# Integral domain and ascending chain condition proof

Show that an integral domain $A$ is a principal ideal domain if every ideal $I$ of $A$ is principal, that is, of the form $I=(a)$. Show directly that the ideals in a PID satisfy the a.c.c.

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I can show that, but since I don't know what definitions you use, I would say the first part of the statement is trivial (a PID is an integral domain where every ideal is prinipical, according to the definition I know). For the second part, just follow the standard proof. PS: If this is homework, it is considered polite to add the homework tag. PPS: The tone of your question is not considered acceptable - you should ask questions, not give commands (thus downvote). – Johannes Kloos May 2 '12 at 14:27
What you asked has pretty much nothing to do with the nullstellensatz...... – BenjaLim May 2 '12 at 14:29
@Johannes To me, it is far more impolite to downvote a question simply due to the choice of language (imperative), than to ask an imperative question. And considering that the OP just joined today, why not be more considerate? Downvotes should be reserved for mathematical matters, not poetic. And to the person who voted to close as "too localized", please do elaborate. I am disappointed to see such nonconstructive unwelcoming behavior towards new members. – Gone May 2 '12 at 15:05
@BillDubuque +1! – rschwieb May 2 '12 at 15:40
@Johannes: few people would admit in public that they were wrong and act accordingly. Congratulations for this exemplary behaviour. – Georges Elencwajg May 2 '12 at 17:37
Let $I_1 \subseteq I_2 \subseteq \ldots$ be an infinite ascending chain of ideals. Now, note that $I := \bigcup_{k \ge 1} I_k$ is again an ideal, so there is some $a \in I$ such that $I = (a)$. But by the definition of $I$, there must be some $k \ge 1$ such that $a \in I_k$. Obviously, $(a) \subseteq I_k \subseteq I = (a)$, so $I_k = (a) = I$. Furthermore, for $j \ge k$, we have $I = I_k \subseteq I_j \subseteq I$, so for all $j \ge k$, $I_j = I_k$.