The Stacks project

Proof. By property (3) of a direct sum dévissage, there is an inclusion $M_{\alpha + 1}/M_{\alpha } \to M$ for each $\alpha \in S$. Consider the map

given by the sum of these inclusions. Further consider the restrictions

for $\beta \in S$. Transfinite induction on $S$ shows that the image of $f_{\beta }$ is $M_{\beta }$. For $\beta =0$ this is true by $(0)$. If $\beta +1$ is a successor ordinal and it is true for $\beta $, then it is true for $\beta + 1$ by (3). And if $\beta $ is a limit ordinal and it is true for $\alpha < \beta $, then it is true for $\beta $ by (2). Hence $f$ is surjective by (1).

Transfinite induction on $S$ also shows that the restrictions $f_{\beta }$ are injective. For $\beta = 0$ it is true. If $\beta +1$ is a successor ordinal and $f_{\beta }$ is injective, then let $x$ be in the kernel and write $x = (x_{\alpha + 1})_{\alpha + 1 \leq \beta + 1}$ in terms of its components $x_{\alpha + 1} \in M_{\alpha + 1}/M_{\alpha }$. By property (3) and the fact that the image of $f_{\beta }$ is $M_{\beta }$ both $(x_{\alpha + 1})_{\alpha + 1 \leq \beta }$ and $x_{\beta + 1}$ map to $0$. Hence $x_{\beta +1} = 0$ and, by the assumption that the restriction $f_{\beta }$ is injective also $x_{\alpha + 1} = 0$ for every $\alpha + 1 \leq \beta $. So $x = 0$ and $f_{\beta +1}$ is injective. If $\beta $ is a limit ordinal consider an element $x$ of the kernel. Then $x$ is already contained in the domain of $f_{\alpha }$ for some $\alpha < \beta $. Thus $x = 0$ which finishes the induction. We conclude that $f$ is injective since $f_{\beta }$ is for each $\beta \in S$. $\square$

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 $M$. $\square$

Proof. Write $M = P \oplus Q$. We are going to construct a Kaplansky dévissage $(M_{\alpha })_{\alpha \in S}$ of $M$ which, in addition to the defining properties (0)-(4), satisfies:

1. Each $M_{\alpha }$ is a direct summand of $M$;

2. $M_{\alpha } = P_{\alpha } \oplus Q_{\alpha }$, where $P_{\alpha } =P \cap M_{\alpha }$ and $Q_\alpha = Q \cap M_{\alpha }$.

(Note: if properties (0)-(2) hold, then in fact property (3) is equivalent to property (5).)

To see how this implies the theorem, it is enough to show that $(P_{\alpha })_{\alpha \in S}$ forms a Kaplansky dévissage of $P$. Properties (0), (1), and (2) are clear. By (5) and (6) for $(M_{\alpha })$, each $P_{\alpha }$ is a direct summand of $M$. Since $P_{\alpha } \subset P_{\alpha + 1}$, this implies $P_{\alpha }$ is a direct summand of $P_{\alpha + 1}$; hence (3) holds for $(P_{\alpha })$. For (4), note that

so $P_{\alpha + 1}/P_{\alpha }$ is countably generated because this is true of $M_{\alpha + 1}/M_{\alpha }$.

It remains to construct the $M_{\alpha }$. Write $M = \bigoplus _{i \in I} N_ i$ where each $N_ i$ is a countably generated $R$-module. Choose a well-ordering of $I$. By transfinite recursion we are going to define an increasing family of submodules $M_{\alpha }$ of $M$, one for each ordinal $\alpha $, such that $M_{\alpha }$ is a direct sum of some subset of the $N_ i$.

For $\alpha = 0$ let $M_{0} = 0$. If $\alpha $ is a limit ordinal and $M_{\beta }$ has been defined for all $\beta < \alpha $, then define $M_{\alpha } = \bigcup _{\beta < \alpha } M_{\beta }$. Since each $M_{\beta }$ for $\beta < \alpha $ is a direct sum of a subset of the $N_ i$, the same will be true of $M_{\alpha }$. If $\alpha + 1$ is a successor ordinal and $M_{\alpha }$ has been defined, then define $M_{\alpha + 1}$ as follows. If $M_{\alpha } = M$, then let $M_{\alpha + 1} = M$. If not, choose the smallest $j \in I$ such that $N_ j$ is not contained in $M_{\alpha }$. We will construct an infinite matrix $(x_{mn}), m, n = 1, 2, 3, \ldots $ such that:

1. $N_ j$ is contained in the submodule of $M$ generated by the entries $x_{mn}$;

2. if we write any entry $x_{k\ell }$ in terms of its $P$- and $Q$-components, $x_{k\ell } = y_{k\ell } + z_{k\ell }$, then the matrix $(x_{mn})$ contains a set of generators for each $N_ i$ for which $y_{k\ell }$ or $z_{k\ell }$ has nonzero component.

Then we define $M_{\alpha + 1}$ to be the submodule of $M$ generated by $M_{\alpha }$ and all $x_{mn}$; by property (2) of the matrix $(x_{mn})$, $M_{\alpha + 1}$ will be a direct sum of some subset of the $N_ i$. To construct the matrix $(x_{mn})$, let $x_{11}, x_{12}, x_{13}, \ldots $ be a countable set of generators for $N_ j$. Then if $x_{11} = y_{11} + z_{11}$ is the decomposition into $P$- and $Q$-components, let $x_{21}, x_{22}, x_{23}, \ldots $ be a countable set of generators for the sum of the $N_ i$ for which $y_{11}$ or $z_{11}$ have nonzero component. Repeat this process on $x_{12}$ to get elements $x_{31}, x_{32}, \ldots $, the third row of our matrix. Repeat on $x_{21}$ to get the fourth row, on $x_{13}$ to get the fifth, and so on, going down along successive anti-diagonals as indicated below:

Transfinite induction on $I$ (using the fact that we constructed $M_{\alpha + 1}$ to contain $N_ j$ for the smallest $j$ such that $N_ j$ is not contained in $M_{\alpha }$) shows that for each $i \in I$, $N_ i$ is contained in some $M_{\alpha }$. Thus, there is some large enough ordinal $S$ satisfying: for each $i \in I$ there is $\alpha \in S$ such that $N_ i$ is contained in $M_{\alpha }$. This means $(M_{\alpha })_{\alpha \in S}$ satisfies property (1) of a Kaplansky dévissage of $M$. The family $(M_{\alpha })_{\alpha \in S}$ moreover satisfies the other defining properties, and also (5) and (6) above: properties (0), (2), (4), and (6) are clear by construction; property (5) is true because each $M_{\alpha }$ is by construction a direct sum of some $N_ i$; and (3) is implied by (5) and the fact that $M_{\alpha } \subset M_{\alpha + 1}$. $\square$

Proof. A module is projective if and only if it is a direct summand of a free module, so this follows from Theorem 10.84.4. $\square$