#  $\mathbb{Z}_{2}$-graded identities of the Grassmann algebra over a finite
field

Luís Felipe Gonçalves Fonseca  Departamento de Matemática, Universidade
Federal de Viçosa - Campus Florestal, Rodovia LMG 818, km 06, Florestal, MG,
Brazil<EMAIL_ADDRESS>
###### Abstract.

Let $F$ be a finite field with characteristic $p>2$ and let $G$ be the unitary
Grassmann algebra generated by an infinite dimensional vector space $V$ over
$F$. In this paper, we determine a basis of the $\mathbb{Z}_{2}$-graded
polynomial identities for any non-trivial $\mathbb{Z}_{2}$-grading such that a
basis of $V$ is homogeneous in this grading.

Mathematics Subject Classification 2000: 16R10, 15A75, 16W50, 12E20.

Keywords: Identities, Grassmann Algebra, Graded rings and modules, Finite
Fields.

To my family.

##  1\. Introduction

In mathematics, the exterior product, or wedge product, of vectors is an
algebraic construction used in Euclidean geometry to study areas and volumes.
In linear algebra, the exterior product provides an abstract algebraic manner
for describing the determinant and minors of a matrix. The exterior algebra,
or Grassmann algebra is the algebraic system whose product is the exterior
product. Mathematically speaking, the Grassmann algebra is the largest algebra
that supports an alternating product on vectors.

The Grassmann algebra is also important for PI-Theory. A celebrated result
obtained by Kemer, depicted in (1987, [10]), shows that any associative PI-
algebra, over a field $F$ of characteristic zero, is PI-equivalent to the
Grassmann envelope of a finite- dimensional associative super-algebra.

An important task in PI-Theory is to describe the identities of the Grassmann
algebra. In 1970s, Regev and Krakovsky (1973,[11]) described the identities of
the Grassmann algebra over a field of characteristic zero. Almost two decades
later, Giambruno and Koshlukov (2001,[9]) found a basis for the identities of
Grassmann algebra over an infinite field of characteristic $p>2$. Briefly,
when the ground field is infinite and its characteristic is not equal to two,
the identities of the Grassmann algebra follow from the triple commutator.

When the ground field is finite, its characteristic is $p>2$, and its size is
$q$, it is necessary to include one further identity in the basis. In this
situation, the identities follow from the triple commutator and the polynomial
$x_{1}^{pq}-x_{1}^{p}$. The researchers C. Bekh-Ochir and S. Rankin worked on
this problem in (2011, [2]).

There have been many studies on graded identities of the Grassmann algebra in
the last 15 years. When the ground field has characteristic zero, Giambruno,
Mishchenko and Zaicev (2001,[8]) described the $\mathbb{Z}_{2}$-graded
identities (respectively $\mathbb{Z}_{2}$-graded co-dimension) of the
Grassmann algebra equipped with its canonical grading. Anisimov (2001, [1])
and Da Silva (2009, [5]) finished the computation of $\mathbb{Z}_{2}$-co-
dimensions of Grassmann algebra in the case of a basis of $V$ being
homogeneous in this grading. Da Silva and Di Vincenzo (2009, [6]) described
the $\mathbb{Z}_{2}$-graded identities of the Grassmann algebra for any non-
trivial $\mathbb{Z}_{2}$-grading such that a basis of $V$ is homogeneous in
this grading.

When the ground field is infinite, with positive characteristic, Centrone
(2011,[3]) provided a framework, that describes the $\mathbb{Z}_{2}$-graded
identities of the Grassmann algebra in the situation explored by Da Silva and
Di Vincenzo. As an important contribution to this study, Centrone investigated
the situation in which it is necessary to include the identity $z^{p}$ in the
basis, given a determined $\mathbb{Z}_{2}$-grading.

In this paper, the ground field is finite. We found a basis for the graded
polynomial identities of the Grassmann algebra, for any non-trivial grading,
such that a basis of $V$ is homogeneous in this grading.

So far, our case same as that explored by Centrone. The inclusion of the
identity $y_{1}^{pq}-y_{1}^{p}$ in the present work.

We are going to use some results of Regev [12] and Ochir- Rankin [2] in the
beginning of this print. The work of Centrone [3] and Ochir-Rankin [2]
provided the basis for the strategy that will be employed to prove the main
theorems. The paper of Siderov and Chiripov [4] motivated the construction of
the SS Total Order. The essay of Da Silva and Di Vincenzo [6] was very
important in the section titled Case 3 and contributed to the majority of the
computational lemmas.

##  2\. Preliminaries

In this paper, $F$ will denote a fixed finite field of $charF=p>2$ and
$|F|=q$. Moreover, all vector spaces and all algebras are going to be
considered over $F$. The Greek letter Lambda will be an element of $F$.

######  Definition 2.1.

The algebra $A$ is $\mathbb{Z}_{2}$-graded when $A$ can be written as a sum of
subspaces $A=A_{0}\oplus A_{1}$ such that for all $i,j\in\mathbb{Z}_{2}$,
$A_{i}A_{j}\subset A_{i+j}$. The decomposition $(A_{0},A_{1})$ is called a
$\mathbb{Z}_{2}$-grading on $A$. We shall call $A_{0}$ the even component and
$A_{1}$ the odd component. The $\mathbb{Z}_{2}$-grading $(A,0)$ is called
trivial. An element $a\in A$ is called a homogeneous element when $a\in
A_{0}\cup A_{1}$ and we denote its $\mathbb{Z}_{2}$-degree (when $a\neq 0$) as
$\alpha(a)$.

######  Remark 2.2.

It is well known that $A$ can be graded by $\mathbb{Z}_{2}$ (in a non-trivial
way) if, and only if, $A$ admits an automorphism of order two.

If $\phi:A\rightarrow A$ is an automorphism of order two, then
$A_{0}=\\{2^{-1}(a+\phi(a))|a\in A\\}$ and $A_{1}=\\{2^{-1}(a-\phi(a))|a\in
A\\}$.

Let $Y=\\{y_{1},\cdots,y_{n},\cdots\\}$ and
$Z=\\{z_{1},\cdots,z_{n},\cdots\\}$ be two countable sets of variables. We
denote by $F\langle X\rangle$ the free algebra freely generated by $X=Y\cup
Z$. We define the $\mathbb{Z}_{2}$-degree of $1$ by $0$. For any variable
$y_{i}\in Y$, we say that $\alpha(y_{i})=0$; similarly for any variable
$z_{i}\in Z$ we say that $\alpha(z_{i})=0$. We define the
$\mathbb{Z}_{2}$-degree of a monomial $m=x_{1}\cdots x_{n}\in F\langle
X\rangle$ by $\alpha(m)=\alpha(x_{1})+\cdots+\alpha(x_{n})$. In this way,
$F\langle X\rangle$ is a $\mathbb{Z}_{2}$-graded algebra, whereas $F\langle
X\rangle_{0}$ is spanned by the monomials of $\mathbb{Z}_{2}$-degree $0$ and
the empty word $1$, and $F\langle X\rangle_{1}$ is spanned by the monomials of
$\mathbb{Z}_{2}$-degree $1$.

Let $A$ be a $\mathbb{Z}_{2}$-graded algebra. A polynomial
$f(x_{1},\cdots,x_{n})\in F\langle X\rangle$ is called a
$\mathbb{Z}_{2}$-graded polynomial identity for $A$ (or a $2$-graded
polynomial identity for $A$) when $f(a_{1},\cdots,a_{n})=0$ for all $a_{i}\in
A_{\alpha(x_{1})}$, $i=1,\cdots,n$. The set of all identities of $A$ is
denoted by $T_{2}(A)$. An endomorphism $\phi$ of $F\langle X\rangle$ is called
a $\mathbb{Z}_{2}$-graded endomorphism when $\phi(F\langle
X\rangle_{i})\subset F\langle X\rangle_{i},i=0,1$. An ideal $I\subset A$ is a
$\mathbb{Z}_{2}$-graded ideal when $I=(I\cap A_{0})\oplus(I\cap A_{1})$. It is
called a $T_{2}$-ideal when $\phi(I)\subset I$ for all $\mathbb{Z}_{2}$\-
graded endomorphisms of $F\langle X\rangle$. It is not hard to see that
$T_{2}(A)$ is a $T_{2}$-ideal. Let $S$ be a non-empty subset of $F\langle
X\rangle$. We define the $T_{2}$-ideal generated by $S$ as the intersection of
all $T_{2}$-ideals that contain $S$, and we denote it by $\langle S\rangle$. A
polynomial $f$ is said to be a consequence of $S$ when $f\in\langle S\rangle$.
We say that $S\subset F\langle X\rangle$ is a basis for the
$\mathbb{Z}_{2}$-graded identities of $A$ when $T_{2}(A)=\langle S\rangle$.

