# 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?

• Suppose you have two elements not in $\mathfrak p$ whose product is. CAn you construct a larger ideal adimitting a monomorphism $R/I\to I$? – Brandon Apr 3 at 19:24
• @Brandon That is exactly what I was trying to do, but I can't – giannispapav Apr 3 at 19:36
• Any associated prime of $R$ does the job for $p$. At least one exists as $R$ is Noetherian. – Youngsu Apr 3 at 20:35
• Suppose $ab\in\mathfrak p$, $a\notin\mathfrak p$ and $b\notin\mathfrak p$. Then $a\in(\mathfrak p:b)$, so $\mathfrak p\subsetneq(\mathfrak p:b)\subsetneq R$. On the other side, $R/(\mathfrak p:b)$ is isomorphic to the cyclic submodule of $R/\mathfrak p$ generated by the residue class of $b$. Now restrict the injective morphism $R/\mathfrak p\to R$ to this submodule and find an injective morphism $R/(\mathfrak p:b)\to R$, a contradiction. – user26857 Apr 3 at 21:13
• @user26857 Thank you very much! – giannispapav Apr 4 at 4:19

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.