中国科学院大学学报  2019, Vol. 36 Issue (1): 1-4   PDF    
On the exponent of NmK2($\mathbb{F}$[Cpn])
ZHANG Hao , TANG Guoping     
School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 100049, China
Abstract: Let Cpn be the cyclic p-group of order pn and $\mathbb{F}$ a finite field of characteristic p. For any integer 1 ≤ ln, we obtain infinitely many non-trivial elements of order pl in NmK2($\mathbb{F}$[Cpn]). In fact, these elements form a generating set of NmK2($\mathbb{F}$[Cpn]) and the exponent of NmK2($\mathbb{F}$[Cpn]) is pn.
Keywords: K-theory     Bass Nil groups     truncated polynomial    
交换群NmK2($\mathbb{F}$[Cpn])的指数
张浩, 唐国平     
中国科学院大学数学科学学院, 北京 100049
摘要: 令Cpn是阶为pn的循环p群,$\mathbb{F}$是特征为p的有限域。对于任何整数1≤ln,得到NmK2$\mathbb{F}$[Cpn])中无限多个非平凡的pl阶元素。事实上,这些元素组成NmK2$\mathbb{F}$[Cpn])的一个生成元集,并且NmK2$\mathbb{F}$[Cpn])的指数为pn
关键词: K理论     Bass Nil群     截断多项式    

Let R be a ring with unit. The Bass Nil groups NmKi(R) are introduced by Bass[1] in order to investigate the relation between Ki(R[x1, …, xm]) and Ki(R). For any i$\mathbb{Z}$, NKi(R) is defined to be the kernel of surjective map Ki(R[x1])→Ki(R) induced by x1 |→0. And NmKi(R) is defined by iteration, i.e., the kernel of the surjection Nm-1Ki(R[xm])→Nm-1Ki(R) induced by xm |→0. When i=0, 1, 2, Ki(R) are the classical algebraic K-groups defined by Grothendieck[2], Bass[1] and Milnor[3], respectively. When i < 0, the negative K-theory is defined by Bass[1]. When i>2, Ki(R)=πi(K(R)) is defined to be the i-th homotopy group of the K-theory space K(R) which was first invented by Quillen[4] via plus-construction or Q-construction. As for the Bass Nil groups, the most known property is the following phenomenon.

Theorem A (See Refs. [5-9]). Let R be a ring. For any i, m$\mathbb{Z}$ and m≥1, if NmKi(R)≠0, then it is not finitely generated as an abelian group.

Let p be a prime number. In some cases, NKi(R) are abelian p-groups[8]. However, the exponents of these abelian p-groups are not completely determined. For example, the exponents of NK0($\mathbb{Z}$[C4]) and NK1($\mathbb{Z}$[C4]) are both 2, but the exponent of NK2($\mathbb{Z}$[C4]) is still unknown[10].

Let $\mathbb{F}$ be a finite field of characteristic p and Cpn the cyclic p-group of order pn. Since K2($\mathbb{F}$[Cpn])=K2($\mathbb{F}$)=0 (see Ref. [11]), we have

$ \begin{array}{*{20}{c}} {{K_2}\left( {\mathbb{F}\left[ {{C_{{p^n}}}} \right]\left[ {{x_1}, \cdots ,{x_m}} \right]} \right) \cong {{\left( {1 + N} \right)}^m}{K_2}\left( {\mathbb{F}\left[ {{C_{{p^n}}}} \right]} \right) = } \\ {\mathop \oplus \limits_{i = 1}^m \left( {\begin{array}{*{20}{c}} m \\ i \end{array}} \right){N^i}{K_2}\left( {\mathbb{F}\left[ {{C_{{p^n}}}} \right]} \right).} \end{array} $

In Ref. [12], Juan-Pineda showed the non-finiteness of NK2($\mathbb{F}$p[Cpn]) by giving one non-trivial element of order p and concluded that NK2($\mathbb{Z}$[C])≠0 for any non-trivial cyclic group C. In this paper, we could give infinitely many non-trivial elements of order pl for any 1≤ln in NmK2($\mathbb{F}$[Cpn]). In fact, we give a presentation of NmK2($\mathbb{F}$[Cpn]) in terms of Dennis-Stein symbols and show that the exponent of NmK2($\mathbb{F}$[Cpn]) is pn.

1 Main result

Let $\mathbb{F}$ be a finite field with pf elements and B={1, b, b2, …, bf-1} a basis of $\mathbb{F}$ as a vector space over the finite field $\mathbb{F}$p of p elements. Let Cpn be the cyclic group of order pn (n≥1) generated by σ. Let J=J($\mathbb{F}$[Cpn]) be the Jacobson radical of the group algebra $\mathbb{F}$[Cpn]. The notation $\oplus $ denotes a countably infinite direct sum, i.e., $\oplus $=$\oplus $sS for some countably infinite set S.

Lemma 1.1 NK2($\mathbb{F}$[Cpn])$\cong $K2($\mathbb{F}$[Cpn][x], J[x]).

Proof Observe that $\mathbb{F}$[Cpn][x]/J[x]$\cong $$\mathbb{F}$[x] and $\mathbb{F}$[x]→$\mathbb{F}$[Cpn][x] is a split inclusion. Since $\mathbb{F}$ is a finite field, K2($\mathbb{F}$[Cpn])=K2($\mathbb{F}$)=0. And K2($\mathbb{F}$[x])=K2($\mathbb{F}$)$\oplus $NK2($\mathbb{F}$)=0 because $\mathbb{F}$ is regular. Hence the result follows from the two exact sequences of K-groups:

$ \begin{array}{*{20}{c}} {0 \to {K_2}\left( {\mathbb{F}\left[ {{C_{{p^n}}}} \right]\left[ x \right],J\left[ x \right]} \right) \to {K_2}\left( {\mathbb{F}\left[ {{C_{{p^n}}}} \right]\left[ x \right]} \right)} \\ { \to {K_2}\left( {\mathbb{F}\left[ {{C_{{p^n}}}} \right]\left[ x \right]/J\left[ x \right]} \right) = 0 \to 0,} \end{array} $
$ \begin{array}{*{20}{c}} {0 \to N{K_2}\left( {\mathbb{F}\left[ {{C_{{p^n}}}} \right]} \right) \to {K_2}\left( {\mathbb{F}\left[ {{C_{{p^n}}}} \right]\left[ x \right]} \right) \to } \\ {{K_2}\left( {\mathbb{F}\left[ {{C_{{p^n}}}} \right]} \right) = 0 \to 0.} \end{array} $

Let I=(t1pn) be a proper ideal in the polynomial ring $\mathbb{F}$[t1, t2, …, tm+1]. Then

$ \mathbb{F}\left[ {{C_{{p^n}}}} \right] \cong \mathbb{F}\left[ {{t_1}} \right]/\left( {t_1^{{p^n}}} \right), $
$ \mathbb{F}\left[ {{C_{{p^n}}}} \right]\left[ {{x_1}, \cdots ,{x_m}} \right] \cong \mathbb{F}\left[ {{t_1}, \cdots ,{t_{m + 1}}} \right]/I, $

via σ-1 |→t1 and xi |→ti+1. Let A=$\mathbb{F}$[t1, …, tm+1]/I and M=($\overline {{t_1}} $) be its nilradical where $\overline {{t_1}} $=t1+I. Then $\mathbb{F}$[x1, …, xm]$\cong $A/M and K2(A)$\cong $K2(A, M). So the above lemma becomes NK2($\mathbb{F}$[Cpn])$\cong $K2($\mathbb{F}$[t1, t2]/I, M) (m=1).

