Ring Theory - University College Dublinmathsci.ucd.ie/~astier/alg-grad/notes.pdf · Chapter 1...

Ring TheoryMATH 40010 – 2008-2009

Vincent Astier

Contents

1 Beginning 51.1 Basic notions . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Hamilton’s quaternions . . . . . . . . . . . . . . . . . . . . . . 13

2 Modules 15

3 Wedderburn’s theorem on finite division rings 25

4 Primitive rings and the density theorem 294.1 A way around this problem . . . . . . . . . . . . . . . . . . . 294.2 The study of α(R) and the density theorem . . . . . . . . . . 304.3 Consequences . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.4 Simple algebras . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5 The Jacobson radical 45

6 Semisimple modules and rings 516.1 Composition series . . . . . . . . . . . . . . . . . . . . . . . . 516.2 Semisimplicity . . . . . . . . . . . . . . . . . . . . . . . . . . . 546.3 The structure of semisimple rings . . . . . . . . . . . . . . . . 58

3

4 CONTENTS

Chapter 1

Beginning

1.1 Basic notions

Definition 1.1 A ring is a nonempty set R containing an element 0 andan element 1 and equipped with two binary operations + and . such that:

1. (R,+, 0) is an Abelian group;

2. . is associative, i.e. for every a, b, c ∈ R a.(b.c) = (a.b).c;

3. Left and right distributivity:For every a, b, c ∈ R a.(b+ c) = a.b+ a.c and (a+ b).c = a.c+ b.c;

4. For all a ∈ R a.1 = 1.a = a.

R is said to be commutative if ab = ba for all a, b ∈ R.

Remarks:

• In some books, a ring is what we will call a “ring without 1”, i.e. aring that does not contain an element 1 satisfying the last condition.

• We will simply write ab for a.b.

If R is a ring, a ∈ R and n ∈ Z, we define:

na =

a+ · · ·+ a (n times) if n > 00 if n = 0−a+ · · ·+ (−a) (n times) if n < 0

.

Definition 1.2 A subset S of a ring R is called a subring if 0, 1 ∈ S andS, equipped with the addition and multiplication form R, is itself a ring (withsame 0 and 1 as R).

5

6 CHAPTER 1. BEGINNING

Remark: S is a subring of R if and only if 1 ∈ S and if a, b ∈ S then−a, a+ b, ab ∈ S.

Corollary 1.3 Let R be a ring. For all a, b ∈ R, n ∈ Z:

1. 0a = a0 = 0.

2. (−1)a = −a = a(−1).

3. (−a)b = a(−b) = −(ab).

4. (−a)(−b) = ab.

5. (na)b = a(nb) = n(ab)

6. If ab = ba then (a+ b)n =∑n

i=0

(ni

)aibn−i.

Proof: See exercise sheet 1. �

Definition 1.4 Let R be a ring.

1. a ∈ R \ {0} is a left (respectively right) zero divisor if there existsb ∈ R \ {0} such that ab = 0 (respectively ba = 0).

2. a ∈ R is said to be left (respectively right) invertible if there existsc ∈ R (respectively b ∈ R) such that ca = 1 (respectively ab = 1).The element c (respectively b) is called a left inverse of a (respectivelyright inverse of a).An element which is both left and right invertible is called invertibleor a unit. The set of units of R will be denoted by U(R).

Corollary 1.5 Let R be a ring.

1. If a ∈ U(R), then the left and right inverses of a coincide, and aredenoted by a−1, the inverse of a.

2. (U(R), ., 1) is a group.

Definition 1.6 A ring R 6= {0} with no (left or right) zero divisor is calleda domain.A ring in which every nonzero element is a unit is called a division ring(or a skew field). A field is a commutative division ring.

Examples:

• (Z,+, ., 0, 1) is a domain.

1.1. BASIC NOTIONS 7

• (C,+, ., 0, 1) is a field.

• (Mn(R),+, ., 0, In) is a ring (a noncommutative ring if n ≥ 2).

• (Z/nZ,+, ., 0, 1) is a commutative ring.

• R[X] is a domain.

• Z/6Z has zero divisors (for example 3.2 = 0).

• Z/pZ is a field if p is prime:Let b ∈ Z with image b ∈ Z/pZ. We must find the inverse of b. Sincep is prime, we have gcd(b, p) = 1 and there exists u, v ∈ Z such thatbu+ pv = 1. In Z/pZ, this gives bu+ pv = 1, and (using that pv = 0):bu = 1.Since Z/pZ is commutative, bu = ub = 1, and u is the inverse of b.

Definition 1.7 Let R and S be rings. A map f : R 7→ S is a homomor-phism (or morphism) of rings if:

1. f(1) = 1.

2. For all a, b ∈ R, f(a+ b) = f(a) + f(b) and f(ab) = f(a)f(b).

If f is injective (resp. surjective, bijective), then f is called a monomor-phism (resp. epimorphism, isomorphism) of rings.A morphism of rings from R to R is called an endomorphism of R.The kernel of f is ker(f) = {x ∈ R | f(x) = 0}.The image of f is Im(f) = {f(x) | x ∈ R}.Two rings R and S are isomorphic if there is an isomorphism of rings ffrom R to S. This is denoted by R ∼= S (or R ' S).

Examples of morphisms: R → Mn(R)x 7→ x.In

, Z → Z/pZx 7→ x+ pZ

.

Z → Zx 7→ 2x

is not a morphism since it does not send 1 to 1.

If f : R 7→ S is a morphism of rings, ker(f) has some obvious properties,for instance: 0 ∈ ker(f), and if a, b ∈ ker(f) and r ∈ R, then a + b, ar, ra ∈ker(f). In other words, ker f is an additive subgroup of R, with the additionalproperty that ar, ra ∈ ker f whenever a ∈ ker f and r ∈ RThis motivates the following definition:

Definition 1.8 Let R be a ring. A subset I of R is called a left (respectivelyright) ideal of R if:


1. I is an additive subgroup of R.

2. For every a ∈ I and every r ∈ R we have ra ∈ I (respectively ar ∈ I).

I is called an ideal (or 2-sided ideal) if I is both a left and right ideal.An ideal which is different from R is called a proper ideal.

Remarks:

• ker(f) is an ideal.

• A left (resp. right) ideal I is proper if and only if 1 6∈ I:If 1 6∈ I, then I is necessarily proper. Conversely, suppose I is proper,but 1 ∈ I. Then for every r ∈ R, r = r.1 ∈ I (resp. r = 1.r ∈ I),which gives I = R, a contradiction.

The role of ideals in ring theory can be compared to the role of normalsubgroups in group theory (this will become clear in the next few results).

If R is a ring and I is an ideal of R, then I is a normal subgroup of(R,+, 0) since (R,+, 0) is a commutative group.In particular, the quotient group (R/I,+, 0 + I) is well-defined with:

(a+ I) + (b+ I) = (a+ b) + I.

We can even define a ring structure on R/I:

Theorem 1.9 Let R be a ring and I and ideal (i.e. 2-sided ideal) of R.Then:

1. The additive group R/I can be turned into a ring with the multiplicationgiven by:

(a+ I).(b+ I) = (ab) + I,

and with identity 1 + I.R/I is called the quotient ring of R by I.If R is commutative, then R/I is commutative.

2. The map π : R → R/Ia 7→ a+ I

is a surjective morphism of rings, and

ker(π) = I. π is called the canonical projection from R to R/I.

Proof:

1. We first check that the multiplication is well-defined, i.e.:

(a+ I = a′ + I and b+ I = b′ + I)⇒ (ab) + I = (a′b′) + I.


We have a′ = a+ i, b′ = b+ j with i, j ∈ I.a′b′ = (a+ i)(b+ j) = ab+ aj + ib+ ij, and the last three terms (andhence their sum) are in I since I is an ideal. This gives a′b′+I = ab+I.The other properties (associativity, distributivity, identity element) areeasy to check.

2. To check that π is a morphism of rings, we use the definitions of sumand product in R/I, namely:

(a+ b) + I = (a+ I) + (b+ I) and (a.b) + I = (a+ I).(b+ I).

With this π(a + b) = (a + b) + I = (a + I) + (b + I) = π(a) + π(b),π(ab) = (ab) + I = (a+ I).(b+ I) = π(a).π(b), and π(1) = 1 + I.

�

Remark: We have seen that if f is a morphism of rings, ker(f) is an ideal.Theorem 1.9 2 gives the converse: every ideal is of the form ker(f) for somemorphism of rings f (the π in the theorem).

Proposition 1.10 Let I be an ideal in a ring R. Then there is a one-to-onecorrespondence between the ideals of R containing I and the ideals of R/I,given by J 7→ J/I.(This describes the ideals of R/I using the ideals of R.)

Proof: We first show that the map is injective: Let J1, J2 be ideals of Rcontaining I, such that J1/I = J2/I.Let a ∈ J1. Then a + I ∈ J1/I = J2/I. This implies that there exist i ∈ Iand b ∈ J2 such that a+ i = j2 i.e. a = j2 − i. But j2 − i ∈ J2 since I ⊆ J2.This proves a ∈ J2 and then J1 ⊆ J2.Similarly we have J2 ⊆ J1, which gives J1 = J2.

We now show that the map is surjective: Let L be an ideal of R/I, anddefine J = π−1(L). We check easily that J is an ideal of R (in fact, if f isa morphism of rings f−1(an ideal) is an ideal), and since π is surjective wehave π(J) = L, i.e. J/I = L. �

Lemma 1.11 If {Ik}k∈K are left (resp. right) ideals of a ring R, then⋂k∈K Ik is a left (resp. right) ideal of R (i.e. any intersection of left (resp.

right) ideals is a left (resp. right) ideal).

Proof: Exercise. �

Definition/Corollary 1.12 Let X be a subset of a ring R. The ideal gen-erated by X, denoted by (X), is the intersection of the ideals containing X,i.e.:

(X) =⋂{I | I ideal, X ⊆ I}.


This is the smallest ideal containing X. The elements of X are called thegenerators.We can also define the left (resp. right) ideal generated by X, by taking theintersection of all the left (resp. right) ideals containing X.If X = {x1, . . . , xn} then (X) is denoted by (x1, . . . , xn) and is called finitelygenerated.An ideal (a) generated by a single element a ∈ R is called principal.

Let A1, . . . An be nonempty subsets of a ring R. We denote by A1 + · · ·+Anthe set:

{a1 + · · ·+ an | ai ∈ Ai, i = 1, . . . , n},and by A1. . . . .An the set:

{m∑k=1

a1,k. . . . .an,k | m ∈ N∗, ai,k ∈ Ai, i = 1, . . . , n, k = 1, . . . ,m}

(the set of finite sums of products of the form a1. . . . .an for ai ∈ Ai).

Definition/Proposition 1.13 Let A1, . . . , An be (left, resp. right) ideals ina ring R. Then A1 + · · ·+An and A1. . . . .An are (left, resp. right) ideals ofR.A1 + · · ·+An is called the sum of A1, . . . , An, and A1. . . . .An is called theproduct of A1, . . . , An.

Proof: For left ideals, and A1 + · · · + An (the other parts of the proof aresimilar):

1. A1 + · · ·+ An is an additive subgroup of R:Let a1 + · · ·+an, b1 + · · ·+bn ∈ A1 + · · ·+An. Then −a1 + · · ·+(−an) ∈A1 + · · · + An (since −ai ∈ Ai), and a1 + · · · + an + b1 + · · · + bn ∈A1 + · · ·+ An (since ai + bi ∈ Ai).

2. Product by elements of R (on the left):Let a1 + · · ·+ an ∈ A1 + · · ·+ An and r ∈ R. Then r(a1 + · · ·+ an) =ra1 + · · · + ran ∈ A1 + · · · + An (because rai ∈ Ai, since Ai is a leftideal).

A1 + · · ·+ An is then a left ideal of R. �

Remark: A1 + · · · + An is the smallest (left, resp. right) ideal containingA1, . . .+ An.This can be seen as follows (for left ideals):The smallest left ideal containing A1, . . . , An is by definition (A1 ∪ · · · ∪An),i.e is ⋂

{I | I left ideal, A1, . . . , An ⊆ I}.


We now show that A1 + · · · + An is contained in any left ideal containingA1, . . . , An.Let I be an ideal containing A1, . . . , An. Then I 3 a1 + · · · an, for everyai ∈ Ai, i = 1, · · · , n. This means I ⊇ A1 + · · ·+An, which implies (A1∪· · ·∪An) ⊇ A1 + · · ·An. The first one is the smallest ideal containing A1, . . . , An,and the second one is an even smaller ideal containing A1, . . . , An. Thisimplies (A1 ∪ · · · ∪ An) = A1 + · · ·An.

Theorem 1.14 (Isomorphism theorems) We have the following:

1. Let f : R→ S be a morphism of rings. Then f induces an isomorphismof rings:

f : R/ ker(f) → Im(f)x+ ker(f) 7→ f(x)

.

This result is also true for rings without 1.

2. Let I and J be ideals in a ring R. Then:

(a) There is an isomorphism of rings:

I/(I ∩ J) → (I + J)/Jx+ I ∩ J 7→ x+ J

.

(b) If I ⊆ J , then J/I is an ideal of R/I and there is an isomorphismof rings:

(R/I)/(J/I) → R/J(x+ I) + J/I 7→ x+ J

.

Proof:

1. This is true for the additive Abelian groups (R,+), (S,+) and (ker(f),+),i.e.:

f : R/ ker(f) → Im(f)x+ ker(f) 7→ f(x)

,

is an isomorphism of additive groups.We only have to check that it is a morphism of rings. Let a, b ∈ R.Then f((a+ker(f))(b+ker(f)) = f(ab+ker(f)) = f(ab) = f(a)f(b) =f(a+ ker(f))f(b+ ker(f)), and we also have f(1 + ker(f)) = f(1) = 1.

2. (a) Apply the first part for f : I → (I + J)/Jx 7→ (x+ 0) + J

.

(b) Apply the first part for f : R/I → R/Jx+ I 7→ x+ J

.

�


Definition 1.15 A ring R is simple if the only ideals of R are {0} and R.

Example: A division ring is a simple ring (we will see other examples later).

We know consider rings of matrices:

Definition/Proposition 1.16 Let R be a ring, and let n ∈ N. The ringof n× n matrices over R is the set of n× n matrices with coefficients in R,equipped with the usual sum and product of matrices. Its identity element isIn.It is denoted by Mn(R).

No proof (not hard, but long). �

Examples: Mn(Z), Mn(H), Mn(Z/nZ).

Proposition 1.17 Let R be a ring. Then J is an ideal of Mn(R) if and onlyif J is of the form Mn(I) (the set of n × n matrices with coefficients in I)for some ideal I of R.

Proof: “⇐” Let I be an ideal of R. Then Mn(I) is an additive subgroup ofMn(R) (since I is an additive subgroup of R).Let (aij)1≤i,j≤n ∈Mn(I), and (bij)1≤i,j≤n ∈Mn(R).Then (aij)1≤i,j≤n(bij)1≤i,j≤n = (cij)1≤i,j≤n, with cij =

∑nk=1 aikbkj. The coef-

ficients of the sum are in I (since I is an ideal and aik ∈ I), and then cij ∈ I.This proves (cij)1≤i,j≤n ∈Mn(I).Similarly, (bij)1≤i,j≤n(aij)1≤i,j≤n ∈Mn(I).Mn(I) is then an ideal of Mn(R).

”⇒” Let J be an ideal of Mn(R) and define:

I = {a ∈ R | a appears in the top left corner of some matrix in J}.

I is clearly an ideal of R: take a, b ∈ I and r ∈ R. Then a (resp. b) appearsin the top left corner of some matrix A ∈ J (resp. B ∈ J), and we see thata + b is given by the matrix A + B (which is in J since J is an ideal), andra, ar are given by the matrices (rIn).A,A.(rIn) (which also are in J , sinceJ is an ideal).

We know check that J = Mn(I) and for this we use the following lemma:

Lemma 1.18 For 1 ≤ r, s ≤ n, let Er,s be the matrix in Mn(R) with 1 asthe row r-column s entry and 0 elsewhere.Then for every matrix A = (aij)1≤i,j≤n ∈Mn(R), the matrix Ep,rAEs,q is thematrix with ars as row r-column s entry and 0 elsewhere.

1.2. HAMILTON’S QUATERNIONS 13

No proof (direct and easy computation). �

We prove that Mn(I) ⊆ J :Let A = (aij)1≤i,j≤n ∈ Mn(I), i.e. there exist matrices Aij ∈ J such that aijappears in the top left corner of Aij.Since Aij ∈ J , we will express A using the matrices Aij. Using the previouslemma, we know that the matrix Ei,1AijE1,j contains only zeroes, exceptfor the coordinate (i, j) which is aij. Moreover, this matrix is in J becauseAij ∈ J and J is an ideal.Then A =

∑1≤i,j≤nEi,1AijE1,j ∈ J .

To conclude, we check that J ⊆Mn(I):Let A = (aij)1≤i,j≤n ∈ J . We have to show that aij ∈ I for every 1 ≤ i, j ≤ n.But using lemma 1.18 we have:

aij 0 · · · 00 0 · · · 0. . . . . .0 · · · · · · 0

= E1,iAEj,1,

with E1,iAEj,1 ∈ J since A ∈ J and J ideal. This gives aij ∈ I for everyi, j ∈ {1, . . . , n}, and thus A ∈Mn(I). �

1.2 Hamilton’s quaternions

We know that the ring C is completely determined by:

• C is a ring.

• C is R⊕ Ri as R-vector space.

• i2 = −1 and ia = ai for every a ∈ R.

This has been generalized (by Hamilton):

Definition 1.19 The ring H of (real) quaternions is the ring determinedby:

• H is a ring.

• H = R⊕ Ri⊕ Rj ⊕ Rk as R-vector space.

• i2 = j2 = −1, ij = −ji = k, and i, j, k commute with the elements ofR.


One can check that the definition of ring is verified.We have:

• ik = i(ij) = i2j = −j.

• jk = j(ij) = j(−ji) = −j2i = i.

• ki = (ij)i = (−ji)i = −ji2 = j.

• kj = (ij)j = ij2 = −i.

• k2 = (ij)(ij) = (ij)(−ji) = −ij2i = i2 = −1.

For α = a+ bi+ cj + dk ∈ H we define α= a− bi− cj − dk, the conjugateof α, and we have:αα = (a+ bi+ cj + dk)(a− bi− cj − dk)

= a2 − abi− acj − adk + bia− (bi)(bi)− (bi)(cj)− (bi)(dk)+cja− (cj)(bi)− (cj)(cj)− (cj)(dk) + dka− (dk)(bi)−(dk)(cj)− (dk)(dk)

= a2 − abi− acj − adk + abi+ b2 − bck + bdj+caj + bck + c2 − cdi+ adk − dbj + cdi+ d2

= a2 + b2 + c2 + d2.Applying this to α we get α ¯α = αα = a2 + b2 + c2 + d2, and we then define:

n(α) = αα = αα = a2 + b2 + c2 + d2,

the norm of α.Since the norm is a sum of squares in R, we get the following importantconsequence:If α 6= 0 then n(α) 6= 0 and has an inverse in R. This gives:

(n(α)−1α)α = n(α)−1(αα) = n(α)−1n(α) = 1, and

α(n(α)−1α) = (αα)n(α)−1 = n(α)n(α)−1 = 1

In other words, every nonzero element α ∈ H has an inverse, and this inverseis α−1 = n(α)−1α.H is then a division ring (but H is not commutative since ij = −ji).

Chapter 2

Modules

Definition 2.1 Let R be a ring. A (left) R-module is an additive abeliangroup (M, 0,+) together with a product R×M → M

(r,m) 7→ r.m(multiplication

on the left by the elements of R), such that, for every r, s ∈ R and a, b ∈M :

1. r(a+ b) = ra+ rb;

2. (r + s)a = ra+ sa;

3. r(sa) = (rs)a;

4. 1.a = a.

If R is a field, then a left R-module is called a vector space (over R).

Remarks:

1. There is also a notion of right R-module, with multiplication by ele-ments of R on the right: M ×R → M

(m, r) 7→ m.r.

2. This is a generalization of the notion of vector space.

Examples:

1. Every vector space.

2. Let R be a ring, and R[X] = the ring of polynomials in one variableover R. Then R[X] is an R-module, with:

R×R[X] → R[X](r, a0 + a1X + · · ·+ anX

n) 7→ ra0 + ra1X + · · ·+ ranXn

.

15

16 CHAPTER 2. MODULES

3. Let (G, 0,+) be an Abelian group. Then G is a Z-module with:

Z×G → G(n, g) 7→ ng

.

4. Let R be a ring, and let I be a left ideal of R. Then I is a (left)R-module with:

R× I → I(r, a) 7→ ra

.

Definition 2.2 Let M be an R-module. A subset N of M is called a sub-module if N is an additive subgroup of M and rb ∈ N for all r ∈ R, b ∈ N .

Examples:

1. R is an R-module. The left ideals of R are exactly the submodules ofR (compare the definitions).

2. A vector space is a module. A sub-vector space is a submodule.

Remarks:

1. A submodule is always a module (with the product given by the mod-ule).

2. Any intersection of submodules is a submodule.

Definition 2.3 Let M be an R-module, and let X ⊆ M . The intersectionof all submodules of M containing X is a submodule which is called thesubmodule generated by X (it is the smallest submodule of M containingX).

Proposition 2.4 Let M be an R-module, and let X ⊆ M . The submoduleN generated by X is:

{n∑i=1

riai | n ∈ N, ti ∈ R, ai ∈ X for i = 1, · · · , n}

(it is the set of linear combinations of elements of X, with coefficients in R).

Proof: Let L = {∑n

i=1 riai | n ∈ N, ti ∈ R, ai ∈ X for i = 1, · · · , n}. L isclearly a submodule of M containing X, so N ⊆ L.We now show that any submodule containing X must contain L. Let K bea submodule containing X. Let x =

∑ni=1 riai with ri ∈ R and ai ∈ X.

17

Since K ⊇ X and K is a submodule we have successively riai ∈ K and∑ni=1 riai ∈ K. This proves that L is included in any submodule containing

X, and in particular is included in N .We then have L = N . �.

The notion of basis is central for vector spaces. What happens for modules?

Definition 2.5 Let M be an R-module, and let X ⊆M . We say that:

1. X is linearly independent if

∀n ∈ N ∀x1, . . . , xn ∈ X ∀r1, . . . rn ∈ Rr1x1 + · · ·+ rnxn = 0⇒ r1 = · · · = rn = 0.

2. X generates M (or X is a set of generators of M) if

∀a ∈M ∃n ∈ N ∃x1, . . . , xn ∈ X ∃r1, . . . rn ∈ Ra = r1x1 + · · ·+ rnxn.

(i.e the submodule generated by X is M).

3. X is a basis of M if X is linearly independent and X generates M .

Remark: It is possible to define the sum and direct sum of modules, exactlyas for vector spaces.

DANGER:In general you cannot work with bases in modules as you do with bases invector spaces, as the following example shows:

Consider Z as a Z-module (with usual sum and product in Z). Let a ∈ Z,a > 1.Then the submodule generated by A is Z.a = {na | n ∈ Z} 6= Z. We seethat {a} does not generate Z.We also clearly see that {a} is linearly independent.

Consider now b ∈ Z, b 6= 0. We will show that {a, b} is not linearlyindependent. Let d = gcd(a, b). Then there exist u, v ∈ Z such that au+bv =d. Multiplying on both sides by a we get a(au − d) + b(av) = 0. We nowhave two cases:

• If au− d 6= 0 or av 6= 0. Then {a, b} is not linearly independent.

• If au − d = 0 and av = 0. Then au = d, i.e. a divides d. Since (bydefinition of d) d divides a and a > 0 we have d = a. Using that ddivides b we get b = k.a for some k ∈ Z, which gives b− ka = 0. {a, b}is not linearly independent.


We can also check that the submodule generated by {a, b} is:

Z.d = {nd | n ∈ Z},

(see exercise sheet 2) which is different from Z if d > 1.

Recapitulating, we have:

• {a} is linearly independent but cannot be completed to a basisof Z.

• If gcd(a, b) = 1 with a, b > 1, then {a, b} generates Z, {a, b}is not linearly independent, but no proper subset of {a, b}generates Z.

However, modules over a division ring behave correctly:

Lemma 2.6 Let M be an R-module, with R a division ring. Then a maximallinearly independent subset X of M is a basis.

