Let's reformulate this problem in terms of commutative algebra (its second tag): for an arbitrary field $K$ the ring $K[z_{11},\dots,z_{mn}]/\ker\phi$ is isomorphic to $\text{Im}\ \phi$ which is obviously the subring of $K[x_1,\dots,x_m,y_1,\dots,y_n]$ generated by all the monomials $x_iy_j$.
Now think in terms of affine semigroup rings:
$K[x_1,\dots,x_m,y_1,\dots,y_n]=K[\mathbb{N}^{m+n}]$ and $K[x_iy_j: 1\le i\le m, 1\le j\le n]=K[S],$ where $S\subset\mathbb{N}^{m+n}$ is the subsemigroup generated by the elements $(e_i,f_j)$. (Here we consider $e_i=(0,\dots,1,\dots,0)\in\mathbb{N}^m$ with $1$ on the place $i$, and $f_j=(0,\dots,1,\dots,0)\in\mathbb{N}^n$ with $1$ on the place $j$.) At this moment I leave you the pleasure to prove that $S=\{(a_1,\dots,a_m,b_1,\dots,b_n)\in\mathbb{N}^{m+n}:a_1+\cdots+a_m=b_1+\cdots+b_n\}.$
Theorem 6.1.4 from Bruns and Herzog, Cohen-Macaulay Rings, provides a criterion for the normality of affine semigroup rings. It says that $K[S]$ is normal if and only if $S$ is a normal semigroup, that is, if $n\in\mathbb{N}$, $n>0$, and $x\in\mathbb{Z}S$ (the subgroup of $\mathbb{Z}^{m+n}$ generated by $S$), then $nx\in S$ implies $x\in S$. In our case it is pretty clear that $S$ is normal.
Remark. Even simpler, we can think of $K[x_iy_j: 1\le i\le m, 1\le j\le n]$ as being the Segre product of two polynomial rings and use the fact that the Segre product of two normal rings is normal (why?).