Genus of the curve $y^2=x^3+x^2$ via Riemann-Hurwitz How should one apply the Riemann-Hurwirz formula to calculate the genus of the curve $y^2=x^3+x^2$? If I project the x-coordinate to the projective line, I get
$2g-2=2(2(0)-2)+3$
since there are 3 ramification points at 0, -1, and infinity, each with multiplicity 2. But this does not give an integer value for g. What am I doing wrong?
 A: The Riemann-Hurwitz formula refers to the geometric genus (which is defined as $h^0(X, \Omega_X^1)$). 
However, the usual Riemann-Hurwitz formula assumes that the curves in question are smooth. However, there is a Riemann-Hurwitz formula for singular curves:
$$
2p_a(X) -2 = d(2p_a(Y)-2)+deg R+2 \sum_{P \in X} \delta_P -2d \sum_{Q \in Y} \delta_Q, 
$$
where $\delta_P,\delta_Q$ are the orders - 1 of the singularities on $X$ and $Y$. I found this formula here (it is not difficult to prove). $p_a(X)$ is the arithmetic genus, which is defined as $1-P_X(0)$, where $P_X$ is the Hilbert polynomial of $Y$.
In your case, $Y=\mathbb P^1$ is smooth. The curve $X$ has a single singular point, which is a node, hence of order $2$, so $\delta_P=1$.
The ramification locus is of degree $2$: for singular curves, the ramification divisor is defined for the normalization of $X$, in which the double point has been separated into two points. Hence the only ramification comes from $x=-1$ and at infinity. See the linked article for details. Since $\mathbb P^1$ has genus $0$, the formula says
$$
2p_a(X) - 2 = 2(0-2)+3+2,
$$
hence $p_a(X)=1$, which is expected for a degree $3$ curve.
