Let $C$ be a projective regular integer curve over a (finite) field $k$ and $K:=K(C)$ the function field. Let $f\colon Y\rightarrow C$ be a projective variety with generic fiber $j\colon Y_\eta\rightarrow Y$ and $i\colon X\hookrightarrow Y_\eta$ a closed immersion. Define $X'$ as the zariski-closure of $j(i(X))$ in $Y$ with the reduced structure.

Now I want to show, that $X'\rightarrow C$ is a projective variety with generic fiber $X$.

As $X'\rightarrow Y$ should be a closed immersion it is clear, that $X'\rightarrow C$ is projectiv. Because $X$ is irreducible it follows, that $X'$ is irreducible, too. And because it is reduced, we get integer.

Now to the generic fiber of $X'$. That's the point, I'm no longer sure, if my argumentation holds: Topologically, the fiber $X'\times_C\text{Spec} K$ is $\{x\in X'\mid x\mapsto\eta$ under $X'\rightarrow C\}$, and because $Y_\eta=Y\times_C\text{Spec }K$ the fiber $X'\times_C\text{Spec }K$ should be the smallest closed set in $Y_\eta$, that contains $i(X)$. But $i$ is a closed immersion, and therefore $i(X)=X=X'\times_C\text{Spec }K$ topologically. If I could show that the fiber is reduced, it should follow, that $X\cong X'\times_C\text{Spec }K$. But here's the point I'm stuck.

Thanks for your help in advance.

EDIT Local ring on generic fiber Here is the proof, that the local rings of the fiber are reduced (because $X'$ is reduced). Now the only question is, if my topological argumentation is right.


Your Answer

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

Browse other questions tagged or ask your own question.