LECTURE V EDWIN SPARK 1. More on the Chnese Remander Theorem We begn by recallng ths theorem, proven n the preceedng lecture. Theorem 1.1 (Chnese Remander Theorem). Let R be a rng wth deals I 1, I 2,..., I r, and defne a map ϕ :R R I1 R I2 R Ir, x (x + I 1, x + I 2,..., x + I r ). Then ϕ s an somorphsm f and only f: (a) I + I j = R j and (b) I 1 I 2 I r = (0). Note that (a) ensures surjectvty, whle (b) ensures njectvty. We wll now consder an example, whch recovers the usual statement of the Chnese Remander Theorem. Example 1.2. Take n 1, n 2,..., n r Z, such that gcd(n, n j ) = 1 j, and set n = n 1 n 2... n r and n = n + (n) Z (n). Then, workng n Z (n), consder ( n ) for each. Now, we have: (a) ( n ) + ( n j ) = Z/(n) for j and (b) ( n 1 ) ( n 2 ) ( n r ) = (0). The frst s true because gcd(n, n j ) = 1 mples that (n ) + (n j ) = Z, whch can be reduced mod n. The second s true because anythng n the ntersecton s surely a multple of n. Hence, the Chnese Remander Theorem clams that: Z (n) = Z (n1 ) Z (n2 ) Z (nr ). Now, gven any system of smultaneous congruences x a (mod n ), wth a Z, we can consder (ā 1,..., ā r ) Z (n1 ) Z (n2 ) and from the somorphsm produce a unque x Z (n) that s ts premage. Then x Z s a soluton to the system of congruences, and furthermore, any other soluton dffers from x by a multple of n. Date: 15 March, 2016. 1
2 EDWIN SPARK Ths means that we have recovered the standard Chnese Remander Theorem from the much more general result proven n the prevous lecture. 2. More operatons on deals Here we consder the acton of rng homorphsms on deals. For any rng homomorphsm f : R S and deals I < R and J < S, f 1 (J) s an deal n R, but f(i) s not necessarly one n S. Ths motvates the followng par of defntons: Defnton 2.1. Let f : R S be a rng homomorphsm and I < R and J < S be deals. Then we call f 1 (J) the contracton of J, denoted J c, and the extenson of I s (f(i)) < S, for whch we wrte I e. In general, contractons are ncer and better behaved than extensons. Ths s already mplct n the defnton, but as an example, we consder how these operatons affect prmeness. Exercse 2.2. Suppose p < S s prme. Then p c s prme n R. Proof. Suppose ab p c. Ths means that f(ab) p, so f(a)f(b) p. Because p s prme, t follows that ether f(a) p, n whch case a p c, or f(b) p, whch would mean that b p c. Extensons, however, do not preserve prmeness, as the followng example demonstrates. Example 2.3. Consder the ncluson map f : Z Z[]. We know that the non-zero prme deals of Z are those generated by prme ntegers. However, not all of ther extensons are prme n Z[]. In fact, we have the followng result: Proposton 2.4. Let p Z be prme. Then (p) e s a prme deal n Z[] f and only f p 3 (mod 4). Proof. Note that Z[] s a Unque Factorsaton Doman so every deal s prncpal, and the deal (x) s prme f and only f x s prme, whch s true f and only f x s rreducble. We proceed wth three cases: Case 1 (p = 2). (2) < Z[] s not prme, because 2 = (1 + ) 2 s not rreducble.
LECTURE V 3 Case 2 (p 1 (mod 4)). By Gauss 108, there exsts a Z such that a 2 1 (mod p) f and only f p 1 (mod 4) or p = 2. 1 As such, p a 2 + 1 = (a + )(a ) (n Z[]). Now, f p s prme, then ether p (a + ) or p (a ). However, n ether case, that would requre p 1, whch s clearly absurd. As such, (p) < Z[] s not prme. Case 3 (p 3 (mod 4)). Suppose, for a contradcton, that p s not rreducble. Then we can fnd non-unts α, β Z[] such that p = αβ. Ths means we have α, β > 1 and α 2 β 2 = p 2. These are statements about non-negatve ntegers, so the only possble soluton s α = β = p. But then, f say α = a + b, we must have a 2 + b 2 = p 3 (mod 4), whch s mpossble, as all squares are congruent to ether 1 or 2 (mod 4). It follows that p s rreducble and (p) < Z[] s prme. 3. Modules Defnton 3.1. Let R be a rng. An R-module M s an abelan group equpped wth an R-acton, that s, a map: R M M, (r, m) r m, that s compatble wth both the rng and group structures. That s, the acton sastsfes: r (m + m ) = r m + r m (r + s) m = r m + s m (rs) m = r (s m) 1 m = m. Remark 3.2 (Notatng the acton). It s sometmes useful to wrte the acton as a map µ : R End(M), r µ(r) :M M, m r m. 1 From the Dsqustones. Here we need only the f half and a = ( p 1 2 )! works.
