Zachary Scherr Math 503 HW 3 Due Friday, Feb 12

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1 Zachary Scherr Math 503 HW 3 Due Friay, Feb 1 1 Reaing 1. Rea sections 7.5, 7.6, 8.1 of Dummit an Foote Problems 1. DF 7.5. Solution: This problem is trivial knowing how to work with universal properties. Let F enote the quotient fiel of R, then there is an injective map ϕ: R F. Let D R be any nonempty set of nonzero elements of R which is close uner multiplication. Since ϕ is injective an F is a fiel, we know that ϕ() F for every D. Thus by the universal property of the ring of fractions, there is a unique injective homomorphism ϕ: D 1 R F making the iagram R ϕ F ψ D 1 R ϕ commute (meaning that ϕ = ϕ ψ). Since ϕ is injective, we see that F contains a subring isomorphic to D 1 R.. DF Solution: We prove part of this assertion in class. Let F be a fiel, an consier the map ϕ: Z F efine in problem Since F is a fiel, an hence an integral omain, we know that the characteristic of F is either p, for some prime p, or 0. Assume first that F has characteristic p. Then ker(ϕ) = (p), an by the first isomorphism theorem F 0 := ϕ(z) = Z/pZ is a subfiel of F. Any other subfiel K F contains the multiplicative ientity 1, an hence F 0 = ϕ(z) K proving that F 0 is the smallest subfiel of F. If, on the other han, the characteristic of F is 0 then ϕ is an injection. Let ψ : Z Q be the injection of Z into its fiel of fractions. Since every nonzero element of Z must map to a unit in F, there exists a unique injective ϕ: Q F so that the iagram Z ϕ F ψ Q ϕ commutes. Since ϕ is injective, we have F 0 := ϕ(q) = Q is a subfiel of F. If K F is any subfiel of F then surely 1 K implying that ϕ(z) K. Thus ϕ(ψ(z)) = ϕ(z) K proving that F 0 K since K is close uner inverses. 3. DF 7.6.3

2 Solution: Let R an S be rings with ientities, an let K R S be an ieal. Consier the projection maps π R : R S R π S : R S S given by π R (r, s) = r an π S (r, s) = s. These maps are surjective ring homomorphisms an so by exercise 7.3.4, the sets I = π R (K) R an J = π S (K) S are ieals. By efinition of I an J we have that K I J. For the other inclusion, consier an element (i, j) I J. Since i π R (K), we know that there exists s S with (i, s) K. Similarly, there exists r R with (r, j) K. Thus an so inee I J K an we get equality. (i, j) = (1, 0)(i, s) + (0, 1)(r, j) K 4. DF 7.6.5c Solution: In the notation of the previous parts of this problem we let n 1 = 8, n = 5, n 3 = 81. Then n 1 = 05, n = 648, n 3 = 00. We have n 1 1 (mo n 1 ) n 3 (mo n ) n 3 38 (mo n 3 ). Immeiately we see that t 1 = 1. To fin t we use the Eucliean algorithm so we can back substitute to get 5 = = , 1 = 3 11 = 3 (5 3 1) 11 = Thus (mo n ) an so we take t = 1. For t 3 we have We back solve as follows 1 = 3 81 = = = = = 3 (5 3) = 3 5 = (38 5 7) 5 = = 38 (81 38 ) 15 = Page

3 Thus take t 3 = 3 since (mo 81). This allows us to solve x = (1)(1)(05) + ()(1)(648) + (3)(3)(00) = (mo ) an y = (5)(1)(05) + (1)(1)(648) + (47)(3)(00) (mo ). 5. DF Solution: (a) Let = gc(a, b) an suppose that a bc. Since Z is a Eucliean Domain, there exist integers x, y Z with ax + by =. Thus acx + bcy = c implying that a c. Since a, we also get a c in Z. (b) Suppose we have ax 0 + by 0 = N, an let x, y be another solution to ax + by = N. Subtracting these two equations yiels a(x 0 x) + b(y 0 y) = 0 which we may rearrange to a(x 0 x) = b(y 0 y). Thus a b(y0 y) an if = gc(a, b), then part a) implies that a an integer m with y 0 y = a m or Substituting into y = y 0 m a. a(x 0 x) = b(y 0 y). yiels a(x 0 x) = b a m. Using cancellation an rearranging gives x = x 0 + m b. y0 y. Thus there exists For completeness we shoul actually show that ax + by = N for the x an y escribe above. We can o this since ( ax + by = a x 0 + m b ) ( + b y 0 m a ) = ax 0 + m ab + by 0 m ab = ax 0 + by 0 = N. 6. DF Page 3

