FIELD EXTENSIONS AND K-HOMOMORPHISMS

Let F be a field containing K. Then F is a K-algebra. If E is another field containing K we let Alg_{K}(F, E) denote the set of morphisms of K-algebras from F to E.

Note that if σ ∈ Alg_{K}(F, E) then σ ≠ 0 since σ(1) = 1. Since F is a field, it follows that σ is monic.

(1) F is a purely transcendental extension of K with transcendence degree n if and only for every field E containing K, there there exist u_{1}, …, u_{n} ∈ F (not unique) such that there is a one-to-one correspondence between Alg_{K}(F, E) and the cartesian power E^{n} given by σ ↦ (σ(u_{1}), …, σ(u_{n})).

(2) Let u_{1}, …, u_{n} ∈ F. Then u_{1}, …, u_{n} are algebraically independent if and only the sub-K-algebra of F generated by u_{1}, …, u_{n} is a purely transcendental extension of K with transcendence degree n.

(3) F is an algebraic extension of K if and only if for every u ∈ F and every field E containing K, there exist at most finitely many e ∈ E such that there is a K-homomorphism σ: F → E with σ(u) = e.

(4) F is a purely inseparable extension of K if for every field E containing K there is at most one K-homomorphism from F into E.

(5) F is a finite dimensional separable extension of K if dim_{K} F = n < ∞ and there exists a field E containing K such that there are n K-homomorphisms from F to E.

(6) More generally, F is a separable extension of K if F is algebraic over K and every finite dimensional subfield of F is separable over K.

(7) If F is a field with dimension n over K, then F is a Galois extension of K if Alg_{K}(F, F) contains n homomorphisms. In this case, we write Gal(F/K) = Alg_{K}(F, F).

(8) F is a normal extension of K with for every field E containing K and σ ∈ Alg_{K}(F, E), σ(F) ⊆ F.

Proposition [Hungerford, Lemma 7.5, p 291]. Let E and F be fields containing E. Hom_{K}(F, E) is a vector space with dimension dim_{K} F over E. If σ_{1}, …, σ_{n} are distinct non-trivial mappings in Alg_{K}(F, E), then σ_{1} …, σ_{n} are linearly independent elements of Hom_{K}(F, E). Consequently, n ≤ dim_{K} F.

Definition. The separable degree of F over K is

[F: K]_{s} = sup_{E} | Alg_{K}(F, E) |,

where the supremum is taken over all fields E containing K. (In fact, it suffices to consider any extension E of F which is normal over K.)

We write [F: K] = dim_{K} F. Thus the preceding Proposition can be rephrased as asserting that [F: K]_{s} ≤ [F: K]. Clearly if [F: K] < ∞ then F is separable over K if and only if [F: K]_{s} = [F: K].

Lemma. (1) If K ⊆ F ⊆ F_{1} with F_{1} algebraic over K and σ ∈ Alg_{K}(F, E) for some field E containing K, then σ extends to a K-homomorphism σ_{1} : F_{1} → E_{1} for some field E_{1} containing E.

(2) If F_{1} above is normal over K then σ extends to a K-automorphism of F_{1} .

Corollary. If K ⊆ F ⊆ F_{1} (with F_{1} algebraic over K), then [F_{1} : K]_{s} = [F_{1} : F]_{s} [F: K]_{s}.

Proposition. (1) If f ∈ K[X] is an irreducible polynomial and u, v are roots of f in some extension field F of K, then there exists an extension E of F and a K-homomorphism σ: F → E such that σ(u) = v.

(2) If f is a polynomial in K[X] with distinct roots and F is a normal extension of K in which f has a root, then f is irreducible if and only if Aut_{K} F acts transitively on the roots of f. proof: (1) K[u] ≈ K[X]/(f) ≈ K[v]. By a previous Corollary, a K-homomorphism K[u] → K[v] extends to a K-homomorphism from F into some field E. ✓

Proposition. Let F be an algebraic extension of K.

(1) F is separable over K if and only if the minimal polynomial in K[X] for every element u ∈ F has distinct roots (in some possibly larger extension of K).

(2) F is normal over K if and only if for every u ∈ F, the minimal polynomial in K[X] for u splits completely in F [X].

