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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?

Someone help please? Thanks.

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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 $$ g(Y) = 2g(X) - 1 + \frac12 \deg(D'') = 2\cdot 6 - 1 + 8 = 19. $$

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  • $\begingroup$ Thank you for your answer. Another question is: The curve as a complex manifold, how can i define the local coordinate of infinity?Are there two infinity with branch number 1? @Sasha $\endgroup$ – Jia Minxin Nov 22 '19 at 5:49
  • $\begingroup$ @JiaMinxin: That's right, the preimage of $\infty$ consists of two points with simple ramification at each. $\endgroup$ – Sasha Nov 22 '19 at 6:43
  • $\begingroup$ Thank you so much.@Sasha. $\endgroup$ – Jia Minxin Nov 22 '19 at 9:53

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