On the set of ideals in a ring, divisibility is antisymmetric

Let $A$ and $B$ be ideals in a commutative unital ring $R$. Show that if $A|B$ and $B|A$ then $A = B$. Characterize the ideals which divide $(1)$.

If $A|B$, then $B = AC$ for some $C$. So $B \subseteq A$. Likewise, if $B|A$ then $A \subseteq B$. Thus if $A|B$ and $B|A$, then $A = B$.

Suppose $A$ is an ideal such that $(1) = AB$ for some ideal $B$. Now $(1) \subseteq A$, so that $A = (1)$. That is, the only unit in the semigroup of ideals in $R$ under ideal multiplication is $(1) = R$ itself.