MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 35


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1 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES Modules over PID This week we re proving the fundmentl theorem for finitely generted modules over PID, nmely tht they re ll direct sums of cyclic modules. The proof will be in stges. On the first dy I decomposed module into torsion nd torsionfree prt. The presenttion ws little disorgnized so tht the steps do not follow one fter the other but rther the other wy round, i.e., to prove (1) we need to prove (2) nd to prove (2) we need to prove (3), etc. I cll this the motivtionl order. At the end we will go over the lemms nd put them in correct logicl order torsion nd torsionfree. Suppose tht R is PID nd M is finitely generted Rmodule. The min exmple I tlked bout ws R = Z in which cse M = G is f.g. belin group. Definition 9.1. M is torsionfree if nn(x) = 0 for ll x 0 in M. Definition 9.2. M is torsion if nn(x) 0 for ll x 0 in M. For exmple, R itself is torsionfree nd R/() is torsion. In the cse R = Z, Z n nd Q re torsionfree dditive groups. However, Q is not finitely generted. The finitely generted torsion belin groups re exctly the finite belin groups. The first decomposition theorem is the following. Theorem 9.3. Every f.g. module over PID is direct summnd of torsion module nd torsionfree module: M = tm fm where tm is torsion nd fm is torsion free. The second theorem tells us wht the torsionfree prt looks like: Theorem 9.4. A f.g. module over PID is torsionfree if nd only if it is free: fm = R n torsion submodule. I used two lemms to show tht the second theorem implies the first theorem. During the clss we decided tht these two lemms hold over ny domin. First I need definition. Definition 9.5. Suppose tht M is module over domin R. Then the torsion submodule of M is defined to be the set of ll elements of M with nonzero nnihiltor idel: tm := {x M nn(x) 0}
2 36 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES Lemm 9.6. tm is submodule of M provided tht R is domin. Proof. We need to show tht tm contins 0 nd is closed under ddition nd sclr multipliction. (1) 0 tm since nn(0) = R. (2) If x, y tm then there re nonzero elements, b R so tht x = by = 0. Then b(x + y) = 0. So, x + y tm. (3) If x tm nd r R then nn(rx) nn(x) is nonzero. In the second step we need to know tht b 0. Lemm 9.7. The quotient M/tM is torsionfree provided tht R is domin. Proof. Suppose not. Then there is nonzero element x + tm in M/tM nd 0 in R so tht x + tm is zero, i.e., x tm. But this mens there is nonzero element b R so tht bx = 0. But b 0. So, x tm which is contrdiction. It ws in this proof tht I mentioned the frction nottion: (tm : x) := { R x tm}. Next, I showed tht the second theorem (Theorem 9.4) implies the first (Theorem 9.3). Proof of Theorem 9.4. Let fm = M/tM. Since this is free, there is section s : fm M of the projection mp M M/tM. Then j s : tm fm M is n isomorphism where j : tm M is the inclusion mp. Next, we need to prove tht f.g. torsionfree modules re free nd tht f.g. torsion modules re direct sums of cyclic modules. I intend to use purity to do both pure submodules. Definition 9.8. We sy tht submodule N M is pure if whenever x M nd R with x N there exists z N so tht z = x. In other words: If n element of N is divisible by R in M then it is divisible by in N. The point is tht pure submodules re direct summnds in f.g. modules over PID s. However, using this fct is little tricky s we sw the next dy.
3 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 37 On the second dy I pointed out tht if P is pure submodule of f.g. module M over PID, then P is direct summnd. However, we cnnot use this fct to prove the min theorem becuse it uses the min theorem, nmely tht f.g. modules re direct sums of cyclic modules. However, we use the min theorem on module with fewer number of genertors thn M. So, this cn be used in n inductive proof of the min theorem. The theorem is: Theorem 9.9. Suppose tht P is pure submodule of f.g. module M over PID R nd M/P is direct sum of cyclic modules. Then P is direct summnd of M. Proof. Suppose tht M/P is direct summnd of cyclic modules. Then ech summnd is generted by some element x i + P with nnihiltor ( i ). This mens tht (x i : P ) = ( i ). In other words, i x i P. Since P is pure, there is n element z i P so tht i z i = i x i. But then x i + P = (x i z i ) + P nd i (x i z i ) = 0. So s i (x i +P ) = x i z i gives lifting s i : R/( i ) M of the direct summnd R/( i ) of M/P to M nd, together, these give n isomorphism ( si ) j : R/( i ) P M where j : P M is the inclusion mp. To find the pure submodule, I need to tke mximl cyclic submodule of M. This exists becuse M is Noetherin. So tht is next submodules of free modules. Theorem Suppose tht M = R n is free Rmodule with n genertors where R is PID. Then ny submodule of M is free with n or fewer genertors. Proof. This is by induction on n. If n = 1 then M = R nd the submodules re either R or n idel Rx. But R is domin. So, either x = 0 or nn(x) = 0. So, Rx is free with 0 or 1 genertor. Now suppose tht n 2 nd N is submodule of R n. Then we wnt to show tht N = R m where m n. Let e 1,, e n be the free genertors of M = R n. Let R n 1 denote the free submodule of M generted by e 1,, e n 1. Then, by induction on n, we hve: N R n 1 = R m 1
4 38 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES where m n. If N = N R n 1 we re done. Otherwise, let J be the set of ll e n coordintes of ll elements of N. I.e., { } n 1 J = R ( x N) x = e n + i e i We get t lest one nonzero element in J since N is not contined in R n 1. Then it is esy to see tht J is n idel (or ll of R) since it is closed under ddition nd sclr multipliction nd is nonempty. Therefore, J = (b) is generted by one element b 0. (So, every J hs the form = rb where r R is unique.) By definition, there is n element x 0 N so tht n 1 x 0 = be n + i e i. This cn be used to define homomorphism φ : (N R n 1 ) R N by φ(y, r) = y + rx 0. I clim tht φ is n isomorphism. To see tht φ is onto, tke ny element x N. Then x = e n + (some element of R n 1 ) where = rb. So, x rx 0 N R n 1 nd x = φ(x rx 0, r). To see tht φ is 11 suppose tht φ(y, r) = 0. Then y = rx 0. Looking t the lst coordinte this gives rb = 0 r = 0 y = 0. Therefore φ is n isomorphism. Its inverse gives: N = (N R n 1 ) R = R m 1 R = R m where m n. Corollry If M is n Rmodule generted by n elements then every submodule of M is generted by n elements. In prticulr, M is Noetherin. Proof. If M is generted by x 1,, x n then we hve n epimorphism φ : R n M given by φ( 1,, n ) = i x i. If N M then N is quotient of φ 1 N which is free on n genertors by the theorem.
5 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES mximl cyclic submodules. Since f.g. modules re Noetherin, they hve mximl cyclic submodules. This is becuse ny sequence of cyclic submodules: Rx 1 Rx 2 Rx 3 M must eventully stop. (If there were no mximl cyclic submodule, I could keep going forever.) Lemm Suppose tht M is torsionfree nd Rx M is mximl cyclic submodule. Then x / M for ny nonunit R. (I.e., x = z is unit.) Proof. Suppose tht x = z. Then Rx Rz. Since Rx is mximl cyclic, Rx = Rz which implies tht z = bx for some b R. So, x = z = bx which implies (b 1)x = 0. Since M is torsionfree, this implies tht b = 1, i.e., is unit. Lemm Every mximl cyclic submodule of torsionfree module is pure. Proof. Suppose tht Rx M is mximl cyclic submodule. Suppose there re elements y M, R so tht y Rx. Then y = bx. We need to show tht this is times n element of Rx. In other words, we wnt to show tht b ( divides b). This is the sme s sying tht b R. At this point I explined this equivlent formultion of divisibility. Since R is domin, it is contined in its field of frctions: R Q(R). The frction b is n element of Q(R). If b, then b = r for some r R nd b = r 1 R nd conversely. So, b b R. Let c R be the gretest common divisor of, b. I.e., (, b) = (c). Then c nd c b. I.e.,, b R nd c c c y = b c x since M is torsionfree. (c times the difference is zero. So, the difference is zero.) But (, b ) = (1), i.e., there exist s, t R so tht c c This implies tht 1 = c s + b c t. x = c sx + b c tx = c sx + c ty = (sx + ty) c
