# Are quotients of polynomial rings almost UFDs?

If $K$ is a field then the polynomial ring $K[x_1,\ldots, x_n]$ is a UFD. On the other hand, quotients of such a polynomial ring usually don't enjoy unique factorization: consider, for instance, $\Bbb R[x,y]$ modulo the ideal $(x^2+y^2-1)$. Then $x^2=(1-y)(1+y)$ and likewise $y^2=(1-x)(1+x)$. (Over the complex numbers we also have $1=(x+iy)(x-iy)$, and, as Georges points out, the quotient ring is in fact a UFD.)

My question is: are these (in some sense) the only examples which can be factored in different ways? Let me explain what I mean: The above quotient ring (over the reals), call it $A$, is Noetherian so every element can be factored into irreducible ones. I'd be interested to see further elements (not a multiple of the above ones) which don't factorize uniquely. Can something interesting be said about those elements?

What happens in more general quotient rings (let's assume they are domains)?

Thanks for any help and pointers (in particular, the ones already received!), as well as for your patience if this is trivial.

• I can't extrapolate what you mean by "are these (in some sense) the only examples." – Qiaochu Yuan Nov 15 '09 at 0:31
• Any commutative ring is a quotient of a polynomial ring. – Fernando Muro Sep 2 '17 at 17:37

No,quotients of polynomial rings are definitely not "almost UFDs".

Any finitely generated ring over K is such a quotient and this means a lot of non UFDs. Said differently, any algebraic variety in affine space over K has as ring of regular functions one of your quotients and in general (as your own example over $\mathbb R$ states) it will not be a UFD.

By the way, your parenthetical remark about the complex case is a bit ambiguous: $\mathbb C[x,y]/(x^2+y^2-1)$ IS a UFD: by the change of variables u=x+iy,v=x-iy this ring becomes $\mathbb C[u,v]/(uv-1)=\mathbb C[u,1/u]$, which is factorial and even a PID ( Reason: If $A$ is a PID, so is every ring of fractions $A_S$ [...unless it is a field, you Bourbakistas])

• Dear @Georges, I do not understand your final parenthetical remark. Do you mean that Bourbaki does not call a (commutative!) field principal? – Fred Rohrer Sep 2 '17 at 20:21
• Dear @Fred, Bourbaki considers a field to be a PID : cf. definition on first page of Algebra, Chapter VII). My (not very clear) parenthetical remark alludes to my belief that fields are generally not considred PID's, But maybe that belief is mistaken? – Georges Elencwajg Sep 3 '17 at 7:24

I'll try to give a basic answer, although I'm still learning about this stuff myself. The kind of non-unique factorization you've identified is due to the fact that $\mathbb{R}$ isn't algebraically closed, and isn't as interesting as another kind of non-unique factorization, which I'll exemplify using $\mathbb{C}[x, y]/(y^2 - x^3)$. Such rings arise as rings of functions on algebraic curves, and in that case one can pinpoint exactly what causes unique factorization to fail, which is the existence of singularities (here, at the point $(x, y) = (0, 0)$).

More precisely, it's known that the ring of functions on an algebraic curve $f(x, y) = 0$ has unique factorization of ideals if and only if every point is nonsingular in the sense that the partial derivatives $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ never simultaneously vanish. The geometric intuition here is that locally, at a singular point a variety looks like the intersection of some number of lines, i.e. it is "locally reducible," so the maximal ideal associated to that point isn't generated by one element. For $\mathbb{C}[x, y]/(y^2 - x^3)$ the singular point at $(0, 0)$ is a cusp where two lines meet and the corresponding ideal is generated by $x$ and $y$ but satisfies a nontrivial relation, and this is precisely non-unique factorization. More generally I believe one can characterize the ideals without unique factorization precisely as the ideals vanishing on singular points.

Anyway, the upshot of all this is that as Greg indicates, it is possible for varieties to have lots of nasty singularities. On the other hand, it's relatively easy to fix this problem for the algebraic curve case: the integral closure of the ring of functions will have unique factorization.

Or maybe you just wanted to know something about your specific case. Make the substitution $x = \frac{1 - t^2}{1 + t^2}, y = \frac{2t}{1 + t^2}$; then for example $x^2 = (1 + y)(1 - y)$ can be written as $(1 - t^2)^2 = (1 + 2t + t^2)(1 - 2t + t^2)$. So here the failure of unique factorization is quite simple: certain polynomials in $t$ are being treated as prime which "shouldn't be." On the other hand any polynomial in $x, y$ which, when written as a rational function in $t$, avoids these anomalous primes, will have the usual prime factorization properties as a polynomial in $t$, but these prime factors will not necessarily always come from polynomials in $x, y$.

I would say that it is not true that quotients of polynomials rings are "almost UFDs".

For starters, being a quotient of $k[x_1,\ldots,x_n]$ for some $n$ just says that the ring is finitely generated over $k$. If $k$ is a field these rings are reasonably nice but they can still be quite "badly behaved" and a long way away from having unique factorization.

For instance one can view the class group of a Dedekind domain $A$ as measuring how badly unique factorization fails in $A$. This group can be very large even when $A$ is finitely generated over a field - taking the ring corresponding to an elliptic curve with a point deleted gives examples with infinite class group (the class group is pretty much the underlying elliptic curve in this case).

In fact it is a theorem of Claborn that any abelian group occurs as the class group of some Dedekind domain. I am not sure how far one can get working with finitely generated algebras over a field, although there are other results in this direction that allow one to construct such examples by taking integral closures in quadratic extensions I think, or via rings of functions on elliptic curves (this second being work of Rosen originally).

And all of this is just in dimension 1.

I understand the other part of your question now and I like Ben's answer. In higher dimension trying to understand this will get worse - I think about the best general statement one can make about unique factorization is that every regular local ring is a UFD so if the geometric incarnation of your algebra is singularity free it is locally a UFD.

As I interpret it, the answer to your question is "yes"; all ways that unique factorization fails come from writing an element of your ideal as the difference of two products.

The point the commenters above are making is that you should not think of this as an at all interesting or useful property. In general, understanding all elements of an ideal is hard and all ways of writing them a product is way too much for a human brain to take in.