TITLE: Zerodivisors and nilpotents in $A/I$
QUESTION [5 upvotes]: I'm studying primary decomposition in the case of polynomial rings with coefficients in a field. I have defined associate prime ideals of an ideal I as the radicals of the primary ideals appearing in a decomposition. Now, I find everywhere that when I consider the quotient $A/I$ ($A$ being the polynomial ring) the set of zero-divisors in $A/I$ is given by the union of all associated primes and I can't find a way to prove it. Also, I would need a way to compute the nilradical of $A/I$ in terms of the associated primes. Could someone help me, please? Thanks.

REPLY [3 votes]: If $I=\cap_{i=1}^n Q_i$ is a (reduced) primary decomposition and $P_i=\sqrt{Q_i}$, then $\sqrt{I}=\cap_{i=1}^n P_i$. The nilradical of $A/I$ is $\sqrt{I}/I$. 
Let $a\in A$ be a zerodivisor on $A/I$. Then there exists $\hat b\in A/I$, $\hat b\neq \hat 0$, such that $a\hat b=\hat 0$, that is, $ab\in I$. Since $b\notin I$, there exists $i$ such that $b\notin Q_i$. But $ab\in Q_i$ and $Q_i$ is $P_i$-primary, so $a\in P_i$. Conversely, let $a\in P_i$. Since the primary decomposition is reduced there exists $b\in\cap_{j\neq i}Q_j$, $b\notin Q_i.$ Then $(I:b)=(Q_i:b)$ and therefore $\sqrt{(I:b)}=P_i$. Let $k\ge 1$ be minimal with the property that $a^k\in (I:b)$. Then $a^kb\in I$, and $a^{k-1}b\notin I$. Thus $ax\in I$ and $x\notin I$, where $x=a^{k-1}b$. This shows that $a$ is a zerodivisor on $A/I$.
Example. Take $I=(X^2,XY)$ in $K[X,Y]$. Then $I=(X)\cap (X^2,XY,Y^2)$ is a reduced primary decomposition. The associated prime ideals are $P_1=(X)$ and $P_2=(X,Y)$. Thus $\sqrt{I}=(X)$ and the nilradical of $K[X,Y]/(X^2,XY)$ is $(X)/(X^2,XY)$. The set of zerodivisors of $K[X,Y]/(X^2,XY)$ is $(X,Y)$.