# Ideal with zero localizations at prime ideals containing it

Let $$R$$ be a commutative unital ring. I know that if an $$R$$-module has zero localizations at all prime ideals of $$R$$, then it is the zero module.

Consider a proper ideal $$I\subset R$$ as an $$R$$-module. Is it true that if localizations of this module at all prime ideals containing $$I$$ are zero, then it is the zero module?

This is true for $$R$$ a DVR (because there is only one non-zero prime ideal which contains all proper ideals).

• I just wanted to mention here that the following two statements for a commutative ring $R$ are equivalent: (1) $R$ has a non-trivial idempotent i.e. $R$ is not connected. (2) $R$ has a non-zero finitely generated ideal $J$ such that $J_P=0$ for every prime ideal $P$ containing $J$ .
– user
Apr 19 '19 at 12:56

Counterexamples exist. In fact:

Proposition. TFAE:

1. $$I_{\mathfrak p} = 0$$ for all primes $$\mathfrak p \supset I$$
2. $$\forall x \in I : Ann(x) + I = R$$
3. $$\forall x \in I : \exists y \in I : xy=x$$

In particular, a nonzero proper ideal in a boolean ring is a counterexample.

Proof. $$1 \implies 2$$: Take $$x \in I$$. Suppose there is a prime ideal $$\mathfrak p$$ containing $$Ann(x) + I$$. Localizing at $$\mathfrak p$$, we find $$r \in R - \mathfrak p$$ with $$rx= 0$$. This contradicts $$Ann(x) \subset \mathfrak p$$.

$$2 \implies 3$$: Take $$a \in R$$ and $$y \in I$$ with $$a+y = 1$$ and $$ax = 0$$. Then $$xy = x$$.

$$3 \implies 1$$: Take $$x \in I$$, $$y \in I$$ with $$xy = x$$ and take $$\mathfrak p \supset I$$. Because $$(1-y)x = 0$$ and $$1-y \notin \mathfrak p$$, $$x$$ becomes $$0$$ in $$I_{\mathfrak p}$$. $$\square$$

Milking this, we find:

• such $$I$$ consists of zero divisors.
• There are no counterexamples when $$R$$ is an integral domain.
• If $$I$$ is nonzero, proper and finitely generated, it contains a nontrivial idempotent. Proof: From 2. it follows that $$Ann(I) + I = R$$. Write $$a + x = 1$$ with $$aI = 0$$ and $$x \in I$$. Then $$x^2=x$$.

Since you asked about $$Spec(R)$$ connected and reduced, here's an example (necessarily not Noetherian). Take a field $$k$$, let $$R = k[X_1, X_2, \ldots]$$ modulo the ideal generated by the $$X_iX_j-X_i$$ for $$j>i$$, and let $$I$$ be the ideal generated by the $$X_i$$.

• is there an example when $\mathrm{Spec}\,R$ is irreducible, but not reduced?
– user251240
Apr 10 '19 at 15:59
• but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain).
– user251240
Apr 10 '19 at 17:51
• I do not completely understand this. All this proves is that tensoring with $\mathbb{Z}[x]/(x^2)$ is not going to work. There still might be an irreducible, non-reduced example constructed in another way. Or did you mean something else?
– user251240
Apr 10 '19 at 17:57
• I see. You can forget about my earlier comment. So your question is really, does there exist an irreducible example (which implies not reduced) Apr 10 '19 at 17:57

Consider the ring $$R= \mathbb Z_2 \times \mathbb Z_2 \times... \times \mathbb Z_2 \times.....$$ countable number of times .

Look at the ideal $$I= \bigoplus _{\mathbb N} \mathbb Z_2$$

I claim $$I_p=0 \forall p \in Spec \ R$$ containing $$I$$. Say $$I \subset p$$ Consider $$a\in I$$. Observe that $$a^2=a$$ and hence $$a= 0$$ in $$A_p$$ since $$1-a$$ is not in $$p$$ ( since $$a\in I \subset p$$)

Thus every element of $$I$$ becomes $$0$$ in $$A_p$$ and hence $$I_p=0 \forall p \in Spec \ A$$

• do you think there is an example with $R$ having no idempotents and no nilpotents? Or preferably, an example with $R$ an integral domain?
– user251240
Apr 10 '19 at 14:09
• localization in integral domains can never give you $0$ unless you invert $0$ itself. Apr 10 '19 at 14:10
• then what about an example with $\mathrm{Spec}\,R$ connected reduced? or is that also impossible?
– user251240
Apr 10 '19 at 14:15
• My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring. Apr 10 '19 at 14:17
• but your example is not connected
– user251240
Apr 10 '19 at 14:17