On a conjecture of V. V. Shchigolev

TL;DR

This paper confirms Shchigolev's conjecture that any T-space generated by an infinite subset of a specific set G over fields of characteristic p>2 is finitely based, providing explicit bounds on the basis size.

Contribution

The paper proves Shchigolev's conjecture, demonstrating that all T-spaces generated by infinite subsets of G are finitely based with bounded basis size.

Findings

Shchigolev's conjecture is true for fields of characteristic p>2.

T-spaces generated by infinite subsets of G have bases of size at most i_2 - i_1 + 1.

Provides explicit bounds on the basis size for these T-spaces.

Abstract

V. V. Shchigolev has proven that over any infinite field k of characteristic p>2, the T-space generated by G={x_1^p,x_1^px_2^p,...} is finitely based, which answered a question raised by A. V. Grishin. Shchigolev went on to conjecture that every infinite subset of G generated a finitely based T-space. In this paper, we prove that Shchigolev's conjecture was correct by showing that for any field of characteristic p>2, the T-space generated by any subset {x_1^px_2^p...x_{i_1}^p, x_1^px_2^p...x_{i_2}^p,...}, i_1<i_2<i_3<..., of G has a T-space basis of size at most i_2-i_1+1.

Equations

$$S^{(d)}=S^{(d)}(x_1,x_2,\dots ,x_d)=\sum _{\sigma \in {\Sigma }_d}\prod _{i=1}^d{x}_{\sigma (i)},$$

$$J=\{1,2\}\cup \{r+j-2\mid j\in J^{\prime },j\ge 2\}.$$

$$V_{i_2+tl}+\sum _{j=r}^{\infty }{V}_{i_j}=\sum _{j=1}^{\infty }{V}_{i_j^{\prime }}=\sum _{j\in J^{\prime }}{V}_{i_j^{\prime }}=V_{i_2+tl}+\sum _{\begin{array}{c}j\in J^{\prime }\\ j\ge 2\end{array}}{V}_{i_j^{\prime }}=V_{i_2+tl}+\sum _{\begin{array}{c}j\in J\\ j\ge 3\end{array}}{V}_{i_j}$$

$$V_{i_1}+V_{i_2}+\sum _{j=r}^{\infty }{V}_{i_j}$$

$$=V_{i_1}+V_{i_2}+\sum _{\begin{array}{c}j\in J\\ j\ge 3\end{array}}{V}_{i_j}.$$

$$\sum _{j\in J}{V}_{i_j}=V_{i_1}+V_{i_2}+\sum _{\begin{array}{c}j\in J\\ j\ge 3\end{array}}{V}_{i_j}=V_{i_1}+V_{i_2}+\sum _{j=r}^{\infty }{V}_{i_j}=\sum _{j=1}^{\infty }{V}_{i_j}.$$

$$S^{(d+1)}(x_1,x_2,\dots ,x_{d+1})$$

$$=S^{(d)}(x_1,x_2,\dots ,x_d)x_{d+1}+\sum _{i=1}^d{S}^{(d)}(x_1,x_2,\dots ,x_{d+1}{x}_i,\dots ,x_d)$$

$$=x_{d+1}{S}^{(d)}(x_1,x_2,\dots ,x_d)+\sum _{i=1}^d{S}^{(d)}(x_1,x_2,\dots ,x_i{x}_{d+1},\dots ,x_d).$$

$$S^{(d+1)}(x_1,x_2,\dots ,x_{d+1})\equiv S^{(d)}(x_1,\dots ,x_d)x_{d+1}\equiv x_{d+1}{S}^{(d)}(x_1,\dots ,x_d).$$

$$\sum _{i=1}^d{S}^{(d)}(x_1,\dots ,{\hat{x}}_i,\dots ,x_{d+1})x_i=S^{(d+1)}(x_1,\dots ,x_{d+1})-S^{(d)}(x_1,\dots ,x_d)x_{d+1}$$

$$S^{(d)}(x_2,\dots ,x_{d+1})x_{1}v+$$

$$=\sum _{i=1}^d{S}^{(d)}(x_1,x_2,\dots ,{\hat{x}}_i,\dots ,x_{d+1})x_{i}v\in (S_1^{(d)}v{)}^{S}to0.0pt.\text{hss}$$