Two $\mathbb{Z}_{2}$-graded algebras $A$ and $B$ are called isomorphic (as
super-algebras) when there exists a bijective homomorphism $\rho:A\rightarrow
B$ such that $\rho(A_{i})\subset B_{i}\ \ i=1,2$.

Consider $[x_{1},x_{2}]:=x_{1}x_{2}-x_{2}x_{1}$ the commutator of $x_{1}$ and
$x_{2}$. Inductively, one defines the left normed higher commutator by:

$[x_{1},\cdots,x_{n-1},x_{n}]:=[[x_{1},\cdots,x_{n-1}],x_{n}]\ \
n=3,4,\cdots$.

Subsequently, we shall use the shortened term “commutators” for left normed
higher commutators.

Let
$B=\\{y_{1},\cdots,y_{n},\cdots,z_{1},\cdots,z_{n},\cdots,\cdots,[x_{1},x_{2}],[x_{1},x_{3}],\cdots,\newline
[x_{2},x_{3}],[x_{2},x_{4}],\cdots,[x_{j_{1}},x_{j_{2}},x_{j_{3}}],\cdots,\cdots,[x_{j_{1}},\cdots,x_{j_{n}}],\cdots\\}$
be an ordered linear basis for the subspace generated by $X$ and commutators
$[x_{i_{1}},\cdots,x_{i_{n}}],n=2,3,\cdots$.

Due to the Poincaré-Birkhoff-Witt Theorem (Theorem 1.3.2, of [7]) and one of
Witt’s theorems (namely, Theorem 1.3.5, of [7], which states that the Lie
subalgebra $L(X)$ of $F\langle X\rangle^{(-)}$, the Lie algebra of $F\langle
X\rangle$, which is generated by $X$, is isomorphic to the free Lie algebra
with $X$ as a set of free generators. Moreover, the universal enveloping
algebra of $L(X)$ is $F\langle X\rangle$.), we have the following elements
form a linear basis for $F\langle X\rangle$ (we will denote this linear basis
by $Pr(X)$):

$x_{i_{1}}^{a_{1}}\cdots
x_{i_{n_{1}}}^{a_{n_{1}}}[x_{j_{1}},\cdots,x_{j_{l}}]^{b_{1}}\cdots[x_{r_{1}},\cdots,x_{r_{t}}]^{b_{n_{2}}},$

$a_{1},\cdots,a_{n_{1}},b_{1},\cdots,b_{n_{2}}$ are non-negative integers,
$x_{i_{1}},\cdots,x_{i_{n_{1}}},[x_{j_{1}},\cdots,x_{j_{l}}],\cdots,\newline
[x_{r_{1}},\cdots,x_{r_{t}}]\in B$, and
$x_{i_{1}}<\cdots<x_{i_{n_{1}}}<[x_{j_{1}},\cdots,x_{j_{l}}]<\cdots<[x_{r_{1}},\cdots,x_{r_{t}}]$.

##  3\. $\mathbb{Z}_{2}$-gradings on Grassmann algebra

Let $G$ be the (unitary) infinite dimensional Grassmann algebra generated by a
vector space $V$ with basis $\\{e_{1},e_{2},\cdots,e_{n},\cdots\\}$.

