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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?

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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?

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    $\begingroup$ This argument shows in particular that every Artinian domain is a field. $\endgroup$
    – Crostul
    Sep 9, 2015 at 16:30
  • $\begingroup$ that $\alpha$ is invertible? $\endgroup$ Sep 9, 2015 at 16:59
  • $\begingroup$ Indeed, since (WLOG taking $k_{1} < k_{2}$) we have $\alpha^{k_{1}} \in \langle \alpha^{k_{2}} \rangle)$, so $\alpha^{k_{1}} = \alpha^{nk_{2}}$ for some $n \in \mathbb{N}$, so we have $\alpha^{nk_{2}} - \alpha^{k_{1}} = 0$. Now factor and use the fact that you're in an integral domain. $\endgroup$ Sep 9, 2015 at 20:24
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    $\begingroup$ @AlexWertheim, $\alpha^{k_{1}} = \alpha^{nk_{2}} b$, for some $b$. $\endgroup$
    – lhf
    Sep 9, 2015 at 21:56
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    $\begingroup$ @PrinceKumar: It does not. I am operating under the (standard) convention that an integral domain is assumed to be commutative. $\endgroup$ Aug 16, 2017 at 20:08
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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 $I=J$.

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