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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 Lin Jun 2 at 0:53

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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.

  • $\begingroup$ +1 This is a nice proof that avoids the route suggested in Lang's hint! $\endgroup$ – 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 $n \geq 1$.


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