not equivalent to any matrix in Smith normal form

hbghlyj

Show that if$$A:=\pmatrix{2&0\\0&X}∈M_{2}(ℤ[X])\text{ and }B:=\pmatrix{2&X\\0&2}∈M_{2}(ℤ[X])$$then $A$ is not equivalent to any matrix in Smith normal form, and $B$ is not equivalent to any diagonal matrix.

Czhang271828

Remember that

• Matrices in Smith's normal form scale the set of normal basis in an "ascending order", whose ideals generated by blocks of entries (ideal blocks) are always invariant under transformations. That is why Smith's normal form is only seen in PID;
• Diagonalisable matrices scale the set of normal basis (aka, pairwise orthogonal eigenvectors) in any order, whose Jordan blocks are always invariant under transformations.
  

Here $(2,X)$ is not principal. $\binom{2\,\,X}{0\,\,2}$ has exactly one eigenvector (in scaling sense).

hbghlyj

Czhang271828 发表于 2023-3-4 07:04
$\binom{2\,\,X}{0\,\,2}$ has exactly one eigenvector

This proves $\binom{2\,\,X}{0\,\,2}$ is not similar to any diagonal matrix,
but doesn't prove it's not equivalent to any diagonal matrix

Czhang271828

hbghlyj 发表于 2023-3-5 20:42
This proves $\binom{2\,\,X}{0\,\,2}$ is not similar to any diagonal matrix,
but doesn't prove it's  ...

Question. $A:=\begin{pmatrix}2&X\\0&2\end{pmatrix}$ is not equivalent to any diagonal matrix.

If not, then it is equivalent to some $B:=\mathrm{diag}(p(x),q(x))$. When $A\sim B$ in $\mathbb Z[X]$, the equivalence relation holds in $\mathbb Z[X]/(2,X)$. Thus $B=O$ in $\mathbb Z[X]/(2,X)$, that is, $p(x)q(x)\in (2,X)^2\setminus (0)$. Since $\det (B)$ is a divisor of $\det (A)$,  $p(x)=q(x)=2$ is the only solution. Therefore, the only candidate is $B=2I$. However, $A\sim B$ in $\mathbb Z[X]/(2)$, a contradiction!

hbghlyj

Is it true that: if $A\sim B$ in $M_2(ℤ[X])$ then $\det A=\pm\det B$