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

1. Apr 8, 2007

### *melinda*

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. Apr 8, 2007

### Hurkyl

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