Processing math: 100%

The Stacks project

Lemma 9.14.2. Let p be a prime number. Let F be a field of characteristic p. Let t \in F be an element which does not have a pth root in F. Then the polynomial x^ p - t is irreducible over F.

Proof. To see this, suppose that we have a factorization x^ p - t = f g. Taking derivatives we get f' g + f g' = 0. Note that neither f' = 0 nor g' = 0 as the degrees of f and g are smaller than p. Moreover, \deg (f') < \deg (f) and \deg (g') < \deg (g). We conclude that f and g have a factor in common. Thus if x^ p - t is reducible, then it is of the form x^ p - t = c f^ n for some irreducible f, c \in F^*, and n > 1. Since p is a prime number this implies n = p and f linear, which would imply x^ p - t has a root in F. Contradiction. \square


Comments (0)

There are also:

  • 3 comment(s) on Section 9.14: Purely inseparable extensions

Your email address will not be published.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.