FINITELY-GENERATED MODULES OVER A PRINCIPAL IDEAL DOMAIN

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1 FINITELY-GENERTED MODULES OVER PRINCIPL IDEL DOMIN EMMNUEL KOWLSKI Throughout ths note, s a prncpal deal doman. We recall the classfcaton theorem: Theorem 1. Let M be a fntely-generated -module. (1) There exsts an nteger n 0 and an somorphsm M n M tors where M tors = m M am = 0 for some a 0} s the torson submodule of M. (2) There exst m 0 and rreducble elements r 1,..., r m, such that the deals r are parwse coprme, M tors (r ) 0 and m M tors = M tors (r ) where we denote N(r) = n N r k n = 0 for some k 0} the r-prmary submodule of any -module N, for any rreducble element r. (3) For each, there exst s 1 and a sequence and an somorphsm M tors (r ) = M(r ) =1 1 ν,1 ν,s 1 s /r ν,. We gve here the proof, fnshng the mssng steps from the lecture. Unless ndcated otherwse, all modules below are over, all lnear maps are -lnear, and all somorphsms are somorphsms of -modules. Proposton 2. (1) Let M be a free module of rank n 0. Then any submodule of M s free of rank n. (2) Let M be a fntely-generated module, wth a generatng set contanng n elements. Then any submodule of M s fntely generated and has a generatng set wth m n elements. Proof. (1) Let x = (x 1,..., x n ) be a bass of M, and let N M be a submodule of M. 1 n, we denote M = x M, N = N M N, 1 whch are submodules of M and N respectvely. We have 0 M 1 M n = M, 0 N 1 N n = N, and M s free of rank wth bass (x 1,..., x ). We denote also λ : M For Date: November 24, 2014, 8:23. 1

2 the -th coordnate -lnear map wth respect to the bass x. We wll show by nducton on, 1 n, that N s free of rank. It wll follow that N n = N s free of rank n. Frst for = 1, we have N 1 = N x 1. The restrcton of λ 1 to N 1 s an -lnear map l 1 : N 1. It s nectve, because λ 1 s an somorphsm M 1, and hence nduces an somorphsm N 1 Im(l1 ). The mage of l 1 s an deal n, and snce s prncpal, there exsts a 1 such that Im(l 1 ) = a 1. Ths s a free module, ether somorphc to f a 1 0, or to 0} f a 1 = 0. In any case, N 1 s a free module of rank 1. Now assume that N 1 s free of rank 1. We study N by consderng the restrcton of λ to N. Its mage λ (N ) s agan an deal n, of the form λ (N ) = a for some a. If a = 0, then λ (N ) = 0}, whch means that any element of N has zero -th coordnate, or n other words that N = N 1. Then N s free of rank 1 by nducton. Suppose on the other hand that a 0. Let then y N be such that λ (y ) = a. We defne a lnear map N 1 N T 1 (x, a) x + ay, and we clam that ths s an somorphsm, whch wll prove that N s free of rank, by nducton agan. Let (x, a) ker(t 1 ) N 1. We have x+ay = 0, hence we get 0 = λ (x) = aλ (y ) = aa, and hence a = 0 (snce s an ntegral doman and a 0). Then x = ay = 0, and therefore ker(t 1 ) = 0}, so T 1 s nectve. Fnally, T 1 s surectve, snce for any x N, we have λ (x) = ba for some b (by defnton of a ), and therefore λ (x by ) = ba ba = 0, hence x = x by N ker(λ ) = N 1, showng that x = T 1 (x, b). (2) Let M be a fntely generated -module and N a submodule. If (x 1,..., x n ) s a generatng set of M, then we have a surectve -lnear map n M T 2 (a ) a x. Let N = T2 1 (N) n. By (1), ths s a free module of rank m n. But then f (y 1,..., y m ) s any bass of N, the elements (T (y 1 ),..., T (y m )) generate T 2 (N ) = N (we have T 2 (N ) = N because T 2 s surectve: any x N s of the form T 2 (x ) for some x n, but then x T2 1 (N) = N ). So N s fntely generated. The next step s: Proposton 3. (1) Let M be a fntely generated -module. Then M s free f and only f M s torson-free,.e, f and only f M tors = 0}. (2) The module M/M tors s free of fnte rank n 0, and there exsts an somorphsm M M tors n. Proof. (1) If M s free then M tors = 0}: ndeed, f (x ) s a bass and x M tors, then we can wrte a = a x 2

