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I have a homework question from Artin's Algebra that asks

Is $i \in \mathbb{Q}[\sqrt[4]{-2}]$?

I suspect that this is not true because $i \sqrt{2} \in \mathbb{Q}[\sqrt[4]{-2}]$ and $\sqrt{2}$ is of course not rational, but I am having a hard time proving it. Perhaps I could consider $\mathbb{Q}[\sqrt[4]{-2}] = \mathbb{Q}[x] / (x^4 + 2)$. Now if $i \in \mathbb{Q}[\sqrt[4]{-2}]$, then $\mathbb{Q}[i] = \mathbb{Q}[x] / (x^2 + 1) \leq \mathbb{Q}[x] / (x^4 + 2)$ and $x^2 + 1 \mid x^4 + 2$, a contradiction? I have a feeling this is not right, but I'm stuck. Also, we haven't covered any Galois theory. Any thoughts would be appreciated!

user131576
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  • Your feeling is right, you can not show this in that way. As an example $2\sqrt 2\in Q(\sqrt 2)$ but $x^2-8$ does not divides $x^2-2$. – mesel Apr 12 '14 at 15:51

2 Answers2

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Notice that $\sqrt[4]{-2}$ is a root of $x^4+2$ and by Eisenstein's creteria it is irreducable over $\mathbb Q$.

So $[\mathbb Q(\sqrt[4]{-2}):\mathbb Q]=4$ and

$$x_1=\sqrt[4]{2}\ \operatorname{cis}(\pi/4)$$ $$x_2=\sqrt[4]{2}\ \operatorname{cis}(3\pi/4)$$ $$x_3=\sqrt[4]{2}\ \operatorname{cis}(5\pi/4)$$ $$x_4=\sqrt[4]{2}\ \operatorname{cis}(7\pi/4)$$ are all roots. Notice that $x_{k+1}/x_k=i$ so if $i\in \mathbb Q(\sqrt[4]{-2})$ then it must includes all other roots as well and under this assuption you can say that $\mathbb{Q}(\sqrt[4]{-2})=\mathbb{Q}(\sqrt[4]{2},i)$.

But clearly $[\mathbb{Q}(\sqrt[4]{2},i):\mathbb{Q}]=8$ which is a contradiction.

Steven Stadnicki
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mesel
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  • Nice answer, but could you clarify how $\sqrt[4]{-2} \in \mathbb{Q}(\sqrt[4]{2},i)$? It is not clear to me. Thanks a bunch! Not that it is really necessary by the way, your reasoning shows $\mathbb{Q}(\sqrt[4]{2},i) \subseteq \mathbb{Q}(\sqrt[4]{-2})$ which is enough to reach your conclusion. – Joachim Apr 12 '14 at 16:38
  • @Joachim : $(\sqrt[4] {2})^2=\sqrt 2 \in Q(\sqrt[4] {2},i)$ so $cis(\pi/4)\in Q(\sqrt[4] {2},i)$ as a result $\sqrt[4] {2} cis(\pi/4)\in Q(\sqrt[4] {2},i)$ – mesel Apr 12 '14 at 16:46
  • @Joachim: you are welcome, yes $Q(\sqrt[4] {2},i)\subseteq Q(\sqrt[4] {-2}) $ is enough. – mesel Apr 12 '14 at 17:06
  • Ah, writing out $\text{cis}(\pi/4)$ explicitly escaped me. Thanks for clearing that up! – Joachim Apr 12 '14 at 17:26
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Hint

Let $\omega = \sqrt[4]{-2}$, then $\mathbb{Q}(\omega)$ is degree 4 over $\mathbb{Q}$, and as a vector space it has basis $\{1,\omega,\omega^2,\omega^3\}$. To show that $i$ is not in $\mathbb{Q}(\omega)$, one only need to show that $\{1,\omega,\omega^2,\omega^3,i\}$ is linearly independent over $\mathbb{Q}$. Write them out explicitly, then it should be easy to see.

mez
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