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I am trying to solve this exercise in Artin.

(a) Define the greatest common divisor of a set $\{a_1, \ldots, a_n\}$ of $n$ integers. Prove that it exists, and that it is an integer combination of $a_1, \ldots, a_n$.

(b) Prove that if the greatest common divisor of $\{a_1, \ldots, a_n\}$ is $d$, then greatest common divisor of $\{a_1/d, \ldots, a_n/d\}$ is $1$.

I am having difficulty getting started, and am really just looking for a hint and some clarifications on interpretation. My thoughts on this at the moment are:

1) My first instinct is to induct on $n$. I don't think we can define the $\gcd$ of an empty set or of a single element, so the base case would would have to be $n = 2$.

2) It doesn't seem that I have any information on these integers. Can we define the $\gcd$ of $0$ or negative integers? Surely the $\gcd$ itself is greaeter than or equal to $1$.

3) I cannot figure out how to 'prove' existence. If we have two integers, the $\gcd$ "exists by definition." If the integers are relatively prime, their $\gcd$ is $1$. If not, there exists some integer $d > 1$ that divides both. There's certainly a finite number of such elements because the distance between any two integers is finite, so taking the maximum gives the $\gcd$.

4) The $n \implies n + 1$ inductive step seems to just be invoking this same $n = 2$ argument I just made.

John P.
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    The gcd of an empty set is $1$, and $\gcd({n})$ is $n$. – MJD Mar 18 '20 at 20:35
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    @MJD If I were to define $\gcd\emptyset$, I would set it to be $0$. This is to make sure that $\gcd (A\cup B)=\gcd(\gcd A,\gcd B)$ for all sets $A,B\subseteq\Bbb Z$. After all, the ideal generated by $\emptyset$ of $\Bbb Z$ is the zero ideal. (Likewise, $\operatorname{lcm}\emptyset$ should be defined to be $1$.) – Batominovski Mar 18 '20 at 20:54
  • Yes, you're absolutely right. – MJD Mar 19 '20 at 00:42

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Hint:Take the subset generated $S$ by $a_1,..,a_n$ and the smallest positive element is $d$.

$d$ divides $a_i$ if not $a_i=b_id+r, r<d$, $ r=d-b_ia_i\in S$ contradiction. Let $c$ be a common divisor, $a_i=c_ic$, since $d\in S, d=b_1a_1+...+b_na_n=b_1c_1c+...+b_nc_nc=(b_1c_1+..+b_nc_n)c$ and $c$ divides $d, c\leq d$.

Tsemo Aristide
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