# 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.

• 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. Apr 17, 2020 at 20:58
• 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. Sep 25, 2023 at 15:16

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 $$Z_R$$.