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Given an algebraically closed field $k$ and a scheme $Z$ over $k$. Consider the projective scheme $X=\mathbb{P}^1_Z=\mathbb{P}_k^1\times_kZ$ and $Y\subset X$ a closed subscheme. Assume that $Y$ does not contain any closed fibre of the projection $p:X\rightarrow Z$, try to show that $\pi:=p|_Y:Y\rightarrow Z$ is an affine morphism, i.e. one can find an affine covering $\{U_i\}$ of $Z$ such that each $\pi^{-1}(U_i)$ is affine.

My attempt:

In the extremely special case $Z=\text{Spec}(k)$, one has $X=\mathbb{P}_k^1$ and $X_z\cong\mathbb{P}_k^1$. Here any closed subscheme $Y$ of $X$ satisfies the condition that $X_z\not\subseteq Y$. So we want the canonical map $Y\hookrightarrow\mathbb{P}_k^1$ to be affine, which is clearly true.

Now let $Z=\text{Spec}(A)$ over $k$, we also have $X_z\cong\mathbb{P}_k^1$ and now $X=\mathbb{P}_A^1$. We know that any closed subscheme of $\mathbb{P}_A^1=\text{Proj}(A[x_0,x_1])$ is determined by a homogeneous ideal $I\subset A[x_0,x_1]$, i.e. $$Y=V_+(I)=\text{Proj}(A[x_0,x_1]/(I))$$ Then I want to define a scheme $Y_k$ such that $Y=Y_k\times_kZ$. If I can do it, them I am done. However, I have no ideal how to apply the setting $X_z\not\subseteq Y$ here.

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    The map $\pi:Y \to Z$ is proper because $X \to Z$ is proper and $Y$ is closed in $X$. The fiber of $\pi$ over $z \in Z(k)$ is a closed subset of $\mathbb{P}^1_k$ that is not equal to $\mathbb{P}^1_k$, and therefore must be finite. So we've shown that $\pi$ is proper and quasi-finite hence finite, and finite morphisms are affine. – Pol van Hoften Mar 24 '21 at 04:28
  • @PolvanHoften Hi Pol, why the fibre of $\pi$ over $z\in Z$ must be finite? –  Mar 24 '21 at 04:56

2 Answers2

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Answer: For any closed point $t\in Z$ it follows the fiber $Y_t$ by definition is a strict closed subscheme $Y_t \subsetneq \mathbb{P}^1_{\kappa(t)} \subseteq \mathbb{P}^1_Z$, and since $dim(\mathbb{P}^1_{\kappa(t)})=1$ it follows $dim(Y_t)=0$ and hence $Y_t$ is finite.

hm2020
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In order to make it complete, I combine Pol van Hoften's comment and hm2020's answer here.

The map $\pi:Y\rightarrow X\rightarrow Z$ is proper because the projection $p:X=\mathbb{P}^1_Z\rightarrow Z$ is proper and $Y$ is closed in $X$ (the closed inclusion $i:Y\subset X$ is proper).

Pick any closed point $z\in Z$, one can see that the fibre $X_z\cong\mathbb{P}_k^1$. By our setting, $Y$ does not contain $X_z$, so the fiber $Y_z$ of $\pi$ over $z$ is a real closed subset of $X_z\cong\mathbb{P}_k^1$. It follows that $\dim(Y_z)=0$. It also follows that $Y_z$ is noetherian (closed subscheme of a noetherian scheme is noetherian). Therefore, $Y_z$ must be finite and thus the morphism $\pi:Y\rightarrow Z$ is quasi-finite.

A proper and quasi-finite morphism is finite, and finite morphisms are affine.

  • quote - zero dimensional noetherian scheme is finite: https://math.stackexchange.com/questions/474873/why-is-every-noetherian-zero-dimensional-scheme-finite-discrete –  Mar 24 '21 at 13:23