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.
Best Answer
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].)