Help with this proof in Fulton book

Question

I'm really confused with the definitions of coordinate rings and field of rational functions. I'm trying to understand this proof which I was stuck in the very beginning:

First I didn't understand the definition of $J_f$. we have $\overline G=G+I(V)$ and $f=f_1+I(V)$, where $f_1$ is the residue of $f$ in $\Gamma(V)$. The $\overline Gf=\bigg(g+I(V)\bigg)\bigg(f_1+I(V)\bigg)=\bigg(gf_1+I(V)\bigg)$, so this multiplication is not always in $\Gamma(V)$, since $\Gamma(V)$ is by definition $k[x_1,...,x_n]/I(V)$?

Second I didn't understand why the points of $V(I_f)$ are exactly those points where $f$ is not defined.

I really need help to understand this proof.

Thanks in advance.

Answer 1

Your first question doesn't exactly hold any water. Of course not every $G\in k[x_1,\ldots,x_n]$ is in $J_f$. The product is always in the fraction field of $k[V]$, and it's actually in $k[V]$ sometimes. That set of things where it actually lands in $k[V]$ is $J_f$.

The reason that the pole set is the zero set of this ideal is that $f$ is the quotient of two elements of $k[V]$ with the denominator nonzero, precisely when there exists an element of $J_f$ (the denominator of the representation) which doesn't vanish at that point. So, you see that the complement of $Z(J_f)$ is the points where $f$ is defined.

EDIT: The below only applies to the case of curves--which is what I thought the sole focus of Fulton was [I know the title, I haven't read the book]. So, keep this in mind.

Also, just as an FYI, this is a strange way to define poles (at least in my experience). Usually the poles of a rational function on a non-singular curve are the points $p$ where it has negative valuation in the canonical valuation of $\mathcal{O}_p$. I don't know if this is something Fulton talks about, but it makes more sense to me, and maybe will to you.