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?

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    $\begingroup$ to correct this proof, replace "Zariski continuous" by "polynomial". Just Zariski-continuity is a very weak property $\endgroup$ – user49822 Aug 23 '14 at 15:47

Yes; just use the analogue of "Hausdorff" for schemes: separatedness (i.e. that the diagonal morphism is a closed immersion). This implies, by the same proof, the fact that you want.

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    $\begingroup$ Or more concretely (for people not comfortable with this), two morphisms (maps defined by polynomials) between irreducible algebraic varieties (e.g. $k^N$) agree if they agree on a Zariski dense set. $\endgroup$ – Donu Arapura Aug 23 '14 at 14:58
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    $\begingroup$ To elaborate, the point is that the locus of equality ("equaliser") is the inverse image of the diagonal, viz. $\{ x : f (x) = g (x) \} = (f, g)^{-1} \Delta$. Of course, one uses the Zariski topology on the product rather than the product topology. $\endgroup$ – Zhen Lin Aug 23 '14 at 18:59
  • $\begingroup$ This seems to do the trick, but I need to think about it a little more. $\endgroup$ – Paul Siegel Aug 23 '14 at 19:49
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    $\begingroup$ Here is a reference that might be useful: Corollary 9.9, page. no. 229 in Gortz and Wedhorn's book on algebraic geometry. Let $S$ be a scheme, $X$ and $Y$ two $S$-shemes. Assume $Y$ is separated over $S$. Let $U\subset X$ be an open dense subscheme, and $f,g:X\rightarrow Y$ two $S$-morphisms such that $f|_{U}=g|_{U}$. Then we have $f|_{X_{\text{red}}}=g|_{X_{\text{red}}}$. $\endgroup$ – Yahoo Aug 23 '14 at 20:52

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.

  • $\begingroup$ Closed points is not enough: the line with the double point (see my example above) is T1. $\endgroup$ – Paul Siegel Aug 23 '14 at 19:50
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    $\begingroup$ Yes, but there is no "subtraction" possible in the space $Y$ above. Let me rephrase the argument: the set of matrices $A$ with $p_A(A)=0$ is closed, (because it is the pre-image of $0$ under the map $ A \mapsto p_A(A)$ that $\{ 0 \}$). Since it contains a dense set, it has to be the entire space. You do not need to use the diagonal. $\endgroup$ – Keivan Karai Aug 23 '14 at 20:53
  • $\begingroup$ I see, that works too - it also helps deal with some of the other Zariski dense arguments that I had in mind. But others might require the diagonal condition. $\endgroup$ – Paul Siegel Aug 23 '14 at 21:26

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