PRIMES 2015 reading project: Problem set #3

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1 PRIMES 2015 readng project: Problem set #3 page 1 PRIMES 2015 readng project: Problem set #3 posted 31 May 2015, to be submtted around 15 June 2015 Darj Grnberg The purpose of ths problem set s to replace an argument n [BFZ-CA3, proof of Proposton 1.8] wth a more elementary argument whch does not use Newton polytopes Some lnear algebra bascs Let us start by dong some lnear algebra over the rng of ntegers. You are probably most used to dong lnear algebra over felds, snce ths s the case most suted to dong lnear algebra: For example, solvng a system of lnear equatons usng Gaussan elmnaton works over felds but not say over Z or over a polynomal rng. 2 We shall do lnear algebra over the rng of ntegers here, because we wll want to use the entres of our vectors as exponents n monomals, and exponents n monomals should be ntegers we don t want to work wth non-nteger powers here. Fortunately, the lnear algebra that we wll do wll be smple enough that t can be made to work over the ntegers. If F s a commutatve rng 3 and f n and m are two nonnegatve ntegers, then F n m denotes the set of all n m-matrces over F. These are the matrces whose entres are n F and whch have n rows and m columns. We regard Z n m as a subset of Q n m for all n and m, because a matrx wth nteger entres can always be regarded as a matrx wth ratonal entres. If F s a commutatve rng and f n N, then GL n F denotes the group of all nvertble n n-matrces A F n n. Here, nvertble means nvertble n F n n ; that s, a matrx A s sad to be nvertble f there exsts a matrx B F n n such that AB = BA = I n where I n s the n n dentty matrx. The meanng of nvertble thus depends on F so we wll clarfy t by sayng nvertble n F n n unless F s clear from the context. For example, the fve matrces , , , , Ths s not the only place where [BFZ-CA3] uses Newton polytopes; but the other place n the proof of Lemma 7.3 s really tangental to what we are dong t proves the only f part of Theorem 1.20, whch I would not call partcularly nterestng. 2 Ths s not to say that the stuaton over Z and over polynomal rngs s hopeless. Some more complcated versons of Gaussan elmnaton do ther job over some of these rngs, and lead to certan weaker and more complcated but also more nterestng! versons of row-reduced matrces. 3 Recall that a commutatve rng s roughly speakng a set S whose elements can be added, subtracted and multpled, and all of these operatons gve results nsde S and satsfy the usual laws commutatvty, assocatvty, dstrbutvty, exstence of 0 and 1.

2 PRIMES 2015 readng project: Problem set #3 page 2 are nvertble n Q n n, but only the frst two of them are nvertble n Z n n. The ffth one, of course, does not le n Z n n to begn wth. We shall study GL n Z n partcular. Recall that a matrx A Q n n s nvertble f and only f ts determnant det A s nonzero. The analogous crteron for Z s a lot more restrctve: A matrx A Z n n s nvertble f and only f ts determnant det A has an nverse n Z. The only ntegers whch have an nverse n Z are 1 and 1, so ths leaves only two possbltes for the determnant. We wll not use ths fact, however, but t s mportant. 4 We start wth some smpler thngs. For any n N, the sets Z n and Q n consst of column vectors of length n at least accordng to my conventon, and we consder Z n as a subset of Q n. I wll usually v 1 v 2 wrte a column vector. n the form v 1, v 2,..., v n T, because the former v n vertcal notaton takes too much space. Of course, the notaton v 1, v 2,..., v n T s just a partcular case of the general notaton A T for the transpose of a matrx A at least f we follow the usual conventon to thnk of column vectors as n 1-matrces and of row vectors as 1 n-matrces. The column vector of length n whose all entres are 0 wll be denoted by 0, or when n s not clear from the context by 0 n. It s called the zero vector. For every n N and {1, 2,..., n}, we let e be the column vector 0, 0,..., 0, 1, 0, 0,..., 0 T Z n, where the 1 s the -th entry of the vector. Ths vector depends on both and n; we just leave the n out of the notaton e because t wll always be clear from the context. Exercse 1. Let n N, and let F be a commutatve rng. Feel free to take F = Z or F = Q here. Show that GL n F s a group where the bnary operaton s multplcaton of matrces Gcds of nteger vectors The greatest common dvsor gcd v of a vector v Z n s defned as the gcd of all the n entres of v. When v = 0, ths s understood to be 0; otherwse, t s a postve nteger. Exercse 2. Let n N, let A Z n n and v Z n. a Prove that gcd v gcd Av. b If A GL n Z, then prove that gcd Av = gcd v. 4 We mght get to explore t later.

