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Suppose that $f:X\rightarrow Y$ is a morphism of finite type schemes over an algebraically closed field $k$. Assume that for every closed point $y\in Y$ the fiber $X_y$ of $f$ in is isomorphic to $\mathbb{P}^n_k$. Now let $y_0\in Y$ be any(nonclosed) point. Is it true that: $X_{y_0}\cong \mathbb{P}^n_{k(y_0)}$?

Suppose now that $\Gamma$ is a scheme over $\mathrm{Spec} (\mathbb{Z})$. For any field $L$ let $\Gamma_L$ be its base change to $L$. How about the previous question with $\mathbb{P}^n$ replaced by $\Gamma$?

asker
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1 Answers1

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The answer is no for the first question, hence the second since the first was a special case.

Let $Y = \mathbb{A}^2_k \setminus V(st)$, the affine plane minus the two axes, and $X = V(x^2 + sy^2 = tz^2) \subset Y \times_k \mathbb{P}^2_{k}$. Let $f \colon X \to Y$ be the projection onto the first coordinate. Then, every fiber $V(x^2 + ay^2 = bz^2) \subset \mathbb{P}^2_k$ over a closed point $(a,b)$ is a non-degenerate conic in $\mathbb{P}^2_k$, hence isomorphic to $\mathbb{P}^1_k$. Over the generic point $\xi \in \mathbb{A}^2_k$, however, the fiber is $$V(x^2 + sy^2 = tz^2) \subset \operatorname{Spec}k(s,t) \times_k \mathbb{P}^2_k = \mathbb{P}^2_{k(s,t)}$$ which has no rational points, hence is not isomorphic to $\mathbb{P}^1_{k(s,t)}$. For if $x_0,y_0,z_0 \in k(s,t)$ satisfied $x_0^2 + sy_0^2 - tz_0^2 = 0$, then the terms of lowest degree in $x_0,y_0,z_0$ cannot cancel each other out.

Takumi Murayama
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