[proofplan]
We prove that the displayed coordinate expansion defines a well-defined homomorphism from the [direct sum](/page/Direct%20Sum) to $M$. The module operations in the direct sum are coordinatewise, so additivity and compatibility with left scalar multiplication follow by restricting to a finite union of supports. The spanning property of the basis gives surjectivity, and the [linear independence](/page/Linear%20Independence) of the basis gives injectivity. Therefore $\Phi$ is a bijective $R$-[linear map](/page/Linear%20Map), hence an isomorphism of left $R$-modules.
[/proofplan]
[step:Define the finite support used in the coordinate sum]
Let $x = (r_i)_{i \in I} \in \bigoplus_{i \in I} R$. Define the support of $x$ by
\begin{align*}
\operatorname{supp}(x) := \{i \in I : r_i \neq 0_R\}.
\end{align*}
By the definition of the direct sum, $\operatorname{supp}(x)$ is finite. Hence
\begin{align*}
\sum_{i \in I} r_i e_i
\end{align*}
means the finite sum
\begin{align*}
\sum_{i \in \operatorname{supp}(x)} r_i e_i.
\end{align*}
If $T \subset I$ is any finite set containing $\operatorname{supp}(x)$, then every term with $i \in T \setminus \operatorname{supp}(x)$ equals $0_R e_i = 0_M$, so the same element of $M$ is obtained by summing over $T$. Thus $\Phi$ is well-defined.
[/step]
[step:Verify that $\Phi$ preserves addition and left scalar multiplication]
Let $x = (r_i)_{i \in I}$ and $y = (s_i)_{i \in I}$ be elements of $\bigoplus_{i \in I} R$. Define the finite set
\begin{align*}
T := \operatorname{supp}(x) \cup \operatorname{supp}(y).
\end{align*}
Since $T$ contains the supports of $x$, $y$, and $x+y$, coordinatewise addition in the direct sum gives
\begin{align*}
\Phi(x+y) = \sum_{i \in T} (r_i+s_i)e_i.
\end{align*}
Using distributivity in the left $R$-module $M$,
\begin{align*}
\sum_{i \in T} (r_i+s_i)e_i = \sum_{i \in T} r_i e_i + \sum_{i \in T} s_i e_i.
\end{align*}
Therefore
\begin{align*}
\Phi(x+y) = \Phi(x)+\Phi(y).
\end{align*}
Now let $a \in R$. Coordinatewise scalar multiplication in $\bigoplus_{i \in I} R$ gives
\begin{align*}
ax = (a r_i)_{i \in I}.
\end{align*}
Since $\operatorname{supp}(ax) \subseteq \operatorname{supp}(x)$, we may sum over $\operatorname{supp}(x)$ and obtain
\begin{align*}
\Phi(ax) = \sum_{i \in \operatorname{supp}(x)} (a r_i)e_i.
\end{align*}
By associativity of scalar multiplication in the left $R$-module $M$,
\begin{align*}
(a r_i)e_i = a(r_i e_i)
\end{align*}
for every $i \in \operatorname{supp}(x)$. Hence
\begin{align*}
\Phi(ax) = a \sum_{i \in \operatorname{supp}(x)} r_i e_i = a\Phi(x).
\end{align*}
Thus $\Phi$ is an $R$-[module homomorphism](/page/Module%20Homomorphism).
[/step]
[step:Use spanning of the basis to prove surjectivity]
Let $m \in M$. Since $(e_i)_{i \in I}$ is a basis of $M$, it spans $M$, so there exist a finite subset $J \subset I$ and coefficients $a_j \in R$ for $j \in J$ such that
\begin{align*}
m = \sum_{j \in J} a_j e_j.
\end{align*}
Define $x = (r_i)_{i \in I} \in \bigoplus_{i \in I} R$ by setting $r_j = a_j$ for $j \in J$ and $r_i = 0_R$ for $i \in I \setminus J$. The support of $x$ is contained in the finite set $J$, so $x$ belongs to $\bigoplus_{i \in I} R$. By the definition of $\Phi$,
\begin{align*}
\Phi(x) = \sum_{j \in J} a_j e_j = m.
\end{align*}
Since $m$ was arbitrary, $\Phi$ is surjective.
[/step]
[step:Use linear independence of the basis to prove injectivity]
Let $x = (r_i)_{i \in I} \in \bigoplus_{i \in I} R$ and assume $\Phi(x)=0_M$. Set
\begin{align*}
S := \operatorname{supp}(x).
\end{align*}
Then $S$ is finite and
\begin{align*}
0_M = \Phi(x) = \sum_{i \in S} r_i e_i.
\end{align*}
Since $(e_i)_{i \in I}$ is a basis, it is $R$-linearly independent. Therefore every coefficient in this finite linear relation is zero, so $r_i = 0_R$ for every $i \in S$. By definition of $S$, also $r_i = 0_R$ for every $i \in I \setminus S$. Hence $x$ is the zero element of $\bigoplus_{i \in I} R$.
[guided]
We must show that no nonzero finitely supported coordinate family can map to $0_M$. Let
\begin{align*}
x = (r_i)_{i \in I} \in \bigoplus_{i \in I} R
\end{align*}
and suppose that $\Phi(x)=0_M$. Define
\begin{align*}
S := \operatorname{supp}(x) = \{i \in I : r_i \neq 0_R\}.
\end{align*}
The defining property of the direct sum is that $S$ is finite. Therefore the equation $\Phi(x)=0_M$ is the finite linear relation
\begin{align*}
\sum_{i \in S} r_i e_i = 0_M.
\end{align*}
Now we use the independence part of the basis hypothesis. Linear independence of $(e_i)_{i \in I}$ means that whenever a finite linear combination of the $e_i$ equals $0_M$, every coefficient in that finite combination must be $0_R$. Applying this to the finite set $S$ and the coefficients $(r_i)_{i \in S}$ gives
\begin{align*}
r_i = 0_R
\end{align*}
for every $i \in S$.
This conclusion says that every coordinate indexed by the support is zero. For indices outside the support, the definition of support already gives $r_i = 0_R$. Hence $r_i = 0_R$ for every $i \in I$, so
\begin{align*}
x = (0_R)_{i \in I}.
\end{align*}
Thus the kernel of $\Phi$ contains only the zero element, and $\Phi$ is injective.
[/guided]
Thus $\Phi$ is injective.
[/step]
[step:Conclude that the coordinate map is an isomorphism]
We have shown that $\Phi$ is an $R$-module homomorphism, that $\Phi$ is surjective, and that $\Phi$ is injective. Therefore $\Phi$ is a bijective $R$-module homomorphism. Hence $\Phi$ is an isomorphism of left $R$-modules from $\bigoplus_{i \in I} R$ to $M$.
[/step]
