On the generators of Nil $K$-groups

TL;DR

This paper investigates the structure of higher Nil K-groups by explicitly describing their generators using binary complexes, advancing the understanding of their algebraic properties.

Contribution

It provides an explicit description of generators for higher Nil K-groups through the use of binary complexes, a novel approach in the field.

Findings

Explicit generators for higher Nil K-groups are constructed.

Binary complexes serve as a tool to understand Nil K-groups.

The approach clarifies the algebraic structure of Nil K-groups.

Abstract

In this article, we study higher Nil $K$-groups via binary complexes. More particularly, we exhibit an explicit form of generators of higher Nil $K$-groups in terms of binary complexes.

Equations

$$\Delta :C\mathcal{N}\to B\mathcal{N},\mathrm{defined}\mathrm{by}\Delta ((N_{\ast },d))=(N_{\ast },d,d).$$

$$0\to K_{n-1}{C}^q\mathcal{N}\stackrel{\Delta }{\to }{K}_{n-1}{B}^q\mathcal{N}\to K_n\mathcal{N}\to 0,$$

$$0\to {\mathrm{Nil}}_n(R)\to K_0(B^q{)}^n\mathbf{Nil}(R)/T_{\mathbf{Nil}(R)}^n\to K_0(B^q{)}^n\mathbf{P}(R)/T_{\mathbf{P}(R)}^n\to 0.$$

$$0\to {\tilde{T}}_R^n\to {\widetilde{\mathrm{Nil}}}_n(R)\to {\mathrm{Nil}}_n(R)\to 0.$$

$$\begin{array}{ccccccccc}000& & & & & & & & \\ \downarrow & & \downarrow & & \downarrow & & & & \\ 0& \to & {\tilde{T}}_R^n& \to & T_{\mathbf{Nil}(R)}^n& \to & T_{\mathbf{P}(R)}^n& \to & 0\\ \downarrow & & \downarrow & & \downarrow & & & & \\ 0& \to & {\widetilde{\mathrm{Nil}}}_n(R)& \to & K_0(B^q{)}^n\mathbf{Nil}(R)& \to & K_0(B^q{)}^n\mathbf{P}(R)& \to & 0\\ \downarrow & & \downarrow & & \downarrow & & & & \\ 0& \to & {\mathrm{Nil}}_n(R)& \to & K_n\mathbf{Nil}(R)& \to & K_n(R)& \to & 0\\ \downarrow & & \downarrow & & \downarrow & & & & \\ 000,& & & & & & & & \end{array}$$

$$0\subseteq \ker (\nu )\subseteq \ker ({\nu }^2)\subseteq \cdots \subseteq \ker ({\nu }^{m-1})\subseteq \ker ({\nu }^m)=R^n.$$

$$0\to \ker ({\nu }^{m-1})\to R^n\to \frac{R^n}{\ker ({\nu }^{m-1})}\to 0.$$

$$\begin{array}{ccccccccc}0& \to & \ker ({\nu }^i)& \to & \ker ({\nu }^{i+1})& \to & \frac{\ker ({\nu }^{i+1})}{\ker ({\nu }^i)}& \to & 0\\ \nu \downarrow & & \nu \downarrow & & \bar{\nu }=0\downarrow & & & & \\ 0& \to & \ker ({\nu }^i)& \to & \ker ({\nu }^{i+1})& \to & \frac{\ker ({\nu }^{i+1})}{\ker ({\nu }^i)}& \to & 0\end{array}$$

$$0\to (\ker ({\nu }^i),\nu )\to (\ker ({\nu }^{i+1}),\nu )\to (\frac{\ker ({\nu }^{i+1})}{\ker ({\nu }^i)},0)\to 0$$

$$[(R^n,\nu )]$$

$$=[(\ker ({\nu }^{m-2}),\nu )]+[(\frac{\ker ({\nu }^{m-1})}{\ker ({\nu }^{m-2})},0)]+[(\frac{R^n}{\ker ({\nu }^{m-1})},0)]$$

$$=[(\ker ({\nu }^2),\nu )]+\cdots +[(\frac{\ker ({\nu }^{m-1})}{\ker ({\nu }^{m-2})},0)]+[(\frac{R^n}{\ker ({\nu }^{m-1})},0)]$$

$$=[(\ker (\nu )\oplus \frac{\ker ({\nu }^2)}{\ker (\nu )}\oplus \cdots \oplus \frac{R^n}{\ker ({\nu }^{m-1})},0)]$$

$$=[(R^n,0)].$$

$$0\to (P_1,{\nu }_1)\to (P,\nu )\to (P_2,{\nu }_2)\to 0$$

$$\begin{array}{ccccccccc}0& \to & P_1& \to & P& \to & P_2& \to & 0\\ {\nu }_1\downarrow & & \nu \downarrow & & {\nu }_2\downarrow & & & & \\ 0& \to & P_1& \to & P& \to & P_2& \to & 0\end{array}$$

$$0\to (C_{\ast },c,\tilde{c})\to (A_{\ast },a,\tilde{a})\to (D_{\ast },d,\tilde{d})\to 0$$

$$0\to (C_{\ast },c,\tilde{c})\to (B_{\ast },b,\tilde{b})\to (D_{\ast },d,\tilde{d})\to 0$$

$$(P_{\ast },p,p^{{}^{\prime }})\oplus (A_{\ast },a,\tilde{a})\cong (Q_{\ast },q,q^{{}^{\prime }})\oplus (B_{\ast },b,\tilde{b}).$$

