29
$\begingroup$

Prove that in a Dedekind domain every ideal is either principal or generated by two elements.

Help me some hints.

Thanks a lot!

$\endgroup$

2 Answers 2

25
$\begingroup$

Hints:

Let $R$ be a Dedekind domain, and let $I\subseteq R$ be any ideal. Factor $I$ as $\mathfrak{p}_1^{e_1}\cdots\mathfrak{p}_n^{e_n}$. Take ANY $\alpha\in I-\{0\}$ and factor $(\alpha)$ as $\mathfrak{p}_1^{f_1}\cdots\mathfrak{p}_n^{f_n}\mathfrak{q}_1^{g_1}\cdots\mathfrak{q}_m^{g_m}$. We know that $f_i\geqslant e_i$ since $(\alpha)\subseteq I$ so that $I\mid (\alpha)$. Now explain why there exists some $\beta\in R$ such that $v_{\mathfrak{p}_i}(\beta)=e_i$ and $v_{\mathfrak{q}_i}(\beta)=0$. Explain then why

$$(\alpha,\beta)=\text{gcd}((\alpha),(\beta))=\mathfrak{p}_1^{e_1}\cdots\mathfrak{p}_n^{e_n}=I$$

This, in fact, shows that for any $I$ and any $\alpha\in I-\{0\}$ you can always find a complementary generator.

$\endgroup$
3
  • $\begingroup$ Corollary 7 said that every ideal of a Dedekind domain can be generated by two elements. But if an ideal isn't generated by two elements, how can show that it is principal? math.stanford.edu/~conrad/210BPage/handouts/… $\endgroup$
    – Truong
    Commented Dec 8, 2013 at 1:05
  • 3
    $\begingroup$ @chuyenvien94 I showed that it requires at most two elements. So, either it's principal or it's two-generated. $\endgroup$ Commented Dec 8, 2013 at 1:06
  • $\begingroup$ Since the link I provide is no longer accessible, I update the link I mentions here: math.stanford.edu/~conrad/210BPage/handouts/dedekind.pdf $\endgroup$
    – Truong
    Commented Oct 7, 2017 at 8:46
9
$\begingroup$

It is enough to show that R/(x) is a PIR for each nonzero nonunit x. If you can use the structure theorem for rings in which every ideal is a product of prime ideals (called general ZPI-rings), this is immediate, since any zero-dimensional such ring is a PIR. (A general ZPI-ring is a finite direct product of Dedekind domains and PIR's with one prime.)

Alternatively, it is not hard to directly prove that R/(x) is a PIR. (This is a much easier special case of the theorem). Being a zero-dimensional semi-quasi-local general ZPI-ring, it is a finite direct product of (necessarily quasi-local) connected such rings. These factors are PIRs since each principal ideal is a power of the maximal ideal, so R/(x) is a PIR.

$\endgroup$
1
  • 22
    $\begingroup$ Although this is an extremely good question, it seems to me unlikely that anyone asking it will simultaneously have the algebraic background to follow, for example, what a "zero-dimensional semi-quasi-local general ZPI-ring" is :) ..... $\endgroup$
    – GaryMak
    Commented Jul 4, 2016 at 9:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .