We next consider semisimple modules in more detail.
Lemma 3.1. Let R be a ring, let M be a left R-module, and let (Si)i∈I be a finite family of simple submodules with sum M = P
i∈ISi. Then there exists a subsetJ ⊂I such that M =L
j∈JSj.
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. Since Si is simple, we conclude thatSi∩P
j∈JSj=Si. It follows that P
j∈JSj =M as desired.
Proposition 3.2. Let R be a ring and letM be a semisimple left R-module.
(i) Let Q be a left R-module and let p: M → Q be a surjective R-linear map.
ThenQis semisimple and there exists an R-linear map s: Q→M such that p◦s:Q→Qis the identity map.
(ii) Let N be a left R-module and let i: N → M be an injective R-linear map.
ThenN is semisimple and there exists anR-linear mapr:M →N such that r◦i:N →N is the identity map.
Proof. (i) We writeM =L
i∈ISias a finite direct sum of simple submodules.
Let J ⊂I be the subset of indices i such that p(Si) 6= {0}. By Lemma 3.1, we can find a subsetK⊂J such thatL
i∈Kp(Si) =Q. Letj: L
i∈KSi→M be the canonical inclusion. Thenp◦jis an isomorphism which shows thatQis semisimple.
Moreover, the composite maps=j◦(p◦j)−1:Q→M has the desired property thatp◦s= idQ.
(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 semisimple and that the mapr= (q◦i)−1◦q:M →N satisfies thatr◦i= idN. We fix a ringRand define Λ(R) be the set of isomorphism classes of the simple leftR-modules that are of the formS=R/IwithI⊂Ra left ideal.1 LetSbe any simple leftR-module. To define thetypeofS, we choose a non-zero elementx∈S and consider theR-linear mapp:R→Sgiven by p(a) =ax. It is surjective, since Sis simple, and hence, induces an isomorphism ¯p:R/I→S, whereI= AnnR(x) is the kernel ofp. We now define the type ofSto be the isomorphism classλ∈Λ(R) of R/I. (Exercise: Show that the type ofSis well-defined.) We prove that semisimple leftR-modules admit the following canonicalisotypic decomposition.
Proposition 3.3. Let R be a ring.
(i) LetM be a semisimple leftR-module, and letMλ⊂M be the submodule given by the sum of all simple submodules of typeλ∈Λ(R). Then
M = M
λ∈Λ(R)
Mλ
and Mλ is a direct sum of simple submodules of type λ. In addition, Mλ is zero for all but finitely manyλ∈Λ(R).
1It is not possible, within standard ZFC set theory, to speak of the isomorphism classes of all simpleR-modules or the set thereof. This is the reason that we define Λ(R) in this way.
10
(ii) LetM andN be semisimple leftR-modules and letf:M →N be anR-linear map. Then for everyλ∈Λ(R),f(Mλ)⊂Nλ.
Proof. We first prove (i) SinceM is semisimple, we can writeM as a finite direct sumM =L
i∈ISi of simple submodules. IfMλ0 =L
i∈IλSi, whereIλ⊂Iis the subset ofi∈Isuch thatSiis of typeλ, thenM =L
λ∈Λ(R)Mλ0 andMλ0 ⊂Mλ. We must show that Mλ ⊂ Mλ0. So let S ⊂ M be a simple submodule of type λ and leti∈I. The compositionfi:S→M →Si of the canonical inclusion and the canonical projection is an R-linear map, and sinceS and Si are both simple left R-modules, the map fi is either zero or an isomorphism. If it is an isomorphism, then we have i∈Iλ, which shows thatS ⊂Mλ0, and hence,Mλ⊂Mλ0 as desired.
Finally, the finite setIis a the disjoint union of the subsetsIλ withλ∈Λ(R), and hence, all but finitely many of these subsets must be empty.
Next, to prove (ii), we letS ⊂M be a simple submodule of type λ. Since S is simple, eitherf(S)⊂N is zero or elsef|S:S→f(S) is an isomorphism of left R-modules. Therefore, f(Mλ)⊂Nλas stated.
Definition3.4. A ringR issemisimple if it semisimple as a left module over itself. A ringR issimple if it is semisimple and if it has exactly one type of simple modules.
We proceed to prove two theorems that, taken together, constitute a structure theorem for semisimple rings.