$$0\to (C_{\ast },c,\tilde{c},0)\to (A_{\ast },a,\tilde{a},0)\to (D_{\ast },d,\tilde{d},0)\to 0$$

$$0\to (C_{\ast },c,\tilde{c},0)\to (B_{\ast },b,\tilde{b},0)\to (D_{\ast },d,\tilde{d},0)\to 0$$

$$K_0{B}^q\mathbf{Nil}(R)\to K_0{B}^q\mathbf{P}(R),[(P_{\ast },p,p^{{}^{\prime }},\nu )]\mapsto [(P_{\ast },p,p^{{}^{\prime }})]$$

$$[(P_{\ast },p,p^{{}^{\prime }})]=[(F_{\ast },f,f^{{}^{\prime }})].$$

$$[(P_{\ast },p,p^{{}^{\prime }},\nu )]+[(Q_{\ast },q,q,0)]=[({\tilde{F}}_{\ast },\tilde{f},{\tilde{f}}^{{}^{\prime }},{\nu }^{{}^{\prime }})].$$

$$[(P_{\ast },p,p^{{}^{\prime }},\nu )]=[({\tilde{F}}_{\ast },\tilde{f},{\tilde{f}}^{{}^{\prime }},{\nu }^{{}^{\prime }})]-[({\tilde{F}}_{\ast },\tilde{f},{\tilde{f}}^{{}^{\prime }},0)]+[(F_{\ast },f,f^{{}^{\prime }},0)].$$

$$x=([({\tilde{F}}_{\ast },\tilde{f},{\tilde{f}}^{{}^{\prime }},{\nu }^{{}^{\prime }})]-[({\tilde{F}}_{\ast },\tilde{f},{\tilde{f}}^{{}^{\prime }},0)])-([(F_{\ast },f,f^{{}^{\prime }},\tilde{\nu })]-[(F_{\ast },f,f^{{}^{\prime }},0)]).$$

$$[(F_{\ast },f,f^{{}^{\prime }},\nu )]-[(F_{\ast },f,f^{{}^{\prime }},0)],$$

$$[(P_{\ast },(p_1,p_1^{{}^{\prime }}),(p_2,p_2^{{}^{\prime }}),\dots ,(p_n,p_n^{{}^{\prime }}),0)]=[(Q_{\ast },(q_1,q_1^{{}^{\prime }}),(q_2,q_2^{{}^{\prime }}),\dots ,(q_n,q_n^{{}^{\prime }}),0)]$$

$$(P_{\ast },(p_1,p_1^{{}^{\prime }}),(p_2,p_2^{{}^{\prime }}),\dots ,(p_n,p_n^{{}^{\prime }}),\nu )\oplus (Q_{\ast },(q,q),(q_2,q_2^{{}^{\prime }}),\dots ,(q_n,q_n^{{}^{\prime }}),0)\cong (F_{\ast },(f_1,f_1^{{}^{\prime }}),(f_2,f_2^{{}^{\prime }}),\dots ,(f_n,f_n^{{}^{\prime }}),{\nu }^{{}^{\prime }}),$$

$$(P_{\ast },(p_1,p_1^{{}^{\prime }}),(p_2,p_2^{{}^{\prime }}),\dots ,(p_n,p_n^{{}^{\prime }}))\oplus (Q_{\ast },(q,q),(q_2,q_2^{{}^{\prime }}),\dots ,(q_n,q_n^{{}^{\prime }}))\stackrel{\alpha }{\cong }(F_{\ast },(f_1,f_1^{{}^{\prime }}),(f_2,f_2^{{}^{\prime }}),\dots ,(f_n,f_n^{{}^{\prime }})),$$

$$K_0(B^q{)}^n\mathbf{Nil}(R)\to K_0(B^q{)}^n\mathbf{P}(R),$$

$$[(P_{\ast },(p_1,p_1^{{}^{\prime }}),(p_2,p_2^{{}^{\prime }}),\dots ,(p_n,p_n^{{}^{\prime }}),\nu )]\mapsto [(P_{\ast },(p_1,p_1^{{}^{\prime }}),(p_2,p_2^{{}^{\prime }}),\dots ,(p_n,p_n^{{}^{\prime }})].$$

Full text

1. Introduction

Given a commutative ring $R,$ let ${\bf P}(R)$ denote the category of finitely generated projective $R$-modules. Let ${\bf Nil}(R)$ be a category consisting of all pairs $(P,\nu),$ where $P$ is a finitely generated $R$-module and $\nu$ is a nilpotent endomorphism of $P.$ Let ${\rm Nil}_{0}(R)$ denote the kernel of the forgetful map $K_{0}({\bf Nil}(R))\to K_{0}({\bf P}(R))=:K_{0}(R).$ The group ${\rm Nil}_{0}(R)$ is generated by elements of the form $[(R^{n},\nu)]-[(R^{n},0)]$ for some $n$ and some nilpotent endomorphism $\nu.$ Using these generators, we show that ${\rm Nil}_{0}(R)=0$ provided every finitely generated torsion free $R$-module is projective (see Theorem 3.3). The hypothesis on $R$ in the above-mentioned result holds for many well-known classes of rings, such as PID, valuation rings, Dedekind domains, and so on. In fact, in case of integral domains, the hypothesis on $R$ is equivalent to $R$ being a Prüfer domain (i.e., a ring which is locally a valuation ring). One of the goals of this article is to determine the generators of higher Nil $K$-groups.

