The Rabinowitz Trick - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T10:29:29Z http://mathoverflow.net/feeds/question/90661 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90661/the-rabinowitz-trick The Rabinowitz Trick Grant Rotskoff 2012-03-09T05:15:11Z 2012-03-09T07:15:00Z <p>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 $J.$</p> <p>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. </p> <p>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?</p> http://mathoverflow.net/questions/90661/the-rabinowitz-trick/90666#90666 Answer by Martin Brandenburg for The Rabinowitz Trick Martin Brandenburg 2012-03-09T07:15:00Z 2012-03-09T07:15:00Z <p>Perhaps the "Rabinowitz trick" is more clear if one writes down the proof backwards in the following way:</p> <p>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.</p>