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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 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).

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  • $\begingroup$ 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 $J$ . $\endgroup$
    – user
    Apr 19, 2019 at 12:56

2 Answers 2

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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 $X_i$.

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  • $\begingroup$ is there an example when $\mathrm{Spec}\,R$ is irreducible, but not reduced? $\endgroup$
    – user251240
    Apr 10, 2019 at 15:59
  • $\begingroup$ but you say that there is no counterexample that is both irreducible and reduced (i.e. integral domain). $\endgroup$
    – user251240
    Apr 10, 2019 at 17:51
  • $\begingroup$ 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? $\endgroup$
    – user251240
    Apr 10, 2019 at 17:57
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    $\begingroup$ I see. You can forget about my earlier comment. So your question is really, does there exist an irreducible example (which implies not reduced) $\endgroup$
    – Bart Michels
    Apr 10, 2019 at 17:57
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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 $

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  • $\begingroup$ do you think there is an example with $R$ having no idempotents and no nilpotents? Or preferably, an example with $R$ an integral domain? $\endgroup$
    – user251240
    Apr 10, 2019 at 14:09
  • $\begingroup$ localization in integral domains can never give you $0$ unless you invert $0$ itself. $\endgroup$
    – user6
    Apr 10, 2019 at 14:10
  • $\begingroup$ then what about an example with $\mathrm{Spec}\,R$ connected reduced? or is that also impossible? $\endgroup$
    – user251240
    Apr 10, 2019 at 14:15
  • $\begingroup$ My example gives you a reduced affine scheme. There are non non trivial nilpotents in the ring. $\endgroup$
    – user6
    Apr 10, 2019 at 14:17
  • $\begingroup$ but your example is not connected $\endgroup$
    – user251240
    Apr 10, 2019 at 14:17

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