Advanced Algebra II structure of rings Definition 0.1. For M 2 RM, we said that M is simple if RM 6= 0 and M has no proper submodule. A ring is R said to be simple if R2 6= 0 and R has no proper ideal. Example 0.2. A simple ring R maybe not a simple left RM. Let R := Mat2(k), the 2 £ 2 matrices over a field k. One can check that ideal of R is of the form Mat2(J) for some J C k. Hence R has no proper ideal, hence a simple ring. However, R has two left ideals Ii := f(ajk)jajk = 08k 6= ig, for i = 1; 2. Therefore, as an left RM, R has proper submodule Ii. Example 0.3. A division ring is a simple ring. Proposition 0.4. A simple RM is cyclic, i.e. M = Ra for some a 2 M. Proof. For each a 2 M, we consider Ra < M. Since M is simple, Ra = 0 or Ra = M. Since RM 6= 0, it follows that Ra 6= 0 hence Ra = M for some a 2 M. We can actually prove more. Let B = fa 2 MjRa = 0g. It’s easy to check that B < M. As we have seen that B 6= M. Therefore, B = 0. That is, M = Ra for all a 6= 0 2 M. ¤ » Proposition 0.5. Let M be a simple RM, then M = R=I for some maximal left ideal I. (Note that R=I is not necessarily a ring, we just consider it as a left RM.) Proof. Since M is cyclic, one has that the natural map ' : R ! M by '(r) = ra for a fixed a 6= 0 2 M is surjective. Consider all these as » left RM, then by the first isomorphism theorem, M = R=I for some submodule I := Ker' = A(a) ½ R. It’s clear that I is a left ideal. M is simple implies that I is maximal. ¤ Note that M = Ra, then a = ea for some e 2 R. e play the role as an ”idempotent”. It follows that ra = rea, i.e r ¡ re 2 A(a) = I for all r 2 R. A left ideal I with the property that there is e 2 R with r ¡ re 2 I for all r 2 R is called regular. We can actually prove the following: Theorem 0.6. A left R-module M is simple if and only if M »= R=I for some regular maximal left ideal I. Proof. We have seen one direction (only if part). Suppose now that M »= R=I for some regular maximal left ideal I. Since I is maximal, there is no proper submodule in M. We still need to check that RM »= R(R=I) 6= 0. Since there is e 2 R with r ¡ re 2 I for all r 2 R. We 1 2 have r(e + I) = r + I 6= 0 if r 62 I. Hence R(R=I) 6= 0. Thus R=I is simple. ¤ A similar and important notion is primitive. We need some more notions before we get into primitive. Proposition 0.7. Let A(M) := fr 2 RjrM = 0g be the annihilator of M. Then A(M) C R is an ideal. Moreover, M is a faithful R=A(M) module, i.e. AR=A(M) = 0 as R=A(M) module. Proof. For r 2 R; s 2 A(M), we have (rs)M = r(sM) = 0, hence rs 2 A(M). Also sr(M) = s(rM) < sM = 0, hence sr 2 A(M). The map R £ M ! M induces a natural map R=A(M) £ M ! M. If for some r 2 R such thatr ¯ 2 AR=A(M) ½ R=A(M).rM ¯ = rM = 0, r 2 A(M) and thusr ¯ = 0. ¤ For M 2 RM, we can consider End(M) the endomorphims of the abelian group M. It’s clear that End(M) is a ring. There is a natural map Φ : R ! End(M) define by Φ(r) = Tr such that Tr(x) := rx for x 2 M. Then KerΦ = A(M). By the isomorphism theorem, we have R=A(M) is isomorphic to a subring of End(M). Definition 0.8. A ring R is primitive if there exists a simple faithful left R-module. If R is primitive, let M be a simple faithful R-module, then R is isomorphic to a subring of End(M). Before we proceed to the study of the structure of primitive rings, we take a break by looking at some examples. Proposition 0.9. A simple ring with identity is primitive. Proof. Since R has identity, R has maximal left ideal, say I. Again, R with identity implies that every left ideal of R is regular. Thus R=I is a simple R-module. It remains to show that A(R=I) = 0. This is the case since 1 62 A(R=I) and A(R=I) C R which is simple. ¤ Proposition 0.10. A commutative ring R is primitive if and only if R is a field. Therefore, if R is a commutative ring with identity, then being sim- ple, primitive and being a field are equivalent. However, if R is com- mutative without identity, then it’s possible to be simple but not a field. Proof. A field is simple with identity, hence is primitive. Conversely, let M be a faithful simple R-module. Then M »= R=I for some regular maximal left ideal I. R is commutative, so I is in fact an ideal. M is faithful, thus A(M) = A(R=I) = 0. It’s clear that I ½ A(R=I) = 0. Moreover, I is regular, that is, there exist e 2 R such 3 that r ¡ re = 0. It follows that e is the identity. We have that R is a commutative ring with identity such that 0 is a maximal ideal. Thus R is a field. ¤ The following example also exhibit the difference between simple and primitive. Example 0.11. Let V be a infinite dimensional vector space over a di- vision ring D. Let R := EndD(V ) be the ring of linear transformations. Id clear that V is a left R-module. Claim 1. V has no proper submodule. And thus V is a simple R-module. Claim 2. A(V ) = 0, thus R is primitive. Claim 3. R is not simple. This is because that we have an ideal I = f½ 2 Rjdim(Im½) < 1g. Remark 0.12. If V is finite dimensional, then R is simple and prim- itive. One notice hat End(M) plays an important role here. Inside End(M), there is yet another interesting object Definition 0.13. The commuting ring of R on M is C(M) = fà 2 End(M)jTrà = ÃTr; 8r 2 Rg: One notice that if à 2 C(M), then Ã(rx) = ÃTr(x) = TrÃ(x) = rÃ(x); for all x 2 M. In other words, à is an RM-homomorphism. So C(M) is actually EndR(M). Theorem 0.14 (Schur’s Lemma). If M is a simple module over R, then EndR(M) is a division ring. Proof. Suppose that à 6= 0 2 EndR(M). Let W := ÃM. One sees that W 6= 0 < M. By the simplicity of M, one has W = M. Thus à is surjective. Similarly, one checks that Kerà M. Thus Kerà = 0, à is injective. ¡1 ¡1 Hence à 2 End(M). It’s direct to check that à 2 EndR(M). ¤ Example 0.15. Let k be a field and V be a n-dimensional vector space ¯ over k. Let R = Matn(k) = Endk(V ). If S ½ R, let S be the subalgebra generated by S in R. Then we can view V as a faithful S¯-module. We say that S is irreducible if V is an simple S¯-module. Then by ¯ Schur’s Lemma, C(S) is a divisionµ ring.¶ 0 ¡1 1. Let n = 2, and S = . One can check that S is 1 0 irreducible and C(S¯) »= C. 4 80 1 0 19 > 0 ¡1 0 0 0 0 ¡1 0 > <B C B C= B 1 0 0 0 C B 0 0 0 1 C 2. Let n = 4, and S = @ A ; @ A . > 0 0 0 ¡1 1 0 0 0 > : 0 0 1 0 0 ¡1 0 0 ; One can check that S is irreducible and C(S¯) »= H the real quaternions..