This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Radical Extensions
The most inseparable extension we can think of is, of course, a \(p\)-radical extension. For this reason, a \(p\)-radical extension is often called a purely inseparable extension. In this post, we first examine the relationship between \(p\)-radical extensions and separable extensions, and explain this using the separable degree.
Let us first prove the following lemma.
Lemma 1 For a field \(\mathbb{K}\) of characteristic \(p\neq 0\), a finite degree \(\mathbb{K}\)-algebra \(A\) is étale if and only if \(A=\mathbb{K}[A^p]\). Moreover, in this case, if \((a_i)\) is a \(\mathbb{K}\)-basis of \(A\), then so are \((a_i^p)\).
Proof
Let us consider an algebraic closure \(\overline{\mathbb{K}}\) of \(\mathbb{K}\). Then for any \(u,v\in \Hom_\Alg{\mathbb{K}}(A, \overline{\mathbb{K}})\), if the restrictions of \(u,v\) to the subalgebra \(\mathbb{K}[A^p]\) are equal, then the equation
\[u(x)^p=u(x^p)=v(x^p)=v(x)^p\]holds for all \(x\in A\), so by definition we know that the inequality
\[[A:\mathbb{K}]_s\leq[\mathbb{K}[A^p]:\mathbb{K}]_s\]holds. If \(A\) is an étale \(\mathbb{K}\)-algebra, then by the inequality and its equality condition in §Étale Algebras, ⁋Proposition 13, the equation
\[[A:\mathbb{K}]=[A:\mathbb{K}]_s\leq[\mathbb{K}[A^p]:\mathbb{K}]_s\leq [\mathbb{K}[A^p]:\mathbb{K}]\]holds, and since \(\mathbb{K}[A^p]\subset A\) trivially holds in general, we obtain \(A=\mathbb{K}[A^p]\).
Now conversely, assume that \(A=\mathbb{K}[A^p]\) and let us show that \(A\) is étale. For this, it suffices to prove the latter condition. Assuming that \((a_i)\) is a \(\mathbb{K}\)-basis of \(A\), the fact that \((a_i^p)\) generate \(\mathbb{K}[A^p]\) as a \(\mathbb{K}\)-vector space is a consequence of §Fields, ⁋Proposition 12, and then from the given assumption \(A=\mathbb{K}[A^p]\), the \((a_i^p)\) also generate \(A\).
Now, to use the result of §Separable Extensions, ⁋Theorem 7, let us show that \(\overline{\mathbb{K}}\otimes_\mathbb{K}A\) is reduced. If we assume that \(u^2=0\) for some \(u\in\overline{\mathbb{K}}\otimes_\mathbb{K}A\), then \(u^p=0\), and hence if we write \(u=\sum \lambda_i\otimes a_i\),
\[0=u^p=\sum_{i\in I} (\lambda_i\otimes a_i)^p=\sum_{i\in I} \lambda_i^p\otimes a_i^p\]and therefore each \(\lambda_i^p\) must be \(0\). Since \(\overline{\mathbb{K}}\) is (of course) reduced, the \(\lambda_i\) must all be \(0\), and hence \(u=0\), from which we obtain the desired result.
Proposition 2 Let \(\mathbb{K}\) be a field with characteristic exponent \(p\), and let \(\mathbb{L}/\mathbb{K}\) be an algebraic extension. Also, let \(S\) be a subset such that \(\mathbb{L}=\mathbb{K}(S)\). If \(\mathbb{L}/\mathbb{K}\) is separable, then \(\mathbb{L}=\mathbb{K}(S^{p^n})\) holds for any \(n\geq 0\). Conversely, if \(\mathbb{L}/\mathbb{K}\) is a finite degree extension and \(\mathbb{L}=\mathbb{K}(S^p)\), then \(\mathbb{L}/\mathbb{K}\) is separable.
Proof
As always, there is nothing to prove when \(p=1\). Thus it suffices to consider the case \(p\neq 1\).
