A left R-module M is an Abelian group M together with a map

φ: R × M → M , (r, m) ↦ φ(r, m) =: r · m ,

satisfying the equations (i)–(iv) in Remark 1.4. Similarly, we define a right R-module as an Abelian group M together with a map

ψ: M × R → M , (m, r) ↦ ψ(m, r) =: m · r ,

satisfying

(i’)  (m + n) · r = m · r + n · r
(ii’)  m · (r + s) = m · r + m · s
(iii’)  m · (r · s) = (m · r) · s
(iv’)  m · 1 = m

for all m ∈ M and r, s ∈ R. The map φ (resp. ψ) is called a left (resp. right) R-module structure on M. We usually write just r · m or rm (resp. m · r or mr) instead of φ(r, m) (resp. ψ(m, r)).

1.6 Remark Let M be an Abelian group and let R be a ring.

(a) By Remark 1.4, each ring homomorphism ρ: R → End(M) gives rise to a left R-module structure φ: R × M → M on M by setting φ(r, m) := (ρ(r))(m). Conversely, if φ: R×M → M is a left R-module structure on M we can associate a ring homomorphism ρ: R → End(M) by setting (ρ(r))(m) := φ(r, m), i.e., ρ(r) is just multiplication by r. These two constructions are inverse bijections between the set of left R-module structures φ: R × M → M and the set of ring homomorphisms ρ: R → End(M).

(b) If φ: R × M → M is a left R-module structure on M, then

φ^{∘}: M × R^{∘} → M , (m, r) ↦ φ(r, m) ,

is a right R^{∘}-module structure on M. (Recall that R^{∘} is the ring with the same elements and addition as in R, but with reversed multiplication: r · s (in R ) = s · r (in R^{∘}).) If R is commutative, we do not need to distinguish between left and right R-module structures.

(c) Similarly as in (a), the right R-module structures ψ: M × R → M correspond bijectively to the ring antihomomorphisms σ: R → End(M). Recall that an antihomomorphism switches the two factors of a product.

(d) If M is a left R-module, then 0 · m = 0 = r · 0 for all m ∈ M and all r ∈ R, since 0 · m = (0 + 0) · m = 0 · m + 0 · m and r · 0 = r · (0 + 0) = r · 0 + r · 0. Moreover, it is easy to show that (−r) · m = −(r · m) = r · (−m) for all r ∈ R and m ∈ M.

(e) If R is a field, then an R-module is the same as an R-vector space.

(f) When we say ‘R-module’ in the sequel, we will always mean ‘left Rmodule’.

1.7 Examples (a) Let R be a ring. Then the ring multiplication R × R → R, (r, s) ↦ rs, defines a left and also a right R-module structure on R. We denote these two modules by _{R}R and R_{R}. More generally, if S ⊆ R is a subring (this always means with the same 1-element), then R is a left and right S-module just by the multiplication defined inside R.

(b) Each Abelian group is a left module over its endomorphism ring.

(c) Each Abelian group is automatically a ℤ-module in a unique way. Therefore, ℤ-modules and Abelian groups are the same mathematical notions.

(d) Let F be a field and let V be an F -vector space. For each F -linear map f: V → V and each polynomial

p(λ) = a_{n}λ^{n} + ⋯ + a_{1}λ + a_{0} ∈ F [λ]

we set

p(f) := a_{n}f^{n} + ⋯ + a_{1}f + a_{0} · id_{V} .

Obviously, p(f) is again an F -linear map from V to V and we obtain for fixed f a ring homomorphism

F [λ] → End_{F} (V) ⊆ End(V) , p ↦ p(f) .

Therefore, V becomes an F [λ]-module (whose structure depends on f) by p·v = (p(f))(v), for p ∈ F [λ] and v ∈ V.

1.8 Definition Let M and N be R-modules. A map f: M → N is called an R-module homomorphism (or R-linear) if it is a group homomorphism and respects the R-module structure, i.e., f(rm) = rf(m) for all r ∈ R and m ∈ M. We denote the set of R-module homomorphisms f: M → N by Hom_{R}(M, N). Note that Hom_{R}(M, N) is a subgroup of Hom(M, N) and that the composition of module homomorphisms is again a module homomorphism. Similarly, we define homomorphisms for right R-modules. As for groups we define the notions of monomorphisms, epimorphisms, isomorphisms, endomorphisms, and automorphisms of modules in the obvious way. We call R-modules M and N isomorphic (and write M ≅ N), if there exists an isomorphism between M and N. Note that if f: M → N is a bijective R-linear map, then also f^{−1} is R-linear. Note also that if R = ℤ, then an R-linear map is just a group homomorphism.

1.9 Definition Let R be a ring and let M be an R-module. An R-submodule of M is a subgroup U of M such that ru ∈ U for all r ∈ R and u ∈ U.

1.10 Examples Let R be a ring and let M be an R-module.

(a) The subgroups {0} and M of M are always R-submodules. If these are the only submodules and if M ≠ {0}, then we call M an irreducible (or simple) R-module.

(b) Let N be an R-module and let f ∈ Hom_{R}(M, N). Then the image im(f) = f(M) of f is an R-submodule of N. More generally, the image f(U) of any R-submodule U of M is again an R-submodule of N. Similarly, the kernel ker(f) is an R-submodule of M. More generally, the preimage f^{−1}(V) of any R-submodule V of N is again an R-submodule of M.

(c) If U is an R-submodule of M, then the factor group M/U carries again a natural R-module structure which is defined by

r · (m + U) := r · m + U

for r ∈ R, m ∈ M. This is well-defined, since for m′ ∈ M with m − m′ ∈ U we have r · m − r · m′ = r · (m − m′) ∈ U and therefore r · m + U = r · m′ + U. The map

ν: M → M/U , m ↦ m + U ,

is R-linear and is called the natural epimorphism.

(d) Let X ⊆ M be a subset of M. Then, the set

⟨X⟩_{R} := {r_{1}x_{1} + ⋯ + r_{n}x_{n} | n ∈ ℕ_{0}, r_{1}, …, r_{n} ∈ R, x_{1}, …, x_{n} ∈ X}

is an R-submodule of M and is called the R-submodule of M generated by X. It is easy to see that this is the smallest R-submodule of M containing X, i.e.,

⟨X⟩_{R} = ⋂_{X⊆U⊆M U is R-submodule of M} U .

We call M a finitely generated R-module, if M is generated by a finite set. We call M cyclic, if M = ⟨m⟩_{R} = Rm for some m ∈ M.

(e) If U_{i}, i ∈ I, are R-submodules of M, then also their intersection ⋂_{i∈I} U_{i} is an R-submodule of M. Moreover, we define the sum ∑_{i∈I} U_{i} of U_{i}, i ∈ I, as the R-submodule generated by ⋃_{i∈I} U_{i}. This is the smallest R-submodule of M containing every U_{i}, i ∈ I.

(f) If M_{i}, i ∈ M, are R-modules, then their direct product ∏_{i∈I} M_{i} is again an R-module with the componentwise group structure and the R-module structure given by r · (m_{i})_{i∈I} := (r · m_{i})_{i∈I}. Moreover, the external direct sum ⨁_{i∈I} M_{i} is defined as the R-submodule of ∏_{i∈I} M_{i} consisting of those families (m_{i})_{i∈I} which have only finitely many non-zero entries. If I is finite, then ∏_{i∈I} M_{i} = ⨁_{i∈I} M_{i}. We abbreviate the direct sum of n copies of _{R}R by R^{(n)}.

(g) The R-submodules of _{R}R are just the left ideals and those of R_{R} are the right ideals of R.

(h) For any subset X of M, the annihilator

ann_{R}(X) := {r ∈ R | rx = 0 for all x ∈ X}

of X is a left ideal of R. If U is an R-submodule of M, then ann_{R}(U) is a (two-sided) ideal of R.

1.11 Theorem (Fundamental theorem of homomorphisms for modules) Let M and N be R-modules and let f ∈ Hom_{R}(M, N). For each R-submodule U of M with U ⊆ ker(f), there exists a unique f̄ ∈ Hom_{R}(M/U, N) such that the diagram

[figure: M; f; N; ν; f̄; M/U]

commutes, where ν: M → M/U is the natural epimorphism. Moreover, f̄ has the same image as f and ker(f̄) = ker(f)/U. The map

Hom_{R}(M/U, N) → {f ∈ Hom_{R}(M, N) | f(U) = 0} , g ↦ g ∘ ν ,

is a group isomorphism.

Proof Everything follows from the fundamental theorem of group homomorphisms; except that one has to verify that the group homomorphism f̄ is Rlinear. But, for r ∈ R and m ∈ M, we have

f̄(r(m + U)) = f̄(rm + U) = (f̄∘ ν)(rm) =
f(rm) = rf(m) = rf̄(ν(m)) = rf̄(m + U) ,

and the proof is complete.

1.12 Remark If R is a ring, M is an R-module, and U is an R-submodule, then the natural epimorphism ν: M → M/U induces a bijection V ↦ V/U between the set of R-submodules of M containing U and the set of all R-submodules of M/U. Moreover, the isomorphism theorems hold similar as for groups, i.e., there are natural R-linear isomorphisms

(M/U)/(V/U) ≅ M/V and (W + U)/U ≅ W/(W ∩ U)

for all R-submodules U ⊆ V ⊆ M and W ⊆ M.

1.13 Definition Let M be an R-module. A subset X of M is called an R-basis of M, if ⟨X⟩_{R} = M and if X is an R-linearly independent set, i.e., any R-linear combination of finitely many pairwise distinct elements in X which equals 0 has 0-coefficients. We call M a free R-module, if M has a basis. We call M a finitely generated free R-module, if M has an R-basis of finite length. Note that this is the same as saying that M is free and finitely generated. Free ℤ-modules are also called free Abelian groups.

1.14 Remark Let R be a ring and let M be an R-module.

(a) A subset X ⊆ M is an R-basis of M if and only if each element of M can be written in a unique way as an R-linear combination of pairwise distinct elements of X.

(b) For each n ∈ ℕ, the R-module R^{(n)} is finitely generated free with R-basis e_{i} = (0, …, 0, 1, 0, …, 0), the 1 being at the i-th position, i = 1, …, n.

1.15 Theorem (Universal property of an R-basis) Let M be a free R-module with R-basis X. Let N be another R-module and let g: X → N be a function. Then there exists a unique homomorphism f ∈ Hom_{R}(M, N) such that f|_{X} = g.

Proof Existence: We write each m ∈ M as R-linear combination m = ∑_{x∈X} r_{x}x with unique coefficients r_{x} ∈ R, x ∈ X, all but finitely many being 0. For m as above we set f(m) := ∑_{x∈X} r_{x}g(x). Then, for another element n = ∑_{x∈X} s_{x}x of M and for r ∈ R, we have m + n = ∑_{x∈X}(r_{x} + s_{x})x and rm = ∑_{x∈X}(rr_{x})x. Hence,

f(m + n) = ∑_{x∈X}(r_{x} + s_{x})g(x) = ∑_{x∈X} r_{x}g(x) + ∑_{x∈X} s_{x}g(x) = f(m) + f(n)

and

f(rm) = ∑_{x∈X}(rr_{x})g(x) = r ∑_{x∈X} r_{x}g(x) = rf(m) .

Therefore, f is R-linear, and obviously f|_{X} = g.

The uniqueness of f is clear, since X generates M as R-module.

1.16 Corollary Any finitely generated free R-module is isomorphic to R^{(n)} for some n ∈ ℕ_{0}.

Proof Let X = {x_{1}, …, x_{n}} be an R-basis of M. Let e_{1}, …, e_{n} be the canonical R-basis of R^{(n)}. Then by the universal property of R-bases there exist f_{1} ∈ Hom_{R}(M, R^{(n)}) and f_{2} ∈ Hom_{R}(R^{(n)}, M) such that f_{1}(x_{i}) = e_{i} and f_{2}(e_{i}) = x_{i} for all i = 1, …, n. This implies f_{2} ∘ f_{1} = id_{M} and f_{1} ∘ f_{2} = id_{R^{(n)}}, so that f_{1} and f_{2} are inverse isomorphisms.

1.17 Example Is x_{1} = (1, 2), x_{2} = (2, 1) a ℤ-basis of ℤ^{(2)}? We check linear independence: If a(1, 2)+b(2, 1) = 0 for a, b ∈ ℤ, then a+2b = 0 and 2a+b = 0. This implies a = −2b and substitution in the second equation gives b = 0 and then we also have a = 0. Thus x_{1}, x_{2} are ℤ-linearly independent. In order to show that ⟨x_{1}, x_{2}⟩_{R} = ℤ^{(2)} we have to verify that (1, 0), (0, 1) ∈ ⟨x_{1}, x_{2}⟩_{R}. But this turns out to be impossible: Suppose (1, 0) = a(1, 2) + b(2, 1) with a, b ∈ ℤ. Then 1 = a + 2b and 0 = 2a + b. So b = −2a and substituting this in the first equation we obtain 1 = −3a which is not possible for a ∈ ℤ. Thus, x_{1}, x_{2} is not a ℤ-basis of ℤ^{(2)}.

1.18 Theorem Let R be a commutative ring and let M be a finitely generated free R-module. Then any two R-bases of M have the same length. This length is called the R-rank of M and is denoted by rk_{R}(M).

Proof Let x_{1}, …, x_{m} and y_{1}, …, y_{n} be R-bases of M and assume that m < n. We can write

y_{i} = ∑_{j=1}^{m} a_{ji}x_{j} and x_{k} = ∑_{l=1}^{n} b_{lk}y_{l}

for all i = 1, …, n and all k = 1, …, m with elements a_{ji}, b_{lk} ∈ R for 1 ⩽ j, k ⩽ m and 1 ⩽ i, l ⩽ n. Substitution yields

y_{i} = ∑_{j=1}^{m} a_{ji} ∑_{l=1}^{n} b_{lj}y_{l} = ∑_{j=1}^{m} ∑_{l=1}^{n} a_{ji}b_{lj}y_{l} = ∑_{l=1}^{n}(∑_{j=1}^{m} b_{lj}a_{ji})y_{l}

for all i = 1, …, n. Since y_{1}, …, y_{n} are R-linearly independent, we obtain

∑_{j=1}^{m} b_{lj}a_{ji} = δ_{li}

for all 1 ⩽ i, l ⩽ n. The n × n-matrices

A := (\begin{matrix}a_{11} & ⋯ & a_{1n} \\ ⋮ & & ⋮ \\ a_{m1} & ⋯ & a_{mn} \\ 0 & ⋯ & 0 \\ ⋮ & & ⋮ \\ 0 & ⋯ & 0 \end{matrix}) and B := (\begin{matrix}b_{11} & ⋯ & b_{1m} & 0 & ⋯ & 0 \\ ⋮ & & ⋮ & ⋮ & & ⋮ \\ b_{n1} & ⋯ & b_{nm} & 0 & ⋯ & 0 \end{matrix})

then satisfy the relation BA = 1_{n}. Therefore, det(B) det(A) = 1. On the other hand, obviously det(A) = 0. This is a contradiction.

1.19 Theorem (Universal property of direct sums) Let M_{i}, i ∈ I, be Rmodules and let M = ⨁_{i∈I} M_{i}. We denote by ι_{i}: M_{i} → M the canonical embedding which maps m_{i} ∈ M_{i} to the family which has the entry m_{i} at i and entries 0 everywhere else. Let also N be an R-module. Then, for each family of homomorphisms f_{i} ∈ Hom_{R}(M_{i}, N), i ∈ I, there exists a unique homomorphism f ∈ Hom_{R}(M, N) such that f_{i} = f ∘ ι_{i} for each i ∈ I.

Proof Let m := (m_{i})_{i∈I} ∈ ⨁_{i∈I} M_{i}. We set f(m) := ∑_{i∈I} f_{i}(m_{i}). Then it is easy to verify that f ∈ Hom_{R}(M, N) and that f_{i} = f ∘ ι_{i} for each i ∈ I. Moreover, f is unique with this property, since M is the sum of the submodules ι_{i}(M_{i}).

1.20 Definition Let M be an R-module and let M_{i},i ∈ I, be submodules of M. M is called the internal direct sum, if the unique homomorphism f: ⨁_{i∈I} M_{i} → M, such that f ∘ ι_{i} is the embedding M_{i} ⊆ M, is an isomorphism. In this case we also write M = ⨁_{i∈I} M_{i} and identify M with the external direct sum via f. Note that, in this case, if g_{i}: M_{i} → N are R-linear maps to some R-module N, then there exists a unique R-linear map g: M → N such that g|_{M_{i}} = g_{i}.

1.21 Theorem Let R be a ring, let M be an R-module, and let U_{i}, i ∈ I, be R- submodules of M. Then the following are equivalent:

(i) M is the internal direct sum of the U_{i}, i ∈ I.

(ii) (α) M = ∑_{i∈I} U_{i}, and

(β) U_{i_{0}} ∩ ∑_{i∈I∖{i_{0}}} U_{i} = 0 for all i_{0} ∈ I. (iii) Each element m ∈ M can be written as m = ∑_{i∈I} u_{i} with unique

elements u_{i} ∈ U_{i}, i ∈ I (only finitely many of which are non-zero).

Proof (i) ⇒ (ii): If f: ⨁_{i∈I} U_{i} → M as in Definition 1.20 is an isomorphism, then ι_{i}(U_{i}) ⊆ ⨁_{i∈I} U_{i} and U_{i} ⊆ M correspond under f. Since ∑_{i∈I} ι_{i}(U_{i}) = ⨁_{i∈I} U_{i}, (α) holds, and since ι_{i_{0}}(U_{i_{0}}) ∩ ∑_{i∈I∖{i_{0}}} ι_{i}(U_{i}) = 0 for all i_{0} ∈ I, (β) holds.

(ii) ⇒ (iii): By (α), each m ∈ M can be written as m = ∑_{i∈I} u_{i} with u_{i} ∈ U_{i} for i ∈ I. By (β), the elements u_{i} ∈ U_{i} are uniquely determined by m. In fact, if also v_{i} ∈ U_{i}, i ∈ I, satisfy m = ∑_{i∈I} v_{i}, and if u_{i_{0}} ≠ v_{i_{0}} for some i_{0} ∈ I, then subtraction yields 0 ≠ v_{i_{0}} − u_{i_{0}} = ∑_{i∈I∖{i_{0}}} u_{i} − v_{i} ∈ U_{i_{0}} ∩ ∑_{i∈I∖{i_{0}}} U_{i}.

(iii) ⇒ (i): The map f: ⨁_{i∈I} U_{i} → M defined on the external direct sum ⨁_{i∈I} U_{i} as in Definition 1.20 is surjective, since each m ∈ M can be expressed as sum of elements in U_{i}, i ∈ I. Moreover, the uniqueness of such an expression implies the injectivity of f.

1.22 Examples (a) Consider the ℤ-submodules U = {(a, b, 0) | a, b ∈ ℤ} and V = {(c, c, c) | c ∈ ℤ} of M = ℤ^{(3)}. Then M = U ⊕ V. In fact, each element (a, b, c) ∈ M can be written as (a, b, c) = (a − c, b − c, 0) + (c, c, c) ∈ U + V, and if (a, b, 0) = (c, c, c) for integers a, b, c, then a = c, b = c, and 0 = c imply a = b = c = 0. But note that M is not the internal direct sum of U and 2V = {(2c, 2c, 2c) | c ∈ ℤ}, since (0, 0, 1) ∉ U + 2V.

(b) The ℤ-module M = ℤ/6ℤ is the internal direct sum of the submodules U = 3ℤ/6ℤ = {0 + 6ℤ, 3 + 6ℤ} and V = 2ℤ/6ℤ = {0 + 6ℤ, 2 + 6ℤ, 4 + 6ℤ} by Theorem 1.21 (iii). (This can also be seen by applying the Chinese Remainder Theorem.)

(c) Let V be a vector space over a field F and let f ∈ End_{F} (V). Consider the F [λ]-module structure of V given by extending λ · v := f(v) to F [λ], cf. Example 1.7 (d). The F [λ]-submodules of V are precisely those F -subspaces U of V with f(U) ⊆ U. We call such a subspace f-stable.

(d) As a more concrete example, let V be a 3-dimensional vector space over ℝ with ℝ-basis x, y, z, and let f ∈ End_{ℝ}(V) be given by f(x) = 3x, f(y) = y+z, and f(z) = −y+2z. Then the subspaces U = ℝx and V = ℝy+ℝz are f-stable. Therefore, with the F [λ]-module structure of V defined as in (c), we have an internal direct sum decomposition M = U ⊕ V of M into F [λ]-submodules.

1.23 Remark (Homomorphisms between free modules and matrices) Throughout this remark let R be a commutative ring.

(a) Let M be an R-module. Since R is commutative we may consider M as left and as right R-module. Using these module structures we can define the matrix product

A · Q and Q · B

for any matrices A and B with entries in R and any matrix Q with entries in M, as long as their sizes are compatible for multiplication. The usual rules for matrix addition and multiplication still hold in this situation.

