# Canonical Divisor of Product of Smooth Curves is Ample

Let $C_1, C_2$ be two curves (so a $1$-dimensional, proper k-scheme) wich are

• smooth

• from genus $h^0(\mathcal{O}_{C_i}) \ge 2$

Consider the surface $S:=C_1 \times C_2$.

The main goal is to see that the canonical / dualising sheaf $\omega_S$ is ample.

Following argument isn't clear to me:

Since $C_i$ are both smooth, their canonical / dualising sheaves are Kaehler bundles $${\displaystyle \omega _{C_i}:=\bigwedge ^{1}\Omega _{C_i}}$$

Consider the canonical projections

$$pr_i: S:=C_1 \times C_2 \to C_i$$

$$\omega_S = pr_1^*(\omega _{C_1}) \otimes pr_2^*(\omega _{C_2})$$

Now to my problem:

Why Segre embedding implies that $\omega_S$ is already ample?

My thoughts:

Since $\omega _{C_i}$ are ample they induce after tensoring with $n$ large enough morphisms $\phi_i: C_i \to \mathbb{P}^{n_i}$.

Segre provides a closed embedding

$$\mathbb{P}^{n_1} \times \mathbb{P}^{n_2} \to \mathbb{P}^m$$

How could this fact imply that $\omega _S$ is ample?

My first idea using concatenation argument failed since $pr_i ^*(\omega _{C_i})$ aren't ample since $pr_i$ aren't finite.

Does anybody have an idea how to use here Sigre to prove the claim?

Let be $f:E \to \mathbb{P}^1$ be a birational morphism.

Why it is then an isomorphism?

Does it something to do with the fact that $\mathbb{P}^1$ is normal?

Since $\omega_{C_1}$ is ample, there exists a positive integer $m_1$ such that $\omega_{C_1}^{\otimes m_1}$ is very ample, i.e. there exists a positive integer $m_1$ and a closed embedding $\phi_1 : C_1 \hookrightarrow\mathbb P^{n_1}$ such that $\omega_{C_1}^{\otimes m_1} =\phi_1^\star \mathcal O_{\mathbb P^{n_1}} (1)$.
Similarly, since $\omega_{C_2}$ is ample, there exists a positive integer $m_2$ such that $\omega_{C_2}^{\otimes m_2}$ is very ample, and there exists a closed embedding $\phi_2 : C_2 \hookrightarrow \mathbb P^{n_2}$ such that $\omega_{C_2}^{\otimes m_2} = \phi_2^\star \mathcal O_{\mathbb P^{n_2}}(1)$.
Hence $$\omega_{C_1 \times C_2}^{\otimes (m_1m_2)}= (\phi_1 \times \phi_2)^\star \left( \pi_1^\star \mathcal O_{\mathbb P^{n_1}}(m_2)\otimes \pi_2^\star \mathcal O_{\mathbb P^{n_2}} (m_1)\right),$$ where $\pi_1 : \mathbb P^{n_1} \times \mathbb P^{n_2} \to \mathbb P^{n_1}$ and $\pi_2 : \mathbb P^{n_1} \times \mathbb P^{n_2} \to \mathbb P^{n_2}$ are the canonical projections.
Now define $N: = \binom{n_1 + m_2}{n_1} + \binom{n_2 + m_1}{n_2}-1$, and consider the map $\varphi : \mathbb P^{n_1} \times \mathbb P^{n_2} \to \mathbb P^{N}$ defined as the composition of the $m_2$th and $m_1$th Veronese embeddings $$\mathbb P^{n_1} \hookrightarrow \mathbb P^{\binom{n_1 + m_2}{n_1} - 1}, \ \ \ \ \mathbb P^{n_2} \hookrightarrow \mathbb P^{\binom{n_2 + m_1}{n_2} - 1}$$ with the Segre embedding $$\mathbb P^{\binom{n_1 + m_2}{n_1} - 1} \times \mathbb P^{\binom{n_2 + m_1}{n_2} - 1} \hookrightarrow \mathbb P^{\binom{n_1 + m_2}{n_1} + \binom{n_2 + m_1}{n_2}-1} = \mathbb P^{N}.$$ It is well known that $$\pi_1^\star \mathcal O_{\mathbb P^{n_1}}(m_2)\otimes \pi_2^\star \mathcal O_{\mathbb P^{n_2}} (m_1) =\varphi^\star\mathcal O_{N}(1).$$
Thus we have exhibited a closed embedding $\psi : = \varphi \circ (\phi_1 \times \phi_2) : C_1 \times C_2 \hookrightarrow \mathbb P^N$, such that $$\omega_{C_1 \times C_2}^{\otimes (m_1m_2)}=\psi^\star \mathcal O_{N}(1).$$ Hence $\omega_{C_1 \times C_2}^{\otimes (m_1 m_2)}$ is very ample, and $\omega_{c_1 \times C_2}$ is ample.