2024-08-08

Theorem (Cayley-Hamilton)

Let be a ring and an ideal. Let be a finitely-generated -module and be an -module endomorphism such that . Then there exist for such that

in .

Proof

Make an -module via . Since is a finitely-generated -module, it is a finitely-generated -module. Localising at , we obtain a finitely generated -module . Let be the subset of elements of the form

is closed under addition and multiplication. If then, by assumption that ,

This implies that in ,

Hence as -modules. Nakayama’s Lemma implies that there exists such that . Write

Then

and implies that #TODO: check this

for some . Setting , then

in .

Remark 2024.1.

Note that this is a monic polynomial!

Proposition

Let be a ring. Consider a short exact sequence of -modules

  1. If is finitely-generated then is finitely-generated.
  2. If and are finitely-generated then is finitely-presented.
  3. If and are finitely-generated (or finitely-presented) then is finitely-generated (or finitely-presented).
  4. ! (Schanuel’s Lemma) If is finitely-presented and is finitely-generated then is finitely-generated.

Remark (Schanuel’s Lemma)

  • Finitely-generated: there exists an exact sequence
  • Finitely-presented: there exists an exact sequence
  • Schanuel’s Lemma implies that for any short exact sequence
    for finitely-presented , is necessarily finitely-generated, i.e., there is a finite number of relations no matter which (finite) collections of generators we choose.

Proof

1.

The composition

gives a finite generation.

3. (finitely-generated case)

Because and are finitely-generated, we can choose surjections and . Because is free, we can lift the map to :

In a commutative diagram, this looks like

Combining the two maps and , we obtain a diagram with exact rows:

By the Snake Lemma, we obtain an exact sequence

But because and are surjective, and . So because it sits between two modules in an exact sequence, i.e., is surjective.

4.

Case:

First, suppose . Since is finitely-presented, we know that there is a surjection with finitely-generated kernel . We now form the following diagram with exact rows by lifting to :

By the Snake Lemma, we get an exact sequence

i.e., there is an exact sequence

But because is finitely-generated, is also finitely-generated by 1.. Similarly, is finitely-generated by 1. because is a surjection. So by 3. (finitely-generated case), is finitely-generated.

General case

In general, we choose a surjection and form the diagram

Here, . By the snake lemma, sits in an exact sequence

and therefore , i.e., is surjective. By the the case already considered, the top horizontal short exact sequence tells us that is finitely-generated. Therefore, is finitely-generated by 1..

Tensor products

Let be a field and an -matrix. Then this gives rise to a bilinear form

In other words, is linear in each variable. From every such map, we get a matrix whose entries are . That is, we get an association of bilinear maps to a matrix, which is a linear object.

Definition (Bilinear map)

Let be a ring and , , be -modules. An -bilinear homomorphism

is a map of sets that is -linear in each argument. That is, for each and , the two maps

Lemma (Properties of bilinear maps)

Let be the set of all -bilinear maps .

  1. is naturally an -module
  2. Switching the arguments is a natural isomorphism of -modules
  3. There is a natural isomorphism of -modules

Proof

Omitted. Read the notes #TODO.

Proposition (Universal property of tensor products)

There exists an -module and an -bilinear map

with the following universal property: if is an -bilinear map then there exists a unique -linear map such that the following diagram commutes:

In other words, and are left and right adjoints respectively: