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.


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 $2^$. 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 $1+\alpha$.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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