MTH 819 Algebra I S13. Homework 1/ Solutions. 1 if p n b and p n+1 b 0 otherwise ) = 0 if p q or n m. W i = rw i

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1 MTH 819 Algebra I S13 Homework 1/ Solutons Defnton A. Let R be PID and V a untary R-module. Let p be a prme n R and n Z +. Then d p,n (V) = dm R/Rp p n 1 Ann V (p n )/p n Ann V (p n+1 ) Note here that p n 1 Ann V (p n )/p n Ann V (p n+1 ) s a module over the feld R/Rp. # 1. Let R be PID. Let p be a prme n R and n Z +. (a) Let (V ) be a famly of untary R-modules. Show that (b) Let b R. Show that (c) Let q be a prme n R and m N. Show that d p,n ( V ) = d p,n (V ) 1 f p n b and p n+1 b d p,n (R/Rb) = 0 otherwse d p,n (R/Rq m 1 f p q and n = m ) = 0 f p q or n m (d) Let V and W be bounded untary R-modules. Show that V s somorphc to W f and only f for all prmes p n R and all n Z +. d p,n (V) = d p,n (W) Observe frst that for any famly of R-modules (W ) and any r R: (1) r and W = rw (2) Ann W (r) = Ann W (r) Also f (U ) s a famly of R-module wth U W, then (3) Thus W / U W /U 1

2 p n 1 Ann V (p n )/p n Ann V (p n+1 ) (2) = p n 1 Ann V (p n )/p n Ann V (p n+1 ) (1) = (3) p n 1 Ann V (p n )/ p n Ann V (p n+1 ) p n 1 Ann V (p n )/p n Ann V (p n+1 ) Snce the dmensons of a drect sum of modules s the sum of the dmensons, ths gves (a). (b) Let r, s, t, v R. s Note r + Rs Ann R/Rs (t) f and only f tr Rs, f and only f s tr and f and only f r. Thus gcd(s,t) s (4) Ann R/Rs (t) = R /Rs R/R gcd (s, t) gcd (s, t) Also s (5) vr/rs = Rv + Rs/Ru = R gcd (v, s) /Rs R/R gcd (s, v) and so (6) vann R/Rs (t) (4) v (R/R gcd (s, t)) (5) gcd (s, t) R/R gcd (gcd (s, t), v) Let b = p k c for some k N and c R wth p c. Then = R/R gcd (s, t) gcd(v, s, t) So f n k then and f k n 1, Thus by (6) and so Snce gcd (b, p n ) = p mn(n,k) and gcd(p n 1, b, p n ) = p mn(n 1,k). gcd (b, p n ) gcd(p n 1, b, p n ) = pn p n 1 = p gcd (b, p n ) gcd(p n 1, b, p n ) = pk p k = 1 p n 1 Ann R/Rb (p n R/Rp f n k ) 0 f k > n dm R/Rp p n 1 Ann R/Rb (p n 1 f k n 1 ) = 0 f k > n 2

3 dm R/Rp p n 1 Ann R/Rb (p n )/p n Ann R/Rb (p n+1 ) = (dm R/Rp p n 1 Ann R/Rb (p n )) (dm R/Rp p n Ann R/Rb (p n+1 )) ths gves Thus (b) holds. 1 1 = 0 f n + 1 k dm R/Rp p n 1 Ann R/Rb (p n )/p n Ann R/Rb (p n+1 ) = 1 0 = 1 f n = k 0 0 = 0 f k n 1 (c) follows mmedately from (b). (d): Let P be set of representatves for the assocated classes of prmes n R. For p P and n N let D p,n be a set of cardnalty d p,n. We wll show (7) V R/Rp n p P D p,n Snce V s bounded, rv = 0 for some 0 r V. In partcular, V s torson module. So by n N (8) V = V p p P where V p s the set of p-elements on V. Let p P and v V p. Then exp R (p) r and exp R (p) p k for some k N. Let e p be maxmal wth p ep r. Then k e p and so p ep V p = 0. Then by Theorem V p = p A p where (A p ) p s a set non cyclc R-submodules of V p. Let A p = Ra p for some a p A p. Then a p s a p-element and so exp R (a p ) = p ep for some e p Z +. Hence by Lemma A p R/Rp ep and so Put I p,n = { I p e,p = n}. Then and so (7) gves V p p R/Rp ep V p R/Rp n p,n n N (10) V R/Rp n p P p,n let q P and m N. Applyng (a) to the drect sum n (10) we get n N d q,m (V) = d q,m (R/Rp m ) p P p,n n N 3

