# Zerodivisors and nilpotents in $A/I$

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.

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Using the projection $A \to A/I$ it's enough to show that for the minimal primary decomposition of the zero ideal $I=(0)=\cap_{i=1}^n I_i$ the zero-divisors of $A$ is the union of all associated primes $P_i:=\sqrt{I_i}.$ – Ehsan M. Kermani Feb 3 '13 at 0:27
Also then the nilradical of $A$ will be $\cap_{i=1}^n P_i.$ – Ehsan M. Kermani Feb 3 '13 at 0:34
Thanks a lot for all the answers. – NeuWolferl Feb 3 '13 at 15:01

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 $(X,Y)$.
You want to show that $xy\in I=Q_1\cap \cdots \cap Q_n$ iff $x,y$ are in the radical $P_i$ of some $Q_i$. For one direction, note that the complement of the union of a set of prime ideals is multiplicative closed, i.e. if $x,y\notin P_1\cup \cdots \cup P_n$ then $xy\notin P_1\cup\cdots\cup P_n$, so $xy\notin I$. For the other, suppose $x\in P_1\cup \cdots \cup P_n$, so $x\in P_i$ for some $i$. Then we have $x^n$ in $Q_i$ for some $n$. Since the intersection $Q_1\cap \cdots \cap Q_n$ is not redundant, we have some $y\in \left(Q_1\cap \cdots\cap \widehat{Q_i}\cap\cdots \cap Q_n\right)\setminus I$, so $y\notin I$ but $x(x^{n-1}y)=x^ny\in I$.