ALGEBRA HW 7 CLAY SHONKWILER

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1 ALGEBRA HW 7 CLAY SHONKWILER Prove, or isprove a salvage: If K is a fiel, a f(x) K[x] has o roots, the K[x]/(f(x)) is a fiel. Couter-example: Cosier the fiel K = Q a the polyomial f(x) = x 4 + 3x The f(x) = (x 2 + )(x 2 + 2), which has roots ±i, ±2i, oe of which, certaily, is i Q. However, sice f(x) is reucible (with the factorizatio just give), we kow that (f(x)) is ot a maximal ieal a, thus, that Q/(f(x)) is ot a fiel. To salvage this claim, let us cosier oly polyomials f K[x] for some fiel K that are of egree 2 or 3. The, if f has o roots i K, the K[x]/(f(x)) is a fiel. To see why, ote that, sice f has egree 2 or 3, f must have a liear factor if f is reucible. However, a liear factor x a has a as a root so, if f has o roots i K, f caot be factore i K[x] by a liear factor. Hece, we coclue that such a f is irreucible a, thus, that (f(x)) is maximal, so K[x]/(f(x)) is a fiel. 2 For each positive iteger, let U = (Z/), the group of uits moulo. Fi a geerator of U 2, a etermie the group structure of U 27 a U 2 explicitly. Cojectures? Proofs? Aswer: The uits of Z/2 will be precisely those itegers less tha 2 that are relatively prime to 2; hece, sice 2 = 2, the o-uits of Z/2 are simply the multiples of. Sice there are such (amely 0,, 22, 33, 44, 55, 66, 77, 88, 99, 0), the orer of U 2 is 2 = 0. Now, cosier 2 U 2. Now, the possible orers of ay elemet i U 2 are 2, 5, 0,, 22, 55. Now, 2 2 = 4 4 mo = mo = mo 2 2 = mo = mo = (2 ) 5 ( 9) 5 = mo 2.

2 2 CLAY SHONKWILER Hece, we see that the orer of 2 i U 2 must be 0, so 2 geerates U 2. Note that the o-uits i Z/27 are precisely the multiples of 3; sice there are 9 such, we see that #(U 27 ) = 8. Now, 2 2 = 4 4 mo = 8 8 mo = 6 6 mo = 32 5 mo = 0 0 mo = mo = 40 3 mo = 26 mo 27, so we see that 2 U 27 has orer 8, so we see that U 27 C 8. Note that the o-uits of Z/2 are the multiples of 3 a 7; amely 0, 3, 6, 7, 9, 2, 4, 5, 8. There are 9 such, so we see that #(U 2 ) = 2. Now, U 2 is certaily abelia, so either U 2 C 2 or U 2 C 2 C 6. Now, 2 2 = 4 4 mo = 8 8 mo = 6 6 mo = 32 mo 2 2 = 22 mo = 69 mo = 26 5 mo = 0 0 mo = mo 2 3 = 43 7 mo = mo 2; hece, we see that 2 has orer 6 a 3 has orer 2 a that, furthermore, U 2 = 2, 3, which is to say that U 2 C 2 C 6. Cojecture: If p prime, the U p k C p k pk. If m a are relatively prime, the U m U m U. Proof. Of seco part. If m, relatively prime, efie f : U m U m U by f(x) = (x, x) where we iterpret the terms o the right moulo m a, respectively. The, for x, y U m, f(xy) = (xy, xy) = (x, x)(y, y), so f is a homomorphism. If x, y U m such that f(x) = f(y), the (x, x) = (y, y) implies x y mo m a x y mo. Hece, there exist i, j such that x = y + im a x = y + j. Thus, y + im = y + j, so im = j is a

3 ALGEBRA HW 7 3 commo multiple of m a. Sice m a are relatively prime, their l.c.m. is m; sice the l.c.m. geerates all multiples, there exists k such that im = k(m) = j, so x x = y + im = y + k(m) = y + j = y + k(m). That is to say, x y mo m, so x = y i U m. Thus, f is ijective. O the other ha, if (a, b) U m U, the, by the Chiese Remaier Theorem, there exists x Z such that x a mo m a x b mo ; hece, f(x) = (a, b), so f is surjective. Therefore, we coclue that f is a isomorphism. 3 (a): Defie the Eucliea algorithm as follows. Give o-zero itegers a a b, write a = bq 0 + r 0 as i the ivisio algorithm (i.e. 0 r 0 b ); the cotiue: b = r 0 q + r, r = r 2 q 3 + r 3, etc. (with 0 r i+ r i ). Show that evetually some r + = 0, a that r is the g.c.. of a a b. Proof. Note that r i is a esceig sequece of o-egative itegers; sice there are o ifiite esceig chais i N, we see that r k = 0 for some k N. Let + be the least iteger such that r + = 0. The r, r are o-zero a r = r q + + r + = r q +, so we see that r r. Now, suppose r r i for all i > j. The r j = r j+ q +2 + r j+2 ; sice r r j+ a r r j+, we ca factor r from the right sie, which guaratees that r r j. Hece, by iuctio, r r i for all 0 i. Hece, sice b = r 0 q + r, r b. Further, sice a = bq 0 + r 0, r a, so we see that r is a commo ivisor of a a b. Sice (a, b) is pricipal a is geerate by the greatest commo ivisor of a a b, if r (a, b), the this suffices to show that r is the g.c.. of a a b. To that e, ote that r 0 = a bq 0 (a, b). I geeral, if r i (a, b) for all i < k, the r k = r k 2 r k q k 2 (r k 2, r k ) (a, b),

