Showing that $(m)\cap (n)=(\operatorname{lcm}(m,n))$ and $(m)+(n)=(\gcd(m,n))$ for any $m,n\in\mathbb{Z}$

Suppose that $I$ and $J$ are ideals of $\mathbb{Z}$, with $I=(m)$ and $J=(n)$. This question has two parts:

1) Let r be the least common multiple of $m$ and $n$. Show that $I\cap J = (r)$.

2) Let $d=(m,n)$. Show that $I+J = (d)$.

Remember that $(a) \subseteq (b)$ if and only if $b$ divides $a$.

1) $(\ell) = (m) \cap (n) \subseteq (m)$ and $(n)$. Thus $m$ and $n$ divide $\ell$ (so $\ell$ is common multiple). What if $k$ is divisible by $m$ and $n$? What would imply that $\ell$ divides $k$ so that $\ell$ is the least common multiple?

2) $(d)=(m,n)=(m)+(n)$. Then $(m)$ and $(n) \subseteq (d)$. Thus $d$ divides $m$ and $n$ (so $d$ is a common divisor). What if $k$ divides $m$ and $n$? What would imply that $k$ divides $d$ so that $d$ is the greatest common divisor?

Hint $\ $ It is straightforward if one employs the universal definitions of lcm and gcd:

$\begin{eqnarray} \rm\ \ (m)\cap(n) = (k) &\iff&\rm [\ (m),(n)\supset (j)\!\! &\iff&\rm (k)\supset (j)\ ] \iff &\rm[\ m,n\ |\ j &\iff&\rm k\ |\ j\ ] \\[.3em] \rm\ \ (m)+(n) = (k) &\iff&\rm [\ (j)\supset (m),(n)\!\! &\iff&\rm (j)\supset (k)\ ] \iff &\rm[\ j\ |\ m,n &\iff&\rm j\ |\ k\ ] \end{eqnarray}$

Notice how the above makes crystal-clear the innate duality between lcm and gcd.