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.



In a commutative Koszul algebra, does every ideal generated by a subset of variables have linear resolution?
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 $I$.


