Failure of finite isolated generation for an irreducible centralizer over a polycyclic group

Problem

For a torsion-free linear group $G$ , is every irreducible system of equations equivalent to a finite system $T$ whose radical is its normal isolated closure?

V_G(S)=V_G(T),\quad |T|<\infty,\qquad \operatorname{Rad}_G(T)\stackrel{?}{=}\sqrt T.

The closure is taken in the ordinary coefficient free product $G*F(X)$ . Equational Noetherianity alone only supplies a finite equivalent system; it does not supply this equality of radicals.

Setup and result

For a group $G$ , write $G[x]=G*\langle x\rangle$ . Each $g\in G$ determines a homomorphism

\operatorname{ev}_g:G[x]\longrightarrow G

that fixes $G$ and sends $x$ to $g$ . For $S\subseteq G[x]$ , put

V_G(S)=\{g\in G:\operatorname{ev}_g(w)=1\text{ for all }w\in S\}, \qquad \operatorname{Rad}_G(S)= \bigcap_{g\in V_G(S)}\ker\operatorname{ev}_g.

A subgroup $N$ of a group $F$ is isolated if

u^m\in N,\quad u\in F,\quad m\geq1 \quad\Longrightarrow\quad u\in N.

We denote by $\sqrt[F]{T}$ the smallest normal isolated subgroup of $F$ containing $T\subseteq F$ , and omit $F$ when the ambient group is clear.

Problem 14.22 of the Kourovka Notebook asks whether every irreducible system of equations over a torsion-free linear group is equivalent to a finite system $T$ satisfying $\operatorname{Rad}_G(T)=\sqrt T$ (3, Problem 14.22). Here equivalence means equality of solution sets, and the normal isolated closure is taken in the ordinary coefficient free product.

We use the commutator convention $[u,v]=u^{-1}v^{-1}uv$ . Set

\tag{4} M=\begin{pmatrix}2&1\\1&1\end{pmatrix},\qquad G=\mathbb Z^2\rtimes_M\mathbb Z,\qquad a=((1,0),0),\qquad t=((0,0),1).

The multiplication in $G$ is $(v,j)(w,k)=(v+M^jw,j+k)$ .

Theorem 1. The group $G$ in (4) is generated by $a,t$ , is torsion-free and polycyclic, and admits an injective homomorphism into $\mathrm{SL}_3(\mathbb Z)$ . Let

F=G[x],\qquad S=\{[x,t]\},\qquad R=\operatorname{Rad}_G(S).

Then $V_G(S)=\langle t\rangle$ is infinite and irreducible, and

\tag{6} \sqrt[F]{T}\subsetneq R \qquad\text{for every finite }T\subseteq R.

In particular, every finite $T\subseteq F$ with $V_G(T)=V_G(S)$ satisfies $\operatorname{Rad}_G(T)\ne\sqrt[F]{T}$ .

The obstruction in (6) is witnessed by homomorphisms that preserve the coefficient group.

Theorem 2. For every finite $T\subseteq R$ , there exist an integer $h>0$ , a torsion-free group $E_h$ , and a homomorphism $\varphi_h:F\to E_h$ whose restriction to $G$ is injective, such that

\varphi_h(w)=1\quad(w\in T),\qquad r_h=[a,x^hax^{-h}]\in R,\qquad \varphi_h(r_h)\ne1.

Consequently $r_h\notin\sqrt[F]{T}$ .

Linearity over $\mathbb Z$ already implies that $G$ is equationally Noetherian (1, Theorem B1). This asserts that systems have finite subsystems with the same solutions. The additional requirement in Problem 14.22 concerns the operation generating their radicals. Over free groups, the finite Nullstellensatz for irreducible varieties follows from the finite-presentation results of Kharlampovich and Myasnikov (2, Theorem 5).

There is also a general positive result relative to the quasivariety generated by $G$ , whose defining laws are all quasi-identities of $G$ with coefficients. Myasnikov and Remeslennikov prove finite presentation in that quasivariety for finitely generated coordinate groups over an equationally Noetherian group (4, Theorem D1). Normal isolated closure permits fewer consequences. The separating groups in Theorem 2 are required to be torsion-free; they are not required to satisfy every quasi-identity of $G$ . Romanovskiy’s Nullstellensatz for divisible rigid soluble groups uses positive formulas and additional inference rules (5). It therefore concerns a different closure operation. We return to the rigidity hypothesis in Remark 6.

