Zero Divisors and Associated Primes of the zero ideal in a Noetherian ring

I have the following question:

Let $P_1, \dots, P_k$ be the associated prime ideals of the zero ideal in the Noetherian ring $R$. Show that $P_1 \cup \dots \cup P_k$ is the set of zero divisors in $R$.

I'm denoting the zero ideal $(0)$, and the set of zero divisors $Z_R$. Let $(0) = Q_1 \cap \dots \cap Q_n$ be a primary decomposition, so that $P_i$ is the associated prime for $Q_i$, i.e. $P_i = \mbox{Rad}(Q_i)$.

Clearly, we want to show both inclusions.

First, let $x \in Z_R$, then $xy = 0$ for some $y \in R$. Then, $xy \in (0) \implies xy \in Q_1 \cap \cdots \cap Q_k \implies xy \in P_1 \cap \cdots \cap P_k$. Since each $P_i$ is prime, $xy \in P_i \implies x \in P_i$ or $y \in P_i$. However, this is where I get stuck, since I need explicitly that $x \in P_i$.

For the other inclusion, I'm not quite sure where to start. If we let $z \in P_1 \cup \cdots \cup P_k$, then $z \in P_i$ for some $i$. Then, $z^m \in Q_i$ for some power $m$. But I don't know how to proceed from here.

Any help would be appreciated.

Solution 1:

From $xy\in Q_{1}\cap \cdots\cap Q_{k}$, we have $xy\in Q_{i}$ for all $i$. Since $Q_{i}$ is $P_{i}$-primary, either $x\in P_{i}$ or $y\in Q_{i}$. If $x\in P_{j}$ for some $j$, then we are done. If not, then $y\in Q_{i}$ for all $i$. So $y\in Q_{1}\cap\cdots\cap Q_{k}$. Then $y=0$. This is a contradiction. So this case does not happen.

The other direction requires more results. My professor said that the associated prime $P_{i}$ has the form $(0:Ra)$ for some $a\in R$. Then $a\neq 0$ because $P_{i}\neq R$. Since $z\in P_{i}=(0:Ra)$, $zRa\subset (0)$, so $za=0$. Then $z$ is a zero divisor.