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).
On
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 $
Counterexamples exist. In fact:
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:
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 $X_i$.