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I have found two different definitions and I wonder if they are equivalent:

Choose $(M_i)_{i\in I}$ a family of submodules of an $R$-Module $M$.

  1. The sum of the $M_i$ over all $i\in I$ is derect.
  2. For every finite subset $F\subset I$ the sum of all $M_i$ with $i\in F$ is direct.

So, is this 'the same'? What is the idea of the proof and/or any references?

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    I don't understand your question. It seems that your two definitions are incomplete. Is what you wrote the only difference? And what does "What is the idea of the proof" mean? One doesn't prove definitions. – 2017-01-06
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    The question is if 1 is equivalent to 2 and if so, why? :) – 2017-01-06
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    What is your definition of "The sum of the $M_i$ is direct" though? – 2017-01-06
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    What you're claiming is true btw – 2017-01-06
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    I am confused by the word direct sum. We had the direct sum of modules defined as being the subspace of the direct product with only finitely many entrys not equal to zero. Besides a sum of vectorspaces is called direct if the intersection is trivial. – 2017-01-06
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    Well there are two notions of direct sum. An (external) direct sum of modules $M_i$ is what you mentioned, the subspace of the direct product with finitely many nonzero entries. If the $M_i$ are submodules of a larger module $M$, you say that the submodule $\sum\limits_i M_i$ is the direct sum of the $M_i$ if the natural map $$\bigoplus\limits_i M_i \rightarrow \sum\limits_i M_i$$ $$(m_i) \mapsto \sum\limits_i m_i$$ is an isomorphism. – 2017-01-06
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    Doesn't 1. mean that $M_j\cap \sum_{i\in I\setminus\{j\}} M_i=\{0\}$? – 2017-01-06
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    Yes, that is equivalent to the definition I gave. This is why I wanted to know what definition you were using – 2017-01-06
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    Maybe we can work with this one? :) This is the inner direct sum, right? I guess this makes sense since in my question we have a 'larger' module $M$ including all $M_i$. – 2017-01-06

2 Answers 2

1

The sum of the submodules is direct if the obvious homomorphism $$ \sigma_I\colon\bigoplus_{i\in I}M_i\to \sum_{i\in I}M_i $$ is an isomorphism or, in other words, an element $x=\sum_{i\in I}M_i$ can be written in a unique way as $$ x=\sum_{i\in I}x_i \quad\text{with}\quad x_i\in M_i,\text{ for } i\in I $$ Note that $\sigma_I$ is surjective by construction, so directness is the same as saying that $\sigma_I$ is injective.

Suppose the sum is direct and consider the diagram $$\require{AMScd} \begin{CD} \bigoplus_{i\in F}M_i @>\sigma_F>> \sum_{i\in F}M_i \\ @VVV @VVV \\ \bigoplus_{i\in I}M_i @>\sigma_I>> \sum_{i\in I}M_i \end{CD} $$ Since the vertical arrows are injective and $\sigma_I$ is an isomorphism, also $\sigma_F$ is injective.

Conversely, if the sum is not direct, take a nonzero element $z$ in the kernel of $\sigma_I$; then this element belongs to the image of some injection $$ \bigoplus_{i\in F}M_i\to \bigoplus_{i\in I}M_i $$ so also the sum restricted to $F$ is not direct. Say that $$ z=(x_i)_{i\in I} $$ and consider $F=\{i\in I:x_i\ne0\}$, which is a finite set. Then $z'=(x_i)_{i\in F}$ is sent to $z$ by the canonical embedding. Since $\sigma_I(z)=0$, also $\sigma_F(z')=0$.

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    Thank you a lot! The vertical arrows are not arbitrary, are they? You mean the obvious embeddings? Besindes I don't get you point in the other direction, more detailed you last statement. – 2017-01-06
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    @mathmarseille Yes, of course, they're the canonical embeddings. – 2017-01-06
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For $J \subseteq I$, let $M_J$ be defined as the direct sum

$$\bigoplus\limits_{i \in J} M_i$$

By definition, this is the $R$-submodule of the cartesian product $\prod\limits_{i \in J}M_i$ consisting of all elements $(m_i)$ such that $m_i = 0$ for all but finitely many $i \in J$.

Since I'm not sure of your definition, let me use the use the following definition: The sum of the $M_i : i \in J$ is direct if and only if the $R$-module homomorphism $\phi_J: M_J \rightarrow M$ given by

$$(m_i) \mapsto \sum\limits_i m_i$$

is injective (equivalently, is an isomorphism onto its image $\sum\limits_i M_i$, the submodule generated by the $M_i$).

Assume that for each finite subset $F \subseteq I$, the sum of the $M_i : i \in F$ is direct, i.e. $\phi_F$ is injective. Let $m = (m_i) \in M_I$, and suppose that $\phi_I(m) = 0$. We want to show that $m$ is zero, i.e. all the $m_i$ are zero. Already, we know that all the $m_i$ are zero except for those $i$ in a finite set $F = \{i_1, ... , i_r\}$.

Let $$n = (m_{i_1}, ... , m_{i_r})$$ which is an element of $M_F$. Our hypothesis is that

$$0 = \phi_I(m) = \sum\limits_i m_i = \sum\limits_{j=1}^r m_{i_j} = \phi_J(n)$$

and since $\phi_J$ is injective, this means that all the $m_{i_j}$ are zero, as required.