I feel like this question is probably easy, but I'm just missing something obvious. I usually pick things up quickly (especially in algebra) but algebraic geometry continues to give me fits.
We will work over a fixed algebraically closed field .
If is an affine or quasi-affine variety, denote by the ring of regular functions on (i.e., an open neighborhood of such that is given by a rational function .
There are two questions here:
1. Show that (where denotes the zero set of which, over , is simply ).
2. Show that . (It is already known that ).
I really feel like I'm just missing something blatantly obvious. I can intuitively believe #1: if we only consider those points where is nonzero, then is defined everywhere, and obviously a rational function. But to actually PROVE that the given ring is exactly equal to the ring of regular functions on the set where is a tougher matter... I think...
Any help would be appreciated. I really, really like this subject, but I'm having so much trouble understanding it fully, or at least to the point where I can competently work with it...