I'm working on a lemma concerning some Galois theory and arithmetics. Let $p$ be an odd prime and $K/F$ be a finite Galois extension of number fields of order prime to $p$ with Galois group $H$. Let $S$ be a finite set of places of $K$ and denote by $K_S(p)$ the maximal $p$-extension of $K$ that is unramified outside the places of $S$ (compositum of all Galois extensions of $K$ that are of $p$-power-degree). I know that $H$ acts transitively on the set of all places of $K$, but my claim is now

Claim: If the set $S$ is invariant under the action of $H$, then $K_S(p)/F$ is normal.

Attempt of a proof: Let $\sigma \in H$ be an F-automorphism of $K$. This gives a F-homomorphismus $\tilde\sigma: K_S(p) \longrightarrow \bar K$, where $\bar K$ is a fixed algebraic closure (this is an algebraic closure for $K_S(p)$ as well). If the image of $K_S(p)$ under $\tilde\sigma$, $M$, was not $K_S(p)$, then the field compositum $M \cdot K_S(p)$ would have the property of being unramified outside of $S$ again, because $\sigma$ and hence $\tilde\sigma$ leaves the places of $S$ invariant. That contradicts the fact that $K_S(p)$ is maximale. Therefore $M$ has to be $K_S(p)$ and $\tilde\sigma$ is in fact an $F$-automorphism of $K_S(p)$.

Is that proof correct?

Thank you for sharing your experience and knowledge with me..

Greetings, Tom

  • $\begingroup$ Ok, I was wrong with my question about the prolongation $\tilde\sigma$. Of course - for showing that $K_S(p) / F$ is normal, it is necessary for an $F$-homomorphism $K_S(p) \longrightarrow \bar{K_S(p)}$ to be an $F$-automorphism of $K_S(p)$. So we just take an arbitrary $F$-homomorphism as above, which naturally can be restricted to $K$ and on that level is an $F$-automorphism, since $K_S(p)$ is normal over $K$, right? $\endgroup$ – BIS HD Aug 13 '13 at 12:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.