# Understanding Rabinowitsch's trick

Rabinowitsch’s trick is a clever argument to prove the Nullstellensatz from its weak form, by introducing an extra variable. Explictly, having proved that every maximal ideal in a polynomial ring over an algebraically closed field $$k$$ has a zero, the trick allows you to prove that if you start with any ideal, take the corresponding algebraic set and then go back, you get the radical of the ideal you started with.

I have always found this argument quite confusing and mysterious. However, as it turns out, abstracting things out slightly and organising the proofs carefully makes the trick seem less magical and more like something anyone could have come up with.

So, let $$k$$ be an algebraically closed field. For any commutative ring $$R$$, let $$\mathsf{mSpec}(R)$$ be the maximal spectrum of $$R$$, i.e. the set of the maximal ideals of $$R$$.

The weak Nullstellensatz as stated above can easily be generalised to any finitely generated $$k$$-algebra $$A$$ as follows:

Theorem 1. The map $$k\text{-}\mathsf{Alg}(A, k) \to \mathsf{mSpec}(A)$$ that maps a homomorphism to its kernel is a bijection.

Proof. The map is obviously injective. The weak Nullstellensatz is exactly the statement that the map is surjective in the case when $$A$$ is a polynomial ring over $$k$$. In fact, the points of the affine space $$\mathbb{A}^n$$ are exactly the $$k$$-algebra homomorphisms $$k[x_1, \ldots, x_n] \to k$$.

For a general finitely generated $$A$$, write $$A$$ as a quotient of a polynomial ring $$k[x_1, \ldots, x_n]$$. Given any maximal ideal $$\mathfrak m$$ of $$A$$, there is a maximal ideal $$\mathfrak m'$$ of $$k[x_1, \ldots, x_n]$$ that contains the pullback of $$\mathfrak m$$. Find a homomorphism $$\phi$$ whose kernel is $$\mathfrak m'$$. Then clearly $$\phi$$ descends to the quotient, and its kernel is a maximal ideal containing $$\mathfrak m$$, so it must be equal to it.$$\square$$

And now we can apply the trick itself. Let $$\mathfrak a$$ be any ideal of $$k[x_1, \ldots, x_n]$$. Denote by $$V(\mathfrak a)$$ the corresponding algebraic set. This can be thought of as the set of $$k$$-algebra homomorphisms to $$k$$ that are zero on $$\mathfrak a$$. Suppose $$f$$ vanishes on $$V(\mathfrak a)$$, i.e. $$f$$ is in the kernel of all the corresponding homomorphisms. Denote by $$\sqrt{\mathfrak a}$$ the radical of $$\mathfrak a$$.

Proposition 2. For $$f$$ as above, $$f \in \sqrt{\mathfrak a}$$.

Proof. Let $$A$$ be the coordinate ring of $$V(\mathfrak a)$$, i.e $$A = k[x_1, \ldots, x_n] / \mathfrak a$$. If $$f$$ is not in the radical of $$\mathfrak a$$, then it is not nilpotent in $$A$$. Therefore, the localisation $$A[f^{-1}]$$ is not the zero ring. Now the key observation is that $$A[f^{-1}]$$ is finitely generated, since it is isomorphic to $$A[t]/(tf - 1)$$. Therefore, by Theorem 1 there exists at least one homomorphism $$\phi: A[f^{-1}] \to k$$, since $$A[f^{-1}]$$, being non-trivial, has at least one maximal ideal. But pulling back $$\phi$$ to $$k[x_1, \ldots, x_n]$$ we get a homomorphism that is zero on $$\mathfrak a$$, but nonzero on $$f$$, a contradiction.$$\square$$

I like how in this reformulation the whole idea of introducing a new variable is now a completely natural operation, since the variable basically appears by itself by unfolding one of the explicit definitions of localisation. The extra variable is only used to prove that the localisation is finitely generated, and does not play any role in the actual argument.

Also, the argument goes through unchanged if we replace the polynomial ring with an arbitrary finitely generated $$k$$-algebra, and define $$V$$ directly in terms of homomorphisms to $$k$$.

In fact, the essence of Proposition 2 is really the following statement, which has essentially the same proof.

Proposition 3. In finitely generated $$k$$-algebra, the nilradical is equal to the Jacobson radical, i.e. it is the intersection of all the maximal ideals.

Proof. Let $$A$$ be a finitely generated $$k$$-algebra. All nilpotent elements of $$A$$ are contained in every prime ideal, and in particular in every maximal ideal. Conversely, if $$f \in A$$ is not nilpotent, the localisation $$A[f^{-1}]$$ is a non-trivial finitely generated algebra, so it admits a homomorphism $$\phi$$ to $$k$$, by Theorem 1. Pulling back $$\phi$$ gives a homomorphism of $$A$$ that maps $$f$$ to a unit. In particular, its kernel is a maximal ideal that does not contain $$f$$.$$\square$$

If $$X$$ is a set of homomorphisms $$A \to k$$, denote with $$I(X)$$ the ideal of $$A$$ of elements that are sent to zero by every homomorphism in $$X$$. So $$V$$ and $$I$$ generalise the usual Galois connection between algebraic sets and ideals of a polynomial ring to arbitrary finitely generated $$k$$-algebras.

We can now make a one-line calculation to prove Proposition 2: $I(V(\mathfrak a)) = \bigcap_{\phi: A/\mathfrak a \to k} \pi^{-1}(\mathsf{ker}(\phi)) = \bigcap_{\mathfrak m \in \mathsf{mSpec}(A/\mathfrak a)} \pi^{-1}(\mathfrak m) = \pi^{-1}\bigcap_{\mathfrak m \in \mathsf{mSpec}(A/\mathfrak a)} \mathfrak m = \pi^{-1}(\sqrt 0) = \sqrt{\mathfrak a},$

where $$\mathfrak a$$ is any ideal of $$A$$, and $$\pi: A \to A/\mathfrak a$$ is the projection to the quotient.