Prove that a subset of an integral domain is relatively prime

Question

This is a statement that appears in the text of D. J. H. Garling's book A Course in Galois Theory (page 30) without proof. I am trying to understand why it is true.

Assume that $R$ is an integral domain, and that $B$ is a non-empty subset of $R$. An element $a\in R$ is called a greatest common divisor of $B$ if $a$ is a common divisor of every element in $B$, and is divisible by every common divisor of all the elements in $B$. Assume that $a$ is the greatest common divisor of $B$, and define $$C=\{c\in{R}:ca\in B\}.$$ Prove that the greatest common divisor of $C$ is equal to $1$.

Here is what I have so far. Since $a$ is a greatest common divisor of $B$, it follows that $B\subset (a)$ where $(a)$ is the ideal generated by $a$. If $d$ is the greatest common divisor of $C$, then since every element of $B$ is a multiple of $a$, $d$ must be a common divisor of $B$, and must therefore divide $a$. If I can prove that there exists $c\in R$ that is prime to $a$ such that $ca\in R$, then I could conclude that $d$ must be a common divisor of two coprime elements and must therefore be equal to $1$. I haven't used the fact that $R$ is an integral domain so far. Wondering whether this helps me find such an element $c$, such that $c$ and $a$ are coprime and $ca\in B$.

Answer 1

Let $d$ be a common divisor of the elements of $C$. By definition, for all $b\in B$, there exists $c_b\in R$ such that $b=c_b a$. Hence $c_b\in C$ for all $b\in B$.

Now, $d$ is a common divisor of the lements of $c$, so we may write $c_b=d e_b$ for some $e_b\in R$, for all $b$. Putting things together, $ad$ is a common divisor of the elements of $b$, so $ad\mid a$, hence $a=ade$ for some $e\in R$. Since $R$ is an integral domain, $1=de$ and $d$ is a unit, and we are done.

Answer 2

Hint: $\,\ \forall i\!:\ d\mid b_i/a\iff \forall i\!:\ ad\mid b_i\!\!\!\overset{\ \,\rm\color{#90f} U}\iff ad\mid\overbrace{\gcd(b_1,b_2,\ldots)}^{\textstyle \gcd(B)=\color{#0a0}a}\iff ad\mid\color{#0a0} a\iff d\mid\color{#c00} 1$

therefore $\, \gcd\{b_i/a\} =\color{#c00}1,\ $ by $\,\rm \color{#90f} U = $ GCD Universal Property.