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Let $k$ be a field and $S=k[x,y,z]$. Let $m=(x,y,z)$ and $I\subseteq m$ a proper homogeneous ideal in $S$. Is this true that we always have:

$$[Im:(x)][Im:(y,z)]\subseteq Im \ ?$$

In a paper we needed this statement for monomial ideals and it is easy to prove, but I can not see either way for general ideals. Has anyone seen this kind of statements before?

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    $\begingroup$ sorry, what does a colon denote? $\endgroup$
    – Fedor Petrov
    Commented Jan 15, 2019 at 17:01
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    $\begingroup$ It is the colon ideal. $I:J = \{f, fJ\subseteq I\}$. $\endgroup$
    – Hailong Dao
    Commented Jan 15, 2019 at 17:05

1 Answer 1

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Perhaps it is not so bad once we are willing to get our hand dirty a little. Let $f\in Im:(x)$. Then $xf = xf_1+yf_2+zf_3$ with $f_i \in I$. Rewriting, we have $x(f-f_1) = yf_2+zf_3$. Since $x,y,z$ form a regular sequence we must have $f=f_1+h$, with $h\in (y,z)$.

Now let $g\in Im:(y,z)$. Since $I$ is proper, $g\in m$. Then $fg = f_1g + hg \in Im$, as desired.

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