2024-08-01

Localisation (continued)

Proposition (from last time)

  1. and are ideals of and respectively.
  2. and .
  3. () and () restrict to bijections
    i.e., is injective with image .

Proof of 3.

)

If is prime then is prime. Observe that if then there exists such that . But . So contains a unit (), which contradicts primality of . Hence, and the map is well-defined.

)

Conversely, let be prime. If , then we claim that

is prime. First, if , then , where and . Hence there exists such that in by definition of . That is,

But , so we have a contradiction. Therefore . This shows that is a proper ideal. Showing that is prime is similar #TODO (exercise).

By 2., is always true, so it remains to prove that if prime satisfies . Let , then . Hence, there exist and such that

That is, there exists such that

But if , then primality of implies that . But that is impossible as as well (by multiplicativity of ) but .

Corollary 1

Let be a ring. Then is nilpotent, i.e., for some , if and only if for all primes .

Proof

)

Suppose that for some . Let be a prime. Then . If , then . Hence either or . Repeating this, we eventually conclude that .

)

Conversely, if for all primes , then for all primes . By the Proposition part 3., there is a bijection

Hence, , and therefore is the -ring. Then in , that is, there exists such that

Hence in .

Corollary 2

Let be a ring and be an ideal. Then

Proof

Apply Corollary 1 to and use the Correspondence Theorem.

Remark

We call radical if .

Corollary 3

Let be a ring and . Then the image of in is a unit if and only if there exists and such that (i.e., ).

Proof

Omitted. Read the notes #TODO.

Corollary 4

Let be a ring.

  1. Let and set . Then is a multiplicative set and
    In other words, we don’t get any additional units, unlike in Corollary 3.
  2. Let be a multiplicative set. Set
    Take , called the saturation of . Then is the largest multiplicative subset of such that
    is an isomorphism.

Remark

Note that . is formed by removing from all the prime ideals which don’t intersect . So there is a well-defined map since we are just adding more denominators.

Proof

Omitted. Read the notes #TODO.

Functoriality

Let be a ring homomorphism. Let be a multiplicative set. Then is a multiplicative set, and the image of under is invertible in . So there exists a unique map such that the following diagram commutes:

Now, applying to this diagram, we find that

By the proposition above, the horizontal maps are injective. So we can identify and with their images under (contraction), and therefore it makes sense to say

Claim

i.e., .

Proof

If () then . So . Therefore

Therefore .

Example

#TODO2
Let be an integral domain and . Then the isomorphism

leads to the diagram (via the inclusion )

Now primes with are of the form where is irreducible:

Zariski topology

Let be a ring. An element is in bijection with an evaluation homomorphism

Then, we can look at the induced map on :

We want to put a topology on such that this map is continuous. If , then

where is the residue field (not a field!?). Further, if , then .