This question has been asked before, but I am stuck on a different part. If $\phi: X \rightarrow Y$ is a morphism of varieties over an algebraically closed field $k$, I'm trying to understand why $\phi(X)$ contains a nonempty open subset of its closure. Here a variety is a ringed space of $k$-algebras which has a finite open cover by maximal spectra of reduced finitely generated $k$-algebras.

I understand the proof when $X$ and $Y$ are affine, $X$ is irreducible, and $\phi(X)$ is dense in $Y$. What I'm having trouble understanding is how we can reduce to that case. In (1.9.5), Springer Linear Algebraic Groups, the proof says we first reduce to the case where $Y$ is affine, then to the case where $X$ is affine, then to the case where $X$ is irreducible.

Reducing to the case where $Y$ is affine:

I'm trying to understand first how we can reduce to the case where $Y$ is affine. I suppose we should start with finite affine open cover $W_i$ of $Y$, and let $U_i = \phi^{-1}W_i$. Then $\phi$ restricts to a morphism of varieties $U_i \rightarrow W_i$. The closure of $\phi(U_i)$ in $W_i$ is $\overline{\phi(U_i)} \cap W_i$. Suppose for each $i$ with $U_i \neq \emptyset$, we have shown that $\phi(U_i)$ contains a nonempty open subset $V_i$ of $\overline{\phi(U_i)} \cap W_i$. That intersection itself is open in $\overline{\phi(U_i)}$. Hence $V_i$ is open in $\overline{\phi(U_i)}$. I want to say that the union of $V_i$ is open in the union of the $\overline{\phi(U_i)}$, which is $\overline{\phi(X)}$. But in general if $A_1$ is open in $B_1$, and $A_2$ is open in $B_2$, then $A_1 \cup A_2$ need not be open in $B_1 \cup B_2$. Even if $B_1$ and $B_2$ are closed in some larger space.


You need an open affine cover of $\overline{\phi(X)}$, not of $Y$. So the trick is to assume without loss of generality from the beginning that $\phi(X)$ is dense in $Y$. Then $\phi(X) \cap W_i$ is dense in $W_i$. Now let $U_i = \phi^{-1}(W_i)$. The claim is that $$\overline{\phi(U_i)} \cap W_i = W_i$$ Once this is proved, we are done, because the $V_i$ will then be open in $Y$. We have $\phi(U_i) = \phi(X) \cap W_i$, so the closure of $\phi(U_i)$ in $W_i$, i.e. $\overline{\phi(U_i)} \cap W_i$, is the closure of $\phi(X) \cap W_i$ in $W_i$. But $\phi(X) \cap W_i$ is dense in $W_i$, so this closure is $W_i$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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