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Chapter 4
Algebras
4.1
Definition
It is time to introduce the notion of an algebra over a commutative ring. So let R be a commutative
ring. An R-algebra is a ring A (unital as always) that is an R-module (left, say) such that
r(ab) = (ra)b = a(rb)
for all r ∈ R, a, b ∈ A. You can say it the other way round: an R-algebra is an R-module equipped
with an R-bilinear multiplication R × A → A making it into a ring. Thus, this multiplication
satisfies the conditions
(A1) 1R a = a;
(A2) (r + s)a = ra + sa;
(A3) r(a + b) = ra + rb;
(A4) (rs)a = r(sa);
(A5) r(ab) = (ra)b = a(rb)
for all r, s ∈ R, a, b ∈ A. (Note strictly speaking I should call this an “associative, unital R-algebra”
– there are other important sorts of algebra which are not associative – but since we won’t meet
them in this course I’ll just stick to algebra.)
Note a Z-algebra is just the old definition of ring: “Z-algebras = rings” just as “Z-modules =
Abelian groups”. So you should view the passage from rings to R-algebras as analogous to the
passage from Abelian groups to R-modules! This is the idea of studying objects (e.g. Abelian
groups, rings) relative to a fixed commutative base ring R.
There is an equivalent formulation of the definition of R-algebra: an R-algebra is a ring A
together with a distinguished ring homomorphism (the “structure map”)
s:R→A
such that the image of s lies in the center Z(A) = {a ∈ A | ab = ba for all b ∈ A}. Indeed, given
such a ring homomorphism, define a multiplication R × A → A by (r, a) $→ s(r)a. Now check
this satisfies the above axioms (A1)–(A5) (the last one being because im s ⊆ Z(A)). Conversely,
given an R-algebra as defined originally, one obtains a ring homomorphism s : R → A by defining
s(r) = r1A , and the image lies in Z(A) by (A5).
Let A be an R-algebra. Then, given an A-module M , we can in particular think of M as just
an R-module, defining rm = s(r)m for r ∈ R, m ∈ M . So you can hope to exploit the additional
structure of the base ring R in studying A-modules. In particular, if R = F is a field and A is
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CHAPTER 4. ALGEBRAS
an F -algebra, A and any A-module M is in particular a vector space over F . So we can talk
about finite dimensional F -algebras and finite dimensional modules over an F -algebra, meaning
their underlying dimension as vector spaces over F .
Note given A-modules M, N , an A-module homomorphism between them is automatically an Rmodule homomorphism (for the underlying R-module structure). However, it is not necessarily the
case that a ring homomorphism between two different R-algebras is an R-module homomorphism.
So one defines an R-algebra homomorphism f : A → B between two R-algebras A and B to
be a ring homomorphism in the old sense that is in addition R-linear (i.e. it is an R-module
homomorphism too). Ring homomorphisms and R-algebra homomorphisms are different things!
Now for examples. Actually, we already know plenty. Let R be a commutative ring. Then, the
polynomial ring R[X1 , . . . , Xn ] is evidently an R-algebra: indeed, R[X1 , . . . , Xn ] contains a copy
of R as the subring consisting of polynomials of degree zero.
The ring Mn (R) of n × n matrices over R is an R-algebra: again, it contains a copy of R
as the subring consisting of the scalar matrices. But note there is a big difference between this
and the previous example: Mn (R) is finitely generated as an R-module (indeed, it is free of rank
n2 ) whereas R[X1 , . . . , Xn ] is not. In case F is a field, Mn (F ) is a finite dimensional F -algebra,
F [X1 , . . . , Xn ] is not.
For the next example, let M be any (left) R-module. Consider the Abelian group
EndR (M ).
We make it into a ring by defining the product of two endomorphisms of M simply to be their
composition. Now I claim that EndR (M ) is in fact an R-algebra: indeed, we define rθ for r ∈
R, θ ∈ EndR (M ) by setting
(rθ)(m) = r(θ(m))(= θ(rm))
for all m ∈ M . Let us check that rθ really is an R-endomorphism of M . Take another s ∈ R.
Then,
s((rθ)(m)) = sr(θ(m)) = θ(srm) = θ(rsm) = (rθ)(sm).
Note we really did use the commutativity of R!
We can generalize the previous two examples. Suppose now that A is a (not necessarily commutative) R-algebra and M is a left A-module. Then, the ring D = EndA (M ) is also an R-algebra,
defining (rd)(m) = r(d(m)) for all r ∈ R, d ∈ D, m ∈ M .
For the final example of an algebra, let G be any group. Define the group algebra RG to be the
free R-module on basis the elements of G. Thus an element of RG looks like
!
rg g
g∈G
for coefficients rg ∈ R all but finitely many of which are zero. Multiplication is defined by the rule

&
'
!
!
!