Given an exact category $\mathcal{N},$ let $i\mathcal{N}$ denote the subcategory of $\mathcal{N}$ whose arrows are isomorphisms. Consider $i\mathcal{N}$ as a category of weak equivalences. Then the $K$-theory spectrum of $\mathcal{N}$ defined as $K\mathcal{N}:=Ki\mathcal{N}.$ The $n$-th $K$-group of $\mathcal{N}$ is defined as $K_{n}\mathcal{N}:=\pi_{n}(Ki\mathcal{N})=K_{n}i\mathcal{N},$ where $n\geq 0$ (see Appendix A of [2]). Let ${\rm Nil}(R)$ be the homotopy fibre of the forgetful functor $K{\bf Nil}(R)\to K{\bf P}(R)=:K(R).$ The $n$-th Nil group ${\rm Nil}_{n}(R)$ is $\pi_{n}{\rm Nil}(R).$ Since the forgetful functor splits, $K{\bf Nil}(R)\simeq{\rm Nil}(R)\times K(R).$ This implies that $K_{n}{\bf Nil}(R)\cong{\rm Nil}_{n}(R)\bigoplus K_{n}(R)$ for every ring $R.$ We also have a natural decomposition $K_{n}(R[t])\cong K_{n}(R)\oplus NK_{n}(R),$ where $NK_{n}(R)=\operatorname{ker}[K_{n}(R[t])\stackrel{{\scriptstyle t\mapsto 0}}{{\to}}K_{n}(R)].$ There is an isomorphism ${\rm Nil}_{n}(R)\cong NK_{n+1}(R)$ for all $n$ and $R$ (see Theorem V.8.1 of [9]). The Nil $K$-groups measure the failure of algebraic $K$-theory to be homotopy invariant. The group ${\rm Nil}_{n}(R)$ is not finitely generated unless it is trivial (for instance, see Proposition IV.6.7.4 of [9]). It is natural to wonder:

How do the generators of the group ${\rm Nil}_{n}(R)$ for $n>0$ look like?

In this article, we are able to figure out generators of ${\rm Nil}_{n}(R)$ for $n>0$ using Grayson’s technique (see [2]) of binary complexes. We refer to section 5 (Theorems 5.4 and 5.8) for the precise results. We hope the explicit form of generators of Nil $K$-groups obtained in this article might be useful for further research.

2. Preliminaries and Grayson’s definition

In order to define Grayson’s $K$-groups, we need an idea of binary complexes. Let us recall the notion from [2] for exact categories.

Binary chain complexes

Let $\mathcal{N}$ denote an exact category. A bounded chain complex $N$ in $\mathcal{N}$ is said to be an acylic chain complex if each differential $d_{i}:N_{i}\to N_{i-1}$ can be factored as $N_{i}\to Z_{i-1}\to N_{i-1}$ such that each $0\to Z_{i}\to N_{i}\to Z_{i-1}\to 0$ is a short exact sequence of $\mathcal{N}.$ Let $C\mathcal{N}$ denote the category of bounded chain complexes in $\mathcal{N}.$ The full subcategory of $C\mathcal{N}$ consisting of bounded acylic complexes in $C\mathcal{N}$ is denoted by $C^{q}\mathcal{N}.$ The category $C^{q}\mathcal{N}$ is exact.

A chain complex in $\mathcal{N}$ with two differentials (not necessarily distinct) is called a binary chain complex in $\mathcal{N}.$ In other words, it is a triple $(N_{*},d,d^{{}^{\prime}})$ with $(N_{*},d)$ and $(N_{*},d^{{}^{\prime}})$ are in $C\mathcal{N}.$ If $d=d^{{}^{\prime}}$ then we call a diagonal binary complex. A morphism between two binary complexes $(N_{*},d,d^{{}^{\prime}})$ and $(\tilde{N}_{*},\tilde{d},\tilde{d}^{{}^{\prime}})$ is a morphism between the underlying graded objects $N$ and $\tilde{N}$ that commutes with both differentials. The category of bounded binary complexes in $\mathcal{N}$ is denoted by $B\mathcal{N}.$ There is always a diagonal functor (see Definition 3.1 of [2])

[TABLE]

As before, let $B^{q}\mathcal{N}$ denote the full subcategory of $B\mathcal{N}$ whose objects are bounded acylic binary complexes. This is also an exact category.

By iterating, one can define exact category $(B^{q})^{n}\mathcal{N}=B^{q}B^{q}\cdots B^{q}\mathcal{N}$ for each $n\geq 0.$ An object of the exact category $(B^{q})^{n}\mathcal{N}$ of bounded acylic binary multicomplexes of dimension $n$ in $\mathcal{N}$ is a bounded $\mathbb{Z}^{n}$- graded collection of objects of $\mathcal{N},$ together with a pair of acyclic differentials $d^{i}$ and $\tilde{d^{i}}$ in each direction $1\leq i\leq n,$ where the differentials $d^{i}$ and $\tilde{d^{i}}$ commute with $d^{j}$ and $\tilde{d^{j}}$ whenever $i\neq j.$ Thus, a typical object looks like $(N_{*},(d^{1},\tilde{d^{1}}),(d^{2},\tilde{d^{2}}),\dots,(d^{n},\tilde{d^{n}})),$ where $N_{*}$ is a bounded $\mathbb{Z}^{n}$- graded collection of objects of $\mathcal{N}.$ We say that an acyclic binary multicomplex $(N_{*},(d^{1},\tilde{d^{1}}),(d^{2},\tilde{d^{2}}),\dots,(d^{n},\tilde{d^{n}}))$ is diagonal if $d^{i}=\tilde{d^{i}}$ for at least one $i.$

Grayson’s Definition

In [7], Nenashev gave a description of $K_{1}$-group in terms of generator and relations using the notion of double exact sequences. Motivated by [7], Grayson defined higher $K$-groups in terms of generators and relations using binary complexes (see [2]). However, Nenashev’s $K_{1}$-group agrees with Grayson’s $K_{1}$-group (see Corollary 4.2 of [4]). In the rest of the paper, we assume the following as the definition of higher $K$-groups.