(3) F is a galois extension of K if and only if the minimal polynomial over K for every u ∈ F splits completely into distinct linear factors over K. proof: (1) It suffices to deal with the case where [F: K] is finite. In this case it is easy to see that there is a maximal subfield F′ of F which is separable over K. If F′ ⊊ F let u ∈ F with u ∉ F′. The minimal polynomial g for u over F′ is a factor of the minimal polynomial for u over K, hence has distinct roots. By the preceding proposition, for each of these roots there is a F′-homomorphism σ defined on F′[u] with taking u to the specified root. Thus [F′[u]: F′]_{s} ≥ deg g = [F′[u]: F′]. Then [F′[u]: K]_{s} = [F′[u]: F′]_{s} [F′: K]_{s} ≥ [F′[u]: F′] [F′: K] = [F′[u]: K], so that F′[u] is separable over K, contradicting the maximality of F′. Thus F′ = F.

(2) If E is a field containing F and σ ∈ Alg_{K}(F, E) and u ∈ F, then σ(u) is a root of the minimal polynomial for u over K. But this root must belong to F, since by hypothesis this polynomial splits completely in F [X]. Hence σ(F) ⊆ F, thus showing that F/K is normal.

Corollary. If F is a galois extension of K then for every u ∈ F with u ∉ K there exists σ ∈ Alg_{K}(F, F) with σ(u) ≠ u. proof: If F is separable over K and u ∉ K, then the minimal polynomial for u over K has more than one root so there exists σ: F → E with σ(u) ≠ u for some extension field E of F. If F/K is normal, we may take E = F. ✓

Corollary. If F is a normal [separable/galois] extension of K and K ⊆ L ⊆ F, then F is a normal [separable/galois] extension of L.

Notation. (1) If G is a group and X is a set/module/ring/field/K-algebra which G acts on, then we write

X^{G} = {x ∈ X | (∀σ ∈ G) σx = x}.

(X^{G} is the subobject of X fixed by G.)

If x ∈ X then the stabilizer (or isotropy subgroup) of x is G_{x} = {σ ∈ G | σx = x}. The orbit of x is Gx = {σx | σ ∈ G}. Notice that distinct orbits are disjoint and that Gx = Gy if and only if y ∈ Gx. We say that G acts transitively on X if X itself is the orbit of one (and therefore all) of its elements, i. e. if for all x, y ∈ X there exists at least one σ ∈ G with y = σx.

If G acts on X then G acts on the set of subobjects of X in the obvious way, i. e. for σ ∈ G, σ(Y) = {σ(y) | y ∈ Y }. If Y is a subobject of X then we say that Y is invariant under G if σY ⊆ Y for all g ∈ G. (Because the elements of G have inverses, it is easy to see that if σ(Y) ⊆ Y then σ(Y) = Y . Clearly a subobject which is fixed under G is invariant, but not necessarily conversely.

If Y is a subobject of X then we will write G_{Y} = {σ ∈ G | (∀y ∈ Y) σ(y) = y}. In other words, G_{Y} is the largest subgroup of G which fixes Y . (This is not standard notation.)

We will write G_{{Y }} = {σ ∈ G | σ(Y) = Y }. Thus G_{{Y }} is the largest subgroup of G under which Y is invariant and is the stabilizer of Y under the action of G on the set of subobjects of X. Clearly G_{{Y }} ⊆ G_{Y} .

Proposition. Let G act on a set X and let x ∈ X. If σ, σ′ ∈ G then σx = σ′x if and only if σ and σ′ are in the same left coset modulo G_{x}. Thus there is a one-to-one correspondence between the elements in the orbit Gx and the space of left cosets G/G_{x}.

Proposition. (1) For Y ⊆ X and τ ∈ G,

G_{τY} = τ^{−1}(G_{Y})τ.

(2) If Y is invariant under G then G_{Y} is a normal subgroup of G.

(3) If Y = X^{G_{Y}} then Y is invariant under G if and only if G_{Y} is a normal subgroup of G. proof: (1) If σ ∈ G then σ ∈ G_{τY} and and only if for all y ∈ Y , σ(τy) = τy, or equivalently τ^{−1}στy = y, i. e. if and only if τ^{−1}στ ∈ G_{Y} .

(2) For τ ∈ G, if τY = Y then τ^{−1}G_{Y} τ = G_{τY} = G_{Y} . Thus if τY = Y for all τ, then G_{Y} is a normal subgroup of G.

(3) Conversely, suppose that Y = X^{G_{Y}} and G_{Y} is a normal subgroup of G. Then for all τ ∈ G, G_{τY} = G_{Y} . Thus if y ∈ τY then for all σ ∈ G_{Y} , σ(y) = y, so that y ∈ X^{G_{Y}} = Y . Thus τY ⊆ Y for all τ ∈ G, showing that Y is invariant under G. ✓

Lemma. A finite subgroup of the multiplicative group of a field is cyclic.

Lemma. If V is a vector field over an infinite field K, then V is not a (set-theoretic) union of a finite number of proper subspaces. proof: Suppose that V = V_{1} ∪ ⋯ ∪ V_{n} with V_{i} ⊊ V . We may assume wlog V is not a union of any proper subset of {V_{1}, …, V_{n}}. Thus we may choose x ∈ V_{1} with x ∉ V_{2} ∪ ⋯ ∪ V_{n} and y ∈ V_{2} with y ∉ V_{1} . Then for any k ≠ 0 ∈ K, kx + y ∉ V_{1} ∪ V_{2} (why?) so kx + y ∈ V_{i} for i ≥ 3. Since K is infinite and there are only finitely many V_{i}, it follows that there exist k, k′ ∈ K such that kx + y, k′x + y ∈ V_{i} for some i ≥ 3. But it then follows that x ∈ V_{i}, a contradiction. ✓ Lemma. If there are only a finite number of fields E with K ⊆ E ⊆ F, then F = K(u) for some u ∈ F. proof: If K is finite then F must also be finite and so the set of non-trivial elements of F is a cyclic group under multiplication and we can choose u to be a generator of this group.

If K is infinite then by the preceding lemma, there exists u ∈ F with u ∉ ⋃ E, where the union is taken over the proper subfields of F containing K. Then K(u) must be all of F. ✓

Lemma. If K ⊆ F ⊆ E and σ_{1}, …, σ_{n} ∈ Alg_{K}(F, E) are such that no σ_{i} fixes F, then there exists u ∈ F such that σ_{i}(u) ≠ u for all i. proof: If K is finite, we may assume that F is finite and choose u to be a generator for its multiplicative group. If K is infinite then, for each i, let F_{i} = {f ∈ F | σ_{i}(f) = f}. By hypothesis, each F_{i} is a proper subfield of F. By a previous lemma, F cannot be a union of this set of proper subfields. Thus there exists u ∈ F with σ_{i}(u) ≠ u for all i. ✓

Lemma. Let F be a finite separable extension of K and K ⊆ L ⊆ F. Let E be a normal extension of K containing F. If the only σ ∈ Alg_{K}(F, E) which fixes L is the inclusion F ↪ E, then L = F. proof: Since F is separable over K, the number of maps in Alg_{K}(F, E) is [F: K]. Now if σ ≠ σ′ ∈ Alg_{K}(F, E) then σ^{−1}σ′ does not fix F, and hence by hypothesis does not fix L. I. e. σ and σ′ have different restrictions to L. Thus there are at least [F: K] distinct K-homomorphisms from L to E so that [L: K] ≥ [L: K]_{s} ≥ [F: K] ≥ [L: K]. It follows that F = L. ✓

Theorem of the Primitive Element. (1) If F is a finite separable extension of K then F = K(u) for some u ∈ F.

(2) If F is a separable extension of K and there exists n such that for every u ∈ F, [K[u]: K] ≤ n, then [F: K] is finite. In fact, [F: K] ≤ n. proof: (1) Let E be a normal extension of K containing F. By a previous lemma there exists u ∈ F such that σ(u) ≠ u for every σ ∈ Alg_{K}(F, E) which does not fix F. Thus K(u) is not fixed by any σ which does not fix F. By the preceding lemma, K(u) = F.

(2) Choose u ∈ F such that [K[u]: K] is maximal. If K[u] ≠ F then there exists v ∈ F with K[u] ⊊ K[u, v]. But by (1) K[u, v] = K[w] for some w, so this yields a contradiction to the maximality of K[u]. Thus F = K[u] and [F: K] ≤ n. ✓ More Notation. If F is a galois extension of K and L a subfield of F containing K, then we will see below that F is a galois extension of L. Gal(F/L) can be identified as a subgroup of Gal(F/K), namely

Gal(F/L) = {σ ∈ Gal(F/K) | (∀ℓ ∈ L) σ(ℓ) = ℓ}.

In other words, Gal(F/L) is the set of elements in Gal(F/K) which fix L. (Hungerford uses the notation L′ for this subgroup.)

