Localization and common zero set (basic algebraic geometry) I have two questions about understanding localization and basic algebraic geometry. 


*

*In class, the professor mentioned that given the basic open set 
$$D(f) = \mathbb{A}^n - Z(f)$$
where $Z(f)$ is the vanishing set of $f\in k[x_1, \cdots x_n] =:A$, then the polynomials on $D(f)$ forms a ring and is isomorphic to the localization $A_f$. 


I know the definition of localization and basic properties, but I can not quite get the picture here. And what does "polynomials on $D(f)$" mean?


*Give an ideal $\mathfrak{a} \subset A$, and $f\in I(Z(\mathfrak{a}))$ where $I(V)$ is the ideal of polynomials that vanishes on the algebraic set $V$, the professor said $\mathfrak{a} A_f = A_f$, some how $Z(\mathfrak{a})$ is contained in the zero set of $f$, and $A_f$ is like the complement of the zero set of $f$. 
I can not get the intuition behind this. 


Thank you very much!
 A: If $U$ is an open set in $k^n$, define $\mathcal O(U)$ to be the $k$-algebra of functions $F: U \rightarrow k$ of the following form: for each $p \in U$, there exists a smaller open set $p \in V \subseteq U$, and polynomials $f, g \in A$, such that for every $q \in V$, $g(q) \neq 0$ and
$$F(q) = \frac{f(q)}{g(q)}$$
In other words, $\mathcal O(U)$ is the $k$-algebra of functions $U \rightarrow k$ which locally look like a quotient of polynomial functions.  So "polynomials on $D(f)$" means $\mathcal O(D(f))$.
Let $a = \frac{h}{f^m}$ be an element of the localization $A_f$.  By definition, $f(q) \neq 0$ for all $q \in D(f)$.  Thus we have a well defined $k$-algebra homomorphism
$$A_f \rightarrow \mathcal O(D(f))$$
which sends $a$ to the function $D(f) \rightarrow k$ given by $q \mapsto \frac{h(q)}{f(q)^m}$.  The result you mentioned in (1) is that this is actually an isomorphism.  The hard part is surjectivity.  A proof is given in Springer, Linear Algebraic Groups, chapter one.  
If $f \in I(Z(\mathfrak a)) = \sqrt{\mathfrak a}$, then $Z(\mathfrak a) \subseteq Z(f)$ (by order reversing correspondence).  Consequently, every $g \in \mathfrak a$ satisfies $g(q) \neq 0$ for all $q \in D(f)$ (if $g(q) = 0$, then $q$ must lie in $Z(f)$, the complement of $D(f)$).  Hence the image of $g$ in $\mathcal O(D(f))$ is a unit, since it now has a well defined inverse there, $\frac{1}{g}$.  But identifying $\mathcal O(D(f))$ with $A_f$, this just says that $\mathfrak a A_f = A_f$.
