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I have to show that the degree $[\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3}):\mathbb{Q}(\sqrt[3]{2})]$ is $\gt1$.

I know that for this purpose it is enought to show that $\sqrt[3]{3} \notin \mathbb{Q}(\sqrt[3]{2})$, but how can I show this?

Zev Chonoles
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Legolas
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    Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level – Elaqqad Aug 05 '15 at 15:43
  • Also, a dupe of this or this. – Jyrki Lahtonen Jul 31 '18 at 05:26

2 Answers2

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You can use traces.

Suppose that $K=\mathbb{Q}(\sqrt[3]2)=\mathbb{Q}(\sqrt[3]3)$, and write $$ \sqrt[3]3= a+b\sqrt[3]2+c\sqrt[3]4, $$ with $a,b,c\in\mathbb{Q}$.

Using the idea in a note by K. Conrad, we have $$0 = Tr_Q^K \sqrt[3]3 = Tr_Q^K (a+b\sqrt[3]2+c\sqrt[3]4)= 3a,$$ $$0 = Tr_Q^K \sqrt[3]6 = Tr_Q^K (a\sqrt[3]2+b\sqrt[3]4+c\sqrt[3]8)= 6c, \textrm{ (this is a typo on the note saying this is $6b$.)}$$ and $$0 = Tr_Q^K \sqrt[3]{12} = Tr_Q^K (a\sqrt[3]4+b\sqrt[3]8+c\sqrt[3]{16})= 6b.$$

Then we have $a=b=c=0$, so that $\sqrt[3]3 = 0$. This is a contradiction.

Sungjin Kim
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  • This method works for i.e. $K=\mathbb{Q}(\sqrt[p]2)=\mathbb{Q}(\sqrt[q]3)$? In other words if we had any prime numbers $p$ and $q$ in the roots and not the $3$ – Legolas Aug 06 '15 at 09:06
  • For distinct prime $p, q$, it is obvious because they have degree $p, q$ respectively. For $p=q$, the same method works. – Sungjin Kim Aug 06 '15 at 18:41
  • i.e. if we had this$[\mathbb{Q}(\sqrt[2]{2},\sqrt[3]{3}):\mathbb{Q}(\sqrt[2]{2})]$ how i can show it? – Legolas Aug 06 '15 at 18:46
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    $[\mathbb{Q}(\sqrt[2]{2},\sqrt[3]{3}):\mathbb{Q}(\sqrt[2]{2})] \leq [\mathbb{Q}(\sqrt[3]3):\mathbb{Q}] = 3$, and $[\mathbb{Q}(\sqrt[2]{2},\sqrt[3]{3}):\mathbb{Q}]$ must have a factor $[\mathbb{Q}(\sqrt[3]3):\mathbb{Q}]=3$. But, $[\mathbb{Q}(\sqrt[2]2):\mathbb{Q}]=2$. So, we must have $[\mathbb{Q}(\sqrt[2]{2},\sqrt[3]{3}):\mathbb{Q}(\sqrt[2]{2})] =3$. – Sungjin Kim Aug 06 '15 at 18:58
  • with the way we said can i take that $$\sqrt[3]3=a+b\sqrt(2)$$ for some $a,b \in Q$ and show this with o contradiction? – Legolas Aug 06 '15 at 19:06
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    For different $p, q$, the way I showed using degree is much easier. In fact, the way with traces, the problem is how you get $\sqrt[3]3\sqrt 2$ has trace $0$. That is not easy if $p\neq q$. – Sungjin Kim Aug 06 '15 at 19:31
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Assume that $$\sqrt[3]3=a+b\sqrt[3]2+c\sqrt[3]4$$ for some $a,b,c\in\Bbb Q$ and try to get a contradiction.

ajotatxe
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  • First, i will assume that$\sqrt[3]{3} \in \mathbb{Q}(\sqrt[3]{2})$ then we have $$\sqrt[3]3=a+b\sqrt[3]2+c\sqrt[3]4$$ for some $a,b,c\in\Bbb Q$.But how i ill conclude that these $a,b,c \notin Q$? – Legolas Aug 05 '15 at 17:08
  • from this equation of $a,b,c$? – Legolas Aug 05 '15 at 17:09