Algebra of linear recurrence relations in arbitrary characteristic

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1 Algebra of lnear recurrence relatons n arbtrary characterstc Nkola V. Ivanov Contents Preface 1. Dvded dervatves of polynomals 2. Sequences and dualty 3. Adjonts of the left shft and of dvded dervatves 4. Endomorphsms and ther egenvalues 5. Torson modules and a property of free modules 6. Polynomals and ther roots 7. The man theorems Note bblographque Bblographe c Nkola V. Ivanov, 2014, Nether the work reported n ths paper, nor ts preparaton were supported by any governmental or non-governmental agency, foundaton, or nsttuton. 1

2 Preface The goal of ths paper s to present an algebrac approach to the basc results of the theory of lnear recurrence relatons. Ths approach s based on the deas from the theory of representatons of one endomorphsms a specal case of whch may be better known to the reader as the theory of the Jordan normal form of matrces). The noton of the dvded dervatves, an analogue of the dvded powers, turned out to be crucal for provng the results n a natural way and n ther natural generalty. The fnal form of our methods was nfluenced by the the umbral calculus of G.-C. Rota. Nether the theory of representaton of one endomorphsm, nor the theory of dvded powers, nor the umbral calculus apply drectly to our stuaton. For each of these theores we need only a modfed verson of a fragment of t. Ths s one of the reasons for presentng all proofs from the scratch. Both these fragments and our modfcatons of them are completely elementary and beautful by themselves. Ths s another reason for presentng proofs ndependent of any advanced sources. Fnally, the theory of the lnear recurrence relaton s an essentally elementary theory, and as such t deserves a self-contaned exposton. The prerequstes for readng ths paper are rather modest. Only the famlarty wth the most basc notons of the abstract algebra, such as the notons of a commutatve rng, of a module over a commutatve rng, and of endomorphsms and homomorphsms are needed. No substantal results from the abstract algebra are used. A taste for the abstract algebra and a superfcal famlarty wth t should be suffcent for the readng of ths paper. The standard expostons of the theory of lnear recurrence relatons present ths theory over algebracally closed felds of characterstc 0, or even only over the feld of complex numbers. In contrast, such restrctons are very unnatural from our pont of vew. The methods of ths paper apply equally well to all commutatve rngs wth unt and wthout zero dvsors; no assumptons about the characterstc are needed. Of course, a form of the condton of beng algebracally closed s needed. We assume only that all roots of the characterstc polynomal of the lnear recurrence relaton n queston are contaned n the rng under consderaton ths can be done usng any of the standard approaches to the theory also). Structure of the paper. The man results are stated and proved n Secton 7, whch depends on all prevous ones. Sectons 1 6 are ndependent wth only one excepton: Secton 3 depends on both Secton 1 and 2. All references are relegated to the Note bblographque at the end. The reasons are the same as N. Bourbak s reasons for not ncludng any references wth the excepton of hs Notes hstorques. We denote by Z the rng of ntegers and by N the set of non-negatve ntegers. We denote by k a fxed entre rng,.e. a commutatve rng wth a unt wthout zero dvsors and such that ts unt s not equal to ts zero. There s a canoncal rng homomorphsm Z k takng 0, 1 Z to the zero and the unt of k respectvely, makng k nto a Z-algebra, and every k-module nto a Z-module. We dentfy 0, 1 Z wth ther mages n k. 2

3 1. Dvded dervatves of polynomals Polynomals n two varables. Let x be a varable, and let k[x] be the k -algebra of polynomals n x wth coeffcents n k. Let y be some other varable, and let k[x, y] be the k -algebra of polynomals n two varables x, y wth coeffcents n k. As s well known, k[x, y] s canoncally somorphc to the k -algebra k[x][y] of polynomals n y wth coeffcents n k[x]. We wll dentfy these two algebras. Ths allows us to wrte any polynomal fx, y) n two varables x, y n the form fx, y) = n = 0 g nx)y n. In fact, ths sum s obvously fnte. Equvalently, the polynomals g n x) are equal to 0 for all suffcent bg n N. The polynomals g n x) are unquely determned by fx, y). The defnton of the dvded dervatves. Let px) k[x]. Then px hence px y) has the form y) k[x, y], and 1) px y) = n = 0 δn p)x)y n, for some polynomals δ n p)x) unquely determned by px). The sum n 1) s actually fnte. Equvalently, δ n p)x) = 0 for all suffcent bg n N. The coeffcent δ n p)x) n front of y n n the sum n 1) s called the n-th dvded dervatve of the polynomal px). We wll also denote δ n p)x) by δ n px) ) or δ n px). Operators δ n. Let n N. By assgnng δ n px) k[x] to px) k[x] we get a map δ n : px) δ n px). Clearly, δ n s a k-lnear operator k[x] k[x]. After substtuton y = 0 the equaton 1) reduces to px) = δ 0 px). Therefore, δ 0 = d = d k[x] Theorem Lebnz formula). Let fx), gx) k[x], and let n N. Then δ n fx)gx)) = j = n δ fx)δ j gx). 3

