According to page 158 of Andreas Gathmann's notes on Algebraic Geometry, http://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2002/main.pdf, if we have some projective plane curve $X$ defined by an homogeneous equation of degree $d$ with coordinates $x_0,x_1,x_2$ not containing the point $(0:0:1)$, it states that $$H^1(X,\mathcal{O}_X)=\left\{\dfrac{x_2^i}{x_0^jx_1^k}:0\le i \le d-1 , j > 0,k > 0, \textrm{ and }i=j+k\right\}$$ and that the arithmetic genus is $\displaystyle{\dfrac{(d-1)(d-2)}{2}}$.

In general, how would one go about finding $H^1(X,\mathcal{O}_X)$ if $X$ is not defined by the equation of a curve (for example, say I want to find $H^1(X,\mathcal{O}_X)$ for the smooth projective model of the singular affine curve $y^2=x^4+x^5$)?

If we let $\bar{X}$ be the smooth projective model of $X$, can we use $H^1(X,\mathcal{O}_X)$ to find a basis for $H^1(\bar{X},\mathcal{O}_\bar{X})$? I suspect such a basis should be a subset of the basis of $H^1(X,\mathcal{O}_X)$, but I'm not too sure.

In a sense $H^i(X, \mathcal O_X)$ only computes "arithmetical information" about $X$, meaning that it doesn't care whether the curve is singular (even reducible). If $X$ is a projective hypersurface (or a curve), all that matters is the degree of the defining equation. In some sense, $H^i(X,\mathcal O_X)$ captures the combinatorics of the defining equations.

What you're asking about is how to compute $H^1(X, \mathcal O_X)$ when $X$ is the desingularization of a singular curve. In general it is hard to use Cech cohomology to do this, but there are formulas relating the geni of singular curves with their resolutions. They depend on the number of singularities and of which kind they are.

In this case, the homogenization of the equation is $z^3y^2=x^4z+x^5$. By using the Jacobian criterion, we find that the projectivization has two singularities. One at the origin, and one at infinity.

Then there is a formula saying that $$g = \frac{(d-1)(d-2)}{2} - \delta_0 -\delta_\infty,$$ where the $\delta_i$ are the "delta invariants", see this MathOverflow answer for more details. What we need to know, is that $\mu=2 \delta-b+1$, where $\mu$ is the Milnor number of the singularity and $b$ is the number of branches. This formula is known as the Milnor-Jung formula.

The Milnor number can be computed by the formula $$\dim_{\mathbb C} \mathbb C[[x,y]]/(f_x,f_y),$$ where $f_x,f_y$ are the partials of the defining local equation. The number $b$ of branches is the number of irreducible components of $f$ in the completion $\mathbb C[[x,y]]$.

So let us compute these numbers for the two singularities. For the singularity at the origin, the local equation is $y^2-x^4-x^5=0$. But notice that $y^2=x^4(1+x)$. Near the origin, the term $1+x$ is a unit, so that in the local ring we can divide by $1+x$. Hence the singularity locally looks like $y^2-x^4=(y-x^2)(y+x^2)$. Since $f_x=-4x^3$ and $f_y=2y$, the Milnor number is easily seen to be $3$. By the factorization, the number of branches is $2$. Hence by the Milnor-Jung formula, we have $3=2\delta_0-2+1$, so that $\delta_0=2$.

To compute $\delta_\infty$, we use the equation $z^3=x^4z+x^5$. This is much harder to do directly, so I used another definition of $\delta$ instead: it is the length of the module $\mathcal O_{\overline X,0}/\mathcal O_{X,0}$, wher the first ring is the local ring of the normalization of the curve at the singularity. This can be computed with Macaulay2 by finding the integral closure. Then one counts which basis elements come from the singular ring. In this case, we find that the degree is $\delta_\infty = 4$.

Hence $g=6-2-4=0$ (as can be confirmed by using Maple for example).

• If I'm not mistaken, you're saying that $H^1(X,\mathcal{O}_X)$ doesn't "care" whether the curve is singular or not, so $H^1(X,\mathcal{O}_X)$ should be the same as $H^1(\bar{X},\mathcal{O}_\bar{X})$. Does this mean that the dimension of $H^1(X,\mathcal{O}_X)$ is always of the form $\dfrac{(d-1)(d-2)}{2}$ (if $X$ corresponds to any projective planar curve)? Aug 3, 2016 at 14:53
• @whetham The answer to second question is "yes" (the genus-degree formula always holds). However, $H^1(X,\mathcal O_X)$ is not necessarily the same as $H^1(\overline X, \mathcal O_{\overline X})$. If $X$ is a cusp, then $H^1=1$, but its normalization is a rational curve, so $H^1=0$. Aug 3, 2016 at 15:17

Let $\overline{X}$ be the normalization of $X$, and let $\pi: \overline{X} \to X$ be the natural map. This map is finite, hence both proper and affine, and so $\pi_*\mathcal O_{\overline{X}}$ is a coherent sheaf of algebras on $X$, which I'll denote by $\mathcal A$. The general theory of affine morphisms shows that we may recover $\overline{X}$ as the relative Spec of $\mathcal A$ over $X$, and also that $H^1(\overline{X}, \mathcal O_{\overline{X}}) = H^1(X,\mathcal A)$.

Thus we may restrict attention to $\mathcal A$. Somewhat more intrinsically, the sections of $\mathcal A$ over an open subset $U$ of $X$ will be the normalization of $\mathcal O_X(U)$ in $K(X)$.

There is a short exact sequence $0 \to \mathcal O_X \to \mathcal A \to \Delta \to 0,$ where $\Delta$ is a sum of sky-scraper sheaves living at the various singular points of $X$. In particular, the higher cohomology of $\Delta$ vanishes, and so there is a surjection $$H^1(X,\mathcal{O}_X) \to H^1(X,\mathcal A)$$ (rather than an inclusion in the opposite direction, as you had guessed), whose kernel is equal to $H^0(\Delta)$.

Computing $\Delta$ is a local problem around the singular points, and the length of the sky-scraper at a singular point $P$ depends on how bad the singularity at $P$ is.