# What are the irreducible elements of the ring of integer matrices?

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.

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.

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 $$i;
• the Smith normal form of $$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.

• Is $2I_n$ reducible?
– lhf
Oct 11, 2021 at 14:12
• @lhf Yes, for instance $\begin{bmatrix}2&0\\0&2\end{bmatrix}=\begin{bmatrix}2&0\\0&1\end{bmatrix}\begin{bmatrix}1&0\\0&2\end{bmatrix}$.
– user562983
Oct 11, 2021 at 14:15
• 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. Oct 11, 2021 at 20:57
• Your argument should work from every principal ideal domain $R$, which means that the irreducible elements of $M_n(R)$ are essentially those of $R$. Oct 12, 2021 at 2:13
• @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.
– user562983
Oct 12, 2021 at 6:17