First, assuming \(\mathbb{L}=\mathbb{K}(S)\), we have
\[\mathbb{K}(\mathbb{L}^p)=\mathbb{K}(\mathbb{K}(S)^p)=\mathbb{K}(\mathbb{K}^p(S^p))=\mathbb{K}(S^p)\]so that \(\mathbb{K}(S^p)=\mathbb{K}(\mathbb{L}^p)=\mathbb{K}[\mathbb{L}^p]\). Applying the above Lemma 1 to \(A=\mathbb{L}\), the case where \(\mathbb{L}/\mathbb{K}\) is a finite degree extension is trivial since this is equivalent to \(\mathbb{L}\) being an étale \(\mathbb{K}\)-algebra. Now even when \(\mathbb{L}/\mathbb{K}\) is of infinite degree, any finite degree subextension \(\mathbb{L}'/\mathbb{K}\) of \(\mathbb{L}/\mathbb{K}\) is separable, so by the preceding discussion,
\[\mathbb{L}'=\mathbb{K}[(\mathbb{L}')^p]=\subseteq \mathbb{K}[\mathbb{L}^p]\]holds, and since \(\mathbb{K}[\mathbb{L}^p]\) is the union of the \(\mathbb{L}'/\mathbb{K}\), we obtain the desired equality. The equality for arbitrary \(n\) follows by a simple induction.
From this we obtain the following two corollaries.
Corollary 3 Any algebraic extension of a perfect field \(\mathbb{K}\) is perfect.
The proof of this is almost obvious from Proposition 2.
Moreover, the following holds.
Corollary 4 For a field \(\mathbb{K}\) with algebraic closure \(\overline{\mathbb{K}}\) and perfect closure \(\mathbb{K}^{p^{-\infty}}\) inside it, a subextension \(\mathbb{L}/\mathbb{K}\) of \(\overline{\mathbb{K}}/\mathbb{K}\) is separable if and only if it is linearly disjoint from \(\mathbb{K}^{p^{-\infty}}\).
Proof
As always, it suffices to consider the situation where \(\mathbb{L}/\mathbb{K}\) is of finite degree. Given a basis \((x_i)\) of \(\mathbb{L}/\mathbb{K}\), that \(\mathbb{L}\) is linearly disjoint from \(\mathbb{K}^{p^{-\infty}}\) is equivalent to \((x_i)\) being linearly disjoint from every \(\mathbb{K}^{p^{-n}}\), which is the same as saying that for any family \((a_i)\) and any \(n\),
\[\sum x_i a_i^{p^{-n}}=0\implies a_i=0\]must always hold. Now taking the \(p^n\)-th power of both sides, we see from this that the \(x_i^{p^n}\) must be free, and therefore they must form a basis of \(\mathbb{L}\), and the converse also holds. Considering dimensions, this is equivalent to \(\mathbb{L}=\mathbb{K}(\mathbb{L}^p)\), so we obtain the desired result from Proposition 2.
Separable Closure
On the other hand, just as with algebraic closures or perfect closures, we can also define the separable closure. Then, just as the perfect closure in §Fields, ⁋Theorem 15 is the relative perfect closure with respect to the algebraic closure, it would be natural to first define the relative separable closure, and then define the relative separable closure inside an algebraic closure as the (absolute) separable closure.
Proposition 5 For a field extension \(\mathbb{L}/\mathbb{K}\), let \(\mathbb{L}_s\) be the set of elements that are algebraic and separable over \(\mathbb{K}\). Then \(\mathbb{L}_s\) is a subextension of \(\mathbb{L}/\mathbb{K}\), and moreover it is the largest separable algebraic extension contained in \(\mathbb{L}\).
