I'm reading 'Arithmetically Cohen-Macaulay Sets of Points in $\mathbb P^1\times\mathbb P^1$' by Elena Guardo and Adam Van Tuyl. (One can read it partially on Google books.)
I doubt whether the 'Bigraded Nullstellensatz' is true or not; they say:
Theorem 2.11 (Bigraded Nullstellensatz)
If $I\subset R=k[x_1,x_2,y_1,y_2]$ is a bihomogeneous ideal and if $F\in R$ is a bihomogeneous polynomial with $\deg F\neq (0,0)$ such that $F(P)=0$ for all $P\in V(I)\subset \mathbb P^1\times \mathbb P^1$, then $F^t\in I$ for some $t>0$.
($k$ is an algebraically closed field.)
Here by a bihomogeneous polynomial they mean a polynomial F which is a homogeneous polynomial in $x_1,x_2$ (resp. $y_1,y_2$) with coefficients in $k[y_1,y_2]$ (resp. $k[x_1,x_2]$).
A bihomogeneous ideal is an ideal in $k[x_1,x_2,y_1,y_2]$ generated by bihomogeneous polynomials. (Equivalently, one can define it to be an ideal I with the following condition:
If $F=\sum F_{p,q}\in I$, where $F_{p,q}$ is a bihomogeneous polynomial of degree $(p,q)$, then each $F_{p,q}$ belongs to $I$.)
They don't give a proof in the textbook; they say the proof is the same as in the homogeneous case. (In the homogeneous case one must assume that $V(I)\neq \emptyset$, but this is a trivial matter.)
I've tried to prove it, but I can't. After the long grappling,
I perhaps found a counterexample.
Let $I=(x_1 y_1 ,x_1 y_2)$. Then $I$ is bihomogeneous and $$V(I) = \{[0:1]\}\times \mathbb P^1.$$ Thus $x_1 \in I(V(I))$, i.e., $x_1$ vanishes on $V(I)$.
However, any powers of $x_1$ are not in I.
Is there something wrong? Thank you.