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Every PID $R$ is Noetherian, since if $a_1R\subsetneq a_2R\subsetneq \cdots$ is an infinite increasing chain of ideals, then the union $I=\cup_i a_iR$ is also an ideal, hence $I=bR$ for some $b\in R$. But then $b\in a_iR$ for some $i$ and we obtain the contradiction $bR\subseteq a_iR\subsetneq a_{i+1}R\subseteq I=bR$.

Question: Does there exist a supremum on the length of chains above a given ideal? $$\sup\{d\in\mathbb{N}: \exists b_1,\ldots,b_d\in R, aR\subsetneq b_1R\subsetneq \cdots \subsetneq b_dR\}<\infty?$$ If not, then what further condition on $R$ is needed to guarantee this?

user26857
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Drew Armstrong
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2 Answers2

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TL;DR: your $d$ is the sum $\alpha$ of the exponents of the primes in the decomposition of $a$.

Elementary version:

If you have an increasing chain $aR \subset b_1R \subset b_2R \subset \ldots \subset b_dR$, then let $a=b_0$, then $b_{i+1}|b_i$ for all $0 < i < d$, and $b_i$ doesn’t divide $b_{i+1}$. So $a=\prod_{0 \leq i < d}{\frac{b_i}{b_{i+1}}}$ is a product of $d$ nonunits, so considering the prime decompositions, we find $d \leq \alpha$.

Conversely, we can weite $a=p_1\ldots p_{\alpha}$ for some primes, not necessarily distinct. Define, for $1 \leq i \leq \alpha$, $b_i=p_{i+1}\ldots p_{\alpha}$, then we have an increasing chain $aR \subset b_1R \subset \ldots \subset b_{\alpha}R$.

Longer/sophisticated version:

You are asking (almost by definition) for what is called the length of the $R$-module $R/aR$.

If $M$ is a $R$-module, its length is the maximal (or $\infty$ if it’s unbounded) length of an increasing chain of submodules.

When a $R$-module has finite length, it can be shown (it isn’t obvious but it’s not too hard either) that its length is the length of any increasing sequence of submodules that is maximal, that is, when there is no submodule which you can “add” to the chain.

The length can be shown to be additive (and all of the claims above hold in any commutative ring with unity, only now we start using the fact that $R$ is a PID), and the PID version of the CRT shows that $R/aR$ is isomorphic to a direct sum of finitely many $R/p_i^{k_i}R$, where $a=\prod_i{p_i^{k_i}}$ is the prime decomposition up to a unit.

We claim that each $R/p^kR$ has length $k$, for $k \geq 1$, and every prime $p$, which shows that your $d$ is the sum of the $k_i$.

Indeed, it’s easy to see that the chain $\{0\}=p^kR/p^kR \subset p^{k-1}R/p^kR \ldots \subset pR/p^kR \subset R/p^kR$ is increasing of length $k$ and contains all the submodules, so that concludes.

Aphelli
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Yes, there is a bound for each $a$, but there need not be a uniform bound for all $R$.

Every PID is a UFD. If we factor $a$ into irreducibles, $$a = um_1^{\alpha_1}\cdots m_r^{\alpha_r},$$ where $u$ is a unit, $m_1,\ldots,m_r$ are pairwise non-associate irreducibles, and $\alpha_i\gt 0$ for all $i$, then $b$ divides $a$ if and only if there exists a unit $v$ and integers $\beta_1,\ldots,\beta_r$, $0\leq \beta_i\leq\alpha_i$, such that $$vb = m_1^{\beta_1}\cdots m_r^{\beta_r},$$ by unique factorization. Thus, there are exactly $\prod_{i=1}^r(\alpha_i+1)$ associate classes of divisors of $a$. Therefore, there are only that many ideals of $R$ that contain $aR$. In particular, every strictly ascending chain starting at $a$ is of length at most that number.

In fact, the bound can be improved, since each step takes away at least one irreducible factor, so in fact the chain is of length at most $\alpha_1+\cdots+\alpha_r$.

However, this also shows that there is no uniform bound for all elements ideals of $R$. In $\mathbb{Z}$, even though every ascending chain is finite, there is no upper bound for how long such an ascending chain can be.

Arturo Magidin
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