(1)We next consider semi-simple modules in more detail

We next consider semi-simple modules in more detail.

Lemma 3.1. Let R be a ring, let M be a left R-module, and let {S_i}_i∈I be a finite family of simple submodules the union of which generates M. Then there exists a subsetJ ⊂I such thatM =L

i∈JSi.

Proof. We consider a subsetJ ⊂Iwhich is maximal among subsets with the property that the sum of submodules P

j∈JSj ⊂M is direct. Now, if i∈IrJ, then Si∩P

j∈JSj 6={0} or else J would not be maximal. SinceSi is simple, we conclude thatSi is contained in the submoduleP

j∈JSj ⊂M. It follows that this

submodule is all ofM. This completes the proof.

Proposition 3.2. Let R be a ring and letM be a semi-simple left R-module.

(i) LetQbe a leftR-module and letp:M →Qbe a surjectiveR-linear map.

ThenQis semi-simple and there exists anR-linear maps:Q→M such thatp◦s:Q→Qis the identity map.

(ii) LetN be a leftR-module and leti:N →M be an injectiveR-linear map.

ThenN is semi-simple and there exists anR-linear mapr:M →N such thatr◦i:N→N is the identity map.

Proof. (i) We writeM =L

i∈IS_ias a finite direct sum of simple submodules.

Let J ⊂I be the subset of indices i such thatp(S_i) is non-zero. By Lemma 3.1, we can find a subset K ⊂J such that L

i∈Kp(S_i) = Q. Letj: L

i∈KS_i → M be the canonical inclusion. Then p◦j is an isomorphism which shows that Q is semi-simple. Moreover, the composite maps=j◦(p◦j)⁻¹:Q→M has the desired property thatp◦sis the identity map ofQ.

(ii) It follows from (i) that there exists a submoduleP ⊂M such that the com- position P →M →M/N of the canonical inclusion and the canonical projection is an isomorphism. Now, if q:M →M/P is the projection onto the quotient by P, thenq◦i:N →M/P is an isomorphism. This shows thatN is semi-simple and that the mapr= (q◦i)⁻¹◦q:M →N satisfies thatr◦i= idN. Let M be a semi-simple left R-module and let Λ be the set of isomorphism classes of simple leftR-modules. If the simple submoduleS ⊂M belongs to the classλ∈Λ, we say thatS has type λ. We prove that semi-simple leftR-modules admit the following canonicalisotypic decomposition.

Proposition 3.3. Let R be a ring.

(i) LetM be a semi-simple left R-module, and letMλ⊂M be the submodule generated by the union of all simple submodules of typeλ. Then

M =M

λ∈Λ

M_λ

andMλ is a direct sum of simple submodules of typeλ.

(ii) Let M and N be semi-simple left R-modules and let f:M → N be an R-linear map. Then f(Mλ)⊂Nλ.

Proof. (i) SinceM is semi-simple, we can writeM =L

i∈ISi as a direct sum of simple submodules. Let M_λ⁰ = L

i∈IλSi where Iλ ⊂ I is the subset of i ∈ I such thatS_i is of typeλ. We haveM =L

λ∈ΛM_λ⁰ andM_λ⁰ ⊂M_λ. We must prove

9

that M_λ⁰ = Mλ. So let S ⊂ M be a simple submodule of type λand let i ∈ I.

The composition fi: S → M → Si of the canonical inclusion and the canonical projection is anR-linear map. SinceS andSiare both simple leftR-modules, the mapf_i is either zero or an isomorphism. If it is an isomorphism, we havei∈I_λby definition. This shows thatS⊂M_λ⁰, and hence,M_λ⊂M_λ⁰ as desired.

(ii) LetS⊂M be a simple submodule of typeλ. Thenf(S)⊂N is either zero or a simple submodule of typeλ. Therefore,f(M_λ)⊂N_λ as stated.

Definition3.4. (i) The ringRissemi-simpleif it semi-simple as a left module over itself.

(ii) The ring R issimple if it is semi-simple and if it has exactly one type of simple modules.

Theorem 3.5. Let R be a semi-simple ring and let R = L

λ∈ΛRλ be the isotypic decomposition ofR as a left R-module.

(i) For every λ∈Λ, the left ideal Rλ⊂R is non-zero. In particular,Λ is a finite set.

(ii) For everyλ∈Λ, the left idealRλ⊂R is also a right ideal.

