# Fun ring theory problem

Right now I’m teaching a class on unique factorization and its failure. One of my hobbies is finding fun problems like “Find a domain where every finitely generated ideal is principle, but not every ideal is finitely generated.” Today one of my students found an answer to the following fun question (which was cute enough that I thought I’d pass it on to all of you).

A weaker statement than unique factorization is just that the number of irreducible factors is independent of factorization. For example, in $\mathbb{Z}[\sqrt{-5}]$, even though 6 factors nonuniquely as $2 \cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})$ both of these factorizations have 2 irreducible factors. How badly can this fail? It turns out the answer is “really badly.” Find a domain R where every nonzero element factors as a unit times a product irreducibles, but which has a fixed element that can be written as a product of more than N irreducibles for any arbitrarily large N.

## 5 thoughts on “Fun ring theory problem”

1. Oh good, I like problems like this.

First of all, it reduces to a question about monoids (or semigroups). If we can find a cancellative monoid S in which:
– every element factors as a product of irreducibles, i.e. the “divides” relation is well-founded when restricted to non-unit elements;
– there is a fixed element that can be written as a product of more than N irreducibles for any arbitrarily large N;
– x^n = y^n implies x=y, for any positive integer n,

then the semigroup ring R := Z[X^{S}] is an integral domain [1], and also it satisfies your criterion, since the “divides” relation of R inherits its well-foundedness from that of S, and every irreducible of S remains irreducible when embedded into R.

To find such a thing, we can just take the monoid with generators x_p, where p ranges over the prime natural numbers, and relations (x_p)^p = (x_q)^q for all prime p and q. Then each generator is irreducible, and the element (x_p)^p is the “fixed element” we seek.

Does that work?

Robin

[1] A Note on Semigroup Rings, Robert Gilmer, The American Mathematical Monthly, Vol. 76, No. 1. (Jan., 1969), pp. 36-37.

2. James says:

For your class on the failure of unique factorization, are you using any publicly available texts or notes? If so, can you say which ones? I’m about to teach something similar, but I haven’t been able to find any notes that I like, and I’m afraid I’m going to have to write them myself, which would be a pain.

3. Allen Knutson says:

Nonmathematical mystery: why did this show up on the RSS feed about 12 hours before it was actuallly on the page?

4. 1. Yeah, that’s basically the example, but there’s an easier way to state that example. It is just k[x^{1/p} for every prime p].

2. No, I wasn’t using any available texts. I’m drawing from experience teaching at the Ross programs, plus many days at tea this year trying to cook up interesting ways to break unique factorization. Today we’re starting out with the fun example Z[2i].

3. Because Ben thought that I’d accidentally posted without completing the post, so he returned it to drafts. I didn’t get his email until 12 hours later, when I cleaned it up a bit so that it was clear that I was just posing the question, not intending to answer it.

5. David Savitt says:

You probably already know that in a number ring, the number of irreducible factors is independent of the factorization if and only if the class number is <= 2?