# Complementation in an extension field

If $E$ is an extension field of $F$, is $F$ necessarily (without assuming the axiom of choice) complemented as a vector subspace of $E$? (Of course the answer is easily yes if the extension is finite.)

The motivation for the question is pedagogical. I'm looking for as elementary as possible (hence no AC) a way of addressing the following basic linear algebra question: if a linear system $A x = b$, with $A$ an $m\times n$ matrix over $F$ and $b \in F^m$, has a solution $x \in E^n$, does it necessarily have a solution in $F^n$?

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What's wrong with Gaussian elimination? – Asaf Feb 10 '13 at 18:39
@Asaf: "Elementary" wasn't quite the right word; clearly one can't get more elementary than Gaussian elimination here. I was hoping for an alternative argument that was still elementary but felt cleaner or more intuitive than Gaussian elimination -- I always feel like Gaussian elimination makes it hard to see what's "really" going on. In any case, never mind the original motivation; now I'm curious about the question in the first sentence. – Mark Meckes Feb 10 '13 at 18:51
@Mark, you can do it with Galois' theory, say move into a (bigger-) Galois extension if necessary, and the solution vector $x$ is fixed by the whole Galois group. Anyway, the second statement is completely algebraic in its nature and does not require any sort of AC (when dealing with finite dimensional extensions) besides the usual assumptions in math. – Asaf Feb 10 '13 at 19:55
It is well known that if $\mathbb{Q}$ is complemented in $\mathbb{R}$ you get nonmeasurable sets in $\mathbb{R}$. I don't think you can get this without some form of AC. – Laurent Moret-Bailly Feb 10 '13 at 20:56
@wccanard: Indeed, it's even easier to forget that once you've lectured several courses on basic vector space theory without once finding a need to mention Gaussian elimination! @Asaf: Of course finite dimensional extensions are trivial to deal with here anyway. The prototypical case that is interesting to think about here is $E = \mathbb{Q}$ and $F = \mathbb{R}$. Finally, @Laurent: Thanks, that's exactly the kind of answer I was looking for. – Mark Meckes Feb 10 '13 at 21:27