Let $L|K$ be a finite Galois extension. Let $G = \text {Gal} \left (L|K \right ).$ Let $H$ be a normal subgroup of $G$ and $M = \text {Fix}_H L.$ Can we say that for every $\tau \in \text {Gal} \left (M|K \right )$ $\exists$ $\sigma \in G$ such that $\sigma|_M = \tau$?

If we drop the condition $H$ a normal subgroup of $G$ and instead if we take any subgroup what happens to the above case?

Any help in this regard will be highly appreciated. Thanks for reading.

  • $\begingroup$ I think that it can be done by using primitive element theorem for finite Galois extensions. Because $L|M$ is a galois extension whenever $L|K$ is a Galois extension. So it has a primitive element say $\zeta$ i.e. $L=M[\zeta].$ $\endgroup$ Nov 25, 2019 at 16:51
  • $\begingroup$ Now if $\tau \in \text {Gal} \left ( M|K \right )$ then if allow $\tau$ to send $\zeta$ to some $\zeta'$ where $\zeta'$ is some zero of the minimal polynomial $\mu_{\zeta,M}$ of $\zeta$ over $M$ then $\tau$ extends to an automorphism of $\text {Gal} \left (L|K \right ).$ $\endgroup$ Nov 25, 2019 at 17:01
  • $\begingroup$ But then every $\tau \in \text {Gal} \left (M|K \right )$ extends to $[L:M]$ many automorphisms of $ \text {Gal} \left (L|K \right ).$ $\endgroup$ Nov 25, 2019 at 17:04
  • $\begingroup$ That will imply $\#\ \text {Gal} \left (M|K \right ) = \frac {[L:K]} {[L:M]} = [M:K].$ But this is not always true unless $\text {Gal} \left (L|M \right )$ is a normal subgroup of $ \text {Gal} \left (L|K \right ).$ Where am I going wrong then? $\endgroup$ Nov 25, 2019 at 17:10

1 Answer 1


Let $\tau\in Gal(M/K)$. Then it can be viewed as a $K$-embedding $M\to K_{alg}$. The theorem of extension of embeddings implies that there exists $\sigma:L\to K_{alg}$ such that $\sigma_M=\tau$ (you can use the fact that $L=K(\alpha)$ for some $\alpha\in L$ if you like). But $L/K$ is a Galois, hence normal, so the image of $\sigma$ is contained in $L$ and we have in fact $\sigma\in Gal(L/K)$.

Notice this is true even if $H$ is not normal. There is no contradiction, because the equality $\sharp Gal(L/K)=\dfrac{[L:K]}{[L:M]}$ is false: ok, any $K$-automorphism of $M$ extends to a $K$-automorphism of $L$, but conversely, a $K$-automorphism of $L$ does not restrict necessarily to a $K$-automorphism of $M$, but only to a $K$-embedding. It would be true only if $H$ is normal!! So an element of $Gal(L/K)$ is not necessarily an extension of an element of $Gal(M/K)$, and you cannot count like you do.

For a specific example, think about $L=\mathbb{Q}(j,\sqrt[3]{2})$ and $M=\mathbb{Q}(\sqrt[3]{2})$. Take $\sigma:L\to L$ which sends $j$ to $j$ and $\sqrt[3]{2}$ to $j\sqrt[3]{2}$. Its restriction to $M$ is NOT a $\mathbb{Q}$-automorphism of $M$.

  • $\begingroup$ What do you mean by "... your last implication is false..."? Which of my implication is not correct. Can you please help me so that I can find my mistake? $\endgroup$ Nov 25, 2019 at 19:10
  • $\begingroup$ The equality $\sharp Gal(L/K)=\dfrac{[L:K]}{[L:M]}$ is not true. An element of $Gal(L/K)$ is not necessarily an extension of an element of $Gal(M/K)$, so the way you count is not correct (I(ve modified my answer to be more clear) $\endgroup$
    – GreginGre
    Nov 26, 2019 at 9:00

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.