3 PRIMES 2015 readng project: Problem set #3 page 3 Exercse 3. Let n be a postve nteger, and let v Z n. Let g = gcd v. Then, there exsts a matrx A GL n Z such that v = gae 1. Ths exercse s crucal, so let me comment on t a bt more. Frst of all, e 1 s the column vector 1, 0, 0, 0,... Z n whch has 1 as ts frst entry and 0 s everywhere else. Thus, Ae 1 s the frst column of the matrx A check ths f you don t know why t holds, and so gae 1 s the result of multplyng ths frst column by g = gcd v. So Exercse 3 says that f n s a postve nteger and f v Z n, then there exsts an nvertble matrx A GL n Z whose frst column, multpled by gcd v, s the vector v. In partcular, f gcd v = 1, then v tself s the frst column of an nvertble matrx A GL n Z. For nstance, the vector 6, 15, 10 T Z 3 s the frst column of the nvertble matrx and, of course, of many other such matrces. Let me gve a few hnts for Exercse 3. We want to prove the exstence of an nvertble matrx A GL n Z such that v = gae 1. Ths s tantamount to constructng an nvertble matrx B GL n Z such that Bv = ge 1 the relaton between A and B s that of beng mutually nverse. A way to do ths s to fnd a sequence of nvertble matrces whch, when multpled onto v from the left one after the other, wll make the vector v smpler step by step, untl v becomes a vector wth only one nonzero entry 5. Then, the entry wll have to be ether g or g why?. A further nvertble matrx namely, I n wll get rd of the mnus f t s a g; a further multplcaton wth a permutaton matrx permutaton matrces are nvertble wll push the nonzero entry nto the frst spot, and we wll have the vector ge 1 n front of us. So the man dffculty s to fnd that sequence of nvertble matrces whch smplfy v. To do so, keep n mnd that f v Z n s a vector and and j are two dstnct elements of {1, 2,..., n}, then the matrx F,j whch has 1 s on the man dagonal, a 1 n cell, j, and 0 s everywhere else s nvertble what s ts nverse?, and the vector F,j v s obtaned from v by subtractng the j-th entry from the -th entry. How can you smplfy v usng such subtracton operatons? 0.3. Remnders on polynomals Fx n N from now on. Let x denote the n-tuple x 1, x 2,..., x n of dstnct ndetermnates. We can use ths n-tuple to defne the polynomal rng Z [x] = Z [x 1, x 2,..., x n ], as well as the Laurent polynomal rng Z [ x ±1] [ ] = Z x 1 ±1, x±1 2,..., x±1 n. The elements of the polynomal rng Z [x] are polynomals,.e., Z-lnear combnatons of monomals whch are expressons of the form x v 1 1 xv 2 2 xv n n where 5 unless v = 0, n whch case there s no nonzero entry at all and we are done

