Kernel of the map from a module to the product of all fibers Let $A$ be a commutative ring with nilradical $\mathfrak{n}$, $M$ an $A$-module. Then it's easy to see that $\mathfrak{n}M$ is contained in the kernel of the composite
$$M\hookrightarrow\prod_{\mathfrak{p}\in{\rm Spec}(A)}M_{\mathfrak{p}}\to\prod_{\mathfrak{p}\in{\rm Spec}(A)}M(\mathfrak{p}),$$
where $M(\mathfrak{p}):=M_{\mathfrak{p}}/{\mathfrak{p}}M_{\mathfrak{p}}$ and the maps are the obvious ones. My question is, do we have an equality here? Or under what mild condition this will be true, e.g. $A$ is noetherian and $M$ is finitely generated?
Another related question is the following: do we have $\bigcap_{\mathfrak{p}\in{\rm Spec}(A)}\mathfrak{p}M=\mathfrak{n}M$? (Again RHS is clearly contained in LHS.)
 A: The answers to both questions are negative in general, even with noetherian/finite generation hypotheses.
Let $M_{\text{nil}}$ denote the kernel of the composite
$$
M\hookrightarrow \prod_{\mathfrak{p}\in \operatorname{Spec}(A)} M_{\mathfrak{p}}\to \prod_{\mathfrak{p}\in \operatorname{Spec}(A)}M(\mathfrak{p}).
$$
Since this map is equal to the composite
$$
M\to \prod_{\mathfrak{p}\in \operatorname{Spec}(A)}M/\mathfrak{p}\to \prod_{\mathfrak{p}\in \operatorname{Spec}(A)}M(\mathfrak{p}),
$$
we have $\mathfrak{n}M\subset \bigcap_{\mathfrak{p}\in \operatorname{Spec}(A)}\mathfrak{p}M\subset M_{\text{nil}}$.
I will give an example of an $M$ with $\mathfrak{n}M\subsetneq \bigcap_{\mathfrak{p}\in \operatorname{Spec}(A)}\mathfrak{p}M$.
Let $k$ be a field and let $A=k[x,y]/(xy)$.
This is a reduced ring and hence $\mathfrak{n}=0$.
Note that $A$ has two minimal prime ideals $(x)$ and $(y)$, and any other prime ideal contains either of them.
Define an $A$-module $M$ by
$$
M=\operatorname{Coker}(A\xrightarrow{(x,-y)}A\oplus A).
$$
Then $[(x,0)]=[(0,y)]\in M$ is contained in $xM\cap yM=\bigcap_{\mathfrak{p}\in \operatorname{Spec}(A)}\mathfrak{p}M$, but it is not contained in $\mathfrak{n}M=0$.
