Timeline for Finding divisors on a curve

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Mar 15, 2019 at 10:14	comment	added	Georges Elencwajg		Dear @isildur: you are absolutely right, the formulation of the answer is a bit sloppy. The correct formulation is that the canonical divisor class of a smooth hypersurface in affine space $\mathbb A^n$ is trivial. In your case the divisor of your differential form $xdx$ is the divisor $D=1.[0]$, which is non-zero but is linearly equivalent to zero since $D= \operatorname {div}(x)$. Hence its class $D]$ is zero in the divisor class group of your curve $y=0$.
Jun 20, 2011 at 5:33	comment	added	isildur		I am confused, take the line $y = 0$ ( x-axis) in $\mathbb{A}^{2}$. The differential form $x\;dx$ will have a divisor $[0]$?
Dec 24, 2009 at 16:47	history	answered	Maharana	CC BY-SA 2.5