Proper ideals generated by central ideals

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?

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Dear Per: I think Parsa's answer is incorrect. Here is my argument: Assume that $I\subset Z:=Z(R)$ is principal, generated by $z\in Z$, and that $RI=R$, that is $Rz=R$. Then $z$ is invertible in $R$, and thus also in $Z$. –  Pierre-Yves Gaillard Feb 19 '12 at 6:57

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 $XZ+YZ=1$.
The only claim that's not clear to me is $Z(R)=k[Z]$. –  Ewan Delanoy Feb 12 '12 at 20:51
Parsa, thanks! This is a nice counterexample! Ewan, from the relations it is clear that $k[Z] \subseteq Z(R)$. Again, using the relations, it is not difficult to verify the other inclusion. –  Per Feb 13 '12 at 10:34
@Per : How do you show the other inclusion ? Since we may rewrite the relations as $YZ=1-XZ,ZX=XZ,ZY=1-XZ$, we see that the monomials not containing $YZ,ZX$ or $ZY$ form a basis of $R$. –  Ewan Delanoy Feb 13 '12 at 20:22
If $f \in Z(R)$, then $fg=gf$ for every $g \in R$. Since $XY \neq YX$ and $Xf=fX$, $Yf=fY$ we see that $f$ cannot have any $X$'s or $Y$'s. –  Parsa Feb 14 '12 at 2:04