Lemma 9.25.1 (Artin-Schreier extensions). Let $L/K$ be a Galois extension of fields of characteristic $p > 0$ with Galois group $\mathbf{Z}/p\mathbf{Z}$. Then $L = K[z]$ with $z^ p - z \in K$.
9.25 Artin-Schreier extensions
Let $K$ be a field of characteristic $p > 0$. Let $a \in K$. Let $L$ be an extension of $K$ obtained by adjoining a root $b$ of the equation $x^ p - x = a$. Then $L/K$ is Galois. If $G = \text{Gal}(L/K)$ is the Galois group, then the map
\[ G \longrightarrow \mathbf{Z}/p\mathbf{Z},\quad \sigma \longmapsto \sigma (b) - b \]
is an injective homomorphism of groups. In particular, $G$ is cyclic of order dividing $p$ as a subgroup of $\mathbf{Z}/p\mathbf{Z}$. The theory of Artin-Schreier extensions gives a converse.
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. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #4892 by Spencer Dembner on
Comment #5168 by Johan on