3 for some a n. If a 0 satsfes ax = 0, we get ax = (aa )x = 0 whch mples that aa = 0 for all, and hence that a = 0 for all, and therefore that x = 0. Conversely, we assume that M tors = 0}, and also that M 0} (snce that other case s obvous). We wll fnd a lnear map from M to a free module, so that Proposton 2 (1) wll mply that M s somorphc to a free module. We let (x 1,..., x n ) be a generatng set of M, wth n 1. We let (y 1,..., y m ), wth m n, be a famly of elements taken from x 1,..., x n } wth m maxmal such that (y 1,..., y m ) s -lnearly ndependent. That such a famly exsts wth m 1 follows from the assumpton that M tors = 0}, snce for any x 0 (whch must exst snce M 0}), the one-element famly (x ) s lnearly ndependent. The submodule N generated by (y 1,..., y m ) s free of rank m n. Let I = a ax N for all x M}. Ths s agan an deal n. We clam that t s non-zero. If ths s so, then for any non-zero a I, we get an nectve lnear map M N T 3 x ax (nectve because a 0, hence ker(t 3 ) M tors = 0}) and therefore M s free by Proposton 2 (2), beng somorphc to the submodule Im(T 3 ) of the fntely generated free module N. To prove the clam, t suffces to show that for any, there exsts a non-zero a wth a x N, snce then a = a 1 a n wll have the desred property. But ths s the case because ether x s one of the bass elements (y 1,..., y m ), and then x N, or else the famly (y 1,..., y m, x ) must be lnearly dependent, by defnton of m (snce t contans more than m elements from (x 1,..., x n )). In that case there exst a, b 1,..., b m, not all zero, wth a x + b y = 0, hence a x N, and a 0 (snce a = 0 would mply mply that all b vansh, n vew of the lnear ndependence of (y 1,..., y m ), contradctng the fact that not all coeffcents are zero). (2) It s easy to see that N = M/M tors has zero torson submodule: f x N tors, then there s a 0 n such that ax = 0; wrtng x as the class of some y M, ths means that ay M tors, and hence there exsts b 0 wth bay = 0. Snce ab 0, we get y M tors, whch means x = 0 n N. Hence, by (1), the module N s a free module. Now consder the proecton π : M N. Pck a bass (x 1,..., x n ) of N, and let y M be any element wth π(y ) = x. We defne a lnear map M tors n M T 4 (t, (a )) t + a y, and we clam that t s an somorphsm, whch wll fnsh the proof. Frst, T 4 s nectve because f (t, (a )) ker(t 4 ), we get a y = t M tors 3

4 hence (applyng π) we derve a x = 0, so that a = 0 for all (snce (x 1,..., x n ) s a bass of N) and then also t = 0. Next, T 4 s surectve, because f x M, and (a ) s defned by we have π(x) = a y, ( π(x) = π a x ), whch means that t = x a x ker π = M tors, and hence x = T 4 (t, (a )) Im(T 4 ). The next step determnes the decomposton of M tors n prmary parts. Proposton 4. Let M be a fntely generated module wth M tors 0}. There exsts m 1 and there exst fntely many rreducble elements r, 1 m, generatng parwse dstnct deals, such that (1) M tors = and M(r ) 0} for 1 m. M(r ) Note that snce M(r ) M tors, we have M(r ) = M tors (r ), and we could also wrte M tors = M tors (r ). Proof. We frst descrbe the set of rreducbles, as those occurng n the factorzaton of some nonzero element a. We wll then prove that they satsfy the desred decomposton. We let I = a ax = 0 for all x M tors }. Ths s once more an deal n. We clam that I s non-zero, and I. Indeed, to fnd a non-zero element n I, we observe that t s enough (by lnearty) to fnd a 0 such that ax = 0 for all elements x n a generatng set of M tors. By Proposton 2 (2), there exsts a fnte such generatng set x 1,..., x n }, and by defnton of M tors, for any such x M tors, there s some a 0 for whch a x = 0. Then a = a 1 a n s a non-zero element of I. To see that I, note that f 1, we would get x = 1 x = 0 for all x M tors, so that M tors = 0}, whch we excluded. Snce s a prncpal deal doman, there exsts consequently some non-zero a wth I = a, and snce I, ths a s not n. Then by exstence of factorzaton, we can wrte a = r n 1 1 rnm m for some m 1, where the r are parwse coprme rreducble elements of. We wll show that these rreducble elements satsfy (1). We frst prove that M tors s spanned by the submodules M(r ). Snce the r are parwse coprme, we have (2) a = 4 r n,