rg g
sh h =
rg sh (gh).
g∈G
h∈G
g,h∈G
In other words, the multiplication in RG is induced by the multiplication in G and R-bilinearity.
Note the RG contains a copy of R as a subring, namely, R1G , so is certainly an R-algebra. The
construction of the group algebra RG allows the possibility of studying the abstract group G by
studying the category of RG-modules – so module theory can be applied to group theory. Note
finally in case G is a finite group and F is a field, the group algebra F G is a finite dimensional
F -algebra.
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4.2. SOME MULTILINEAR ALGEBRA
4.2
Some multilinear algebra
Let F be a field (actually, everything here could be done more generally over an arbitrary commutative ring, e.g. Z – but let’s stick to the case of a field for simplicity). Given a vector space V
over F , we have defined V ⊗ V (meaning V ⊗F V ). Hence, we have (V ⊗ V ) ⊗ V and V ⊗ (V ⊗ V ).
But these two are canonically isomorphic, the isomorphism mapping a generator (u ⊗ v) ⊗ w of the
first to u ⊗ (v ⊗ w) in the second. From now on we’ll identify the two, and write simply V ⊗ V ⊗ V
for either. More generally, we’ll write T n (V ) for V ⊗ · · · ⊗ V (n times), where the tensor is put
together in any order you like, all different ways giving canonically isomorphic vector spaces. Set
(
T (V ) =
T n (V )
n≥0
and call it the tensor algebra on V . Note T 0 (V ) = F by convention, just a copy of the ground
field, and T 1 (V ) = V . Then, T (V ) is an F -vector space given, but we can define a multiplication
on it making it into an F -algebra as follows: define a map
T m (V ) × T n (V ) → T m+n (V ) = T m (V ) ⊗ T n (V )
by (x, y) $→ x ⊗ y. Then extend linearly to give a map from T (V ) × T (V ) → T (V ). Since ⊗ is
associative, the resulting multiplication is associative, and T (V ) is an F -algebra.
To be quite explicit, suppose that ei (i ∈ I) is a basis for V . Then,
{ei1 ⊗ ei2 ⊗ · · · ⊗ ein | n ≥ 0, i1 , . . . , in ∈ I}
gives a basis of “monomials” for T (V ). Multiplication of the basis elements is just by joining
tensors together. In other words, T (V ) looks like “non-commutative polynomials” in the ei (i ∈ I)
– an arbitrary element being a linear combination of finitely many non-commuting monomials.
The importance of the tensor algebra is as follows:
Universal property of the tensor algebra. Given any F -linear map f : V → A to an F algebra A, there is a unique F -linear ring homomorphism f¯ : T (V ) → A such that f = f¯ ◦ i, where
i : V → T (V ) is the inclusion of V into T 1 (V ) ⊂ T (V ).
(Note: this universal property could be taken as the definition of T (V )!)
Proof. The vector space T n (V ) is defined by the following universal property: given a multilinear
map fn : V × · · · × V (n copies) to a vector space W , there exists a unique F -linear map f¯n :
T n (V ) → W such that fn = f¯n ◦ in , where in is the obvious map V × · · · × V → T n (V ). Now define
the map fn so that fn (v1 , v2 , . . . , vn ) = f (v1 )f (v2 ) . . . f (vn ) (multiplication in the ring A). Check
that it is multilinear in the vi , since A is an F -algebra. Hence it induces a unique f¯n : T n (V ) → A.
Now define f¯ : T (V ) → A by glueing all the f¯n together. In other words, using the basis notation
above, we have that
f¯(ei1 ⊗ · · · ⊗ ein ) = f (ei1 )f (ei2 ) . . . f (ein ).
This is an F -linear ring homomorphism, and is clearly the only option to extend f to such a thing.
We can restate the theorem as follows: Let X be a basis of V . Then, given any set map f from
X to an F -algebra A, there exists a unique F -linear homomorphism f¯ : T (V ) → A extending f
(proof: extend the map X → A to a linear map V → A in the unique way then apply the theorem).
In other words, T (V ) plays the role of the “free F -algebra on the set X”. Then you can define
algebras by “generators and relations” but we won’t go into that...
Next, we come to the symmetric algebra on V . Continue with V being a vector space. We’ll
be working with V × · · · × V (n times). Call a map
f : V × ··· × V → W
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CHAPTER 4. ALGEBRAS
a symmetric multilinear map if its multilinear, i.e.
f (v1 , . . . , cvi + c# vi# , . . . , vn ) = cf (v1 , . . . , vi , . . . , vn ) + c# f (v1 , . . . , vi# , . . . , vn )
for each i, and moreover
f (v1 , . . . , vi , vi+1 , . . . , vn ) = f (v1 , . . . , vi+1 , vi , . . . , vn )
for each i (so its symmetric). The symmetric power S n (V ), together with a distinguished map
i : V × · · · × V → S n (V ), is characterized by the following universal property: given any symmetric
multilinear map f : V × · · · × V → W to a vector space W , there exists a unique linear map
f¯ : S n (V ) → W such that f = f¯ ◦ i. Note this is exactly the same universal property as defined
T n (V ) in the proof of the universal property of the tensor algebra, but we’ve added the word
symmetric too.
As usual with universal properties, S n (V ), if it exists, is unique up to canonical isomorphism (so
we call it “the” symmetric algebra). Still, we need to prove existence with some sort of construction.
So now define
S n (V ) = T n (V )/In
where
In = *v1 ⊗ · · · ⊗ vi ⊗ vi+1 ⊗ · · · ⊗ vn − v1 ⊗ · · · ⊗ vi+1 ⊗ vi ⊗ · · · ⊗ vn ,
is the subspace spanned by all such terms for all v1 , . . . , vn ∈ V . The distinguished map i :
V × · · · × V → S n (V ) is the obvious map V × · · · × V → T n (V ) followed by the quotient map
T n (V ) → S n (V ). Now take a symmetric multilinear map f : V × · · · × V → W . By the universal
property defining T n (V ), there is a unique F -linear map f¯1 : T n (V ) → W extending f . Now since
f is symmetric, we have that
f¯1 (v1 ⊗ · · · ⊗ vi ⊗ vi+1 ⊗ · · · ⊗ vn − v1 ⊗ · · · ⊗ vi+1 ⊗ vi ⊗ · · · ⊗ vn ) = 0.
Hence, f¯1 annihilates all generators of In , hence all of In . So f¯1 factors through the quotient S n (V )
of T n (V ) to induce a unique map f¯ : S n (V ) → W with f = f¯ ◦ i. Note we have really just used
the univeral property of T n followed by the universal property of quotients!
So now we’ve defined the nth symmetric power of a vector space. Note its customary to denote
the image in S n (V ) of (v1 , . . . , vn ) ∈ V × · · · × V under the map i by v1 · · · · · vn . So v1 · · · · · vn is
the image of v1 ⊗ · · · ⊗ vn under the quotient map T n (V ) → S n (V ).
We can glue all S n (V ) together to define the symmetric algebra on V :
(
S(V ) =
S n (V )
n≥0
where again S 0 (V ) = F, S 1 (V ) = V . Since each S n (V ) is by construction a quotient of T n (V ), we
see that S(V ) is a quotient of T (V ), i.e.
S(V ) = T (V )/I
)
where I = n≥0 In . I claim in fact that I is an ideal of T (V ), so that S(V ) is actually a quotient
algebra of T (V ). Actually, I claim even more, namely:
4.2.1. Lemma. I is the two-sided ideal of T (V ) generated by the elements v ⊗ w − w ⊗ v for all
v, w ∈ V .
Proof. Let J be the two-sided ideal generated by the given elements. By definition, I is spanned
as an F -vector space by terms like
v1 ⊗ · · · ⊗ vi ⊗ vi+1 ⊗ · · · ⊗ vn − v1 ⊗ · · · ⊗ vi+1 ⊗ vi ⊗ · · · ⊗ vn =
v1 ⊗ · · · ⊗ vi−1 ⊗ (vi ⊗ vi+1 − vi+1 ⊗ vi ) ⊗ vi+2 ⊗ · · · ⊗ vn .
4.2. SOME MULTILINEAR ALGEBRA
101
This clearly lies in J, hence I ⊆ J. On the other hand, the generators of the ideal J lie in I, so it
just remains to show that I is a two-sided ideal of T (V ). Since T (V ) is spanned by monomials of
the form u = u1 ⊗ · · · ⊗ um , it suffices to check that for any generator
v = v1 ⊗ · · · ⊗ vi ⊗ vi+1 ⊗ · · · ⊗ vn − v1 ⊗ · · · ⊗ vi+1 ⊗ vi ⊗ · · · ⊗ vn
of I, we have that uv and vu both lie in I. But that’s obvious!
The lemma shows that S(V ) really is an F -algebra, as a quotient of T (V ) by an ideal. Multiplication of two monomials v1 · · · · · vm and w1 · · · · · wn in S(V ) is just by concatenation, giving
v1 · · · · · vm · w1 · · · · · wn . Moreover, since · is symmetric, we can reorder this as we like in S(V ) to
see that
v1 · · · · · vm · w1 · · · · · wn = w1 · · · · · wn · v1 · · · · · vn .
Hence, since these “pure” elements (the images of the pure tensors which generate T (V )) span
S(V ) as an F -vector space, we see that S(V ) is a commutative F -algebra. Note not all elements
of S(V ) can be written as v1 · · · · · vm , just as not all tensors are pure tensors.
The symmetric algebra S(V ), together with the inclusion map i : V → S(V ), is characterized
by the following universal property (compare with the universal property of the tensor algebra):
Universal property of the symmetric algebra. Given an F -linear map f : V → A, where A
is a commutative F -algebra, there exists a unique F -linear homomorphism f¯ : S(V ) → A such
that f = f¯ ◦ i.
Proof. Since A is commutative, the map f¯1 : T (V ) → A given by the universal property of the
tensor algebra annihilates all elements v ⊗ w − w ⊗ v ∈ T 2 (V ). These generate the ideal I by the
preceeding lemma. Hence, f¯1 factors through the quotient S(V ) of T (V ).
Now suppose that V is finite dimensional on basis e1 , . . . , em . Then, we’ve seen S(V ) before!
Indeed, let F [x1 , . . . , xm ] be the polynomial ring over F in indeterminates x1 , . . . , xm . The map
ei $→ xi extends to a unique F -linear map V → F [x1 , . . . , xm ], hence since the polynomial ring is
commutative, the universal property of symmetric algebras gives us a unique F -linear homomorphism S(V ) → F [x1 , . . . , xm ].
Now, S(V ) is spanned by the images of pure tensors of the form ei1 ⊗ · · · ⊗ ein . Moreover, any
such can be reordered in S(V ) using the symmetric property to assume that i1 ≤ · · · ≤ in . Hence,
S(V ) is spanned by the ordered monomials of the form ei1 . . . ein for all i1 ≤ · · · ≤ in and all n ≥ 0.
Clearly, such a monomial maps to xi1 . . . xin in the polynomial ring F [x1 , . . . , xm ]. But we know (by
definition) that the ordered monomials give a basis for the F -vector space F [x1 , . . . , xm ]. Hence,
they must in fact be linearly independent in S(V ) too, and we’ve constructed an isomorphism:
Basis theorem for symmetric powers. Let V be an F -vector space of dimension m. Then,
S(V ) is isomorphic to F [x1 , . . . , xm ], the isomorphism mapping a basis element ei of V to the
indeterminate xi in the polynomial ring. In particular, S n (V ) has basis given by all ordered monomials in the basis of the form
ei1 · · · · · ein
with 1 ≤ i1 ≤ · · · ≤ in ≤ m.
In this language, the universal property of symmetric algebras gives that the polynomial algebra
F [x1 , . . . , xm ] is the free commutative F -algebra on the generators x1 , . . . , xm . Note in particular
that if V is finite dimensional, say of dimenion m, the basis theorem implies
*
+
m+n−1
n
dim S (V ) =
n
(exercise!).
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CHAPTER 4. ALGEBRAS
The last important topic in multilinear algebra I want to cover is the exterior algebra. The
construction goes in much the same way as the symmetric algebra, however unlike there we do not
have a known object like the polynomial algebra to compare it with – so we have to work harder
to get the analogous basis theorem to the basis theorem for symmetric powers.
Start with V being any vector space. Define K to be the two-sided ideal of the tensor algebra
T (V ) generated by the elements
{x ⊗ x | x ∈ V }.
Note that
(v + w) ⊗ (v + w) = v ⊗ v + w ⊗ v + v ⊗ w + w ⊗ w.
So K also contains all the
, elements {v ⊗ w + w ⊗ v | v, w ∈ V } automatically. Define the exterior
(or Grassmann) algebra (V ) to be the quotient algebra
,
(V ) = T (V )/K.
,
We’ll write v1 ∧ · · · ∧ vn for the image in (V ) of the pure tensor v1 ⊗ · · · ⊗ vn ∈ T (V ). So, now
we have “anti-symmetric” properties like:
v1 ∧ · · · ∧ vi ∧ vi ∧ . . . vn = 0
and
v1 ∧ · · · ∧ vi ∧ vi+1 ∧ . . . vn = −v1 ∧ · · · ∧ vi+1 ∧ vi ∧ . . . vn .
)
Since the ideal K is generated by homogeneous elements, K = n≥0 Kn where Kn = K ∩ T n (V ).
It follows that
( ,n
,
(V ) =
(V )
n≥0
,n
∼
where
(V
,n) = T (V )/Kn is the subspace spanned by the images of all pure tensors of degree n.
We’ll call
(V ) the nth
,nexterior power.
The exterior power
(V ), together with the distinguished map
,n
i : V × ··· × V →
(V ),
(v1 , . . . , vn ) $→ v1 ∧ · · · ∧ vn ,
n
is characterized by the following universal property. First, call a multlinear map f : V ×· · ·×V → W
to a vector space W alternating if we have that
f (v1 , . . . , vj , vj+1 , . . . , vn ) = 0
,n
whenever vj = vj+1 for some j. The map i : V × · · · × V →
V is multilinear and alternating.
Moreover, given
any
other
multilinear,
alternating
map
f
:
V
×
·
·
· × V → W , there exists a unique
,n
linear map f¯ :
V → W such that f = f¯ ◦ i. The proof is the same as for the symmetric powers,
you should compare the two constructions! Using this all important universal property, we can
now prove:
Basis ,
