# $C\times \mathbb{P}^1$ birational to $C'\times \mathbb{P}^1$ implies that $C$ and $C'$ are isomorphic.

Let $C$ and $C'$ be algebraic curves (complete and nonsingular over an algebraically closed field). Let us suppose that $C\times \mathbb{P}^1$ is birational to $C'\times \mathbb{P}^1$. I want to prove that $C$ and $C'$ are isomorphic.

Birational algebraic curves are isomorphic. Therefore, it suffices to prove that $C$ and $C'$ are birational.

Let $U\subset C\times \mathbb{P}^1$ and $U'\subset C'\times \mathbb{P}^1$ be such that $U\simeq U'$. I denote $\pi:C\times \mathbb{P}^1\rightarrow C$ and $\pi':C'\times \mathbb{P}^1\rightarrow C'$ to the first projections. I guess that $$\pi(U)\rightarrow U\rightarrow U'\rightarrow\pi'(U')$$ gives a birrational map between open subsets of $C$ and $C'$, but I am not able to prove it.

If $g(C)=g(C')=0$, there is nothing to prove. So, assume that $g(C')>0$. Then, we have the map $U\to U'\to C'$. The composite is dominant. For any point $p\in C$, we get a morphism $U\cap (\{p\}\times\mathbb{P}^1)\to C'$. Since there are no non-constant map from a genus zero curve to a positive genus curve, we see that this map is constant. This means, the map factors through a rational dominant map $C\to C'$. Thus $g(C)\geq g(C')>0$ and by reversing roles we get $g(C')\geq g(C)$. So, $g(C)=g(C')$. This immediately finishes the proof when $g\geq 2$, since any map of such curves is always an isomorphism.
Finally to the $g=1$ case. Then, the induced map $C\to C'$ is finite and thus etale. We want to show its degree is one. If not, take a general point $p\in C'$ and let $P\neq Q$ be inverse images. Then, we get a birational morphism $\{P,Q\}\times \mathbb{P}^1\cap U\to \{p\}\times \mathbb{P}^1\cap U'$. Easy to check that this map can not be an injection, since both $\{P\}\times\mathbb{P}^1$ and $\{Q\}\times\mathbb{P}^1$ maps dominantly to the image.
• could you elaborate out the argument why if we know that for every $p \in C$ the induced morphism $U\cap (\{p\}\times\mathbb{P}^1)\to C'$ is constant then the map $U\to U'\to C'$ factors through rational map $C \to C'$ as scheme morphism? The existence of such factorization between underlying sets is clear by set theoretical quotient property, but I don't know an argument why this factorization exactly also happens in category of schemes. sorry, if I oversee an obvious reason Mar 31, 2021 at 0:40