$$S^{(d)}(x_1,\dots ,{\hat{x}}_i,\dots ,x_{d+1})x_{i}v$$

$$\equiv S^{(d)}(x_2,\dots ,{\hat{x}}_i,\dots ,x_{d+1},x_{i}v)x_1\mathrm{mod}{S}_1^{(d)}.$$

$$S^{(d)}(x_{2},\ldots,x_{d+1})x_{1}v+\bigl{(}(\sum_{i=2}^{d}S^{(d)}(x_{2},\ldots,\hat{x}_{i},\ldots,x_{i}v)\bigr{)}x_{1}\in(S_{1}^{(d)}v)^{S}+S_{1}^{(d)}.$$

$$(S_1^{(d)}uv{)}^S\subseteq (S_1^{(d)}u{)}^S+(S_1^{(d)}v{)}^S+S_1^{(d)},$$

$$[x^p,v]=\sum _{i=0}^p{x}^i[x,v]x^{p-i}=\frac{1}{(p-1)!}{S}^{(p)}(x,x,\dots ,x,[x,v]),$$

$$[x_1^p{x}_2^p\cdots x_m^p{x}_{m+1}^p,v]=[x_1^p{x}_2^p\cdots x_m^p,v]x_{m+1}^p+x_1^p{x}_2^p\cdots x_m^p[x_{m+1}^p,v].$$

$$(x+y{)}^p=\sum _{i=0}^p\sum _{\begin{array}{c}J\subseteq J_p\\ |J|=i\end{array}}{P}_J=y^p+x^p+\sum _{i=1}^{p-1}\frac{1}{i!(p-i)!}{S}^{(p)}(x,y;i).$$

$$(x+y{)}^{2p}=y^{2p}+x^{2p}+2x^p{y}^p+[y^p,x^p]+u^2+(x^p+y^p)u+u(x^p+y^p).$$

$$x_1^p\prod _{i=1}^m(2x_{2i}^p{x}_{2i+1}^p)\in H_1(H_1+H_1{S}_1^{(p)}+S_1^{(p)}{H}_1{)}^m.$$

Full text

1 Introduction

In [2] (and later in [3], the survey paper with V. V. Shchigolev), A. V. Grishin proved that in the free associative algebra with countably infinite generating set $\{\,x_{1},x_{2},\ldots\,\}$ over an infinite field of characteristic 2, the $T$-space generated by the set $\{\,x_{1}^{2},x_{1}^{2}x_{2}^{2},\ldots\,\}$ is not finitely based, and he raised the question as to whether or not over a field of characteristic $p>2$, the $T$-space generated by $\{\,x_{1}^{p},x_{1}^{p}x_{2}^{p},\ldots\,\}$ is finitely based. This was resolved by V. V. Shchigolev in [4], wherein he proved that over an infinite field of characteristic $p>2$, this $T$-space is finitely based. Shchigolev then raised the question in [4] as to whether every infinite subset of $\{\,x_{1}^{p},x_{1}^{p}x_{2}^{p},\ldots\,\}$ generates a finitely based $T$-space. In this paper, we prove that over an arbitrary field of characteristic $p>2$, every subset of $\{\,x_{1}^{p},x_{1}^{p}x_{2}^{p},\ldots\,\}$ generates a $T$-space that can be generated as a $T$-space by finitely many elements, and we give an upper bound for the size of a minimal generating set.

Let $p$ be a prime (not necessarily greater than 2) and let $k$ denote an arbitrary field of characteristic $p$. Let $X=\{\,x_{1},x_{2},\ldots\,\}$ be a countably infinite set, and let $k_{0}\langle X\rangle$ denote the free associative $k$-algebra over the set $X$.

Definition 1.1**.**

For any positive integer $d$, let

[TABLE]

where $\Sigma_{d}$ is the symmetric group on $d$ letters. Then define $S_{1}^{(d)}=\{\,S^{(d)}\,\}^{S}$, the $T$-space generated by $\{\,S^{(d)}\,\}$, and for all $n\geq 1$, $S_{n+1}^{(d)}=(S_{n}^{(d)}S_{1}^{(d)})^{S}$.

