Let $A = k[x_1 , \dots , x_n] / I$ be a commutative Koszul algebra; that is, the ideal $(x_1 , \dots , x_n)$ has linear minimal free resolution. Does it follow that the ideal generated by any subset of variables $(x_{i_1} , \dots , x_{i_\ell})$ also has a linear minimal free resolution?

The answer seems to be yes. Indeed, it seems like the resolution of the subset of variables is obtained as a direct summand of the Priddy (generalized Koszul) complex, which is acyclic by the Koszul assumption on $A$. Probably this subcomplex is realized as a Tate construction, and I was looking for a reference (or a quick proof/counterexample) of the question in the title.

Edit: As provided by Hailong, there is indeed a counterexample to the above claim. I would like to mention that there is a notion of "strongly Koszul" introduced by Herzog, Hibi, and Restuccia. This imposes the additional assumption that all colon ideals of the form $(x_{i_1} , \dots , x_{i_k} ) : x_{i_{k+1}}$ (for $i_1 < \cdots < i_{k+1}$) are generated by a subset of the variables. It turns out that being strongly Koszul does imply that the ideal generated by any subset of the variables has linear resolution.


1 Answer 1


The ring $R= K[a,b,c,d]/(ac,ad,ab-bd,a^2+bc,b^2)$ is Koszul but the ideal $I=(b)$ is not Koszul as $bc^2=0$, and indeed $c^2$ appears in the presentation matrix of $I$.

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    $\begingroup$ Thank you, I have been trying to prove this for like 2 days now and it is actually false. No wonder I had a hard time. $\endgroup$
    – Rellek
    Feb 17, 2021 at 19:00
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    $\begingroup$ You are welcome, this is precisely what MO is for. $\endgroup$ Feb 17, 2021 at 19:01

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