MTH 310-3 Astract Algera I and Numer Theory S18 Review for Eam 2/Solutions 1. Let n e an integer. (a) For 0 n 10, verify that n 2 n + 11 is a prime. () If n 0, 1 and n 0, 1 (mod 11), show that n 2 n + 11 is not a prime. (c) Prove or disprove: If n / 0, 1 (mod 11), then n 2 n + 11 is a prime. (a) Define f(n) n 2 n + 11 and oserve that (1) f(n) n(n 1) + 11 Hence f(n + 1) (n + 1)n + 11 n 2 + n + 11 (n 2 n + 11) + 2n f(n) + 2n So f(n + 1) f(n) + 2n. Using this formula we can easily compute f(n) for 0 n 10. n 2 3 4 5 6 7 8 9 10 2n 0 2 4 6 8 12 14 16 18 20 n 2 n + 13 17 23 31 41 53 67 83 101 All the numers in the last row are primes and so (a) holds. () Assume that n 0, 1 (mod 11) and n 0, 1. We need to show that n 2 n + 11 is not a prime. We will first show that 11 divides n 2 n + 11. Since n 0, 1 (mod 11) we have [n] 11 [0] 11 or [n] 11 [1] 11 y Theorem In Z 11 we have n 0 or n 1, 0 2 0 + 11 0 and 1 2 1 + 11 11 0 Thus [n 2 n + 11] 11 [0] 11 and so 1.4.9 shows that 11 divides (n 2 n + 11) 0. So (1) 11 (n 2 n + 11). Since n 0, 1, we have n < 0 or n > 1. If n < 0, then n 2 n n 2 > 0 and if n > 1, then n 2 n n(n 1) > 0. So in either case n 2 n > 0 and so n 2 n + 11 > 11. Thus 11 ±(n 2 n + 11). From (1) we know that 11 divides n 2 n + 11 and so n 2 n + 11 is not a prime. (c) This is false. Indeed 15 / 0, 1 (mod 11) and f(15) 15 14 + 11 210 + 11 221 13 17. So f(15) is not a prime. 2. Let R e an integral domain and a, R. Show that the following statements are equivalent: (a) a and a. () a is an associate of in R. (a) (): Suppose that a and a. Then y definition of divide a u and av for some u, v R. Case 1: a 0 R. Then 0 R and so 0 R y (1.4.3)(). By 1.12.6 is an equivalence relation and so is refleive. It follows that 0 R 0 R. As a 0 R and 0 R t we conclude that a in this case. Case 2: a 0 R. Sustituting av into a u gives a (av)u. Hence y Aiom 2, a a(vu). By definition of identity (A 10), a a1 R. Thus a1 R a(vu). By assumption of Case 2, a 0 R. Since R is an integral domain we can apply the Multiplicative Cancellation Law 1.8.7 and conclude that 1 R uv. Thus y 1.12.9 u is a unit. Since a u this shows that a is associate to. Thus (a) implies () () (a): If a then 1.12.8 shows that a and a.
3. Let n and m e a positive integer and suppose f Z n Z m is a surjective homomorphism. (a) Show that f([k] n ) [k] m for all k N. (Hints: What is f([1] n )? Use induction on k.) () Show that m n in Z.(Hint: Consider f([n] n )). (a) For k N let P (k) e the statement P (k) f([k] n ) [k] m By (1.11.7)(a), f([0] n ) [0] m and so P (0) holds. Since f is a surjective homomorphism (1.11.7)(a) implies that f([1] n ) [1] m. Thus also P (1) holds. Suppose now that P (k) is true. We compute: f([k + 1] n ) f([k] n + [1] n ) Definition of Addition in Z n f([k] n ) + f([1] n ) since f is a homomorphism [k] m + [1] m P (k) and P (1) [k + 1] m Definition of Addition in Z, Thus P (k + 1) holds. So y the principal of induction, P (k) holds for all k N. () Since n (n 0) we have [n] n [0] n y 1.4.9. Thus also f([n] n ]) f([0] n ) and (a) implies that [n] m [0] m. Hence 1.4.9 shows that m (n 0) and so m n. 4. Which of the following functions are homomorphisms? (a) f Z Z, n 2n. () g R Z, a, where R a Z. a 0 a 0 (c) h Z 2 [] Z 2 [], f f 2. (a) We have f(1 1) f(1) 2 1 2 and f(1) f(1) 2 2. Thus f(1 1) f(1) f(1). Hence f does not respect multiplication and so f is not a homomorphism. () Thus g g 0 and g g 1 1 1. g g g. So g is not a homomorphism. (c) We will show that h is a homomorphism. Let d Z 2 []. Since Z 2 is a commutative ring with identity 1, 2.2.4 shows that Z 2 [] is a commutative ring with identity 1. So 1d d. Also 1 + 1 0 in Z 2. We compute We proved (1) d + d 0 for all d Z 2 []. Now let a, Z 2 []. Then d + d 1d + 1d A 8 (1 + 1)d 0d (1.2.9)(c) 0. h(a + ) (a + ) 2 (a + )(a + ) A 8 a(a + ) + (a + ) A 8 (a 2 + a) + (a + 2 ) GAL (a 2 + (a + a)) + 2 (1) (a 2 + 0) + 2 A 4 a 2 + 2 h(a) + h()
