Separable Extensions

Properties of Étale Algebras

We now define the notion of a separable extension. First, we will examine the properties of étale algebras discussed in the previous post.

Lemma 1 For an algebraically closed field \(\mathbb{K}\), suppose that a commutative \(\mathbb{K}\)-algebra \(A\) of finite degree satisfies \(\Omega_{A/\mathbb{K}}=0\). Then in \(A\), for any maximal ideal \(\mathfrak{m}\), we have \(\mathfrak{m}=\mathfrak{m}^2\).

Proof

The field \(A/\mathfrak{m}\) is a finite degree extension of the algebraically closed field \(\mathbb{K}\), so it must be an algebraic extension, and therefore \([A/\mathfrak{m}:\mathbb{K}]=1\). Thus for each \(a\in A\), to have \(a-\lambda\in \mathfrak{m}\), we can find \(\lambda\in \mathbb{K}\). Then the map

\[A \rightarrow \mathfrak{m}/\mathfrak{m}^2;\qquad a\mapsto a-\lambda\]

defines a function \(D:A \rightarrow \mathfrak{m}/\mathfrak{m}^2\) that is evidently a \(\mathbb{K}\)-derivation. Now from the universal property of [Multilinear Algebra] §Differential Modules, ⁋Proposition 8, \(D\) must factor through \(\Omega_{A/\mathbb{K}}\), which is \(0\) by assumption, so \(D=0\).

To establish our characterization of étale algebras, we need to show the following lemma, which starts from the result of the above lemma.

Lemma 2 For a commutative ring \(A\), if a finitely generated ideal \(\mathfrak{a}\) satisfies \(\mathfrak{a}=\mathfrak{a}^2\), then there exists an idempotent \(e\in A\) such that \(\mathfrak{a}=Ae\).

Proof

First, let the generators of \(\mathfrak{a}\) be \(a_1,\ldots,a_r\). Then, since \(\mathfrak{a}=\mathfrak{a}^2\), for each \(i\), the equation

\[a_i=\sum_{j=1}^r x_{ij}a_j\]

holds for some \(x_{ij}\). For the resulting \(r\times r\) matrix, let \(I_r-(x_{ij})\) be \(M\). Then, through the adjoint matrix of \(M\), the equation

\[NM=(\det M) I_r\]

holds for some \(r\times r\) matrix \(N\). Now calculating where the column vector corresponding to \(a_j\) goes on both sides, we know that \((\det M)a_j=0\) must hold for all \(j\), and therefore \((\det M)\mathfrak{a}=0\) holds.

On the other hand, by definition the matrix \(M\), after reduction modulo \(\mathfrak{a}\), becomes \(I_r\), so its determinant \(\det M\) is also (modulo \(\mathfrak{a}\)) \(1\). Now setting \(e=1-\det M\), \(e\) is the desired idempotent.

Then the differential characterization of étale algebras is given as follows.

Theorem 3 For a finite degree commutative \(\mathbb{K}\)-algebra \(A\), \(A\) is étale if and only if \(\Omega_{A/\mathbb{K}}=0\).

Proof

First, since \(\Omega_{A/\mathbb{K}}\) and the notion of an étale algebra behave well under base change, we may assume that \(\mathbb{K}\) is algebraically closed. More precisely, \(A\) being étale is equivalent to the \(\overline{\mathbb{K}}\)-algebra \(A_{(\overline{\mathbb{K}})}=A\otimes_\mathbb{K}\overline{\mathbb{K}}\) being diagonalizable for the algebraic closure \(\overline{\mathbb{K}}\) of \(\mathbb{K}\) (§Étale Algebras, ⁋Proposition 8), while by the canonical isomorphism of [Multilinear Algebra] §Differential Modules, ⁋Proposition 12 and the definition of the \(\overline{\mathbb{K}}\)-algebra \(A_{(\overline{\mathbb{K}})}\) we obtain the isomorphism