Definition 2.1**.**

(see Corollary 7.4 of [2]) Let $\mathcal{N}$ be an exact category. For $n\geq 1,$ $K_{n}\mathcal{N}$ is the abelian group having generators $[N],$ one for each object $N$ of $(B^{q})^{n}\mathcal{N}$ and the relations are:

1. (1)

$[N^{{}^{\prime}}]+[N^{{}^{\prime\prime}}]=[N]$* for every short exact sequence $0\to N^{{}^{\prime}}\to N\to N^{{}^{\prime\prime}}\to 0$ in $(B^{q})^{n}\mathcal{N};$* 2. (2)

$[T]=0$* if $T$ is a diagonal acyclic binary multicomplex.*

Note that if we only consider the relation (1) then it is just $K_{0}(B^{q})^{n}\mathcal{N}.$

Definition 2.2**.**

We denote $T_{\mathcal{N}}^{n}$ as the subgroup of $K_{0}(B^{q})^{n}\mathcal{N}$ generated by the $K_{0}$-classes of all diagonal acyclic binary multicomplexes in $(B^{q})^{n}\mathcal{N}.$

Remark 2.3**.**

In view of Definition 2.1, we have $K_{n}\mathcal{N}\cong K_{0}(B^{q})^{n}\mathcal{N}/T_{\mathcal{N}}^{n}.$**

Lemma 2.4**.**

For each $n\geq 1$, there is a split short exact sequence

[TABLE]

which is functorial in $\mathcal{N}.$

Proof.

See Lemma 2.7 of [3]. ∎

Remark 2.5**.**

For $n=1,$ $K_{0}C^{q}\mathcal{N}\cong{\rm im}(\Delta)=T_{\mathcal{N}}^{1}.$ Thus, the above lemma implies that $K_{0}B^{q}\mathcal{N}\cong T_{\mathcal{N}}^{1}\oplus K_{1}\mathcal{N}.$ **

Higher Nil $K$-groups via binary complexes

The category ${\bf Nil}(R)$ is exact. By Remark 2.3, we can view the $n$-th Nil $K$-group as a quotient of the Grothendieck group of the category $(B^{q})^{n}{\bf Nil}(R).$ More precisely, $K_{n}{\bf Nil}(R)\cong K_{0}(B^{q})^{n}{\bf Nil}(R)/T_{{\bf Nil}(R)}^{n}$ for $n\geq 0.$ Here, $T_{{\bf Nil}(R)}^{n}$ is a subgroup of $K_{0}(B^{q})^{n}{\bf Nil}(R)$, as described in Definition 2.2. Moreover, there is a split exact sequence

[TABLE]

Let $\widetilde{{\rm Nil}}_{n}(R)$(resp. $\widetilde{T}_{R}^{n}$) denote the kernel of the map $K_{0}(B^{q})^{n}{\bf Nil}(R)\to K_{0}(B^{q})^{n}{\bf P}(R)$ (resp. $T_{{\bf Nil}(R)}^{n}\to T_{{\bf P}(R)}^{n}$). Clearly, the map $T_{{\bf Nil}(R)}^{n}\to T_{{\bf P}(R)}^{n}$ is surjective. We have the following result, which will be used in section 5.

Lemma 2.6**.**

For $n\geq 1,$ there is a canonical short exact sequence

[TABLE]

If $n=1$ then this sequence splits, i.e., $\widetilde{{\rm Nil}}_{1}(R)\cong{\rm Nil}_{1}(R)\oplus\widetilde{T}_{R}^{1}.$

Proof.

The assertion follows by chasing the following commutative diagram

[TABLE]

where rows are split exact sequences, the second and third columns are exact sequences. For $n=1,$ the second column is split exact (see Remark 2.5). This forces that the first column is split exact provided $n=1.$ ∎

3. Vanishing of zeroth Nil $K$-groups

We know ${\rm Nil}_{0}(R)\cong NK_{1}(R)=\operatorname{ker}[K_{1}(R[t])\stackrel{{\scriptstyle t\mapsto 0}}{{\to}}K_{1}(R)]$ (for instance, see Proposition III.3.5.3 of [9]). The homotopy invariance of $K$-theory is known for regular noetherian rings and valuation rings (see [5]). Thus, ${\rm Nil}_{0}(R)=0$ provided $R$ is a regular noetherian or valuation ring. In this section, we discuss a condition on $R$ under which ${\rm Nil}_{0}(R)$ is trivial.

For a fix $n,$ $(R^{n},\nu)$ is an object in ${\bf Nil}(R).$ Assume that $\nu$ is non-zero nilpotent. Since $\nu$ is nilpotent, there exist a least $m\in\mathbb{N}$ such that $\nu^{m}=0$ and $\nu^{r}\neq 0$ for $r<m.$ Then we have a chain of $R$-modules

[TABLE]

Lemma 3.1**.**

$\frac{\operatorname{ker}(\nu^{i+1})}{ker(\nu^{i})}$* is a torsion free $R$-module for $1\leq i\leq m-1.$*

Proof.