(b) Let M and N be free R-modules with ordered R-bases X = (x_{1}, …, x_{m}) and Y = (y_{1}, …, y_{n}), respectively. Let f ∈ Hom_{R}(M, N) and, for i = 1, …, m, write

f(x_{i}) = ∑_{j=1}^{n} a_{ji}y_{j}  (1.23.a)

with a_{ji} ∈ R for 1 ⩽ j ⩽ n, 1 ⩽ i ⩽ m. Then, f is uniquely determined by the elements a_{ji} ∈ R, and conversely, for any given set of elements a_{ji} ∈ R, 1 ⩽ j ⩽ n, 1 ⩽ i ⩽ m, we may define an R-linear map f: M → N by the Equations (1.23.a) using the universal property of a basis. If we arrange the elements a_{ji} in the form of an n × m-matrix

A = (\begin{matrix}a_{11} & ⋯ & a_{1m} \\ ⋮ & & ⋮ \\ a_{n1} & ⋯ & a_{nm} \end{matrix}) ,

then the above arguments show that we obtain a bijection between Hom_{R}(M, N) and M_{n×m}(R). This is a group isomorphisms, since if the matrix B is attached to a further map g ∈ Hom_{R}(M, N) then the matrix A + B is attached to f + g. We say that A represents f with respect to the bases X and Y, and we denote A by _{Y} M_{X}(f). The i-th column of _{Y} M_{X}(f) gives the coefficients of f(x_{i}) with respect to the basis Y. If we write the elements R^{(m)} and R^{(n)} as columns and identify them with M and N, respectively, by

( r_{1} ; ⋮ ; r_{m}) ↦ ∑_{i=1}^{m} r_{i}x_{i} and (r_{1} ; ⋮ ; r_{n}) ↦ ∑_{j=1}^{n} r_{j}y_{j} ,

then f translates to an R-linear map R^{(m)} → R^{(n)} which is given as the multiplication by A:

( r_{1} ; ⋮ ; r_{m}) ↦ A ( r_{1} ; ⋮ ; r_{m}) .

Note that with the notation in (a) we can also write the Equations (1.23.a) in the form

(f(x_{1}), …, f(x_{m})) = (y_{1}, …, y_{n})A .

(c) If P is a further free R-module with finite R-basis Z = (z_{1}, …, z_{p}) and if g ∈ Hom_{R}(N, P), then we have

ZM_{X}(g ∘ f) = _{Z}M_{Y} (g) ·_{Y} M_{X}(f) .

In fact, if _{Z}M_{Y} (g) = (g_{kj}), then the i-th column of _{Z}M_{X}(g ∘ f) is determined by

g(f(x_{i})) = ∑_{j=1}^{n} a_{ji}g(y_{j}) = ∑_{j=1}^{n} a_{ji} ∑_{k=1}^{p} b_{kj}z_{k} = ∑_{k=1}^{p}(∑_{j=1}^{n} b_{kj}a_{ji})z_{k} ,

and the coefficient of z_{k} is precisely the (k, i)-entry of B · A.

(d) If we use other R-bases X′ = (x′_{1}, …, x′_{m}) and Y′ = (y′_{1}, …, y′_{n}) for M and N, respectively, then by (c) we have

Y′M_{X′}(f) = _{Y′}M_{Y} (id_{N}) ·_{Y} M_{X}(f) ·_{X}M_{X′}(id_{M}) .

More explicitly, if

id_{M}(x′_{i}) = x′_{i} = ∑_{j=1}^{m} s_{ji}x_{j} and id_{N}(y′_{k}) = y′_{k} = ∑_{l=1}^{n} t_{lk}y_{l}

and if we define the matrices S = (s_{ji}) and T = (t_{lk}), then S = _{X}M_{X′}(id_{M}) and T = _{Y} M_{Y′}(id_{N}). These matrices are invertible, since by part (c) we have

XM_{X′}(id_{M}) ·_{X′}M_{X}(id_{M}) = _{X}M_{X}(id_{M}) = 1_{m} ,

and similar in the reversed order and for Y and Y′. This also shows that S^{−1} = _{X′}M_{X}(id_{M}) and T^{−1} = _{Y′}M_{Y} (id_{N}) so that

Y′M_{X′}(f) = T^{−1}AS .

(e) If x′_{1}, …, x′_{m} are arbitrary elements of M and if we write

x′_{i} = ∑_{j=1}^{m} s_{ji} · x_{j} ,

for each i = 1, …, m, then the matrix S = (s_{ji}) is invertible in M_{m}(R) if and only if x′_{1}, …, x′_{m} is again an R-basis of M. In fact, if x′_{1}, …, x′_{m} is an R-basis, then S is invertible by part (d). Conversely, assume that S is invertible. The above equations can be written as

(x′_{1}, …, x′_{m}) = (x_{1}, …, x_{m})S ,

so that we have

(x_{1}, …, x_{m}) = (x′_{1}, …, x′_{m})S^{−1} .

This shows that x_{1}, …, x_{m} ∈ ⟨x′_{1}, … x′_{m}⟩_{R} and therefore, M = ⟨x′_{1}, …, x′_{m}⟩_{R}. Also, if r_{1}x′_{1} + ⋯ + r_{m}x′_{m} = 0 for elements a_{1}, …, a_{m} ∈ R, then

(r_{1}, …, r_{m}) ( x′_{1} ; ⋮ ; x′_{m}) = 0

and so

(r_{1}, …, r_{m})S^{t} ( x_{1} ; ⋮ ; x_{m}) = 0 .

Since x_{1}, …, x_{m} are linearly independent, we obtain (r_{1}, …, r_{m})S^{t} = 0, and, since S^{t} is invertible, also (r_{1}, …, r_{m}) = 0.

Recall that, for commutative R, a square matrix S with coefficients in R is invertible if and only if det(S) is invertible in R, cf. [J, Theorem 2.1, p. 96].

1.24 Examples (a) Let V be an ℝ-vector space with basis x, y, z and let f ∈ End_{ℝ}(V) be given by f(x) = 3x, f(y) = y + z, f(z) = −y + 2z as in Example 1.22 (c). Then the matrix representing f with respect to the basis (x, y, z) is

(\begin{matrix}3 & 0 & 0 \\ 0 & 1 & −1 \\ 0 & 1 & 2 \end{matrix}) .

If we want to represent f with respect to the basis (x, y, y + z), we have to find S ∈ GL_{3}(ℝ) such that

(x, y, y + z) = (x, y, z)S .

This implies

S = (\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}) and S^{−1} = (\begin{matrix}1 & 0 & 0 \\ 0 & 1 & −1 \\ 0 & 0 & 1 \end{matrix})

so that f is represented with respect to the basis (x, y, y + z) by

(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & −1 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix}3 & 0 & 0 \\ 0 & 1 & −1 \\ 0 & 1 & 2 \end{matrix}) (\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}) =

(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & −1 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix}3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 3 \end{matrix}) = (\begin{matrix}3 & 0 & 0 \\ 0 & 0 & −3 \\ 0 & 1 & 3 \end{matrix}) .

(b) Consider the elements (0, 1, 2), (2, 3, 1), (−1, −1, 0) ∈ ℤ^{(3)}. Do they form a ℤ-basis of ℤ^{(3)}? We have to determine, if det(S) is invertible in ℤ for

S = (\begin{matrix}0 & 2 & −1 \\ 1 & 3 & −1 \\ 2 & 1 & 0 \end{matrix}) .

Now, det(S) = −4 − 1 + 6 = 1. Therefore, the above elements form a ℤ-basis of ℤ^{(3)}.

2 Modules over Principal Ideal Domains

2.1 Definition Let R be a domain (i.e., a commutative ring without zero divisors except 0 and with 1 ≠ 0) and let M be an R-module. An element m ∈ M is called an R-torsion element, if there exists r ∈ R ∖ {0} such that rm = 0. The set tor_{R}(M) of torsion elements is an R-submodule of M, called the R-torsion submodule of M. M is called R-torsion free, if tor_{R}(M) = 0. Note that R-linear maps map R-torsion elements to R-torsion elements, and that a free R-module is R-torsion free. If R = ℤ, we just write tor(M) and call this the torsion subgroup of M. Obviously, tor(M) consists of the elements of finite order.

2.2 Theorem Let M be a finitely generated free R-module over a principal ideal domain R and let U be an R-submodule of M. Then also U is R-free and rk_{R}(U) ⩽ rk_{R}(M).

Proof By induction on n = rk_{R}(M). If n = 0, then M = 0, U = 0, and U is R-free with basis ∅. Let n > 0. Choose an R-basis f_{1}, …, f_{n} of M and set M′ := ⟨f_{2}, …, f_{n}⟩_{R} ⊆ M. Then M is R-free of rank n − 1. If U is contained in M′, then by induction U is R-free with rk_{R}(U) ⩽ n − 1. So we can assume that U ⊈ M′. Consider the R-linear map π: M → R, r_{1}f_{1} + ⋯ + r_{n}f_{n} ↦ r_{1}. Since R is a principal ideal domain and U ⊈ M′, there exists 0 ≠ d ∈ R with π(U) = dR. Let u_{1} ∈ U with π(u_{1}) = d. We claim that U = Ru_{1} ⊕ (M′ ∩ U). In fact, Ru_{1} + (U ∩ M′) ⊆ U and Ru_{1} ∩ U ∩ M′ ⊆ Ru_{1} ∩ M′ = 0. Moreover, let u ∈ U be arbitrary and let π(u) = rd with r ∈ R. Then u = ru_{1} + (u − ru_{1}) with π(u − ru_{1}) = 0, so u − ru_{1} ∈ U ∩ M′. By induction, U ∩ M′ is R-free of rank at most n − 1. Let e_{2}, …, e_{k} be an R-basis of U ∩ M′ with 1 ⩽ k ⩽ n. Then obviously e_{1} := u_{1}, e_{2}, …, e_{k} is an R-basis of U = Ru_{1} ⊕ (U ∩ M′).

2.3 Definition Let R be a commutative ring. Tow matrices A, B ∈ M_{m×n}(R) are called R-equivalent, if there exist P ∈ GL_{m}(R) and Q ∈ GL_{n}(R) such that B = P AQ. This defines an equivalence relation on M_{m×n}(R).

2.4 Remark Let R be a principal ideal domain.

(a) Let a_{1}, …, a_{n} be elements in R. Recall that an element d ∈ R is called a greatest common divisor (or short a gcd) of a_{1}, …, a_{n} if d divides each a_{i} (1 ⩽ i ⩽ n) and if for any other element e ∈ R which also divides each a_{i} (1 ⩽ i ⩽ n) one has e | d. Recall that d is a gcd of a_{1}, …, a_{n} if and only if Ra_{1}+⋯ Ra_{n} = dR. Therefore, two gcds d, d′ of a_{1}, …, a_{n} are always associates in R, i.e., there exists a unit u of R such that d′ = du.

(b) Let A_{1} ∈ M_{m×n}(R) and A_{2} ∈ M_{n×p}(R). Recall that, for i ∈ ℕ, an iminor of A_{1} is the determinant of an i × i-submatrix of A arising from choosing any i rows and i columns of A_{1}. For i ∈ ℕ, let Δ_{i}(A_{1}) be a gcd of the i-minors of A_{1}. If i is bigger than m or n, then we set Δ_{i}(A_{1}) = 0. Similarly we define Δ_{i}(A_{2}) and Δ_{i}(A_{1}A_{2}). Since the columns of A_{1}A_{2} are R-linear combinations of those of A_{1}, also the columns of any (i × i)-submatrix S of A_{1}A_{2} are R-linear combinations of i-subcolumns of A_{1} (using the same i rows). Since

det(x_{1}, …, x_{j−1}, rx_{j} + sy_{j}, x_{j+1}, …, x_{i}) =

r det(x_{1}, …, x_{j−1}, x_{j}, x_{j+1}, …, x_{i}) + s det(x_{1}, …, x_{j_{1}}, y_{j}, x_{j+1}, …, x_{i}) ,

for any columns x_{1}, …, x_{n}, y_{j} with i entries in R (1 ⩽ j ⩽ i) and any r, s ∈ R, we can express det(S) as R-linear combination of i-minors of A_{1}. Therefore, Δ(A_{1}) | Δ(A_{1}A_{2}). Similarly, using row arguments, we have Δ_{i}(A_{2}) | Δ(A_{1}A_{2}). In particular, if A, B ∈ M_{m×n}(R) are R-equivalent, then Δ_{i}(A) and Δ_{i}(B) are associates in R. Note that the 1-minors of A are just the entries of A.

(c) Consider the (n × n)-matrices

T_{ij}(a) = (\begin{matrix}1 \\ & ... & a \\ . & . & . \\ & & & 1 \end{matrix}) ,

with a ∈ R in the i-th row and j-th column 1 ⩽ i, j ⩽ n, i ≠ j,

D_{i}(u) = (\begin{matrix}1 \\ & . & . & . \\ & & & & u \\ & & & & & . & . & . \\ & & & & & & & & 1 \end{matrix}) ,

with a unit u of R, and

P_{ij} = (\begin{matrix}1 \\ & . & . & . \\ & & & & 0 & & & & 1 \\ & & & & & . & . & . \\ & & & & 1 & & & & 0 \\ & & & & & & & & & . & . & . \\ & & & & & & & & & & & & 1 \end{matrix}) ,

with 1 ⩽ i, j ⩽ n, i ≠ j. The diagonal entries of P_{ij} are all equal to 1 except the two 0-entries at the i-th and j-th position. Multiplication by T_{ij}(a) from the left (resp. right) to a matrix A adds the j-th row (resp. i-th column) of A multiplied by a to the i-th row (resp. j-th column) of A. Multiplication by D_{i}(u) from the left (resp. right) multiplies the i-th row (resp. column) of A by u. Multiplication by P_{ij} from the left (resp. right) interchanges the i-th and j-th row (resp. column) of A. Obviously,

T_{ij}(a)T_{ij}(−a) = 1_{n} ,
D_{i}(u)D_{i}(u^{−1}) = 1_{n} ,
P_{ij}P_{ij} = 1_{n} ,

so that all these matrices lie in GL_{n}(R).

2.5 Theorem Let R be a principal ideal domain. Then each matrix A = (a_{ij}) ∈ M_{m×n}(R) is equivalent to a matrix of the form

(\begin{matrix}d_{1} \\ & . & . & . \\ & & & & d_{k} \\ & & & & & 0 \\ & & & & & & . & . & . \end{matrix} )

with non-zero elements d_{1}, …, d_{k} ∈ R such that d_{1} | d_{2} | ⋯ | d_{k}. Moreover, k is uniquely determined by A and d_{1}, …, d_{k} are unique up to multiplication with units of R. The elements d_{1}, …, d_{k} are called invariant factors of A.

Proof We define the length l(r) of an element r ∈ R as the number of factors occurring in a prime decomposition of R. If R is a unit we set l(r) = 0, and for r = 0 we set l(0) = ∞. We will first show that A is equivalent to a matrix of the above form. If A = 0 or if A is a 1 × 1-matrix, this is obvious. So we assume that A ≠ 0 and that A has at least two entries.

Step 1: We show that A is similar to a matrix of the form

B = (\begin{matrix}b_{11} & 0 & ⋯ & 0 \\ 0 \\ ⋮ & & C \\ 0 \end{matrix})  (2.5.a)

with an (m−1)×(n−1)-matrix C = (c_{ij}) such that b_{11} | c_{ij} for all 1 ⩽ i ⩽ m−1, 1 ⩽ j ⩽ n − 1. We will prove this by induction on l(A) := min{l(a_{ij}) | 1 ⩽ i ⩽ m, 1 ⩽ j ⩽ n}.

If l(A) = 0, then there is an element a_{ij} in A which is a unit. Using permutations of rows and columns we may assume that a_{11} is a unit. Then we can add the first row multiplied by −a_{i1}a_{11}^{−1} to the i-th row (i = 2, …, m) and the first column multiplied by −a_{1j}a_{11}^{−1} to the j-th column (j = 2, …, n) and obtain the desired matrix B.

If l > 0, then we can again assume that l(a_{11}) is minimal among the l(a_{ij}). If every other entry of A is divisible by a_{11}, we set b_{11} = a_{11} and can add appropriate multiples of the first row and column to the other rows and columns to obtain the desired matrix B. Note that all the entries which are not in the first row or first column remain divisible by a_{11} = b_{11}, since a_{11} is a gcd of the 1-minors of A and so of B.

If there exists an entry a_{ij} in A which is not divisible by a_{11}, then we may assume that a_{ij} = a_{12} by permutations of rows and columns if necessary (or a_{ij} = a_{21} if there is only a single column; but this case is handled similarly). We set a := a_{11}, b := a_{12}, and choose a gcd d of a and b. Then l(d) < l(a_{11}). Moreover, there exist x, y ∈ R with ax + by = d and we can write a = rd and b = sd with r, s ∈ R. We set

Q := (\begin{matrix}x & s & 0 & ⋯ & 0 \\ y & −r & 0 & ⋯ & ⋮ \\ 0 & 0 & 1 & ... & ⋮ \\ ⋮ & & ... & ... & 0 \\ 0 & ⋯ & ⋯ & 0 & 1 \end{matrix}) ∈ M_{n}(R) .

Note that Q is invertible, since

(\begin{matrix}x & s \\ y & −r \end{matrix}) (\begin{matrix}r & s \\ y & −x \end{matrix}) = (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) = (\begin{matrix}r & s \\ y & −x \end{matrix}) (\begin{matrix}x & s \\ y & −r \end{matrix}) .

Now the first row of AQ has the form (d, 0, ∗, …, ∗), and since l(d) < l(a_{11}) = l(A), we can now apply induction to this matrix.

Step 2: We next prove the assertion by induction on min{m, n}. If m = 1 or n = 1, we are finished by Step 1. So we assume that m, n ⩾ 2. Then, by Step 1, A is equivalent to a matrix B as in (2.5.a) with d_{1} := b_{11} | c_{ij} for all 1 ⩽ i ⩽ m − 1 and 1 ⩽ j ⩽ n − 1. By induction, there exist P′ ∈ GL_{m−1}(R) and Q′ ∈ GL_{n−1}(R) such that

P′CQ′ = (\begin{matrix}d_{2} \\ & . & . & . \\ & & & & d_{k} \\ & & & & & 0 \\ & & & & & & . & . & . \end{matrix} ) =: D′

with non-zero d_{2} | ⋯ | d_{k}. Therefore, with

P := (\begin{matrix}1 & 0 \\ 0 & P′ \end{matrix}) and Q := (\begin{matrix}1 & 0 \\ 0 & Q′ \end{matrix}) ,

we obtain

P BQ = (\begin{matrix}d_{1} \\ & d_{2} \\ & & . & . & . \\ & & & & & d_{k} \\ & & & & & & 0 \\ & & & & & & & . & . & . \end{matrix} ) .

Now, by Remark 2.4 (b), any two gcds of the entries of C and D′, respectively, are associates in R, so that d_{1} also divides d_{2}. This finishes the first part of the proof.

Next we assume that two matrices

D = (\begin{matrix}d_{1} \\ & d_{2} \\ & & . & . & . \\ & & & & & d_{k} \\ & & & & & & 0 \\ & & & & & & & . & . & . \end{matrix} ) and E = (\begin{matrix}e_{1} \\ & e_{2} \\ & & . & . & . \\ & & & & & e_{l} \\ & & & & & & 0 \\ & & & & & & & . & . & . \end{matrix} )

in M_{m×n}(R) with non-zero d_{1}, …, d_{k}, e_{1}, …, e_{l} satisfying d_{1} | ⋯ | d_{k} and e_{1} | ⋯ | e_{l} are equivalent. It will be convenient to set d_{k+1} = d_{k+2} = ⋯ = 0 and e_{l+1} = e_{l+2} = ⋯ = 0. Then r_{i} = d_{1}d_{2} ⋯ d_{i} and s_{i} = e_{1}e_{2} ⋯ e_{i} are gcds of the i-minors of D and E, respectively, and they are associates in R by Remark 2.4 (b). In particular, r_{i} = 0 if and only if s_{i} = 0. This implies k = l. Moreover, there exist units u_{1}, …, u_{k} of R with s_{i} = r_{i}u_{i}. Now it is easy to show by induction on i that e_{1} = d_{1}u_{1} and e_{i+1} = d_{i+1}u_{i+1}u_{i}^{−1} for i ⩾ 2. This concludes the proof of the Theorem.

