Let $A\in Mat(m,n;R)$ for $m\le n$ and $R$ a local ring. Consider the $mn\times(m^2+n^2)$ matrix $A\otimes 1_{nn}\oplus 1_{mm}\otimes A^T$, here $1_{mm}$, $1_{nn}$ are identity matrices. I'd like to know as much as possible about the annihilator of cokernel for this matrix.

One can write this matrix explicitly: $\begin{pmatrix} A&0_{mn}&\cdots&0_{mn}\\0_{mn}&A&\cdots&0_{mn}\\\cdots&\cdots&&\cdots\\0_{mn}&\cdots&0_{mn}&A\end{pmatrix}\oplus\begin{pmatrix} a_{11}1_{mm}&a_{21}1_{mm}&\cdots&a_{m1}1_{mm}\\a_{12}1_{mm}&\cdots&\cdots&a_{m2}1_{mm}\\\cdots&\cdots&&\cdots\\a_{1n}1_{mm}&\cdots&&a_{mn}1_{mm}\end{pmatrix}$.

Here $0_{mn}$ is the block of zeroes, $1_{mm}$ is the identity matrix. By $\oplus$ one means the juxtaposition of the $mn\times m^2$ block and the $mn\times n^2$ block.

As is seen from the explicit form: $ann.coker(A\otimes 1_{nn}\oplus 1_{mm}\otimes A^T)\supseteq ann.coker(A)$. In the simplest case, $m=1$, one actually has the equality. I guess for $m>1$ the two ideals are close, but not equal.

How to understand/bound/(compute??) $ann.coker(A\otimes 1_{nn}\oplus 1_{mm}\otimes A^T)$?

There are lots of tricks for computing the determinants of tricky matrices (and hence the fitting ideals) but I do not know any tricks to compute the annihilator of cokernel. Any suggestion?

Let $A\in Mat(m,n;R)$ for $m\le n$ and $R$ a local ring. Consider the $mn\times(m^2+n^2)$ matrix $A\otimes 1_{nn}\oplus 1_{mm}\otimes A^T$, here $1_{mm}$, $1_{nn}$ are identity matrices. I'd like to know as much as possible about the annihilator of cokernel for this matrix.

One can write this matrix explicitly: $\begin{pmatrix} A&0_{mn}&\cdots&0_{mn}\\0_{mn}&A&\cdots&0_{mn}\\\cdots&\cdots&&\cdots\\0_{mn}&\cdots&0_{mn}&A\end{pmatrix}\oplus\begin{pmatrix} a_{11}1_{mm}&a_{21}1_{mm}&\cdots&a_{m1}1_{mm}\\a_{12}1_{mm}&\cdots&\cdots&a_{m2}1_{mm}\\\cdots&\cdots&&\cdots\\a_{1n}1_{mm}&\cdots&&a_{mn}1_{mm}\end{pmatrix}$.

Here $0_{mn}$ is the block of zeros, $1_{mm}$ is the identity matrix. By $\oplus$ one means the juxtaposition of the $mn\times m^2$ block and the $mn\times n^2$ block.

As is seen from the explicit form: $ann.coker(A\otimes 1_{nn}\oplus 1_{mm}\otimes A^T)\supseteq ann.coker(A)$. In the simplest case, $m=1$, one actually has the equality. I guess for $m>1$ the two ideals are close, but not equal.

How to understand/bound/(compute??) $ann.coker(A\otimes 1_{nn}\oplus 1_{mm}\otimes A^T)$?

There are lots of tricks for computing the determinants of tricky matrices (and hence the fitting ideals) but I do not know any tricks to compute the annihilator of cokernel. Any suggestion?

upd. It seems that the integral closures of the ideals $ann.coker(A\otimes 1_{nn}\oplus 1_{mm}\otimes A^T)$ and $ann.coker(A)$ coincide. Further, if $n>m$ and $A$ is left-right equivalent to a matrix with a column of zeros, then $ann.coker(A\otimes 1_{nn}\oplus 1_{mm}\otimes A^T)=ann.coker(A)$. Still, I'd like some general methods (beyond the particular tricks)