Let R = ab | a ∈ Z, b = 3n for some n ≥ 0, gcd(a, b) = 1 ⊂ Q. Find all sub-rings of R. Because rings as dened in the book have the additive identity and multiplicative identity, we immediately know any subring S of R must contain Z. In particular, Z is a subring of R. Suppose, however, that the subring S had an a element n for some n ≥ 1 and a 6= 0. If necessary we can take its additive 3 inverse and assume a > 0. Then under addition, we'd have a a a = n + · · · + n ∈ S. 3 |3 {z 3 } 3n−1 times Now take the greatest integer less than or equal to jak 3 jak a . Then , denoted 3 3 ∈ Z ⊂ S and hence a jak ∈ S. − 3 3 It follows that either 1 2 ∈ S or ∈ S. 3 3 If 13 ∈ S then S = R. Suppose 23 ∈ S . Then − 13 = 23 − 13 ∈ S and hence 13 ∈ S (closed under additive inverses). Thus, S = R. In conclusion, subrings of R are just Z and R itself. 3.2.14: Suppose there was a eld F of order 6, then the additive subgroup is isomorphic to C6 . Thus there is an element of (additive) order 6 which generates the additive subgroup, call it g . Thus we have F = {0, g, 2g, 3g, 4g, 5g}. But then 2g · 3g = (g + g)(g + g + g) = g2 + g2 + g2 + g2 + g2 + g2 = 6g 2 = 0 where the last equality comes from Lagrange's theorem: the (additive) order of element divides the order of the (additive) group. 3.2.16b An additive subgroup of F must have order 2, 4, 8. If it is 2, then it must be {0, 1}. If it is 8 it must be F . Assume it is 4. Then the multiplicative part is order 3 and a subgroup of the multiplicative group of F which is order 7. But 3 doesn't divide 7, so no such subeld of order 4 may exist. 3.2.16c Then we may have 4 or 8 and respectively multiplicative groups of order 3 or 7. 3 does divide 15, whereas 7 does not. Thus the same argument says we cannot have a subeld of order 8, but we wouldn't know about a subeld of order 4. 3.2.29c 1 We show every nonzero element a + bω has an inverse. Formally, we might want to write 1 1 a − bω = a + bω a + bω a − bω a − bω = 2 a + b2 = a(a2 + b2 )−1 − b(a2 + b2 )−1 . Thus, it suces to show that a2 + b2 6= 0. If, however, we could nd a, b ∈ Zp such that a2 + b2 = 0 then if a = 0 then b = 0, contradicting that a + bω is nonzero, if b = 0 then a = 0, contradicting that a + bω is nonzero, and nally if both are nonzero then (ab−1 )2 + 1 = a2 (b−1 )2 + 1 = 0 which contradicts the corollary that x2 = −1 has no solutions in Zp for prime p with p ≡ 3 (mod 4). 3.3.17 a) When R is commutative, because the center is a subring and hence contains 1. Thus R = Z . If R = Z , then R is a commutative ring. b) Z is a commutative subring of R. Then for a ∈ Z , a 6= 0, consider the principal ideal generated by a. Because a ∈ Ra, then Ra is not the 0 ideal and hence Ra = R, because R is simple. It follows that a has an inverse in R: ba = ab = 1 for some b ∈ R. We conclude that Z is a eld (when R is simple). c) Suppose R/Z = ha + Zi for some a. Thus given r, s ∈ R/Z we have rs = (ja + u)(ka + v) for some integers j, k and some elements u, v ∈ Z . Expanding we have (ja)(ka) + (ja)v + u(ka) + uv . Using the property of Z and the commutativity of Z we have (ja)(ka) + (ja)v + u(ka) + uv = (ka)(ja) + v(ja) + (ka)u + vu and the right-hand side equals sr. 2
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