Localization and Spectrum
Recall that the ideals of correspond to a subset of the ideals of A. If we restrict ourselves to prime ideals, we get the following nice bijection.
The above gives a bijection between
If is an ideal, then implies for some . Proof: exercise.
First we show is prime in . Suppose , satisfy . Then by the above trick, there exists such that . But so either or and hence or .
Note that is not the whole ring since if it contains 1, then for some , ,
Continuing the previous paragraph, it remains to show .
(⊆) Suppose so . Thus .
(⊇) Suppose so that . Then for some . Since we have . ♦
The above map identifies as a subspace of .
Let . Then is injective and continuous. On the other hand, suppose is a closed subset of for an ideal . Let , an ideal of A. It remains to show .
(⊆) Suppose , a prime ideal of . Then so .
(⊇) Suppose is a prime of A containing . By theorem 1, from the previous article. ♦
Philosophically, localization removes prime ideals which are “too big”.
Basic Open Sets
Let . By the above result, is identified with the subspace . But this is exactly the basic open set . Hence,
as a homeomorphism.
Let be the coordinate ring of variety V over an algebraically closed field k. Pick and let .
Concretely, if is a closed subspace, then where
because of the following result.
Prove that if A is a reduced ring, so is .
As a consequence, we see that if is a regular function on affine variety V, the open set is also an affine variety. In summary, a basic open subset of an affine variety is an affine variety.
The simplest example is , which is isomorphic to the “hyperbola” .
Consider the case for a prime ideal so . The first result implies:
The prime ideals of correspond bijectively with the prime ideals of A contained in .
In particular, has exactly one maximal ideal .
Thus the spectrum of looks something like this.
If and , then
A ring is called a local ring if it has a unique maximal ideal.
We often write for the local ring, where is the maximal ideal of A.
The following is all you need to know about local rings.
A ring is local if and only if its set of non-units forms an ideal, in which case this ideal is the unique maximal ideal of .
We may assume .
(⇒) Let be local. If is not a unit it generates a proper ideal , which must be contained in a maximal ideal of A. But is the only maximal ideal so . Thus is the set of non-units of .
(⇐) Suppose the set of non-units is an ideal . Then any maximal ideal cannot contain any units so . Equality must hold by maximality. ♦
Local Rings in Geometry
Let again, where V is an affine k-variety with k algebraically closed. Fix a point and consider its associated maximal ideal . If where , , then so (defined above) contains P. Hence defines a function
for some open neighbourhood U of P.
A function of this form is said to be regular at P. Unfortunately this is dependent on our choice of representatives.
Let , corresponding to and . Prove that if and only if there is an open set W,
Hence we define the following notion.
Let be a point on the topological space . For a set S, an S-valued local function at P is a function for some open neighbourhood U of P.
If , are local functions at P, we say they have the same germ at P if there is an open W,
Note that this is an equivalence relation.
Hence classifies the germs of regular local functions at P.
Let where . Take .
where represent their images in . We take the functions:
which are well-defined on . Now since and but they define the same germ at P:
in . Indeed, if we take out the component , then is an open subset such that .
Prove the preceding statement.
When V is an irreducible variety, is an integral domain and this case is particularly nice since S has no zero-divisors (as long as it does not contain 0).
A rational function on V is a regular local function f at any point of V. Thus a rational function is of the form for some non-empty open .
We identify two rational functions
if there exists a non-empty open such that .
1. Prove that this is an equivalence relation. [Hint: use the fact that in an irreducible space, any two non-empty open subsets intersect.]
2. Prove that in fact, if there exists such that , then it holds for .
Algebraically, a rational function is of the form with and for some point . The latter condition just means so we have:
The set of rational functions on V is exactly the field of fractions of .
We call this the rational function field of V and denote it by .
Take the curve with coordinate ring . Prove that , where is the field of rational functions in X.
Question to Ponder
Let be a prime ideal (not maximal in general) corresponding to the irreducible closed subset . Explain in geometric terms.