Let $x+ker(\nu^{i})$ be a torsion element of $\frac{\operatorname{ker}(\nu^{i+1})}{ker(\nu^{i})}.$ Then there exist a non-zero-divisor $r\in R$ such that $rx\in ker(\nu^{i}).$ This implies that $r\nu^{i}(x)=0$ in $R^{n}.$ So, $x\in ker(\nu^{i}).$ ∎

Lemma 3.2**.**

$[(R^{n},\nu)]=[(R^{n},0)]$* in $K_{0}({\bf Nil}(R))$ provided every finitely generated torsion free $R$-module is projective.*

Proof.

We have an exact sequence of $R$-modules

[TABLE]

By Lemma 3.1, $\frac{R^{n}}{\operatorname{ker}(\nu^{m-1})}$ is a finitely generated projective $R$-modules. Thus the sequence splits, and we get that $\operatorname{ker}(\nu^{m-1})$ is also a finitely generated projective $R$-module. By considering the exact sequence $0\to\operatorname{ker}(\nu^{m-2})\to\operatorname{ker}(\nu^{m-1})\to\frac{\operatorname{ker}(\nu^{m-1})}{\operatorname{ker}(\nu^{m-2})}\to 0$ and using Lemma 3.1, we obtain $\operatorname{ker}(\nu^{m-2})$ is a finitely generated projective $R$-module. Continuing this way, each $\operatorname{ker}(\nu^{i})$ is a finitely generated projective $R$-module. Observe that the following diagram of exact sequences

[TABLE]

is commutative for each $i,$ $1\leq i\leq m-1.$ Thus, we have an exact sequence

[TABLE]

in ${\bf Nil}(R)$ for $1\leq i\leq m-1.$ In $K_{0}({\bf Nil}(R))$, we get

[TABLE]

∎

Theorem 3.3**.**

Let $R$ be a commutative ring with unity. Assume that every finitely generated torsion free $R$-module is projective. Then ${\rm Nil}_{0}(R)=0.$

Proof.

Recall that ${\rm Nil}_{0}(R)$ is generated by elements of the form $[(R^{n},\nu)]-[(R^{n},0)]$ for some $n$ and some nilpotent endomorphism $\nu.$ Lemma 3.2 yields the result. ∎

Remark 3.4**.**

The hypothesis on $R$ in the above theorem holds for any Prüfer domain. We say that a ring $R$ is a Prüfer domain if $R_{\mathfrak{p}}$ is a valuation domain for all prime ideals $\mathfrak{p}$ of $R.$ Clearly, a valuation ring is Prüfer. A domain $R$ is Prüfer if and only if every finitely generated torsion free $R$-module is projective. The ring of integer-valued polynomials ${\rm Int}(\mathbb{Z})=\{f\in\mathbb{Q}[x]|f(\mathbb{Z})\subset\mathbb{Z}\}$ is a Prüfer domain. In fact, it is a non-noetherian Prüfer domain (see [1]). One can see [8] for $K$-theory of Prüfer domains. **

4. Cofinality Lemma

Let $\mathcal{N}$ be an exact category. An exact subcategory $\mathcal{M}$ of $\mathcal{N}$ is called closed under extensions whenever there exists a short exact sequence $0\to N_{1}\to N\to N_{2}\to 0$ in $\mathcal{N}$ with $N_{1}$ and $N_{2}$ in $\mathcal{M}$ then $N$ is isomorphic to an object of $\mathcal{M}.$ We say that an exact subcategory $\mathcal{M}$ of $\mathcal{N}$ is cofinal in $\mathcal{N}$ if for every object $N_{1}\in\mathcal{N}$ there exists $N_{2}\in\mathcal{N}$ such that $N_{1}\oplus N_{2}$ is isomorphic to an object of $\mathcal{M}.$ Let ${\bf Free}(R)$ denote the category of finitely generated free $R$-modules. Clearly, ${\bf Free}(R)$ is an exact subcategory of ${\bf P}(R)$ which is cofinal and closed under extensions.

We can define a category ${\bf Nil(Free}(R))$ whose object are pairs $(F,\nu),$ where $F$ is in ${\bf Free}(R)$ and $\nu$ is a nilpotent endomorphism. A morphism $f:(F_{1},\nu_{1})\to(F_{2},\nu_{2})$ is a $R$-module map $f:F_{1}\to F_{2}$ such that $f\nu_{1}=\nu_{2}f.$ One can check that ${\bf Nil(Free}(R))$ is an exact category.

Lemma 4.1**.**

The category ${\bf Nil(Free}(R))$ is an exact full subcategory of ${\bf Nil}(R)$ which is cofinal and closed under extensions.

Proof.

Let $(P,\nu)\in{\bf Nil}(R).$ Then there exists a $Q$ in ${\bf P}(R)$ such that $\alpha:P\oplus Q\cong R^{n}$ for some $n>0.$ Note that $(Q,0)\in{\bf Nil}(R).$ So, we get $(P,\nu)\oplus(Q,0)\cong(R^{n},\nu^{{}^{\prime}}),$ where $\nu^{{}^{\prime}}=\alpha(\nu,0)\alpha^{-1}.$ This implies that ${\bf Nil(Free}(R))\subseteq{\bf Nil}(R)$ is cofinal.