\[\Omega_{A_{(\overline{\mathbb{K}})}/\overline{\mathbb{K}}}\cong \Omega_{A/\mathbb{K}}\otimes_A A_{(\overline{\mathbb{K}})}=\Omega_{A/\mathbb{K}}\otimes_A A\otimes_\mathbb{K}\overline{\mathbb{K}}\cong \Omega_{A/\mathbb{K}}\otimes_\mathbb{K}\overline{\mathbb{K}}\]

and therefore \(\Omega_{A/\mathbb{K}}=0\) is equivalent to \(\Omega_{A_{(\overline{\mathbb{K}})}/\overline{\mathbb{K}}}=0\). That is, to prove the given theorem, it suffices to assume that \(\mathbb{K}\) is algebraically closed and show that \(A\) being diagonalizable is equivalent to \(\Omega_{A/\mathbb{K}}=0\).

First, assuming that \(A\) is diagonalizable, \(A\) is generated by idempotents by the second condition of §Étale Algebras, ⁋Proposition 6. But for any idempotent \(e\), considering the Leibniz rule,

\[\dd{(e)}=\dd{(e^2)}=e\dd{(e)}+e\dd{(e)}=2e\dd{(e)}\]

and multiplying both sides by \(e\) gives \(e\dd{(e)}=0\), so substituting this back into the above shows that \(\dd{e}=0\). On the other hand, since \(\Omega_{A/\mathbb{K}}\) is explicitly presented as the free module generated by such elements \(\dd{e}\) modulo appropriate relations ([Multilinear Algebra] §Differential Modules, ⁋Example 10), from this we see that \(\Omega_{A/\mathbb{K}}=0\).

Now conversely, assume \(\Omega_{A/\mathbb{K}}=0\) and let us show that \(A\) is diagonalizable. We proceed by induction on the degree of \(A\), and the case of degree \(1\) is trivial. Now in the general case, choose a maximal ideal \(\mathfrak{m}\) of \(A\). Then by Lemma 1 we have \(\mathfrak{m}=\mathfrak{m}^2\), and therefore applying Lemma 2 we can find an idempotent \(e\) such that \(\mathfrak{m}=Ae\). Meanwhile, from this, if we decompose \(A\) into the direct sum of \(\mathfrak{a}=(1-e)A\) and \(\mathfrak{m}\), from the assumption that \(\mathbb{K}\) is algebraically closed we see that the extension \(A/\mathfrak{m}\) must have degree \(1\). That is, we can write this direct sum as

\[A\cong \mathfrak{a}\oplus\mathfrak{m}\cong \mathbb{K}\times A/\mathfrak{a}\]

Now since \(\Omega\) is a right exact functor ([Multilinear Algebra] §Differential Modules, ⁋Proposition 13), \(\Omega_{(A/\mathfrak{a})/\mathbb{K}}\) is a quotient of \(\Omega_{A/\mathbb{K}}=0\) and hence is \(0\). Therefore, \(A/\mathfrak{a}\) is diagonalizable by the induction hypothesis, and so is \(A\), which is its product with \(\mathbb{K}\).

Earlier, in §Purely Inseparable Extensions, ⁋Example 9, we examined a situation that could cause problems when developing Galois theory; based on this example, let us examine Theorem 3.

Example 4 A favorable case, as we saw in the introduction to §Purely Inseparable Extensions, is \(\mathbb{Q}(\sqrt{2})/\mathbb{Q}\). From the computation in [Multilinear Algebra] §Differential Modules, ⁋Example 10, we know that \(\Omega_{\mathbb{Q}[\x]/\mathbb{Q}}\) is generated by \(\dd{\x}\) as a free \(\mathbb{Q}[\x]\)-module. On the other hand, considering the ideal of \(\mathbb{Q}[\x]\) given by \(\mathfrak{I}=(\x^2-2)\), from [Multilinear Algebra] §Differential Modules, ⁋Proposition 14 we have the following exact sequence