Proof: Let N be the submodule generated by X. By definition, X is a basisof N , and the proof is finished if N = M .Suppose N 6= M , take a ∈M \N and consider X ∪ {a}:If we prove that X ∪ {a} is linearly independent, this will contradict “Xmaximal” and we will have N = M .Suppose ra+ r1x1 + · · ·+ rnxn = 0, with r, r1, . . . rn ∈ R, x1, . . . .xn ∈ X. Wedistinguish two cases:

• If r = 0, then r1x1 + · · · + rnxn = 0, which gives r1 = · · · = rn = 0since X is linearly independent.

• If r 6= 0, we get a = r−1(r1x1 + · · · + rnxn). In particular we havea ∈ N , a contradiction. This case is impossible. �

Theorem 2.7 Let M be a module over a division ring R. Then:

1. Every linearly independent subset of M is contained in a basis of M(in particular, every non-zero module has a basis).

2. Every set of generators of M contains a basis of M .

3. Any two bases of M have the same cardinality.

Proof:

19

1. Let S = {mi}i∈I be a linearly independent subset of M , and let A ={T ⊆M | T is linearly independent}. A, with the (partial) order givenby the inclusion (T1 ≤ T2 iff T1 ⊆ T2) is a partially ordered set. Wewill apply Zorn’s lemma to A:A is non-empty since S ∈ A.Let B ⊆ A be a chain in A We want to find an upper bound of B inA. Let TB =

⋃T∈B T . We first check that TB ∈ A: Let n ∈ N and

x1, . . . xn ∈ TB. By definition of TB there exist T1, . . . , Tn ∈ B suchthat xi ∈ Ti for i = 1, . . . , n.Since B is a chain, one of the Ti contains the others, say T1. We thenhave x1, . . . , xn ∈ T1, and since T1 is linearly independent, {x1, · · · , xn}is linearly independent.This shows that TB is linearly independent, i.e. TB ∈ A. And bydefinition of TB, it is necessarilly an upper bound of B.

Zorn’s lemma then applies and gives a maximal element Tm in A: Tmis linearly independent and is not contained in any bigger linearly in-dependent subset of M .

We now show that Tm generates M :Let N be the submodule generated by Tm, and suppose N 6= M .Choose x ∈M \N . We show that Tm ∪ {x} is linearly independent:Let n ∈ N and x1, . . . , xn ∈ Tm, r, r1, . . . , rn ∈ R such that rx+ r1x1 +· · · + rnxn = 0. We deduce rx = (−r1)x1 + · · · + (−rnxn), and wedistinguish between 2 cases:

• If r = 0 then r1x1+· · ·+rnxn = 0, which gives (since Tm is linearlyindependent): r1 = · · · = rn = 0.

• If r 6= 0 then x = r−1(−r1)x1 + · · · + r−1(−rn)xn ∈ M , a contra-diction. This case is impossible.

Thus the first case is the only possibility, we have r = r1 = · · · = rn = 0.Tm ∪ {x} is linearly independent, which contradicts the fact that Tm ismaximal.

2. Exercise. The idea of the proof goes as follows: Let X be a set of gener-ators of M . Use Zorn’s lemma to find a maximal linearly independentsubset of X. This is a basis.

3. No proof. �

Definition 2.8 Let M and N be R-modules over a ring R. A map f : M →N is a R-module homomorphism (or morphism of R-modules) if for


all a, b ∈M , for all r ∈ R:

f(a+ b) = f(a) + f(b) and f(ra) = rf(a).

f is a monomorphism (resp. epimorphism, isomorphism) if f is in-jective (resp. surjective, bijective).The kernel of f is ker(f) = {x ∈M | f(x) = 0}.The image of f is Im(f) = {f(x) | x ∈M}.EndR M denotes the set of all morphisms of R-modules from M to M (sucha morphism is called an endomorphism).Two modules M and N are isomorphic if there is an isomorphism of modulesf from M to N . This is denoted by M ∼= N (or M ' N).

Remarks:

• ker(f), Im(f) are submodules of M , N .

• If L is a submodule of M , then f(L) is a submodule of N . If P is asubmodule of N , then f−1(P ) is a submodule of M .

• ker(f) = {0} ⇔ f injective .

Theorem 2.9 Let N be a submodule of an R-module M . Then the quotientgroup M/N is an R-module with the action of R on M/N defined by:

R×M/N → M/N(r, a+N) 7→ ra+N.

The map π : M → M/Na 7→ a+N

is a surjective morphism of R-modules, with

kernel N . π is usually called the canonical projection and M/N is called thequotient of M by N .

Proof: Similar to the corresponding one for rings. �

Theorem 2.10 (Isomorphism theorems) We have the following:

1. Let f : M → N be a morphism of modules. Then f induces an isomor-phism of modules:

f : M/ ker(f) → Im(f)a+ ker(f) 7→ f(a)

.

2. Let A and B be submodules of M . Then:

21

(a) There is an isomorphism of modules:

A/(A ∩B) → (A+B)/B.x+ A ∩B 7→ x+B

.

(b) If A ⊆ B, then B/A is a submodule of M/A and there is anisomorphism of modules:

(M/A)/(B/A) → M/B(x+ A) +B/A 7→ x+B

.

Proof: Similar to the corresponding one for rings. �

Proposition 2.11 Let M be an R-module and let N be a submodule of M .There is a one-to-one correspondance between the submodules of M contain-ing N and the submodules of M/N , given by (if π : M → M/N is thecanonical projection):

{sunmodules of L containing N} 7→ {submodules of L/N}π−1(P ) ←[ P .

Proof: Similar to the corresponding one for rings (proposition 1.10). �

Definition 2.12 A nonzero module M is simple (or irreducible) if the onlysubmodules of M are {0} and M .

Remark: If M is simple, then M is generated by any nonzero element:If m ∈ M \ {0}, then the submodule generated by m is non-zero, and thenmust be M (this submodule is R.m).

Definition 2.13 Let R be a ring. An ideal (resp. left, right ideal) I in R issaid to be maximal if I is different from R and for every ideal (resp. left,right ideal) J of R:

I ⊆ J ⇒ (J = I or J = R).

An ideal (resp. left, right ideal) I in R is said to be minimal if I is differentfrom {0} and for every ideal (resp. left, right ideal) J of R:

J ⊆ I ⇒ (J = {0} or J = I).

A ring may have several maximal (or minimal) ideals. Examples:

• LetR = Z. For a, b ∈ Z, the ideal (a, b) generated by a, b is (gcd(a, b)) =Z.gcd(a, b), the ideal generated by gcd(a, b).Then, if p is a prime, (p) = Z.p is maximal (so Z has a lot of maximalideals):Let I be an ideal such that Z.p $ I. Take a ∈ I \Z.p. Then (a, p) ⊆ I.But (a, p) = (gcd(a, p)) = (1) = Z. This proves that I = Z.


• Let S = Z/2Z be the ring of integers modulo 2, and let R = S × S,with product and sum coordinate by coordinate. S is a ring, and thesubsets {0} × S and S × {0} are ideals of R. They are clearly bothminimal (since they only contain 2 elements.)

Lemma 2.14 A ring R has at least one maximal ideal (resp. maximal left,maximal right ideal).

Proof: Consider E = {I proper ideal of R} = {I ideal of R | 1 6∈ I}, withthe order given by the inclusion, and use Zorn’s lemma (easy exercise). �

Proposition 2.15 A left R-module M is simple if and only if M ∼= R/L forsome maximal left ideal L of R.

Proof: “⇒” We have M = R.m for some m ∈M , m 6= 0.Define φ : R → M

r 7→ r.m. This is a morphism of R-modules. φ is surjective, so

induces an isomorphism from R/ ker f to M (by the isomorphism theorem).We show that ker f is a maximal left ideal of R: Let J be a left ideal of Rwith ker f ⊆ J . Then f(J) is a submodule of M , so is equal to {0} or M . Inthe first case we get J = ker f , in the second case J = R, which shows thatker f is a maximal left ideal of R.“⇐” Let φ be the isomorphism fromM to R/L, and letN be an R-submoduleof M . Then φ(N) is a left ideal of R/L. In particular φ(N) is of the formJ/L for some left ideal J of R, J ⊇ L (see proposition 1.10).Since L is maximal we have J = L or J = R. If J = L then φ(N) = {0},i.e. N = {0}. If J = R then φ(N) = R/L, i.e. N = M . �

Proposition 2.16 (Schur’s lemma) If M is a simple R-module, thenEndRM is a division ring.

Remark (exercise): EndRM is always a ring, under the following operations,for f, g ∈ EndRM :

• (f + g)(m) = f(m) + g(m) ∀m ∈M .

• fg = f ◦ g, i.e. (fg)(m) = (f ◦ g)(m) ∀m ∈M .

The identity is the identity map from M to M .Proof: We have to show that every non-zero element f ∈ EndRM has aninverse. Let f ∈ EndRM \ {0}.Im f is a submodule of M , and so is {0} or M (since M is simple). The firstcase is impossible because f 6= 0. We then have Im f = M , i.e. f surjective.ker f is a submodule of M , and so is {0} or M (since M is simple). The

23

second case is impossible because f 6= 0. We then have ker f = {0}, i.e. finjective.f is then bijective, and so has an inverse, which is easily checked to be anelement of EndRM . �

Remarks:

1. Rn is an R-module.

2. If M has a basis {u1, . . . , un} then M is isomorphic to Rn by:

Rn φ→ M(r1, . . . , rn) 7→ r1u1 + · · ·+ rnun

.

And EndRRn ∼→ EndRMf 7→ φ ◦ f ◦ φ−1

.

We conclude this part on modules by recalling the link between endomor-phisms and square matrices.

Definition 2.17 Let R be a ring. The opposite ringRop is the ring definedby:

• Rop has the same elements as R and the same addition.

• The multiplication ×op in Rop is defined by a×op b = ba (product com-puted in R).

Remarks:

• If R is a division ring, then Rop is a division ring.

• If R is commutative, we have Rop = R.

Theorem 2.18 If M is an R-module isomorphic to Rn, then EndRM ∼=Mn(Rop).

We skip the proof, since it is similar to the one for vector spaces, but brieflyconsider the case n = 2 to justify the introduction of Rop:

Let f ∈ EndR(R2). Say f(

(10

)) =

(a1

a2

)and f(

(01

)) =

(b1

b2

). The isomor-

phism is the usual map sending f to the matrix

(a1 b1

a2 b2

).

However, for this matrix to work “properly”, it must be considered as a


matrix with coefficients in Rop. Indeed, by the R-linearity of f we have:

f(

(uv

)) = f(u

(10

)+ v

(01

)) = uf(

(10

)) + vf(

(01

)) =

(ua1 + vb1

ua2 + vb2

),

while computing with matrices with coefficients in R gives:(a1 b1

a2 b2

)(uv

)=

(a1u+ b1va2u+ b2v

),

which is not what we want.On the other hand, computing with matrices with coefficients in Rop givesthe desired result:(

a1 b1

a2 b2

)(uv

)=

(a1 ×op u+ b1 ×op va2 ×op u+ b2 ×op v

)=

(ua1 + vb1

ua2 + vb2

).

Chapter 3

Wedderburn’s theorem onfinite division rings

Definition 3.1 Let R be a ring. The center of R is:

C(R) = {x ∈ R | xy = yx ∀y ∈ R}

(it is also sometimes denoted by Z(R)).

C(R) is a subring of R but is not (in general) an ideal of R:

C(M2(R)) = {diagonal matrices in M2(R)}= R.I2,

which is not an ideal.

Remark: R is a (left) C(R)-module, with:

C(R)×R → R(x, r) 7→ xr

.

The module properties are verified because R is a ring and C(R) is a subringof R.

Theorem 3.2 (Wedderburn) A finite division ring is commutative (i.e. isa field).

Proof: Let R be a finite division ring. We assume that R 6= {0}. We easilycheck that the inverse of a nonzero an element of C(R) is in C(R). C(R) isthen also a division ring. Since C(R) is commutative, it is a field.We have seen that R is a C(R)-module. Since in our case C(R) is a field,R is a vector space over C(R), and is then isomorphic (as vector space) to

25

26CHAPTER 3. WEDDERBURN’S THEOREM ON FINITE DIVISION RINGS

C(R)n for some integer n.Let q = card(C(R)). We have q ≥ 2 since 0, 1 ∈ C(R) (and 0 6= 1 sinceR 6= {0}), and R has qn elements.We will prove that C(R) = R, or, equivalently, that n = 1.

