Smith normal form and last invariant factor of certain matrices

Question

I've copied over this question from what I asked on Mathematics Stack Exchange, in the hope that some experts can provide some relevant insight.

Suppose we have row vectors $x_1$, $x_2$ , $y_1$ , $y_2 \in \mathbb{Z}^2$, consider the $4$ by $4$ matrix $\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \\ x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix}$.

Let $M_1 = \begin{bmatrix} x_1 & y_1 & (0, 0) \\ -y_1 & x_1 & y_1 \\ x_2 & y_2 &(0, 0) \\ -y_2 & x_2 & y_2 \end{bmatrix}$ and $M_2 =\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \\ x_2 & y_2 \\ -y_2 & x_2 \\ (0, 0) &y_1 \\ (0, 0) &y_2 \end{bmatrix}$.

The last invariant factor of $M_i$ is defined to be the $\gcd$ of the determinants of all $4$ by $4$ minors of $M_i$.

Question:If the last invariant factor of $M_2$ is $1$, does this implies that the last invariant factor of $M_1$ is $1$?

My idea was to apply some elementary row and column operations on $M_1$ and $M_2$, call the resulting matrix $M_1'$ and $M_2'$, and show that all of the $4$ by $4$ minors in $M_2'$ can be found in $M_1'$. However, it is not obvious to me how to find $M_1'$ and $M_2'$.

There is also another potential idea, let $r_1, \ldots, r_4$ denotes the rows of $M_1$, and $r'_1, \ldots r'_6$ denote the rows of $M_2$, then an eqvaiblent statement to the question would be: $$ \text{ $\frac{\mathbb{Z}^4 }{ \langle r'_1, \ldots, r'_6 \rangle}$ is torsion free implies that $\frac{\mathbb{Z}^6 }{ \langle r_1, \ldots, r_4 \rangle}$ is torsion free. } $$

Thank you for reading, any idea for this would be really appreciated.

Answer 1

To mark the question answered, I copied below my accepted answer from Mathematics Stack Exchange.

There are no matrices $M_1$ and $M_2$ which you are trying to find because of the following proposition.

Proposition. Let $p$ be any prime. Then $p$ divides the last determinant divisor of $M_1$ iff $p$ divides the last determinant divisor of $M_2$. Note that we can relax the independence condition.

Proof. By elementary transformations which do not change the last determinant divisors, we can transform the matrices $M_1$ to and $M_2$, to $M_1’= \begin{bmatrix} x_1 & y_1 & (0, 0) \\ x_2 & y_2 &(0, 0) \\ (0, 0) & x_1 & y_1 \\ (0, 0) & x_2 & y_2 \end{bmatrix}$ and $M_2’= \begin{bmatrix} x_1 & (0, 0) \\ x_2 & (0, 0) \\ y_1 & x_1 \\ y_2 & x_2 \\ (0, 0) &y_1 \\ (0, 0) &y_2 \end{bmatrix}$, respectively.

For any matrix $M$ over $\mathbb Z$ let $\overline{M}$ be the matrix $M$ with its entries replaced by their residues modulo $p$. We consider the following matrices over the field $\mathbb Z_p$ of residues modulo $p$. Put $X=\overline{\begin{bmatrix} x_1 \\ x_2\end{bmatrix}}$, and $Y=\overline{\begin{bmatrix} y_1 \\ y_2\end{bmatrix}}$, $N_1=\overline{M’_1}=\begin{bmatrix} X & Y & 0\\ 0 & X & Y\end{bmatrix}$, and $N_2=\overline{M’_2}=\begin{bmatrix} X & 0\\ Y & X \\ 0 & Y\end{bmatrix}$. Then $p$ divides the last determinant divisor of $M_1$ (resp. $M_2$) iff the rank of $N_1$ (resp. $N_2$) is at most $3$. If any of matrices $X$ and $Y$ is nonsingular then both $N_1$ and $N_2$ have rank $4$ and we are done. If any of matrices $X$ and $Y$ is zero then the rank $N_1$ equals the rank of $N_2$ and we are done. So it remains to consider the case when both ranks of $X$ and $Y$ are $1$. Simultaneously applying to $X$ and $Y$ the elementary transformations which do not change the ranks of $N_1$ and $N_2$, we can transform $X$ to $\begin{bmatrix} \overline{1} & 0 \\ 0 & 0\end{bmatrix}$ and $Y$ to some matrix $Z$. If the last row of $Z$ is zero then both matrices $N_1$ and $N_2$ are singular and we ore done. Otherwise by the elementary row transformations which do not change the ranks of $N_1$ and $N_2$ we can keep the matrix $X$ and annulate the first row of the matrix $Z$. Then we can easily see that matrices $N_1$ and $N_2$ have equal rank, so and we are done. $\square$