\[\mathfrak{I}/\mathfrak{I}^2\overset{\overline{d}}{\longrightarrow}\Omega_{\mathbb{Q}[\x]/\mathbb{Q}}\otimes_{\mathbb{Q}[\x]}\mathbb{Q}(\sqrt{2})\overset{\Omega_0(u)}{\longrightarrow}\Omega_{\mathbb{Q}(\sqrt{2})/\mathbb{Q}}\longrightarrow 0\]

from which we know that \(\Omega_{\mathbb{Q}(\sqrt{2})/\mathbb{Q}}\) is generated by the element \(\dd{(\sqrt{2})}\) as a \(\mathbb{Q}(\sqrt{2})\)-module. But from the computation

\[0=\dd{(2)}=\dd{((\sqrt{2})^2)}=2\sqrt{2}\dd{(\sqrt{2})}\]

and the fact that \(2\sqrt{2}\) is invertible in \(\mathbb{Q}(\sqrt{2})\), we know that \(\dd{(\sqrt{2})}=0\), and therefore \(\Omega_{\mathbb{Q}(\sqrt{2})/\mathbb{Q}}=0\).

On the other hand, for the algebraic extension of \(\mathbb{K}=\mathbb{F}_p(t)\) given by \(\mathbb{K}(t^{1/p})=\mathbb{K}[\x]/(\x^p-t)\) examined in §Purely Inseparable Extensions, ⁋Example 9, the computation above breaks down; just as in the computation above, \(\Omega_{\mathbb{K}(t^{1/p})/\mathbb{K}}\) is generated by \(\dd{(t^{1/p})}\) as a \(\mathbb{K}(t^{1/p})\)-module, but the computation

\[0=\dd{(t)}=\dd{((t^{1/p})^p)}=p(t^{1/p})^{p-1}\dd{(t^{1/p})}\]

does not yield the result \(\dd{(t^{1/p})}=0\). Since \(\mathbb{K}(t^{1/p})\) has characteristic \(p\) anyway, \(p(t^{1/p})^{p-1}\) is \(0\), and thus the above equation gives no relation whatsoever on \(\dd{(t^{1/p})}\). Indeed, explicitly considering any field \(\mathbb{K}\) and algebraic extension \(\mathbb{K}(\alpha)=\mathbb{K}[\x]/(f)\), we have

\[\Omega_{(\mathbb{K}[\x]/(f))/\mathbb{K}}\cong\frac{\Omega_{\mathbb{K}[\x]/\mathbb{K}}\otimes_{\mathbb{K}[\x]}\mathbb{K}(\alpha)}{\mathfrak{I}/\mathfrak{I}^2}\cong\frac{ {\mathbb{K}[\x]\dd{\x}}\otimes_{\mathbb{K}[\x]}\mathbb{K}[\x]/(f)}{(\dd{f})}\cong \frac{\mathbb{K}[\x]}{(f, f')}\dd{\x}\]

so the two computations above follow from this.

The case we wish to exclude is precisely when the minimal polynomial \(f\) has a multiple root, that is, when \(\dd{f}=0\); thus the notion of an étale algebra will be useful. We shall soon define a separable extension as a field extension in which every finite degree subextension is étale. (Definition 8) Then, as we saw in Example 4, any algebraic extension of \(\mathbb{Q}\) is a separable extension. Furthermore, we will show that any algebraic extension of a perfect field is separable. (Proposition 9) Let us examine a few more properties of étale algebras that we will use in this process.

Proposition 5 For any field \(\mathbb{K}\), a finite degree, commutative \(\mathbb{K}\)-algebra \(A\) is reduced if and only if there exist suitable finite degree field extensions of \(\mathbb{K}\), \(\mathbb{L}_1,\ldots, \mathbb{L}_n\), such that \(A\) is isomorphic to \(\mathbb{L}_1\times\cdots\times \mathbb{L}_n\) as a \(\mathbb{K}\)-algebra.

