MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a complex irreducible quasi-projective variety, $f:X\longrightarrow\mathbb{P}^N$ a morphism, $H\subset\mathbb{P}^N$ a hyperplane, $Z:=f^{-1}(H)$ which is irreducible, $Y\subset X$ a irriducible closed subset. Clearly we have $f(Y)\cap H=f(Y\cap Z)$, is it true that $\overline{f(Y)}\cap H=\overline{f(Y\cap Z)}$?

EDIT: Moreover $Y\cap Z$ is irreducible (hence not empty).


share|cite|improve this question
How does $X$ enter into your question? As far as I can see, everything only depends on $f|_Y$. – user2035 Jan 29 '12 at 10:46
Usually "irreducible" is compatible with being empty, I think, so "hence not empty" is unclear. – Jim Humphreys Jan 29 '12 at 14:54
Several standard sources (Bourbaki, EGA, Hartshorne) require irreducible spaces to be nonempty. – user2035 Jan 29 '12 at 15:12
up vote 2 down vote accepted

Actually, $\overline{f(Y)}\cap H$ and $\overline{f(Y\cap Z)}$ don't even have to be of the same dimension:

Let $X=Y=\mathbb A^2$ with coordinates $x,y$ and $f:X\to \mathbb P^2$ the morphism $(x,y)\mapsto [x:xy:1]$. Further let $x_0,x_1,x_2$ denote the homogenous coordinates on $\mathbb P^2$ and let $H=Z(x_0)$. Then $Z=f^{-1}H=Z(x)\subset Y=\mathbb A^2$. Now, as you observe, $f(Y\cap Z)=f(Y)\cap H=\{[0:0:1]\}$, a single (closed!) point and hence the same holds for its closure: $\overline{f(Y\cap Z)}=\{[0:0:1]\}$. At the same time, $f$ is clearly dominant, i.e., $f(Y)$ is dense in $\mathbb P^2$ and hence $\overline{f(Y)}\cap H=H$.

The problem is that quasi-projective varieties are missing some pieces. In this example if you compactify $Y$ to start with, and resolve the indeterminacies of the morphism, then the image is the entire $\mathbb P^2$ and everything is dandy. Of course, if you assumed that $Y$ was projective, then the statement would be trivially true, since in that case $f(Y)$ is closed.

To salvage the situation and get a condition for a quasi-projective variety to have this condition, you can do the following:

In addition to your setup, assume that:
$\bullet$ $f(Y)\subset \mathbb P^n$ is a quasi-projective variety. (I.e., it's open in its closure).
$\bullet$ $H\cap \overline{f(Y)}$ is irreducible (you can take this instead of assuming that $Y\cap Z$ is irreducible).

In this case it follows that $f(Y\cap Z)=f(Y)\cap H=f(Y)\cap \big(H\cap \overline{f(Y)}\big)$ is a non-empty open subset of the irreducible set $H\cap \overline{f(Y)}$ and hence it is dense in it.

The above example shows that the first condition ($f(Y)$ being quasi-projective) is necessary and Dustin's example shows that the second ($H\cap \overline{f(Y)}$ being irreducible) is. This seems to suggest that this statement is the best you can hope for. (Actually, instead of irreducibility in the second condition you could assume that $f(Y)$ intersects non-trivially all the components of $\overline{f(Y)}\cap H$, but that seems harder to check).

share|cite|improve this answer
Thank you very much. – gio Jan 30 '12 at 9:25

Even with the additional assumption in your edit, the answer is still no. Take $X=Y=\mathbb A^1$ and let $f\colon X \rightarrow \mathbb P^2$ be given by sending $x$ to $(1:x:x^2)$. Let $(a:b:c)$ denote the coordinates on $\mathbb P^2$ and take $H$ to be the hyperplane defined by $b=0$. Then $Z = f^{-1}(H)$ is the single point $x=0$ and $f(Z)$ is the single closed point $(1:0:0)$. On the other hand, $\overline {f(Y)}$ is the variety defined by $b^2=ac$, so $H \cap \overline{f(Y)}$ consists of two points $(1:0:0)$ and $(0:0:1)$.

share|cite|improve this answer
Thank you very much. – gio Jan 30 '12 at 9:21

Let $X=Y=\mathbb A^{1}$, and consider the natural inclusion $\mathbb A^1 \to \mathbb P^1$. Let $H$ be the missing point, then $Y \cap Z$ is trivial but $\overline{f(Y)}=\mathbb P^1$.

If you want to demand $Z$ nonempty, choose 3 lines, $A$, $B$, and $C$ intersecting at a point in $\mathbb P^2$. Let $X=\mathbb P^2 - A$, let $H=B$, and let $Y = X \cap C$. $f$ is just the inclusion. Then $Y \cap Z$ is again zero, but $\overline{f(Y)}=C$, which intersects $H$ at one point.

share|cite|improve this answer
Thanks, but I had forgotten to write that $Y\cap Z$ is irreducible... – gio Jan 29 '12 at 10:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.