f = ae b e , i.e., ru + P = (r + P )(u + P ) = (s + P )(t + P ) = st + P. Then since ru st P and su P we conclude that r s t u = ru st

Transcription

1 Mth 662 Spring 204 Homewor 2 Drew Armstrong Problem 0. (Drwing Pictures) The eqution y 2 = x 3 x defines curve in the complex plne C 2. Wht does it loo lie? Unfortuntely we cn only see rel things, so we substitute x = + ib nd y = c + id with, b, c, d R. Equting rel nd imginry prts then gives us two simultneous equtions: () (2) 3 3b = c 2 d 2, b 3 + b 3 2 b = 2cd. These equtions define rel 2-dimensionl surfce in rel 4-dimensionl spce R 4 = C 2. Unfortuntely we cn only see 3-dimensionl spce so we will interpret the b coordinte s time. Setch the curve in rel (, c, d)-spce t time b = 0. [Hint: It will loo -dimensionl to you.] Cn you imgine wht it loos lie t other times b? At time b = 0 eqution (2) becomes 0 = cd which implies tht c = 0 or d = 0. When d = 0 eqution () becomes 3 = c 2. This is curve in the (, c)-plne which we setched on HW. When c = 0 eqution () becomes 3 = d 2, or ( ) 3 ( ) = d 2. This curve lives in the (, d)-plne. It loos just lie the curve in the (, c)-plne but it is reflected cross the (c, d)-plne nd rotted 90. The full solution is the disjoint union of these two:

2 It loos lie four circles glued together in chin. If you cn imgine, the two outer circles meet t the point t infinity, nd the four circles together form the seleton of torus. The other times sweep out the surfce of the torus. For exmple, here I hve plotted the curves in (, c, d)-spce for b {.5,.4,.3,.2,., 0,.,.2,.3,.4,.5}: Problem. (Locl Rings) Let R be ring. We sy R is locl if it contins unique (nontrivil) mximl idel. () Prove tht R is locl if nd only if its set of non-units is n idel. (b) Given prime idel P R, prove tht the locliztion { } R P := b :, b R, b P is locl ring. [Hint: The mximl idel is clled P R P.] (c) Consider prime idel P R. By prt (b) we cn define the residue field R P /P R P. Prove tht we hve n isomorphism of fields: Frc(R/P ) R P /P R P. Proof. For prt (), let M R denote the set of non-units. Note tht M R becuse M. First we ssume tht M is n idel. In this cse, let I R be ny idel of R not contined in M. Then by definition I contins unit nd hence I = R. (If u I is unit then u I nd u R imply = uu I. Then for ll r R we hve r = r I.) We conclude tht M is the unique mximl idel of R, hence R is locl. Conversely, ssume tht R is locl with unique mximl idel m < R. Since m R we now tht m contins no units, hence m M. On the other hnd, we will show tht M m. Suppose for contrdiction tht there exists x M with x m. Since x m nd R/m is field (m is mximl) there exists y R such tht xy + m = (x + m)(y + m) = + m. This implies tht xy = + for some m. But then + m since otherwise = (+) is in m (this is contrdiction becuse m R). Hence the idel (xy) = (+) strictly contins m nd since m is mximl this implies (xy) = R. We conclude tht xy is unit, hence x is unit: x(y(xy) ) = (xy)(xy) =. This contrdicts the fct tht x M nd we conclude tht M = m is n idel.

3 [Remr: I could hve given shorter proof of M m s follows. Consider ny x M. Since (x) < R is proper idel, it is contined in some proper mximl idel, hence (x) m. We conclude tht x m. But this rgument implicitly uses the Axiom of Choice. The proof I gve bove shows tht the Axiom of Choice is not necessry.] For prt (b), let P R be prime nd consider the locliztion { } R P := b :, b R, b P. I will show tht the nonunits of R P form n idel. We cn thin of R P s subring of Frc(R). Let b R P. Since b 0 this frction hs inverse b Frc(R). This inverse will be in R P if nd only if P. In other words, b R P is nonunit if nd only if P. Let { } P R P := b :, b R, P, b P denote the set of nonunits. This is n idel becuse given b, c d P R P nd e f R P (i.e. with, c, e P nd b, d, f P ) we hve b c d bc = P R P d bd becuse d bc P nd bd P, nd b e f = e bf P R P becuse e P nd bf P. We conclude tht R P is locl with mximl idel P R P. For prt (c), note tht R P /P R P is field becuse P R P < R P is mximl idel. Note lso tht R/P is domin becuse P < R is prime idel, thus we cn form the field of frctions Frc(R/P ). I clim tht these two fields re isomorphic. To see this, note first tht s + P 0 + P implies s P. Thus we cn define mp from Frc(R/P ) to R P /P R P by (3) r + P s + P r s + P R P. To see tht this is well-defined, consider s, u P nd suppose tht r+p s+p = t+p u+p, i.e., ru + P = (r + P )(u + P ) = (s + P )(t + P ) = st + P. Then since ru st P nd su P we conclude tht r s t u = ru st su P R P. It is esy to see tht the mp is surjective ring homomorphism (detils omitted). Finlly we will show tht the mp is injective by showing tht the ernel is trivil. Consider r s R P (i.e. with s P ) nd suppose tht r s + P R P = P R P, i.e., tht r s P R P. This mens tht r P nd hence r+p s+p is the zero element of Frc(R/P ). We conclude tht Frc(R/P ) R P /P R P. Problem 2. (Forml Power Series) Let K be field nd consider the ring of forml power series: K[[x]] := { 0 + x + 2 x x 3 + : i K for ll i N }. The degree of power series does not necessrily exist. However, for ll nonzero f(x) = x we cn define the order ord(f) := the minimum such tht 0. () Prove tht K[[x]] is domin.