Note that the set $\mathcal{B}=\\{e_{i_{1}}.\cdots.e_{i_{n}}|n\geq
0,i_{1}<\cdots<i_{n}\\}$ forms a linear basis for $G$.

We denote by $1_{G}$ the unity of $G$. The subalgebra of $G$ with linear basis
$\mathcal{B}-1_{G}$ is the infinite dimensional non-unitary Grassmann algebra
over $F$, and is denoted by $G^{*}$.

######  Definition 3.1.

For $a=e_{i_{1}}\cdots e_{i_{n}}\in\mathcal{B}-\\{1_{G}\\}$, let
$supp(a)=\\{e_{i_{1}},\cdots,e_{i_{n}}\\}$ (support of $a$) and
$wt(a):=|supp(a)|$, while $supp(1_{G})=\emptyset$ and $wt(1_{G})=0$. Now, for
any $g=\sum_{i=1}^{n}\lambda_{i}a_{i}\in G-\\{0\\}$ (where
$a_{i}\in\mathcal{B}$ and $\lambda_{i}\in F-\\{0\\}$). Let
$supp(g):=\cup_{i=1}^{n}supp(a_{i})$ (support of $g$) and
$wt(g):=max\\{wt(a_{i})|i=1,\cdots,n\\}$ (support-length of $g$) and
$dom(g):=\sum_{wt(a_{i})=wt(g)}\lambda_{i}a_{i}$ (dominant part of $g$), while
we define $supp(0)=\emptyset$ and $|supp(0)|=0$.

Consider the following automorphisms of order $2$ on $G$:

$\left\\{\begin{array}[]{l}\phi_{0}:G\rightarrow G\\\ e_{i}\mapsto-
e_{i},i=1,2,\cdots,n,\cdots\\\ \end{array}\right.$

$\left\\{\begin{array}[]{l}\phi_{\infty}:G\rightarrow G\\\ e_{i}\mapsto
e_{i},\ \mbox{if}\ i\ \mbox{is even}\\\ e_{i}\mapsto-e_{i},\ \mbox{if}\ i\
\mbox{is odd}\end{array}\right.$

$\left\\{\begin{array}[]{l}\phi_{k^{*}}:G\rightarrow G\\\ e_{i}\mapsto-
e_{i},i=1,\cdots,k\\\ e_{i}\mapsto e_{i},i=k+1,\cdots,\end{array}\right.$

$\left\\{\begin{array}[]{l}\phi_{k}:G\rightarrow G\\\ e_{i}\mapsto
e_{i},i=1,\cdots,k\\\ e_{i}\mapsto-e_{i},i=k+1,\cdots,\end{array}\right.$

Each one of those four automorphisms induces a non-trivial
$\mathbb{Z}_{2}$-grading on $G$.

:

Induced by $\phi_{0}$: $G_{can}$ (or $G_{0}$). In this grading,
$V\subset(G_{can})_{1}$;

:

Induced by $\phi_{\infty}$: $G_{\infty}$. In this grading,
$\\{e_{2},\cdots,e_{2n},\cdots\\}\subset(G_{\infty})_{0}$ and
$\\{e_{1},e_{3},\cdots,e_{2n+1},\cdots\\}\subset(G_{\infty})_{1}$;

:

Induced by $\phi_{k^{*}}$: $G_{k^{*}}$. In this grading,
$\\{e_{1},\cdots,e_{k}\\}\subset(G_{k^{*}})_{1}$ and
$\\{e_{k+1},e_{k+2},\cdots,\\}\subset(G_{k^{*}})_{0}$;

:

Induced by $\phi_{k}$: $G_{k}$. In this grading,
$\\{e_{1},\cdots,e_{k}\\}\subset(G_{k})_{0}$ and $\newline
\\{e_{k+1},e_{k+2},\cdots,\\}\subset(G_{k})_{1}$.

From now on, $G$ is going to be owned with a grading induced by one of the
four automorphisms reported above.

##  4\. $\mathbb{Z}_{2}$-graded identities for the Grassmann algebra

In this section, we are going to show some graded identities for $G$.