6 40 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES since bx = y. By the previous lemm, this implies tht c i.e., c R. So b = b c c R s desired. We need one more lemm. is unit, Lemm If P is pure submodule of torsionfree module M then M/P is torsionfree. Proof. Suppose not. Then there is nonzero element x + P M/P so tht (x + P ) = 0 + P. I.e., x P. But P is pure. So, there is z P so tht z = x. Then (x z) = 0 in M which implies tht x = z since M is torsion free. So, x + P = z + P = P is the zero element of M/P which is contrdiction. Now I m redy to prove Theorem 9.4: Finitely generted torsionfree modules over PID s re free, completing the proof tht M = tm fm = tm R n. Proof of Theorem 9.4. The proof is by induction on the number of genertors. If M hs one genertor x then M = Rx = R is free. So, suppose M hs n genertors x 1,, x n where n 2. Since Rx 1 is cyclic submodule of M it is contined in mximl cyclic submodule Rx. By Lemm 9.13, Rx is pure. By Lemm 9.14, this implies tht M/Rx is torsionfree. But M/Rx is generted by n 1 elements (the imges of the genertors x 2,, x n ). So, it is free, sy, M/Rx = R m. But this implies tht is lso free. M = Rx R m = R R m It remins to show tht the torsion submodule tm is lso direct sum of cyclic modules. I wnt to do the sme proof, nmely, since M is Noetherin, we cn find mximl cyclic submodule Rx. Since M/Rx will be generted by n 1 elements, it is sum of cyclic modules. After some stumbling, I decided I need more precise construction of mximl cyclic submodule. I took genertor with miniml nnihiltor. And this works the best for pprimry modules.
7 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES pprimry decomposition. pprimry modules re generliztions of belin pgroups. We showed tht finite nilpotent groups re products of their psylow subgroups. For finite belin groups nd, more generlly, for torsion modules over PID s, this is very esy to prove. It follows from the unique fctoriztion theorem. (I.e., every PID is UFD.) Definition An element x M is clled pprimry if it hs nnihiltor nn(x) = (p n ) for some n 0 where p R is irreducible. A module M is clled p primry if every element is pprimry. (Note tht 0 M is pprimry for every p.) I should hve prove the following lemm first: Lemm If p n x = 0 where n 0 then x is pprimry. Proof. Suppose nn(x) = (). Then p n x = 0 implies tht p n = b for some b R. By unique fctoriztion this implies tht = up k where u is unit in R nd k n. But then nn(x) = (up k ) = (p k ). For modules over PID, the Sylow theorems re very esy to prove: Lemm The set of pprimry elements of ny module M over PID is submodule. Definition If p R is irreducible, let M p be the submodule of M consisting of ll pprimry elements of M. Proof. If x, y re nonzero pprimry elements of M then p n x = 0 nd p m y = 0. Then p n rx = 0 for ny r R nd p n+m (x+y) = 0. Therefore, rx nd x + y re pprimry by the lemm. Theorem Every torsion module M over PID is direct sum of pprimry modules: M = M p. p Proof. First choose irreducible elements p i so tht every irreducible element of R is up i for some unit u nd some unique p i. By the lemm, for ech of these irreducibles we hve submodule M pi M. By the universl property this gives homomorphism φ : M pi M I clim tht this is n isomorphism.
8 42 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES It follows from unique fctoriztion tht this mp is onto: Suppose x 0 M with nnihiltor nn(x) = (). Since M is torsion, 0. So = u where u is unit. Then, for ech i, i k i = up n 1 1 i p n k k R. There is no irreducible element of R which divides ech of these elements. This implies tht there re elements r i R so tht Apply this to x to get: where x = 1 = k r i k i r i x i = x = i i x. k r i x i is p i primry since i x i = x = 0. This shows tht x = φ(r i x i ) i is in the imge of φ nd therefore φ is onto. To show tht φ is 11 suppose tht (x i ) i is in the kernel of φ. Then k x i = 0 where x i is p i primry. Thus sum is finite since the direct sum is equl to the wek product. Suppose tht nn(x i ) = ( i ). Then the product p n 2 2 p n k k nnihiltes x 2,, x k nd therefore nnihiltes their sum x x k = x 1 This implies tht p n 2 2 p n k k (p n 1 1 ), i.e., p n 2 2 p n k k = p n 1 1 for some R. By unique fctoriztion this is only possible if n 1 = 0, i.e., x 1 = 0. The sme rgument shows tht x i = 0 for ll i. So, φ is monomorphism nd thus n isomorphism s climed.
