Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
How do you know the basis will have 9 vectors? – Chris Eagle Feb 22 '12 at 19:02
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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.