Let $R$ be a commutative ring, when $\mathfrak p$ is a prime ideal, there is the localization $M_{\mathfrak p}:=S^{-1}M$, where $S=R\setminus\mathfrak p$. Show:

If $M$ is a nonzero $R$-module, then there exists a prime ideal $\mathfrak p$ such that $M_{\mathfrak p}\neq 0$.

I don't understand this exercise, in which case is $M_{\mathfrak p}$ zero ? $M$ is nonzero, this is good. So does it remain to prove that the set $R\setminus\mathfrak p$ is nonempty ?

$\textbf{EDIT:}$ I think I got it:

If $M$ is nonzero $\implies\exists m\in M,m\neq 0$

let $I=\{a\in R:aM=0\}$

since $1\in R\Rightarrow1\cdot m=m\neq 0\implies I\neq R$

and $I$ is an ideal, so $\exists$ maximal ideal $\mathfrak p$, s.t. $I\subseteq\mathfrak p$

but every maximal ideal is prime, so suppose $M_{\mathfrak p}=0$ then ;

$m/1=0$ in $M_{\mathfrak p}\implies m/1=0/1$

by definition $\exists s\in R\setminus\mathfrak p$ such that $0=s(1\cdot m-1\cdot 0)=sm$

Hence $s\in I\subseteq\mathfrak p$ but $s\in R\setminus\mathfrak p\implies s\notin\mathfrak p$ which is absurd.

  • $\begingroup$ Your proof is wrong starting from the last line "Hence $s∈I$..." (why $s∈I$?). $\endgroup$
    – user26857
    Oct 27, 2014 at 9:42
  • $\begingroup$ @inequal : Related: math.stackexchange.com/questions/2001936 $\endgroup$
    – Watson
    Jan 31, 2017 at 17:34
  • $\begingroup$ You have to change the definition of $I$: by the fact that $sm=0$ you can only say that $s\in J:=\{a\in R\mid am=0\}$. Instead of $I$ you can use $J$ and I think that the proof works. $\endgroup$
    – marco
    Dec 22, 2021 at 13:16

2 Answers 2


$M_p=0$ if and only if $x/s=0/1$ for all $x\in M$ and $s\in R-p$ if and only if for all $x\in M$ there is $t_p\in R-p$ such that $t_px=0$.

If all localizations $M_p$ are $0$, then $M=0$.

Pick an element $x\in M$. Since $x/s=0/1$ in $M_p$ there is $t_p\in R-p$ such that $t_px=0$. Now notice that the ideal generated by all $t_p$, when $p$ runs through the set of prime ideals of $R$, equals the whole ring. Write $1$ as a (finite) linear combination of $t_p$ and multiplying by $x$ get $x=0$.

Edit. An alternative proof (using annihilators):

If $M\ne 0$ pick a non-zero element $x\in M$ and consider its annihilator $\operatorname{Ann}(x)=\{a\in R\mid ax=0\}$. This is an ideal of $R$ and clearly $\operatorname{Ann}(x)\ne R$. There is a prime ideal $p$ of $R$ such that $\operatorname{Ann}(x)\subseteq p$. Then $x/1\ne0/1$ in $M_p$, otherwise there is $t_p\in R-p$ such that $t_px=0$, a contradiction. This proves that $M_p\ne0$.

  • 1
    $\begingroup$ May I ask how do you conclude that "ideal generated by all $t_p$, when $p$ runs through the set of prime ideals of $R$, equals the whole ring"? $\endgroup$
    – MathEric
    Aug 6, 2021 at 23:24
  • 1
    $\begingroup$ It is not contained in any prime ideal. $\endgroup$
    – user26857
    Aug 7, 2021 at 12:18

Since the question now has an answer, I'll just add an illustrative example.

The $\mathbb Z$-module $M=\mathbb Z/2\mathbb Z$ has non-zero localization $M_{2\mathbb Z}\neq 0$ at the prime $2\mathbb Z=(2)\subset \mathbb Z$ .
But all its other localizations are zero: $M_{\mathfrak p}=0$ for all primes $\mathfrak p\neq (2)$ namely for $\mathfrak p=(0),(3),(5),(7),(11),\cdots,(4999),\cdots\subset \mathbb Z$ .

  • $\begingroup$ Thanks for your example. but why is this true for example for $M_{(3)}= 0$ because if I consider $\frac12\in M_{(3)}$, but $\frac12=\frac24$ and $2=0$ in $M$ implies $\frac12=0$. and if we have $M_{2\mathbb Z}$ then a nonzero element looks like $\frac{1}{odd number}$, and since $S$ has only odd numbers, we can't get $0$. Is my explanation correct ? $\endgroup$
    – inequal
    Oct 26, 2014 at 8:28

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