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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. 1 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: 2 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 ∗ 3 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. 4