On a finitely generated submodule of a direct sum of modules....

[steenis]

I am new on this forum, this is my gift for you.

Suppose $(M_i)_{i \in I}$ is a family of left $R$-modules and $M = \bigoplus_{i \in I} M_i$ (external direct sum).

Suppose $N = \langle x_1, \cdots ,x_m \rangle$ is a finitely generated submodule of $M$.

Then for each $j = 1, \cdots ,m$, there is a finite $I_j \subset I$ such that $x_j \in \bigoplus_{i \in I_j} M_i$.

Can anyone help me with the proof of this?
This is from a book of "Ribenboim – Rings and modules (1969)". This is part of the proof of (d) on p.21 (chapter I, section 6).

 

[fresh_42]

I am new on this forum, this is my gift for you.

Suppose $(M_i)_{i \in I}$ is a family of left $R$-modules and $M = \bigoplus_{i \in I} M_i$ (external direct sum).

Suppose $N = \langle x_1, \cdots ,x_m \rangle$ is a finitely generated submodule of $M$.

Then for each $j = 1, \cdots ,m$, there is a finite $I_j \subset I$ such that $x_j \in \bigoplus_{i \in I_j} M_i$.

Can anyone help me with the proof of this?
This is from a book of "Ribenboim – Rings and modules (1969)". This is part of the proof of (d) on p.21 (chapter I, section 6).

How do you write the elements $m \in M$, i.e. what do we know about the sums $m=\sum_{\iota \in I}m_\iota$ or simply: how is the direct sum defined?

 

[steenis]

$M$ is defined as the external direct sum of the modules $(M_i)_{i \in I}$, so "write" the elements $m \in M$ with "coordinates", $m = (m_i)_{i \in I}$.

In case $M$ is an internal direct sum, the answer is easy, can you see that?

 

[fresh_42]

As far as I have learned it, a specific $m$ must always be written as a family $m=(m_\iota)_{\iota \in I}= \sum_{\iota \in I} m_\iota$ with almost all $m_\iota =0$ and then the statement follows automatically.

 

[steenis]

$(m_\iota)_{\iota \in I}$ is never equal to $\sum_{\iota \in I} m_\iota$, because these are two different "things": for instance, the point $(a,b)$ in the real plane is not equal to the sum of its coordinates $a+b$.
There is, however, an isomorphism between the external direct sum $\bigoplus M_i$ and the internal direct sum $\Sigma M_i$ that maps $(m_\iota)_{\iota \in I}$ onto $\sum_{\iota \in I} m_\iota$

 

[fresh_42]

$(m_\iota)_{\iota \in I}$ is never equal to $\sum_{\iota \in I} m_\iota$, because these are two different "things": for instance, the point $(a,b)$ in the real plane is not equal to the sum of its coordinates $a+b$.
There is, however, an isomorphism between the external direct sum $\bigoplus M_i$ and the internal direct sum $\Sigma M_i$ that maps $(m_\iota)_{\iota \in I}$ onto $\sum_{\iota \in I} m_\iota$

That's formally correct, but doesn't mean anything. Whether you write $(a,b)$ or $a+b$ where $a$ and $b$ are from different sets, namely $M_1=\mathbb{R}$ and $M_2=\mathbb{R}$ makes no difference. It only matters if you identify $M_1=M_2$ which I nowhere had. Sorry, I didn't know you simply wanted to argue about notation.

I fold.

 

[steenis]

It is not an argument about notations. The external direct sum $\bigoplus M_i$ and the internal direct sum $\Sigma M_i$ are different concepts. They are only isomorphic if the modules $M_i$ are submodules of one module: $M_i \leq X$ for all $i$, otherwise the internal direct product is not defined. And that is the case in my question, there is no $X$ given such that $M_i \leq X$.
There is a way to make the two direct sums isomorphic, and, I think, the way to do that is the key to the solution of my problem.

 

[steenis]

This was my question:

Suppose $(M_i)_{i \in I}$ is a family of left $R$-modules and $M = \bigoplus_{i \in I} M_i$ (external direct sum).

Suppose $N = \langle x_1, \cdots ,x_m \rangle$ is a finitely generated submodule of $M$.

Then for each $j = 1, \cdots ,m$, there is a finite $I_j \subset I$ such that $x_j \in \bigoplus_{i \in I_j} M_i$.

It took me a while, but the solution is:

Each $x_j$ is an element of $M = \bigoplus_{ i \in I} M_i$.

Therefore $x_j = (m_{ij})_{I \in I}$ where $m_{ij}$ is the i-th component of $x_j$ in $\bigoplus_{ i \in I} M_i$; $m_{ij} \in M_i$

M is an external direct sum, so only finitely many $m_{ij}$ are nonzero.

Let $I_j = \{i \in I | m_{ij} \neq 0 \}$, $I_j$ is finite.

Then $x_j \in \bigoplus_{ i \in I_j} M_i$.