Let $I:i_{1}<i_{2}<\cdots$ be a sequence of positive integers (finite or infinite), and then for each $n\geq 1$, let $R_{{n,I}}^{(d)}=\sum_{j=1}^{n}S_{i_{j}}^{(d)}$. When the sequence $I$ is understood, we shall usually write $R_{{n}}^{(d)}$ instead of $R_{{n,I}}^{(d)}$. Finally, let $R_{{\infty,I}}^{(d)}$ (even if the sequence is finite) denote the $T$-space generated by $\{\,S_{i}^{(d)}\mid i\in I\,\}$. We shall prove that $R_{{\infty,I}}^{(d)}$ has a $T$-space basis of size at most $i_{2}-i_{1}+1$.

Definition 1.2**.**

Let $H_{1}=\{\,x_{1}^{p}\,\}^{S}$, and for each $n\geq 1$, let $H_{n+1}=(H_{n}H_{1})^{S}$.

Then for any positive integer $n$, let $L_{n,I}=\sum_{j=1}^{n}H_{i_{j}}$, and let $L_{\infty,I}$ denote the $T$-space generated by $\{\,h_{i}\mid i\in I\,\}$. We prove that $L_{\infty,I}$ is finitely generated as a $T$-space, with a $T$-space basis of size at most $i_{2}-i_{1}+1$. In particular, this proves that Shchigolev’s conjecture is valid.

2 Preliminaries

In this section, $k$ denotes an arbitrary field of characteristic an arbitrary prime $p$, and $V_{i}$, $i\geq 1$, denotes a sequence of $T$-spaces of $k_{0}\langle X\rangle$ satisfying the following two properties:

• (i)

$(V_{i}V_{j})^{S}=V_{i+j}$;

• (ii)

for all $m\geq 1$, $V_{2m+1}\subseteq V_{m+1}+V_{1}$.

Lemma 2.1**.**

For any integers $r$ and $s$ with $0<r<s$, $V_{s+t(s-r)}\subseteq V_{r}+V_{s}$ for all $t\geq 0$.

Proof.

The proof is by induction on $t$. There is nothing to show for $t=0$. For $t=1$, let $m=s-r$ in (ii) to obtain that $V_{2s-2r+1}\subseteq V_{s-r+1}+V_{1}$, then multiply by $V_{r-1}$ to obtain $V_{r-1}V_{2s-2r+1}\subseteq V_{r-1}V_{s-r+1}+V_{r-1}V_{1}\subseteq(V_{r-1}V_{s-r+1})^{S}+(V_{r-1}V_{1})^{S}=V_{s}+V_{r}$. But then $V_{2s-r}=(V_{r-1}V_{2s-2r+1})^{S}\subseteq V_{s}+V_{r}$, as required.

Suppose now that $t\geq 1$ is such that the result holds. Then $V_{s+(t+1)(s-r)}=(V_{s+t(s-r)}V_{s-r})^{S}\subseteq((V_{s}+V_{r})V_{s-r})^{S}=V_{2s-r}+V_{s}\subseteq V_{r}+V_{s}+V_{s}=V_{r}+V_{s}$. The result follows now by induction. ∎

For any increasing sequence $I:i_{1}<i_{2}<\cdots$ of positive integers, we shall refer to $i_{2}-i_{1}$ as the initial gap of $I$.

Proposition 2.1**.**

For any increasing sequence $I=\{\,i_{j}\,\}_{j\geq 1}$ of positive integers, there exists a set $J$ of size at most $i_{2}-i_{1}+1$ with entries positive integers such that the following hold:

• (i)

$1,2\in J$;

• (ii)

$\sum_{j=1}^{\infty}V_{i_{j}}=\sum_{j\in J}V_{i_{j}}$.

Proof.

