Help in a question of Hartshorne's algebraic geometry book I'm trying to solve this question:

Let $\psi:\mathbb P^1\times \mathbb P^1\to \mathbb P^3$ be the Segre embedding given by $\psi([a_0:a_1],[b_0:b_1])\to [a_0b_0:a_0b_1:a_1b_0:a_1b_1]$. This corresponding to the ring homomorphism $\theta:k[z_{00},z_{01},z_{10},z_{11}]\to k[x_0,x_1,y_0,y_1]$ which maps $z_{ij}$ to $x_iy_j$ (see question 2.14 Hartshorne).
I need help to prove that $\ker\theta=\langle z_{00}z_{11}-z_{01}z_{10}\rangle$. If I prove that, I think I solve the question.
Thanks a lot.
 A: Let $I:= \langle z_{00} z_{11} - z_{01} z_{10} \rangle.$ It is clear that $I\subseteq \ker \theta.$ Note that for any element $p$ of $k[z_{00}, z_{01}, z_{10}, z_{11} ]$ we can find a coset representative of $p+I$ which has the form $$ \sum A_{i,j,k} z_{00}^i z_{01}^j z_{11}^k + \sum B_{i',j',k'} z_{00}^{i'} z_{10}^{j'} z_{11}^{k'}.$$ This is because in any term of the form $z_{00}^{a_1} z_{01}^{a_2} z_{10}^{a_3} z_{11}^{a_4}+I$ we can use $z_{00}z_{11} +I = z_{01} z_{10}+I$ to reduce the term until either $a_3=0$ or $a_4=0.$ 
Now suppose $p= \sum A_{i,j,k} z_{00}^i z_{01}^j z_{11}^k + \sum B_{i', j' , k'} z_{00}^{i'} z_{10}^{j'} z_{11}^{k'}$ is in the kernel, so 
$$0=\theta(p) = \sum A_{i,j,k} x_0^{i+j} x_1^k y_0^i y_1^{j+k} + \sum B_{i',j',k'} x_0^{i'} x_1^{j'+k'} y_0^{i'+j'} y_1^{k'}.$$
Hence the coefficients $A_{i,j,k}, B_{i',j',k'}$ vanish except possibly when we have the same quad-nomial term appearing in each sum i.e. when $i+j=i', k=j'+k', i=i'+j'$ and $j+k=k',$ in which case we have $A_{i,j,k} = -B_{i',j',k}.$ These conditions imply that $j=-j'$ which must then both be $0,$ and so $i=i', k=k'.$ This means that $p\in I$ and hence $\ker \theta \subseteq I.$
