# Automorphisms of composite fields of finite extensions

If $\phi:\mathbb{Q}(a_1\ldots,a_n)\rightarrow\mathbb{Q}(b_1,\ldots,b_n)$ is an isomorphism of finite extensions of $\mathbb{Q}$ such that $\phi(a_i)=b_i$, can one extend $\phi$ to an automorphism $\sigma$ of the composite $\mathbb{Q}(a_1,\ldots,a_n,b_1,\ldots,b_n)$? Is ther an obvious counter-example?

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The composite is not well-defined without fixing a common extension of both fields, but I assume that you want $\mathbb{C}$ or a fixed closure $\overline{\mathbb{Q}}$ for that. –  Martin Brandenburg Dec 7 '12 at 20:15
Yeah I was thinking $\mathbb{C}$ –  greg Dec 7 '12 at 20:17
@MartinBrandenburg: when talking about algebraic extensions, you would usually assume that they're all contained in some fixed algebraic closure... –  tomasz Dec 7 '12 at 20:36
If they were transcendental, they wouldn't be finite extensions. –  Hurkyl Dec 7 '12 at 22:05
I want to say $a = \sqrt[u]{1 + \sqrt[v]{2}}$ and $b = \sqrt[u]{1 + \zeta_v \sqrt[v]{2}}$ where $u,v$ are appropriately chosen form a counter-example. But they aren't counter-examples for very small values of $u$ and $v$. :( –  Hurkyl Dec 8 '12 at 10:44

I've not found a definite answer yet, but here are some partial results. I can't guarantee that I have not overthought it and that there is a much simpler solution. :)

Note that by Abel's theorem on primitive element there is an $\alpha$ such that ${\bf Q}(a_1,\ldots,a_n)={\bf Q}[\alpha]$. In this case, it will also be true that ${\bf Q}[\phi(\alpha)]={\bf Q}(b_1,\ldots,b_n)$ (because $\phi$ is an isomorphism) and $\alpha$ and $\phi(\alpha)$ are conjugate.

Consider $K$, the splitting field of $\alpha$'s minimal polynomial $f_\alpha$. What you ask is equivalent to asking if there's an automorphism $\Phi$ of $K$ such that $\Phi(\alpha)=\phi(\alpha)$, and such that $\Phi(\Phi(\alpha))\in{\bf Q}[\alpha,\phi(\alpha)]$.

This is true, for example, if $\alpha$ and $\phi(\alpha)$ already generate $K$ (e.g. if $\alpha$ is a root of unity, or $\alpha=\sqrt[4]{2},\phi(\alpha)=i\sqrt[4]{2}$, or if $K$ is of prime degree), or if $\phi$ is the restriction of an involutive automorphism (e.g. if $\alpha$ and $\phi(\alpha)$ are complex conjugate), it's also true if the degree of $K$ is the product of at most two (maybe not distinct) primes (so in particular if $f_\alpha$ is at most cubic), and also if the Galois group of $K$ is bitransitive (in particular, if it is the full symmetric group, which I think is the "generic" situation), but I don't think it has to be the case.

Ideally, a counterexample would have $K$ of degree $24$ (anything less won't work, as pointed out by Hurkyl), with $f_\alpha$ of degree $4$. But then the Galois group is the full permutation group, so there's no counterexample with $f_\alpha$ of degree four. This does seem to make the hypothesis viable... but hypotheses that fail only for polynomials of degree five or more are not unheard of. ;)

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If $f$ has degree $4$ and doesn't have a degree 24 splitting field, then either $\phi(\alpha) \in \mathbb{Q}(\alpha)$, or $\mathbb{Q}(\alpha, \phi(\alpha)) = K$ because $\mathbb{Q}(\alpha)$ is already degree 4, so any further extension is the whole thing. –  Hurkyl Dec 7 '12 at 22:11
@Hurkyl: good point. :) –  tomasz Dec 7 '12 at 22:14