Why for every prime ideal there exists a submodule such that annihilator of submodule is this prime ideal? Let $R$ be a commutative ring with identity and $M$ a faithful and finitely generated $R$-module. For any prime ideal $P$ of $R$, there exists a prime $R$-submodule $K$ of $M$ such that $P=(K:M)$. Why?
(A submodule $K\subsetneq M$ is called prime if $am\in K$ implies $aM\subseteq K$ or $m\in K$.) 
 A: 
Let $R$ be a commutative ring with identity and $M$ a finitely generated faithful $R$-module. Then for any prime ideal $P$ of $R$ there exists an $R$-submodule $K$ of $M$ such that $P=(K:M)$.

Set $$K(P)=\{x\in M:\exists\, s\in R\setminus P, sx\in PM\}.$$ ($K(P)$ is the inverse image of $PM_P$ under the canonical map $M\to M_P$.)
Let's first prove that whenever $K(P)\ne M$ we have $(K(P):M)=P$. It's obvious that $PM\subseteq K(P)$. Conversely, let $a\in(K(P):M)$, equivalently $aM\subseteq K(P)$. Since $K(P)\ne M$ there is $x\in M\setminus K(P)$. In particular, $ax\in K(P)$, and therefore there is $s\in R\setminus P$ such that $sax\in PM$. If $a\notin P$ then $sa\in R\setminus P$ and from the definition of $K(P)$ we get $x\in K(P)$, a contradiction.
It remains to prove that under the hypothesis $K(P)\ne M$. Suppose the contrary. Then $M=Rx_1+\cdots+Rx_n$ with $x_i\in K(P)$. For each $i$ there is $s_i\in R\setminus P$ such that $s_ix_i\in PM$, and then we can write $s_ix_i=\sum_{j=1}^np_{ij}x_i$ with $p_{ij}\in P$. Let $d=\det(p_{ij}-\delta_{ij}s_i)$. We have $dM=0$, and since $M$ is faithful $d=0$. On the other side, $d=p+(-1)^ns_1\cdots s_n$ with $p\in P$, and therefore $d\ne 0$, a contradiction.
