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I am asked to find the degree and basis for a given field extension $\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{6},\sqrt[3]{24}) $

Now I know that the degree for each vector is $3$, and that the basis will have $9$ vectors. I found the answer in the back of the book as $\{1,\sqrt[3]{ 2},\sqrt[3]{ 4},\sqrt[3]{ 3},\sqrt[3]{ 6},\sqrt[3]{ 12},\sqrt[3]{9},\sqrt[3]{ 18},\sqrt[3]{ 36}\}$ but I would like to know how you find them. Thanks!

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How do you know the basis will have 9 vectors? –  Chris Eagle Feb 22 '12 at 19:02
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Chris is right: this is not trivial at all! Equivalently, it is a real problem to show that $[\mathbb Q( \sqrt[3]{2},\sqrt[3]{3}):\mathbb Q]=9$ –  Georges Elencwajg Feb 22 '12 at 20:27

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Let $\mathbb{F}=\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{6}, \sqrt[3]{24})$

$\sqrt[3]{24}=\sqrt[3]{2^33}=2\sqrt[3]{3}$. So $\sqrt[3]{3} \in \mathbb{F}$. But $\sqrt[3]{2}\sqrt[3]{3}=\sqrt[3]{6}$ so it's redundant. Therefore, $\mathbb{F} = \mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$.

Then $\sqrt[3]{2}$ raised to $1,2,3$ powers yields $\sqrt[3]{2},\sqrt[3]{4},2$. $\sqrt[3]{3}$ raised to $1,2,3$ powers yields $\sqrt[3]{3},\sqrt[3]{9},3$.

So far we've found $1,\sqrt[3]{2},\sqrt[3]{4},\sqrt[3]{3},\sqrt[3]{9}$ (I've replaced $2$ and $3$ with the equivalent basis member $1$).

Now you just need to worry about products among these elements. These products will complete your list.

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This is, of course, after you have proved that the nine numbers you get are linearly independent over the rationals - which takes a bit of work. –  Gerry Myerson May 15 '12 at 7:42

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