A tensor $t \in V \otimes W$ is called decomposable if $t = v \otimes w$ for some $v \in V$ and $w \in W$. The set of decomposable tensors is the image of the map \begin{equation} f : \mathbb{R}^2 \oplus \mathbb{R}^2 \to \mathbb{R}^2 \otimes \mathbb{R}^2, \hspace{2mm} f(v,w) = v \otimes w. \end{equation} First I showed that this map is not injective in general.
Now I want to show the following non-sense: The image of this map $f$ (decomposable tensors) is the zero set of a certain quadratic polynomial.
I tried to approach this problem the following way:
First I proved that $\mathbb{R}^2 \otimes \mathbb{R}^2 \cong \mathbb{R}^{2 \times 2}$ and with $\lbrace e_1,e_2 \rbrace$ denoting the standard basis of $\mathbb{R}^2$ the basis of $\mathbb{R}^2 \otimes \mathbb{R}^2$ is given by $\lbrace e_1 \otimes e_1, e_1 \otimes e_2, e_2 \otimes e_1, e_2 \otimes e_2 \rbrace$. From the association with matrices in $\mathbb{R}^{2 \times 2}$ the tensor products $e_i \otimes e_j$ for $i,j = 1,2$ can be seen as matrices.
Now I want to find a polynomial $\chi \in \mathbb{R}_{}[x]$ given by
$\chi(x) = a x^2 + bx + c$ where $a \neq 0$ such that
\begin{equation}
\chi(f(v,w)) = \chi(v \otimes w) = a(v \otimes w)^2 + b(v \otimes w) + cI = 0
\end{equation}
for all $v \in \mathbb{R}^2$ and $w \in \mathbb{R}^2$ where the above polynomial is a matrix polynom and $I$ is the $2 \times 2$ identity matrix.
Since $v = v_1 e_1 + v_2 e_2$ and $w = w_1 e_1 + w_2 e_2$ for some real coefficients $v_1,v_2,w_1,w_2 \in \mathbb{R}$ then
\begin{equation}
v \otimes w = v_1 w_1 e_1 \otimes e_2 + v_1 w_2 e_1 \otimes e_2 + v_2 w_1 e_2 \otimes e_1 + v_2 w_2 e_2 \otimes e_2.
\end{equation}
If I now try to find the coefficients such that the zero set condition of the polynomial $\chi$ holds I get
\begin{equation}
\chi(v \otimes w) = \begin{pmatrix}
c + v_1 w_1 (b + a v_1 w_1 + a v_2 w_2) & v_1 w_2 (b + a v_1 w_1 + a v_2 w_2) \\
v_2 w_1 (b + a v_1 w_1 + a v_2 w_2) & c + v_2 w_2 (b + a v_1 w_1 + a v_2 w_2)
\end{pmatrix} \stackrel{!}{=} 0
\end{equation}
therefore $b = -a(v_1 w_1 + v_2 w_2)$ and $c = 0$ for all $v,w$ means $a = 0$ and $b = 0$ so the polynom vanishes which is not a quadratic polynomial.