# Kelly's theorem for quadratic polynomials

Let $f_1, \ldots, f_m$ be homogeneous irreducible quadratic polynomials in $\mathbb{C}[x_1, \ldots, x_n]$.

Assume that these polynomials are pairwise coprime.

Denote $P:= f_1 \cdot f_2 \ldots \cdot f_m$.

Assume that these polynomials have the following property: for every $i$ and for every $j \not= i$ the ideal $\langle f_i, f_j \rangle$ contatns $\frac{P}{f_i \cdot f_j}$.

Question. Is it true that $\text{trdeg}_{\mathbb{C}}(f_1, \ldots, f_m ) = O(1)$?

Background for the question.

If the word "quadratic" changes to the word "linear" then my question has an affirmative answer by Kelly's theorem. Indeed, points in $\mathbb{CP}^n$ corresponds to linear forms in $\mathbb{C}[x_1, \ldots, x_n]$. Statement "point $A$ belongs to a line BC" can be rewritten as "linear form $l_A$ belongs to ideal $\langle l_B, l_C \rangle$".

Finally, the statement "the ideal $\langle f_i, f_j \rangle$ contatns $\frac{P}{f_i \cdot f_j}$" is equivalent to "there exists $k \not= i, j$ such that $\langle f_i, f_j \rangle$ contains $f_k$" if all $f_i$ are linear. This is true since ideals $\langle f_i, f_j \rangle$ are prime in this case.

So, my question would has an affirmative answer if all ideal $\langle f_i, f_j \rangle$ were prime for quadratic $f_i$. However, this is not true. Nevertheless the statement "the ideal $\langle f_i, f_j \rangle$ contatns $\frac{P}{f_i \cdot f_j}$" can be changed to $\langle f_i, f_j \rangle$ contatns $f_{k_1} \cdot f_{k_2} ... \cdot f_{k_s}$", where $s$ is some constant. This is notion of Hailong Dao.

• What does $O(1)$ mean in the question? – Mohan Jun 21 '18 at 17:49
• @Mohan trdeg is bounded by some constant – Alexey Milovanov Jun 21 '18 at 18:03
• So, if I understand, you are letting both $n,m$ vary? – Mohan Jun 21 '18 at 18:55
• @Mohan exactly. – Alexey Milovanov Jun 21 '18 at 19:02