4 Proof. By applyng 1) to px) = fx) and to px) = gx), we get fx y) = = 0 δ fx)y, gx y) = j = 0 δj gx)y j. By multplyng these two denttes, we get fx y)gx y) =,j = 0 δ fx)y δ j gx)y, and hence fx y)gx y) =,j = 0 δ fx) δ j gx)y j = ) n = 0 j = n δ fx) δ j gx) y n. The theorem follows Corollary Lebnz formula for δ 1 ). Let fx), gx) k[x]. Then δ 1 fx)gx)) = δ 1 fx)gx) + fx)δ 1 gx). In other terms, δ 1 s a dervaton of the rng k[x] Lemma. ) δ 0 1) = 1 and δ n 1) = 0 for all n 1. ) δ 0 x = x, δ 1 x = 1, and δ n x = 0 for all n 2. Proof. As we noted above, δ 0 = d. In partcular, δ 0 1) = 1. For px) = 1, the formula 1) takes the form 2) 1 = δ 0 1)y 0 + n = 1 δn 1)y n. Snce δ 0 1)y 0 = 1 y 0 = 1, the formula 2) mples that 0 = n = 1 δn 1)y n. It follows that δ n 1 = 0 for n 1. Ths proves the part ) of the lemma. For px) = x, the formula 1) takes the form x + y = δ 0 x)y 0 + δ 1 x)y 1 + It follows that δ 0 x) = x, ) of the lemma. n = 2 δn x)y n. δ 1 x) = 1, and δ n x) = 0 for all n 2. Ths proves the part 4

5 1.4. Lemma. δ 1 x n ) = nx n 1 for all n N, n 1. Proof. The case n = 1 was proved n Lemma 1.3. Suppose that n N, n 1, and we already know that δ 1 x n ) = nx n 1. By Corollary 1.2, δ 1 x n 1 ) = δ 1 x x n ) = δ 1 x)x n + xδ 1 x n ) = 1 x n + xnx n 1 ) = x n + nx n = n 1)x n. An applcaton of nducton completes the proof. Remark. By Lemma 1.4, the operator δ 1 : k[x] k[x] agrees on the powers x n k[x] wth the operator d: fx) f x) of takng the usual formal dervatve. Snce both these operators are k-lnear, δ 1 = d. But f N, 2, then the operator δ s not equal to the operator of takng the -th dervatve. Ths mmedately follows ether from Lemma 1.5 or from Theorem 1.7 below. Bnomal coeffcents. Let n N. For N, n n) N, by the bnomal formula n, we defne the bnomal coeffcents 3) x + y) = n = 0 n n) x n y n. Gven arbtrary numbers a, b N, we defne a b) as n b b), where n = a b. Gven arbtrary ntegers a, b Z, we set a b) = 0 f at least one of the numbers a, b s not n N. We prefer the notaton a b) to the classcal one by the typographcal reason, and because the new notaton helps to brng to the lght the fact that we do not use any propertes of a b) except the above defnton Lemma. δ n x ) = n n) x n. Proof. It s suffcent to compare 3) wth the defnton 1) of dvded dervatves. The left shft operator. The left shft operator λ: k[x] k[x] s just the operator of multplcaton by the polynomal x: λpx) = λp)x) = xpx). 5

6 The reasons for callng λ the left shft operator wll be clear later. The man property of the left shft operator and the dvded dervatves s the followng commutaton relaton Theorem. Let us set δ 1 = 0. Then 4) δ n λ λ δ n = δ n 1. for all n N. Proof. Let px) k[x], and let y be a varable dfferent from x. By the Lebnz formula from Theorem 1.1, δ n xpx) ) = j = n δ x) δ j px) ). But by Lemma 1.3, δ 0 x = x, δ 1 x = 1, and δ x = 0 for 2. Therefore δ n xpx) ) = xδ n px) ) + δ n 1 px) ), or, what s the same, δ n xpx)) xδ n px) ) = δ n 1 px) ). Rewrtng the last dentty n terms of λ, we get δ n λp)x) ) λ δ n px) )) = δ n 1 px) ),.e. δ n λ px) ) λ δ n px) ) = δ n 1 px) ). Snce px) k[x] was arbtrary, ths proves the theorem. The followng theorem wll be not used n the rest of the paper Theorem Composton of dvded dervatves). Let n, m N. Then δ n δ m = m n) δ n+m. Proof. Let px) k[x]. Let u, z be two new varables dfferent from both x and y. If we apply 1) to u, z n the role of x, y respectvely and use m nstead of n), we get pu z) = m = 0 δm p)u)z m. Let us set u = x y and apply 1) to each polynomal δ m p)x y): 6

7 5) px y) z) = m = 0 δm p)x y) z m = m = 0 n = 0 δn δ m p)x) y n) z m = m, n = 0 δn δ m p)x) y n z m Alternatvely, we can apply 1) to y z n the role of y and then apply 3): 6) px y z)) = k = 0 δk px) y z) k = m+n=k m n)ym z n) k = 0 δk px) = m, n = 0 δm+n px) m n)y m z n = m m, n = 0 n)δm+n px) y m z n. By the assocatvty of the addton, x y) z = x y z) and hence px y) z) = px y z)). By combnng ths equalty wth 5) and 6) we conclude that δ n δ m px)) = m n)δ m+n px)) for all px) k[z] and n, m N. The theorem follows. 2. Sequences and dualty Sequences. A sequence of elements of a set X s defned as a map N X. For a sequence s we usually denote the value s), N by s and often call t the -th term of s. The set of all sequences of elements of X wll be denoted by S X. We are, frst of all, nterested n the case when X s a k-module, and especally n the case when X s equal to k consdered as a k-module. When t s clear from the context to what set X the terms of the consdered sequences belong, we call the sequences of elements of X smply sequences. Let M be a k-module. Then the set S M has a canoncal structure of a k-module. The k-module operatons on S M are the term-wse addton of sequences and the term-wse multplcaton of sequences by elements of k, defned n the followng obvous way. 7

8 The term-wse sum r s of sequences r, s S M s defned by r s) = r s, and the term-wse product cs of c k and s S M s defned by cs) = cs. Modules of homomorphsms. Let M, M be k-modules. Then the set HomM, M ) of k-homomorphsms M M has a canoncal structure of a k-module. The addton s defned as the addton of k-homomorphsms, and the product af of an element a k and a k-homomorphsms F: M M s defned by af)m) = afm), where m M. Note that the obvous) verfcaton of the fact that af s a k-homomorphsm uses the commutatvty of k. We are mostly nterested n the case of M = k[x], and especally n the case of M = k[x] and M = k, where k[x] s consdered as a k-module by forgettng about the multplcaton of elements of k[x], and the rng k s consdered as a module over tself. For the rest of ths secton M denotes a fxed k-module. A parng between S M and k[x]. Consder a sequence s S M and a polynomal px) = = 0 c x k[x]. Of course, the sum here s actually fnte,.e. c = 0 for all suffcently large. Let s, px) = = 0 c s M. Snce c = 0 for all suffcently large, the sum n the rght hand sde of ths formula s well defned. The map defned by, : S M k[x] M, : s, px)) s, px) s our parng between k[x] and S M. Obvously, t s a k-blnear map and hence ndeed deserves to be called a parng). The parng, defnes a k-lnear map D M : S M Hom k[x], M ) by the usual rule D M s): px) s, px). 8

9 Note that, obvously, s, x = s for every s S M, 7) s = D M s) x ) n N. Therefore for every N and every s S M Theorem Dualty). The parng, s non-degenerate n the sense that the map D M : S M Hom k[x], M ) s an somorphsm of k-modules. Proof. Note that k-homomorphsm F: k[x] M s determned by ts values Fx ) on the monomals x, N because every polynomal px) k[x] s a fnte sum of powers x, N wth coeffcents n k ). By 7) the terms s of a sequence s S M are equal to the values D M s)x ). It follows that all terms of s, and hence the sequence s are determned by the homomorphsm D M s). Therefore, the map D M s njectve. In order to prove that D M s surjectve, let us consder an arbtrary k-homomorphsm F: k[x] M. Let s S M be the sequence wth s = Fx ). Then k-homomorphsms F and D M s) take the same values at all powers x. It follows that F = D M s) cf. the prevous paragraph). Therefore, the map D M s surjectve. The theorem follows. Each k-endomorphsm E: k[x] k[x] defnes ts dual endomor- Dual endomorphsms. phsm E : Hom k[x], M ) Hom k[x], M ) by the usual formula E h) = h E, for all k-homomorphsms h: k[x] M. Obvously, f E, F are two k-endomorphsms k[x] k[x], then E F) = F E. Adjont endomorphsms. Let M be a k-module. Snce D M s an somorphsm by Theorem 2.1, we can use D M to turn the dual map E : Homk[x], M) Homk[x], M) of an endomorphsm E: k[x] k[x] nto a map S M S M. Namely, let E = D M ) 1 E D M : S M S M. Then D M E = E D M. We wll call E the adjont endomorphsm of E. 9

10 For every par E, F: k[x] k[x] of k-endomorphsms E F) = F E. Ths mmedately follows from the correspondng property E F) = F E of dual endomorphsms Lemma. Let M be a k-module, and let E: k[x] k[x] be a k-endomorphsm. The adjont map S M S M s the unque map E such that 8) p, E s) = Ep), s for all p = px) k[x], s S M. Proof. Let p = px) k[x] and let s S M. By the defnton of D M we have: p, E s) = D M E s) ) p ) = D M E s) ) p ) ; Ep), s = D M s) Ep) ) = E D M s) ) p ) = E D M s) ) p ). Therefore, 8) s equvalent to D M E s) ) p ) = E D M s) ) p ). It follows that 8) holds for all p = px) k[x], s S M f and only f D M E = E D M. The lemma follows. 3. Adjonts of the left shft and of dvded dervatves As n the prevous secton, M denotes a fxed k-module. The adjont of the left shft operator. Let L = λ, where λ s the left shft operator from Secton 1. In vew of the followng lemma call L also the left shft operator Lemma. For every sequence s S M the terms of the sequence Ls) are Ls) ) = s 1. Proof. Recall that s = s, x for any sequence s S M. Therefore by Lemma 2.2 Ls) ) = Ls), x = λ s), x = s, λx ) = s, x 1 = s 1. The lemma follows. 10

11 3.2. Corollary. For every n N and every sequence s S M the terms of the sequence L n s) are L n s) ) = s n. Proof. For n = 0 the corollary s trval, because L 0 = d. For n 1 the corollary follows from Lemma 3.1, f we use an nducton by n. The adjonts of the dvded dervatves. Let D n = δ n ), where n N and δ n s the n-th dvded dervatve operator from Secton 1. Recall see Secton 1) that δ 0 : k[x] k[x] s the dentty of k[x]. Therefore D 0 : S M S M s also the dentty of S M. Recall that n Theorem 1.6 we also ntroduced operator δ 1. Let D 1 = δ 1 ). Snce δ 1 = 0 by the defnton, we have D 1 = 0. The followng commutaton relatons are the most mportant for our purposes propertes of the adjont operators L = λ and D n = δ n ) Theorem. For every α k and every n N L D n D n L = D n 1 and L α) D n D n L α) = D n 1, where we nterpret α as the operator S M S M of multplcaton by α k. Proof. By takng the adjont dentty of the dentty 4) from Theorem 1.6, we get δ n λ ) λ δ n ) = δ n 1 ), and hence λ δ n) δ n ) λ = δ n 1). In vew of the defntons of L and D n, ths mples the frst dentty of the theorem. Snce D n s a k-lnear operator, we have α D n = D n α, where α s nterpreted as the multplcaton operator. Clearly, the frst dentty of the theorem together wth α D n = D n α mples the second one. The sequences sα) and sα, n). Let α k and n N. Let us defne sequences sα) and sα, n) by sα) = α and sα, n) = D n sα) ). 11

12 Obvously, sα, 0) = sα). Note that sα) 0 even f α = 0, because sα) 0 = α 0 = 1 by the defnton for all α k. For explct formulas for sequences sα, n) wth n 1 the reader s referred to Theorem 3.6 below. No such formulas are used n ths paper Lemma. Let α k and s S M. Then L α ) s) = 0 f and only f s has the form s = βsα), where β k. Proof. The condton L α ) s) s equvalent to Ls) = αs. The latter condton holds f and only f Ls) ) n = αs n for all n N. By Lemma 3.1 Ls) ) n = s n 1. Therefore, ) L α s) = 0 f and only f sn 1 = αs n for all n N. An applcaton of the nducton completes the proof Lemma. Suppose that a N, α k, and s S M. If n a, then L α ) a sα, n) ) = sα, n a). Proof. The lemma s trval f a = 0. Let us prove the lemma for a = 1. In vew of the defnton of sequences sα, n), we need to prove that L α ) D n sα) )) = D n 1 sα) ). By applyng the second dentty of Theorem 3.3 to sα), we get L α) D n sα) ) D n L α) sα) ) = D n 1 sα) ), whch s equvalent to L α) D n sα) )) D n L α) sα) )) = D n 1 sα) ). Snce L α ) sα) ) = 0 by Lemma 3.4, we see that L α) D n sα) )) = D n 1 sα) ), ) ).e. L α sα, n) = sα, n 1). Ths proves the lemma for a = 1. The general case follows from ths one by nducton. The followng theorem s not used n the rest of the paper. 12

13 3.6. Theorem. Let n N. For every sequence s S M the terms of the sequence D n s) are D n s) ) = n n)s n. In addton, for every α k the terms of the sequence sα, n) are sα, n) ) = n n)α n. Proof. Recall that s = s, x for any sequence s S M. Together wth Lemma 2.2 ths fact mples that D n s) ) = D n s), x = δ n ) s), x = s, δ n x ). Snce δ n x ) = n n) x n by Lemma 1.5, we have s, D n x ) = s, n n) x n = n n) s, D n x n ) = n n)s n. The frst part of the theorem follows. Let us apply the frst part to s = sα). We get sα, n) = D n sα) )) = n n)sα) n = n n)α n. Ths proves the second part of the theorem. 4. Endomorphsms and ther egenvalues Representaton of the polynomal algebra defned by an endomorphsm. Let x be a varable, and let k[x] be the k -algebra of polynomals n x wth coeffcents n k. Let M be a k-module. The k-endomorphsms M M form a k -algebra End M wth the composton as the multplcaton. For every k-endomorphsms E: M M and every a N we wll denote by E a the a-fold composton E E... E. As usual, we nterpret the 0-fold composton E 0 as the dentty endomorphsm d End M. 13

14 For a k-module endomorphsm E: M M and a polynomal fx) = c 0 x n + c 1 x n 1 + c 2 x n c n k[x] one can defne an endomorphsm fe): M M by the formula fe) = c 0 E n + c 1 E n 1 + c 2 E n c n. The map fx) fe) s a homomorphsm k[x] End M of k-algebras. Ths follows from the obvous denttes x a x b = x a b and E a E b = E a b. Ths homomorphsm defnes a structure of k[x]-module on M. Of course, ths structure depends on E. We wll denote by k[e] the mage of the homomorphsm fx) fe). commutatve, the mage k[e] s a commutatve subalgebra of End M. Snce k[x] s Egenvalues. Suppose that a k-module endomorphsm E: M M s fxed. Let α k. The kernel KerE α) s called the egenmodule of E correspondng to α and s denoted also by E α. Clearly, E α s a k-submodule of M. An element α k s called an egenvalue of E f the kernel E α = KerE α) 0. The set of elements v M such than E α) v) = 0 for some N s called the extended egenmodule of E correspondng to α and s denoted by Nlα). Clearly, Nlα) s a k- submodule of M Lemma. Let α k. Then the followng statements hold. ) The submodules E α and Nlα) are E-nvarant. ) E α and Nlα) are k[x]-submodules of M. ) The submodule Nlα) s non-zero f and only f α s an egenvalue. Proof. Let us prove ), ) frst. Note that E α) E = E E α) for every N, because k[e] s a commutatve subalgebra of End M. Therefore, f E α) v) = 0, then E α) Ev)) = ) E α) E v) = E E α) ) ) v) = E E α) v) In the case = 1 ths mples that EE α ) E α. In general, ths mples that EKerE α) ) KerE α), and hence ENlα)) Nlα). Ths proves ), and ) mmedately follows. = 0. 14

