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Localization and the strong Nullstellensatz

A basic idea in topology and analysis is to study a space by restricting attention to arbitrarily small neighborhoods of a point. It is desirable, therefore, to have a notion of looking at small neighborhoods of a point which can be stated in entirely ring-theoretic terms. More generally, we’d like to have a way to ignore some points and only think about others. The tool that allows us to do this is called localization, and it offers a conceptual proof of the strong Nullstellensatz from the weak Nullstellensatz, which, as you’ll recall, is the tool that allows us to describe the category of affine varieties as the opposite of a category of algebras.

Rational functions

It would be nice if everywhere-defined functions told you everything in a geometric situation. However, it’s a common situation in complex geometry (for example) that this is not enough. For example, there are no nonconstant holomorphic functions on a compact Riemann surface such as the Riemann sphere, so if we want to think about compact Riemann surfaces from a ring-of-functions perspective we need to expand our notion of a function. The standard solution is to work instead with the field of meromorphic functions, which are functions holomorphic almost everywhere except at an isolated set of points, where they have at worst poles.

If we think of meromorphic functions as quotients of holomorphic functions, this suggests that the corresponding move in the affine variety case is to replace the affine coordinate ring $\mathbb{C}[V]$ by its ring of fractions $\mathbb{C}(V)$, the function field of $V$. For example, the function field of $\mathbb{C}$ is precisely the field of rational functions in one variable, which is in turn precisely the field of meromorphic functions on the Riemann sphere. The Riemann sphere is the one-point compactification of $\mathbb{C}$, and remarkably this story generalizes: it turns out that any compact Riemann surface is essentially the compactification – though an algebraic geometer would say the projectivization – of an affine complex variety. We haven’t introduced projective varieties yet, but they are the natural setting in which to think about isomorphism classes of function fields.

Function fields should be compared with number fields such as $\mathbb{Q}$ and its finite extensions. The elements of $\mathbb{Q}$ may be regarded as “meromorphic functions” on the set of primes – although they cannot be reduced $\bmod p$ for every prime $p$, they have only a finite set of “poles.” Moreover, they have a well-defined notion of order at every prime. As we will see, this is an important way to think about prime factorization, and one way to think about this series of posts is as an exploration of the question of to what extent number fields can be regarded as meromorphic functions in this manner.

Localization for affine varieties

Given a variety $V$, say that an element of $\mathbb{C}(V)$ is regular at $p \in V$ if its denominator does not vanish there, i.e. it can be evaluated at $p$. Given a subset $W \subseteq V$, the localization of $\mathbb{C}[V]$ at $W$ is the ring of regular functions on $W$. For example, if $W$ is empty, the localization is just the entire function field. If $W = V$, then any function $f = \frac{g}{h}$ has the property that if $h$ doesn’t generate the unit ideal (hence $f$ is a polynomial) then it vanishes somewhere, since it is contained in a maximal ideal, so the localization is just $\mathbb{C}[V]$. (Hence this definition of regular is consistent with our previous one.)

Theorem (strong Nullstellensatz): $\mathbb{C}[x_1, ... x_n]$ is Jacobson.

Proof. It suffices to show that if $J$ is an ideal, then $I(V(J)) = \text{rad}(J)$. It’s clear that the RHS contains the LHS, so we only have to show the other inclusion. By Noetherianness, $J$ is finitely generated, say by elements $f_1, ... f_k$. Let $g$ be an element of $I(V(J))$, hence $g$ vanishes on all of $V(J)$. Now we are going to allow $g$ to be invertible, which should be thought of as passing to the localization at $V(J) - V(g)$, which should be empty. Formally, we will now consider $\mathbb{C}[x_1, ... x_n, y]/(f_1, ... f_k, 1 - gy)$, where $y$ is the new inverse of $g$. The claim that $g$ vanishes on all of $V(J)$ is equivalent to the claim that this new ring has no maximal ideal by the weak Nullstellensatz, which is possible if and only if $(f_1, ..., f_k, 1 - gy)$ is the unit ideal. So we have a linear combination

$\displaystyle h_0 (1 - gy) + \sum_{i=1}^{k} h_i f_i = 1$

where $h_i \in \mathbb{C}[x_1, ... x_n, y]$. The result now follows by substituting $y = \frac{1}{g}$ and multiplying by a sufficiently large power of $g$ to clear denominators.

This proof technique is usually known as the Rabinowitsch trick, but I think calling it a trick undermines its conceptual content. The key point here is that some open subsets of an affine variety are themselves varieties in one higher dimension and we proved the weak Nullstellensatz in all dimensions.

Localization for commutative rings

The abstract content of localization is that we pass from a ring $R$ to a larger ring where more of the elements of $R$ are invertible than they were before because we no longer care about the “poles” their inverses have in $\text{Spec } R$. A basic observation is that if we adjoin inverses of two functions, we necessarily get an inverse of their product. In other words, the set of new inverses we get is closed under multiplication.

A set $S$ of elements of $R$ is said to be a multiplicative set if it is closed under multiplication and doesn’t contain zero. As in the formal definition of the quotient field, we can define the localization of $R$ with respect to $S$ as the ring of elements of the form $\frac{r}{s}, r \in R, s \in S$ with addition, multiplication, and equivalence defined in the usual way, and we denote this new ring by $S^{-1} R$. An important special case is when $S$ is the complement of a prime ideal $P$, in which case the localization is denoted $R_P$. Confusingly, another important special case (as we just saw) is when $S = \{ f, f^2, f^3, ... \}$ for some non-nilpotent element $f$, in which case the localization is denoted $R_f$.

There is a canonical injection $R \to S^{-1} R$ which gives rise to a map $\text{Spec } S^{-1} R \to \text{Spec } R$. When $R$ is an integral domain, the localization at $(0)$ is precisely the fraction field and the above map is precisely the inclusion of the generic point in $\text{Spec } R$. This picture generalizes. Given a map of rings $\phi : A \to B$ and an ideal $I$ of $A$, the image $\phi(I)$ is not necessarily an ideal of $B$. It generates an ideal called the extended ideal $I^e$; this construction is adjoint to the construction of the contracted ideal.

Proposition: The prime ideals of $S^{-1} R$ are precisely the extensions of the prime ideals of $R$ disjoint from $S$. Hence the map $\text{Spec } S^{-1} R \to \text{Spec } R$ is injective.

Proof. Given a prime ideal $P$ disjoint from $S$, every element of $S$ is invertible in $R/P$, so the canonical map $R \to R/P$ factors through $S^{-1} R$. (This is essentially a universal property.) The preimage of this ideal is precisely the extended ideal. In the other direction, if $Q$ is a prime ideal of $S^{-1} R$ then the contraction of this ideal is a prime ideal of $R$ which must be disjoint from $S$, and the map $R \to S^{-1} R / Q$ must be surjective because the inverses of the elements of $S$ exist, so $Q$ is unique.

Corollary: If $P$ is a prime ideal, $R_P$ is a local ring with unique maximal ideal the extension of $P$, and $R_P/P^e \simeq R/P$.

At some point in the history of commutative algebra it was recognized that many important questions could be resolved by reducing to the local case. Part of this recognition was the realization that certain properties of a ring are local, i.e. they held if and only if they held at all localizations at prime ideals, and certain other properties are not. For example, it turns out that neither being a PID nor being a UFD are local properties, which is part of the reason why they are considered “delicate” rather than “robust.” We’ll discuss this more later.