Quotient modules

Let R be a ring with 1, and let N ďM be R-modules.Since M is an abelian group, any subgroup is normal, so the(additive) quotient group

M{N “ tm`N | m PMu is well-defined.

TheoremThe group M{N is also an R-module under the action

r ¨m “ r ¨m where m “ m`N

Moreover, the natural projection π : M ÑM{N is an R-modulehomomorphism.

Proof. We already know M{N is well-defined as a group and thatπ is a group homomorphism (with kernel N). So since therelations of the ring action are inherited from the action on M , weneed only check (1) the action is well-defined, and (2) theR-action factors out of π (i.e. πpr ¨mq “ r ¨ πpmq).

Quotient modules

Let R be a ring with 1, and let N ďM be R-modules.Since M is an abelian group, any subgroup is normal, so the(additive) quotient group

M{N “ tm`N | m PMu is well-defined.

TheoremThe group M{N is also an R-module under the action

r ¨m “ r ¨m where m “ m`N

Moreover, the natural projection π : M ÑM{N is an R-modulehomomorphism.

Proof. We already know M{N is well-defined as a group and thatπ is a group homomorphism (with kernel N). So since therelations of the ring action are inherited from the action on M , weneed only check (1) the action is well-defined, and (2) theR-action factors out of π (i.e. πpr ¨mq “ r ¨ πpmq).

Isomorphism theorems

Let R be a ring with 1, and M an R-module, with submodules Aand B.

1. If ϕ : M Ñ N is an R-module homomorphism, then kerpϕq isa submodule of M and

M{kerpϕq – ϕpMq.

2. We have

pA`Bq{B – A{pAXBq.

3. Suppose A ď B. Then B{A is a submodule of M{A and

pM{Aq{pB{Aq – pM{Bq.

4. For a submodule N ďM , the natural projectionπ : M ÑM{N gives a bijection

tA | A submod, N ď A ďMu ÐÑ tA | A submod M{Nu

where A “ πpAq “ A{N . This bijection additionally preservescontainment, sums, and intersections.

Direct sums and productsJust like for vector spaces, the direct sum A1 ‘ ¨ ¨ ¨ ‘An of a finitecollection of R-modules is equal to A1 ˆ ¨ ¨ ¨ ˆAn an (additive)group, along with the R-action (See also [D & F, §10.3, ex 20])

r ¨ pa1, . . . , anq “ pr ¨ a1, . . . , r ¨ anq.

Proposition (D & F, §10.3, Prop 5)

Let M be an R-module, and let A1, . . . , An be a collection ofsubmodules. Writing A “ A1 ` ¨ ¨ ¨ `An, the following areequivalent:

1. The map

ρ :n

à

i“1

Ai Ñ A pa1, . . . , anq ÞÑ a1 ` ¨ ¨ ¨ ` an

is an R-module isomorphism.

2. For all j “ 1, . . . , n, we haveAj X pA1 ` ¨ ¨ ¨ `Aj´1 `Aj`1 ` ¨ ¨ ¨ `Anq “ 0.

3. Every a P A can be written uniquely in the forma “ a1 ` ¨ ¨ ¨ ` an.

Direct sums and productsJust like for vector spaces, the direct sum A1 ‘ ¨ ¨ ¨ ‘An of a finitecollection of R-modules is equal to A1 ˆ ¨ ¨ ¨ ˆAn an (additive)group, along with the R-action (See also [D & F, §10.3, ex 20])

r ¨ pa1, . . . , anq “ pr ¨ a1, . . . , r ¨ anq.

Proposition (D & F, §10.3, Prop 5)

Let M be an R-module, and let A1, . . . , An be a collection ofsubmodules. Writing A “ A1 ` ¨ ¨ ¨ `An, the following areequivalent:

1. The map

ρ :n

à

i“1

Ai Ñ A pa1, . . . , anq ÞÑ a1 ` ¨ ¨ ¨ ` an

is an R-module isomorphism.

2. For all j “ 1, . . . , n, we haveAj X pA1 ` ¨ ¨ ¨ `Aj´1 `Aj`1 ` ¨ ¨ ¨ `Anq “ 0.

3. Every a P A can be written uniquely in the forma “ a1 ` ¨ ¨ ¨ ` an.

Proposition (See also D & F, Prop 5.1.2 & Ex. 5.1.14.)

Let M1, . . . ,Mn be R-modules, and let M “M1 ‘ ¨ ¨ ¨ ‘Mn.