4 PRIMES 2015 readng project: Problem set #3 page 4 v 1, v 2,..., v n N. The elements of the Laurent polynomal rng Z [ x ±1] are Laurent polynomals,.e., Z-lnear combnatons of Laurent monomals whch are expressons of the form x v 1 1 xv 2 2 xv n n where v 1, v 2,..., v n Z. We can vew Z [x] as a subrng of Z [ x ±1]. Let us recall how the noton of an rreducble element of a commutatve rng s defned: If R s a commutatve rng, then an element a R s sad to be rreducble f and only f the element a s not nvertble n R, and there are no two elements b and c of R whch are not nvertble n R and satsfy bc = a. For example, the rreducble elements of Z are the prme numbers and the negatves of the prme numbers. The elements 1 and 1 are not rreducble they fal the frst condton, as they are nvertble, and the composte postve ntegers are not rreducble ether they fal the second condton. On the other hand, Q has no rreducble elements ndeed, 0 Q fals the second condton, whle all other elements fal the frst one. The rreducble elements of Z [x] are the rreducble polynomals, n the approprate sense. For example, 2, x 1, x2 2 + x2 3 and x2 1 2 are rreducble, whereas 6, x 1x 2, x2 2 x2 3 and x2 1 4 are not. The rreducble elements of Z [ x ±1] are more or less the same, but only more or less. The stuaton s slghtly dfferent because all Laurent monomals have become nvertble n Z [ x ±1]. Thus, for example, x 1, although rreducble n Z [x], s not rreducble n Z [ x ±1], because t s nvertble n the latter rng. Exercse 4 c below shows that almost-monomals ±x 1, ±x 2,..., ±x n are essentally the only exceptons. Exercse 4. a Show that a polynomal P Z [x] s nvertble n Z [x] f and only f t equals 1 or 1. b Show that a Laurent polynomal P Z [ x ±1] s nvertble n Z [ x ±1] f and only f ether P or P s a sngle Laurent monomal. c Let P Z [x] be a polynomal such that none of the varables x 1, x 2,..., x n dvdes P n Z [x]. Then, show that P s rreducble as an element of Z [x] f and only f P s rreducble as an element of Z [ x ±1]. We say that two elements a and b of a commutatve rng R are coprme f every common dvsor of a and b s nvertble. When R = Z, ths noton of coprmalty s exactly the classcal noton of coprmalty for ntegers Monomal substtutons

5 PRIMES 2015 readng project: Problem set #3 page 5 Defnton 0.1. Let A Z n n. If u = u 1, u 2,..., u n s an n-tuple of nvertble elements of a commutatve rng R, then we let A u denote the n-tuple defned as a 1,1 a 1,2 a 1,n follows: Wrte A n the form A = a,j 1,j n = a 2,1 a 2,2 a 2,n Then, a n,1 a n,2 a n,n set A u = u a 1,1 1 u a 1,2 2 u a 1,n n, u a 2,1 1 u a 2,2 2 u a 2,n n,..., u a n,1 1 u a n,2 2 u a n,n n. Ths s the n-tuple whose j-th entry s n-tuple of nvertble elements of R. n u a j,. Notce that A u s agan an =1 For a trval example, I n u = u for every n-tuple u. Other examples are 1 0 u 2 3 1, u 2 = u 1 1 u0 2, u2 1 u3 2 = u 1, u 2 1 u3 2 ; 1 2 u 0 3 1, u 2 = u 1 1 u2 2, u0 1 u3 2 = u 1 u 2 2, u3 2. Exercse 5. Let A Z n n and B Z n n. Prove that A B u = AB u whenever u s an n-tuple of nvertble elements of a commutatve rng R. A consequence of Exercse 5 s that every A GL n Z and every n-tuple u of nvertble elements of a commutatve rng R satsfy A 1 A u = A 1 A u = I n u = u. 1 =I n Now, f u s an n-tuple of nvertble elements of a commutatve rng R, and f P Z [ x ±1] s a Laurent polynomal, then we can substtute the entres of u for the varables x 1, x 2,..., x n n P, and obtan an element of R, whch we denote by P [u]. It s more commonly denoted by P u, but we prefer to wrte P [u] due to a lesser chance of confusng t wth a product when u s a 1-tuple. For nstance, f n = 2, R = Q, u = 1, 3 and P = x 1 x 2, then P [u] = 1 x 2 x = 8. Of course, every 3 P Z [ x ±1] satsfes P [x] = P, because substtutng the varables x 1, x 2,..., x n for themselves does not change P. We can combne the notatons we have now ntroduced to descrbe certan substtutons n Laurent polynomals. Namely, f A Z n n s a matrx, then A x s an n-tuple of Laurent monomals recall that x = x 1, x 2,..., x n, and thus P [A x] s defned for every Laurent polynomal P Z [ x ±1]. For nstance, [ ] 1 0 [ ] P x = P x 2 3 1, x1 2 x3 2.

