3
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As part of a larger problem I need to show that

$i$ is not contained in the field extension $\mathbb Q(\sqrt[3]{2},\zeta)$, where $\zeta$ is the third root of unity.

I understand that the third root of unity is equal to $${-1}/{2} + i\sqrt{3}/2.$$ I'm unsure how to procure a contradiction now though. Do I have to consider the degree? How would I do that?

Thanks

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  • $\begingroup$ What is $\sqrt{(3)}$ ? $\endgroup$ – Dietrich Burde Oct 9 '13 at 17:27
  • $\begingroup$ He wrote \sqrt(3) $\endgroup$ – Martin Argerami Oct 9 '13 at 17:28
  • $\begingroup$ I'm very sorry. I wrote it incorrectly. Please see my edits in a second. It's not sqrt(3), it's the third root of 2. $\endgroup$ – Flabbergasted in Fields Oct 9 '13 at 17:31
  • $\begingroup$ Also, I've considered attempting to derive a contradiction by stating that i is not in Q(third root of 2) or Q(zeta), but that doesn't tell me enough, I suspect. $\endgroup$ – Flabbergasted in Fields Oct 9 '13 at 17:34
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Both $K=\mathbb{Q}(\sqrt[3]{2},\zeta)$ and $L=\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$ have degree $6$ over $\mathbb{Q}$. But the latter is obviously real, so $K\neq L$, and in particular $\sqrt{3}\notin K$.

Now, if $i\in K$, can we show that $\sqrt{3}\in K$ for a contradiction?

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  • $\begingroup$ Thank you, that's an interesting method. I have a question, though. To show sqrt(3) is not in K, is it sufficient to note that because $a+b\sqrt[3]{2}+c\sqrt{3}i/2 = \sqrt{3}$ implies c is complex, and c cannot be complex as the base field is real, that $\sqrt{3}$ cannot be in the field? $\endgroup$ – Flabbergasted in Fields Oct 9 '13 at 18:14
  • $\begingroup$ You could do it this way, though I think you need more coefficients on the left. But the point is that if $\sqrt{3} \in K$, then $L\subset K$, and since the two fields have the same degree over $\mathbb{Q}$, this means $L=K$. But $L\subset\mathbb{R}$, while $\zeta\notin\mathbb{R}$. $\endgroup$ – Slade Oct 9 '13 at 18:18
3
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Also, I've considered attempting to derive a contradiction by stating that $i$ is not in $\mathbb{Q}(\sqrt[3]{2})$ or $\mathbb{Q}(\zeta)$, but that doesn't tell me enough, I suspect.

It does. Since $i^2 + 1 = 0$, we know that $[\mathbb{Q}(i):\mathbb{Q}] = 2$. Supposing we already know that $i \notin \mathbb{Q}(\zeta)$, we see that $[\mathbb{Q}(i,\zeta):\mathbb{Q}] = 4$, so every finite extension containing $i$ and $\zeta$ must have degree divisible by $4$.

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  • $\begingroup$ I see. So because Q(\sqrt[3]{2}) has degree 3, Q(\sqrt[3]{2},\zeta) has degree 6, and 4 doesn't divide 6, which is a contradiction? $\endgroup$ – Flabbergasted in Fields Oct 9 '13 at 17:50
  • $\begingroup$ Right. That's the contradiction if you assume you had $i \in \mathbb{Q}(\sqrt[3]{2},\zeta)$. $\endgroup$ – Daniel Fischer Oct 9 '13 at 17:50
  • $\begingroup$ Thank you for your help. I have one further question, because this is actually a concept I've been rocky on (the tower law and related issues). I see that Q(\sqrt[3]{2}) has degree 3, clearly; but what allows me to say that I can simply multiply the degree of Q(\sqrt[3]{2}) and the degree of Q(\zeta) to get the degree of the extension? What concept allows me to prove that Q(\sqrt[3]{2},zeta) over Q(\sqrt[3]{2}) equals 2? $\endgroup$ – Flabbergasted in Fields Oct 9 '13 at 17:53
  • $\begingroup$ The minimal polynomial of $\zeta$ over $\mathbb{Q}$ is $X^2 + X + 1$. So the minimal polynomial of $\zeta$ over $\mathbb{Q}(\sqrt[3]{2})$ has degree either $2$ or $1$. It doesn't have degree $1$ - most simply seen because $\sqrt[3]{2} \in \mathbb{R}$ - so $\zeta$ has the same minimal polynomial over both fields, and $[\mathbb{Q}(\sqrt[3]{2},\zeta) : \mathbb{Q}(\sqrt[3]{2})] = 2$. $\endgroup$ – Daniel Fischer Oct 9 '13 at 17:57
  • $\begingroup$ Thank you for your help, Daniel. I accepted the other answer because it was a bit simpler and given first but I really appreciate the assistance. $\endgroup$ – Flabbergasted in Fields Oct 9 '13 at 20:37

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