# Genus of $y^4=x^{14}+x$?

I want to calculate the genus of the algebraic curve $$y^4=x^{14}+x$$ by Riemann-Hurwitz formula.

I know there are $$42$$ branch numbers over all finite points by the discriminant of the curve. But i don't know how to calculate the branch number of infinity, because infinity is a singular point. I want to know how to resolve the singularities of infinity.

What's more, how can i use puiseux series to distinct the regular points, branch points and singular points?

The curve is a smooth compactification the cyclic covering of $$\mathbb{A}^1$$ branched over the divisor $$D \subset \mathbb{A}^1$$ defined by the equation $$x^{14} + x = 0$$. First, consider the double covering $$f \colon X \to \mathbb{P}^1$$ branched over $$D$$. This is a hyperelliptic curve of genus $$g$$, where $$2g + 2 = \deg(D) = 14$$, i.e., $$g = 6$$. Note that $$f^{-1}(D) = 2D'$$, where $$D' \subset X$$ is a divisor of degree $$14$$. Note also that $$\mathcal{O}_X(D') \cong f^*\mathcal{O}_{\mathbb{P}^1}(7)$$ is not divisible by 2 in $$\mathrm{Pic}(X)$$. However, the divisor $$D'' = D' + f^{-1}(\infty)$$ of degree 16 is divisoble by 2, because $$\mathcal{O}_X(D'') \cong f^*\mathcal{O}_{\mathbb{P}^1}(8).$$ Therefore, we can consider the double covering $$Y \to X$$ branched over $$D''$$. Clearly, over $$\mathbb{A}^1$$ the curve $$Y$$ gives the required cyclic covering. Moreover, it is smooth. Hence its genus can be computed by the Hurwitz formula as $$g(Y) = 2g(X) - 1 + \frac12 \deg(D'') = 2\cdot 6 - 1 + 8 = 19.$$
• @JiaMinxin: That's right, the preimage of $\infty$ consists of two points with simple ramification at each. – Sasha Nov 22 '19 at 6:43