Allow $k$ be a field. Define $\phi: k[x,y,z,w] \to k[s_0,s_1,t_0,t_1]$ by \begin{align*} &x\mapsto s_0t_0, &y\mapsto s_1t_1,\\ &z\mapsto s_1t_0, &w\mapsto s_0t_1 \end{align*} and for any $\alpha \in k$, $\alpha \mapsto \alpha$. Show that $\ker \phi = (xy-zw)$.
Showing $(xy-zw) \subset \ker\phi$ is easy. But my problem comes from the reverse direction. How do we start to show that any arbitrary polynomial in $\ker\phi$ factors into $g(x,y,z,w)(xy-zw)$. The factorization doesn't seem trivial. Is there some other fact that helps this?
Again, this question is supposed to only be done in the classic algebraic sense with no machinery from algebraic geometry.
Hints, but not complete answers, are encouraged.
Hint: if you perform a division algorithm type operation on elements of $f \in k[x,y,z,w]$, you can eventually get $g \in k[x,y,z,w]$ which has eliminated any terms with positive powers of both $x$ and $y$, yet still $f \equiv g \pmod{xy-zw}$. Thus, $g$ will only involve monomials of the form $x^a z^b w^c$ and $y^a z^b w^c$. Now, if you can show that $\phi(g) = 0$ implies $g = 0$, you will be done.