This old MO thread and its comments contains a discussion of the Zariski density proof of Cayley-Hamilton (I have also asked a separate question about the proof Victor gives in the comments here). Victor Protsak argues that it is circular:
Awk, this is so backwards. Zariski density of semisimple matrices in all matrices logically depends on the Cayley-Hamilton theorem
As someone who has given this proof on a few occasions I'd like to clarify my understanding of this!
Here is what I think Victor means. For an $n \times n$ matrix $A \in M_n$, with characteristic polynomial $p_A(t) = \det(tI - A)$, we want to prove $p_A(A) = 0$. This is a collection of $n^2$ polynomial identities in the $n^2$ matrix coefficients $a_{ij}$ of $A$, so it suffices to prove it on a Zariski dense subscheme of $M_n \cong \mathbb{A}^{n^2} \cong \text{Spec } \mathbb{Z}[a_{ij}]$. Now:
- CH is clearly true for diagonalizable matrices $A$ (over any field).
- Write $\Delta$ for the discriminant of the characteristic polynomial $p_A$. The complement $\Delta \neq 0$ is a non-empty open subscheme of $M_n$ (because there exists a diagonal matrix with distinct diagonal entries!), hence (since $M_n$ is irreducible) Zariski dense.
- Matrices in the complement $\Delta \neq 0$ of the discriminant locus are diagonalizable (over an algebraically closed field), hence the diagonalizable matrices are also Zariski dense, so CH is true identically (equivalently, over any commutative ring).
We could perform this argument over a fixed base field $K$, passing to its algebraic closure $\overline{K}$ as necessary, to be more concrete and avoid scheme language. But I want to run this argument over $\mathbb{Z}$ to conclude Cayley-Hamilton over any commutative ring.
Now, I think Victor is pointing out that there is a potential circularity in step 3:
I don't see a direct proof of Zariski density of diag matrices w/distinct eigenvalues (geom pts), because some form of CH is needed to relate them to the condition that the char polynomial has distinct roots.
That is, without CH, Victor is saying we don't necessarily know that $\Delta \neq 0$ implies that a matrix is diagonalizable.
However, it seems to me that we do: we can straightforwardly prove without CH that if $\lambda$ is a root of $p_A$ over a field $K$ then it's an eigenvalue of $A$ over $K$, so if $\Delta_A \neq 0$ then $p_A$ is separable over $K$, hence has distinct roots over $\overline{K}$, so $A$ has $n$ distinct eigenvalues over $\overline{K}$, hence a basis of eigenvectors over $\overline{K}$.
We could even work directly over the generic point by taking $K = Q(\mathcal{O}_{M_n}) \cong \mathbb{Q}(a_{ij})$ to be the fraction field of the ring of functions on $M_n$; in other words, we can reduce to the observation that the generic matrix has distinct eigenvalues and hence is diagonalizable (over $\overline{K}$, or just the finite extension of $\mathbb{Q}(a_{ij})$ given by the splitting field of the generic characteristic polynomial). This genericity argument can also be done using no scheme language at all, just standard facts about polynomials and fields; all we need to observe is that we're trying to prove that a bunch of elements of $\mathbb{Z}[a_{ij}]$ vanish, and that $\mathbb{Z}[a_{ij}]$ is an integral domain (of course this is the key fact we need to conclude that non-empty open subschemes are dense), so it embeds into its field of fractions $K$ and hence into $\overline{K}$.
Question: Is my understanding of the situation accurate?
I am not super familiar with scheme language regarding geometric points, subschemes, and generic points, although I think I have some idea, so any general clarification on how this circle of ideas works would also be much appreciated. Thanks!
