I'm trying to calculate the genus of a curve over a non-algebraically closed field via Riemann-Hurwitz.
Let $K = \mathbb Q(t)$ with $t$ transcendental, and let $F$ be the extension of $K$ obtained by adjoining a root of
$$f(x) = x^2 – (t^2 – 10t – 5)$$
Since $K$ is associated to projective space, it has genus zero. Let $g$ be the genus of a smooth curve with function field isomorphic to $F$. Then Riemann-Hurwitz says that
$$2g -2 = 2*(-2) + \sum_{P} e_P – 1$$
$$g = -1 + \frac 12 \sum_{P} e_P – 1$$
From the discriminant, it looks like the curve is ramified at two points, $\infty$ and $(t^2 – 10t – 5)$ with ramification index $2$ at each. This gives $g=0$.
If I base change to $\mathbb Q(\alpha)$ where $\alpha$ is a root of $t^2 – 10t – 5$, it seems that the map would now be ramified at three points: $\infty$, $t-\alpha$, and $t-\alpha'$, the conjugate of $\alpha$, all with index $2$. But that would make the genus $1/2$ which is absurd, on top of the fact that I thought genus was a geometric invariant.
Why do the two seem to work out differently, and especially what is going wrong with the last computation?
Best Answer
Here is the statement of the Riemann-Hurwitz Theorem I alluded to in the comments. (Rosen, Number Theory in Function Fields, Theorem 7.16, p. 90).
Theorem. Let $L/K$ be a finite, separable, geometric extension of function fields. Then $$ \DeclareMathOperator{\D}{\mathfrak{D}} \DeclareMathOperator{\P}{\mathfrak{P}} 2g_L - 2 = [L:K] (2g_K - 2) + \deg_L(\D_{L/K}) $$ where $\mathfrak{D}_{L/K}$ is the different ideal.
If all the ramified primes of $L$ are tamely ramified (which is the case here since the ground field has characteristic $0$), then $\D_{L/K} = \sum_{\P} (e(\P/P) - 1) \P$, so the formula becomes $$ 2g_L - 2 = [L:K] (2g_K - 2) + \sum_{\P} (e(\P/P) - 1) \deg_L(\P) \, . $$
Turning to your example, your mistake is that $F$ is not ramified above $\infty$. A geometric way to see this is the following. Homogenizing the curve defining $F$, we obtain the curve $C: X^2 - (Y^2 - 10YZ - 5Z^2) = 0$, where $x = X/Z$ and $t = Y/Z$, and we are considering the map $\pi: C \to \mathbb{P}^1$, $[X:Y:Z] \mapsto [Y:Z]$. To compute $\pi^{-1}([1:0])$, we plug $Z = 0$ into the equation for $C$, obtaining $$ 0 = X^2 - Y^2 = (X-Y)(X+Y) $$ so $\pi^{-1}([1:0]) = \{[1:1:0], [1:-1:0]\}$. Since $\sum_i e_i f_i = 2$ by the fundamental identity, then $f_i = e_i = 1$, so $\pi$ is unramified above $\infty$.
For a more function field theoretic approach, let $s = 1/t$ and $r = x/t = xs$. Then maximal order of $F$ at infinity is $R := \frac{\mathbb{Q}[r,s]}{(r^2 - (1 - 10s - 5s^2))}$. To determine the splitting above $\infty$, we examine how $sR$ factors. Using the equation defining $R$, we find $sR = (r-1,s)(r+1,s)$, and these primes are distinct, so $F$ is unramified above $(s)$.
Let $\P = (x)$ and $P = (t^2 - 10t - 5)$. The residue field of $\P = (x)$ is \begin{align*} \frac{\mathbb{Q}[t,x]/(x^2 - (t^2 - 10t - 5))}{(x)} \cong \frac{\mathbb{Q}[t]}{(t^2 - 10t - 5)} \end{align*} which has dimension $2$ as a $\mathbb{Q}$-vector space, so $\deg_L(\P) = 2$.
Applying Riemann-Hurwitz, we have \begin{align*} 2g_L - 2 = 2 (2 \cdot 0 - 2) + (e(\P/P) - 1) \deg_L(\P) = -4 + (2-1) \cdot 2 = -2 \end{align*} so $g_L = 0$, as we had hoped.