The Stacks project

Lemma 9.12.4. Let $F$ be a field. An irreducible polynomial $P$ over $F$ is separable if and only if $P$ has pairwise distinct roots in an algebraic closure of $F$.

Proof. Suppose that $\alpha \in \overline{F}$ is a root of both $P$ and $P'$. Then $P = (x - \alpha )Q$ for some polynomial $Q$. Taking derivatives we obtain $P' = Q + (x - \alpha )Q'$. Thus $\alpha $ is a root of $Q$. Hence we see that if $P$ and $P'$ have a common root, then $P$ does not have pairwise distinct roots. Conversely, if $P$ has a repeated root, i.e., $(x - \alpha )^2$ divides $P$, then $\alpha $ is a root of both $P$ and $P'$. Combined with Lemma 9.11.2 this proves the lemma. $\square$

Comments (2)

Comment #2963 by xd on

should be in the first sentence of the proof?

There are also:

  • 7 comment(s) on Section 9.12: Separable extensions

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Your email address will not be published.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 09H3. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 9.12.4 as pdf