Why det is a natural transformation?

The aim of this blog is proide a categorical approach to $M_n(-),U,G{L}_n,\det (-)$$M_n(-),U,GL_n,\det(-)$.

Let $R$$R$ be a ring, $M_n(R)$$M_n(R)$ means the $n\times n$$n\times n$ matrix ring over $R$$R$, $U(R)$$U(R)$ means the unit group of $R$$R$.

Lemma. $M_n,U,G{L}_n$$M_n,U,GL_n$ are functors.

It is not hard to see that $M_n(-):\mathrm{R}\mathrm{i}\mathrm{n}\mathrm{g}\to \mathrm{R}\mathrm{i}\mathrm{n}\mathrm{g}$$M_n(-):\rm Ring\to Ring$​.

For object

$$(R,+,\cdot )\mapsto (M_n(R),+,\circ )$$

For morphism

$$(f:R\to S)⟼(M_N(f):M_n(R)\to M_n(S))$$

For each element

$$M_n(f):\left(\begin{array}{c}{r}_{i,j}\end{array}\right)\mapsto \left(\begin{array}{c}f(r_{i,j})\end{array}\right)$$

The unit group functor $U(-):\mathrm{R}\mathrm{i}\mathrm{n}\mathrm{g}\to \mathrm{G}\mathrm{r}\mathrm{p}$$U(-):\rm Ring\to Grp$ maps $R$$R$ to its unit group, and naturally ring homomorphism becomes group homomorphism.

Then $G{L}_n(-):=U\circ M_n(-)$$GL_n(-):=U\circ M_n(-)$.

Proposition. $\det (-)$$\det(-)$ is a natural transformation between $G{L}_n(-)$$GL_n(-)$ and $U(-)$$U(-)$.

Proof.

We already discuss the definition and property of $\det (-)$$\det(-)$ over $R$$R$ in here via the exterior power functor $\begin{array}{c}\underset{R}{\overset{n}{\bigwedge }}(-)\end{array}$$\bigwedge^n_R(-)\\$.

By the functorial property of $\begin{array}{c}\underset{R}{\overset{n}{\bigwedge }}(-)\end{array}$$\bigwedge^n_R(-)\\$, we see that for each $R$$R$,

$$\det \left(AB\right)=\det \left(A\right)\det B,\det \left(A\right)\in U(R)⟺A\in G{L}_n(R).$$

Hence ${\det }_R:G{L}_n(R)\to U(R)$$\det_R :GL_n(R)\to U(R)$ gives us a family of group homomorphism.

Then the proposition is claimed by $U(f)({\det }_{R}A))={\det }_S(G{L}_n(f)(A))$$U(f)(\det_R A))=\det_S(GL_n(f)(A))$. For any ring homomorphism $f:R\to S$$f:R\to S$.

You should draw the square by you own, and see the diagram commute is equivalent to

$$U(f)({\det }_{R}A))={\det }_S(G{L}_n(f)(A))$$

The equation holds since $f$$f$ is a ring homomorphism. $◻$$\square$