# Non-Noetherian domain but finitely generated ideals principal

Posted by Jason Polak on 02. January 2018 · Write a comment · Categories: commutative-algebra · Tags: ,

A finitely-generated module over a principal ideal domain is always isomorphic to $R^n\oplus R/a_1\oplus\cdots\oplus R/a_n$ where $n$ is a nonnegative integer and $a_i\in R$ for $i=1,\dots,n$. This is called the structure theorem for modules over a principal ideal domain. Examples of principal ideal domains include fields, $\Z$, $\Z[\sqrt{2}]$, and the polynomial ring $k[x]$ when $k$ is a field.

If $a\in R$ is not a unit, then $R/a$ is not projective, since $a$ annihilates any element of $R/a$ and therefore $R/a$ cannot be the direct summand of any free module. Therefore, we can conclude from the structure theorem that any finitely-generated projective module over a principal ideal domain is a free module. Don’t get your hopes up though: there are many examples of non-free projective modules.

But let’s stick with principal ideal domains. It is actually true that every projective module over a principal ideal domain is free. Kaplansky in [1] proved the following even stronger theorem:

Theorem. If $R$ is an integral domain in which every finitely generated ideal is principal, then every projective $R$-module is free.

It seems rather strange that an integral domain might have infinitely generated ideals, but yet have all of its finitely generated ideals principal. So what is an example of such a ring? Kaplansky doesn’t give one.

One answer lies in…valuation rings! This is rather amusing because when I started writing this post I wasn’t even thinking of valuation rings.

Recall that a ring $R$ is a valuation ring if it is a domain and for every $a,b\in R$, either $a\mid b$ or $b\mid a$. In other words, a valuation ring is a domain for which the principal ideals are linearly ordered. I talked a little bit about totally ordered sets of ideals in the last post. The ring of integers $\Z$ is not a valuation ring, but the ring $k[[x]]$ of formal power series over a field is a valuation ring! However, both of these are principal ideal domains and hence not the oddities that we desire. It follows directly from the definition of valuation ring that every finitely-generated ideal in a valuation ring is principal.

What we need is a valuation ring that is not Noetherian. Will such valuation rings please stand up?

Here’s an example. Take the monoid of nonnegative rational numbers, ${\Q}^+ = \{ a\in \Q : a\geq 0\}$ whose binary operation is addition, and take the monoid ring $\Q[\Q^+]$. It is by definition the vector space whose basis consists of the symbols $[a]$ for every $a\in \Q^+$. Multiplication is defined on symbols by $[a][b] = [a + b]$ and this defines a multiplication on all of $\Q[\Q^+]$ by extending symbol multiplication linearly.

Then $\Q[\Q^+]$ is a integral domain with vector space addition and the multiplication just defined, and it is in fact a valuation ring and so every finitely generated ideal is in fact principal. However:

The commutative ring $\Q[\Q^+]$ is not Noetherian.
Proof. Consider the ideal $I = ( [1/n] : n =1,2,3,\dots)$. That is, $I$ is the ideal generated by the set of symbols of reciprocals of the positive natural numbers. We claim that $I$ is not finitely-generated. If it were finitely-generated, it would be principal and generated by a single element of the form $[a]$ where $a$ is a nonnegative rational number. So let us assume that $I = ([a])$ and attempt a contradiction.

Now, this assumption of $I = ([a])$ means that for each $n=1,2,\dots$ there exists an element $x\in \Q[\Q^+]$ such that $x[a] = [1/n]$. It is easy to see that $x$ itself must be of the form $[b]$ for some $b\in \Q[\Q^+]$. So in other words, for each $n$ there exists a nonnegative rational $b$ such that $a + b = 1/n$. Since $n$ is arbitrary, this implies that $a = 0$, which is impossible as $[0]\not\in I$.

Going back to Kaplansky’s theorem, we see that every projective module over the monoid ring $\Q[\Q^+]$ is free! Now how’s that for a cool example? Can you think of another way to show that every projective over this ring is free?

[1] Kaplansky, “Projective Modules”. The Annals of Mathematics, Second Series. 68 (2), September 1958, pp.372-377.