It is well-known that $[x_{1},x_{2},x_{3}]\in T_{2}(G)$. Another well-known
fact is:

######  Lemma 4.1.

$[x_{1}^{p},x_{2}],[x_{1},x_{2}][x_{2},x_{3}]\in\langle[x_{1},x_{2},x_{3}]\rangle$.

######  Lemma 4.2.

Let $F$ be a field of $charF=p>2$ and $|F|=q$. Then $\newline z_{1}^{p}\in
T_{2}(G)$. Moreover, if $\lambda.1_{G}+a\in G$ ($a\in G\cap G^{*},\lambda\in
F$), then $a^{p}=\lambda^{p}.1_{G}$.

###### Proof.

These results are respectively consequences of Lemma 1.2-b and Corollary 1.5-a
by [12]. ∎

######  Corollary 4.3.

In the notation of Lemma 4.2, we have $y_{1}^{pq}-y_{1}^{p}\in T_{2}(G)$.

######  Definition 4.4.

Let $f=\sum_{j=1}^{l}\lambda_{j}m_{j}\in F\langle y_{1},\cdots,y_{n}\rangle$
be a linear combination of elements from $Pr(X)$, where
$\psi(m_{1})=\cdots=\psi(m_{l})=1$. That polynomial is called a $p$-polynomial
when $Deg_{y_{i}}m_{j}\equiv 0$ mod $p$ and $Deg_{y_{i}}m_{j}<qp$, for all
$i\in\\{1,\cdots,n\\}$ and $j\in\\{1,\cdots,l\\}$.

######  Proposition 4.5.

Let $F$ be a field of $charF=p>2$ and $|F|=q$. Let $f(y_{1},\cdots,y_{n})$ be
a $p$-polynomial. If $f\in T_{2}(G)$ is a $p$-polynomial, then $f$ is the zero
polynomial.

###### Proof.

It is enough to repeat word for word Lemma 3.4 and Corollary 3.1 by [2]. ∎

######  Corollary 4.6.

In the notation of Proposition 4.5, if $f(y_{1},\cdots,y_{m})$ is a non-zero
$p$-polynomial, then there exist $\alpha_{1},\cdots,\alpha_{m}\in F$ such that
$f(\alpha_{1}.1_{G},\cdots,\alpha_{m}.1_{G})\neq 0$.

###### Proof.

According to Proposition 4.5, $f(y_{1},\cdots,y_{m})\notin T_{2}(G)$.

On the other hand, if $\lambda.1_{G}+a\in G$ ($a\in G\cap G^{*}$), then
$a^{p}=\lambda^{p}.1_{G}$. So, there exist $\alpha_{1},\cdots,\alpha_{m}\in F$
such that $f(\alpha_{1}.1_{G},\cdots,\alpha_{m}.1_{G})\neq 0$ and we are done.
∎

##  5\. $SS$ and $SS$ Total Order

Let $u=z_{i_{1}}^{a_{1}}\cdots
z_{i_{n_{1}}}^{a_{n_{1}}}[x_{j_{1}},\cdots,x_{j_{l}}]^{c_{1}}\cdots[x_{r_{1}},\cdots,x_{r_{t}}]^{c_{n_{2}}}$
be an element of $Pr(X)$. For modulo
$\langle[x_{1},x_{2},x_{3}],z_{1}^{p}\rangle$, $u$ can be written as:

$(\prod_{r=1}^{n_{1}}z_{i_{r}}^{b_{i_{r}}})[x_{t_{1}},x_{t_{2}}]\cdots[x_{t_{2l-1}},x_{t_{2l}}]$,

where $0\leq b_{i_{1}},\cdots,b_{i_{m}}<p-1;\
x_{t_{1}}<\cdots<x_{t_{2l}};\lambda\in\\{-1,0,1\\}$.

######  Definition 5.1.

Let
$a=(\prod_{r=1}^{n}y_{j_{r}}^{a_{j_{r}}})(\prod_{r=1}^{m}z_{i_{r}}^{b_{i_{r}}})[x_{t_{1}},x_{t_{2}}]\cdots[x_{t_{2l-1}},x_{t_{2l}}]\in
Pr(X)$. We say that $a\in SS$ when $0\leq
a_{j_{1}},\cdots,a_{j_{n}},b_{i_{1}},\cdots,b_{i_{m}}\leq p-1$, and $\psi(a)$
is multilinear or $\psi(a)=1$.

######  Definition 5.2.

Let
$a=(\prod_{r=1}^{n}y_{j_{r}}^{a_{j_{r}}})(\prod_{r=1}^{m}z_{i_{r}}^{b_{i_{r}}})[x_{t_{1}},x_{t_{2}}]\cdots[x_{t_{2l-1}},x_{t_{2l}}]\in
SS$. We define:

:

$beg(a):=(\prod_{r=1}^{n}y_{j_{r}}^{a_{j_{r}}})(\prod_{r=1}^{m}z_{i_{r}}^{b_{i_{r}}})$
and $\psi(a):=x_{t_{1}}\cdots x_{t_{2l}}$;

:

$\Pi(Y)(a):=(\prod_{r=1}^{n}y_{j_{r}}^{a_{j_{r}}})$ and
$\Pi(Z)(a):=(\prod_{r=1}^{m}z_{i_{r}}^{b_{i_{r}}})$;

:

$pr(z)(a)=z_{i_{1}}$,
$\Pi_{1}(Z)(a):=(z_{i_{1}}^{b_{i_{1}}-1}\prod_{r=2}^{m}z_{i_{r}}^{b_{i_{r}}})$
(if $b_{i_{1}}\geq 1$);

:

$Deg_{x_{i}}a$: the number of times that the variable $x_{i}$ appears in
$beg(a)\psi(a)$;

:

$deg_{Y}a:=\sum_{y\in Y}Deg_{y}(a)$, $deg_{Z}a:=\sum_{z\in Z}Deg_{z}(a)$ and
$dega:=deg_{Z}a+deg_{Y}a$.

######  Definition 5.3.

Let $u,v\in SS$ such that $\psi(u)=\psi(v)=1$. We say that $u<_{lex-rig}v$
when $Deg_{x_{1}}u<Deg_{x_{1}}v$ for some $x_{1}\in X$. Moreover,
$Deg_{x}u=Deg_{x}v$ for every $x>x_{1}$ (with respect to ordered basis of
$L(X)$).

######  Definition 5.4 (SS Total Order).

Given two elements $u,v\in SS$, we say that $u<v$ when:

:

$degu<degv$ or;

:

$degu=degv$, but $beg(u)<_{lex-rig}beg(v)$ or;

:

$degu=degv,beg(u)=beg(v)$, but $\psi(u)<_{lex-rig}\psi(v)$.

######  Remark 5.5.

Notice that if $u\in SS-\\{1\\}$, then $1<u$.

##  6\. Test polynomials

The next proposition is immediate consequence of Lemmas 4.1, 4.2, and
Corollary 4.3.

######  Proposition 6.1.

Let $F$ be a field of $charF=p>2$ and $|F|=q$. Let
$f=\sum_{i=1}^{n}\lambda_{i}v_{i}$ be a linear combination from $Pr(X)$.

For modulo $\langle[x_{1},x_{2},x_{3}],z_{1}^{p},y_{1}^{pq}-y_{1}^{p}\rangle$,
$f$ can be written as:

$\sum_{i=1}^{m}f_{i}u_{i}$,

where $f_{1},\cdots,f_{m}$ are $p$-polynomials and $u_{1},\cdots,u_{m}\in SS$
is (are) distinct.

######  Definition 6.2.

A test polynomial is an expression of the form:

$f=f_{0}+\sum_{i=1}^{n}f_{i}u_{i}$,

where $f_{0}$ is a $p$-polynomial and $f_{1},\cdots,f_{n}$ are non-zero
$p$-polynomials. Moreover $u_{1},\cdots,u_{n}\in SS-\\{1\\}$ is (are)
distinct. The Leading Term of $f$ ($LT(f)$) is the greatest element of
$\\{u_{1},\cdots,u_{n}\\}$.

