Let $R$ be a unital ring and denote its center by $Z(R)$. If $I$ is an ideal of $Z(R)$, then the set $RI$ (consisting of finite sums of elements of the form ra where $r\in R$ and $a\in I$) is clearly an ideal of $R$.
My question is the following:
If $I$ is a proper ideal of $Z(R)$, is $RI$ necessarily a proper ideal of $R$?
The proof that I had in mind does not seem to work out, and I am now suspecting that the answer is negative. Are there any nice and intuitive counter-examples?
I'm new to non-commutative algebra, but maybe the following counterexample works. Take the noncommutative polynomial ring $k<X,Y,Z>$ over a field $k$, and mod out by the following relations: $XZ-ZX, YZ-ZY,$ and $XZ+YZ-1$. I guess by mod out by these relations I mean go modulo the two-sided ideal generated by these three elements, and denote the quotient ring $R$. Then the center is $Z(R)=k[Z]$. Take $I=(Z) \subsetneq Z(R)$. However, we have $RI=R$ because $XZ+YZ=1$.