Theorem 3.5. Let R be a semisimple ring and let R = L
λ∈Λ(R)Rλ be the isotypic decomposition ofR as a left R-module.
(i) For everyλ∈Λ(R), the left idealRλ⊂R is non-zero. In particular, the set of typesΛ(R)is finite.
(ii) For everyλ∈Λ(R), the left ideal Rλ⊂R is also a right ideal.
(iii) Leta, b∈R and writea=P
λ∈Λ(R)aλ andb=P
λ∈Λ(R)bλ with aλ, bλ∈Rλ. Thenab=P
λ∈Λ(R)aλbλ andaλbλ∈Rλ.
(iv) For everyλ∈Λ(R), the subsetRλ⊂Ris a ring with respect to the restriction of the addition and multiplication onR, and the identity element is the unique elementeλ∈Rλ such thatP
λ∈Λ(R)eλ= 1.
(v) For everyλ∈Λ(R), the ring Rλ is simple.
Proof. (i) LetS be a simple leftR-module of typeλ. We choose a non-zero elementx∈Sand consider again the surjectiveR-linear mapp:R→Sdefined by p(a) =ax. By Proposition 3.2 there exists an R-linear map s:S →R such that p◦s= idS. But thens(S)⊂Ris a simple submodule of type λ, and hence,Rλ is non-zero. Finally, it follows from Proposition 3.3 (i) that Λ(R) is a finite set.
(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. This shows thataλbµ ∈Rλ∩Rµ, and since
Rλ∩Rµ=
(Rλ ifλ=µ, {0} ifλ6=µ,
the claim follows.
(iv) We have already proved in (iii) that the multiplication onR restricts to a multiplication onRλ. Now, for allaλ∈Rλ, we have
aλ=aλ·1 =aλ·(X
µ∈Λ
eµ) =X
µ∈Λ
aλ·eµ =aλ·eλ
and the identityaλ=eλ·aλ is proved analogously. It follows thatRλis a ring and thateλ∈Rλ is its identity element.
(v) Let Sλ be a simple left R-module of type λ. Since Rλ ⊂ R, the left multiplication byRonSλdefines a left multiplication byRλ onSλ. To prove that this defines a left Rλ-module structure onSλ, 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
fλ(eλ·x) =eλ·fλ(x) =fλ(x),
we conclude thateλ·x=x, for allx∈Sλ, as desired. We further note thatSλ is a simple left Rλ-module. Indeed, it follows from (iii) that a subsetN ⊂Sλ is an R-submodule if and only if it is an Rλ-submodule. Finally, by Proposition 3.3 (i), the left R-module Rλ is a direct sum Sλ,1⊕ · · · ⊕Sλ,r of simple submodules, all of which are isomorphic to the simple left R-moduleSλ. Therefore, also as a left Rλ-module,Rλ is the direct sumSλ,1⊕ · · · ⊕Sλ,r of submodules, all of which are isomorphic to the simple leftRλ-moduleSλ. This shows that Rλ is a semisimple ring, and (i) shows that every simple leftRλ-module is isomorphic toSλ. SoRλ is
a simple ring.
Remark 3.6. The inclusion map iλ: Rλ → R is not a ring homomorphism unless R = Rλ. Indeed, the map iλ takes the identity element eλ ∈ Rλ to the elementeλ ∈R, which is not equal to the identity element 1∈R, unlessR=Rλ. However, the projection map
pλ:R→Rλ
that takesa=P
µ∈Λaµ withaµ∈Rµ toaλ is a ring homomorphism. In general, theproduct ring of the family of rings (Rλ)λ∈Λ is the defined to be the set
Y
λ∈Λ
Rλ={(aλ)λ∈Λ |aλ∈Rλ}
with componentwise addition and multiplication. The identity element in the prod- uct ring is the tuple (eλ)λ∈Λ, where eλ∈Rλ is the identity element. We may now restate Theorem 3.5 (ii)–(v) as saying that the map
p:R→ Y
λ∈Λ(R)
Rλ
defined by p(a) = (pλ(a))λ∈Λ is an isomorphism of rings, and that each of the component ringsRλ is a simple ring.
Theorem 3.7. The following statements holds.
