I have n Vector spaces $V_1,...,V_n$ and would like to show that $\dim(V_1\times...\times V_n)=dimV_1+...+dimV_n$
Is it possible to show this relation using somehow that the dimension of the tensor products $\dim(V_1\bigotimes...\bigotimes V_n)=dim V_1...dimV_n$
If M is set of all function with finite support on $V_1\times...\times V_n$ then $V_1\bigotimes...\bigotimes V_n=M/M_0$ i.e a quotient space so I thought there might be a way to show the dimension formula above.
There's no need to use the tensor product. The proof that the dimension of a (finite) product of vector spaces is equal to the sum of their dimensions can be done elementarily.
Suppose $V_i$ has basis $\mathcal{B}_i$, and let $$\mathcal{B} = \{ (0, \cdots, 0, v_i, 0, \cdots, 0)\, :\, v_i \in \mathcal{B}_i ,\ 1 \le i \le n\}$$ That is, $\mathcal{B}$ is the set of vectors in $V_1 \times \cdots \times V_n$ whose $i$th components are the basis vectors of $V_i$.
Prove that $\mathcal{B}$ is a basis for $V_1 \times \cdots \times V_n$, and that $\mathcal{B}$ has $|\mathcal{B}_1| + \cdots + |\mathcal{B}_n|$ elements, i.e. there is a bijection between $\mathcal{B}$ and the disjoint union of the $\mathcal{B}_i$s.
This works even when the vector spaces are not finite-dimensional.