I was researching $\mathcal{M}_n(\mathbb{Q})$, the set of square $n\times n$ matrices with rational entries, as a semigroup with matrix multiplication. For $A,B\in\mathcal{M}_n(\mathbb{Q})$, the equivalence relation $A\sim B$ will be true when $A=cB$ for $c\in\mathbb{Q}$, non-zero. The quotient, $\mathcal{M}_n(\mathbb{Q})/\sim$, is also a semigroup (looks like lines in $\mathbb{R}^{n^2}$, when $\mathcal{M}_n(\mathbb{Q})$ is interpretted in Euclidean space).
I was wondering if this quotient semigroup is finitely generated (the set if finitely generated and one only has to hit a point on each line to generate it). I tried to prove by contradiction with finite generators and trying to find a non-generated element but to no avail.
Actually, the semigroup $S = \mathcal{M}_2(\mathbb{Q})/{\sim}$ is not finitely generated. Indeed, let $\cal A$ be a finite set of matrices and let $T$ be the semigroup generated by the matrices of the form $cA$, with $c \in {\Bbb Q}$ and $A \in {\cal A}$. Let us show that $T \not= {\cal M}_2({\Bbb Q})$. Consider the finite set $F$ of irreducible rational numbers equal to the determinant of some matrix in $\cal A$. Let $P$ be the finite set of prime numbers dividing the numerator or the denominator of some element of $F$ and let $p$ be a prime number not in $P$.
I claim that the matrix $\pmatrix{p&0 \\ 0&1}$ is not in $T$. Otherwise, its determinant, which is equal to $p$, would be the determinant of a matrix of $T$, and hence of the form $c^2\frac{n}{m}$ where $p$ neither divides $n$ nor $m$. Writing $c$ as $\frac{a}{b}$, one would get $pb^2m = a^2n$, which is not possible.