Proof

Since \(\mathbb{L}_1\times\cdots\times \mathbb{L}_n\) is reduced, one direction is trivial. Conversely, let \(A\) be a finite degree reduced commutative \(\mathbb{K}\)-algebra. If \(A\) is a field there is nothing more to show, so it suffices to prove the case where \(A\) is not a field, and as usual by induction on the degree of \(A\) it is enough to show that any (non-field) reduced algebra \(A\) can always be written as a nontrivial product \(A_1\times A_2\).

To see this, we show that \(A\) has an idempotent other than \(0,1\). Since any ideal of \(A\) is a finite-dimensional \(\mathbb{K}\)-vector space, among the ideals of \(A\) other than \(0\), we can choose an ideal of smallest dimension as a \(\mathbb{K}\)-vector space, \(\mathfrak{a}\). Then by the minimality of \(\mathfrak{a}\) and the assumption that \(A\) is reduced, we have \(\mathfrak{a}^2=\mathfrak{a}\), and therefore Lemma 2 can be applied.

Then the claim of Proposition 9 mentioned above is essentially contained in the following lemma.

Lemma 6 For a perfect field \(\mathbb{K}\), any finite degree, reduced \(\mathbb{K}\)-algebra is étale.

Proof

Consider a \(\mathbb{K}\)-algebra \(A\) satisfying the conditions of the claim. First, by Proposition 5, there exist extensions such that \(A\cong \mathbb{L}_1\times\cdots \times\mathbb{L}_n\). On the other hand, since the product of étale algebras is étale, for the given claim it suffices to show that any finite degree field extension over a perfect field \(\mathbb{K}\) is étale. Therefore, applying Theorem 3, it suffices to show that for any \(\alpha\in A\), \(\dd{\alpha}=0\). By the computation in Example 4, we know that the equation \(f'(\alpha)\dd{\alpha}=0\) must hold, and by similar logic we must show that \(f'(\alpha)\neq 0\).

Suppose for contradiction that \(f'(\alpha)=0\). In any case, if \(f\) is constant there is nothing to prove, so we may assume \(f\) has degree at least one; then by the result of §Fields, ⁋Proposition 19, we have \(\ch(\mathbb{K})=p\neq 0\) and \(f\in \mathbb{K}[\x^p]=\mathbb{K}[\x]^p\). But \(f\) is a minimal polynomial, hence irreducible, and an irreducible polynomial cannot belong to \(\mathbb{K}[\x]^p\) by definition, which is a contradiction. Therefore \(f'(\alpha)\neq 0\).

Using the above lemma, we can now produce another characterization of étale algebras as follows.

Theorem 7 For any finite degree commutative \(\mathbb{K}\)-algebra \(A\), the following are all equivalent.

1. \(A\) is étale.
2. For any extension \(\mathbb{L}/\mathbb{K}\), \(A_{(\mathbb{L})}=\mathbb{L}\otimes_\mathbb{K}A\) is reduced.
3. There exists an extension \(\mathbb{L}/\mathbb{K}\) such that \(\mathbb{L}\) is perfect and \(A_{(\mathbb{L})}\) is reduced.

Proof

That the second condition implies the third is trivial from the existence of a perfect closure, and that the first condition implies the second is the result of §Étale Algebras, ⁋Proposition 8. Thus it suffices to show that the third condition implies the first. Now this third result follows again from Lemma 6 and §Étale Algebras, ⁋Proposition 8.

Separable Extensions

The time has finally come to state the following definition.

Definition 8 An algebraic extension \(\mathbb{L}/\mathbb{K}\) is called a separable extension if every finite degree subextension of \(\mathbb{L}/\mathbb{K}\) is an étale \(\mathbb{K}\)-algebra.

We can define this notion for non-algebraic field extensions as well, but for now we only consider separable algebraic extensions.

By definition, for any finite degree \(\mathbb{K}\)-algebra \(\mathbb{L}\) that is itself a field, being étale and being separable are the same thing. From the definition, if \(\mathbb{L}/\mathbb{K}\) is separable then any subextension is separable, and conversely if all (finite degree) subextensions are separable then \(\mathbb{L}/\mathbb{K}\) is also separable; both are trivial.

