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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?

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  • $\begingroup$ Suppose you have two elements not in $\mathfrak p$ whose product is. CAn you construct a larger ideal adimitting a monomorphism $R/I\to I$? $\endgroup$ – Brandon Apr 3 at 19:24
  • $\begingroup$ @Brandon That is exactly what I was trying to do, but I can't $\endgroup$ – giannispapav Apr 3 at 19:36
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    $\begingroup$ Any associated prime of $R$ does the job for $p$. At least one exists as $R$ is Noetherian. $\endgroup$ – Youngsu Apr 3 at 20:35
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    $\begingroup$ 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. $\endgroup$ – user26857 Apr 3 at 21:13
  • $\begingroup$ @user26857 Thank you very much! $\endgroup$ – giannispapav Apr 4 at 4:19
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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.

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