# If $F^p = F$ and $E/F$ is algebraic, then $E/F$ is separable and $E^p = E$ : Corollary V.6.12 from Lang's *Algebra* [duplicate]

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This is from Lang's Algebra (page 251)

Proposition 6.11 Let $$E/F$$ be a normal field extension. Let $$E^G$$ be the fixed field of $$\operatorname{Aut}(E/F)$$. Then, $$E^G$$ is purely inseparable over $$F$$ and $$E$$ is separable over $$E^G$$.

And below is a corollary of this theorem:

Corollary 6.12. Let $$F$$ be a field with characteristic $$p\neq 0$$ such that $$F^p=F$$. Then, every algebraic extension $$E$$ of $$F$$ is separable and $$E^p=E$$.

How this is a corollary of the above theorem?

Lang states that "Every algebraic extension is contained in a normal extension, so Proposition 6.11 can be applied to get this", but how?

Let $$E$$ be an algebraic extension of $$F$$. Then, there is a field extension $$L$$ of $$E$$ such that $$L/F$$ is normal.

Let $$\phi\colon F\to F:a\mapsto a^p$$.

Then, by hypothesis, $$\phi$$ is a field automorphism of $$F$$.

Then, $$\phi$$ can be extended to a field monomorphism $$\sigma \colon \bar F \to \bar F$$, but since $$\phi$$ is not fixing $$F$$, I don't get what this has to do with Proposition 6.11.

These are what all I know. How do I proceed to prove the corollary?

## marked as duplicate by Brahadeesh, YuiTo Cheng, Eric Wofsey, José Carlos Santos, Lee David Chung LinJun 2 at 0:53

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## 2 Answers

Let $$a \in \overline{F}$$, $$F(a) \cong F[x]/(h(x))$$.

If $$F(a)/F$$ is not separable then $$\gcd(h ,h') \ne 1$$, thus $$h' = 0$$ (since otherwise $$\gcd(h,h')$$ would divide $$h$$ which wouldn't be irreducible) and $$h(x) = g(x^p)$$ for some $$g \in F[x]$$.

If also $$F = F^p$$, we can take $$f(x) \in F[x]$$ such that $$f^p(x) = g(x)$$ ($$p$$-th power of the coefficients) so that $$h(x) =f^p(x^p)=( f(x))^p$$.

But then $$h(a)=(f(a))^p = 0$$ implies $$f(a) = 0$$, and since $$\deg(f) = \frac{\deg(h)}{p}$$ it contradicts that $$h$$ was the minimal polynomial of $$a$$.

Whence $$F(a)/F$$ had to be separable.

• +1 This is a nice proof that avoids the route suggested in Lang's hint! – Brahadeesh May 31 at 14:15

Let $$\operatorname{char}(k) = p > 0$$. Suppose that $$k$$ is perfect, that is, $$k^p = k$$.

We prove the first part of the corollary for finite extensions of $$k$$.

Let $$E/k$$ be a finite extension. There is a normal extension $$K/k$$ such that $$E \subset K \subset E^a$$. Let $$G = \operatorname{Aut}(K/k)$$ and let $$K^G$$ be the fixed field of $$G$$. Then $$K/K^G$$ is separable and $$K^G/k$$ is purely inseparable, by Proposition 6.11.

It suffices to show that $$K^G=k$$. For then $$K/K^G = K/k$$, so $$K/k$$ is separable. Hence, $$E/k$$ is also separable.

Now we show that $$K^G=k$$. Since $$K^G / k$$ is purely inseparable, every $$\alpha \in K^G$$ is purely inseparable over $$k$$, that is, for each $$\alpha \in K^G$$ there exists $$n \geq 0$$ such that $$\alpha^{p^n} \in k$$. To show that $$\alpha \in K^G \implies \alpha \in k$$, we need to show that $$n = 0$$ works for every $$\alpha \in K^G$$. For the sake of contradiction, suppose that $$\alpha \in K^G$$ such that $$\alpha \not\in k$$. Let $$n \geq 1$$ be the least positive integer such that $$\alpha^{p^n} \in k$$. Since $$k^p = k$$, there exists $$a \in k$$ such that $$\alpha^{p^n} = a^p$$. Hence, $$\alpha^{p^{n-1}} = a \in k$$, which contradicts the minimality of $$n$$. Hence, proved.

Next, we prove the first part of the corollary for algebraic extensions of $$k$$ of infinite degree.

Let $$E/k$$ be an algebraic extension such that $$[E:k] = \infty$$. To show that $$E/k$$ is separable, we show equivalently that every $$\alpha \in E$$ is separable over $$k$$. So, let $$\alpha \in E$$ and consider the extension $$k(\alpha)/k$$. This is a finite extension, so, by what we have proved earlier, $$k(\alpha)/k$$ is separable. Hence, $$\alpha$$ is separable over $$k$$.

Now, we prove the second part of the corollary for finite extensions of $$k$$.

Let $$E/k$$ be a finite extension. We have proved that $$E/k$$ is separable. So, by Corollary 6.10 (see below), $$E^{p^n}k = E$$ for all $$n \geq 1$$. In particular, $$E^p k = E$$. Since $$k \subset E \implies k^p \subset E^p$$ and since $$k$$ is perfect, we have that $$k \subset E^p$$. So, $$E^p k = E^p$$. Thus, $$E^p = E$$.

Next, we prove the second part of the corollary for algebraic extensions of $$k$$ of infinite degree.

Let $$E/k$$ be an algebraic extension such that $$[E:k] = \infty$$. Then, $$E$$ is the union of all its finitely generated subextensions containing $$k$$, that is, $$E = \bigcup k(\alpha_1,\dots,\alpha_n),$$ where the union is taken over all finite subsets $$\{ \alpha_1,\dots,\alpha_n \}$$ of $$E$$. Each subextension $$k(\alpha_1,\dots,\alpha_n)$$ is a finite extension of $$k$$, so, by what we have shown above, each such subextension is perfect. Since $$E^p = \bigcup k(\alpha_1,\dots,\alpha_n)^p,$$ we have that $$E^p = E$$. Hence, proved.

Corollary 6.10. Let $$E^p$$ denote the field of all elements $$x^p$$, $$x \in E$$. Let $$E$$ be a finite extension of $$k$$. If $$E^p k = E$$, then $$E$$ is separable over $$k$$. If $$E$$ is separable over $$k$$, then $$E^{p^n} k = E$$ for all $$n \geq 1$$.