The claim promised earlier can now also be easily shown.

Proposition 9 A field \(\mathbb{K}\) is perfect if and only if every algebraic extension \(\mathbb{L}/\mathbb{K}\) is separable.

Proof

Since every finite degree subextension of any algebraic extension \(\mathbb{L}/\mathbb{K}\) is a finite degree reduced \(\mathbb{K}\)-algebra, if \(\mathbb{K}\) is perfect then one direction is trivial from Lemma 6. Thus it suffices to show the converse.

Suppose for contradiction that \(\mathbb{K}\) is not perfect, and thus has characteristic \(p\neq 0\). Then from the assumption that \(\mathbb{K}\) is not perfect, we can consider a (relative in an algebraic closure \(\overline{\mathbb{K}}/\mathbb{K}\)) \(p\)-radical extension \(\mathbb{K}(a)/\mathbb{K}\). On the other hand, from the embedding \(\mathbb{K}\hookrightarrow\overline{\mathbb{K}}\), the resulting map \(\mathbb{K}(a)\rightarrow\overline{\mathbb{K}}\) is unique by §Purely Inseparable Extensions, ⁋Proposition 6. In other words, the set \(\Hom_{\Alg{\mathbb{K}}}(\mathbb{K}(a), \overline{\mathbb{K}})\) is a singleton, and therefore

\[1=[\mathbb{K}(a):\mathbb{K}]_s\lneq [\mathbb{K}(a):\mathbb{K}]=p^e\]

so \(\mathbb{K}(a)\) is not étale and therefore not separable.

The proof of the above proposition justifies calling \([\mathbb{K}(a):\mathbb{K}]_s\) the separable degree. On the other hand, even if \(\mathbb{K}\) is not perfect, a particular polynomial \(f\) may not have multiple roots. The following proposition is merely a compilation of what we have observed so far.

Proposition 10 For an irreducible polynomial \(f\in \mathbb{K}[\x]\), the following are all equivalent.

1. \(f'\) is not \(0\).
2. \(f\) and \(f'\) are coprime in \(\mathbb{K}[\x]\).
3. \(f\) has a root in \(\mathbb{L}[\x]\) that is not a multiple root for some extension \(\mathbb{L}/\mathbb{K}\).
4. \(f\) splits into a product of distinct linear factors in \(\mathbb{L}[\x]\) for some extension \(\mathbb{L}/\mathbb{K}\).
5. \(f\) has no multiple roots in \(\overline{\mathbb{K}}/\mathbb{K}\).
6. \(\mathbb{K}[\x]/(f)\) is an étale \(\mathbb{K}\)-algebra.
7. \(\mathbb{K}\) has characteristic \(0\), or \(\ch(\mathbb{K})=p\) and \(f\not\in\mathbb{K}[\x^p]\).

A polynomial \(f\) satisfying these conditions is called a separable polynomial. Then restating Proposition 9 once more, we see that \(\mathbb{K}\) being perfect is equivalent to every irreducible polynomial in \(\mathbb{K}[\x]\) being separable. Focusing on the element added through \(f\), we can define as follows.

Definition 11 For a field extension \(\mathbb{L}/\mathbb{K}\), an algebraic element \(x\in \mathbb{L}\) is called a separable element if \(\mathbb{K}(x)/\mathbb{K}\) is a separable extension.

By definition, if \(f\) is the minimal polynomial of \(x\), then \(f\) must be separable, and in this case \(x\) is a simple root of \(f\). These concepts all (of course) mean the same thing. That is, for an extension \(\mathbb{L}/\mathbb{K}\), an element \(x\in\mathbb{L}\), and the minimal polynomial \(f\) of \(x\), the following are all equivalent:

1. \(x\) is separable.
2. \(f\) is separable.
3. \(x\) is a simple root of \(f\).

Moreover, just as an extension in which every element is algebraic is an algebraic extension, the following holds.

Proposition 12 For an algebraic extension \(\mathbb{L}/\mathbb{K}\), the following hold.