5 and the Chnese Remander Theorem gves an somorphsm of rngs /a f /r n. Let g denote the proecton /a. Snce f s an somorphsm and g s surectve, there exst elements e for 1 m such that f g(e ) = (0,..., 0, 1, 0,...), where the 1 s at the -th poston. The sum e = e e m then satsfes for some α, snce e 1 ker(f g) = a. Now for any x M tors, we have e = 1 + αa x = 1 x = (e αa)x = ex = e 1 x + + e n x, (snce ax = 0 by defnton of I). Furthermore, we clam that r n e a for all, from whch t follows that r n e x = 0, and hence that e x M(r ). Hence x = e 1 x + + e n x and we deduce that (3) M tors = 1 m M(r ). M(r ), To check the clam, we use (2). Note frst that r n e r n. Then, from the defnton of e, we see that e r n for all, and hence we conclude that r n e r n = a. There only remans to prove that the decomposton (3) s a drect sum decomposton. Ths means that, for any gven, we must show that M(r ) M(r ) = 0}. But f x belongs to ths ntersecton, on the one hand there exsts k 0 such that r k x = 0, and on the other hand, wrtng x = x wth x M(r ), we see that there exsts k 0, for, such that r k x = 0, and therefore ( ) x = 0. Thus the deal contans the deal r k as well as the deal ( r k I x = a ax = 0} ) r k. Snce these are coprme (because ther generators nvolve dfferent rreducble elements), we have I x =, and therefore 1 I x, whch means that x = 0. 5

6 The fnal step s the decomposton of M(r ). In fact, we state ths step for a arbtrary fntely generated -module M whch s p-prmary for some rreducble element p,.e., such that M(p) = M. Ths fnshes the proof of the classfcaton theorem because M(p) always has ths property, for any fntely-generated module M. Proposton 5. Let p be an rreducble element and M a fntely-generated module such that M(p) = M. There exst s 0, a sequence of ntegers, and an somorphsm 0 ν 1 ν s (4) M /p ν 1 /p νs. Proof. We frst observe that the -module V = M/pM s n a natural way a vector space over the feld K = /p, wth multplcaton nduced by the -module structure (.e., for λ K and v V, we put λv = ax for arbtrary a wth class λ n K, and arbtrary x M wth class v n V ; one checks by a smple computaton that ax s ndependent of the choces of a and x). The mage n V of a generatng set of M generates V, frst as an -module and then (from the defnton) as a K-vector space. It follows that the dmenson dm K V s fnte. We wll prove the proposton by nducton on ths nteger dm K V. Frst, for any x M, we denote I x = a ax = 0}, whch s an deal n. It s generated by a power p ν(x) of p, snce I x p k for some k 0 by defnton, whch means that a generator must dvde p k. Let ν = maxν(x) x M}. Ths maxmum s well-defned, because f (x 1,..., x n ) s a generatng set of M, we have I x p max(ν(x)) for all x M, from whch (by lnearty) we get ν(x) max(ν(x ))). We now begn the nducton. Let s = dm K V. If s = 0, we obtan M = pm, and therefore M = p k M for all k 1. But then takng k = ν, t follows that M = 0}, and the statement s then obvous (wth s = 0). We now assume that s = dm K V 1, and that the proposton holds for any fntely generated p-prmary module N where dm K N/pN s 1. We then frst note that ν 1: otherwse, we would get ν(x) = 0 for all x, and ths means that M = 0}, contradctng that s 1. Let y 1 M be an element wth ν(y 1 ) = ν, and let N = M/y 1 be the quotent of M by the submodule generated by y 1. We denote by π the proecton π : M N. We wll show that N has a decomposton (4) wth s 1 summands, and that M s somorphc to /p ν N, whch wll gve the decomposton clamed for M. We frst check that N s p-prmary and satsfes dm K N/pN s 1, so that we can apply the nducton assumpton to N. The frst part s easy: for any x N, wth x = π(y) for some y M, we deduce p ν x = π(π ν y) = 0. For the second, we observe that the composton M π N N/pN s a lnear map wth kernel contanng pm, hence nduces a lnear map S : M/pM N/pN. When both sdes are seen as K-vector spaces, ths map s K-lnear (ust from the defnton of the vector space structure). It s surectve, snce S s obtaned by passng to a quotent (whch does not change the mage) a composte of surectve maps. Hence s = dm K M/pM dm K N/pN. 6