4 4 CLAY SHONKWILER so r k (a, b). Thus, by iuctio, we see that r i (a, b) for 0 i. Specifically, r (a, b), so we coclue that r is the g.c.. of a a b. (b): Use this to fi the g.c.. of 55 a 65. Aswer: The steps escribe i (a) above lea to the followig series of equatios: 55 = (65)() = (504)() = (47)(3) = (63)(2) = (2)(3) + 0 so we see that the g.c.. of 55 a 65 is 2. (c): Verify, i the calculatios of part (b), that (i the otatio of (a)), = q 0 + q + q q Also verify i these calculatios that if we write q 0 + q + = q q i lowest terms, the x, y form a solutio to the Diophatie equatio 65x 55y =, where = gc(55, 65). Ca solutios to other equatios be fou i this way? Explore. Proof. Note that x y = = = = = 55 3 = Now, turig to the seco equatio, a = = = (6) 55(9) = 0, 46 0, 395 = 2 = gc(65, 55).

5 ALGEBRA HW 7 5 I geeral, cosier the Diophatie equatio ax + by =, where is the g.c.. of a a b. The we ca use the Eucliea algorithm to fi a, traversig backwars, we see that, sice = r a r i+2 = r i r i+ q i+2, = r = r 2 r q = r 2 [r 3 r 2 q ]q = r 3 q + r 2 [ + q q ] = r 3 q + [r 4 r 3 q 2 ][ + q q ] = r 4 [ + q q ] + r 3 [q q 2 ( + q q ]. = ar + br for r, r i terms of the r i s a q j s. This proceure always gives a solutio of the Diophatie Equatio ax + by = where is the g.c.. of a a b. 4 Do the aalog of problem 3 with Z replace by k[x], where k is a fiel. I parts (b) a (c), replace 55 a 65 with x 3 + x 2 + x a x 2 +. (a): Give f(x), g(x) k[x], efie the algorithm give by f(x) g(x) r 0 (x). = g(x)q 0 (x) + r 0 (x) = r 0 (x)q (x) + r (x) = r (x)q 2 (x) + r 2 (x) where r i 0 or eg(r i ) < eg(r i ). The eg(r i ) is a ecreasig sequece of o-egative itegers, so it must reach zero i fiitely may steps. Let + be the smallest iteger such that eg(r + ) = 0. The r (x) = r (x)q + (x) + r + (x) = r (x)q + (x), so we see that r (x) r (x). Now, suppose r (x) r i (x) for all i > j. The r j (x) = r j+ (x)q +2 (x) + r j+2 (x); sice r (x) r j+ (x) a r (x) r j+ (x), we ca factor r (x) from the right sie, which guaratees that r (x) r j (x). Hece, by iuctio, r (x) r i (x) for all 0 i. Hece, sice r (x) g(x). Further, sice g(x) = r 0 (x)q (x) + r (x), f(x) = b(x)q 0 (x) + r 0 (x), r (x) f(x), so we see that r (x) is a commo ivisor of f(x) a g(x). Sice (f(x), g(x)) is pricipal a is geerate by the greatest

