How do we show that prime C* algebras have trivial center A prime C* algebra is a C* algebra with the property that the product of any two of its non zero ideals is non zero. The claim is that it has trivial center, i.e., the only central elements are multiplies of the identity. I would be grateful for a hint on how to prove this. Thanks.
 A: It is more or less similar to how we show division Banach algebra over complex number is $\mathbb{C}$.
Suppose we can find a nonscalar element $a$ in the center of the prime C$^*$-algebra, say $A$, consider the ring homomorphism from $A$ to $A(a-\lambda)$ by sending $x$ to $x(a-\lambda)$, where $\lambda$ is one spectrum point of $a$, then if $ker$ is nonzero, primeness implies $ker\times A(a-\lambda)=0$ Which is wrong by definition of the homomorphism. so $A\cong A(a-\lambda)$, so especially $a-\lambda$ is invertible, which can not be true, so $a=\lambda$.
Edit: this answer has problems, that I don't think can be fixed (if they can I'll be happy to be proven wrong):
The map $\pi:x\longmapsto x(a-\lambda)$ is linear, but it is not a homomorphism. Its kernel is still an ideal (because $a$ is central), so one gets $A\simeq A(a-\lambda)$ as vector spaces. But this does not imply that $a-\lambda $ is invertible from here. As I mentioned in a comment below, one can have $A\simeq Ax$ (as C$^*$-algebras!) and even that is not enough to show that $x$ is invertible. Concretely, if $A=\ell^\infty(\mathbb N)$ and $x=(1,0,1,0,1,0,\cdots)$, and $Ax\simeq A$. 
I have posted an answer with a completely different argument. 
A: The proof in ougao's answer is wrong; I've made a few comments there. Below is a different argument.
Suppose that the centre $Z(A)$ of $A$ has an element that is not a scalar multiple of the identity. Then $A$ has a selfadjoint central element $a\in Z(A)$ that is not a scalar multiple of the identity. So the spectrum of $a$ has two or more points. Let $\alpha=\min\sigma(a)$, $\beta=\max\sigma(a)$; we have $\alpha<\beta$. 
Let $f\in C[\alpha,\beta]$ with $f(\alpha)=1$, and $f(t)=0$ for $t\geq (\alpha+\beta)/2$; similarly, let $g\in C[\alpha,\beta]$ with $g(\beta)=1$ and $g(t)=0$ for $t\leq(\alpha+\beta)/2$. Put $b=f(a)$, $c=g(a)$. Then $b,c\in Z(A)$ (since $a\in Z(A)$ and $Z(A)$ is a C$^*$-algebra) and, as $fg=0$, we have $bc=0$. Then $bAc=Abc=0$, so $A$ is not prime (or, consider the nonzero ideals $A_b=\overline{Ab}$ and $A_c=\overline{Ac}$ generated by $b$ and $c$ respectively, and note that $A_bA_c=0$). 