The proof of the proposition shall be by induction on the initial gap. By Lemma 2.1, for a sequence with initial gap 1, we may take $J=\{\,i_{1},i_{2}\,\}$ . Suppose now that $l>1$ is an integer for which the result holds for all increasing sequences with initial gap less than $l$, and let $i_{1}<i_{2}<\cdots$ be a sequence with initial gap $i_{2}-i_{1}=l$. If for all $j\geq 3$, $V_{i_{j}}\subseteq V_{i_{1}}+V_{i_{2}}$, then $J=\{\,1,2\,\}$ meets the requirements, so we may suppose that there exists $j\geq 3$ such that $V_{i_{j}}$ is not contained in $V_{i_{1}}+V_{i_{2}}$. By Lemma 2.1, this means that there exists $j\geq 3$ such that $i_{j}\notin\{\,i_{2}+ql\mid q\geq 0\,\}$. Let $r$ be least such that $i_{r}\notin\{\,i_{2}+ql\mid q\geq 0\,\}$, so that there exists $t$ such that $i_{2}+tl<i_{r}<i_{2}+(t+1)l$. Form a sequence $I^{\prime}$ from $I$ by first removing all entries of $I$ up to (but not including) $i_{r}$, then prepend the integer $i_{2}+tl$. Thus $i_{1}^{\prime}$, the first entry of $I^{\prime}$, is $i_{2}+tl$, while for all $j\geq 2$, $i_{j}^{\prime}=i_{r+j-2}$. Note that $i_{2}^{\prime}-i_{1}^{\prime}=i_{r}-(i_{2}+tl)\leq l-1$. By hypothesis, there exists a subset $J^{\prime}$ of size at most $i_{2}^{\prime}-i_{1}^{\prime}+1\leq l=i_{2}-i_{1}$ that contains 1 and 2 and is such that $\sum_{j=1}^{\infty}V_{i_{j}^{\prime}}=\sum_{j\in J^{\prime}}V_{i_{j}^{\prime}}$. Set

[TABLE]

Then $|J|=|J^{\prime}|+1\leq i_{2}-i_{1}+1$ and

[TABLE]

and by Lemma 2.1, $V_{i_{2}+tl}\subseteq V_{i_{2}}+V_{i_{2}}$, so

[TABLE]

Finally, the choice of $r$ implies that

[TABLE]

This completes the proof of the inductive step. ∎

We remark that in Proposition 2.1, it is possible to improve the bound from $i_{2}-i_{1}+1$ to $2(\log_{2}(2(i_{2}-i_{1}))$.

In the sections to come, we shall examine some important situations of the kind described above.

3 The $R_{{n}}^{(d)}$ sequence

We shall have need of certain results that first appeared in [1]. For completeness, we include them with proofs where necessary. In this section, $p$ denotes an arbitrary prime, $k$ an arbitrary field of characteristic $p$, and $d$ an arbitrary positive integer.

The proof of the first result is immediate.

Lemma 3.1**.**

Let $d$ be a positive integer. Then

[TABLE]

Corollary 3.1**.**

Let $d$ be any positive integer. Then modulo $S_{1}^{(d)}$,

[TABLE]

Proof.

This is immediate from (2) and (3) of Lemma 3.1. ∎

We remark that Corollary 3.1 implies that for every $u\in S^{(d)}_{1}$ and $v\in k_{0}\langle X\rangle$, $[\,{u},{v}\,]\in S^{(d)}_{1}$. While we shall not have need of this fact, we note that in [4], Shchigolev proves that if the field is infinite, then for any $T$-space $V$, if $v\in V$, then $[\,{v},{u}\,]\in V$ for any $u\in k_{0}\langle X\rangle$.

The next proposition is a strengthened version of Proposition 2.1 of [1].

Proposition 3.1**.**

For any $u,v\in k_{0}\langle X\rangle$,

• (i)

$(S_{1}^{(d)}uv)^{S}\subseteq S_{1}^{(d)}+(S_{1}^{(d)}u)^{S}+(S_{1}^{(d)}v)^{S}$; and

• (ii)

$(uvS_{1}^{(d)})^{S}\subseteq S_{1}^{(d)}+(uS_{1}^{(d)})^{S}+(vS_{1}^{(d)})^{S}$.

Proof.

We shall prove the first statement; the proof of the second is similar and will be omitted. By (1) of Lemma 3.1,

[TABLE]

and by (2) of Lemma 3.1, $S^{(d+1)}(x_{1},\ldots,x_{d+1})-S^{(d)}(x_{1},\ldots,x_{d})x_{d+1}\in S_{1}^{(d)}$. Let $v\in k_{0}\langle X\rangle$. Then

[TABLE]

Now for each $i=2,\ldots,d$, we use two applications of Corollary 3.1 to obtain

[TABLE]

Thus

[TABLE]

Thus for $u\in k_{0}\langle X\rangle$, we obtain $S^{(d)}(x_{2},\ldots,x_{d+1})uv\in(S_{1}^{(d)}u)^{S}+(S_{1}^{(d)}v)^{S}+S_{1}^{(d)}$, and so

[TABLE]

as required. ∎

Corollary 3.2**.**

Let $d$ be any positive integer. Then the sequence $S_{n}^{(d)}$, $n\geq 1$, satisfies

• (i)

For all $m,n\geq 1$, $(S_{m}^{(d)}S_{n}^{(d)})^{S}=S_{m+n}^{(d)}$;

• (ii)

For all $m\geq 1$, $S_{2m+1}^{(d)}\subseteq S_{m+1}^{(d)}+S_{1}^{(d)}$.

