Theorem 10.83.4. Suppose $M$ is a direct sum of countably generated $R$-modules. If $P$ is a direct summand of $M$, then $P$ is also a direct sum of countably generated $R$-modules.

**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:

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

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

(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 induction 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:

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

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 $M_{\alpha } \subset M_{\alpha + 1}$. $\square$

## 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)

There are also: