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Let $R$ be a PID. If $(a)$ is a nonzero ideal, then there are finitely many ideals containing $(a)$.

I know that this question has already been asked/answered here, but I wanted to write a more explicit solution. Is the following correct?

Since $R$ is a UFD, let $a = a_1a_2\cdots a_n$ where each $a_i$ is irreducible/prime. Let $X$ be the set of all divisors of $a$ and let $x \in X$. Since $R$ is a UFD, let $x = x_1x_2\cdots x_k$ where each $x_i$ is irreducible/prime. Since $x \mid a$, $k <n$ and each $x_i$ is associate to an $a_i$, call it $\bar{a}_i$. So, $x = u\bar{a}_1\cdots \bar{a}_k$ where $u$ is a unit in $R$. Hence, $(x) = (\bar{a}_1\cdots \bar{a}_k)$.

Thus, the number of ideals containing $(a)$ is $\leq nC1+ nC2 + nC3+\cdots+nCn$ (where $nCr$ is $n$ choose $r$) and hence is finite.

Community
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user2640211
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    Hint:: $\ (b)\supseteq (a)\iff b\mid a,,$ so it is equivalent to $,a,$ has finitely many divisors (up to associates). – Bill Dubuque Dec 12 '16 at 18:55

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Observe with respect to Bill Dubuque's reply, one can generalize then that it must be the case that if $R$ is a UFD, it satisfies the Ascending Chain Condition among its principal ideals.

Dylan Baldridge
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