ALGEBRA HW 7 CLAY SHONKWILER

ALGEBRA HW 7 CLAY SHONKWILER 1 Whch of the followng rngs R are dscrete valuaton rngs? For those that are, fnd the fracton feld K = frac R, the resdue feld k = R/m (where m) s the maxmal deal), and a unformzer π. For the others, explan why not. Z, Z (5), Z[1/5], R[x], R[x] (x 2), R[x, 1/(x 2)], Q[x] (x 2 +1), C[x, y] (x,y), ( R[x, y]/(x 2 + y 2 1) ) (x 1,y), ( R[x, y]/(y 2 x 3 ) ) (x,y). Answer: Z: Z s not a d.v.r., snce Z has maxmal deals (p) for p prme; that s, Z s not a local rng. Z (5) : Z (5) s a d.v.r., wth maxmal deal m = (5). The unformzer s π = 5, snce every deal s of the form (5 a ). The fracton feld K = frac Z (5) s { a } K = b : a, b Z (5), b 0 = Q. Fnally, the resdue feld s k = Z (5) /(5) = Z/5. Z[1/5]: Z[1/5] s not a d.v.r., snce Z[1/5] has maxmal deals (p) where p prme and p 5; that s, Z[1/5] s not local. R[x]: R[x] s not a d.v.r., snce (x a) s a maxmal deal of R[x] for all a; that s, R[x] s not a local rng. R[x] (x 2) : R[x] (x 2) s a d.v.r., wth maxmal deal m = (x 2). The unformzer s π = x 2. The fracton feld K = frac R[x] (x 2) s { } f(x) K = g(x) f(x), g(x) R[x] (x 2), g(x) 0 = R(x). The resdue feld s k = R[x] (x 2) /(x 2) = R. R[x, 1/(x 2)]: Ths s not a d.v.r., snce, n R[x, 1/(x 2)], (x a) s a maxmal deal for a 2; that s, R[x, 1/(x 2)] s not a local rng. Q[x] (x 2 +1): Ths s a d.v.r., wth maxmal deal m = (x 2 + 1). The unformzer s π = x 2 + 1. The fracton feld K = frac Q[x] (x 2 +1) s K = { f g : f, g Q[x] (x 2 +1), g 0 The resdue feld s } = Q[x] (x 2 +1) k = Q[x] (x 2 +1)/(x 2 + 1) = Q[]. 1 ( 1 x 2 + 1 ) = Q(x).

2 CLAY SHONKWILER C[x, y] (x,y) : C[x, y] (x,y) s not a d.v.r., snce t s not dmenson 1; specfcally, ( t s dmenson 2. R[x, y]/(x 2 + y 2 1) ) : Ths rng s a d.v.r., wth maxmal deal (x 1,y) m = (y). Ths s because x + 1 / (x 1, y), so 1 x+1 R and ( ) ( (x + 1)(x 1) x 2 ) ( ) 1 x 1 = = = y2 y x + 1 x + 1 x + 1 = y (y). x + 1 Hence, (x 1, y) = (y), so R s a d.v.r. wth unformzer π = y. Now, R(y) njects nto R and R(y)[x]/(x 2 + y 2 1) contans R. Snce x 2 + (y 2 1) s rreducble n R(y)[x], R(y)[x]/(x 2 + y 2 1) s a feld and so contans the fracton feld K = frac R. Now, K R(y) so, snce R(y)[x]/(x 2 + y 2 1) s a quadratc extenson of R(y), we see that ether K = R(y) or K = R(y)[x]/(x 2 + y 2 1). Clearly, R R(y), so we conclude that The resdue feld s K = R(y)[x]/(x 2 + y 2 1). k = ( R[x, y]/(x 2 + y 2 1) ) /(y) = R[x]/(x 1) = R. (x 1,y) ( R[x, y]/(y 2 x 3 ) ) (x,y) : As we saw n class, z = x y s n the ntegral closure of R, but s not contaned n R, so R s not ntegrally closed and, therefore, not a d.v.r. 2 Let R be a dscrete valuaton rng wth fracton feld K, maxmal deal m, and dscrete valuaton v. If a, b K, defne ρ(a, b) = 2 v(a b) f a b, and defne ρ(a, a) = 0. (a): Show that ρ defnes a metrc on K. Proof. Suppose ρ(a, b) = 0. Then 2 v(a b) = 0 or a = b. However, snce 2 c > 0 for all c Z, only the latter case obtans, so a = b. On the other hand, by defnton, ρ(a, a) = 0. To show symmetry, note frst that symmetry s trval n the case that a = b. Thus, suppose π s the unformzer of the dscrete valuaton. Then, for a, b K, a b, a b = uπ n where u R and v(a b) = n. Then b a = (a b) = uπ n ; snce u R, v(b a) = n = v(a b). Hence, so ρ s symmetrc. ρ(a, b) = 2 v(a b) = 2 v(b a) = ρ(b, a),