Lemma 1.2 Let 1≤tn-1 and 1≤h < p be integers. If pn-t-1 < kpn-t, $\left\lceil {{\rm{lo}}{{\rm{g}}_p}\frac{{{p^n} + 1}}{{pk}}} \right\rceil = \left\lceil {{\rm{lo}}{{\rm{g}}_p}\frac{{{p_n}}}{{pk-h}}} \right\rceil = t$, and if k=1, $\left\lceil {{\rm{lo}}{{\rm{g}}_p}\frac{{{p^n} + 1}}{{pk}}} \right\rceil = \left\lceil {{\rm{lo}}{{\rm{g}}_p}\frac{{{p_n}}}{{pk-h}}} \right\rceil = n$, where $\left\lceil a \right\rceil $=min{s$\mathbb{Z}$|sa} denotes the smallest integer no less than a.

Proof If k=1, the computation is easy. Suppose pn-t-1+1≤kpn-t, the result follows from the inequalities

$ {p^{t - 1}} < \frac{{{p^n}}}{{{p^{n - t + 1}} - h}} \le \frac{{{p^n}}}{{pk - h}} \le \frac{{{p^n}}}{{{p^{n - t}} + p - h}} < {p^t}, $
$ {p^{t - 1}} = \frac{{{p^n}}}{{{p^{n - t + 1}}}} < \frac{{{p^n} + 1}}{{pk}} \le \frac{{{p^n} + 1}}{{{p^{n - t}} + p}} < {p^t}. $

Theorem 1.1 Let Cpn be the cyclic group of order pn (n≥1) generated by σ. For any integer m≥1, NmK2($\mathbb{F}$[Cpn])$\cong $$\oplus $($\mathop \oplus \limits_{i = 1}^n $$\mathbb{Z}$/pi$\mathbb{Z}$) can be generated by these elements:

the generators of order pl (1≤ln-1) are 〈b(σ-1)pk-1$\prod\limits_{j = 1}^m {} $xjβj, (σ-1)〉, where pn-l-1 < kpn-l and gcd(p, β1, …, βm)=1, 〈b(σ-1)pk($\prod\limits_{j = 1}^m {} $xjβj)/xi, xi〉, where 1≤im, pn-l-1k < pn-l, gcd(p, β1, …, βm)=1 and i≠min{j|βj0 mod p}, 〈b(σ-1)pk-h($\prod\limits_{j = 1}^m {} $xjβj)/xi, xi〉, where 1≤im, pn-l-1 < kpn-l, 1≤h < p; and generators of order pn are 〈b(σ-1)p-1$\prod\limits_{j = 1}^m {} $xjβj, (σ-1)〉, where gcd(p, β1, …, βm)=1, 〈b(σ-1)h($\prod\limits_{j = 1}^m {} $xjβj)/xi, xi〉, where 1≤im and 1≤h < p.

For all the above symbols, bB and β=(β1, …, βm)∈$\mathbb{N}$+m, where $\mathbb{N}$+ is the set of positive integers.

Proof Suppose we get a generating set of K2($\mathbb{F}$[Cpn][x1, …, xm])$\cong $K2(A, M) in terms of Dennis-Stein symbols. Fix j different indeterminates in {x1, …, xm}, say {xi1, …, xij}. The elements of the direct summand NjK2($\mathbb{F}$[Cpn])K2($\mathbb{F}$[Cpn][xi1, …, xij])K2($\mathbb{F}$[Cpn][x1, …, xm]) can be represented by using those Dennis-Stein symbols containing these j different indeterminates. Hence K2($\mathbb{F}$[Cpn][x1, …, xm]) contains mj pieces of NjK2($\mathbb{F}$[Cpn]). So the elements of NmK2($\mathbb{F}$[Cpn]) can be represented by using those Dennis-Stein symbols in K2($\mathbb{F}$[Cpn][x1, …, xm]) containing all the m indeterminates.

