# Calculating quotient ring over polynomial ring without guessing

Suppose we have a quotient ring over a polynomial ring, i.e. we have an ideal $I$ and a ring, $K[X_1,...,X_n]$, then when we can we identify $K[X_1,...,X_n]/I$?

What do I mean by this? Well, for example, we have $\mathbb{R}[x]/(x^2+1) \cong \mathbb{C}$, and $\mathbb{C}[x,y]/(x-y) \cong \mathbb{C}[x]$.

So given an ideal, and a ring, is there any way of seeing what $K[X_1,...,X_n)/I$ is "naturally" isomorphic to, in a sense?

I've come across this when trying to identify prime ideals. I.e., given an ideal, how can we quickly check that it's prime?

• Checking an ideal is prime is a very hard task in general. Dec 3, 2015 at 7:35

In the first question you are asking for a classification of quotients of the polynomial ring, right?

This is (nearly) impossible. Let me give a few reasons why.

If we relax the conditions of polynomial ring a bit, i.e. we allow polynomial rings over $\mathbb{Z}$ with infinite variables, then any ring is of this form. Classifying all rings is a too hard problem.

But even if you only allow $K$ to be a field and finitely many variables, this task seems impossible. The basic objects of study in algebraic geometry are varieties. If you restrict to affine varieties over a field $K$ they are up to an equivalence of categories the same as finitely generated $K$-algebras, which are quotients of a polynomial ring over $K$. There are partial results, but there are no complete classification results.

What we do know is that if we take the quotient of a ring by a prime ideal (resp. maximal ideal), then we get a domain (resp. field).

In one variable you can use the structure theorem for finitely generated modules over a principal ideal domain. For two variables the prime ideals are "known" see here.

There is no general criterion to check if an ideal is prime. In one variable this is the same as finding irreducible polynomials. There are partial results as Gauss's Lemma and Eisenstein's criterion among others, but even in one variable there is no easy way.