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Determine the Galois group of the splitting field of the minimal polynomial of the following algebraic numbers $\sqrt{2}+\sqrt{3}+\sqrt{6}$ over $\Bbb{Q}$.

It is clear that $\mathbf{Q}(\sqrt{2},\sqrt{3})$ is a splitting field of the minimal polynomial with degree 4 as $\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \mid a,b,c,d\in\mathbf{Q}\}$, and it is a normal extension(as already splitting field). Hence the degree of the extension field equals the degree of its Galois group. But here after I couldn't determine its Galois group. Could somebody please give me a hint?

Thanks

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  • $\begingroup$ @ Dietrich Burde... how can we show whether its Galois is $(\mathbb{Z}/2\mathbb{Z})^3$? Thanks! $\endgroup$ – UserAb Mar 31 '16 at 20:32
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    $\begingroup$ @ Dietrich Burde: isn't $\mathbb{Q}(\sqrt{2}+\sqrt{3}+\sqrt{6})=\mathbb{Q}(\sqrt{2},\sqrt{3})$ of degree 4 over $\mathbb{Q}$? $\endgroup$ – carmichael561 Mar 31 '16 at 20:32
  • $\begingroup$ Yes, of course (I misread $\sqrt{5}$). $\endgroup$ – Dietrich Burde Mar 31 '16 at 20:33
  • $\begingroup$ $\mathbf Q(\sqrt2,\sqrt3,\sqrt6)=\mathbf Q(\sqrt2,\sqrt3$ has degree $4$ over $\mathbf Q$, hence its Galois group has order $4$. $\endgroup$ – Bernard Mar 31 '16 at 20:34
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With a little work, you can show that $\mathbb{Q}(\sqrt{2}+\sqrt{3}+\sqrt{6})=\mathbb{Q}(\sqrt{2},\sqrt{3})$, which has degree 4 over $\mathbb{Q}$. Therefore its Galois group is either $C_4$ or $C_2\times C_2$.

And $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has two quadratic subfields, hence the Galois group is $C_2\times C_2$.

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  • $\begingroup$ (In fact, it has three quadratic subfields: $\mathbb Q(\sqrt 2)$, $\mathbb Q(\sqrt 3)$, and $\mathbb Q(\sqrt 6)$.) $\endgroup$ – Ravi Fernando Mar 31 '16 at 22:25
  • $\begingroup$ Yes, good point. Of course anything more than one rules out $C_4$. $\endgroup$ – carmichael561 Mar 31 '16 at 22:49
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There are better ways of showing that $\sqrt2+\sqrt3$ is of degree four over $\Bbb Q$, but its minimal polynomial is $f(X)=X^4-10X^2+1$, for which $F(X+1)=X^4+4X^3-4X^2-16X-8$, clearly irreducible by a suitable variant of Eisenstein (its Newton polygon has one segment of slope $-3/4$).

As a result, we have $\Bbb Q(\sqrt2+\sqrt3\,)\subset\Bbb Q(\sqrt2,\sqrt3\,)$, both of them of degree four over $\Bbb Q$, and therefore equal. But $\sqrt6\in\Bbb Q(\sqrt2,\sqrt3\,)$, so that $\Bbb Q(\sqrt2+\sqrt3+\sqrt6\,)\subset\Bbb Q(\sqrt2,\sqrt3,\sqrt6)=\Bbb Q(\sqrt2,\sqrt3\,)$.

Thus the degree of your field is four, not eight. (I see @carmichael561 has beat me to it with a more efficient answer.)

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  • $\begingroup$ I see... thanks. I edited this. $\endgroup$ – UserAb Mar 31 '16 at 22:14

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