Set 1 - Ma 5b - Solutions Luciena Xiao January 16, 2017 7.1.7. Let R be a ring and let Z(R) denote its center. First note that 1 2 Z(R) because 1r _ = r = r1_ for any r 2 R: Now, to show Z(R) is a subring, it suffices to show closure under subtraction and multiplication: for any a; b 2 Z(R) and r 2 R; we have (a − b)r = ar − br = ra − rb = r(a − b); and abr = arb = rab: Thus Z(R) is indeed a subring that contains the identity. Suppose R is a division ring. For any a 2 Z(R), a 2 R and a−1 exists because R is a division ring. For any r 2 R; a−1r = (r−1a)−1 = (ar−1)−1 = a−1r; which implies a−1 2 Z(R): This proves that Z(R) is a division ring. Clearly, Z(R) is commutative, so Z(R) is a field. 7.1.13. (a) Clearly, akbk = (ab)k and n = akbjakbk; so (ab)k ≡ 0 mod n; i.e. ab is nilpotent in Z=nZ: m (b) Suppose a 2 Z=nZ is nilpotent, then nja for some positive integer m: Thus if p is any prime divisor of n; then pjam; which implies pja: Qk mi Conversely, suppose any prime divisor p of n also divides a: Write n = i=1 pi where pi are m primes. Let m = maxfm1; ··· ; mkg; then nja ; so a is nilpotent in Z=nZ: 3 3 By the above statement, all nilpotent elements of Z=72Z must be multiples of 6 since 72 = 2 3 : Therefore they are: 0; 6; 12; 18; 24; 30; 36; 42; 48; 54; 60; 66: (c) Suppose f 2 R is nilpotent. Then there exists some positive integer m such that for any x 2 X; (f(x))m = 0: Since F is a field and f(x) 2 F; f(x) cannot be a zero divisor, so f(x) = 0: But this holds for any x 2 X; so f is the zero function. Therefore R has no non-zero nilpotent element. 7.1.14. (a) Since x is nilpotent, there exists some minimal positive integer m with xm = 0: If m = 1; x = 0; otherwise xx_ m−1 = 0 where xm−1 6= 0; which implies x is a zero divisor. (b) gain, since x is nilpotent, there exists m such that xm = 0: Since R is commutative, for any r 2 R we have (rx)m = rmxm = rm0_ = 0; so rx is nilpotent. (c) Let m be as above. 1 = 1 + xm = (1 + x)(1 − x + ··· + (−x)m−1); so 1 + x is a unit. (d) For any unit a 2 R; a + x = a(1 + a−1x): By parts (b) and (c), 1 + a−1x is a unit. A product of units is a unit, so a + X is a unit. 1 7.1.23. We first prove Of is a subring containing the identity. 1 = 1 + 0f!; so 1 2 Of : For any x = a + bf!; y = c + df! 2 Of ; we have x − y = (a − c) + (b − d)f! 2 Of : Also, xy = 2 2 2 (ac + bdf ! ) + (ac + bd)f!: If Dp6≡ 1 mod 4; then ! = D and xy is clearly in mathcalOf ; if D ≡ 1 mod 4; then !2 = (1 + D)2=4 = (D − 1)=4 + ! and 4jD − 1; so we can write xy = 2 (ac + bdf (D − 1)=4) + (ac + bd + bdf)f!; which is in Of : Thus Of is a subring. 0 0 One can define a map φ : O!Of by a + b! 7! a + b f! where f is the remainder of b modulo f: It is easy to check that this map is a f-to-1 group homomorphism, so [O : Of ] = jker(φ)j = f: Now conversely suppose R ⊂ O is a subring with the stated properties. Then there exists a smallest integer f 0 such that f 0! 2 R since otherwise R wouldn't have finite index. Now if there 0 0 0 0 exists n! 2 R with f - n; then n ! 2 R where n is the remainder of n modulo f ; contradicting 0 0 the minimality of f : Therefore R must be of the form Z[f !]: By the previous result we conclude 0 f = f; i.e. R = Of : 7.3.17. (a) Suppose φ(1) 6= 1: Then φ(1)2 = φ(12) = φ(1); so φ(1)(φ(1) − 1) = 0; but φ(1) − 1 6= 0; so we conclude φ(1) is a zero divisor. If S is an integral domain, then S contains no zero divisor, so φ(1) must be the identity. (b) If u is a unit, then 1 = φ(1) = φ(uu_ −1) = φ(u)φ(u−1); so φ(u) is a unit and φ(u−1) = φ(u)−1: 7.3.22. (a) Write I = fx 2 Rjax = 0g: Clearly 0 2 I; so I is nonempty. For any x; y 2 I, we have a(x − y) = ax − ay = 0 − 0 = 0 and a(xy) = (ax)y = 0y _ = 0; so I is closed under subtraction and multiplication. Thus I is a subring of R: For any x 2 I and r 2 R; a(xr) = (ax)r = 0; so Is ⊂ R for all s 2 R: Thus I is a right ideal. Similarly, the left annihilator is a left ideal. (b) Write I = fx 2 Rjxa = 0 for all a 2 Lg: By an argument analogous to part (a) we see I is a subring. For any r 2 R; x 2 I; a 2 L we have (rx)a = r(xa) = r0_ = 0 and (xr)a = x(ra) = 0 since ra 2 L: Thus I is a two-sided ideal. 2.