2.6 Corollary Let R be a principal ideal domain and let M be a finitely generated R-module. Then there exist s ∈ ℕ_{0}, k ∈ ℕ_{0}, and non-units d_{1}, …, d_{s} ∈ R ∖ {0} with d_{1} | ⋯ | d_{s} such that

M ≅ R/d_{1}R ⊕ ⋯ ⊕ R/d_{s}R ⊕ R^{(k)} .  (2.6.a)

Proof Let M be generated by y_{1}, …, y_{m} ∈ M as R-module and let f ∈ Hom_{R}(R^{(m)}, M) be defined by f(e_{i}) = y_{i}. Then U := ker(f) ⊆ R^{(m)} is again R-free with n := rk_{R}(U) ⩽ m (by Theorem 2.2), and f is surjective, so M ≅ R^{(m)}/U. Choose an ordered R-basis X of U and consider the matrix A ∈ M_{m×n}(R) representing the inclusion U ⊆ R^{(m)} with respect to X and (e_{1}, …, e_{m}). By Theorem 2.5 and since f is injective, there exist d′_{1}, …, d′_{n} ∈ R ∖ {0} with d′_{1} | ⋯ | d′_{n} such that A is equivalent to the (m × n)- matrix

(\begin{matrix}d′_{1} \\ . & . & . \\ & & d′_{n} \\ 0 & ⋯ & 0 \\ ⋮ & & ⋮ \end{matrix}) .

This means that there exists an R-basis f_{1}, …, f_{m} of R^{(m)} such that the elements d′_{1}f_{1}, …, d′_{n}f_{n} form an R-basis of U. We set d′_{n+1} = ⋯ = d′_{m} := 0 and obtain

M ≅ R^{(m)}/U ≅ (Rf_{1} ⊕ ⋯ ⊕ Rf_{m})/(Rd′_{1}f_{1} ⊕ ⋯ ⊕ Rd′_{m}f_{m})
≅ R/d′_{1}R ⊕ ⋯ ⊕ R/d′_{m}R ≅ R/d′_{1}R ⊕ ⋯ ⊕ R/d′_{n}R ⊕ R^{(m−n)} .

Omitting those d′_{i}s which are units, we obtain

M ≅ R/d_{1}R ⊕ ⋯ ⊕ R/d_{s}R ⊕ R^{(m−n)}

with d_{1} | ⋯ | d_{s}.

2.7 Remark Let M be as in Corollary 2.6 and (2.6.a). Then tor_{R}(M) ≅ R/d_{1}R ⊕ ⋯ ⊕ R/d_{s}R. Moreover, there exists a free submodule U of M of R-rank k such that M = tor_{R}(M) ⊕ U.

2.8 Theorem (Chinese Remainder Theorem) Let R be an arbitrary ring and let I_{1}, …, I_{n} be two-sided ideals of R such that I_{k} + I_{l} = R for any k ≠ l from {1, …, n}. Then the map

R/I_{1} ∩ … ∩ I_{n} → R/I_{1} × … × R/I_{n} ,
r + I_{1} ∩ ⋯ ∩ I_{n} ↦ (r + I_{1}, …, r + I_{n}) ,

is a well-defined isomorphism of rings and R-modules.

Proof Consider the ring and R-module homomorphism

ρ: R → R/I_{1} × ⋯ × R/I_{n} , r ↦ (r + I_{1}, …, r + I_{n}) .

Its kernel is obviously I_{1} ∩ … ∩ I_{n}. Therefore, it suffices to show that ρ is surjective. Fix k ∈ {1, …, n}. It suffices to find b ∈ R such that b + I_{k} = 1 + I_{k} and b + I_{l} = 0 + I_{l} for all l ≠ k. But for each l ≠ k there exist elements a_{l} ∈ I_{k} and b_{l} ∈ I_{l} such that a_{l} + b_{l} = 1. Therefore, b := b_{1} ⋯ b_{k−1}b_{k+1}b_{n} ∈ I_{l} for all l ≠ k. Moreover,

1 + I_{k} = (a_{1} + b_{1}) ⋯ (a_{n} + b_{n}) + I_{k} = b_{1} ⋯ b_{n} + I_{k} ,

which completes the proof.

2.9 Remark If R is a principal ideal domain and d_{1}, …, d_{n} ∈ R are pairwise coprime, then the ideals I_{1} := d_{1}R, …, I_{n} := d_{n}R satisfy the hypothesis of Theorem 2.8 and I_{1}∩…∩I_{n} = d_{1} ⋯ d_{n}R. Therefore, we obtain an isomorphism

ρ: R/d_{1} ⋯ d_{n}R → R/d_{1}R × ⋯ × R/d_{n}R ,
r + d_{1} ⋯ d_{n}R ↦ (r + d_{1}R, …, r + d_{n}R) ,

of rings and R-modules.

2.10 Corollary Let R be a principal ideal domain and let M be a finitely generated R-module. If P denotes a set of representatives for the primes of R up to associates, then

M ≅ R/p_{1}^{α_{1}}R ⊕ ⋯ ⊕ R/p_{t}^{α_{t}}R ⊕ R^{(l)}  (2.10.a)

for appropriate t ∈ ℕ_{0}, l ∈ ℕ_{0}, p_{1}, …, p_{t} ∈ P and α_{1}, …, α_{t} ∈ ℕ.

Proof This follows immediately from Corollary 2.6 and Remark 2.9, by decomposing each d_{i} into a product of powers of primes.

2.11 Definition Let R be a principal ideal domain and let M be an R-module. For any prime p of R we set

M_{p} := {m ∈ M | p^{α}m = 0 for some α ∈ ℕ}

and call M_{p} the p-component or the p-primary part of M. We call M a p-primary module if M = M_{p}. Obviously, f(M_{p}) ⊆ N_{p} for each map f ∈ Hom_{R}(M, N) to some R-module N. Note that M_{p} ⊆ tor_{R}(M) is an R-submodule and that M_{p} = M_{p′} if p and p′ are associates in R.

2.12 Theorem Let R be a principal ideal domain, let M be an R-module, and let P be as in Corollary 2.10. Then

tor_{R}(M) = ⨁_{p∈P} M_{p} .

Proof Obviously, M_{p} ⊆ tor_{R}(M). Conversely, let 0 ≠ m ∈ tor_{R}(M) and let r ∈ R ∖ {0} such that rm = 0. Then r is not a unit, and we can write r = up_{1}^{α_{1}} ⋯ p_{n}^{α_{n}} with a unit u of R, p_{1}, …, p_{n} ∈ P, and α_{1}, …, α_{n} ∈ ℕ. We may assume that u = 1 by choosing u^{−}1r instead of r if necessary. Set r_{i} := r/p_{i}^{α_{i}} for i = 1, …, n. Then 1 is a gcd of r_{1}, …, r_{n} and there exist a_{1}, …, a_{n} ∈ R such that a_{1}r_{1} + ⋯ + a_{n}r_{n} = 1. Therefore, m = a_{1}r_{1}m + ⋯ + a_{n}r_{n}m with a_{i}r_{i}m ∈ M_{p_{i}} for each i = 1, …, n. This implies that tor_{R}(M) ⊆ ∑_{p∈P} M_{p}.

Finally we show that the sum ∑_{p∈P} M_{p} is a direct sum. It suffices to show that if m_{1} + ⋯ + m_{n} = 0 with m_{i} ∈ M_{p_{i}} for distinct p_{1}, …, p_{n} ∈ P, then m_{1} = ⋯ = m_{n} = 0. Let α_{1}, …, α_{n} ∈ ℕ be such that p_{i}^{α_{i}}m_{i} = 0 and define r := p_{1}^{α_{1}} ⋯ p_{n}^{α_{n}} and r_{i} := r/p_{i}α_{i} as above. Then r_{i}m_{j} = 0 for all i ≠ j. As above, there exist a_{1}, …, a_{n} ∈ R with a_{1}r_{1} + ⋯ + a_{n}r_{n} = 1. Therefore, m_{i} = a_{1}r_{1}m_{i} + ⋯ + a_{n}r_{n}m_{i} = a_{i}r_{i}m_{i} = a_{i}r_{i}(m_{1} + ⋯ + m_{n}) = 0.

2.13 Lemma Let R be a principal ideal domain and let p be a prime of R. Assume that t, u ∈ ℕ_{0} and 1 ⩽ α_{1} ⩽ ⋯ ⩽ α_{t} and 1 ⩽ β_{1} ⩽ ⋯ ⩽ β_{u} are integers such that

R/p^{α_{1}}R ⊕ ⋯ ⊕ R/p^{α_{t}}R ≅ R/p^{β_{1}}R ⊕ ⋯ ⊕ R/p^{β_{u}}R  (2.13.a)

as R-modules. Then t = u and α_{i} = β_{i} for all i = 1, …, t.

Proof Let M be an R-module and let k ∈ ℕ_{0}. Then p^{k+1}M ⊆ p^{k}M ⊆ M are R-submodules and p^{k}M/p^{k+1}M is an R-module which is annihilated by pR. Therefore, p^{k}M/p^{k+1}M is a vector space over the field F := R/pR by setting (r + pR)(m + p^{k+1}M) := rm + p^{k+1}M for r ∈ R and m ∈ p^{k}M. If M = R/p^{α}R for some α ∈ ℕ, we claim that

dim_{F} (p^{k}M/p^{k+1}M) = \begin{cases}0, if k ⩾ α, \\ 1, if k < α. \end{cases}

In fact, p^{k}M = 0 if k ⩾ α, and p^{k}M ≅ p^{k}R/p^{α}R if k ⩽ α. Thus, for k < α, we have

p^{k}M/p^{k+1}M ≅ (p^{k}R/p^{α}R)/(p^{k+1}R/p^{α}R) ≅ p^{k}R/p^{k+1}R ≅ R/pR = F

as R-modules and therefore also as F -modules so that the claim is proven.

If we denote the left and right hand side of (2.13.a) by M and N, respectively, then for each k ∈ ℕ_{0}, the dimension dim_{F} (p^{k}M/p^{k+1}M) (resp. dim_{F} (p^{k}N/p^{k+1}N)) counts the number of indices i = 1, …, t (resp. j = 1, …, u) such that α_{i} > k (resp. β_{j} > k). But since M ≅ N, this implies t = u (by setting k = 0) and α_{1} = β_{1}, …, α_{t} = β_{t} (by increasing k).

2.14 Corollary Let R be a principal ideal domain and let D be the set of products of elements in P. Then each non-zero element of R which is not a unit has a unique associate in D. If we write a finitely generated R-module M as

M ≅ R/d_{1}R ⊕ ⋯ ⊕ R/d_{s}R ⊕ R^{(k)}

as in Corollary 2.6 with non-units d_{1}, …, d_{s} ∈ R ∖ {0} such that d_{1} | ⋯ | d_{s}, and as

M ≅ R/p_{1}^{α_{1}} ⊕ ⋯ ⊕ R/p_{t}^{α_{t}}R ⊕ R^{(l)}

as in Corollary 2.10 with prime powers p_{1}^{α_{1}}, …, p_{t}^{α_{t}} of primes in P, then k = l and k, d_{1}, …, d_{s}, p_{1}^{α_{1}}, …, p_{t}^{α_{t}} are uniquely determined (the prime powers up to permutations). The elements d_{1}, …, d_{s} are called the invariant factors of M and p_{1}^{α_{1}}, …, p_{t}^{α_{t}} are called the elementary divisors of M.

Proof We have

R/d_{1}R ⊕ ⋯ ⊕ R/d_{s}R ≅ tor_{R}(M) ≅ R/p_{1}^{α_{1}} ⊕ ⋯ ⊕ R/p_{t}^{α_{t}} ,

and therefore,

R^{(k)} ≅ M/tor_{R}(M) ≅ R^{(l)}

which implies k = l and the uniqueness of k by Theorem 1.18. Moreover, by decomposing tor_{R}(M) into its p-primary parts (for p ∈ P) we obtain the uniqueness of the prime powers p_{1}^{α_{1}}, …, p_{t}^{α_{t}} by Lemma 2.13, since

M_{p} ≅ ⨁_{{i∈\{1,... ,t\} \atop p_{i}=p}} R/p_{i}^{α_{i}}R .

Moreover, for fixed p ∈ P, the Chinese Remainder Theorem implies (R/d_{i}R)_{p} ≅ R/p^{β_{i}}R, if d_{i} decomposes as d_{i} = d′_{i}p^{β_{i}} and p does not divide d′_{i}. Therefore, M_{p} ≅ R/p^{β_{1}} ⊕ ⋯ ⊕ R/p^{β_{s}} with β_{1} ⩽ ⋯ ⩽ β_{s}. Now Lemma 2.13 shows that β_{1}, …, β_{s} are uniquely determined by M. Therefore, d_{1}, …, d_{s} are uniquely determined. This completes the proof.

2.15 Remark If R = ℤ one usually chooses the set of primes for P. Then D = {2, 3, 4, 5, 6, … }. If R = F [λ] for a field F one usually chooses the set of monic irreducible polynomials for P. Then D is the set of monic polynomials of degree ⩾ 1.

2.16 Example (a) By Corollary 2.14, there are exactly two Abelian groups of order 20 (up to isomorphism), namely ℤ/4ℤ ⊕ ℤ/5ℤ (which is isomorphic to ℤ/20ℤ) and ℤ/2ℤ ⊕ ℤ/2ℤ ⊕ ℤ/5ℤ (which is isomorphic to ℤ/2ℤ ⊕ ℤ/10ℤ).

(b) Let

u_{1} = (\begin{matrix}−2 \\ 0 \\ 2 \end{matrix}) , u_{2} = (\begin{matrix}1 \\ 2 \\ 3 \end{matrix}) ,

and set U := ⟨u_{1}, u_{2}⟩_{ℤ} ⊆ ℤ^{(3)}. Then X := (u_{1}, u_{2}) is an ordered ℤ-basis of U. We will try to find the invariant factors of M := ℤ^{(3)}/U. We have to proceed as in the proof of Corollary 2.6. For this reason we have to start with the matrix A representing the inclusion U ⊆ ℤ^{(3)} with respect to some basis, say X and E = (e_{1}, e_{2}, e_{3}). This gives

A = (\begin{matrix}−2 & 1 \\ 0 & 2 \\ 2 & 3 \end{matrix}) .

Next we have to transform A as in Theorem 2.5 into a diagonal matrix:

A = (\begin{matrix}−2 & 1 \\ 0 & 2 \\ 2 & 3 \end{matrix}) ⟶^{•P_{12}} (\begin{matrix}1 & −2 \\ 2 & 0 \\ 3 & 2 \end{matrix}) ⟶^{T_{21}(−2)•} (\begin{matrix}1 & −2 \\ 0 & 4 \\ 3 & 2 \end{matrix}) ⟶^{T_{31}(−3)•}

(\begin{matrix}1 & −2 \\ 0 & 4 \\ 0 & 8 \end{matrix}) ⟶^{•T_{12}(2)} (\begin{matrix}1 & 0 \\ 0 & 4 \\ 0 & 8 \end{matrix}) ⟶^{T_{32}(−2)•} (\begin{matrix}1 & 0 \\ 0 & 4 \\ 0 & 0 \end{matrix}) =: B .

This already shows that

M ≅ ℤ/1ℤ ⊕ ℤ/4ℤ ⊕ ℤ ≅ ℤ/4ℤ ⊕ ℤ

and that the invariant factors of M consist just of a single one, namely d_{1} = 4. But we can derive more information from what we computed: B represents still the inclusion U ⊆ ℤ^{(3)} but with respect to different bases Y = (v_{1}, v_{2}) and E′ = (e′_{1}, e′_{2}, e′_{3}) of U and ℤ^{(3)}, respectively. In order to compute Y and E′ we have to write B in the form B = T^{−1}AS, and then have

(v_{1}, v_{2}) = (u_{1}, u_{2})S = (u_{1}, u_{2})P_{12}T_{12}(2) =
(u_{2}, u_{1})T_{12}(2) = (u_{2}, u_{1} + 2u_{2}) .

and

(e′_{1}, e′_{2}, e′_{3}) = (e_{1}, e_{2}, e_{3})T = (e_{1}, e_{2}, e_{3})T_{21}(2)T_{31}(3)T_{32}(2) =
(e_{1} + 2e_{2}, e_{2}, e_{3})T_{31}(3)T_{32}(2) = (e_{1} + 2e_{2} + 3e_{3}, e_{2}, e_{3})T_{32}(2) =
(e_{1} + 2e_{2} + 3e_{3}, e_{2} + 2e_{3}, e_{3}) .

This also shows that in the above isomorphism M ≅ ℤ/4ℤ⊕ℤ the element in M corresponding to (1 + 4ℤ, 0) is e′_{2} + U and the one corresponding to (0 + 4ℤ, 1) is e′_{3} + U.

3 Normal Forms of Matrices

3.1 Notation Throughout this section we fix a field F, a finite dimensional F - vector space V, an ordered F -basis X = (v_{1}, …, v_{n}) of V, and an endomorphism f ∈ End_{F} (V). We set A := _{X}M_{X}(f) = (a_{ij}) ∈ M_{n}(F), i.e.,

f(v_{i}) = ∑_{j=1}^{n} a_{ji}v_{j} , i = 1, …, n .

We put R := F [λ] and view V a an R-module via g(λ) · v := (g(f))(v) for g(λ) ∈ R and v ∈ V.

3.2 Remark (a) If g(λ) = a_{m}λ^{m} + ⋯ + a_{1}λ + a_{0} ∈ R with a_{m} ≠ 0, then R/g(λ)R is an F -vector space with F -basis

1 + g(λ)R, λ + g(λ)R, …, λ^{m−1} + g(λ)R .

In fact, dividing an arbitrary h(λ) ∈ R by g(λ) gives h(λ) = q(λ)g(λ) + r(λ) with q(λ), r(λ) ∈ R and deg(r(λ)) < m, so that h(λ) + g(λ)R = r(λ) + g(λ)R. Moreover, if b_{0}(1 + g(λ)R) + ⋯ + b_{m−1}(λ^{m−1} + g(λ)R) = 0, then b_{0} + b_{1}λ + ⋯ b_{m−1}λ^{m−1} ∈ g(λ)R which implies b_{0} = ⋯ = b_{m−1} = 0, since deg(g(λ)) = m.

(b) If g(λ) = (λ − r)^{m} for some r ∈ F, then also

1 + g(λ)R, (λ − r) + g(λ)R, …, (λ − r)^{m−1} + g(λ)R

form an F -basis of R/g(λ)R, since these m elements of R/g(λ)R are F -linear independent by a similar argument as in (a).

(c) V is a finitely generated R-torsion module, since it is finite dimensional over F. Suppose that V ≅ R/p_{1}(λ)^{α_{1}}R ⊕ ⋯ ⊕ R/p_{t}(λ)^{α_{t}}R with monic irreducible p_{i}(λ) ∈ R and α_{i} ∈ ℕ. Then n = dim_{F} (V) = n_{1} + ⋯ + n_{t} with n_{i} = α_{i} deg(p_{i}(λ)), and V (as R-module) has generators z_{1}, …, z_{t} ∈ V corresponding to the elements (0, …, 0, 1+p_{i}(λ)^{α_{i}}R, 0, …, 0) for i = 1, …, t. Then, by (a),

Z = (z_{1}, f(z_{1}), …, f^{n_{1}−1}(z_{1}), z_{2}, f(z_{2}), …)

is an F -basis of V and we have

ZM_{Z}(f) = (\begin{matrix}B_{1} \\ & . & . & . \\ & & & & B_{t} \end{matrix})

with

B_{i} = (\begin{matrix}0 & ⋯ & ⋯ & 0 & −a_{0}^{(i)} \\ 1 & ... & & ⋮ & ⋮ \\ 0 & ... & ... & ⋮ & ⋮ \\ ⋮ & ... & ... & 0 & ⋮ \\ 0 & ⋯ & 0 & 1 & −a_{n_{i}−1}^{(i)} \end{matrix}) ∈ M_{n_{i}}(F) , i = 1, …, t,

if p_{i}(λ)^{α_{i}} = λ^{n_{i}} + a_{n_{i}−1}^{(i)} + ⋯ + a_{0}^{(i)}. Any matrix like the B_{i}s, such that the polynomial formed from the last column is a power of an irreducible polynomial of R is called an F -rational block. Any block diagonal matrix with F -rational blocks is called F -rational. The matrix _{Z}M_{Z}(f) is called the rational normal form of f or A. One can show that up to reordering the blocks B_{i} it is uniquely determined by f or A. That means, if S^{−1}AS is again F -rational for some S ∈ GL_{n}(F), then the blocks of S^{−1}AS coincide with the B_{1}, …, B_{t} above.

