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In the proof of the finite generation of the invariant ring of a finite group acting on $k[x_1,\dots,x_n]$, at one time there is a symbol I don't understand. The situation is as follows.

$k$ is a field of characteristic $p<\infty$, and $P$ its prime field. Suppose that $k$ arose from $P$ by adjoining finitely many (algebraic or transcendental) elements. Then she writes:

Because the prime field $P$ is perfect, it follows that $k^{1/p}$ is finite with respect to $k$.

I would have suspected that $k^{1/p}=\{x\in\bar k:x^p\in k\}$. This was called "Wurzelkörper" earlier in the paper. What is it, and why does the above follow from $P$ being perfect?

Thanks!

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$P$ being perfect means that exponentiation by $p$ is an isomorphism of $P$, so we can take $p$-th roots of any element in $P$ . You are right with the definition of $k^{1/p}$. To prove the statement suppose $k$ is generated by $x_1,\ldots,x_n$ over $P$. I claim that $k^{1/p}$ is (finitely) generated by $x_1^{1/p},\ldots, x_n^{1/p}$. Indeed let $y\in k$ be of the form $$ y=a_1x_1+\ldots+a_nx_n,\quad a_i\in P. $$ If $b_i\in P$ is the $p$-th root of $a_i$ then the $p$-th root of $y$ in $\overline{k}$ is $$ y^{1/p}=b_1x_1^{1/p}+\ldots+b_nx_n^{1/p} $$

Remark: $p$-th roots of elements are unique in characteristic $p$. If $a^p=b^p$ then $a^p-b^p=(a-b)^p=0$, so $a=b$.

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  • $\begingroup$ Great answer, thanks! $\endgroup$ – InvisiblePanda Jun 27 '13 at 14:07

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