# Is every $\tau \in \text {Gal} \left (M|K \right ) = \sigma|_M$ for some $\sigma \in \text {Gal} \left (L|K \right )$?

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.

• 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. $L=M[\zeta].$ Nov 25, 2019 at 16:51
• 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 $\text {Gal} \left (L|K \right ).$ Nov 25, 2019 at 17:01
• But then every $\tau \in \text {Gal} \left (M|K \right )$ extends to $[L:M]$ many automorphisms of $\text {Gal} \left (L|K \right ).$ Nov 25, 2019 at 17:04
• 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? Nov 25, 2019 at 17:10

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 $$M$$.

• 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? Nov 25, 2019 at 19:10
• 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) Nov 26, 2019 at 9:00