(d) If p_{i}(λ)^{α_{i}} = (λ − r_{i})^{n_{i}} for each i = 1, …, t and some r_{1}, …, r_{t} ∈ F (note that this always is the case for F = ℂ), and if z_{1}, …, z_{t} ∈ V are as in (c), then we choose the F -basis

Y = (z_{1}, (λ − r_{1})z_{1}, (λ − r_{1})^{2}z_{1}, …, (λ − r_{1})^{n_{1}−1}z_{1}, z_{2}, (λ − r_{2})z_{2}, …) = (z_{1}, (f − r_{1}id_{V})(z_{1}), (f − r_{1}id_{V})(z_{1}), …)

of V and obtain

Y M_{Y} (f) = (\begin{matrix}J_{1} \\ & . & . & . \\ & & & & J_{t} \end{matrix})

with

J_{i} = (\begin{matrix}r_{i} \\ 1 & ... \\ & ... & ... \\ & & 1 & r_{i} \end{matrix}) ∈ M_{n_{i}}(F) , i = 1, …, t.

Matrices of the form J_{i} are called Jordan blocks, matrices of the form _{Y} M_{Y} (f) are called Jordan matrices and _{Y} M_{Y} (f) is called the Jordan normal form of f or A. Again, the Jordan normal form of A or f is unique up to permutation of the blocks in the same sense as the rational normal form. Note that the rational and Jordan (if existent) normal forms are already determined by the R-module structure of V.

3.3 Theorem The invariant factors of the R-module V are the monic non-unit invariant factors of λ · 1_{n} − A ∈ M_{n}(R).

Proof According to the proof of Corollary 2.6, the invariant factors of V are the non-unit invariant factors of the matrix representing the inclusion ker(ρ) ⊆ R^{(m)}, where ρ: R^{(m)} → V is any surjective R-linear map. We choose

ρ: R^{(n)} → V , e_{i} ↦ v_{i} ,

and claim that the elements

f_{i} := λe_{i} − ∑_{j=1}^{n} a_{ji}e_{j} ∈ R^{(n)} , i = 1, …, n ,

form an R-basis of ker(ρ). Then, the matrix representing the inclusion ker(ρ) ⊆ R^{(n)} with respect to (f_{1}, …, f_{n}) and (e_{1}, …, e_{n}) is just λ · 1_{n} − A, and the proof would be complete.

Since ρ(f_{i}) = f(v_{i}) − ∑_{j=1}^{n} a_{ji}v_{j} = 0, we have f_{1}, …, f_{n} ∈ ker(ρ). Since λe_{i} = f_{i}+∑_{j=1}^{n} a_{ji}e_{j}, we have Re_{i} ⊆ Rf_{i}+∑_{j=1}^{n} F e_{j}, and so R^{(n)} = ∑_{i=1}^{n} Rf_{i}+ ∑_{i=1}^{n} F e_{i}. If u ∈ ker(ρ) and u = ∑_{i=1}^{n} h_{i}(λ)f_{i} + ∑_{i=1}^{n} b_{i}e_{i}, then applying ρ to this equation gives 0 = ρ(∑_{i=1}^{n} b_{i}e_{i}) = ∑_{i=1}^{n} b_{i}v_{i}. But this implies b_{1} = ⋯ = b_{n} = 0, and so u ∈ ⟨f_{1}, …, f_{n}⟩_{R}. Moreover, if ∑_{i=1}^{n} h_{i}(λ)f_{i} = 0 with h_{i}(λ) ∈ R, then

∑_{i=1}^{n} λh_{i}(λ)e_{i} = ∑_{i=1}^{n} h_{i}(λ) ∑_{j=1}^{n} a_{ji}e_{j} ,

and so

λh_{k}(λ) = ∑_{i=1}^{n} h_{i}(λ)a_{ki} ,

for each k = 1, …, n. If one of the h_{k}(λ)s is non-zero and if we choose l ∈ {1, …, n} such that h_{l}(λ) has maximal degree, then we obtain λh_{l}(λ) = ∑_{i=1}^{n} h_{i}(λ)a_{li} which is impossible, since the two sides have different degrees.

3.4 Remark In order to find an F -basis W = (w_{1}, …, w_{n}) of V such that W M_{W} (f) is a Jordan matrix or an F -rational matrix, we have to determine T in

T^{−1}(λ1_{n} − A)S = D = (\begin{matrix}1 \\ & . & . & . \\ & & & & 1 \\ & & & & & d_{1}(λ) \\ & & & & & & . & . & . \\ & & & & & & & & & d_{s}λ \end{matrix})

with non-constant polynomials d_{1}(λ) | ⋯ | d_{s}(λ) in R, and with S, T ∈ GL_{n}(R). Then the elements u_{1}, … u_{s} ∈ V which are determined by

(v_{1}, …, v_{n})T = (∗, …, ∗, u_{1}, …, u_{s})

generate V as R-module such that V = Ru_{1}⊕⋯⊕Ru_{s} with ann_{R}(u_{i}) = d_{i}(λ)R, so that Ru_{i} ≅ R/d_{i}(λ)R under u_{i} ↔ 1 + d_{i}(λ)R. Going through the proof of the Chinese Remainder Theorem for each R/d_{i}(λ)R then gives the elements z_{1}, …, z_{t} ∈ V (t ⩾ s) as in Remark 3.2 (c), and we can continue as in Remark 3.2.

3.5 Remark (a) Let d_{1}(λ) | ⋯ | d_{s}(λ) be the invariant factors of λ1_{n} − A ∈ M_{n}(R). Then d_{s}(λ) is the minimal polynomial m_{A}(λ) of A ( since g(A) = 0 ⟺ g(f)v_{i} = 0 for all i = 1, …, n ⟺ g(λ)V = 0 ⟺ g(λ) ∈ ann_{R}(V) = d_{s}(λ)R ⟺ d_{s}(λ) | g(λ)). Moreover, the product d_{1}(λ) ⋯ d_{s}(λ) is the characteristic polynomial ch_{A}(λ) of A (since det(diag(1, …, 1, d_{1}(λ), …, d_{s}(λ))) and det(λ1_{n} − A) are associates by Theorem 3.3 and both are monic). In particular this shows that m_{A}(λ) divides ch_{A}(λ) and that they have the same irreducible factors.

(b) We call two matrices A, B ∈ M_{n}(F) similar if there exists S ∈ GL_{n}(F) such that B = S^{−1}AS. It is not difficult to show that the following are equivalent:

(i) A and B are similar.

(ii) A and B have the same rational normal form (up to reordering the blocks).

(iii) λ1_{n} − A and λ1_{n} − B have the same monic invariant factors.

If F = ℂ, the rational normal form in (ii) can be replaced by the Jordan normal form.

II. Field Extensions and Galois Theory

1 Questions

We will address the following questions in the course of this chapter:

• Are there formulae giving the solutions of a polynomial equation X^{n} + a_{n−1}X^{n−1} + ⋯ a_{1}X + a_{0} (with a_{0}, …, a_{n−1} from a field) in terms of the operations +, −, ·, /, and ^{k} √. Note that we can do this in the case n = 2: The solutions of X^{2} + aX + b are

\frac{1}{2}(−a ± \sqrt{a^{2} − 4b}).

• Let F ⊆ E be fields. Are there finitely or infinitely many intermediate fields between E and F? How can we determine them?

• Geometric problems: Suppose we have only a straight edge and a compass.

– Can we square the circle?

– Can we trisect any given angle?

– Can we construct a regular n-gon?

– Can we double a cube?

– What numbers in the complex plane can we construct, starting from 0 and 1?

2 Basic Definitions and Properties

2.1 Definition Let F ⊆ E be fields (we also say that E/F is a field extension) and let x ∈ E. If there exists 0 ≠ f ∈ F [X] such that f(x) = 0, we call x algebraic over F; otherwise we call x transcendental over F. If x is algebraic over F, the polynomials g ∈ F [X] with g(x) = 0 form a non-zero ideal of F [X]. The monic generator min_{x/F} ∈ F [X] of this ideal is called the minimal polynomial of x over F. It has the property that min_{x/F} (x) = 0 and for every g ∈ F [X] with g(x) = 0 one has min_{x/F} | g. Obviously, min_{x/F} is irreducible in F [X].

2.2 Example The element \sqrt{2} is algebraic over ℚ with minimal polynomial min_{\sqrt{2}/ℚ} = X^{2} − 2. The minimal polynomial of 5 over ℚ is X − 5.

2.3 Remark Let E/F be a field extension.

(a) Recall that if there exists an n ∈ ℕ such that n · 1_{F} = 1 + ⋯ + 1 = 0, then the minimal such n is a prime p, called the characteristic of F and denoted by char(F). If no such n exists, then we set char(F) = 0.

(b) Note that E is an F -vector space. We set [E : F] := dim_{F} E and call [E : F] the degree of the extension E/F. It is easy to see that if E′/E is a further field extension, if {x_{i} | i ∈ I} is an F -basis of E, and if {y_{j} | j ∈ J} is an E-basis of E′, then {x_{i}y_{j} | (i, j) ∈ I × J} is an F -basis of E′. In particular, E′/F is of finite degree if and only if E′/E and E/F are of finite degree. In this case we have [E′ : F] = [E′ : E][E : F].

(c) For x ∈ E we write F [x] for the smallest subring of E containing F and x. It is easy to see that this is the set

{f(x) | f ∈ F [X]} .

Moreover, we write F (x) for the smallest subfield of E containing F and x. It is easy to see that this is the set

{f(x)/g(x) | f, g ∈ F [X], g(x) ≠ 0} .

More generally, for any subset {x_{i} | i ∈ I} of E we define F [x_{i} | i ∈ I] (resp. F (x_{i} | i ∈ I)) to be the smallest subring (resp. subfield) of E containing F and {x_{i} | i ∈ I}. These are given by

{f(x_{i_{1}}, …, x_{i_{n}}) | n ∈ ℕ, i_{1}, …, i_{n} ∈ I, f ∈ F [X_{1}, …, X_{n}]}

and

{f(x_{i_{1}}, …, x_{i_{n}})/g(x_{j_{1}}, …, x_{j_{m}}) | m, n ∈ ℕ, i_{1}, …, i_{n}, j_{1}, …, j_{m} ∈ I, f ∈ F [X_{1}, …, X_{n}], g ∈ F [X_{1}, …, X_{m}], g(x_{j_{1}}, …, x_{j_{m}}) ≠ 0} .

(d) Note that a field F has only the ideals 0 and F. Therefore, each ring homomorphism f: F → R to any ring R is injective (f(1) = 1).

2.4 Theorem Let E/F be a field extension and let x ∈ E be algebraic with minimal polynomial m(X) = X^{n} + a_{n−1}X^{n−1} + … + a_{1}X + a_{0}.

(a) The map ρ: F [X] → F [x], f ↦ f(x), is a surjective ring homomorphism with kernel (m(X)).

(b) The ring F [x] is a field; in particular, F [x] = F (x).

(c) We have [F (x) : F] = n, and 1, x, x^{2}, …, x^{n−1} is an F -basis of F (x).

Proof (a) It is easy to check that ρ is a ring homomorphism. It is surjective by the description of F [x] in Remark 2.3 (c).

(b) By (a) we have a ring isomorphism ρ̄: F [X]/(m(X)) → F [x]. Since m(X) is irreducible, both rings are fields.

(c) We know that 1+(m(X)), X +(m(X)), …, X^{n−1}+(m(X)) is an F -basis of F [X]/(m(X)). Since ρ̄ is also F -linear, ρ̄(X^{i}+(m(X))) = x^{i}, i = 0, …, n−1, form an F -basis of F (x).

2.5 Definition A field extension E/F is called algebraic, if every element of E is algebraic over F.

2.6 Theorem Let E/F be a field extension.

(a) For x ∈ E we have: x is algebraic over F ⟺ [F (x) : F] < ∞.

(b) If [E : F] < ∞, then E is algebraic over F.

(c) If E = F (x_{i} | i ∈ I) with elements x_{i} ∈ E, i ∈ I, which are algebraic over F, then E is algebraic over F.

Proof (a) ⇒: This follows from Theorem 2.4. ⇐: If [F (x) : F] = n, then the set 1, x, x^{2}, …, x^{n} is F -linearly dependent and we obtain a relation a_{n}x^{n} + ⋯ a_{1}x + a_{0} = 0 with a_{i} ∈ F, not all of them being zero.

(b) This is clear with (a).

(c) Let y ∈ E = F (x_{i} | i ∈ I). Then y ∈ F (y_{1}, …, y_{k}) for some finite subset {y_{1}, …, y_{k}} of {x_{i} | i ∈ I}. Since y_{i} is algebraic over F it is also algebraic over F (y_{1}, …, y_{i−1}), i = 1, …, k. Therefore, [F (y_{1}, …, y_{i}) : F (y_{1}, …, y_{i−1})] is finite for all i = 1, …, k by (a). Now Remark 2.3 (b) implies that [F (y_{1}, …, y_{k}) : F] is finite. Hence, y is algebraic over F by (b).

3 The Field of Constructible Elements

3.1 Definition Let z_{1}, …, z_{k} ∈ ℂ. We set C_{0}(z_{1}, …, z_{k}) := {0, 1, z_{1}, …, z_{k}} and define C_{n+1}(z_{1}, …, z_{k}) ⊆ ℂ, for n ∈ ℕ_{0}, recursively as the union of C_{n}(z_{1}, …, z_{k}) and the complex numbers arising as intersections of

(a) two different lines, each one drawn through two elements of C_{n}(z_{1}, …, z_{k}), or

(b) two different circles, each one having an element of C_{n}(z_{1}, …, z_{k}) as center and the distance between two elements of C_{n}(z_{1}, …, z_{k}) as radius, or (c) a line and a circle as in (a) and (b).

We call C(z_{1}, …, z_{k}) := ⋃_{n∈ℕ_{0}} C_{n}(z_{1}, …, z_{k}) the set of elements constructible from {z_{1}, …, z_{k}}. We will write just C_{n} and C if k = 0 (i.e., if we start with C_{0} = {0, 1}).

3.2 Theorem Let z_{1}, …, z_{k} ∈ ℂ. The set C(z_{1}, …, z_{k}) is the smallest subfield E of ℂ which contains z_{1}, …, z_{k} and is closed under complex conjugation (z ∈ E ⇒ z̄ ∈ ℂ) and closed under taking square roots (z ∈ ℂ, z^{2} ∈ E ⇒ z ∈ E).

Proof (a) C(z_{1}, …, z_{k}) is a field: Let x, y ∈ C(z_{1}, …, z_{k}). It is not difficult to show that x + y, −x, x · y, and x^{−1} (if x ≠ 0) arise after finitely many of the steps (a)-(c) of Definition 3.1. For details see [J, pp. 217–218].

(b) C(z_{1}, …, z_{k}) is closed under complex conjugation, since we can construct the reflection of a given point at the real axis.

(c) C(z_{1}, …, z_{k}) is closed under taking square roots: Suppose z ∈ ℂ with z^{2} = x ∈ C(z_{1}, …, z_{k}). Then z arises from bisecting one of the two angles between the positive real ray and the ray with origin in 0 through x, and by taking the square root of the length of x. Both can be constructed.

(d) Let E ⊆ ℂ be a subfield which contains z_{1}, …, z_{k} and is closed under complex conjugation and under taking square roots. Then i ∈ E, since i^{2} = −1 ∈ E. Moreover, for a = α_{1} + α_{2}i ∈ ℂ with α_{1}, α_{2} ∈ ℝ, we have: a ∈ E ⟺ α_{1} ∈ E and α_{2} ∈ E. In fact α_{1} = (a + ā)/2 and α_{2} = (a − ā)/2i. We will show by induction on n ∈ ℕ_{0} that C_{n}(z_{1}, …, z_{k}) ⊆ E. Obviously, C_{0}(z_{1}, …, z_{k}) = {0, 1, z_{1}, …, z_{k}} ⊆ E. Assume that C_{n}(z_{1}, …, z_{k}) ⊆ E. Any line through elements of C_{n}(z_{1}, …, z_{k}) is given as the set of solutions x = ξ_{1} + ξ_{2}i of the equation ξ_{2} = αξ_{1} + β or as the set of solutions of the equation ξ_{1} = γ for some α, β, γ ∈ E ∩ ℝ. Note that the distance ρ between any two elements of C_{n}(z_{1}, …, z_{k}) is in E. Therefore, an arbitrary circle with center at m = μ_{1} + μ_{2}i ∈ C_{n}(z_{1}, …, z_{k}) and radius ρ which is the distance between two points of C_{n}(z_{1}, …, z_{k}) is the set of solutions x = ξ_{1} + ξ_{2}i of the equation (ξ_{1} − μ_{1})^{2} + (ξ_{2} − μ_{2})^{2} = ρ^{2}. Therefore, the intersections of any two such lines or circles are solutions of linear or quadratic equations with coefficients from E. Therefore, its solutions are still elements in E and we have C_{n+1}(z_{1}, …, z_{k}) ⊆ E.

3.3 Definition Let E/F be a field extension. A square root tower in E over F is a finite series of field extensions

F = F_{0} ⊆ F_{1} ⊆ ⋯ ⊆ F_{l} ⊆ E

such that there exist elements u_{i} ∈ F_{i} with F_{i} = F_{i−1}(u_{i}) and u_{i}^{2} ∈ F_{i−1} for all i = 1, …, l. In particular, F_{i} = F (u_{1}, …, u_{i}) and [F_{i} : F] | 2^{i} for all i = 1, …, l.

3.4 Theorem Let z_{1}, …, z_{k} ∈ ℂ and set F := ℚ(z_{1}, …, z_{k}, z̄_{1}, …, z̄_{k}). For z ∈ ℂ the following are equivalent:

(i) z ∈ C(z_{1}, …, z_{k}).

(ii) There exists a square root tower F = F_{0} ⊆ … ⊆ F_{l} in ℂ over F such that z ∈ F_{l}.

Proof (ii) ⇒ (i): Let F = F_{0} ⊆ … ⊆ F_{l} be a square root tower as in (ii). Then F_{l} is contained in every subfield E of ℂ which contains z_{1}, …, z_{k} and which is closed under complex conjugation and under taking square roots. Hence, by Theorem 3.2, F_{l} is contained in C(z_{1}, …, z_{k}).

(i) ⇒ (ii): Let E denote the set of elements z ∈ ℂ which satisfy (ii). By Theorem 3.2, it suffices to show that E is a field which contains z_{1}, …, z_{k} and which is closed under complex conjugation and under taking square roots. Obviously, E contains z_{1}, …, z_{k}. Moreover, if x, y ∈ E and if x ∈ F_{l}, y ∈ E_{m}, with square root towers F = F_{0} ⊆ ⋯ ⊆ F_{l} and F = E_{0} ⊆ ⋯ ⊆ E_{m} such that F_{i} = F_{i−1}(u_{i}), E_{j} = E_{j−1}(v_{j}) for elements u_{i}, v_{j} ∈ E satisfying u_{i}^{2} ∈ F_{i−1} and v_{j}^{2} ∈ E_{j−1} for all 1 ⩽ i ⩽ l and 1 ⩽ j ⩽ m, then F = F_{0} ⊆ ⋯ ⊆ F_{l} ⊆ F_{l}(v_{1}) ⊆ F_{l}(v_{1}, v_{2}) ⊆ ⋯ ⊆ F_{l}(v_{1}, …, v_{m}) is a square root tower containing x+y, xy, −x, and x^{−1} (if x ≠ 0). This shows that E is a field. E is also closed under taking square roots, since we just have to extend a square root tower by one step. To prove that E is closed under taking complex conjugation, assume that z ∈ F_{l} and that F = F_{0} ⊆ ⋯ ⊆ F_{l} ⊆ ℂ is a square root tower with F_{i} = F_{i−1}(u_{i}) and u_{i}^{2} ∈ F_{i−1} for all i = 1, …, l. Then, z̄ ∈ F̄_{l} and F = F̄ = F̄_{0} ⊆ ⋯ ⊆ F̄_{l} ⊆ ℂ is a square root tower with F̄_{i} = F̄_{i−1}(ū_{i}) and ū_{i}^{2} ∈ F̄_{i−1} for all i = 1, …, l.

3.5 Corollary Let z_{1}, …, z_{k} ∈ ℂ and assume that z ∈ C(z_{1}, …, z_{k}). Then [F (z) : F] is a power of 2, where F = ℚ(z_{1}, …, z_{k}, z̄_{1}, …, z̄_{k}).

Proof This follows immediately from Theorem 3.4.

3.6 Theorem The duplication of the cube is impossible.

Proof Assume that it is possible. Then \sqrt[3]{2} ∈ C and [ℚ(\sqrt[3]{2}) : ℚ] = deg(min_{\sqrt[3]{2}/ℚ}(X)) is be a power of 2 by Corollary 3.5. But min_{\sqrt[3]{2}/ℚ}(X) = X^{3}−2 since it is irreducible in ℤ[X] (by Eisenstein’s criterion), and then also in ℚ[X] (cf. [J, 2.16, Lemma 3]). This is a contradiction.