6 6 CLAY SHONKWILER commo ivisor of f(x) a g(x), if r (x) (f(x), g(x)), the this suffices to show that r (x) is the g.c.. of f(x) a g(x). To that e, ote that r 0 (x) = f(x) g(x)q 0 (x) (f(x), g(x)). I geeral, if r i (x) (f(x), g(x)) for all i < k, the r k (x) = r k 2 (x) r k (x)q k 2 (x) (r k 2 (x), r k (x)) (f(x), g(x)), so r k (x) (f(x), g(x)). Thus, by iuctio, we see that r i (x) (f(x), g(x)) for 0 i. Specifically, r (x) (f(x), g(x)), so we coclue that r (x) is the g.c.. of f(x) a g(x). (b): The steps escribe i (a) above yiel the followig sequece of equatios: x 3 + x 2 + x = (x 2 + )(x + ) x 2 + = ( )( x 2 ) + 0, so we see that x 3 + x 2 + x a x 2 + are relatively prime. (c): Note that (x + ) + x 2 = (x + )(x2 + ) x 2 + x 2 + = x3 + x 2 + x x 2. + Furthermore, i the seco equatio, x + = x + a (x 2 + )(x + ) (x 3 + x 2 + x)() = = gc(x 2 +, x 3 + x 2 + x). Now, i our emostratio i 3(c) that this geeral proceure will solve equatios of the form ax + by = i ot epe o ay properties of the itegers other tha the fact that we ca o the Eucliea algorithm i Z a that Z is a itegral omai. As such, the same argumet emostrates that we ca use this proceure to fi solutios of the equatio f(x)k(x)+g(x)l(x) = (x), where (x) is the g.c.. of f(x) a g(x). (a): Show that 2 is irratioal. 5 Proof. Suppose 2 Q. The there exist a, b Z such that a b = 2 where a a b are relatively prime. The ( a ) 2 a 2 2 = = b b 2,

7 ALGEBRA HW 7 7 or a 2 = 2b 2 = (2b)b. However, this implies that a is a multiple of b; sice a a b are relatively prime, this is impossible. From this cotraictio, the, we coclue that, i fact, 2 is irratioal. (b): More geerally, show that if m Z a x 2 m has o root i Z, the x 2 m has o root i Q. Proof. Suppose x 2 m has a root i Q but oe i Z. The there exist a, b Z relatively prime such that ( a ) 2 a 2 0 = m = b b 2 m a2 b 2 = m. The a 2 = mb 2 = (mb)b, so a is a multiple of b; sice a a b are relatively prime, this is impossible, so we coclue that there are o such a a b a, therefore, that x 2 m has o roots i Q. (c): Still more geerally, show that if a 0, a,..., a Z, a if the polyomial f(x) = x + a x + + a x + a 0 has o root i Z, the it has o root i Q. Proof. Suppose f has a root i Q but o roots i Z. The there exist c, Z such that c a are relatively prime a ( c ) ( c ) ( c + a a + a 0 = 0. ) Multiplyig both sies by b, we see that Hece, c + a c a c + a 0 = 0. c = ( a c a 2 c a c a 0 ), so we see that c. However, sice c a are relatively prime, this is impossible. From this cotraictio, the, we coclue that if f has o roots i Z the it has o roots i Q. (): What if, i part (c), the polyomial a x +a x + +a x+ a 0 (for some itegers a 0, a,..., a ) is cosiere istea? Aswer: Clearly, the roots of the polyomial 9x 2 4 are x = ± 2 3 ; amely, this polyomial has o roots i Z but oes have roots i Q. That havig bee sai, there are coitios o the ratioal roots of such a polyomial. If f(x) = a x + a x a x + a 0 has ratioal root c Q such that c, Z a c, are relatively prime, the ( c ) ( c ) ( c a + a a + a 0 = 0. ) Thus, multiplyig by b, we see that a c + a c a c + a 0 = 0.

8 8 CLAY SHONKWILER Hece, a c = (a c + a 2 c a c + a 0 ) a a 0 = c(a c + a c a 2 c + a ). Thus, a c a c a a 0 ; sice c a are relatively prime, we see that a a c a 0. This fact the puts sigificat restrictios o the possibilities for ratioal roots of such a polyomial. 6 (a): Describe the maximal ieals i each of the followig rigs: (Z/2)[x], C[x, y, z, t], R[[x]], Z (2), Z[/5], Z/5, C[x, y]/(y 2 x 3 ), Q[i], R R, C[x]/(x 2 ). (b): Describe all the uits i these rigs. Which have oly fiitely may uits? 7 Let p be a prime umber a let be a positive iteger such that p mo. (a): Show that the map φ : (Z/p) (Z/p), give by φ (x) = x, is exactly -to-oe. Proof. Sice (Z/p) C p, we kow that (Z/p) = a for some a (Z/p). Hece, for all b (Z/p), b = a m for some m. Further, b p = for all b (Z/p) a, sice p mo, p is ivisible by. Now, ote that φ (a k(p ) ) = (a k(p ) ) = a k(p ) = (a p ) k = k =. Sice a k(p ) are istict for k =,...,, we see that φ is at least -to-oe, sice each of the elemets of the form a k(p ) is i the kerel of φ. O the other ha, if b = a j is i the kerel of φ, the a p = = φ (b) = b = (a j ) = a j, so j = k(p ) for some iteger k a, hece, j = k(p ). If k = + i for some i 0, the a so j = ( + i)(p ) a j i(p ) (p )+ = a = (p ) + = a p a i(p ) i(p ) = a i(p ),

