I am going over a proof in the Fundamental Theorem of Galois Theory and I need a little clarification. I hope you can help.

Let $L/K$ be a finite extension with Galois group $G$ and let $M$ be an intermediate field. Denote by $M^*$ the group of all $M$-automorphisms of $L$. Then a part of the Fundamental Theorem of Galois Theory states:

If an intermediate field $M$ is a normal extension of $K$ then the Galois group of $M/K$ is isomorphic to the quotient group $G/M^*$.

Now the proof is rather simple just defining the map $\phi: G \to G'$ by $\phi(\tau)=\tau \mid_M$ where $\tau \in G$ and $G'$ is the Galois group of $M/K$. Then $\phi$ is a group homomorphism and surjective. The deciding point in the proof is the claim that the kernel of $\phi$ is $M^*$, i.e., the kernel of $\phi$ is the group of all $M$-automorphisms of $L$. Why is this so? It seems to me that if $\tau \in M^*$, $\phi(\tau)=\tau \mid_M$ and $\tau \mid_M$ is just the identity on $M$ (since it is an $M$-automorphism)?

If I can get the kernel to be $M^*$ then $G' \cong G/\text{ker}(\phi) = G/M^*$ and that would be the end of it.


If $\varphi$ is an automorphism of $L$ that fixes $M$, then it will restrict to the identity automorphism on the extension $M \subset K$ (since it fixes every element of M). Likewise, any element of the Galois group of $L$ over $K$ that restricts to the identity on $M$ must fix $M$ pointwise (mostly by definition here), so it follows that $\ker \phi = $ Gal$(L/M)$ (where $\phi$ is the restriction map).

As you said in your post, this implies that Gal$(M/K) \cong G/M^{\ast}$, where $G = $Gal($L/K$) and $M^{\ast}$ = Gal($L/M$).

  • $\begingroup$ Thanks for your quick reply! But I still don't see how the automorphisms in $\text{Gal}(L/M)$ are mapped to '$0$'? Ah, but perhaps the $0$-element in the restriction to $M$ are those that are the identity in the first place? $\endgroup$ – Numbersandsoon Aug 5 '14 at 15:41
  • $\begingroup$ Remember that your map $\phi$ is a homomorphism between Galois groups, so the "$0$" element in question is the identity map $I$ with $I(x) = x$ for all $x$ in the given field. By definition every automorphism in Gal$(L/M)$ fixes $M$ elementwise, so it follows that these automorphisms act like the identity map on $M$. So when you restrict their domain to $M$ they all become the identity map, hence $\phi$ maps everything in Gal$(L/M)$ to the "$0$" element of Gal($M/K$). $\endgroup$ – A. Barron Aug 5 '14 at 15:47
  • $\begingroup$ Okay! The ''o'' element is the identity map. Then I'm following. Thanks! $\endgroup$ – Numbersandsoon Aug 5 '14 at 15:50

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