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Let $\phi: X \rightarrow Y$ be a morphism be varieties over an algebraically closed field. I'm trying to prove that $\phi(X)$ contains a nonempty open subset of $\overline{\phi(X)}$. I know how to solve the problem when $X$ and $Y$ are affine and irreducible with $\phi(X)$ dense in $Y$, and I'm trying to understand how we can reduce the problem to this case. I am generally having trouble with problems like this, I don't understand how people can reduce to the affine case so quickly.

So far I have reduced to the case where $Y$ is affine, and I'm currently trying to understand how to reduce further.

By a variety, I mean a locally ringed space of $k$-valued functions with a finite open cover by open affines. Here an affine variety is the maximal spectrum of a reduced finitely generated algebra over the (algebraically closed) field.

D_S
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  • Well if $V_0$ is a subset of $\phi(U)$ which is open in $\overline{\phi(U)}$, what would be the required subset of $\phi(X)$ which is open in $\overline{\phi(X)}$? – D_S Apr 02 '16 at 02:30
  • You're right, that's a good point. If I recall correctly, one first reduces to the case that $X,Y$ are irreducible, $Y$ is affine, and $\phi$ is dominant. To reduce down to $X$ affine, if $\bigcup_i U_i$ is an open affine cover of $X$, then the $U_i$ are dense in $X$, so their images $\phi(U_i)$ are dense in $Y$, hence the restrictions $U_i\to Y$ are dominant morphisms of irreducible affines. Then you can take the union of opens of $\overline{\phi(U_i)}$ which are in $\phi(U_i)$. – Ben West Apr 02 '16 at 03:11
  • Digging back through my old notes on Springer I think this bothered me as well. I made a note to myself there that Proposition 2.2.13 of Algebraic Groups and Differential Galois Theory by Crespo proves this in more detail. – Ben West Apr 02 '16 at 03:20
  • Thank you, that was very helpful. Feel free to post what you said as an answer, I will accept it. – D_S Apr 02 '16 at 05:27
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    I saw your other comments, and I agree, I'm dubious too now. If you unaccept, I will deleted the post for now (the site won't allow me to otherwise). I'll think about it if I have time, or hopefully someone more well-versed in AG will step in. – Ben West Apr 02 '16 at 20:52

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To start, we can certainly assume that $\overline{\phi(X)}=Y$ (if this isn't true, just replace $Y$ by $\overline{\phi(X)}$). Now let $C_1,\dots,C_n$ be the irreducible components of $Y$. Then $U=Y\setminus (C_2\cup \dots\cup C_n)$ is a nonempty open subset of $Y$. Furthermore, $U\subseteq C_1$ and hence $U$ is irreducible since it is an open subset of an irreducible variety. Replacing $Y$ by $U$ and $X$ by $\phi^{-1}(U)$, we may assume $Y$ is irreducible (if $\phi(X)$ contains a nonempty open subset of $U$, then $\phi(X)$ contains a nonempty open subset of $Y$ since $U$ is open in $Y$). Replacing $Y$ by a nonempty affine open subset of $Y$ and replacing $X$ by its inverse image, we may also assume $Y$ is affine.

Now let $D_1,\dots, D_m$ be the irreducible components of $X$. The sets $\overline{\phi(D_i)}$ are closed sets whose union is $\overline{\phi(X)}=Y$, so by irreducibility of $Y$ one of them is all of $Y$. Replacing $X$ by the corresponding component, we may assume $X$ is irreducible. Now replace $X$ by a nonempty affine open subset (which is dense in $X$ and hence does not disturb the condition $\overline{\phi(X)}=Y$), and we've reduced to the case you know how to do.

Eric Wofsey
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