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.
Integral domain and ascending chain condition proof
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commutative-algebra
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0I 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. – 2012-05-02
1 Answers
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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 $j \ge k$, $I_j = I_k$.
Edit: This seems answer the question (cf. comments).
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0yes it is just the ACC, thanks a lot for you. – 2012-05-05