If a ∈ R, define:C(a) = {x ∈ R | xa = ax}.

We have C(R) ⊆ C(a), and we can check that C(a) is a ring (even a divisionring). C(a) is also a C(R)-vector-space, and this gives C(a) ∼= C(R)n(a) forsome integer n(a). In particular, C(a) has qn(a) elements.

We show that n(a) | n:The first step is to note that R is a C(a)-module. Since C(a) is a divisionring, R has a basis as C(a)-module, say {u1, . . . , uk}.We also know that C(a) is a C(R)-vector space of dimension n(a). Let{v1, . . . , vn(a)} be a basis of C(a) as C(R)-vector space.Since we have a sequence of inclusions:

C(R) ⊆ C(a) ⊆ R,

it is easy to check that the set {viuj | i = 1, . . . , n(a), j = 1, . . . , k} is a basisof R as C(R)-vector space.The dimension of R as C(R)-vector space is then k.n(a). But it is also n. Sowe have n = k.n(a), and n(a) divides n.

We now consider the multiplicative group R? of non-zero elements of R.Then:

C(a) \ {0} = {x ∈ R? | xa = ax}= N(a) the normalizer of a in R?.

Recall from group theory that if G is a group and if N(a) is the normalizerof an element of G:

1. We say that an element b is a conjugate of a if there exists c such thatb = c−1ac.

2. Conjugacy is an equivalence relation.

3. When G is a finite group, the number of conjugates of a is [G : N(a)] =|G|/|N(a)|.

4. An immediate consequence is:

|G| =∑

one a in each conjugacy class

|G||N(a)|

(because G is the disjoint union of all conjugacy classes).

27

Applying this to our case:G = R?, |G| = qn − 1, N(A) = C(a) \ {0}, |N(a)| = qn(a) − 1,we have:

qn − 1 =∑

one a in each conjugacy class

qn − 1

qn(a) − 1. (3.1)

We now consider the conjugacy classes of elements in C(R), the center of R:

If a ∈ C(R), we have N(a) = R?, |N(a)| = qn − 1, and |G||N(a)| = 1.

Moreover, if a, b ∈ C(R) with a 6= b, then a and b are in different conjugacyclasses (suppose they are in the same. Then there exists c ∈ R? such thata = c−1bc = b (since b ∈ C(R)), a contradiction).Then, in the sum (3.1), every element a of R? which is in C(R) represents a

new conjugacy class, and adds 1 to the sum (since |G||N(a)| = 1). We also know

that there are |C(R) \ {0}| = q − 1 such elements. With this, the equality(3.1) can be written:

qn − 1 = (q − 1) +∑ qn − 1

qn(a) − 1, (3.2)

where the sum is over one a in each conjugacy class not meeting the center.Note that we have n(a)|n.

We will show that the equality (3.2) is impossible with n(a)|n, unless n = 1.This is the content of the rest of the section. �

We do this by considering cyclotomic polynomials over R, and we will startby some of their properties:Let Un be the multiplicative group of n-th roots of 1 in C. This is cyclicgroup, and if ξ is a n-th root of 1, ξ is called primitive if it generates Un:Un = {ξk | k ∈ Z}.The n-th cyclotomic polynomial is:

gn(x) = (x− ξ1)(x− ξ2) · · · (x− ξr),

where ξ1, · · · , ξr are all the distinct primitive n-th roots of 1 in C.Here are (without proof) some of the properties of cyclotomic polynomials:

1.∏

d|n gd(x) = xn − 1.

2. gn(x) ∈ Z[x]; the leading coefficient of gn is 1 (we say that g is monic).

3. gn(x) is the minimal polynomial in Q[x] for the primitive n-th roots of1. In particular, gn(x) is irreducible in Q[x].

28CHAPTER 3. WEDDERBURN’S THEOREM ON FINITE DIVISION RINGS

Using the first property and since n(a)|n (and n(a) < n), we have:

xn − 1

xn(a) − 1=

∏d|n, d-n(a)

gd(x) ∈ Z[x].

We also see that this polynomial is divisible by gn(x) (since gn(x) appears inthe product on the right).Hence qn − 1 and qn−1

qn(a)−1are divisible by gn(q). Using this in equation (3.2):

qn − 1 = (q − 1) +∑ qn − 1

qn(a) − 1,

we get that gn(q) | q − 1 in Z.

We are done if we can prove that this is false for n > 1, and it is clearlyenough to show that |gn(q)| > q − 1.We have:

gn(q) =∏

ξ primitive n-th root of 1

(q − ξ).

If n > 1, then every ξ primitive n-th root of 1 is a complex number of norm 1and is different from 1, while q is on the real axis and is greater that 1. Thedistance between q and ξ is then greater than the distance between q and 1.In other words |q − ξ| > q − 1.Then |gn(q)| > (q − 1)k for some k > 1, and since q − 1 ≥ 1 we conclude|gn(q)| > q − 1.

Chapter 4

Primitive rings and the densitytheorem

This chapter starts with the following idea:If R is a ring and M is an R-module, every element r ∈ R defines a mapfrom M to M as follows

f : R → MM ,r 7→ fr

where fr : M → Mm 7→ rm

.

The problem here is that fr is not in general an endomorphism of M , so notas nice as one would wish (from an algebraic point of view): If it were so, wewould have, for every m ∈M and s ∈ R:

fr(sm) = sfr(m), i.e. (rs)m = (sr)m,

which may not hold if R is not commutative.

4.1 A way around this problem

We first see that we can consider M as an EndRM -module. For this we haveto define a product from EndRM ×M to M . We do this as follows:

EndRM ×M → M

(f,m) 7→ f.mdefinition

= f(m) .

This operation gives a structure of EndRM -module over M , because, forf, g ∈ EndRM and a, b ∈M :

29

30 CHAPTER 4. PRIMITIVE RINGS AND THE DENSITY THEOREM

f.(a+ b) = f(a+ b) = f(a) + f(b) = f.a+ f.b,(f + g).a = (f + g)(a) = f(a) + g(a) = f.a+ g.a,f.(g.a) = f.(g(a)) = f(g(a)) = (f ◦ g)(a) = (fg).a,Id.a = Id(a) = a.

So M can be seen as an EndRM -module. To keep notation simple we writeD for EndRM .

We now consider a map very similar to the map f introduced above:

α : R → EndDM,r 7→ αr

where αr : M → Mm 7→ rm

.

We first have to check that αr is an element of EndDM : Let m1,m2 ∈ M ,and θ ∈ D. We have:

αr(m1 +m2) = r(m1 +m2) = rm1 + rm2 = αr(m1) + αr(m2),αr(θm1) = r(θ(m1)) = rθ(m1) = θ(rm1) = θ.rm1 = θαr(m1)

(note that this now works even if R is not commutative).We can even show that α is a morphism of rings: Let r, s ∈ R. We have,

for every m ∈M :

α(r + s)(m) = αr+s(m) = (r + s)m = rm+ sm = α(r)(m) + α(s)(m),α(rs)(m) = αrs(m) = (rs)m = r(sm) = α(r)(α(s)(m)) = (α(r) ◦ α(s))(m),α(1)(m) = 1.m = m.

Since this is true for every m ∈ M , we have α(r + s) = α(r) + α(s),α(rs) = α(r)◦α(s) and α(1) = Id, which shows that α is a morphism of rings.

In the rest of this chapter we will study this map α in some particularcases.

4.2 The study of α(R) and the density theo-

rem

We begin with some preliminary notions.

Definition 4.1 Let M be an R-module, and let B ⊆M . The (left) annihi-lator of B in R is:

AnnRB = {r ∈ R | rb = 0 ∀b ∈ B}.

This is a left ideal of R (easy exercise).

4.2. THE STUDY OF α(R) AND THE DENSITY THEOREM 31

Proposition 4.2 Let M be an R-module, and let N be a submodule of M .Then AnnRN is an ideal of R.

Proof: We already know that it is a left ideal.We have to show: ∀a ∈ AnnRN ∀r ∈ R ar ∈ AnnRN . Let m ∈ N . Then:(ar)m = a(rm) = 0 since rm ∈ N (N is a submodule). �.

Definition 4.3 A left R-module M is faithful if AnnRM = {0}. A ring Ris primitive if there exists a simple faithful R-module.

Remarks:

1. An R-module M is faithful if for every r ∈ R:(rm = 0 ∀m ∈M)⇒ r = 0.

2. R is a faithful R-module:If rm = 0 ∀m ∈ R, then r = r.1 = 0.

Proposition 4.4 Let R be a simple ring. Then:

1. Every nonzero R-module M is faithful.

2. R is primitive.

Proof:

1. Let M be a nonzero R-module. AnnRM is an ideal of R and is differentfrom R because 1.m = m ∀m ∈M (so 1 6∈ AnnRM).Since R is simple, the only possibility is AnnRM = {0}.

2. Using the first part, we just have to find a simple R-module, i.e anR-module M whose only submodules are {0} and M .Let L be a maximal left ideal of R (see lemma 2.14). Since the left idealsof R are exactly the submodules of R, we know that L is a maximalsubmodule of R.We consider the module R/L. We see that since L is maximal, themodule R/L does not contain any proper nonzero submodule: Letπ : R → R/L be the canonical projection, and let N be a submoduleof R/L. Then π−1(N) is a submodule of R, which contains L. Since Lis maximal we have π−1(N) = R or π−1(N) = L. This gives N = R/Lor N = 0.R/L is then a simple R-module. �

We now introduce the ideas that lead to the density theorem.


• LetM be a simple R-module. By Schur’s lemma, D = EndRM is a divi-sion ring (so M , when seen as a D-module, is nothing more complicatedthan a vector space, except that the “field” D is not commutative).

• Let M be a faithful R-module. Then the map α is injective, i.e. R canbe identified with a subring of EndDM :Let r ∈ R, and suppose α(r) = 0, i.e. αr(m) = 0 for every m ∈M .Then rm = 0 for every m ∈ M , which means r ∈ AnnRM . SinceM is faithful, we have AnnRM = {0}, which gives r = 0. We haveker(α) = {0} i.e. α is injective.In particular α defines an isomorphism of rings from R to α(R).

We place ourselves in this nice situation an assume from now on that Mis a simple faithful R-module, and we indentify R with its image α(R) inEndDM .

The next question that we now address is to determine how R “sits” inEndDM . It turns out that R takes a lot of “space” in EndDM . To expressthis, we introduce the notion of density:

Warning: If you don’t have some knowledge of topology, godirectly to definition 4.6, you will not lose anything.

If you know some topology, this short section is here to give you a richerdefinition of the notion of density, which makes in particular more intuitivethe claim that R takes a lot of space in EndDM .

Let M be an D-module, for some division ring D. Let T be the topology onEndDM defined by taking as a basis the sets of the form:

B(a1, . . . , an; b1, . . . , bn) = {f ∈ EndDM | f(ai) = bi for i = 1, . . . , n},

where n is any natural number, and ai, bi are elements of M .

Definition 4.5 (Density) Let D be a division ring, and let M be a D-module. Define T as above.

1. T is called the finite topology on EndDM .

2. We say that a subring R of EndDM is dense if R is dense in EndDMfor the finite topology.

(For the same definition without topology, see definition 4.6.)

We now take some time to translate what “R dense in EndDM” means:It says that every nonempty open set of the finite topology contains an ele-ment of R.


In other words, using the basis B(a1, . . . , an; b1, . . . , bn) of the finite topology,we get:

∀n ∈ N ∀a1, . . . , an, b1, . . . , bn ∈MB(a1, . . . , an; b1, . . . , bn) 6= ∅ ⇒ ∃r ∈ R r ∈ B(a1, . . . , an; b1, . . . , bn).

Since we are working with a module over a division ring, it is easy to see thatif B(a1, . . . , an; b1, . . . , bn) 6= ∅, then:

B(a1, . . . , an; b1, . . . , bn) = B(ai1 , . . . , aik ; bi1 , . . . , bin),

for some linearly independant subset {ai1 , . . . , aik} of {a1, . . . , an}.And of course, if a1, . . . , an are linearly independant, thenB(a1, . . . , an; b1, . . . , bn) is nonempty (for every b1, . . . , bn).

