A field is rigid iff its automorphism group is trivial.

A field $F$ is perfect iff all irreducibles in $F[x]$ are separable.

Is every rigid field perfect?

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    $\begingroup$ An explicit counter-example is $\mathbb{F}_3(x)[y]/(y^3+y^2=x^4-x^3+1)$ (following Case 1 of Bjorn Poonen's paper). $\endgroup$ – Martin Brandenburg Oct 27 '13 at 8:08

The answer is no.

The natural place to look for a rigid field is in the function field of a generic, high genus smooth curve. For instance, over the complex numbers, Hurwiz's automorphism theorem guarantees that a compact Riemann surface of genus $g>1$ has at most $84(g-1)$ automorphisms. Having at least one nontrivial automorphism, for a Riemann surface of genus $g>2$, should be viewed as a very special property (genus $2$ is not sufficient, because genus $2$ curves are all hyperelliptic, and therefore admit an involution).

This paper of Bjorn Poonen shows that for any field $k$ and any integer $g \geq 3$, there exists a smooth curve $X$ of genus $g$ over $k$ such that $X$ has no nontrivial automorphisms over $\overline{k}$ (so, a fortiori it has no automorphisms over $k$).

Taking $k=\mathbf F_p$, it follows that the function field $\mathbf F_p(X)$ of such a curve $X/\mathbf F_p$ is rigid. However, being a finite extension of the rational function field $\mathbf F_p(T)$, it cannot be perfect, or else it would contain all $p$-power roots of $T$, and would have infinite degree over $\mathbf F_p(T)$.

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  • $\begingroup$ The construction of such a field $\mathbb{F}_p(X)$ is too advanced for me at the moment. However, assuming what you've said is correct, we can then construct a field $K$ such that $K$ is pointwise fixed by all automorphisms of $K$, such that $K$ is not perfect. This contradicts the statement of a problem I am working on at the moment. So either there is a mistake in the problem I am working on, or another mistake in the reasoning here. $\endgroup$ – pre-kidney Oct 27 '13 at 1:19
  • $\begingroup$ Wait a minute. We know that $X$ has no nontrivial automorphisms over $\overline{k}$. But does this imply that it is rigid over $k$? For instance, there may be non-trivial automorphisms of $k$ which do not extend to $\overline{k}$. Can this occur? $\endgroup$ – pre-kidney Oct 27 '13 at 1:36
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    $\begingroup$ @Hurkyl It does! And if the curve $X/k$ had an automorphism over $k$, it would yield, by base-change to $\overline{k}$, an automorphism of $X\times_k \overline{k}/\overline{k}$, i.e. "an automorphism of $X$ over $\overline{k}$". Base-change gives an injective map $\text{Aut}_k(X) \to \text{Aut}_\overline{k}(X\times_k \overline{k})$. $\endgroup$ – Bruno Joyal Oct 27 '13 at 1:48
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    $\begingroup$ Since there seems to be some doubt, let me just say: yes, this solution is entirely correct. (And it is the way I would have solved the problem. I'm not sure that one could do something substantially simpler.) $\endgroup$ – Pete L. Clark Oct 27 '13 at 6:47
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    $\begingroup$ Dear @Hurkyl, We're not extending maps from $k$ to $\bar{k}$, we are base-changing along $k\rightarrow\bar{k}$. For example, a map $\varphi:A\rightarrow B$ of $k$-algebras (meaning $\varphi$ is the identity on $k\hookrightarrow A,B$) yields a map $\varphi_{\bar{k}}:A\otimes_k\bar{k}\rightarrow B\otimes_k\bar{k}$ by base change which is a $\bar{k}$-algebra map, i.e., that is the identity on $\bar{k}\hookrightarrow A\otimes_k\bar{k},B\otimes_k\bar{k}$. $\endgroup$ – Keenan Kidwell Oct 30 '13 at 19:19

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