theorem for exterior powers. Suppose that V is finite dimensional, with basis e1 , . . . , em .
n
Then,
(V ) has basis given by all monomials
ei1 ∧ · · · ∧ ein
for all sequences 1 ≤ i1 < · · · < in ≤ m. In particular,
* +
,n
m
dim
(V ) =
,
n
and is zero for n > m.
103
4.2. SOME MULTILINEAR ALGEBRA
,n
,n
Proof. Since
(V ) is a quotient of T n (V ), we certainly have that
(V ) is spanned by all
monomials ei1 ∧ · · · ∧ ein . Now using the antisymmetric properties of ∧, we get that it is spanned
by the given strictly ordered monomials. The problem is to prove that the given elements are
linearly
,1independent. For this, we proceed by induction on n, the case n = 1 being immediate
since
(V ) = V . Now consider n > 1.
∗
Let f1 , . . . , fn be the basis for
the given basis e1 , . . . , en for V . For each i = 1, . . . , n,
,nV dual to
,n−1
we wish to define a map Fi :
(V ) →
(V ). Start with the multilinear, alternating map
,n−1
f¯i : V × · · · × V →
(V ) defined by
f¯i (v1 , . . . , vn ) =
n
!
j=1
(−1)j fi (vj )v1 ∧ · · · ∧ vj−1 ∧ vj+1 ∧ · · · ∧ vn .
Its clearly multilinear, and to check its alternating, we just need to see that if vj = vj+1 then:
f¯i (v1 , . . . , vj , vj+1 , . . . , vn ) = (−1)j fi (vj )v1 ∧ · · · ∧ vj+1 ∧ · · · ∧ vn +
(−1)j+1 fi (vj+1 )v1 ∧ · · · ∧ vj ∧ · · · ∧ vn
,
gives that f¯i induces a unique linear map Fi :
which is zero. Now the univeral property of
,n
,n−1
V →
V.
Now we show that the given elements are linearly independent. Take a linear relation
!
ai1 ,...,in ei1 ∧ · · · ∧ ein = 0.
i1 <···<in
Apply the linear map F1 , which annihilates all ej except for e1 by definition. We get:
!
a1,i2 ,...,in ei2 ∧ · · · ∧ ein = 0.
1<i2 <···<in
By the induction hypothesis, all such monomials are linearly independent, hence all a1,i2 ,...,in are
zero. Now apply F2 in the same way to get that all a2,i2 ,...,in are zero, etc...
,n
Now let’s focus -on.a special case. Suppose that dim V = n and consider
V . Its dimension,
n
by the theorem, is n = 1, and it has basis just the element e1 ∧ · · · ∧ en . Let f : V → V be a
linear map. Define
,n
f¯ : V × · · · × V →
(V )
by f¯(v1 , . . . , vn ) = f (v1 ) ∧ · · · ∧ f (vn ). One easily checks that this is multilinear and alternating.
Hence, it induces a unique linear map we’ll denote
,n
,n
∧n f :
(V ) →
(V ).
,n
But
(V ) is one dimensional, so such a linear map is just multiplication by a scalar. In other
words, there is a unique scalar which we’ll call det f determined by the equation
f (e1 ) ∧ · · · ∧ f (en ) = (det f )e1 ∧ · · · ∧ en .
Of course, det f is exactly the determinant of the linear transformation f . Note our definition of
determinant is basis free, always a good thing.
You can see what we’re calling det f is the same as usual, as follows. Suppose the matrix of f
in our fixed basis is A, defined from
!
f (ej ) =
Ai,j ei .
i
104
CHAPTER 4. ALGEBRAS
Then,
f (e1 ) ∧ · · · ∧ f (en ) =
!
i1 ,...,in
Ai1 ,1 Ai2 ,2 . . . Ain ,n ei1 ∧ · · · ∧ ein .
Now terms on the right hand side are zero unless all i1 , . . . , in are distinct, i.e. (i1 , . . . , in ) =
(g1, g2, . . . , gn) for some permutation g ∈ Sn . So:
f (e1 ) ∧ · · · ∧ f (en ) =
!
g∈Sn
Ag1,1 Ag2,2 . . . Agn,n eg1 ∧ · · · ∧ egn .
Finally, eg1 ∧ · · · ∧ egn = sgn(g)e1 ∧ · · · ∧ en where sgn denotes the sign of a permutation. So:
f (e1 ) ∧ · · · ∧ f (en ) =
This shows that
det f =
!
sgn(g)Ag1,1 Ag2,2 . . . Agn,n e1 ∧ · · · ∧ en .
!
sgn(g)Ag1,1 Ag2,2 . . . Agn,n
g∈Sn
g∈Sn
which is exactly the usual “Laplace expansion” definition of determinant.
Here’s a final result to illustrate how nicely the universal property definition of determinant
gives its properties:
Multiplicativity of determinant. For linear transformations f, g : V → V , we have that
det(f ◦ g) = det f det g.
Proof.
By definition, ∧n [f ◦ g] is the map uniquely determined by
(∧n [f ◦ g])(v1 ∧ · · · ∧ vn ) = f (g(v1 )) ∧ · · · ∧ f (g(vn )).
But ∧n f ◦ ∧n g also satisfies this equation. So, ∧n (f ◦ g) = ∧n f ◦ ∧n g. The left hand side is scalar
multiplication by det(f ◦ g), the right hand side is scalar multiplication by det f det g.
4.3
Chain conditions
Let M be a (left or right) R-module. Then, M is called Noetherian if it satisfies the ascending
chain condition (ACC) on submodules. This means that every ascending chain
M1 ⊆ M2 ⊆ M3 ⊆ . . .
of submodules of M eventually stabilizes, i.e. Mn = Mn+1 = Mn+2 = . . . for sufficiently large n.
Similarly, M is called Artinian if it satisfies the descending chain condition (DCC) on submodules. So every descending chain
M1 ⊇ M2 ⊇ M3 ⊇ . . .
of submodules of M eventually stabilizes, i.e. Mn = Mn+1 = Mn+2 = . . . for sufficiently large n.
A ring R is called left (resp. right) Noetherian if it is Noetherian viewed as a left (resp. right)
R-module. Similarly, R is called left (resp. right) Artinian if it is Artinian viewed as a left (resp.
right) R-module. In the case of commutative rings, we can omit the left or right here, but we
cannot in general as the pathological examples on the homework show.
In this chapter, we’ll mainly be concerned with the Artinian property, but it doesn’t make sense
to introduce one chain condition without the other. We will discuss Noetherian rings in detail in
the chapter on commutative algebra. By the way, you should think of the Noetherian property
105
4.3. CHAIN CONDITIONS
as a quite weak finiteness property on a ring, whereas the Artinian property is rather strong (see
Hopkin’s theorem below for the justification of this).
We already know plenty of examples of Noetherian and Artinian rings. For instance, any PID is
Noetherian (Lemma 2.4.1), but it need not be Artinian (e.g. Z is not Artinian: (p) ⊃ (p2 ) ⊃ . . . is
an infinitely descending chain). The ring Zpn for p prime is both Noetherian and Artinian: indeed,
here there are only finitely many ideals in total, the (pi ) for 0 ≤ i ≤ n.
The main source of Artinian rings is as follows. Let F be a field and suppose that R is an
F -algebra. Then I claim that every R-module M which is finite dimensional as an F -vector space
is both Artinian and Noetherian. Well, R-submodules of M are in particular F -vector subspaces.
And clearly in a finite dimensional vector space, you cannot have infinite chains of proper subspaces.
So finite dimensionality of M does the job immediately!
In particular, if the F -algebra R is finite dimensional as a vector space over F , then R is both
left and right Artinian and Noetherian. Now you see why finite dimensional algebras over a field
(e.g. Mn (F ), the group algebra F G for G a finite group, etc...) are particularly nice things!
i
π
4.3.1. Lemma. Let 0 −→ K −→ M −→ Q −→ 0 be a short exact sequence of R-modules. Then
M is Noetherian (resp. Artinian) if and only if both K and Q are Noetherian (resp. Artinian).
Proof. I just prove the result for the Noetherian property, the Artinian case being analogous.
First, suppose M satisfies ACC. Then obviously K does as it is isomorphic to a submodule of M .
Similarly Q does by the lattice isomorphism theorem for modules.
Conversely, suppose K and Q both satisfy ACC. Let M1 ⊆ M2 ⊆ . . . be an ascending chain of
R-submodules of M . Set
Ki = i−1 (i(K) ∩ Mi ),
Qi = π(Mi ).
Then, K1 ⊆ K2 ⊆ . . . is an ascending chain of submodules of K, and Q1 ⊆ Q2 ⊆ . . . is an
ascending chain of submodules of Q. So by assumption, there exists n ≥ 1 such that KN = Kn
and QN = Qn for all N ≥ n. Now, we have evident short exact sequences
0 −→ Kn −→ Mn −→ Qn −→ 0
and
0 −→ KN −→ MN −→ QN −→ 0
for each N ≥ n. Letting α : Kn → KN , β : Mn → MN and γ : Qn → QN be the inclusions, one
obtains a commutative diagram with α and γ being isomorphisms. Then the five lemma implies
that β is surjective, hence MN = Mn for all N ≥ n as required.
4.3.2. Theorem. If R is left (resp. right) Noetherian, then every finitely generated left (resp.
right) R-module is Noetherian. Similarly, if R is left (resp. right) Artinian, then every finitely
generated left (resp. right) R-module is Artinian.
Proof. Let’s do this for the Artinian case, the Noetherian case being similar. So assume that
R is left Artinian and M is a finitely generated left R-module. Then, M is a quotient of a free
R-module with a finite basis. So it suffices to show that R⊕n is an Artinian left R-module. We
proceed by induction on n, the case n = 1 being given. For n > 1, the submodule of R⊕n spanned
by the first (n − 1) basis elements is isomorphic to R⊕(n−1) , and the quotient by this submodule
is isomorphic to R. By induction both R⊕(n−1) and R are Artinian. Hence, R⊕n is Artinian by
Lemma 4.3.1.
The next results are special ones about Noetherian rings (but see Hopkin’s theorem below!)
4.3.3. Lemma. Let R be a left Noetherian ring and M be a finitely generated left R-module. Then
every R-submodule of M is also finitely generated.
106
CHAPTER 4. ALGEBRAS
Proof. By Theorem 4.3.2, M is Noetherian. Let N be an arbitrary R-submodule of M and let
A be the set of finitely generated R-submodules of M contained in N . Note A is non-empty as
the zero module is certainly there!
Now I claim that A contains a maximal element. Well, pick M1 ∈ A . If M1 is maximal in A ,
we are done. Else, we can find M2 ∈ A with M1 ⊃ M2 . Repeat. The process must terminate, else
we construct an infinite ascending chain of submodules of N ⊆ M , contradicting the fact that M
is Noetherian. (Warning: as with all such arguments we have secretly appealed to the axiom of
choice here!
So now let N # be a maximal element of A . Say N # is generated by m1 , . . . , mn . Now take
any m ∈ N . Then, (m1 , . . . , mn , m) is a finitely generated submodule of M contained in N and
containing N # . By maximality of N # , we therefore have that
N # = (m1 , . . . , mn , m),
i.e. m ∈ N # . This shows that N = N # , hence N is finitely generated.
4.3.4. Corollary. Let R be a ring. Then, The following are equivalent:
(i) R is left Noetherian;
(ii) every left ideal of R is finitely generated;
(Similar statements hold on the right, of course).
Proof. (i)⇒(ii). This is a special case of Lemma 4.3.3, taking M = R.
/
(ii)⇒(i). Let I1 ⊆ I2 ⊆ . . . be an ascending chain of left ideals of R. Then, I = n≥1 In is also
a left ideal of R, hence finitely generated, by a1 , . . . , am say. Then for some sufficiently large n, all
of a1 , . . . , am lie in In . Hence In = I and R is Noetherian.
Now you see that – for commutative rings – Noetherian is an obvious generalization of a PID.
Instead of insisting all ideals are generated by a single element, one has that every ideal is generated
by finitely many elements.
4.4
Wedderburn structure theorems
Now we have the basic language of algebras and of Artinian rings and modules, we can begin to
discuss the structure of rings. The first step is to understand the structure of semisimple rings.
Recall that an R-module M is simple or irreducible if it is non-zero and has no submodules other
than M and (0).
Schur’s lemma. Let M be a simple R-module. Then, EndR (M ) is a division ring.
Proof. Let f : M → M be a non-zero R-endomorphism of M . Then, ker f and im f are both
R-submodules of M , with ker f 4= M and im f 4= (0) since f is non-zero. Hence, since M is simple,
ker f = (0) and im f = M . This shows that f is a bijection, hence invertible.
Throughout the section, we will also develop parallel but stronger versions of the results dealing
with finite dimensional algebras over algebraically closed fields (recall a field F is algebraically
closed if every monic f (X) ∈ F [X] has a root in F ).
Relative Schur’s lemma. Let F be an algebraically closed field and A be an F -algebra. Let M
be a simple A-module which is finite dimensional as an F -vector space. Then, EndR (M ) = F .
Proof. Let f : M → M be an A-endomorphism of M . Then in particular, f : M → M is an
F -linear endomorphism of a finite dimensional vector space. Since F is algebraically closed, f has
an eigenvector vλ ∈ M of eigenvalue λ ∈ F (because the characteristic polynomial of f has a root
in F ).
107
4.4. WEDDERBURN STRUCTURE THEOREMS
Now consider f −λ idM . It is also an A-endomorphism of M , and moreover its kernel is non-zero
as vλ is annihilated by f − λ idM . Hence since M is irreducible, the kernel of f − λ idM is all of M ,
i.e. f = λ idM . This shows that the only R-endomorphisms of M are the scalars, as required.
Now, let R be a non-zero ring. Call R left semisimple if R R is semisimple as a left R-module
(recall section 3.4). Similarly, R is right semisimple if RR is semisimple as a right R-module.
Finally, R is simple if it has no two-sided ideals other than R and (0). Here’s a lemma from the
previous chapter that should be quite familiar by now.
4.4.1. Lemma. Let A be an R-algebra, M a left A-module and D = EndA (M ). Then,
EndA (M ⊕n ) ∼
= Mn (D)
as R-algebras.
Proof.
Let us write elements of M ⊕n as column vectors