9 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES decomposition of pprimry modules. Now we come to the finl step in the proof of the min theorem, nmely, I will prove tht every f.g. pprimry module is direct sum of cyclic modules. But first some lemms. Lemm Any quotient of pprimry module is pprimry. Proof. This is obvious. If x M then p n x = 0 for some n. But then p n (x + N) = 0 in M/N. So, M/N is pprimry. Lemm Suppose M is f.g. pprimry module nd R so tht p /. Then, there is b R so tht multipliction by b is the identity on M. Proof. Let x 1,, x k be genertors for M nd suppose tht nn(x i ) =. Let n = mx(n i ). Then p n M = 0. Since / (p), (, p n ) = (1). So, there exist b, c R so tht b + p n c = 1 But p n = 0 on M. So, b = 1 on M. Theorem If M is f.g. pprimry module over PID R then M is direct sum of cyclic pprimry modules. Proof. Let x 1,, x k be genertors for M where k is miniml. Then we will show by induction on k tht M is direct sum of k or fewer cyclic summnds. Since M/Rx 1 is generted by k 1 elements, it is direct sum of k 1cyclic pprimry modules by induction. So, it suffices to show tht Rx 1 is pure submodule of M. Let nn(x i ) = nd let n = mx(n i ). We will ssume tht n = n 1. Then p n M = 0 nd p n 1 x 1 0. Clim Rx 1 is pure submodule of M. proof : Suppose tht y M nd R so tht y Rx 1. If y = 0 then y = 0 Rx 1. So, we my ssume tht y 0. Then y = bx 1 for some b R. Write, b s = p k s, b = p m t where s, t re not divisible by p. The condition y = p m tx 1 0 mens tht m < n. So, n m 1 0. By the lemm, there is n element r R so tht sr = 1 on M. This gives: (9.1) y = p k sy = p m tx 1 = p m srtx 1 So, it suffices to show tht k m since this would give y = (p m k rtx 1 ). To prove tht k m multiply both sides of Eqution (9.1) by p n m 1.
10 44 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES This gives p k+(n m 1) sy = p m+(n m 1) tx 1 = p n 1 tx 1 0 since p n 1 x 1 0 nd multipliction by t is n isomorphism on M. But this implies tht the left hnd side is lso nonzero. Since p n M = 0 this implies tht k + n m 1 < n In other words, k < m + 1 or k m. This proves the clim nd the theorem.
11 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES Structure theorem for f.g. modules over PID. Now we hve the complete proof of the following existence theorem. Theorem Every f.g. module over PID is direct sum of cyclic primry modules M = R r R/( i ). Proof. Here is n outline of the entire proof. (1) First, we defined the torsion submodule tm of M. This exists since R is domin. (2) The quotient M/tM is torsionfree. Agin this holds over ny domin. (3) Since R is PID, every submodule of f.g. free module is f.g. free. This implies: () M is Noetherin (ll submodules re finitely generted) nd (b) Every f.g. torsionfree module is free. (4) The conclusion ws tht M is direct sum of tm nd f.g. free module: M = tm R r (5) If p R is irreducible, the set of pprimry elements of ny module forms submodule M p. (6) It follows from unique fctoriztion tht every torsion module is direct sum of pprimry modules: tm = M pi (7) In pprimry module M, genertor with miniml nnihiltor genertes pure cyclic submodule. Then we use the lemm: (8) If N is pure submodule of M nd M/N is direct sum of cyclic modules then N is direct summnd of M. (9) Every f.g. pprimry module is direct sum of cyclic pprimry modules. The structure theorem for finitely generted modules over PID lso sys the decomposition is unique: Theorem For ny f.g. module M over PID, the numbers r nd the sequence of pirs (p i, n i ) re uniquely determined up to permuttion of indices. We will prove this lter using tensor products. The number r = rk(m) is clled the rnk of M. It is the dimension of Q(R) M.
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