"Evident" functor from CRng to Grp

Question

In Categories for the Working Mathematician, p.14:

Functors arise naturally in algebra. To any commutative ring $K$, the set of all non-singular $n \times n$ matrices with entries in $K$ is the usual general linear group $GL_n(K)$; moreover, each homomorphism $f: K \rightarrow K'$ of rings produces in the evident way a homomorphism $GL_n f: GL_n(K) \rightarrow GL_n(K')$ of groups.

The most evident functor for me would be applying $f$ to each coefficient of a matrix in $GL_n(K)$. But applying the zero homomorphism for example would yield the zero matrix, which is singular.

What is the evident functor the author refers to ?

Answer 1

Your guess is correct. $GL_{n,f} $ applies $f$ to the entries. But this is not a problem, since the zero homomorphism is not a morphism in the category Crng. Namely each morphism $$g:R\to S$$ in this category, by definition fulfills $$ g(1_R)=1_S.$$ To see that $GL_{n,f}$ is well-defined, you can use the following:

1. An element $A\in GL_n(R)$ is invertible if and only if $det(A)\in R^*$
2. $det(GL_{n,f}(A))= f(det(A))$
3. Each ring homomorphism sends units to units

If you need more elaboration on any point, let me know.

Answer 2

Two comments: first, it's unnecessary to require commutativity of the rings. Second, it's also unnecessary to know anything about determinants. If $f : R \to S$ is a ring homomorphism and $X \in M_n(R)$ is a square matrix with inverse $Y \in M_n(R)$, meaning that $XY = YX = I$, then applying $f$ respects multiplication, so we get

$$f(XY) = f(X) f(Y) = f(I) = I$$

and similarly for $f(YX)$, so $f(X)$ has inverse $f(Y)$.

Said another way: a ring homomorphism $f : R \to S$ applied componentwise produces a ring homomorphism $M_n(f) : M_n(R) \to M_n(S)$. Now we can apply the "group of units" functor $(-)^{\times} : \text{Ring} \to \text{Grp}$ to this homomorphism, using that $GL_n(R) = M_n(R)^{\times}$.