Order and residue of 1-form $x^{-1}dx$ Consider  the rational 1-form $x^{−1}dx$ on $\mathbb{P}^1$. I am asked to compute its order and residue at all $P \in \mathbb{P}^1$. 
Could somebody help me with this? I do not really how to start here. I would also be happy with a hint where I can find good examples. 
 A: If the coordinate on the other patch is $u$, with the relationship $1/x = u$, then $du = d (1/x) = -1/x^2 dx$, so $- x^2 du = dx$, $-1/u^2 du = dx$. So computing more: $x^{-1} dx = -1 u /u^2 du = - u^{-1} du$.
Now, on the affine $x$-chart $dx$ is a nonvanishing regular differential. Hence the divisor associated to some $f(x)/g(x) dx$ on that chart is the divisor associated to the rational function $f(x) / g(x)$ (on that chart!). (If you don't like the word divisor, then all I am saying is that the zeros and poles of $f(x)/g(x) dx$ on the $x$-chart are the zeros and poles of the rational function $f(x) / g(x)$ - counted with multiplicity - but done only inside the $x$-chart.)
We have also computed an expression for our rational function on the other chart, as an expression $h(u) du$, where $h(u)$ is a rational function. This lets you compute the order of the vanishing or order of the pole at infinity.
(Note that in your case, you get a total sum #ZEROS - #POLES to be -2, when you count with multiplicity. This will be the same for any rational differential form.)
As for the residues at these poles - I just looked up what this meant, so be a little more skeptical than usual. You can read the residues from the local coordinate descriptions given above, e.g. $(\ldots a_{-1} 1/t + \ldots )dt$ has residue $a_{-1}$ (why is it independent of the coordinate chosen? I don't really understand why.) So, unless I lost a sign, it should be 1 and -1. The sum is zero, which is what it should be, according to a theorem stating that the sum of residues of a rational differential form on a projective curve should be zero.
