Finite homological dimension of the finite graded Khovanov-Seidel module category

The positive-length ideal $J$ and its powers. Let $J\le A_m$ be the $\mathbb{Z}$-span of the classes of the paths of positive length. By [L2] every positive-length basis element is one of the arrows $u_i$, $d_i$ for $0\le i<m$ or the returns $(i\mathrm{\mid }i-1\mathrm{\mid }i)$ for $1\le i\le m$, while the monotone and reverse monotone length-two paths have class $0$ in $I_m$ by [F1]; hence $J={\sum }_{i=0}^{m-1}(\mathbb{Z}{u}_i+\mathbb{Z}{d}_i)+{\sum }_{i=1}^m\mathbb{Z}(i\mathrm{\mid }i-1\mathrm{\mid }i)$ and $A_m=\left({⨁}_{i=0}^m\mathbb{Z}{e}_i\right)\oplus J$ as graded abelian groups, and $J$ is a two-sided ideal because multiplying a positive-length class on either side either concatenates to a longer path or gives $0$ by [F3]. Multiplying two positive-length basis classes gives either a noncomposable product, hence $0$ by [F3], a path of length at least three, hence $0$ by [L2], or one of the products $u_i{d}_i=(i\mathrm{\mid }i+1\mathrm{\mid }i)$ and $d_i{u}_i=(i+1\mathrm{\mid }i\mathrm{\mid }i+1)$ for $0\le i<m$, the product $u_0{d}_0=(0\mathrm{\mid }1\mathrm{\mid }0)$ having class $0$ by [L2]; at each interior vertex the two returns agree by [F1], so $J^2$ is the $\mathbb{Z}$-span of the returns $(1\mathrm{\mid }0\mathrm{\mid }1),\dots ,(m\mathrm{\mid }m-1\mathrm{\mid }m)$, and every product of three positive-length classes is a path of length at least three or $0$ and hence vanishes by [L2], so $J^3=0$.

The graded quotient ring $E=A_m\mathrm{/}J$. By step 1.1 the composite ${⨁}_{i=0}^m\mathbb{Z}{e}_i\to A_m\to A_m\mathrm{/}J$ is a $\mathbb{Z}$-module isomorphism, and by [F3] the vertex classes satisfy $e_i{e}_j={\delta }_{ij}{e}_i$ and $1={\sum }_i{e}_i$, so $E=A_m\mathrm{/}J$ is isomorphic as a unital ring to the product ${\prod }_{i=0}^m\mathbb{Z}$ with coordinate idempotents the images of the $e_i$; these images have degree $0$ by [F1], so $E$ is concentrated in internal degree $0$. In particular $E$ is a free abelian group of rank $m+1$, and each component $e_{i}E$ is $\mathbb{Z}{e}_i$.

The three-step filtration of $M$ and its finitely generated quotients. Let $M$ be a finitely generated graded left $A_m$-module with homogeneous generators $x_1,\dots ,x_N$. The sets $JM\supseteq J^{2}M\supseteq J^{3}M=0$ are graded $A_m$-submodules, and the quotients $Q_1:=M\mathrm{/}JM$, $Q_2:=JM\mathrm{/}{J}^{2}M$, $Q_3:=J^{2}M$ are annihilated by $J$ since $J{Q}_1=0$, $J{Q}_2=J^{2}M\mathrm{/}{J}^{2}M=0$ and $J{Q}_3=J^{3}M=0$ by step 1.1, so each $Q_k$ is a graded $E$-module by step 2.1. Each $Q_k$ is finitely generated as an $E$-module: $JM$ is the $A_m$-submodule generated by the finitely many products $g{x}_j$ with $g$ in the finite basis set $\{u_i,d_i:0\le i<m\}\cup \{(i\mathrm{\mid }i-1\mathrm{\mid }i):1\le i\le m\}$, and $J^{2}M$ by the finitely many $gh{x}_j$ with $g,h$ in that set, so both are finitely generated, and $Q_1$ is a quotient of $M$.

Finitely generated graded $E$-modules decompose into abelian groups. Let $N$ be a finitely generated graded left $E$-module with homogeneous generators $y_1,\dots ,y_s$ of degrees $d_1,\dots ,d_s$. Since $E$ is concentrated in degree $0$ by step 2.1, $N_d={\sum }_{j:d_j=d}\mathbb{E}{y}_j$, so $N_d=0$ for $d\mathrm{\notin }\{d_1,\dots ,d_s\}$; writing $N_{d,i}:=e_i{N}_d$ gives $N_d={⨁}_{i=0}^m{N}_{d,i}$ by the idempotent decomposition of step 2.1, and each $N_{d,i}$ is the image of ${\mathbb{Z}}^{s_d}$ under the map that reads the $i$-th coordinate, where $s_d$ is the number of generators of degree $d$, so $N_{d,i}$ is a finitely generated abelian group. Hence $N$ is the finite direct sum of the graded $E$-modules obtained by placing $N_{d,i}$ in degree $d$, on which $e_i$ acts as the identity of $N_{d,i}$ and $e_j$ with $j\mathrm{\ne }i$ acts as $0$.