3.7 Theorem It is not possible to trisect an arbitrary angle.

Proof Assume that this is possible. Then ζ_{18} := cos 20^{∘} + i sin 20^{∘} ∈ C, since ζ_{6} = cos 60^{∘} + i sin 60^{∘} ∈ C. But with ζ_{18} also a := cos 20^{∘} = (ζ_{18} − ζ̄_{18})/2 ∈ C. Since cos(3θ) = 4 cos^{3} θ − 3 cos θ, we have 4a^{3} − 3a = cos 60^{∘} = 1/2. With a also b := 2a − 1 is in C and we have

0 = 4(\frac{b + 1}{2})^{3} − 3(\frac{b + 1}{2}) − \frac{1}{2} = \frac{1}{2}(b^{3} + 3b^{2} − 3) ,

and so b^{3} + 3b^{2} − 3 = 0. But X^{3} + 3X^{2} − 3 is irreducible in ℤ[X] by Eisenstein’s criterion, and then also in ℚ[X] (cf. [J, 2.16, Lemma 3]). Therefore, this is the minimal polynomial of b ∈ C over ℚ and [ℚ(b) : ℚ] = 3. This contradicts Corollary 3.5.

4 Splitting Fields and Normal Field Extensions

4.1 Definition Let F be a field. An F -algebra (R, i) is a ring R together with a ring homomorphism i: F → R such that i(F) ⊆ Z(R). An F -algebra homomorphism between two F -algebras (R, i) and (S, j) is a ring homomorphism φ: R → S such that f(i(α)) = j(α) for all α ∈ F. Note that R carries the structure of an F -vector space via α · r := i(α)r for α ∈ F and r ∈ R. Often we identify F via i with its image i(F) = F · 1_{R} in R. Then we have φ(α) = α for any F -algebra homomorphism φ and α ∈ F. Often we only write R instead of (R, i) if the map i is obvious from the context. We denote the set of F -algebra homomorphisms between R and S by Hom_{F −alg}(R, S). If (E, i) is an F -algebra and also a field, then we may consider E/F as a field extension by identifying F with i(F).

4.2 Examples (a) If E/F is a field extension, then E is an F -algebra.

(b) The polynomial ring F [X_{i} | i ∈ I] in the indeterminate X_{i}, i ∈ I, is an F -algebra.

(c) The matrix ring M_{n}(F) is an F -algebra with i(α) = α · 1_{n} for α ∈ F.

4.3 Theorem Let F be a field and let X_{i}, i ∈ I, be a family of indeterminates. If R is a commutative F -algebra and r_{i}, i ∈ I, is a family of elements in R, then there exists a unique φ: Hom_{F −alg}(F [X_{i} | i ∈ I], R) such that φ(X_{i}) = r_{i}.

Proof The uniqueness part follows from the fact that each element of F [X_{i} | i ∈ I] is an F -linear combination of products of the elements X_{i}, i ∈ I. In order to show the existence of φ, let K be the set of all functions k: I → ℕ_{0} such that k(i) ≠ 0 for only finitely many i ∈ I. Then the elements X^{k} := ∏_{i∈I} X_{i}^{k(i)}, k ∈ K form an F -basis of F [X_{i} | i ∈ I]. Therefore, there exists an F -linear map φ: F [X_{i} | i ∈ I] → R such that φ(X^{k}) = ∏_{i∈I} r_{i}^{k(i)}, for all k ∈ K. Now it is easy to see that

φ(X^{k}X^{l}) = ∏_{i∈I} r_{i}^{k(i)+l(i)} = φ(X^{k})φ(X^{l})

for all k, l ∈ K. Note that φ(1) = 1. For arbitrary elements ∑_{p=1}^{n} α_{p}X^{k_{p}} and ∑_{q=1}^{m} β_{q}X^{l_{q}} of F [X_{i} | i ∈ I] with k_{p}, l_{q} ∈ K and α_{p}, β_{q} ∈ F for 1 ⩽ p ⩽ n and 1 ⩽ q ⩽ m, we have:

φ((∑_{p=1}^{n} α_{p}X^{k_{p}})(∑_{q=1}^{m} β_{q}X^{l_{q}})) = ∑_{p=1}^{n} ∑_{q=1}^{m} α_{p}β_{q}φ(X^{k_{p}}X^{l_{q}})

= ∑_{p=1}^{n} ∑_{q=1}^{m} α_{p}β_{q}φ(X^{k_{p}})(φ(X^{l_{q}}) = φ(∑_{p=1}^{n} α_{p}X^{k_{p}})φ(∑_{q=1}^{m} β_{q}X^{l_{q}}) ,

which shows that φ is a ring homomorphism. Moreover, φ(α) = φ(α · 1) = αφ(1) = α1_{R}, for each α ∈ F. This implies that φ ∈ Hom_{F −alg}(F [X_{i} | i ∈ I], R).

4.4 Definition Let F be a field and let f ∈ F [X] with deg(f) = n > 0. An extension field E of F is called a splitting field of f over F if f decomposes as f(X) = α(X − r_{1}) ⋯ (X_{r} − n) in E[X] (with α ∈ F and r_{1}, …, r_{n} ∈ E) and if E = F (r_{1}, …, r_{n}). Note that then E/F is a finite (i.e., [E : F] < ∞) and algebraic field extension.

4.5 Remark Let F be a field and let f ∈ F [X] with deg(f) > 0. Then there exists an extension field E of F of finite degree such that f has a root r ∈ E. In fact, let p(X) = α_{n}X^{n} + … + α_{1}X + α_{0} be an irreducible factor of f in F [X]. Then E := F [X]/(p) can be considered as an extension field of F and for r := X + (p) ∈ E we have:

p(r) = α_{n}(X + (p))^{n} + … + α_{1}(X + (p)) + α_{0}
= α_{n}X^{n} + … + α_{0} + (p) = p(X) + (p) = 0 ∈ E .

Therefore, f(X) = g(X)(X − r) with some g ∈ E[X].

4.6 Theorem Let F be a field and let f ∈ F [X] with deg(f) > 0. Then there exists a splitting field E of f over F.

Proof Induction on n = deg(f). If n = 1 we may take E = F. So let deg(f) ⩾ 2. By Remark 4.5, there exists a finite extension field E_{1} of F such that f(X) = (X −r_{1})g(X) with g ∈ E_{1}[X] and r_{1} ∈ E_{1}. Since deg(g) < n, by induction there exists a finite field extension E_{2}/E_{1} such that g(X) = α(X − r_{2}) ⋯ (X − r_{n−1}) with r_{2}, …, r_{n} ∈ E_{2} and α ∈ F. Therefore, f(X) = α(X − r_{1}) ⋯ (X − r_{n}) in E_{2}(X) and E := F (r_{1}, …, r_{n}) is a splitting field of f over F.

4.7 Example A splitting field of f(X) = X^{3} − 2 over ℚ is E := ℚ(\sqrt[3]{2}, ζ_{3} \sqrt[3]{2}, ζ_{3}^{2} \sqrt[3]{2}) with ζ_{3} = e^{2πi/3} = (−1 + \sqrt{3}i)/2, since these are the roots of f(X). We have E = ℚ(\sqrt[3]{2}, \sqrt{3}i) and so [E : ℚ] = [E : ℚ(\sqrt[3]{2})][ℚ(\sqrt[3]{2}) : ℚ] = 2 · 3 = 6.

4.8 Remark Let E_{1}/F and E_{2}/F be two extension fields of F and let a ∈ E_{1}. Moreover, let φ ∈ Hom_{F −alg}(E_{1}, E_{2}). If f(X) := min_{a/F} (X) = X^{n} + α_{n−1}X^{n−1} + ⋯ + α_{0}, then

0 = φ(0) = φ(a^{n} + α_{n−1}a^{n−1} + ⋯ + α_{0})
= φ(a)^{n} + α_{n−1}φ(a)^{n−1} + ⋯ + α_{0} = f(φ(a)) .

Therefore, φ(a) ∈ E_{2} is also a root of f. In particular, if there exists an irreducible f ∈ F [X] which has a root in E_{1} but not in E_{2}, then Hom_{F −alg}(E_{1}, E_{2}) = ∅.

4.9 Lemma Let F and F̃ be fields and let φ: F → F̃ be a field isomorphism. Let E/F and Ẽ/F̃ be field extensions, let a ∈ E be algebraic over F with f(X) := min_{a/F} (X) = X^{n} + α_{n−1}X^{n−1} + ⋯ + α_{0}, and let ã ∈ Ẽ be algebraic over F̃ with min_{ã/F̃}(X) = f̃(X) := X^{n} + φ(α_{n−1})X^{n−1} + ⋯ + φ(α_{0}). Then there exists a unique ring homomorphism ψ: F (a) → F̃(ã) such that ψ(a) = ã and ψ(α) = φ(α) for all α ∈ F. Moreover, ψ is an isomorphisms.

Proof We extend φ: F → F̃ to the ring isomorphism

Φ: F [X] → F̃[X] , α_{n}X^{n} + ⋯ + α_{0} ↦ φ(α_{n})X^{n} + ⋯ + φ(α_{0}) .

Then f̃ = Φ(f), and Φ induces an isomorphism

Φ̄: F [X]/(f) → F̃[X]/(f̃) .

Composing Φ̄ with the ring isomorphisms ρ̄: F [X]/(f) ≅ F (a) and ¯ρ̃: F̃[X]/(f̃) ≅ F̃(ã) from Theorem 2.4, we obtain a ring isomorphism with the desired properties.

Since each element in F (a) is an F -linear combination of the elements 1, a, …, a^{n−1}, there exists at most one ring homomorphism ψ: F (a) → F̃(ã) such that ψ(a) = ã and ψ(α) = φ(α) for all α ∈ F.

4.10 Corollary Let E/F and Ẽ/F be field extensions with E = F (a) for some a ∈ E which is algebraic over F. Let f := min_{a/F} ∈ F [X] and let ã_{1}, …, ã_{n} be the roots of f in Ẽ (without repetition). Then, there exist exactly n F -algebra homomorphisms φ_{1}, …, φ_{n} : E → Ẽ and they are given by φ_{i}(a) = ã_{i}, for i = 1, …, n.

Proof This is immediate from Lemma 4.9 and Remark 4.8.

4.11 Theorem Let F, F̃ be fields, let φ: F → F̃ be a field isomorphism, let f(X) = α_{n}X^{n} + ⋯ + α_{1}X + α_{0} ∈ F [X] with α_{n} ≠ 0 and n > 0, and set f̃(X) := φ(α_{n})X^{n} + ⋯ + φ(α_{0}) ∈ F̃[X]. If E is a splitting field of f over F and if Ẽ is a splitting field of f̃ over F̃, then there exists a field isomorphism ψ: E → Ẽ such that ψ(α) = φ(α) for all α ∈ F. In particular, if F = F̃ and φ = id_{F}, we obtain that any two splitting fields of f over F are isomorphic as F -algebras.

Proof We proof the assertion by induction on m := [E : F]. If m = 1, then f(X) = α_{n}(X − r_{1}) ⋯ (X − r_{n}) with r_{1}, …, r_{n} ∈ F, f̃(X) = φ(α_{n})(X − φ(r_{1})) ⋯ (X − φ(r_{n})), and Ẽ = F̃. Hence, we may take ψ := φ. Now let m = [E : F] ⩾ 2. Then f(X) has an irreducible factor p(X) with deg(p) ⩾ 2, and p̃ := Φ(p) ∈ F̃[X] is an irreducible factor of f̃ (with Φ as in the proof of Lemma 4.9). With f also p decomposes into linear factors in E[X] and there exists a ∈ E with p(a) = 0. Similarly there exists ã ∈ Ẽ such that p̃(ã) = 0. By Lemma 4.9, there exists a field isomorphism σ: F (a) → F̃(ã) such that σ(α) = φ(α) for all α ∈ F. Obviously, E (resp. Ẽ) is still a splitting field of f(X) (resp. f̃) over F (a) (resp. F̃(ã)). By induction we obtain a field isomorphism ψ: E → Ẽ such that ψ(β) = σ(β) for each β ∈ F (a). Hence, ψ(α) = φ(α) for each α ∈ F.

4.12 Definition An algebraic field extension E/F is called normal, if each irreducible f ∈ F [X] which has a root in E splits completely into linear factors in E[X].

4.13 Theorem Let E/F be a finite field extension. The following are equivalent:

(i) E/F is normal.

(ii) E is a splitting field of some f ∈ F [X] over F.

(iii) For each field extension E′/E and each F -algebra homomorphism φ: E → E′, one has φ(E) ⊆ E.

Proof (i) ⇒ (ii): Let a_{1}, …, a_{n} ∈ E with E = F (a_{1}, …, a_{n}) and set f(X) := min_{a_{1}/F} (X) ⋯ min_{a_{n}/F} (X). Since E/F is normal, f decomposes in E[X] into linear factors. Moreover, E is obtained by adjoining the roots of f in E to F. Hence, E is a splitting field of f over F.

(ii) ⇒ (iii): Let f ∈ F [X] be as in (ii) and write

f(X) = α_{(}X − a_{1}) ⋯ (X − a_{n})

with α ∈ F and a_{1}, …, a_{n} ∈ E. Then E = F (a_{1}, …, a_{n}). Let E′/E be a field extension and let φ ∈ Hom_{F −alg}(E, E′). By Remark 4.8, for each i = 1, …, n, there exists π(i) ∈ {1, …, n} such that φ(a_{i}) = a_{π(i)}. In particular, φ(E) = φ(F (a_{1}, …, a_{n})) ⊆ F (a_{1}, …, a_{n}) = E.

(iii) ⇒ (i): Let a ∈ E and let f := min_{a/F} (X) ∈ F [X]. Let a_{1}, …, a_{n} ∈ E such that E = F (a_{1}, …, a_{n}) and let E′ be a splitting field of

g(X) := f(X) · min_{a_{1}/F} (X) ⋯ min_{a_{n}/F} (X)

over E. Let b ∈ E′ with f(b) = 0. Then, by Lemma 4.9, there exists an F - algebra isomorphism σ: F (a) → F (b) with σ(a) = b. Note that E′ is also a splitting field of g(X) over F, F (a), and F (b). By Theorem 4.11, σ can be extended to an F -algebra isomorphism ψ: E′ → E′. Let φ: E → E′ be the restriction of ψ to E. By (iii) we then have b = ψ(a) ∈ ψ(E) ⊆ E, so that f(X) splits in E[X] into linear factors.

4.14 Corollary Let E/F be a finite field extension. Then there exists a finite field extension E′/E such that E′/F is normal.

Proof Let a_{1}, …, a_{n} ∈ E such that E = F (a_{1}, …, a_{n}) and let E′ be a splitting field of f(X) := min_{a_{1}/F} (X) ⋯ min_{a_{n}/F} (X) over E. Then E′ is also a splitting field of f over F. Therefore, E′/F is normal by Theorem 4.13.

4.15 Example ℚ(\sqrt[3]{2})/ℚ is not normal, since \sqrt[3]{2} ∈ ℚ(\sqrt[3]{2}) is a root of the irreducible polynomial X^{3} − 2 ∈ ℚ[X], but the other roots ζ_{3} \sqrt[3]{2} and ζ_{3} \sqrt[3]{2} are not in ℚ(\sqrt[3]{2}). However, ℚ(\sqrt[3]{2}, \sqrt{3}i) is normal over ℚ, since it is a splitting field of X^{3} − 2 over ℚ (see Example 4.7).

5 Separable Field Extensions

5.1 Definition Let F be a field.

(a) An irreducible polynomial f ∈ F [X] is called separable, if f has deg(f) distinct roots in a splitting field of f over F. An arbitrary polynomial f ∈ F [X] ∖ F is called separable over F, if every irreducible factor of f in F [X] is separable.

(b) Let E/F be an algebraic field extension. An element a ∈ E is called separable over F, if min_{a/F} (X) is separable. The field extension E/F is called separable, if every element a ∈ E is separable over F. We call F perfect, if every algebraic extension E/F is separable.

(c) We call the F -linear map

\frac{d}{dX} : F [X] → F [X] , X^{n} ↦ nX^{n−1} ,

the formal derivation and f′(X) := \frac{d}{dX} (f(X)) the formal derivative of f(X) ∈ F [X]. It is straight forward to verify that (f + g)′ = f′ + g′, (αf)′ = α(f′), (f · g)′ = f′ · g + f · g′, and (f ∘ g)′ = (f′ ∘ g) · g′, for α ∈ F and f, g ∈ F [X]. Note that \frac{d}{dX} (X^{p}) = 0 if F has characteristic p.

5.2 Lemma Let F be a field and let f ∈ F [X] be irreducible. Then the following are equivalent:

(i) f is separable.

(ii) For every field extension E/F, f(X) decomposes in E[X] into distinct irreducible factors.

(iii) f and f′ are coprime in F [X].

(iv f′ ≠ 0.

Proof (i) ⇒ (ii): Let E/F be a field extension and suppose that f(X) = p(X)^{2}g(X) with p, g ∈ E[X] and p irreducible in E[X]. Then f(X) has less than deg(f) distinct roots in a splitting field of f over E, and then also in a splitting field over F, which is a contradiction.

(ii) ⇒ (iii): Let E be a splitting field of f over F and let f(X) = α(X − r_{1}) ⋯ (X − r_{n}) be a decomposition in E[X]. We have (by induction on n)

f′(X) = α ∑_{i=1}^{n}(X − r_{1}) ⋯ (X − r_{i−1})(X − r_{i+1}) ⋯ (X − r_{n}) .

Assume that f and f′ have a common irreducible factor p ∈ F [X]. Then p(r_{i}) = 0 for some i = 1 …, n and we obtain

0 = f′(r_{i}) = (r_{i} − r_{1}) ⋯ (r_{i} − r_{i−1})(r_{i} − r_{i+1}) ⋯ (r_{i} − r_{n}) .

This implies r_{i} = r_{j} for some j ∈ {1, …, n} with i ≠ j, which contradicts (ii). (iii) ⇒ (iv): Assume that f′ = 0. Then f is a common divisor of f and f′. (iv) ⇒ (i): Let E be a splitting field of f over F and let f = α(X−r_{1}) ⋯ (X−

r_{n}) with α ∈ F and r_{1}, …, r_{n} ∈ E. Assume that r_{i} = r_{j} for some i ≠ j from {1, …, n}. Then (X − r_{i}) is a common divisor of f and f′ in E[X], hence f′(r_{i}) = 0. But this implies that f = min_{r_{i}/F} (X) | f′. This is a contradiction, since deg(f′) < deg(f) and f′ ≠ 0.

5.3 Proposition Let F be a field and let f ∈ F [X].

(a) If char(F) = 0, then one has: f′ = 0 ⟺ f ∈ F.

(b) If char(F) = p > 0, then one has: f′ = 0 ⟺ f(X) = g(X^{p}) for some g ∈ F [X].

Proof Let f = ∑_{i=0}^{n} α_{i}X^{i}, so f′ = ∑_{i=1}^{n} α_{i}iX^{i−1}. Therefore, f′ = 0 if and only if α_{i}i = 0 for all i ⩾ 1. If char(F) = 0, then the last statement is equivalent to α_{i} = 0 for all i ⩾ 1. In the case of positive characteristic it is equivalent to α_{i} = 0 for all i ⩾ 1 which are not divisible by p, and this implies

f = α_{0} + α_{p}X^{p} + α_{2p}X^{2p} + ⋯ = g(X^{p})

with g = α_{0} + α_{p}X + α_{2p}X^{2} + · · ·. Conversely, if g ∈ F [X] is arbitrary, then g(X^{p}) ∈ F [X] has vanishing derivative by Definition 5.1.

5.4 Corollary Every field F of characteristic 0 is perfect.

Proof By Lemma 5.2 and Proposition 5.3 every f ∈ F [X] which is not constant is separable over F.

5.5 Definition Let F be a field of characteristic p > 0. Note that for α, β ∈ F we have

(α + β)^{p} = ∑_{i=0}^{p} (\begin{matrix}p \\ i \end{matrix}) α^{i} + β^{p−i} = α^{p}β^{p} ,

since (\begin{matrix}p \\ i \end{matrix}) is divisible by p for all i = 1, …, p−1. Therefore, the map φ: F → F,

α ↦ α^{p}, is a ring homomorphism which we call the Frobenius homomorphism of F.

5.6 Theorem Let F be a field of characteristic p > 0. Then F is perfect if and only if the Frobenius homomorphism is surjective.

Proof ⇒: Let α ∈ F. We have to show that α = β^{p} for some β ∈ F. Let E be a splitting field of f(X) := X^{p} − α over F, and let b ∈ E be a zero of f(X), i.e., α = b^{p}. Then min_{b/F} (X) divides X^{p} − α = X^{p} − b^{p} = (X − b)^{p}. Therefore, min_{b/F} (X) = (X − b)^{r} for some r ∈ {1, …, p}. But since b is separable over F, r = 1 and b ∈ F.

