Cayley–Hamilton Theorem via Yoneda Lemma

A Concise Proof of the Cayley–Hamilton Theorem

Once you have proved the Cayley–Hamilton theorem over algebraically closed fields (by AG),you know it holds for integral domains, and in particular for the integral domain

$$\mathcal{U}=\mathbb{Z}[x_1,\dots ,x_n,n]$$

This ring is the representing object for the functor $M_n$$M_n$, with universal element the generic matrix $X=(x_{i,j})$$X = (x_{i,j})$.
For any commutative ring $A$$A$, a matrix in $M_n(A)$$M_n(A)$ corresponds to a ring homomorphism $\phi :\mathcal{U}\to A$$\varphi : \mathcal{U} \to A$.

We already know that $p_X(X)=0$$p_X(X) = 0$; therefore

$$p_M(M)=\phi (p_X(X))=\phi (0)=0.$$

In abstract terms, the assignment ${\alpha }_R:M\mapsto p_M(M)$$\alpha_R : M \mapsto p_M(M)$ is actually a natural transformation, and it is the zero transformation.
Consider the following diagram; it is easy to verify that it commutes:

$$\begin{array}{ccc}{M}_n(R)& \stackrel{{\alpha }_R}{\to }& M_n(R)\\ \downarrow M_n(\varphi )& & \downarrow M_n(\varphi )\\ M_n(S)& \stackrel{{\alpha }_S}{\to }& M_n(S)\end{array}$$

Yoneda’s lemma tells us that

$$\mathrm{Nat}(M_n,M_n)\cong \mathrm{Nat}({\mathrm{Hom}}_{\mathsf{CRing}}(\mathcal{U},-),{\mathrm{Hom}}_{\mathsf{CRing}}(\mathcal{U},-))\cong M_n(\mathcal{U})$$

Thus $\alpha$$\alpha$ is completely determined by its action on the universal element. That is, we only need to consider...

Proof

Since $\mathrm{\Phi }$$\Phi$ is a natural transformation, we have the following commutative diagram:

This diagram shows that the natural transformation $\mathrm{\Phi }$$\Phi$ is completely determined by ${\mathrm{\Phi }}_A({\mathrm{id}}_A)=u$$\Phi_A(\mathrm{id}_A) = u$ since for each morphism $f:A\to X$$f: A \to X$ one has...

$${\alpha }_u(X)=0_u(X)⟹\alpha =0$$

Here $0$$0$ is the zero natural transformation, sending every matrix to the zero matrix.