Proof.

The first statement follows immediately from Definition 1.1 by an elementary induction argument. For the second statement, let $m\geq 1$. Then by Proposition 3.1, for any $u,v\in S_{m}^{(d)}$, $(S_{1}^{(d)}uv)^{S}\subseteq S_{1}^{(d)}+(S_{1}^{(d)}u)^{S}+(S_{1}^{(d)}v)^{S}$, which implies that $(S_{1}^{(d)}S_{m}^{(d)}S_{m}^{(d)})^{S}\subseteq S_{1}^{(d)}+(S_{1}^{(d)}S_{m}^{(d)})^{S}$. By (i), this yields $S_{2m+1}^{(d)}\subseteq S_{1}^{(d)}+S_{m+1}^{(d)}$, as required. ∎

Theorem 3.1**.**

Let $I$ denote any increasing sequence of positive integers with initial gap $g$. Then $R_{{\infty,I}}^{(d)}$ is finitely based, with a $T$-space basis of size at most $g+1$.

Proof.

Denote the entries of $I$ in increasing order by $i_{j}$, $j\geq 1$. By Corollary 3.2 and Proposition 2.1, there exists a set $J$ of positive integers with $|J|\leq i_{2}-i_{1}+1$ and $R_{{\infty,I}}^{(d)}=R_{{n,I}}^{(d)}=\sum_{j\in J}S_{i_{j}}^{(d)}$. Since for each $i$, the $T$-space $S_{i}^{(d)}$ has a basis consisting of a single element, the result follows. ∎

4 The $L_{n}$ sequence

We shall make use of the following well known result. An element $u\in k_{0}\langle X\rangle$ is said to be essential if $u$ is a linear combination of monomials with the property that each variable that appears in any monomial appears in every monomial.

Lemma 4.1**.**

Let $V$ be a $T$-space and let $f\in V$. If $f=\sum f_{i}$ denotes the decomposition of $f$ into its essential components, then $f_{i}\in V$ for every $i$.

Proof.

We induct on the number of essential components, with obvious base case. Suppose that $n>1$ is an integer such that if $f\in V$ has fewer than $n$ essential components, then each belongs to $V$, and let $f\in V$ have $n$ essential components. Since $n>1$, there is a variable $x$ that appears in some but not all essential components of $f$. Let $z_{x}$ and $f_{x}$ denote the sum of the essential components of $f$ in which $x$ appears, respectively, does not appear. Then evaluate at $x=0$ to obtain that $f_{x}=f\mkern-5.0mu\mid_{{}_{x=0}}\in V$, and thus $z_{x}=f-f_{x}\in V$ as well. By hypothesis, each essential component of $f_{x}$ and of $z_{x}$ belongs to $V$, and thus every essential component of $f$ belongs to $V$, as required. ∎

Corollary 4.1**.**

$S_{1}^{(p)}\subseteq H_{1}$.

Proof.

$S^{(p)}$ is one of the essential components of $(x_{1}+x_{2}+\cdots+x_{p})^{p}$, and since $(x_{1}+x_{2}+\cdots+x_{p})^{p}\in H_{1}$, it follows from Lemma 4.1 that $S^{(p)}\in H_{1}$. Thus $S_{1}^{(p)}\subseteq H_{1}$. ∎

Corollary 4.2**.**

For every $m\geq 1$, $S_{m}^{(p)}\subseteq H_{m}$.

Proof.

The proof is an elementary induction, with Corollary 4.1 providing the base case. ∎

Corollary 4.3**.**

For any $u\in H_{1}$ and any $v\in k_{0}\langle X\rangle$, $[\,{u},{v}\,]\in H_{1}$.

Proof.

It suffices to observe that

[TABLE]

which belongs to $H_{1}$ by virtue of Corollary 4.1. ∎

We remark again that in [3], Shchigolev proves that if $k$ is infinite, then every $T$-space in $k_{0}\langle X\rangle$ is closed under commutator in the sense of Corollary 4.3. Since we have not required that $k$ be infinite, we have provided this closure result (see also Lemma 4.4 below).

