One answer to this Lemma on infinitely generated projective modules shows that every finitely generated module of a non-countably generated projective module is contained in a countably generated direct summand. Now I would like to ask : When can every countably generated submodule of a a non-countably generated projective module be contained in a countably generated direct summand ?

My rings are commutative with unity. You my assume the ring to be Noetherian if the general case is not easy to handle.

  • $\begingroup$ As far as I can see, the argument in the answer of the link also works for countably generated submodules (if one assumes the axiom of countable choice that ensure that the countable union of countable sets is again countable). $\endgroup$
    – tj_
    May 21, 2018 at 18:27
  • $\begingroup$ @tj_: could you please elaborate and write it as an answer ? I would be happy to accept ... $\endgroup$
    – user111524
    May 21, 2018 at 18:35

1 Answer 1


This is an elaboration of Ralph's answer to the question linked by the OP. I use the notation from there. Moreover, I assume the axiom of countable choice, i.e. countable unions of countable sets are countable.

Let $P$ be a non-countably generated projective $R$-module with projective base $(x_i, f_i)_{i\in I}$. A projective base has the two properties:

For all $x \in P$: $$\text{supp}(x) := \{i \in I \mid f_i(x) \neq 0\} \text{ is finite}\tag{1}$$ $$x = \sum_{i \in\text{supp}(x)}f_i(x)x_i \tag{2}$$

Let $M$ be a submodule of $P$ with countably (finite or infinite) many generators $y_k$. Define inductively $$I_0 := \bigcup_k \text{supp}(y_k)$$ $$I_{n+1} := \bigcup_{i \in I_n} \text{supp}(x_i)$$ and let $J = \bigcup_{n \ge 0}I_n$. By the axiom of countable choice, the sets $I_n, J$ are countable. Let $Q$ be the submodule of $P$ generated by (the countable many) $x_i,\,i \in J$.

By $(2)$, $y_k$ is an $R$-linear combination of $x_i\,(i \in \text{supp}(y_k)\subseteq J)$ and hence $M \subseteq Q$.

The essential step is to show that $Q$ is a direct summand of $P$. This is done by showing that the following $R$-linear map is a projection (i.e. fixes $Q$ pointwise):
$$\kappa: P \to Q,\,\,\,x \mapsto \sum_{i \in \text{supp}(x)\cap J}f_i(x)x_i$$

Let $x_j$ be a generator of $Q$. That means $j \in J$, i.e. there is $n \ge 0$ such that $j \in I_n$. Hence $\text{supp}(x_j)\subseteq I_{n+1}\subseteq J$ and
$$\kappa(x_j) \overset{\text{def}}{=}\sum_{i \in \text{supp}(x_j)\cap J}f_i(x_j)x_i = \sum_{i \in \text{supp}(x_j)}f_i(x_j)x_i\overset{(2)}{=}x_j.$$ So $\kappa$ is the identity on $Q$ and the inclusion $Q \hookrightarrow P$ is a splitting of $\kappa$.

Finally, $Q$ is a proper submodule of $P$, since $Q$ is countably generated and $P$ is not countably generated.$\blacksquare$

  • $\begingroup$ where exactly is the uncountability of the projective base is needed ? $\endgroup$
    – user111524
    May 21, 2018 at 23:37
  • $\begingroup$ It's needed to ensure that the direct summand is proper (last sentence). Without that assumption $Q$ could be the whole $P$. $\endgroup$
    – tj_
    May 21, 2018 at 23:44
  • $\begingroup$ oh so in any case, countably generated submodule of projective is always in countably generated summand , right ? $\endgroup$
    – user111524
    May 21, 2018 at 23:48
  • $\begingroup$ Yes, but if the projective module itself is countably generated, the statement is trivial. $\endgroup$
    – tj_
    May 21, 2018 at 23:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy