Math 307 Abstract Algebra Homework 10 Sample solution n 1. An element a of a ring R is nilpotent if a = 0 for some n 2 N. (a) Show that if a and b are nilpotent elements of a commutative ring, then a + b is also nilpotent. (b) Show that a ring R has no nonzero nilpotent element if and only if 0 is the only solution of x2 = 0 in R. n m Solution. (a) Suppose a = 0 = b with n; m 2 N. Because R is commutative, the Binomial theorem applies and n+m X n + m (a + b)n+m = ajbm+n−j = 0 j j=0 by the fact that aj = 0 or bn+m−j = 0 depending on j ≥ n or j < n. (b) If there is a nonzero x 2 R satisfies x2 = 0, then x is a nilpotent. If y 2 R is a nonzero nilpotent and k > 1 is the smallest positive integer such that yk = 0, then x = yk−1 satisfies x2 = y2k−2 = ykyk−2 = 0. 2. Show that the set S of all nilpotent elements of a communtative ring R is an ideal, i.e., S is a subring satisfying ax 2 S for every a 2 S and x 2 R. Solution. Let A be the set of nilpotent elements of a commutative ring R. First, 0 2 A; if x; y 2 A so that xn = 0 = ym, then (x−y)m+n = 0 by the same proof as in (a) of the previous question. Thus, x − y 2 A. Moreover, if z 2 R, then (xz)n = xnzn = 0. So, A is an ideal. 3. Suppose R is a commutative ring with unity and charR = p, where p is a prime. Show that φ : R ! R defined by φ(x) = xp is a ring homomorphism. p Solution. Note that for k = 1; : : : ; p−1, k = p!=(k!(p−k)!) is divisible by p. [To see this, note that if m = p!=(k!(p − k)!), then p! = mk!(p − k)! and p cannot be a prime factor of k!(p − k)!. p Pp p j p−j p p So, p is a factor of m.] Thus, φ(x + y) = (x + y) = j=0 j x y = x + y = φ(x) + φ(y), and φ(xy) = (xy)p = xpyp = φ(x)φ(y). So, φ is a ring homomophism. 0 4. Let R1 and R2 be rings, and φ : R1 ! R2 be a ring homomorphism such that φ(R1) 6= f0 g, 0 where 0 is the additive identity of R2. (a) Show that if R1 has a unity and R2 has no zero-divisors, then φ(1) is a unity of φ(R1). (b) Show that the conclusion in (a) may fail if R2 has zero-divisors. 0 Solution. (a) We prove the stronger result that φ(1) is the unity in R2. Since φ(R) 6= f0 g, 0 there is x 2 R such that φ(x) 6= 0 . Now for any z 2 R2, φ(x)φ(1)z = φ(x1)z = φ(x)z so that φ(1)z = z by left cancellation, and zφ(1)φ(x) = zφ(1x) = zφ(x) so that zφ(1) = z by right cancellation. Thus, φ(1)z = zφ(1) = z for all z 2 R2; φ(1) is the identity in R2. (b) Suppose φ : Z ! Z ⊕ Z such that φ(n) = (n; 0). Then φ(1) = (1; 0) is not the unity in Z ⊕ Z. 1 5. Find an multiplicative inverse of 2x + 1 in Z4[x]. Is the inverse unique? Solution. (2x + 1)(2x + 1) = 4x2 + 4x + 1 = 1. So, (2x + 1) is its own inverse. Note that in a ring with unity, if b; c are the inverses of a, then b = b(ac) = (ab)c = c. So, the inverse if always unique if it exists. 6. (a) Given an example to show that a factor ring of an integral domain may have zero-divisors. (b) Give an example to show that a factor ring of a ring with zero-divisors may be an integral domain. ∼ Solution. (a) Let R = Z, A = 4Z. Then R=A = Z4 has a zero divisor. ∼ (b) Let R = Z4 and A = h2i. Then R=A = f0 + A; 1 + Ag = Z2 has no zero divisors. 7. Let R1 and R2 be rings, and φ : R1 ! R2 be a ring homomorphism. (a) Show that if A is an ideal of R1, then φ(A) is an ideal of φ(R1). (b) Give an example to show that φ(A) may not be an ideal of R2. Solution. (a) Note that φ(A) is a subgroup of R2 using the group theory result. So, it is a subgroup of φ(R1). For any b 2 φ(A) and y 2 φ(R1), there are a 2 A, x 2 R1 such that φ(a) = b and φ(x) = y. Because ax; xa 2 A, we have by = φ(a)φ(x) = φ(ax) 2 φ(A) and yb = φ(x)φ(a) = φ(xa) 2 φ(A). (b) Suppose A = R1 = Z, R2 = Q, and φ(x) = x. Then φ is a ring homomorphism, φ(A) = φ(R1) = Z so that φ(A) is an ideal in φ(R1). But φ(A) = Z is not an ideal in R2 = Q, say, 1 2 Z, 1=2 2 Q and 1 · (1=2) 2= Z. 2 2 8. (8 points) Let A = hx + x + 1i = f(x + x + 1)f(x): f(x) 2 Z2[x]g ⊆ Z2[x]. (a) Show that Z2[x]=A = fa + bx + A : a; b 2 Z2g has 4 elements. Proof: Note that Z2[x]=A = ff(x) + A : f(x) 2 Z2[x]g: Now for every f(x) 2 Z2[x], f(x) = a + bx + (x2 + x + 1)q(x) so that f(x) + A = a + bx + (x2 + x + 1)q(x) + A = a + bx + A. Thus, Z2[x]=A = fa + bx + A : a; b 2 Z2g = fA; 1 + A; x + A; 1 + x + Ag has four distinct elements. (b) Show that (a + bx + A)(c + dx + A) = (ac + bd) + (ad + bc + bd)x + A. Proof: (a + bx + A)(c + dx + A) = ac + adx + bcx + bdx2 + A = ac + adx + bcx + bdx2 + bd + bd + bdx + bdx + A(ac + bd) + (ad + bc + bd)x + bd(x2 + x + 1)A = (ac + bd) + (ad + bc + bd)xA. (c) For each nonzero element a + bx + A 2 Z2[x]=A, show that there is c + dx + A 2 Z2[x]=A such that (a + bx + A)(c + dx + A) = 1 + A, and deduce that F = Z2[x]=A is a field. Proof: Because Z2[x]=A is a commutative ring with unity, we only need to show that every nonzero element has an inverse. Then Z2[x]=A is a field. Now, (1 + A)(1 + A) = 1 + A and (x + A)(1 + x + A) = 1 + A. The result follows. (d) Show that the nonzero elements in F form a cyclic group under multiplication. Proof: Note that (x + A)1 = x + A,(x + A)2 = x2 + A = 1 + x + A,(x + A)3 = x3 + A = x(x2 + x + 1) + (x2 + x + 1) + 1 + A = 1 + A. The result follows. 2.