The proof proceeds by collecting words into $\mathbb R[z,z^{-1}]\rtimes\mathbb Z^2$ . Specialization at $z=\lambda^n$ , where $\lambda=(3+\sqrt5)/2$ , detects evaluation at $x=t^n$ . The same collection data lift to torsion-free central extensions. Each fixed word has vanishing central coordinate when the interaction distance is sufficiently large, while a commutator at that distance remains nontrivial.

The coefficient group and isolated closure

Lemma 3. Let $F,H$ be groups and let $T\subseteq F$ . The subgroup

\sqrt[F]{T} = \bigcap_{\substack{N\trianglelefteq F,\;T\subseteq N\\ N\text{ isolated}}}N

is normal and isolated. If $H$ is torsion-free and $f:F\to H$ is a homomorphism with $T\subseteq\ker f$ , then $\sqrt[F]{T}\subseteq\ker f$ . If $G$ is torsion-free, then $\operatorname{Rad}_G(S)$ is normal and isolated for every $S\subseteq G[x]$ .

Proof. Intersections preserve normality and the defining implication for isolation. The family in the intersection is nonempty because it contains $F$ , and every member contains $T$ . If $u^m\in\ker f$ with $m\geq1$ , then $f(u)^m=1$ ; torsion-freeness gives $f(u)=1$ . Thus $\ker f$ is normal and isolated. The claimed containment follows from the intersection definition. Apply this observation to each coefficient-fixing evaluation into $G$ to obtain the assertion about radicals. ◻

We next record the spectral facts used throughout the proof. Define

\tag{9} \lambda=\frac{3+\sqrt5}{2},\qquad \ell(u,v)=u+(\lambda-2)v \quad((u,v)\in\mathbb Z^2).

Lemma 4. The number $\lambda$ is irrational, satisfies $\lambda^2-3\lambda+1=0$ , and is greater than $1$ . The additive map $\ell:\mathbb Z^2\to\mathbb R$ is injective, and

\tag{10} \ell(M^jv)=\lambda^j\ell(v) \qquad(j\in\mathbb Z,\ v\in\mathbb Z^2).

Proof. The first assertions follow from (9) and the irrationality of $\sqrt5$ . If $\ell(u,v)=0$ and $v\ne0$ , then $\lambda=2-u/v$ , a contradiction. Hence $v=0$ , and then $u=0$ ; additivity now implies injectivity. Direct substitution, using the quadratic identity for $\lambda$ , gives

\ell(2u+v,u+v)=\lambda\ell(u,v).

Iteration proves (10) for nonnegative $j$ . Since $M$ is invertible and $\lambda\ne0$ , the same identity applied to $M^{-1}v$ proves the negative powers. ◻

Proposition 5. The group $G$ has the algebraic properties asserted in Theorem 1. Moreover,

\tag{12} C_G(t)=\langle t\rangle.

Proof. The inverse matrix

M^{-1}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}

has integer entries, so the action is defined for every integral power. If $(v,j)^m=1$ for $m\geq1$ , projection to the cyclic factor gives $mj=0$ , hence $j=0$ . Then $mv=0$ , whence $v=0$ . Thus $G$ is torsion-free.

Let $A=\mathbb Z^2\times\{0\}$ , and put $b=((0,1),0)$ . The subnormal series

1\ \trianglelefteq\ \langle a\rangle\ \trianglelefteq\ A\ \trianglelefteq\ G

has three infinite cyclic factors. The identity

tat^{-1}=a^2b

shows that $a,t$ generate $b$ , hence $A$ and $G$ .

Regarding $v$ as a column vector, define

\tag{16} \rho(v,j)= \begin{pmatrix}M^j&v\\0&1\end{pmatrix}.

Block multiplication agrees with the multiplication in $G$ . Every matrix in (16) has integral entries and determinant $(\det M)^j=1$ . If $\rho(v,j)$ is the identity, then $v=0$ and $M^j=I$ . Applying $\ell$ to $M^j(1,0)=(1,0)$ gives $\lambda^j=1$ . Since $\lambda>1$ , this forces $j=0$ . Thus $\rho$ is injective. In particular,

\rho(a)=\begin{pmatrix}1&0&1\\0&1&0\\0&0&1\end{pmatrix}, \qquad \rho(t)=\begin{pmatrix}2&1&0\\1&1&0\\0&0&1\end{pmatrix}.

Finally, $(v,j)$ commutes with $t$ precisely when $Mv=v$ . The matrix $M-I$ has determinant $-1$ , so this is equivalent to $v=0$ . Since $t^j=(0,j)$ , equation (12) follows. ◻

Remark 6. The natural abelian normal module in this example satisfies

(M^2-3M+I)v=0\qquad(v\in\mathbb Z^2).

Equivalently, the nonzero Laurent polynomial $t^2-3t+1$ annihilates that module. Torsion-freeness of $G$ therefore does not supply the module torsion-freeness required by the definition of a rigid series in (5). This distinction is separate from the choice of closure operation.

Laurent collection and irreducibility

Let $\mathcal A=\mathbb R[z,z^{-1}]$ , regarded as an additive group, and let $Q=\mathbb Z^2$ . For $q=(j,k)\in Q$ , define

\tag{19} \sigma_q(P)=\lambda^jz^kP\qquad(P\in\mathcal A).

Then $\sigma_0=1$ and $\sigma_q\sigma_r=\sigma_{q+r}$ . Thus $Q$ acts by additive automorphisms on $\mathcal A$ . Write

L=\mathcal A\rtimes_\sigma Q.

We denote its elements by $(P,j,k)$ , so that

\tag{21} (P,j,k)(P',j',k') = \bigl(P+\lambda^jz^kP',\,j+j',\,k+k'\bigr).

By (10), the map

\iota:G\longrightarrow L,\qquad \iota(v,j)=(\ell(v),j,0)

is a homomorphism, where a scalar is identified with its constant Laurent polynomial. It is injective by Lemma 4. The universal property of the free product gives a homomorphism

\tag{23} \chi:F\longrightarrow L,\qquad \chi|_G=\iota,\qquad \chi(x)=(0,0,1).

For $w\in F$ , write $\chi(w)=(P_w,j_w,k_w)$ . No surjectivity assertion about $\chi$ is needed.

Lemma 7. Let $w\in F$ and $n\in\mathbb Z$ . If $\operatorname{ev}_{t^n}(w)=(v_n,m_n)$ , then

\tag{24} \ell(v_n)=P_w(\lambda^n),\qquad m_n=j_w+nk_w.

In particular,

\tag{25} \operatorname{ev}_{t^n}(w)=1 \quad\Longleftrightarrow\quad P_w(\lambda^n)=0\ \text{ and }\ j_w+nk_w=0.

Proof. For a coefficient $g=(v,j)$ , the collection data are $(\ell(v),j,0)$ , and the assertion is immediate. For $x^m$ , $m\in\mathbb Z$ , they are $(0,0,m)$ , while $\operatorname{ev}_{t^n}(x^m)=t^{nm}=(0,nm)$ .

Suppose the assertion holds for $u,w$ . Write their evaluations as $(v_n,m_n)$ and $(v'_n,m'_n)$ . The lattice coordinate of the evaluation of $uw$ is $v_n+M^{m_n}v'_n$ . Therefore

\begin{align*} \ell(v_n+M^{m_n}v'_n) &=P_u(\lambda^n)+\lambda^{j_u+nk_u}P_w(\lambda^n)\\ &=\bigl(P_u+\lambda^{j_u}z^{k_u}P_w\bigr)(\lambda^n). \end{align*}

The cyclic coordinate is $(j_u+j_w)+n(k_u+k_w)$ , as required by (21). Every element of $F$ is a finite product of elements of $G$ and integral powers of $x$ ; their inverses are already included among these factors. Induction on such a product proves (24). Equation (25) follows from injectivity of $\ell$ . ◻

Lemma 8. If $P\in\mathcal A$ is nonzero, then $\{n\in\mathbb Z:P(\lambda^n)=0\}$ is finite. For each $w\in F$ , the set

Z_w=\{n\in\mathbb Z:\operatorname{ev}_{t^n}(w)=1\}

is either finite or all of $\mathbb Z$ . Moreover,

\tag{28} Z_w=\mathbb Z \quad\Longleftrightarrow\quad P_w=0,\quad j_w=0,\quad k_w=0.

Proof. Multiplying $P$ by a suitable power of $z$ gives a nonzero ordinary polynomial. Since $\lambda^n\ne0$ , this multiplication does not change its zeros at the points $\lambda^n$ . These points are pairwise distinct because $\lambda>1$ . A nonzero polynomial has finitely many roots, proving the first assertion.

Apply (25). If $P_w\ne0$ , then $Z_w$ is finite. If $P_w=0$ and $k_w\ne0$ , the equation $j_w+nk_w=0$ has at most one integral solution. If $P_w=0=k_w$ , then $Z_w$ is empty or all of $\mathbb Z$ , according as $j_w\ne0$ or $j_w=0$ . Alternatively, when $Z_w=\mathbb Z$ , evaluation at $n=0,1$ forces $j_w=k_w=0$ , and the first assertion forces $P_w=0$ . The converse in (28) follows directly from (25). ◻

Equip $G$ with the group Zariski topology: the complements of the single-equation sets $V_G(\{w\})$ , $w\in F$ , form an open subbasis. Thus every $V_G(T)$ is closed. An algebraic set is irreducible if it is nonempty and cannot be covered by two proper relatively closed subsets.

Proposition 9. The algebraic set $V_G(S)=\langle t\rangle$ is infinite and irreducible, and

\tag{29} R=\ker\chi.

Proof. The centralizer computation (12) gives $V_G(S)=\{t^n:n\in\mathbb Z\}$ . Projection to the cyclic factor shows that $n\mapsto t^n$ is injective, so this set is infinite.

Give $\mathbb Z$ the cofinite topology. The inverse image of $V_G(\{w\})$ under $n\mapsto t^n$ is $Z_w$ , which is closed in that topology by Lemma 8. Hence the parametrization is continuous. The cofinite topology on an infinite set is irreducible: two nonempty open subsets have finite complements and must intersect. A continuous image of an irreducible space is irreducible, since the inverse images of a cover by two proper relatively closed subsets would give such a cover of the domain. This proves irreducibility of $V_G(S)$ .

Finally, $w\in R$ means $Z_w=\mathbb Z$ . By (28), this is equivalent to $\chi(w)=(0,0,0)=1$ , proving (29). ◻

Torsion-free groups with distant interactions

Write $P_i$ for the coefficient of $z^i$ in $P\in\mathcal A$ . Thus $\operatorname{supp}P$ is a finite subset of $\mathbb Z$ . Let $\mathcal C=\mathbb R^{\mathbb Z}$ , the additive group of all real functions on $\mathbb Z$ . For an integer $h\geq1$ , define

\tag{30} \beta_h:\mathcal A\times\mathcal A\longrightarrow\mathcal C, \qquad \beta_h(P,P')(i)=P_iP'_{i+h}-P'_iP_{i+h}.

The map is biadditive and alternating.

Lemma 10. The operation

\tag{31} (P,c)(P',c') = (P+P',\,c+c'+\beta_h(P,P'))

makes $\mathcal A\times\mathcal C$ a torsion-free group $N_h$ . Its identity is $(0,0)$ , its inverse operation is $(P,c)^{-1}=(-P,-c)$ , and

\begin{align*} (P,c)^m&=(mP,mc) \qquad(m\geq0),\tag{32a}\\ [(P,c),(P',c')]&=(0,2\beta_h(P,P')).\tag{32b} \end{align*}

Proof. Biadditivity gives the cocycle identity

\beta_h(P,P')+\beta_h(P+P',P'') = \beta_h(P',P'')+\beta_h(P,P'+P''),

which is exactly associativity of the central coordinate in (31). The vector coordinate is additive. The identity and inverse assertions follow from $\beta_h(P,0)=0$ and $\beta_h(P,-P)=0$ . Induction gives (32a), because $\beta_h(mP,P)=m\beta_h(P,P)=0$ . Both $\mathcal A$ and $\mathcal C$ are torsion-free additive groups, so (32a) proves torsion-freeness of $N_h$ .

Alternation and biadditivity imply $\beta_h(P',P)=-\beta_h(P,P')$ . Applying (31) to the commutator gives

(-P,-c)(-P',-c')(P,c)(P',c') =\bigl(0,\beta_h(P,P')-\beta_h(P',P)\bigr) =\bigl(0,2\beta_h(P,P')\bigr).

This proves (32b). ◻

For $q=(j,k)\in Q$ , define an additive automorphism of $\mathcal C$ by

\tag{35} (\tau_qc)(i)=\lambda^{2j}c(i-k).

Then $\tau_q\tau_r=\tau_{q+r}$ , with inverse $\tau_{-q}$ . The coefficient form of (19) is

(\sigma_qP)_i=\lambda^jP_{i-k}.

Consequently

\begin{align*} \beta_h(\sigma_qP,\sigma_qP')(i) &=\lambda^{2j} \bigl(P_{i-k}P'_{i-k+h}-P'_{i-k}P_{i-k+h}\bigr)\notag\\ &=\bigl(\tau_q\beta_h(P,P')\bigr)(i). \tag{37} \end{align*}

It follows that

(P,c)\longmapsto(\sigma_qP,\tau_qc)

is an automorphism of $N_h$ , and these automorphisms define a $Q$ -action. Let

\tag{39} E_h=N_h\rtimes Q.

Writing elements as $(P,c,q)$ , its multiplication is

\tag{40} (P,c,q)(P',c',q') = \bigl(P+\sigma_qP',\, c+\tau_qc'+\beta_h(P,\sigma_qP'),\, q+q'\bigr).

Proposition 11. The group $E_h$ is torsion-free. The map

\tag{41} \iota_h:G\longrightarrow E_h,\qquad \iota_h(v,j)=(\ell(v),0,(j,0))

is an injective homomorphism. Setting

\xi_h=(0,0,(0,1))

defines a homomorphism

\tag{43} \varphi_h:F\longrightarrow E_h,\qquad \varphi_h|_G=\iota_h,\qquad \varphi_h(x)=\xi_h.

The homomorphism

\pi_h:E_h\longrightarrow L,\qquad \pi_h(P,c,(j,k))=(P,j,k)

satisfies $\pi_h\varphi_h=\chi$ .

Proof. If $y^m=1$ in $E_h$ , where $m\geq1$ , projection to $Q$ shows that its $Q$ -coordinate is zero. Thus $y\in N_h$ , and Lemma 10 gives $y=1$ .

Two constant Laurent polynomials have zero $\beta_h$ -interaction: their supports are contained in $\{0\}$ , and $h>0$ . Also, $\sigma_{(j,0)}\ell(v)=\lambda^j\ell(v)$ . Equations (40) and (10) now give

\iota_h(v,j)\iota_h(w,k) = \bigl(\ell(v+M^jw),0,(j+k,0)\bigr).

This proves that $\iota_h$ is a homomorphism. Its coordinates recover $j$ and $\ell(v)$ , so injectivity follows from that of $\ell$ . Equation (43) follows from the free product universal property.

Forgetting the central coordinate in (40) gives (21), so $\pi_h$ is a homomorphism. The maps $\pi_h\varphi_h$ and $\chi$ agree on the coefficient group and on $x$ , and therefore agree on $F$ . ◻

Finite families and the separating commutator

For $w\in F$ , write

\tag{46} \varphi_h(w)=(P_w,c_h(w),(j_w,k_w)).

The Laurent and exponent coordinates here are independent of $h$ by Proposition 11.

Lemma 12. For any $P,P'\in\mathcal A$ , there is an integer $B\geq0$ such that

h>2B\quad\Longrightarrow\quad \beta_h(P,P')=0.

Proof. Choose $B$ with $\operatorname{supp}P\cup\operatorname{supp}P'\subseteq[-B,B]$ . If $P_iP'_{i+h}\ne0$ , then both $i$ and $i+h$ lie in $[-B,B]$ , which implies $h\leq2B$ . Thus this product is zero when $h>2B$ . The same argument applies to $P'_iP_{i+h}$ , so (30) vanishes at every $i$ . ◻

Lemma 13. For every $w\in F$ , there is an integer $N_w\geq0$ such that

c_h(w)=0\qquad\text{for every integer }h>N_w.

Proof. The assertion holds for every coefficient $g\in G$ by (41). It also holds for every $x^m$ , $m\in\mathbb Z$ , since

\varphi_h(x^m)=(0,0,(0,m)).

Suppose it holds for $u,w$ . Equations (40) and (46) give

\tag{50} c_h(uw)= c_h(u)+\tau_{(j_u,k_u)}c_h(w) +\beta_h\bigl(P_u,\sigma_{(j_u,k_u)}P_w\bigr).

The first two terms vanish for all sufficiently large $h$ . The Laurent polynomials in the last term do not depend on $h$ , so Lemma 12 makes that term zero for all sufficiently large $h$ as well. Take the maximum of these three bounds. Induction on a product of coefficient elements and integral powers of $x$ completes the proof. ◻

Proposition 14. For every finite $T\subseteq R$ , there is an integer $N\geq0$ such that

\varphi_h(w)=1 \qquad(h>N,\ w\in T).

Proof. If $w\in R=\ker\chi$ , then $P_w=0$ and $j_w=k_w=0$ . Thus (46) reduces to $\varphi_h(w)=(0,c_h(w),(0,0))$ . Lemma 13 makes this element the identity for all sufficiently large $h$ . Take the maximum of the finitely many bounds for $w\in T$ , or take $N=0$ if $T$ is empty. ◻

Lemma 15. For every integer $h>0$ , the word $r_h=[a,x^hax^{-h}]$ belongs to $R$ , while $\varphi_h(r_h)\ne1$ .

Proof. Since $\ell(1,0)=1$ , equations (41) and (43) give

\varphi_h(a)=(1,0,(0,0)),\qquad \varphi_h(x^hax^{-h})=(z^h,0,(0,0)).

For the second identity, conjugation by $\xi_h^h=(0,0,(0,h))$ acts on $N_h$ by $(\sigma_{(0,h)},\tau_{(0,h)})$ , sending $(1,0)$ to $(z^h,0)$ . Both images have zero $Q$ -coordinate, so (32b) yields

\tag{53} \varphi_h(r_h)= \bigl(0,\,2\beta_h(1,z^h),\,(0,0)\bigr).

The constant polynomial $1$ has coefficient $1$ at $0$ and coefficient $0$ at $h$ ; the polynomial $z^h$ has coefficient $1$ at $h$ and coefficient $0$ at $0$ , because $h>0$ . Hence $\beta_h(1,z^h)(0)=1$ , and the central coordinate of (53) at $0$ is $2\ne0$ . This proves $\varphi_h(r_h)\ne1$ .

Applying $\pi_h$ to (53) gives the identity of $L$ . Since $\pi_h\varphi_h=\chi$ , we obtain $\chi(r_h)=1$ , and therefore $r_h\in R$ by (29). ◻

Proof of Theorems 1 and 2. The algebraic properties of $G$ are proved in Proposition 5, and the infinitude and irreducibility of $V_G(S)$ in Proposition 9.

Let $T\subseteq R$ be finite. Choose $N$ as in Proposition 14, and take $h=N+1$ . Then $\varphi_h$ kills $T$ . Its target $E_h$ is torsion-free and its coefficient restriction is injective by Proposition 11. Lemma 15 gives $r_h\in R$ with $\varphi_h(r_h)\ne1$ . By Lemma 3,

\sqrt[F]{T}\subseteq\ker\varphi_h, \qquad r_h\notin\sqrt[F]{T}.

The same lemma makes $R$ normal and isolated. Since $T\subseteq R$ , we have $\sqrt[F]{T}\subseteq R$ , and the omitted element $r_h$ makes the containment strict.

Finally, suppose that $T\subseteq F$ is finite and $V_G(T)=V_G(S)$ . Every equation in $T$ vanishes on $V_G(S)$ , so $T\subseteq R$ . Equality of solution sets also gives $\operatorname{Rad}_G(T)=R$ . Thus (6) implies $\sqrt[F]{T}\subsetneq\operatorname{Rad}_G(T)$ , as required. ◻

References

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