and so (2) h(a + ) h(a) + h(). Also h(a) (a) 2 (a)(a) GCL (aa)() a 2 2 h(a)h(). and so (3) h(a) h(a)h() By (2) and (3), h is a homomorphism. 5. Let F a a + a, Z 2. Given that F is a suring of M 2 (Z 2 ). Show that F is a field. By Theorem 1.5.6 Z 2 {0, 1} and so y definition of F : F,,, Put 1 F. Since 1 F is an identity in M 2 (Z 2 ), 1 F is also an identity in F. In particular, 1 F is a unit. Moreover, and so oth and are units in F. Thus all non-zero elements in F are units. It remains to show that F is commutative. For a, Z 2 define M(a, ) a a +. Let, y F. Then y definition of F, M(a, ) and y M(c, d) for some some a,, c, d Z 2. We compute y M(a, )M(c, d) a a + c d d c + d ac + d ad + (c + d) c + (a + )d d + (a + )(c + d) ac + d ad + (c + d) c + (ad + d) d + ((ac + c) + (ad + d)) ac + d ad + (c + d) ad + (c + d) (ac + d) + (ad + (c + d)) M(ac + d, ad + c + d) Thus y M(ac + d, ad + c + d). This formula yields a formula for y y interchanging a and c and also interchanging and d. So
Since Z 2 is commutative and associative y M(ca + d, c + da + d) ac + d ca + d and ad + c + d c + da + d Thus So F is commutative. y M(ac + d, ad + c + d) M(ca + d, c + da + d) y. 6. Let f R S e a homomorphism of rings. Let A e a suring of R and put B {f(a) a A}. Show that B is a suring of S. We need to verify the four conditions of the Suring Theorem hold for the suset B of S. Let s S. Then y definition of B: ( ) s B s f(r) for some r A Let, y B. Then y (*) ( ) f(u) and y f(v) for some u, v A. Since A is a suring of R, the suring theorem implies that ( ) 0 R, u + v, uv and u all are in A (I) 0 S (1.11.7)(a) f(0 R ). By (***), 0 R A and so y (*), 0 S B. (II) + y f(u) + f(v) f hom f(u + v). By (***), u + v A and so y (*), + y B. (III) y f(u)f(v) f hom f(uv). By (***), uv A and so y (*), y B. (IV) f(u) (1.11.7)() f( u). By (***), u A and so y (*), B. Thus the four conditions of the Suring Theorem hold for B and so B is a suring of S. 7. Let p e a prime integer. Show that Z p is a field. By 1.6.4 Z p is a commutative ring with identity [1] p. Since p is a prime, we have p ±1 and so p 1. Thus [1] p [0] p y 1.4.9. It remains to show that any nonzero element of Z p is a unit in Z p. So let a Z p with a [0] p. Then 1.5.6 shows a [k] p for some k Z with 0 < k < p. Suppose that p k. Then 1.9.1 implies that p k. But k > 0 and so k k < p, a contradiction. Thus p k. As p is a prime we can apply (1.10.2)().. Thus either ( p k and p gcd(a, p) ) or ( p k and 1 gcd(k, p) ) Since p k this shows that 1 gcd(k, p). Hence 1.12.4 shows that [k] p is a unit in Z p. As a [k] p this shows that a is a unit in Z p. We proved that Z p is a commutative ring with non-zero identity and that every non-zero elements of Z p is a unit in Z p. Thus Z p is a field (y definition of a field). 8. Show that the function f C C, a + i a i, is an isomorphism. Let, y C, then a + i and y c + di for some a, R. Suppose that f() f(y). Then a i c id and so a c and d. Thus a + i c + id y. Hence f is injective. Note that f(a i) f(a + i( )) a + i( ( )) a + i and so f surjective. As f is injective and surjective, f is ijective.