4 Solution: We have 85 85(1 13i) = = i i. Thus for our first quotient we can take q = (1 6i) an we have Next, we look at r = a bq = 85 (1 + 13i)(1 6i) = 6 7i i 6 7i = (1 + 13i)(6 + 7i) = 1 + i. Here there is no remainer, an so the full Eucliean algorithm is 85 = (1 + 13i)(1 6i) + (6 7i) i = (6 7i)( 1 + i) + 0. Thus we have the equality (85, i) = (6 7i). Since the units of Z[i] are ±1, ±i you might also get 6 + 7i, 7 + 6i or 7 6i. Next we look at We roun this to (1 + i) an get i 47 13i = i. ( i) (47 13i)(1 + i) = 7 + i. Repeating, we have We roun to 1 i an get 47 13i 7 + i = i. Repeating, we have (47 13i) ( 7 + i)( 1 i) = 4 5i. 7 + i 4 5i = 3i so we can stop. The full Eucliean algorithm looks like i = (47 13i)(1 + i) + ( 7 + i) 47 13i = ( 7 + i)( 1 i) + ( 4 5i) 7 + i = ( 4 5i)( 3i) + 0. Therefore ( i, 47 13i) = ( 4 5i) = (4 + 5i). 7. DF Solution: Let I be a nonzero ieal of Z[i]. Since Z[i] is a Eucliean Domain, there exists a non-zero Page 4

5 α Z[i] with I = (α). For every β Z[i] we know that there exists q, r Z[i] with β = αq + r where either r = 0 or N(r) < N(α). This tells us that moulo I, every element of Z[i] is equivalent to some element whose norm is smaller than the norm of α. Since N(a + bi) = a + b, we have that there are only finitely many elements of Z[i] of boune norm, so in fact the ring Z[i]/I must be finite. 8. DF Solution: (a) We can rephrase the efinition of the least common multiple as follows. It is an element e R satisfying 1. e (a) an e (b).. if e (a) an e (b) for some e R then e (e). Property 1 says that (e) (a) (b), while property says that if (e ) (a) (b) then (e ) (e). Thus the lcm, if it exists, is a generator for the largest principal ieal containe in (a) (b). Notice that if such an ieal exists then property says in fact that it must be unique. (b) Let a, b R be nonzero elements of a Eucliean omain R. Then since every ieal of R is principal we have that (a) (b) = (e) for some e R. Therefore (e) is the largest principal ieal of R containe in (a) (b) an so e is a least common multiple of a an b. Moreover, since the ieal is unique an (α) = (β) if an only if β = αu for some u R, we see that e is unique up to multiplication by a unit. (c) Let () = (a, b) an e = ab. Since a an b, we have e = a b = b a (a) (b) an so (e) (a) (b). For the converse, let r (a) (b). Then there exists x, y R with r = ax = by. This equality shows us that a by an so problem 4 tells says that a y. If y = a y then r = by = b ( a y ) = ey. Thus r (e) an hence (a) (b) (e). We have just prove that (a) (b) = (e) an therefore e = ab is a least common multiple for a an b. 9. Consier the Eucliean Domain R = F 3 [x]. (a) Fin gc(x 3 + x + x +, x 6 + x 5 + x 4 + x + x + 1) in R. Solution: We perform the Eucliean algorithm, which requires oing polynomial long ivision. We have x 6 + x 5 + x 4 + x + x + 1 = (x 3 + x + x + )(x 3 + x + x) + (x + 1) x 3 + x + x + = (x + 1)(x + ) + 0. Page 5