(iii) Let a, b∈R and write a=P

λ∈Λaλ andb=P

λ∈Λbλ with aλ, bλ∈Rλ. Thenab=P

λaλbλ andaλbλ∈Rλ.

(iv) For everyλ∈Λ,Rλ is a ring with respect to the restriction of the mul- tiplication on R and the identity element is the unique element e_λ ∈R_λ such thatP

λe_λ= 1.

(v) For everyλ∈Λ, the ringR_λ is simple.

Proof. (i) LetS be a simple leftR-module of typeλ. We choose a non-zero element x∈ S and consider the R-linear map p: R →S defined by p(a) =a·x.

The imagep(S)⊂S, which is a non-zero submodule of a simple leftR-module, is necessarily all ofS, sopis surjective. We conclude from Proposition 3.2 that there exists an R-linear maps:S →R such that p◦s= id_S. But thens(S)⊂ R is a simple submodule of typeλ.

(ii) Let a ∈ R and let ρa: R → R be the map ρa(b) = ba defined by right multiplication bya. It is an R-linear map from the leftR-moduleR to itself. By Proposition 3.3 (ii), we conclude thatρa(Rλ)⊂Rλwhich is precisely the statement thatRλ⊂R is a right ideal.

(iii) SinceRµ ⊂R is a left ideal, we haveaλbµ ∈Rµ, and since Rλ ⊂R is a right ideal, we haveaλbµ ∈Rλ. It follows thataλbµ ∈Rλ∩Rµ which is equal to Rλ and{0}, respectively, asλ=µandλ6=µ.

(iv) We have already proved in (iii) that the multiplication onR restricts to a multiplication onR_λ. Now, for alla_λ∈R_λ, we have

a_λ=a_λ·1 =a_λ·(X

µ∈Λ

e_µ) =X

µ∈Λ

a_λ·e_µ =a_λ·e_λ

and the identitya_λ=e_λ·a_λ is proved analogously. It follows thatR_λ is a ring.

(v) Let Sλ be a simple left R-module of type λ. Since Rλ ⊂ R, the left multiplication ofR onSλ defines a left multiplication ofRλ onSλ. To prove that this defines a left Rλ-module structure on Sλ, we must show that eλ·x=x, for all x ∈ Sλ. We have just proved that eλ·y = y, for all y ∈ Rλ. Moreover, by Proposition 3.3 (i), we can find an injectiveR-linear mapfλ:Sλ→Rλ. Since

f_λ(e_λ·x) =e_λ·f_λ(x) =f_λ(x),

we conclude thateλ·x=x, for allx∈Sλ, as desired. We further note thatSλ is a simple leftRλ-module. Indeed, it follows from (iii) that the subsetN⊂Sλis anR- submodule if and only if it is anRλ-submodule. Finally, by Proposition 3.3 (i), the leftR-moduleR_λis isomorphic to a direct sumS_λ,1+· · ·+Sλ,rof simple submodules, all of which are isomorphic to the simple left R-module S_λ. Therefore, as a left R_λ-module,R_λ is isomorphic to the direct sumS_λ,1+· · ·+S_λ,r of submodules, all of which are isomorphic to the simple leftR_λ-moduleS_λ. This shows that R_λ is a semi-simple ring, and we conclude from (i) that every simple leftR_λ-module is

isomorphic toSλ. SoRλ is a simple ring.

Remark 3.6. The inclusion map i_λ: R_λ → R is not a ring homomorphism unlessR =R_λ. Indeed, the mapi_λ takes the multiplicative identity elemente_λ∈ Rλ to the elementeλ∈Rwhich is not equal to the multiplicative identity element 1∈RunlessR=Rλ. However, the projection map

p_λ:R→R_λ that takesa=P

µ∈Λaµ withaµ∈Rµ toaλ is a ring homomorphism. In general, theproduct ring of the family of rings{R_λ∈Λ} is the defined to be the set

Y

λ∈Λ

R_λ={(aλ)_λ∈Λ |a_λ∈R_λ}

with componentwise addition and multiplication. The multiplicative identity ele- ment in the product ring is the tuple (eλ)_λ∈Λ where eλ∈Rλ is the multiplicative unit element. We may now restate Theorem 3.5 (ii)–(v) as saying that the map

p:R→ Y

λ∈Λ

R_λ

defined by p(a) = (pλ(a))λ∈Λ is an isomorphism of rings, and that each of the component ringsR_λ is a simple ring.

