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Let $F=\mathbb{F}_p(X,Y)$ be the field of rational functions in variables $X,Y$ over the finite field of $p$ elements. Let $K=\mathbb{F}_p(X^p,Y^p)$ be a subfield. Note that for any $f\in F$, $f^p\in K$. Deduce from this that $F/K$ is not a simple extension.

Spook
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  • I tried contradiction methods. So assuming $F=K(f)$, $f^p=g\in K$ then $f$ is not separable over $K$. That's all I could think of, and I don't think this is taking me anywhere. – Spook Jan 12 '13 at 23:51
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    Maybe if you can prove that the degree of the extension is $p^2$ all is done. Try it! –  Jan 13 '13 at 00:01
  • I couldn't show the set ${X,Y,XY,X^2,Y,XY^2,...X^pY^p}$ is a linearly independent set. I took that this was what you meant by $p^2$. But this is an exercise in a section on separability. – Spook Jan 13 '13 at 00:21
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    Presumably the reason it's in a section on separability is that this section (or maybe the next section) contains the theorem of the primitive element, and the point of the exercise is to show that this theorem really needs the hypothesis of separability. – Andreas Blass Jan 13 '13 at 02:11

1 Answers1

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First of all $\mathbb{F}_p(X^p,Y^p)$ is a function field, which means that we may treat $X^p,Y^p$ as variables. The polynomial $(T-X)^p = T^p-X^p$ is irreducible over $\mathbb{F}_p(Y^p)[X^p]$ by Eisenstein's criterion, but then also over the field of fractions $\mathbb{F}_p(X^p,Y^p)$. Hence, $\mathbb{F}_p(X,Y^p)$ has degree $p$ over $\mathbb{F}_p(X^p,Y^p)$. Similarily one proves that $\mathbb{F}_p(X,Y)$ has degree $p$ over $\mathbb{F}_p(X,Y^p)$. Hence, $F=\mathbb{F}_p(X,Y)$ has degree $p^2$ over $K=\mathbb{F}_p(X^p,Y^p)$. But since $F^p \subseteq K$, every element of $F$ has degree $\leq p$ over $K$, so that it cannot generate $F/K$.

Martin Brandenburg
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