######  Definition 6.3.

Let $f=f_{0}+\sum_{j=1}^{n}f_{j}u_{j}$ be a test polynomial, where $n\geq 2$
and $deg_{Z}beg(LT(f))>0$. An element $u_{i}\in\\{u_{1},\cdots,u_{n}\\}$ is
called a bad term of $f$ when the following assertions hold true:

  1. (1)

$Deg_{x}(u_{i})=Deg_{x}(LT(f))$ for all $x\in X$, i.e, $u_{i}$ and $LT(f)$
have the same multi-degree;

  2. (2)

If $z\neq pr(z)(LT(f))$, then $Deg_{z}beg(LT(f))=Deg_{z}beg(u_{i})$;

  3. (3)

If $z=pr(z)(LT(f))$, then $Deg_{z}beg(u_{i})+1=Deg_{z}beg(LT(f))$;

  4. (4)

If $x\in Y$, then $Deg_{x}beg(LT(f))\leq Deg_{x}beg(u_{i})$.

We denote by $LBT(f)$ the greatest bad term of $f$ (leading bad term of $f$).

######  Lemma 6.4.

In the notation of Definition 6.3, if $u_{i}$ is a bad term of $f$, we have:

  1. (1)

$\Pi_{1}(Z)(LT(f))=\Pi(Z)(u_{i})$;

  2. (2)

If $z_{1}=pr(z)(LT(f))$, then $Deg_{z_{1}}(\psi(u_{i}))=1$;

  3. (3)

There exists a variable $x\in Y$ such that
$Deg_{x}beg(LT(f))<Deg_{x}beg(u_{i})$;

  4. (4)

$deg_{Z}(beg(u_{i}))+deg_{Y}(\psi(u_{i}))<deg_{Z}(beg(LT(f)))+deg_{Y}(\psi(LT(f)))$;

  5. (5)

$\Pi(Y)(LT(f))<_{lex-rig}\Pi(Y)(LBT(f))$.

###### Proof.

Statements 1 and 2 are immediate consequences of the definition of bad term.

Statement 3: suppose on the contrary that
$Deg_{y}beg(LT(f))=Deg_{y}beg(u_{i})$ for all $y\in Y$. So
$deg_{Y}\psi(u_{i})=deg_{Y}\psi(LT(f))$. Moreover, notice that
$deg_{Z}\psi(u_{i})=deg_{Z}\psi(LT(f))+1$. However,
$deg(\psi(LT(f)))-deg(\psi(u_{i}))$ is an even integer. This fact proves the
statement 4.

Statements 4 and 5 are immediate consequences of statement 3 and the
definition of bad term. ∎

######  Definition 6.5.

Let $f(x_{1},\cdots,x_{m})=f_{0}+\sum_{t=1}^{n}f_{t}u_{t}$ be a test
polynomial and $\newline u_{i}\in\\{u_{1},\cdots,u_{n}\\}$. A
$T_{u_{i}}$-sequence is an $m$-tuple, $(a_{1},\cdots,a_{j},\cdots,a_{m})\in
G^{m}$, whose elements $a_{j}$ satisfy:

:

1) If $Deg_{x_{j}}u_{i}=0$, $a_{j}=0$.

:

2) If $Deg_{x_{j}}u_{i}>0$, $a_{j}$ is a sum of homogeneous elements (each
homogeneous element has coefficient $1$) of $\mathbb{Z}_{2}$-degree
$\alpha(x_{j}).$ Moreover:

2.1) The number of summands of $a_{j}$ is $Deg_{x_{j}}u_{i}$.

2.2) If $a_{j1}$ and $a_{j2}$ are summands of $a_{j}$, $supp(a_{j1})\cap
supp(a_{j2})=\emptyset$.

2.3) If $Deg_{x_{j}}(\psi(u_{i}))=0$, the summand (summands) of $a_{j}$ has
(have) support-length $2$.

