I’m having a hard time understanding why can’t we see tensors as simple products on the base field. See, if i select two subfields of a field $(\mathbb{F}; +; *)$ in such a way that they form a finite field extension $K=\mathbf{V}(\mathbb{F}^n)$, $Dim(\mathbf{V})=n$ for some natural $n$ over the original field, with $B=\{e_0;e_1;…;e_n\}$ being the base of the field extension $K$, then, since $\mathbf{V}$ is a subfield of the field itself, multiplication is automatically defined.
Let $\vec{v}$ and $\vec{w}$ be elements of $\mathbf{V}$. Then there is a trivial natural isomorphism:
$$φ:\mathbf{V}\otimes\mathbf{V}\rightarrow\mathbb{F}$$
Given by the relation
$$φ(\vec{v}\otimes\vec{w})=a_v*b_w$$
Where $a_v$ and $b_w$ are the corresponding values of each vector in the main field. So, in this sense the “tensor product” between two (or more) vectors is just a multiplication. As an example. Take a function over $\mathbf{V}(\mathbb{F}^2)$; $$f(x;y)=(xe_0+ye_1)^2$$ Evidently, if we have $f(a;b)+f(c;d)$, then $$f(a;b)+f(c;d) \\ =(ae_0)^2+2abe_0e_1+(be_1)^2+(ce_0)^2+2cde_0e_1+(de_1)^2 \\ =(a^2+c^2)e_0^2+2(ab+cd)e_0e_1+(b^2+d^2)e_1^2$$ so $f$ is (kinda) linear, hence, it acts like a map from $\mathbf{V}$ to $\mathbf{V}\otimes\mathbf{V}$… i could even define the composition of $f$ with the previous isomorphism $φ$ And claim $(f\circ φ)(a;b)$ is a bilinear map! I’m probably wrong about something, so a little clarification would be very helpful. Specifically, i would like to know if: is $(f\circ φ)$ a bilinear map? Can $f(x;y)$ be regarded as a tensor product of a vector by itself? Is there any situation in wich tensor products have exactly the same components as the expanded form of a multiplication? Thanks for the help. Maybe i’m just mixing concepts.