We next prove the following structure theorem for simple rings. We recall from Schur’s lemma that the endomorphism ring of a simple module is a division ring.

Theorem 3.7. The following statements holds.

(i) LetDbe a division ring and letR=Mn(D)be the ring ofn×n-matrices.

ThenR is a simple ring with the leftR-moduleS =M_n,1(D) of column n-vectors as its simple module, and the map

ρ:D→EndR(S)^op defined byρ(a)(x) =xa is a ring isomorphism.

(ii) Let R be a simple ring and let S be a simple left R-module. ThenS is a finite dimensional right vector space over the division ringD= EndR(S)^op opposite of the ring ofR-linear endomorphisms ofS, and the map

λ:R→End_D(S) defined byλ(a)(x) =ax is a ring isomorphism.

Proof. (i) We have proved in Lemma 2.12 thatSis a simpleR-module. Now, letei ∈M1,n(D) be the row vector whoseith entry is 1 and whose remaining entries are 0. Then the mapf:S⊕ · · · ⊕S →R, where there arensummandsS, defined byf(v₁, . . . , v_n) =v₁e₁+· · ·+v_ne_n is an isomorphism of leftR-modules. Indeed, in then×n-matrixv_ie_i, theith column isv_i and the remaining columns are zero.

This shows that R is a semi-simple ring. By Theorem 3.5 (i), we conclude that every simple leftR-module is isomorphic toS. Hence, the ringR is simple.

It is readily verified that the mapρis a ring homomorphism. Now, the kernel ofρis a two-sided ideal in the division ringD, and hence, is either zero or all ofD.

Butρ(1) = id_S is not zero, so the kernel is zero, and hence the mapρis injective.

It remains to show that ρis surjective. So letf:S →S be anR-linear map. We must show that there existsa∈D such that for ally∈S,f(y) =ya. To this end, we fix a non-zero elementx∈S and choose a matrixP∈Rsuch thatP x=xand such thatP S=xD⊂S. Sincef isR-linear, we have

f(x) =f(P x) =P f(x)∈xD

which shows that f(x) = xa with a ∈ D. Now, given any y ∈ S, we can find a matrixA∈R such thatAx=y. Again, sincef isR-linear, we have

f(y) =f(Ax) =Af(x) =Axa=ya

as desired. This shows thatρis surjective, and hence, and isomorphism.

(ii) Since R is a simple ring with simple left R-module S, there exists an isomorphism of leftR-modulesf: Sⁿ→R from the direct sum of finite numbern copies ofS ontoR. We now have ring isomorphisms

R^op−^∼→End_R(R)−^∼→End_R(Sⁿ)−^∼→M_n(End_R(S)) =M_n(D^op)

where the left-hand isomorphism is given by Remark 2.6, the middle isomorphism is induced by the chosen isomorphismf, and the right-hand isomorphism takes the endomorphismg to the matrix of endomorphisms (gij) with the endomorphismgij

defined to be the composition gij =pi◦g◦ij of the inclusion ij:S → Sⁿ of the jth summand, the endomorphismg: Sⁿ→Sⁿ, and the projectionpi: Sⁿ→S on theith summand. It follows that we have a ring isomorphism

R−^∼→Mn(D^op)^op−^∼→Mn((D^op)^op) =Mn(D)

given by the composition of the isomorphism above and the isomorphism that takes the matrix Ato its transpose^tA. This shows that the simple ring Ris isomorphic to the simple ring Mn(D) we considered in (i). Therefore, it suffices to show that the map λis an isomorphism in this case. But this is precisely the statement of

Corollary 2.5.

Exercise3.8. LetDbe a division ring, letR=Mn(D), and letS =Mn,1(D).

We viewS as a leftR-module and as a rightD-vector space.

(1) Let x∈S be a non-zero vector. Show that there exists a matrix P ∈R such thatP S=xD⊂S. (Hint: Tryx=e₁ first.)

(2) Let x, y∈S be non-zero vectors. Show that there exists a matrixA∈R such thatAx=y.

Remark 3.9. The center of a ring R is the subringZ(R)⊂R of all elements a∈R with the property that for allb∈R, ab=ba; it is a commutative ring. The center k =Z(D) of the division ring D clearly is a field, and it is not difficult to show that alsoZ(Mn(D)) =k. It is possible for a division ring Dto be of infinite dimension over the centerk. However, one can show that ifDis of finite dimension doverk, thend=m²is a square and every maximal subfieldE⊂Dhas dimension moverk. For example, the center of the division ring of quarternionsHis the field of real numbersRand the complex numbersC⊂His a maximal subfield.

