I want to see if there is a well defined map $$\int_0^\theta\text{d}\theta:\frac{\mathbb{R}[x,y]}{\langle x^2+y^2-1\rangle}\rightarrow\frac{\mathbb{R}[x,y]}{\langle x^2+y^2-1\rangle}.$$ I am starting my study of algebraic geometry and I have found this puzzling. This is related to another post Polar coordinates for polynomials on the circle.
I have tried going into polar coordinates but there is no simple expression for an integral of the form $$\int_0^\theta \text{d}\theta\,\cos(\theta)^n\sin(\theta)^m.$$ On the other hand, in complex coordinates one has $$\int_\mathcal{C}\text{d}z\,z^n\bar{z}^m=\begin{cases}i r^{n+m}\theta, & 1+n-m=0\\ \frac{z^{n+1/2}\bar{z}^{m-1/2}-(z\bar{z})^{(n+m)/2}}{1+n+m},&\text{otherwise.}\end{cases}$$ In here $\mathcal{C}$ is the arc in the circle of radius $r$ between $0$ and $\theta$. Thus, in $\mathbb{C}[z,\bar{z}]$ this doesn't even seem to be well defined. For example, in the case $1+n-m=0$ it does not even yield a function.
It at least does induce a well defined operator on some polynomials. The key is that I did the computation in polar coordinates wrong. The relevant computation is $$\int_0^\theta\text{d}\theta\,z^n\bar{z}^m=\frac{z^n\bar{z}^m}{i(n-m)}\in\mathbb{C}[z,\bar{z}],$$ as long as $n\neq m$. Since taking the real part of a complex number is linear and every real polynomial can be obtained by taking the real part of a complex one, this proofs what we needed. The problem of $n\neq m$ in the above computation is solved because on the circle the polynomials $z^n\bar{z}^n$ are in the same equivalence class than $1$.