4 and so by (c) d q,m (V) = q,m 1 = I q,m Thus I q,m s a set of cardnalty d q,m and so (10) mples (7). Suppose now that V W. Then also and so also p n 1 Ann V (p n )/p n Ann V (p n+1 ) p n 1 Ann W (p n )/p n Ann W (p n+1 ) for all prmes p n R and all n N. d p,n (V) = d p,n (W) Conversely suppose that d p,n (V) = d p,n (V) for all prmes p n R and all n N. Then (7) appled the V and W shows that both V and W are somorphc to So V W. R/Rp n p P D p,n n N Lemma B. Let R be a PID and 0 b R. Let r = p d pdk k prmes and d 1,... d k Z +. Then where p 1,..., p n are parwse non-assocate R/Rb k R/Rp d =1 Let P be set of representatve for the assocate classes of prmes wth p P for all 1 k. For p P let d p N wth p dp b and p dp+1 b. So d p = d f p = p for some 1 k and d p = 0 f p p for all 1 k. p P and n Z +. Then by Exercse 1(c) 1 f n = d p d p,n (R/Rb) 0 f n d p hence the Lemma follows from equaton (7) n the proof of Exercse 1. # 2. Let R be a PID and V a untary, torson R-module. Show that V s a locally cyclc R-module f and only f for all prmes p of R, Ann V (p) s a cyclc R-module. (Recall that an R-module s called locally cyclc f all fntely-generated R-submodules of V are cyclc.) Suppose frst that V s locally cyclc R-module and let p be a prme n R. Ann V (p) = 0, Ann V (p) s cyclc. So suppose 0 Ann V (p) and pck 0 u Ann V (p). Let v Ann V (p). Snce V s locally cyclc, u, v R = w for some w V. Then w u, v Ann V (p) and u = rw for some r R. Snce u 0 and pw = 0, p r. Snce R s a PID ths gves gcd (p, r) = 1 and 1 = xp + yr for some x, y R. Thus w = 1w = (xp + yr)w = x(pw) + y(rw) = x0 + yv = yv. So v Rw = Ryv Ru. Sne ths holds for all v Ann V (p), Ann V (p) = Ru and so Ann V (p) s cyclc, Suppose next that Ann V (p) s cyclc for all prmes p V. Let M be fntely generated R-submodule of V. By Theorem

5 M R R... R R/Rp k R/Rpkn n k tmes for some n, k N, k 1..., k n Z = and prmes p 1,..., p n n R. Snce V s a torson module and R s not, k = 0. Suppose that p p j for some 1 < j n. Put p = p. By equatons (2) and (4) n the proof of Exercse 1 Ann k R/Rp R/Rp k j j (p) Ann R/Rp k (p) Ann R/Rp k j (p) R/R gcd (p k, p) R/R gcd (p k j, p) = R/Rp R/Rp Thus dm R/Rp Ann V (p) dm R/Rp R/Rp R/Rp = 2. On the other hand Ann V (p) s cyclc and so Ann V (p) = Rv for some v Ann V (p). Then v 0, exp R (v) p and Ann V (p) R/Rp. Hence dm Ann V (p) = 1, a contradcton. Thus p 1,..., p k are parwse non-assocate prmes and Lemma B shows that So M s cyclc and V s locally cyclc. M R/Rp k R/Rpkn n R/Rp k pkn n # 3. Let F be a feld and a F. Consder the matrces A = a and B = 0 0 a 0 over F. Determne the Jordan Canoncal Form of A and B over F. (Note that the answers wll depend on F and a). Vew F 4 as an F[x]-module va gv = vg(a) for all g F[x] and v F 4. Put f = x Note that A = M( f ), where M( f ) s the matrx defned n Lemma So lemma shows that F 4 F[x]/F[x] f an F[x]-module. Suppose that 2 = 0 n F. Then x = (x 1) 4 and so F 4 F[x]/(x 1) 4. Thus the Jordan canoncal form or A s Case 2 = 0 M(x 1, 4) = Suppose now that 1 1. Assume that x s rreducble. Then A s n Jordan canoncal form s 5

