Monomorphism of noetherian R-modules Suppose $R$ is a noetherian ring. Then there exists a $R$-module monomorphism:
$R/I \rightarrow R$
with $I$ being a prime ideal of $R$.
I found a proof but I cannot really understand the intuition behind it. Can anyone elaborate why we would expect something like ths to happen ?
EDIT: I understand that I should show that the maximal ideal of the set of ideals $S=\{I:\text{there exists monomorphism from } R/I\rightarrow R\}$ is in fact prime. The proof I have found (here) uses a contradiction argument but I can't really follow it. I was wandering if there exists something more elementary and/or intuitive. 
 A: The statement should be:

Let $R$ be a noetherian (commutative) ring; then there exist a prime ideal $I$ of $R$ and a monomorphism of $R$-modules $R/I\to R$.

In general, you cannot find a monomorphism $R/I\to R$, for an arbitrary prime ideal $I$; think for instance to $R=\mathbb{Z}$, where the only ideals $I$ with the property are $\{0\}$ and $\mathbb{Z}$, the former of which is indeed prime.
The link you provide shows nothing for me, so I'll give a proof, possibly quite similar to the one in the book.
Let $\mathcal{A}$ be the set of proper ideals $I$ of $R$ such that there exists a monomorphism $R/I\to R$. Then $\mathcal{A}$ is not empty, because $\{0\}\in\mathcal{A}$. Since $R$ is noetherian, there exists a maximal element $P$ in $\mathcal{A}$. I claim that $P$ is prime.
If it is not prime, there exist $a,b\in R$, $a,b\notin P$, such that $ab\in P$. Note that $J=\{x\in R:ax\in P\}$ is an ideal properly containing $P$, because $b\in J$. Moreover $1\notin J$, because $a\notin P$.
By maximality, $J\notin\mathcal{A}$ and is a proper ideal, so no $R$-module homomorphism $R/J\to R$ is injective. On the other hand, there exists a monomorphism $f\colon R/P\to R$, by construction.
Consider $g\colon R/J\to R$ defined by $g(r+J)=f(ar+P)$. This is well defined, because if $r\in J$, then $ar\in P$ by definition. It is also an $R$-module homomorphism, by direct verification. On the other hand, $g(r+J)=0$ if and only if $f(ar+P)=0$, which means $ar\in P$, because $f$ is injective, so $r\in J$. Hence also $g$ is injective: a contradiction.
