All $\mathbb{Q}$-automorphism of $\mathbb{Q}(\sqrt{3},\sqrt{5})$ I have a question related to count the number of all $\mathbb{Q}$-automorphisms of $\mathbb{Q}(\sqrt{3},\sqrt{5})$. Since $\mathbb{Q}(\sqrt{3},\sqrt{5})$ is a splitting field of $(x^2-3)(x^2-5)$ over $\mathbb{Q}$. There is a theorem guarantee that (such theorem require $\mathbb{Q}(\sqrt{3},\sqrt{5})$ to be a splitting field of some polynomial):


*

*There is an automorphism maps $\sqrt{3}$ to $\sqrt{3}$ and fix $\Bbb Q$.

*There is an automorphism maps $\sqrt{3}$ to $-\sqrt{3}$ and fix $\Bbb Q$.

*There is an automorphism maps $\sqrt{5}$ to $\sqrt{5}$ and fix $\Bbb Q$.

*There is an automorphism maps $\sqrt{5}$ to $-\sqrt{5}$ and fix $\Bbb Q$.


So there are $2\times 2=4$ possible automorphisms, $|\textbf{Aut}_{\Bbb Q}\mathbb{Q}(\sqrt{3},\sqrt{5})|=4$. But my question is, is there indeed an automorphism maps $\sqrt{3}$ to $\sqrt{3}$ and meanwhile maps $\sqrt{5}$ to $-\sqrt{5}$? I know the answer is yes, but is it definitely true? Or is it so just by trivial reason? Or should we need a theorem to state this result? PS: I found that Hungerford explicit construct such four automorphisms, to check these are all indeed exist.
 A: Your descriptions of the four automorphisms ar lacking. For 1. and 2. you describe what happens to $\sqrt 3$, but not what happens to $\sqrt5$, and vice versa for 3. and 4.
Given an arbitrary element $a+b\sqrt3+c\sqrt5+d\sqrt{15}$ in your field, where $a, b, c, d$ are rational, the four automorphisms are:
$$
a+b\sqrt3+c\sqrt5+d\sqrt{15}\mapsto a+b\sqrt3+c\sqrt5+d\sqrt{15}\\
a+b\sqrt3+c\sqrt5+d\sqrt{15}\mapsto a-b\sqrt3+c\sqrt5-d\sqrt{15}\\
a+b\sqrt3+c\sqrt5+d\sqrt{15}\mapsto a+b\sqrt3-c\sqrt5-d\sqrt{15}\\
a+b\sqrt3+c\sqrt5+d\sqrt{15}\mapsto a-b\sqrt3-c\sqrt5+d\sqrt{15}
$$
Or, in words:


*

*One automorphism that maps $\sqrt3$ to $\sqrt3$ and $\sqrt5$ to $\sqrt5$

*One automorphism that maps $\sqrt3$ to $-\sqrt3$ and $\sqrt 5$ to $\sqrt5$

*One automorphism that maps $\sqrt3$ to $\sqrt3$ and $\sqrt 5$ to $-\sqrt5$

*One automorphism that maps $\sqrt3$ to $-\sqrt3$ and $\sqrt5$ to $-\sqrt5$
with the implicit understanding that the image of all other elements are defined by the properties of homomorphisms (as well as by fixing $\Bbb Q$, although that's superfluous in this specific case).
You should, of course, check that these are indeed automorphisms by showing that they are bijective homomorphisms.
A: You don't really have to check all the computations suggested by @Arthur 's answer.
The degree of the extension $\mathbb{Q}(\sqrt{3},\sqrt{5})$ is 4 (because $1,\sqrt{3},\sqrt{5}$ and $\sqrt{15}$ are linearly independent, for example). Also, the field being a splitting field of a separable polynomial(over Q any poly with simple roots is separable) implies that the extension is Galois. So the group of automorphisms will be either $\mathbb{Z}_2 \times \mathbb{Z}_2$ or $\mathbb{Z}_4$. But since $\sqrt{3}$ $\mapsto -\sqrt{3}$ (and leave Q fixed) and $\sqrt{5} \mapsto -\sqrt{5}$(and leave Q fixed again) are distinct automorphisms of $\mathbb{Q}(\sqrt{3})$ and $\mathbb{Q}(\sqrt{5})$ respectively, that can be extended to automorphisms of $\mathbb{Q}(\sqrt{3},\sqrt{5})$ (any automorphism of a smaller field may be extended to an automorphism of the larger field) and generate subgroups of order two in our Galois group G, it follows that the group cannot be $\mathbb{Z}_4$ (since $\mathbb{Z}_4$ has only one subgroup of order 2).
A: Yes. You have an automorphism for every election of signs. But you are no counting your morphisms well. For example, the morphisms 1. and 3. could be the identity.
