# Prove that a subset of an integral domain is relatively prime

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 $$ca\in B$$.

## 2 Answers

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.

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.