4 (b) Prove tht K[[x]] is Eucliden domin with norm function ord : K[[x]] {0} N. (You cn define ord(0) = if you wnt.) [Hint: Given f, g K[[x]] we hve f g if nd only if ord(f) ord(g), so the reminder is lwys zero.] (c) Prove tht the units of K[[x]] re just the power series with nonzero constnt term. (d) Conclude tht K[[x]] is locl ring. (e) Prove tht Frc (K[[x]]) is isomorphic to the ring of forml Lurent series: K((x)) := { n x n + n+ x n+ + n+2 x n+2 + : i K for ll i n }. Proof. Given power series f(x) = x nd g(x) = l b lx l recll tht the coefficient of x m in f(x)g(x) is given by +l=m b l. I clim tht ord(fg) = ord(f) + ord(g). Indeed, if m < ord(f)+ord(g) then +l = m implies tht either < ord(f) or l < ord(g) (otherwise we hve +l ord(f)+ord(g) > m). Thus every term in the sum +l=m b l is zero. However, if m = ord(f)+ord(g) then the coefficient of x m in f(x)g(x) is +l=m b l = ord(f) b ord(g) 0 becuse ord(f) 0 nd b ord(g) 0 by ssumption (nd K is domin). We conclude tht ord(fg) = ord(f) + ord(g). For prt (), ssume tht f, g K[[x]] re nonzero. This implies tht ord(f), ord(g) < nd hence ord(fg) = ord(f) + ord(g) <. We conclude tht K[[x]] is domin. For prt (b), consider f(x) = x nd g(x) = l b lx l with g 0 (i.e. with ord(g) < ).We wnt to prove tht there exist q, r K[[x]] with f = qg + r nd either r = 0 or ord(r) < ord(g). Indeed, if ord(f) < ord(g) then we cn simply te q(x) = 0 nd r(x) = f(x). If ord(f) ord(g) then we cn perform long division s follows. Let b be the lowest coefficient of g(x). Then let f = f nd for ll n such tht f n 0 define f n+ (x) := f n (x) n b xord(fn) ord(g) g(x). where n is the lowest coefficient of f n (x). By construction we hve ord(g) ord(f ) < ord(f 2 ) < ord(f 3 ) < so this is lwys defined. If the lgorithm termintes with f N = 0 then we set n = 0 for ll n N, otherwise we let the lgorithm run forever (i.e. use induction). In the end we obtin forml power series q(x) := n b xord(fn) ord(g) n with the property tht f(x) = q(x)g(x) (the reminder is lwys zero!). This proves tht K[[x]] is Eucliden. [Probbly proof by exmple would hve been better, but I didn t feel lie typesetting n infinite long division in L A TEX. I encourge you to compute n exmple yourself.] In the proof of (b) note tht we ctully showed tht given two power series f, g K[[x]] we hve g f if nd only if ord(g) ord(f). For prt (c), note tht g K[[x]] is unit if nd only if g divides. By the bove remr this hppens if nd only if ord(g) ord() = 0, i.e., if nd only if ord(g) = 0. Finlly, note tht ord(g) = 0 if nd only if g hs nonzero constnt term. For prt (d), note tht the set of nonunits of K[[x]] re just the power series with zero constnt term, i.e., the power series divisible by x: (x) := {xf(x) : f(x) K[[x]]}. Since this is n idel we conclude tht K[[x]] is locl ring. For prt (e), we sy tht f(x) = x is forml Lurent series if there exists minimum r Z (possibly negtive) such tht r 0. In this cse we define ord(f) = r. Let K((x)) denote the ring of forml Lurent series with ddition nd multipliction defined just s for power series. Then K[[x]] K((x)) is the subring or Lurent series with nonnegtive order.

