Completion and endomorphism ring of injective envelope Let $(R, m ,k)$ be a commutative Noetherian local ring. We denote by $E$ the injective envelope of $k$ and by $R^～$ the $m$-adic completion of $R$. For any module $M$ over $R$, we let $M^*=Hom_R(M,E)$. 
Let $E_n=\{x\in E\mid m^nx=0\}$. Then $E_n\cong (R/m^n)^*$ and $E=\varinjlim E_n$, that is, the union of $E_n$ is $E$.
We have the following isomorphism chain:
$$Hom_R(E,E)\cong (\varinjlim E_n)^*\cong lim_{\leftarrow}(E_n)^*\cong lim_{\leftarrow}(R/m^n)^{**}\cong lim_{\leftarrow}R/m^n=R^～$$ Here, we have used that $(R/m^n)^{**}\cong R/m^n$ since $R/m^n$ has finite length.
Question 1: Is the isomorphism $Hom_R(E,E)\cong R^～$ of $R$-modules obtained above a ring isomorphism? If so, how to prove it?
I have found an interesting thing.
Let $\alpha: R\rightarrow R^～$ be the natural embedding. Let
$\beta :R\rightarrow End_R(E)$ be the $R$-homomorphism sending $r$ to $x \mapsto rx$. Denote by $\delta:End_R(E)\rightarrow R^～$ the isomorphism of Question 1. I think that, in general, $\delta\beta $ is not equal to $\alpha$. Let $ass(E)$ denote the set of associate primes of $E$. Since $ass(E)=\{m\}$, we have $ass(E)=Supp(E)=V(ann(E)) = \{m\}$. So, if $\dim_{Krull}(R) > 0$, then $ann(E)$ is not equal to $\{0\}$.
Question 2: When is $\beta$ injective?
Question 3: When does the identity $\delta\beta=\alpha$ hold true?
 A: 

Claim 1. The $R$-isomorphism $Hom_R(E,E) \cong R^～$ of Question 1 is also a ring isomorphism.


We will use


Lemma. Let $M = Rm$ be a cyclic $R$-module. Then there is a unique ring structure on $M$ such that multiplication is an $R$-bilinear map with identity $m$.
Proof. Let $\pi: R \rightarrow M$ the $R$-homomorphism defined by $\pi(r) = rx$. Assume there is an $R$-bilinear multiplication $\cdot$ on $M$ that turns $M$ into a ring. Then $\pi(x) \cdot \pi(y) = xy (\pi(1) \cdot \pi(1))$, from which we infer unicity. As $M \cong R/\ker(\pi)$ as an $R$-module, the existence is also guaranteed.


Now we can prove Claim 1.


Proof of Claim 1. Let us inspect the composition of the first two isomorphisms in the chain, namely 
    $\Psi: Hom_R(E,E)\cong lim_{\leftarrow}(E_n)^* = lim_{\leftarrow} Hom_R(E_n, E)$.
    The $R$-isomorphism $\Psi$ is given by $\Psi(f) = (f_{\vert E_n})_n$ where $f_{\vert E_n}$ denotes the restriction of $f \in Hom_R(E,E)$ to $E_n \subseteq E$. Since $f(E_n) \subseteq E_n$ for every $f \in Hom_R(E_n, E)$, we have $Hom_R(E_n, E) = Hom_R(E_n, E_n)$, so that $lim_{\leftarrow} Hom_R(E_n, E) = lim_{\leftarrow} Hom_R(E_n, E_n)$ is endowed with a natural ring structure, that is, componentwise addition and multiplication, where multiplication is the composition of homomorphisms. Clearly, $\Psi$ is a ring isomorphism with respect to this structure. Since $Hom_R(E_n, E_n) \cong R/m^n$ as $R$-modules, with the identy to the indentity and the two rings are cyclic $R$-modules for which multiplication is $R$-bilinear, they are isomorphic as rings by the above lemma. It follows that the composition of the last two isomorphisms of the chain is a ring isomorphism. As result, $Hom_R(E,E)$ and $R^～$ are also isomorphic as rings.


The next questions also have simple answers.


Claim 2. The $R$-homomorphism $\beta$ is always injective.
Proof. By definition, the $R$-homomorphism $\beta$ is injective if and only if $E$ is a faithful $R$-module. This always holds (even if $R$ is not Noetherian), see [1, Exercise 18.5].
Claim 3. The identity $\delta \beta = \alpha$ always holds true.
Proof. The maps $\alpha$ and $\delta \beta$ are $R$-homomorphisms defined on a cyclic $R$-module and both map $1$ to $1$. Therefore they coincide.


So, what's wrong with the following claim by OP?
$$ ass(E)=Supp(E)=V(ann(E)) = \{m\} ?$$
The identity $ass(E) = \{m \}$ holds true by [1, Theorem 18.4.v] and the inclusions $ass(M) \subseteq Supp(M) \subseteq V(ann(M))$ are valid for any commutative ring $R$ and any $R$-module $M$. If $M$ is finitely generated, then we have $Supp(M) = V(ann(M))$, see [1, Remark of §4], but equality doesn't hold in general and this is the only problem. Indeed, it fails for instance if $R$ is a discrete valuation ring and $E$ is the injective envelop of its residue field $k$ ($E = Frac(R)/R$ is not finitely generated in this case). Regarding the inclusion, $ass(M) \subseteq Supp(M)$ we also know that the minimal primes of $ass(M)$ are those of $Supp(M)$ if $R$ is Noetherian [1, Theorem 6.5]. In the context of the question, it is not difficult to show that, indeed, $ass(E) = Supp(E)$ (one can use, e.g., [1, Theorem 18.4.v]).

[1] H. Matsumura, "Commutative Ring Theory", 1989.
