Finitely generated injective module implies ring is Artinian?

Question

If $R$ is a commutative Noetherian ring that admits a non-zero finitely generated injective module $E$, is $R$ Artinian?

Some Observations: This is true if $R$ is local, as seen here. In the general case, one can decompose $E$ as a direct sum of injective hulls $\displaystyle\bigoplus_{\mathfrak p\in\mathrm{Ass}_R(M)} E_R\left(R/\mathfrak p\right)^{a_{\mathfrak p}}$; clearly, this direct sum must be finite.

As a consequence, $E_R(R/\mathfrak p)$ is a finitely generated $R$-module for every $\mathfrak p\in\mathrm{Ass}_R(M)$. But since $E_R(R/\mathfrak p)\cong E_{R_{\mathfrak p}}(\kappa(\mathfrak p))$, we have that $E_{R_{\mathfrak p}}(\kappa(\mathfrak p))$ is a finitely generated $R_{\mathfrak p}$-module. Hence, $R_{\mathfrak p}$ is Artinian, that is, $\mathrm{height}~\mathfrak p = 0$ for all $\mathfrak p\in\mathrm{Ass}_R(M)$.

On the other hand, in search of a counterexample, it suffices to find a Noetherian ring $R$ of positive dimension with a minimal prime $\mathfrak p$ such that $R_{\mathfrak p}$ is a finitely generated $R$-module. Indeed, if such a ring $R$ existed, then $E_R(R/\mathfrak p)\cong E_{R_{\mathfrak p}}(\kappa(\mathfrak p))$ would be a finitely generated $R_{\mathfrak p}$-module, and hence, a finitely generated $R$-module.

I have been rather unsuccessful in finding such a ring $R$. The only thing to note is that if $R\to R_{\mathfrak p}$ is module-finite, then it is integral, whence $\mathrm{Spec}\left(R_{\mathfrak p}\right)\to\mathrm{Spec}(R)$ is a closed map; and as a result, $\mathfrak p$ is maximal in $R$.

I would greatly appreciate any advice on how to proceed, either in proving the statement or finding a counterexample.

Answer 1

Let $R=k[x,y]/(x-x^2,y-xy)$, with $k$ a field, be the example given in the comment. Then $\mathfrak{p}$ $=$ $(\overline{x},\overline{y})$ is both a minimal prime and a maximal ideal. Note that $\mathfrak{p}$ $=$ $\overline{x}\mathfrak{p}$, so, by (5) $\Rightarrow$ (1) of Lemma 10.108.2 in 04PQ, $\mathfrak{p}$ is a pure ideal of $R$. That is, $k$ $=$ $R/\mathfrak{p}$ is flat over $R$. Clearly $k$ is injective over $k$, so, by Lemma 47.3.2 in 08XN, $k$ is also injective over $R$. And $k$ is finitely generated. The Krull dimension of $R$ is $1$, so $R$ is not Artinian.

As noted by Aphelli, this ring $R$ is not connected: $\overline{x}$ is a nontrivial idempotent. (Indeed, by Lemma 10.108.5 in 04PQ, every finitely generated pure ideal is generated by an idempotent...)

Now let $R$ be a connected Noetherian ring of dimension $>0$, and $\mathfrak{m}$ a maximal ideal. Suppose the injective hull $E$ $=$ $\mathrm{E}(R/\mathfrak{m})$ is a finitely generated $R$-module. Then $E$ has finite length over $R$ (e.g. Cor 3.85 in Lam, "Lectures on modules and rings"). Hence by Th. 5 in Rosenberg & Zelinsky - Finiteness of the injective hull, $R_{\mathfrak{m}}$ is Artinian. So $\mathfrak{m}$ is a minimal prime ideal of $R$. As $\mathfrak{m}R_{\mathfrak{m}}$ is the only prime ideal of $R_{\mathfrak{m}}$, it is the nilradical of $R_{\mathfrak{m}}$. Since $\mathfrak{m}$ is finitely generated, we have $\mathfrak{m}^nR_{\mathfrak{m}}$ $=$ $0$ for some $n$. It follows that $s\mathfrak{m}^n$ $=$ $0$ in $R$ for some $s\in R\setminus\mathfrak{m}$. Then $Rs+\mathfrak{m}^n$ $=$ $R$, so, if $x_1,\cdots,x_t$ generate the ideal $\mathfrak{m}^n$, we have $as+\Sigma_{1\le i\le t}\,b_ix_i$ $=$ $1$ for suitable $a,b_1,\cdots,b_t\in R$. Putting $y$ $=$ $as$ and $z$ $=$ $\Sigma_{1\le i\le t}\,b_ix_i$, one has $y+z$ $=$ $1$ and $yz$ $=$ $0$, and hence $z$ $=$ $(y+z)z$ $=$ $yz+z^2$ $=$ $z^2$. So $z$ is an idempotent. If $z$ $=$ $0$ then $as$ $=$ $1$, so $s$ is a unit, so $\mathfrak{m}^n$ $=$ $0$ because $s\mathfrak{m}^n$ $=$ $0$. But then $R$ is Artinian ($\mathfrak{q}$ prime $\Rightarrow$ $\mathfrak{m}^n\subseteq\mathfrak{q}$ $\Rightarrow$ $\mathfrak{m}\subseteq\mathfrak{q}$), contradiction. Since $z\in\mathfrak{m}^n$, necessarily $z$ $\ne$ $1$. Thus $z$ is a nontrivial idempotent, contradicting connectedness of $R$.

This leaves the case of finitely generated $\mathrm{E}(R/\mathfrak{q})$ where $\mathfrak{q}$ is a non-maximal prime.

Edit The general case has been settled by Aphelli.

Answer 2

As Matthé van der Lee’s counterexample shows, we need to assume that $R$ is connected.

In this case, the claim is true by a refinement of tj_’s argument. Indeed, if $p$ is a prime ideal, $E_p$ is a finitely generated injective $R_p$-module (Stacks, Lemma 0A6I), so, by the local case, either $E_p\neq 0$ or $p$ has height zero.

In particular, if $p$ has height zero and $q\supset p$ is a prime ideal, then $E_q\neq 0$, so $q$ has height zero and $q=p$.

Thus $\{p\}$ is a closed subspace of $Spec(R)$ stable under generization, so it contains an open neighborhoood of $p$ by Stacks, Lemma 05LW. Thus $\{p\}$ is a closed open subspace of $Spec(R)$. Since $R$ is connected, $Spec(R)=\{p\}$ thus $R$ is Artinian.