Primary ideals of the polynomial ring

Question

Is it possible to classify all the primary ideals of the polynomial ring $K[X_1,\ldots ,X_n]$ where $K$ is a field.

Or, give a big class of examples of primary ideals which are not prime ideals.

Answer 1

When you restrict to special classes like monomial or binomial ideals (those generated by polynomials with one (monomial) or two (binomial) terms) then combinatorial characterizations exist. For instance, a monomial ideal $I\subset K[X_1,\dots,X_n] =:S $ is primary if and only if in the quotient $S/I$ every image of a variable is either regular or nilpotent. For binomial ideals the story is more complicated but things can be said. See Eisenbud/Sturmfels "Binomial ideals", Dickenstein/Matusevich/Miller, "Combinatorics of binomial primary decomposition", and Kahle/Miller "Decompositions of commutative monoid congruences and binomial ideals".

Answer 2

Symbolic powers give big classes of primary ideals which are not prime. Recall that given a prime $P$, the $n$th symbolic power of $P$ (denoted by $P^{(n)}$) is the $P$-primary ideal in the minimal prime decomposition $P^n$. It can also be described as $(P^n R_P) \cap R$.

It is perhaps surprising that $P^{(n)} \neq P^n$, but the only real large class of ideals where that always happens is for prime ideals which are cut out by regular sequences (ie, complete intersections).

In algebraic geometry, $P$-primary ideals show up frequently when $P$ is a height one ideal. Indeed, suppose that $D$ is a prime divisor on a normal variety $X = \text{Spec} R$. Then $$\Gamma(X, \mathcal{O}_X(-nD)) = P^{(n)}$$ where $P$ is the prime ideal defining $D$. For example, consider the prime $P = (x,y) \subseteq k[x,y,z]/(x^2 - yz)$, it is a good exercise to verify that $P^{(2)} \neq P^2$ in this example.

A poor classification: Based upon the symbolic power idea, one can classify $P$-primary ideals as follows (for any prime $P$). The set of $P$-primary ideals is equal to the set of all $$ Q \cap R $$ where $Q$ runs over all ideals of $R_P$ such that $\sqrt{Q} = P R_P$.