The statement “R is dense in EndDM” can then be expressed as follows:

For every n ∈ N, for every linearly independant elements a1, . . . , anof M , for every b1, . . . , bn ∈ M , there exists r ∈ R such that r ∈B(a1, . . . , an; b1, . . . , bn).

We give two more reformulations, the first of which can be used as an alter-nate definition of density:

∀n ∈ N, ∀a1, . . . , an ∈ M linearly independent, ∀b1, . . . , bn ∈ M ,there exists r ∈ R such that r(a1) = b1, . . . , r(an) = bn.

The part of the course containing topology ends here. You cannow safely go on even if you have not yet seen any topology.

Definition 4.6 (Density) Let D be a division ring and let M be a D-module. Let R be a subring of EndDM . We say that R is dense in EndDMif

∀n ∈ N, ∀a1, . . . , an ∈ M linearly independant, ∀b1, . . . , bn ∈ M ,there exists r ∈ R such that r(a1) = b1, . . . , r(an) = bn.

(In other words, you can find an element of R which sends the members ofa finite linearly independent subset to some given values.)

We can reformulate the definition as follows:

∀n ∈ N, ∀a1, . . . , an ∈M linearly independant, R.(a1, . . . , an) = Mn

(where R.(a1, . . . , an) = {(r(a1), . . . , r(an)) | r ∈ R}).The special case for n = 1 is forth mentionning:

∀a ∈M R.a = M.


Proposition 4.7 Let D be a division ring and let M be a D-module. Let Rbe a dense subring of EndDM . Then R is primitive.

Proof: We have to find a simple faithful R-module, and we will show thatM is such a module:We first check that M is an R-module. We have seen that M is an EndDM -module, with product defined by:

EndD ×M → M


= f(m) .

Since R ⊆ EndDM , we can define:

R×M → M


= f(m) ,

and it is easy to check that it verifies the definition of an R-module (sinceM is an EndDM -module).M is a faithful R-module:Let f ∈ AnnRM , i.e. fm = 0 ∀m ∈ M , i.e. f(m) = 0 ∀m ∈ M . Sincef ∈ EndDM , this means f = 0.M is a simple R-module:Let N be an R-submodule of M , and suppose N 6= {0}. Take a ∈ N \ {0}.We have R.a ⊆ N , and by the last line before the proposition (case n = 1)we know that R.a = M . This implies N = M . M is then a simple R-module.

�

Remark: Dense subrings of EndDM do of course exist: For instance EndDMitself.

The main result is in fact the converse of proposition 4.7:

Theorem 4.8 (Jacobson’s density theorem) Let R be a primitive ringand let M be a simple faithful R-module. Consider M as a module over thedivision ring D = EndDM , and consider R as a subring of EndDM .Then R is dense in EndDM .

Proof: To prove the theorem, it suffices to show that for every n ∈ N,given m1, . . . ,mn ∈ M and m ∈ M such that m is not in the submodule Ngenerated by m1, . . . ,mn, we can find r ∈ R such that:

r.N = {0} and r.m 6= 0.

To see this, suppose that the above property is true.Then we can prove a slightly better result:


Since rm 6= 0, we have R.rm 6= {0}. Since R.rm is the R-submodule gener-ated by rm and M is a simple R-module, we have R.rm = M . This gives:

∀b ∈M ∃s ∈ R srm = b and sr.N = {0}. (4.1)

We now prove the density:Let a1, . . . , an ∈M be linearly independent over D, and let b1, . . . , bn ∈M .Let Vi be the D-submodule of M generated by {a1, . . . , an} \ {ai}.Since ai 6∈ Vi we can find ti ∈ R such that tiai = bi and tiVi = {0} by (4.1).Define t = t1 + · · ·+ tn. We have:

ait = ai(t1 + · · ·+ tn)= aiti= bi,

for every i = 1, . . . , n, and R is dense in EndDM . �

We now prove the property used in the proof:

Lemma 4.9 Let M be a simple faithful R-module. Consider M as a D =EndRM-module (D is a division ring). If N is a finite-dimensional D-submodule of M and m ∈ M \N , then there exists r ∈ R such that rm 6= 0and r.N = {0}.

Proof: We do it by induction on n = dimN :

• n = 0, i.e N = {0}. Take r = 1.

• Suppose dimDN = n > 0. Let {u1, . . . , un} be a D-basis of N , and letW be the D-submodule generated by {u1, . . . , un−1}.We have N = W ⊕D.un.Consider I = AnnRW . This is a left ideal of R, and then I.un is aR-submodule of M . Since M is a simple R-module, we only have twopossibilities: I.un = {0} or I.un = M .We check that the first one is impossible: We have un 6∈ W anddimDW = n−1. So by induction, there exists r ∈ R such that run 6= 0and r.W = {0}. In other words, run 6= 0 and r ∈ AnnRW = I. Thisimplies run ∈ I.un and since run 6= 0, this proves I.un 6= {0}.So Iun = M , which gives that every element of M is of the form runfor some r ∈ I.

We are looking for r ∈ R such that rm 6= 0 and rN = {0}.Assume that there is no such r, i.e.:

∀r ∈ R rN = {0} ⇒ rm = 0. (4.2)


We then look for a contradiction.We can then define θ : M = Iun → M

run 7→ rm(with r ∈ I).

We first check that θ is well-defined (if r1un = r2un we must haveθ(r1un) = θ(r2un)): Let r1un = r2un with r1, r2 ∈ I. Then (r1−r2)un =0. Moreover, using r1 − r2 ∈ I = AnnRW , we have (r1 − r2)W = {0}.This implies (r1 − r2).(W ⊕Dun) = {0}, i.e. (r1 − r2).N = {0}. Thehypothesis (4.2) yields (r1 − r2)m = 0, and then r1m = r2m. θ is well-defined. It is then easy to check that θ is a morphism of R-modules,which gives θ ∈ EndRM = D.

Then, for every r ∈ I, θ(run) = rm. This gives:

0 = θ(run)− rm = r(θ(un)−m) for every r ∈ R. (4.3)

This in turns implies that θ(un) − m ∈ W , because otherwise, byinduction we could find s ∈ R such that sW = {0} (i.e. s ∈ I =AnnRW ) and s(θ(un)−m) 6= 0, a contradiction to (4.3).And we have:

m = θ(un)− (θ(un)−m)= θ.un − (θ(un)−m),

(the last line is because when we see M as a D-module, the product ofm ∈M by an element f ∈ D is defined by f.m = f(m)).Since θ.un ∈ D.un and θ(un)−m ∈ W , we get m ∈ Dum ⊕W = N , acontradiction. �

4.3 Consequences

We will now see some consequences of the density theorem:

Proposition 4.10 Let R be a primitive ring. Then for some division ring∆, one of the following holds:

1. R is isomorphic to Mn(∆).

2. For every n ∈ N there is a subring Sn of R and a surjective morphismof rings Sn →Mn(∆).

Proof: Let M be a simple faithful R-module, and let D = EndRM . By thedensity theorem, we know that R is a dense subring of EndDM , if we identifyit with it image by α (see page 32).

4.3. CONSEQUENCES 37

1. Suppose dimDM is finite, so equal to some n ∈ N (the dimension existsbecause D is a division ring, and M has a basis over D)Let {v1, . . . , vn} be a basis of M over D. We show that α is surjective:Let f ∈ EndDM . f is completely determined by f(v1), . . . , f(vn). Butsince v1, . . . , vn are linearly independent, by the density ofR in EndDM ,there exists r ∈ R such that rv1 = f(v1), . . . , rvn = f(vn), i.e. αr(v1) =f(v1), . . . , αr(vn) = f(vn), i.e. αr = f .So α is surjective, and R ∼= EndDM ∼= Mn(Dop) (the last isomorphismbecause dimDM = n; see theorem 2.18).

2. —No proof yet, since it is an exercise in the third exercise sheet— �

Before seeing a more important consequence of the density theorem, we in-troduce the “chain conditions”.

Definition 4.11 A familly of subsets {Ci}i∈I in a set C is said to satisfythe ascending chain condition (ACC for short) if there does not exist aninfinite strictly increasing chain:

Ci1 & Ci2 & · · · & Cik & · · ·

of elements of {Ci}i∈I .An equivalent condition is:For any ascending chain Ci1 ⊆ Ci2 ⊆ · · · ⊆ Cik ⊆ · · · , of elements of {Ci}i∈I ,there exists n ∈ N such that Cin = Cin+1 = Cin+2 = · · · .

There is a corresponding notion for descending chains:

Definition 4.12 A familly of subsets {Ci}i∈I in a set C is said to satisfythe descending chain condition (DCC for short) if there does not existan infinite strictly descending chain:

Ci1 ' Ci2 ' · · · ' Cik ' · · ·

of elements of {Ci}i∈I .An equivalent condition is:For any descending chain Ci1 ⊇ Ci2 ⊇ · · · ⊇ Cik ⊇ · · · , of elements of{Ci}i∈I , there exists n ∈ N such that Cin = Cin+1 = Cin+2 = · · · .

Examples:

• If C is a finite set, every familly {Ci}i∈I of subsets of C satisfies boththe ACC and the DCC.


• The ideals of Z are of the form a.Z for a ∈ Z, and a.Z ⊆ b.Z if and onlyif b divides a. This implies that the set of ideals of Z satisfies the ACC(you can only divide a given element of Z a finite number of times).Remark that the set of ideals of Z does not satisfy the DCC, because(for example) Z ' 2.Z ' 22.Z ' 23.Z · · · .

Definition 4.13 1. A ring R is left (resp. right) Noetherian if theset of left (resp. right) ideals of R satisfies the ACC (i.e. there is noinfinite strictly increasing chain of left (resp. right) ideals).R is Noetherian if it is both left and right Noetherian.

2. A ring R is left (resp. right) Artinian if the set of left (resp. right)ideals of R satisfies the DCC (i.e. there is no infinite strictly descendingchain of left (resp. right) ideals).R is Artinian if it is both left and right Artinian.

There is also a notion of Noetherian and Artinian modules (where Noetherianmeans the set of submodules satifies the ACC, and Artinian means the setof submodules satisfies the DCC).

Examples:

1. Using the example after definition 4.12, we know that Z is Noetherianbut not Artinian.

2. A division ring D is both Noetherian and Artinian, because the only(left, right) ideals of D are {0} and D.

3. A finite ring is both Noetherian and Artinian.

We give the next two results without proof, as examples of constructions thatpreserve the “Noetherian” and/or “Artinian” properties:

Proposition 4.14 If R is left (resp. right) Noetherian, then Mn(R) is left(resp. right) Noetherian.If R is left (resp. right) Artinian, then Mn(R) is left (resp. right) Artinian.

Remark: The proof of this proposition cannot be done using proposition1.17, because proposition 1.17 gives informations about ideals (i.e. two-sidedideals), but the “Artinian” and “Noetherian” properties deal with left or rightideals.

Theorem 4.15 (Hilbert basis theorem) If R is a commutative Noethe-rian ring, then so is R[x] (and then R[x1, . . . , xn]).

We now go back to the consequences of the density theorem:

4.3. CONSEQUENCES 39

Theorem 4.16 (Wedderburn-Artin) The following conditions on a leftArtinian ring (i.e. a ring satisfying the DCC on left ideals) are equivalent:

1. R is simple.

2. R is primitive.

3. R is isomorphic to Mn(∆), for some n ∈ N and some division ring ∆.

This theorem applies in particular to finite rings (since a finite ring is neces-sarily left Artinian), and gives the following special case:

Theorem 4.17 The following conditions on a finite ring are equivalent:

1. R is simple.

2. R is primitive.

3. R is isomorphic to Mn(F ), for some n ∈ N and some finite field F .

Proof of theorem 4.17: We apply theorem 4.16. Since R is finite, the divisionring ∆ is necessarily finite, and is then a field by Wedderburn’s theorem. �

We need two preliminary results before the proof of the Wedderburn-Artintheorem:

Proposition 4.18 If R is a ring satisfying the DCC on left ideals (i.e. R isleft Artinian), then R has a minimal left ideal.

Proof: We use Zorn’s lemma on the set:

E = {I ⊆ R | I is a left ideal I 6= {0}},

with order given by I ≤ I ′ if and only if I ′ ⊆ I (the order given by the reverseinclusion).We check the hypotheses for Zorn:

• E is clearly a partially orderd set (this is always the case if the orderon E is given by the inclusion or the reverse inclusion).

• Let C = {Ji}i∈A be a chain in E. We want to find an upper bound forC in E. Suppose that no such upper bound exists (?), and take Ji0 ∈ E.By (?), Ji0 is not an upper bound of C, so there exists Ji1 ∈ C suchthat Ji1 6≤ Ji0 . Since Ji0 , Ji1 are in a chain (C), we deduce Ji0 < Ji1 .By (?), Ji1 is not an upper bound of C, and applying the same reason-ning we find Ji2 ∈ C such that Ji0 < Ji1 < Ji2 . We continue and find astrictly increasing chain of elements of C

Ji0 < Ji1 < · · · < Jin < · · ·


Using the definition of ≤ we get an infinite strictly descending chain ofleft ideals of R:

Ji0 ' Ji1 ' · · · ' Jin ' · · ·

which contradicts the hypothesis on R.C has then an upper bound in E.

Zorn’s lemma then gives a maximal element in E for the order ≤, which isthen a minimal left ideal of R. �

Proposition 4.19 Let D be a division ring, M be a D-module, and let Rbe a dense subring of EndDM .Then R is left Artinian (i.e satisfies the DCC on left ideals) if and only ifdimDM is finite (in which case R = EndDM ∼= Mn(Dop)).

Proof: “⇒” Suppose dimDM is infinite, and let {u0, u1, u2, . . .} be an infinitelinearly independent subset of M .The idea of the proof is to consider AnnR{u1, . . . , uk} for k ≥ 0, to get adescending chain of ideals in R.To do this, we first have to be able to consider M as an R-module:Since R ⊆ EndDM , we can define:

R×M → M

(r,m) 7→ r.mdefinition

= r(m) .

This turns M into an R-module (the properties are easily verified).We can then define Ik = AnnR{u0, . . . , uk}. This a left ideal of R. Since{u0, . . . , uk} ⊆ {u0, . . . , uk+1} we have Ik+1 ⊆ Ik. Let m ∈ M \ {0}. Since{u0, . . . , uk+1} is linearly independent and R is dense in EndDM , there existsr ∈ R such that:

r(u0) = r(u1) = · · · = r(uk) = 0, and r(uk+1) = m 6= 0.

So r ∈ AnnR{u0, . . . , uk} = Ik, but r 6∈ AnnR{u0, . . . , uk+1} = Ik+1, whichgives a stricktly descending chain:

I0 ' I1 ' I2 · · · In ' In+1 · · ·

contradicting the hypothesis “R Artinian”.“⇐” Suppose dimDM = n, and let {v1, . . . , vn} be a basis of M . Let f ∈EndDM . f is completely determined by f(v1), · · · , f(vn).Since R is dense in EndDM and {v1, . . . , vn} is linearly independent, thereexists r ∈ R such that r(v1) = f(v1), · · · , r(vn) = f(vn). Since r and f are

4.4. SIMPLE ALGEBRAS 41

both in EndDM this implies r = f .So R = EndDM ∼= Mn(Dop), which is Artinian since D (hence Dop) is adivision ring (see proposition 4.14). �

We can now prove Wedderburn-Artin theorem:Proof of theorem 4.16:“1 . ⇒ 2 .” We have to find a simple faithful R-module. By proposition4.18, we know that R has a minimal left ideal J (remark that in particularJ 6= {0}). J is an R-module (with the product given by R). We check thatJ is a simple R-module:Let N be an R-submodule of J . Then N is an R-submodule of the R-moduleR, and is then a left ideal of R. Since N ⊆ J and J is a minimal left ideal,we have N = {0} or N = J .

We check that J is faithful:AnnR J is an ideal of R (since J is a submodule of R; see proposition 4.2).Since R is simple, we have AnnR J = {0} or AnnR J = R. Suppose AnnR J =R. This means that for every u ∈ J , R.u = {0}. In particular, J ⊆ {r ∈R | R.r = {0}} = L, which is a proper ideal of R (exercise; it is properbecause it does not contain 1). Since R is simple, it implies L = {0} andthen J = {0}, a contradiction.Then AnnR J = {0}, i.e. J is faithful.“2 . ⇒ 3 .” Let M be a simple faithful R-module. By the density theorem,R is dense in EndDM , where D = EndRM is a division ring. Then applyproposition 4.19.“3 .⇒ 1 .” This is a very easy consequence of proposition 1.17. �

4.4 Simple algebras

Definition 4.20 An algebra A over a field F is a ring with the followingproperties

1. F is a subring of A;

2. F ⊆ C(R).

Examples:

1. Mn(F ) is an F -algebra, where F is identified with F · In.

2. H is an R-algebra, but not a C-algebra, since C 6⊆ C(H).

Let A be an F -algebra. Since F ⊆ A, the sum and product of A allow us toconsider A as an F -vector space. In particular dimF A, the dimension of Aas F -vector space, is well-defined.


Definition 4.21 Let A be an F -algebra (in this formulation, F is alwaysassumed to be a field).

1. The dimension of A (over F ) is dimF A, and A is called finite dimen-sional if dimF A is finite.

2. A is simple if it is simple as a ring (i.e. has no proper ideals).

3. A is a division algebra if it is also a division ring.

The following theorem will be used in the course on quadratic forms.

Theorem 4.22 (Wedderburn structure theorem) Let A be a finite di-mensional simple F -algebra. Then A ∼= Mn(∆) (as ring) for some n ∈ Nand some finite dimensional division algebra ∆ over F .Moreover, n is unique and ∆ is unique up to isomorphism.

Proof: We first show that A satisfies the DCC on left ideals: Let I be a leftideal of A. Then I can also be considered as an F -vector space. So every leftideal of A is a left vector space. Since dimF A is finite, there cannot be aninfinite strictly decreasing chain of subspaces of A. In particular A satisfiesthe DCC on left ideals.

Since A is also simple, theorem 4.16 tells us A ∼= Mn(∆) for some divisionring ∆. We then have ∆ ⊆ A via the identification of ∆ and ∆·In. If we showthat F ⊆ ∆, then ∆ will be an F -vector space and will be finite dimensionalsince it is included in A, which is finite dimensional as F -vector space.

The unicity of ∆ and n follows from the next lemma. �

Lemma 4.23 Let R = Mn(D) and R′ = Mn′(D′), with n, n′ ∈ N and D,D′

division rings. Assume the rings R and R′ are isomorphic. Then n = n′ andD ∼= D′.

Proof: The result follows from a series of simple facts:Fact 1: Let θ : R′ → R be an isomorphism, and let S be an R-module. ThenS can be seen as an R′-module when equipped with the following product:

R′ × S → S(r′, s) 7→ θ(r′)s

Fact 2: If S is a simple faithful R-module, then S, seen as an R′-module (asin Fact 1), is also simple and faithful.Proof of fact 2: Let N be a nonzero R′-submodule of S. Then N can beseen as an R-submodule of the R-module S when equipped with the productR×N → N , (r, n) 7→ rn. Since S is simple we get N = S. The faithfulness

4.4. SIMPLE ALGEBRAS 43

is immediate from the definition of the product. End of proof of fact 2.Fact 3: The R-endomorphisms of S are exactly the R′-endomorphisms of S.Proof of fact 3: Straightforward.

If S is a simple faithful R-module, using the proof of theorem 4.16 weget Dop ∼= EndRS = EndR′S ∼= D′op. So D and D′ are isomorphic and letρ : D → D′ be an isomorphism of rings.

We then have two isomorphism of rings:

θ : R′ → R and ρ : D → D′.

Step 1a: Mn′(D′), equipped with the product D × Mn′(D

′) → Mn′(D′),

(dIn′ ,M) 7→ θ−1(dIn)M , is a D-module.Step 1b: θ−1 is then an isomorpshim of D-modules between Mn(D) andMn′(D

′).Step 2a: Mn′(D

′), equipped with the product D × Mn′(D′) → Mn′(D

′),(dIn′ ,M) 7→ ρ(d)In′M , is a D-module.Step 2b: The natural extension of ρ: ρ+ : Mn′(D) → Mn′(D

′) is then anisomorphism of D-modules from Mn′(D) to Mn′(D

′).

Combining the results of steps 1b and 2b, we get an isomorphism of D-modules from Mn′(D) to Mn(D). Comparing the dimensions gives us n = n′.

�

We close this chapter with an observation on some simple modules:Let D be a division ring, let V = Dn for some n ∈ N, and let R = EndDV .

Then V is a left R-module with the action

R× V → V(f, v) 7→ f(v),

and

Proposition 4.24 1. V is a simple R-module;

2. Any simple R-module is isomorphic to V .

Proof:

1. Let M be a non-zero R-submodule of Dn, and let m ∈ M , m 6= 0.Then R ·m ⊆M , but R ·m = Dn. So M = Dn.

2. Let {e1, . . . , en} be any basis of Dn and define

ξ : R = EndDV → V ⊕ · · · ⊕ Vf 7→ (f(e1), . . . , f(en)).


It is clear that ξ is an isomorphism of R-modules.

Let M be a simple R-module, and let m ∈ M , m 6= 0. We defineλ : R→M , λ(r) = rm, and we define

τ : V ⊕ · · · ⊕ V ξ−1

→ Rλ→M.

We obtain in this way n morphism of R-modules τi (for i = 1, . . . , n):

Vid→ 0⊕ · · · ⊕ 0⊕ V ⊕ 0⊕ · · · ⊕ 0→ R

λ→M

(where the second map is the restriction of ξ−1), with τ = τ1 + · · ·+ τn.

Since τ 6= 0, one of the τi is non zero, say τ1 6= 0. It follows that τ1 isan isomorphism since both V and M are simple R-modules. �

Chapter 5

The Jacobson radical

Definition 5.1 Let R be a ring. An ideal I of R is primitive if R/I is aprimitive ring.

Proposition 5.2 Let I be an ideal of a ring R. Then I is primitive if andonly if I = AnnRM for some simple R-module M .

Proof: “⇒” Since R/I is primitive, there exists a simple faithful R/I-moduleM . We can see M as a R-module with the product (exercise):

R×M → M

(r,m) 7→ rmdefinition

= (r + I)m . (5.1)

We have to check that I = AnnRM and that M is a simple R-module.1) I = AnnRM :“⊆” Let i ∈ I. Then im = (i+ I)m = (0 + I)m = 0m = 0, for every m ∈M .So I ⊆ AnnRM .“⊇” Let r ∈ AnnRM , i.e. rm = 0 for every m ∈ M . Then, by definitionof the product in (5.1) we have (r + I)m = 0 for every m ∈ M . This givesr+I ∈ AnnR/IM , which is {0} since M is a faithful R/I-module. So r+I = 0in R/I, i.e. r ∈ I.

2) M is a simple R-module:Let N be an R-submodule of M . We check that N is also a R/I-submoduleof M with the product:

R/I ×N → N

(r + I,m) 7→ (r + I)mdefinition

= rm

(check it; you have to check that the product is well defined, i.e. does notdepend on the representative chosen in the equivalence class r + I. This

45

46 CHAPTER 5. THE JACOBSON RADICAL

works because I = AnnRM). Since M is a simple R/I-module, we haveN = {0} or N = M .

“⇐” Let I = AnnRM , with M a simple R-module. We can see M as aR/I-module with the product:

R/I ×M → M

(r + I,m) 7→ (r + I)mdefinition

= rm

(the justification and the fact that it is well-defined are as above).An R/I-submodule of M can also be seen as an R-submodule of M (usingthe idea in (5.1)), so M is a simple R/I-module.M is a faithful R/I-module:Let r+I ∈ AnnR/IM , i.e. (r+I)m = 0 for every m ∈M , which gives, usingthe definition of the product: rm = 0 for every m ∈M . So r ∈ AnnRM = I,and r + I = 0 + I = 0 in R/I. �

Definition 5.3 The Jacobson radical of a ring R is

J(R) =⋂{I | I primitive ideal of R}.

If R has no primitive ideal, we take J(R) = R.

Remark: J(R) is an ideal of R.

Lemma 5.4 Let R be a ring. Every left (resp. right, 2-sided) ideal of R iscontained in a maximal left (resp. right, 2-sided) ideal of R.

