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

  • #1
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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).
 
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Answers and Replies

  • #2
14,202
11,485
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?
 
  • #3
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##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?
 
  • #4
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As far as I have learnt 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.
 
  • #5
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##(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##
 
  • #6
14,202
11,485
##(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.
 
  • #7
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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.
 
  • #8
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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 ##.
 

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