a. No. 1 A, A ≠ R.

c. Yes. ,

e. No. 1 A.

a. (1 + A)(r + A) = (1 · r + A) = r + A and (r + A)(1 + A) = (r · 1 + A) = r + A.

c. If R is commutative then, for all r + A, s + A in R/A :

a. mr + ms = m(r + s) mR, −(mr) = m(− r) mR; s(mr) = m(sr) mR; (mr)s = m(rs) mR. If mr = 0 and mt = 0 then m(r + t) = mr + mt = 0; m(− r) = − mr = 0; m(rs) = (mr)s = 0; and m(sr) = s(mr) = 0 for all s R.

a. A × B is clearly an additive subgroup and (r, s)(a, b) = (ra, sb) A × B for all (r, s) R × S and (a, b) A × B. Similarly (a, b)(r, s) A × B.

c. By (b) let A × B be a maximal ideal of R × S. Then either A = R or B = S (otherwise A × B ⊂ A × S ⊂ R × S). If B = S then A is maximal in R [if A ⊆ B ⊆ R then A × S ⊆ B × S ⊆ R × S so B = A or B = R]. Similarly, if A = R then B is maximal in S. Conversely A × S is maximal in R × S if A is maximal in R, with a similar statement for R × B.

a. Since i A = Ri, and since i is a unit in R, A = R. So R/A is the zero ring.

c. A = R(1 + 2i). By the Hint 3 + i A so i + A = − 3 + A. Hence (m + ni) + A = (m − 3n) + A, so each coset in R/A has the form k + A, . Moreover, 5 = 2(3 + i) − (1 + 2i) A so (as in Example 6)

a. Assume nR ≠ 0 and let A = {r R nr = 0}. Then A is an ideal of R and A ≠ R because nR ≠ 0. So A = 0.

a. If a, b ann X then (a ± b)x = ax ± bx = 0 and (ra)x = 0 for all r R.

c. a ann (X ∪ Y) at = 0 for all t X ∪ Y at = 0 for all t X and at = 0 for all t Y.

e. By (d) and (b), ann (X) ⊇ ann { ann [ann (X)]}. Let b ann (X). If y ann [ann (X)] then by = 0. Hence b ann { ann [ann (X)]}.

a. X + Y is a subgroup because 0 = 0 + 0 X + Y and, if r = x + y and r′ = x′ + y′ are in X + Y then r + r′ = (x + x′) + (y + y′) X + Y and −r = (− x) + (− y) X + Y. We have X ⊆ X + Y because x = x + 0 X + Y for all x X. Similarly Y ⊆ X + Y.

c. S + A is an additive subgroup because S and A are; and 1 = 1 + 0 S + A. Finally (s + a)(s′ + a′) = ss′ + (sa′ + as′ + aa′) S + A because sa′ + as′ + aa′ A.

a. If Z = Z(R) is an ideal then 1 Z forces Z = R, and R is commutative. Conversely, R commutative implies Z = R is an ideal.

c. Write Z = Z(R) and let as an additive group. Let r, s R. Then r + Z = m(b + Z) and s + Z = n(b + Z) where m, , say r = mb + z and s = nb + z′ where z, z′ Z. Hence

a. We have by (3) of Theorem 4. If let x = b + A and x = c + A where b B and c C. Hence b + A = c + A, so b − c A ⊆ C. But then b c + C = C, so b B ∩ C. Finally then, proving that

a. If M is a maximal ideal then R/M is a field, so R/M is an integral domain, so M is prime.

c. No. 0 is a prime ideal of because is a domain, but 0 is not maximal because is not a field.

a. (r + A)² = r² + A = r + A for all r R.

c. Given r + A, if rs = 1 = sr then (r + A)(s + A) = 1 + A = (s + A)(r + A). If rⁿ = 0 then (r + A)ⁿ = rⁿ + A = 0 + A.

a. Suppose e² = e in R. Then (e + A)² = e + A in R/A so e + A = 0 + A or e + A = 1 + A by hypothesis. Thus e A or 1 − e A. But e A means e is nilpotent (hypothesis) so eⁿ = 0, n ≥ 1. But e² = e ⇒ eⁿ = e for all n ≥ 1, so e = 0. If 1 − e A then 1 − e = 0 in the same way because (1 − e)² = 1 − e.

c. Suppose the only unit in R/A is 1 + A. If u R^∗ then u + A is a unit in R/A, so u + A = 1 + A by hypothesis. Hence u − 1 A, that is u 1 + A. This shows that R^∗⊆ 1 + A ; the other inclusion holds because A consists of nilpotents.

a. is a field so 0 is the only maximal ideal.

c. The divisors of 10 are 1, 2, 5, 10. So the lattice of additive subgroups is as shown. They are clearly ideals and 2 and 5 are the maximal ones,

a. Since R is commutative, rab = (ra)b Rb and rab = (rb)a Ra for all r R.

c. If u is a unit then u Ru implies Ru = R by Theorem 2. Conversely, if Ru = R then 1 Ru, say 1 = vu, . Hence u is a unit (R is commutative).

e. If a = ub, u R^∗, then a Rb so Ra ⊆ Rb. But b = u⁻¹a Ra, so Rb ⊆ Ra too. Conversely, if Ra = Rb then a Rb, say a = ub, u R. Similarly b = va, , so a = u(va) = (uv)a. If a = 0 then b = va = 0 so a = 1b. If a ≠ 0 then cancellation (R is a domain) gives 1 = uv.

a. AB is clearly an additive subgroup and r(∑ _i=1a_ib_i) = ∑ _i=1(ra_i)b_i AB for all r. Similarly . Now AB ⊆ A because A is an ideal. Similarly AB ⊆ B, so AB ⊆ A ∩ B.

c. AR ⊆ A because A is an ideal; A ⊆ AR because a = a · 1 for all a A. Thus A = AR; similarly A = RA.

a. If in let . Then xx^∗ = a² − 2b² = a² + b² in . Now 0² = 0, 1² = 2² = 1, in so a² + b² ≠ 0 if a ≠ 0 or b ≠ 0. Thus .

a. If aⁿ = 0 then (ra)ⁿ = rⁿaⁿ = 0 for all r. If also b^m = 0 consider

c. Let . Then and are nilpotent, but is not .

a. Here J(R) = {0} is an ideal.

c. Write If is an ideal of then so k pⁿ. Hence k = p^t for t ≤ n, so A ⊆ M where It follows that M is the unique maximal ideal of so is local and

e. Write J = J(R) and J/A = {r + A r J}. If r + A is a nonunit in R/A then r is a nonunit in R [rs = 1 ⇒ (r + A)(s + A) = 1 + A]. Hence J(R/A) ⊆ J/A. Conversely, let r + A J/A, so r J. We must show r + A is a nonunit in R/A. Suppose not, and write (r + A)⁻¹ = s + A. Then rs − 1 A so rs − 1 J. But r J and J is an ideal, so rs J. Thus −1 J, 1 J, a contradiction.

a. If a, b M then ab M because ab P ⇒ a P or b P (P is prime). Clearly 1 M because P ≠ R.

c. Define . This is an ideal of R_P because since av + bu + P if a, b P; and because ra P if a P. If , a P, then ar = uv so uv P. P is prime so u P or , a contradiction. Thus is a nonunit if a P so J ⊆ J(R_P). Conversely, let . Then we must show r P. But so in R_P, a contradiction. So r P and . Hence J = J(R_P) so J(R_P) is an ideal.

a. We show that J(R) = A. Clearly A ≠ R so A consists of nonunits; that is A ⊆ J(R). Let r J(R). If r A then r + A is a unit in R/A, say (r + A)⁻¹ = s + A. Thus rs − 1 A, say rs − 1 = a. Hence rs = 1 + a so rs is a unit in R because a is nilpotent by Example 17 §3.1. Similarly sr is a unit in R. If , we see that r is a unit, contrary to r J(R).

c. The power series for e^x is . If a A then a is nilpotent so is an element of R which is nilpotent (R is commutative; see Exercise 34(a)). Thus σ : A → 1 + A given by is a map. Similarly so if a A, . Thus τ : 1 + A → A is well defined by τ(u) = ln u. Now τσ(a) = ln [e^a] = a and στ(u) = e^lnu = u hold, so σ and τ are inverses. Hence σ is a bijection. Finally σ(a + b) = e^a+b = e^ae^b = σ(a)σ(b) for ab A. Thus σ is a group isomorphism.