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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? $$y=x^4 + x^3 + x^2 + x + 1$$

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    $\begingroup$ 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 ... $\endgroup$ Commented Nov 8, 2020 at 0:03
  • $\begingroup$ Is the singularity at infinity of multiplicity 3? It seems it would need to be to reduce the genus by 3. $\endgroup$ Commented Nov 8, 2020 at 0:08
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    $\begingroup$ Yes, you need to learn homogeneous coordinates are work out simple examples. $\endgroup$ Commented Nov 8, 2020 at 0:13
  • $\begingroup$ 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? $\endgroup$ Commented Nov 8, 2020 at 0:15

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The clarification I needed was provided by Ted Shifrin in the comments. The mistake in my thinking was in the notion that:

"it seems that I can easily choose P(x) such that Q(x,y)=0 has no nodes of multiplicity 2 or more."

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 $0$.

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