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Let $A$ be a ring and consider the ring generated by $A$ and $kn^2$ indeterminates $X_{lij}$ (with $1 \leq l \leq k $ and $1 \leq i,j \leq n$). If $M_l$ is the matrix $( X_{lij} )_{i,j}$, then one can consider the ideal generated by the coefficients of all commutants $M_{l_1}M_{l_2} - M_{l_2} M_{l_1}$. Now, let $B_{n,k}$ be the quotient ring.

If $A$ is an integral domain, is it true that all $B_{n,k}$ are also integral domains ?

For $n=1$ or $k=1$, the ring $B_{n,k}$ is simply a polynomial algebra over $A$ - hence is an integral domain. For $n=k=2$, $B_{2,2}$ is isomorphic to $C[X,Y]$ where $C$ is the $A$-algebra corresponding to the vanishing of all $2$-minors of a generic $2 \times 3$ matrix - so we are done. These are the only cases for which I know the answer to the question above.

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    $\begingroup$ For $k=2$, this is an open problem. arxiv.org/abs/math/0306275 I don't know whether it is also open for larger $k$, or whether counter-examples are known. $\endgroup$ Commented Sep 28, 2013 at 11:38
  • $\begingroup$ It is known that $B_{n,2}$ is irreducible. The part that is open is whether $B_{n,2}$ is reduced. It is also conjectured (M. Artin and M. Hochster) that $B_{n,2}$ is in fact Cohen-Macaulay, but this is open for $n>4$. $\endgroup$ Commented Sep 29, 2013 at 0:22
  • $\begingroup$ I must add that for k>2 this paper might be of interest: csun.edu/~asethura/papers/CommPairsTriples.pdf $\endgroup$ Commented Sep 29, 2013 at 0:29

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