m1
 .. 
 . 
mn
with mi ∈ M . Suppose we have f ∈ EndA (M ⊕n ). Let fi,j : M → M be the map sending m ∈ M
to the ith coordinate of the vector


0
 .. 
 . 


 0 




f
 m 
 0 


 . 
 .. 
0
where the m is in the jth row. Then, fi,j is an
Moreover,
 

f1,1 . . . f1,n
m1
 ..   ..
..
..
f  .  =  .
.
.
fn,1 . . . fn,n
mn
A-module homomorphism, so is an element of D.

m1
  .. 
 . 
mn

(matrix multiplication!),
so that f is uniquely determined by the fi,j . Thus, we obtain an R-algebra isomorphism f $→ (fi,j )
between EndA (M ⊕n ) and Mn (D).
Wedderburn’s first structure theorem. The following conditions on a non-zero ring R are
equivalent:
(i) R is simple and left Artinian;
(i# ) R is simple and right Artinian;
(ii) R is left semisimple and all simple left R-modules are isomorphic;
(ii# ) R is right semisimple and all simple right R-modules are isomorphic;
(iii) R is isomorphic to the matrix ring Mn (D) for some n ≥ 1 and D a division ring.
Moreover, the integer n and the division ring D in (iii) are determined uniquely by R (up to
isomorphism).
Proof. First, note the condition (iii) is left-right symmetric. So let us just prove the equivalence
of (i),(ii) and (iii).
108
CHAPTER 4. ALGEBRAS
(i)⇒(ii). Let U be a minimal left ideal of R. This exists because R is left Artinian so we have
DCC on left ideals! Then, U = Rx for any non-zero x ∈ U , and Rx is a simple left R-module.
Since R is a simple ring, the non-zero two-sided ideal RxR must be all of R. Hence,
!
Rxa.
R = RxR =
a∈R
Note each Rxa is a homomorphic image of the simple left R-module Rx, so is either isomorphic
to Rx or zero. Thus we have written R R as a sum of simple R-modules, so by Lemma 3.4.1, there
exists a subset S ⊆ R such that
(
R=
Rxa
a∈S
with each Rxa for a ∈ S being isomorphic to U . Hence, R is left semisimple.
Moreover, if M is any simple left R-module, then M ∼
= R/J for a maximal left ideal J of R. We
can pick a ∈ S such that xa ∈
/ J. Then, the quotient map R → R/J maps the simple submodule
Rxa of R to a non-zero submodule of M . Hence using simplicity, M ∼
= Rxa. This shows all simple
R-modules are isomorphic to U .
(ii)⇒(iii). Let U be a simple left R-module. We have that
(
R=
Ra
a∈S
for some subset S )
of R, where each left R-module Ra is isomorphic to U . I claim that S is finite.
Indeed, 1R lies in a∈S Ra hence in Ra1 ⊕ · · · ⊕ Ran for some finite subset {a1 , . . . , an } of S. But
1R generates R as a left R-module, so in fact {a1 , . . . , an } = S.
Thus,
R∼
= U ⊕n
for some n, uniquely determined as the composition length of R R. Now, by Schur’s lemma,
EndR (U ) ∼
= D, where D a Division ring uniquely determined up to isomorphism as the endomorphism ring of a simple left R-module. Hence applying Lemma 4.4.1,
EndR (R R) ∼
= EndR (U ⊕n ) ∼
= Mn (D).
Now finally, we observe that
EndR (R R) ∼
= Rop ,
the isomorphism in the forward direction being determined by evaluation at 1R . To see that this is
an isomorphism, one constructs the inverse map, namely, the map sending r ∈ R to fr :R R →R R
with fr (s) = sr for each s ∈ R (“fr is right multiplication by r”). Now we have shown that
Rop ∼
= Mn (D).
Hence, noting that matrix transposition is an isomorphism between Mn (D)op and Mn (Dop ), we
get that
R∼
= Mn (Dop )
and Dop is also a division ring.
(iii)⇒(i). We know that Mn (D) is simple as it is Morita equivalent to D, which is a division
ring (or you can prove this directly!). It is left Artinian for instance because Mn (D) is finite
dimensional over the division ring D. This gives (i).
Corollary. Every simple left (or right) Artinian ring R is an algebra over a field.
Proof.
Observe Mn (D) is an algebra over Z(D), which is a field.
The relative version for finite dimensional algebras over algebraically closed fields is as follows:
4.4. WEDDERBURN STRUCTURE THEOREMS
109
Relative first structure theorem. Let F be an algebraically closed field and A be a finite dimensional F -algebra (hence automatically left and right Artinian).
(i) A is simple;
(ii) A is left semisimple and all simple left R-modules are isomorphic;
(ii# ) A is right semisimple and all simple right R-modules are isomorphic;
(iii) A is isomorphic to the matrix ring Mn (F ) for some n (indeed, n2 = dim A).
Proof. The proof is the same, except that one gets that the division ring D equals F since one
can use the stronger relative Schur’s lemma.
Wedderburn’s second structure theorem. Every left (or right) semisimple ring R is isomorphic to a finite product of matrix rings over division rings:
R∼
= Mn1 (D1 ) × · · · × Mnr (Dr )
for uniquely determined ni ≥ 1 and division rings Di . Conversely, any such ring is both left and
right semisimple.
)
Proof. Since R is left semisimple, we can decompose R R as
i∈I Ui where each Ui is simple.
Note 1R already lies in a sum of finitely many of the Ui , hence the index set I is actually finite.
Now gather together isomorphic Ui ’s to write
RR
where
= H1 ⊕ · · · ⊕ Hm
Hi ∼
= Ui⊕ni
for irreducible R-modules Ui and integers ni ≥ 1, with Ui ∼
4 Uj for i 4= j. By Schur’s lemma,
=
Di = EndR (Ui ) is a division ring. Moreover, there are no R-module homomorphisms between Hi
and Hj for i 4= j, because none of their composition factors are isomorphic. Hence:
∼ EndR (H1 ⊕ · · · ⊕ Hm )op ∼
R∼
= EndR (R R)op =
= EndR (H1 )op × · · · × EndR (Hm )op
op
op
∼
).
= Mn1 (D1 )op × · · · × Mnm (Dm )op ∼
= Mn1 (D1 ) × · · · × Mnm (Dm
The integers ni are uniquely determined as the multiplicities of the irreducible left R-modules in a
composition series of R R, while the division rings Di are uniquely determined up to isomorphism
as the endomorphism rings of the simple left R-modules.
For the converse, we need to show that a product of matrix algebras over division rings is left
and right semisimple. This follows because a single matrix algebra Mn (D) over a division ring is
both left and right semisimple according to the first Wedderburn structure theorem.
The theorem shows:
Corollary. A ring R is left semisimple if and only if it is right semisimple.
So we can now just call R semisimple if it is either left or right semisimple in the old sense.
Also:
Corollary. Any semisimple ring is both left and right Artinian.
Proof. A product of finitely many matrix rings over division algebras is left and right Artinian.
Finally, we need to give the relative version:
110
CHAPTER 4. ALGEBRAS
Relative second structure theorem. Let F be an algebraically closed field and A be a finite
dimensional F -algebra that is semisimple. Then,
A∼
= Mn1 (F ) × · · · × Mnr (F )
for uniquely determined integers ni ≥ 1.
Proof.
Just use the relative Schur’s lemma in the same proof.
This theorem lays bare the structure of finite dimensional semisimple algebras over algebraically
closed fields: A is uniquely determined up to isomorphism by the integers ni , which are precisely
the dimensions of the simple A-modules.
4.5
The Jacobson radical
I want to explain in this section how understanding of the case of semisimple rings gives information
about more general rings. The key notion is that of the Jacobson radical, which will be defined
using the following equivalent properties:
Jacobson radical theorem. Let R be a ring, a ∈ R. The following are equivalent:
(i) a annihilates every simple left R-module;
(i)# a annihilates every simple right R-module;
(ii) a lies in every maximal left ideal of R;
(ii)# a lies in every maximal right ideal of R;
(iii) 1 − xa has a left inverse for every x ∈ R;
(iii)# 1 − ay has a right inverse for every y ∈ R;
(iv) 1 − xay is a unit for every x, y ∈ R.
Proof. Since (iv) is left-right symmetric, I will only prove equivalence of (i)–(iv).
(i)⇒(ii). If I is a maximal left ideal of R, then R/I is a simple left R-module. So, a(R/I) = 0,
i.e. a ∈ I.
(ii)⇒(iii). Assume (ii) holds but 1 − xa does not have a left inverse for some x ∈ R. Then,
R(1 − xa) 4= R. So, there exists a maximal left ideal I with
R(1 − xa) ≤ I < R.
But then, 1 − xa ∈ I, and a ∈ I by assumption. Hence, 1 ∈ I so I = R, a contradiction.
(iii)⇒(i). Let M be a simple left R-module. Take u ∈ M . If au 4= 0, then Rau = M as M is
simple, so u = rau for some r ∈ R. But this implies that (1 − ra)u = 0, hence since (1 − ra) has a
left inverse by assumption, we get that u = 0. This contradiction shows that in fact au = 0 for all
u ∈ M , i.e. aM = 0.
(iv)⇒(iii). This is trivial (take y = 1).
(i),(iii)⇒(iv). For any y ∈ R and any simple left R-module M , ayM ⊆ aM = 0. So ay also
satisfies the condition in (i), hence (iii). So for every x ∈ R, 1 − xay has a left inverse, 1 − b
say. The equation (1 − b)(1 − xay) = 1 implies b = (b − 1)xay. Hence bM = 0 for all simple left
R-modules M , so b satisfies (i) hence (iii). So, 1 − b has a left inverse, 1 − c, say. Then,
1 − c = (1 − c)(1 − b)(1 − xay) = 1 − xay,
hence
c = xay.
Now we have that 1 − b is a left inverse to 1 − xay, and 1 − xay is a left inverse to 1 − b. Hence,
1 − xay is a unit with inverse 1 − b.
111
4.5. THE JACOBSON RADICAL
Now define the Jacobson radical J(R) to be
J(R) = {a ∈ R | a satisfies the equivalent conditions in the theorem}.
Thus for instance using (i), J(R) is the intersection of the annihilators in R of all the simple left
R-modules, or using (ii)
8
J(R) =
I
I
where I runs over all maximal left ideals of R. This implies that J(R) is a left ideal of R, but
equally well using (ii)# ,
8
I
J(R) =
I
where I runs over all maximal right ideals of R so that J is a right ideal of R. Hence, J(R) is a
two-sided ideal of R.
The following result is a basic trick in ring theory and maybe gives a first clue as to why the
Jacobson radical is so important.
Nakayama’s lemma. Let R be a ring and M be a finitely generated left R-module. If J(R)M =
M then M = 0.
Proof. Suppose that M is non-zero and set J = J(R) for short. Let X = {m1 , . . . , mn } be a
minimal set of generators of M , so m1 4= 0. Since JM = M , we can write
m1 = j1 m1 + · · · + jn mn
for some ji ∈ J. So,
(1R − j1 )m1 = j2 m2 + · · · + jn mn .
But 1R − j1 is a unit by the definition of J(R). So we get that
m1 = (1R − j1 )−1 j2 m2 + · · · + (1R − j1 )−1 jn mn .
This contradicts the minimality of the initial set of generators chosen.
We will apply Nakayama’s lemma on the next chapter, but actually never in this chapter. The
remainder of the section is concerned with the Jacobson radical in an Artinian ring!!! All these
results are false if the ring is not Artinian...
First characterization of the Jacobson radical for Artinian rings. Suppose that R is left
(or right) Artinian. Then, J(R) is the unique smallest two-sided ideal of R such that R/J(R) is a
semisimple algebra.
Proof. Pick a maximal left ideal I1 of R. Then (if possible) pick a maximal left ideal I2 such that
I2 4⊇ I1 (hence I1 ∩ I2 is strictly smaller than I1 ). Then (if possible) pick a maximal left ideal I3
such that I3 4⊇ I1 ∩ I2 (hence I1 ∩ I2 ∩ I3 is strictly smaller than I1 ∩ I2 ). Keep going! The process
must terminate after finitely many steps, else you construct an infinite descending chain
R ⊃ I1 ⊃ I1 ∩ I2 ⊃ I1 ∩ I2 ∩ I3 ⊃ . . .
of left ideals of R, contradicting the fact that R is left Artinian. We thus obtain finitely many
maximal left ideals I1 , I2 , . . . , Ir of R such that I1 ∩ · · · ∩ Ir is contained in every maximal left ideal
of R. In other words,
J(R) = I1 ∩ · · · ∩ Ir ,
so the Jacobson radical of an Artinian ring is the intersection of finitely many maximal left ideals.
112
CHAPTER 4. ALGEBRAS
Consider the map
a $→ (a + I1 , . . . , a + Ir ).
R → R/I1 ⊕ · · · ⊕ R/Ir ,
It is an R-module map with kernel I1 ∩ · · · ∩ Ir = J(R). So it induces an embedding
R/J '→ R/I1 ⊕ · · · ⊕ R/Ir .
The right hand side is a semisimple R-module, as it is a direct sum of simples, hence R/J is also
a semisimple R-module. This shows that the quotient R/J is a semisimple ring.
Now let K be another two-sided ideal of R such that R/K is a semisimple ring. By the lattice
isomorphism theorem, we can write
(
Bi /K
R/K =
where9
Bi is a left ideal of R containing K,
Ci = j'=i Bj for short. Then,
9
i∈I
i∈I
9
Bi = R and Bi ∩ ( j'=i Bj ) = K for each i. Set
R/Ci = (Ci + Bi )/Ci ∼
= Bi /(Bi ∩ Ci ) = Bi /K.
