I am reading something on the web talking about free modules over a ring $R$. At the very beginning there is this identity:
$$\displaystyle\oplus_{\alpha\in A}R=\{s:A\to R: s\text{ is finitely supported}\}$$
Here $A$ is any set. The text says that this is an example of free $R$-module. What I can't understand is:
1) What is the left hand side of the identity? Maybe the direct sum of finitely many copies of $R$?
2) Why left hand side is the same as right hand side? I mean, I have checked that RHS is an $R$-module, with obvious pointwise operations, but why is the same as LHS, whatever LHS is? I mean, that "=" is a strict equal or maybe an isomorphism?
3) Assuming that this is free, what is a basis for it?
1) the left hand side is the direct sum of $|A|$ many copies of $R$. And it is not assumed that $A$ is finite.
One definition of the direct sum which you may be familiar with is that $A_1\oplus A_2 \oplus A_3 \oplus \dotsb$ is the set of all finitary sums of the form $a_1+a_2+a_3+\dotsb + a_{i_n}$ with each $a_i\in A_i$. Another definition is as tuples of elements (so it's a subset of $A_1\times A_2\times A_3\times\dotsb$) which are only finitely often nonzero.
The direct sum $A_1\oplus A_2 \oplus A_3 \oplus \dotsb$ comprises tuples, so it is a subset of the direct product $A_1\times A_2 \times A_3 \times \dotsb$ of all tuples. The direct sum should be thought of as actual formal sums, and summation is built inductively out of addition. Addition is a binary operation, so you can only inductively build up finite sums, so in an algebraic setting, only finite summations are considered. And taking infinite sums is not usually considered an algebraic operation; for example that operation often requires topological data (take a limit of partial sums, etc). This is why the direct sum only includes tuples which are finitely often nonzero, and the tuples in the direct product should not be thought of as sums of its components, unless you want to consider formal infinite sums.
2) The left-hand side is the same as the right-hand side because tuples and functions (or sequences) are the same thing. Why is this?
Consider a sequence $\{a_n\}_{n\in\mathbb{N}}$ such as the Fibonacci sequence with $a_1=1, a_2=1, a_3=2, a_4=3, a_5=5, a_6=8,\dotsc$. Just like a tuple it has a first, second, third, etc element, and this property characterizes tuples. A tuple is an ordered list like $(1,1,2,3,5,8,\dotsc)\in R\times R\times R\times\dotsb$. It can also be thought of as a function $a\colon \mathbb{N}\to R$ with $a(1)=1, a(2)=1, a(3)=2, a(4)=3, a(5)=5, a(6)=8,\dotsc$.
So we may think of our direct sum as a finite length sequence, or as a tuple, or as functions from the indexing set. They are the same.
I had written "a tuple is the same as a sequence is the same as a function", but that's not entirely true. A tuple like $(4,7)$ may live in a direct product like $2\mathbb{Z}\times(2\mathbb{Z}+1)$, where the two factors are not the same. Elements of such a product are not functions, because functions always take values in a fixed codomain. We cannot think of this tuple as a function with $f(1)=4, f(2)=7$, because what would be the codomain of this function?
However when all the factors (or summands, in case of direct sum) are the same (as in your example), then it is valid to state that tuples, sequences, and functions are the same thing.
3) yes it's free. A natural basis is the set of functions $\{e_a\}_{a\in A}$ where each $e_a$ is defined as the function which sends $a$ to $1$ and all other elements to $0$.
$$e_a(b)= \left\{ \begin{array}{cc} 1 & a=b \\ \\ 0 & a\neq b \\ \end{array} \right.$$
Viewing elements as tuples, this basis is written as $(1,0,0,0,\dotsc), (0,1,0,0,\dotsc), (0,0,1,0,\dotsc),(0,0,0,1,\dotsc),\dotsc$. It is sometimes called the standard basis.
Note that the standard basis does not span the direct product. You need a strictly larger cardinality basis and some nonconstructive mathematics to get a basis for the direct product.