ALGEBRA HW 7 3 To see that ρ satsfes the trangle nequalty, suppose a, b, c K. If b = a or b = c, the trangle nequalty s trval, so suppose that b a and b c. Smlarly f a = c, then the trangle nequalty s trval, so assume a c. Then v(a c) = v(a b + b c) mn{v(a b), v(b c)}, so 2 v(a c) max { 2 v(a b), 2 v(b c)}. Hence, snce 2 n > 0 for all n Z, ρ(a, c) = 2 v(a c) 2 v(a b) + 2 v(b c) = ρ(a, b) + ρ(b, c). (b): Show that ρ s an ultrametrc (=non-archmedan metrc);.e., t satsfes the strong trangle nequalty ρ(a, c) max(ρ(a, b), ρ(b, c)). Proof. Let a, b, c K. If a = c, then ρ(a, c) = 0 max{ρ(a, b), ρ(b, c)} trvally. If b = a or b = c then ths statement s trval, so assume b a, b c, a c. Then n part (a) we saw that { ρ(a, c) = 2 v(a c) max 2 v(a b), 2 v(b c)} = max{ρ(a, b), ρ(b, c)}. (c): Show that (K, ρ) s a topologcal feld,.e. that t s a topologcal space n whch addton and multplcaton defne contnuous maps K K K. Proof. K s certanly a topologcal space under the metrc topology, and K K s a topologcal space under the product topology. Let (a, b) K K and let ɛ > 0. If (c, d) K K such that 0 < ρ(a, c) < ɛ and 0 < ρ(b, d) < ɛ, then, f a + b c + d, ρ(a + b, c + d) = 2 v((a+b) (c+d)) = 2 v((a c)+(b d)) max {2 v(a c), 2 v(b d)} = max{ρ(a, c), ρ(b, d)} < ɛ. Obvously f a + b = c + d, then ρ(a + b, c + d) = 0 < ɛ. Therefore, snce our choce of ɛ > 0 was arbtrary, we see that the functon defned by addton s contnuous at (a, b). Snce our choce of (a, b) was arbtrary, ths mples that addton, as a map K K K, s contnuous. Turnng to multplcaton, agan let (a, b) K K and let ɛ > 0. ɛ Let δ 1 = max{1,ρ(b,0)} and let δ ɛ 2 = max{δ 1,ρ(a,0)}. Now, f (c, d) K K such that ρ(a, c) < δ 1 and ρ(b, d) < δ 2, then, so long as ab cd (f ab = cd then ρ(ab, cd) = 0 < ɛ), (1) ρ(ab, cd) max {ρ(ab, cb), ρ(bc, cd)}.

4 CLAY SHONKWILER (2) (3) Now, f ab = cb, then ρ(ab, cb) = 0 < ɛ. If ab cb, then b 0 and a c. Hence, ρ(ab, cb) = 2 v(ab cb) = 2 v(b(a c)) = 2 (v(b)+v(a c)) = 2 v(b) 2 v(a c) = ρ(b, 0) ρ(a, c) < ρ(b, 0)δ 1 ɛ < ρ(b, 0) ρ(b, 0) = ɛ. Thus, ρ(ab, cb) < ɛ. On the other hand, f bc = cd, then ρ(bc, cd) = 0 < ɛ. If bc cd, then c 0 and b d. Hence, ρ(bc, cd) = 2 v(bc cd) = 2 v(c(b d)) = 2 (v(c)+v(b d)) = 2 v(c) 2 v(b d) = ρ(c, 0) ρ(b, d) (max{ρ(c, a), ρ(a, 0)}) ρ(b, d) = (max{δ 1, ρ(a, 0)}) ρ(b, d) ɛ < (max{δ 1, ρ(a, 0)}) max{δ 1, ρ(a, 0)} = ɛ. Thus, ρ(bc, cd) < ɛ. Therefore, combnng (1), (2) and (3), we see that ρ(ab, cd) max {ρ(ab, cb), ρ(bc, cd)} < ɛ. Snce our choce of ɛ was arbtrary, we see that the functon from K K K defned by multplcaton s contnuous at (a, b); snce our choce of (a, b) was arbtrary, we see that multplcaton s contnuous on all of K K. Thus, havng shown that K s a topologcal space and that addton and multplcaton defned contnuous functons K K K, we conclude that K s a topologcal feld. (d): Show that n K, the closed unt dsc about 0 s R and the open unt dsc about 0 s m. Proof. Recall that R = {a K : v(a) 0}. Then, for any a R, v(a) 0 and so 2 v(a) 1. ρ(0, a) = ρ(a, 0) = 2 v(a 0) = 2 v(a) 1.

