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Let $P_1,\ldots, P_d, Q_1, \ldots, Q_k \in \mathbb{C}[x_0,\ldots, x_n]$ be homogenous polynomials of degree at most $r$.

Assume that $P_1 \cdot P_2 \cdots P_{d-1} \cdot P_d \in \langle Q_1, \ldots, Q_k \rangle$. Here $\langle h_1, \ldots, h_s \rangle$ is the ideal with the generators $h_1, \ldots, h_s$.

Is it true that for some $\{ i_1, \ldots, i_f \} \subseteq \{1, \ldots, d\}$ the polynomial $P_{i_1} \cdots P_{i_f} \in \langle Q_1, \ldots, Q_k \rangle$, where $f= f(k,r)$?

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  • $\begingroup$ It is trivially true if you allow $\{ i_1, \ldots, i_f \} = \{1, \ldots, d\}$. Isn't it? $\endgroup$
    – Luc Guyot
    Dec 16, 2017 at 21:06
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    $\begingroup$ @LucGuyot I think the OP wants $f$ to only depend on $k,r$. $\endgroup$
    – Gjergji Zaimi
    Dec 16, 2017 at 21:08

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Yes. The point is that many invariants of the ideal $I=(Q_1,\dots,Q_k)$ can be bounded depending only on $k$ and $r$. It is pure luck that the following paper has collected many of them in a very convenient Proposition 4.6. http://www-personal.umich.edu/~asnowden/papers/genstillman-071517.pdf

In particular, you can find a primary decomposition of $I = I_1\cap\dots \cap I_l$ such that each of the $I_i$ is $\mathfrak p_i$-primary and the number of generators of $I_i$ as well as degrees and $l$ itself are bounded by some function of $k$ and $r$ (using part 6 of the Prop. cited above).

So the problem reduces to the case when $I$ is $\mathfrak p$-primary. But there is a $B$ such that $\mathfrak p^B\subseteq I$, and this number can also be bounded on the degrees and number of generators of $I$ (part 10 of loc. cit.). Such $B$ works.

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    $\begingroup$ Very nice answer! In addition $\mathbb{C}$ can be replaced by any algebraically closed field, according to your source. $\endgroup$
    – Luc Guyot
    Dec 16, 2017 at 22:33
  • $\begingroup$ Thank you! I have a question. Do you mean in your solution that some of the indexes $i_1, \ldots i_f$ can be the same? I understand your solution in this case: if $P_1 \cdots P_k$ belongs to $\mathfrak p$-primary ideal $I$ then some $P_i$ belongs to $\mathfrak p$ and hence $P_i^B \in I$. Did you mean something else? $\endgroup$
    – Alexey Milovanov
    Jan 11, 2018 at 0:48
  • $\begingroup$ @AlexeyMilovanov: no, I don't assume the indexes can be the same, otherwise it is too easy. Once you reduced to the case of one prime ideal $P$, just remove all the $f$ that is not in $P$. The product of the rest is still in $I$. If there are at most $B$ elements remain, we are done. If not, then choose $B$ of them, the product is in $P^B\subseteq I$. $\endgroup$
    – Hailong Dao
    Jan 11, 2018 at 5:13
  • $\begingroup$ > The product of the rest is still in $I$. Could you explain it? I understand only that the rest in $P$. $\endgroup$
    – Alexey Milovanov
    Jan 11, 2018 at 9:28
  • $\begingroup$ @AlexeyMilovanov: if $f \notin P$ then it is a nonzero divisor mod $I$, as $P$ is the only associated prime. So if $fx \in I$ then $x\in I$. $\endgroup$
    – Hailong Dao
    Jan 11, 2018 at 14:55

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