Suppose that the sequence

[TABLE]

is exact with $(P_{1},\nu_{1}),(P_{2},\nu_{2})\in{\bf Nil(Free}(R)),$ i.e., the following diagram

[TABLE]

is commutative with exact rows and $P_{1},P_{2}\in{\bf Free}(R).$ Let $\beta$ denote the isomorphism $P\cong P_{1}\oplus P_{2}.$ We define a nilpotent endomorphism of $P_{1}\oplus P_{2}$ as $\beta\nu\beta^{-1}.$ Therefore, $(P,\nu)\cong(P_{1}\oplus P_{2},\beta\nu\beta^{-1})$ in ${\bf Nil(Free}(R)).$ Hence the lemma. ∎

We now consider categories $B^{q}{\bf Nil(Free}(R))$ and $B^{q}{\bf Nil}(R).$ Note that one can identify $B^{q}{\bf Nil(Free}(R))$ with ${\bf Nil}(B^{q}{\bf Free}(R))$ and $B^{q}{\bf Nil}(R)$ with ${\bf Nil}(B^{q}{\bf P}(R))$ because nilpotent endomorphism commutes with each differential.

Lemma 4.2**.**

The categories $C^{q}{\bf Nil(Free}(R))\subseteq C^{q}{\bf Nil}(R)$ and $B^{q}{\bf Nil(Free}(R))\subseteq B^{q}{\bf Nil}(R)$ both are cofinal and closed under extensions.

Proof.

By Lemma 6.2 of [2], we know that if $\mathcal{M}$ is an exact full subcategory of $\mathcal{N}$ which is cofinal and closed under extensions then for every $(N_{*},d,d^{{}^{\prime}})\in B^{q}\mathcal{N}$ there is an object $(L_{*},e)\in C^{q}\mathcal{N}$ such that $(N_{*},d,d^{{}^{\prime}})\oplus(L_{*},e,e)\cong(M_{*},f,f^{{}^{\prime}})\in B^{q}\mathcal{M}.$ The assertion now follows from Lemma 4.1. ∎

5. Generators of ${\rm Nil_{n>0}}(R)$

We first describe generators for ${\rm Nil_{1}}(R)(\cong NK_{2}(R))$. Before that we need some preparations. Let us begin with the following Lemma.

Lemma 5.1**.**

If $[(P_{*},p,p^{{}^{\prime}})]=[(Q_{*},q,q^{{}^{\prime}})]$ in $K_{0}B^{q}{\bf P}(R)$ then $[(P_{*},p,p^{{}^{\prime}},0)]=[(Q_{*},q,q^{{}^{\prime}},0)]$ in $K_{0}B^{q}{\bf Nil}(R).$

Proof.

Since $[(P_{*},p,p^{{}^{\prime}})]=[(Q_{*},q,q^{{}^{\prime}})]$ in $K_{0}B^{q}{\bf P}(R),$ there are short exact sequences (see Exercise II.7.2 of [9])

[TABLE]

and

[TABLE]

in $B^{q}{\bf P}(R)$ such that

[TABLE]

Note that

[TABLE]

and

[TABLE]

both are short exact sequences in $B^{q}{\bf Nil}(R).$ Thus, $[(A_{*},a,\tilde{a},0)]=[(B_{*},b,\tilde{b},0)]=[(C_{*},c,\tilde{c},0)]+[(D_{*},d,\tilde{d},0)]$ in $K_{0}B^{q}{\bf Nil}(R).$ By using the isomorphism (5.1), we get $(P_{*},p,p^{{}^{\prime}},0)\oplus(A_{*},a,\tilde{a},0)\cong(Q_{*},q,q^{{}^{\prime}},0)\oplus(B_{*},b,\tilde{b},0).$ Hence the assertion. ∎

Lemma 5.2**.**

For every $(P_{*},p,p^{{}^{\prime}},\nu)$ in $B^{q}{\bf Nil}(R)$ there exists a $(Q_{*},q,q,0)$ in $B^{q}{\bf Nil}(R)$ such that $(P_{*},p,p^{{}^{\prime}},\nu)\oplus(Q_{*},q,q,0)\cong(F_{*},f,f^{{}^{\prime}},\nu^{{}^{\prime}}),$ where $(F_{*},f,f^{{}^{\prime}},\nu^{{}^{\prime}})\in B^{q}{\bf Nil(Free}(R)).$

Proof.

Note that $(P_{*},p,p^{{}^{\prime}})\in B^{q}{\bf P}(R).$ By Lemma 6.2 of [2], there exists a $(Q_{*},q,q)$ such that $(P_{*},p,p^{{}^{\prime}})\oplus(Q_{*},q,q)\cong(F_{*},f,f^{{}^{\prime}}),$ where $(F_{*},f,f^{{}^{\prime}})\in B^{q}{\bf Free}(R).$ Let $\alpha$ denote the above isomorphism. Define $\nu^{{}^{\prime}}:=\alpha(\nu,0)\alpha^{-1}.$ The isomorphisms $\alpha$ and $\alpha^{-1}$ commute with differentials. We need to check that $\nu^{{}^{\prime}}$ is a nilpotent endomorphism on $(F_{*},f,f^{{}^{\prime}})$. Let us just consider one case. Checking for other case is similar. We have $\alpha(p,q)=f\alpha$ and $(p,q)\alpha^{-1}=\alpha^{-1}f.$ Moreover, $\nu$ commutes with $p$ and $p^{{}^{\prime}}.$ Then $f\nu^{{}^{\prime}}=f\alpha(\nu,0)\alpha^{-1}=\alpha(p,q)(\nu,0)\alpha^{-1}=\alpha(\nu,0)(p,q)\alpha^{-1}=\alpha(\nu,0)\alpha^{-1}f=\nu^{{}^{\prime}}f.$ This shows that $\nu^{{}^{\prime}}$ is a nilpotent endomorphism on $(F_{*},f,f^{{}^{\prime}})$ and we get the desired isomorphism $(P_{*},p,p^{{}^{\prime}},\nu)\oplus(Q_{*},q,q,0)\cong(F_{*},f,f^{{}^{\prime}},\nu^{{}^{\prime}}).$ ∎

Recall that we have the forgetful map

[TABLE]

and $\widetilde{{\rm Nil}}_{1}(R)$ denotes the kernel of the above map.

