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A basis for $\mathbb{Q}(\sqrt 2 + \sqrt 3)$ over $\mathbb{Q}$ is $\{1,\sqrt 2 , \sqrt 3 , \sqrt 6 \}$

The roots of $x^2 -2$ are $\pm \sqrt 2$ and the roots of $x^2 -3$ are $\pm \sqrt 3$ so to find automorphisms of $Gal$ we have to map roots of the same polynomial to another root so $\theta(\sqrt 2)= \pm \sqrt 2$and $\theta \sqrt 3 = \pm \sqrt 3$

These are possible automorphisms:

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My question is How do I get out of calculating that these automorphisms preserve addition and multiplication? In other words a shortcut :)

I know that $|Gal(\mathbb{Q}(\sqrt 2 + \sqrt 3)/\mathbb{Q})| = [\mathbb{Q}(\sqrt 2 + \sqrt 3): \mathbb{Q}] = 4$ and according to my book this fact allows me to omit details that these mappings preserve $+/*$. I don't get why.

I did one them, but multiplying 2 four entry polynomials is not fun :(.

Thanks

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  • $\begingroup$ I don't think the accepted answer fully answers your question. Don't forget that there is a very useful theorem that you can always find an automorphism fixing $\mathbb{Q}$ and mapping an element to a root of its minimal polynomial over $\mathbb{Q}$. This shows immediately that sending $\sqrt{2}$ to $-\sqrt{2}$ gives an automorphism, and likewise for $\sqrt{3}$. Composing gives the 4th automorphism. $\endgroup$ – user21820 Dec 22 '14 at 14:33
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Here is a shortcut. You already know that the extension is Galois, and that it is of degree 4. Thus there must be four automorphisms - but you have presented all of the four possibilities. Therefore...

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For addition, it's obvious. As for multiplication, think of the four-entry polynomial $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ as the two-entry polynomial $A+B\sqrt{2}$ with $A,B\in\mathbb{Q}(\sqrt{3})$. Thus $\theta_2(A+B\sqrt{2})=A-B\sqrt{2}$ and that should make things easier.

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