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 $f= f(k,r)$?

  • $\begingroup$ It is trivially true if you allow $\{ i_1, \ldots, i_f \} = \{1, \ldots, d\}$. Isn't it? $\endgroup$
    – Luc Guyot
    Commented Dec 16, 2017 at 21:06
  • 3
    $\begingroup$ @LucGuyot I think the OP wants $f$ to only depend on $k,r$. $\endgroup$ Commented Dec 16, 2017 at 21:08

1 Answer 1


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.

  • 1
    $\begingroup$ Very nice answer! In addition $\mathbb{C}$ can be replaced by any algebraically closed field, according to your source. $\endgroup$
    – Luc Guyot
    Commented 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? $\endgroup$ Commented 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 $P^B\subseteq I$. $\endgroup$ Commented 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 $P$. $\endgroup$ Commented 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 $x\in I$. $\endgroup$ Commented Jan 11, 2018 at 14:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.