6 Thus gc(x 3 + x + x +, x 6 + x 5 + x 4 + x + x + 1) = x + 1. Since GCDs are unique up to multiplication by a unit, if we wante a monic GCD then we can multiply by to get gc(x 3 + x + x +, x 6 + x 5 + x 4 + x + x + 1) = x +. (b) Fin a polynomial f(x) R with f(x) x + 1 (mo x 3 + x + x + ) f(x) x 3 + x (mo x 6 + x 5 + x 4 + x + x + 1). Solution: From the previous part we have that (x + 1) = (x 6 + x 5 + x 4 + x + x + 1) (x 3 + x + x + )(x 3 + x + x) = (x 6 + x 5 + x 4 + x + x + 1) + (x 3 + x + x + )(x 3 + x + x). Let a(x) = (x 6 + x 5 + x 4 + x + x + 1) an b(x) = (x 3 + x + x + )(x 3 + x + x) an f(x) = a(x) + xb(x) = x 7 + x 5 + x 4 + x + 1. I claim that f(x) is a solution to our system of congruences. The easiest way to see this is from the equality x + 1 = a(x) + b(x). Since (x 6 + x 5 + x 4 + x + x + 1) ivies a(x) we get x + 1 b(x) (mo x 6 + x 5 + x 4 + x + x + 1). Similarly, since (x 3 + x + x + ) ivies b(x) we get Thus while x + 1 a(x) (mo x 3 + x + x + ). f(x) a(x) + xb(x) (mo x 6 + x 5 + x 4 + x + x + 1) xb(x) (mo x 6 + x 5 + x 4 + x + x + 1) x(x + 1) (mo x 6 + x 5 + x 4 + x + x + 1) x 3 + x (mo x 6 + x 5 + x 4 + x + x + 1) f(x) a(x) + xb(x) (mo x 3 + x + x + ) a(x) (mo x 3 + x + x + ) x + 1 (mo x 3 + x + x + ). 10. Moify the proof that Z[i] is a Eucliean Domain to prove that Z[ ] is a Eucliean Domain with respect to N(a+b ) = a +b, an that Z[ ] is a Eucliean Domain with respect to N(a+b ) = a b. Page 6

7 Solution: Let s begin with Z[ ]. Let α, β Z[ ] with β 0. If α = a + b an β = c + then in C we have α β = a + b c + = (a + b )(c ) a + b bc a c + = c + +. c + Let r = a+b c + an s = bc a c +. Then r, s Q so we can fin integers m, n Z with r m 1 an s n 1. Let γ = α β(m + n ). Since b, n Z we have γ Z[ ]. In C we have γ β = α β (m + n ) = (r m) + (s n). Taking the norm gives N ( ) γ = (r m) + (s n) = 1 β = 3 4. Since N is multiplicative, we get N(γ) = 3 4 N(β) an hence we see that Z[ ] is a Eucliean omain. Next we consier Z[ ]. Here the proof is nearly ientical to the other two, we just have to eal with the complication of the absolute value symbol. Let α, β Z[ ] with β 0. As in the solution above, we can rationalize the enominator to write α β = r + s where r, s Q. Again, choose integers m, n with m r 1 an n s 1. Set γ = α β(m + n ) Z[ ] then Thus N ( ) γ = N((r m) + (s n) ) = (r m) (s n). β ( ) γ N = (r m) (s n) β max{(r m), (s n) } { 1 max 4, 1 } = 1. By multiplicatively we therefore get N(γ) 1 N(β). REMARK: The above proof also shows that Z[ 3] is a Eucliean omain. Unfortunately the proof oes Page 7

8 not work for Z[ 3]. The reason is that you en up with = 1 an you cannot guarantee strict ecrease in the norm. However, the above proof oes go through for the larger ring [ ] { ( ) } Z = a + b a, b Z. In this ring, we have ( ( )) N a + b = N ( a + b + b ) ( 3 = a + b + 3 ) ( ) b = a + ab + b. 11. Let R be a commutative ring an let D R be a set of non-zero elements which oesn t contain any zero-ivisors an is close uner multiplication. In class we constructe the ring of fractions D 1 R an an injection ψ : R D 1 R. (a) Let J D 1 R be an ieal. Prove that there exists an ieal I R with { } i J = D 1 I := i I, D. Solution: We know that I = ψ 1 (J) R is an ieal of R. I claim in fact that J = D 1 I. To make the notation less cumbersome, ientify R with its image ψ(r) D 1 R via ψ(r) = r 1. By efinition of I we know for any i I that i 1 J. Since J is close uner multiplication by elements of D 1 R we also have i = 1 i 1 J for any D. Thus D 1 I J. For the other inclusion note that every element in J is of the form r for some D an r R. Since J is an ieal, we have r 1 = r J as well. Since ψ(r) = r 1 we have that r ψ 1 (J) = I an hence r D 1 I. We ve prove that J D 1 I an therefore we get equality. (b) Show that the map I D 1 I gives a one-to-one corresponence between prime ieals in D 1 R an prime ieals in R which intersect trivially with D. Solution: We prove above that every ieal of D 1 R is of the form D 1 I for some ieal I in R. In fact, for every ieal I R it is equally easy to check that D 1 I is an ieal of D 1 R. To see this, let x 1, y D 1 I. Then x y = x y 1 D 1 I 1 1 since x y 1 I. This proves that D 1 I is a subgroup of D 1 R, an for any r D 1 R we have r x = rx D 1 I 1 1 since rx I. Assume first that J D 1 R is prime. Then we prove on the last homework that I = ψ 1 (J) is a prime ieal of R. I claim that I D =. To prove this we assume for the sake of Page 8

