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G.c.d.'s and PID's

  1. Apr 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Prove: If a, b are nonzero elements in a PID, then there are elements s, t in the domain such that sa + tb = g.c.d.(a,b).

    2. Relevant equations

    g.c.d.(a,b) = sa + tb if sa + tb is an element of the domain such that,
    (i) (sa + tb)|a and (sa + tb)|b and
    (ii) If f|a and f|b then f|(sa + tb)

    3. The attempt at a solution

    Since a, b, s, t, are all elements of the PID, so is sa + tb by properties of rings.

    I also know that if f|a and f|b then a = xf and b = yf. So f|(sa + tb) can be written as f|(sx + ty)f, which shows that f|(sa + tb) as desired.

    I'm just not sure how to satisfy criteria (i) of the definition of g.c.d.

    I know that if d and d' are g.c.d.'s of a and b, then d and d' are associates, but I'm not sure how to use this to my advantage.

    Any suggestions would be appreciated!
  2. jcsd
  3. Apr 8, 2007 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    You don't seem to have invoked the fact you are in a PID, rather than in an arbitrary ring...
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