Genus of extension $\mathbb{C}(T)(\sqrt{T^n + 1})$

Question

Let $k = \mathbb{C}$ and $K$ is the extension $\mathbb{C}(T)(\sqrt{T^n + 1})$ of $\mathbb{C}(T)$ with $n \ge 2$ an even integer. I suspect that the genus of $K$ is $(n - 2)/2$, but all attempts at showing this have not really gone anywhere. Could anybody tell me of an easy way to see this, assuming it is true?

Answer 1

Your extension field $K$ is the field of rational functions $\operatorname {Rat}(C)$ of the smooth affine algebraic curve $C\subset \mathbb A^2_k$ given by the equation $y^2=x^n+1$ .
Since $n$ is even, the closure $\bar C \subset \mathbb P^2_k$ given by the equation $Y^2Z^{n-2}=X^n+Z^{n}$.
This is not a smooth curve, but its normalization is smooth and yields a ramified covering space $$ f:(\bar C)^{\operatorname {nor}} \to \mathbb P^1$$ which is unramified over $\infty \in \mathbb P^1$.
The ramification points of $f$ are over the $n$ distinct roots of $T^n+1$ and the Riemann-Hurwitz theorem then yields the required genus $g$ : $$2g-2=2(2.0-2)+n$$ so that $g=\frac {n-2}{2}$, as you correctly predicted.
(Remember that the field of rational functions of $(\bar C)^{\operatorname {nor}}$ is the same as that of its open affine piece $C$, namely $K$))