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There is a theorem in Lang which says that if $L/k$ is Galois and $k\subseteq K$ is any field extension, if $L,K$ are subfields of a larger field, then $LK$ is Galois over $K$. I was wondering if $LK$ is also Galois over $L$?

I think this would be true if $K\subseteq L$ since then we can apply the same theorem to the extension $K\subseteq LK$ and use that $(LK)L=LK$. Is that right?

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I don't understand your second question... If $K\subseteq L$, then $LK=L$ which is trivially Galois over $L$. –  Álvaro Lozano-Robledo Mar 9 '12 at 19:03

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up vote 3 down vote accepted

Counterexample: let $k=\mathbb{Q}$, $L=\mathbb{Q}[\sqrt{2}]$, and $K=\mathbb{Q}[\sqrt[3]{2}]$.

Then $LK=\mathbb{Q}[\sqrt{2},\sqrt[3]{2}]$ is not Galois over $L=\mathbb{Q}[\sqrt{2}]$, because $x^3-2$ is irreducible over $L$, $LK$ contains one of its roots, but it does not contain the complex roots of $x^3-2$.

(If $K\subseteq L$, then you aren't doing anything...)

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No, in general $LK$ is not Galois over $L$.
Indeed, take $L=k$. Then $LK=K$ and the question becomes :

$$\text {Is an arbitrary field extension } \;K/k \; \text {Galois?} $$
Need I answer?

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