# A Fixed Field of $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$

I was working on a review problem from Dummit and Foote and came across the following issue. It is clear that the Galois group of the splitting field for the polynomial $(x^2-2)(x^2-3)(x^2-5)$ has order $8$ and one of the automorphisms is $$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5} \mapsto a-b\sqrt{2}-c\sqrt{3}+d\sqrt{5}$$ which I would think has associated fixed field $\mathbb{Q}(\sqrt{5})$. However, after completing the problem I found in the solutions that this automorphism has fixed field $\mathbb{Q}(\sqrt{6},\sqrt{5})$. Why is the $\sqrt{6}$ there? I would think that by setting $$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5} =a-b\sqrt{2}-c\sqrt{3}+d\sqrt{5}$$ we find $$b\sqrt{2}+c\sqrt{3}=-b\sqrt{2}-c\sqrt{3}$$ so that $b\sqrt{2}+c\sqrt{3}=0$ so that $(b/c)^2=3/2$, impossible as $b,c \in \mathbb{Q}$. How how is it that this automorphism is fixing $\mathbb{Q}(\sqrt{6})$?

• The automorphism fixes the square root of 6, what's the problem? Recall not all elements of your field are linear combination of those four numbers, $1$ and the three roots, there are also combinations of products. You can tell, because the dimension is 8, not 4. – Adam Hughes Jul 29 '14 at 0:06
• @AdamHughes I didn't observe that until catching rogerl's answer. I thought it was sufficient to check the image of the generating elements alone. Is it always the case that one need check all the possible products of images of the generating set? – Kyle L Jul 29 '14 at 0:09
• Fixed fields are fixed sub vector spaces, so you need to check on a linear combination of basis elements. In particular, you are looking for fixed subspaces for the linear transformations that are the automorphisms in question. – Adam Hughes Jul 29 '14 at 0:13
• @AdamHughes Thank you, that makes it very clear. I had been too focused on the action on the generating set and not enough on the action on the basis. – Kyle L Jul 29 '14 at 0:15

## 1 Answer

Under this automorphism, $\sqrt{2}\mapsto -\sqrt{2}$ and $\sqrt{3}\mapsto -\sqrt{3}$, so that $\sqrt{6}\mapsto\sqrt{6}$.

• Thanks! That clears up the question so easily! It is always the case that one need consider not only the image of the generating elements but all their possible products as well? – Kyle L Jul 29 '14 at 0:08
• Yes, of course, because you are talking about field homomorphisms. – rogerl Jul 29 '14 at 0:09
• @KyleL Does that make sense now? – rogerl Jul 29 '14 at 0:18
• I believe so. Just to be sure, if it were enough for the fixed field to be a vector subspace we would only need to check what I did, check the image of the basis elements. However, since a fixed subfield also has products (it's a field so it has multiplication and addition) we need to check products of the basis elements under the automorphisms as well. Correct? – Kyle L Jul 29 '14 at 0:53
• That's essentially correct. As you point out, $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ is both a $\mathbb{Q}$-vector space and a field extension; the two are very different algebraic structures. – rogerl Jul 29 '14 at 1:04