It is high time that we see an example of a semi-simple ring. In general, if k is a commutative ring and ifGis a group, the group ringk[G] is defined to be the freek-module with basisGand with multiplication

(X

g∈G

agg)·(X

g∈G

bgg) =X

g∈G

( X

h,k∈G hk=g

ahbk)g.

We note thatG⊂k[G] as the set of basis elements; the unit element e∈Gis also the multiplicative unit element in the ringk[G]. Moreover, the map η:k →k[G]

defined by η(a) =a·eis ring homomorphism. IfM is a leftk[G]-module, we also say thatM is a k-linear representation of the groupG.

Let k be a field and let η:Z → k be the unique ring homomorphism. We define the characteristic of k to be the unique non-negative integer char(k) such that ker(η) = char(k)Z. For example, the fields Q, R, and C have characteristic zero while, for every prime numberp, the field Z/pZhas characteristicp.

Exercise 3.10. Letk be a field. Show that char(k) is either zero or a prime number, and that every integernnot divisible by char(k) is invertible ink.

Theorem3.11 (Maschke’s theorem). Let kbe a field and letGbe finite group whose order is not divisible by the characteristic ofk. Then the group ringk[G]is a semi-simple ring.

Proof. We show that every leftk[G]-moduleM of finite dimensionmoverk is a semi-simple leftk[G]-module. The proof is by induction onm; the basic case m= 1 follows from Example 2.11, since a left k[G]-module of dimension 1 over k is simple as a left k-module, and hence, also as a left k[G]-module. So we let M be a left k[G]-module of dimension m > 1 over k and assume, inductively, that every left k[G]-module of smaller dimension is semi-simple. We must show that M is semi-simple. IfM is simple, we are done. If M is not simple, there exists a non-zero proper submoduleN⊂M. We leti:N→M be the inclusion and choose a k-linear mapσ: M → N such that σ◦i = id_N. The map σ is not necessarily k[G]-linear. However, we claim that the maps:M →N defined by

s(x) = 1

|G|

X

g∈G

gσ(g⁻¹x)

isk[G]-linear and satisfiess◦i= idN. Indeed,sisk-linear and ifh∈G, then s(hx) = 1

|G|

X

g∈G

gσ(g⁻¹hx) = 1

|G|

X

g∈G

hh⁻¹gσ(g⁻¹hx)

= 1

|G|

X

k∈G

hkσ(k⁻¹x) =hs(x)

which shows thatsisk[G]-linear. Moreover, we have (s◦i)(x) = 1

|G|

X

g∈G

gσ(g⁻¹i(x)) = 1

|G|

X

g∈G

gσ(i(g⁻¹x))

= 1

|G|

X

g∈G

gg⁻¹x=x

which shows thats◦i= id_N. This proves the claim. Now, letP be the kernel ofs.

The claim shows that M is equal to the direct sum of the submoduleN, P ⊂M.

ButN andP both have dimension less thanmoverk, and hence, are semi-simple by the inductive hypothesis. This shows thatM is semi-simple as desired.

Example 3.12 (Cyclic groups). To illustrate the theory above, we determine the structure of the group ringsC[C_n], R[C_n], and Q[C_n], where C_n is the cyclic group of ordern. Theorem 3.11 shows that the three rings are semi-simple rings, and their structure are given by Theorems 3.5 and 3.7 once we identify their isomorphism classes of simple modules; we proceed to do so. We fix choices of a generatorg∈C_n and of a primitiventh root of unity ζn∈C.

We first consider the complex group ringC[Cn]. For every 06k < n, we define the leftC[Cn]-moduleC(ζ_n^k) to be the sub-C-vector space C(ζ_n^k)⊂Cspanned by the elementsζ_n^ki with 06i < n and with the module structure defined by

(

n−1

X

i=0

a_igⁱ)·z=

n−1

X

i=0

a_iζ_n^kiz.

