I'm reading Stewart's Galois Theory. In Chapter 8, we define that:

Let $L:K$ be a field extension, so that $K$ is a subfield of the subfield $L$ of $\mathbb{C}$. A $K$-automorphism of $L$ is an automorphism $\alpha$ of $L$ such that $\alpha(k) = k$ for all $k \in K$.

Then in example 8.4.3, we are discussing the $\mathbb{Q}$ automorphism of $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})$. We know that $t^2 -5$ is irreducible on $\mathbb{Q}(\sqrt{2},\sqrt{3})$,$t^2 -2$ is irreducible on $\mathbb{Q}(\sqrt{3},\sqrt{5})$, $t^2 -3$ is irreducible on $\mathbb{Q}(\sqrt{2},\sqrt{5})$. Then he simply says "Thus there are three $\mathbb{Q}$-automorphisms of $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})$:

$\rho_1 : \sqrt{2} \rightarrow -\sqrt{2}, \sqrt{3} \rightarrow \sqrt{3}, \sqrt{5} \rightarrow \sqrt{5}$

$\rho_2 : \sqrt{2} \rightarrow \sqrt{2}, \sqrt{3} \rightarrow -\sqrt{3}, \sqrt{5} \rightarrow \sqrt{5}$

$\rho_3 : \sqrt{2} \rightarrow \sqrt{2}, \sqrt{3} \rightarrow \sqrt{3}, \sqrt{5} \rightarrow -\sqrt{5}$

I'm not quite sure how he got these three $\mathbb{Q}$-automorphisms, can anybody explain to me the reasoning behind? Thank you very much!!


Since any $\Bbb Q$-morphism must preserve the coefficients of the polynomial $x^2 -2$, it must permute the roots of that polynomial. I.e., it must send $\sqrt{2}$ to $\pm \sqrt{2}$. Similarly for $\sqrt{3}, \sqrt{5}$. Thus in total, there are eight possible $\Bbb Q$-morphisms. The author has pointed out three of them, but of course there are others. In fact, all eight possibilities are indeed morphisms, and they are generated by the three the author lists.

  • $\begingroup$ This is super clear. Thank you! $\endgroup$ – nekodesu Jan 6 '16 at 3:41

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