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Let $K$ be the splitting field of $x^2 + 2$ over $\mathbb{Q}$.

Prove or disprove that $ i = \sqrt{-1}$ is an element of $K$.

Q. How can I prove that? And, in general, how can I prove an element belongs to a splitting field?

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    $\begingroup$ In fact $K=\mathbb Q(\sqrt{-2})$. So you can try to show that $i$ can/cannot be written as $a+b\sqrt{-2}$ for $a,b\in\mathbb Q$. $\endgroup$ – awllower Jun 19 '17 at 15:49
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If $i\in \mathbb Q(\sqrt{-2})$, there are $a,b\in\mathbb Q$ s.t. $$a+b\sqrt{-2}=i\implies a^2+2\sqrt{-2}ab-2b^2=-1\implies \frac{2b^2-1-a^2}{2ab}=\sqrt{-2}$$$$\implies -2=\left(\frac{2b^2-1-a^2}{2ab}\right)^2,$$ which is a contradiction.

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    $\begingroup$ Of course, we might have $ab=0$ but then either $a=\pm 1$ or $b=\pm \frac 1{\sqrt 2}$ $\endgroup$ – lulu Jun 19 '17 at 16:11
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    $\begingroup$ You must let the OP think a little bit ;-) First, $b=0$ is not possible. If $ab=0$ then $a=0$, and $b\notin\mathbb Q$. Contradiction ! @lulu $\endgroup$ – Surb Jun 19 '17 at 16:17
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    $\begingroup$ Oh, I agree. I'm just pointing out, for the OP really, that you need to eliminate the case $ab=0$. $\endgroup$ – lulu Jun 19 '17 at 16:18
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Since $x^2+2$ has degree $2$, its splitting field $K$ is obtained by just adding a root of it, so $K=\mathbb{Q}(\sqrt{-2})=\mathbb{Q}(i \sqrt{2})$.

If $i$ were in $K$ the same would be true for $\sqrt{2}$, so $\mathbb{Q}(\sqrt{2}, \, i) \subseteq K$ which is absurd since the first extension has degree $4$ over $\mathbb{Q}$. In fact, we have a tower of extensions $$\mathbb{Q} \subset \mathbb{Q}(i) \subset \mathbb{Q}(i, \, \sqrt{2}),$$ the second strict inclusion coming from the fact that $$\sqrt{2} \notin \mathbb{Q}(i)= \{z \in \mathbb{C}\, | \, \mathrm{Re}(z), \, \mathrm{Im}(z) \in \mathbb{Q} \}.$$

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    $\begingroup$ That is circular reasoning. $\endgroup$ – lulu Jun 19 '17 at 15:55
  • $\begingroup$ Where is the circularity? $\endgroup$ – Francesco Polizzi Jun 19 '17 at 15:57
  • $\begingroup$ How do you know that the degree of $\mathbb Q (\sqrt 2,i)$ has degree $4$? Knowing that is entirely equivalent to the statement the OP wants. $\endgroup$ – lulu Jun 19 '17 at 15:59
  • $\begingroup$ I do not agree. Clearly $\sqrt{2}$ is not contained in $\mathbb{Q}(i)$, just because $\sqrt{2}$ is not rational. In other words, $x^2-2$ is trivially irreducible on $\mathbb{Q}(i)$. $\endgroup$ – Francesco Polizzi Jun 19 '17 at 16:01
  • $\begingroup$ You could, I suppose, argue that $\mathbb Q(i)$ can't be a subfield of $\mathbb R$. But that wouldn't help if the OP asked "show that $\sqrt 3 \notin \mathbb Q(\sqrt 2)$" say. At some point you need to argue that the desired element can't be written as an algebraic expression in the generator of your splitting field. $\endgroup$ – lulu Jun 19 '17 at 16:04

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