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I am trying to find $[\mathbb{Q}(\sqrt [3] {3},\sqrt [3] {2}):\mathbb{Q}]$. My guess is it is $9$. There are 3 possibilities-3,6,9. If it is not 9 then $X^{3}-3$ is not irreducible over $\mathbb{Q}(\sqrt [3] {2})$ hence it has a root in $\mathbb{Q}(\sqrt [3] {2})$. But $\mathbb{Q}(\sqrt [3] {2})$ is real field so the only possibility is that it contains $\sqrt [3] {3}$. To show this is not the case I assumed that $\sqrt [3] {3}=a+b\sqrt [3] {2}+c\sqrt [3] {4}$ and cubed both side. I have not yet completed it , but seems this methode is very long. Is there a better approach?

I WANT A SOLUTION WHICH DOES NOT USE ANY GALOIS THEORY/ALGEBRAIC NUMBER THEORY.

Bingo
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  • Hint: Find a primitive element which generates $\mathbb{Q}(\sqrt [3] {3},\sqrt [3] {2})$. – Celsius Sep 20 '14 at 18:28
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    This question is shown in the list of related questions on the right: http://math.stackexchange.com/questions/367013/prove-sqrt33-notin-mathbbq-sqrt32 – Martin Sleziak Sep 20 '14 at 18:41
  • I want some elementary solution (using basic field theory) , not using any Galois theory/algebraic number theory. – Bingo Sep 21 '14 at 04:17
  • Note that your edit automatically put the post in the reopen review queue. You can also ask for reopening here. (If you post your request in that thread, you should explain the reason why the post should be reopened.) – Martin Sleziak Sep 21 '14 at 05:19
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    BTW $[\mathbb{Q}(\sqrt [3] {3},\sqrt [3] {2}):\mathbb{Q}]=[\mathbb{Q}(\sqrt [3] {3},\sqrt [3] {2}):\mathbb{Q}(\sqrt[3]2)] \cdot [\mathbb{Q}(\sqrt [3] {2}):\mathbb{Q}]$ implies that the possibilities are 3 and 9. – Martin Sleziak Sep 21 '14 at 05:20

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