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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?

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Sure. You introduce a new "dimension" to "spread out" things. This is also a common trick in algebraic K-theory. But others will explain this more properly ... By the way: 1+ for your question because in my opinion learning mathematics is not about verification of the proofs, but rather to understand what the inventor had in mind and thereby understand the original idea (which often gets lost during the process of writing-up things, unfortunately). –  Martin Brandenburg Mar 9 '12 at 7:04
The Rabinowitsch trick isn't a trick; it's localization. –  Qiaochu Yuan Mar 9 '12 at 7:59
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up vote 15 down vote accepted

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.

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