Does there exist a field $k$ and a subring $R$ of $S = M\_2(k)$ such that $R$ is not finitely generated over its center, $S=kR$ and $1\_R = 1\_S$? ($S$ is the algebra of $2 \times 2$ matrices over $k$.) 



M_2(k) as a central extension
I think the answer is "yes". Let $A$ be a non-Noetherian integral domain (for example a polynomial ring in infinitely many variables over a field), let $I$ denote a non-finitely-generated ideal, and let $k$ be the field of fractions of $A$. Let $R$ denote the ring of $2\times 2$ matrices with coefficients in $A$ and with bottom left hand entry in $I$. 


I think this ticks all the boxes. For example $kR=M\_2(k)$ because I can scale any element of $M\_2(k)$ until it's in $M\_2(R)$ and then again so that all entries are in $I$.


However, I don't think $R$ can be finitely-generated over its centre (which is easily checked to be $A$). For if $r\_1,r\_2,\ldots,r\_n$ are finitely many elements of $R$ then the ring they generate over $A$ will be contained in the $2\times 2$ matrices with coefficients in $A$ and bottom left hand entry in $J$, the finitely-generated ideal generated by the bottom left hand entries of the $r\_i$, and this is a proper subset of $I$.