⇐: Let E/F be an algebraic field extension. Assume that some a ∈ E is not separable over F. Then min′_{a/F} (X) = 0 and so min_{a/F} (X) = α_{0} + α_{1}X^{p} + ⋯ + X^{np}. Let β_{i} ∈ F with α_{i} = β_{i}^{p} for each i = 0, …, n − 1. Then

min_{a/F} (X) = β_{0}^{p} + β_{1}^{p}X^{p} + ⋯ β_{n−1}^{p}X^{(n−1)p} + X^{np} =
(β_{0} + β_{1}X + ⋯ + β_{n−1}X^{n−1} + X^{n})^{p}

contradicting the irreducibility of min_{a/F} (X).

5.7 Corollary Every finite field F is perfect.

Proof The Frobenius homomorphism φ: F → F is injective and then also surjective, since F is finite.

5.8 Example Let F := 𝔽_{2}(T) be the function field over the field with 2 elements in the indeterminate T. Then f(X) := X^{2} − T ∈ F [X] is irreducible. In fact, if not, then there exists α ∈ F with T = α^{2} = p^{2}/q^{2} for some p, q ∈ 𝔽_{2}[X], q ≠ 0. But then, T q(T)^{2} = p(T)^{2}, which is impossible, since T q(T)^{2} has odd degree and p(T)^{2} has even degree. Therefore, the splitting field of f(X) over F is not separable, i.e., F (\sqrt{T})/F is not separable.

5.9 Theorem Let F be a field and let G ⩽ F^{×} be a finite subgroup of the unit group F^{×} := F ∖ {0} of F. Then G is cyclic. In particular, if F is finite, then F^{×} is cyclic.

Proof Let |G| = p_{1}^{α_{1}} ⋯ p_{t}^{α_{t}} be the prime factor decomposition of |G| and let P_{i} ⩽ G be the Sylow p_{i}-subgroups of G. Since G is Abelian, we have G ≅ P_{1} × ⋯ × P_{t}, and by the Chinese Remainder Theorem it suffices to show that each P_{i} is cyclic. Assume that P_{i} is not cyclic. Then g^{p_{i}^{α_{i}−1}} = 1 for each g ∈ P_{i}. But then the polynomial X^{p_{i}^{α_{i}−1}} − 1 ∈ F [X] has at least p_{i}^{α_{i}} different roots in F, which is a contradiction.

5.10 Theorem Let E/F be a finite separable field extension. Then there exists a primitive element for E/F, i.e. an element a ∈ E such that E = F (a).

Proof If F is finite, also E is finite and E = F (a) for every a ∈ E^{×} with ⟨a⟩ = E^{×}. Therefore, we assume that |F | = ∞ and write E = F (x_{1}, …, x_{n}) with x_{1}, …, x_{n} ∈ E and minimal n ∈ ℕ. Assume that n > 1. Set a := x_{1} and b := x_{2}, let E_{1} be a splitting field of min_{a/F} · min_{b/F} over E, and write

m_{a/F} (X) = ∏_{i=1}^{r}(X − a_{i}), m_{b/F} (X) = ∏_{j=1}^{s}(X − b_{j}),

with a = a_{1}, …, a_{r}, b = b_{1}, …, b_{s} ∈ E_{1}. Since b is separable over F, the elements b_{1}, …, b_{s} are pairwise distinct, and there exists γ ∈ F such that γ ≠ \frac{a_{i}−a}{b−b_{j}} for each i = 1, …, r and j = 2, …, s. Set c := a + γb = a_{1} + γb_{1}. Then F (c) ⊆ F (a, b). Since min_{a/F} (c − γb) = 0, the polynomial (X − b) divides min_{a/F} (c − γX) and min_{b/F} (X) in F (a, b)[X]. By the choice of γ, c − γb_{j} = a + γ(b − b_{j}) ≠ a_{i} for any i = 1, …, r and j = 1, …, s. Therefore, min_{a/F} (c − γb_{j}) ≠ 0 for all j = 2, …, s, and X − b is a gcd of min_{b/F} (X) and min_{a/F} (c − γX) in F (a, b)[X]. But min_{b/F} (X) and min_{a/F} (c − γX) are contained in F (c)[X], and so X − b ∈ F (c)[X] (use the Euclidean algorithm or see Exercise 39). Therefore, b ∈ F (c) and also a = c − γb ∈ F (c). This implies F (a, b) ⊆ F (c) and E = F (c, x_{3}, …, x_{n}), contradicting the minimality of n.

5.11 Lemma Let E/F be an algebraic field extension, assume that char(F) = p > 0, and let a ∈ E. Then F (a^{p}) ⊆ F (a), and one have:

(a) If a is separable over F, then a ∈ F (a^{p}) and F (a) = F (a^{p}).

(b) If a is not separable over F, then [F (a) : F (a^{p})] = p.

Proof Obviously F (a^{p}) ⊆ F (a).

(a) Assume that a is separable over F. Set E′ := F (a^{p}) ⊆ F (a). Since min_{a/E′} | min_{a/F} and since min_{a/F} is separable, also min_{a/E′} is separable. Clearly, min_{a/E′}(X) | X^{p} − a^{p} ∈ F (a)[X]. This implies that min_{a/E′} = X − a and a ∈ E′.

(b) Assume that a is not separable over F. Then min′_{a/F} (X) = 0 by Lemma 5.2, and min_{a/F} (X) = g(X^{p}) for some monic irreducible g ∈ F [X] by Proposition 5.3. Since g(a^{p}) = min_{a/F} (a) = 0, we have g = min_{a^{p}/F} (X). This implies that

[F (a) : F (a^{p})][F (a^{p}) : F] = [F (a) : F] = deg(min_{a/F})
= p deg(min_{a^{p}/F}) = p[F (a^{p}) : F] ,

and so [F (a) : F (a^{p})] = p.

5.12 Theorem Let E/F be an algebraic field extension and let a_{1}, …, a_{n} ∈ E. Assume that a_{i} is separable over F (a_{1}, …, a_{i−1}) for each i = 1, …, n. Then F (a_{1}, …, a_{n})/F is separable.

Proof We may assume that char(F) = p > 0. Set F_{0} := F and F_{i} := F_{i−1}(a_{i}) for i = 1, …, n. Then F_{i} = F_{i−1}(a_{i}^{p}) for each i = 1, …, n by Lemma 5.10. Therefore, each element of

F_{n} = F_{n−1}(a_{n}^{p}) = F_{n−2}(a_{n−1}^{p}, a_{n}^{p}) = ⋯ = F (a_{1}^{p}, …, a_{n}^{p})

is an F -linear combination of p-powers of elements in F_{n}. Let b ∈ F_{n} and set k := deg(min_{b/F} (X)) = [F (b) : F]. We will show that F (b) = F (b^{p}) which implies that b is separable over F by Lemma 5.11. For that purpose we extend the F -basis b_{1} := 1, b_{2} := b, …, b_{k} := b^{k−1} of F (b) to an F -basis b_{1}, …, b_{l} of F_{n}. Let x ∈ F_{n} be arbitrary. Then there exist ξ_{1}, …, ξ_{l} ∈ F with x = ξ_{1}b_{1} + ⋯ + ξ_{l}b_{l} and we have x^{p} = ξ_{1}^{p}b_{1}^{p} + ⋯ + ξ_{l}^{p}b_{l}^{p}. Since every element of F_{n} is an F -linear combination of elements of the form x^{p}, the elements b_{1}^{p}, …, b_{l}^{p} span F_{n} as an F -vector space, Therefore they form an F -basis of F_{n}. In particular, 1, b^{p}, …, b^{p(k−1)} are F -linearly independent and we obtain k = [F (b) : F] ⩾ [F (b^{p}) : F] ⩾ k, which implies [F (b) : F] = [F (b^{p}) : F] and F (b) = F (b^{p}).

5.13 Corollary Let E/F be an algebraic field extension. If E = F (a_{1}, …, a_{n}) for elements a_{1}, …, a_{n} ∈ E which are separable over F, then E/F is separable.

Proof If a_{i} is separable over F, so also over F (a_{1}, …, a_{i−1}), since min_{a_{i}/F (a_{1},... ,a_{i−1})} | min_{a_{i}/F}. Now Theorem 5.12 applies.

6 Galois Extensions

6.1 Definition For a field extension E/F we denote by Aut(E/F) the set of F -algebra isomorphisms φ: E → E. Note that Aut(E/F) is a group under composition.

6.2 Remark In the above situation, Aut(E/F) is contained in Hom_{F} (E, E), the group of F -linear homomorphisms from E to E. Note that Hom_{F} (E, E) is an E-vector space by setting (x · f)(y) := x(f(y)) for f ∈ Hom_{F} (E, E) and x, y ∈ E. It is not difficult to see that if E/F has finite degree n, then dim_{E}(Hom_{F} (E, E)) = n (see Exercise 42).

6.3 Lemma (Dedekind) For every field extension E/F, the set Aut(E/F) is E-linearly independent in the E-vector space Hom_{F} (E, E). In particular, if [E : F] is finite, then |Aut(E/F)| ⩽ [E : F].

Proof Assume that Aut(E/F) is not E-linearly independent and let a_{1}φ_{1} + ⋯ + a_{n}φ_{n} = 0 be a non-trivial relation with a_{1}, … a_{n} ∈ E and φ_{1}, …, φ_{n} ∈ Aut(E/F) such that n is minimal. Then n ⩾ 2, a_{1} ≠ 0, and there exists y ∈ E with φ_{1}(y) ≠ φ_{n}(y). We have for each x ∈ E:

0 = a_{1}φ_{1}(xy) + ⋯ + a_{n}φ_{n}(xy) = a_{1}φ_{1}(y)φ_{1}(x) + ⋯ + a_{n}φ_{n}(y)φ_{n}(x)

and

0 = φ_{n}(y)(a_{1}φ_{1}(x) + ⋯ + a_{n}φ_{n}(x)) = a_{1}φ_{n}(y)φ_{1}(x) + ⋯ + a_{n}φ_{n}(y)φ_{n}(x) .

Subtraction yields a shorter relation

0 = (b_{1}φ_{1} + ⋯ + b_{n−1}φ_{n−1})(x)

for all x ∈ E with b_{i} := a_{i}(φ(y) − φ_{n}(y)) for i = 1, …, n − 1 and b_{1} ≠ 0. This contradicts the minimality of n.

6.4 Remark Let E/F be a field extension and let X ⊆ Aut(E/F) be a subset. Then the set

Fix_{E}X := {a ∈ E | φ(a) = a for all φ ∈ X}

is a subfield of E containing F. Conversely, if F′ is an intermediate field in E/F, then Aut(E/F′) is a subgroup of Aut(E/F). Therefore we obtain maps between the set of intermediate fields F′ in E/F and the set of subgroups H of G := Aut(E/F):

Fix_{E}: H ↦ Fix_{E}(H) ,
Aut(E/−): F′ ↦ Aut(E/F′) .

Obviously, we have

F′ ⊆ Fix_{E}(Aut(E/F′)) , H ⩽ Aut(E/F ix_{E}(H)) ,  (6.4.a)

for each intermediate field F′ of E/F and each subgroup H ⩽ G. Moreover, if F′ ⊆ F′′ are intermediate fields of E/F and if H ⩽ K are subgroups of G, then

Aut(E/F′′) ⩽ Aut(E/F′) and Fix_{E}(K) ⊆ Fix_{E}(H) .  (6.4.b)

Note that (6.4.a) and (6.4.b) imply

Fix_{E}(Aut(E/F ix_{E}(H))) = Fix_{E}(H)  (6.4.c)

and

Aut(E/F ix_{E}(Aut(E/F))) = Aut(E/F) .  (6.4.d)

6.5 Definition A finite field extension E/F is called a Galois extension if F = Fix_{E}(Aut(E/F)). In this case, we call Aut(E/F) the Galois group of E/F and denote it by Gal(E/F).

6.6 Example (a) ℚ(\sqrt{2})/ℚ is a Galois extension, since Aut(ℚ(\sqrt{2})/ℚ) = \{1, τ\} with τ(\sqrt{2}) = −\sqrt{2}, and since Fix_{ℚ(\sqrt{2})}(\{1, τ\}) = \{a + b\sqrt{2} | a, b ∈ ℚ, a + b\sqrt{2} = a − b\sqrt{2}\} = ℚ.

(b) ℚ(\sqrt[3]{2})/ℚ is not a Galois extension, since Aut(ℚ(\sqrt[3]{2})/ℚ) = \{1\} and since Fix_{ℚ( \sqrt[3]{2})}(\{1\}) = ℚ(\sqrt[3]{2}) ≠ ℚ.

(c) ℂ/ℝ is a Galois extension, since Aut(ℂ/ℝ) = {1,¯·} and Fix_{ℂ}({1,¯·}) = ℝ.

6.7 Theorem For a finite field extension E/F the following are equivalent: (i) E/F is a Galois extension. (ii) E/F is separable and normal. (iii) E is a splitting field of some polynomial f ∈ F [X] which is separable

over F.

(iv) There exists a ∈ E such that E = F (a) and min_{a/F} (X) has [E : F] distinct roots in E.

(v) |Aut(E/F)| = [E : F].

Proof (i)⇒(ii): First we show that E/F is separable. Let a ∈ E and let {σ(a) | σ ∈ Aut(E/F)} = {a_{1}, …, a_{n}} such that a = a_{1}, …, a_{n} are pairwise distinct. Set f(X) := (X − a_{1}) ⋯ (X − a_{n}) ∈ E[X] and write f(X) = X^{n} + b_{n−1}X^{n−1} + ⋯ + b_{0} with b_{0}, …, b_{n−1} ∈ E. For σ ∈ Aut(E/F) denote by σ̃: E[X] → E[X] the induced F -algebra isomorphism. Since each σ ∈ Aut(E/F) just permutes a_{1}, …, a_{n}, we have σ̃(f) = f and therefore, σ(b_{i}) = b_{i} for all i = 1, …, n−1. By (i), this implies b_{0}, …, b_{n−1} ∈ F and min_{a/F} (X) | f. Therefore, a is separable over F.

By Theorem 5.10, there exists a ∈ E such that E = F (a). With the above considerations for this a, we see that E is a splitting field of min_{a/F} over F, and therefore, E/F is normal by Theorem 4.13.

(ii)⇒(iii): Since E/F is normal, it is a splitting field for some monic polynomial f(X) ∈ F [X]. Since E/F is separable, each of the roots of f(X) in E is separable over F. Therefore, their minimal polynomials (which are just the irreducible factors of f(X) in F [X]) are separable. By definition, also f(X) is separable over F.

(iii)⇒(iv): E/F is normal by Theorem 4.13. All the roots a_{1}, …, a_{n} ∈ E of f are separable over F and E = F (a_{1}, …, a_{n}). Therefore, E/F is separable by Corollary 5.12. By Theorem 5.10, there exists a ∈ E with E = F (a). Since E/F is normal, min_{a/F} (X) splits completely into linear factors in E[X]. They are pairwise distinct, since a is separable over F. Therefore, min_{a/F} (X) has deg(min_{a/F}) = [E : F] pairwise distinct roots in E.

(iv)⇒(v): Let a = a_{1}, …, a_{n} ∈ E be the distinct roots of min_{a/F} (X). Note that n = [E : F]. By Corollary 4.10, there exist exactly n F -algebra homomorphisms φ_{1}, …, φ_{n}: E → E, given by φ_{i}(a) = a_{i}, for i = 1, …, n. Since they are injective, they are also surjective, and |Aut(E/F)| = n = [E : F].

(v)⇒(i): (6.4.d), Lemma 6.3, and (6.4.a) imply

|Aut(E/F)| = |Aut(E/Fix_{E}(Aut(E/F)))| ⩽
⩽ [E : Fix_{E}(Aut(E/F))] ⩽ [E : F] = |Aut(E/F)| .

Therefore, [E : Fix_{E}(Aut(E/F))] = [E : F] and F = Fix_{E}(Aut(E/F)).

6.8 Corollary Let E/F be a Galois extension and let F ⊆ F′ ⊆ E be an intermediate field. Then also E/F′ is a Galois extension.

Proof This is immediate from part (iv) of Theorem 6.6.

6.9 Lemma Let E/F be a field extension and let H ⩽ Aut(E/F) be a finite subgroup. Then [E : Fix_{E}(H)] ⩽ |H|.

Proof Assume that n := |H| < [E : Fix_{E}(H)], choose elements a_{1}, …, a_{n+1} ∈ E which are linearly independent over Fix_{E}(H), and write H = {φ_{1}, …, φ_{n}}. Then there exists a solution (x_{1}, …, x_{n+1}) ∈ E^{n+1} ∖ {0} of the linear equation system

∑_{j=1}^{n+1} φ_{i}^{−1}(a_{j})x_{j} = 0 , i = 1, …, n.

Let l ∈ {1, …, n + 1} with x_{l} ≠ 0. By Lemma 6.3, there exists y ∈ E with φ_{1}(y) + ⋯ + φ_{n}(y) ≠ 0. Then also (z_{1}, …, z_{n+1}) := yx_{l}^{−1}(x_{1}, …, x_{n+1}) is a solution of the above equation system with z_{l} = y ≠ 0. Applying φ_{i} to the i-th equation we obtain

∑_{j=1}^{n+1} φ_{i}(z_{j})a_{j} = 0 , for i = 1, …, n.

Addition of these equations yields

0 = ∑_{i=1}^{n} ∑_{j=1}^{n+1} φ_{i}(z_{j})a_{j} = ∑_{j=1}^{n+1}(∑_{i=1}^{n} φ_{i}(z_{j}))a_{j} .

But ∑_{i=1}^{n} φ_{i}(z_{j}) ∈ Fix_{E}(H), for all j = 1, …, n + 1, since

σ(∑_{i=1}^{n} φ_{i}(z_{j})) = ∑_{i=1}^{n}(σφ_{i})(z_{j})

and {σφ_{1}, …, σφ_{n}} = {φ_{1}, …, φ_{n}} for all σ ∈ H. Since a_{1}, …, a_{n+1} are linearly independent over Fix_{E}(H), we obtain ∑_{i=1}^{n} φ_{i}(z_{j}) = 0 for all j = 1, …, n + 1, in particular for j = l and z_{l} = y. This is a contradiction.

6.10 Corollary Let E/F be a field extension and let H ⩽ Aut(E/F) be a finite subgroup. Then H = Aut(E/Fix_{E}(H)) and E/Fix_{E}(H) is a Galois extension with Gal(E/Fix_{E}(H)) = H.

Proof By (6.4.a), Lemma 6.3, and Lemma 6.9, be have

|H| ⩽ |Aut(E/Fix_{E}(H))| ⩽ [E : Fix_{E}(H)] ⩽ |H| ,

and we have equality everywhere. In particular, H = Aut(E/Fix_{E}(H)), and [E : Fix_{E}(H)] = |H| < ∞, so that E/Fix_{E}(H) is a Galois extension.

6.11 Theorem (Main theorem of Galois theory) Let E/F be a Galois extension. Set G := Gal(E/F). Then the maps Fix_{E} and Aut(E/−) are inverse bijections between the set of subgroups H of G and the set of intermediate fields F′ of E/F. If F′ and H are in correspondence, then [E : F′] = |H| and [F′ : F] = [G : H].

Proof Let F′ be an intermediate field of E/F. Then E/F′ is a Galois extension by Corollary 6.8, and so F′ = Fix_{E}(Aut(E/F′)). Moreover, for each H ⩽ G we have H = Aut(E/Fix_{E}(H)) by Corollary 6.10. If H and F′ correspond to each other, then |H| = |Aut(E/F′)| = [E : F′] by Corollary 6.7. Moreover,

[F′ : F] = \frac{[E : F]}{[E : F′]} = \frac{|G|}{|H|} = [G : H] .

6.12 Remark Let E/F be a Galois extension and set G := Gal(E/F). Let H ⩽ G and σ ∈ G. Then

Fix_{E}(σHσ^{−1}) = σ(Fix_{E}(H)) .

In fact, for a ∈ E we have:

a ∈ Fix_{E}(σHσ^{−1}) ⟺ στσ^{−1}(a) = a for all τ ∈ H
⟺ τ(σ^{−1}(a)) = σ^{−1}(a) for all τ ∈ H
⟺ σ^{−1}(a) ∈ Fix_{E}(H)
⟺ a ∈ σ(Fix_{E}(H)) .

6.13 Theorem Let E/F be a Galois extension, let H ⩽ G := Aut(E/F), and set F′ := Fix_{E}(H). Then the following are equivalent:

(i) F′/F is a Galois extension.

(ii) F′/F is normal.

(iii) σ(F′) ⊆ F′ for all σ ∈ G.

(iv) H is a normal subgroup of G.