Proof
First, since any element of a separable extension is separable (§Separable Extensions, ⁋Proposition 12), any subextension of \(\mathbb{L}\) that is separable is always contained in \(\mathbb{L}_s\). On the other hand, conversely, an algebraic extension generated only by separable elements is also separable by §Separable Extensions, ⁋Proposition 12, so \(\mathbb{K}(\mathbb{L}_s)\) is itself a separable extension, and again by the above claim \(\mathbb{K}(\mathbb{L}_s)\subseteq \mathbb{L}_s\) holds. That is, \(\mathbb{K}(\mathbb{L}_s)=\mathbb{L}_s\) holds, and thus we obtain the desired result from Proposition 2.
As mentioned above, \(\mathbb{L}_s\) is called the relative separable algebraic closure (in \(\mathbb{L}/\mathbb{K}\)).
The main result of this post is that any algebraic extension always splits completely into a separable part (corresponding to the relative separable closure) and an inseparable part. This can be checked as follows.
Theorem 6 For an algebraic extension \(\mathbb{L}/\mathbb{K}\) and \(\mathbb{L}_s\) defined in Proposition 5, the following hold.
- \(\mathbb{L}/\mathbb{L}_s\) is a \(p\)-radical extension.
- For a subextension \(\mathbb{M}/\mathbb{K}\), suppose that \(\mathbb{L}/\mathbb{M}\) is \(p\)-radical. Then \(\mathbb{M}\) contains \(\mathbb{L}_s\).
- \(\mathbb{L}_s\) is the unique subextension of \(\mathbb{L}\) that is separable over \(\mathbb{K}\) and such that \(\mathbb{L}\) is \(p\)-radical over it.
Proof
First, for the first claim, the case \(\ch(\mathbb{K})=0\) is trivial, so let us consider the case \(\ch(\mathbb{K})=p>0\). Now, given \(x\in \mathbb{L}\) and its minimal polynomial \(f\), there exists a suitable \(m\geq 0\) such that \(f\in\mathbb{K}[\x^{p^m}]\) but \(f\not\in \mathbb{K}[\x^{p^{m+1}}]\), and in this case we can choose a polynomial \(g\) such that \(f(\x)=g(\x^{p^m})\). However, since \(f\) is irreducible, \(g\) is as well, and therefore \(g\) becomes the minimal polynomial of the element \(x^{p^m}\) of \(\mathbb{K}\). Now, from the last equivalent condition of §Separable Extensions, ⁋Proposition 10, \(g\) is separable, and therefore \(x^{p^m}\) belongs to \(\mathbb{L}_s\). Hence \(x\) is \(p\)-radical over \(\mathbb{L}/\mathbb{L}_s\), and from this we obtain the desired result.
On the other hand, given a subextension \(\mathbb{M}/\mathbb{K}\) satisfying the second assumption, let \(x\in \mathbb{L}_s\). Then \(x\) is separable over \(\mathbb{K}\), and hence also over \(\mathbb{M}\). However, since \(\mathbb{L}/\mathbb{M}\) is \(p\)-radical, \(x\) is \(p\)-radical over \(\mathbb{M}\). Again, from the last equivalent condition of §Separable Extensions, ⁋Proposition 10, the minimal polynomial of \(x\) must lie in \(\mathbb{K}[\x^p]\), but at the same time, from the condition that \(x\) is \(p\)-radical, for the height \(e\) of \(x\), \(\x^{p^e}-x^{p^e}\) must be the minimal polynomial of \(x\). Therefore \(e=0\) and \(\x-x\) must be the minimal polynomial of \(x\), so \(x\) must belong to \(\mathbb{M}\).
The last claim follows from the uniqueness in Proposition 5.
Since the property of an algebraic extension being separable is stable under base change, if we take two subextensions \(\mathbb{L}/\mathbb{K}\), \(\mathbb{K}'/\mathbb{K}\) of some extension, and the relative separable closure \(\mathbb{L}_s\) of \(\mathbb{K}\) in \(\mathbb{L}\), then we can verify that \(\mathbb{K}'(\mathbb{L}_s)\) becomes the relative separable closure of \(\mathbb{K}'\) in \(\mathbb{K}'(\mathbb{L})\). Also, from the uniqueness of the relative separable closure, for a finite degree extension \(\mathbb{L}/\mathbb{K}\) we can verify that the formula
\[\mathbb{L}_s=\bigcap_{n\geq 0} \mathbb{K}(\mathbb{L}^{p^n})\]holds.
