gcd in principal ideal domain

Let $R$ be a principal ideal domain. Show that any pair of nonzero elements $a, b\in R$ have a greatest common divisor and that for any greatest common divisor $d$, we have $d$ in $aR + bR$. Show also that $a, b$ are relatively prime if and only if $1\in aR + bR$.

My attempt: $aR + bR$ is an ideal of R. Let $(aR+bR) =(d)$. That is $(aR+bR)$ is generated by element $d$ in $R$. $a\in aR+bR$ , hence $a = dm$ for some $m\in R$. Now, $b\in aR+bR$, hence $b = dn$ for some $n\in R$. That shows $a$ and $b$ has a common divisor. But I don't know how to go further to turn $d$ into the greatest common divisor.

For b) Suppose $\gcd(a,b)=1$, then $1$ is in $aR+bR=R=(1)$. Conversely, if $1$ is in $aR + bR$, then $aR+bR=(1)=R$, $\gcd(a,b)=1$.

Solution 1:

Let $\,(a,b) = (d).\,$ Then $\,a,b \in (d)\,\Rightarrow\, d\mid a,b,\,$ so $\,d\,$ is a common divisor of $\,a,b.\,$

Conversely $\,d\in (a,b)\,$ so $\,d = r a + sb,\ r,s\in R,\ $ so $\ c\mid a,b\,\Rightarrow\, c\mid d = ra+sb.\,$

Hence $\,d\,$ is a common divisor of $\,a,b\,$ that is divisible by every common divisor. Therefore, by definition $\,d\,$ is a greatest common divisor of $\,a,b.$

Hint for $(b):\ $ $\,1\in (a,b)\iff (1) = (a,b) = (\gcd(a,b))\ $ by above