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?



$P$ 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 $a=b$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.