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In point set topology, we have the following result, which is easily proved.

Theorem. Let $Y$ be Hausdorff space and $f,g:X \to Y$ be continuous functions. If there exists a set $A\subset X$ such that $\bar{A} = X$ and $f|_A = g|_A$, then $f=g$.

I was trying to understand what would be the natural generalization of this fact in the category of schemes. We know that the correct analogous of a Hausdorff space is a separated scheme. So I was thinking in a statement like this:

"Let $Y$ be a separated scheme and $f,g:X \to Y$ be morphisms of schemes. If there exists a set $A\subset X$ such that $\bar{A} = X$ and $f|_A = g|_A$, then $f=g$."

The first problem is that we must be careful with this restriction "$f|_A$", since $f$ is a morphism and I want to consider the morphism of sheaves also. Then I saw that Liu's book on Algebraic Geometry has the following statement:

"Let $Y$ be a separated scheme, $X$ a reduced scheme, and $f,g:X \to Y$ morphisms of schemes. If there exists a dense open subset $U$ such that $f|_U=g|_U$, then $f=g$."

Now this makes sense, since we are dealing with open subsets now. But I still find this result too restrictive. So I came up with this:

"Let $Y$ be a separated scheme, $X$ a reduced scheme, and $f,g:X \to Y$ morphisms of schemes. If there exists a morphism $\varphi:S \to X$ such that $\varphi(S)$ is dense in $X$ and $f\circ \varphi = g\circ \varphi$, then $f=g$."

It's easy to see that Liu's proof of the result concerning only the open set also applies to this context. Finally, let's go to the questions:

  1. Is this really the best generalization? Is there any other results in this direction that are at least slightly different?

  2. I can see where the "reduced" hypothesis enters in the proof, but I found it a little strange. Is it just a technical point or can be "understanded" in some sense? Maybe counterexamples of this fact when this hypothesis isn't valid would help to clarify , but I didn't think of any.

P.S. Sorry for the bad english.

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To answer your question (1): Let $X=\mathrm{Spec}(B)$ be such that $B\to O_X(U)$ is not injective for some dense open subset $U$ (this can't happen if $B$ is reduced), let $Y=\mathrm{Spec}\mathbb Z[t]$. Fix an element $b\in B$ non-zero such that $b|_U=0$.

Let $\varphi: \mathbb Z[t] \to B$ be any ring homomorphism and let $\psi : \mathbb Z[t] \to B$ be defined by $\psi(t)=\varphi(t)+b$. Then the corresponding morphisms $f, g : X\to Y$ coincide on $U$ but are not equal.

Standard example of such $B$: $B=k[x,y]/(x^2, xy)$ and $U=X\setminus \{ (0,0)\}=D(y)$.

For your question (1), you can replace the hypothesis $X$ reduced by $S\to X$ schematically dominant. When $X$ is noetherian, this means that the image of $S\to X$ contains the associated points of $X$ (= maximal points of $X$ if the latter is reduced).

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    $\begingroup$ Thank you! I didn't know what is a schematically dominant morphism, and so I ended up discovering that EGA IV 11.10 deals with this kind of morphisms and prove the equivalence with the property of unique extension. If anyone is interested, it worths looking at it. $\endgroup$
    – user1971
    Jul 15, 2012 at 0:21
  • $\begingroup$ @John, sorry I didn't give a reference for schematically dominant, but you did the right thing. $\endgroup$
    – user18119
    Jul 16, 2012 at 9:02

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