1. If \(\mathbb{L}/\mathbb{K}\) is separable, then every element of \(\mathbb{L}\) is separable.
2. Conversely, if \(A\) is a set of separable elements of \(\mathbb{L}\) and \(\mathbb{L}=\mathbb{K}(A)\), then \(\mathbb{L}\) is a separable extension.

Proof

The first result is trivial, so it suffices to show the second. That is, we must take an arbitrary finite degree subextension \(\mathbb{M}\) of \(\mathbb{L}\) and show that it is étale. First, since \(\mathbb{L}=\mathbb{K}(A)\) and \(\mathbb{M}\) is a finite degree subextension of \(\mathbb{L}\), we can choose finitely many elements \(x_1,\ldots, x_m\) from \(A\) such that

\[\mathbb{M}\subseteq \mathbb{K}(x_1,\ldots, x_m)=\mathbb{K}[x_1,\ldots, x_m]\]

Here, that each \(\mathbb{K}[x_i]\) is a separable extension is trivial from the assumption on \(A\), so they are étale; then \(\mathbb{K}[x_1,\ldots, x_m]\) is obtained from their tensor product \(\mathbb{K}[x_1]\otimes\cdots\otimes \mathbb{K}[x_m]\) by quotienting by the relations expressing their associativity and commutativity, and since \(\mathbb{M}\) is a subalgebra of this, \(\mathbb{M}\) is étale by §Étale Algebras, ⁋Corollary 14.

Furthermore, a finite degree separable extension can be generated by a single element. That is, if \(\mathbb{L}/\mathbb{K}\) is a finite degree separable extension, we can choose an appropriate \(x\in \mathbb{L}\) such that \(\mathbb{L}=\mathbb{K}[x]\). Such an element is called a primitive element.

Theorem 14 shows that a primitive element can always be found for finite degree separable extensions. For this, we need the following lemma.

Lemma 13 For an infinite field \(\mathbb{K}\), fix a commutative \(\mathbb{K}\)-algebra \(A\). If \(A\) has only finitely many subalgebras, and \(V\) is a subspace generating \(A\), then there exists \(x\in V\) such that \(A=\mathbb{K}[x]\).

Proof

Suppose that the subalgebras of \(A\) other than \(A\) itself are only \(A_1,\ldots, A_n\). Then, first, since \(V\) generates \(A\), it cannot generate any of the \(A_i\); that is, \(V\not\subseteq A_i\). If we show from this that \(V\not\subseteq A_1\cup\cdots\cup A_n\), then for \(x\in V\setminus A_1\cup\cdots \cup A_n\), the subalgebra of \(A\), \(\mathbb{K}[x]\), cannot equal any \(A_i\), and therefore we must have \(A=\mathbb{K}[x]\), which will prove what is desired.

Suppose for contradiction that \(V\subseteq A_1\cup\cdots\cup A_n\). From the given assumption, \(V\not\subseteq A_n\), so there exists some \(x\in V\setminus A_n\). Then for any \(y\in V\), we can consider the infinite set

\[\{x\}\cup \{y+\lambda x\mid \lambda\in\mathbb{K}\}\]

This set is a subset of \(V\) and therefore a subset of \(A_1\cup\cdots\cup A_n\). Hence, by the pigeonhole principle, at least two of these must belong to the same \(A_i\), and from this fact we know that \(x,y\) are simultaneously contained in some \(A_i\). By assumption, \(x\not\in A_n\), so \(i\) cannot be \(n\), and therefore \(y\) (just like \(x\)) must belong to one of \(A_1,\ldots, A_{n-1}\). From this we see that \(y\) must belong to \(A_1\cup\cdots \cup A_{n-1}\), and since \(y\) is an arbitrary element of \(V\), we have \(V\subseteq A_1\cup\cdots\cup A_{n-1}\).

Now, repeating this process inductively, we obtain \(V\subseteq A_1\), which is a contradiction; therefore we must have \(V\not\subseteq A_1\cup\cdots\cup A_n\).

