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I am trying to follow this question and response... Integral multiples of g.c.d.(m,n)

let $J_m$ be the set of all integral multiples of the integer $m$. 1) Prove that $J_{g.c.d(m,n)}$ contains all the elements of $J_m$ and $J_n$.

2) Prove that if $J_r$ contains all the elements of $J_m$ and $J_n$, then $J_r$ contains all the elements of $J_{g.c.d(m,n)}$.

for part 1) the elements of $J_{g.c.d(m,n)}$ are of the form $c*g.c.d(m,n)$, where $c\in \mathbb{Z}$.

Now if $x$ is in $J_m$, then ∃ $a\in \mathbb{Z}$ | $am=x$. So we want $x=am=b*g.c.d(m,n)$ for some $b\in \mathbb Z$. If we pick $b=am/g.c.d(m,n)$, then our task is done.

I am having trouble seeing how $$J_m\cup J_n \subseteq J_{gcd(m,n)} \rightarrow J_{gcd(m,n)} \subseteq J_r$$ because $J_m\cup J_n$ is smaller than $J_{gcd(m,n)}$ yet it is saying $J_{gcd(m,n)}$ is contained in $J_r$ because of the fact that $J_m\cup J_n \subseteq J_{gcd(m,n)}$ which I don't buy...

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MathIsHard
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If $J_m \cup J_n ⊆ J_r$, then $r|m$ and $r|n$. Since, $r|m$ and $r|n$ then it must be the case that $r|gcd(m,n)$. So, $r = gcd(m,n)$ or $r|gcd(m,n)$. In either case, $J_{gcd(m,n)} \subseteq J_r$.

Note, the critical step in this proof is the fact that $gcd(m,n)$ is the largest number number that divides both $m$ and $n$. However, any distinct number that divides $gcd(m,n)$ will generate a larger 'integral multiple ideal'.

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  • Thank you for the help. I am a bit stuck still because don't we know that the element only divides m or n? Not necessarily both since it is a $\cup$? – MathIsHard Feb 14 '17 at 15:55
  • Oh but maybe I see what you mean.. That r would have to divide both if they are both contained in it.. Thank you! – MathIsHard Feb 14 '17 at 16:02