ALGEBRA HW 7 5 So R = {a K : ρ(0, a) 1}, the closed unt dsc n K. Now, the maxmal deal m = {a R : v(a) > 0}. Hence, f a m, v(a) > 0, so 2 v(a) < 1, so ρ(0, a) = ρ(a, 0) = 2 v(a 0) = 2 v(a) < 1. Thus, m = {a K : ρ(0, a) < 1}, the open unt dsc n K. 3 Let K be a feld and let f(x) K[x] be a non-zero polynomal of degree n. (a): Show that f a K s a root of f, then (x a) dvdes f(x) n K[x]. Proof. We prove ths by nducton on the degree of f. Note that, snce we re n a feld, we may as well assume f s monc. If the degree of f s 1, then f(x) = x b, so f a s a root of f, a b = 0, so a = b. Now, suppose that, f a s a root of f mples (x a) dvdes f(x) for all polynomals such that deg(f) = k. Then, f f(x) K[x] such that deg(f) = k + 1 and a s a root of f, then f(x) = x k+1 + c k x k +... + c 1 x + c 0. Consder the frst step of the dvson algorthm: x a Now, let ) x n 1 x n + c n 1 x n 1 + c n 2 x n 2... + c 1 x + c 0 x n ax n 1 (c n 1 a)x n 1 + c n 2 x n 2 +... + c 1 x + c 0 g(x) = f(x) x k (x a) = (c k + a)x k + c k 1 x k 1 +... + c 1 x + c 0. Then, f a s a root of f, g(a) = (c k +a)a k +c k 1 a k 1 +... c 1 (a)+c 0 = a k+1 +c k a k +...+c 1 (a)+c 0 = f(a) = 0, so a s also a root of g. Now, snce deg(g) = k, we know, by the nducton hypothess, that (x a) dvdes g(x), so g(x) = (x a)h(x). Therefore, f(x) = x k (x a) + g(x) = x k (x a) + h(x)(x a) = (x k + h(x))(x a), so x a dvdes f(x). (b): Deduce that f has at most n roots n K.

6 CLAY SHONKWILER Proof. Let a 1 be a root of f. Then, by part (a), f(x) = (x a 1 )f 1 (x), for some f 1 (x) K[x] wth deg(f 1 ) = n 1. Then, f a 2 s another root of f, then a 2 must also be a root of f 1, so f(x) = (x a 1 )(x a 2 )f 2 (x) where deg(f 2 ) = n 2. Iteratng ths process, we see that we eventually factor f nto a product of rreducbles; n ths factorzaton, there can be at most n lnear factors, and, thus, at most n assocated roots of f. Snce K[x] s a UFD, ths factorzaton s unque up to multplcaton by a constant, so f s not dvsble by any lnear factors not appearng n ths factorzaton. Hence, f b s a root of f, then, by (a), x b must dvde f(x), so t must already appear n ths factorzaton. Therefore, we see that there are at most n roots of f. (c): Wll the argument and concluson of part (b) stll hold f K s replaced by a dvson algebra? Explan. Answer: If K s replaced by a dvson algebra, the result n part (b) no longer holds. To see why, consder the dvson algebra H and the polynomal x 2 + 1 n H[x]. Then ±, ±j and ±k are all roots of x 2 + 1, so ths polynomal has at least sx roots n H despte only beng of degree 2. The reason (b) fals f K s replaced by a dvson algebra s that K[x] s not necessarly a UFD f K s a dvson algebra, and we defntely needed unque factorzaton n part (b). 4 Let R be a commutatve rng of characterstc p (where p s prme) and defne F : R R by a a p. (a): Show that F s a rng endomorphsm. Proof. Clearly, F (0) = 0 p = 0 and F (1) = 1 p = 1. Now, f a, b R, then F (ab) = (ab) p = a p b p = F (a)f (b). Also, ( ) ( ) p p F (a+b) = (a+b) p = a p + a p 1 b+...+ ab p 1 +b p = a p +b p = F (a)+f (b), 1 p 1 snce ( p k) s a multple of p for 1 k p 1. Therefore, we see that F preserves the denttes, addton and multplcaton, so F : R R s an endomorphsm.

