# Functors from rings to groups

For any commutative ring $$K$$, the set of all non-singular $$n\times n$$ matrices with entries in $$K$$ is the usual general linear group $$\operatorname{GL}_n(K)$$; moreover, each homomorphism $$f:K\to K'$$ of rings produces in the evident way a homomorphism $$\operatorname{GL}_nf:\operatorname{GL}_n(K)\to \operatorname{GL}_n(K')$$ of groups. These data define for each natural number $$n$$ a functor $$\operatorname{GL}_n:\operatorname{CRng}\to \operatorname{Grp}$$.

These lines come from MacLane, Categories for the working mathematician. What happens if I remove "commutative"? In other words, $$\operatorname{GL}_n:\operatorname{Rng}\to \operatorname{Grp}$$ is still a functor?

• what is the evident way to define $GL_nf$?
– john
Commented May 2, 2020 at 20:53
• @john I suppose $A\to B$, with $b_{ij}=f(a_{ij})$ Commented May 2, 2020 at 20:55
• yeah right that works.
– john
Commented May 2, 2020 at 20:56
• I see no reason to demand commutativity. $GL_n(K)$ is still a group for a non commutative ring and $GL_nf$ still a homomorphism. If you want to prove it just check the axioms ;)
– john
Commented May 2, 2020 at 20:58
• Perhaps MacLane focused on nonsingular because determinant and therefore non-singular become more complicated. However if you define $\operatorname{GL}_n(K)$ to be the units group of $M_n(K)$, then you're ok.
– jgon
Commented May 2, 2020 at 22:15

No commutativity of multiplication is necessary, and by the way also no subtraction. If $$R$$ is any semiring, we can form the semiring of matrices $$M_n(R)$$ over $$R$$. Clearly, this defines a functor $$M_n : \mathbf{SemiRing} \to \mathbf{SemiRing}$$. We also have the forgetful functor $$U : \mathbf{SemiRing} \to \mathbf{Mon}$$ (forgets addition) and the group of units functor $$(-)^{\times} : \mathbf{Mon} \to \mathbf{Grp}$$. The composition of these functors is a therefore also a functor, namely $$\mathrm{GL}_n : \mathbf{SemiRing} \to \mathbf{Grp}$$. You can also restrict it to the subcategory of rings.