If $A=K(H)$, where $H$ is an infinite dimensional separable Hilbert space, then $A$ is simple and nuclear, and the multiplier algebra $M(A)$ of $A$ is not nuclear.


My question is: can we find a non-unital simple nuclear $\mathbb C$ $\*$- algebra $B$ such that the multiplier algebra of $B$ is also nuclear?



multiplier algebra of a simple $C^*$ algebra
If $A$ is a $\sigma$-unital, simple, non-unital $C^\ast$-algebra, then $M(A)$ is non-exact (in particular, it is non-nuclear). The following argument is modelled after Yemon's idea in the comments above. Also, it uses the very deep theorem of Kirchberg, that quotients of exact $C^\ast$-algebras are exact (this can probably be avoided).


If $A$ is type I, this is the fact that $B(H)$ is not exact, so we assume that $A$ is not type I. Assume for contradiction that $M(A)$ is exact.


Let $h\in A$ be strictly positive, $\| h \|=1$, let $t\_1 > t\_2 > \dots$ be in the spectrum of $h$, with $\lim t\_n =0$, and let $f\_n\colon [0,1] \to [0,1]$ be continuous functions of norm 1, supported only close to $t\_n$ (so that all supports are mutually disjoint), such that $f\_n(t\_n) = 1$. Let $d\_n = f\_n(h)$. 
It is easily seen that the map $(a\_n)\_{n\in \mathbb N} \mapsto \sum a\_n$ (strict convergence) induces an embedding $\prod\_{n\in \mathbb N} \overline{d\_n A d\_n} \hookrightarrow M(A)$. As exactness passes to $C^\ast$-subalgebras, it follows by our assumption that $M(A)$ is exact, that $\prod \overline{d\_n A d\_n}$ is exact.


As each $\overline{d\_n A d\_n}$ is not type I, a theorem of Glimm (Cor. 6.7.4 in Pedersen's book "$C^\ast$-algebras and their automorphism groups") implies that there are $C^\ast$-subalgebras $B\_n \subseteq \overline{d\_n A d\_n}$ such that $B\_n$ has a quotient isomorphic to the UHF algebra $M\_{n^\infty}$. Hence $\prod\_{n\in \mathbb N} B\_n$, which is exact since it embeds into $\prod \overline{d\_n A d\_n}$, has a quotient isomorphic to $\prod\_{n\in \mathbb N} M\_{n^\infty}$. As quotients of exact $C^\ast$-algebras are exact (by a deep theorem of Kirchberg), $\prod M\_{n^\infty}$ is exact. However, this has a $C^\ast$-subalgebra $\prod\_{n\in \mathbb N} M\_n$ which is then exact, hence a contradiction.