We follow the notations in Ref. [13]. Let $\mathbb{N}$={0, 1, 2…} be the set of non-negative integers and $\mathbb{N}$+={1, 2, 3, …} the set of positive integers. Let εi=(0, …, 0, 1, 0, …, 0)∈$\mathbb{N}$m+1 be the i-th basis vector. For α$\mathbb{N}$m+1, write tα=t1α1tαm+1m+1 where t1=σ-1 and ti+1=xi for 1≤im. Define

$ \Delta ' = \left\{ {\alpha = \left( {{\alpha _1}, \cdots ,{\alpha _{m + 1}}} \right) \in \mathbb{N}_ + ^{m + 1}\left| {{\alpha _1} \geqslant {p^n}} \right.} \right\}, $
$ \Lambda ' = \left\{ {\left( {\alpha ,i} \right) \in \mathbb{N}_ + ^{m + 1} \times \left\{ {1,2, \cdots ,m + 1} \right\}} \right\}. $

For (α, i)∈Λ′, let [α, i]=min{k$\mathbb{Z}$|-εi∈Δ′} and w(α, i)=min{w$\mathbb{N}$|pw≥[α, i]}. Then [α, 1]=$\left\lceil {\frac{{{p^n} + 1}}{{{\alpha _1}}}} \right\rceil $, [α, j]=$\left\lceil {\frac{{{p^n}}}{{{\alpha _1}}}} \right\rceil $ for any j≠1. If gcd(p, α1, …, αm+1)=1, put [α]=max{[α, i]|αi0 mod p}. Set Λ′0={(α, i)∈Λ′| gcd(p, α1, …, αm+1)=1, i≠min{j|αj0 mod p, [α, j]=[α]}}, and let Λ′>10={(α, i)∈Λ′0|[α, i]>1}.

Then by Corollary 2.7 in Ref. [13] and the above discussion, NmK2($\mathbb{F}$[Cpn]) has a presentation with

generators: 〈btα-εi, ti〉, where bB, (α, i)∈Λ′>10;

relations: pw(α, i)btα-εi, ti〉=0, where w(α, i)=「logp[α, i].

It is sufficient to determine the set Λ′>10.

If α1=pn and at least one of αj with pαj, then only (α, 1)∈Λ′>10 and [α, 1]=2.

If α1=pk for some 1≤k < pn-1 and j is the smallest number such that pαj, i.e., p|α1, …, p|αj-1 and pαj, then all (α, i) except (α, j) are in Λ′>10.

If pα1, i.e., α1=pk-h for some 1≤kpn-1 and 1≤h < p, then for each j≥2, (α, j)∈Λ′>10.

So one gets

$ \begin{array}{*{20}{c}} {\Lambda {'}_{ > 1}^0 = \left\{ {\left( {\alpha ,1} \right)\left| {{\alpha _1} = {p^n},\gcd } \right.\left( {p,{\alpha _2}, \cdots ,{\alpha _{m + 1}}} \right) = 1} \right\}}\\ { \cup \left\{ {\left( {\alpha ,i} \right)\left| {\begin{array}{*{20}{c}} {{\alpha _1} = pk,1 \le k < {p^{n - 1}},}\\ {\gcd \left( {p,{\alpha _2}, \cdots ,{\alpha _{m + 1}}} \right) = 1,}\\ {i \ne \min \left\{ {j\left| {{\alpha _j} ≢ 0\bmod p} \right.} \right\}} \end{array}} \right.} \right\}}\\ { \cup \left( {\mathop \cup \limits_{j = 2}^{m + 1} \left\{ {\left( {\alpha ,j} \right)\left| {\begin{array}{*{20}{c}} {{\alpha _1} = pk - h,}\\ {1 \le k \le {p^{n - 1}},}\\ {1 \le h < \rho } \end{array}} \right.} \right\}} \right).} \end{array} $

Let bB. For any β$\mathbb{N}$+m, write xβ=x1β1x2β2xmβm. We can get a presentation of NmK2($\mathbb{F}$[Cpn]) in terms of the following Dennis-Stein symbols with generators

·〈b(σ-1)pk-1xβ, (σ-1)〉 where 1≤kpn-1 and gcd (p, β1, …, βm)=1;

·〈b(σ-1)pkxβ-εi, xi〉 where 1≤k < pn-1, gcd (p, β1, …, βm)=1 and i≠min{j|βj0 mod p};

·〈b(σ-1)pk-hxβ-εi, xi〉 where 1≤kpn-1, 1≤h < p, 1≤im.

