4
$\begingroup$

I am stuck with problem 22, chapter 3 in Marcus' book Number Fields which says:

Suppose $\alpha^5=2\alpha+2$. Prove that the ring of integers of $\mathbb Q[\alpha]$ is $\mathbb Z[\alpha]$. Prove the same thing also if $\alpha^5+2\alpha^4=2$.

Try: Discriminant for both of them is not square free.

$\endgroup$
4
$\begingroup$

Let's look at the first part. Let $R$ be the ring of integers of $\mathbb Q[\alpha]$. Let $\mathfrak p$ be a maximal ideal of $R$ containing $2$. Then $\alpha^5=2(1+\alpha)\in \mathfrak p$, hence $\alpha\in \mathfrak p$. Moreover $1+\alpha\notin \mathfrak p$, so $2\in \mathfrak p^5R_\mathfrak p$. This implies that the ramification index at $\mathfrak p$ is at least $5$. But $\mathbb Q[\alpha]$ has degree $5$, so the ramification index is exactly $5$, the residue extension is trivial and there is only one prime above $2$. By Problem 21 just before the one you are considering, $2^4$ divides the discriminant of $R$.

Now the discriminant of $X^5-2X-2$ is $2^4.3.13.67$. So it is equal to the discriminant of $R$ and therefore $\mathbb Z[\alpha]=R$.

The seconde part should be solved similarly.

Edit To see the ramification index at $\mathfrak p$ is at least $5$ without using localization (I don't know how things are organized in Marcus), we can just compare ideal decompositions of $\alpha^5 R$ and $2(1+\alpha)R$, noticing that no prime ideal can contains at the same time $\alpha$ and $1+\alpha$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But what is $R_\mathfrak{p}$ and how to prove sum no $21$? $\endgroup$ – user12345 Feb 21 '13 at 19:36
  • $\begingroup$ @QiL'8 +1 For your answer. Also, can you expand a bit on how the localization thing came in? Thanks. $\endgroup$ – user38268 Feb 21 '13 at 21:36
  • $\begingroup$ $R_{\mathfrak p}$ is the localization of $R$ at $\mathfrak p$. The resulting ring is a discrete valuation ring. I don't understand what do you mean by sum no 21. $\endgroup$ – user18119 Feb 21 '13 at 21:37
  • 1
    $\begingroup$ @BenjaLim: as $1+\alpha\notin \mathfrak p$, it is invertible in $R_{\mathfrak p}$, so $2=\alpha^5 (1+\alpha)^{-1}\in \mathfrak p^5R_\mathfrak p$ in $R_\mathfrak p$. $\endgroup$ – user18119 Feb 21 '13 at 21:44
  • $\begingroup$ @QiL'8 I understand that fact, however what I don't understand is why does this imply that the ramification index is at least $5$. Thanks, $\endgroup$ – user38268 Feb 21 '13 at 22:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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