Here I asked a question about the tensor products- Construction of tensor product over module and apparently I do not understand what a free module is.
More specifically, in page 24 of Atiyah's Commutative Algebra there is considered the free $A$-module $A^{(M\times N) }$ where $M,N$ are $A$-modules. Then they say that the elements of this thing are formal linear combinations of elements from $M\times N$ with coefficients in $A$, i.e. expressions of the form $\displaystyle\sum_{i=1}^{n} a_i (m_i,n_i)$.
I cannot wrap my head around the difference between this (so called free) $A$-module and the one given by the additive group $M\times N$ where addition is defined by $(x,y)+(z,t)=(x+z,y+t)$ and which is made into an $A$-module by $a(m,n)=(am,an)$. What exactly is this free $A$-module $A^{(M\times N) }$?
I would appreciate a dumbed-down, step by step explanation because apparently I have a hard time understanding this concept.
Formally, it is the set of maps from $M\times N$ with finite support, i.e. the set of $f:M\times N\to A$ such that $f(m,n)=0$, but for a finite number of $(m,n)\in M\times N$.
If $I$ is the finite set of $(m,n)\in M\times N$ with a non-zero image,we can denote $a_i=f(i)$ and identifying a map with the set of its values, write the map as $$\sum_{i=(m,n)\in I} a_{i} \mathbf 1_{(m,n)},$$ where $\mathbf 1_{(m,n)}$ is the map which takes the value $1$ at $i=(m,n)$, $0$ elsewhere.
One conventionally denotes this sum in a shorter way $$\sum_{i\in I}a_i(m_i,n_i),$$ but one must not forget that in this context of formal linear combinations, the module structure of $M\times N$ is forgotten, so that,say, $$(m,n)+(m',n')\ne (m+m',n+n'),$$ since the former is the map which takes the value $1$ both at $(m,n)$ and $(m',n')$, $0$ elsewhere, and the latter takes the value $1$ at the sole point $(m+m',n+n')$.