6 PRIMES 2015 readng project: Problem set #3 page 6 Exercse 6. Let A GL n Z. a If P Z [ x ±1] s a Laurent polynomal, then show that P s rreducble f and only f P [A x] s rreducble. b If P Z [ x ±1] and Q Z [ x ±1] are two Laurent polynomals, then show that P and Q are coprme f and only f P [A x] and Q [A x] are coprme. Exercse 6 a shows that, for example, f n = 2, then a Laurent polynomal P = P [x 1, x 2 ] Z [ x ±1] s rreducble f and only f P [x 1, x 1 x 2 ] Z [ x ±1] s rreducble 1 0 because x 1, x 1 x 2 = x. 1 1 If v = v 1, v 2,..., v n T Z n s any vector, then we defne a Laurent monomal x v Z [ x ±1] by x v = x v 1 1 xv 2 2 xv n n. Ths s a monomal n Z [x] when all entres of v are nonnegatve. Exercse 7. Let n be a postve nteger. Let v Z n and let g = gcd v. Prove that there exsts an A GL n Z such that x v = x g 1 [A x]. To make sense of the notaton x g 1 [A x], recall that x g 1 s a Laurent monomal and thus a Laurent polynomal, and that A x s an n-tuple of Laurent monomals; thus x g 1 [A x] means the result of substtutng A x for x n x g 1. Ths exercse s useful: It shows that every monomal x v can be obtaned from the monomal x g 1 for g = gcd v by an nvertble substtuton of Laurent monomals for ts varables. Why nvertble? Because 1 shows that t s undone by a smlar substtuton usng the matrx A 1. Let us get some mleage out of ths: Exercse 8. Let v Z n be such that gcd v = 1. Prove that the Laurent polynomal x v + 1 s rreducble n Z [ x ±1]. Here are some examples for Exercse 8: For n = 4 and v = 2, 3, 0, 4 T, we have x v = x1 2x3 2 x 4 4, and thus Exercse 8 yelds that the Laurent polynomal x1 2x3 2 x s rreducble n Z [ x ±1]. For n = 2 and v = 6, 9 T, we have x v = x1 6x9 2, and Exercse 8 cannot be appled because gcd v = 3 = 1. Nor does ts consequent hold: The Laurent polynomal x1 6x factors as x6 1 x = x1 2x x1 4x6 2 x2 1 x , and thus s not rreducble. For n = 2 and v = 2, 4 T, we have x v = x1 2x4 2. Here, Exercse 8 cannot be appled ether, but the Laurent polynomal x1 2x s nevertheless rreducble. So the converse of Exercse 8 does not hold! The followng exercse mght help solvng Exercse 8:

7 PRIMES 2015 readng project: Problem set #3 page 7 Exercse 9. Let n be a postve nteger. Let h be a postve nteger. Let P Z [ x ±1] be a Laurent polynomal whch dvdes x1 h + 1. Show that P has the form x v Q x 1, where x v s a Laurent monomal and Q Z [t] s a polynomal n a sngle varable t. Exercse 8 dscussed rreducblty. Let us now study coprmalty: Exercse 10. Let v Z n and w Z n. Prove that f the Laurent polynomals x v + 1 and x w + 1 are not coprme n Z [ x ±1], then the vectors v and w are proportonal.e., there exsts some ntegers b and c whch are not both 0 and whch satsfy bv = cw. Note that more can be sad: Namely, the Laurent polynomals x v + 1 and x w + 1 are coprme n Z [ x ±1] f and only f there exst no odd coprme ntegers b and c such that bv = cw. Ths s what Fomn and Zelevnsky need to prove [BFZ-CA3, Lemma 3.1]; but t s not needed for provng [BFZ-CA3, Proposton 1.8], and t s harder to show, so you can just as well gnore t Replacng [BFZ-CA3, Lemma 3.1] Now, let us show how to use Exercse 10 n place of [BFZ-CA3, Lemma 3.1] n the proof of [BFZ-CA3, Proposton 1.8]. We use the notatons of [BFZ-CA3]. Let Σ = x, B be a seed of geometrc type. Assume that the matrx B has full rank. We need to check that all seeds mutaton equvalent to Σ are coprme. We shall prove that the seed Σ s coprme. 2 Why s ths suffcent for provng [BFZ-CA3, Proposton 1.8]? Well, assume that we have proven 2. Now, [BFZ-CA3, Lemma 3.2] 6 yelds that every seed mutaton equvalent to Σ has full rank. Then, ths seed must be coprme by 2, appled to ths seed nstead of Σ 7. Therefore, [BFZ-CA3, Proposton 1.8] s proven under the assumpton that 2 holds. So t remans to prove 2. Indeed, recall what t means for the seed Σ to be coprme: It means that the polynomals P 1, P 2,..., P n appearng n the exchange relatons [BFZ-CA3, 1.3] are coprme n ZP [x]. 8 Snce our seed s of geometrc 6 whch, by the way, s a generalzaton of [Lampe, Exercse 2.4 a] 7 Here, we are usng the fact that coprmalty of a seed s defned n terms of the seed alone. Thus, f y, C s a seed n F obtaned by some mutatons from an orgnal seed x, B, then coprmalty of the seed y, C means that the polynomals appearng n ts exchange relatons are parwse coprme n ZP [y], not n ZP [x]. [ ] 8 Recall that ZP s the Laurent polynomal rng Z x n+1 ±1, x±1 n+2,..., x±1 m here.

8 PRIMES 2015 readng project: Problem set #3 page 8 type, we have p + j = n< m; b,j >0 x b,j and p j = n< m; b,j <0 x b,j for all j. follows: Ths allows us to smplfy the formula [BFZ-CA3, 1.3] for P j x as P j x = = p + j = n< m; b,j >0 n< m; b,j >0 x b,j x b,j 1 n; b,j >0 x b,j + p j 1 n; b,j >0 = n< m; b,j <0 x b,j = 1 m; b,j >0 = 1 m; b,j >0 x b,j + x b,j 1 m; b,j <0 x b,j + 1 n; b,j <0 n< m; b,j <0 x b,j x b,j 1 n; b,j <0 x b,j = 1 m; b,j <0 x b,j x b,j. 3 We can further transform the rght hand sde by notcng that 1 m; b,j 0 Thus, 3 becomes x b,j = = 1 m; b,j <0 1 m; b,j <0 x b,j x b,j x b,j 1 m; b,j =0 =x 0 =x 0=1 1 m; b,j =0 } {{ } =1 1 = 1 m; b,j <0 x b,j. P j x = 1 m; b,j >0 x b,j + 1 m; b,j <0 x b,j = 1 m; b,j 0 x b,j = 1 m; b,j >0 x b,j + 1 m; b,j 0 x b,j. 4

9 PRIMES 2015 readng project: Problem set #3 page 9 So far we have been workng n Z [x]. Indeed, all exponents on the rght hand sde of 4 are nonnegatve. Let us now work n Z [ x ±1] ; ths allows us to get an even smpler form for P j x: Namely, P j x = 1 m; b,j >0 x b,j = x b,j / 1 m x b,j 1 m; b,j 0 snce the s whch satsfy b,j >0 are exactly the s whch do not satsfy b,j 0 = x b,j / 1 m = x b,j 1 m 1 m; b,j 0 1 m; b,j 0 = x b,j m x b,j x b,j 1 m; b,j x b,j + 1 m; b,j 0 1 m; b,j 0 1 m; b,j 0 x b,j x b,j x b,j. 5 For every j {1, 2,..., n}, let b j be the j-th column of the matrx B. b j = b 1,j, b 2,j,..., b m,j T Z m, and Then, x b j = Hence, 5 smplfes as follows: P j x = x b,j +1 1 m Hence, because 1 m; b,j 0 =x b j 1 m; b,j 0 x b,j. 1 m x b,j x = b j m; b,j 0 x b,j. 6 [ P j x x b j + 1 n Z x ±1] 7 x b,j s just a Laurent monomal, and thus s nvertble n Z [ x ±1].