Lemma 5.3**.**

The group $\widetilde{{\rm Nil}}_{1}(R)$ is generated by elements of the form $[(F_{*},f,f^{{}^{\prime}},\nu)]-[(F_{*},f,f^{{}^{\prime}},0)],$ where $(F_{*},f,f^{{}^{\prime}},\nu)$ and $(F_{*},f,f^{{}^{\prime}},0)$ both are in $B^{q}{\bf Nil(Free}(R)).$

Proof.

By Lemma 4.2, $B^{q}{\bf Nil(Free}(R))\subseteq B^{q}{\bf Nil}(R)$ is cofinal with closed under extensions. Let $x\in K_{0}(B^{q}){\bf Nil}(R).$ Then $x$ is of the form $[(P_{*},p,p^{{}^{\prime}},\nu)]-[(F_{*},f,f^{{}^{\prime}},\tilde{\nu})],$ where $(P_{*},p,p^{{}^{\prime}},\nu)\in B^{q}{\bf Nil}(R)$ and $(F_{*},f,f^{{}^{\prime}},\tilde{\nu})\in B^{q}{\bf Nil(Free}(R))$ (see Remark II.7.2.1 of [9]). If $x\in\widetilde{{\rm Nil}}_{1}(R)$ then

[TABLE]

Since $(P_{*},p,p^{{}^{\prime}})\in B^{q}{\bf P}(R),$ there exists a $(Q_{*},q,q)\in B^{q}{\bf P}(R)$ such that $[(P_{*},p,p^{{}^{\prime}})]+[(Q_{*},q,q)]=[(\tilde{F}_{*},\tilde{f},\tilde{f}^{{}^{\prime}})]$ in $K_{0}B^{q}{\bf P}(R)$ (by Lemma 6.2 of [2]). Here $(\tilde{F}_{*},\tilde{f},\tilde{f}^{{}^{\prime}})\in B^{q}{\bf Free}(R).$ We also have (see Lemma 5.2)

[TABLE]

By (5.2), $[(F_{*},f,f^{{}^{\prime}})]+[(Q_{*},q,q)]=[(\tilde{F}_{*},\tilde{f},\tilde{f}^{{}^{\prime}})]$ in $K_{0}B^{q}{\bf P}(R).$ Using Lemma 5.1, we get $[(F_{*},f,f^{{}^{\prime}},0)]+[(Q_{*},q,q,0)]=[(\tilde{F}_{*},\tilde{f},\tilde{f}^{{}^{\prime}},0)]$ in $K_{0}B^{q}{\bf Nil}(R).$ Now, (5.3) implies that

[TABLE]

Therefore,

[TABLE]

This shows that $\widetilde{{\rm Nil}}_{1}(R)$ is generated by elements of the form $[(F_{*},f,f^{{}^{\prime}},\nu)]-[(F_{*},f,f^{{}^{\prime}},0)],$ where $(F_{*},f,f^{{}^{\prime}},\nu)$ and $(F_{*},f,f^{{}^{\prime}},0)$ both are in $B^{q}{\bf Nil(Free}(R)).$ ∎

Theorem 5.4**.**

The group ${\rm Nil}_{1}(R)$ is generated by elements of the form

[TABLE]

where $(F_{*},f,f^{{}^{\prime}},\nu),(F_{*},f,f^{{}^{\prime}},0)\in B^{q}{\bf Nil(Free}(R))$ and $f\neq f^{{}^{\prime}}.$

Proof.

We know $\widetilde{{\rm Nil}}_{1}(R)\cong{\rm Nil}_{1}(R)\oplus\widetilde{T}_{R}^{1}$ (see Lemma 2.6). Hence the assertion by Lemma 5.3. ∎

Generators of ${\rm Nil}_{n}(R)$ for $n>1$

We consider the case ${\rm Nil}_{n}(R)(\cong NK_{n+1}(R))$ for $n>1.$

Lemma 5.5**.**

If $[(P_{*},(p_{1},p_{1}^{{}^{\prime}}),(p_{2},p_{2}^{{}^{\prime}}),\dots,(p_{n},p_{n}^{{}^{\prime}}))]=[(Q_{*},(q_{1},q_{1}^{{}^{\prime}}),(q_{2},q_{2}^{{}^{\prime}}),\dots,(q_{n},q_{n}^{{}^{\prime}}))]$ in $K_{0}(B^{q})^{n}{\bf P}(R)$ then

[TABLE]

in $K_{0}(B^{q})^{n}{\bf Nil}(R).$

Proof.

The proof is similar to Lemma 5.1. More precisely, just rewrite the proof of Lemma 5.1 for $(B^{q})^{n}{\bf P}(R)$. ∎

Lemma 5.6**.**

For every $(P_{*},(p_{1},p_{1}^{{}^{\prime}}),(p_{2},p_{2}^{{}^{\prime}}),\dots,(p_{n},p_{n}^{{}^{\prime}}),\nu)$ in $(B^{q})^{n}{\bf Nil}(R)$ there exists a $(Q_{*},(q,q),(q_{2},q_{2}^{{}^{\prime}}),\dots,(q_{n},q_{n}^{{}^{\prime}}),0)$ in $(B^{q})^{n}{\bf Nil}(R)$ such that

[TABLE]

where $(F_{*},(f_{1},f_{1}^{{}^{\prime}}),(f_{2},f_{2}^{{}^{\prime}}),\dots,(f_{n},f_{n}^{{}^{\prime}}),\nu^{{}^{\prime}})$ is in $(B^{q})^{n}{\bf Nil(Free}(R)).$

Proof.