The relations are

$ \cdot \;\;{p^{\left\lceil {{{\log }_p}\frac{{{p^n} + 1}}{{pk}}} \right\rceil }}\left\langle {b{{\left( {\sigma - 1} \right)}^{pk - 1}}{x^\beta },\left( {\sigma - 1} \right)} \right\rangle = 0, $
$ \cdot \;\;{p^{\left\lceil {{{\log }_p}\frac{{{p^n}}}{{pk}}} \right\rceil }}\left\langle {b{{\left( {\sigma - 1} \right)}^{pk}}{x^{\beta - {\varepsilon ^i}}},{x_i}} \right\rangle = 0, $
$ \cdot \;\;{p^{\left\lceil {{{\log }_p}\frac{{{p^n}}}{{pk - h}}} \right\rceil }}\left\langle {b{{\left( {\sigma - 1} \right)}^{pk - h}}{x^{\beta - {\varepsilon ^i}}},{x_i}} \right\rangle = 0. $

Then by Lemma 1.2, the result follows.

2 Examples

Example 2.1 Let C4 be the cyclic group of order 4 generated by σ. Then NK2($\mathbb{F}$2[C4])$\cong $$\oplus $($\mathbb{Z}$/2$\mathbb{Z}$$\oplus $$\mathbb{Z}$/4$\mathbb{Z}$) can be generated by these elements: the generators of order 4 are

$ \left\{ {\left\langle {\left( {\sigma - 1} \right){x^{i - 1}},x} \right\rangle \left| {i \ge 1} \right.} \right\}, $
$ \left\{ {\left\langle {\left( {\sigma - 1} \right){x^{2i - 1}},\left( {\sigma - 1} \right)} \right\rangle \left| {i \ge 1} \right.} \right\}, $

and the generators of order 2 are

$ \left\{ {\left\langle {{{\left( {\sigma - 1} \right)}^3}{x^{i - 1}},x} \right\rangle \left| {i \ge 1} \right.} \right\}, $
$ \left\{ {\left\langle {{{\left( {\sigma - 1} \right)}^3}{x^{2i - 1}},\left( {\sigma - 1} \right)} \right\rangle \left| {i \ge 1} \right.} \right\}. $
$ {N^2}{K_2}\left( {{\mathbb{F}_2}\left[ {{C_4}} \right]} \right) \cong { \oplus _\infty }\left( {\mathbb{Z}/2\mathbb{Z}\; \oplus \;\mathbb{Z}/4\mathbb{Z}} \right) $

N2K2($\mathbb{F}$2[C4])$\cong $$\oplus $($\mathbb{Z}$/2$\mathbb{Z}$$\oplus $$\mathbb{Z}$/4$\mathbb{Z}$) can be generated by these elements: the generators of order 4 are

$ \left\{ {\left\langle {\left( {\sigma - 1} \right){x^{i - 1}}{y^j},x} \right\rangle \left| {i \ge 1} \right.,j \ge 1} \right\}, $
$ \left\{ {\left\langle {\left( {\sigma - 1} \right){x^i}{y^{j - 1}},y} \right\rangle \left| {i \ge 1} \right.,j \ge 1} \right\}, $
$ \left\{ {\left\langle {\left( {\sigma - 1} \right){x^{2i - 1}}{y^j},\left( {\sigma - 1} \right)} \right\rangle \left| {i \ge 1} \right.,j \ge 1} \right\}, $
$ \left\{ {\left\langle {\left( {\sigma - 1} \right){x^{2i}}{y^{2j - 1}},\left( {\sigma - 1} \right)} \right\rangle \left| {i \ge 1} \right.,j \ge 1} \right\}, $