9 ALGEBRA HW 7 9 so we see there are at most elemets i the kerel. Thus, #(ker φ ) =, so φ is exactly -to-oe. (b): Deuce that there are exactly p th powers. elemets of (Z/p) that are Proof. Sice (Z/p) /ker φ Image φ a Image φ is exactly the set of th powers i (Z/p), we see that #(Image φ ) = #((Z/p) /ker φ ) = #((Z/p) ) #(ker φ ) = p. Hece, there are exactly p elemets of (Z/p) that are th powers. (c): What happes if istea the cogruece hypothesis is roppe? Aswer: Let = (, p ). Let m = /. The, if b = a j ker φ, a p = = φ (b) = b = (a j ) = a j, so j = k(p ) for some k. Thus, j = k(p ) iteger i a j = i(p ). Obviously, a (+l)(p ) = a (p ) + l(p ) = a p a i(p ) = k m = a i(p ), p. So k m so there are at most possible choices for i. O the other ha, (a i(p ) ) = (a i(p ) ) m = (a i(p ) ) m =, is a so a i(p ) ker φ for all i =,...,. Hece, we coclue that ker φ has orer, so φ is -to-oe a thus, by the same argumet give i (b), there are p elemets of (Z/p) that are th powers. 8 (a): Which of the followig elemets of Z[i] ca be factore otrivially? For each oe that ca be, o so explicitly. 2, 3, 5, 7,, 3, 5, 3i, 5i, 2 + i, 3 + i. Aswer: Let N(a) eote the orm of a Z[i]. Recall that i C (a, therefore, i Z[i] C), N(ab) = N(a)N(b). Furthermore, sice N(ab) = c for some c N, we see that N(a) 2 a N(b) 2 must ivie c. 2 = ( + i)( i) = i 2 = 2, so we see that 2 ca be factore o-trivially. N(3) = 3 so, if ab = 3, N(a) 2 9 a N(b) 2 9. Sice the oly otrivial factorizatio of 9 is 3 3, we see that N(a) 2 = N(b) 2 = 3 or

10 0 CLAY SHONKWILER either a or b is a uit. I the latter case, the factorizatio is trivial. Furthermore, if a = α + βi, the 3 = N(a) 2 = α 2 + β 2 ; sice 3 is ot the sum of ay two squares, we see that this is impossible, so 3 ca oly be factore trivially. 5 = ( + 2i)( 2i), so 5 ca be factore o-trivially. N(7) = 7, so, if ab = 7, N(a) 2 49 a N(b) The oly otrivial factorizatio of 49 is as 7 7, so we see that either a or b is a uit or N(a) 2 = N(b) 2 = 7. However, if a = α + βi, the 7 = N(a) 2 = α 2 + β 2 ; sice 7 is ot the sum of two squares, this is impossible, so 7 ca oly be factore trivially. N() =, so, if ab =, N(a) 2 2 a N(b) 2 2. The oly o-trivial factorizatio of 2 is as, so we see that either a or b is a uit or N(a) 2 = N(b) 2 =. However, if a = α + βi, the = N(a) 2 = α 2 + β 2 ; sice is ot the sum of two squares, this is impossible, so ca oly be factore trivially. 3 = (2 + 3i)(2 3i), so 3 ca be factore o-trivially. 5 = 3 5, so 5 ca be factore o-trivially. N(3i) = 3, so, if ab = 3i, N(a) 2 9 a N(b) 2 9. The oly otrivial factorizatio of 9 is as 3 3, so we see that either a or b is a uit or N(a) 2 = N(b) 2 = 3. However, if a = α + βi, the 3 = N(a) 2 = α 2 + β 2 ; sice 3 is ot the sum of two squares, this is impossible, so 3i ca oly be factore trivially. 5i = (2 + i)( + 2i), so 5i ca be factore o-trivially. N(2 + i) = 5, so, if ab = 2 + i, the N(a) 2 N(b) 2 = 5. Sice 5 caot be factore o-trivially, we see that either N(a) = or N(b) = ; either way, either a or b is a uit, so 2 + i caot be factore o-trivially. 3 + i = ( + i)(2 i), so 3 + i ca be factore o-trivially. (b): Make a cojecture about which Gaussia itegers ca be factore o-trivially. Cojecture: a Z[i] caot be factore o-trivially if a oly if either N(a) is a prime cogruet to 3 moulo 4 or N(a) = b for some prime b. DRL 3E3A, Uiversity of Pesylvaia aress: shokwil@math.upe.eu