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Im soooooo close to solving this problem (Rings)

  1. Apr 27, 2007 #1
    Let R be a ring of characteristic m > 0, and let n be any
    integer. Show that:

    if 1 < gcd(n,m) < m, then n · 1R is a zero divisor

    heres what i got out of this:

    Let gcd(n,m) = b

    1< d < m so m/d = b < m
    and d | n

    Also, m * 1_R = 0

    can someone please offer some insight?
  2. jcsd
  3. Apr 27, 2007 #2
    You know [itex]m 1_R=0_R[/itex]. You need to show that there are some x, y in R with [itex]x\cdot (n 1_R) = 0_R[/itex] and [itex](n 1_R) \cdot y = 0_R[/itex].

    I suggest trying [itex]y=x = \frac{m}{(n,m)}1_R[/itex]. :smile:
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