Doubt in proof of existence of tensor product I have a question regarding the proof of the proposition 2.12 in Atiyah and Macdonald's book.
They say:
" Let $C$ denote the free $A$-module $A^{M \times N}$. The elements of $C$ are formal linear combinations of elements of $M \times N$ with coefficients in $A$, i.e they are expressions of the form $ \sum_{i=1}^n a_i \cdot (x_i,y_i) (a_i \in A, x_i \in M, y_i \in N).$
Let $D$ be the submodule of $C$ genearated by all elements of $C$ of the following types:
$(x+x',y)-(x,y) -(x',y) \\
(x,y+y')-(x,y) -(x,y') \\
(ax,y)- a \cdot(x,y) \\
(x,ay)- a \cdot(x,y) \\$
Let $T= C/D.$ For each basis element $(x,y)$ of $C$, let $ x \otimes y$ denote its image in $T$. Then $T$ is generated by the elements of the form $x \otimes y$ "
My doubt is why $C$ is a free  module. C can be think of sequence of coordinates of each $(x,y)$ but in such case $A(x,y)$ need not be isomorphic to $A$
Definition: a free $A$ module is of form $\oplus_{i\in I} M_i$ such that each $M_i $ is isomorphic to $A$
Please help me with this
 A: Let me try to provide a kind of answer here by talking about the free abelian group on a set $X = \{A,B,C\}$ of three generators.
You'll see things like 'this is a formal sum of terms like $7A -2B + 6C$, with coefficients in the integers. $0A + 0B + 0C$ serves as an identity element, and the inverse of $aA + bB + cC$ is just $-aA -bB -cC$.'
That's all very well, but where is this "addition" and scalar multiplication taking place? What the hell does "formal sum" mean?
The answer is to define it in terms of things you already know. The "formal sum" $7A -2B + 6C$ is actually a function $f$ on the set $X$, one for which $f(A) = 7, f(B) = -2, f(C) = 6.$ Addition of formal sums is defined as ... addition of integer-valued functions, which you already know about. Inverses? That's just negation of integer-valued functions. Once you see this, it's easy to see that all the axioms for a group are satisfied, and it's pretty easy to do the conversion from "formal sum of elements of $X$" to "integer-valued function on $X$" and back again.
Now let's look at the quotation above:

Let D be the submodule of C generated by all elements of C of the following types:
$(x+x',y)-(x,y) -(x',y) \\
(x,y+y')-(x,y) -(x,y')\\
(ax, y) - a \cdot (x, y) \ldots$

and think about that third one in the case where $a$ is the element $2 \in A$. And let's suppose that $x_0$ and $y_0$ are particular elements of $X$ and $Y$. Then we can define a function $h$ on $X \times Y$ by saying that  $h(x, y) = 0$ for almost every pair $(x, y)$ except that
$$
h(2x_0, y_0) = 1\\
h(x_0, y_0) = -2.
$$
Written out in formal-sum terms, this is just
$$
(2x_0, y_0) - 2(x_0, y_0).
$$
Now $D$ contains all such formal sums, for every possible $x_0 \in X$ and $y_0 \in Y$, and lots of others (corresponding to the other rows in the description) as well.
In $C/D$, the equivalence class of our function $h$ will be the same as the equivalence class of the everywhere-zero function, so we can say that
$$
[(2x_0, y_0) - 2(x_0, y_0)] = [0]
$$
But just as when we talk of the integers mod 3, we often use the symbols $0, 1, 2$ rather than the more proper $[0], [1], [2]$, to denote the three residue classes, folks sometimes leave off the brackets and say that in $C/D$, we have
$$
(2x_0, y_0) - 2(x_0, y_0) = 0 \tag{1}
$$
When we write it like this, it's hard to tell whether this is supposed to be an equality in $C$ or in $C/D$. (It's the latter, but you can't tell that by looking!)
Now $V$ is "generated by" all single-term formal sums, i.e., functions that are $1$ on some particular element of $X \times Y$ and $0$ everywhere else. (You should think through why this is true.)
That means that $V/D$ is also generated by the equivalence classes of these "basic" formal sums; as it happens, they "overgenerate" -- there are far too many of them to be a generating set "without relations". But to help distinguish between $C$ and $C/D$, we make up a new notation:
$$
x_0 \otimes y_0
$$
is used to denote the equivalence class of $(x_0, y_0)$ in $C/D$. And now formula 1 above becomes
$$
(2x_0) \otimes y_0 - 2 (x_0 \otimes y_0) = 0
$$
or more often, that's written
$$
(2x_0) \otimes y_0 = 2 (x_0 \otimes y_0). 
$$
I hope this helps a bit.