This is simple, so Ci is a maximal left ideal of R. Hence, each Ci ⊇ J(R). So K =
contains J(R). Thus J(R) is the unique smallest such ideal.
:
Ci also
Thus you see the first step to understanding the structure of an Arinian ring: understand the
semisimple ring R/J(R) in the sense of Wedderburn’s structure theorem.
4.5.1. Corollary. Let R be a left Artinian ring and M be a left R-module. Then, M is semisimple
if and only if J(R)M = 0.
Proof. If M is semisimple, it is a direct sum of simples. Now, J(R) annihilates all simple left
R-modules by definition, hence J(R) annihilates M . Conversely, if J(R) annihilates M , then we
can view M as an R/J(R)-module by defining (a + J(R))m = am for all a ∈ R, m ∈ M (one needs
J(R) to annihilate M for this to be well-defined!). Since R/J(R) is semisimple by the previous
theorem, M is a semisimple R/J(R)-module. But the R-module structure on M is just obtained
by lifting the R/J(R)-module structure, so this means that M is semisimple as an R-module too.
Second characterization of the Jacobson radical for Artinian rings. Let R be a left (or
right) Artinian ring. Then, J(R) is a nilpotent two-sided ideal of R (i.e. J(R)n = 0 for some
n > 0) and is equal to the sum of all nilpotent left ideals of R.
Proof.
Suppose x ∈ R is nilpotent, say xn = 0. Then, (1 − x) is a unit, indeed,
(1 − x)(1 + x + x2 + · · · + xn−1 ) = 1.
So if I is a nilpotent left ideal of R, then every x ∈ I satisfies condition (iii) of the Jacobson radical
theorem. This shows every nilpotent ideal of R is contained in J(R). It therefore just remains to
prove that J(R) is itself a nilpotent ideal.
Set J = J(R). Consider the chain
J ⊇ J2 ⊇ J3 ⊇ . . .
of two-sided ideals of R. Since R is left Artinian, the chain stabilizes, so J k = J k+1 = . . . for some
k. Set I = J k , so I 2 = I. We need to prove that I = 0.
Well, suppose for a contradiction that I 4= 0. Choose a left ideal K of R minimal such that
IK 4= 0 (use the fact that R is left Artinian). Take any a ∈ K with Ia 4= 0. Then, I 2 a = Ia 4= 0,
so the left ideal Ia of R coincides with K by the minimality of K. Hence, a ∈ K lies in Ia, so we
can write a = xa for some x ∈ I. So, (1 − x)a = 0. But x ∈ J, so 1 − x is a unit, hence a = 0,
which is a contradiction.
4.6. CHARACTER THEORY OF FINITE GROUPS
113
4.5.2. Corollary. In particular, a left (or right) Artinian ring R is semisimple if and only if it
has no non-zero nilpotent ideals.
We end with an important application:
Hopkin’s theorem. Let R be a left Artinian ring and M be a left R-module. The following are
equivalent:
(i) M is Artinian;
(ii) M is Noetherian;
(iii) M has a composition series;
(iv) M is finitely generated.
Proof. (i)⇒(iii) and (ii)⇒(iii). Let J = J(R). Then, J is nilpotent by the second characterization
of the Jacobson radical. So, J n = 0 for some n. Consider
M ⊇ JM ⊇ J 2 M ⊇ · · · ⊇ J n M = 0.
It is a descending chain of R-submodules of M . Set Fi = J i M/J i+1 M . Then, Fi is annihilated by
J, hence is a semisimple left R-module. Now, if M is Artinian (resp. Noetherian), so is each Fi ,
so each Fi is in fact a direct sum of finitely many simple R-modules. Thus, each Fi obviously has
a composition series, so M does too by the lattice isomorphism theorem.
(iii)⇒(iv). Let M = M1 ⊃ · · · ⊃ Mn = 0 be a composition series of M . Pick mi ∈ Mi − Mi+1
for each i = 1, . . . , n − 1. Then, the image of mi in Mi /Mi+1 generates Mi /Mi+1 as it is a simple
R-module. It follows that the mi generate M . Hence, M is finitely generated.
(iv)⇒(i) is Theorem 4.3.2.
(iii)⇒(ii) and (iii)⇒(i). These both follow immediately from the Jordan-Hölder theorem (actually, the Schreier refinement lemma using that any refinement of a composition series is trivial).
Now we can show that “Artinian implies Noetherian”:
4.5.3. Corollary. If R is left (resp. right) Artinian, then R is left (resp. right) Noetherian.
Proof.
Apply Hopkin’s theorem to the Artinian R-module R R.
4.6
Character theory of finite groups
Wedderburn’s theorem has a very important application to the study of finite groups, indeed, it
is the starting point for the study of character theory of finite groups. In this section, I’ll follow
Rotman section 8.5 fairly closely.
Maschke’s theorem. Let G be a finite group and F be a field either of characteristic 0 or of
characteristic 0 < p ! |G|. Then, the group algebra F G is a semisimple algebra.
Proof. It is certainly enough to show that every left F G-module M is semisimple. We just need
to show that every short exact sequence
π
0 −→ K −→ M −→ Q −→ 0
of F G-modules splits. Well, pick any F -linear map θ : Q → M such that π ◦ θ = idQ . The problem
is that θ need not be an F G-module map.
Well, for g ∈ G, define a new F -linear map
g
θ : Q → M,
x $→ gθ(g −1 x).
114
Then,
CHAPTER 4. ALGEBRAS
(π ◦ g θ)(x) = π(gθ(g −1 x)) = g(π ◦ θ)(g −1 x) = gg −1 x = x
so each g θ also satisfies π ◦ g θ = idQ . Now define
Avθ =
1 !g
θ
|G|
g∈G
(note |G| is invertible in F by assumption on characteristic). This is another F -linear map such
that π ◦ Avθ = idQ . Moreover, Avθ is even an F G-module map:
hAvθ(x) =
1 !
1 !
hgθ(g −1 x) =
gθ(g −1 hx) = Avθ(hx).
|G|
|G|
g∈G
g∈G
Hence, Avθ defines the required splitting.
4.6.1. Remark. Maschke’s theorem is actually if and only if... F G is a semisimple algebra if and
only if char F = 0 or char F ! |G|. Let me just give one example. Take F of characteristic p > 0
and G = Cp = *x,. The group algebra F G is commutative and artinian. Now recall a commutative
artinian ring is semisimple if and only if it has no non-zero nilpotent elements. But The corollary
can be applied in particular to commutative rings. In that case, if x ∈ R is a nilpotent element,
the ideal (x) it generates is a nilpotent ideal. So, a commutative Artinian ring is semisimple if
and only if it has no non-zero nilpotent elements. Now let G = Cp , the cyclic group of order p.
Maschke’s theorem shows that if F is any field of characteristic different from p, then F G is a
semisimple. Conversely, suppose F is a field of characterisitic p. The group algebra F G is finite
dimensional, hence is a commutative Artinian ring. Consider the element
x − 1 ∈ FG
where x is a generator of G. We have that
(x − 1)p = xp + (−1)p = 1 + (−1)p = 0.
Hence, it is a non-zero nilpotent element. Thus, F G is not semisimple in this case.
We are going to be interested from now on in the following situation:
• G is a finite group (so that F G is artinian);
• the field F is of characteristic 0 (so that F G is semisimple);
• the field F is algebraically closed (so that the souped-up version of Wedderburn’s theorem
holds).
Actually I’ll just assume F = C from now on. Then, we have shown that
(1)
CG ∼
= Mn1 (C) × · · · × Mnr (C)
where r is the number of isomorphism classes of irreducible CG-modules and n1 , . . . , nr are their
dimensions. These give some interesting invariants of the group G defined using modules... Note
first (considering dimension of each side as a C-vector space) that:
|G| = (n1 )2 + · · · + (nr )2 .
The number r has a simply group theoretic meaning too:
4.6.2. Lemma. The number r in (1) is equal to the number of conjugacy classes in the group G.
4.6. CHARACTER THEORY OF FINITE GROUPS
115
Proof. Let us compute dim Z(CG) in two different ways. First, since the center of Mn (C) is one
dimensional spanned by the identity matrix, dim Z(CG) = r (one for each matrix algebra in the
Wedderburn decomposition). On the other hand if
!
ag g ∈ CG
g∈G
is central then conjugating by h ∈ G you see that
ag = ahgh−1
for all h ∈ G. Hence the coefficients ag are constant
on conjugacy classes. Hence if C1 , . . . , Cs are
9
the conjugacy classes of G, the elements zi = g∈Ci g form a basis {z1 , . . . , zs } for Z(CG). Hence
r = dim Z(CG) = s.
4.6.3. Example. Say G = S3 . There are three conjugacy classes. Hence r = 3, i.e. there are three
isomorphism classes of irreducible CS3 -modules. Moreover n21 + n22 + n23 = 6 so the dimensions can
only be 1, 1 and 2.
4.6.4. Example. Say G is abelian. Then there are r = |G| conjugacy classes, and n21 +· · ·+n2r = r
hence each ni = 1. This proves that there are n isomorphism classes of irreducible CG-module,
and all of these modules are one dimensional.
Before going any further I want to introduce a little language. A (finite dimensional complex)
representation of G means a pair (V, ρ) where V is a finite dimensional complex vector space
and ρ : G → GL(V ) is a group homomorphism. The representation is called faithful if this map
ρ is injective. Note having a representation (V, ρ) of G is exactly the same as having a finite
dimensional CG-module V : given a representation (V, ρ) we get a CG-module structure on V by
defining gv := ρ(g)(v); conversely given a CG-module structure on V we get a representation
ρ : G → GL(V ) by defining ρ(g) to be the linear map v $→ gv. This is all one big tautology. But I
will often switch between calling things CG-modules and calling them representations. Sorry.
Say ρ : G → GL(V ) is a representation of G. Pick a basis for V , v1 , . . . , vn . Then each element
of GL(V ) becomes an invertible n × n matrix, and in this way you can regard ρ instead as a group
homomorphism ρ : G → GLn (C). This is a matrix representation: ρ maps the group G to the
group of n × n invertible matrices. You should compare the notion of matrix representation with
that of permutation representation from before: that was a map from G to the symmetric group
Sn for some n. It corresponded to having a G-set X of size n... Its all very analogous to what
we’re doing now, but sets have been replaced by vector spaces. Before we called a permutation
representation faithful if the map G → Sn was injective (then it embedded G as a subgroup of a
symmetric group); now we are calling a (linear) representation faithful if the map G → GLn (C) is
injective (then it embeds G as a subgroup of the matrix group)...
4.6.5. Example. Recall that GL(C) ∼
= GL1 (C) is just the multiplicative group C× . For any group
G, there is always the trivial representation, namely, the homomorphism mapping every g ∈ G to
1 ∈ GL1 (C). This corresponds to the trivial CG-module equal to C as a vector space with every
g ∈ G acting as 1.
4.6.6. Example. For the symmetric group Sn , there is a group homomorphism sgn : Sn → {±1}.
The latter group sits inside C× , so you can view this as another 1-dimensional representation, the
sign representation. The corresponding module is not isomorphic to the trivial module (providing
n > 1). Now we’ve constructed both of the one dimensonal CS3 -modules: one is trivial, one is
sign. What about the two dimensional CS3 -module?
There is a simple way to construct (linear) representations of G out of permutation representations. Suppose X = {x1 , . . . , xn } is a finite G-set. Let CX be the C-vector space on basis X. Each
116
CHAPTER 4. ALGEBRAS
g ∈ G defines an automorphism of the set X. Extending linearly, this means each g ∈ G can be
viewed as a vector space automorphism of CX, i.e. an element of GL(CX). Thus we have defined
a map ρ : g → GL(CX), i.e. we have turned the set X into a linear representation CX of G. In
module language, CX is a CG-module. Identify GL(CX) with GLn (C) via the choice x1 , . . . , xn
of basis. Then ρ(g) becomes an n × n matrix for each g ∈ G. Note the ij-entry of this matrix is
1 if gxj = xi and it is zero otherwise. Since X was a G-set this means that the matrix ρ(g) is a
permutation matrix: all its entries are zeros and ones, and there is just one non-zero entry in every
row and column. So amongst all matrix representations of G, the ones coming from permutation
representations are very simple...
4.6.7. Example. Let’s just go back to S3 again. It IS a permutation group, so it has a natural
permutation representation on the set X = {1, 2, 3}. The corresponding matrix representation
S3 → GL3 (C) is one you all know perfectly well. For instance, the image of the three cycle (1 2 3)
with respect to the standard basis v1 , v2 , v3 of CX labelled by the elements of the set X (I’d better
not just call them 1, 2, 3!) is the matrix


