Let $R$ be a commutative, unital ring and $M$ an $R$-module. It is well-known that if $M$ is nonzero, there is some prime $\mathfrak{p}\in \operatorname{Spec} R$ with $M\otimes_R R_\mathfrak{p}$ nonzero.

My question is, broadly, can we replace $R_\mathfrak{p}$ with the residue field $\kappa(\mathfrak{p})$ in this statement? (A more detailed form of the question is below.)

If $M$ is finitely generated over $R$, then the answer is yes. In this case, if $M$ is nonzero, take $\mathfrak{p}$ be some prime of $R$ such that $M_\mathfrak{p}$ is nonzero. Since it is finitely generated and nonzero, Nakayama's lemma gives $\mathfrak{p}_\mathfrak{p}M_\mathfrak{p} \subsetneq M_\mathfrak{p}$. Then $M\otimes_R\kappa(\mathfrak{p}) = M_\mathfrak{p}/\mathfrak{p}_\mathfrak{p}M_\mathfrak{p}$ is also nonzero.

This argument shows that in the f.g. case, any time $M\otimes R_\mathfrak{p}$ is nonzero, $M\otimes\kappa(\mathfrak{p})$ will be nonzero as well. This ceases to be true when we drop the finite generation hypothesis: take $R=\mathbb{Z}$ and $M=\mathbb{Q}$. Then $M\otimes R_\mathfrak{p}$ is never zero, and yet $M\otimes\kappa(\mathfrak{p})$ is zero for all $\mathfrak{p}$'s of $\operatorname{Spec}\mathbb{Z}$ except $(0)$.

Nonethelss, in this case, the nonzeroness of the $\mathbb{Z}$-module $\mathbb{Q}$ is still "detected" over the residue field at the zero ideal. So:

1) Is it true even in the non-f.g. case that nonzeroness of $M$ is detected by nonzeroness of $M\otimes\kappa(\mathfrak{p})$ for some $\mathfrak{p}\in\operatorname{Spec}R$? If "yes", what's the argument?

2) If "no", what's a counterexample, and is there a natural class of rings $R$ such that the answer becomes "yes"?


The standard counterexample here is $M=\mathbb{Q}/\mathbb{Z}$ with $R=\mathbb{Z}$. More generally, if $R$ has dimension $>0$, you can take any nonmaximal prime $\mathfrak{p}\subset R$ and consider $M=\kappa(\mathfrak{p})/(R/\mathfrak{p})$.

On the other hand, suppose $(R,\mathfrak{p})$ is local and zero-dimensional and $\mathfrak{p}^n=0$ for some $n$ (in particular, this holds if $\mathfrak{p}$ is finitely generated). Then for any $R$-module $M$, $M=\mathfrak{p}M$ implies $M=\mathfrak{p}M=\mathfrak{p}^2M=\dots=\mathfrak{p^n}M=0$. It follows that residue fields detect nonzeroness of modules over any locally Noetherian $0$-dimensional ring. Some hypothesis like "Noetherian" is necessary here--for instance, if $\mathfrak{p}$ contains an infinite sequence of elements $a_1,a_2,\dots$ such that $a_1a_2\dots a_n\neq 0$ for all $n$, you can take $M$ to be the colimit of the sequence $R\stackrel{a_1}\to R\stackrel{a_2}\to R\stackrel{a_3}{\to}\dots$, and then $M$ will be a nonzero module such that $\mathfrak{p}M=M$.


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