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.
Did you mean in the definition in the section “Local rings in Geometry” . Also, can you describe what S is in -valued local function at .
A typo in the lemma in the section “Irreducible varieties” : “We denote call this the ….”
Thanks, corrected the parts. On your first comment, one can set , but the point is that one really cares about what goes on “close to P”. Hopefully, this edit is clearer.
Would you please share your personal email I’d with me? I have a few questions regarding measure theory and I believe you may answer them. Thank you.
Maybe post the question here? I’ll try to help but can’t guarantee. My measure theory is a little rusty.
I wonder why there are many theories of integration; I feel that such a state of affairs seems contradictory to the spirit of Mathematics, namely exactitude, proceeding from a set of axioms through logical deduction.
Progress in algebra is in isolating important subsets of sets and study their behaviour; this approach eventually leads one to answer a question related to number theory. But, in analysis, the approach seems otherwise.Lebesgue’s theory, unlike Riemann’s, can integrate larger class of functions. What does that mean? That I change my theory of integration as and when I like in order to integrate a special class of functions of my interest; for instance Dirichlet’s function on [0,1] in the case of Lebesgue’s theory. Please clarify these issues. Is it the case that adopting Lebesgue’s theory enables one to make sense of several phenomenon, like in Fourier analysis, which eventually helps one to explain many physical phenomenon. Or is it just that Lebesgue’s theory was adopted by mathematicians, because of its simplicity, generality, and applicability as a theory of integration.
I’m no expert in analysis but my understanding is that Lebesgue integration is definitely preferred to Riemann integration because it can be generalized to integration on measure spaces (e.g. Haar measure on locally compact groups, including p-adic groups and Lie groups). Once you learn Lebesgue integration, you should probably forget Riemann integration.
That being said, Riemann integration has the advantage of being easier to understand so it is not leaving the university curriculum.
What is the need to show that “show since we already knew that is continuous?
Sorry, it is in Proposition 1.
Hi. What we’re showing is that identifies as a subspace of .
For an injective map , it is not enough to show that is continuous (i.e. if is closed, then so is ). We also need to show that any closed subset of is of the form for some closed .
I can’t understand why we need the “useful trick”. Because in previous post by a lemma we know that what are the elements of P(S-1A), i mean a/s € S-1A, where a€ P and s€S. Please clarify this!!!!
Sorry, a/s € P(S-1A) where a € P and s€ S
It’s a little subtle since does not imply here. The “useful trick” is really easy to prove but will be repeatedly used again and again that’s why we mention it here.