How to define the (absolute) separable algebraic closure is obvious.
Definition 7 A field \(\mathbb{K}\) is separably closed if every separable algebraic extension of \(\mathbb{K}\) is trivial. If an extension \(\mathbb{L}\) of a field \(\mathbb{K}\) is separable algebraic and \(\mathbb{L}\) is separably closed, then it is called a separable algebraic closure of \(\mathbb{K}\).
Then, since any separable algebraic extension is algebraic, an algebraically closed field is always separably closed. Conversely, since any algebraic extension of a perfect field \(\mathbb{K}\) is separable, a separably closed perfect field is algebraically closed.
Then the following holds.
Proposition 8 Fix an algebraic closure \(\overline{\mathbb{K}}\) of a field \(\mathbb{K}\).
- The relative separable algebraic closure \(\overline{\mathbb{K}}_s\) in \(\overline{\mathbb{K}}\) is a separable closure of \(\mathbb{K}\).
- The separable algebraic closure of \(\mathbb{K}\) is uniquely determined up to isomorphism.
Proof
- \(\overline{\mathbb{K}}_s\) is separable by Proposition 5, and since it is a subextension of the algebraic closure \(\overline{\mathbb{K}}\), it is algebraic. Therefore, the claim is proved if we show that any separable algebraic extension \(\mathbb{L}/\overline{\mathbb{K}}_s\) is a trivial extension of \(\overline{\mathbb{K}}_s\). Since the extension \(\mathbb{L}/\overline{\mathbb{K}}_s\) is algebraic, there exists a unique \(\overline{\mathbb{K}}_s\)-homomorphism \(u:\mathbb{L}\rightarrow\overline{\mathbb{K}}\) (§Algebraic Closures, ⁋Theorem 5), and its image \(u(\mathbb{L})\) is separable algebraic by §Separable Extensions, ⁋Proposition 15, and therefore \(u(\mathbb{L})=\overline{\mathbb{K}}_s\) holds.
- Similarly, this follows by using §Algebraic Closures, ⁋Theorem 5.
From this, we know that the separable algebraic closure is the smallest separably closed algebraic extension. That is, if \(\mathbb{L}/\mathbb{K}\) is a separable algebraic closure of \(\mathbb{K}\), and for a field extension \(\mathbb{M}/\mathbb{K}\) the field \(\mathbb{M}\) is separably closed, then there exists a unique morphism \(\mathbb{L}\rightarrow\mathbb{M}\).
Separable Degree
By Theorem 6, any finite degree extension \(\mathbb{L}/\mathbb{K}\) can be split into a separable part and a non-separable part, written as \(\mathbb{L}/\mathbb{L}_s/\mathbb{K}\).
Proposition 9 In the above situation, \([\mathbb{L}:\mathbb{K}]_s=[\mathbb{L}_s:\mathbb{K}]\).
Proof
By definition, \([\mathbb{L}:\mathbb{K}]_s\) is defined as the number of \(\mathbb{K}\)-algebra homomorphisms from \(\mathbb{L}\) to \(\overline{\mathbb{K}}\), where \(\overline{\mathbb{K}}\) is an algebraic closure of \(\mathbb{K}\). However, since \(\overline{\mathbb{K}}\) is an algebraically closed field, it is a perfect field, and therefore by §Radical Extensions, ⁋Proposition 6, for any given \(\mathbb{K}\)-algebra homomorphism \(\mathbb{L} \rightarrow \overline{\mathbb{K}}\) there is a unique \(\mathbb{L}_s\rightarrow \overline{\mathbb{K}}\) defined, and conversely, for any given \(\mathbb{K}\)-algebra homomorphism \(\mathbb{L}_s \rightarrow \overline{\mathbb{K}}\) we can restrict it to \(\mathbb{L}\) to obtain \(\mathbb{L}\rightarrow\overline{\mathbb{K}}\). From this we obtain the equality
\[[\mathbb{L}:\mathbb{K}]_s=[\mathbb{L}_s:\mathbb{K}]_s\]On the other hand, since \(\mathbb{L}_s/\mathbb{K}\) is a finite degree separable extension, it is an étale algebra, and therefore by §Étale Algebras, ⁋Proposition 13 we have \([\mathbb{L}_s:\mathbb{K}]_s=[\mathbb{L}_s:\mathbb{K}]\), obtaining the desired result.
