We have this theorem on page 62 of Shafarevich's Basic Algebraic Geometry I:

Theorem: If $f : X \to Y$ is a regular map (of quasi-projective varieties) and $f(X)$ is dense in $Y$ then $f(X)$ contains an (non-empty) open set of $Y$.

The proof to this theorem starts with ``The assertion of the theorem reduces at once to the case that both $X$ and $Y$ are irreducible and affine''.

I'd like to know how to reach this reduction?

I have some feelings for the reduction to the irreducible case: If $X,Y$ are not irreducible, they can be decomposed into irreducible components as $X = \bigcup_{i=1}^n X_i$ and $Y = \bigcup_{j=1}^m Y_j$ with $X_i$ irreducible and closed in $X$ and $Y_j$ irreducible and closed in $Y$. Then it follows that $f(X_i)$ must fall completely in some $Y_j$, otherwise, it would contradicts to that $X_i$ is irreducible. Now, suppose we may find an open subset $U_{ij}$ of each $Y_j$ contained in $f(X_i)$, but how can we construct an open subset of $Y$ from all the scratches?

As for the reduction to the affine case, I have completely no idea.

Any help will be appreciated. Thank you in advance.


1 Answer 1


You're on the right track with your ideas.

First, we finish the reduction to the irreducible case. Suppose we know the assertion for $X,Y$ irreducible. We first make the reduction to $X$ not necessarily irreducible while maintaining $Y$ irreducible. If $f(X)\subset Y$ is dense, then at least one of $f(X_i)$ must be dense in $Y$ - if not, then $f(X)$ isn't dense (as a union of a finite number of non-dense sets is again non-dense). So there's at least one irreducible component $X_i$ so that $f(X_i)\subset Y$ is dense and thus $f(X)\supset f(X_i)$ must contain a nonempty open by our assumption.

If $Y$ isn't necessarily irreducible, then we can apply our logic from the previous paragraph to each irreducible component in turn - for any irreducible component $Y_j$, there's an irreducible component $X_i$ so that $f(X_i)$ is dense in $Y_j$, so $f(X_i)$ contains a nonempty open subset of $Y_j$. We may assume that this open subset misses all the other irreducible components of $Y_j$, since $Y_j\cap Y_k$ is closed and nowhere dense in both $Y_j$ and $Y_k$ for $j\neq k$. It's then straightforwards to check that the union of all of these open sets over all $Y_j$ is again open and contained in the image of $X$.

For the affine case, we can apply the reduction to the irreducible case first, so that we have $f:X\to Y$ with $f(X)\subset Y$ dense and $X,Y$ irreducible. There are three key ingredients here, all from point-set topology:

Lemma 1: in an irreducible topological space, any nonempty open subset is dense.

Lemma 2: If $A\subset B$ is dense and we have a continuous map of topological spaces $f:B\to C$, then $f(A)\subset f(B)$ is also dense.

Lemma 3: If $A$ is dense in $B$ and $B$ is dense in $C$, then $A$ is dense in $C$.

Exercise: prove each of these yourself. Now that you've done that, do you see how to piece these together? More details in the spoiler text below:

Pick an open affine $V\subset Y$ and then pick an open affine $U\subset f^{-1}(V)\subset X$. As $U$ is dense in $X$ by lemma 1, $f(U)$ is dense in $f(X)$ by lemma 2 so it must be dense in $Y$ by lemma 3. But we know that $f(U)\subset V$, so $f(U)\subset V$ must be dense. By our assumption that we know the affine case, $f(U)$ must contain a nonempty open of $V$. But this is also a nonempty open of $Y$, and we're done.


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