Let $R$ be a ring (not necessarily commutative). A non-unit and non-zero-divisor element $r \in R$ will be called irreducible if for all $a,b \in R$ such that $r=ab$, then $a$ or $b$ is a unit.

The group of units in $M_n(\mathbb{Z})$ is the group $SL^{\pm}_n(\mathbb{Z})$ of matrices of determinant $\pm 1$.

Question 1: What are the irreducible elements of $M_n(\mathbb{Z})$?
If it is too hard, let us restrict to $n=2$.

Note that a sufficient condition for an element $M \in M_n(\mathbb{Z})$ to be an irreducible element is that $\det(M)$ is a prime element of $\mathbb{Z}$. About the converse, let us ask the following:

Question 2: Let $M,N \in M_n(\mathbb{Z})$ such that $\det(M)=\det(N) \neq 0$. Is there $A,B \in SL^{\pm}_n(\mathbb{Z})$ such that $N=AMB$?

A positive answer to Question 2 implies that above sufficient condition is also necessary, and so would answer Question 1.


1 Answer 1


Notice that if a matrix has determinant plus or minus a prime, then by Binet it's irreducible.

For the converse, notice that given any matrix $A$ there are $S, T\in SL^\pm(n,\Bbb Z)$ such that $SAT$ is in Smith normal form. Now, a diagonal matrix $D$ (and therefore a matrix in Smith normal form) is irreducible if and only if all its diagonal entries are invertible except exactly one, which is irreducible. If $SAT=D_1D_2$, then $A=(S^{-1}D_1)(D_2T^{-1})$, so the Smith normal form of $A$ being irreducible is a necessary condition.

The description of the elementary divisors in terms of the determinant divisors tells us that the irreducible matrices in $M_n(\Bbb Z)$ must have determinant equal to (plus or minus) a prime, and $d_i(A)=1$ for all $i<n$.

Therefore, the following are equivalent:

  • $A$ is irreducible in $M_n(\Bbb Z)$;
  • $\det(A)$ is $\pm$ a prime;
  • $\det(A)$ is $\pm$ a prime and $d_i(A)=1$ for all $i<n$;
  • the Smith normal form of $A$ is $\begin{bmatrix}I_{n-1}&0\\ 0& p\end{bmatrix}$ for some prime $p$.

Added: As for the second question, the page on Smith normal form clairifes when two matrices $A,B\in \Bbb Z^{m\times n}$ are left-right equivalent. Call $d_i(X)$ the greatest common divisor of all the $i\times i$ minors of $X$, $\alpha_1=d_1(X)$ and $\alpha_i(X)=\frac{d_i(X)}{d_{i-1}(X)}$ for all $2\le i\le \operatorname{rk}X$. Then, there are invertible matrices $S\in SL^\pm(m,\Bbb Z)$, $T\in SL^\pm(n,\Bbb Z) $ such that $A=SBT$ if and only if $\operatorname{rk}A=\operatorname{rk}B$ and $\alpha_i(A)\Bbb Z=\alpha_i(B)\Bbb Z$ for all $1\le i\le \operatorname{rk}A$. This invariant realises exactly the (weakly) decreasing sequences of at most $\min\{m,n\}$ non-zero ideals.

  • $\begingroup$ Is $2I_n$ reducible? $\endgroup$
    – lhf
    Oct 11, 2021 at 14:12
  • 1
    $\begingroup$ @lhf Yes, for instance $\begin{bmatrix}2&0\\0&2\end{bmatrix}=\begin{bmatrix}2&0\\0&1\end{bmatrix}\begin{bmatrix}1&0\\0&2\end{bmatrix}$. $\endgroup$
    – user562983
    Oct 11, 2021 at 14:15
  • 2
    $\begingroup$ Recently, after having read some more category theory, this really makes me wonder how much of the argument can be done by using $\det$ as a functor between preordered monoids, with an adjoint given by the map you gave that replaces an entry of the identity matrix. I thought about it a bit and thought these should be mutually adjoint, but I wasn't sure if that was enough alone to make the maximal principal right ideals correspond. I think maybe the SNF is enough to ensure this is an equivalence of categories, and then I think they would definitely correspond. $\endgroup$
    – rschwieb
    Oct 11, 2021 at 20:57
  • $\begingroup$ Your argument should work from every principal ideal domain $R$, which means that the irreducible elements of $M_n(R)$ are essentially those of $R$. $\endgroup$ Oct 12, 2021 at 2:13
  • $\begingroup$ @SebastienPalcoux Yes, it's the same for all PIDs (in fact, all those $\pm$s are quite artificial). When I started writing I decided to stick to $\Bbb Z$ because I wasn't too sure of the answer and I thought you might not know much ring theory. When I realised that you certainly knew I decided not to write everything all over again: I figured it out, so I thought that you would too (which you did). The word essentially is ok, but to me, as a layman, the surprising part was that $\det$ is enough. $\endgroup$
    – user562983
    Oct 12, 2021 at 6:17

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