<s>[INST] 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 $\mathbb{Q}(\sqrt{6})$?
 [/INST] Under this automorphism, $\sqrt{2}\mapsto -\sqrt{2}$ and $\sqrt{3}\mapsto -\sqrt{3}$, so that $\sqrt{6}\mapsto\sqrt{6}$.
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