(1) Fix i P rns. The setMi “ tp0, . . . , 0, m

loomoon

i

, 0, . . . , 0q | m PMiu ĎM

is a submodule isomorphic to Mi, and

M{Mi –M1 ˆ ¨ ¨ ¨ ˆMi´1 ˆMi`1 ˆMn.

(2) For each fixed i define the projectionπi : M ÑMi by pm1, . . . ,mnq ÞÑ mi.

Then πi is a surjective homomorphism withkerpπiq “ tpm1, . . .mi´1, 0,mi`1, . . .mnq|mj PMju –M{Mi.

(3) For each i P rns, let Ni is a submodule of Mi. ThenN “ N1 ‘ ¨ ¨ ¨ ‘Nn is a submodule of M , and

M{N – pM1{N1q ‘ ¨ ¨ ¨ ‘ pMn{Nnq.

Schur’s lemma

Recall that a M is simple or irreducible if its only submodules areM and 0.

Theorem (Schur’s lemma)

If M and M 1 are simple R-modules and ϕ : M ÑM 1 is anR-module homomorphism, then ϕ is either the 0 map, or it is anisomorphism.

Proof.Consider the kernel and image of ϕ, recalling that both arethemselves modules.

Schur’s lemma

If M and M 1 are simple R-modules and ϕ : M ÑM 1 is anR-module homomorphism, then ϕ is either the 0 map, or it is anisomorphism.

Some consequences:

1. If M is simple, then

EndRpMq “ tR-module homomorphisms ϕ : M ÑMu

is a division ring, i.e. EndRpMq “ AutR Y t0u.

2. If N is a simple submodule of an R-module M , andϕ : M Ñ L is an R-module homomorphism, then N Ă kerpϕqor L contains a submodule isomorphic to N .

3. If M – N1 ‘N2, where N1 and N2 are simple nonisomorphicmodules, then

EndRpMq “ EndRpN1q ˆ EndRpN2q,

and, again, EndRpNiq are division rings.

Schur’s lemma

If M and M 1 are simple R-modules and ϕ : M ÑM 1 is anR-module homomorphism, then ϕ is either the 0 map, or it is anisomorphism.

Some consequences:

1. If M is simple, then

EndRpMq “ tR-module homomorphisms ϕ : M ÑMu

is a division ring, i.e. EndRpMq “ AutR Y t0u.

2. If N is a simple submodule of an R-module M , andϕ : M Ñ L is an R-module homomorphism, then N Ă kerpϕqor L contains a submodule isomorphic to N .

3. If M – N1 ‘N2, where N1 and N2 are simple nonisomorphicmodules, then

EndRpMq “ EndRpN1q ˆ EndRpN2q,

and, again, EndRpNiq are division rings.

Schur’s lemma

If M and M 1 are simple R-modules and ϕ : M ÑM 1 is anR-module homomorphism, then ϕ is either the 0 map, or it is anisomorphism.

Some consequences:

1. If M is simple, then

EndRpMq “ tR-module homomorphisms ϕ : M ÑMu

is a division ring, i.e. EndRpMq “ AutR Y t0u.

2. If N is a simple submodule of an R-module M , andϕ : M Ñ L is an R-module homomorphism, then N Ă kerpϕqor L contains a submodule isomorphic to N .

3. If M – N1 ‘N2, where N1 and N2 are simple nonisomorphicmodules, then

EndRpMq “ EndRpN1q ˆ EndRpN2q,

and, again, EndRpNiq are division rings.

Schur’s lemma

If M and M 1 are simple R-modules and ϕ : M ÑM 1 is anR-module homomorphism, then ϕ is either the 0 map, or it is anisomorphism.

Some consequences:

4. Homework: If M is simple and N is a nontrivial propersubmodule of M , then N –M .

In general, every submodule of M‘n is isomorphic to M‘k forsome k “ 0, . . . , n.Consequently, if A and B are completely decomposableR-modules with

A –à

λPΛ1

M‘cλλ , B –

à

λPΛ2

M‘dλλ

with Mλ being distinct (up to isomorphism) simpleR-modules, we have

HomRpA,Bq –à

λPΛ1XΛ2

HomRpM‘cλλ ,M‘dλ

λ q.

Schur’s lemma

If M and M 1 are simple R-modules and ϕ : M ÑM 1 is anR-module homomorphism, then ϕ is either the 0 map, or it is anisomorphism.

Some consequences:

4. Homework: If M is simple and N is a nontrivial propersubmodule of M , then N –M .In general, every submodule of M‘n is isomorphic to M‘k forsome k “ 0, . . . , n.

Consequently, if A and B are completely decomposableR-modules with

A –à

λPΛ1

M‘cλλ , B –

à

λPΛ2

M‘dλλ

with Mλ being distinct (up to isomorphism) simpleR-modules, we have

HomRpA,Bq –à

λPΛ1XΛ2

HomRpM‘cλλ ,M‘dλ

λ q.

Schur’s lemma

If M and M 1 are simple R-modules and ϕ : M ÑM 1 is anR-module homomorphism, then ϕ is either the 0 map, or it is anisomorphism.

Some consequences:

4. Homework: If M is simple and N is a nontrivial propersubmodule of M , then N –M .In general, every submodule of M‘n is isomorphic to M‘k forsome k “ 0, . . . , n.Consequently, if A and B are completely decomposableR-modules with

A –à

λPΛ1

M‘cλλ , B –

à

λPΛ2

M‘dλλ

with Mλ being distinct (up to isomorphism) simpleR-modules, we have

HomRpA,Bq –à

λPΛ1XΛ2

HomRpM‘cλλ ,M‘dλ

λ q.

Free modules

An R-module M is said to be free on a subset S ĂM if for everynon-zero element m PM , there is a unique set of non-zeroelements r1, . . . , rn P R and a1, . . . , an P S such that

m “ r1a1 ` ¨ ¨ ¨ ` rnan (n depends on m).

We say S is a basis of free generators for M . If R is commutative,then |S| is called the rank of M (over R).

Idea: Free R-modules are the “most like vector spaces”.

Ex: The free abelian group of rank r is Zr, and is a free Z-moduleof rank r.

Note that the uniqueness must come on both sides (from the ringand the module).

Non-ex: The trivial module T for CS3 is not free: for any v P T ,we have 1 ¨ v “ p12q ¨ v “ p2p12q ´ p132qqv “ ¨ ¨ ¨ , and so on.

Note: If R ‰ 0 and M is free with basis S, then 0 R S.

Free modules

An R-module M is said to be free on a subset S ĂM if for everynon-zero element m PM , there is a unique set of non-zeroelements r1, . . . , rn P R and a1, . . . , an P S such that

m “ r1a1 ` ¨ ¨ ¨ ` rnan (n depends on m).

We say S is a basis of free generators for M . If R is commutative,then |S| is called the rank of M (over R).

Idea: Free R-modules are the “most like vector spaces”.

Ex: The free abelian group of rank r is Zr, and is a free Z-moduleof rank r.

Note that the uniqueness must come on both sides (from the ringand the module).

Non-ex: The trivial module T for CS3 is not free: for any v P T ,we have 1 ¨ v “ p12q ¨ v “ p2p12q ´ p132qqv “ ¨ ¨ ¨ , and so on.

Note: If R ‰ 0 and M is free with basis S, then 0 R S.

Free modules

An R-module M is said to be free on a subset S ĂM if for everynon-zero element m PM , there is a unique set of non-zeroelements r1, . . . , rn P R and a1, . . . , an P S such that

m “ r1a1 ` ¨ ¨ ¨ ` rnan (n depends on m).

We say S is a basis of free generators for M . If R is commutative,then |S| is called the rank of M (over R).

Idea: Free R-modules are the “most like vector spaces”.

Ex: The free abelian group of rank r is Zr, and is a free Z-moduleof rank r.

Note that the uniqueness must come on both sides (from the ringand the module).

Non-ex: The trivial module T for CS3 is not free: for any v P T ,we have 1 ¨ v “ p12q ¨ v “ p2p12q ´ p132qqv “ ¨ ¨ ¨ , and so on.

Note: If R ‰ 0 and M is free with basis S, then 0 R S.

Free modules

An R-module M is said to be free on a subset S ĂM if for everynon-zero element m PM , there is a unique set of non-zeroelements r1, . . . , rn P R and a1, . . . , an P S such that

m “ r1a1 ` ¨ ¨ ¨ ` rnan (n depends on m).

We say S is a basis of free generators for M . If R is commutative,then |S| is called the rank of M (over R).

Idea: Free R-modules are the “most like vector spaces”.

Ex: The free abelian group of rank r is Zr, and is a free Z-moduleof rank r.

Note that the uniqueness must come on both sides (from the ringand the module).

Non-ex: The trivial module T for CS3 is not free: for any v P T ,we have 1 ¨ v “ p12q ¨ v “ p2p12q ´ p132qqv “ ¨ ¨ ¨ , and so on.

Note: If R ‰ 0 and M is free with basis S, then 0 R S.

Free modules

An R-module M is said to be free on a subset S ĂM if for everynon-zero element m PM , there is a unique set of non-zeroelements r1, . . . , rn P R and a1, . . . , an P S such that

m “ r1a1 ` ¨ ¨ ¨ ` rnan (n depends on m).

We say S is a basis of free generators for M . If R is commutative,then |S| is called the rank of M (over R).

Idea: Free R-modules are the “most like vector spaces”.

Ex: The free abelian group of rank r is Zr, and is a free Z-moduleof rank r.

Note that the uniqueness must come on both sides (from the ringand the module).

Non-ex: The trivial module T for CS3 is not free: for any v P T ,we have 1 ¨ v “ p12q ¨ v “ p2p12q ´ p132qqv “ ¨ ¨ ¨ , and so on.

Note: If R ‰ 0 and M is free with basis S, then 0 R S.

Free modules

An R-module M is said to be free on a subset S ĂM if for everynon-zero element m PM , there is a unique set of non-zeroelements r1, . . . , rn P R and a1, . . . , an P S such that

m “ r1a1 ` ¨ ¨ ¨ ` rnan (n depends on m).

We say S is a basis of free generators for M . If R is commutative,then |S| is called the rank of M (over R).

Idea: Free R-modules are the “most like vector spaces”.

Ex: The free abelian group of rank r is Zr, and is a free Z-moduleof rank r.

Note that the uniqueness must come on both sides (from the ringand the module).

Non-ex: The trivial module T for CS3 is not free: for any v P T ,we have 1 ¨ v “ p12q ¨ v “ p2p12q ´ p132qqv “ ¨ ¨ ¨ , and so on.

Note: If R ‰ 0 and M is free with basis S, then 0 R S.

Free modules

An R-module M is said to be free on a subset S ĂM if for everynon-zero element m PM , there is a unique set of non-zeroelements r1, . . . , rn P R and a1, . . . , an P S such that

m “ r1a1 ` ¨ ¨ ¨ ` rnan (n depends on m).

We say S is a basis of free generators for M . If R is commutative,then |S| is called the rank of M (over R).

Idea: Free R-modules are the “most like vector spaces”.

Ex: The free abelian group of rank r is Zr, and is a free Z-moduleof rank r.

Note that the uniqueness must come on both sides (from the ringand the module).

Non-ex: The trivial module T for CS3 is not free: for any v P T ,we have 1 ¨ v “ p12q ¨ v “ p2p12q ´ p132qqv “ ¨ ¨ ¨ , and so on.

Note: If R ‰ 0 and M is free with basis S, then 0 R S.

Free modulesFree (over S): For every 0 ‰ m PM , there are uniquer1, . . . , rn P R´ 0 and a1, . . . , an P S such thatm “ r1a1 ` ¨ ¨ ¨ ` rnan.

TheoremFor any set S, there is a free R-module FrpSq on the set S.

Moreover, FrpSq satisfies the following “universal property”: If Mis any R module and ϕ : S ÑM is any map (of sets), then thereis a unique R-module homomorphism Φ : FrpSq ÑM such thatΦ|S “ ϕ:

S FrpSq

M

incl

ϕ ! Φö

Finally, if |S| “ n ă 8, then FrpSq –à

aPS

Ra – Rn.

Free modulesFree (over S): For every 0 ‰ m PM , there are uniquer1, . . . , rn P R´ 0 and a1, . . . , an P S such thatm “ r1a1 ` ¨ ¨ ¨ ` rnan.

TheoremFor any set S, there is a free R-module FrpSq on the set S.

Moreover, FrpSq satisfies the following “universal property”: If Mis any R module and ϕ : S ÑM is any map (of sets), then thereis a unique R-module homomorphism Φ : FrpSq ÑM such thatΦ|S “ ϕ:

S FrpSq

M

incl

ϕ ! Φö

Finally, if |S| “ n ă 8, then FrpSq –à

aPS

Ra – Rn.

Free modulesFree (over S): For every 0 ‰ m PM , there are uniquer1, . . . , rn P R´ 0 and a1, . . . , an P S such thatm “ r1a1 ` ¨ ¨ ¨ ` rnan.

TheoremFor any set S, there is a free R-module FrpSq on the set S.

Moreover, FrpSq satisfies the following “universal property”: If Mis any R module and ϕ : S ÑM is any map (of sets), then thereis a unique R-module homomorphism Φ : FrpSq ÑM such thatΦ|S “ ϕ:

S FrpSq

M

incl

ϕ ! Φö

Finally, if |S| “ n ă 8, then FrpSq –à

aPS

Ra – Rn.