# On a variety of right-symmetric algebras

Nurlan Ismailov  Astana IT University, Mangilik El avenue, 55/11, Business
center EXPO, block C1, Astana, 010000, Kazakhstan<EMAIL_ADDRESS>and
Ualbai Umirbaev  Department of Mathematics, Wayne State University, Detroit,
MI 48202, USA; and Institute of Mathematics and Mathematical Modeling, Almaty,
050010, Kazakhstan<EMAIL_ADDRESS>
###### Abstract.

We construct a finite-dimensional metabelian right-symmetric algebra over an
arbitrary field that does not have a finite basis of identities.

Mathematics Subject Classification (2020): Primary 17D25, 17A50. Secondary
15A24, 16R10.

Key words: right-symmetric algebras, identities, Specht property.

##  1\.  Inroduction.

We say that a variety of algebras has the Specht property or is Spechtian if
any of its subvarieties has a finite basis of identities. In other words, a
variety of algebras is Spechtian if the set of all its subvarieties satisfies
the descending chain condition with respect to inclusion. In 1950 Specht [31]
formulated a problem on the Specht property for the variety of all associative
algebras over a field of characteristic zero.

Specialists extended the study of this problem for any varieties of algebras
over fields of any characteristic. In 1970 Vaughan-Lee [36] constructed an
example of a finite-dimensional Lie algebra over a field of characteristic
$p=2$ that does not have a finite basis of identities. In 1974 Drensky [8]
extended this result to fields of any positive characteristic $p>0$. In 1978
Medvedev [25] showed that varieties of metabelian Malcev, Jordan, alternative,
and $(-1,1)$ algebras are Spechtian. In 1984 Umirbaev [33] proved that the
variety of metabelian binary Lie algebras over a field of characteristic $\neq
3$ has the Specht property. In 1980 Medvedev [26] also constructed an example
of a variety of solvable alternative algebras over a field of characteristic
$2$ with an infinite basis of identities. In 1985 Umirbaev [34] proved that
the varieties of solvable alternative algebras over a field of characteristic
$\neq 2,3$ have the Specht property. Pchelintsev [27] constructed an almost
Spechtian variety of alternative algebras over a field of characteristic 3.
The Specht property of so-called bicommutative algebras is proven in [9].

In 1976 Belkin [1] proved that the variety of metabelian right-alternative
algebras does not have the Specht property. In 1978 L’vov [24] constructed a
six-dimensional nonassociative algebra over an arbitrary field satisfying the
identity $x(yz)=0$ with an infinite basis of identities. In 1986 Isaev [15]
adapted L’vov’s methods for right-alternative algebras and constructed a
finite-dimensional metabelian right-alternative algebra over an arbitrary
field with an infinite basis of identities. In 2008 Kuz’min [22] gave a
sufficient condition for the varieties of metabelian right-alternative
algebras over a field of characteristic $\neq 2$ to be Spechtian.

In 1988 Kemer [16, 17] positively solved the famous Specht problem [31] and
proved that every variety of associative algebras over a field of
characteristic zero has a finite basis of identities. Later the Specht problem
was negatively solved for the variety of associative algebras over fields of
positive characteristic $p>0$ [2, 13, 28]. It is also known that the varieties
of Lie algebras generated by a finite-dimensional algebra over a field of
characteristic zero have the Specht property [14, 18]. Despite the efforts of
many specialists in this field, the question of whether the variety of Lie
algebras over a field of characteristic zero has the Specht property remains
open.

This paper is devoted to the study of the Specht property for the variety of
right-symmetric algebras. Recall that an algebra $A$ over a field $\mathbb{F}$
is called right-symmetric if it satisfies the identity

(1) $(a,b,c)=(a,c,b),$

where $(a,b,c)=(ab)c-a(bc)$ is the associator of $a,b,c\in A$.

Right-symmetric algebras are Lie admissible, that is, any right-symmetric
algebra with respect to the commutator $[x,y]=xy-yx$ is a Lie algebra. Very
often right-symmetric (or left-symmetric) algebras are called pre-Lie algebras
and play an important role in the theory of operads [23]. Right-symmetric
algebras arise in many different areas of mathematics and physics [3].

In 1994 Segal [30] constructed a basis of free right-symmetric algebras.
Chapoton and Livernet [5] and, independently, Löfwall and Dzhumadil’daev [11]
gave other bases of free right-symmetric algebras in terms of rooted trees.
The identities of right-symmetric algebras were studied by Filippov [12], and
he proved that any right-nil right-symmetric algebra over a field of
characteristic zero is right nilpotent. An analogue of the PBW basis Theorem
for the universal (multiplicative) enveloping algebra of a right-symmetric
algebra was given in [19]. The Freiheitssatz and the decidability of the word
problem for one-relator right-symmetric algebras were proven in [20].
Recently, Dotsenko and Umirbaev [7] determined that the variety of right-
symmetric algebras over a field of characteristic zero is Nielson-Schreier,
that is, every subalgebra of a free right-symmetric algebra is free.

A right-symmetric algebra with an additional identity

$a(bc)=b(ac)$

is called a Novikov algebra. The class of Novikov algebras is an important and
well-studied subclass of right-symmetric algebras. Recently there was great
progress in the study of identities, solvability, and nilpotency [38, 12, 10,
29, 35, 32]. In 2022 Dotsenko, Ismailov, and Umirbaev [6] proved that (a)
every Novikov algebra satisfying a nontrivial polynomial identity over a field
of characteristic zero is right-associator nilpotent and (b) the variety of
Novikov algebras over a field of characteristic zero has the Specht property.

In this paper, we continue the study of the identities of right-symmetric
algebras. Namely, using the constructions and methods of L’vov [24] and Isaev
[15], we construct a finite-dimensional metabelian right-symmetric algebra
over an arbitrary field that does not have a finite basis of identities. In
fact, our algebra belongs to the variety of algebras $\mathcal{R}$ defined by
the identities

(2) $[[a,b],c]=0,$ (3) $(ab)a=0,$

and

(4) $(ab)(cd)=0.$

We determine some identities and operator identities of the variety
$\mathcal{R}$ in Section 2. In Section 3, a series of algebras $P_{n}$ of this
variety is constructed. A linear basis of free algebras of the variety
$\mathcal{R}$ is constructed in Section 4. Section 5 is devoted to the study
of the relationships between the polynomial identities and the operator
identities of the algebras $P_{n}$. The main result of the paper is given in
Section 6 and says that the algebra $P_{2}$ does not have a finite basis of
identities.

##  2\.  A variety of right-symmetric algebras

Let $\mathbb{F}$ be an arbitrary fixed field. In what follows, all vector
spaces are considered over $\mathbb{F}$. As above, $\mathcal{R}$ denotes the
variety of algebras defined by the identities (2), (3), and (4).

######  Lemma 2.1.

Every algebra of the variety $\mathcal{R}$ is right-symmetric and right
nilpotent of index $4$.

###### Proof.

The linearization of (3) gives

(5) $(ab)c+(cb)a=0.$

This identity and (4) imply that

(6) $((ab)c)d=-(dc)(ab)=0.$

Using (3) and(2) one can also get

$(a,b,c)-(a,c,b)=(ab)c-a(bc)-(ac)b+a(cb)$ $=-(cb)a-a(bc)+(bc)a+a(cb)$
$=[b,c]a-a[b,c]=[[b,c],a]=0,$

i.e., $\mathcal{R}$ is a variety of right-symmetric algebras. ∎

Let $A$ be an arbitrary algebra of the variety $\mathcal{R}$. Recall that for
any $x\in A$ the operators of right multiplication $R_{x}$ and left
multiplication $L_{x}$ on $A$ are defined by

$aR_{x}=ax\quad\text{and}\quad aL_{x}=xa,$

respectively. Set also $V_{x,y}=L_{x}R_{y}$.

######  Lemma 2.2.

(7) $V_{x,x}=0,\quad\quad V_{x,y}=-V_{y,x}.$ (8)
$xR_{y}L_{z}L_{t}=yV_{x,z}L_{t}-xR_{y}V_{t,z}.$ (9)
$xR_{y}L_{z}=xV_{z,y}+yR_{x}L_{z}-yV_{z,x}.$ (10)
$xR_{y}V_{z,t}=yR_{x}V_{z,t}.$ (11) $V_{x,y}R_{z}=0.$ (12)
$V_{x,y}(L_{z}L_{t}+V_{t,z})=0.$

###### Proof.

The identities (3) and (5) immediately imply (7). By (1) and (4) we get

$xR_{y}L_{z}L_{t}=t(z(xy))$
$=(tz)(xy)+t((xy)z)-(t(xy))z=yV_{x,z}L_{t}-xR_{y}V_{t,z}.$

From the identity (2) follows (9).

Then (1) and (6) give that

$xR_{y}V_{z,t}=(z(xy))t$ $=((zx)y)t+(z(yx))t-((zy)x)t=yR_{x}V_{z,t}.$

By (6) we obtain $tV_{x,z}R_{t}=((xt)z)t=0$, and, therefore, $V_{x,y}R_{z}=0$.
Set $v=uV_{x,y}$. Then (6), (1), and (4) imply that

$vL_{z}L_{t}=t(zv)-t(vz)=(tz)v-(tv)z=-vV_{t,z}.$

∎

##  3\.  Algebras $P_{n}$

For each natural $n$ we define the algebra $P_{n}$ with a linear basis

$a_{ij},\,b_{ij},\,c_{i},\,d_{ij},\,e_{ij},$

where $i,j\in\\{1,2,\ldots,n\\}$, and with the product defined by

$a_{ij}c_{i}=d_{ij},\quad b_{ij}c_{i}=e_{ij},$
$a_{ij}e_{ij}=e_{ij}a_{ij}=-b_{ij}d_{ij}=-d_{ij}b_{ij}=c_{j},$

where all zero products are omitted.

Set

$A_{n}=\mathrm{Span}\\{a_{ij},\,b_{ij}\,|\,1\leq i,j\leq n\\}$

and

$D_{n}=\mathrm{Span}\\{c_{i},\,d_{ij},\,e_{ij}\,|\,1\leq i,j\leq n\\},$

where $\mathrm{Span}\,X$ denotes the linear span of $X$. Then $A_{n}$ is a
subalgebra of $P_{n}$ and $D_{n}$ is an ideal of $P_{n}$. Moreover, $P_{n}$ is
a direct sum of the vector spaces $A_{n}$ and $D_{n}$. Set also

$C_{n}=\mathrm{Span}\\{c_{i}\,|\,1\leq i\leq n\\},\ \
\overline{C}_{n}=\mathrm{Span}\\{d_{ij},\,e_{ij}\,|\,1\leq i,j\leq n\\}.$

Then

(13) $P_{n}^{2}=D_{n},\quad A_{n}^{2}=D_{n}^{2}=0,\quad
D_{n}=C_{n}\oplus\overline{C}_{n},$ $D_{n}P_{n}=C_{n},\quad
P_{n}A_{n}=C_{n},\quad P_{n}C_{n}=\overline{C}_{n},\quad C_{n}P_{n}=0,\quad
P_{n}\overline{C}_{n}=C_{n}.$

######  Lemma 3.1.

The algebra $P_{n}$ belongs to the variety $\mathcal{R}$.

###### Proof.

Obviously the space of commutators $[P_{n},P_{n}]$ coincides with
$\overline{C}_{n}$, which is in the center of $P_{n}$, i.e., (2) holds.

In order to verify the identity (3), it is sufficient to check the identities
(3) and (5) for all elements of the basis of $P_{n}$. Let us begin with (3).
Since $A_{n}^{2}=D_{n}^{2}=(D_{n}A_{n})D_{n}=0$, we may assume that $a\in
A_{n}$ and $b\in D_{n}$. Consider all nonzero products of the space
$A_{n}D_{n}$. If $a=a_{ij}$ and $b=c_{i}$, then

$(a_{ij}c_{i})a_{ij}=d_{ij}a_{ij}=0.$

If $a=a_{ij}$ and $b=e_{ij}$, then

$(a_{ij}e_{ij})a_{ij}=c_{j}a_{ij}=0.$

The other cases can be verified similarly.

Now let’s verify (5). Since $(D_{n}P_{n})P_{n}=0$, the product $(ab)c$ is
nonzero only if $a=a_{ij},\,b=c_{i},\,c=b_{ij}$ or
$a=b_{ij},\,b=c_{i},\,c=a_{ij}$. Thus,

$(ab)c+(cb)a=-c_{j}+c_{j}=0.$

The identity (4) follows from the relations $P^{2}_{n}=D_{n}$ and
$D^{2}_{n}=0$. ∎

######  Lemma 3.2.

For all $x,y\in P_{n}$, $d\in D_{n}$ we have

$(A_{n}+\overline{C}_{n})V_{x,y}=0,\quad V_{d,y}=V_{y,d}=0.$

###### Proof.

The relations (13) give that $(P_{n}A_{n})P_{n}\subseteq C_{n}P_{n}=0$ and
$(P_{n}\overline{C}_{n})P_{n}\subseteq C_{n}P_{n}=0$, i.e., the first equality
of the lemma holds. Similarly, by noting that $(D_{n}P_{n})P_{n}\subseteq
C_{n}P_{n}=0$ and $(P_{n}P_{n})D_{n}\subseteq D_{n}D_{n}=0$, we can deduce the
second equality of the lemma. ∎

Denote by $\mathrm{Ann}_{l}P_{n}$ the space of left annihilators of $P_{n}$.

######  Lemma 3.3.

$\mathrm{Ann}_{l}P_{n}=C_{n}$.

###### Proof.

Assume that $x\in(A_{n}+\overline{C}_{n}+C_{n})\cap\mathrm{Ann}_{l}P_{n}$ and
express it as

$x=\sum_{i,j}(\alpha_{ij}a_{ij}+\beta_{ij}b_{ij}+\gamma_{ij}d_{ij}+\delta_{ij}e_{ij}+\epsilon_{i}c_{i}),$

where
$\alpha_{ij},\beta_{ij},\gamma_{ij},\delta_{ij},\epsilon_{i}\in\mathbb{F}$.
Then we have

$xc_{i}=\sum_{j}(\alpha_{ij}d_{ij}+\beta_{ij}e_{ij}),\quad
xa_{ij}=\delta_{ij}c_{j},\quad xb_{ij}=-\gamma_{ij}c_{j}.$

From these equations, it can be deduced that
$\alpha_{ij}=\beta_{ij}=\gamma_{ij}=\delta_{ij}=0$. Therefore, we can conclude
that $x=\sum_{i}\epsilon_{i}c_{i}$. Consequently,
$\mathrm{Ann}_{l}P_{n}=C_{n}$. ∎

##  4\.  Structure of free algebras of $\mathcal{R}$

Let $F(X)$ be the free algebra of the variety $\mathcal{R}$ generated by an
infinite countable set $X=\\{x_{1},x_{2},\ldots,x_{n},\ldots\\}$.

######  Proposition 4.1.

The set of elements $\mathcal{B}$ of $F(X)$ of the forms

$x_{i}$, $x_{i}\hat{R}_{x_{j}}L_{x_{s}}$,
$x_{i}\hat{R}_{x_{j}}V_{x_{p_{1}},x_{q_{1}}}\cdots
V_{x_{p_{k}},x_{q_{k}}}\hat{L}_{x_{s}},$

where $i<j$ and $p_{r}<q_{r}$ for all $r=1,2,\ldots,k$, $k\geq 1$, and
$\hat{T}_{x}$ denotes that the operator $T_{x}$ might not occur, is a basis of
$F(X)$.

###### Proof.

In order to show that $\mathcal{B}$ linearly spans $F(X)$ it is sufficient to
verify that, for any $v\in\mathcal{B}$, the elements $vR_{x_{i}}$ and
$vL_{x_{i}}$ belong to the linear span of $\mathcal{B}$. This is easy to do
using the identities (1), (4), and Lemma 2.2. For example, let

$v=x_{i}\hat{R}_{x_{j}}V_{x_{p_{1}},x_{q_{1}}}\cdots
V_{x_{p_{k}},x_{q_{k}}}L_{x_{s}}.$

Then

$vR_{x_{r}}=x_{i}\hat{R}_{x_{j}}V_{x_{p_{1}},x_{q_{1}}}\cdots
V_{x_{p_{k}},x_{q_{k}}}L_{x_{s}}R_{x_{r}}=x_{i}\hat{R}_{x_{j}}V_{x_{p_{1}},x_{q_{1}}}\cdots
V_{x_{p_{k}},x_{q_{k}}}V_{x_{s},x_{r}}.$

By (12), we get

$vL_{x_{r}}=x_{i}\hat{R}_{x_{j}}V_{x_{p_{1}},x_{q_{1}}}\cdots
V_{x_{p_{k}},x_{q_{k}}}L_{x_{s}}L_{x_{r}}=-x_{i}\hat{R}_{x_{j}}V_{x_{p_{1}},x_{q_{1}}}\cdots
V_{x_{p_{k}},x_{q_{k}}}V_{x_{r},x_{s}}.$

Applying (7) we can express $vR_{x_{r}}$ and $vL_{x_{r}}$ as a linear
combination of elements of $\mathcal{B}$.

It remains to prove the linear independence of elements of $\mathcal{B}$.
Suppose that $f=f(x_{1},x_{2},\ldots,x_{n})\in F(X)$ is a nontrivial linear
combination of elements of $\mathcal{B}$. Suppose that $v\in\mathcal{B}$ and
$\deg_{x_{i}}(v)=k$. Let’s write $v=v(x_{i},\ldots,x_{i})$ in order to differ
the presence of $x_{i}$ in different places. To linearize $v$ in $x_{i}$ we
use new variables $y_{1},\ldots,y_{k}\in X$ and, after renumeration, we can
assume that $y_{r}<x_{j}$ if $i<j$ and $x_{j}<y_{r}$ if $j<i$ for all $1\leq
r\leq k$. Notice that every word $v(y_{\sigma(1)},\ldots,y_{\sigma(k)})$,
where $\sigma\in S_{k}$ and $S_{k}$ is the symmetric group in $k$ symbols, is
an element of $\mathcal{B}$. Then the full linearization of $v$ in $x_{i}$ is
a linear combination of basis elements
$v(y_{\sigma(1)},\ldots,y_{\sigma(k)})$. Therefore, by linearizing a
nontrivial element $f$, we obtain a nontrivial element that is a linear
combination of multilinear elements from $\mathcal{B}$. Substituting zeroes
instead of some variables, if necessary, we can make $f$ linear in each
variable. Therefore, we can assume that $f$ is a multilinear nontrivial
identity in the variables $x_{1},\ldots,x_{n}$. Let

$f=\sum_{i=1}\alpha_{i}u_{i},$

where $\alpha_{i}\in\mathbb{F}$ and $u_{i}\in\mathcal{B}$. Suppose, for
example, that

$u_{1}=x_{i}R_{x_{j}}V_{x_{p_{1}},x_{q_{1}}}\cdots
V_{x_{p_{k}},x_{q_{k}}}L_{x_{s}}.$

Set $x_{i}=d_{1,2}$, $x_{j}=-b_{1,2}$, $x_{p_{r}}=a_{r+1,r+2}$,
$x_{q_{r}}=-b_{r+1,r+2}$ for all $r=1,2,\ldots k$, $x_{s}=a_{k+2,k+3}$. We
have $c_{i}V_{a_{ij},-b_{ij}}=c_{j}$ for all $i,j$. Then the value of $u_{1}$
under this substitution is $d_{k+2,k+3}$ and the value of any other $u_{i}$ is
$0$. Consequently, the value of $f$ is $\alpha_{1}d_{k+2,k+3}\neq 0$. Thus,
$f$ is not an identity for $\mathcal{R}$.

If $L_{x_{s}}$ does not appear in $u_{1}$, then we perform the same
substitutions for the variables. If $R_{x_{j}}$ does not appear in $u_{1}$,
then we simply set $x_{i}=c_{2}$ and perform the same substitutions for the
rest of the variables as described above. In both cases the value of $f$ is
nonzero. This completes our proof. ∎

Let $M=M(F(X))$ be the multiplication algebra of the algebra $F(X)$. Denote by
$E_{0}$ the subalgebra (without identity) of $M$ generated by the operators
$V_{x_{i},x_{j}}$ with $i<j$ for all $i,j=1,2,\ldots$. Set also

$E_{1}=\sum_{j\geq 1}E_{0}L_{x_{j}},\quad E_{2}=\sum_{i\geq
1}R_{x_{i}}E_{0},\quad E_{3}=\sum_{i,j\geq 1}R_{x_{i}}E_{0}L_{x_{j}},$

and

$R_{k}=\sum_{i\geq 1}x_{i}E_{k},\quad\mbox{for }k=0,1,2,3.$

According to Proposition 4.1, the space $F(X)$ is the direct sum of the
subspaces $R_{k}$ and the linear span of the elements of $\mathcal{B}$ of
degree less than or equal to 3.

######  Lemma 4.2.

An identity $zf(x_{1},\ldots,x_{m})=0$, where $f\in E_{0}$, is a consequence
of a system of identities

(14) $tg_{j}(x_{1},\ldots,x_{l})=0,\quad g_{j}\in E_{0},\quad
j\in\mathcal{J},$

in the variety $\mathcal{R}$, where $\mathcal{J}$ is any set of indices, if
and only if the operator $f(x_{1},\ldots,x_{m})$ belongs to the ideal of the
associative algebra $E_{0}$ generated by the set $G$ of all operators
$\varphi(g_{j})$, where $\varphi$ runs over the set of all linear
endomorphisms $\varphi:X\rightarrow\mathbb{F}X=\sum_{i\geq 1}\mathbb{F}x_{i}$
and $j\in\mathcal{J}$.

###### Proof.

Suppose that $f$ belongs to the ideal of $E_{0}$ generated by $G$. Then

$f=\sum_{r=1}^{t}u_{r}g_{j_{r}}^{\varphi_{r}}v_{r},$

for some linear endomorphisms $\varphi_{r}$ and $u_{r},v_{r}\in E_{0}$.
Therefore,

$zf=\sum_{r=1}^{t}(zu_{r})g_{j_{r}}^{\varphi_{r}}v_{r}$

and $zf=0$ is a consequence of the system of identities (14).

Let’s describe all the consequences of the identities (14). Let
$\varphi:F(X)\to F(X)$ be an arbitrary endomorphism and set
$\varphi(x_{i})=y_{i}+h_{i}$, where $y_{i}\in\mathbb{F}X$ and $h_{i}\in
F(X)^{2}$ for all $i$. Since $g_{j}\in E_{0}$, using (4) and (6), we get

$t\varphi(g_{j})=tg_{j}(y_{1},\ldots,y_{l})=0.$

Thus, a general form of consequences of the identities (14) can be expressed
as

$\sum_{r=1}^{t}u_{r}g_{j_{r}}^{\varphi_{r}}v_{r},$

where $u_{r}\in F(X)$, $v_{r}\in M(F(X))$, and $\varphi_{r}$ are linear
endomorphisms. We know that $g_{j_{r}}^{\varphi_{r}}\in E_{0}$. We also claim
that $u_{r}$ and $v_{r}$ can be represented in the forms

$x_{i}\hat{R}_{x_{j}}V_{x_{p_{1}},x_{q_{1}}}\cdots
V_{x_{p_{k}},x_{q_{k}}},\quad\quad
V_{x_{p^{\prime}_{1}},x_{q^{\prime}_{1}}}\cdots,V_{x_{p^{\prime}_{k}},x_{q^{\prime}_{k}}}\hat{L}_{x_{s}},$

respectively, where $i<j$, $p_{l}<q_{l}$, $p^{\prime}_{l}<q^{\prime}_{l}$ and
$k=0,1,\ldots$.

Suppose that $u_{r}$ is a basis element that ends with $L_{x_{s}}$. Then, by
(11) and (12), we can derive that

$V_{x_{i},x_{j}}L_{x_{s}}V_{x_{k},x_{t}}=V_{x_{i},x_{j}}L_{x_{s}}L_{x_{k}}R_{x_{t}}=-V_{x_{i},x_{j}}V_{x_{k},x_{s}}R_{x_{t}}=0.$

Consequently, we have $V_{x_{i},x_{j}}L_{x_{s}}E_{0}=0$.

If $u_{r}=x_{i}R_{x_{j}}L_{x_{k}}$, then by (8) and (11) we get

$u_{r}V_{x_{s},x_{t}}=x_{i}R_{x_{j}}L_{x_{k}}L_{x_{s}}R_{x_{t}}=x_{j}V_{x_{i},x_{k}}L_{x_{s}}R_{x_{t}}-x_{i}R_{x_{j}}V_{x_{s},x_{k}}R_{x_{t}}=x_{j}V_{x_{i},x_{k}}V_{x_{s},x_{t}}.$

So, we can conclude that $u_{r}$ has the claimed form.

Now, let’s consider the case when $v_{r}$ is a basis element that starts with
$R_{y}$. According to (11), we have $E_{0}R_{y}=0$. If $v_{r}$ starts with
$L_{x_{i}}L_{x_{j}}$, then by using (12), we find

$V_{x_{k},x_{s}}L_{x_{i}}L_{x_{j}}=-V_{x_{k},x_{s}}V_{x_{j},x_{i}}.$

Hence, $v_{r}$ also has the claimed form.

If $zf=0$ is a consequence of the identities (14), then we get an equality of
the form

$x_{m+1}f(x_{1},\ldots,x_{m})=\sum_{r=1}^{t}\lambda_{r}x_{i_{r}}w_{r}g_{j_{r}}^{\varphi_{r}}v_{r},$

where $x_{i_{r}}w_{r}=u_{r}$, $w_{r}\in E_{0}+E_{2}$ and $v_{r}\in
E_{0}+E_{1}$. Notice that every element
$x_{i_{r}}w_{r}g_{j_{r}}^{\varphi_{r}}v_{r}$ belongs to $\mathcal{B}$.
Consequently, we may assume that $x_{i_{r}}=x_{m+1}$, $w_{r},v_{r}\in E_{0}$,
and

$f(x_{1},\ldots,x_{m})=\sum_{r=1}^{t}\lambda_{r}w_{r}g_{j_{r}}^{\varphi_{r}}v_{r}.$

∎

##  5\.  Identities of $P_{n}$.

In this section, we study the connections between the identities and the
operator identities of $P_{n}$ for $n\geq 2$.

######  Lemma 5.1.

If $f=f(x_{1},\ldots,x_{m})\in F(X)$ and $f=0$ is an identity of $P_{n}$ for
$n\geq 2$, then

(15) $f=f_{0}+f_{1}+f_{2}+f_{3}\in F(X),\quad f_{k}\in R_{k},$

and $f_{k}=0$ is an identity of $P_{n}$ for all $k=0,1,2,3$.

###### Proof.

Let

$f=\sum_{i=1}^{m}\lambda_{i}x_{i}+\sum_{i,j=1}^{m}\lambda_{ij}x_{i}x_{j}+\sum_{i,j,k=1,i<j}^{m}\lambda_{ijk}x_{i}R_{x_{j}}L_{x_{k}}+f^{\prime},$

where $f^{\prime}$ is a linear combination of elements from $\mathcal{B}$ of
degree $\geq 4$.

We first show that $\lambda_{i}=\lambda_{ij}=\lambda_{ijk}=0$ for all
$i,j,k=1,\ldots,m$. For any fixed $i$ the substitution $x_{i}=c_{1}$ and
$x_{j}=0$ for all $j\neq i$ gives that $\lambda_{i}c_{1}=0$, which implies
$\lambda_{i}=0$.

If $i\neq j$, then the substitution $x_{i}=a_{11}$, $x_{j}=c_{1}$, and
$x_{k}=0$ for all $k\neq i,j$, makes the value of $f$ equal to
$\lambda_{ij}d_{11}=0$. We get the same value if $i=j$ under the substitution
$x_{i}=x_{j}=a_{11}+c_{1}$ and $x_{k}=0$ for all $k\neq i,j$. This gives
$\lambda_{ij}=0$ in both cases.

Assume that $i<j>k$. If $i\neq k$, then the substitution $x_{i}=b_{11}$,
$x_{j}=d_{11}$, $x_{k}=a_{12}$, and $x_{t}=0$ for all $t\neq i,j,k$, makes the
value of $f$ equal to $-\lambda_{ijk}d_{12}$. This gives that
$\lambda_{ijk}=0$. If $i=k$, then the substitution $x_{i}=b_{11}$,
$x_{j}=d_{11}$, and $x_{t}=0$ for all $t\neq i,j$, gives that
$-\lambda_{iji}e_{11}=0$ and $\lambda_{iji}=0$. If $i<j=k$, then the
substitution $x_{i}=d_{11}$, $x_{j}=b_{11}$, and $x_{t}=0$ for all $t\neq
i,j$, gives that $-\lambda_{ijj}e_{11}=0$ and $\lambda_{ijj}=0$. Finally, if
$i<j<k$, then the substitution $x_{i}=d_{11}$, $x_{j}=x_{k}=b_{11}$, and
$x_{t}=0$ for all $t\neq i,j$, gives that
$-\lambda_{ijk}e_{11}-\lambda_{ikj}e_{11}=0$, i.e.,
$\lambda_{ijk}=-\lambda_{ikj}=0$.

Thus, $f$ is a linear combination of elements of $\mathcal{B}$ of degree $\geq
4$. Suppose that $f$ is written as in (15). Taking into account the relations
$D_{n}P_{n}\subseteq C_{n}$ and $P_{n}C_{n}\subseteq\overline{C}_{n}$ it can
be observed that the images of $F_{0}=f_{0}+f_{2}$ and $F_{1}=f_{1}+f_{3}$
belong to $C_{n}$ and $\overline{C}_{n}$, respectively. Therefore, if $f=0$ is
an identity of $P_{n}$, then $F_{0}=0$ and $F_{1}=0$ are also identities of
$P_{n}$.

Suppose that

$f_{k}(x_{1},\ldots,x_{m})=\sum_{i=1}^{m}x_{i}g_{i}^{(k)}(x_{1},\ldots,x_{m}),$

where $g_{i}^{(k)}\in E_{k}$ and $k=0,1,2,3$. Let $p_{1},\ldots,p_{m}\in
P_{n}$ with $p_{s}=v_{s}+\overline{v}_{s}+a_{s}$, where $v_{s}\in C_{n}$,
$\overline{v}_{s}\in\overline{C}_{n}$, $a_{s}\in A_{n}$. By Lemma 3.2 we can
obtain that

$p_{i}V_{p_{j},p_{k}}=v_{i}V_{p_{j},p_{k}}=v_{i}V_{v_{j}+\overline{v}_{j}+a_{j},v_{k}+\overline{v}_{k}}+v_{i}V_{v_{j}+\overline{v}_{j},a_{k}}+v_{i}V_{a_{j},a_{k}}=v_{i}V_{a_{j},a_{k}}.$

Then

$f_{0}(p_{1},\ldots,p_{m})=\sum_{i=1}^{m}v_{i}g_{i}^{(0)}(a_{1},\ldots,a_{m})=f_{0}(v_{1}+a_{1},\ldots,v_{m}+a_{m}).$

It is easy to see that
$(A_{n}+C_{n})R_{a_{i}+v_{i}}V_{a_{j}+v_{j},a_{s}+v_{s}}=0$ for all
$a_{i},a_{j},a_{s}\in A_{n}$ and $v_{i},v_{j},v_{s}\in C_{n}$. It follows that

$f_{2}(v_{1}+a_{1},\ldots,v_{m}+a_{m})=0.$

Thus,

$f_{0}(p_{1},\ldots,p_{m})$
$=f_{0}(v_{1}+a_{1},\ldots,v_{m}+a_{m})+f_{2}(v_{1}+a_{1},\ldots,v_{m}+a_{m})$
$=F_{0}(v_{1}+a_{1},\ldots,v_{m}+a_{m})=0.$

Therefore, we can conclude that $f_{0}=0$ and $f_{2}=0$ are identities of
$P_{n}$. Similarly, we can establish that $f_{1}=0$ and $f_{3}=0$ are also
identities of $P_{n}$. ∎

######  Lemma 5.2.

If $f=f(x_{1},\ldots,x_{m})\in R_{1}+R_{3}$, then $fx_{m+1}\in R_{0}+R_{2}$
and if
$f(x_{1},\ldots,x_{m})x_{m+1}=0$ is an identity of $P_{n}$, then $f=0$ is an
identity of $P_{n}$ as well.

###### Proof.

We have $fx_{m+1}\in R_{0}+R_{2}$ by the definition of the spaces $R_{i}$,
where $0\leq i\leq 3$. If $fx_{m+1}=0$ is an identity of $P_{n}$, then all
values of $f$ in $P_{n}$ belong to $C_{n}=\mathrm{Ann}_{l}(P_{n})$ by Lemma
3.3. However, since $f$ is an element of $R_{1}+R_{3}$, the values of $f$ must
belong to $\overline{C}_{n}$. Consequently, $f=0$ is an identity of $P_{n}$. ∎

Recall an exact formal definition of the linearization of identities [37,
Chapter 1]. Let $\mathcal{V}$ be an arbitrary variety of algebras and
$\mathbb{F}\langle X\rangle$ be its free algebra over $\mathbb{F}$ generated
by $X=\\{x_{1},x_{2},\ldots\\}$. Let $y\in\mathbb{F}\langle X\rangle$ be an
arbitrary fixed element. For a nonnegative integer $k$, we define the linear
mapping $\Delta^{k}_{x_{i}}(y)$ on $\mathbb{F}\langle X\rangle$ as follows:

  * •

$\Delta^{0}_{x_{i}}(y)$ is the identity mapping;

  * •

$x_{s}\Delta^{k}_{x_{i}}(y)=0$, if either $k>1$ or $k=1$, $i\neq s$;

  * •

$x_{i}\Delta^{1}_{x_{i}}(y)=y$;

  * •

$(uv)\Delta^{k}_{x_{i}}(y)=\sum_{r+s=k}(u\Delta^{r}_{x_{i}}(y))(v\Delta^{s}_{x_{i}}(y))$,

where $x_{i}\in X$ and $u,v$ are any monomials in $\mathbb{F}\langle
X\rangle$. We also write $\Delta_{x_{i}}(y)$ instead of
$\Delta^{1}_{x_{i}}(y)$.

######  Lemma 5.3.

Suppose that $f=f(x_{1},\ldots,x_{m})\in R_{2}$. Then
$f\Delta_{i}(x_{m+1}x_{m+2})\in R_{0}$ for all $1\leq i\leq m$. Moreover,
$f=0$ is an identity of $P_{n}$ if and only if $P_{n}$ satisfies the following
system of identities

(16) $f(x_{1},\ldots,x_{m})\Delta_{i}(x_{m+1}x_{m+2})=0,\quad 1\leq i\leq m.$

###### Proof.

Let $w=xR_{y}V_{z_{1},t_{1}}\cdots V_{z_{r},t_{r}}\in\mathcal{B}$ and $u,v\in
X$. We have

$(xR_{y})\Delta_{x}(uv)=(uv)R_{y}=vV_{u,y}.$

By (1), (4) and (6), we get

$(xR_{y}V_{z_{1},t_{1}})\Delta_{y}(uv)=(z_{1}(x(uv)))t_{1}=((z_{1}x)(uv)-(z_{1}(uv))x+z_{1}((uv)x))t_{1}$
$=-((z_{1}(uv))x)t_{1}+(z_{1}((uv)x))t_{1}=(z_{1}((uv)x))t_{1}=vV_{u,x}V_{z_{1},t_{1}}.$

By (4) we get that $w\Delta_{z_{i}}(uv)=w\Delta_{t_{i}}(uv)=0$ for any
$i=1,\ldots,r$. Thus, if $f(x_{1},\ldots,x_{m})\in R_{2}$, then
$f(x_{1},\ldots,x_{m})\Delta_{i}(x_{m+1}x_{m+2})\in R_{0}$.

If $p_{1},\ldots,p_{m}\in P_{n}$ and $v_{1},\ldots,v_{m}\in D_{n}$, then we
have

(17)
$f(p_{1}+v_{1},\ldots,p_{m}+v_{m})=f(p_{1},\ldots,p_{m})+\sum_{i=1}^{m}f(p_{1},\ldots,p_{m})\Delta_{i}(v_{i}).$

In fact, by Lemma 1.3 from [37], the relation

$f(x_{1}+y_{1},\ldots,x_{m}+y_{m})=\sum_{i_{1},\ldots,i_{m}\geq
0}f\Delta_{1}^{i_{1}}(y_{1})\cdots\Delta_{m}^{i_{m}}(y_{m})$
$=f(x_{1},\ldots,x_{m})+\sum_{i=1}^{m}f(x_{1},\ldots,x_{m})\Delta_{i}(y_{i})+g,$

where $y_{1},\ldots,y_{m}\notin\\{x_{1},\ldots,x_{m}\\}$ are distinct
variables and the degree of $g$ in the variables $y_{1},\ldots,y_{m}$ is
greater than one, holds in $\mathbb{F}\langle X\rangle$. By substituting
$x_{i}=p_{i},y_{i}=v_{i}$ and using the fact that $D_{n}^{2}=0$, we can obtain
the relation (17).

If $f=0$ is an identity of $P_{n}$, then the relation (17) implies that

$f(p_{1},\ldots,p_{m})\Delta_{i}(v)=f(p_{1},\ldots,p_{i}+v,\ldots,p_{m})-f(p_{1},\ldots,p_{m})=0$

for all $p_{i}\in P_{n}$ and $v\in D_{n}$. In other words, the algebra $P_{n}$
satisfies the system of identities (16).

Conversely, suppose that the system of identities (16) holds in $P_{n}$.
Assume that $p_{1},\ldots,p_{m}\in P_{n}$ of the form $p_{i}=a_{i}+v_{i}$,
where $a_{i}\in A_{n}$ and $v_{i}\in D_{n}$. Then using the relation (17), we
have

$f(p_{1},\ldots,p_{m})=f(a_{1}+v_{1},\ldots,a_{m}+v_{m})$
$=f(a_{1},\ldots,a_{m})+\sum_{i=1}^{m}f(p_{1},\ldots,p_{m})\Delta_{i}(v_{i})=f(a_{1},\ldots,a_{m}).$

Considering $A^{2}_{n}=0$ and $f\in R_{2}\subseteq F(X)^{2}$, we can conclude
that $f(a_{1},\ldots,a_{m})=0$. Consequently, $f(p_{1},\ldots,p_{m})=0$. ∎

######  Lemma 5.4.

If $f=f(x_{1},\ldots,x_{m})\in R_{0}$ and $f=0$ is an identity of $P_{n}$ of
the form

$f=\sum_{i=1}^{m}x_{i}g_{i},$

where $g_{i}\in E_{0}$, then $x_{m+1}g_{i}=0$ is an identity of $P_{n}$.

###### Proof.

For a fixed $i$ set $x_{i}=v+a_{i}$ and $x_{j}=a_{j}$ for all $j\neq i$, where
$v\in D_{n}$ and $a_{j}\in A_{n}$. Taking into account the relations
$A_{n}^{2}=D_{n}^{2}=0$ and Lemma 3.2, one can have

$f(x_{1},\ldots,x_{m})=vg_{i}(a_{1},\ldots,a_{m})=0.$

Hence, $x_{m+1}g_{i}=0$ is an identity of $P_{n}$. ∎

######  Proposition 5.5.

For an arbitrary polynomial $f=f(x_{1},\ldots,x_{m})\in F(X)$ there exist
$t(m)=2m(m+3)$ polynomials $g_{i}(x_{1},\ldots,x_{m+3})\in E_{0}$, where
$i=1,\ldots,t(m)$, such that $f(x_{1},\ldots,x_{m})=0$ is an identity of
$P_{n}$ for $n\geq 2$ if and only if $P_{n}$ satisfies the system of
identities

$zg_{i}(x_{1},\ldots,x_{m+3})=0,\quad 1\leq i\leq t(m).$

###### Proof.

Let $f=f(x_{1},\ldots,x_{m})\in F(X)$ and suppose that $f=0$ is an identity of
$P_{n}$. Then by Lemma 5.1 we obtain

$f=f_{0}+f_{1}+f_{2}+f_{3},\quad f_{k}\in R_{k},$

and $f_{k}=0$ is an identity of the algebra $P_{n}$.

By Lemma 5.4, the identity $f_{0}=\sum_{i=1}^{m}x_{i}g_{i}=0$ is equivalent to
the system of $m$ identities $x_{m+1}g_{i}=0$ of $P_{n}$, where $1\leq i\leq
m$.

By Lemma 5.2, the identity $f_{1}=0$ is equivalent to $f_{1}x_{m+1}=0$ and
$f_{1}x_{m+1}\in R_{0}$. Moreover, if

$f_{1}x_{m+1}=\sum_{i=1}^{m}x_{i}g_{i},\quad g_{i}\in E_{0},$

then, by Lemma 5.4, the identity $f_{1}x_{m+1}=0$ is equivalent to the system
of $m$ identities $x_{m+2}g_{i}=0$ of $P_{n}$, where $1\leq i\leq m$.

By Lemma 5.3, the identity $f_{2}=0$ is equivalent to the system of $m$
identities
$f_{2}(x_{1},\ldots,x_{m})\Delta_{i}(x_{m+1}x_{m+2})=0$, where $i=1,\ldots,m$,
and we have $f_{2}\Delta_{i}(x_{m+1}x_{m+2})\in R_{0}$. Hence, by Lemma 5.4,
it is equivalent to a system of $m(m+2)$ identities of the form
$x_{m+3}g_{i}=0$, where $g_{i}(x_{1},\ldots,x_{m+2})\in E_{0}$ and
$i=1,\ldots,m(m+2)$.

By Lemma 5.2, the identity $f_{3}=0$ is equivalent to $f_{3}x_{m+1}=0$ and
$f_{3}x_{m+1}\in R_{2}$. The identity (4) implies that
$(f_{3}x_{m+1})\Delta_{m+1}(x_{m+2}x_{m+3})=0$. Then, by Lemma 5.3, $f_{3}=0$
is equivalent to the system of $m$ identities
$0=(f_{3}x_{m+1})\Delta_{i}(x_{m+2}x_{m+3})\in R_{0}$, where $i=1,\ldots,m$.
Moreover, by Lemma 5.4, it is equivalent to a system of $m(m+2)$ identities of
the form $x_{m+4}g_{j}=0$, where $g_{j}(x_{1},\ldots,x_{m+3})\in E_{0}$ and
$1\leq j\leq m(m+2)$.

Thus, $f=0$ is equivalent to a system of $t(m)=2m(m+3)$ identities of the form
$zg_{i}(x_{1},\ldots,x_{m+3})=0$, where $g_{i}(x_{1},\ldots,x_{m+3})\in E_{0}$
and $i=1,\ldots,t(m)$. ∎

Let $B$ be an arbitrary algebra in $\mathcal{R}$. We define $E_{0}(B)$ as the
algebra of operators generated by $V_{b_{1},b_{2}}$ for all $b_{1},b_{2}\in
B$, that acts on the algebra $B$. Denote by $T(E_{0}(B))$ the ideal of $E_{0}$
defined as the intersection of the kernels of all possible homomorphisms from
$F(X)$ to $B$. The elements of $T(E_{0}(B))$ are called $V$-identities of $B$.

######  Lemma 5.6.

$E_{0}(P_{n})\cong M_{n}(\mathbb{F})$, where $M_{n}(\mathbb{F})$ is algebra of
$n\times n$ matrices.

###### Proof.

According to Lemma 3.2, $E_{0}(P_{n})$ annihilates the subspace
$A_{n}+\overline{C}_{n}$, and $C_{n}$ is an invariant subspace of $P_{n}$
under its action. Consequently, $E_{0}(P_{n})$ is isomorphic to a subalgebra
$L$ of the algebra $End_{\mathbb{F}}C_{n}$. Furthermore, the operator
$V_{b_{ij},a_{ij}}\in E_{0}(P_{n})$ sends the element $c_{i}$ to $c_{j}$, and
$c_{k}$ to zero if $k\neq i$, resembling the action of a unit matrix.
Therefore, the subalgebra $L$ coincides with the entire algebra
$End_{\mathbb{F}}(C_{n})\cong M_{n}(\mathbb{F})$. ∎

######  Proposition 5.7.

If the algebra $P_{n}$ has a finite basis of identities for $n\geq 2$, then
the ideal $T=T(E_{0}(P_{n}))$ is generated by polynomials of bounded degrees.

###### Proof.

Suppose that $P_{n}$ has a finite basis of identities for $n\geq 2$. By
Proposition 5.5, modulo (1), (3), and (4), every identity is equivalent to a
finite system of identities of (14). Consequently, by Lemma 4.2, there exists
a finite set of elements $G\subseteq T$ such that the identities $tg=0$, where
$g\in G$, form a basis of identities of $P_{n}$. Let $m$ be the maximum of the
degrees of polynomials in $G$. By the same Lemma 4.2, the ideal $T$ is
generated by all $\varphi(g)$, where $g\in G$ and $\varphi$ is linear.
Consequently, $T$ is generated by elements of degrees $\leq m$. ∎

##  6\.  Identities of $P_{2}$.

We are going to prove that $P_{2}$ does not have a finite basis of identities.
First, let’s construct some important examples of algebras.

######  Proposition 6.1.

For any $s>5$ there exists an algebra $B\in\mathcal{R}$ with the following two
properties:

  1. (1)

$B$ is generated by a set $Q=\\{q_{1},\ldots,q_{s+3}\\}$ such that
$T\nsubseteq T(E_{0}(B))$.

  2. (2)

Let $C$ be a subalgebra of $B$ generated by any subset $Q^{\prime}$ of $Q$
with $s$ elements. Then

$tg(c_{1},\ldots,c_{k})=0$

for all $g(x_{1},\ldots,x_{k})\in T$, $c_{1},\ldots,c_{k}\in C$, and $t\in B$.

###### Proof.

Set $n=s-5\geq 1$. Let $H$ be the free algebra with identity in the variety of
algebras generated by the field $\mathbb{F}$ with free generators
$\\{h_{1},\ldots,h_{n}\\}$. Denote by $W$ the subspace of $H$, spanned by all
words in $h_{1},\ldots,h_{n}$, including the identity $1$, that do not contain
at least one $h_{i}$. Then $W\neq H$. By Theorem 1.6 from [37], the algebra
$A=H\otimes_{\mathbb{F}}P_{3}$ belongs to $\mathcal{R}$. Consider the
subalgebra $L$ of $A$ generated by the following set of elements:

(18) $\\{1\otimes c_{1},1\otimes a_{11},1\otimes b_{11},1\otimes
a_{12},1\otimes b_{12},h_{i}\otimes a_{22},1\otimes b_{22},1\otimes
a_{23},1\otimes b_{23}\\}$

where $i=1,\ldots,n$.

We note that

$1\otimes c_{2}=-(1\otimes b_{12})((1\otimes a_{12})(1\otimes c_{1})),$
$h_{j}\otimes c_{2}=-(1\otimes b_{22})((h_{j}\otimes a_{22})(1\otimes
c_{2})),$ $h_{i}h_{j}\otimes c_{2}=-(1\otimes b_{22})((h_{i}\otimes
a_{22})(h_{j}\otimes c_{2})).$

Thus, by induction on the length of $h$, one can derive that $h\otimes
c_{2}\in L$ for any word $h$ in $h_{1},\ldots,h_{n}$. In addition, $h\otimes
c_{3}\in L$ since

$h\otimes c_{3}=-(1\otimes b_{23})((1\otimes a_{23})(h\otimes c_{2})).$

Note that $h\otimes c_{3}$ is a two-sided annihilator of $L$ since

$L\subseteq H\otimes(D_{3}+\sum_{i\leq j\leq
3,(i,j)\neq(3,3)}(\mathbb{F}a_{ij}+\mathbb{F}b_{ij}).$

Consequently, $N=H\otimes c_{3}$ and $N^{\prime}=W\otimes c_{3}$ are ideals of
$L$. Set $B=L/N^{\prime}$ and let’s show that it satisfies the properties (1)
and (2) of the proposition.

Verification of Property (1). Denote by $q_{1},\ldots,q_{s+3}$ the images of
the generators of (18) under the natural projection $L\rightarrow
L/N^{\prime}$. By Lemma 5.6, the algebra $E_{0}(P_{2})$ satisfies the well-
known Hall’s identity

$[[\overline{f}_{1},\overline{f}_{2}]\circ[\overline{f}_{3},\overline{f}_{4}],\overline{f}_{5}]=0,$

for all $\overline{f}_{i}\in E_{0}(P_{2})$, where $1\leq i\leq 5$ and $a\circ
b=ab+ba$. It follows that $S=[[f_{1},f_{2}]\circ[f_{3},f_{4}],f_{5}]\in T$ for
all $f_{i}\in E_{0}$.

It is easy to choose $f_{1},\ldots,f_{5}\in E_{0}$ and $\varphi:F(X)\to L$
such that

$f^{\varphi}_{1}=V_{1\otimes b_{12},1\otimes a_{12}}\prod_{i=1}^{n}V_{1\otimes
b_{22},h_{i}\otimes a_{22}},\quad f^{\varphi}_{2}=f^{\varphi}_{5}=V_{1\otimes
b_{11},1\otimes a_{11}},$ $f^{\varphi}_{3}=V_{1\otimes b_{22},1\otimes
a_{22}},\quad f^{\varphi}_{4}=V_{1\otimes b_{23},1\otimes a_{23}}.$

The actions of the operators
$f^{\varphi}_{1},f^{\varphi}_{2},f^{\varphi}_{3},f^{\varphi}_{4}$ on $L$ give
us

$f^{\varphi}_{1}f^{\varphi}_{2}=0,\quad
f^{\varphi}_{2}f^{\varphi}_{1}=V_{1\otimes b_{12},v\otimes a_{12}},$
$f^{\varphi}_{3}f^{\varphi}_{4}=V_{1\otimes b_{23},1\otimes a_{23}},\quad
f^{\varphi}_{4}f^{\varphi}_{3}=0,$

and we have

$S^{\varphi}=[-V_{1\otimes b_{12},v\otimes a_{12}}\circ V_{1\otimes
b_{23},1\otimes a_{23}},V_{1\otimes b_{11},1\otimes a_{11}}]=V_{1\otimes
b_{12},v\otimes a_{12}}V_{1\otimes b_{23},1\otimes a_{23}},$

where $v=h_{1}\cdots h_{n}$. Since

$(1\otimes c_{1})S^{\varphi}=v\otimes c_{3}\neq 0(mod\,N^{\prime}),$

we obtain $S\notin T(E_{0}(B))$ and therefore $T\nsubseteq T(E_{0}(B))$.

Verification of Property (2). Let $L^{\prime}$ be a subalgebra of $L$
generated by a subset of the set (18) that contains no more than $s$ elements.
Assume that $f(x_{1},\ldots,x_{k})\in T$. Let $M$ be the set of all elements
of the form $(1\otimes c_{1})f(l_{1},\ldots,l_{k})$, where $l_{i}\in
L^{\prime}$. We claim that

(19) $M\cap N\subseteq N^{\prime}.$

Let’s assume that (19) does not hold. In other words, there is an element

$g=(h_{1}\cdots h_{n}\otimes c_{3})(h^{\prime}\otimes
c_{3})+h^{\prime\prime}\otimes c_{3}\in M\cap N$

for some nonzero $h^{\prime}\in H$ and some $h^{{}^{\prime\prime}}\in W$.

Note that

$(1\otimes c_{1})V_{1\otimes b_{12},1\otimes a_{12}}=1\otimes
c_{2},\quad(1\otimes c_{2})V_{1\otimes b_{22},h_{i}\otimes
a_{22}}=h_{i}\otimes c_{2},$ $(h_{i}\otimes c_{2})V_{1\otimes
b_{22},h_{j}\otimes a_{22}}=h_{i}h_{j}\otimes c_{2},\,(h_{1}\cdots
h_{n}\otimes c_{2})V_{1\otimes b_{23},1\otimes a_{23}}=h_{1}\cdots
h_{n}\otimes c_{3}.$

Without using $1\otimes c_{1}$ and all of the operators

$V_{1\otimes b_{12},1\otimes a_{12}},\quad V_{1\otimes b_{22},h_{i}\otimes
a_{22}},\quad V_{1\otimes b_{23},1\otimes a_{23}},$

we cannot get elements containing the product $h_{1}\ldots h_{n}$. This means
that $M\cap N$ contains $g$ if and only if the following $s+1$ elements appear
in our calculations:

$1\otimes c_{1},1\otimes a_{12},1\otimes b_{12},h_{i}\otimes a_{22},1\otimes
b_{22},1\otimes a_{23},1\otimes b_{23}\quad(i=1,\ldots,n).$

It is impossible since $L^{\prime}$ is generated by $s$ elements. This
contradiction establishes the inclusion (19).

Set $\overline{f}=f(l_{1},\ldots,l_{k})$ for some fixed $l_{1},\ldots,l_{k}\in
L^{\prime}$. We show that $L\overline{f}=0\,(mod\,N^{\prime})$. By Lemma 3.2,
one can have that
$(H\otimes(\overline{C}_{3}+A_{3}+\mathbb{F}c_{3}))E_{0}(L)=0$. Then, because
of (19), it is sufficient to prove that

(20) $(1\otimes c_{1})\overline{f}=0(mod\,N),\quad(H\otimes
c_{2})\overline{f}=0.$

If $l_{i}=l_{i}^{\prime}+l_{i}^{\prime\prime}$, where

$l_{i}^{\prime}\in H\otimes\sum_{i\leq j\leq
2}(\mathbb{F}a_{ij}+\mathbb{F}b_{ij}),\quad l_{i}^{\prime\prime}\in
H\otimes(\mathbb{F}a_{23}+\mathbb{F}b_{23}),$

then

(21) $(1\otimes c_{1})\overline{f}=(1\otimes
c_{1})f(l_{1}^{\prime},\ldots,l_{k}^{\prime})(mod\,N).$

We have

$f(x_{1}+y_{1},\ldots,x_{k}+y_{k})=f(x_{1},\ldots,x_{k})+g(x_{1},\ldots,x_{k},y_{1},\ldots,y_{k}),$

where every monomial of $g\in E_{0}$ involve at least one variable from
$y_{1},\ldots,y_{k}$. Since $LV_{l^{\prime\prime}_{i},L}\subseteq N$, we
obtain

$(1\otimes
c_{1})g(l^{\prime}_{1},\ldots,l^{\prime}_{k},l^{\prime\prime}_{1}\ldots,l^{\prime\prime}_{k})=0(mod\,N),$

which implies (21).

Consequently, to prove the first relation of (20), without loss of generality
we can assume that $l_{i}\in H\otimes\sum_{i\leq j\leq
2}(\mathbb{F}a_{ij}+\mathbb{F}b_{ij})$. In this case, the elements
$c_{1},l_{1},\ldots,l_{k}$ generate a subalgebra of $H\otimes P_{2}$. The
algebra $H$ can be embedded into the Cartesian product $\mathbb{F}^{\alpha}$
of the algebra $\mathbb{F}$. So, $H\otimes P_{2}$ can be embedded into
$\mathbb{F}^{\alpha}\otimes P_{2}\cong P^{\alpha}_{2}$ and satisfies all the
identities of $P_{2}$. Consequently, it satisfies all $V$-identities from $T$.
Thus, the first relation of (20) is established. The second relation of (20)
can be established similarly using the equality

$H\otimes(\sum_{j\leq 2}(\mathbb{F}a_{1j}+\mathbb{F}b_{1j}))c_{2}=0.$

In this case, we can assume that

$l_{i}\in H\otimes(\sum_{2\leq i,\,j\leq
3}(\mathbb{F}a_{ij}+\mathbb{F}b_{ij})).$

Then the elements $c_{2},l_{1},\ldots,l_{k}$ generate an algebra isomorphic to
a subalgebra of $H\otimes P_{2}$. Thus, we have
$L\overline{f}=0(mod\,N^{\prime})$. The factorization by $N^{\prime}$
completes the proof of Proposition 6.1. ∎

######  Lemma 6.2.

Let $\Sigma$ be a set of generators of the ideal $T=T(E_{0}(P_{2}))$ of
$E_{0}$. Then for any natural number $s$, there exists a polynomial
$f_{s}\in\Sigma$ that depends on more than $s$ variables.

###### Proof.

Suppose, contrary, that $\Sigma$ consists of polynomials that depend on $\leq
s$ variables. Let $B$ be the algebra satisfying the conditions of Proposition
6.1. Consider the epimorphism $\tau:F(X)\rightarrow B$ defined by

$\tau(x_{i})=\left\\{\begin{array}[]{rcl}q_{i}&\mbox{if}&i\leq s+3,\\\
0&\mbox{if}&i>s+3.\end{array}\right.$

This induces the epimorphism $\tilde{\tau}:E_{0}\rightarrow E_{0}(B)$ defined
by

$\tilde{\tau}g(x_{1},\ldots,x_{k})=g(\tau(x_{1}),\ldots,\tau(x_{k})).$

If $g(x_{1},\ldots,x_{k})\in\Sigma$, then $k\leq s$ and $c_{i}=\tau(x_{i})$
belong to a subalgebra of $B$ generated by $\leq s$ elements. By Proposition
6.1(2), $g(c_{1},\ldots,c_{k})=0$. Thus, $g\in Ker\,\tilde{\tau}$. So
$Ker\tilde{\tau}$ contains $\Sigma$ and, consequently, $T$ as well.

Now let $f(x_{1},\ldots,x_{m})\in T$. For any $b_{1},\ldots,b_{m}\in B$ there
exist $r_{i}\in F(X)$ such that $b_{i}=\tau(r_{i})$ for all $i=1,\ldots,m$.
Since $f(r_{1},\ldots,r_{m})\in T$, we have
$f(b_{1},\ldots,b_{m})=\tilde{\tau}f(r_{1},\ldots,r_{m})=0$. This proves that
every element of $T$ is a $V$-identity of $B$. This contradicts Proposition
6.1(1). ∎

######  Theorem 6.3.

Algebra $P_{2}$ over an arbitrary field $\mathbb{F}$ does not have a finite
basis of identities.

###### Proof.

If $P_{2}$ has a finite basis of identities, then the ideal
$T=T(E_{0}(P_{2}))$ is generated by polynomials of bounded degree by
Proposition 5.7. This contradicts Lemma 6.2. ∎

## Acknowledgments

The authors would like to thank the Max Planck Institute für Mathematik and
the first author would like to thank Wayne State University for their
hospitality and excellent working conditions, where some part of this work has
been done.

The second author is supported by the grant AP09261086 of the Ministry of
Science and Higher Education of the Republic of Kazakhstan.

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