The Stacks project

Lemma 9.28.2. Let $K$ be a field of characteristic $p > 0$. Let $L/K$ be a separable algebraic extension. Let $\alpha \in L$.

  1. If the coefficients of the minimal polynomial of $\alpha $ over $K$ are $p$th powers in $K$ then $\alpha $ is a $p$th power in $L$.

  2. More generally, if $P \in K[T]$ is a polynomial such that (a) $\alpha $ is a root of $P$, (b) $P$ has pairwise distinct roots in an algebraic closure, and (c) all coefficients of $P$ are $p$th powers, then $\alpha $ is a $p$th power in $L$.

Proof. It follows from the definitions that (2) implies (1). Assume $P$ is as in (2). Write $P(T) = \sum \nolimits _{i = 0}^ d a_ i T^{d - i}$ and $a_ i = b_ i^ p$. The polynomial $Q(T) = \sum \nolimits _{i = 0}^ d b_ i T^{d - i}$ has distinct roots in an algebraic closure as well, because the roots of $Q$ are the $p$th roots of the roots of $P$. If $\alpha $ is not a $p$th power, then $T^ p - \alpha $ is an irreducible polynomial over $L$ (Lemma 9.14.2). Moreover $Q$ and $T^ p - \alpha $ have a root in common in an algebraic closure $\overline{L}$. Thus $Q$ and $T^ p - \alpha $ are not relatively prime, which implies $T^ p - \alpha | Q$ in $L[T]$. This contradicts the fact that the roots of $Q$ are pairwise distinct. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 031V. Beware of the difference between the letter 'O' and the digit '0'.