Question 32844
Prove that the set S of rational numbers (in lowest term) with odd denominators is
a subring of Q.
(b) Let I be the set of elements of S with even numerators. Prove that I is an ideal in S.
(c) Show that S/I consists of exactly two elements.

 proof: a) S = { p/q| p, q in Z,(p,q) = 1 & q is odd}
   p1/q1, p2/q2 in S --> p1/q1- p2/q2 = (p1q2 - p2q1)/q1 q2 is in S
  and p1/q1* p2/q2 = p1p2/q1q2 is in S
 So, S is a subring of Q.
 
 b) I = {p/q in S| p even}
  As in a) test I is a subring of S (for you).

  for r in S, p/q in I, so p is even and so
  r * pq = pp'/qq' if r = p'/q' is in I since pp' is even.
  Hence, rI < I  and so I is an ideof Q.

 c) S/I ={ s + I | s in S}
   for any s in S, let s = p/q 
 s is in I if p is even.(i.e s+I = I)
 if p is odd, then s - 1/3 = p/q - 1/3 = (3p-q)/3q belonging to I
  (since 3p-q is even)
 Hence, s + I = 1/3 for any s = p/q (p odd) in I.
 This shows S/I  = {I, 1/3 + I} with two elements.

 Kenny