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.

  • $\begingroup$ The primes $P_i$ are the maximal elements among the ideals that are the zero-divisors of elements of $R.$ A word of caution.: you started with $x \in Z_R$ and concluded either $x \in P_i$ or $y\in P_i.$ You wanted to show $x \in P_i.$ You might want to argue that if $xy=o,$ then $x$ is in the annihilator of $y$ and consider what I said at the beginning. $\endgroup$ Apr 17, 2020 at 20:58
  • $\begingroup$ More generally, if $R$ is Noetherian and $M$ is an $R$-module, then the set of zero-divisors of $R$ for $M$, $\{r\in R\mid\exists m\in M\setminus\{0\},rm=0\}$, equals the union of the associated primes for $M$. See Matsumura's Commutative Algebra, Corollary 7.2. $\endgroup$ Sep 25, 2023 at 15:16

4 Answers 4


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.


Here is a complete answer following your approach. Let $0 = Q_1 \cap Q_2 \cap \dots \cap Q_k$ be a reduced primary decomposition with $\mbox{rad}(Q_i) = P_i$.

First suppose $x \in Z_R$ and pick nonzero $y \in R$ such that $xy = 0$. If $y \in Q_i$ for all $1 \leq i \leq k$, then $y \in Q_1 \cap \cdots \cap Q_k =0$ yields a contradiction. Thus $y \notin Q_i$ for some $i$. Since $xy \in Q_i$ and $Q_i$ is $P_i$-primary, $x \in \mbox{rad}(Q_i) = P_i$.

For the other direction, let $z \in P_1 \cup \cdots \cup P_k$. Then there exists $P_i$ containing $z$, say $P_1$. Now consider $Q_2 \cap \cdots \cap Q_k$. This is nonzero ideal because $0 = Q_1 \cap Q_2 \cap \dots \cap Q_k$ is reduced. Pick nonzero $w \in Q_2 \cap \cdots \cap Q_k$ and a positive integer $m$ such that $z^m \in Q_1$. Then $0 = z^m w \in Q_1 \cap \cdots \cap Q_k$. Pick minimal $n \geq 1$ such that $z^n w = 0$. If $n=1$, then we are done. If $n>1$, $z^{n-1}w \neq 0 $ by the minimality. Hence $0 = z \cdot z ^{n-1}w$ with $z^{n-1}w \neq 0 $.


This is a complement of @Delong's answer. We shall show that each $P_i$ is given by annihilators, which is a consequence of the first uniqueness theorem of primary decomposition(e.g. Atiyah-Macdonald, Theorem 4.5.) But we can go as follows:

First, verify that if $B \neq 0$ is a unitary $R$-module and $P$ is maximal in the set of ideals $\mathfrak{X}_B= \{\mbox{ann}(x) : 0 \neq x \in B\}$, then $P$ is prime (This is the exercise 8.3.8. of the Hungerford's algebra book.)

Fix $i$ and consider $B = \cap_{j \neq i} Q_j$. Let $x \in B$ be given. If $a \in \mbox{ann}(x)$, then $ax =0 \in Q_i$. Since $Q_i$ is $P_i$-primary and $x \notin Q_i$, we have $a \in P_i$. This shows $\mbox{ann}(x) \subset P_i$. On the other hand $Q_i \subset \mbox{ann}(x)$ because $\cap_{j=1}^{n} Q_j =0$. Taking radicals, we have $P_i \subset \mbox{Rad} \left(\mbox{ann}(x)\right)$.

Now take a maximal element $\mbox{ann}(y)$ of $\mathfrak{X}_B$. This is possible because $R$ is Noetherian. Now $\mbox{ann}(y) \subset P_i \subset \mbox{Rad} \left(\mbox{ann}(y)\right) = \mbox{ann}(y) $, which completes the proof.


If we use the First uniqueness theorem from Atyah MacDonald book, then the proof is straightforward.

First Uniqueness Theorem Let $I$ be a decomposable ideal in the ring $R$ and $I=\underset{1\leq i\leq n}\bigcap Q_i$ be a minimal primary decomposition of $I$. Let $P_i=\mathrm{rad}(Q_i)$, ${1\leq i\leq n}$. Then the $P_i$ are precisely the prime ideals which occur in the set of ideals $\mathrm{rad}(I:x)$, $x\in R$, and hence are independent of the particular decomposition of $I$.

In our case, $I=(0)$ ideal. Therefore, {$P_i\scriptsize \,({1\leq i\leq n})$}={$\,\mathrm{rad}(0:x)$ prime | $x\in R\,$}

So, $\underset{1\leq i\leq n}\cup$ {$P_i$} =$\underset{x\neq 0}\cup ${$\,\mathrm{rad}(0:x)$ prime | $x\in R\,$}.

Now, $\underset{x \neq 0}\cup 0${$\,\mathrm{rad}(0:x)$ prime | $x\in R \,$}=$\underset{x \neq 0}\cup${$\,\mathrm(0:x)$ prime | $x\in R \,$}

is easy to see because if $y$ is in the $\underset{x \neq 0}\cup${$\,\mathrm{rad}(0:x)$ prime | $x\in R\,$},

then $y^n$ $\in$ $\underset{x \neq 0}\cup${$\,\mathrm(0:x)$ prime | $x\in R\,$} for some integer $n\gt0$ which means that there exists some $z\neq0$ such that $y^nz=0$ and therefore $yy^{n-1}z=0$, so $y\in$$\underset{x \neq 0}\cup${$\,\mathrm(0:x)$ prime | $x\in R\,$}.

But $\underset{x \neq 0}\cup${$\,\mathrm(0:x)$ | $x\in R \,$} is $\mathrm Ann(x)$ where $x\neq0$ which means that this set is a set of all zero divisors or in your case $Z_R$.


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