# a computation in Galois theory

I am confused with some computation in Galois theory (this is not homework, just my weird curiosity).

Let $k$ be a field of positive characteristic $p\neq 2$ that contains all roots of unity (e.g. algebraically closed), let $K$ be the Artin-Schreier extension $k(x)[\mu]/(\mu^p-\mu-x)$ and let $L$ be the extension of $K$ obtained by adjoining the square root of $\mu$, call it $\alpha$. Now $K$ is Galois over $k(x)$ with Galois group cyclic of order $p$, as it is an Artin-Schreir extension. $L$ is Galois over $K$ with Galois group cyclic of order 2, as a Kummer extension. The Galois group $G$ of $L$ over $k(x)$ must be an extension of $C_p$ by $C_2$ $$1 \to C_2 \to G \to C_p \to 1$$ By Schur-Zassenhaus (most likely there is a simpler way to see it) since the orders of $C_p$ and $C_2$ are coprime, $G$ is a semiderect product, and since there is no non-trivial $C_p$-module structure on $C_2$, it is a direct product.

I wonder what is the action of the generator of $C_p$ on $L$ and what is the subextension of $L$ of degree 2 fixed by the action of $C_p$.

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Are you sure that $L$ is Galois over $k(x)$? The generator of $Gal(K/k(x))$ maps $\mu\mapsto\mu+1$. If $L/k(x)$ is Galois, this would need to extend to an automorphism of $L$. Such an automorphism needs to map $\alpha$ to a square root of $\mu+1$, but I don't see, why $\sqrt{\mu+1}$ should exist in $L$? –  Jyrki Lahtonen May 1 '12 at 19:06
you are right; question closed. Thank you! The smallest galois extension containing $L$ then is the one that contains all square roots of $\mu+k$, $0\leq k < p$, and the Galois group is semidirect product of $(C_2)^p$ with $C_p$ permuting the factors cyclically. –  Dima Sustretov May 1 '12 at 19:33