Let d=GCD(a,b)
Then aR < dR, bR < dR. So, aR+bR < dR.
As R is
principal ideal ring, then aR+bR=cR, for some c in R.
Then a*1+b*0=cu, for
some u, hence c|a. Analogously, c|b. So c|d by definition.
Then, aR+bR = cR
> dR > aR+bR . Result: aR+bR=dR, where d=GCD(a,b).
Then there are such
elements u,v in R, that au+bv=d.
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