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Consider the polynomial $P(X) = X^5 - X + 1 \in \mathbb{Q}[X]$, and let $x \in \mathbb{C}$ be a root of $P(X)$. Let $K = \mathbb{Q}(x)$. How can you prove that the ring of integers $\mathcal{O}_K$ is a principal ideal domain?

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Compute its class group. –  Qiaochu Yuan Jul 17 '13 at 19:21
    
I tried but I found some problems: it's easy to compute the Minkowski bound, but I don't know any factorization result for the ideals of the form $p\mathcal{O}_k$ (I know there is such a result for quadratic fields). Is there a not-too-difficult way to do this computation without using a software? –  Mattia Coluccia Jul 17 '13 at 19:30
    
@MattiaColuccia: factorization results for ideals $p{\mathcal O}_k$ are nearly the same for any number field $k$ as they are for quadratic fields. A good term to look up in this direction is "Kummer-Dedekind theorem". One thing that changes for non-quadratic $k$ is that the ring of integers of $k$ might not have the form ${\mathbf Z}[\alpha]$ for some $\alpha$ (examples already occur for cubic $k$). –  KCd Jul 18 '13 at 6:27

1 Answer 1

First, $P(X)$ is irreducible with discriminant $2869=19\cdot 151$, which is squarefree. Hence the ring of integers of $K$ is $\mathbb{Z}[x]$. Using the Minkowski bound, we see that every class of ideals for K contains an integral ideal $\frak{a}$ with norm $$ N(\frak{a})<0.062\cdot \sqrt{2869}<4. $$ Now we have that there are no prime ideals $\frak{p}$ with norm $2$ or $3$ (because then the residue field would be $\mathbb{F}_2$ or $\mathbb{F}_3$, so that $P(X)$ would have a root modulo $2$ (or $3$), which is not the case). From this we see that the ring of integers is a PID, i.e., the class number of $K$ is $1$. The Galois group of the spiltting field of $P(X)$ is the full symmetric group $S_5$.

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Thank you very much! –  Mattia Coluccia Jul 17 '13 at 21:25
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@MattiaColuccia: if you're satisfied with the answer, accept it. –  tomasz Jul 17 '13 at 22:25

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