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I am trying to determine $[\Bbb Q(\sqrt2,\sqrt{-3}):\Bbb Q]$ and $[\Bbb Q(\sqrt[3]2,\zeta_3\sqrt[3]{2}):\Bbb Q]$? $(\zeta_3$ is the cubic root of unity).

For the first one, it seems to me that $[\Bbb Q(\sqrt2):\Bbb Q]=2$ and then $[\Bbb Q(\sqrt2,\sqrt{-3}):\Bbb Q(\sqrt2)]=3.$ Therefore $[\Bbb Q(\sqrt2,\sqrt{-3}):\Bbb Q]=6$. But to prove that $[\Bbb Q(\sqrt2,\sqrt{-3}):\Bbb Q(\sqrt2)]=3.$ I need to prove that the polynomial $x^2+3$ is irreducible over $\Bbb Q(\sqrt2).$It is easy to prove that it is irreducible over $\Bbb Q$ using Eisenstein. But how to prove it is irreducible over $\Bbb Q(\sqrt2)?$

For the second one, I really don't know how to do that.

1 Answers1

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For the first one you can show that $i\sqrt{3}=\sqrt{-3}$ is indipendent from $\sqrt{2}$ so the two extension are not intersected and the final degree is $4$ because the minimal polynomial of $\sqrt{-3}$ is $x^2+3$ and the minimal polinomyal of $\sqrt{2}$ is $x^2-2$. These two polynoms are relatively prime, so you have that the degree of the extension is $4$.

For the second one the minimal polynomial of $\sqrt[3]{2}$ and of $\zeta_3\sqrt[3]{2}$ is $x^3-2$, which is irriducible for Eisenstein. So the degree of the extension is 6, because you have to add to the field both $\sqrt[3]{2}$ and $\zeta_3$: $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$ and $[\mathbb{Q}(\zeta_3):\mathbb{Q}]=2.$

Alberto Andrenucci
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    May I please ask for an argument for why the polynomial $x^2+3$ is irreducible over $\Bbb Q(\sqrt2)$? –  Mar 26 '17 at 01:45
  • It is irreducible because his roots are $\pm i \sqrt{3} $ and there is no way to obtain them as a $\sqrt {2} $ linear combination. Remember that you can always see an extension like a bigger vectorial space. – Alberto Andrenucci Mar 26 '17 at 08:57