Filtrations of the pieces with vertex-module quotients. By [F9] the finitely generated abelian group $N_{d,i}$ of step 3.2 is isomorphic to ${\mathbb{Z}}^r\oplus {⨁}_j\mathbb{Z}\mathrm{/}(p_j^{e_j})$, so it carries a finite $\mathbb{Z}$-submodule filtration with successive quotients isomorphic to $\mathbb{Z}$ or to $\mathbb{Z}\mathrm{/}p$: add the $r$ free cyclic summands one at a time, and for each cyclic summand $C=\mathbb{Z}\mathrm{/}{p}^e$ use the chain $0\subseteq p^{e-1}C\subseteq \cdots \subseteq pC\subseteq C$, whose successive quotients $p^{j}C\mathrm{/}{p}^{j+1}C$ are cyclic of order $p$. Each of these submodules is stable under the $E$-action of step 3.2, because $e_i$ acts as the identity and every $e_j$ with $j\mathrm{\ne }i$ as $0$ on the piece placed in degree $d$, so the filtration is a filtration by $E$-submodules and, by [L5], its successive quotients are the internal shifts $S_i\{d\}$ in the free case and $S_i\mathrm{/}p{S}_i\{d\}$ in the cyclic torsion case. Splicing the finitely many such filtrations over the finite index set of step 3.2 into the three-step filtration $0\subseteq J^{2}M\subseteq JM\subseteq M$ of step 3.1 by taking preimages of filter pieces under the displayed quotient maps yields a finite filtration $0=F_0\subseteq F_1\subseteq \cdots \subseteq F_t=M$ of $M$ by graded $A_m$-submodules whose successive quotients $F_k\mathrm{/}{F}_{k-1}$ are internal shifts of $S_i$ or of $S_i\mathrm{/}p{S}_i$ for some $0\le i\le m$ and some prime $p$.

The uniform bound on the pieces. Let $0\le i\le m$. By [L6] the resolution of $S_i$ is the total complex of the staircase double complex with nonzero entries $C_{p,q}=P_{i+p-q}\{p\}$ on $0\le p\le m-i$, $0\le q\le i+p$, so its nonzero terms lie in total degrees $n=p+q$ with $0\le n\le (m-i)+m=2m-i$; hence $S_i$ has a finite graded projective resolution of length at most $2m-i\le 2m$, and the mapping cone of [L6] adds exactly one homological degree, so $S_i\mathrm{/}p{S}_i$ has a finite graded projective resolution of length at most $2m-i+1\le 2m+1$. Since the internal shift is an automorphism of $A_m\text{-mod}$ by [L4] that carries finite graded projectives to finite graded projectives — a finite graded projective is a degree-zero direct summand of a finite direct sum of shifts $A_m\{r_j\}$ by [L10], and shifting that decomposition by $d$ presents $P\{d\}$ as a degree-zero direct summand of the finite direct sum of the shifts $A_m\{r_j+d\}$ — every internal shift of $S_i$ or of $S_i\mathrm{/}p{S}_i$ admits a finite graded projective resolution of length at most $2m+1$.

Extension-stability over the finite filtration. Let $L:=2m+1$. We show by induction on $k$ that every member $F_k$ of the filtration of step 4.1 has a finite graded projective resolution of length at most $L$. For $k=0$ the zero module has the zero resolution $0\to F_0\to 0$ of length $0\le L$. For $k\ge 1$ the sequence $0\to F_{k-1}\to F_k\to F_k\mathrm{/}{F}_{k-1}\to 0$ is degreewise exact in $A_m\text{-mod}$; step 5.1 gives the quotient $F_k\mathrm{/}{F}_{k-1}$ a finite graded projective resolution of length at most $L$, the inductive hypothesis gives one for $F_{k-1}$, and [L8] then gives one for $F_k$ of length at most $\max (L,L)=L$, using only the finitely many lifts of that construction.

Conclusion. Every finitely generated graded left $A_m$-module $M$ has a finite graded projective resolution of length at most $L=2m+1$, namely the one built in step 6.1 for $k=t$; by [L7] this is a projective resolution in the sense of the statement, its length bound depends only on $m$, and the successive quotients of the filtration producing it are internal shifts of the vertex modules $S_i$ and their prime quotients, so $\mathrm{pd}M\le 2m+1$ uniformly on $A_m\text{-mod}$ and the finite graded module category has finite homological dimension. The proof used only the finite path basis, the structure theory of a finitely generated abelian group, the finitely many lifts of [L8] and the explicit resolutions of [L6], and it selected no family indexed by an infinite set, so no choice principle is assumed.