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Let D be a domain and $\emptyset \subset A \subseteq D^*$

If $x \in D^*$ and $GCD(xA)\neq \emptyset$ then $GCD(A)\neq\emptyset$ and $GCD(xA) = xGCD(A)$.

I've already figured out how to show that $GCD(A)\neq\emptyset$ and that there exists an element $r \in GCD(A)$ but I'm not sure how to show that $GCD(xA) = xGCD(A)$.

Any advice would be great!

user2553807
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1 Answers1

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The proof is essentially the same as in the classic case $\rm\,|A| = 2,\,$ namely

Theorem $\rm\ \ \gcd A\ =\ \gcd(xA)/x\ \ $ if $\rm\ \gcd(xA)\ $ exists in $\rm\:D.$

Proof $\rm\quad\: c\mid A \iff xc \mid xA \iff xc\mid \gcd(xA) \iff c\mid \gcd(xA)/x\ \ \ $ QED

Above we used the universal definition of the GCD.

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Bill Dubuque
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