Moreover, if (i)–(iv) hold, then the restriction map r: Gal(E/F) → Gal(F′/F), σ ↦ σ|_{F′}, is an epimorphism with kernel H. In particular, r induces an isomorphism

r̄: G/H → Gal(F′/F) .

Proof (i)⇒(ii): This is immediate from Theorem 6.7.

(ii)⇒(iii): This follows from Theorem 4.13(iii).

(iii)⇒(iv): This follows from Remark 6.12, since

σHσ^{−1} = Aut(E/σ(F′)) ⩾ Aut(E/F′) = H

for all σ ∈ G implies σ^{−1}Hσ ⩽ H for all σ ∈ G.

(iv)⇒(iii): If σHσ^{−1} = H for all σ ∈ G, then σ(F′) = F′ for all σ ∈ G by Remark 6.12.

(iii)⇒(i): If (iii) holds, then r: Aut(E/F) → Aut(F′/F), σ ↦ σ|_{F′}, is a well-defined group homomorphism. For σ ∈ Aut(E/F) we have: σ ∈ ker(r) ⟺ σ|_{F′} = id_{F′} ⟺ σ ∈ Aut(E/F′). Therefore, r induces a monomorphism r̄: Aut(E/F)/Aut(E/F′) → Aut(F′/F). But, by Theorem 6.11 and Lemma 6.3, we also have

|Aut(E/F)/Aut(E/F′)| = [G : H] = [F : F′] ⩾ |Aut(F′/F)| .

This implies, that we have equality and that r̄ and r are surjective. Moreover, since [F : F′] = |Aut(F′/F)|, F′/F is a Galois extension by Theorem 6.7.

6.14 Definition Let F be a field and let f(X) ∈ F [X] be a separable irreducible polynomial. We define the Galois group of f over F as the Galois group G of a splitting field E of f(X) over F. By Theorem 4.11, two different such splitting fields have isomorphic Galois groups. Note that, by Remark 4.8, G acts on the set {r_{1}, …, r_{n}} of roots of f(X) in E. Moreover, if σ ∈ G acts as the identity on {r_{1}, …, r_{n}}, then σ = id_{E}, since E = F (r_{1}, …, r_{n}). Therefore, we can regard G as a subgroup of the symmetric group Sym(r_{1}, …, r_{n}) ≅ S_{n}.

6.15 Theorem Assume the notation of Definition 6.14. Then G acts transitively on R = {r_{1}, …, r_{n}}.

Proof Let R_{1} be the G-orbit of r_{1} and set R_{2} := R ∖ R_{1}. We will show that R_{2} is empty. Write f(X) = α(X − r_{1}) ⋯ (X − r_{n}) with α ∈ F. If we define

g(X) := α ∏_{r∈R_{1}}(X − r) and h(X) := ∏_{r∈R_{2}}(X − r) .

then f = gh. For σ ∈ G, let σ̃: E[X] → E[X] denote the induced ring isomorphism. Then

σ̃(g(X)) = α ∏_{r∈R_{1}}(X − σ(r)) = g(X) ,

since {σ(r) | r ∈ R_{1}} = R_{1}. Similarly, σ̃(h(X)) = h(X). This implies that g(X) and h(X) have coefficients in Fix_{E}(G) = F. Since f(X) is irreducible and g(X) is not constant, h(X) = 1 and R_{2} = ∅.

6.16 Example Let E := ℚ(\sqrt[3]{2}, ζ) with ζ := e^{2πi/3}. Then E is a splitting field of the separable polynomial f(X) = X^{3} − 2. Therefore, E/F is a Galois extension with G := Gal(E/ℚ) of order [E : ℚ] = 6. Since min_{\sqrt[3]{2}/ℚ}(X) = X^{3} − 2 = (X − \sqrt[3]{2})(X − ζ \sqrt[3]{2})(X − ζ^{2} \sqrt[3]{2}) and min_{ζ/ℚ}(X) = X^{2} + X + 1 = (X − ζ)(X − ζ^{2}) (we have X^{3} − 1 = (X − 1)(X^{2} + X + 1)), Remark 4.8 only allows the following 6 possibilities of ℚ-algebra homomorphisms from E to E:

id: \sqrt[3]{2} ↦ \sqrt[3]{2} , ζ ↦ ζ ,
σ: \sqrt[3]{2} ↦ ζ \sqrt[3]{2} , ζ ↦ ζ ,
σ^{2} : \sqrt[3]{2} ↦ ζ^{2} \sqrt[3]{2} , ζ ↦ ζ ,
τ: \sqrt[3]{2} ↦ \sqrt[3]{2} , ζ ↦ ζ^{2} ,
στ: \sqrt[3]{2} ↦ ζ \sqrt[3]{2} , ζ ↦ ζ^{2} ,
σ^{2}τ: \sqrt[3]{2} ↦ ζ^{2} \sqrt[3]{2} , ζ ↦ ζ^{2} .

Since |G| = 6, these must be the elements of G. Note that in this example, G considered as a permutation group of the roots \{\sqrt[3]{2}, ζ \sqrt[3]{2}, ζ^{2} \sqrt[3]{2}\} of X^{3} − 2, is the full symmetric group.

The subgroups of G are G, H_{1} = {1, σ, σ^{2}}, H_{2} = {1, τ}, H_{3} = {1, στ}, H_{4} = {1, σ^{2}τ}, and {1}. Its fixed fields F = ℚ, F_{1}, F_{2}, F_{3}, F_{4}, and E satisfy [F_{1} : ℚ] = 2 and [F_{i} : ℚ] = 3 for i = 2, 3, 4. Since ζ ∈ F_{1}, \sqrt[3]{2} ∈ F_{2}, ζ^{2} \sqrt[3]{2} ∈ F_{3}, and ζ \sqrt[3]{2} ∈ F_{4}, we have F_{1} = ℚ(ζ), F_{2} = ℚ(\sqrt[3]{2}), F_{3} = ℚ(ζ^{2} \sqrt[3]{2}), and F_{4} = ℚ(ζ \sqrt[3]{2}). These are all the intermediate fields of ℚ(\sqrt[3]{2}, ζ)/ℚ.

7 First Applications of Galois Theory

7.1 Definition A field F is called algebraically closed, if F/F is the only algebraic field extension of F.

Obviously, the following are equivalent:

(i) F is algebraically closed.

(ii) Every f ∈ F [X] ∖ F has a root in F.

(iii) Every f ∈ F [X] ∖ F is a product of linear factors in F [X].

(iv) Every irreducible polynomial in F [X] has degree 1.

7.2 Theorem (Fundamental theorem of algebra, Gauß 1777-1855) ℂ is algebraically closed.

Proof Let f ∈ ℂ[X]∖ℂ and let E/ℂ be a splitting field of f over ℂ. We have to show that E = ℂ. Since ℂ/ℝ is finite, there exists a finite normal field extension E′/ℝ by Corollary 4.4. Since char(ℝ) = 0, E′/ℝ is also separable and therefore a Galois extension. Let P be a Sylow 2-subgroup of G := Gal(E′/ℝ) and set E_{2} := Fix_{E′}(P). Then [E_{2} : ℝ] = [G : P] is odd. Since E_{2}/ℝ is separable, there exists a ∈ E_{2} with E_{2} = ℝ(a). Since deg(min_{a/ℝ}) = [E_{2} : ℝ] is odd, min_{a/ℝ} has a root in ℝ by Analysis. Therefore, deg(min_{a/ℝ}) = 1, E_{2} = ℝ, and G = P. This implies that E′/ℂ is a Galois extension with Q := Gal(E′/ℂ) of order 2^{r} for some r ∈ ℕ_{0}. Assume that r ⩾ 1. Then, by the Sylow theorems, Q has subgroup Q_{1} of index 2, and Q_{1} is normal in Q. Therefore E_{1} := Fix_{E′}(Q_{1}) is a Galois extension of ℂ of degree 2. Let b ∈ E_{1} such that ℂ(b) = E_{1} and let min_{b/ℂ} = X^{2} +αX +β with α, β ∈ ℂ. But if γ ∈ ℂ is a square root of α^{2} − 4β (square roots exist in ℂ), then (−α ± γ)/2 ∈ ℂ is a root of min_{b/ℂ}(X) = X^{2} + αX + β. This is a contradiction. Therefore, r = 0, E′ = ℂ, and E = ℂ.

7.3 Remark (Summary of the basic theory of finite fields) Let F be a finite field. Then char(F) = p for some prime p and F is a vector space over the subfield P := {0, 1, 1+1, …, } which has p elements and is isomorphic to ℤ/pℤ. Therefore, |F | = p^{n} for n := [F : P]. Moreover, F^{×} = ⟨a⟩ for some a ∈ F by Theorem 5.9. Obviously, every element in F is a root of X^{p^{n}} − X ∈ P [X]. Since F = P (a), F is a splitting field of X^{p^{n}} − X over P. Now Theorem 4.11 shows that any two fields with p^{n} elements are isomorphic. We know that F/P is separable, since P is perfect, and also normal, since F is a splitting field X^{p^{n}} − X. Therefore, F/P is a Galois extension. Let σ: F → F, x ↦ x^{p} be the Frobenius automorphism. Then σ ∈ G := Gal(F/P) and for k ∈ ℕ we have:

σ^{k} = id_{F} ⟺ σ^{k}(a) = a ⟺ a^{p^{k}} = a .

Therefore, the order of σ is n = [F : P] = G and G = ⟨σ⟩ is cyclic. By the main theorem of Galois theory, the subfields of F are given as Fix_{F} (⟨σ^{m}⟩), with m | n. We have [Fix_{F} (⟨σ^{m}⟩) : P] = [G : ⟨σ^{m}⟩] = m. Therefore, |Fix_{F} (⟨σ^{m}⟩)| = p^{m} and Fix_{F} (⟨σ^{m}⟩) = P (a^{p^{n/m}}).

We still do not know, if for every prime power p^{n}, with n ∈ ℕ, there exists a finite field with p^{n} elements. For that purpose, let E be a splitting field of f(X) = X^{p^{n}} − X ∈ ℤ/pℤ[X]. Then certainly E is a finite field. Let again σ: E → E denote the Frobenius automorphism of E. Since E = ℤ/pℤ(R), where R is the set of roots of X_{X}^{p^{n}} in E, and since σ^{n}(r) = r for all r ∈ R and σ(α) = α for all α ∈ ℤ/pℤ, we have σ^{n}(a) = a for all a ∈ E. Therefore, every element of E is a root of f(X) and |E| ⩽ p^{n}. On the other hand, f(X) has no multiple roots in E, since f′(X) = −1 and f are coprime (cf. Exercise 49). This shows that |E| ⩾ p^{n}. Hence, E is a field with p^{n} elements.

In order to construct a field with p^{n} elements it is usually easier to find a polynomial f(X) ∈ ℤ/pℤ[Z] of degree n which is irreducible. Then ℤ/pℤ[X]/(f(X)) is a field with p^{n} elements.

7.4 Definition Let F be a field and let n ∈ ℕ.

(a) An element ζ ∈ F with ζ^{n} is called an n-th root of unity. If ζ has order n in F^{×}, then ζ is called a primitive n-th root of unity. We write μ_{n}(F) for the multiplicative group of n-th roots of unity in F. Every ζ ∈ μ_{n}(F) is a primitive d-th root of unity for a unique divisor d of n. Conversely μ_{d}(F) ⩽ μ_{n}(F) for each divisor d of n. Therefore, if we write μ̃_{n}(F) ⊆ μ_{n}(F) for the subset of primitive n-th roots of unity in F, the group μ_{n}(F) is the disjoint union of the subsets μ̃_{d}(F), where d runs through all positive divisors of n.

(b) Every splitting field of X^{n} − 1 over F is called the n-th cyclotomic field over F and is denoted by F_{n}. The polynomial

Φ_{n}(X) := ∏_{ζ∈μ̃_{n}(F_{n})}(X − ζ) ∈ F_{n}[X]

is called the n-th cyclotomic polynomial over F.

7.5 Remark Let F be a field and let n ∈ ℕ.

(a) Since X^{n} − 1 can have at most n roots in F, we have |μ_{n}(F)| ⩽ n, and μ_{n}(F) is cyclic by Theorem 5.9.

(b) If F contains a primitive n-th root of unity ζ, then |⟨ζ⟩| = n and therefore, μ_{n}(F) = ⟨ζ⟩. For i ∈ ℤ we then have: ζ^{i} is again a primitive n-th root of unity if and only if gcd(i, n) = 1. Therefore, |μ̃_{n}(F)| equals the number of elements i + nℤ ∈ ℤ/nℤ with gcd(i, n) = 1, which are exactly the units in ℤ/nℤ. Their number is φ(n), where φ denotes the Euler φ-function. In particular, for the n-th cyclotomic polynomial Φ_{n} ∈ F_{n}[X] we have either Φ_{n} = 1 or Φ_{n} has degree φ(n), depending on whether there exists a primitive n-th root of unity in F_{n} or not.

(c) If char(F) = p > 0 and n = pm, then there are no primitive n-th roots of unity in F, since 0 = ζ^{pm} − 1 = (ζ^{m} − 1)^{p} for every ζ ∈ μ_{n}(F), and so ζ has order at most m. In particular, this holds for F_{n} instead of F, so that Φ_{n}(X) = 1 in this case.

(d) If char(F) ∤ n (these are the remaining cases, i.e., char(F) = 0 or char(F) = p > 0 with p ∤ n), then (X^{n} − 1)′ = nX^{n−1} is coprime to X^{n} − 1. Therefore, by Exercise 49, the roots of X^{n} − 1 in F_{n} are pairwise distinct, and they form a cyclic group ⟨ζ⟩ of order n. This implies that F_{n} contains a primitive n-th root of unity ζ. Moreover, F_{n} = F (ζ), since X^{n} − 1 = ∏_{i=1}^{n}(X − ζ^{i}).

(e) Assume that char(F) ∤ n. Then deg(Φ_{n}(X)) = φ(n), and since μ_{n}(F_{n}) is the disjoint union of μ̃_{d}(F_{n}), where d | n, we obtain

X^{n} − 1 = ∏_{ζ∈μ_{n}(F_{n})}(X − ζ) = ∏_{d|n} Φ_{d}(X) .

In particular, Φ_{n}(X) can be determined recursively by

Φ_{n}(X) = \frac{X^{n} − 1}{∏_{d|n, d≠n} Φ_{d}(X)} .

Note that since each Φ_{d}(X) has leading coefficient 1, and since Φ_{1}(X) = X − 1 has coefficients in ℤ · 1_{F}, also Φ_{n} has coefficients in ℤ · 1_{F} by an inductive argument. If F has characteristic 0, the coefficients lie in ℤ, and if F has prime characteristic p, then the coefficients lie in the unique subfield of F of order p. In particular, if we compute Φ_{n}(X) ∈ ℤ[X] for F = ℚ, then Φ_{n} ∈ ℤ[X] and we obtain Φ_{n}(X) for an arbitrary field F with char(F) ∤ n by using the unique ring homomorphism ℤ → F and extending it to a ring homomorphism ℤ[X] → F [X] in the obvious way. In the sequel we will usually mean the cyclotomic polynomial with integer coefficients when we use the symbol Φ_{n}(X).

7.6 Example (a) μ_{n}(ℝ) = {1, −1} for every even n.

(b) ζ := e^{2πi/n} ∈ ℂ is a primitive n-th root of unity in ℂ.

(c) If F is a finite field with p^{n} elements, then μ_{p^{n}−1}(F) = F^{×} and F contains a primitive p^{n} − 1-th root of unity.

(d) If p is a prime, then

Φ_{p} = \frac{X^{p} − 1}{X − 1} = X^{p−1} + X^{p−2} + ⋯ + X + 1 .

7.7 Theorem Let F be a field and let n ∈ ℕ with char(F) ∤ n. Then F_{n}/F is a Galois extension with G := Gal(F_{n}/F) isomorphic to a subgroup of (ℤ/nℤ)^{×}. In particular, G is abelian.

Proof By Remark 7.5(d), X^{n} − 1 is separable over F. Therefore, F_{n}/F is a Galois extension. Also, by Remark 7.5(d), F_{n} = F (ζ) with a primitive n-th root of unity ζ. Clearly, σ(ζ) is again a primitive n-th root of unity for all σ ∈ G. Therefore, σ(ζ) = ζ^{i} for a unique i + nℤ ∈ (ℤ/nℤ)^{×}. Moreover, if τ ∈ G and τ(ζ) = ζ^{j} with j + nℤ ∈ (ℤ/nℤ)^{×}, then (στ)(ζ) = σ(ζ^{j}) = σ(ζ)^{j} = ζ^{ij}. Therefore, the resulting map G → (ℤ/nℤ)^{×} is a group homomorphism. It is injective, since σ ∈ G is uniquely determined by σ(ζ).

7.8 Remark Let F and n ∈ ℕ be as in Theorem 7.7. Since G := Gal(F_{n}/F) is abelian, every subgroup of G is normal in G. Therefore, every intermediate field E in F_{n}/F is a Galois extension of F and its Galois group is a factor group of G and hence abelian. A kind of converse for the base field ℚ to that was proved by Kronnecker and Weber using techniques from algebraic number theory: Let F ⊂ ℂ, then F/ℚ is a Galois extension with abelian Galois group if and only if F ⊆ ℚ(ζ) for some root of unity ζ ∈ ℂ.

7.9 Theorem For each n ∈ ℕ, Φ_{n}(X) ∈ ℤ[X] is irreducible in ℤ[X] and in ℚ[X]. Moreover, if ζ ∈ ℂ is a primitive n-th root of unity, then ℚ_{n} := ℚ(ζ) is a Galois extension of ℚ of degree φ(n) and (ℤ/nℤ)^{×} → Gal(ℚ_{n}(ζ)/ℚ), i + nℤ ↦ σ_{i}, with σ_{i}(ζ) = ζ^{i}, is a group isomorphism.

Proof It suffices to show that Φ_{n}(X) is irreducible in ℤ[X]. The rest follows from Gauß’ Lemma and from Theorem 7.7 and its proof.

Assume that f ∈ ℤ[X] is an irreducible monic factor of Φ_{n}, and let X^{n} −1 = f(X)g(X) with g(X) ∈ ℚ[X]. Then g(X) ∈ ℤ[X], since f(X) is monic. Let η ∈ ℚ_{n} be a root of f. Then f = min_{η/ℚ} and η ∈ μ̃_{n}(ℚ_{n}).

Claim: For every prime p with p ∤ n we have f(η^{p}) = 0. The claim is proved as follows. Assume that f(η^{p}) ≠ 0 for some prime p ∤ n. Then f(η^{p})g(η^{p}) = (η^{p})^{n} − 1 = 1 − 1 = 0 and g(η^{p}) = 0. Since f = min_{η/ℚ}, there exists h ∈ ℚ[X] such that g(X^{p}) = f(X)h(X). Since f is monic, g ∈ ℤ[X] implies h ∈ ℤ[X]. Let φ: ℤ[X] → ℤ/pℤ[X] be the ring homomorphism with φ(a_{n}X^{n} + ⋯ + a_{0}) = ā_{n}X^{n} + ⋯ + ā_{0}. Since f is monic, φ(f) is not constant and has an irreducible factor q ∈ ℤ/pℤ[X]. Write g = X^{k} +a_{k−1}X^{k−1} +⋯+a_{0} with a_{k−1}, …, a_{0} ∈ ℤ. Since ā^{p} = ā for all a ∈ ℤ, we have

φ(f)φ(h) = φ(g(X^{p})) = X^{kp} + ā_{k−1}X^{(k−1)p} + ⋯ + ā_{0}
= (X^{k} + ā_{k−1}X^{k−1} + ⋯ + ā_{0})^{p} = φ(g)^{p} .

Since q | φ(f), also q | φ(g) and q^{2} | φ(f · g) = φ(X^{n} − 1) = X^{n} − ¯1. This contradicts Remark 7.5(d), and the claim is proved.

Now, η^{p} is a root of f for all primes p with p ∤ n. By induction on r we obtain that η^{p_{1}···p_{r}} is a root of f for arbitrary (not necessarily distinct) primes p_{1}, …, p_{r} not dividing n. Therefore, f(η^{m}) = 0 for all m ∈ ℕ with gcd(m, n) = 1, which shows that each primitive n-th root of unity is a root of f. This implies f = Φ_{n}.

7.10 Theorem Let n = 2^{α}p_{1}^{α_{1}} ⋯ p_{t}^{α_{t}} ∈ ℕ with α ∈ ℕ_{0}, α_{1}, …, α_{t} ∈ ℕ, and with primes 3 ⩽ p_{1} < p_{2} < ⋯ < p_{t}. The regular n-gon is constructible from 0 and 1 if and only if p_{1}, …, p_{t} are Fermat primes (these are primes p of the form p = 2^{2^{r}} + 1 for some r ∈ ℕ_{0}) and α_{1} = ⋯ = α_{t} = 1.

