Return to Answer

Let $k$ be an algebraically closed field.

Let $f:X\to \mathbb{P}^1$ be a smooth proper morphism with fibres of dimension one. Note that the fibres of $f$ are geometrically connected by Stein factorization and the fact that $\mathbb{P}^1$ is simply connected (Riemann-Hurwitz).

Theorem. The morphism $f$ is isotrivial.

Proof. Let $g$ be the genus of the fibres. Clearly, if $g=0$, then all geometric fibres are isomorphic. Thus, if $g=0$, then $f$ is isotrivial. Next, assume that $g=1$. Then, as the moduli space of elliptic curves is affine, the moduli map associated to the Jacobian of $f$ is constant, so that $Jac(f)$ has constant $j$-invariant. It follows that $f$ is isotrivial. Finally, it is a theorem of Moret-Bailly that any genus $g>1$ curve over $\mathbb{P}^1$ is isotrivial; see Lemme 5 in [1]. QED

[1] Laurent Moret-Bailly. Un théorème de pureté pour les familles de courbes lisses. C. R. Acad. Sc. Paris, t. 300, Serie I, n. 14, 1985.

Remark. If $k$ has characteristic zero, then Moret-Bailly's theorem is due to Parshin. Parshin's theorem follows from the "hyperbolicity" of the moduli stack of genus $g$ ($g>1$) curves. There are many notions of "hyperbolicity" lurking around there, and all of them imply the statement you want (in characteristic zero). (More generally, if $k$ is of characteristic zero, any smooth proper morphism $X\to \mathbb{P}^1$ whose fibres are smooth proper connected varieties with ample canonical bundle is isotrivial by work of Kovács, Migliorini, Viehweg-Zuo.)

Let $F$ be a fibre of $f$ over a closed point of $\mathbb{P}^1_k$.

If $g>1$, then the Isom-scheme $Isom(F\times \mathbb{P}^1, X)\to \mathbb{P}^1$ is finite etale and thus trivial. Thus, $f$ is trivial.

If $g=1$, then probably the family $X\to \mathbb{P}^1$ is trivial Will Sawin's argument below.

If $g=0$, as Daniel Loughran explains in the comments below, the (isotrivial) morphism $f$ $f$ has a section by Tsen's theorem, so that $X$ is a Hirzebruch surface.

Let $k$ be an algebraically closed field.

Let $f:X\to \mathbb{P}^1$ be a smooth proper morphism with fibres of dimension one. Note that the fibres of $f$ are geometrically connected by Stein factorization and the fact that $\mathbb{P}^1$ is simply connected (Riemann-Hurwitz).

Theorem. The morphism $f$ is isotrivial.

Proof. Let $g$ be the genus of the fibres. Clearly, if $g=0$, then all geometric fibres are isomorphic. Thus, if $g=0$, then $f$ is isotrivial. Next, assume that $g=1$. Then, as the moduli space of elliptic curves is affine, the moduli map associated to the Jacobian of $f$ is constant, so that $Jac(f)$ has constant $j$-invariant. It follows that $f$ is isotrivial. Finally, it is a theorem of Moret-Bailly that any genus $g>1$ curve over $\mathbb{P}^1$ is isotrivial; see Lemme 5 in [1]. QED

[1] Laurent Moret-Bailly. Un théorème de pureté pour les familles de courbes lisses. C. R. Acad. Sc. Paris, t. 300, Serie I, n. 14, 1985.

Remark. If $k$ has characteristic zero, then Moret-Bailly's theorem is due to Parshin. Parshin's theorem follows from the "hyperbolicity" of the moduli stack of genus $g$ ($g>1$) curves. There are many notions of "hyperbolicity" lurking around there, and all of them imply the statement you want (in characteristic zero). (More generally, if $k$ is of characteristic zero, any smooth proper morphism $X\to \mathbb{P}^1$ whose fibres are smooth proper connected varieties with ample canonical bundle is isotrivial by work of Kovács, Migliorini, Viehweg-Zuo.)

Let $F$ be a fibre of $f$ over a closed point of $\mathbb{P}^1_k$.

If $g>1$, then the Isom-scheme $Isom(F\times \mathbb{P}^1, X)\to \mathbb{P}^1$ is finite etale and thus trivial. Thus, $f$ is trivial.

