Let us start with $V,W$ to $R$-modules. In order to define their tensor product we first introduce something called the free R-module $Free(V \times W)$ whose basis is given by the set of all ordered pairs $(v,w) \in V \times W$. This means that an element of $Free(V\times W)$ is a unique finite linear finite combination of elements of $V \times W$. Now we consider the $R$-submodule $S$ spanned by all elements that have a kind of linearity.. that is $S = \{(v1 + v2, w) - (v1,w) - (v2,w), (rv,w) - r(v,w), \text{and the same for the second entry}\}$. Then we define the tensor product $V \otimes_R W$ as the quotient $Free(V \times W)/S$. An element (hence a class) of this space is written as $v \otimes w$ and is called the tensor product.
Questions:
- What exactly is the relation for say $(a,b)$ be an element of $[(v,w)]$ ?
- In the proof of $(v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w$ we want to show that one class is the same as the other one. That is $[(v_1+v_2,w)] = [(v_1,w) + (v_2,w)]$.
Now, I believe this "follows" by the quotient that we make since each equivalence class is supposed to satisfy the relations of $S$. But I am not being able to properly express this.
I noticed another post where there's a comment saying that by doing this quotient, we are essentially forcing that the defining relations on $S$ are all equal to zero. For instance, we would have $(v_1+v_2,w) - (v_1,w) - (v_2,2) = 0$ in $Free(V \times W)/S$. Is this the case? I can see that this solves que issue since then automatically we would have $$[(v_1+v_2,w)] = [(v_1,w)+(v_2,w)].$$But this is also not clear to me.
Thank you.
Thank you for the answer @Mark.
We say that $a \sim b$ in $Free(V \times W)/S$ iff $b-a \in S$. In this case, one of the elements that generate $S$ is $(v_1+v_2,w) - (v_1,w)-(v_2,w)$. So yes, $$(v_1,w) + (v_2,w) \sim (v_1+v_2,w),$$showing that both classes are equivalent.