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.

  • $\begingroup$ 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 $F$? $\endgroup$ Feb 16 '18 at 4:31
  • $\begingroup$ 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. $\endgroup$
    – Hiro Wat
    Feb 16 '18 at 13:31
  • $\begingroup$ Thus we may conclude that $$I(V(I))=(\sqrt{I}:(x_1,x_2)) + (\sqrt{I}:(y_1,y_2)) !!!!!!$$ $\endgroup$
    – Hiro Wat
    Feb 16 '18 at 14:44

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