4
$\begingroup$

Possible Duplicate:
Easy way to show that $\mathbb{Z}[\sqrt[3]{2}]$ is the ring of integers of $\mathbb{Q}[\sqrt[3]{2}]$

Is the integral domain $\mathbb{Z}[\sqrt[3]{2}]$ equal to its integral closure in its quotient field $\mathbb{Q}(\sqrt[3]{n})$?

I believe it would be enough to show that $\mathbb{Z}[\sqrt[3]{2}]$ is a UFD, but I'm not sure if it's true. I approached this by attempting to show it is a Euclidean domain, but with a cube root, I couldn't think of a good norm.

Otherwise, I would take $\alpha/\beta$ in the quotient field, and multiply the equation $ (\alpha/\beta)^m+a_1(\alpha/\beta)^{m-1}+\cdots+a_m=0\qquad a_i\in\mathbb{Z}[\sqrt[3]{2}] $ through by $\beta^m$. So if $p$ where any prime dividing $\beta$ and not $\alpha$ I would get a contradiction, so $\alpha/\beta\in\mathbb{Z}[\sqrt[3]{2}]$.

I just don't know if this is the right approach though, since I can't show that $\mathbb{Z}[\sqrt[3]{2}]$ is an Euclidean domain, PID, or UFD. Is there an obvious integral element in $\mathbb{Q}(\sqrt[3]{2})\setminus\mathbb{Z}[\sqrt[3]{2}]$ that I'm missing to prove otherwise?\

Edit: Is it possible to conclude this without resorting to a lot of machinery from algebraic number theory and strictly using tools from commutative algebra? I didn't follow the answers in the duplicate question.

  • 0
    @Arturo I flagged for moderators' help for the same reason you edited the post. I couldn't edit it for some reason! Thanks anyway!2012-01-20

0 Answers 0