TITLE: A question about "Zariski dense" arguments
QUESTION [30 upvotes]: This question is a little basic, but I think it is consistent with the goals of MO. 
My question is about a certain type of argument in algebraic geometry which exploits the abundance of dense sets in the Zariski topology.  A classical example is the Cayley-Hamilton theorem, and I will frame my question by sketching a false proof.

Theorem: Let $A$ be a $n \times n$ matrix with complex entries.  Then $p_A(A) = 0$ where $p_A$ is the characteristic polynomial of $A$.
False proof:
Step 1: The theorem is trivial for diagonalizable matrices.
Step 2: The set of diagonalizable matrices is Zariski dense in $\mathbb{C}^{n^2}$ because it contains the complement of the zero locus of the discriminant polynomial.
Step 3: The function $\mathbb{C}^{n^2} \to \mathbb{C}^{n^2}$ given by $A \mapsto p_A(A)$ is Zariski continuous, so it vanishes everywhere since it vanishes on a Zariski dense set.

The Flaw: Step 3 uses the following "fact" (which perhaps belongs as an answer to the "common false beliefs" question): if $f$ and $g$ are continuous functions between topological spaces $X$ and $Y$ which agree on a dense subset of $X$ then they agree everywhere.  But this need not be the case if $Y$ is not Hausdorff (and the Zariski topology certainly is not): Let $X = \mathbb{R}$, let $Y$ be the line with the double origin, and let $f, g \colon X \to Y$ be the maps which restrict to the identity on $\mathbb{R} - \{0\}$ and which satisfy $f(0) = 0_1$, $g(0) = 0_2$.  I have seen this error in textbooks.
In the case of the Cayley-Hamilton theorem, there is an easy fix: if you give $\mathbb{C}^{n^2}$ the norm topology then the diagonalizable matrices are still dense, $A \mapsto p_A(A)$ is still continuous, and now everything is Hausdorff.  But Zariski dense arguments come up a lot in algebraic geometry, sometimes in contexts where there isn't a norm topology conveniently lying around.  So my question is: can the problem be fixed in some sort of uniform way?

REPLY [16 votes]: At least in the proof of Cayley-Hamilton you mentioned, you do not need the Hausdorff condition. All you need is for the points to be closed, which is the case for Zariski topology. The point is that you can reformulate the condition $f(x)=g(x)$ as $f(x)-g(x)=0$, so the argument goes through without any issues.