# Help Understanding Genus of a Plane Algebraic Curve

I have been attempting to teach myself algebraic geometry, and I am unfortunately confused about the geometric genus of an algebraic curve. My understanding is that from the degree of the curve, $$d$$, one computes $$(d-1)(d-2)/2$$ Then one subtracts $$r(r-1)/2$$ for each node, where $$r$$ is the multiplicity of that node.

Here is where I get confused. Suppose I have a polynomial $$P(x)$$ of degree $$d$$. Then the curve defined by $$Q(x,y) = P(x)-y = 0$$ is also of degree $$d$$. $$Q$$ is furthermore rational, because it can be parameterized by $$x=t$$ $$y=P(t)$$

Further, it seems that I can easily choose $$P(x)$$ such that $$Q(x,y)=0$$ has no nodes of multiplicity 2 or more, (am I mistaken?) and hence the genus of $$Q$$ would be $$(d-1)(d-2)/2$$. However, I also understand (incorrectly?) that curves are rational if and only if their geometric genus is $$0$$. Can someone help me with what I am missing?

Perhaps it would be useful to deal with a concrete example. What is the genus of this curve? $$y=x^4 + x^3 + x^2 + x + 1$$

• Your curve is singular at infinity. The concepts you're talking about are for compact curves, not affine curves, so you take the closure in $\Bbb P^2$ ... and ... Commented Nov 8, 2020 at 0:03
• Is the singularity at infinity of multiplicity 3? It seems it would need to be to reduce the genus by 3. Commented Nov 8, 2020 at 0:08
• Yes, you need to learn homogeneous coordinates are work out simple examples. Commented Nov 8, 2020 at 0:13
• Thank you. I just worked out that $$yz^3 = x^4+x^3z+x^2z^2+xz^3+z^4$$ does indeed have a singularity of multiplicity $3$ at $(0:1:0)$. Would you like to fill in the answer for credit? Commented Nov 8, 2020 at 0:15

If one extends the curve $$Q(x,y)=P(x)-y=-y+\sum_{i=0}^dc_ix^i=0$$ to the projective plane, one has $$Q^h(x,y,z)=-yz^{d-1}+\sum_{i=0}^dc_ix^iz^{d-i}=0$$ If one then dehomogenizes onto the $$xz$$ plane, one gets $$\hat{Q}(x,z) = -z^{d-1}+\sum_{i=0}^dc_ix^iz^{d-i}=0$$ for which $$\hat{Q}(0,0)=0$$ is clearly a solution, and for which the lowest degree term is $$z^{d-1}$$. Thus, $$Q^h(x,y,z)=0$$ has a singular point at $$(0:1:0)$$ of multiplicity $$d-1$$ and so, the curve defined by $$Q$$ will, of necessity, be of genus $$0$$.