Given an ideal of a ring $R$, is there any way by which the associated primes of $R/I$ can be computed without knowing a primary decomposition of $I$?

Question

Suppose I've been given an ideal $I$ of a commutative ring $R$ and I don't know the primary decomposition of $I$. How do I find the associated primes of $R/I$? Please give some approach if possible. More specifically, I've been trying to find $Ass(R/I)$ in the cases where (i) $R=k[X,Y]$ and $I=(X^2,XY)$ and (ii) $R=k[X,Y,Z]$ and $I=(X,Y)(X,Z)$. In both these cases, I actually know the primary decomposition of $I$ (it's $(X,Y)^2\cap (X)$ in (i) and $(X,Y)\cap (X,Z)\cap (X,Y,Z)^2$ in (ii)) and hence I can find the associated primes to be $(X,Y)$ and $(X)$ in (i) and ($X,Y),(X,Z)$ and $(X,Y,Z)$ in (ii). But supposing the primary decomposition is not known, how can I possibly find the associated primes in these two problems? Any such approach is welcome.

Answer 1

You can find them if you know the resolution of the ideal. This is Theorem 8.3.2 in Hal Schenck's "Computational Algebraic Geomtry" (see here)

Let $M$ be a finitely generated, graded module over a polynomial ring $R$ and $P$ a prime of codimension $c$. Then $P \in Ass(M) \Leftrightarrow P \in Ass(Ext^c(M,R))$.

Now all you need to know to compute the Ext groups is the resolution of the ideal.

In the case $I=(xy,x^2)$ the resolution is easy: $$ 0 \to R(-3) \xrightarrow{(-y, x)} R(-2)^2 \xrightarrow{(x², xy)^t} R \to R/I \to 0. $$ Thus Ext is the homology of $$ 0 \to R \xrightarrow{(xy, x^2)} R(2)^2 \xrightarrow{(x,-y)} R(3) \to 0 $$

The left map is injective, so $Ext^0(M,R)=0$. The middle map has kernel $im([y,x])$ and the image of the left map makes $Ext^1(M,R)=R(1)/\langle x \rangle$.

Thus the associated primes in codimension $1$ is precisely $\langle x \rangle$.

We compute $Ext^2(M,R)$ to be $R(3)/(x,y)$, and the associated prime is $\langle x,y \rangle$, just as you found out! f In conclusion, the answer is yes!