Show that an integral domain with finitely many ideals is a field [duplicate]

Hint: let $R$ be your integral domain, and suppose $\alpha \in R$ is nonzero. Consider the ideals $\langle \alpha \rangle, \langle \alpha^{2} \rangle, \langle \alpha^{3} \rangle, \ldots$. What must be true if any of these two ideals $\langle \alpha^{k_{1}} \rangle$ and $\langle \alpha^{k_{2}} \rangle$ are equal?

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 $I=J$.