Applying Riemann-Roch theorem prove that the Riemann sphere has no genus.

Question

I have recently studied Riemann-Roch theorem. So how can I prove that the Riemann sphere has no genus by applying Riemann-Roch theorem? Thanks in advance for any help...

Answer 1

Riemann-Roch implies that the degree of a canonical divisor on a compact Riemann surface of genus $g$ is $2g-2$.

On the other hand, a direct computation using differential forms shows that any canonical divisor on the Riemann sphere has degree $-2$, hence $g=0$.

Answer 2

Hint : show that if $p \sim q$ (as divisor) then $X \cong \Bbb P^1$. Show then that Riemann Roch implies that there is $p,q \in X$ with $p \sim q$.