2024-08-06

Corollary

Let be a ring. There is a bijection

Proof

Omitted. Read the notes #TODO.

Corollary

Let be a domain. Then is irreducible.

Proof

.

Definition (Module)

Let be a ring (assumed to be commutative with unity). An -module is is an abelian group with an action of . That is, it is a pair where is an abelian group and

is a ring homomorphism. Note that is not necessarily commutative.

Definition (Module homomorphism)

An -module homomorphism is a map such that for all , .

Examples

  1. -modules, i.e., abelian groups.
  2. For a field, -modules are the same as -vector spaces.
  3. Let be a field (or any commutative ring) and , the polynomial ring in variable. Then -modules are -vector spaces plus the data of an endomorphism, the action of . To see this, a map out of is just the data of
    Hence, given a -module, we get , where is the underlying -module. A -module homomorphism must satisfy

Kernels and cokernels

Kernels and cokernels of -module homomorphisms as abelian groups are canonically -modules, and they inherit the appropriate universal properties.

Lattice isomorphism theorem

Let be a ring. Let be an -module and be an -submodule. Then there is a bijection

⟻

where is the projection.

Definitions (Free, finitely-generated, finitely-presented)

Let be a ring. An -module is:

  • free if for some set .
  • finitely-generated if there exists a surjection for some .
  • finitely-presented if there exists a surjection for some with finitely-generated.

Example: bases of -vector spaces

Let be a field and be a -module. Then the following are equivalent:

  • is a finite-dimensional -vector space.
  • is free and finitely-presented.
  • is finitely-presented.
  • is finitely-generated.

Remark

Let be a surjection of rings. If is a finitely-generated -module, then is a finitely-generated -module. To see this, consider

If is finitely-generated and is a finitely-presented -module, then is a finitely-presented -module.

Lemma (Nakayama)

Let be a ring and be a finitely-generated -module. Let be a subset closed under addition and multiplication. If , then there exists such that (or for all ). In particular, if , then .

Definition (Jacobson radical)

The Jacobson radical of a ring is the ideal

Proof

Case:

First, assume that . We will prove that . Let generate (e.g., choose and set where is the th basis element). Now the condition that implies that there exist such that

But (from the above definition). So we can write

Repeating this, we get .

General case

In general, set . Then is a multiplicative subset. Also, , and . This means that by the case above which we have already considered. If generate , then generate , i.e.,

So there exist such that for all . Now setting

Then for all . So .

Example

Let be a local ring and be a finitely-generated -module. Then if and only if . By Lemma (Nakayama), since , .

Let be a ring and be a finitely-generated -module. For , we can construct:

  • ,
  • , a (finitely-generated) -module; and
  • is a -vector space.

So for every (finitely-generated) -module, we get a family of vector spaces over each point of . This is kind of like a bundle. Nakayama says that if the vector space is at a point, then is actually at the local ring . One can conclude from this that will be in a neighbourhood of , i.e., the support of is closed. (It is true that .)

Example

Let be a ring, and be a finitely-generated -module. Let be an -module endomorphism. If is surjective, then it is an isomorphism.

Remark

In this sense, a finitely-generated module is similar to a vector space. If we have a finite-dimensional vector space , and a surjective endomorphism , then it must be an isomorphism by rank-nullity.

Proof

We view as an -module where . Therefore is a finitely-generated -module. Surjectivity of tells us that . By Lemma (Nakayama), there exists such that . That is, there exists such that . Set . If , then

That is, is an inverse to ( and commute).