Theorem. Let H be a finite group acting faithfully on a field F and let u ∈ F. Let H_{u} be the stabilizer of u, i. e. H_{u} = {σ ∈ H | σ(u) = u}. Let

g(x) = ∏_{H/H_{u}}(X − σ(u)),

where the product is taken over a set of representatives σ for the space of left cosets H/H_{u}. Then

(1) g has distinct roots and is the minimal polynomial for u over F^{H}.

(2) F is a galois extension of F^{H}.

(3) [F: F^{H}] = |H|.

(4) Gal(F/F^{H}) = H.

(5) If F = K(u) for some subfield K of F^{H}, then the coefficients of g generate

F^{H} over K. proof: Since H acts faithfully on F and fixes F^{H}, we will think of H as a subgroup of Aut_{F^{H}} (F).

(1) Restated, one sees that g(x) = ∏(X − σ(u)), where the σ(u) (which constitute the roots of g) range over the orbit of u under H. Thus g has distinct roots. Furthermore, each τ ∈ H permutes the orbit of u, so τ(g(X)) = ∏(X − τσ(u)) = g(X) so that g ∈ F^{H}[X]. And g must be irreducible as a polynomial in F^{H}[X] since H acts transitively on its roots. Since g(u) = 0, g is thus the minimal polynomial for u over F^{H}.

(2) From (1) we see that the minimal polynomial in F^{H}[X] for every element u ∈ F has distinct roots and splits completely in F [X]. Thus F is a galois extension of F^{H}.

(3) By (1), for any u ∈ F, the minimal polynomial for u over F^{H} has degree at most |H|, so [F^{H}[u]: F^{H}] ≤ |H|. Since F/F^{H} is separable, it follows that [F: F^{H}] ≤ |H| (see the theorem of the primitive element). On the other hand, the elements of H are distinct F^{H}-homomorphism of F so |H| ≤ [F: F^{H}]_{s} = [F: F^{H}].

(4) H ⊆ Gal(F/F^{H}) and | Gal(F/F^{H})| ≤ [F: F^{H}]_{s} = [F: F^{H}] = |H|.

(5) Suppose that F = K(u) and let g as above be the minimal polynomial for u over F^{H}. Then F = F^{H}(u) so by (3), deg g = |H|. Let L be the K-subalgebra of F generated by the coefficients of g. (L is in fact a field, since it’s an integral domain with finite dimension over K.) Then g ∈ L[X] and g(u) = 0, so [F: L] = [K[u]: L] ≤ deg g = [F: F^{H}]. Since L ⊆ F^{H} by (1), it follows that

F^{H} = L. ✓

Theorem. Let F be a galois extension of K.

(1) If H is a subgroup of Gal(F/K) then Gal(F/F^{H}) = H.

(2) If K ⊆ L ⊆ F then F^{Gal(F/L)} = L.

(3) If H is a subgroup of Gal(F/K) then H is a normal subgroup if and only if F^{H} is a normal extension of K. In this case, Gal(F^{H}/K) ≈ Gal(F/K) / Gal(F/F^{H}). proof: (1) See the preceding proposition.

(2) F is a galois extension of L. Now suppose that u ∈ F and u ∉ L. Then there exists a L-homomorphism σ: F → F such that σ(u) ≠ u. Then σ ∈ Gal(F/L), so u ∉ F^{Gal(F/L)} . Thus F^{Gal(F/L)} ⊆ L. But the converse inclusion is trivial.

(3) See a previous proposition. ✓

Hilbert Theorem 90 [Hungerford, p. 292]. If F is a galois extension of K and Gal(F/K) is cyclic generated by σ, then an element u ∈ F has norm 1 if and only if u = v/σ(v) for some v ∈ F.

Corollary. If K contains a primitive dth root of unity and F is a galois extension of K degree d such that Gal(F/K) is cyclic, then F is generated over K by a dth root of some element of K. proof: let ζ ∈ K be a primitive dth root of unity. Then ζ has norm 1 so by the Hilbert Theorem 90, ζ = v/σ(v) for some v ∈ F. Then σ(v) = ζ^{−1}v and so σ(v^{d}) = ζ^{−d}v^{d} = v^{d}. Since σ generates Gal(F/K), thus v^{d} ∈ K. Furthermore, since σ^{i}(v) = ζ^{−i}v ≠ v for 0 < i < d, Gal(F/K[v]) = {1}. Thus F = K[v]. ✓

For a little bit of historical backgroud, see Scientific American, April, 1982, pp. 136–49, The Short Life of Evariste Galois.