The Stacks project

Proof. Let $\sigma $ be a generator of $\text{Gal}(L/K)$. Consider $\sigma : L \to L$ as a $K$-linear operator. Observe that $\sigma ^ p - 1 = 0$ as a linear operator. Applying linear independence of characters (Lemma 9.13.1), there cannot be a polynomial of degree $< p$ annihilating $\sigma $. We conclude that the minimal polynomial of $\sigma $ is $x^ p - 1 = (x - 1)^ p$. This implies that there exists $w \in L$ such that $(\sigma - 1)^{p - 1}(w) = y$ is nonzero. Then $\sigma (y) = y$, i.e., $y \in K$. Thus $z = y^{-1}(\sigma - 1)^{p - 2}(w)$ satisfies $\sigma (z) = z + 1$. Since $z \not\in K$ we have $L = K[z]$. Moreover since $\sigma (z^ p - z) = (z + 1)^ p - (z + 1) = z^ p - z$ we see that $z^ p - z \in K$ and the proof is complete. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 9.25: Artin-Schreier 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Your email address will not be published.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 09DY. Beware of the difference between the letter 'O' and the digit '0'.