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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?

Bruno Joyal
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pre-kidney
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1 Answers1

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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 $\mathbf F_p(T)$.

Bruno Joyal
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  • 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. – pre-kidney Oct 27 '13 at 01:19
  • 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? – pre-kidney Oct 27 '13 at 01:36
  • @pre-kidney, No: the condition of being defined over $k$ is stronger than that of being defined over $\overline{k}$. In essence, "defined over $k$" means that the polynomial map which describes the morphism has coefficients in $k$, so a morphism defined over $k$ is also defined over $\overline{k}$. – Bruno Joyal Oct 27 '13 at 01:39
  • @Marie: Doesn't "automorphism over $\overline{k}$" imply that $\overline{k}$ is fixed by such an automorphism? Another loophole, then, is automorphisms that, when extended to $\overline{k}$, don't actually fix $\overline{k}$, such as the Frobenius map. –  Oct 27 '13 at 01:40
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    @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 $\text{Aut}k(X) \to \text{Aut}\overline{k}(X\times_k \overline{k})$. – Bruno Joyal Oct 27 '13 at 01:48
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    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.) – Pete L. Clark Oct 27 '13 at 06:47
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    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 $\bar{k}\hookrightarrow A\otimes_k\bar{k},B\otimes_k\bar{k}$. – Keenan Kidwell Oct 30 '13 at 19:19