(i) LetDbe a division ring and letR=Mn(D)be the ring ofn×n-matrices. Then Ris a simple ring with the leftR-moduleS=Mn,1(D)of columnn-vectors as its simple module, and the map
ρ:D→EndR(S)op defined byρ(a)(x) =xais a ring isomorphism.
(ii) LetR be a simple ring and letS 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 of S, and the map
λ:R→EndD(S) defined byλ(a)(x) =axis a ring isomorphism.
Here, in (ii), the ring EndR(S)opis a division ring by Schur’s lemma, which we proved last time.
Proof. (i) We have proved in Lemma 2.11 thatS is a simple left R-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 are nsummandsS, defined byf(v1, . . . ,vn) =v1e1+· · ·+vnen is an isomorphism of leftR-modules.
Indeed, in then×n-matrixviei, theith column isvi and the remaining columns are zero. This shows thatR is a semisimple ring. By Theorem 3.5 (i), we conclude that every simple leftR-module is isomorphic toS. Hence, the ringRis 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) = idS 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 thatPx=xand such thatP S=xD⊂S. (The existence of such a matrixP will be shown on the problem set.) Sincef isR-linear, we have
f(x) =f(Px) =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, an isomorphism.
(ii) Since R is a simple ring with simple left R-module S, there exists an isomorphism of left R-modules f: S ⊕ · · · ⊕S → R from the direct sum of a finite number, sayn, of copies ofS ontoR. We now have ring isomorphisms
Rop−∼→EndR(R)−∼→EndR(Sn)−∼→Mn(EndR(S)) =Mn(Dop)
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 compositiongij=pi◦g◦ij of the inclusionij:S →Snof thejth summand, the endomorphism g:Sn → Sn, and the projectionpi: Sn → S onto theith summand. It follows that we have a ring isomorphism
R−∼→Mn(Dop)op−∼→Mn((Dop)op) =Mn(D)
given by the composition of the isomorphism above and the isomorphism that takes the matrixAto its transpose matrixAt. This shows that the simple ringR is isomorphic to the simple ringMn(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, so the proof is complete.
Remark 3.8. 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 is a field, and it is not difficult to show that also Z(Mn(D)) = k·In. It is possible for a division ring D to 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 now high time that we see an example of a semisimple ring. In general, if k is a commutative ring andGa group, then the group ringk[G] is defined to be the freek-module with basisGand with the “convolution” 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 byη(a) =a·eis ring homomorphism. IfM is a leftk[G]-module, then 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 fieldsQ,R, andChave characteristic 0, while for every prime numberp, the fieldZ/pZhas characteristicp.
Exercise 3.9. Let k 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.
Theorem 3.10 (Maschke’s theorem). Let k be a field and let G be a finite group, whose order is not divisible by the characteristic of k. Then the group ring k[G] is a semisimple ring.
Proof. We show that every leftk[G]-moduleM of finite dimensionmoverk is a semisimple leftk[G]-module. The proof is by induction on m; 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 leftk-module, and hence, also as a leftk[G]-module. So we letn >1 and assume, inductively, that every leftk[G]-module of dimension m < noverkis semisimple. We must show that ifM is a leftk[G]-module of dimensionm=nover k, then M is semisimple. 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 map ρ: M → N such that σ◦i = idN. The map ρis not necessarilyk[G]-linear. However, we claim that the mapr:M →N defined by
r(x) = 1
|G|
X
g∈G
gρ(g−1x)
isk[G]-linear and satisfiesr◦i= idN. Indeed,risk-linear and ifh∈G, then r(hx) = 1
|G|
X
g∈G
gρ(g−1hx) = 1
|G|
X
g∈G
hh−1gρ(g−1hx)
= 1
|G|
X
k∈G
hkρ(k−1x) =hr(x)
which shows thatrisk[G]-linear. Moreover, we have (r◦i)(x) = 1
|G|
X
g∈G
gρ(g−1i(x)) = 1
|G|
X
g∈G
gρ(i(g−1x))
= 1
|G|
X
g∈G
gg−1x=x
which shows thatr◦i= idN. This proves the claim. Now, letP be the kernel ofr.
The claim shows that M is equal to the direct sum of the submodulesN, P ⊂M. ButN and P both have dimension less thann overk, and hence, are semisimple by the induction hypothesis. This shows thatM is semisimple as desired.
