Question: Let C be an nxn matrix, then the polynomials in C (with appropriate coefficients) form an algebra of commuting matrices. I feel that I should know if the converse is true but I do not. So my first question (adding several conditions) is: Suppose A and B are symmetric integer matrices with AB=BA, must there be a matrix C such that both A and B are polynomials in C? This would imply that the algebra which A and B generate has dimension n or less. In general: Let **A** be an algebra of pairwise commuting nxn matrices over an integral domain D, Under what conditions (if any) on D and **A** can we conclude that there is a matrix C so that **A** is contained in the D-algebra generated by C? I am thinking of conditions such as D is an algebraically closed field, D has characteristic zero, the matrices are symmetric. etc.

Answer: 
Consider the subalgebra of $M\_{2n}(k)$ spanned by the matrices of the form $\left(\begin{smallmatrix}0&A\\0&0\end{smallmatrix}\right)$ (all blocks are $n\times n$) together with the identity, which is commutative. Its dimension is larger than $2n$ when $n$ is sufficiently large, so it is not generated by a single matrix.
The open problem that surprised me was on the lower bound of dimensions of maximal commutative subalgebras of $M\_n(\mathbf C)$. A maximal commutative sublagebra of $M\_n(\mathbf C)$ can have dimension strictly lass than $n$. See for example [Courter's article](http://ams.rice.edu/mathscinet/pdf/174583.pdf?pg1=IID&s1=52245&vfpref=html&r=9), where he gives a 13 dimensional maximal commutative subalgebra of $M\_{14}(\mathbf C)$. I think, In general the problem of finding lower bound for the dimension of maximal commutative subalgebras in $M\_n(\mathbf C)$ is still open.