6 Case 2 0, x rreducble M(x ) = A = M(x 1, 4) = Assume that x has root a n F. So a 4 = 1 1. Also a 8 = 1 and so a has multplcatve order 8. Thus a, = 1, 3, 5, 7 are parwse dstnct. If s an odd nteger, then (a ) 4 = (a 4 ) = ( 1) = 1 and so a s root of x 4 = 1. So a, a 3, a 5 and a 7 are the roots of x and Lemma B now shows that x 4 = 1 = (x a)(x a 3 )(x a 5 )(x a 7 ) F 4 F[x]/F[x](x 4 + 1) F[x]/F[x](x a) F[x]/F[x](x a 3 )F[x]/F[x](x a 5 )F[x]/F[x](x a 7 ) Thus the Jordan canoncal form of A s a Case 2 0, a 4 = 1 M(x a, 1 x a 3, 1 x a 5, 1 x a 7 0 a 3 0 0, a) = 0 0 a a 7 Assume that f s nether rreducble nor has a root n F. Then f = gh where g and h are monc rreducble polynomal of degree 2 n F[x]. Let g = x 2 + ax + b and h = x 2 + cx + d wth a, b, c, d F. Then and so x = (x 2 + ax + b)(x 2 + cx + d) = x 4 + (a + c)x 3 + (ac + b + d)x 2 + (bc + ad)x + bd (1) a + c = 0, ac + b + d = 0, bc + ad = 0 andbd = 1 So c = a and 0 = bc + ad = ab + ad = a(d b) Hence a = 0 or b = d. Suppose that a = 0. Then also c = a = 0 and (1) gves b + d = 0 and bd = 1. So d = b and b 2 = 1. Note that x 2 + b and x 2 b are rreducble f and only f x 2 + b and x 2 b have no roots and f and only f x has no roots. Also Snce 1 1, also b b and so x = (x 2 + b)(x 2 b) F 4 F[x]/F[x](x 4 + 1) F[x]/F[x](x 2 + b) F[x]/F[x](x 2 b) 6

7 Thus the Jordan canoncal form of A s Case b 2 = 1, x has no roots M(x 2 b, 1 x 2 b b, 1) = 0 0 b 0 Suppose next that a 0. Then b = d and c = a. So (1) becomes a 2 + 2b = 0 and b 2 = 1. Thus b = ±1, a 2 = 2b and x = (x 2 + ax + b)(x 2 ax + b) By the quadratc formula x 2 ± ax + b has root n F f and only f b + a2 4 square and so f and only f x has a root. Snce a a, these are non-assocate rreducble polynomals, and F 4 F[x]/F[x](x 4 + 1) F[x]/F[x](x 2 + ax + b) F[x]/F[x](x 2 ax + b) If b = 1 then the Jordan canoncal form of A s 2b = b + 4 = 2b 4 = ( a 2 ) s a Case 2 0, a 2 = 2 and x has no roots M(x 2 + ax + 1, 1 x 2 1 a 0 0 ax + 1, 1) = a and f b = 1, Case 2 0, a 2 = 2 and x has no roots M(x 2 + ax 1, 1 x 2 1 a 0 0 ax 1, 1) = a To compute the Jordan canoncal form of B vew F 4 as an F[x]-module va gv = vg(b) for all g F[x] and v F 4. Let m F[x] be a monc polynomal wth m exp F[x] (F 4 ). So m generates the deal Ann F[x]] (F 4 ) and so m s the unque monc polynomal of mnmal degree wth respect to m(b) = 0. Note that m(b) = m 0 B 0 + m 1 B m k 1 B k 1 + B k, where k = deg m. So we can compute m by fndng the smallest k such that B 0, B, B 2,..., B k s lnearly ndependent over F. 7

