I am reading Miles Reid Undergraduate Commutative Algebra and I am working through some of the exercises in chapter 6.
6.9: For $A=k[V]$ the coordinate ring of a variety $V\subset k^n$, and $f\in A$, prove that $A[1/f]$ is the coordinate ring of a variety $V_f\subset k^{n+1}$ which is a natural one to one correspondence with the (Zariski) open set $V\setminus V(f)$
I believe I want to use the previous problem (6.8) If $S=\{1,f,f^2,f^3,\ldots \}$, is a multiplicative set of $A$, then $Spec(A_f)\subset Spec(A)$ is the complement of the closed set $\mathcal{V}(f)$.
My proof: By definition, $\mathcal{V}(f):=\{P\in Spec(A) : f\in P \}$. Hence, $Spec(A) \setminus \mathcal{V}(f)= \{P\in Spec(A) : f \notin P\}= \{P \in Spec(A): 1,f,f^2,\ldots \notin P\}$ as each $P$ is prime. Thus, $Spec(A) \setminus \mathcal{V}(f) = \{ P \in Spec(A) : P\cap S= \varnothing \}$. As per the hint, by Corollarry 6.3(iv), we can identify $Spec(A_f)$ with $Spec(A) \setminus \mathcal{V}(f)$
Using this lemma, and the definition of $A=k[V]=k[X_1,\ldots X_n]/I$ for $I=rad(I)=\{\text{ polynomials that vanish on all points }P\in V\}$, I can view any $f\in A$ as a polynomial function $f:V\to k$ meaning I can write $f$ as polynomial coordinates $x_i$ then $A[1/f]=k[V,1/f]$ I have coordinates for the first $n$ components but I am lost as to how $1/f$ changes it. Intuitively, it makes sense that this will be the set of points that don't vanish on $f$, but I am unsure where to go from here.