The Stacks project

Remark 10.93.2. Lemma 10.93.1 does not hold without the countable generation assumption. For example, the $\mathbf Z$-module $M = \mathbf{Z}[[x]]$ is flat and Mittag-Leffler but not projective. It is Mittag-Leffler by Lemma 10.91.4. Subgroups of free abelian groups are free, hence a projective $\mathbf Z$-module is in fact free and so are its submodules. Thus to show $M$ is not projective it suffices to produce a non-free submodule. Fix a prime $p$ and consider the submodule $N$ consisting of power series $f(x) = \sum a_ i x^ i$ such that for every integer $m \geq 1$, $p^ m$ divides $a_ i$ for all but finitely many $i$. Then $\sum a_ i p^ i x^ i$ is in $N$ for all $a_ i \in \mathbf{Z}$, so $N$ is uncountable. Thus if $N$ were free it would have uncountable rank and the dimension of $N/pN$ over $\mathbf{Z}/p$ would be uncountable. This is not true as the elements $x^ i \in N/pN$ for $i \geq 0$ span $N/pN$.


Comments (0)


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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 059Y. Beware of the difference between the letter 'O' and the digit '0'.