10 PRIMES 2015 readng project: Problem set #3 page 10 Now, let us fnally prove that the seed Σ s coprme. To do so, we need to show that the polynomals P 1, P 2,..., P n are parwse coprme n ZP [x]. Indeed, let and j be two dstnct elements of {1, 2,..., n}. We need to show that P and P j are coprme n ZP [x]. Assume the contrary. Then, there exsts a non-nvertble polynomal g ZP [x] whch dvdes each of P and P j n ZP [x]. Consder ths g. We call g a polynomal, but n truth t s a polynomal over ZP, so t can have nverses of the frozen varables x n+1, x n+2,..., x m but not of x 1, x 2,..., x n. We have g P j = P j x x b j + 1 n Z [ x ±1] because of 7. Smlarly, g x b + 1 n Z [ x ±1]. The matrx B has full rank. Thus, ts columns are lnearly ndependent. In partcular, no two of ts columns are proportonal. Thus, the vectors b and b j beng two dstnct columns of B are not proportonal. Thus, the Laurent polynomals x b + 1 and x b j + 1 must be coprme n Z [ x ±1] because otherwse, Exercse 10 appled to m, b and b j nstead of n, v and w would yeld that the vectors b and b j are proportonal. In other words, every Laurent polynomal h Z [ x ±1] whch dvdes both x b + 1 and x b j + 1 must be nvertble n Z [ x ±1]. Thus, g s nvertble n Z [ x ±1] snce g s a Laurent polynomal whch dvdes both x b + 1 and x b j + 1. We are almost done: We know that g s nvertble n Z [ x ±1] but non-nvertble n ZP [x]. These two facts almost make a contradcton, but not qute, because Z [ x ±1] = ZP [x] n general. But Exercse 4 b appled to P = g yelds that ether g or g s a sngle Laurent monomal. We WLOG assume that g s a Laurent monomal because otherwse, we can just replace g by g and get back nto ths case. In other words, g = x k 1 1 xk 2 2 xk m m for some k 1, k 2,..., k m Z. Consder these k 1, k 2,..., k m. Snce x k 1 1 xk 2 2 xk m m = g ZP [x], we must have k l 0 for all l n because the only ndetermnates whch are nvertble n ZP [x] are x n+1, x n+2,..., x m. But f we have k l = 0 for all l n, then the monomal g s nvertble n ZP [x] because t s a Laurent monomal n the ndetermnates x n+1, x n+2,..., x m only, and such monomals are already nvertble n ZP, whch s mpossble because g s nonnvertble n ZP [x] by defnton. Therefore, there exsts some l n such that k l > 0. Consder ths l. Clearly, k l > 0 shows that x l g n ZP [x]. Thus, x l g P j = P j x = 1 m; b,j >0 x b,j + 1 m; b,j <0 x b,j by 3 n ZP [x]. Hence, the term x l must appear n both monomals 1 m; b,j <0 x b,j 1 m; b,j >0 x b,j and because these two monomals surely cannot cancel each other, and because x l s not nvertble n ZP [x]. Therefore, l must both satsfy b,l > 0 so

11 PRIMES 2015 readng project: Problem set #3 page 11 that t can appear n the frst monomal and satsfy b,l < 0 so that t can appear n the second monomal. But ths s clearly absurd. Ths gves us the contradcton we wanted, and thus 2 s proven. Hence, [BFZ-CA3, Proposton 1.8] s proven. References [BFZ-CA3] Arkady Berensten, Sergey Fomn, Andre Zelevnsky, Cluster algebras III: Upper bounds and double Bruhat cells, arxv preprnt arxv:math/ v3. Ths was later publshed n: Duke Mathematcal Journal, Vol. 126, No. 1, [Lampe] Phlpp Lampe, Cluster algebras, A verson wth my correctons: - verson 4 dec 2013 EV.pdf?dl=0

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