9 contraiction that there exists I with D. Then ψ() = 1 J, but 1 is a unit in D 1 R. This implies that J = D 1 R which is a contraiction since J is prime. Therefore we see that I is a prime ieal of R which intersects trivially with D an by part a), J = D 1 I. For the converse, let P R be a prime ieal which intersects trivially with D. We alreay know that K = D 1 P is an ieal of D 1 R. For any D we know that 1 K since P, an hence K is a proper ieal of D 1 R. Let a 1, b be elements of D 1 R with ab = a b K. 1 1 By efinition of K we must have ab P an so a P or b P. Therefore a 1 an hence K is a prime ieal of D 1 R. K or b K 3 Challenge Problems Challenge Problems ten to be harer than the rest of the problems (an sometimes more interesting). You o not nee to turn these in, but you shoul get something out of thinking about these. 1. Let R be a commutative ring, an let N R be the ieal of nilpotent elements of R. Prove that N equals the intersection of all prime ieals in R. (Hint: if R is not nilpotent then one can form the ring D 1 R for D = { k k 1}. Use properties of this ring.) Solution: Let x N. Then there exists n 1 so that x n = 0. Thus for any prime ieal P R we have x n = 0 P, an so we must have x P proving that N is containe in the intersection of all primes in R. Conversely, let R be an element which is containe in the intersection of all prime ieals of R. We will argue by contraiction that is nilpotent. If is not nilpotent then the D = { k k 1} is a multiplicatively close subset of R which oesn t contain 0. We can therefore form the ring D 1 R an we have a map ψ : D 1 R. The ring D 1 R is a ring with 1 0 so it contains a maximal ieal M. Consequently, M is also prime in D 1 R an hence P = ψ 1 (M) is prime in R. By the last problem above, P D = implying that P which contraictions the hypothesis that is containe in all prime ieals of R. REMARK: As I was writing up this proof I realize that there is a gap in the above argument since might be a zero-ivisor an we on t know how to form D 1 R if D contains zero-ivisors. It s true that one can form D 1 R if D contains zero-ivisors, but the efinition has to be moifie slightly. When we construct D 1 R as equivalence classes in R D, we say that two pairs (a, b) an (c, ) are equivalent if a bc = 0 OR a bc is a zero-ivisor in R. This efinition will turn D 1 R into a ring with 1 0, but the map ψ : R D 1 R may no longer be injective. The above proof still works since the statement of the last problem is still vali in this new setting.. Dummit an Foote Page 9

10 Solution: Let n = ab a b. First we will prove that there is no solution to ax + by = n with x, y Z +. Since gc(a, b) = 1 an n = ab a b = a(b 1) + b( 1), we know that all solutions to ax + by = n are of the form n = a(b 1 tb) + b( 1 + ta) for t Z. If t 1 then b 1 tb < 0, an if t < 1 then 1 + ta < 0. Thus we see that no solution to ax + by = n can have x, y 0. Next, we must show if m 1 then there is a solution to ax + by = m + n with x, y 0. Since gc(a, b) = 1 we know that for every m 1 there is a solution an that all solutions are given by ax 0 + by 0 = m a(x 0 + bt) + b(y 0 at) = m where t Z. Ajusting t if necessary, we can guarantee that (b 1) x 0 0. Since m 1 an x 0 0, we must then also have y 0 1. Thus m + n = (ax 0 + by 0 ) + (ab a b) = a(x 0 + (b 1)) + b(y 0 1). We have our esire solution since both x 0 + (b 1) an y 0 1 are non-negative. Page 10