15 Fnally, let us prove ). Suppose that v 0 and E α) v) = 0. Let be the smallest nteger such that E α) v) = 0. Note that > 0 because v 0. Let w = E α) 1 v). Then w 0 and E α)w) = 0. Therefore E α = KerE α) 0. Ths proves ). Torson free modules. A k-module M s called torson-free, f αm = 0 mples that ether α = 0, or m = 0, where α k and m M. Snce k s assumed to be a rng wthout zero dvsors, k n s a torson free module for any non-zero n N. For the rest of ths secton we wll assume that M s a torson-free module Lemma. Let α 1, α 2,..., α n be dstnct egenvalues of an endomorphsm E: M M. Let Ker 1, Ker 2,..., Ker n be the correspondng egenmodules,.e. Ker = KerE α ) for each = 1, 2,..., n. Then the sum of these egenmodules s a drect sum,.e. an element v Ker 1 + Ker Ker n admts only one presentaton v = v 1 + v v n wth v Ker for all = 1, 2,..., n. Proof. It s suffcent to prove that f v 1 v 2... v n = 0 and v Ker for all, then v 1 = v 2 =... = v n = 0. Suppose that v 1 v 2... v n = 0, v Ker for all, and not all v are equal to 0. Consder the maxmal nteger m such that 9) v 1 + v v m = 0 for some elements v Ker such that v m 0. Note that n ths case v 0 also for some m 1, n vew of 9). By applyng E α m to 9), we get E α m )v 1 ) E α m )v m 1 ) + E α m )v m ) = 0. Snce v Ker = KerE α ) and therefore Ev ) = α v for all, we see that 10) α 1 α m )v α m 1 α m )v m 1 + α m α m )v m = 0, 11) α 1 α m )v α m 1 α m )v m 1 = 0. Snce the egenvalues α are dstnct, α α m 0 for m 1. Snce our module M s assumed to be torson-free, ths mples that α α m )v 0 f m 1 and v 0. As we noted above, v 0 for some m 1. Therefore, the equalty 11) contradcts to the choce of m. Ths contradcton proves the lemma. 15

16 4.3. Lemma. Let α 1, α 2,..., α n be dstnct egenvalues of an endomorphsm E: M M. Let Nl 1, Nl 2,..., Nl n be the correspondng extended egenmodules,.e. Nl = Nlα ) for = 1, 2,..., n. Then the sum of these extended egenmodules s a drect sum,.e. an element v Nl 1 + Nl Nl n admts only one presentaton v = v 1 + v v n wth v Nl for all = 1, 2,..., n. Proof. It s suffcent to prove that f v 1 v 2... v n = 0 and v Nl for all, then v 1 = v 2 =... = v n = 0. Suppose that v 1 v 2... v n = 0, v Nl for all, and not all v are equal to 0. The proof proceeds by replacng, n several steps no more than n), the orgnal elements v by new ones n such a way that eventually not only v Nl, but, moreover, v Ker = KerE α ), and stll not all v are equal to 0. Obvously, ths wll contradct to Lemma 4.2. Let E = E α for all = 1, 2,..., n. If = 1, 2,..., n 1, or n, then E 0 v ) v and E a v ) = 0 for some nteger a 1. If v 0, then we defne a as the largest nteger a 0 such E a v ) 0. Then E a v ) 0 and E a 1 v ) = 0. In partcular, E α )E a v )) = E E a v )) = E a 1 v ) = 0, and hence E a v ) Ker. If v = 0, then we set a = 0 and E a v ) Ker s stll true. Let us fx an nteger k between 1 and n. Let w = E a k k v ), where = 1, 2,..., n. By applyng E a k k to v 1 v 2... v n = 0, we conclude w 1 w 2... w n = 0. Note that snce the submodules Nl are E-nvarant by Lemma 4.1, w Nl for every. Clam 1. If v 0, then w = 0. Proof of Clam 1. If = k and v = v k 0, a k. Suppose that k and v 0. Then E a w ) = E a E a k k v ) ) = E a k k then w = w k = E a k k v k) 0 by the choce of E a v ) ) But E a v ) Ker and E a v ) 0 by the choce of a. Snce E acts of Ker as the multplcaton by α, we have E a k k E a v ) ) = E α k ) a k E a v ) ) = α α k ) a k E a v ) ). Snce α α k and k s a rng wthout zero dvsors, α α k ) a k 0. Snce M s a torson free k-module and E a v ) 0, ths mples that α α k ) a k E a v ) ) 0. It follows that E a w ) = α α k ) a k E a v ) ) 0, 16

17 and hence w 0. Ths completes the proof of the clam. Clam 2. If v Ker, then w Ker. Proof of Clam 2. Suppose that v Ker,.e. E v ) = 0. Snce E = E α and E k = E α k obvously commute, t follows that E w ) = E E a k k v )) = E a k k E v )) = 0. Ths proves the clam. To sum up, we see that by applyng E a k k to the equalty v 1 v 2... v n = 0 wth v Nl for all we get another equalty w 1 w 2... w n = 0 such that for all : ) w Nl ; ) f v 0, then w 0; ) f v Ker, then w Ker. In addton, w k = E a k k v k) Ker k even f v k dd not belonged to the egenmodule Ker k. Therefore, we can take w 1, w 2,..., w n as the new elements v 1, v 2,..., v n, ncreasng the number of elements belongng to the correspondng egenmodules by an approprate choce of k f some v dd not belonged to egenmodules yet). It follows that by startng wth the equalty v 1 v 2... v n = 0 and consecutvely applyng endomorphsms E a k k for k = 1, 2,..., n, we wll eventually prove the equalty v 1 v 2... v n = 0 for some new vectors v such that v Ker for all, and stll not all v are equal to 0. The contradcton wth Lemma 4.2 completes the proof Lemma. Let E: M M be an endomorphsm of M and let α be an egenvalue of E. Suppose that v Nlα). Let a 0 be the largest nteger such that E α) a v) 0, and let v = E α) v) for = 0, 1,..., a. Then the homomorphsm k a 1 M defned by x 0, x 1,..., x a ) x 0 v 0 + x 1 v x a v a s an somorphsm onto ts mage. In partcular, v 0, v 1,..., v a k-submodule of M. are free generators of a free Proof. It s suffcent to prove that our homomorphsm s njectve. In other terms, t s suffcent to prove that f 12) x 0 v 0 + x 1 v x a v a = 0 17

