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I'm trying to understand the proof of the statement that, if $L\mathop{:}K$ is a finite,normal and separable with intermediate field $M$ such that $M\mathop:K$ is a normal extension, then the quotient group $\Gamma(L:K)/\Gamma(L:M)$ is isomorphic to $\Gamma(M:K)$. Where $\Gamma$ is used to denote the Galois group of a field extension. The proof is quite simple and involves constructing a function $$\phi:\Gamma(L:K)\to\Gamma(M:K)$$ where for any $K$ automorphism $\alpha \in \Gamma(L:K)$ the definition is $\phi(\alpha)=\alpha|_M$. My text states that because $\alpha|_M$ is a $K$ automorphism on $M$ (by a previous theorem which is fine) it follows that $\phi$ is a group homomorphism. If I accept this fact the rest of the proof is trivial but it seems to me that it does not matter that $\alpha|_M$ being a $K$ automorphism as it does not come into the proof at all. It seems to me that for any $m\in M$ and $\gamma,\alpha \in \Gamma(L:K)$ that simply by how function restriction is defined that $(\gamma\circ\alpha)|_M(m)=(\gamma|_M\circ\alpha|_M)(m)$. What am I missing that makes the fact of the restriction of $\alpha$ being a automorphism necessary?

$\textbf{Edit:}$ After further ruminations in the shower I believe that the crux of the matter lies in the fact if $\alpha|_M$ wasn't an automorphism on $M$ then we're not guaranteed that $a|_M(m)$ is even in the domain of $\gamma|_M$, which is rather problematic. Without $\alpha|_M$ being a automorphism on $M$ we couldn't use the trivial argument that I thought we could, as in my argument I assumed that the composition is always well defined; so in essence I used the very thing I thought I wasn't using. Thus I believe I have pretty much answered my own question, but I'd still appreciate any other views.

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  • $\begingroup$ Take $m\in M$. Then $\phi(\gamma \circ \alpha)(m)=(\gamma \circ \alpha )(m)$ since the image of the two map are indentical on $M$. But then it is equal to $\phi(\gamma)\circ \phi(\alpha)(m)$. $\endgroup$
    – user966
    Commented Jun 4, 2016 at 17:52

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Just evaluate, let $m\in M$, $\phi(\gamma\circ\alpha)(m)=(\gamma\circ\alpha)\mid_M(m)=(\gamma\circ\alpha)(m)=\gamma(\alpha(m))$, and $(\phi(\gamma)\circ\phi(\alpha))(m)=((\gamma\mid_M)\circ(\alpha\mid_M))(m)=\gamma\mid_M(\alpha\mid_M(m))=\gamma\mid_M(\alpha(m))=\gamma(\alpha(m))$, here you use that $\alpha(m)\in M $, which means that $\phi(\gamma\circ\alpha)=\phi(\gamma)\circ\phi(\alpha)$.

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