Lemma 10.93.1. Let $M$ be an $R$-module. If $M$ is flat, Mittag-Leffler, and countably generated, then $M$ is projective.

## 10.93 Characterizing projective modules

The goal of this section is to prove that a module is projective if and only if it is flat, Mittag-Leffler, and a direct sum of countably generated modules (Theorem 10.93.3 below).

**Proof.**
By Lazard's theorem (Theorem 10.81.4), we can write $M = \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i$ for a directed system of finite free $R$-modules $(M_ i, f_{ij})$ indexed by a set $I$. By Lemma 10.92.1, we may assume $I$ is countable. Now let

be an exact sequence of $R$-modules. We must show that applying $\mathop{\mathrm{Hom}}\nolimits _ R(M, -)$ preserves exactness. Since $M_ i$ is finite free,

is exact for each $i$. Since $M$ is Mittag-Leffler, $(\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, N_{1}))$ is a Mittag-Leffler inverse system. So by Lemma 10.86.4,

is exact. But for any $R$-module $N$ there is a functorial isomorphism $\mathop{\mathrm{Hom}}\nolimits _ R(M, N) \cong \mathop{\mathrm{lim}}\nolimits _{i \in I} \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, N)$, so

is exact. $\square$

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 $N/pN$.

Theorem 10.93.3. Let $M$ be an $R$-module. Then $M$ is projective if and only it satisfies:

$M$ is flat,

$M$ is Mittag-Leffler,

$M$ is a direct sum of countably generated $R$-modules.

**Proof.**
First suppose $M$ is projective. Then $M$ is a direct summand of a free module, so $M$ is flat and Mittag-Leffler since these properties pass to direct summands. By Kaplansky's theorem (Theorem 10.84.5), $M$ satisfies (3).

Conversely, suppose $M$ satisfies (1)-(3). Since being flat and Mittag-Leffler passes to direct summands, $M$ is a direct sum of flat, Mittag-Leffler, countably generated $R$-modules. Lemma 10.93.1 implies $M$ is a direct sum of projective modules. Hence $M$ is projective. $\square$

Lemma 10.93.4. Let $f: M \to N$ be universally injective map of $R$-modules. Suppose $M$ is a direct sum of countably generated $R$-modules, and suppose $N$ is flat and Mittag-Leffler. Then $M$ is projective.

**Proof.**
By Lemmas 10.82.7 and 10.89.7, $M$ is flat and Mittag-Leffler, so the conclusion follows from Theorem 10.93.3.
$\square$

Lemma 10.93.5. Let $R$ be a Noetherian ring and let $M$ be a $R$-module. Suppose $M$ is a direct sum of countably generated $R$-modules, and suppose there is a universally injective map $M \to R[[t_1, \ldots , t_ n]]$ for some $n$. Then $M$ is projective.

## 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.

## Comments (0)