p.95 in Eisenbud:
If $M$ is any module and $P$ is a minimal prime over $\operatorname{Ann}(M)$, then the submodule $M'\subset M$ defined by $$M'=\ker(M\to M_P)$$ is $P$-primary because $M/M'$ injects into $(M/M')_P=M_P$. In this situation, $M'$ is called the $P$-primary component of $0$ in $M$.
I don't understand this claim. $M/M'$ is $P$-primary iff (by definition) $\operatorname{Ass}(M/M')=\{P\}$. By an earlier result, a module $N$ is $\mathfrak p$-coprimary (meaning that $\{0\}\subset N$ is $\mathfrak p$-primary) iff $\mathfrak p$ is minimal over $\operatorname{Ann}(N)$ and $N$ injects into $N_\mathfrak p$. How does it follow that $M'$ is $P$-primary? And why does $M/M'$ inject into $(M/M')_P$? (In particular, how come that $M'_P=0$?)
Consider the localization map $M/M'\to(M/M')_P, \bar m\mapsto \bar m/1$. By a general result, $(M/M')_P\simeq M_P/M'_P$. Note that $M'_P=0$: let $m/s\in M'_P$; then $m/s=(m/1)(1/s)$; since $m\in M'=\ker (M\to M_P)$, it follows that $m/1=0$, so $m/s=0$. Now $M'\subset M$ is $P$-primary iff $Ass(M/M')=\{P\}$ iff $M/M'$ is $P$-coprimary iff $P$ is a prime minimal over $Ann(M)$ and $M/M'$ injects into $(M/M')_P$.