Projective modules: Wedderburn rings April 10, 2008 8 Wedderburn rings A Wedderburn ring is an artinian ring which has no nonzero nilpotent left ideals. Note that if R has no left ideals I such that I 2 = 0 then I has no nonzero nilpotent left ideals at all. Also, if R has no nonzero nilpotent two-sided ideals then R has no nonzero nilpotent left ideals. The name Wedderburn ring honors the seminal work of JHM Wedderburn, who proved that a finite-dimensional C-algebra which has no nilpotent left ideals must be a finite direct product of of full matrix algebras over C. This was later generalized by E Artin, who showed that a ring which has the descending chain condition on left ideals and which has no nonzero nilpotent left ideals that is, what we now call a Wedderburn ring must be a finite direct product of matrix algebras, each factor defined over some (possibly noncommutative) field. In particular, every simple artinian ring is a matrix ring over a field. Further work by N Jacobson, A Mal cev and others refined this structure theorem to encompass arbitrary artinian rings, using a tool we now call the Jacobson radical. First, we link the structure of Wedderburn rings to our previous work on projective modules. The following theorem says, in essence, that Wedderburn rings are precisely those for which the category of modules is trivial. (That is not to say that the individual modules are trivial!) Theorem 8.1. Suppose that R is ring. The following are equivalent. 1. Every R-module is projective. 2. Every R-module is completely reducible. 3. R is completely reducible. 4. R is a Wedderburn ring. Proof. 1 2: If every module is projective and N < M then the projectivity of M/N implies that the sequence 0 N M M/N 0 splits, whence N is a direct summand of M. 1
Conversely, suppose that every module is completely reducible. Given a module M choose a free module F and a surjection 0 K F M 0. The complete reducibility of F implies that F = K M, whence M is projective. 2 3: If every module is completely reducible then R is. Conversely, if R is completely reducible, then it is a sum of simple modules. Hence every free module is also a sum of simple modules, whence also completely reducible. But then every module is a homomorphic image of a free module, whence completely reducible. 3 4: If R is completely reducible then R is a direct sum of minimal left ideals. Since 1 lies in a finite sum, R = R 1 is in fact a finite sum. Since a simple module is artinian a finite sum of simple modules is artinian. Finally, if I is a minimal left ideal then I is a direct summand, hence I = Ie for some idempotent. In particular, I 2 0. Hence R is a Wedderburn ring. Conversely, since R is artinian it contains at least one minimal left ideal I 1, say. Since R contains no nilpotent left ideals I 1 is a direct summand. Say R = I 1 J 1. Again, if J 1 0 then it contains a minimal left ideal I 2, say. Since I 2 is a direct summand of R then it is a direct summand of J 1, by the Modular Law. Say, R = I 1 I 2 J 2. Continuing this way we obtain a descending chain J 1 > J 2 > which can terminate only when some J k = 0. At this point we have decomposed R as a finite direct sum of minimal left ideals. This theorem is the key to proving the Wedderburn-Artin Theorem. To exploit it letr be any ring and consider the ring End R (R). Each element φ End R (R) is determined by its value on 1, since for every x R we have that φ(x) = φ(x 1) = xφ(1). On the other hand, any choice of r R determines an φ End R (R) with φ(1) = r by defining φ(x) = xr. This gives a bijection between R and End R (R). It is straightforward to check that this bijection is an additive. However, composition of endomorphisms corresponds to the reverse of multiplication: if φ r and ψ s then φψ sr, since (φψ)(x) = (xs)r = x(sr). Said another way, End R (R) = R op, where the opposite ring R op is the same as R, except that multiplication is reversed: if is the multiplication in R op then a b = ba, where the latter multiplication is the one defined in R. Left modules for R become right modules for R op, and the other way around. Now consider a Wedderburn ring R. As a left R-module, R is a finite direct sum of simple submodules (minimal left ideals) R j. Number these so that S 1 = R 1 = R2 = = Rk1 = S2 = R k1+1 = R k1+2 = = R k1+k 2 = Thus The S kj j R = S k1 1 Sk2 2 Skn n. are called the isotypic components. 2
Lemma 8.2 (Schur s Lemma). If S and T are simple R-modules then any nonzero homomorphism φ Hom R (S, T ) is an isomorphism. Hence if S = T then Hom R (S, T ) = 0, and End R (S) is a (possibly noncommutative) field. Proof. If 0 φ Hom R (S, T ) then ker φ S and im φ 0. Since S and T are simple this implies that ker φ = 0 and im φ = T. That is, φ is an isomorphism. Lemma 8.3. Suppose that M = S k and N = T l are isotypic, with S and T simple. 1. If S = T then HomR (M, N) = 0. 2. End R (M) = M k (F op ), where F = End R (S). Proof. If φ Hom R (M, N) then φ(x 1,..., x k ) = (φ 1 (x 1,..., x k ),..., φ l (x 1,..., x k )), where φ j (x 1,..., x k ) = i φ ij (x i ), where φ ij Hom R (S, T ). Slightly more explicitly, φ ij is the composition of the inclusion of S into the i-th coordinate of M, followed by φ, followed by projection onto the j-th coordinate of N. In particular, if S = T then each φij = 0, whence φ = 0. On the other hand, if S = T and k = l then φ has been represented by a k k matrix with entries in F = End R (S). However, as before when R is not commutative we must multiply matrices on the right: that is to say, we regard the multiplication in F op. Note that if I is a minimal left ideal in a ring R, then the two-sided ideal IR is isotypic, since it is the sum of the left ideals Ix. If Ix 0 then the map y yx is an R-module isomorphism I = Ix. Suppose moreover that I is a direct summand of R, and that J is any left ideal isomorphic to I. Say R = I K and φ is an R-isomorphism from I to J. The composition R R/K φ = I J < R is in End R (R), and so must equal right multiplication by some x R. That is, J = Ix < IR. In particular, if R is a Wedderburn ring then the isotypic components are the 2-sided ideals generated by the minimal left ideals. 3
Theorem 8.4 (Wedderburn-Artin Theorem). If R is a Wedderburn ring then R = M k1 (F 1 ) M kn (F n ), where the F j are (possibly noncommutative) fields. More precisely, if S 1,..., S n are the pairwise nonisomorphic minimal left ideals of R, and S j R = S kj j, as R-modules, then as rings we have the isomorphisms R = S 1 R S n R, where S j R = M kj (F j ) and F j = End R (S j ). In particular, a simple artinian ring is a matrix algebra. Conversely, any matrix algebra is a simple artinian ring, and any finite product of matrix algebras is a Wedderburn ring. Proof. We have proven that R op = End R (R) has such a decomposition, with fields F op j. It remains only to observe that M n (F op ) op = M n (F ), under the map which sends each matrix to its transpose. Note a consequence of this theorem: left Wedderburn is the same as right Wedderburn. Let s examine one of the most important examples of a Wedderburn ring, namely the group algebra F G of a finite group G over a commutative field F of characteristic 0. I claim that every F G-module V is completely reducible. This result is called Maschke s Theorem. First of all, since F < F G, V is an F -vector space. Hence if W is a submodule then there is a vector-space splitting f : V W that is, a linear transformation such that f(x) = x for every x W. We use an averaging process to produce an F G-module splitting f: f(x) = 1 g 1 f(gx). This is a F G-module map since if h G and x V f(hx) = 1 g 1 f(ghx) = 1 hh 1 g 1 f(ghx) = h 1 (gh) 1 f(ghx) = h 1 g 1 f(gx) = h f(x). 4
Hence if r = h c hh F G then f(rx) = f( h c h hx) = h c h f(hx) = c h h f(x) = r f(x). h Second, it is a splitting, since if x W then so is each gx, whence f(x) = 1 g 1 f(gx) = 1 g 1 gx = x = x. By contrast, if p = char(f ) then F G is never a Wedderburn ring. Indeed, if we set ζ = g then it is straightforward to check that ζ Z(F G) and ζ 2 = 0. Hence I = ζf G is a nonzero nilpotent ideal. 5