Theorem 14 Fix an infinite field \(\mathbb{K}\). For an algebraic extension \(\mathbb{L}/\mathbb{K}\), the following are equivalent.

1. \(\mathbb{L}\) has a primitive element.
2. \(\mathbb{L}\) has only finitely many subextensions.

Proof

First, suppose \(\mathbb{L}\) has a primitive element \(x\), and let \(f(\x)\in \mathbb{K}[\x]\) be the minimal polynomial of \(x\). If we consider a monic polynomial dividing the polynomial \(f(\x)\) in \(\mathbb{L}[\x]\), namely \(g(\x)\in \mathbb{L}[\x]\), then whenever such a polynomial \(g\) is given, we can consider the subextension generated by the coefficients of \(g\) in \(\mathbb{L}\). Denote this by \(\mathbb{K}_g\). Then our claim is that these \(\mathbb{K}_g\) are precisely the subextensions of \(\mathbb{L}\). In particular, if \(f\) has degree \(d\), there are at most \(2^d\) such extensions, so the second condition will hold.

To prove the claim, let \(\mathbb{M}\) be an arbitrary subextension. Then since \(x\) is a primitive element, \(\mathbb{M}[x]=\mathbb{L}\) holds. That is, \(x\) is an algebraic element of the extension \(\mathbb{L}/\mathbb{M}\), so we can choose a minimal polynomial \(h(\x)\in\mathbb{M}[\x]\), which by §Algebraic Extensions, ⁋Theorem 15 is a monic polynomial in \(\mathbb{L}[\x]\) dividing \(f\). Thus \(\mathbb{K}_h\) is defined as above, and by its definition \(\mathbb{K}_h\subseteq\mathbb{M}\). On the other hand, since \(x\) is a primitive element of \(\mathbb{L}/\mathbb{K}\), we have the equality

\[\mathbb{K}_h[x]=\mathbb{M}[x]=\mathbb{L}\]

But by definition \([\mathbb{L}:\mathbb{M}]=\deg h\), and since \(h(\x)\in \mathbb{K}_h[\x]\) satisfies \(h(x)=0\), we have \([\mathbb{L}:\mathbb{K}_h]\leq\deg h\). From this we conclude that \(\mathbb{K}_h=\mathbb{M}\) must hold.

Now assuming the second condition, since \(\mathbb{L}/\mathbb{K}\) has only finitely many subextensions, setting \(A=\mathbb{L}\) satisfies the hypotheses of Lemma 13, and we obtain the desired result.

In particular, if \(\mathbb{L}/\mathbb{K}\) is a finite degree separable extension, then it is in particular a finite degree étale \(\mathbb{K}\)-algebra, and therefore by §Étale Algebras, ⁋Proposition 9 we know that the second condition of the above theorem holds.

Theorem 14 always holds even when \(\mathbb{K}\) is a finite field, but to prove this requires a somewhat more delicate counting argument than Lemma 13, so we postpone it.

On the other hand, separability is essentially (almost) an étale algebra, and since étale algebras behave well under base change (§Étale Algebras, ⁋Corollary 14), with slight modifications as in the proof of Proposition 12, we can show that separability also behaves well under base change in the following two cases.

Proposition 15 For an algebraic extension \(\mathbb{M}/\mathbb{L}/\mathbb{K}\), \(\mathbb{M}/\mathbb{K}\) is separable if and only if both \(\mathbb{M}/\mathbb{L}\) and \(\mathbb{L}/\mathbb{K}\) are separable.

Proof

If \(\mathbb{M}/\mathbb{K}\) is separable, then that its subextension \(\mathbb{L}/\mathbb{K}\) is separable is clear from the definition, and since any element of \(\mathbb{M}\) is separable over \(\mathbb{K}\) (Proposition 12), viewing it in the extension \(\mathbb{M}/\mathbb{L}\), it is also separable over \(\mathbb{L}\) (Proposition 10). That is, since every element of \(\mathbb{M}\) is separable, we know again by Proposition 12 that \(\mathbb{M}/\mathbb{L}\) is separable.

