image of Segre embedding is cut out (scheme-theoretically) by vanishing minors

Question

I am doing exercise 10.6.B of Vakil's FOAG (July 31, 2023 version).
Basically, I need to show that the image of the Segre embedding is cut out by equations so that the matrix \begin{bmatrix} a_{00}&\cdots&a_{0n}\\ \cdots&\cdots&\cdots\\ a_{m0}&\cdots&a_{mn} \end{bmatrix} has rank 1, i.e. all $2\times 2$ minors vanish. So we need the ideal generated by $(a_{ij}a_{pq}-a_{pj}a_{iq}).$
Note that the Segre embedding is defined in the following way: $$\phi:\mathbb{P}_A^m\times_A\mathbb{P}_A^n\to \mathbb{P}_A^{mn+m+n}$$ sending $([x_0:\cdots:x_m],[y_0,\cdots,y_n])$ to $[x_0y_0:\cdots:x_my_n]$
Here is what I tried:
Consider the ring homomorphism $\sigma:A[a_{00},\cdots,a_{mn}]\to A[x_0y_0,\cdots,x_my_n]$ mapping $a_{ij}$ to $x_iy_j$. This corresponds to the substitution. Then I wish to establish the isomorphism $$im(\phi)\cong \operatorname{Proj}A[a_{00},\cdots,a_{mn}/\alpha$$ where $\alpha:=ker(\sigma)$. Here is where I find it confusing: I do not really know how to describe $[\mathfrak{p}]\in im(\phi)$. The image of this morphism is confusing for me. I tried to solve the affine case $$\phi|_{D_+(x_i)\times D_+(y_j)}:D_+(x_i)\times D_+(y_j) \to D_+(z_{ij})$$ but also felt lost when dealing with the image. Just in case, the variety case is clear to me.
Any help is sincerely appreciated! Thank you.