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Consider the following proposition with its relative proof:


Let $k$ be an algebraically closed field of characteristic $0$.

a) If $L$ is a subfield of $k$, then every elements of $\operatorname {Aut} (L)$ extends to an element of $\operatorname{Aut} (k)$

b) $\operatorname{Fix}_k\left(\operatorname{Gal}(k/L)\right)=L$

Proof: $a)$ Let $S$ be a transcendence basis for $k/L$, then every element $\sigma\in\operatorname{Aut}(L)$ extends naturally to an element $\widetilde\sigma\in\operatorname{Aut}(L(S))$; since $k$ is an algebraic closure of $L(S)$, we use the isomorphism extension theorem to conclude that $\widetilde\sigma$ extends to an element of $\operatorname{Aut}(k)$.

$b)$ Obviously it is enough to prove that for every $x\in k\setminus L$ there is an element $\sigma\in\operatorname{Gal}(k/L)$ such that $\sigma(x)\neq x$. We distinguish two cases:

  1. $x$ is transcendental over $L$. Consider the field $L(x)$, then the assignment $x\mapsto-x$ induces a unique element $\sigma\in\operatorname{Gal}(L(x)/L)$ that moves $x$. By the point $a)$ this $\sigma$ extends to an element of $\operatorname{Gal}(k/L)$.
  2. $x$ is algebraic over $L$. Since in characteristic $0$ every irreducible polynomial is separable, if $f=\textrm{min}\left(x,L\right)$ then there exists in $k$ (remember that $f$ splits over $k$) a root $y$ of $f$ such that $x\neq y$. Let $M$ be the splitting field of $f$ over $L$ and look at the inclusions
    $$L\subseteq L(x)\subseteq M\subseteq k$$ $M$ is normal over $L$ and the canonical $L$-isomorphism $\sigma:L(x)\longrightarrow L(y)$ can be viewed as an immersion $\sigma: L(x)\longrightarrow k$. By the characterization of normal extensions there is an element $\tau\in\operatorname{Gal}(M/L)$ such that $\tau_{|L(x)}=\sigma$ (in particular $\tau(x)=y$), therefore by the point $a)$ $\tau$ extends to an element of $\operatorname{Gal}(k/L)$ which moves $x$.

Now my question: Is the point $b)$ of the proposition true also when the characteristic of $k$ is $p\neq 0$? The above proof uses the separability of all irreducible polynomials!

Edit: Moreover I'd like to know if there are shorter or more elegant proofs of the proposition.

Thanks in advance.

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    $\begingroup$ I don't think so. The following is not a counterexample, because the bigger field is not algebraically closed, but I bet it may be extended. Take $\alpha\in\mathbb{F_p}\setminus\mathbb{F}_p^p$, and consider $\beta\in\overline{\mathbb{F}_P}$ such that $\beta^p=\alpha$. Now take $L=\mathbb{F}_p$ and $k=L(\beta)$, the only automorphism of $k$ over $L$ is the identity, so the fixed field is $k$. What you need is some hypothesis to guarantee the separability of $k/L$: $\operatorname{char}(k)=0$ is only used to ensure it. $\endgroup$ – Giulio Bresciani Mar 19 '14 at 23:50
  • $\begingroup$ @Giulio Every element of $\Bbb F_p$ is a $p$th power, perhaps you mean $\Bbb F_p(T)$? $\endgroup$ – blue Mar 20 '14 at 19:19
  • $\begingroup$ Uh, yes, sorry. Replace $\mathbb{F}_p$ with some field with characteristic $p$ and without $p$-roots. $\endgroup$ – Giulio Bresciani Mar 20 '14 at 19:41
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Let $k=\Bbb F_p(T)$, denote by $K$ its algebraic closure, and $G={\rm Aut}_k(K)$ the automorphisms of $K$ which fix $k$ pointwise. Let $l=\Bbb F_p(T^{1/p})$ be the splitting field of $X^p-T\in\Bbb F_p(T)[X]$. Since it is a splitting field it must be the case that $\sigma l=l$ for all $\sigma \in G$. An action $\sigma$ must send $T^{1/p}$ to a root of the polynomial $X^p-T=(X-T^{1/p})^p$, but this polynomial's only root is $T^{1/p}$, so $\sigma$ fixes $T^{1/p}$ and hence fixes $\Bbb F_p(T^{1/p})$ pointwise. Therefore, $\Bbb F_p(T^{1/p})$ is contained in the fixed field $K^G$, which means that $K^G=k$ is impossible. In fact, $\Bbb F_p(T^{1/p^\infty})\subseteq K^G$.

So we see why separability is important: inseparability of an element means a Galois action can't move it anywhere else (it has no conjugates), thus exhibiting a nontrivial element fixed by $G$.

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