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Morandi's Field and Galois Theory, exercise 5.19b

Let $K$ and $L$ be Galois extensions of $F$. The restriction of function map, namely, $\sigma\mapsto(\sigma\vert_K,\sigma\vert_L)$ induces an injective group homomorphism $\varphi\colon\operatorname{Gal}(KL/F)\to\operatorname{Gal}(K/F)\times\operatorname{Gal}(L/F)$. Show that $\varphi$ is surjective if and only if $K\cap L=F$.

It's not hard to show that $\varphi$ is a monomorphism. If it's surjective, it's not hard to show that $K\cap L=F$ as follow:

Fix $\alpha\in K\cap L$, let $\beta$ be a root of the minimal polynomial of $\alpha$ over $F$. Since $K,L$ are normal, $\beta\in K\cap L$. By isomorphism extension theorem, we can choose $\tau_1\in\operatorname{Gal}(K/F)$ such that $\tau_1(\alpha)=\beta$. For surjectivity of the map, there's $\sigma$ such that $\sigma\vert_K=\tau_1$ and $\sigma_L=\mathrm{id}$, which forces $\alpha=\beta$, therefore $\alpha\in F$, since $K,L$ are separable over $F$.

The converse seems hard. I cannot show that when $K,L$ are arbitrary Galois extensions. If they are both finite dimensional, the statement follows from natural irrationality: $\operatorname{Gal}(KL/L)\cong\operatorname{Gal}(K/K\cap L)$, which implies that $[KL:L]=[K:K\cap L]=[K:F]$, therefore $[KL:F]=[K:F][L:F]$, and note that $\varphi$ is injective, thus surjective.

Any help? Thanks!

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  • $\begingroup$ If there intersection is $F$, then $KL\cong K\otimes_F L$. It seems as though you can use this to your advantage. For, then if $(\sigma,\tau)$ is in the codomain, you get a map on $K\otimes_F L$ by $x\otimes y\mapsto \sigma(x)\otimes \tau(y)$. $\endgroup$ Sep 29, 2013 at 0:47
  • $\begingroup$ @AlexYoucis How did you show the disjointness? I don't know how to make use of the condition that $K/F$ and $L/F$ are Galois. Apparently, if there were no any condition, it should be false. We'd only need to consider $F=\mathbb Q,K=F(\sqrt[3]2),L=F(\sqrt[3]2\omega)$ where $\omega=\exp(2\pi/3)$. $\endgroup$
    – Yai0Phah
    Oct 6, 2013 at 5:09
  • $\begingroup$ @AlexYoucis, I'm not following your argument; shouldn't it be their instead of there? $\endgroup$ May 4, 2015 at 22:52

2 Answers 2

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We need the following lemma.

Lemma Let $K/F$ be a (not necessarily finite dimensional) Galois extension, $L/F$ an arbitrary extension. Clearly $KL/L$ is Galois. Then the restriction map, namely, $\sigma\mapsto \sigma\mid K$ induces an isomorphism $\psi\colon \mathrm{Gal}(KL/L) \rightarrow \mathrm{Gal}(K/K\cap L)$.

Proof: We regard $\mathrm{Gal}(KL/L)$ and $\operatorname{Gal}(K/K\cap L)$ as topological groups with Krull topologies. Clearly $\psi$ is continuous and injective. Let $H = \psi(\mathrm{Gal}(KL/L))$. Since $\mathrm{Gal}(KL/L)$ is compact, $H$ is also compact. Since $\mathrm{Gal}(K/K\cap L)$ is Hausdorff, $H$ is closed. Clearly the fixed subfield of $K$ by $H$ is $K \cap L$. Hence $H = \mathrm{Gal}(K/K\cap L)$ by the fundamental theorem of (not necessarily finite dimensional) Galois theory. This completes the proof.

Now we prove the following proposition with which the OP had a problem.

Proposition Let $K$ and $L$ be Galois extensions of $F$. The restriction of function map, namely, $\sigma\mapsto(\sigma\vert_K,\sigma\vert_L)$ induces a group homomorphism $\varphi\colon\operatorname{Gal}(KL/F)\to\operatorname{Gal}(K/F)\times\operatorname{Gal}(L/F)$. Suppose $K\cap L=F$. Then $\varphi$ is an isomorphism.

