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

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.

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 $$I$$.