8 a a 3a 0 a B a =, B =, B a 2a 0 = B 0 0 a 0 3 a a =, B a 4 0 a 2 4a 0 = 0 0 a a a 0 0 a a 2 B 0 and B 1 are clearly lnearly ndependent. If 2 0, also B 0, B 1 and B 2 are lnearly ndependent. If 2 = 0, then B 2 + ab 0 = 0 and so m = x 2 + a n ths case. Suppose next that 2 0. Suppose that B 3 = a 0 B 0 + a 1 B + a 2 B 2. The (1, 2) and (1, 3) coeffcent mples a 1 = a and a 1 = 3a. Thus a = 3a, 2a = 0 and snce 2 0, a = 0. If a = 0, then B 3 = 0 and so m = x 3 f a = 0 and 2 0. Note that B 4 = 2aB 2 a 2 B 0 and so m = x 4 2ax 2 + a = (x 2 a) 2 f 2 0 and a 0. To summarze x 2 + a f 2 = 0 (2) m = x 3 f 2 0, a = 0 (x 2 a) 2 f 2 0, a 0 Note that F 4 =1 n F[x]/F[x]p k for some rreducble monc polynomals p 1,..., p n n F[x] and postve ntegers k 1,..., k n. Note that Thus dm F F[x]/F[x]p k = k deg p and exp F[x] F[x]/F[x]p k = p k (3) n =1 k deg p = dm F 4 = 4 Also by Lemma the exponent of a sum of modules s the least common multpler the exponents of the summands and so (4) m = exp F[x] (F 4 ) = lcm n =1p k Suppose 2 = 0 and x 2 + a s rreducble. Then m = x 2 + a s rreducble and (4) shows that p k = x 2 + a for all 1 k. (3) mples k = 2 and so the Jordan canoncal form of A s Case 2 = 0, x 2 + a rreducble M(x 2 + a, 1 x 2 a a, 1) = 0 0 a 0 Suppose next that 2 = 0 and x 2 +a s not rreducble. Then a = b 2 for some b F and m = x 2 +a = (x+b) 2. 8

9 Hence (4) shows that p k = (x + b) k for some 1 k 2 and there exsts wth k = 1. Usng (3) and reorderng the p k1 f necessary we get n = 2, p k1 1 = pk2 2 = (x + b)2 or n = 2, p k1 1 = (x + b)2, p k2 2 = pk2 3 = x + b Let p = x + b and V = F 4. Then p 2 = m and so p 2 V and Ann V (p 2 ) = V = Ann V (p 3 ). Thus Also F[x]/F[x]p F and so pann V (p 2 )/p 2 Ann V (p 3 ) = pv/p 2 V = pv/0 pv d p,2 (B) = dm F ppv Note that pv = F 4 p(b) s just the Column space of the matrx b b a b 0 1 b 2 b 0 1 p(b) = B + bi = = 0 0 b b a b 0 0 b 2 b Note the the second columns f b tmes the frst plus the thrd, and the last columns f b tmes the thrd. Thus Ths shows that exactly two of the p k pv = (b, 1, 1, 0), (0, 0, b, 1) F and d p,2 = dm F pv = 2 are equal to p 2 = (x + b) 2. Thus the Jordan canoncal form of B s b Case 2 = 0, b 2 0 b 0 0 = a M(x + b, 2 x + b, 2) = 0 0 b b Suppose next that 2 0 and a = 0. Then m = x 3. So (4) mples that p k = x k, and k 3 for all 1 n and k = 3 for at least one. Thus (3) shows that (possble after reorderng) n2, p k1 1 = x3 and p k2 2 = x.thus the Jordan canoncal form of B s Case 2 0, a = 0 M(x, 3 x, 1) = Suppose next that 2 0 and x 2 a s rreducble. (and so a 0). Then m = (x 2 a) 2. From (3) and (4) we get n = 1 and p k1 1 = m = (x2 a) 2. Thus the Jordan canoncal form of B s 9

