Math 594. Composites of radical towers

Let k be a field, and K/k an extension field. Let k_{1}, k_{2} ⊆ K be two subfields containing k, and assume each k_{j} is “solvable in radicals” in the sense that there exists a k-embedding k_{j} ↪ L_{j} where L_{j}/k is a radical tower (i.e., L_{j}/k can be expressed as a finite tower of extensions, each level generated over the level below by a single element which admits some power that lies in the level below). We wish to prove that the composite field k_{1}k_{2} ⊆ K is also solvable in radicals. In other words, we want to construct an extension L/k into which k_{1}k_{2} embeds with L/k a radical tower.

A natural idea is to create a “composite” L of L_{1} and L_{2} over k (note that, a priori, L_{1} and L_{2} are not given inside of a common extension of k) and to somehow put k_{1}k_{2} inside of here compatibly with the mechanism by which each k_{j} is k-embedded into L_{j}. Such an L/k seems likely to be a radical tower over k, and hence would solve our problem. This is not quite what we will do (it may be impossible to get compatibility with the initial k-embeddings of the k_{j}’s if k_{1} ∩ k_{2} ⊆ K is larger than k), but it gives the general idea.

We emphasize that the concept of composite field k_{1}k_{2} of the k_{j}’s over k inside of K depends very much (in general) on how the k_{j}’s sit in relation to each other inside of K as extensions of k. That is, k_{1}k_{2} is not usually intrinsic to the k_{j}’s as abstract extensions of k (there is an important condition called linear disjointness which does render this compositum intrinsic to the k_{j}’s over k, but we won’t address this issue here).

1. Some embedding lemmas

The first step is to put the L_{j}’s into a common normal radical tower over k. First, we need to construct an abstract compositum of the L_{j}’s over k, which is to say an extension L/k into which each L_{j} embeds via some k-embedding ι_{j} : L_{j} ↪ L, with L generated over k by the subfields ι_{j}(L_{j}). Rather more generally, we have the following lemma which has nothing to do with radical towers.

Lemma 1.1. Let L_{1} and L_{2} be two finite extensions of a field k. Then there exists a finite extension L/k and k-embeddings ι_{j} : L_{j} ↪ L such that L is the compositum of the subfields ι_{j}(L_{j}) over k.

This lemma is true without finiteness (or even algebraicity) conditions, and is easily proven using Zorn’s Lemma and tensor products: let L be the quotient of the nonzero k-algebra L_{1} ⊗_{k} L_{2} by a maximal ideal. The finite case can be handled without using such tools, and this case is all we need, so there seems to be some merit in handling it directly.

Proof. If L_{1} = k(α) and α has minimal polynomial f over k, then let L = L_{2}[T]/g where g is an irreducible factor of f in L[T]. Define ι_{2} in the obvious way, and define

ι_{1} : L_{1} ≃ k[T]/f → L_{2}[T]/g = L

in the obvious way (via T ↦ T). Concretely, we are sending α ∈ L_{1} to the residue class T mod g ∈ L.

The general case is now easily handled by expressing L_{1} is a finite tower of primitive extensions and inducting on the number of steps in the tower. More precisely, we may suppose L_{1} = K(α) where K is a subfield over k for which there exists an “abstract compositum” F of K and L_{2} over k (i.e., there exist kembeddings of K and L_{2} into F which generate F over k), and then we take L to be an abstract compositum of F and L_{1} over K. ■

Lemma 1.2. Let L/k be an extension which is a compositum over k of two subextensions L_{1}, L_{2}/k, with each L_{j}/k a radical tower. Then L/k is a radical tower.