0 0 1
 1 0 0 .
0 1 0
Now clearly the vector v1 + v2 + v3 is fixed by all of G so it spans a one dimensional submodule
(isomorphic to the trivial module). By Maschke’s theorem that had better have a complement.
For instance {v1 − v2 , v2 − v3 } is a complement (all a1 v1 + a2 v2 + a3 v3 with a1 + a2 + a3 = 0).
Let’s write down matrices with respect to the new basis v1 + v2 + v3 , v1 − v2 , v2 − v3 instead: we
get the following matrices


1 0 0
ρ(1) =  0 1 0  ,
0 0 1


1 0
0
ρ((1 2)) =  0 −1 1  ,
0 0
1


1 0 0
ρ((2 3)) =  0 1 0  ,
0 1 −1


1 0
0
−1  ,
ρ((1 3)) =  0 0
0 −1 0


1 0 0
ρ((1 2 3)) =  0 0 −1  ,
0 1 −1


1 0
0
ρ((1 3 2)) =  0 −1 1  .
0 −1 0
Note all these matrices are block diagonal matrices. The top 1×1 block is the trivial representation
of G on V1 = C(v1 + v2 + v3 ), the bottom 2 × 2 block is a two dimensional representation of G
on V2 = C(v1 − v2 ) ⊕ C(v2 − v3 ). Of course V2 is irreducible. The decomposition V = V1 ⊕ V2 of
V into irreducibles corresponds in matrix language to choosing a basis so that each ρ(g) is block
diagonal... and since the blocks are irreducible representations you cannot do any better.
In other words, if V and W are two finite dimensional CG-modules, so is V ⊕ W . Pick
basis, to view V as a matrix representation ρ : G → GLn (C) and W as a matrix representation
σ : G → GLm (C). With respect to the basis for V ⊕ W obtained by concatenating the two basis,
4.6. CHARACTER THEORY OF FINITE GROUPS
117
the matrix representation ρ ⊕ σ : G → GLm+n (C) corresponding to the module V ⊕ W has all g
mapping to block diagonal matrices diag(ρ(g), σ(g))... This is how you should think of direct sums
of CG-modules in terms of matrices.
Let V be a finite dimensional CG-module. Equivalently, let (V, ρ) be a representation of G. Define its character χV : G → C to be the function with χV (g) equal to the trace of the endomorphism
ρ(g). (Recall that tr(AB) = tr(BA), hence if A is invertible tr(ABA−1 ) = tr(A−1 AB) = tr(B),
so the trace of a matrix depends only on its similarity class, so it makes sense to define trace of an
endomorphism as the trace of the matrix of the endomorphism in any basis...) Note χV extends
linearly to a map also denoted χV : CG → C, sometimes we’ll apply χV to elements of the group
algebra, but usually I won’t think of χV in this way.
4.6.8. Example. The character χ1 of the trivial CG-module is χ1 (g) = 1 for all g ∈ G. The
character ψ of the regular CG-module CG is ψ(g) = |G| if g = 1 and ψ(g) = 0 for all other
1 4= g ∈ G.
A class function is a function f from G to C that is constant on conjugacy classes, i.e.
f (hgh−1 ) = f (g). For example, the character χV of any finite dimensional CG-module is a class
function:
(2)
tr(ρ(hgh−1 )) = tr(ρ(h)ρ(g)ρ(h)−1 ) = tr(ρ(h)−1 ρ(h)ρ(g)) = tr(ρ(g))
hence χV (hgh−1 = chiV (g). Let C(G) denote the vector space of all class functions on G. There
is an obvious basis for C(G): let C1 , . . . , Cr be the conjugacy classes of G. Let δi : G → C be the
function with δi (g) = 1 if g ∈ Ci , 0 otherwise. Then C(G) has basis δ1 , . . . , δr . There is a less
obvious basis for C(G): let L1 , . . . , Lr be a complete set of pairwise non-isomorphic irreducible
CG-modules. (Recall the number of them is the number of conjugacy classes in G too...) Let
χi = χLi be the corresponding characters.
4.6.9. Theorem. χ1 , . . . , χr is a basis for C(G).
Proof.
Consider Wedderburn:
CG = Mn1 (C) × · · · × Mnr (C).
In the product on the right hand side, let ei = (0, · · · , 0, Ini , 0, · · · , 0), i.e. the identity matrix of
the ith matrix algebra. The irreducible representations on the right hand side are the modules
L1 , . . . , Lr where Li is the module of column vectors for Mni (C) on which ei acts as 1 and all other
ej act as zero. So χi (x) is just the trace of the ith matrix in the Wedderburn decomposition of an
element of x ∈ CG... But χi (ej ) = δi,j nj . This proves that χ1 , . . . , χr are linearly independent.
Hence they form a basis by dimension considerations.
4.6.10. Corollary. Two finite dimensional CG-modules V and W are isomorphic if and only if
χV = χW , i.e. they have the same characters.
)r
)r
i
i
Proof. Note V is a direct sum of irreducibles, say V ∼
. Similarly W ∼
.
= i=1 L⊕a
= 9i=1 L⊕b
i
i
r
∼
Now V = W
if
and
only
if
a
=
b
for
all
i
(say
by
Jordan-Holder
theorem).
But
χ
=
a
χ
i
i
V
i
i
i=1
9r
and χW = i=1 bi χi . So χV = χW if and only if ai = bi for all i by the theorem...
In the proof of the theorem, we exploited the idempotents ei ∈ CG coming from the identity
matrices in the Wedderburn decomposition. Let us repeat this, because these are important: there
are mutually orthogonal central idempotents e1 , . . . , er ∈ CG summing to the identity with the
property that ei acts on the jth irreducible module Lj (of dimension nj ) as δi,j . Sorry there is a
lot of fixed notation: r is the number of conjugacy classes, C1 , . . . , Cr are the conjugacy classes,
L1 , . . . , Lr are representatives of the isomorphism classes of irreducible modules, χi is the character
118
CHAPTER 4. ALGEBRAS
of Li , ni is the dimension of Li or χi (1), ei is the central idempotent corresponding to Li ... To make
things worse I will always assume from now on that C1 = {1} and L1 is the trivial module. Also
let ci = |Ci | be the size of the ith conjugacy class. Also I’ll pick once and for all a representative
gi in the conjugacy class Ci .
The character table of G is the r × r matrix with rows labelled by χ1 , . . . , χr and columns
labelled by C1 , . . . , Cr with ij-entry equal to χi (gj ). Of course this is independent of the particular
representative gj of Cj chosen because χi is a class function. (There are two choices in all of these
that are important: the indexing of the conjugacy classes by 1, . . . , r determines the ordering of
the columns of the character table and the indexing of the isoclasses of irreducible representation
by 1, . . . , r determines the ordering of the rows of the character table).
The character table of G is a crucial invariant of the group G...
4.6.11. Example. Here is the character table of S3 :
χ1
χ2
χ3
1 (1 2) (1 2 3)
1 1
1
1 −1
1
2 0
−1
Here χ2 is the sign representation, χ3 is the two dimensional one sitting in the natural permutation
representation. Note for example that given any finite dimensional CG-module V you can decompose it into9irreducibles given just
χ and its character table: just work out how to
)r its character
r
i
...
In
class
I will apply this to the natural permutation
write χ = i=1 ai χi then V ∼
= i=1 L⊕a
i
representation and the regular representation.
Before we forget all the notation, let me prove a slightly technical lemma.
9
(i)
4.6.12. Lemma. If ei = g∈G ag g then
a(i)
g =
ni χi (g −1 )
.
|G|
Proof. Let ψ be the character of the regular CG-module and g ∈ G. We’ll compute ψ(ei g −1 ) in
two different ways. On the one hand,
! (i)
ei g −1 =
ah hg −1 .
h
So since ψ(1) = |G| and ψ is zero on all other group elements,
On the other hand, ψ =
9r
j=1
ψ(ei g −1 ) = a(i)
g |G|.
nj χj . So
ψ(ei g −1 ) =
!
nj χj (ei g −1 ).
j
But ei acts as zero on all Lj for j 4= i, and it acts as 1 on Li . So we get that
ψ(ei g −1 ) = ni χi (g −1 ).
Comparing these formulas proves the lemma.
Okay, now let’s do something clever. Introduce a Hermitian form on the complex vector space
C(G) by defining the pairing of two class functions χ and ψ from
1 !
(χ, ψ) =
χ(g)ψ(g)
|G|
g∈G
9
1
2
where − denotes complex conjugation. Note that (χ, χ) = |G|
g∈G |χ(g)| which is > 0 if and
only if χ 4= 0. So it is a positive definite Hermitian form. So we can call it an inner product...
119
4.6. CHARACTER THEORY OF FINITE GROUPS
4.6.13. Theorem (Orthogonality relations). (i) With respect to the inner product just defined,
χ1 , . . . , χr are orthonormal. Hence for any character χ,
χ=
r
!
(χ, χi )χi .
i=1
(ii) (row orthogonality relation) With our usual notation for the character table, we have for
any i, j = 1, . . . , r that
;
r
!
0
if i 4= j,
ck χi (gk )χj (gk ) =
|G| if i = j.
k=1
(iii) (column orthonality relation) For any j, k = 1, . . . , r we have that
;
r
!
0
if j 4= k,
χi (gj )χi (gk ) =
|G|/cj if j = k.
i=1
Proof. (i) We know from the preceeding lemma that χi (ej ) = δi,j nj . Also ej =
Hence,
1 !
δi,j nj =
nj χj (g −1 )χi (g).
|G| g
1
|G|
9
g
nj χj (g −1 )g.
You’ll show on the exercises that χj (g −1 ) is χj (g) = χj (g)−1 . Hence the right hand side is
nj
|G| (χi , χj ).
(ii) This is just a restatement of (i) in the language of the character table.
(iii) Let A be the character table, i.e. the matrix with ij-entry χi (gj ). Let B be the matrix
with ij-entry ci χj (gi )/|G|. Let us compute AB: its ij-entry is
r
1 !
ck χi (gk )χj (gk ) = δi,j .
|G|
k=1
Therefore AB = I. Therefore BA = I. Let’s compute its ij-entry:
1 !
ci χk (gi )χk (gj ) = δi,j .
|G|
k
I think that’s it.
This gives some rather strong arithmetic properties that the character table must satisfy. Using
it you can compute character tables using only a relatively small amount of information. Here are
some examples.
4.6.14. Example. Let us first compute the character table of the cyclic group C4 of order 4. Its
abelian, so r = 4 and n1 = n2 = n3 = n4 = 1. Let x be a cyclic generator. A 1 dimensional
representation must map x → ω with ω 4 = 1 in C. Let ω be a primitive 4th root of one. Then the
four choices are x $→ ω, x $→ ω 2 = −1, x $→ ω 3 and x $→ ω 4 = 1. So the character table is:
Ci
ci
χ1
χ2
χ3
χ4
1 x x2 x3
1 1
1
1
1 1
1
1
1 ω −1 ω 3
1 −1 1 −1
1 ω 3 −1 ω
Note χ2 can be thought of as the composite of the map C4 ! C2 mapping x2 to 1, then the
non-trivial character of C2 ...
120
CHAPTER 4. ALGEBRAS
4.6.15. Example. Next, let’s look at V4 = {1, x, y, xy}. There are four obvious one dimensional
representations. For example you can map x to −1, y to 1 then xy has to map to −1. So you get
Ci
ci
χ1
χ2
χ3
χ4
1 x
y
xy
1 1
1
1
1 1
1
1
1 −1 1 −1
1 1 −1 −1
1 −1 −1 1
You should check the orthogonality relations hold in both cases... These two examples give groups
G, H with CG ∼
= CH but with different character tables: character tables are a finer invariant
than just the information in the Wedderburn decomposition.
4.6.16. Example. Next lets look at D4 = {1, x, x2 , x3 , y, xy, x2 y, x3 y}. Recall that x2 is central.
So there is a surjection D4 ! V4 under which x2 maps to 1. Using this the four irreducible
characters of V4 lift to give four irreducible characters of D4 . There are just 5 conjugacy classes,
and you can work out the last irreducible character just using the orthogonality relations.
Ci
ci
χ1
χ2
χ3
χ4
χ5
1 x2
x
y
xy
1 1
2
2
2
1 1
1
1
1
1 1 −1 1 −1
1 1
1 −1 −1
1 1 −1 −1 1
2 −2 0
0
0
4.6.17. Example. Finally lets look at Q3 = {1, 1̄, i, ī, j, j̄, k, k̄}. The conjugacy classes are
{1}, {1̄}, {i, ī}, {j, j̄}, {k, k̄}.
There’s a map Q3 ! V4 , 1̄ $→ 1}. This gives four irreducible characters, the last comes for free.
Ci
ci
χ1
χ2
χ3
χ4
χ5
1 1̄
i
j
k
1 1
2
2
2
1 1
1
1
1
1 1 −1 1 −1
1 1
1 −1 −1
1 1 −1 −1 1
2 −2 0
0
0
Note this is the same as the character table in the preceeding example. So we’ve found groups
G 4∼
= H with the same character tables, just in case you were wondering if character tables separated
groups...
The purpose of the next theorem is to convince you that the character table of a finite group
does actually say quite a bit about the structure of the group. In fact character theory is absolutely
crucial in many parts of finite group theory, e.g. the classification of finite simple groups would
never have been possible without it.
4.6.18. Theorem. Let ρ : G → GL(V ) be a representation of G, and χ = χV be its character. Let
ker χ = {g ∈ G | χ(g) = χ(1)}. Then,
(i) |χ(g)| ≤ χ(1) = dim V ;
(ii) ker ρ = ker χ, hence ker χ is a normal subgroup of G;
121
4.6. CHARACTER THEORY OF FINITE GROUPS
(iii) if χ =
9t
s=1
as χis with as > 0 then ker χ =
:
i
ker χis ;
(iv) if N is
:tany normal subgroup of G, then there are irreducible characters χi1 , . . . , χis such that
N = s=1 ker χi .
Proof. (i) Note that g |G| = 1 for any g ∈ G. Hence the minimal polynomial of ρ(g) divides x|G| −1.
In particular, it has distinct linear factors, so ρ(g) is diagonalizable, and all its eigenvalues are |G|th
roots of unity. Hence, χ(g) is a sum of χ(1) roots of unity. Now the triangle inequality implies
that |χ(g)| ≤ χ(1).
(ii) If g ∈ ker ρ, then |χ(g)| = χ(1), hence g ∈ ker χ. Conversely, if g ∈ ker χ then χ(g) = χ(1) so
equality holds in the triangle inequality, so all of the eigenvalues of ρ(g) are equal. Say ρ(g) = ωI,
a |G|th root of 1. So χ(g) = χ(1)ω. But χ(g) = χ(1) by hypothesis, so in fact ω = 1 and g ∈ ker ρ
as ρ(g) = I.
9t
9
9 (iii) Say χ =<< 9 s=1 as χis <<with9as > 0. Take g ∈ ker χ. Then χ(g) = s as χis (g). Hence χ(1) =
and all
s as χis (g) =
s as |χis (g)|. So equality holds in the triangle inequality,
s as χis (1) =
9
χ
(g)
are
equal
to
some
complex
number
z
(say)
times
χ
(1).
But
then
χ(g)
=
a
χ
(1)z =
i
i
s
i
s
s
s
9s
:
a
χ
(1),
hence
z
=
1.
Therefore
χ
(1)
=
χ
(g)
for
all
s,
so
g
∈
ker
χ
.
The
reverse
s
i
i
i
i
s
s
s
s
s
s
inclusion is clear.
(iv) Let N be a normal subgroup. Lift the regular character of G/N to G. This gives a character
ψ of G with ψ(g) = [G : N ] if g ∈ N , ψ(g) = 0 if g ∈
/ N . For this, ker ψ = N .
Let us now compute the character table of S4 .
C
|C|
χ1
χ2
χ3
χ4
χ5
1
1
1
1
2
3
3
(1 2)
6
1
−1
0
1
−1
(1 2 3)
8
1
1
−1
0
0
(1 2 3 4)
6
1
−1
0
−1
1
(1 2)(3 4)
3
1
1
2
−1
−1
Here’s how you get it. Start with the trivial and sign character. (These are lifts of the two
irreducible characters of the cyclic group C2 via the surjection sgn : S4 ! C2 ). Next look at the
surjection S4 ! S4 /V4 ∼
= S3 . Lifting the two dimensional irreducible character of S3 through this
gives χ3 . Next look at the natural permutation representation χ. Its values on these classes are
the numbers of fixed points: 4, 2, 1, 0, 0. Of course it contains χ1 as a constituent. Subtracting
χ − χ1 gives χ4 (check its irreducible by showing (χ4 , χ4 ) = 1). Finally χ5 is completed by using
the orthogonality relations, or noting χ5 = χ2 χ4 (pointwise multiplication).
I want to end the section by discussing the various tricks around to compute character tables. By
a character of G I mean the character of some CG-module. So characters are special class functions:
the ones which are Z≥0 -linear combinations of irreducible characters. These techniques all give
ways to find some genuine characters. They need not be irreducible (a character is irreducible if
and only if (χ, χ) = 1). But they are a starting point... You can then reduce them bit by bit to
eventually pin down the irreducible characters.
• Characters of finite abelian groups are easy.
• For any group, there’s always the trivial character
χ1 , and the regular character ψ with
9r
ψ(1) = |G|, ψ(g) = 0 for g 4= 1. Note ψ = i=1 ni χi (by Wedderburn’s theorem), so ψ
contains everything else ... but its hard to decompose it without more data.
• Lifts. If N is a normal subgroup of G, then you can lift irreducible characters of G/N to G
through G ! G/N . The character table of G/N gives some but not all rows of the character
table of G...
122
CHAPTER 4. ALGEBRAS
• Permutation characters. If G acts on a set X, the corresponding CG-module CX gives rise
to a character χ of G with χ(g) equal to the number of fixed points of g on X. This is an
easy source of characters... Its always the case that (χ, χ1 ) (where χ1 is the trivial character
of G) is the number of orbits of G on X. (This is the orbit counting lemma: the number of
orbits of G on X is the average number of fixed points ... that is the definition of the inner
product on C(G)).
• Tensor products. There is an obvious pointwise multiplication on the vector space C(G)
making it into a commutative C-algebra: (χ · ψ)(g) = χ(g)ψ(g). In fact the product of two
characters is always a character. Indeed, let V and W be two CG-modules with characters
χV and χW . Make the vector space V ⊗ W (tensor product over C) into a G-module so
that g(v ⊗ w) = (gv) ⊗ (gw) for g ∈ G, v ∈ V, w ∈ W . Then, I claim that χV χW = χV ⊗W .
To see this, use the following fact: suppose α : V → V, β : W → W are diagonalizable
endomorphisms. Then tr(α ⊗ β) = tr(alpha) tr(β). I’ll leave the proof as an exercise: pick
bases in which everything is diagonal then its not too bad. Note this statement is actually true
even if things are not diagonalizable, but its a little harder to prove. But of course we know
for any finite dimensional representation ρ : G → GL(V ) that all ρ(g) ARE diagonalizable
matrices, because their minimal polynomials have distinct linear factors.
• *Induction/restriction*. Let H be a subgroup of G. Then CH is a subalgebra of CG. So
any CG-module is a CH-module by restriction. At the level of characters, any character χ
of G restricts to a character χ ↓H of H... its just the restriction of the function on G to a
function on H. Conversely, any CH-module V gives rise to a CG-module CG ⊗CH V . This
is called the “induced module”. At the level of characters, there is a corresponding notion of
“induced character”, denoted χ ↑G if χ was a character of H. In class, I’ll prove the following
formula for χ ↑G : if t1 , . . . , tk is a system of G/H-coset representatives then
χ ↑ (g) =
G
k
!
χ̇(t−1
i gti )
i=1
where χ̇(g) is χ(g) if g ∈ H, χ̇(g) = 0 otherwise. For example, if G acts transitively on a
set X and you let H be the stabilizer of a point x ∈ X, hence X is identified with the coset
space G/H via the orbit map, then this formula gives that 1H ↑G (g) is the number of fixed
points of g on X. Here 1H denotes the trivial character of H. So permutation characters are
a special case of induced characters: they are precisely the characters induced from trivial
representations of subgroups...
There is one important fact about induced characters which is a good excuse to remind you
some more about tensor products. Let us write (., .)G for the inner product on C(G) and
(., .)H for the inner product on C(H) to avoid confusing the two. The following formula is
known as Frobenius reciprocity:
(χ ↑G , ψ)G = (χ, ψ ↓H )H
for any χ ∈ C(G), ψ ∈ C(H).
Modern proof. First suppose that V and W are two finite dimensional CG-modules. I
want to observe that the inner product (χV , χW )G computes the dimension of the vector
) ⊕ai
) b
space
HomCG (V,
and W = j Ljj . So χV =
9
9W ). To prove this, decompose
9 V = i Li
ai χi , χW =
bj χj . Hence (χV , χW )G = i ai bi . This equals dim HomCG (V, W ), since
dim HomCG (Li , Lj ) = δi,j .
Now to prove the Frobenius reciprocity formula, it suffices to consider the case that χ = χV
for some CH-module V and ψ = χW for some CG-module W . In terms of homomorphisms,
the formula claims that
HomCG (CG ⊗CH V, W ) ∼
= HomCH (V, W ↓H ).
123
4.7. BURNSIDE’S P A QB THEOREM
Now notice that W ↓H is isomorphic as a CH-module to HomCG (CG, W ), where CG is being