Also f( + y) f((a + i) + (c + di)) f((a + c) + ( + d)i) (a + c) ( + d)i (a i) + (c di) f(a + i) + f(c + di) f( + y). and f(y) f((a + i)(c + di)) f((ac d) + (ad + c)i) (ac d) (ad + c)i (ac ( )( d)) + (a( d) + ( )c)i (a i)(c di) f(a + i)f(c + di). So f respects addition and multiplication and so f is a homomorphism. Since f is ijective, f is an isomorphism. 9. Consider the ring R M 2 (R) and the polynomials f 2 + and g in R[]. Compute fg and deg(fg). fg 2 + 5 + + (2.1.5)(c) 5 + + 2 0 The highest non-zero coefficient is the coefficient of 3, so deg fg 3. 10. Consider the polynomials f 3 + 1 and g 4 2 + 2 + 1 in Z 5 []. Find polynomials q and r in Z 5 [] such that f gq + r and deg r < deg g. Note first that (in Z 5 []) we have g 4 2 + 2 + 1 2 + 2 + 1. 2 2 + 2 + 1 3 + 1 3 2 2 2 2 + 4 + 1 2 2 4 2 3 + 3 So q 2 4 + 3 and r 3 + 3. 11. Consider the polynomials f 4 + 3 2 2 + 3 + 1 and g 3 3 2 + + 1 in Z 7 []. Find polynomials u and v in Z 5 [] such that fu + gv gcd(f, g).
E 1 4 + 3 2 2 + 3 + 1 f 1 + g 0 F 0 3 3 2 + + 1 f 0 + g 1 E 0 2 2 2 f 0 + g 2 F 0 3 1 2 G 1 f 0 + g ( 2 4) E 0 ( + 2) F 1 2 2 + 2 f 1 + g (2 + 4) E 1 G 1 E 1 2 + 1 f 3 + g ( + 2) F 1 ( 2) 1 3 G 2 f ( 3) + g ( 2 + 2) E 1 F 2 2 f ( 3) + g ( 2 2 2) E 0 G 2 E 2 + 2 f 3 + g ( 2 + 2 + 2) F 2 ( 1) 1 1 Hence 0 + 2 2 2 2 4 + 3 2 2 + 3 + 1 4 + 3 2 2 2 2 3 3 2 2 + 1 2 2 2 4 4 2 2 + 2 2 2 + 1 2 2 2 2 2 1 + 2 2 + 2 2 + 2 2 2 0 + 2 f 3 + g ( 2 + 2 + 2) gcd(f, g) 12. Find all monic reducile polynomials of degree 2 in Z 5 []. We will first determine all the monic reducile polynomials of degree 2 on Z 5 []. For this let f e a monic polynomial in Z 5 [] of degree 2. By Theorem 2 on the Solutions of Homework 9, f is reducile if and only if f is a product of two non-constant monic polynomial of lower degree in Z 5 []. Hence, since deg f 2, f is reducile if and only if f is the product of two monic polynomials of degree 1. By Theorem 1.5.6, Z 5 {0, 1, 2, 3, 4} {0, 1, 2, 1, 2}. Thus the monic polynomials of degree 1 in Z 3 [] are, + 1, + 2, 2. So the reducile monic polynomials of degree 2 in Z 5 [] are + 1 + 2 2 1 2 2 + 2 + 2 2 2 2 + 1 2 + 2 + 1 2 2 + 2 2 2 2 1 + 2 2 1 2 + 1 2 + 2 2 2 + 1 2 + 2 + 2 1 2 2 + 1 Note here that Z 5 [] is commutative, so fg gf and the entries left lank in lower half of the tale appear in the upper half. The monic polynomials of degree 2 in Z 5 [] are 2, 2 +1, 2 + 2, 2 2, 2 1, 2 +, 2 + + 1, 2 + + 2, 2 + 2, 2 + 1, 2 + 2, 2 + 2 + 1, 2 + 2 + 2, 2 + 2 2, 2 + 2 1, 2 2, 2 2 + 1, 2 2 + 2, 2 2 2, 2 2 1, 2, 2 + 1, 2 + 2, 2 2, 2 1, So the irreducile monic polynomials of degree 2 in Z 5 [] are 2 + + 1, 2 + + 2, 2 + 2, 2 2, 2 + 2 2, 2 + 2 1, 2 2 1, 2 + 1 2 + 2, 2 2, Let f e a polynomial of degree 2 in Z 5 [] which is not monic. Then f has leading coefficient 1. It follows that ˇf f ( 1) f and f ˇf. By Theorem 2 on the Solutions of Homework 2. f is reducile
if and only if ˇf is reducile. Hence f is irreducile if and only if ˇf is irreducile. Hence we can otain a complete list of all irreducile polynomial of degree 2 y multiplying the aove list of 10 monic irreducile polynomials with the four units 1, 2, 2, 1 in Z 5. This will give a list of 40 irreducile polynomials of degree 2 in Z 5 [].