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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. Tags quick Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post bignick79 Algebra 3 June 15th, 2010 10:41 AM peterle1 Algebra 1 March 3rd, 2010 01:08 AM DouglasM New Users 2 February 18th, 2010 11:32 AM Dark Consort Algebra 8 August 22nd, 2008 11:40 AM Helpless13 Algebra 2 July 10th, 2008 09:15 PM

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