The Rabinowitz Trick

2012-03-09T05:15:11Z

The recent question about problems which are solved by generalizations got me thinking about the Rabinowitz trick, which is used to prove a statement of Hilbert's Nullstellensatz, specifically, the inclusion of the ideal generated by an affine variety $V(J)$ over an algebraically closed field into the radical of $J.$

Let $0\neq f\in J,$ as above. In the course of the proof, one extends the given polynomial ring by a single indeterminate and writes its elements as, $$\sum_{i=1}^l h_ig_i + h(X_n\cdot f - 1),$$ where $h_i,h\in k[X_1,\dots,X_{n+1}]$ and $g_i\in k[X_1,\dots,X_n].$ One then applies the weak Nullstellensatz, to see that, indeed, every element of $k[X_1,\dots,X_{n+1}]$ can be written in the above form. Then, mapping back to the smaller polynomial ring, via $X_{n+1} \mapsto \frac{1}{f}$ yields the result, by simply clearing denominators.

My question is this: While the trick uses some exceedingly clever algebra, does it have some sort of deeper geometric meaning? Why does it make sense to try this in the first place?

Answer by Martin Brandenburg for The Rabinowitz Trick

2012-03-09T07:15:00Z

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.

The Rabinowitz Trick - MathOverflow

The Rabinowitz Trick

Answer by Martin Brandenburg for The Rabinowitz Trick