Could you please help me finish the proof below. The only problem is the $(\Leftarrow)$ part of a).
Proposition???: In any domain:
a) $(a_1,\ldots,a_n)\!=\!(a)\;\Leftrightarrow\;a\!=\!\gcd(a_1,\ldots,a_n)$
b) $Ra_1\!\cap\!\ldots\!\cap\!Ra_n\!=\!Ra\;\Leftrightarrow\; a\!=\!\mathrm{lcm}(a_1,\ldots,a_n)$.
Proof:
$(\Rightarrow)\!:$ If $(a_1,\ldots,a_n)\!=\!(a)$, then $a_i\!\in\!(a)$, so $a\mid a_i$. If any other a'\mid a_i, then a_i\!\in\!(a'), so (a')\geq(a_1,\ldots,a_n)\!=\!(a), hence a'\mid a. $\checkmark$
$(\Leftarrow)\!:$ If $a\!=\!\gcd(a_1,\ldots,a_n)$, then $a\mid a_i$, so $a_i\!\in\!(a)$, hence $(a_1,\ldots,a_n)\leq(a)$. If a'\!\in\!(a), then ???
$(\Rightarrow)\!:$ If $Ra_1\!\cap\!\ldots\!\cap\!Ra_n\!=\!Ra$, then $a\!\in\!Ra_i$ so $a_i\mid a$. If any other a_i\mid a', then a'\!\in\!Ra_1\!\cap\!\ldots\!\cap\!Ra_n\!=\!Ra, so a\mid a'. $\checkmark$
$(\Leftarrow)\!:$ If $a\!=\!\mathrm{lcm}(a_1,\ldots,a_n)$, then $a_i\mid a$, so $a\!\in\!Ra_i$, hence $Ra\!\leq\!Ra_1\!\cap\!\ldots\!\cap\!Ra_n$. If a'\!\in\!Ra_1\!\cap\!\ldots\!\cap\!Ra_n, then a_i\mid a', so a\mid a' and a'\!\in\!Ra. $\checkmark$
Is this not true in any domain? I would think it is, since the analogous statement for $\mathrm{lcm}$ is...