5 I clim tht K((x)) is field. Indeed, given ny two Lurent series f, g K((x)) with g 0, the long division process defined bove cn be used to obtin q(x) K((x)) such tht f(x) = q(x)g(x). We we do not require ord(g) ord(f). In fct, becuse ord(q) = ord(f) ord(g) we will hve q K[[x]] if nd only if ord(g) ord(f). If f(x) = then we obtin q(x) = g(x). Since K((x)) is field contining K[[x]] we cn identify Frc(K[[x]]) with the subfield of K((x)) consisting of elements of the form f(x)g(x) for f, g K[[x]] with g 0. But note tht every Lurent series f(x) K((x)) hs this form. Indeed, if ord(f) 0 then f(x) = f(x)() Frc(K[[x]]) nd if ord(f) < 0 then f(x) = (x ord(f) f(x))(x ord(f) ) Frc(K[[x]]) becuse x ord(f) f(x) nd x ord(f) re in K[[x]]. We conclude tht Frc(K[[x]]) = K((x)). [As you my now, ny function f : C C holomorphic in n nnulus hs convergent Lurent series expnsion there. This mes complex nlysis very lgebric subject.] Problem 3. (Prtil Frction Expnsion) To wht extent cn we un-dd frctions? Let R be PID. Consider, b R with b = p e pe 2 2 pe where p,..., p re distinct primes nd e,..., e. () Prove tht there exist,..., R such tht b = p e + 2 p e p e [Hint: First prove it when b = pq with p, q coprime. Use Bézout.] Now ssume tht R is Eucliden domin with norm function N : R {0} N. (b) Prove tht there exist c, r ij R such tht b = c + e i r ij i= j= where for ll i, j we hve either r ij = 0 or N(r ij ) < N(p i ). [Hint: If p is prime, prove tht we cn write p s where either r = 0 or N(r) < N(p). Then use ().] e q p e + r p e Now suppose tht the norm function stisfies N() N(b) nd N( b) mx{n(), N(b)} for ll, b R {0}. (c) Prove tht the prtil frction expnsion from prt (b) is unique. [Hint: Suppose we hve two expnsions c + e i r ij i= j= = b = c + Then we get prtil frction expnsion of zero: 0 b = = (c c ) + b, e i r ij i= j= e i (r ij r ij ). i= j=.

6 For ll i, j define ˆb ij := b/, so tht b(c c) = e i i= j= (r ij r ij)ˆb ij. Suppose for contrdiction tht there exist i, j such tht r ij r ij nd let j be mximl with this property. Use the lst eqution bove to show tht p i divides (r ij r ij ) nd hence N(p i ) N(r ij r ij) mx{n(r ij ), N(r ij)} < N(p i ). Contrdiction.] (d) If K is field nd R = K[x] then the norm function N(f) = deg(f) stisfies the hypotheses of prt (c) so the expnsion is unique. Compute the unique expnsion of x 5 + x + (x + ) 2 (x 2 + ) R(x). (e) If R = Z then the norm function N() = does not stisfy b mx{, b }. However, if we require reminders r, r to be nonnegtive then it is true tht r r mx{ r, r } nd the proof of uniqueness in (c) still goes through. Compute the unique expnsion of 77 2 Q with nonnegtive prmeters r ij 0. Proof. Consider, b R with b = p e pe where p,..., p re distinct primes nd e,..., e. For prt (), note tht since R is PID we must hve (p e, pe 2 2 pe ) = (d) where d is the gretest common divisor. Since p e nd p e 2 2 pe re coprime this implies tht d =, nd hence there exist c, c 2 R such tht Multiplying both sides by b gives = c p e 2 2 pe + c 2 p e. b = c p e + c 2 p e 2 2. pe Now the result follows by induction. For prt (b), let R be Eucliden nd suppose tht we hve written b = p e + + p e. Now consider ny p with, p R nd p prime. We cn divide by p to obtin q, r R such e tht = pq + r nd either r = 0 or N(r) < N(p). In other words, we hve p e = qp + r p e = q p e + r p e where either r = 0 or N(r) < N(p). By induction we obtin p e = q p 0 + r p + r 2 p r e p e, where for ll i we hve r i = 0 or N(r i ) < N(p). Then, combining these expressions for ech summnd i p e i of b gives i b = c + e i r ij i= j= where for ll i, j we hve r ij = 0 or N(r ij ) < N(p i ). This is clled prtil frction expnsion of b.,

