# Prüfer domains are Arithmetical rings

Suppose $R$ be a Prüfer domain. How should I Prove that it is an arithmetical ring?

• Is it really plausible that this is a homework question? The topic is relatively advanced commutative algebra: e.g. I taught a semester graduate course in the subject and did not make it to this material. In the textbook treatments I know of, this result is part of a rather important theorem and is proved in the text, not left as an exercise. Have any of the closers ever encountered this as a homework question in a course? – Pete L. Clark Aug 5 '13 at 19:52
• @YACP: It could be homework -- anything could be homework -- but do you find it plausible? E.g. have you ever seen this question assigned as homework? Does it seem to have the level of difficulty of a homework question? Is it something which students in this field are generally expected to work out for themselves? "Second, the OP didn't show his/her own thoughts about the question." Of course I agree, and it seems that people are voting to close as "homework" for that reason. That seems appropriate for some questions -- especially, for undergraduate level questions -- but not others. – Pete L. Clark Aug 5 '13 at 20:09
• For instance, if this question were asked on MO, it would probably not have been closed there. But this site is supposed to be more all-purpose and inclusive. Doesn't that seem a little strange? Also, I can't quite put my finger on why, but I suspect that some of the people who voted to close the question don't know how to answer it, and that doesn't sit so well with me. – Pete L. Clark Aug 5 '13 at 20:13
• Finally, yes, the OP has not put a lot of effort into her questions and has not responded well to very reasonable followup questions. I don't think this means a lack of respect, necessarily. Maybe there are severe language issues here. If this unresponsiveness means that you don't want to answer her questions, that is more than reasonable. But is it really a good idea to aver that a question can be closed based on the user but would stay open if someone else asked it? – Pete L. Clark Aug 5 '13 at 20:21
• Thanks everyone for reopening the question. – Pete L. Clark Aug 5 '13 at 20:23

Assuming that by an "arithmetical ring" you mean a ring in which for all ideals $\mathfrak{a}, \mathfrak{b}, \mathfrak{c}$ we have $\mathfrak{a} \cap (\mathfrak{b} + \mathfrak{c}) = (\mathfrak{a} \cap \mathfrak{b}) + (\mathfrak{a} \cap \mathfrak{c})$, then this is one of a standard list of equivalent condition for an integral domain to be a Prüfer domain. For instance, Theorem 6.6 in the text Multiplicative Ideal Theory (henceforth MIT) by Larsen and McCarthy contains ten such equivalent conditions, the first being Prüfer domain (i.e., nonzero finitely generated ideals are invertible) and the last being arithmetical ring.

The sequence of logical implications among these 10 conditions is one of the most complicated I have seen -- it is rather far from being a ten-cycle! In fact this result is stated in my commutative algebra notes but I am still working on typing up the proof, which will be the longest in the entire set of notes (at least three pages). They prove Prüfer implies arithmetical in MIT as follows:

Show that (i) Prüfer $\implies$ (iii): nonzero finitely generated ideals are cancellable.
(iii) $\implies$ (iv): For every prime ideal $\mathfrak{p}$ of $R$, the localization $R_{\mathfrak{p}}$ is a valuation ring.
And then they prove (iv) $\implies$ (x): arithmetical ring.

In fact I see that I have typed up some of this already, so if no one else intervenes to give a better answer, perhaps I will take this opportunity to get this transcription done.

Remark: Mr. Bill Dubuque has written several posts here and elsewhere giving much longer lists of equivalent conditions for a domain to be Prüfer. To the best of my recollection his posts do not (understandably!) contain proofs of all of these implications, but they probably contain references, some of which may be freely accessible online (as MIT is not).


Okay, here it goes. I will insert another condition on an integral domain $R$.

(iii$'$): if $\mathfrak{a},\mathfrak{b},\mathfrak{c}$ are ideals of $R$ with $\mathfrak{a}$ nonzero and finitely generated and such that $\mathfrak{a} \mathfrak{b} \subset \mathfrak{a} \mathfrak{c}$, then $\mathfrak{b} \subset \mathfrak{c}$.

(i) $\implies$ (iii): Let $\mathfrak{a}$, $\mathfrak{b}$, $\mathfrak{c}$ be ideals of a Prüfer domain, with $\mathfrak{a}$ finitely generated and nonzero and such that $\mathfrak{a} \mathfrak{c} = \mathfrak{b} \mathfrak{c}$. Then since $\mathfrak{a}$ is invertible, multiply both sides by $\mathfrak{a}^{-1}$ to deduce $\mathfrak{b} = \mathfrak{c}$.

(iii) $\implies$ (iii$'$): If $\aa \bb \subset \aa \c$, then $\aa \c = \aa \bb + \aa \c = \aa(\bb+ \c)$; cancelling $\aa$ gives $\c = \bb + \c$, so $\bb \subset \c$.

(iii$'$) $\implies$ (iv): Let $\pp$ be a prime ideal of $R$. I will make use of the (much easier!) exercise that an integral domain is a valuation ring iff principal ideals are linearly ordered under inclusion. So it suffices to show that for any $\frac{a}{s}, \frac{b}{t} \in R_{\pp}$, we have either $(\frac{a}{s}) \subset (\frac{b}{t})$ or $(\frac{b}{t}) \subset (\frac{a}{s})$. Since $\frac{1}{s}, \frac{1}{t} \in R^{\times}$, it is equivalent to show that $(a) \subset (b)$ or $(b) \subset (a)$: for this we may clearly assume $a,b \neq 0$. Now

$\langle ab \rangle \langle a,b \rangle \subset \langle a^2, b^2 \rangle \langle a,b \rangle$,

so

$ab = xa^2 + yb^2$ for some $x,y \in R$. Then

$\langle y b \rangle \langle a,b \rangle \subset \langle a \rangle \langle a,b \rangle$,

so

$\langle yb \rangle \subset \langle a \rangle$.

Put

$yb = au$ for some $u \in R$. Then

$ab = xa^2 + uab$, or $xa^2 = ab(1-u)$.

Case 1: $u \in \pp$. Then $1-u \notin \pp$, so $b = a\left(\frac{x}{1-u}\right) \in a R_{\pp}$.

Case 2: $u \notin \pp$. Then $a = b\left(\frac{y}{v} \right) \in b R_{\pp}$.

(iv) $\implies$ (x): We always have the inclusion of ideals

$\iota: (\aa \cap \bb) + (\aa \cap \c) \subset \aa \cap (\bb + \c)$. Whether a homomorphism of $R$-modules is an isomorphism (hence the identity, in this case) can be checked "locally", i.e., after pushing forward to $R_{\pp}$ for all maximal ideals $\pp$ of $R$. Thus we reduce to the case of a valuation ring. But a valuation ring is a uniserial ring (or chain ring) -- i.e., given any two ideals, one contains the other -- and this makes the identity almost trivial to check: e.g. if $\aa \subset \bb \subset \c$, then $\aa \cap (\bb + \c) = \aa = (\aa \cap \bb) + (\aa \cap \c)$. The other five cases are no harder.