Lemma 4.2**.**

For any $m,n\geq 1$, $(H_{m}H_{n})^{S}=H_{m+n}$.

Proof.

The proof is by an elementary induction on $n$, with Definition 1.2 providing the base case. ∎

Lemma 4.3**.**

For any $m\geq 1$, $(S_{1}^{(p)}H_{2m})^{S}\subseteq H_{1}+H_{m+1}$ and $(H_{2m}S_{1}^{(p)})^{S}\subseteq H_{1}+H_{m+1}$.

Proof.

By Proposition 3.1 (i), for any $u,v\in H_{m}$, we have $S_{1}^{(p)}uv\subseteq S_{1}^{(p)}+(S_{1}^{(p)}u)^{S}+(S_{1}^{(p)}v)^{S}$. By Corollary 4.2, this gives $S_{1}^{(p)}H_{m}H_{m}\subseteq H_{1}+(H_{1}H_{m})^{S}$, and then from Lemma 4.2, we obtain $S_{1}^{(p)}H_{2m}\subseteq H_{1}+H_{m+1}$. The proof of the second part is similar. ∎

Lemma 4.4**.**

Let $m\geq 1$. For every $u\in H_{m}$ and $v\in k_{0}\langle X\rangle$, $[\,{u},{v}\,]\in H_{m}$.

Proof.

The proof is by induction on $m$, with Corollary 4.3 providing the base case. Suppose that $m\geq 1$ is such that the result holds. It suffices to prove that for any $v\in k_{0}\langle X\rangle$, $[\,{x_{1}^{p}x_{2}^{p}\cdots x_{m}^{p}x_{m+1}^{p}},{v}\,]\in H_{m+1}$. We have

[TABLE]

By hypothesis, $[\,{x_{1}^{p}x_{2}^{p}\cdots x_{m}^{p}},{v}\,]\in H_{m}$, while $x_{m+1}^{p}\in H_{1}$ and thus by Corollary 4.3, $[\,{x_{m+1}^{p}},{v}\,]\in H_{1}$ as well. Now by definition, $[\,{x_{1}^{p}x_{2}^{p}\cdots x_{m}^{p}},{v}\,]x_{m+1}^{p}\in H_{m+1}$ and $x_{1}^{p}x_{2}^{p}\cdots x_{m}^{p}[\,{x_{m+1}^{p}},{v}\,]\in H_{m+1}$, which completes the proof of the inductive step. ∎

Lemma 4.5**.**

Let $m\geq 1$. Then $H_{i}S^{(p)}H_{2m-i}\subseteq H_{1}+H_{m+1}$ for all $i$ with $1\leq i\leq 2m-1$.

Proof.

Let $m\geq 1$. We consider two cases: $2m-i\geq m$ and $2m-i<m$. Suppose that $2m-i\geq m$, and let $u\in H_{i}$, $w\in H_{m-1}$ and $z\in H_{m-i+1}$. Then $uS^{(p)}wz=([\,{u},{S^{(p)}w}\,]+S^{(p)}wu)z=[\,{u},{S^{(p)}w}\,]z+S^{(p)}wuz$. Since $u\in H_{i}$, it follows from Lemma 4.4 that $[\,{u},{S^{(p)}w}\,]\in H_{i}$. But then by Lemma 4.2, $[\,{u},{S^{(p)}w}\,]z\in H_{i+m-i+1}=H_{m+1}$. As well, by Corollary 4.1 and Lemma 4.2, $S^{(p)}wuz\in S_{1}^{(p)}H_{m-1+i+m-i+1}=S_{1}^{(p)}H_{2m}$, and by Lemma 4.3, $S_{1}^{(p)}H_{2m}\subseteq H_{1}+H_{m+1}$. Thus $uS^{(p)}wz\in H_{1}+H_{m+1}$. This proves that $H_{i}S^{(p)}H_{m-1}H_{m-i+1}\subseteq H_{1}+H_{m+1}$, and so by Lemma 4.2, $H_{i}S^{(p)}H_{2m-i}=H_{i}S^{(p)}(H_{m-1}H_{m-i+1})^{S}\subseteq H_{1}+H_{m+1}$. The argument for the case when $2m-i<m$ is similar and is therefore omitted. ∎

Proposition 4.1**.**

Let $p>2$. Then for every $m\geq 1$, $H_{2m+1}\subseteq H_{1}+H_{m+1}$.

