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I would like to know if my solution to the following exercise is correct. If not, then I will be grateful for a correct argument. (I am working with varieties over an algebraically closed field, not schemes)

Exercise: Suppose that $f,g:X \to Y$ are two morphisms, where Y is separated. If $f$ and $g$ agree on some dense open subset $U \subset X$ then $f = g$.

My argument: Let $W = \{x \in X\:|\:f(x) = g(x)\}$. I know by assumption that $U \subset W$, and wish to show that $W = X$. Since $Y$ is separated, the graph of $f$ (call it $\Gamma_f$) is closed in $X \times Y$. So by continuity, the preimage $(\mathrm{id},g)^{-1}(\Gamma_f) = W$ is closed in $X$. But then $X = Cl_X(U) \subset Cl_X(W) = W \Rightarrow W = X$ as required.

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Are there non-separated varieties? –  Matt Oct 23 '12 at 16:03
Yes - one example is 'the affine line with two origins'. The underlying set of this is constructed as follows: you take the disjoint union of two copies of $\mathbb{A}^1$ (call the coordinate on the first one $x$ and the coordinate on the second $y$) and quotient out by the relation $x$ ~ $y$ if and only if $x \neq 0$ and $y \neq 0$ and $x = y$. You give it the variety structure such that the quotient map is a morphism of varieties. –  AKr Oct 23 '12 at 16:37
However, every (quasi-)affine and every (quasi-)projective variety is separated I believe. –  AKr Oct 23 '12 at 16:40
Hmm...then maybe you should give us your definition of variety. If I'm not mistaken Hartshorne's definition is an integral, separated, scheme of finite type (over a field). It isn't that relevant to the question. I was just being annoying because you claimed to be working with varieties and not schemes, but then included the hypothesis separated for $Y$. –  Matt Oct 23 '12 at 17:24
I mean 'variety' in the sense of Kempf's book: a space with functions, $(X,\mathcal{O}_X)$ over a field $k = \overline{k}$, with the property that there exists a finite covering by open sets $X = \bigcup_1^n{U}_i$ and each $(U_i,\mathcal{O}_X|U_i)$ is isomorphic to an affine variety ($\mathrm{maxSpec}(A)$ for some finitely generated $k$-algebra $A$, which has no nilpotents). I believe such objects are sometimes called 'algebraic sets'. I really should have clarified this, since the word 'variety' means different things to different people. –  AKr Oct 23 '12 at 18:51
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2 Answers

up vote 1 down vote accepted

You proved correctly that $f=g$ as maps of the underlying spaces of $X, Y$. That they are equal as morphisms either follows from the definition in Kempf's book (I can't check) or has to be proved (used the fact the $O_X(U)$ is reduced).

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I have added this comment as an answer, since the text keeps overflowing. My apologies, I am new to this. –  AKr Oct 24 '12 at 8:46
@Akr, no problem. You are welcome. –  user18119 Oct 25 '12 at 22:02
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In response to the answer by QiL (it was too long for a comment):

Kempf's definition of a morphism is: a continuous map of the underlying topological spaces, $f:X\to Y$ such that regular functions on Y pull back to regular functions on X: for any $V\subset Y$ open, $r\in \mathcal{O}_Y(U)\Rightarrow f^{∗}r \in \mathcal{O}_X(f^{−1}(U))$.

So my understanding is that I need to add the following extra line:

Moreover, $f,g$ induce the same pullback on regular functions. Indeed, for any open $U \subset Y$ and any $r \in \mathcal{O}_Y(U)$, consider the pullbacks $f^{*}r, g^{*}r$ in $\mathcal{O}_X(f^{-1}(U)) = \mathcal{O}_X(g^{-1}(U))$. Both these regular functions take the same value in $k$ at every point of $f^{-1}(U)$. Since $\mathcal{O}_X(f^{-1}(U))$ is reduced (implying regular functions are determined by their values) it follows that $f^{*}r=g^{*}r$. Since $f,g$ are equal as functions, and induce the same pullback, they are equal as morphisms.

Finally, just to be clear: in Kempf's book, the structure sheaf $\mathcal{O}_X$ of a variety $X$ assigns, by definition, to every open set $U \subset X$, a $k$-algebra of functions $U \to k$. So it will always be reduced (have no nonzero nilpotents).

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