Noetherian ring has an associated prime ideal Question
Let $R$ be a Noetherian ring. Show that there is a prime ideal $\mathfrak{p}$ and a monomorphism of $R$-modules $f:R/\mathfrak{p}\to R$.
Attempt
I considered the set of ideals $\{I< R: \exists \text{ monomorphism }  f:R/I\to R\}$ which is not empty hence has a maximal element, say $\mathfrak{p}$. 
How can I show that $\mathfrak{p}$ is prime?
 A: I would go about this in a different way. For a commutative ring $R$, prime ideal $\mathfrak p \subset R$, and $R$-module $M$, the following are equivalent:
(1) there is an injective $R$-module homomorphism $R/\mathfrak p \to M$;
(2) there is an element $u\in M$ with $\mathrm{Ann}_R(u):= \{r \in R : ru = 0\} = \mathfrak p$.
To see why (2) implies (1), notice that the $R$-module homomorphism $R\to M$ given by $r\mapsto ru$ has kernel $\mathfrak p$. So we'll go about this problem by trying to satisfy (2). Since $\mathrm{Ann}_R(0) = R$, we need to assume $M\ne 0$ if we want there to be such a prime ideal, and we obviously need $R \ne 0$ too (if we want any prime ideals at all). For us, $M=R$, but let's keep it general. $R \ne 0$ is Noetherian and $M \ne 0$ is an $R$-module.

It is in fact true that any nonzero element of $M$ has a nonzero multiple with prime annihilator. Let $0\ne u \in M$. Let $\mathcal S = \{I \subset R : I = \mathrm{Ann}_R(ru) \text{ for some } r\in R \text{ with } ru \ne 0\}$. Notice that if $I\in \mathcal S$, then $I \ne R$, since $1\notin \mathrm{Ann}_R(ru)$ when $ru \ne 0$. Zorn's lemma tells us there's a maximal element $\mathfrak p \in \mathcal S$ under containment, and $\mathfrak p = \mathrm{Ann}_R(ru)$ for some $r\in R$ with $ru\ne 0$. Suppose $a,b \in R-\mathfrak p$ are such that $ab\in \mathfrak p$. Then $bru \ne 0$ and $\mathrm{Ann}_R(bru) \in \mathcal S$ contains $a$ as well as $\mathfrak p$, contradicting maximality of $\mathfrak p$ in $\mathcal S$. So $\mathfrak p$ is prime. 
