Is there a simpler, more abstract proof of the Cayley-Hamilton theorem for matrices?

Question

The Cayley-Hamilton theorem is equivalent to: Let $R$ be a ring and let $M_n(R)$ be $n\times n$ matrices over $R$. Then the minimal polynomial of $A \in M_n(R)$ over $R$ divides the characteristic polynomial of $A$.

For instance. In order to reduce the confusion of having $X = $ a matrix in a polynomial. Let $R'$ be the subring of matrices $\{ a I : a \in R\}$. It's clearly isomorphic to $R$. Now consider the characteristic polynomial as an element of $R'[X]$.

Answer 1

This has no pretense to be "The" answer!

I am no algebraist but I remember a nice proof I was taught when I was I student, I thought I'd share it.

In a nutshell: True for diagonalizable matrices, then use "algebraic continuation".

Let's write down some details.

Lemma ("algebraic continuation"): Let $k$ be an infinite field. Let $P, Q \in k[X_1, \dots, X_n]$ be polynomials of $n$ variables with $Q \neq 0$ . If $P$ vanishes on the set $\{x \in k^n ~\colon~ Q(x) \neq 0\}$, then $P = 0$.

This lemma expresses that "non-empty open sets are dense in the Zariski topology". It's not hard to show (I'll give you a hint if you want).

Now:

Theorem (Cayley-Hamilton): Let $R$ be a commutative unital ring. Let $A \in M_n(R)$ be a square matrix and denote by $\chi_A(X) \in R[X]$ its characteristic polynomial. Then $\chi_A(A)$ is the zero matrix.

Let's give a proof when $R = k$ is an infinite field for the moment. By the "algebraic continuation" lemma, it is enough to show that the theorem is true when $A$ lies in some "dense open set". More precisely, each coefficient of the matrix $\chi_A(A)$ is a polynomial in the $n^2$ coefficients of $A$. It is enough to show that it vanishes on some set $\{Q \neq 0\}$, where $Q$ is a nonzero polynomial in $n^2$ variables. Let's take $Q(A) = \mathrm{Disc}(\chi_A)$ (the discriminant of the polynomial $\chi_A$). The set where $Q \neq 0$ consists precisely of matrices $A$ whose eigenvalues are all distinct in an algebraic closure $\bar{k}$ of $k$. Such matrices are diagonalizable over $\bar{k}$ so it is easy to check that $\chi_A(A)$ = 0 (I'll let you do that).

We're done!

Wait, how does this extend to an arbitrary unital ring? Well, each of the coefficients of the matrix $\chi_A(A)$ is actually a polynomial in the $n^2$ coefficients of $M$ with integer coefficients. These polynomials must be zero because we showed that Cayley-Hamilton holds for $\mathbb{Q}$ hence $\mathbb{Z}$. (NB: I think some people would say something like "$\mathbb{Z}$ is an initial object in the category of unital rings" or whatever).