What is the zero element in the direct sum of modules?

Question

Given an index set $I$ and $R$-modules $\{M_i\}_{i\in I}$, with $R$ being a commutative ring with identity, the direct product of $M_i$ is defined as the set $$\oplus_{i\in I}M_i =\{(m_i)_{i\in I}: m_i \in M_i \text{ such that all except finitely many $m_i = 0$}\}$$

Since for a tuple $(m_i)_{i\in I} \in \oplus_{i\in I}M_i$, not all elements are zero, what will be the zero elements in that group? since the direct sum of modules is a module.

What confuses me also is the fact that if $I$ is a finite set then $\oplus_{i\in I}M_i = \prod_{i\in I} M_i$, where $\prod_{i\in I} M_i$ is the direct product. For instance, Taking $\mathbb{Z}$-module $2\mathbb{Z}$ and $3\mathbb{Z}$, we have $2\mathbb{Z} \oplus 3\mathbb{Z} = \{(0,0), (0,1), (0,2), (1,0), (1,1), (1,2)\}$ and in that case, we have an element, $(0,0)$, where all components are zero.

Answer 1

I think you are misunderstanding the phrase "all except finitely many are zero". It does not mean "not all of them are zero"; rather, it means "the nonzero elements are finite in number". That is to say, there may be 5 nonzero elements, or 100 of them, or 10 billion of them, but there can't be infinitely many of them.

Notice that this definition allows for the possibility that there may be no nonzero elements at all, because $0$ is a finite number.

Answer 2

An element $(m_i)\in\bigoplus_{i\in I} M_i$ must have all but finitely many $m_i=0$. "Finitely many" includes "zero", so you could have all but zero of the $m_i$ equal to $0$, i.e. $m_i=0$ for all $i$. This will be the zero element of the direct sum.

Answer 3

To add to the answers, you mention that the direct sum of modules equals the direct product if the index set $I$ is finite.

The direct product has the same definition as the direct sum except you remove the requirement that all but finitely many $m_i = 0$. This seems to be the source of confusion, so maybe if we focus on the direct product first, things will be clearer.

First, notice the element where all the $m_i = 0$ is in the direct product (since we can choose the $m_i$ as we wish). Does adding the condition that there can only be finitely many non-zero $m_i$ affect this element?

If not, you've just convinced yourself that this element is in the direct sum and consequently found your zero element.