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Find the Galois group of $f(x)=x^4+5x^2+5\in \mathbb{Q}[x]$.

This is solved here, Exersice 3: https://math.berkeley.edu/~serganov/114/solhwg.pdf

I have a question about it (I will not write all the details).

First, he find out that $f(x)$ is irreducible over $\mathbb{Q}$ and has four distinct (irrationals) roots $\alpha_1,\alpha_2,\alpha_3,\alpha_4$. Then he proves $\alpha_2,\alpha_3,\alpha_4\in\mathbb{Q}(\alpha_1)$, hence the splitting field of $f(x)$ is $\mathbb{Q}(\alpha_1)/\mathbb{Q}$ and its Galois group is of order $4$. Finally, he takes $s\in G$ (the Galois group) such that $s(\alpha_1)=\alpha_2$ and proves $s$ has order $4$. So the Galois group is $\mathbb{Z}_4$.

My only question is: how do you know there is $s\in G$ such that $s(\alpha_1)=\alpha_2$?

Added: I know that $s(\alpha_1)\in\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}$ (for every $s$ that is a $\mathbb{Q}$-automorphism).

Thank you.

Talexius
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    I believe (at least in characteristic $0$) that when $f$ is irreducible you can always map one root to another in the splitting field of $f$ via an automorophism of the splitting field. – Gregory Grant Jun 16 '16 at 17:56
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    Yes if $\alpha$ and $\beta$ are two roots of an irreducible $f$ then $\Bbb Q[\alpha]$ is isomorphic to $\Bbb Q[\beta]$ and that isomorphism extends to an automorphism of the splitting field. – Gregory Grant Jun 16 '16 at 17:57
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    Oh, thank you. I see it. Btw in this case $\mathbb{Q}(\alpha_1)=\mathbb{Q}(\alpha_2)$ so we don't have to extend it. @GregoryGrant – Talexius Jun 16 '16 at 18:16
  • Related: https://math.stackexchange.com/questions/204709 – Watson Dec 26 '16 at 13:03

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