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April 25th, 2012, 08:23 AM   #1
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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.
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April 28th, 2012, 03:06 PM   #2
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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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