Lemma 9.12.4 (09H3)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 9: Fields
Section 9.12: Separable extensions
Lemma 9.12.4 (cite)

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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 October 21, 2017 at 12:54

should be $\alpha\in\overline{F}$ in the first sentence of the proof?

Comment #3089 by Johan on January 31, 2018 at 16:29

Thanks, fixed here.

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