Download 31 Semisimple Modules and the radical

31
Semisimple Modules and the radical
If the characteristic of F does not divide the order of G then we saw that
every F G-module is a direct sum of simple modules. In this section we
examine modules over arbitrary algebras A having this property, i.e., that
they are direct sums of simple modules.
We assume as before that A is a finite dimensional algebra over the field
F and that all modules are finitely generated (and thus finite dimensional)
left A-modules. Furthermore, the letter M will always denote an A-module
and the letter S will always denote a simple A-module.
Theorem 31.1. For any f.g. A-module M the following are equivalent.
1. M is a direct sum of simple modules.
2. M is a sum of simple modules.
3. Every submodule of M is a direct summand.
Modules with this property will be called semi-simple (s-s).
Proof. Obviously, (1) implies (2). Also (3) implies (1) since each simple
submodule is a direct summand. It remains to show that (2) implies (3).
Assume by contradiction that (2) holds but (3) doesn’t hold. Let N
be the largest (maximum dimension) submodule of M which is not a direct
summand. Since M is a sum of simple modules, there is a simple submodules
S so that S is not contained in N . Then S ∩ N = 0 so
S + N = S ⊕ N.
But this is larger than N so must be a direct summand of M . Thus M =
K ⊕ (S ⊕ N ) and N is a direct summand of M .
Corollary 31.2. Every submodule and quotient module of a s-s module is
s-s.
Proof. If M ∼
= ⊕Si then any quotient of M is a sum of the images of the Si .
However, every homomorphic image of a simple module is simple. Thus the
quotient satisfies condition (2) in the theorem. For submodules of M note
that by (3), every submodule is also a quotient module.
Semi-simple modules are also characterized by the property that their
radicals are zero.
Definition 31.3. The radical rM of M is defined to be the smallest submodule of M so that M/rM is semi-simple. 31.5 gives another definition.
Theorem 31.4.
1. M/rM is semi-simple.
2. M is semi-simple iff rM = 0.
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Proposition 31.5. The radical of M is equal to the intersection of all kernels
of epimorphisms p : M ³ S where S is simple. I.e.,
rM = ∩ ker p = {x ∈ M | p(x) = 0 ∀p : M ³ S}
Proof. The submodule rM ∩ ker p is the kernel of the homomorphism
M → (M/rM ) ⊕ S
which has s-s image. By minimality of rM we have ker p ∩ rM = rM ,
i.e., rM ⊆ ker p for all p : M ³ S. Thus rM ⊆ ∩ ker p. Conversely,
suppose that x ∈ M is not in rM . Then it maps nontrivially to one of the
simple summands of M/rM ∼
= ⊕Si . So pi (x) 6= 0 for some i showing that
∩ ker p ⊆ rM .
Prove as an exercise:
Corollary 31.6. rM 6= M unless M = 0.
Semisimple algebras
An algebra A is called semi-simple iff all f.g. A-modules are s-s.
Theorem 31.7. A is a s-s algebra iff rA = 0 (i.e., iff it is a s-s A-module).
Proof. A s-s algebra implies all modules s-s implies rA = 0. Conversely,
suppose that A is a s-s module. Then any module M is a quotient of An and
thus s-s.
Theorem 31.8. rA is a two-sided ideal.
Proof. By definition, rA is a submodule and thus a left ideal of A. We need
to show that rA is also a right ideal, i.e., for all x ∈ A we want to show
rAx ⊆ rA. In other words, any homomorphism p : A → S contains rAx in
its kernel. But right multiplication by x is a homomorphism of left A-modules
so the composition:
( )x
p
A −−→ A −
→S
is a homomorphism so it sends rA to zero, i.e., p sends rAx to zero as
required.
Corollary 31.9. Simple algebras (i.e., those with no nontrivial two-sided
ideals) are semi-simple.
Proof. A simple ⇒ rA = 0 ⇒ A s-s.
Example 31.10. A division algebra D is simple and thus semi-simple. (Any
nonzero element is invertible and thus generated the whole ring.)
Another example:
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Theorem 31.11. The n × n matrix algebra Mn (D) over a division algebra
D is semi-simple.
Proof. Suppose that I is a nonzero two sided ideal. Then I will show that I =
Mn (D) by showing that I contains all elementary matrices xij (d) (having d
as ij-entry and 0’s everywhere else). For example:


0 0 0
x32 (d) = 0 0 0
0 d 0
Suppose that t ∈ I, t 6= 0. Then one of the entries tpq = a 6= 0. Thus D
contains a−1 and
xip (a−1 )txqj (d) = xij (d) ∈ I
as promised.
In general we have:
Theorem 31.12. If A is a s-s algebra then every simple module is isomorphic
to a summand of A.
However in the case of Mn (D) the is only one such simple module.
Theorem 31.13. Up to isomorphism the only simple Mn (D)-module is the
space of column vectors Dn .
Proof. Since Mn (D) is the direct sum of the n column vector spaces it suffices
by Theorem 31.12 to show that Dn is simple. The argument is the same as
the proof of Theorem 31.11.
Suppose t ∈ Dn is nonzero. Then one of its coordinates is nonzero, say
tj = a 6= 0. Then:

   
0 0 s1 a−1 0 0
∗
s1
0 0 s2 a−1 0 0 ∗ s2 
n
X

   
−1
 a = s3 
0
0
s
a
0
0
xij (si a−1 )t = 
3

   
0 0 s4 a−1 0 0 ∗ s4 
i=1
0 0 s5 a−1 0 0
∗
s5
is an arbitrary element of Dn .
Other examples
Take the 6-dimensional algebra


 ∗ ∗ ∗ 
A3 := 0 ∗ ∗


0 0 ∗
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As a left A-module this breaks up into a direct sum of the three columns:
P1 = S1 = F ⊕ 0 ⊕ 0
P2 = F ⊕ F ⊕ 0
P3 = I3 = F ⊕ F ⊕ F
However, only the first submodule P1 = S1 is simple. Since S1 ⊆ P2 ⊆ P3 ,
P2 and P3 are not simple. But P2 is generated by any element whose second
coordinate is nonzero. (See the proof of Theorem 31.13.) Consequently, P2 is
not s-s so it must be indecomposable. Similarly, P2 is not a direct summand
of P3 so P3 is also indecomposable. The radical of P2 is S1 with quotient
P2 /S1 ∼
= S2 = 0 ⊕ F ⊕ 0. Similarly the radical of P3 is P2 with quotient S3 .
This means that


 0 ∗ ∗ 
rA3 = rP1 ⊕ rP2 ⊕ rP3 = 0 0 ∗


0 0 0
Note that this is the set of nilpotent elements of A3 . The matrix ring M3 (F )
also contains nilpotent elements (since A3 ⊆ M3 (F )). However, M3 (F )
contains no nontrivial nilpotent two sided ideal. rA3 is a nilpotent ideal in
A3 since the product of any three elements of rA3 is zero.
Another example is F Z/3 where F is a field of characteristic 3. In that
case we have (1 − g)3 = 1 − g 3 = 0 where Z/3 = hgi. This implies that
(1 − g) generates a nilpotent ideal. Since this ideal is 2-dimensional it must
be the radical of the algebra. As a module, F Z/3 is indecomposable, has
radical equal to the ideal generated by (1 − g) and the quotient is the simple
module F with trivial action of Z/3 (gx = x). The analysis of this example
implicitly uses the fact that r(M ⊕ N ) = rM ⊕ rN . Consequently, if M/rM
is 1-dimensional, M is indecomposable. But the converse is not true.
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