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Suppose $F/L$, $F'/L$, $L/K$ finite extensions of fields. If $F$, $F'$ isomorphic over $K$ then does it follow that they are isomorphic over $L$? I think probably not, but I can't come up with a counterexample. I've tried thinking about splitting fields of quartics or about function fields and neither has given me any joy. Obviously finite fields are no good. Could someone give me a hint? Thanks!

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You should be able to come up with an example with $K = \mathbb{Q}$. What kind of things have you tried for $L$? – Qiaochu Yuan May 18 '12 at 15:36
I've tried quadratic and cubic extensions, but can't see anything that works... – Edward Hughes May 18 '12 at 15:41
Well, here's kind of a big hint. I don't claim that this is the only way to do the problem. Choose $L$ so that it embeds into $\mathbb{R}$, and choose $F, F'$ so that if you fix an embedding of $L$ into $\mathbb{R}$, one of $F, F'$ is also embeddable into $\mathbb{R}$ and the other isn't. – Qiaochu Yuan May 18 '12 at 15:43
Is it $L=\mathbb{Q}(\sqrt{2})$,$F=\mathbb{Q}(\sqrt[4]{2})$, $F'=\mathbb{Q}(i\sqrt[4]{2})$? – Edward Hughes May 18 '12 at 16:01
@hgbreton: yep! – Qiaochu Yuan May 18 '12 at 16:05
up vote 6 down vote accepted

This is indeed false. An easy counterexample is $L=\mathbb{Q}(\sqrt{2})$, $F=\mathbb{Q}(\sqrt[4]{2})$, $F'=\mathbb{Q}(i\sqrt[4]{2})$.

Clearly $F$, $F'$ isomorphic over $\mathbb{Q}$ by $\sqrt[4]{2}\mapsto i\sqrt[4]{2}$. Now suppose that $F$, $F'$ isomorphic over $L$ by $\tau$ say. Then $\tau(\sqrt[4]{2})=a+ib\sqrt[4]{2}$ some $a,b\in L$. We must have


This forces $a=0$ so $b^2=-1$ and contradiction.

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Alternatively, $\tau$ must map a root of $X^4 - 2$ to another root, so it pretty much has to send $\sqrt[4]{2}$ to $\pm i\sqrt[4]{2}$, but then $\sqrt{2}$ maps to $-\sqrt{2}$ so we can't fix $L$. (PS. 2011 paper 1 question 18H, right? :P) – Ben Millwood Jun 2 '12 at 17:48
Yep - that's right! – Edward Hughes Jun 2 '12 at 18:29

Both $F = \Bbb{Q}(\sqrt[4]{2})$ and $F' = \Bbb{Q}(i\sqrt[4]{2})$ are isomorphic to $\Bbb{Q}[x]/(x^4 - 2)$, hence $\Bbb{Q}$-isomorphic, ($x^4 - 2$ is irreducible over $\Bbb{Q}$, so these are field extensions. The actual isomorphism is given by $\sqrt[4]{2} \mapsto i\sqrt[4]{2}$). However, they are not isomorphic over $L = \Bbb{Q}(\sqrt{2})$, as $x^2 + \sqrt{2} \in L[x]$ splits in $F'$, but not in $F$.

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