The injective hull for a module always exists, however over certain rings modules may not have projective covers. I have a question.
If $A$ is an Artinian module on a Noetherian local ring $R$ then $A$ has projective cover? If not, please give a counter example.


2 Answers 2


No: as soon as the (local noetherian) ring itself is not artinian, there exists an artinian module with no projective cover.

First recall (over any ring) that a projective cover of $M$ is $N$ projective and an epimorphism $N\to M$ that is non-surjective in restriction to any proper submodule of $N$.

In the current context ($R$ local noetherian), for $k$ the residual field, say that $M$ is aperiodic if $\mathrm{Hom}(M,k)=0$. If $M$ is aperiodic and nonzero then it has no projective cover $L\to M$: indeed, since $N\neq 0$, $M\neq 0$; also every projective $R$-module is free. Hence $L/P\simeq k$ for some submodule $P$; then since $M$ is aperiodic, the image of $P$ in $M$ should be all of $M$, contradicting that $L\to M$ is a projective cover.

Then it suffices to check the existence of a nonzero artinian aperiodic module. Let $I$ be the injective hull of $k$. It is artinian (by an old result of Vámos, see 19.1 in Lam's book "Lectures on modules and rings").

The Matlis dual of $I$ (see remainder below) is the free module of rank one over the completion $\hat{R}$ of $R$. Since $R$ is non-artinian, so is $\hat{R}$. Hence the latter has a quotient that is a non-artinian domain $D$ (e.g., of Krull dimension 1). Hence $\mathrm{Hom}(k,D)=0$. By Matlis duality, $D$ corresponds to a submodule $J$ of $I$ with $\mathrm{Hom}(J,k)=0$. Since $D$ has infinite length, so does $J$. Hence $J$ is an aperiodic module as requested.

(Added) Remainder on Matlis duality. This is a very useful tool, which mostly reduces the understanding of artinian modules to standard knowledge about finitely generated modules.

First observe that any artinian $R$-module is canonically a module over the completion $\hat{R}$ of $R$, and $R$-module homomorphisms between artinian $R$-modules are $\hat{R}$-module homomorphisms. The Matlis dual of an artinian $R$-module $M$ is $T(M)=\mathrm{Hom}_{R\mathrm{-mod}}(M,I)$. The first important result in Matlis duality is that $T(M)$ is a finitely generated $\hat{R}$-module. If $N$ is a finitely generated $\hat{R}$-module, write $T'(N)=\mathrm{Hom}_{\hat{R}\mathrm{-mod}}(N,I)$; this is an artinian $\hat{R}$-module and we can view it as an $R$-module. One can view $T'(T(M)$ as the bidual of $M$; Matlis duality says that the canonical $R$-module homomorphism $M\to T'(T(M))$ is an isomorphism for every artinian $R$-module $M$. Thus $T$ is a contravariant equivalence of categories between artinian $R$-modules and finitely generated $\hat{R}$-modules. (We have $T(k)=k$ and it restricts to a contravariant involutive self-equivalence of categories of the category of $R$-modules of finite length.) A standard reference for Matlis duality is the book by Bruns and Herzog.

  • $\begingroup$ I wrote this an hour ago but without internet, so I post it anyway, although Jeremy already provides the answer in one particular (but quite typical) case. $\endgroup$
    – YCor
    Dec 12, 2017 at 13:43

Take $R=\mathbb{Z}_p$, the $p$-adic integers, and $A=\mathbb{Q}_p/\mathbb{Z}_p$. Then $A$ is an Artinian $R$-module, but doesn't have a projective cover.

[Since $R$ is local, projectives are free. If $\varphi=\pmatrix{\varphi_1\\\varphi_2}:\mathbb{Z}_p\oplus\mathbb{Z}_p\to A$ is a homomorphism then $\varphi_1$ factors through $\varphi_2$ (or vice versa), say $\varphi_1=\varphi_2\theta$. Then $\left\{\left(x,-\theta(x)\right)\vert x\in\mathbb{Z}_p\right\}$ is a direct summand of $\mathbb{Z}_p\oplus\mathbb{Z}_p$ in the kernel of $\varphi$. So the kernel of any map to $A$ from a free module of rank greater than one has a non-zero direct summand in its kernel.]


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.