# If $K,E$ are subfields of $\Omega/F$ then $KE/F$ is a finite Galois imply $K/K\cap E$ is Galois?

Let $\Omega/F$ be a field extension and $K,E$ be two subfields of $\Omega/F$. Assume that $KE/F$ is a finite Galois.

I have a theorem in my lecture notes that claim $\text{Gal}(KE/E)\cong \text{Gal}(K/K\cap E)$, while it is easy to see that $KE/F$ is Galois $\implies KE/E$ is Galois I can not understand why $K/K\cap E$ is Galois (it is separable since $KE/F$ is, but I don't see why it's normal).

Why $K/K\cap E$ is Galois ?

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I think you need to assume that $K$ is Galois over $K\cap E$, in which case you get the isomorphism... – Arturo Magidin Jul 20 '12 at 21:27

You need to assume that $K$ is Galois over $K\cap E$; if that is the case, then the isomorphism drops out of the Galois correspondence and the Isomorphism Theorems.

We can replace $F$ with $K\cap E$, so that we are in the following situation:

• $KE/F$ is finite Galois;
• $K\cap E = F$.

If $G=\mathrm{Gal}(KE/F)$, let $M$ be the subgroup corresponding to $K$ and $N$ the subgroup corresponding to $E$. Then $M\cap N=\{e\}$ (since $KE$ is the field "on top"), and $\langle M,N\rangle = G$.

Now, $K$ is Galois over $F$ if and only if $M$ is normal in $\langle M,N\rangle$.

In particular, if $K$ is Galois over $F$, then $M\triangleleft \langle M,N\rangle$, so $\langle M,N\rangle = MN$. Thus, $N\cap M$ is normal in $N$, and by the isomorphism theorems we have that $$\frac{G}{M} =\frac{MN}{M} \cong \frac{N}{N\cap M} \cong N.$$ Now, $\frac{G}{M}\cong\mathrm{Gal}(K/F)$; and $N=\mathrm{Gal}(KE/E)$; so we get the isomorphism if $K$ is Galois.

For an example showing that the given conditions do not imply that $K$ is Galois over $K\cap E$, let $F=\mathbb{Q}$, $K=\mathbb{Q}[\sqrt[3]{2}]$, and $E=\mathbb{Q}(\zeta)$, where $\zeta$ is a primitive cubic root of unity; then $K\cap E=\mathbb{Q}$. Then $KE$ is the splitting field of $x^3-2$ over $\mathbb{Q}$, hence is Galois. Even though $KE$ is Galois over $E$, $K$ is not Galois over $\mathbb{Q}$, so you cannot have the claimed isomorphism.

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Thank you very much! – Belgi Jul 20 '12 at 21:45