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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 $aR+bR=(1)=R$, $\gcd(a,b)=1$.

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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

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    $\begingroup$ Thanks for respond. My question is why c |d but not d |c? $\endgroup$
    – user73195
    Commented Apr 9, 2015 at 0:38
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    $\begingroup$ @user73195 Above $\,c\,$ is any common divisor of $\,a,b.\,$ To show that $\,d\,$ is a greatest common divisor we must show that it is divisible by every such common divisor $\,c, \ $ Here "greatest" is with respect to divisibility. $\endgroup$
    – Bill Dubuque
    Commented Apr 9, 2015 at 0:41
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    $\begingroup$ if we let d first ,like (a,b) =(d) and argue that d is a common divisor of a, b. Then, if c |a,b , then c | d. Yes. I got it. But if we say c| a,b, then both c and d are divisors of a, b. How to know that c |d but not d |c. Of course, c and d are associate. But if c |d, then d is greatest, if d|c, then d is not greatest. Am I right? Sorry for being dummy. :) $\endgroup$
    – user73195
    Commented Apr 9, 2015 at 0:43
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    $\begingroup$ @user73195 Yes, there are two steps/directions. the inclusion $\,(a,b)\subseteq (d)\,$ yields that $\,d\,$ is a common divisor of $\,a\,b,\,$ and the reverse inclusion $\,(a,b)\supseteq (d)\,$ yields that it is greatest, because it yields the Bezout identity expressing $\,d\,$ as an $R$-linear combination of $\,a,b.\,$ Such linear common divisors are always greatest, as the proof shows. $\endgroup$
    – Bill Dubuque
    Commented Apr 9, 2015 at 0:49

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