10 Case 2 0, x 2 a rreducble M(x 2 a a, 2) = 0 0 a 0 Suppose next that 2 0, a 0 and x 2 a s not rreducble. Then x 2 a = (x b)(x + b) for some b F and snce 2 0, b b. Also m = (x b) 2 (x + b) 2. Hence (4) shows that p k = (x b) 2 for some and = (x + b) 2 for some j. (3) mples n = 2 and so the Jordan canoncal form of B s p k j j b Case 2 0, b 2 0 b 0 0 = a 0 M(x b, 2 x + b, 2) = 0 0 b b # 4. Let R be rng and consder the followng homomorphsm of exact sequences of R-lnear maps: Show that: A 1 f 1 A2 f 2 A3 f 3 A4 f 4 A5 α1 α 2 α 3 α 4 α. 5 B 1 B 2 B 3 B 4 g 1 g 2 g 3 g 4 B 5 (a) If α 1 s onto and α 2 and α 4 are 1-1, then α 3 s 1-1. (b) If α 5 s 1-1 and α 2 and α 4 are onto, then α 3 s onto. (a) Snce α 2 s onto Img 1 = g 1 (B 1 ) = g 1 (α 1 (A 1 )) = Im(g 1 α 1 ). Thus Consder the functon α 2 (ker f 2 ) = α 2 (Im f 1 )Im(α 2 f 1 ) = Im g1 α 1 = Im g1 = ker g 2 Then α 2 = π ker(g2) α 2 A 2 B 2 / ker g 2, a 2 α 2 (a 2 ) + ker g 2 ker α 2 = {a 2 A 2 α 2 (a 2 ) ker g 2 } Snce α 2 s 1-1, α 2 (a 2 ) = α 2 (a) for some a ker f 2 f and only f a 2 ker f 2. As ker α 2 = α 2 (ker f 2 ), ths shows ker α 2 = ker f 2 and so we obtan 1-1 R-lnear functon α 2 A 2 / ker f 2 B 2 / ker g 2, a 2 + ker f 2 α 2 (a 2 ) + ker g 2 Note that we also have 1-1 -R-lnear maps 10

11 and f 2 A 2 / ker f 2 A 3, a 2 + ker f 2, a 2 + ker f 2 f 2 (a 2 ) g 2 B 2 / ker g 2 B 3, b 2 + ker g 2, b 2 + ker g 2 g 2 (b 2 ) and so we obtan the followng homomorphsm of exact sequences of R-lnear maps: Thus (a) follows from the Short Fve Lemma. (b) α 5 s 1-1, 0 A 2 / ker f 2 f 2 A3 f 3 A4 α2 α 3 α. 4 0 B 2 / ker g 2 B 3 g 2 g 3 B 4 Let a 4 A 4. Then ker(g 4 α 4 ) = ker(α 5 f 4 ) = ker f 4 = Im f 3 α 4 (a 4 ) Img 3 α 4 (a 4 ) ker g 4 a 4 ker(g 4 α 4 ) a 4 Im f 3 Snce α 4 s onto, ths mples α 4 (Img 3 ) = Im f 3. So we obtan the followng homomorphsm of exact sequences of R-lnear maps: Thus (b) follows from the Short Fve Lemma. f 2 f 3 A 2 A3 Im f3 0 α2 α 3 α 4 Im f3. B 2 B 3 Img 3 0 g 2 g 3 # 5. Let R be a rng. Show that the sequence of R-modules f g h 0 A B D E 0. s exact f and only f there exst short exact sequences of R-modules 0 A wth g = g 2 g 1. f B Suppose frst that g 1 C 0 and 0 C 0 A f B g D s exact. Put C = Img = ker h. Then g = d C g and the sequences g 2 D h E 0. h E 0 11