18 for some x 0, x 1,..., x a k, then x 0 = x 1 =... = x a = 0. Suppose that 12) holds and x 0 for some. Let b N be the mnmal nteger wth the property x b 0. Let us apply E α) a b to 12). Note that f > b, then ) E α) a b v ) = E α) a b E α) v) E α) a b v) = 0 because a b > a and E α) n v) = 0 for n > a by the choce of a. Therefore, the operator E α) a b takes the left hand sde of 12) to ) x b E α) a b v b ) = x b E α) a b E α) b v) = x b E α) a b b v) = x b E α) a v), and hence the result of applcaton of E α) a b to 12) s 13) x b E α) a v) = 0. But E α) a v) 0 by the choce of a, and x b 0 by the choce of b. Snce the module M s assumed to be torson free, these facts together wth 13) lead to a contradcton. Ths contradctons shows that 12) may be true only f x = 0 for all. 5. Torson modules and a property of free modules Torson modules. An element m M of a k-module M s called a torson element f xm = 0 for some non-zero x k. A k-module M s called a torson module f every element of M s a torson element Lemma. Let n N. If M s a k-submodule of a k-module N and both M and N are somorphc to k n, then the quotent N/M s a torson module. Proof. Suppose that N/M s not a torson module. Then there s an element v N/M such that αv 0 f α 0. For such a v the map α αv s an njectve homomorphsm of k-modules k N/M. Let us lft v N/M to an element v 0 N, so v s the mage of v 0 under the canoncal surjecton N N/M. Then the map α αv 0 s an njectve homomorphsm of k-modules k N. Clearly, f v 1, v 2,..., v n s a bass of M whch exsts because M s somorphc to k n ), then v 0, v 1,..., v n s a bass of kv 0 M. Therefore, kv 0 M s a submodule somorphc to k n 1 of the module N somorphc to k n. In partcular, there exst an njectve k- homomorphsm J: k n 1 k n. 18

19 Snce k has no zero dvsors, t can be embedded nto ts feld of fractons, whch we wll denote by F. Moreover, the k-homomorphsm k n 1 k n extends to an F-lnear map F n 1 F n, whch we wll denote by J F. Clam. J F s njectve. Proof of the clam. Suppose y 0, y 1,..., y n ) F n 1 s non-zero and belongs to the kernel of J F. Snce F s the feld of fractons of k, there s an element z k such that zy 0, zy 1,..., zy n k. For such an element z k the n 1)-tuple zy 0, zy 1,..., zy n ) belongs to k n 1, and Jzy 0, zy 1,..., zy n ) = J F zy 0, zy 1,..., zy n ) = z J F y 0, y 1,..., y n ) = z0 = 0. Snce F s the feld of fractons of k, y 0, y 1,..., y n ) 0 mples that the n 1)-tuple zy 0, zy 1,..., zy n ) 0. At the same tme ths n 1)-tuple belongs to the kernel of J, n contradcton wth the njectvty of J. The clam follows. As s well known, for a feld F there are no njectve F-lnear maps F n 1 F n. The contradcton wth the above clam proves that N/M s ndeed a torson module. 6. Polynomals and ther roots 6.1. Lemma. Let px) k[x] be a polynomal wth leadng coeffcent 1, and let α k. Then α k s a root of px) f and only f 14) px) = x α)qx) for some polynomal qx) k[x] wth leadng coeffcent 1. If α s a root, then qx) s unquely determned by 14). Proof. Suppose that px) = x α)qx) and both px), qx) have the leadng coeffcents 1. Then deg qx) = deg px) 1 and the polynomals px), qx) have the form px) = x n + c 1 x n a n 1 x + c n, qx) = x n 1 + d 1 x n d n 2 x + d n 1, where c 1,..., c n, d 1,..., d n 1 k. Let us compute the product x α)qx) = x α) x n d n 2 x + d n 1 ). 19

20 Obvously, x α)qx) = x n + d 1 x n d n 2 x 2 + d n 1 x αx n 1... αd n 3 x 2 αd n 2 x αd n 1. It follows that px) = x α)qx) f and only f c 1 = d 1 α c 2 = d 2 αd 1... = c n 1 = d n 1 αd n 2 c n = αd n 1, or, equvalently, d 1 = α + c 1 d 2 = αd 1 + c 2... = d n 1 = αd n 2 + c n 1 αd n 1 + c n = 0. These equaltes allow to compute the coeffcents d 1, d 2,..., d n 1 n terms of the coeffcents c 1, c 2,..., c n 1. Namely, d 1 = α + c 1 and d = α + c 1 α 1 + c 2 α c 1 α + c for 2 n 1. Therefore, the last equalty αd n 1 c n = 0 holds f and only f α α n 1 + c 1 α n c n 1 α ) + c n = 0,.e. f and only f pα) = 0. The lemma follows Corollary. Let px) k[x] be a polynomal wth leadng coeffcent 1, and let α k. Then there s a number m N and a polynomal rx) k[x] such that px) = x α) m rx) and α s not a root of rx). The number m and the polynomal rx) are unquely determned by px) and α. Proof. If pα) 0, then, obvously, m = 0 and rx) = px). If pα) = 0, we can apply Lemma 6.1. If qα) 0, then m = 1, rx) = qx) and we are done. If qα) = 0, then we can apply Lemma 6.1 agan. Eventually we wll get a presentaton px) = x α) m rx) such that rα) 0. By consecutvely applyng the unqueness part of Lemma 6.1, we 20

21 see that x α) m 1 rx), x α) m 2 rx),..., and, eventually, m and rx) are unquely determned by px) and α. The multplcty of a root. We wll denote by deg px) the degree of the polynomal px). If α s a root of px), then the number m from Corollary 6.2 s called the multplcty of the root α Corollary. Let px) k[x] be a polynomal wth leadng coeffcent 1. The number k of dstnct roots of px) s fnte and k deg px). If α 1, α 2,..., α k s the lst of all dstnct roots of px), and f µ 1, µ 2,..., µ k are the respectve multplctes of these roots, then 15) px) = x α 1 ) µ1 x α2 ) µ2 x α k ) µk rx), where rx) k[x] has no roots n k. The polynomal rx) s unquely determned by px). Proof. By consecutvely applyng the exstence part of the Corollary 6.2, we see that a factorzaton of the form 15) exsts. Smlarly, the unqueness of rx) follows from the unqueness part of Corollary 6.2. Polynomals wth all roots n k. Agan, let px) k[x] be a polynomal wth leadng coeffcent 1. We say that px) has all roots n k, f n the factorzaton 15) the polynomal rx) = 1. In other words, px) has all roots n k, f px) has the form px) = x α 1 ) µ1 x α2 ) µ2 x α k ) µk, for some α 1, α 2,..., α k k and some non-zero µ 1, µ 2,..., µ k N. Obvously, then deg px) = µ 1 µ 2... µ k. 7. The man theorems Let us fx for the rest of ths secton a polynomal px) = x n + c 1 x n c n 1 x + c n k[x] wth the leadng coeffcent 1. Consder the left shft operator L: S k S k from Secton 2. As t was explaned n Secton 4, the operator L defnes a homomorphsm of k -algebras k[x] End S k by the rule fx) fl). We are nterested n the kernel Ker pl). 21

22 7.1. Lemma. A sequence s S k belongs to Ker pl) f and only f 16) s + c 1 s c n 1 s n 1 + c n s n = 0 for all N, n. Proof. Let us compute the terms of pl)s), usng Corollary 3.2 at the last step: pl)s) ) = L n + c 1 L n c n 1 L + c n )s) ) = L n s) ) + c 1 L n 1 s) ) c ) n 1 Ls) + c ) n s = s n + c 1 s n c n 1 s 1 + c n s. Ths calculaton shows that s Ker fl) f and only f 17) s n + c 1 s n c n 1 s 1 + c n s = 0 for all ntegers 0. Clearly, 17) holds for all ntegers 0 f and only f 16) holds for all ntegers n. The lemma follows. Remark. Classcally, a sequence s S k s called recurrent f ts terms satsfy 16) for all N, n, and the equaton 16) s called a lnear recurrence relaton. Ths explans the ttle of the paper Lemma. The map F: Ker pl) k n defned by F: s s 0, s 1,..., s n 1 ) k n s an somorphsm. In partcular, Ker pl) s a free k-module of rank n. Proof. By Lemma 7.1 the kernel Ker pl) s equal to the k-submodule of S k consstng of sequences s satsfyng the relaton 16) for all N, n. Clearly, 16) allows to compute each term s, n of s as the lnear combnaton of n mmedately precedng terms s 1, s 2,..., s n of s wth coeffcents c 1, c 2,..., c n ndependent of. Therefore, such a sequence s s determned by ts frst n terms s 0, s 1,..., s n 1. Moreover, these n terms can be prescrbed arbtrarly. The lemma follows Theorem. Suppose that α k s a root of px) of multplcty µ. Then pl)sα, a)) = 0 for each a N, 0 a µ 1, where sα, a) are the sequences defned n Secton 3, the paragraph mmedately precedng Lemma

23 Proof. By Corollary 6.2, px) has the form px) = x α) µ qx). Therefore 18) pl) = L α ) µ ) µ. ql) = ql) L α If a µ 1, then µ 1 a 0, and we can present L α ) µ as the followng product: ) µ ) µ 1 a ) ) a. 19) L α = L α L α L α By Lemma ) L α ) a sα, a) ) = sα, a a) = sα, 0) = sα), and by Lemma 3.4, 21) L α ) sα) ) = 0. By combnng 19), 20), and 21), we see that L α ) µ sα, a) ) = 0. By combnng the last equalty wth 18), we get pl) sα, a) ) = ql) L α ) µ sα, a) ) = ql)0) = 0. The theorem follows Theorem. Suppose that px) has all roots n k. Let k be the number of dstnct roots of px), let α 1, α 2,..., α k be these roots, and let µ 1, µ 2,..., µ k be, respectvely, the multplctes of these roots. Then sequences sα u, a), where 1 u k and 0 a µ u 1, are free generators of a free k-submodule of Ker fl). Proof. By Theorem 7.3, all these sequences belong to Ker pl). In partcular, they are generators of a k-submodule of Ker pl). Let us prove that they are free generators. Let 1 u k. By Lemma 3.5, 22) L αu ) µu 1 sαu, µ u 1) ) = sα u, µ u 1) µ u 1)) = sα u, 0) = sα u ). By Lemma 3.4, L α u ) sαu ) ) = 0 and hence L α u ) µ u sα u, µ u 1) = L α u )sα)) = 0. In partcular, sα u, µ u 1) belongs to the extended egenmodule Nlα u ). 23

