Lemma 10.85.2. Let $M$ be a countably generated $R$-module with the following property: if $M = N \oplus N'$ with $N'$ a finite free $R$-module, then any element of $N$ is contained in a free direct summand of $N$. Then $M$ is free.
Proof. Let $x_1, x_2, \ldots $ be a countable set of generators for $M$. We inductively construct finite free direct summands $F_1, F_2, \ldots $ of $M$ such that for all $n$ we have that $F_1 \oplus \ldots \oplus F_ n$ is a direct summand of $M$ which contains $x_1, \ldots , x_ n$. Namely, given $F_1, \ldots , F_ n$ with the desired properties, write
\[ M = F_1 \oplus \ldots \oplus F_ n \oplus N \]
and let $x \in N$ be the image of $x_{n + 1}$. Then we can find a free direct summand $F_{n + 1} \subset N$ containing $x$ by the assumption in the statement of the lemma. Of course we can replace $F_{n + 1}$ by a finite free direct summand of $F_{n + 1}$ and the induction step is complete. Then $M = \bigoplus _{i = 1}^{\infty } F_ i$ is free. $\square$
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Your email address will not be published.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)