If $g=1$, then probably the family $X\to \mathbb{P}^1$ is trivial Will Sawin's argument below.

If $g=0$, as Daniel Loughran explains in the comments below, the (isotrivial) morphism   $f$ has a section by Tsen's theorem, so that $X$ is a Hirzebruch surface.

I assume the base field is of characteristic zero below.

Let $f:X\to \mathbb{P}^1$ be a smooth proper morphism with fibres of dimension one. Note that the fibres of $f$ are geometrically connected by Stein factorization and the fact that $\mathbb{P}^1$ is simply connected.

Theorem.  $f$ is isotrivial.

Proof. Let $g$ be the genus of the fibres. Clearly, if $g=0$, then all geometric fibres are isomorphic, so $f$ is isotrivial by definition. Next, assume that $g=1$. Then, as the moduli space of elliptic curves is affine, the moduli map associated to the Jacobian of $f$ is constant, so that $Jac(f)$ has constant $j$-invariant. It follows that $f$ is isotrivial. Finally, it is a theorem of Parshin that any genus $g>1$ curve over $\mathbb{P}^1$ is isotrivial. (More generally, any smooth proper morphism $X\to \mathbb{P}^1$ whose fibres are smooth proper connected varieties with ample canonical bundle is isotrivial by work of Kovács, Migliorini, Viehweg-Zuo.) Parshin's theorem follows from the "hyperbolicity" of the moduli stack of genus $g$ ($g>1$) curves. There are many notions of "hyperbolicity" lurking around there, and all of them imply the statement you want. QED

Thus, $f$ is isotrivial. Let $F$ be a fibre (over a closed point of $\mathbb{P}^1$).

If $g>1$, then the Isom-scheme $Isom(F\times \mathbb{P}^1, X)\to \mathbb{P}^1$ is finite etale and thus trivial. Thus, $f$ is trivial.

If $g=1$, then probably the family $X\to \mathbb{P}^1$ is trivial. But I can't remember the correct argument at the moment.

Comment: If $k$ is of characteristic $p$, then it follows from a theorem of Moret-Bailly, that for $g>1$, $f$ is isotrivial; ses Lemme 5 in "Un théorème de pureté pour les familles de courbes lisses" in the C. R. Acad. Sc. Paris, t. 300, Serie I, n. 14, 1985. So, the assumption that $k$ is of char. zero is not necessary.

Let $k$ be an algebraically closed field.

Let $f:X\to \mathbb{P}^1$ be a smooth proper morphism with fibres of dimension one. Note that the fibres of $f$ are geometrically connected by Stein factorization and the fact that $\mathbb{P}^1$ is simply connected (Riemann-Hurwitz).

Theorem. The morphism $f$ is isotrivial.

Proof. Let $g$ be the genus of the fibres. Clearly, if $g=0$, then all geometric fibres are isomorphic. Thus, if $g=0$, then $f$ is isotrivial. Next, assume that $g=1$. Then, as the moduli space of elliptic curves is affine, the moduli map associated to the Jacobian of $f$ is constant, so that $Jac(f)$ has constant $j$-invariant. It follows that $f$ is isotrivial. Finally, it is a theorem of Moret-Bailly that any genus $g>1$ curve over $\mathbb{P}^1$ is isotrivial; see Lemme 5 in [1]. QED

[1] Laurent Moret-Bailly. Un théorème de pureté pour les familles de courbes lisses. C. R. Acad. Sc. Paris, t. 300, Serie I, n. 14, 1985.

Remark. If $k$ has characteristic zero, then Moret-Bailly's theorem is due to Parshin. Parshin's theorem follows from the "hyperbolicity" of the moduli stack of genus $g$ ($g>1$) curves. There are many notions of "hyperbolicity" lurking around there, and all of them imply the statement you want (in characteristic zero). (More generally, if $k$ is of characteristic zero, any smooth proper morphism $X\to \mathbb{P}^1$ whose fibres are smooth proper connected varieties with ample canonical bundle is isotrivial by work of Kovács, Migliorini, Viehweg-Zuo.)

Let $F$ be a fibre of $f$ over a closed point of $\mathbb{P}^1_k$.

If $g>1$, then the Isom-scheme $Isom(F\times \mathbb{P}^1, X)\to \mathbb{P}^1$ is finite etale and thus trivial. Thus, $f$ is trivial.

