# Cartesian Product of Vector Spaces as Basis for another space

This doubt arose because of another question of mine: Kostrikin's Definition of Tensor Product. The point is that in the definition posted on this link, we build a set $M$ with functions with finite support on another set, and then we consider some special $\delta$ functions (I'll not post all of the definition again to avoid duplicating content).

Well, then I've got a very good answer, that said something that now is getting me curious: "I start with the $\mathbb{K}$-space $M$ with basis $L_1×L_2$". The author of the answer explained then that this is the construction of the set $M$ with the functions and with those $\delta$'s.

My point is: I have already seem this statement on some books: "some space $V$ with basis $V_1 \times V_2$". What does that really mean ? I think it's more general than just for the definition of tensor product. Is this always the construction of functions with finite support written as linear combinations of those $\delta$ functions ?

If $B$ is an arbitrary set, and $R$ is a ring (in your case a field, but this construction doesn't have anything to do with fields; if you are not familiar with modules just replace them with vector spaces), then there is a canonical free $R$-module $F(B)$ with a basis which can be identified with $B$. The idea is that elements should look like finite linear combinations of elements in $B$ with coefficients in $R$, and that the coefficients can be seen as a function which assigns to every element of $B$ its coefficient in $R$. More formally, the underlying set of $F(B)$ is defined to be the set of functions $B \to R$ with finite support (remember that our linear combinations should be finite). This set becomes a submodule of the module of all functions $B \to R$, equipped with pointwise module structure. There is an injection $i : B \to F(B)$ sending $b \in B$ to the function which is supported at $b$ and has value $1$ there. That is, $i(b)(b')=\delta_{b,b'}$ (Kronecker-delta). For $\lambda \in F(B)$ we have $\lambda = \sum_{b \in B} \lambda(b) \cdot i(b)$. From this it follows that $i(B)$ is a basis of $F(B)$, as desired. Usually one writes $b$ instead of $i(b)$.
This module enjoys a universal property: If $M$ is an arbitrary $R$-module and $j : B \to M$ is a function to the underlying set, then there is a unique homomorphism $\alpha : F(B) \to M$ with $\alpha \circ i = j$. This can be also expressed as the statement that $F(-) : \mathsf{Set} \to \mathsf{Mod}(R)$ is left adjoint to the forgetful functor. This just means that $F(B)$ is the "best" module "approximating" the set $B$.
Now for the tensor product: If $R$ is a commutative ring and $M,N$ are $R$-modules, the idea is that $M \otimes_R N$ also should consist of formal sums of elements of the form $m \otimes n$, but we want these tensors to behave bilinear in each variable. And in the universal property above, we only want to be able to extend bilinear maps. In order to achive this, consider $F(M \times N)$ and consider the submodule generated by all bilinear relations (for example $(m+m',n)-(m,n)-(m',n)$ etc.), and let $M \otimes_R N$ be the quotient. Then, by construction, there is a universal bilinear map $M \times N \to M \otimes_R N$. That is, whenever $T$ is an $R$-module and $M \times N \to T$ is a bilinear map, it extends uniquely to a linear map $M \otimes_R N \to T$. When you really want to understand the tensor product, you should remember this universal property. In fact, this serves as a definition of the tensor product. The explicit structure of the elements is quite irrelevant for most of the problems.