I am studying Tensor product from the book "Commutative Algebra" written by N.S. Gopalakrishnan. Here I am struggling in understanding the existence of tensor product properly. The author has defined tensor product between two $R$-modules in the following way $:$
Let $M$ and $N$ be two $R$-modules. Then the tensor product of $M$ and $N$ is a pair $(T,\theta)$ where $T$ is a $R$-module and $\theta : M \times N \longrightarrow T$ is a $R$-bilinear map such that given any $R$-module $K$ and given any $R$-bilinear map $f: M \times N \longrightarrow K$ there exists a unique $R$-linear map $\bar f : T \longrightarrow K$ such that $\bar f \theta = f$.
Now the author proved the uniqueness of the tensor product as follows $:$
Let $F$ be a free $R$-module on the set $M \times N$, i.e. all formal finite $R$-linear combination of the type $\sum_i a_i (x_ i,y_i)$, $a_i \in R$, $x_i \in M$ and $y_i \in N$, with formal addition and natural scalar multiplication.Let $G$ be the submodule of $F$ genetated by all the elements of the type
$$(x+x',y)-(x,y)-(x',y),\ x,x' \in M,\ y \in N,$$
$$(x,y+y')-(x,y)-(x,y'),\ x \in M,\ y,y' \in N,$$
$$(ax,y)-a(x,y),\ x \in M,\ y \in N,\ a \in R,$$
$$(x,ay)-a(x,y),\ x \in M,\ y \in N,\ a \in R.$$
Here the author considered $T:=F/G$ and showed that it is the tensor product of $M$ and $N$. All the things done looks ok to me except the construction of $G$. From the construction of $G$ it is clear that all the elements of the form $(x+x',y)-(x,y)-(x',y),\ x,x' \in M,\ y \in N$ are inside $G$. If we take $x=x'=0$ then we will find that $(0,y) \in G$ for all $y \in M$. Similarly all the generators of the form $(x,y+y')-(x,y)-(x,y')$,\ $x \in M,\ y,y' \in N$ are inside $G$. So if we take $y=y'=0$ here we find that $(x,0) \in G$, for all $x \in M$ SInce $G$ is a module so we have $(x,y) \in G$ for all $(x,y) \in M \times N$. But this implies $F=G$. Then what is $T$? It contains only zero element! What a mysterious thing it is!
I am in a fix. I don't know whether I am correct or not. Please help me in this regard.
Thank you in advance.
It is not true that $(x,y)=(x,0)+(0,y)$ in $F$. Indeed, $F$ is the free module on the set $M\times N$ (which is very different from the module $M\times N$), so the elements of the form $(a,b)$ are all linearly independent of each other.
So, even though $G$ contains $(x,0)$ and $(0,y)$ for any $x$ and $y$, this does not mean it always contains $(x,y)$ and therefore is all of $F$.