$(a_1,\ldots,a_n)\!=\!(a)\;\Leftrightarrow\;a\!=\!\gcd(a_1,\ldots,a_n)$? 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...

No, this is not true in $\mathbb{Z}[x,y]$, since $gcd(x,y)=1$, but $(x,y)$ is not principal (not to mention that $(x,y)\subsetneq R$).  In particular, part a) doesn't hold in any domain that is not a Bézout domain (ie an integral domain in which every finitely generated ideal is principal--in particular, every PID is Bézout).
Of course, if we're under the assumption that $R$ is a PID, then for your proof of $(\Leftarrow)$ in a), you can use the fact that $(a_1,a_2,\cdots,a_n)=(x)$ for some $x\in R$ and then show that $x$ must necessarily be a greatest common divisor of $a_1,\cdots,a_n$.