12 0 A are exact. f B Suppose next that g C 0 and 0 C d C D h E 0 0 A f B g 1 C 0 and 0 C are short exact sequences of R-modules wth g = g 2 g 1. Snce g 1 s onto, Img = Im(g 2 g 1 ) = Img 2 = ker h. Snce g 2 s 1-1, ker g = ker(g 2 g 1 ) = ker g 1 = Im f. So s exact. 0 A f B g D g 2 D h E 0. # 6. Show that somorphsm of short exact sequences s an equvalence relaton. Obvous. h E 0 Lemma C. Let R be a rng, S a subrng of R and A subset of R wth R = S [A]. Let V be rght R-module and W an left R-module. (a) Let f V W D be an S -balanced map wth f (va, w) = f (v, aw) for all v V, a A and w W. Then f s R-balanced. (b) Put U = va w v aw v V, a A, w W V S W. Then s a tensor product of V and W over R. V W (V S W)/U, (v, w) v w + U Proof. (a) Put T = {t R f (vt, w) = f (v, tw) for all v V, w W}. Snce f s S -balanced S T and by assumpton A T. We clam that T s subrng of R, So let r, t T, v V and w W. Then and v(r + t) w = (vr + vt) w = vr w + vt w = v rw + v tw = v (rw + tw) = v (r + t)w v(rt) w = (vr)t w = vr tw = v r(tw) = v (rt)w So r + t and rt T and T s subrng of R. Snce S A T and R = S [A], ths shows that R = T and so f s R-balanced. (b) By (a) s R-balanced. Let f V W D be R-balanced. Then f s also S balanced and so there exsts a unque Z-lnear functon g V S W wth f = g. Snce f s R-balanced, va w v aw ker g for all v V, a A, w W. Snce g s Z-lnear we conclude that U ker g. Hence g (V S W)/U D, z + U g(z) s the unque Z-lnear functon from (V S W)/U to D wth f = g. So s ndeed the tensor product of V and W over R. 12

13 Corollary D. Let R be rng, V and R-module and α End R (V). Let Φ R End(V) the homomorphsm assocated to the acton of R on V. Note that α commutes a each Φ(r), r R, there exsts a unque homomorphsm Φ α R[x] End(V) wth (Φ α ) R = Φ and Φ α (x) = α and so V s an R[x]-module va f v = (Φ a f )v = ( f α)v. Defne Then s an R[x]-somorphsm wth nverse β R[x] R V R[x] R V, f v f x v f αv (R[x] R V)/Im β V, ( f v) + Imβ ( f α)v V (R[x] R V)/Im β, v 1 v + Imβ Proof. Note that β( f v) = f x v f xv and so snce R[x] R V s generated by the f v, Hence C(b) show that Im β = f x v f xv f R[x], v V R[x] V (R[x] R V)/Im, β s tensor product of R[x] and V over R[x]. By Example (b) also ( f, v) ( f v) + Imβ R[x] V V, ( f, v) f v s tensor product of R[x] and V over R[x]. Thus the lemma follows from the unqueness of the tensor product. Defnton E. Let R be a rng, A an I J-matrx n A and (, j) I J. (a) (, j) s called a non-zero poston f a j 0. (b) (, j) s called a leadng poston f a j 0 but a l = 0 and a kl = 0 for all k I and all j l J. So (, j) s the only non-zero poston n Row and n column j of A. (c) A s called a monomal matrx f all non-zero poston are lead postons. Lemma F. Let R be a rng and V a free R-module wth bass v = (v j ) j J. Let (w ) be famly n V and W = w I. Let A M I J(R) defned by w = j J A j v. Suppose there exsts nvertble matrces B M I I(R) and C M J J such that D = BAC 1 s a monomal matrx. For j J put u j = l J c l j v l and u j = u j + W. Defne d j = d j f there exsts I such that (, j) s lead poston of D, and d j = 0 otherwse. Then (a) V/W = j J Ru j. (b) Ru j R/Rd j for all j J. Proof. Note that W s the mage of the lnear functon α R J V defned by α(e ) = w, where e = (e ) s the standards bass for R I. Also A s the matrx of α wth respect to e and v. Snce B s nvertble, the rows of B for a bass f of R I and snce C s nvertble, u = (u j ) j J s a bass for V. Note that D s the matrx of α wth respect to f and u. Let I. If the -row of D s zero, then α f = 0. Otherwse (, j) s a lead-poston of D for some j J and α( f ) = d j u j = d j u j. Thus 13