Proof (for left ideals): Let I be a left ideal of R, and consider:

E = {J | J left ideal of R, J ⊇ I, J 6= R}= {J | J left ideal of R, J ⊇ I, 1 6∈ J}.

E, with the order given by the inclusion, is a partially ordered set. ApplyZorn’s lemma to find a maximal element of E. �

Lemma 5.5 Let R be a ring. For y ∈ R the following are equivalent:

1. y ∈⋂{M | M maximal left ideal of R}.

2. 1− xy is left-invertible for every x ∈ R.

3. y.M = {0} for every simple R-module M .

47

Remark: 3 . ⇔ y ∈⋂{AnnRM | M simple R-module }

⇔ y ∈⋂{I | I primitive ideal of R} (by proposition 5.2)

⇔ y ∈ J(R).

Proof: 1 .⇒ 2 . Suppose 1− xy is not left-invertible, for some x ∈ R. DefineI = R.(1 − xy). I is different from R since 1 6∈ I, and I is a left ideal. SoI ⊆ L for some L maximal left ideal of R.We have 1− xy ∈ I ⊆ L, i.e. 1− xy = l ∈ L. But y ∈ L by hypothesis, andthen xy ∈ L. This implies 1 = l + xy ∈ L, and then L = R, a contradiction.2 .⇒ 3 . Assume there exists a simple R-module M such that yM 6= {0}. Inparticular there exists m ∈M such that ym 6= 0.Consider R.(ym), the R-submodule generated by ym. It is nonzero sinceym 6= 0. Since M is a simple R-module, we get R.(ym) = M .In particular there exists r ∈ R such that rym = m, which implies (1 −ry)m = 0. But 1− ry is left-invertible by hypothesis, and if we multiply thisequation on the left by its left-inverse, we have m = 0, a contradiction.3 . ⇒ 1 . Let L be a maximal left ideal of R. R/L is an R-module (withproduct R × R/L → R/L, (r, a + L) 7→ ra + L), which is simple (exercise:it is because L is a maximal left ideal of R, i.e. a maximal left submoduleof the R-module R). By hypothesis we have y.R/L = {0}, and in particulary(1 + L) = 0, i.e. y + L = 0, which means y ∈ L. �

Corollary 5.6 Let R be a ring. Then:

J(R) =⋂{L | L maximal left ideal of R}.

Proof: See the remark after lemma 5.5. �

Lemma 5.7 Let R be a ring, and let y ∈ R. The following are equivalent:

1. y ∈ J(R).

2. ∀x, z ∈ R 1− xyz ∈ U(R).

Proof: 2 . ⇒ 1 . By taking z = 1, we have 2 . from lemma 5.5, which means(by part 1 . of lemma 5.5) y ∈ J(R).1 . ⇒ 2 . Let y ∈ J(R) and x, z ∈ R. Since J(R) is an ideal, yz ∈ J(R).So by lemma 5.5, 1 − xyz is left invertible, so there exists u ∈ R such thatu(1− xyz) = 1 (?).Since J(R) is an ideal, xyz ∈ J(R). By 2 . in lemma 5.5, 1 + u(xyz) is left-invertible.But (?) gives u(xyz) = u − 1, so we have 1 + u(xyz) = u is left-invertible.This implies u ∈ U(R) since u is also right invertible by (?). This suffices to


show that 1− xyz = u−1 and then belongs to U(R) (since for an element ofU(R) the right and left inverses coincide, and are in U(R)). �

Corollary 5.8 J(R) is the largest left ideal I ⊆ R such that 1 + I ⊆ U(R).

Proof: We have to prove that every left ideal I satisfying 1 + I ⊆ U(R) is asubset of J(R).Let I be a left ideal such that 1 + I ⊆ U(R). We use the characterization ofJ(R) given by corollary 5.6 and lemma 5.5:Let y ∈ I and x ∈ R. Since I is a left ideal, −xy ∈ I, and then, by hypothesis,1 − xy ∈ U(R). Since this is true for every x ∈ R we have y ∈ J(R). Thisproves I ⊆ J(R). �

Remarks:

1. From lemma 5.5 and 5.7 we have:

∀x ∈ R 1− xy is left invertible⇓

∀x, z ∈ R 1− xyz is invertible,

which looks much stronger.

2. The properties / characterizations of J(R) that we obtained in termsof left ideals can be rephrased in terms of right ideals. In particularJ(R) is the intersection of all right ideals of R. This can be seen forinstance in lemma 5.7 where J(R) is described in terms that do notdepend on left/right.

Definition 5.9 An element of a ring R is nilpotent if an = 0 for somen ∈ N.A left (reps. right, two-sided) ideal I of R is nil if every element of I isnilpotent; I is nilpotent if In = {0} for some n ∈ N.

An easy example of nilpotent element can be found in M2(R) with a =(0 10 0

), because a2 = 0.

Remark: In = I.I. · · · .I (n times)

= {∑m

i=1 ai,1. · · · .ai,n | m ∈ N, ai,k ∈ I}(see definition page 10).

This gives:

I nilpotent ⇔ ∃n ∈ N In = {0}⇒ ∀a1, · · · , an ∈ I a1. · · · .an = 0⇒ ∀a ∈ I an = 0.

49

So I nilpotent implies I nil. (The converse is false! See exercise sheet 4.)

Theorem 5.10 Let R be a ring and let I be a nil left ideal of R. ThenI ⊆ J(R).

Proof: Let a ∈ I. a is nilpotent and there exits n ∈ N such that an = 0.Define r = −a+ a2 − a3 + · · ·+ (−1)n−1an−1. Then:r + a+ ra = a2 − a3 + · · ·+ (−1)n−1an−1 − a2 + a3 + · · ·+ (−1)n−2an−1

= 0.This implies (1 + r)(1 + a) = 1 + r + a + ra = 1. We also get in a similarway: (1 + a)(1 + r) = 1. We have proved 1 + a ∈ U(R) for every a ∈ I, i.e.1 + I ⊆ U(R), and by corollary 5.8 we get I ⊆ J(R). �

Proposition 5.11 If R is a left Artinian ring (i.e. satisfies the descendingchain condition on left ideals), then J(R) is nilpotent.

Corollary 5.12 If R is a left Artinian ring then:

1. If I is a nil left ideal of R, then I is nilpotent.

2. J(R) is the unique maximal nilpotent left ideal of R.

Proof of corollary 5.12: It is an immediate consequence of theorem 5.10 andproposition 5.11. �

Remark: Proposition 5.11 and corollary 5.12 apply in particular to finiterings, since a finite ring as necessarilly left Artinian.

Proof of proposition 5.11: Let J = J(R) and consider:

J ⊇ J2 ⊇ J3 ⊇ · · ·

It is a descending chain of left ideals, and since R is left Artinian, there existsk ∈ N such that J i = Jk for every i ≥ k.We prove that Jk = {0} (which will give J nilpotent):Suppose Jk 6= {0}. Then J2k = Jk 6= {0}. Consider:

E = {L left ideal of R | JkL 6= {0}},

with the order given by the reversed inclusion (i.e. L1 ≥ L2 if and only ifL1 ⊆ L2). E is nonempty because Jk ∈ E. Using that R is left Artinian, wecan verify that E satisfies the hypothesis of Zorn’s lemma (see the proof ofproposition 4.18), and by applying it we get a minimal element I0 of E.Since I0 ∈ E, we have JkI0 6= {0} and there exists a ∈ I0, a 6= 0, such thatJka 6= {0}.Jka is a left ideal of R and Jka ∈ E (since Jk(Jka) = J2ka = Jka 6= {0}).


But we also have Jka ⊆ I0 since a ∈ I0 and I0 is a left ideal. Since I0 isminimal in E, this implies Jka = I0.Using a ∈ I0, we get r ∈ Jk such that ra = a.

We have r ∈ Jk ⊆ J , so by lemmas 5.5 and 5.7, 1− r ∈ U(R). Let α ∈ R besuch that α(1− r) = 1. We then have α(1− r)a = 1.a = a, but we also have:α(1− r)a = (α− αr)a

= αa− αra= αa− αa= 0.

This gives a = 0, a contradiction. �

Chapter 6

Semisimple modules and rings

6.1 Composition series

Definition 6.1 Let M be an R-module.

1. A series S of M is a finite chain of submodules of M :

M0 ⊆M1 ⊆ · · · ⊆Mn.

Its length is the number of proper inclusions.Its factors are the quotient modules Mi/Mi−1.Its set of modules is M(S) = {M0, . . . ,Mn}.

2. Let S and T be two series of M . T is a refinement of S if M(S) ⊆M(T ), and a proper refinement if M(S) (M(T ).

3. A composition series of a module M is a series of M that has noproper refinement (i.e., and it is an easy exercise, a series in whichevery nonzero factor module is simple).

Remark:

1. Not every module has a composition series. For instance, let M be aninfinite dimensional vector space over a field K. Then M is a K-moduleand has no composition series (for dimension reasons).

2. Let M be a finite-dimensional K-vector space, with basis {u1, . . . , un}.Then M has a composition series of length n:

{0} ( 〈u1〉 ( 〈u1, u2〉 ( · · · ( 〈u1, . . . , un−1〉 ( M.

51

52 CHAPTER 6. SEMISIMPLE MODULES AND RINGS

Lemma 6.2 Let R be a ring, let M be an R-module and let M ′ be a propersubmodule of M . Assume that M has a composition series of length n. ThenM ′ has a composition series of length less than n.

Proof: Let {0} ( M1 ( · · · ( Mn = M be a composition series of M . Weconsider the series of M ′:

{0} ⊆M1 ∩M ′ ⊆ · · · ⊆Mn ∩M ′ = M ′,

and we show that one of the inclusions must be an equality.

The quotient (Mi+1 ∩M ′)/(Mi ∩M ′) is isomorphic to (Mi+1 ∩M ′) +Mi/Mi,a submodule of Mi+1/Mi which is simple. So (Mi+1 ∩M ′)/(Mi ∩M ′) = {0}or (Mi+1 ∩M ′)/(Mi ∩M ′) is simple and (Mi+1 ∩M ′) +Mi = Mi+1.But this later case can not happen for every i: We assume that it doesand prove by induction on i that Mi ⊆ M ′, a contradiction for i = n. Ifi = 0 we clearly have M0 ⊆ M ′. Supposing by induction that Mi ⊆ M ′, weobtain M ′ ∩Mi+1 = (M ′ ∩Mi+1) +Mi = Mi+1 (by hypothesis). This meansMi+1 ⊆M ′. �

Theorem 6.3 Let R be a ring and let M be an R-module.

1. M has a composition series if and only if M is Artinian and Noethe-rian.

2. If M has a composition series of length n, then every series of M haslength at most n and can be refined to a composition series.

Proof:

1. “⇐” First observe that if a module satisfying the ACC on submodulesalways contains a maximal proper submodule (the proof is similar tothat of proposition 4.18.We then choose a maximal proper submodule M1 of M , a maximalproper submodule M2 of M1, and so on. . .By the DCC on submodules, there is k ∈ N such that Mk = M` forevery ` ≥ k. But this can only happen if Mk = {0}. Then M ⊇M1 ⊇· · · ⊇Mk−1 ⊇ {0} is a composition series for M .“⇒” It follows from the next item.

2. Let N0 ( N1 ( · · · ( Nk = M be a series of M . We show by inductionon n that k ≤ n. This is clear if n = 0 since in this case M = {0}.By lemma 6.2, Nk−1 has a composition series of length at most n− 1,

6.1. COMPOSITION SERIES 53

so by induction the length of the series N0 ( · · · ( Nk−1 is less thann− 1, so the length of N0 ( N1 ( · · · ( Nk = M is less than n.

Knowing that every series of M has length at most n, it is then possibleto use Zorn’s lemma (on the set of series of M of length at most n) toshow the existence of a maximal series of M , which is then necessarillya composition series of M . �

Remark: In particular if M has a composition series, then any two compo-sition series of M have the same length:Let S1 be a composition series of length n1 and S2 be a composition seriesof length n2. Then n1 ≤ n2 and n2 ≤ n1, so n1 = n2.

Proposition 6.4 The following are equivalent, for an R-module M :

1. M is both Noetherian and Artinian.

2. Every chain of submodules of M is finite.

3. M has a composition series.

Proof: See exercise sheet 5. �

It is actually possible to obtain the following two results (which we willnot need) about composition series.

