Math Overflow offers this explanation of the Rabinowitsch trick in proving the strong Nullstellensatz from the weak Nullstellensatz.
https://mathoverflow.net/questions/90661/the-rabinowitz-trick
"Perhaps the "Rabinowitz trick" is more clear if one writes down the proof backwards in the following way:
Let $I \subseteq k[x_1,\dotsc,x_n]$ be an ideal and $f \in I(V(I))$, we want to prove $f \in \mathrm{rad}(I)$. In other words, we want to prove that $f$ is nilpotent in $k[x_1,\dotsc,x_n]/I$, or in other words, that the localization $(k[x_1,\dotsc,x_n]/I)_f$ vanishes. By general nonsense this algebra is isomorphic to $k[x_1,\dotsc,x_n,y]/(I,fy-1)$. But, clearly $V(I,fy-1)=\emptyset$ and therefore the Weak Nullstellensatz implies that $(I,fy-1)=(1)$, i.e. that the quotient vanishes."
I don't follow the claim that goes "By general nonsense this algebra is isomorphic to..." Should this be immediately obvious when reading the proof? Can you break it down further for me?
$R_f$ is isomorphic to $R[y]/(fy-1)$, which we can show using the universal property of localizations:
For any ring morphism $\varphi: R\to S$ that inverts $f$, there is a unique extension $\overline{\varphi}:R[y]\to S$ sending $y$ to $\varphi(f)^{-1}$. The kernel contains $fy-1$, and so $\overline{\varphi}$ factors through $R[y]/(fy-1)$.
A morphism $R[y]/(fy-1)\to S$ extending $\varphi$ must send $y$ to $\varphi(f)^{-1}$. In other words, any ring morphism $\varphi: R\to S$ that inverts $f$, factors uniquely through $R\to R[y]/(fy-1)$.
It follows that $R[y]/(fy-1)\cong R_f$.