# Artin-Schreier Theorem and Galois Extensions

This is what I know of the Artin-Schreier Theorem for field extensions:

Let $K< L$ be a proper Galois extension such that $L$ has a prime characteristic $p\neq 0$ and $|\mathrm{Aut}(L/K)|=p$ then there is an $\alpha\in L\backslash K$ such that $L=K(\alpha)$ and $\alpha$ has a minimal polynomial (over $K$) of the form $T^p -T+a$ for some $a\in K$.

I tried to prove this by collecting some known results (I think this is how Artin himself proved it) but I ended up using the fact that $K<L$ must be a Galois extension. I have seen this informally stated without requiring the extension to be Galois. So, I ask whether we can just remove this condition that $K<L$ be Galois (can this not be concluded?). I couldn't really do this without assuming Galois extension because I needed the fact that an invariant element for the generator of the automorphism group is an element of the lower field $K$. Can one prove that the extension is Galois if we only assume that the order of the automorphism group is the characteristic? If not, is there a counterexample for the case when this statement does not hold when we remove the requirement that the extension is Galois?

• I suppose that Aut(L/K) means the group of automorphisms of L which fix K. Then one of the classical theorems states that the finite extension L/K is Galois iff the order of Aut(L/K) equals the degree. This has nothing to do with char K. The A-S. theorem absolutely needs 2 hypotheses: char K = p and L/K Galois has degree p. It follows in particular that Aut(L/K) is cyclic and, if I remember well, this allows to use the additive form of Hilbert's theorem 90. Note that even if L/K is cyclic, but not of degree p, the proof does not work. – nguyen quang do May 8 '17 at 7:14
• Yes $\mathrm{Aut}(L/K)$ is just as you described. So, in short, you are writing: In the Theorem I just wrote, I cannot remove "Galois". Because, without assumption it is Galois, I do not know apriori if my field extension has the same degree as the group order (I just know that the group order is the same as the characteristic). If $L/K$ is cyclic then by definition it is Galois. Then the degree of extension becomes $p$ (because the order of the automorphism group fixing $K$ is $p$). – quantum May 8 '17 at 9:13
• Still I would like to see a counterexample that shows how the theorem fails if we do not assume Galois extension. – quantum May 8 '17 at 9:14
• Ah! Now I understand what bothers you. For short, write G= Aut(L/K), which has order p by hypothesis. Adopt Artin's point of view on Galois theory: take the top field as the fundamental object and construct the bottom field as the fixed subfield under a finite group of automorphisms. If M denotes the subfield of L fixed by G , the extension L/M is Galois, with group G, hence [L:M] = p, hence M = K because p is a prime, which means that L/K itself is Galois . Note that char K plays no role at this stage. – nguyen quang do May 8 '17 at 14:27
• As Torsten's answer and the comments by others explain the assumption of the extension being Galois is necessary. In case you can't locate a suitable textbook here is an on-site proof. Sorry about blowing my own trumpet to this extent. The proof idea is certainly not mine (more like standard textbook stuff). It is an additive adaptation of the similar textbook proof for the fact that if $L/K$ is Galois of order $p$, $p$ a prime, and $K$ contains a primitive $p$th root of unity, then $L=K(z)$, where $z^p\in K$, i.e. a root extension. – Jyrki Lahtonen Jan 17 '18 at 11:58

Following the discussion in the comments, setting $q=p^p$,
$$K =\Bbb{F}_p(t), L=\Bbb{F}_{q}(t^{1/p})$$
is a counterexample where $L|K$ is not separable. One has the intermediate field
$$M = \Bbb{F}_p(t^{1/p})$$
with $M|K$ purely inseparable, and $Aut(L|K) = Aut(L|M) = Gal(L|M) \simeq Gal(\Bbb{F}_q|\Bbb{F}_p)$ has order $p$. But $[L:K] = [L:M][M:K] = p^2$.
Again, following the discussion in the comments, this is the only thing that can go wrong. So one could relax the condition "$L|K$ Galois" to "$L|K$ separable". In particular, if $K$ is perfect, one could remove that condition entirely.