Let $\omega = \mathrm d \eta$ be an exact rational $n$-form on $\Bbb P^n$.

It may happen that the polar locus of $\eta$ is not included in the polar locus of $\omega$. But is it true that $\omega = \mathrm d \eta_0$, where $\eta_0$ is an $(n-1)$-form which is regular where $\omega$ is?

In other words (and considering the affine case), assume that $$ F = \partial_1 G_1 + \dotsb + \partial_n G_n $$ with $F$ and $G_i$'s rational functions in $x_1,\dotsc,x_n$.

Is it possible to choose the $G_i$ such that their denominators are a power of the denominator of $F$?

I've been unable to provide a counter example to this question whereas the following formulation seems to indicate that there exists one.

Let $X$ be the open set of $\Bbb P^n$ where $\omega$ is regular, and $Z$ the polar locus, in $X$, of $\eta$, so that $Z$ is a hypersurface of $X$. If I'm not mistaken, a positive answer to my question is equivalent to the injectivity of the restriction map in the de Rham cohomology $$H^n(X)\to H^n(X\setminus Z).$$

And following Hartshorne's *On the De Rham cohomology of algebraic varieties*, we have an exact sequence
$$ \dotsb \to H^{n-1}(X\setminus Z) \to H^{n-2}(Z) \to H^n(X)\to H^n(X\setminus Z), $$
so that the injectivity of the last arrow is equivalent to the nullity of the previous one, or the surjectivity of the one before, the Poincaré residue map. Is there any reason for this residue map to be surjective? What if $Z$ is smooth, or is a hyperplane?

For $n=2$ we have to check that $$0\to H^1(X) \to H^1(X\setminus Z) \to H^0(Z) \to 0$$ is exact. Which seems to be true … although I don't know why.

Any thought will be much appreciated, including on the specific case $n=2$.