Proof. Since it is clear that $\varphi$ is injective, it suffices to prove that it is surjective. Let $G_1 = \mathrm{Gal}(K/F), G_2 = \mathrm{Gal}(L/F), G = \mathrm{Gal}(KL/F)$. By the lemma, given $\sigma_1 \in G_1$, there exists $\sigma \in \mathrm{Gal}(KL/L)$ such that $\sigma\mid K = \sigma_1$. Since $\sigma \in G$ and $\sigma\mid L = 1_L$, $G_1\times 1 \subset \varphi(G)$. Similarly $1\times G_2 \subset \varphi(G)$. Hence $G_1\times G_2 = \varphi(G)$. This completes the proof.

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  • $\begingroup$ It's an ancient post and I feel glad that I eventually receive an answer. It would be better if you can furnish some good reference for infinite Galois theory. It seems that the corresponding chapter of Morandi's book is indigestive to me. $\endgroup$
    – Yai0Phah
    Jun 19, 2014 at 14:39
  • $\begingroup$ @FrankScience I learned the subject from Bourbaki's textbook on algebra. However, I suppose many people think it is too dry. I have found this article on the internet. I think it is easy to understand. math.leidenuniv.nl/~astolk/docs/galois.pdf $\endgroup$
    – user157678
    Jun 19, 2014 at 15:51
  • $\begingroup$ Well, I'm looking for some experts to check this proof, so the acceptance might be postponed. Sorry! $\endgroup$
    – Yai0Phah
    Jun 22, 2014 at 2:31
  • $\begingroup$ @FrankScience Please don't accept the answer unless you understand it. The lemma is proved not only in the article(see Theorem, p.6)(math.leidenuniv.nl/~astolk/docs/galois.pdf) that I referred to, but also in Milne's course note(see Proposition 7.14, p.96)(jmilne.org/math/CourseNotes/ft.html). If you don't understand the proof of the proposition above, please let me know where you are stuck. $\endgroup$
    – user157678
    Jun 22, 2014 at 2:56
  • $\begingroup$ Well, thanks. I've been away from Galois theory since then, and I still find that I don't digest Galois theory well. Many theorems are still nontrivial to me, so I need to refresh the materials. Infinite Galois theory and profinite groups are still left indigestions, which I want to compare with ring completions, also an application of inverse limit. Incidentally, is there any material on (elementary) geometric (especially, algebro-geometric) viewpoint towards Galois theory? $\endgroup$
    – Yai0Phah
    Jun 22, 2014 at 3:18
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I think this solution will give more general intuition.

Let $\varphi : Gal(KL/F) \to Gal(K/F) \times Gal(L/F)$ which sends $\sigma \to (\left.\sigma\right|_{K}, \left.\sigma\right|_{L})$.

Lemma: $Gal(KL/F)$ is isomorphic to the subgroup $H = \{(\sigma, \tau) \}$ of $Gal(K/F) \times Gal(L/F)$ where $\left.\sigma\right|_{K \cap L} = \left.\tau\right|_{K \cap L}$. So $\varphi$ is surjective if and only if $H = Gal(K/F) \times Gal(L/F)$, and this happens if and only if $K \cap L = F$.

Proof. $\varphi$ is clearly injective (any element in kernel must fix $K$ and $L$, therefore, it fixes $KF$ which implies the identity). Let $A = img(\varphi)$. It's obvious that $A \subset H$. To prove the reverse, not that $KL$ is Galois over $F$ and $K \cap L$ is a subfield containing $F$. Then any embedding from $K \cap L$ to $\bar{F}$ is of the form $\left.\phi\right|_{K \cap L}$ for some $\phi \in Gal(KL/F)$ (generally, given field $F$ with Galois extension $K$, if $E$ is subfield of $K$ containing $F$, then any embedding of $E$ which fixes $F$ is of the form $\left.\phi\right|_{E}$ for some $\phi \in Gal(K/F)$. This is a nice property, and I'll let you play with it!). If $(\phi, \tau) \in H$, then $\left.\sigma\right|_{K \cap L} = \left.\tau\right|_{K \cap L}$ and $\left.\sigma\right|_{K \cap L} =\left.\tau\right|_{K \cap L}$ is actually of them form $\left.\phi\right|_{K \cap L}$ for some $\phi \in Gal(KL/F)$ which implies $H \subset A$, hence $H=A$.

Hope this helps!

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