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Clearly $[\mathbb Q(i): \mathbb Q] = 2$. I wanted to prove that $X^2+2$ is irreducible over $\mathbb Q(i)$ in order to get the degree of the whole extension. My attempt:

Assume $X^2+2$ is reducible. Than there must be an $\alpha \in \mathbb Q(i)$ s.t. $a^2+2 =0 $ Write $\alpha = (p+qi)$ where $p,q \in \mathbb Q$. Then $$ p^2 + 2pqi -q^2 +2 = 0 $$ That means $pq=0$. If $p=0$ then $2=q^2$ so $\sqrt 2 \in \mathbb Q$. Contradiction. If $q=0$ then $p^2 = -2$ so $\sqrt{-2} \in \mathbb Q$. Contradiction. So $\alpha$ can not exist and hence $X^2+2$ is irr. over $\mathbb Q(i)$.

Is this correct ?

  • Yep, then you conclude with the multiplicativity formula for degrees – AndreasT Jun 18 '13 at 14:14
  • Note that "then there must be an $\alpha$..." only works because this is a polynomial of degree 2 (or 3). In general, just because a polynomial is reducible over some field does not necessarily imply it has a root in that field. You might already know this, but it's probably best to mention this fact and write it into the solution. – Omnivium Jun 18 '13 at 14:16
  • Yes absolutely. I knew this but its good to mention :) –  Jun 18 '13 at 14:21

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$\mathbb{Q}(i,\sqrt{-2})=\mathbb{Q}(i,\sqrt{2})$. Indeed $i$ belongs to both fields and $\sqrt{-2}=i\sqrt{2}\in\mathbb{Q}(i,\sqrt{2})$, while $\sqrt{2}=-i\sqrt{-2}\in\mathbb{Q}(i,\sqrt{-2})$ (I assume $\sqrt{-2}=i\sqrt{2}$, it would be the same with the other determination).

Since clearly $i\notin\mathbb{Q}(\sqrt{2})$, we have: $$ [\mathbb{Q}(\sqrt{2},i):\mathbb{Q}(\sqrt{2})]=2 $$ and we can use the following dimension formula: $$ [\mathbb{Q}(\sqrt{2},i):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},i):\mathbb{Q}(\sqrt{2})] [\mathbb{Q}(\sqrt{2}):\mathbb{Q}] =2\cdot2=4 $$

Alex Johnson
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