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
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 nonsurjective 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 nonartinian, so is $\hat{R}$. Hence the latter has a quotient that is a nonartinian 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 selfequivalence 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. $\endgroup$– YCorDec 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 nonzero direct summand in its kernel.]