What is the genus of an unramified extension of a function field?

Question

Suppose L/K is a degree $n$m separable, geometric extensions of function fields where both function fields have constants $\mathbb{F}_q$. Suppose also that $L$ is unramified over $K$. Let $g_L$ be the genus of $L$ and let $g_K$ be the genus of $K$.

Riemann-Hurwitz tells us that $$2g_L - 2 = n(2g_K -2) + \text{deg}_L\left(\text{Diff}(L/K)\right)$$ where $\text{Diff}(L/K)$ is the different divisor. So $(\text{Diff}(L/K) = \sum_{\frak{p}}c_{\frak p}\frak p$ where all but finitely many of the $c_{\frak p}$s are zero. Further, since $L/K$ is an unramified extensions here, $c_{\frak p}=0$ for all primes $\frak p$.

Then by Riemann-Hurtiz we have then that $g_L=\frac{n}{2}(2g_K-2)+1$.

Two questions:

1. Is this computation correct?

2. If so, suppose $g_K=0$. If $n>1$ we have $g_L < 0$. Does this mean that genus $0$ function fields don't have unramified extensions?

Answer 1

Yes, this is true.

• This is analogous to the result from algebraic number theory that any algebraic extension of $\Bbb{Q}$ is ramified above some prime $p$.
• The genus zero rational function field $K=\Bbb{F}_q(x)$ has no unramified extensions with $\Bbb{F}_q$ as its field of constants. In many cases this can be seen directly. For example, when $q$ is odd, a quadratic extension $L/K$ is of the form $L=K(y)$ with $y^2=f(x)$, $f(x)\in K\setminus \Bbb{F}_q$ square-free. It is then easy to see that the places associated to factors of the numerator or the denominator of $f(x)$ ramify.
• Another consequence is that an unramified extension of the function field of an elliptic curve (so $g=1$) also has genus one. IIRC this has applications to the study of morphisms between elliptic curves.