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 April 25th, 2012, 08:23 AM #1 Newbie   Joined: Apr 2012 Posts: 16 Thanks: 0 Help quick If m is an integer, show that mR = {mr | r\in R} and A_m {r\in R | mr = 0} are ideals of R.
 April 28th, 2012, 03:06 PM #2 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: Help quick it's easy to do this just from the definition of an ideal. let x,y be in mR. then x = mr, and y = ms, for some r,s in R. thus: x - y = mr - ms = m(r - s), which is in mR. thus (mR,+) is a subgroup of (R,+). now let a be any element of R. then: ax = a(mr) = a(r + r +...+ r) (m times) = ar + ar +...+ ar (m times) = m(ar), which is in mR, and: xa = (r + r +...+ r)a (m times) = ra + ra +...+ ra (m times) = m(ra), which is in mR. thus mR is an ideal of R. ****** now suppose x,y are in A_m. then mx = 0 and my = 0, so m(x - y) = mx - my = 0 - 0 = 0, so x - y is in A_m, thus (A_m,+) is a subgroup of (R,+). again, let a be any element of R, with x in A_m. then m(ax) = ax + ax +...+ ax (m times) = a(x + x +...+ x) (m times) = a(mx) = a0 = 0, so ax is in A_m. also: m(xa) = xa + xa +...+ xa (m times) = (x + x +...+ x)a (m times) = (mx)a = 0a = 0, so xa is in A_m, thus A_m is an ideal of R.

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