## 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. $\square$

Comment #4892 by Spencer Dembner on

Does the sentence beginning "By linear algebra..." get used? It seems like everything used later in the proof follows from the minimal polynomial fact alone.

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