By repeatadly using Lemma 4.2, $(B^{q})^{n-1}{\bf Free}(R)\subseteq(B^{q})^{n-1}{\bf P}(R)$ is cofinal and closed under extensions. By Lemma 6.2 of [2], for $(P_{*},(p_{1},p_{1}^{{}^{\prime}}),(p_{2},p_{2}^{{}^{\prime}}),\dots,(p_{n},p_{n}^{{}^{\prime}}))\in(B^{q})^{n}{\bf P}(R)$ there exists a $(Q_{*},(q,q),(q_{2},q_{2}^{{}^{\prime}}),\dots,(q_{n},q_{n}^{{}^{\prime}}))$ such that

[TABLE]

where $(F_{*},(f_{1},f_{1}^{{}^{\prime}}),(f_{2},f_{2}^{{}^{\prime}}),\dots,(f_{n},f_{n}^{{}^{\prime}}))\in(B^{q})^{n}{\bf Free}(R).$ Define $\nu^{{}^{\prime}}:=\alpha(\nu,0)\alpha^{-1}.$ Note that $\alpha$ and $\alpha^{-1}$ commute with differentials in each direction. The rest of the argument is similar to Lemma 5.2. ∎

Recall that $\widetilde{{\rm Nil}}_{n}(R)$ denotes the kernel of

[TABLE]

[TABLE]

Lemma 5.7**.**

The group $\widetilde{{\rm Nil}}_{n}(R)$ is generated by elements of the form

[TABLE]

where $(F_{*},(f_{1},f_{1}^{{}^{\prime}}),(f_{2},f_{2}^{{}^{\prime}}),\dots,(f_{n},f_{n}^{{}^{\prime}}),\nu)$ and $(F_{*},(f_{1},f_{1}^{{}^{\prime}}),(f_{2},f_{2}^{{}^{\prime}}),\dots,(f_{n},f_{n}^{{}^{\prime}}),0)$ both are in $(B^{q})^{n}{\bf Nil(Free}(R)).$

Proof.

By Lemma 4.2, $(B^{q})^{n}{\bf Nil(Free}(R))\subseteq(B^{q})^{n}{\bf Nil}(R)$ is cofinal with closed under extensions. Let $x\in K_{0}(B^{q})^{n}{\bf Nil}(R).$ By applying Remark II.7.2.1 of [9] for the categories $(B^{q})^{n}{\bf Nil(Free}(R))\subseteq(B^{q})^{n}{\bf Nil}(R),$ we get

[TABLE]

$(P_{*},(p_{1},p_{1}^{{}^{\prime}}),(p_{2},p_{2}^{{}^{\prime}}),\dots,(p_{n},p_{n}^{{}^{\prime}}),\nu)\in(B^{q})^{n}{\bf Nil}(R)$ and $(F_{*},(f_{1},f_{1}^{{}^{\prime}}),(f_{2},f_{2}^{{}^{\prime}}),\dots,(f_{n},f_{n}^{{}^{\prime}}),\tilde{\nu})\\ \in(B^{q})^{n}{\bf Nil(Free}(R)).$ Suppose $x\in\widetilde{{\rm Nil}}_{n}(R)$. Thus,

[TABLE]

in $K_{0}(B^{q})^{n}{\bf P}(R).$ The rest of the argument is similar to the case of $\widetilde{{\rm Nil}}_{1}(R)$. By using Lemmas 5.5 and 5.6, we get

[TABLE]

where all the entries are in $(B^{q})^{n}{\bf Nil(Free}(R)).$ Therefore, $\widetilde{{\rm Nil}}_{n}(R)$ is generated by elements of the form

[TABLE]

∎

By Lemma 2.6, ${\rm Nil}_{n}(R)\cong\widetilde{{\rm Nil}}_{n}(R)/\widetilde{T}_{R}^{n}.$ Hence we get,

Theorem 5.8**.**

The group ${\rm Nil}_{n}(R)$ is generated by elements of the form

[TABLE]

where $(F_{*},(f_{1},f_{1}^{{}^{\prime}}),(f_{2},f_{2}^{{}^{\prime}}),\dots,(f_{n},f_{n}^{{}^{\prime}}),\nu)),(F_{*},(f_{1},f_{1}^{{}^{\prime}}),(f_{2},f_{2}^{{}^{\prime}}),\dots,(f_{n},f_{n}^{{}^{\prime}}),0))$ are objects of $(B^{q})^{n}{\bf Nil(Free}(R)).$

Remark 5.9**.**

D. Grayson in [2, Remark 8.1] remarked that acyclic binary multicomplexes supported on $[0,2]^{n}$ suffice to generate whole group $K_{n}\mathcal{N}.$ Recently, D. Kasprowski and C. Winges establish Grayson’s remark in [6](more precisely, see Theorem 1.3 of [6]). In view of [6], generators of ${\rm Nil}_{n}(R)$ for $n>0$ obtained in Theorems 5.4 and 5.8 can be restricted to acyclic binary multicomplexes supported on $[0,2]^{n}.$**