7 For prt (c), suppose tht the Eucliden norm stisfies N() N(b) nd N( b) mx{n(), N(b)} for ll, b R {0}, nd suppose we hve two prtil frction expnsions c + e i r ij i= j= = b = c + e i r ij i= j= I clim tht r ij = r ij for ll i, j (nd hence lso c = c ). To see this, we subtrct the expnsions: 0 = (c c) + Then we multiply both sides by b to get b(c c ) = e i (r ij r ij ) i= j= e i (r ij r ij)ˆb ij, i= j= where ˆb ij := b/ R. Now ssume for contrdiction tht we hve r mn r mn for some m, n nd let n be mximl with this property. Tht is, suppose tht we lso hve r mj = r mj for ll j > n. In this cse, note tht pem n m divides ˆb ij for every nonzero term in the sum, thus since R is domin we cn cncel it to get (4) b (c c ) = where b = p e pn m p e nd ˆb ij = e i (r ij r ij)ˆb ij, i= j= p e pe i j i p n m p e i < m p e pe i j i p n m p e i > m p e pn j m p e i = m, j n 0 i = m, j > n Finlly, note tht p m divides (r mn r mn)ˆb mn becuse p m divides every other term of the sum (4). Since p m is prime, Euclid sys tht p m (r mn r mn) or p m ˆb mn. But by definition we now tht p m does not divide ˆb mn. We conclude tht p m divides r mn r mn nd then the ssumed properties of the norm imply tht N(p m ) N(r mn r mn) mx { N(r mn ), N(r mn) } < N(p m ). Contrdiction. For prts (d) nd (e) I will nively follow the steps of the proof. I will not use ny trics lie differentition. (You cn get the solution fster with trics.) For prt (d) we first loo for polynomils f(x) nd g(x) such tht = f(x)(x + ) 2 + g(x)(x 2 + ). For this we consider the set of triples f, g, h R[x] with f(x)(x + ) 2 + g(x)(x 2 + ) = h(x) nd pply row reduction: f(x) g(x) h(x) 0 (x + ) 2 0 x 2 + 2x x/2 + x/2.

8 We conclude tht ( x/2)(x + ) 2 + ( + x/2)(x 2 + ) = nd hence (x + ) 2 (x 2 + ) = ( x/2)(x + )2 + ( + x/2)(x 2 + ) (x + ) 2 (x 2 + ) = x/2 x x/2 (x + ) 2. Multiplying both sides by x 5 + x + gives x 5 + x + (x + ) 2 (x 2 + ) = 2 (x6 + x 2 + x) x (x6 + 2x 5 + x 2 + 3x + 2) (x + ) 2. Now we del with both of the summnds seprtely. First we divide 2 (x6 +x 2 +x) by x 2 + to get hence 2 (x6 + x 2 + x) = 2 (x4 x 2 + 2)(x 2 + ) (x 2), 2 2 (x6 + x 2 + x) x 2 + = 2 (x4 x 2 + 2) + 2 (x 2) (x 2 + ). Next we divide 2 (x6 + 2x 5 + x 2 + 3x + 2) by (x + ) to get hence 2 (x6 + 2x 5 + x 2 + 3x + 2) = 2 (x5 + x 4 x 3 + x 2 + 3)(x + ) 2, 2 (x6 + 2x 5 + x 2 + 3x + 2) (x + ) 2 = Finlly, we divide 2 (x5 + x 4 x 3 + x 2 + 3) by (x + ) to get hence 2 (x5 + x 4 x 3 + x 2 + 3) + /2 (x + ) (x + ) 2. 2 (x5 + x 4 x 3 + x 2 + 3) = 2 (x4 x 2 + 2x 2)(x + ) + 5, Putting everything together gives 2 (x5 + x 4 x 3 + x 2 + 3) = (x + ) 2 (x4 x 2 + 2x 2) + 5/2 (x + ). x 5 + x + 5/2 (x + ) 2 (x 2 = (x 2) + + ) (x + ) + /2 (x 2)/2 + (x + ) 2 (x 2. + ) [By doing everything out longhnd I ment to show tht it is possible, not tht it is esy.] For prt (e) we first fctor 2 = 3 4 with 3, 4 coprime. Now we loo for x, y Z with 3x + 4y =. This cn be done by inspection: = 3( ) + 4 [If it couldn t be done by inspection we would use the Eucliden lgorithm.] Dividing by 2 gives 2 = 3( ) + 4 = , nd then multiplying by 77 gives 77 2 =

9 Now we del with both of the summnds seprtely. First we divide 77 by 3 to get Then we divide 77 by 2 to get Finlly, we divide 39 by 2 to get Putting everything together gives 77 = = = 2( 39) = = 2( 20) + 39 = = This result is unique s long s we use positive reminders. [Why did I s you to do this? Becuse I lwys wondered bout prtil frctions. They pper in Clculus to show us tht ll rtionl functions over R cn be integrted in elementry terms. For exmple: x 5 + x + (x + ) 2 (x 2 + ) dx = 2 x2 2x ln(x + ) 2(x + ) 4 ln(x2 + ) + rctn(x). But then prtil frctions mysteriously dispper from the curriculum. Now t lest we now why.]