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Given a field $K$, we consider its algebraic closure $\tilde{K}$ and the separable closure $K^{\text{sep}}$ inside it. Assume nonzero characteristic, because if $\text{char}K=0$, these coincide.

Claim: the restriction map $\text{Aut}_K(\tilde{K}) \to \text{Aut}_K(K^{\text{sep}})=\text{Gal}(K^{\text{sep}}/K)$ is an isomorphism.

Please help me understand why it's true. There's a sketch in the second answer here, but I'm struggling to fill in the details. Assuming $\text{char}K=p$:

  1. $\tilde{K}/K^{\text{sep}}$ is a purely inseparable extension, check.
  2. Therefore for each $a \in \tilde{K}$, $a^{p^n} \in K^{\text{sep}}$ for some $n$ - why? I understand why the minimal polynomial of $a$ is actually a polynomial in $X^p$, but not sure how to continue from that.
  3. Therefore $\tilde{K}$ is obtained from $K^\text{sep}$ by adjoining all the $p^n$-th roots - check, given 2.
  4. Therefore any automorphism of $\tilde{K}$ that is identity on $K^\text{sep}$ must be identity on $\tilde{K}$ - why? Why can't this automorphism permute the roots of every $X^{p^n}-a$ in some clever way?
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It is a standard result that if $f(x) \in K[x]$ is inseparable (ie has a root in $\overline{K}\setminus K^{sep}$) then there is another polynomial $g(x) \in K[x]$ such that $f(x)=g(x^p)$. If $g$ is then separable we are done else you can induct to say that $f(x)=h(x^{p^n})$ for some separable polynomial $h$.

Rephrasing this, if $a$ is a root of $f$, then $a^{p^n}$ is a root of the separable polynomial $h$, ie $a^{p^n} \in K^{sep}$.

For 4. consider the polynomial $x^p-a$ where $a$ is chosen so that this polynomial is irreducible over $K$. Let $t$ be such that $t^p=a$, i.e. $t$ is a root. Then note that $(x-t)^p=x^p-t^p=x^p-a$ since all the cross terms in the binomial expansion carry a factor of $p$ in the binomial coefficient so become 0 in characteristic $p$. (This is precisely why polynomials can only be inseparable in positive characteristic.)

Now you know that any Galois automorphism must permute the roots of $x^p-a$ but there is in fact only one root (which has multiplicity $p$), so it has no choice but to send it to itself.

The same idea applies to $x^{p^n}-a$ so the roots get fixed by every Galois element, which gives the isomorphism you're after.

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  • $\begingroup$ All my confusion is gone. Much obliged! $\endgroup$ – AnatolyVorobey Nov 1 '15 at 16:28

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