# Bihomogeneous Nullstellensatz

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:

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.

• Here’s a thought: what if, in theorem 2.11, instead of saying that $\deg F \neq (0,0)$, we further require that $$\deg F \neq (0, \ast) \text{ or } (\ast, 0)?$$ This seems to prevent counterexamples like yours. Or would that exclude some interesting polynomials $F$? Feb 16 '18 at 4:31
• I think one may show that $$I(V(I))\cap (x_1,x_2)\cap (y_1,y_2) \subset \sqrt{I},$$ reducing the problem to the affine case. Feb 16 '18 at 13:31
• Thus we may conclude that $$I(V(I))=(\sqrt{I}:(x_1,x_2)) + (\sqrt{I}:(y_1,y_2)) !!!!!!$$ Feb 16 '18 at 14:44