14 Let r R J. Then W = Imα = α( f ) I R = d j u j j J R = { s j d j u j s R J } j J r j u j = 0 j J r j u j W j J j J r j u j = s j d j u j j J for some s R J r j u j = s j d j for some s R J and all j J r j Rd j for all j J In partcular r j u j = 0 f and only f r j Rd j and we conclude that (a) and (b) hold. Lemma G. Let (R, deg) be an Eucldean doman, I and J be sets wth J fnte, and A an I J-matrx n R. Defne m N and matrces A (s) ), B (s),c (s), 0 s m nductvely as follows: (1) Put A (0) = A, B (0) = I I, C (0) = I J and t = 0. (2) Put D = A (t). (3) If D s a monomal matrx, put m = t and termnate the algorthm. (4) If D s not a monomal matrx let (k, l) I J be a non-zero, non-leadng poston wth deg(d kl ) mnmal. For k I choose q, r n R wth D l = q D kl + r and deg(r ) < deg(d kl ). Put q k = 0 and defne D = [D q D k ] and B t+1 = [B q B k ] For l j J choose u j, v j n R wth D k j = u j D kl + v j and deg(v j ) < deg(d kl ). Put v l = 0 and defne A (t+1) = [ D, j v j D,l ], j J. Let C (t+1) be the J J-matrx obtaned by addng j J v j C j to row l of C (t). Replace t by t + 1 and contnue wth at Step (2). Then the algorthm termnates n fnally many steps. Put D = A (m), B = B (m) and C = C (m). Then D s a monomal I J matrx. B s an nvertble I I-matrx, C s an nvertble J J-matrx and D = BAC 1. Proof. For X M IJ (R) let z(x) be the number of leadng postons of X and deg(x) = mn{deg X j (, j) I J (, j) a non zero, non-leadng poston} We wll show by nducton that for all 0 t m, (a) B (t) and C (t) are nvertble. (b) A (t) = B (t) AC (t) 1. 14

15 (c) If t > 0, then z(a (t) ) > z(a (t 1) ) or deg A (t) < deg A (t 1). Note that ths holds for t = 0. Assume t holds for t and D = A (t) s not monomal. Let E be the I I matrx wth column k equal to (q ) and all other columns equal to zero. Let F be the J J matrx whose l-row s (v j ) j J and all other rows are zero. Then D = (I I E)D, B (t+1) = (I I E)B (t), A (t+1) = D(I J F), C (t+1) = (I J + F)C (t) Snce both I I E and I J + F are nvertble we conclude that B (t+1) and C (t+1) are nvertble. So (a) holds for t + 1. Moreover, A (t+1) = (I I E)D(I J F) = (I I E)B (t) AC (t) 1 (I J + F) 1 = (I I E)B (t) A((I J + F)C (t) ) 1 = B (t+1) AC (t+1) 1 and (b) holds for t + 1. Suppose that deg A (t+1) deg A (t). Note that for l, the (, l) entry of A (t+1) s equal to v. Snce deg v < deg(d kl ) = deg A (t) and deg A (t+1) deg A (t) we conclude that (k, l) s the only non-zero entry n column l of A (t+1). Smlarly, (k, l) s the only non-zero entry n row k of A (t+1) and thus (k, l) s a leadng poston of A (t+1). Also observe that f (x, y) s leadng poston of D = A (t), then (x, y) s stll leadng poston of A (t+1). Thus z(a (t+1) > z(a (t) and so (c) holds for t + 1. Snce J s fnte and the number of leadng poston of any I J matrx s at most J, (c) mples that the algorthm termnates n fntely many steps. (a) an (b) appled wth t = m now shows that the lemma holds. # 7. Let V be a fnte dmensonal vector space over the feld F and A End F (V). Fnd an algorthm to compute a bass v for V such that M vv (A) s n Jordan Canoncal Form. Step 1: Somehow fnd a bass w = (w ) m =1 for V and method to compute the coordnates u V wth respect to w. Step 2: Compute B = M ww (A) for A wth respect to v. Step 3: Step 1: Let Z + and suppose A 0,..., A 1 have already been computed. and are lnear Fnd a monc polynomal m F[x] of mnmal degree such that m(a) = 0. Step 2: Fnd parwse dstnct monc rreducble polynomals p 1,..., p n and postve ntegers k 1,..., k n wth m = p k pkn n 15