Math 594. Generalized splitting fields and embeddings into a normal extension

Let E/k be an algebraic extension, and L/k a normal extension. Let us assume there exists at least one k-embedding E ↪ L.

Theorem 0.1. Fix a subextension E_{0}/k inside of E/k. Under the above assumptions, any k-embedding i : E_{0} ↪ L lifts to an E_{0}-embedding j : E ↪ L (viewing L as an extension of E_{0} via i). That is, if L/k is normal and the question of the existence of a k-embedding E into L admits an affirmative answer, then the following diagram can always be filled in to be commutative:

\xymatrix{E \ar[r]^{?} & L \\ E_{0} \ar[u] \ar[ur]^{i} \\ k \ar[u] \ar[uur]}

As a special case, if E = L then this says that for any normal extension L/k and any subfield L_{0} ⊆ L containing k, any k-embedding L_{0} ↪ L lifts to a map L ↪ L which must in fact be an automorphism (since we have seen in class that a self-map of an algebraic extension is necessarily an automorphism). For example, by composing a k-automorphism of L_{0} with the canonical inclusion of L_{0} into L, we see that every k-automorphism of L_{0} lifts to a k-automorphism of L (but be careful: not every k-automorphism of L will induce an automorphism of L_{0}, unless L_{0} happens to also be normal over k). The normality assumption in the theorem is crucial: if we let k = Q and L = Q(2^{1/4}) = Q[X]/(X^{4} − 2) with x ∈ L the residue class of X modulo X^{4} − 2, there is no k-automorphism of L which lifts the non-trivial automorphism of the subfield E = Q(x^{2}) ≃ Q(\sqrt{2}) (proof?).

Now we turn to the proof of the theorem. We begin with a definition which generalizes our earlier notion of splitting field.

Definition 0.2. Let Σ be a set of non-constant polynomials in k[T]. We define a splitting field of Σ over k to be an algebraic extension k_{Σ} of k over which all f ∈ Σ split into a product of linears, and which is generated over k by the roots of all such f.

If Σ = {f_{1}, …, f_{n}} is finite, it is clear that a splitting field of Σ is the same concept as that of a splitting field of the polynomial ∏ f_{j} in our earlier sense. Hence, we see existence and uniqueness up to k-isomorphism for finite Σ. In fact, this works in general:

Lemma 0.3. Let Σ be an arbitrary set of non-constant polynomials in k[T]. Then a splitting field of Σ over k exists and is unique up to k-isomorphism. Moreover, an algebraic extension of k is normal if and only if it is the splitting field of some such Σ.

Proof. It is clear that a splitting field of Σ is a compositum over k of splitting fields of each f ∈ Σ, so normality of the latter implies normality of the former (recall that an extension of k which is a compositum of finite normal subextensions is itself a normal extension, ultimately because of the real theorem that the splitting field of a polynomial is normal). Thus, splitting fields in our generalized sense are necessarily normal over k. Conversely, if L/k is a normal extension, then for each a ∈ L we see (by normality) that L contains a splitting field L_{a}/k for the minimal polynomial f_{a} of a over k. Thus, L is a splitting field for the set of f_{a}’s. This settles the second part of the lemma.

To handle the existence and uniqueness (up to isomorphism), note (for those who prefer finitistic proofs) that for finite Σ we already know both aspects from our earlier work with splitting fields (and the remarks preceding the lemma). To handle the general case, let \overline{k} be an algebraic closure of k. Define k_{Σ} to be the subfield of \overline{k} generated over k by all roots of all f ∈ Σ. This is clearly a splitting field of Σ, settling existence. For uniqueness, let k′ be a splitting field of Σ over k. We shall prove that k′ is k-isomorphic to the explicitly constructed subfield k_{Σ} inside of \overline{k}. By algebraicity of k′ over k, there exists a k-embedding i of k′ into \overline{k}.

Since k′ is generated over k by roots of the elements f ∈ Σ, i(k′) ⊆ \overline{k} is also generated over k by such roots, so i(k′) ⊆ k_{Σ}. This yields an injection k′ ↪ k_{Σ} which we must prove is surjective.

Let a ∈ k_{Σ} be the root of some f ∈ Σ. Such a’s generate k_{Σ} over k, so surjectivity of the k-map i will follow if each such a lies in i(k′). But k′ is a splitting field of Σ over k, so f splits into a product of linears over k′. Applying i carries this into a linear factorization of i(f) = f over k_{Σ}. One of these latter linear factors must have a as a root. Thus, some linear factor ℓ ∈ k′[T] of f has the property that i(ℓ) has a as a root. If a′ ∈ k′ is the unique root of the linear ℓ in k′, then i(a′) ∈ i(k′) is the unique root of i(ℓ) in k_{Σ}. Thus, i(a′) = a.

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Now we prove the theorem.

Proof. By the lemma, we can fix a set Σ of non-constant polynomials over k for which the normal L/k is a splitting field of Σ. That is, L is generated over k (and hence over any subfield containing k) by roots of the elements of Σ. Fix a k-embedding j of E into L. This induces a k-embedding j_{0} of E_{0} into L. Fix an arbitrary k-embedding i of E_{0} into L. Viewing Σ as a set of polynomials in E_{0}[T] and using i to embed E_{0} into L over k, L acquires the structure of a splitting field for Σ over E_{0} since L is even generated over k by roots of elements of Σ and i is a k-embedding. Since both i and j_{0} make L into a splitting field of Σ over E_{0}, by the uniqueness aspect of the lemma we conclude that there exists an automorphism σ of L carrying j_{0} into i. Thus, σ ∘ j : E → L lifts i.

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