I have the following problem. Let $B$ be a $C^*$-algebra, $E$ be a Hilbert $B$-module. I want to show that the multiplier algebra of $\mathbb{K}(E)$ is $\mathbb{B}(E)$. In the "ordinary" case of $E=H$ just a Hilbert space (no Hilbert $B$-module) the easiest way in my opinion is to consider the non-degenerate faithful representation given by inclusion $\mathbb{K}(H)\hookrightarrow\mathbb{B}(H)$. In this case the Multiplier algebra is given by the idealizer $Id(\mathbb{K}(H))=\{ T\in\mathbb{B}(H)\ |\ T\mathbb{K}(H)\subset \mathbb{K}(H), \mathbb{K}(H)T\subset\mathbb{K}(H)\}$ and clearly this is $\mathbb{B}(H)$. I want to reduce to this case by considering a faithful non degenerate representation $\rho:\mathbb{B}(E)\to\mathbb{B}(H)$. The Problem is now can I assume $\rho(\mathbb{K}(E))=\mathbb{K}(H)$ and $\rho(\mathbb{B}(E))=\mathbb{B}(H)$? For the construction with Idealizers at least i need to show that the restriction of $\rho$ to $\mathbb{K}(E)$ is also non degenerate, since then $M(\mathbb{K}(E))=Id(\mathbb{K}(E))$ and then there is only left to show that $Id(\mathbb{K}(E))=Id(\mathbb{B}(E))$ (which is also non trivial I think).
The other way around starting with a faithful non-degenrate representation on $\mathbb{K}(E)$ will fail since the extension to a representation on $\mathbb{B}(E)$ is no longer faithful (right?).
I think the more abstract question would be: If $J\subset A$ is an essential ideal in a (possibly non unital?) $C^*$-algebra, then $M(J)=M(A)?
The statement for $M(\mathbb{K}(E))=\mathbb{B}(E)$ is true but I don't understand the proofs since they use a different method (In both of Blackadars books e.g.).
I cannot contribute much to the Hilbert-module part, but I can answer the question about essential ideals.
For starters, if $A$ is unital, then $M(A)=A$, so to have $M(J)=M(A)$ you need $M(J)=A$. This is false, in general. For instance let $A=C[0,2]$ and $$J=\{f\in C[0,2]: f(1)=0\}.$$ It is then well-known that $M(J)=C_b([0,2]\setminus\{1\})$, the continuous bounded functions on $[0,2]\setminus\{1\}$. These algebra cannot be isomorphic to $C[0,2]$ because $C[0,2]$ has no nontrivial projections, while $$1_{[0,1)}\in M(J).$$ The example can be made non-unital and non-abelian by using that the multiplier algebra of a direct sum is the direct sum of the multipliers, and same with esential ideals. So for instance $$ J=C_0([0,2]\setminus\{1\})\oplus K(H)\subset C[0,2]\oplus K(H)=A $$ is an inclusion of non-unital C$^*$-algebras, where $J$ is an essential ideal in $A$ and $M(J)=C_b([0,2]\setminus\{1\})\oplus B(H)$ is not isomorphic to $M(A)=C[0,2]\oplus B(H)$.