Lemma 10.84.3. Let $M$ be an $R$-module. Then $M$ is a direct sum of countably generated $R$-modules if and only if it admits a Kaplansky dévissage.
Proof. The lemma takes care of the “if” direction. Conversely, suppose $M = \bigoplus _{i \in I} N_ i$ where each $N_ i$ is a countably generated $R$-module. Well-order $I$ so that we can think of it as an ordinal. Then setting $M_ i = \bigoplus _{j < i} N_ j$ gives a Kaplansky dévissage $(M_ i)_{i \in I}$ of $M$. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: