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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!