and the generators of order 2 are

$ \left\{ {\left\langle {{{\left( {\sigma - 1} \right)}^3}{x^{i - 1}}{y^j},x} \right\rangle \left| {i \ge 1} \right.,j \ge 1} \right\}, $
$ \left\{ {\left\langle {{{\left( {\sigma - 1} \right)}^3}{x^i}{y^{j - 1}},y} \right\rangle \left| {i \ge 1} \right.,j \ge 1} \right\}, $
$ \left\{ {\left\langle {{{\left( {\sigma - 1} \right)}^2}{x^{2i - 1}}{y^{2j - 1}},x} \right\rangle \left| {i \ge 1} \right.,j \ge 1} \right\}, $
$ \left\{ {\left\langle {{{\left( {\sigma - 1} \right)}^2}{x^{2i - 1}}{y^{j - 1}},y} \right\rangle \left| {i \ge 1} \right.,j \ge 1} \right\}, $
$ \left\{ {\left\langle {{{\left( {\sigma - 1} \right)}^3}{x^{2i - 1}}{y^j},\left( {\sigma - 1} \right)} \right\rangle \left| {i \ge 1} \right.,j \ge 1} \right\}, $
$ \left\{ {\left\langle {{{\left( {\sigma - 1} \right)}^3}{x^{2i}}{y^{2j - 1}},\left( {\sigma - 1} \right)} \right\rangle \left| {i \ge 1} \right.,j \ge 1} \right\}, $

where x, y are indeterminates.

Corollary 2.1 NK1($\mathbb{Z}$[Cp2])$\cong $$\oplus $$\mathbb{Z}$/p$\mathbb{Z}$.

Proof Assume Cp2 is generated by σ. There is a Milnor square,

where ζp2 is a primitive p2-th root of unity and $\mathbb{Z}$[ζp2] is the ring of integers in Q(ζp2). By the Mayer-Vietoris sequence for NK-functors, we get an exact sequence

$ \begin{array}{*{20}{c}} {N{K_2}\left( {\mathbb{Z}\left[ {{C_{{p^2}}}} \right]} \right) \to N{K_2}\left( {\mathbb{Z}\left[ {{\zeta _{{p^2}}}} \right]} \right) \oplus } \\ {N{K_2}\left( {\mathbb{Z}\left[ {{C_p}} \right]} \right) \to N{K_2}\left( {{\mathbb{F}_p}\left[ {{C_p}} \right]} \right) \to } \\ {N{K_1}\left( {\mathbb{Z}\left[ {{C_{{p^2}}}} \right]} \right) \to N{K_1}\left( {\mathbb{Z}\left[ {{\zeta _{{p^2}}}} \right]} \right) \oplus N{K_1}\left( {\mathbb{Z}\left[ {{C_p}} \right]} \right).} \end{array} $

Since $\mathbb{Z}$[ζp2] is regular, NKn($\mathbb{Z}$p2])=0 for all n. The order of Cp is square-free. Hence NK1($\mathbb{Z}$[Cp])=0 (see Ref. [14]). And NK2($\mathbb{Z}$[Cp])$\cong $$\oplus $$\mathbb{Z}$/p$\mathbb{Z}$ (see Ref. [15]). So the above exact sequence becomes

$ \begin{array}{*{20}{c}} {N{K_2}\left( {\mathbb{Z}\left[ {{C_p}} \right]} \right) \to N{K_2}\left( {{\mathbb{F}_p}\left[ {{C_p}} \right]} \right) \to } \\ {N{K_1}\left( {\mathbb{Z}\left[ {{C_{{p^2}}}} \right]} \right) \to 0.} \end{array} $

Moreover, we have NK2($\mathbb{F}$p[Cp])$\cong $$\oplus $$\mathbb{Z}$/p$\mathbb{Z}$ and NK1($\mathbb{Z}$[Cp2])≠0 (see Ref. [14]). Hence NK1($\mathbb{Z}$[Cp2]) is not finitely generated, therefore NK1($\mathbb{Z}$[Cp2])$\cong $$\oplus $$\mathbb{Z}$/p$\mathbb{Z}$.

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