Therefore it is justified to call \([\mathbb{L}:\mathbb{K}]_s\) the separable degree. Moreover, from the formula
\[[\mathbb{L}:\mathbb{K}]=[\mathbb{L}:\mathbb{L}_s][\mathbb{L}_s:\mathbb{K}]=[\mathbb{L}:\mathbb{K}]_s[\mathbb{L}:\mathbb{L}_s]\]we can define the inseparable degree of the extension \(\mathbb{L}/\mathbb{K}\) as \([\mathbb{L}:\mathbb{K}]_i=[\mathbb{L}:\mathbb{L}_s]\).
If \(\ch\mathbb{K}=0\), then \([\mathbb{L}:\mathbb{K}]_i=1\) always holds, and if \(\ch\mathbb{K}=p\), then \([\mathbb{L}:\mathbb{K}]_i\) is always a power of \(p\). (Theorem 6) However, since there are plenty of polynomials of degree \(p^e\) that do not define a \(p\)-radical extension, there is no way to determine the value of \([\mathbb{L}:\mathbb{K}]_i\) from (say) \([\mathbb{L}:\mathbb{K}]\) alone.
However, the following still holds.
Proposition 10 Fix an extension \(\Omega/\mathbb{K}\), and consider two finite degree subextensions \(\mathbb{L}/\mathbb{K}\), \(\mathbb{M}/\mathbb{K}\) of \(\Omega\). The following hold.
- If \(\mathbb{L}\subseteq \mathbb{M}\), then \([\mathbb{M}:\mathbb{K}]_s=[\mathbb{M}:\mathbb{L}]_s[\mathbb{L}:\mathbb{K}]_s\) and \([\mathbb{M}:\mathbb{K}]_i=[\mathbb{M}:\mathbb{L}]_i[\mathbb{L}:\mathbb{K}]_i\).
-
For any subextension \(\mathbb{K}'\) of \(\Omega\), the formulas
\[[\mathbb{K}'(\mathbb{L}):\mathbb{K}']_s\leq [\mathbb{L}:\mathbb{K}]_s,\qquad [\mathbb{K}'(\mathbb{L}):\mathbb{K}']_i\leq [\mathbb{L}:\mathbb{K}]_i\]hold, and equality holds when \(\mathbb{K}'\) and \(\mathbb{L}\) are linearly disjoint.
-
The two inequalities
\[[\mathbb{K}(\mathbb{L}\cup \mathbb{M}):\mathbb{K}]_s\leq [\mathbb{L}:\mathbb{K}]_s[\mathbb{M}:\mathbb{K}]_s,\qquad [\mathbb{K}(\mathbb{L}\cup \mathbb{M}):\mathbb{K}]_i\leq [\mathbb{L}:\mathbb{K}]_i[\mathbb{M}:\mathbb{K}]_i\]hold, and equality holds when \(\mathbb{L}\) and \(\mathbb{M}\) are linearly disjoint.
The proofs of these are almost tautological; for example, the first result follows from §Algebraic Extensions, ⁋Proposition 2 and §Étale Algebras, ⁋Proposition 12. The remaining results also hold naturally for the inseparable degree, since similar results hold for the extension degree and the separable degree respectively.
댓글남기기