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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 a zero module.

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

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

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 $X_i$.

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 $ain I$. Observe that $a^2=a$ and hence $a= 0 $ in $A_p$ since $1-a $ is not in $p$ ( since $ain I subset p$)

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