Krull dimension of a direct limit of modules

Question

Suppose that $\left\{M_{\lambda}\right\}$ is a directed system of $R$-modules, all of them with finite Krull dimension, $n$. Is it true that $\dim\varinjlim M_{\lambda}\leq\sup\left\{\dim{M_{\lambda}}\right\}$?

Thank you.

Answer 1

Set $M := \varinjlim_{\lambda \in \Lambda} M_{\lambda}$. Let $M'_{\lambda} \subset M$ be the image of the canonical maps $M_{\lambda} \to M$; then $\operatorname{Supp} M'_{\lambda} \subseteq \operatorname{Supp} M_{\lambda}$. Thus after replacing the $M_{\lambda}$ by $M'_{\lambda}$, we reduce to the case when $M_{\lambda} \to M$ and the transition maps $M_{\lambda_{1}} \to M_{\lambda_{2}}$ are injective. In this case we show that in fact we have equality $\operatorname{dim} M = \sup_{\lambda \in \Lambda} \operatorname{dim} M_{\lambda}$. Since $M_{\lambda} \to M$ is injective, we have $\operatorname{dim} M_{\lambda} \le \operatorname{dim} M$. Conversely, let $\mathfrak{p}_{0} \subset \dotsb \subset \mathfrak{p}_{m}$ be a chain of primes contained in $\operatorname{Supp} M$; then for every $i=0,\dotsc,m$ there exists $\lambda_{i} \in \Lambda$ such that $\mathfrak{p}_{i} \in \operatorname{Supp} M_{\lambda_{i}}$. Since the index category $\Lambda$ is filtered, we may choose $\lambda' \in \Lambda$ such that $\lambda_{i} \le \lambda'$ for all $i=0,\dotsc,m$; then $\mathfrak{p}_{i} \in \operatorname{Supp} M_{\lambda'}$ as well since $M_{\lambda_{i}} \to M_{\lambda'}$ is injective. Thus $\operatorname{dim} M_{\lambda'} \ge m$.