How does one calculate genus of an algebraic curve?
Hmm. I was hoping someone who actually knows algebraic geometry would write an answer to this question.
First, some intuition. As it turns out, complex projective non-singular algebraic curves are the same thing as (connected) compact Riemann surfaces, which topologically are compact oriented surfaces. By the classification of compact surfaces, such surfaces are uniquely classified by a single number, their genus $g$, which counts how many holes there are in the surface. More precisely, for every $g$ there is an oriented surface $S_g$ which is the connected sum of $g$ tori (that is, it's a "doughnut with $g$ holes"), and every (connected) compact orientable surface is homeomorphic to $S_g$ for a unique $g$.
The genus $g$ has several equivalent definitions, and some of these generalize to algebraic geometry where we do not have direct access to topological information. Unfortunately, none of them are particularly easy to describe. An excellent introduction to this subject can be found in Fulton's Algebraic Curves.
So the idea is clear for non-singular curves. However, the genus turns out to be a birational invariant of curves (in particular, invariant under deletion of finitely many points), so it is possible to extend the definition of the genus to singular curves by declaring the genus of a singular curve to be the genus of a non-singular curve birational to it.
Example. Consider the singular curve $y^2 = x^3$ of degree $3$ in $\mathbb{P}^2(\mathbb{C})$ (equivalently $\mathbb{A}^2(\mathbb{C})$). It has a singular point at the origin of order $2$. Now, a non-singular curve of degree $3$ in $\mathbb{P}^2(\mathbb{C})$ has genus $1$ (see elliptic curve), but this curve doesn't: in fact, using the birational map $t \mapsto (t^2, t^3)$ we see that this curve is birational to $\mathbb{P}^1(\mathbb{C})$, hence has genus $0$.
So we see from the above that, roughly speaking, singularities decrease the "expected" genus of a curve (where "expected" means the number $\frac{(d-1)(d-2)}{2}$ that one gets from the genus-degree formula). Exactly how much singularities decrease the expected genus appears to me to be a somewhat complicated question and I am not the one to discuss it in detail. However, for "ordinary" singular points (I am not sure exactly what this means) of order $r$ it seems that the genus gets decreased by $\frac{r(r-1)}{2}$. So the genus in your first example is $$\frac{9 \cdot 8}{2} - 3 \frac{5 \cdot 4}{2} - \frac{4 \cdot 3}{2} = 36 - 30 - 6 = 0$$
and the genus in your second example is $$\frac{4 \cdot 3}{2} - \frac{3 \cdot 2}{2} - 3 \frac{2 \cdot 1}{2} = 6 - 3 - 3 = 0.$$
I am not an algebraic geometer either, but I tried to figure out how to compute the genus of an algebraic curve to satisfy my own curiosity. Although I have yet to succeed, I would like to share what I did learn.
First, those still getting used to projective space and homogeneous coordinates should read the first two sections of the appendix in Rational Points on Elliptic Curves by Silverman. It provides both motivation and intuition for these concepts.
Below I attempt to explain how to compute the genus by hand. Alternatively, one can use a computer algebra system like Maple to compute the genus.
This answer by Vogler on The Math Forum provided by Hans in a comment is indeed helpful. It explains almost everything in a very accessible way. That answer is based on Algebraic Curves by Walker (see sections 7.1 to 7.5). Another reference is Algebraic Curves: An Introduction to Algebraic Geometry by Fulton, which is freely available online (see section 7.5). These sections I point out in both books deal with the most difficult part of computing the genus, which is how to handle non-ordinary singularities.
With only ordinary singularities, things are much easier. Let $f(x,y) = 0$ define a nonsingular algebraic curve $C$ of degree $d$ with just $n$ ordinary singular points $p_i$ (for $1 \le i \le n$), where $p_i$ has multiplicity $r_i$. Then $$\operatorname{genus}(C) = \frac{(d-1)(d-2)}{2} - \sum_{i=1}^n \frac{r_i (r_i - 1)}{2}.$$
A point (in dehomogenized coordinates) is singular if $$f(x,y) = \partial_x f(x,y) = \partial_y f(x,y) = 0.$$ Let $(a,b)$ be a singular point. To determine the order of $(a,b)$, compute $f(a + x t, b + y t)$. Then the order of $(a,b)$ is the minimum value for $r$ such that $g(x,y) t^r$ is not identically zero. Now write $g(x,y)$ as $y^r h(x/y)$. Then $(a,b)$ is an ordinary singularity if $\gcd(h, h') = 1$ and is non-ordinary otherwise.
(Note that this explanation only mentions the variables $x$ and $y$, but there is another variable $z$ that is implicitly set to 1. Don't forget to consider the points "at infinity", which is when $z=0$. Read the appendix by Silverman if this isn't clear.)
As Mariano said in a comment, an ordinary singularity is a point with distinct tangents (and is non-ordinary if any tangent appears more than once). To get a feel for this, see the example figures by Fulton on page 32 (or the examples by Walker on page 57).
I am completely unsure what to do if the curve is reducible.
To compute the genus of an irreducible algebraic curve with non-ordinary singularities, we transform it into another algebraic curve with the same genus and no non-ordinary singularities using a so-called birational transformation. In contrast to the explanations above, this part is best explained in homogenous coordinates.
This transformation is obtained by repeatedly performing two steps. In the first step, we transform $C$ to a new curve $C'$ satisfying several properties. Vogler states these properties (with my paraphrasing) as follows. Let $p=(a,b,c)$ be a non-ordinary singular point of multiplicity $r$. Then
- $p=(1,0,0)$ in projective coordinates;
- The points $(0,1,0)$ and $(0,0,1)$ are not on $C'$;
- The line $x = 0$ does not intersect $C'$ at any singular point;
- The lines $y = 0$ and $z = 0$ do not intersect $C'$ at any singular point other than $p=(1,0,0)$ of multiplicity $r$.
Fulton says that a curve satisfying these conditions is in excellent position (see page 90).
The first condition is easy to satisfy, as the curve $C_1$ defined by $$f'(x,y,z) = f(a x, y + b, z + c) = 0$$ has this property. However, I am unsure how to systematically obtain further transformations to satisfy the other properties while maintaining the previous ones (even if these last three conditions should typically hold as Vogler points out). Vogler gives one example of such transformations while Fulton and Walker leave this step as an exercise for the reader. If someone could modify my answer by explaining this step, that would be fantastic.
Now given that our curve $C'$ defined by $f'(x,y,z)=0$ satisfies the above properties, we transform to a new curve $C''$ defined by $f''(x,y,z)=0$, where $$f'(yz,xz,yz) = x^r f''(x,y,z).$$ Then we repeat this whole process starting with $C''$ until we obtain a curve with no non-ordinary singularities, at which point we can compute the genus using the formula above.