The leftC[Cn]-module C(ζ_n^k) is simple. For as aC-vector space, C(ζ_n^k) =C, and therefore has no non-trivial proper submodules. Suppose thatf:C(ζ_n^k)→C(ζ_n^l) is aC[C_n]-linear isomorphism. Then we have

ζ_n^kf(1) =f(ζ_n^k) =f(g·1) =g·f(1) =ζ_n^lf(1),

where the first and third equalities follows from C[C_n]-linearity. Since f(1) 6= 0, we conclude thatk=l. So thensimple leftC[C_n]-modulesC(ζ_n^k), 06k < n, are pairwise non-isomorphic. Therefore, Theorem 3.5 (i) implies that

C[Cn] =

n−1

M

k=0

C(ζ_n^k)

as a leftC[C_n]-module. The endomorphism ring End_C[Cn](C(ζ_n^k)) is isomorphic to the fieldCfor all 06k < n.

We next consider the real group ring R[C_n]. Again, for 0 6k < n, we define the leftR[C_n]-moduleR(ζ_n^k) to be the sub-R-vector space R(ζ_n^k)⊂Cspanned by the elementsζ_n^ki with 06i < n and with the module structure defined by

(

n−1

X

i=0

a_igⁱ)·z=

n−1

X

i=0

a_iζ_n^kiz.

The leftR[Cn]-moduleR(ζ_n^k) is simple. For given two elementsz, z⁰∈R(ζ_n^k), there exists ω ∈ R[Cn] with ω·z = z⁰. The dimension of R(ζ_n^k) as an R-vector space is either 1 or 2 according as ζ_n^k ∈ R or ζ_n^k ∈/ R. Moreover, we find that the left R[Cn]-modulesR(ζ_n^k) andR(ζ_n^l) are isomorphic if and only if the complex numbers ζ_n^k andζ_n^l are conjugate. Again, from Theorem 3.5 (i), we conclude that

R[C_n] =

[n/2]

M

k=0

R(ζ_n^k)

as a leftR[Cn]-module. Here [n/2] is the largest integer less than or equal ton/2.

The endomorphism ring End_R[Cn](R(ζ_n^k)) is isomorphic toR, if k= 0 ork=n/2, and toC, otherwise.

Finally, we consider the rational group ringQ[Cn]. For all 06k < n, we define the left Q[Cn]-moduleQ(ζ_n^k) to be the sub-Q-vector spaceQ(ζ_n^k)⊂Cspanned by the elementsζ_n^ki with 06i < n and with the module structure defined by

(

n−1

X

i=0

aigⁱ)·z=

n−1

X

i=0

aiζ_n^kiz.

Again,Q(ζ_n^k) is a simple leftQ[C_n]-module, since givenz, z⁰ ∈Q(ζ_n^k), there exists an elementω∈Q[C_n] withω·z=z⁰. Suppose that

{ζ_n^ki|06i < n}={ζ_n^li|06i < n} ⊂C. Then we may define aQ[Cn]-linear isomorphism

f:Q(ζ_n^k)→Q(ζ_n^l)

to be the uniqueQ-linear map that takesζ_n^kitoζ_n^li, for all 06i < n. Suppose that the set{ζ_n^ki|06i < n}hasdelements. Thenddivides nand

{ζ_n^ki|06i < n}={ζ_dⁱ |06i < d}

withζd∈Ca primitivedth root of unity. LetQ(ζd)⊂Cbe the left Q(ζd)-module defined by the sub-Q-vector spaceQ(ζd)⊂Cspanned by theζ_dⁱ with 06i < dand with the module structure

(

n−1

X

i=0

z_igⁱ)·z=

n−1

X

i=0

a_iζ_dⁱz.

Then we define aQ[C_n]-linear isomorphism f:Q(ζ_d)→Q(ζ_n^k)

to be the uniqueQ-linear map that takesζ_dⁱ toζ_n^ki. It is not difficult to show that the dimension of Q(ζd) as a Q-vector space is equal to the number ϕ(d) of the integers 16i6dthat are prime tod. Moreover, since

X

d|n

ϕ(d) =n

we conclude from Theorem 3.5 (i) that these represent all isomorphism classes of simple leftQ[C_n]-modules. Therefore,

Q[Cn] =M

d|n

Q(ζd)

as a leftQ[C_n]-module. We note thatQ(ζ_d)⊂Cis a subfield, thedth cyclotomic field overQ. The endomorphism ring End_Q[Cn](Q(ζ_d))^opis isomorphic to the field Q(ζ_d) for every divisordofn.

Remark3.13 (Modular representation theory). If the characteristic of the field kdivides the order of the groupG, then the group ringk[G] is not semi-simple, and it is a very difficult problem to understand the structure of this ring. For example, ifFp is the field withpelements andSp is the symmetric group onpletters, then the structure of the ringFp[Sp] is only understood for a few primesp.