Existence of Tensor Product of Modules

Question

I am reading "Commutative Algebra", written by Michael Atiyah. I have a problem about constructing the tensor product of modules. In the book, the tensor product of elements of A-Modules M and N, denoted by $x\otimes y$ for $x \in M$ and $y \in N$, is defined as below:

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
$\sum\nolimits_{i=1}^{n} a_i.(x_i,y_i)(a_i \in A, x_i \in M, y_i \in N).$

Let $D$ be the submodule of $C$ generated by all elements of $C$ of the following types:

$(x+x',y)-(x,y)-(x',y)$

$(x,y+y')-(x,y)-(x,y')$

$(ax,y)-a.(x,y)$

$(x,ay)-a.(x,y)$.

Let $T = C/D$. For each basis element $(x,y)$ of $C$, let $x\otimes y$ denote its image in $T$. The mapping $g: M \times N \rightarrow T$, defined by $g(x,y) = x\otimes y$ is $A$-bilinear.

I proved that g is bilinear: $(x+x')\otimes y = (x+x',y) + D.$ From the definition we have $(x+x',y)-(x,y)-(x',y)$ belongs to $D$. Therefore, $(x+x',y) + D = (x,y) + (x',y) + D = (x,y)+D+(x',y)+D = x\otimes y + x'\otimes y. $ Likewise, we have $x \otimes (y+y')=x\otimes y + x\otimes y'.$

$(ax)\otimes y = (ax,y)+D = a.(x,y) + D = (x,ay) + D. \Rightarrow (ax)\otimes y = a.(x\otimes y) = x \otimes (ay).$

My question is that, what is the operation between elements of A and C? What about multiplication between elements of C? Can the above computations be altered in this way: $(x+x',y) = (x,y) + (x',0)$?

Answer 1

$C$ is a free $A$-module with basis indexed by the set $M\times N$, so the operation between elements of $A$ and $C$ is just the scalar multiplication as $A$-module.

Concretely, if $\sum_i a_i(x_i, y_i)$ is an element of $C$, and $a$ is an element of $A$, then we have $$ a \cdot \sum_i a_i(x_i, y_i) = \sum_i aa_i(x_i, y_i).$$

For your second question, there is no multiplication between elements of $C$ (unless $M$ and $N$ have some extra structure).

For your third question, the notation $(x, y)$ means the basis element index by the element $(x, y) \in M\times N$, hence all the three elements $(x + x', y), (x, y), (x', 0)$ are in general three different basis elements, hence the identity $(x + x', y) = (x, y) + (x', 0)$ is not true in $C$.