Math 594. Uniqueness of algebraic closure

Let k be a field, and \overline{k}/k a choice of algebraic closure. As a first step in the direction of proving that \overline{k} is “unique up to (non-unique) isomorphism”, we prove:

Lemma 0.1. Let L/k be an algebraic extension, and L′/L another algebraic extension. There exists a k-embedding i : L ↪ \overline{k}, and once i is picked there exists a k-embedding L′ ↪ \overline{k} extending i.

Proof. Since an embedding i : L ↪ \overline{k} realizes the algebraically closed \overline{k} as an algebraic extension of L (and hence as an algebraic closure of L), by renaming the base field as L it suffices to just prove the first part: any algebraic extension admits an embedding into a specified algebraic closure.

Define Σ to be the set of pairs (k′, i) where k′ ⊆ L is an intermediate extension over k and i : k′ ↪ \overline{k} is a k-embedding. Using the inclusion i_{0} : k ↪ \overline{k} that comes along with the data of how \overline{k} is realized as an algebraic closure of k, we see that (k, i_{0}) ∈ Σ, so Σ is non-empty. We wish to apply Zorn’s Lemma, where we define a partial ordering on Σ by the condition that (k′, i′) ≤ (k′′, i′′) if k′ ⊆ k′′ inside of L and i′′|_{k′} = i′. It is a simple exercise in gluing set maps to see that the hypothesis of Zorn’s Lemma is satisfied, so there exists a maximal element (K, i) ∈ Σ.

We just have to show K = L. Pick x ∈ L, so x is algebraic over K (as it is algebraic over k). If f_{x} ∈ K[T] is the minimal polynomial of x, then K(x) ≃ K[T]/f_{x}. Using i : K ↪ \overline{k} realizes \overline{k} as an algebraic closure of K, so f_{x} ∈ K[T] has a root in \overline{k}. Pick such a root, say r, and then we define K[T] → \overline{k} by using i on the coefficients K and sending T to r. This map kills f_{x}, and hence factors through the quotient to define a map of fields K[T]/f_{x} ↪ \overline{k} extending i. Composing this with the isomorphism K(x) ≃ K[T]/f_{x} therefore defines an element (K(x), i′) ∈ Σ which dominates (K, i). By maximality, this forces (K(x), i′) = (K, i), or in other words K(x) = K as subfields of L. This holds for all x ∈ L, and says exactly x ∈ K. Thus, L = K, as desired.

■

As an application of the lemma, we get:

Theorem 0.2. Let \overline{k}_{1} and \overline{k}_{2} be two algebraic closures of k. Then there exists an isomorphism \overline{k}_{1} ≃ \overline{k}_{2} over k.

Beware that the isomorphism in the theorem is nearly always highly non-unique (it can be composed with any k-automorphism of \overline{k}_{2}, of which there are many in general). Thus, one should never write \overline{k}_{1} = \overline{k}_{2}; always keep track of the choice of isomorphism. In particular, always speak of an algebraic closure rather than the algebraic closure; there is no “preferred” algebraic closure except in cases when there are no nontrivial automorphisms over k (which happens for fields which have the property of being “separably closed”, a notion we’ll encounter later).

Proof. By the lemma, applied to L = \overline{k}_{1} (algebraic over k) and \overline{k} = \overline{k}_{2} (an algebraically closed field equipped with a structure of algebraic extension of k), there exists a k-embedding i : \overline{k}_{1} ↪ \overline{k}_{2}. Since \overline{k}_{1} is algebraic over k and \overline{k}_{2} is algebraically closed, it follows that the k-embedding i realizes \overline{k}_{2} as an algebraic extension of \overline{k}_{1}. But an algebraically closed field (such as \overline{k}_{1}) admits no non-trivial algebraic extensions (why not?), so the map i is forced to be an isomorphism. More concretely, any y ∈ \overline{k}_{2} is a root of an irreducible monic f ∈ k[T], and f = ∏(T − r_{j}) in \overline{k}_{1}[T] since \overline{k}_{1} is algebraically closed, so applying i shows that the i(r_{j})’s exhaust the roots of f in \overline{k}_{2}. Thus, y = i(r_{j}) for some j, so indeed i is surjective.

■