7 On the other hand, we clam that the mage v 1 of y 1 n M/pM s a non-zero element n the kernel of S, whch has therefore dmenson > 0. It then follows that dm K M/pM > dm K N/pN by lnear algebra. To see the clam, note that t s certanly true that S(v 1 ) = 0, snce π(y 1 ) = 0. nd v 1 0, because v 1 = 0 would mean that y 1 pm, and n that case we would have y 1 = py 2 for some y 2 M, so that p ν y 2 = p ν 1 y 1 0, showng that I y2 s contaned n I y1 and dfferent from I y1, or equvalently that ν(y 2 ) > ν(y 1 ) = ν, whch contradcts the defnton of ν. Thus we may apply the nducton assumpton to N and deduce that there s an somorphsm N U 2 m /p µ where m s (t s convenent here to start the ndex at 2) and µ 2 µ m. The next step wll be to show that there exsts N M such that the restrcton of the proecton π to N gves an somorphsm N N. For ths purpose, for 2 m, we let x = U 1 (0,..., 0, 1, 0,..., 0), where the 1 s n the -th poston. We have I x = p µ for all. We wll now show that there exst elements y M such that π(y ) = x, wth the addtonal property that I y = I x for all. 1 In fact, ths property holds for any x N. To see ths, let µ = ν(x). Let z M be frst any element such that π(z) = x, and let κ = ν(z) 0. Snce p κ x = π(p κ z) = 0, t follows that κ µ. Of course, we also have κ ν by defnton of ν. The possbltes for y wth π(y) = x are gven by y = z + by 1 for some b. Snce any such y has ν(y) µ, t s enough to fnd b such that p µ y = 0 to conclude that I y = I x. We have p µ y = p µ z + p µ by 1. Snce p µ x = π(p µ z) = 0, t follows that p µ z = cy 1 for some c. Further, snce κ µ, we can multply both sdes by p κ µ to deduce that p κ µ cy 1 = 0. Hence p κ µ c I y1, so p κ µ c = p ν d for some d. Ths means that c = p µ+ν κ d, and snce κ ν, that c p µ, say c = p µ d. Therefore p µ y = p µ (d + b)y 1, and hence takng b = d, we fnd the requred element y = z d y 1 wth p µ y = 0. Now, the map /p µ M T 5 2 m (a ) a y s then well-defned, where each a s an element of mappng to a n /p µ. Indeed, f a s replaced by a = a + pµ b for some b, then a y = a y + bp µ y = a y. The map T 5 s then lnear, and furthermore t s nectve: f a y = 0, we obtan (a ) = 0 2 m /p µ 1 Note that the exstence of such elements s a necessary condton to the exstence of a submodule N M as above. 7

8 by applyng U π, whch means (because of the somorphsm U) that a = pµ b for some b, n whch case a = 0 n /p µ for all. Let N M be the mage of T 5. The composton U 1 T5 1 (where T5 1 s defned on N ) s therefore an somorphsm N N. In fact, ths s ust the proecton π restrcted to N : y N s mapped frst by T5 1 to (a ) such that a y = y, and then to U 1 ((a )) = a x = π(y). We can now conclude by checkng that the lnear map N /p ν M (x, a) x + ay 1 s an somorphsm. Frst note that t s well-defned because f a and a are two elements of mappng to a modulo p ν, we get a a = p ν b for some b, and hence a y 1 = a y 1 + p ν by 1 = a y 1. The map s surectve because for any y M, we can fnd y N wth π(y ) = π(y), and then y y ker(π) = y 1. It s fnally also nectve, because f x + ay 1 = 0, then we get π(x) = 0, whch means x = 0 snce π : N N s an somorphsm, and then ay 1 = 0 means that a = 0 n /p ν. We conclude by establshng the unqueness propertes of the classfcaton theorem. Theorem 6. Let M be a fntely-generated -module. Wth notaton as n Theorem 1, the followng unqueness statements hold: (1) The nteger n depends only on M up to somorphsm; (2) The set of deals r } generated by the rreducble elements r depends only on M up to somorphsm; (3) For each rreducble r, the nteger s 0 and the non-decreasng sequence such that depends only on M up to somorphsm. M(r) 1 ν 1 ν s Proof. It suffces to take a module M of the form (5) M = n 1 s /r ν, /r ν, 1 m 1 s wth s 0, 1 ν,1 ν,s and to show how to recover from M the nvarants n, m, the rreducbles r and the sequences (ν, ), usng only defntons that wll extend to an somorphc module. We wll denote by π the proecton M n, and by π, : M /r ν, the proectons on the correspondng factors. 8