4 EDWIN SPARK Here End(M) s the rng of abelan group homomorphsms from M to M. The notaton s due to the fact that such maps are (partcular) endomorphsms. Defnton 3.3. Let M and N be R-modules. Then an R-module homomorphsm s an abelan group homomorphsm ϕ : M N such that ϕ(r m) = r ϕ(m). An R-module somorphsm s an nvertble R-module homomorphsm. If such a morphsm from M to N exsts, we say that M and N are somorphc as R-modules, and can wrte M = N. Remark 3.4. The process through whch we have just gone defnes the category of R-modules. We defned a collecton of objects, and then the maps (category theoretcally, morphsms) between them. Ths then gves us for free a sense of same-ness. For ths reason, ths s the general procedure usually followed by defntons of new thngs. Defnton 3.5. If M s an R-module, a submodule of M s a subgroup N M whch s also closed under the R-acton. That s, t satsfes: r x N, r R, x N. Example 3.6 (R = k, a feld). k-modules are vector spaces over k. Example 3.7 (R = Z). Z-modules are abelan groups. Example 3.8 (R = k[x], k a feld). Let M be a k[x]-module. Then snce k k[x], M s also a k-module, and accordngly a vector space over k. Note that ths dentfcaton means that the acton of α k k[x] s necessarly scalar multplcaton by α. Furthermore, the acton map (µ n Remark 3.2) must assgn x to some lnear operator T End(M). Ths choce, together wth the module axoms, determnes the entre acton, snce we necessarly have µ( n k x ) = =0 n k T. Ths means that each k[x]-module can be assocated wth a k-vector space, together wth a prvleged lnear transformaton. The reversblty of the assocaton allows us to dentfy the two colletons, that of k[x]-modules and that of pars (V, T ), where V s a k-vector space and T End(V ). =0
LECTURE V 5 4. Operatons on Modules Defnton 4.1. Let M be an R-module, wth some submodules {M }. Then defne the sum of submodules, M, by { } M = m m M, and the ntersecton of submodules, M, as M = {m m M }. M and M are both contaned n M as submodules, the closure n each case beng gven by the closure of each M. Defnton 4.2. Let M be an R-module, wth submodule N. Then the module quotent M N s defned by takng the group quotent and defnng the acton by r (m + N) = (r m) + N. From ths defnton, we have the usual somorphsm theorems, whch are proven n the same way as they are for groups and rngs. Theorem 4.3. 1. Let φ : M N be an R-module homomorphsm. Then M ker(φ) = m(φ). 2. Let M 1 and M 2 be submodules of M. Then (M 1 + M 2 ) M1 = M 2 M1 M 2. 3. Let L M N be R-modules. Then ( L N ) ( M N ) = L M. Defnton 4.4. Let M be an R-module, and I < R an deal. Then we defne IM by { } IM = r m r I, m M. Exercse 4.5. IM s a submodule of M. Proof. To show IM s a subgroup, take r, s IM. Then, there exsts r, s I so that r = r m, s = s m and because I s a subgroup of R, r s I for every. Ths means that r s = r m s m = (r s )m IM.
6 EDWIN SPARK To see IM s closed under the acton, take r R and s m IM. Then r ( s m ) = r (s m ) = (rs ) m whch s n IM, because I s closed under multplcaton by all of R. Notce that we needed the entre defnton of I as deal n the proof, so n order for IM to be a submodule, I must be an deal of R. Defnton 4.6. Let M be an R-module. Then the annhlator of M s Ann(M) = {r R r m = 0 m M}. Exercse 4.7. Ann(M) s an deal of R. Proof. Let r, r Ann(M). Then for any m M, (r r ) m = r m r m = 0 0 = 0, so (r r ) Ann(M) and the annhlator s a subgroup. Furthermore, for any s R, (sr) m = s (r m) = s 0 = 0, so (sr) Ann(R). Ths ensures closure under multplcaton by any element of R, so Ann(M) s an deal of R.