ALGEBRA HW 7 7 (b): If R s a feld, determne whch elements le n the set {a R F (a) = a}. Answer: If a {a R F (a) = a}, then a s a root of the polynomal x p x. On the other hand, roots of ths polynomal are certanly n the set, so we see that {a R F (a) = a} = {a R a s a root of x p x}. Now, by 3(b) above, x p p has at most p roots n K. On the other hand, snce K s a feld of characterstc p, t contans the prme feld Z/p. Now, f a Z/p, then a = 1 +... + 1; hence, a p = (1+...+1) p = 1 p +(1+...+1) p =... = 1 p +...+1 p = 1+...+1 = a, so a s a root of x p x. Thus, all elements of Z/p are roots of ths polynomal, so there are at least p roots (snce there are p elements of Z/p). Therefore, we conclude that there are exactly p roots of x p p n R, the elements of Z/p. So Z/p = {a R F (a) = a} where Z/p s consdered as contaned n R. (c): If R s a feld, must F be njectve? surjectve? Answer: If R s a feld, then F : R R s an endomorphsm by (a), so ker F s an deal of R. Certanly F (1) = 1, so ker F R; the only other deal of R s (0), so ker F = (0). Hence, F s njectve. On the other hand, suppose R = Z/p(x). Then R s a feld. Consder the element x R. If x = F (a) = a p for some a R, then a = f(x) g(x) for f(x), g(x) Z/p[x] and ( ) f(x) p x = = (f(x))p g(x) (g(x)) p, so xg(x) p = f(x) p. If a n x n s the hghest-order term of g(x) and b m x m s the hghest-order term of f(x), then ths mples that a n x np+1 = b m x mp, so a n = b m and np + 1 = mp, whch s clearly mpossble, snce p > 1. Therefore, we see that there s no element of R mappng to x, so F : R R s not surjectve. (d): If R s a fnte feld, show that F s an automorphsm. Proof. We showed, n (c), that F : R R s an njecton. Snce an njectve map from a fnte set to tself must be surjectve, ths suffces to show that F : R R s a bjecton. Snce F s also an endomorphsm, ths mples that F s an automorphsm.

8 CLAY SHONKWILER 5 Let K be a feld and let G be a subgroup of the multplcatve group K = K {0}. (a): Show that f a, b K have fnte orders m, n, then there s a c K whose order s the least common multple of m, n. Proof. Frst, suppose m and n are relatvely prme. Then the l.c.m. of m and n s mn. Now, consder ab. Frst, note that (ab) mn = a mn b mn = (a m ) n (b n ) m = 1. Furthermore, f 1 = (ab) k = a k b k, then b k = ( a k) 1 = a k. Snce the order of a k dvdes m and the order of b k dvdes n and m and n are relatvely prme, we see that b k = a k has order 1;.e. a k = b k = 1. Hence, k s a multple of both m and n, and so k s a multple of mn. Therefore, we see that the order of ab s mn, the l.c.m. of m and n. More generally, f a and b have orders m and n, respectvely, let l be the l.c.m. of m and n. Then l = p α 1 1 pα k for p prme and m, then a m/(pα ) has order p α α N. Now, f p α p α n, then a n/(pα c K such that c has order p α k c = =1 k ) has order p α ; smlarly, f. Hence, for each, there exsts. Let c. Then, snce p α and p α j j are relatvely prme for all j, the result proved above demonstrates that c has order k p α = l. =1 (b): Show that f G s fnte then t s cyclc. Proof. Let l = the l.c.m. of the orders of the elements of G. Then #(G) s a multple of l. By 3(b), there are at most l roots of the polynomal x l 1 n K. On the other hand, f a G, then a l = 1 snce l s a multple of the order of a. Hence, each element of G s a root of x l 1, so #(G) = l and the elements of G are all the roots of x l 1. Now, by part (a), there exsts c K such that the order of c s l. Snce c s, thereby, a root of x l 1, c G. Snce all the powers of c are also solutons of x l 1, the l dstnct powers of c are precsely the elements of G, so we see that G = c s cyclc.

ALGEBRA HW 7 9 (c): Conclude that f K L s an extenson of fnte felds, then L = K[a] for some a K. Proof. Snce L s fnte, L s also fnte and so, by part (b) above, L s cyclc. Let a L be a generator of L ; then L = K[a]. DRL 3E3A, Unversty of Pennsylvana E-mal address: shonkwl@math.upenn.edu