24 In addton, 22) together wth the fact that sα u ) 0 mples that µ u nteger a such that ) a L αu sαu, µ u 1) ) 0. 1 s the largest Lemma 4.4 mples that the sequences L α u ) a sα, µ u 1) ) for a = 0, 1,..., µ u 1 form a bass of a free submodule of Nlα u ) Ker pl). Snce L αu ) a sαu, µ u ) ) = sα u, µ u a) by Lemma 3.5, ths mples that the sequences sα u, a) for a = 0, 1,..., µ u 1 form a bass of a free submodule of Nlα u ) Ker pl). By combnng ths result wth Lemma 4.3, we see that the sequences sα u, a) from the theorem form a bass of a free submodule of Ker pl). Ths completes the proof Theorem. Let S Ker pl) be the free k-module generated by the sequences sα u, a) from Theorem 7.4. Then the quotent k-module Ker pl))/s s a torson module. Proof. Let n = deg px). By Lemma 7.2, Ker pl) s a free module of rank n,.e. s somorphc to k n. Snce n = µ 1... µ k, we have exactly n sequences sα u, a). By Theorem 7.4, they are free generators of S. In partcular, S s also somorphc to k n. It remans to apply Lemma Corollary. If k s a feld, then S = Ker pl). Proof. A torson module over a feld s equal to 0. Note bblographque Ths note s concerned only wth the works whch nfluenced the present paper. The author dd not attempted to wrte even an ncomplete account of the hstory of the theory of lnear recurrence relatons. The theory of recurrent sequences s an oblgatory topc for any ntroducton to combnatorcs. But all too often the proofs are presented only for Fbonacc numbers, even f the general case s dscussed. The author stumbled upon ths tradton n P. Cameron s textbook [C]. The dscusson of the general case n [C] s lmted by the followng. 24

25 In ths case, suppose that α s a root of the characterstc equaton wth multplcty d. Then t can be verfed that the the d functons α n, nα n,..., n d 1 α n are all solutons of of the recurrence relaton. Dong ths for every root, we agan fnd enough ndependent solutons that k ntal values can be ftted. The justfcaton of ths s the fact that the solutons clamed can be substtuted n the recurrence relaton and ts truth verfed. Ths dscusson gnores at least two sgnfcant ssues. Frst, the clam that the sequences n α n provde solutons cannot be justfed by the substtuton of them n the recurrence relatons and a routne verfcaton smply because they are solutons only f d 1. Second, one needs to prove that these solutons are lnearly ndependent. The standard approach to the general case s based on the theory of generatng functons and the partal fractons expanson of ratonal functons. Ths method s elegantly presented n Chapter 3 of M. Hall s classcal book [H]. A recent presentaton of ths method can be found n Secton 3.1 of M. Agner s book [A]. When ths approach s chosen, the exstence of partal fracton expansons s smply quoted as a tool external to the theory of lnear recurrence relatons. The theory of partal fractons s an applcaton of the theory of modules over prncpal entre rngs. See N. Bourbak [B], Secton 2.3, or S. Lang [L], Secton IV.5. Hence the theory of lnear recurrence relatons s also an applcaton of the theory of modules. Once ths s realzed, t s only natural to use the theory of modules drectly, wthout usng the generatng functons and partal fractons as an ntermedary. Another applcaton of the same part of the theory of modules s the theory of the Jordan normal form. See N. Bourbak [B], Secton 5, or S. Lang, Chapter XIV. Of course, the theory of the Jordan normal form precedes the theory of modules and s usually presented wthout any references to the latter. It s only natural to adapt drectly the standard arguments from the theory of the Jordan normal form to the theory of lnear recurrence relaton. Ths s done n Secton 4 of the present paper, whch was heavly nfluenced by I.M. Gelfand s classcs [G]. The materal of Sectons 2 and 3 took ts present form under the nfluence of G.-C. Rota s deas about the umbral calculus. See, for example, [RR]. The defnton of dvded dervatves was motvated by the desre to prove the man results wthout any restrctons on the characterstc of the base rng k and at the same tme to avode brute force calculatons wth bnomal coeffcents. After ths work For the purposes of ths paper, any calculaton wth bnomal coeffcents usng the well known expresson n terms of factorals s a brute force calculaton. 25

26 was completed, the author came across the paper [D] by J. Deudonné, from whch he learned that ths noton was frst ntroduced by H. Hasse, F.K. Schmdt, and O. Techmüller n 1936, wth applcatons to varous questons of algebra n mnd. See the references n [D]. Ths was a pleasant surprse, especally because a major part of the author s work s devoted to Techmüller modular groups and Techmüller spaces. Bblographe [A] M. Agner, A course n enumeraton, Sprnger-Verlag, Berln, Hedelberg, [B] [C] N. Bourbak, Algèbre, Chaptre VII, Modules sur les anneaux prncpaux, Masson, Pars, 1981 édton orgnale). Sprnger-Verlag, Berln, Hedelberg, 2007 rémpresson nchangée). P.J. Cameron, Combnatorcs: topcs, technques, algorthms. Cambrdge Unversty Press, Cambrdge, [D] J. Deudonné, Le calcul dfférentel dans les corps de caractérstque p > 0, Proceedngs of the Internatonal Congress of Mathematcans, 1954, V. 1, [G] [Ge] [H] I.M. Gelfand, Lectures on lnear algebra Russan), 4th edton, Nauka, Moscow, pp. A.O. Gel fond, The calculus of fnte dfferences n Russan), Gos. zd-vo tekhnkoteoret. lt-ry, Moscow, pp. M. Hall, Jr., Combnatoral theory, Blasdell Publshng Co., Waltham Massachussetts) - Toronto - London, 1967 Orgnal edton). Wley, Hoboken, NJ, 1986 Second edton). [L] S. Lang, Algebra, Revsed 3rd edton, Sprnger-Verlag, New York, [RR] S.M. Roman, G.-C. Rota, The umbral calculus, Advances n Mathematcs, V. 27, No ), December 30, 2014 Aprl 16, 2016 Mnor edts and Note bblographque ) 26