TITLE: Show that an integral domain with finitely many ideals is a field
QUESTION [10 upvotes]: I know that an integral domain with finite number of elements is a field, but, how do relate this with the finitude of the number of ideals?

REPLY [8 votes]: Let $R$ be your integral domain, and take $a \in R$, $a \ne 0$. We have to prove that $a$ is a unit.
Let $\mathcal I$ be the set of ideals of $R$ and consider the map $\mathcal I \to \mathcal I$ given by $I \mapsto aI$.
This map is injective (*). Since $\mathcal I$ is finite, the map is surjective. Thus, $R=aI$ for some ideal $I$ and so $1=ai$ for some $i\in I$. This means that $a$ is a unit.
(*) If $I$ and $J$ are ideals of $R$ with $aI=aJ$, then for every $i\in I$, there is a $j\in J$ such that $ai=aj$. Since $R$ is a domain and $a\ne0$, this implies that $i=j\in J$. In other words, $aI=aJ$ implies $I \subseteq J$. By symmetry, $J \subseteq I$ and so $I=J$.