Proof. If L_{2} = k(α) with α^{d} = a ∈ k, then L = L_{1}(α) with α^{d} = a ∈ L_{1}, so by expressing L_{1}/k as a radical tower we get an expression for L/k as a radical tower. In general, L_{2} is an n-step radical tower over k and we induct on n (the case n = 1 having just been settled). For n > 1 we write L_{2} = K(α) where α^{d} ∈ K and K/k is a radical tower in n − 1 steps. Then F = L_{1}K is a radical tower over k by the inductive hypothesis and L = F (α) with α^{d} ∈ K ⊆ F, so we conclude that L/k is also a radical tower. ■

Quite generally, if F/k is a finite extension and F′/F is a normal closure over k, then F′ is a compositum over k of subfields σ(F) for σ ∈ Aut(F′/k). Indeed, this compositum is stable under Aut(F′/k) and thus must be normal over k since F′/k is normal, and by definition of normal closure it follows that this normal subextension of F′ which contains F has to be equal to F′. As a special case, if L/k is a radical tower, then a normal closure L′/k is a compositum over k of finitely many subextensions σ(L) each of which is abstractly k-isomorphic to L, so each σ(L) is a radical tower. By Lemma 1.2, we obtain:

Theorem 1.3. If L/k is a finite extension which is a radical tower, then so is a normal closure of L/k.

The interest of this theorem lies in its role in the proof of:

Corollary 1.4. If L_{1} and L_{2} are radical towers over k, then there exist k-embeddings ι_{j} : L_{j} ↪ L with L/k a radical tower which is normal.

Proof. By Lemma 1.1, we can find k-embeddings L_{j} ↪ F for some finite extension F/k. By replacing F with a normal closure over k, we may assume F/k is normal. The compositum K of L_{1} and L_{2} over k inside of F is a radical tower, by Lemma 1.2. Since F/k is normal, it contains a normal closure of K/k. By Theorem 1.3, this normal closure is a radical tower, and hence may be taken as L. ■

2. Main result

We have now gathered everything we need to prove the result of interest. Recall the setup: we give ourselves extensions k_{1}, k_{2}/k with k-embeddings k_{j} ↪ L_{j} where each L_{j}/k is a radical tower. We also assume the k_{j}’s are given to us inside of some extension field K/k. We want to prove that the compositum k_{1}k_{2} over k inside of K admits a k-embedding into some radical tower L/k. By Corollary 1.4, we may enlarge the L_{j}’s to reduce to the case L_{1} = L_{2} with this common extension moreover normal over k. In other words, we may assume there exists a finite normal extension L/k which is a radical tower and into which each k_{j} admits a k-embedding. We shall prove that the compositum k_{1}k_{2} ⊆ K admits a k-embedding into L. Beware that this need not be compatible with the initial k-embeddings of the k_{j}’s into L (e.g., it may happen that k_{1} ∩ k_{2} ⊆ K is strictly bigger than k and the k-embeddings k_{j} ↪ L do not agree on this overlap field).

Our problem has nothing to do with radical towers, and is really a problem concerning normal extensions. Namely, we are given k-embeddings of k_{1} and k_{2} into a normal extension L/k, and we wish to prove that there exists a k-embedding of the subfield k_{1}k_{2} ⊆ K into L. By Lemma 1.1, there exists a finite extension F/L into which K admits a k-embedding. We may even enlarge F so that it is normal over k. We pick a k-embedding ι : K ↪ F, and it suffices to show that each ι(k_{j}) lies in L, for then ι(k_{1}k_{2}) ⊆ L. What we know is that there exists some k-embedding ι_{j} : k_{j} ↪ L. Viewing this as a k-embedding of k_{j} into F, and keeping in mind that we wish to show that ι|_{k_{j}} : k_{j} ↪ F has image inside of L, we are reduced to the following problem: if F/L/k is a tower of fields with F and L normal over k, and κ/k is an extension which admits a k-embedding into L, then we claim that any k-embedding κ ↪ F has image inside of F. For each α ∈ κ, it suffices to solve this problem for the subfield k(α) ⊆ κ. But if f ∈ k[T] is the minimal polynomial of α, then the hypothesis that k(α) admits a k-embedding into L implies that f has a root in L, and then normality of L/k implies that f splits over L. Hence, all roots of f in F lie in L. Thus, any k-embedding of k(α) into F must send α to a root of f in F, and all such roots lie in L. This solves our problem.