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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.

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  • $\begingroup$ There is a huge difference. You should think of the free $A$-module on a set $X$ as a module with basis $X$ (just like for vector spaces). That means that this free module forgets all structure on $X$ itself. So when you consider the free module on $M\times N$, it's simply the $A$-module whose basis is the set $M\times N$. $\endgroup$ – Mathematician 42 Jul 31 '18 at 9:21
  • $\begingroup$ I think I might have an idea, but I am not very sure. Suppose that $M\times N$ is countable and index its elements $(m_1,n_1),(m_2,n_2),...$. The elements of the free $A$-module mentioned in the question are then infinite arrays $(a_1,a_2,a_3,....)$ with $a_i\in A$ in which all but finitely many entries are zero. In particular $\displaystyle\sum_{i=1}^{n} a_i (m_i,n_i)$ is the array which has $a_i$ in position $i$ and $0$ everywhere else. Is this right? $\endgroup$ – Earl Jones Jul 31 '18 at 9:22
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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 $(m+m',n+n')$.

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The module you suggest is called the direct product of $M$ and $N$, and has elements $M\times N$. In contrast, the free module $A^{(M\times N)}$ is a module with basis $M\times N$, and whose elements are formal linear combinations. This is going to be a much bigger module.

In general, given any set $X$ we can form the free module $A^X$ in the way above. This is in some sense the most general module that contains $X$.

The way I think of the free module construction is that even though the set $X$ may have some structure, (for example, if $X=M\times N$ then we may consider it as the product $A$-module), we ignore that structure.

So, even though there is a way to interpret sums $\Sigma_{i=1}^na_i(m_i,n_i)$ as elements of $M \times N$, when we form the free module we forget about this, and just leave the sums unevaluated. The reason this is useful is that we can choose to interpret the sum in a different way later.

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