Definition 6.5 Two series of M are equivalent if there is a bijection betweentheir sets of nonzero factor modules, so that corresponding factor modules areisomorphic.

Theorem 6.6 (Schreier) Any two series of a module have equivalent re-finements.

Theorem 6.7 (Jordan-Holder) Any two composition series of a moduleare equivalent.

The Jordan-Holder theorem is a direct consequence of the Schreier theoremsince a composition series has no proper refinement. Conversely, the Schreiertheorem follows easily out of the Jordan-Holder theorem with the help oftheorem 6.3.


6.2 Semisimplicity

We first need a few concepts.

Definition 6.8 Let M be an R-module, and let Mi, i ∈ I be submodules ofM .

1. M is the sum of the submodules Mi, i ∈ I if each x ∈M can be writtenas a finite sum of elements of the submodules Mi, i ∈ I. We writeM =

∑i∈IMi.

2. M is the direct sum of the submodules Mi, i ∈ I if each x ∈ M canbe written in a unique way as finite sum of elements of the submodulesMi, i ∈ I. We write M =

⊕i∈IMi.

Remark: It M =∑i ∈ IMi, then M =

⊕i∈IMi if and only if for every

finite subset J of I and every k ∈ I \ J we have Mk ∩∑

j∈JMj = {0}. thedirect sum of the Mi, i ∈ I if and only if for ev

Definition 6.9 Semisimple modules are defined by any of the equivalent con-ditions presented in the next proposition.

Proposition 6.10 Let M be an R-module. The following are equivalent:

1. M is a direct sum of simple submodules.

2. M is a sum of simple submodules.

3. Every submodule of M is a direct summand (i.e. for every submoduleN of M , there is a submodule P of M such that M = N ⊕ P ).

Proof: 1. ⇒ 2. is clear.2. ⇒ 1. and 3.: By hypothesis we have M =

∑i∈I Si, where the Si are

simple submodules of M .Let N be a submodule of N , and consider

S = {J ⊆ I | all sums in the following expression are direct sums N +∑

i∈J Si}= {J ⊆ I | N ∩

∑i∈J Si = {0} ∧ [Sj ∩ (N +

∑i∈J,i6=j Si) = {0} ∀j ∈ J ]}.

Applying Zorn’s lemma (exercise: chack that the hypotheses are satisfied),we find a maximal element J . Note that by choice of J the sum N +

∑i∈J Si

is direct, i.e. equal to N ⊕⊕

i∈J Si. In particular we now only have to provethat N +

∑i∈J Si = M .

6.2. SEMISIMPLICITY 55

Assume N +∑

i∈J Si ( M . Then there is k ∈ I such that Sk 6⊆ N +∑

i∈J Si.It follows:

(N +∑i∈J

Si) ∩ Sk ( Sk, and thus

(N +∑i∈J

Si) ∩ Sk

since Sk is simple.This means that the sum N +

∑i∈J∪{k} Si is direct, contradicting the choice

of J .

The case N = {0} yields M =⊕

i∈J Si.

3. ⇒ 2.: The first step consists in showing that any cyclic (=generatedby one element only, so of the form Ra for some a ∈ M) submodule of Mcontains a simple submodule:The map φ : R → Ra, r 7→ ra is module homomorphism from the left R-module R to the left R-module Ra. Its kernel is a left ideal of R, which isthen contained in a maximal left ideal L of R. Then φ(L) = La is a maximalsubmodule of Ra (easy exercise), and therefore the quotient module Ra/La issimple (easy exercise again). We now use the hypothesis: By 3. La is a directsummand in M , i.e. there is a submodule N of M such that M = La ⊕N .It follows easily that Ra = La⊕ (Ra∩N), which yields Ra∩N ∼= Ra/La asR-module, so Ra ∩N is simple.

Let now N be the sum of all simple submodules of M . Then M = N⊕N ′ forsome submodule N ′ of M . We check that N ′ = {0}: Otherwise N ′ containsa cyclic submodule of M (take any a ∈ N \ {0} and consider Ra). By theabove argument, N then contains a simple submodule S, so S ⊆ N , whichcontradicts N ∩N ′ = {0}. �

Proposition 6.11 1. A direct sum of semisimple R-modules is semisim-ple.

2. Every submodule of a semisimple R-module is semisimple.

3. Every quotient module of a semisimple module is semisimple.

Proof: Since it follows rather easily from proposition 6.10, we only present asketch of the proof.

1. This one is obvious.

2. Use the (easy) fact that ifM is a module and S, T and A are submodulesof A, then M = S ⊕ T and S submodule of A imply A = S ⊕ (T ∩A).


3. If M is semisimple and N is a submodule of M , then M = N ⊕ P forsome submodule P of M . P is semisimple and M/N ∼= P . �

Definition 6.12 A ring R is called semisimple if every left R-module issemisimple.

Proposition 6.13 For a ring R, the following statements are equivalent:

1. R is semisimple.

2. The left R-module R is semisimple.

3. R is a direct sum of minimal left ideals.

4. R is a direct sum of finitely many minimal left ideals.

Proof: 1. ⇒ 2. Clear.2. ⇒ 1. By proposition 6.11, since every left R-module is a isomorphic to aquotient of a direct sum of copies of R.1. ⇒ 2. and 3. By definition of semisimplicity, R is a direct sum of simplesubmodules, and a simple submodule of R is a minimal left ideal of R. SoR =

⊕i∈I Li, where the Li are minimal left ideals of R. But such a direct sum

is always finite. Indeed, consider the element 1 ∈ R. We have 1 =∑

j∈J ei,where J is a finite subset of I and ej ∈ Lj, for every j ∈ J .Consider now x ∈ Lk, for some k 6∈ J . Then x · 1 = x, i.e.

∑j∈J xej = x.

Since x ∈ Lk and xej ∈ Lj, the fact that the Li are in direct sum gives usx = 0, so Lk = {0} for every k 6∈ J . �

Corollary 6.14 Every semisimple ring is left Artinian, left Noetherian, rightArtinian, right Noetherian.

Theorem 6.15 Let R be a ring. The following statements are equivalent:

1. R is semisimple.

2. R is left Artinian and J(R) = {0}.

3. R if left Artinian and has no nonzero nilpotent ideal.

Proof: The equivalence of 2 and 3 follows at once from proposition 5.11 andtheorem 5.10.“1 ⇒ 2” Assume R is semisimple, and consider the left ideal J(R). SinceJ(R) is then also an R-submodule of R, there is a submodule P of R suchthat R = J(R)⊕P as R-module. By exercise sheet 4, there are idempotentse, f ∈ R such that e ∈ J(R), f ∈ P , 1 = e+ f , J(R) = J(R)e and P = Pf .

6.2. SEMISIMPLICITY 57

Since e ∈ J(R), by proposition 5.5, f = 1− e has a left inverse u ∈ R. Sincef is an idempotent (f 2 = f) we obtain

1 = uf = uf 2 = (uf)f = f,

from which follows e = 0, and therefore J(R) = J(R)e = {0}.“2⇒ 1” We first show that if I is a minimal left ideal of R, then I is a directsummand of R:Since I 6= {0} we have I 6⊆ J(R). Since J(R) is the intersection of all maxi-mal left ideals of R, there is a maximal left ideal L of R not containing I.Since I is minimal, it is a simple left R-module, so I ∩ L is either {0} or I.Thesecond possibility means I ⊆ L, which is impossible. So I ∩ L = {0}.The maximality of A gives R = I + A, and so we have R = I ⊕ A.We now use the Artinian property to repeat this process until we get thewhole ring R:Choose a minimal left ideal I1 of R (it exists since R is Artinian, see propo-sition 4.18). As we just saw, R = I1⊕B1 for some left ideal B1. In the samewaty, B1 contains a minimal left ideal I2 of R, and B1 = I2 ⊕ B2, where B2

is a left ideal of R. It follows R = I1 ⊕ I2 ⊕B2.We continue this process, which must stop at some point because R is leftArtinian and the sequence of left ideals

B1 ) B2 ) · · ·

is strictly decreasing. Since the only way for this sequence to stop decreasingit for some Bk to be equal to {0}, we get k ∈ N such that Bk = {0}, whichgives

R = I1 ⊕ I2 ⊕ · · · ⊕ Ik,

proving that R is semisimple. �

Proposition 6.16 Let R be a left Artinian ring. Then a left R-module Mis semisimple if and only if J(R)M = {0}.

Proof: Let J = J(R). By definition of J we have JS = {0} for every simpleleft R-module S, since J ⊆ AnnR S. It yields JM = {0} for every semisimpleleft R-module M .Conversely, assume JM = {0}. Then the product R/J ×M → M , (r +J,m) 7→ r +m is well-defined and defines a structure of R/J-module on M .Moreover the R/J-module M and the R-module M have the same submod-ules (easy verification).Now R/J is still Artinian (easy) and we have J(R/J) = {0} (see remark ...).By theorem 6.15 we know that R/J is then semisimple. In particular the


R/J-module M is semisimple, i.e. every submodule of the R/J-module Mis a direct summand, i.e. every submodule of the R-module M is a directsummand, i.e. M is semisimple. �

Theorem 6.17 (Hopkins-Levitzki) Let R be a left Artinian ring. Thenfor a R-module M the following properties are equivalent:

1. M is Noetherian.

2. M is Artinian.

3. M is of finite lenth.

Proof: If M is semisimple then 1, 2 and 3 are equivalent since a Noetherianor Artinian module cannot be direct sum of infinitely many simple modules.The general case: Let J = J(R). Let M be Noetherian (or Artinian). Bycorollary 5.12 there is n ∈ N such that Jn = {0}, which yields the followingdecreasing sequence

M ⊇ JM ⊇ · · · ⊇ JnM = {0}.

Since M is Noetherian (or Artinian), it is easy to check that, for everyi = 0, . . . , n, J iM and J i/J i+1M are Noetherian (or Artinian).But J iM/J i+1M is semisimple by proposition 6.16 (since J · J iM/J i+1M ={0}. So J iM/J i+1M has a composition series, and therefore M has a com-position series. �

Corollary 6.18 (Hopkins) Every left Artinian ring is left Noetherian.

6.3 The structure of semisimple rings

Theorem 6.19 (Wedderburn-Artin) A ring R is semisimple if and onlyif it is isomorphic to a direct product Mn1(D1) × · · · ×Mnk

(Dk), for somek, n1, . . . , nk ∈ N and some division rings D1, . . . , Dk.

Proof: By proposition 6.13 3 we have

R = I1 ⊕ · · · ⊕ Ik,

where the I` are minimal left ideals ofR, and in particular simpleR-submodulesof R.By theorem 2.18 we have a ring ismorphism Rop ∼= EndRR, i.e. a ring iso-morphism Rop ∼= EndR(I1 ⊕ · · · ⊕ Ik). �

Index

ACC, 37annihilator, 30Artinian, 38ascending chain condition, 37

basis, 17

center, 25chain condition

ascending, 37descending, 37

commutative, 5cyclotomic polynomial, 27

DCC, 37dense, 32, 33density theorem, 34descending chain condition, 37division ring, 6domain, 6

endomorphismof rings, 7of modules, 20

epimorphismof modules, 20of rings, 7

faithful, 31field, 6finite topology, 32finitely generated

ideal, 10

generated

submodule, 16generator, 17

homomorphismof modules, 19of rings, 7

ideal, 8left, 7right, 7

image, 7, 20inverse, 6

left, 6right, 6

invertible, 6left, 6right, 6

irreduciblemodule, 21

isomorphic, 7isomorphism

of modules, 20of rings, 7

isomorphism theoremfor modules, 20for rings, 11

kernel, 7, 20

linearly independent, 17

maximalideal, 21

module, 15monomorphism

59

60 INDEX

of modules, 20of rings, 7

morphismof modules, 19of rings, 7

nil, 48nilpotent

element, 48ideal, 48

Noetherian, 38norm, 14

primitiveideal, 45ring, 31

productideal, 10

properideal, 8

quaternions, 13

ring, 5primitive, 31quotient, 8

Schur’s lemma, 22simple

module, 21ring, 12

submodule, 16subring, 5sum

ideal, 10

topologyfinite, 32

unit, 6

vector space, 15

Wedderburn’s theorem, 25

Wedderburn-Artin theorem, 39

zero divisor, 6

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