Proof The regular n-gon is constructible if and only if ζ := e^{2πi/n} is constructible. By Theorem 3.4, this is equivalent to ℚ(ζ) being contained in the top field of a square root tower ℚ = F_{0} ⊆ F_{1} ⊆ ⋯ ⊆ F_{m}. This in turn is equivalent to [ℚ(ζ) : ℚ] being a power of 2. But

[ℚ(ζ) : ℚ] = φ(n) = 2^{α−1}(p_{1} − 1) ⋯ (p_{t} − 1)p_{1}^{α_{1}−1} ⋯ p_{t}^{α_{t}−1} ,

and we obtain the equivalent condition α_{1} = ⋯ α_{t} = 1 and p_{i} = 2^{β_{i}} + 1 with some β_{i} ∈ ℕ, for all i = 1, …, t. But note, that if 2^{s} + 1 is a prime, then necessarily s is a power of 2, because otherwise s = qt with an odd prime q and t ∈ ℕ and we obtain a decomposition 2^{s} + 1 = (2^{t} + 1)((2^{t})^{q−1} − (2^{t})^{q−2} + − ⋯ + 1). This completes the proof.

8 Solvability of Polynomial Equations by Radicals

8.1 Definition A group G is called solvable, if there exist finitely many subgroups

G = G_{0} ⊵ G_{1} ⊵ ⋯ ⊵ G_{n} = 1

such that G_{i−1}/G_{i} is abelian for all i = 1, …, n.

8.2 Lemma Let G be a group, let H ⩽ G and N ⊴ G.

(a) If G is solvable, then also G and G/N are solvable.

(b) G is solvable if and only if G/N and N are solvable.

Proof (a) Let G = G_{0} ⊵ G_{1} ⊵ ⋯ ⊵ G_{n} = 1 with G_{i−1}/G_{i} abelian for all i = 1, …, n. Then H = H ∩ G_{0} ⩾ H ∩ G_{1} ⩾ ⋯ ⩾ H ∩ G_{n} = 1 with H ∩ G_{i} ⊴ H ∩ G_{i−1} and

H ∩ G_{i−1}/H ∩ G_{i} = H ∩ G_{i−1}/(H ∩ G_{i−1}) ∩ G_{i}
≅ (H ∩ G_{i−1})G_{i}/G_{i} ⩽ G_{i−1}/G_{i}

abelian. This shows that H is solvable. Moreover,

G/N = G_{0}N/N ⩾ G_{1}N/N ⩾ ⋯ ⩾ G_{n}N/N = N/N

with G_{i}N/N ⊴ G_{i−1}N/N and

(G_{i−1}N/N)/(G_{i}N/N) ≅ G_{i−1}N/G_{i}N = G_{i−1}(G_{i}N)/G_{i}N
≅ G_{i−1}/G_{i−1} ∩ G_{i}N ≅ (G_{i−1}/G_{i})/(G_{i−1} ∩ G_{i}N/G_{i})

abelian. Therefore, G/N is solvable.

(b) By (a) we only have to show that if N and G/N are solvable, then also G is solvable. Let

G/N = X_{0} ⊵ X_{1} ⊵ ⋯ ⊵ X_{n} = N/N

and

N = N_{0} ⊵ N_{1} ⊵ ⋯ ⊵ N_{m} = 1

such that X_{i−1}/X_{i} and N_{j−1}/N_{j} is abelian for all i = 1, …, n and j = 1, …, m. Then X_{i} = G_{i}/N for a unique subgroup G_{i} with N ⩽ G_{i} ⩽ G for i = 1, …, n. Moreover, since G_{i}/N ⊴ G_{i−1}/N, also G_{i} ⊴ G_{i−1} and G_{i−1}/G_{i} ≅ (G_{i−1}/N)/(G_{i}/N) is abelian. Therefore, the chain

G = G_{0} ⊵ G_{1} ⊵ ⋯ ⊵ G_{0} = N = N_{0} ⊵ N_{1} ⊵ ⋯ ⊵ N_{m} = 1

has the right properties to imply that G is solvable.

8.3 Example The symmetric groups S_{1}, S_{2}, S_{3}, S_{4} are solvable. In fact, S_{1} and S_{2} are abelian, and S_{3} ⊵ A_{3} ⊵ 1 and S_{4} ⊵ A_{4} ⊵ V_{V} ⊵ 1 with V_{4} := {1, (12)(34), (13)(24), (14)(23)} and abelian factor groups S_{3}/A_{3}, A_{3}/1, S_{4}/A_{4}, A_{4}/V_{4}, V_{4}/1.

8.4 Theorem S_{n} is solvable if and only if n ⩽ 4.

Proof By Example 8.3, we only have to show that S_{n} is not solvable for n ⩾ 5. By Lemma 8.2, it suffices to show that the subgroup H ⩽ S_{n} which is generated by the set of 3-cycles (i, j, k), is not solvable. Assume that H is solvable. Then there exists H_{1} ◁ H such that H/H_{1} is abelian. In particular H′ := ⟨aba^{−1}b^{−1} | a, b ∈ H⟩ ⩽ H_{1} < H. But for each 3-cycle (i, j, k) there exist elements l, m ∈ {1, …, n} such that |{i, j, k, l, m}| = 5 and we obtain

(i, j, l)(i, k, m)(i, j, l)^{−1}(i, k, m)^{−1} = (i, j, k)

implying that H′ = H and contradicting H′ < H. Therefore, H and S_{n} are not solvable.

8.5 Theorem Let F be a field and let n ∈ ℕ be such that char(F) ∤ n. Moreover, let α ∈ F and let E be a splitting field of X^{n} − α over F.

(a) E contains a primitive n − th root of unity ζ.

(b) If b ∈ E is such that b^{n} = α, then b, ζb, …, ζ^{n−1}b are the roots of X^{n}−α. In particular, X^{n} − α has n pairwise distinct roots and E = F (b, ζ).

(c) E/F, E/F (ζ), and F (ζ)/F are Galois extensions.

(d) Gal(E/F (ζ)) is isomorphic to a subgroup of ℤ/nℤ. In particular, if F contains a primitive n-th root of unity, then E/F has a cyclic Galois group.

Proof (a) (X^{n} − α)′ = nX^{n−1} and X^{n} − α are coprime. Therefore, X^{n} − α has n pairwise distinct roots b_{1}, …, b_{n} in E by Exercise 49. Then b_{1}/b_{1}, b_{2}/b_{1}, …, b_{n}/b_{1} are pairwise distinct n-th roots of unity, since (b_{i}/b_{1})^{n} = α/α = 1. Therefore, |μ_{n}(E)| = n. Moreover, μ_{n}(E) is cyclic by Theorem 5.9, and hence generated by a primitive n-th root of unity ζ.

(b) This is obvious.

(c) E/F is a Galois extension, since E is the splitting field of the separable polynomial X^{n} − α. F (ζ)/F is a Galois extension by Theorem 7.7. E/F (ζ) is a Galois extension, since E/F is.

(d) Let σ ∈ Gal(E/F (ζ)). Then σ(b) ∈ {b, ζb, ζ^{2}b, …, ζ^{n−1}b} and σ(b)/b ∈ ⟨ζ⟩. This defines a function φ: Gal(E/F (ζ)) → ⟨ζ⟩. For σ, τ ∈ Gal(E/F (ζ)) we have:

φ(στ) = \frac{σ(τ(b))}{b} = \frac{σ(b\frac{τ(b)}{b})}{b} = \frac{σ(b)\frac{τ(b)}{b}}{b} = φ(σ)φ(τ) ,

since τ(b)/b ∈ F (ζ). Therefore, φ is a group homomorphism. If φ(σ) = 1, then σ(b) = b, hence σ = id_{E}, since E = (F (ζ))(b).

8.6 Theorem Let n ∈ ℕ and assume that F is a field containing a primitive n-th root of unity ζ. If E/F is a Galois extension with cyclic Galois group G = ⟨σ⟩ of order n, then there exists α ∈ F such that E is a splitting field of X^{n} − α over F.

Proof By Lemma 6.3, there exists c ∈ E such that

b := c + ζσ(c) + ζ^{2}σ^{2}(c) + ⋯ + ζ^{n−1}σ^{n−1}(c) ≠ 0 .

We have

σ(b) = σ(c) + ζσ^{2}(c) + ⋯ + ζ^{n−2}σ^{n−1}(c) + ζ^{n−1}b = ζ^{−1}b ,

and therefore, σ(F (b) = F (b) and σ^{m}(b) = ζ^{−m}b for all m ∈ ℕ. Therefore, the restrictions of 1, σ, σ^{2}, …, σ^{n−1} form n pairwise distinct elements in Aut(F (b)/F) and we obtain

n ⩽ |Aut(F (b)/F)| ⩽ [F (b) : F] = [E : F] = n ,

which implies equality everywhere and F (b) = E. Since σ(b^{n}) = σ(b)^{n} = (ζb)^{n} = b^{n}, we have α := b^{n} ∈ F. Obviously, E is a splitting field of X^{n} − α over F, since b, ζb, …, ζ^{n−1}b are the roots of X^{n} − α.

8.7 Definition A field extension E/F is called a radical extension if there exists a chain F = F_{0} ⊆ F_{1} ⊆ ⋯ ⊆ F_{r} = E and elements b_{i} ∈ F_{i}, n_{i} ∈ ℕ for i = 1, …, r such that F_{i} = F_{i−1}(b_{i}) and b_{i}^{n_{i}} ∈ F_{i−1} for all i = 1, …, r. In particular, [E : F] is finite. A polynomial equation f(X) = 0 with irreducible f ∈ F [X] is called solvable by radicals over F, if there exists a radical extension E/F such that f(X) splits into linear factors in E[X].

8.8 Lemma Let F be a field of characteristic 0 and let E/F be a radical extension. Then there exists an extension E′/E such that E′/F is a radical Galois extension.

Proof Let F = F_{0} ⊆ F_{1} ⊆ ⋯ ⊆ F_{r} = F, b_{i} ∈ F_{i}, n_{i} ∈ ℕ be as in Definition 8.7. By Corollary 4.14, there exists a finite field extension Ẽ/E such that Ẽ/F is normal and hence a Galois extension. We write G := Gal(Ẽ/F) = {σ_{1} = idẼ, σ_{2}, …, σ_{s}} and set

F_{ij} := F (σ_{1}(b_{1}), …, σ_{s}(b_{1}), σ_{1}(b_{2}) …, σ_{s}(b_{2}), … σ_{1}(b_{i}), …, σ_{j}(b_{i}))

for i = 1, …, r and j = 1, …, s. Then we have a tower of fields

F ⊆ F_{11} ⊆ F_{12} ⊆ ⋯ ⊆ F_{1s} ⊆ F_{21} ⊆ ⋯ ⊆ F_{2s} ⊆ ⋯ ⊆ F_{rs} ⊆ Ẽ .

For each i = 1, …, r, F_{is}/F is a Galois extension. In fact, each τ ∈ G permutes the elements σ_{1}(b_{1}), …, σ_{s}(b_{1}), …, σ_{1}(b_{i}), …, σ_{s}(b_{i}), and therefore, τ(F_{is}) ⊆ F_{is} and Theorem 6.13 applies. In particular, E′ := F_{rs} is a Galois extension of F.

Obviously, E = F (b_{1}, …, b_{r}) ⊆ E′. Moreover, for i = 1, …, r and j = 2, …, s, we have

σ_{j}(b_{i})^{n_{i}} = σ_{j}(b_{i}^{n_{i}}) ∈ σ_{j}(F_{i−1}) ⊆ σ_{j}(F_{i−1,s}) = F_{i−1,s} ⊆ F_{i,j−1} .

Moreover, σ_{1}(b_{1})^{n_{1}} = b_{1}^{n_{1}} ∈ F and for i = 2, …, n we have

σ_{1}(b_{i})^{n_{i}} = b_{i}^{n_{i}} ∈ F_{i−1} ⊆ F_{i−1,1} ⊆ F_{i−1,s} .

Therefore, E′/F is also a radical extension.

8.9 Corollary Let F be a field of characteristic 0 and let f(X) ∈ F [X] be irreducible. Assume that f has a root in a radical field extension E/F. Then there exists a radical field extension E′/F such that f(X) splits completely into linear factors in E′[X].

Proof By Lemma 8.8 there exists a radical Galois extension E′/F with E ⊆ E′. Since E′/F is normal and contains a root of f, f splits completely into linear factors in E′[X].

8.10 Theorem Let F be a field of characteristic 0 and let E/F be a radical Galois extension. Then G := Gal(E/F) is solvable.

Proof Let F = F_{0} ⊆ F_{1} ⊆ ⋯ ⊆ F_{r} = E, b_{i} ∈ F_{i}, n_{i} ∈ ℕ be as in Definition 8.7. Set n = n_{1} ⋯ n_{r} and let Ẽ := E(ζ) with a primitive n-th root of unity ζ. Since E/F and F (ζ)/F are Galois extensions, also Ẽ · F (ζ)/F is a Galois extension, because E · F (ζ)/F is normal by Exercise 37. By Lemma 8.2, it suffices to show that G̃ := Gal(Ẽ/F) is solvable. The chain

F = F_{0} ⊆ F_{0}(ζ) ⊆ F_{1}(ζ) ⊆ ⋯ ⊆ F_{r}(ζ) = E(ζ) = Ẽ

corresponds to a chain of subgroups

G̃ ⩾ G̃_{0} ⩾ G̃_{1} ⩾ ⋯ ⩾ G̃_{r} = 1 .

Since F (ζ)/F is a Galois extension, G̃_{0} ⊴ G̃ and G̃/G̃_{0} = Gal(F (ζ)/F) is abelian by Theorem 7.7. Moreover, for i = 1, …, r, we have F_{i}(ζ) = F_{i−1}(b_{i}, ζ) = (F_{i−1}(ζ))(b_{i}) with b_{i}^{n_{i}} ∈ F_{i−1}(ζ). Therefore, F_{i}(ζ) is a splitting field of X^{n_{i}} − b_{i}^{n_{i}} over F_{i−1}(ζ). By Theorem 8.5(c) and (d), F_{i}(ζ)/F_{i−1}(ζ) is a Galois extension with cyclic Galois group. Therefore, G̃_{i} ⊴ G̃_{i−1} and G_{i−1}/G_{i} = Gal(F_{i}(ζ)/F_{i−1}(ζ)) is abelian. This implies that G̃ is solvable.

8.11 Theorem Let E/F be a Galois extension with solvable Galois group G of order n and assume that char(F) = 0. Then the n-th cyclotomic field E(ζ), where ζ a primitive n-th root of unity, is a radical Galois extension of F.

Proof Since F (ζ)/F and E/F are Galois extensions, also E(ζ) = E · F (ζ)/F is a Galois extension. By Exercise 52, E(ζ)/F (ζ) is a Galois extension with

G := Gal(E(ζ)/F (ζ)) ≅ Gal(E/E ∩ F (ζ)) ⩽ Gal(E/F) .

Therefore, also G is solvable and |G| | n. Let

G = G_{0} ⊵ G_{1} ⊵ ⋯ ⊵ G_{r} = 1

be as in Definition 8.1. Since every finite abelian group is a direct sum of cyclic groups, we may even assume that G_{i−1}/G_{i} is cyclic for i = 1, …, r, by refining the above chain if necessary. Consider the intermediate fields F_{i} := Fix_{E(ζ)}(G_{i}) in E(ζ)/F (ζ) for i = 1, …, r. They form a chain

F ⊆ F (ζ) = F_{0} ⊆ ⋯ ⊆ F_{r} = E(ζ) ,

with F_{i}/F_{i−1} a Galois extension, since G_{i} ⊴ G_{i−1} for all i = 1, …, r. Moreover, Gal(F_{i}/F_{i−1}) ≅ G_{i−1}/G_{i} is cyclic. By Theorem 8.6, there exists b_{i} ∈ F_{i} with b_{i}^{n_{i}} ∈ F_{i−1}. Moreover, ζ^{n} ∈ F. Therefore, E(ζ)/F is a radical extension.

8.12 Theorem Let F be a field of characteristic 0, let f ∈ F [X] be irreducible, let E be a splitting field of f over F, and set G := Gal(E/F), the Galois group of f over F. Then the following are equivalent:

(i) f(X) = 0 is solvable by radicals over F.

(ii) G is solvable.

Proof (i)⇒(ii): Let E′/F be a radical extension such that f(X) splits into linear factors in E′[X]. By Lemma 8.8 we may choose E′ such that E′/F is also a Galois extension. By Theorem 8.10, Gal(E′/F) is solvable. Let r_{1}, …, r_{n} ∈ E′ be the roots of f. Then F (r_{1}, …, r_{n}) ⊆ E′ is a splitting field of f over F, hence G is isomorphic to the factor group Gal(F (r_{1}, …, r_{n})/F) of the solvable group Gal(E′/F). Hence G is solvable by Lemma 8.2.

(ii)⇒(i): If G is solvable, there exists a radical extension E′/F with E ⊆ E′ by Theorem 8.11. Since f splits into linear factors in E[X], it also does in E′[X].

8.13 Lemma Let p be a prime.

(a) Each element σ ∈ S_{p} of order p is a p-cycle.

(b) If σ ∈ S_{p} has order p and τ ∈ S_{p} is a transposition, then ⟨σ, τ⟩ = S_{p}.

Proof (a) Assume that σ is not a p-cycle. Then, in the cycle notation it contains a k-cycle (a_{1}, …, a_{k}) for some k ∈ {2, …, p − 1}. This implies that σ^{k}(a_{1}) = a_{1} = σ^{p}(a_{1}). Since k and p are coprime, it also implies σ(a_{1}) = a_{1} in contradiction to σ(a_{1}) = a_{2}.

(b) Since σ has order p, σ is a p-cycle by (a). Write σ = (a_{1}, …, a_{p}) and τ = (a_{i}, a_{j}). Since also σ = (a_{i}, …, a_{p}, a_{1}, …, a_{i−1}), we can write σ = (b_{1}, …, b_{p}) and τ = (b_{1}, b_{k}) with k ∈ {2, …, p}. Then σ^{k−1} = (b_{1}, b_{k}, …) is again a p-cycle by (a), and we can write σ^{k−1} = (c_{1}, c_{2}, …, c_{p}) and τ = (c_{1}, c_{2}) for certain c_{1}, …, c_{p}. Define ρ ∈ S_{p} by ρ(i) := c_{i} for i = 1, …, p. Then ρ^{−1}σ^{k−1}ρ = (1, 2, 3, …, p) and ρ^{−1}τρ = (1, 2). It is well-known that S_{p} = ⟨(1, 2, …, p), (1, 2)⟩ and we obtain

S_{p} = ρ⟨(1, 2, …, p), (1, 2)⟩ρ^{−1} = ⟨ρ(1, 2, …, p)ρ^{−1}, ρ(1, 2)ρ^{−1}⟩
= ⟨σ^{k−1}, τ⟩ ⩽ ⟨σ, τ⟩ .

8.14 Theorem Let p be a prime and let f(X) ∈ ℚ[X] be an irreducible polynomial of degree p which has exactly p − 2 real roots. Then the Galois group of f is isomorphic to S_{p}. In particular, f(X) = 0 is not solvable by radicals over ℚ, if p ⩾ 5.

Proof Let f = c(X − a_{1}) ⋯ (X − a_{p}) be a decomposition in ℂ[X]. Then E := ℚ(a_{1}, …, a_{p}) is a splitting field of f over ℚ. E/ℚ is a Galois extension and we denote its Galois group by G. For each i = 1, …, p also ā_{i} is a root of f. Therefore, the restriction of the complex conjugation is an element τ ∈ G, which fixes the p − 2 real roots and interchanges the two non-real roots. On the other hand, since E contains ℚ(a_{1}) and since [ℚ(a_{1}) : ℚ] = deg(min_{a_{1}/ℚ}) = deg(f) = p, also p | |G| and G contains an element σ of order p. The action of G on {a_{1}, …, a_{p}} gives rise to a monomorphism φ: G → S_{p} as in Definition 6.14. Then φ(τ) is a transposition and φ(σ) is an element of order p. By the previous lemma, the homomorphism φ is also surjective.

8.15 Example The polynomial f(X) = X^{5} − 4X + 2 is irreducible in ℤ[X] and ℚ[X] by Eisenstein. Since f′(X) = 5X^{4} − 4 has exactly two real roots, namely ±\sqrt[4]{4/5}, and since f(\sqrt[4]{4/5}) < 0 and f(−\sqrt[4]{4/5}) > 0, f has exactly 3 real roots. Therefore, the Galois group of f is isomorphic to S_{5}, and its roots lie in no radical extension E/F.

References

[J] Jacobson: Basic Algebra I, 2nd edition, Freeman 1985.