Recall some definitions: a sub-C*-algebra $A$ of $B(H)$, the algebra of bounded operators on a Hilbert space $H$, is called irreducible if the only closed $A$-invariant subspaces of $H$ are $0$ and $H$ itself. A (general) C*-algebra $A$ is called primitive if there is a irreducible faithful representation of $A$ in some $B(H)$. We denote $M(A)$ the Multiplier Algebra of $A$.
How do we show that if $A$ is primitive then $Z(M(A))$ (the center of $M(A)$) is trivial, that is, $Z=\mathbb{C}\cdot 1$?
I looked for this, and there are in fact some places where this is stated (for example, this answer in MO), but I couldn't find the proof anywhere.
The first problem I encountered is "Given $p\in Z(M(A))$, how do I find a candidate $\lambda\in\mathbb{C}$ for which $p=\lambda1$?". After reading this answer of the question How do we show that prime C* algebras have trivial center, I tought of using Gelfand-Mazur theorem. It is clear that $Z(M(A))$ is a Banach algebra, so the problem is to show, using irreducibility of $A$, that it is a field.
You have $A\subset M(A)\subset B(H)$ (since $A''=B(H)$). So an element of the centre of $M(A)$ commutes with all the elements of $A$. Then $$ Z(M(A))\subset A'=A'''=B(H)'=\mathbb C\ I. $$