Therefore, the core of this proposition is the converse. First, if \(\mathbb{M}/\mathbb{K}\) had finite degree, then separable algebras and étale algebras are the same thing, so the claim is immediate from the third result of §Étale Algebras, ⁋Proposition 12. The idea of the proof, similar to the above, is that even if \(\mathbb{M}/\mathbb{K}\) has infinite degree, looking at a single specific element allows us to reduce this to finite degree.

Suppose an arbitrary \(x\in \mathbb{M}\) is given, and its minimal polynomial \(f\in\mathbb{L}[\x]\) is given; then \(f\) is a separable polynomial. Now for \(\mathbb{L}/\mathbb{K}\), define the subextension \(\mathbb{L}'\) to be the subextension obtained by adjoining the coefficients appearing in \(f\) to \(\mathbb{K}\). Then since \(\mathbb{L}'\) is generated by finitely many algebraic elements, \(\mathbb{L}'/\mathbb{K}\) is a finite degree extension. Now \(f\) is, in both \(\mathbb{L}\) and \(\mathbb{L}'\), the minimal polynomial of \(x\), and since it is separable in \(\mathbb{L}\), it is also separable in \(\mathbb{L}'\). That is, \(x\) is a separable element of the extension \(\mathbb{M}/\mathbb{L}'\), and setting \(\mathbb{M}'=\mathbb{L}'(x)\), \(\mathbb{M}'/\mathbb{L}'\) is a finite degree separable extension, that is, a finite degree étale algebra. On the other hand, since \(\mathbb{L}/\mathbb{K}\) is separable, its subextension \(\mathbb{L}'\) is also separable, and therefore \(\mathbb{L}'/\mathbb{K}\) is also a finite degree étale algebra. Hence their tower \(\mathbb{M}'/\mathbb{K}\) is also a finite degree étale algebra, and therefore a finite degree separable extension. That is, \(x\) is a separable element over \(\mathbb{K}\), and since the choice of \(x\) was arbitrary, we obtain the desired result.

Proposition 16 Fix a suitable extension of \(\mathbb{K}\), and suppose a subextension \(\mathbb{K'}/\mathbb{K}\) and an algebraic subextension \(\mathbb{L}/\mathbb{K}\) of this extension are given. Then for \(\mathbb{L}'=\mathbb{K}'(\mathbb{L})\), the following hold.

1. If \(\mathbb{L}/\mathbb{K}\) is separable, then so is \(\mathbb{L}'/\mathbb{K}'\).
2. Conversely, if \(\mathbb{L}'/\mathbb{K}'\) is separable and \(\mathbb{L}/\mathbb{K}, \mathbb{K}'/\mathbb{K}\) are linearly disjoint, then \(\mathbb{L}/\mathbb{K}\) is also separable.

Proof

1. By assumption, every element of \(\mathbb{L}\) is separable over \(\mathbb{K}\), so every element of \(\mathbb{K}'(\mathbb{L})\) is separable over \(\mathbb{K}'\).
2. To show this, we must show that any finite degree subextension \(\mathbb{M}/\mathbb{K}\) is étale. First, from the given assumption, \(\mathbb{M}\) and \(\mathbb{K}'\) are linearly disjoint, and thus \(\mathbb{M}_{(\mathbb{K}')}=\mathbb{M}\otimes_\mathbb{K}\mathbb{K}'\) is isomorphic to \(\mathbb{K}'(\mathbb{M})\). (§Algebraic Extensions, ⁋Proposition 10) On the other hand, since \(\mathbb{K}'(\mathbb{M})/\mathbb{K}'\) has finite degree, it is étale from the assumption that \(\mathbb{L}'/\mathbb{K}'\) is separable. Now since étale morphisms are stable under base change, we obtain the desired result. (§Étale Algebras, ⁋Corollary 14)

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References

[Bou] N. Bourbaki. Algebra II: Chapters 4–7. Springer, 2003.