Proof.

First, consider the expansion of $(x+y)^{p}$ for any $x,y\in k_{0}\langle X\rangle$. It will be convenient to introduce the following notation. Let $J_{p}=\{\,1,2,\ldots,p\,\}$. For any $J\subseteq J_{p}$, let $P_{J}=\prod_{i=1}^{p}z_{i}$, where for each $i$, $z_{i}=x$ if $i\in J$, otherwise $z_{i}=y$. As well, for each $i$ with $1\leq i\leq p-1$, we shall let $S^{(p)}(x,y;i)=S^{(p)}(\underbrace{x,x,\ldots,x}_{i},\underbrace{y,y,\ldots,y}_{p-i})$. Observe that $S^{(p)}(x,y;i)=i!(p-i)!\sum_{\begin{subarray}{c}J\subseteq J_{p}\\ |J|=i\end{subarray}}P_{J}$. We have

[TABLE]

Let $u=\sum_{i=1}^{p-1}\frac{1}{i!(p-i)!}S^{(p)}(x,y;i)$, so that $(x+y)^{p}=x^{p}+y^{p}+u$, and note that $u\in S_{1}^{(p)}$. Then

[TABLE]

Since $(x+y)^{2p}$, $x^{2p}$, $y^{2p}$, and, by Lemma 4.4, $[\,{y^{p}},{x^{p}}\,]$ all belong to $H_{1}$, it follows (making use of Corollary 4.2 where necessary) that $2x^{p}y^{p}\in H_{1}+H_{1}S_{1}^{(p)}+S_{1}^{(p)}H_{1}$.

Consequently, for any $m\geq 1$,

[TABLE]

By Corollary 4.1, Lemma 4.2, and Lemma 4.5, $H_{1}(H_{1}+H_{1}S_{1}^{(p)}+S_{1}^{(p)}H_{1})^{m}\subseteq H_{1}+H_{m+1}$, and since $p>2$, it follows that $\prod_{i=1}^{2m+1}x_{i}^{p}\in H_{1}+H_{m+1}$. Thus $H_{2m+1}\subseteq H_{1}+H_{m+1}$, as required. ∎

Theorem 4.1** (Shchigolev’s conjecture).**

Let $p>2$ be a prime and $k$ a field of characteristic $p$. For any increasing sequence $I=\{\,i_{j}\,\}_{j\geq 1}$, $L_{\infty,I}$ is a finitely based $T$-space of $k_{0}\langle X\rangle$, with a $T$-space basis of size at most $i_{2}-i_{1}+1$.

Proof.

By Lemma 4.2 and Proposition 4.1, the sequence $H_{n}$ of $T$-spaces of $k_{0}\langle X\rangle$ meets the requirements of Section 2. Thus by Proposition 2.1, for any increasing sequence $I=\{\,i_{j}\,\}_{j\geq 1}$ of positive integers, there exists a set $J$ of positive integers such that $|J|\leq i_{2}-i_{1}+1$ and $L_{\infty,I}=\sum_{j=1}^{\infty}H_{i_{j}}=\sum_{j\in J}H_{i_{j}}$. Since for each $i$, $H_{i}$ has $T$-space basis $\{\,x_{1}^{p}x_{2}^{p}\cdots x_{i}^{p}\,\}$, it follows that $L_{\infty,I}$ has a $T$-space basis of size $|J|\leq i_{2}-i_{1}+1$. ∎

Shchigolev’s original result was that for the sequence $I^{+}$ of all positive integers, $L_{\infty,I^{+}}$ is a finitely-based $T$-space, with a $T$-space basis of size at most $p$. It was then shown in [1], a precursor to this work, that $L_{\infty,I^{+}}$ has in fact a $T$-space basis of size at most 2 (the bound of Theorem 4.1, since $i_{1}=1$ and $i_{2}=2$).

It is also interesting to note that the results in this paper apply to finite sequences. Of course, if $I$ is a finite increasing sequence of positive integers, then $L_{\infty,I}$ has a finite $T$-space basis, but by the preceding work, we know that it has a $T$-space basis of size at most $i_{2}-i_{1}+1$.