If $g=1$, then probably the family $X\to \mathbb{P}^1$ is trivial Will Sawin's argument below.

If $g=0$, as Daniel Loughran explains in the comments below, the (isotrivial) morphism $f$ $f$ has a section by Tsen's theorem, so that $X$ is a Hirzebruch surface.

I assume the base field is of characteristic zero below.

Let $f:X\to \mathbb{P}^1$ be a smooth proper morphism with fibres of dimension one. Note that the fibres of $f$ are geometrically connected by Stein factorization and the fact that $\mathbb{P}^1$ is simply connected.

Theorem. $f$ is isotrivial.

Proof. Let $g$ be the genus of the fibres. Clearly, if $g=0$, then all geometric fibres are isomorphic, so $f$ is isotrivial by definition. Next, assume that $g=1$. Then, as the moduli space of elliptic curves is affine, the moduli map associated to the Jacobian of $f$ is constant, so that $Jac(f)$ has constant $j$-invariant. It follows that $f$ is isotrivial. Finally, it is a theorem of Parshin that any genus $g>1$ curve over $\mathbb{P}^1$ is isotrivial. (More generally, any smooth proper morphism $X\to \mathbb{P}^1$ whose fibres are smooth proper connected varieties with ample canonical bundle is isotrivial by work of Kovács, Migliorini, Viehweg-Zuo.) Parshin's theorem follows from the "hyperbolicity" of the moduli stack of genus $g$ ($g>1$) curves. There are many notions of "hyperbolicity" lurking around there, and all of them imply the statement you want. QED

Thus, $f$ is isotrivial. Let $F$ be a fibre (over a closed point of $\mathbb{P}^1$).

If $g>1$, then the Isom-scheme $Isom(F\times \mathbb{P}^1, X)\to \mathbb{P}^1$ is finite etale and thus trivial. Thus, $f$ is trivial.

If $g=0$, then $f$ is a Brauer-Severi scheme. But all Brauer-Severi schemes over $\mathbb{P}^1$ are trivial, as the Brauer group of $\mathbb{P}^1$ is trivial.

If $g=1$, then probably the family $X\to \mathbb{P}^1$ is trivial. But I can't remember the correct argument at the moment.

I assume the base field is of characteristic zero below.

Let $f:X\to \mathbb{P}^1$ be a smooth proper morphism with fibres of dimension one. Note that the fibres of $f$ are geometrically connected by Stein factorization and the fact that $\mathbb{P}^1$ is simply connected.

Theorem. $f$ is isotrivial.

Proof. Let $g$ be the genus of the fibres. Clearly, if $g=0$, then all geometric fibres are isomorphic, so $f$ is isotrivial by definition. Next, assume that $g=1$. Then, as the moduli space of elliptic curves is affine, the moduli map associated to the Jacobian of $f$ is constant, so that $Jac(f)$ has constant $j$-invariant. It follows that $f$ is isotrivial. Finally, it is a theorem of Parshin that any genus $g>1$ curve over $\mathbb{P}^1$ is isotrivial. (More generally, any smooth proper morphism $X\to \mathbb{P}^1$ whose fibres are smooth proper connected varieties with ample canonical bundle is isotrivial by work of Kovács, Migliorini, Viehweg-Zuo.) Parshin's theorem follows from the "hyperbolicity" of the moduli stack of genus $g$ ($g>1$) curves. There are many notions of "hyperbolicity" lurking around there, and all of them imply the statement you want. QED

Thus, $f$ is isotrivial. Let $F$ be a fibre (over a closed point of $\mathbb{P}^1$).

If $g>1$, then the Isom-scheme $Isom(F\times \mathbb{P}^1, X)\to \mathbb{P}^1$ is finite etale and thus trivial. Thus, $f$ is trivial.

If $g=1$, then probably the family $X\to \mathbb{P}^1$ is trivial. But I can't remember the correct argument at the moment.

Comment: If $k$ is of characteristic $p$, then it follows from a theorem of Moret-Bailly, that for $g>1$, $f$ is isotrivial; ses Lemme 5 in "Un théorème de pureté pour les familles de courbes lisses" in the C. R. Acad. Sc. Paris, t. 300, Serie I, n. 14, 1985. So, the assumption that $k$ is of char. zero is not necessary.