viewed as a (CG, CH)-bimodule by right multiplication for H. Now the formula is exactly
adjointness of tensor and hom!
• *Geometry*. Well we won’t go there. But there are lots of beautiful ways to get characters
from groups acting on topological spaces and their cohomology....
Let’s end this introduction to character theory by computing the characters of the symmetric
group S5 . Remark: the symmetric groups are remarkable in that the characters of all Sn are
integers. Usually character values really are complex numbers, but for a character of degree n we
do at least know that all χ(g)’s are sums of n |G|th roots of unity... In particular all character
values are algebraic integers.
C
|C|
χ1
χ2
χ3
χ4
χ5
χ6
χ7
1 (12)
1 10
1
1
1 −1
4
2
4 −2
5
1
5 −1
6
0
(123) (1234)
20
30
1
1
1
−1
1
0
1
0
−1
−1
−1
1
0
0
(12345)
24
1
1
−1
−1
0
0
1
(12)(34)
15
1
1
0
0
1
1
−2
(123)(45)
20
1
−1
−1
1
1
−1
0
Working: Work out the conjugacy classes and their orders. Then we have the trivial and sign
representations.
Next look at the natural permutation character which has values 5, 3, 2, 1, 0, 1, 0 on the conjugacy classes. It contains the trivial character, so subtract it... Then what is left is irreducible since
its inner product with itself is 1. This gives a degree 4 irreducible character, and tensoring that
with sign gives another one.
The remaining three characters either have degrees (5, 5, 6) or (1, 6, 7). But the degree of a
character divides the order of the group (see next section) so the second thing couldn’t happen. Now
we need some input. Consider the permutation action on the set of pairs {1, 2}, {1, 3}, {1, 4}, {1, 5},
{2, 3}, {2, 4}, {2, 5}, {3, 4}, {3, 5}, {4, 5}. Computing fixed points gives the character values
10, 4, 1, 0, 0, 2, 1
on the conjugacy classes in the order listed. Subtract the trivial character. We’ve got
9, 3, 0, −1, −1, 1, 0.
Now take the inner product with χ3 . You get 1. So subtract χ3 . Then check that is indeed
irreducible.
Now tensor with sign to get χ6 . Finally work out χ7 using orthogonality relations.
4.7
Burnside’s pa q b theorem
You know that simple arithmetic plays an important role in finite group theory. The character
table is the way that “higher arithmetic” comes into play. By “higher arithmetic” I mean algebraic
number theory... Don’t forget: if χ is a character of degree n, then each character value χ(g) is a
sum of n |G|th roots of unity. We need to understand such numbers to make further progress...
In class we’ll digress at this point to cover sections 5.1 and 6.1 from these notes, in particular,
we’ll meet the definition of an algebraic integer: an element of C that is a root of a monic polynomial
in Z[x]. We’ll prove in §6.1 that the sum or the product of two algebraic integers is again an
algebraic integer. Hence all character values χ(g) are algebraic integers. Here are a couple of basic
lemmas.
124
CHAPTER 4. ALGEBRAS
4.7.1. Lemma. Let α be an algebraic number and mα (x) ∈ Q[x] be its minimal polynomial (the
monic polynomial with coefficients in Q of smallest possible degree having α as a root). Then α is
an algebraic integer if and only if mα (x) ∈ Z[x].
Proof. Let α be an algebraic integer. Let p(x) be a monic polynomial of minmal degree in Z[x]
having α as a root. Note p(x) is irreducible in Z[x] (if it factorizes then by looking at the leading
coefficients both factors are monic and one of them already has α as a root). Hence by Gauss’
lemma it is irreducible in Q[x]. Clearly mα (x)|p(x). Together these imply that mα (x) = p(x).
Note we will prove the next lemma again in §6.1 – in fancy language it says that Z is integrally
closed in Q.
4.7.2. Lemma. A rational algebraic integer (i.e. an algebraic integer lying in Q) is actually an
integer (i.e. an element of Z).
Proof. Let α be an algebraic integer that can be written as a rational number p/q in lowest terms.
Say f (x) = xn + a1 xn−1 + · · · + an is a monic polynomial in Z[x] with α as a root. Then,
q n−1 f (α) = 0 =
It follows that
pn
q
pn
+ a1 pn−1 + · · · + an q n−1 .
q
is already an integer, i.e. q|pn . This contradicts that p/q was in its lowest terms.
Finally we need the following lemma about algebraic integers, which I’ll prove again in §6.1
without using the noetherianness of Z (it will make more sense then when it is put into the right
context – right now it looks like magic).
4.7.3. Lemma. Let α ∈ C. Suppose there exists a faithful Z[α]-submodule U of C that is finitely
generated as an abelian group. (Here, faithful means that annZ[α] (U ) = {0}). Then α is an
algebraic integer.
Proof. The main task is to prove that Z[α] is a noetherian Z-module. Once we’ve done that, the
lemma is easy: look at 1, α, α2 , . . . . By ACC there must be some k such that αk ∈ Z + Zα +
· · · + Zαk−1 . Hence α is a root of a monic polynomial in Z[x] of degree k. Hence α is an algebraic
integer.
To prove that Z[α] is a noetherian Z-module, note that since U is a finitely generated Zmodule, so is EndZ (U ). To prove this statement, let F be a free Z-module of finite rank having
U as a quotient. Then EndZ (U ) is a Z-submodule of HomZ (F, U ) (it’s the homomorphism that
factor through the quotient U of F ...). The latter is a direct sum of finitely many copies of
HomZ (Z, U ) ∼
= U , which is finitely generated. But Z is noetherian so submodules of finitely
generated Z-module are finitely generated. Hence the submodule EndZ (U ) is finitely generated
too.
But now the assumption that U is faithful means that the representation Z[α] → EndZ (U )
is injective. So Z[α] is a Z-submodule of a noetherian Z-module, hence its noetherian as Z is a
noetherian ring.
Now I’m going to prove an important theorem which is really just a warm up to proving
Burnside’s pa q b theorem. It is our first application of algebraic integers to character theory.
4.7.4. Theorem. With the notation of the previous section, the degrees ni of the irreducible characters of a finite group G all divide |G|.
9
Proof. Recall that ei = g∈G α1 χi (g −1 )g where α = |G|/ni . Since ei is an idempotent, e2i = ei ,
so
!
αei =
χi (g −1 )gei .
g∈G
125
4.7. BURNSIDE’S P A QB THEOREM
Now for any g, h ∈ G, define ah,g ∈ C from the equation
!
gei =
ah,g h.
h∈G
The above equation shows that
!
!
!
αah,g h = αgei =
χi (k −1 )(gk)ei =
χi (k −1 )ah,gk h.
h∈G
k∈G
h,k∈G
9
Equating coefficients shows that αah,g = k∈G χi (k −1 )ah,gk . Now let M be the Z-submodule of
C generated by all {za
∈ G, z |G| = 1}. Clearly M is finitely generated as a Z-module.
9 h,g | h, g −1
Moreover, αzah,g = k∈G χi (k )zah,gk . The coefficient χi (k −1 )z is a sum of |G|th roots of 1,
because character values are. This shows that αzah,g ∈ M . Hence M is a Z[α]-submodule of C,
clearly faithful as C is a field, that is finitely generated as a Z-module. Hence by the previous
lemma, α is an algebraic integer. But α ∈ Q, hence by Lemma 4.7.2 α is an integer. Hence ni ||G|.
Okay, now we’re going to do a similar trick to prove Burnside’s pa q b theorem. We need a
little more preparation, which needs a little Galois theory. Let me recall: if f (x) ∈ Q[x] is a
polynomial, its splitting field E is the subfield of C obtained by adjoining to Q all the roots of
the polynomial f (x). The Galois group of f (x) over Q is the Galois group Gal(E/Q) of the field
extension Q ⊆ E. That is, it is the group of all field automorphisms of E leaving Q invariant
(that last condition is guaranteed in this case that the small field is Q since Q doesn’t have any
non-trivial automorphisms...). Note the fixed subfield of the Galois group
{α ∈ E | g(α) = α for all g ∈ Gal(E/Q)}
is precisely the field Q again. This follows from the fundamental theorem of Galois theory. Note the
Galois group of f (x) permutes the set of roots of the polynomial f (x), hence Gal(E/Q) embeds
into the symmetric group on the set of roots of f (x). This is where group theory came from
originally. You probably met the following result when you first studied Galois theory.
4.7.5. Lemma. If f (x) ∈ Q[x] is monic and irreducible, then its Galois group acts transitively on
the set of roots of f (x).
#
Proof. Note
= f (x) has distinct roots (else (f (x), f (x)) 4= 1). So if R is the set of roots of f (x),
f (x) == α∈R (x − α). Suppose the Galois group has two orbits on R, R = R1 8 R2 . Then
g(x) = α∈R1 (x − α) divides f (x). But the Galois group fixes g(x), hence all its coefficients lie in
the fixed subfield, namely, Q. Hence f (x) is reducible.
Now here is the main difficult thing that we’ll need about algebraic numbers.
4.7.6. Lemma. Suppose ε1 , . . . , εn are N th roots of unity and α =
integer with |α| < 1. Then α = 0.
1
n (ε1 + · · · + εn )
is an algebraic
Proof. Let mα (x) ∈ Z[x] be the minimal polynomial of α. Its irreducible over Q[x]. Let β be
any root of mα (x). Let G be its Galois group, i.e. G = Gal(E/Q) where E is the splitting field
of mα (x) over Q. By the previous lemma, G acts transitively on the roots of mα (x). Hence there
exists g ∈ G such that g(α) = β. Let F be the splitting field of xN − 1 over E. Since g maps
xN − 1 to itself, the automorphism g of E extends to an automorphism also denoted g of F (e.g.
by Existence and uniqueness of splitting fields in 5.3). Hence
β=
1
(gε1 + · · · gεn )
n
and each gεi is also an N th root of unity, hence has |gεi | ≤ 1. It follows from this and the triangle
inequality that |β| ≤ 1.