9 (1) In (5), the nteger n s the largest non-negatve nteger such that the module M contans n elements whch are -lnearly ndependent. Indeed, t s clear that M contans n such elements (a bass of n for nstance), so t suffces to prove that n + 1 elements (x 1,..., x n+1 ) n M are necessarly lnearly dependent. For ths, take frst a dependency relaton a 1 π(x 1 ) + a n+1 π(x n+1 ) = 0 wth a n the fracton feld K of, wth not all a zero: ths exsts because we can see (x 1,..., x n+1 ) as elements of the vector space K n, and n+1 elements of a vector space are lnearly dependent. Now wrte a = b /c wth c 0. Multplyng both sdes of the dependency relaton wth c = c 1 c n+1, we obtan a relaton ca 1 π(x 1 ) + + ca n+1 π(x n+1 ) = 0 where each ca belongs to and not all are zero. Fnally consder a = r ν,s 0}, and note that aca 1 x aca n+1 x n+1 = 0 (ndeed, applyng π or π, to the left-hand sde gves zero for all and ). Snce not all coeffcents aca are zero, ths s a dependency relaton among (x 1,..., x n+1 ). (2) For any rreducble r n whch s not n the set r 1,..., r m } (whch may be empty s m = 0), the lnear map M M m r x rx s nectve. Indeed, ths s clear for the restrcton to the free part n, and f x /r ν, belongs to ker m r, we get I x r ν, + r = (snce r s coprme wth r ) so that x = 0. The nectvty of m r means n partcular that M(r) = 0}. The same argument shows that for any fxed, M(r ) has zero ntersecton wth the other prmary components, namely wth /r ν k, k. We then deduce that k M(r ) = 1 s /r ν, 0 (snce ν,1 1). So the set of deals generated by the rreducble elements r s the set of deals for whch M(r ) 0}. (3) Consder frst an rreducble element p and a module M = /p ν for some ν 1. The proecton map /p then passes through the quotent to defne a surectve -lnear map M /p (because ν 1 so that p ν s contaned n the kernel). The kernel of ths map, n turn, s pm, and hence t nduces an somorphsm M/pM /p. 9

10 pplyng ths to each component of a p-prmary module /p ν 1 s wth s 0 and 1 ν 1 ν s, we deduce that there s an -lnear somorphsm M/pM (/p) s. If we use the /p-vector space structure on M/pM descrbed n the proof of Proposton 5, ths map s (/p)-lnear, and therefore s an somorphsm of vector spaces. In partcular s = dm /p M/pM s unquely determned by M up to somorphsm. We fnally show how to determne the sequence (ν ) nductvely, by showng frst how to determne the ndex k 0 such that 1 ν 1 ν k < ν k+1 = = ν s (n other words, ν k+1,..., ν s are all equal to the largest exponent among the sequence (ν ); n partcular k = 0 f all exponents are equal). If we can do ths, then replacng M by /p ν k 1 k wll allow us to nductvely determne the next largest exponent ν k, etc. To determne k, we frst note that ν s s determned as the smallest nteger ν 0 such that p ν M = 0. If ν s = 1, then all exponents must be equal to 1, and we are done. Suppose that ν s 2. Then we consder the lnear map M M T 6 x p νs 1 x. Snce p νs 1 x = 0 for any x /p ν f ν < ν s, and snce the kernel of multplcaton by p νs 1 on /p νs s p/p νs, t follows that s ker T 6 = p/p νs. =k+1 By the same arguments as at the begnnng, we then deduce that (because ν s 2). s k = dm /p ker(t 6 )/p ker(t 6 ). ETH Zürch D-MTH, Rämstrasse 101, CH-8092 Zürch, Swtzerland E-mal address: kowalsk@math.ethz.ch 10

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