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

  • Thread starter *melinda*
  • Start date
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1. Homework Statement

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. Homework 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!
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
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You don't seem to have invoked the fact you are in a PID, rather than in an arbitrary ring...
 

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