Let $V = V(J) \subset k^{n}$ be an algebraic variety. Let $K[V] = K[X_{1},\dots,X_{n}]/(I(V))$ be its coordinate ring. For any $f \in K[V]$, the basic open set $V_{f} := V-V(f)$ is in bijective correspondence with the algebraic variety $W := V(I(V),(x_{n+1}f-1)) \subset k^{n+1}$ via the map $$(x_{1},\dots,x_{n}) \longmapsto (x_{1},\dots,x_{n}, \dfrac{1}{f(x_{1},\dots,x_{n})})$$
My question is:
I want to show that the coordinate ring $$K[W] = K[X_{1},\dots,X_{n},X_{n+1}]/(I(V(I(V),(x_{n+1}f-1))))$$ is isomorphic to $K[V][1/f]$. I know that $$\dfrac{(K[X_{1},\dots,X_{n}]/I(V))[X_{n+1}]}{(X_{n+1}f-1)} \cong (K[X_{1},\dots,X_{n}]/I(V))[1/f] = K[V][1/f]$$ but I don't know how to proceed with the left hand side in order to arrive to $K[W]$. For me, an algebraic variety is an irreducible algebraic set.
1) You have to show that $I = J+(X_{n+1}f-1)$ is prime in $k[X_1,\ldots,x_{n+1}]$; equivalently, show that $k[X_1,\ldots,X_{n+1}]$ is an integral domain. For the sake of simplicity call $R=k[X_1,\ldots,X_n]$, $S=k[X_1,\ldots,X_{n+1}]$.
By the isomorphism $A\otimes_R R/J\simeq A/JA$ (here $JA$ is the extension of $J$ in $A$) with $A=S/(X_{n+1}f-1)$ we get that
$$\frac{S/(X_{n+1}f-1)}{(J+(X_{n+1}f-1))/(X_{n+1}f-1)}\simeq \frac{S}{(X_{n+1}f-1)}\otimes_R \frac{R}{J}\simeq R[1/f]\otimes_R \frac{R}{J} \simeq \frac{R[1/f]}{J}\simeq (R/J)[1/f]$$
Now you have your results: you can deduce it in the following way: note that $(R/J)[1/f] = (R/J)_f \simeq R_f / J_f$ where $R_f$ is the localisation with respect to the multiplicative part $S=\{f^n:n\in \mathbb{N}\} $. Since the ideal $J$ does not intersect $S$, the localised ideal $S^{-1}J =:J_f$ is prime. Hence $(R/J)[1/f]$ is a domain and it follows that $J+(X_{n+1}f-1)$ is a prime ideal.