Center of an ideal of a $C^\ast$-algebra.

Question

If $A$ is a $C^\ast$-algebra and $I$ is an ideal of $A$, then $Z(I)=Z(A)\cap I$? where Z(.) is the center.

I have read in a paper https://doi.org/10.1093/imrn/rnaa133 that it is well known fact that $Z(I)=Z(A)\cap I$ for a $C^\ast$-algebra $A$ and $I$ two sided closed ideal of $A$.

I have done that $Z(A)\cap I$ trivially contained in $Z(I)$. But I am getting no idea how to do converse containment. Is there any idea or hint how to do it. Is some property of $C^\ast$-algebra will be used to prove that?

Answer 1

Let $x\in Z(I)$. Let $\{u_k\}$ be an approximate unit for $I$. If $a\in A$, then $$ax = \lim_k a (u_k x) = \lim_k (au_k)x = \lim_k x(au_k) = \lim_k (xa)u_k = xa$$ so $x\in Z(A)$.