2.4) If $Deg_{x_{j}}(\psi(u_{i}))=1$ and $Deg_{x_{j}}u_{i}\geq 2$, only one
summand of $a_{j}$ has support-length equal to $1$, but the other summand
(summands) of $a_{j}$ has (have) support-length $2$.

2.5) If $Deg_{x_{j}}(\psi(u_{i}))=1$ and $Deg_{x_{j}}u_{i}=1$, $a_{j}$ has
support-length $1$.

Furthermore, if $r\neq l$, $supp(a_{r})\cap supp(a_{l})=\\{\\}$.

######  Definition 6.6.

Let $f(x_{1},\cdots,x_{m})=f_{0}+\sum_{t=1}^{n}f_{t}u_{t}$ be a test
polynomial, where $deg_{Z}(beg(u_{i}))>0$ for some
$u_{i}\in\\{u_{1},\cdots,u_{n}\\}$. An Almost Type-$u_{i}$ sequence is an
$m$-tuple, $AT_{u_{i}}=(a_{1},\cdots,a_{j},\cdots,a_{m})\in G^{m}$ that is
defined like a Type-$u_{i}$ sequence, with the exception of property 2.3.
Instead, the following properties hold:

:

I) If $x_{j}\neq pr(z)(u_{i})$ and $Deg_{x_{j}}(\psi(u_{i}))=0$, the summand
(summands) of $a_{j}$ has (have) support-length $2$;

:

II) If $x_{j}=pr(z)(u_{i})$, $Deg_{x_{j}}(\psi(u_{i}))=0$ and
$Deg_{x_{j}}(u_{i})\geq 2$, only one summand of $a_{j}$ has support-length
$1$, but the other has (have) support-length $2$;

:

III) If $x_{j}=pr(z)(u_{i})$, $Deg_{x_{j}}(\psi(u_{i}))=0$ and
$Deg_{x_{j}}(u_{i})=1$, then $a_{j}$ has support-length $1$.

In the notation of Definition 6.5 (respectively Definition 6.6), we say that
an element $g\in\mathcal{B}$ is complete with respect to $T_{u_{i}}$
(respectively $AT_{u_{i}}$) when:

$supp(g)=(\bigcup_{i=1}^{m}supp(a_{i}))$.

######  Definition 6.7.

Let $f(x_{1},\cdots,x_{m})=f_{0}+\sum_{t=1}^{n}f_{t}u_{t}$ be a test
polynomial. A Scalar-$f$ sequence is an $m$-tuple
$(\lambda_{1}.1_{G},\cdots,\lambda_{m}.1_{G})\in(F.1_{G})^{m}$, where
$\lambda_{i}=0$ if $x_{i}\in Z$.

The next two lemmas may readily be seen from an adaptation of the Binomial
Theorem.

######  Lemma 6.8.

Let $u_{1}=x_{1}^{m}$ be a test polynomial. If $(\sum_{i=1}^{m}a_{i})$ is a
Type-$u$ sequence and $(\lambda_{1}.1_{G})$ is a Scalar-$u$ sequence, then:

:

$dom\ \ u(\sum_{i=1}^{m}a_{i})=m!a_{1}.\cdots.a_{m}$;

:

$dom\ \ u(\lambda_{1}.1_{G}+\sum_{i=1}^{m}a_{i})=m!a_{1}.\cdots.a_{m}$

######  Lemma 6.9.

Let $u=z_{1}^{m}$ be a test polynomial. If $(\sum_{i=1}^{m}a_{i})$ is an
Almost Type-$u$ sequence, then:

:

$dom\ \ u(\sum_{i=1}^{m}a_{i})=m!a_{1}.\cdots.a_{m}$.

######  Lemma 6.10.

Let $f(x_{1},\cdots,x_{m})=f_{0}+\sum_{t=1}^{n}f_{t}u_{t}$ be a test
polynomial. Consider $u_{i}\in\\{u_{1},\cdots,u_{n}\\}$. If a Type-$u_{i}$
sequence $T_{u_{i}}=(A_{1},\cdots,A_{m})$ exists, then:

:

$dom(u_{i}(A_{1},\ldots,A_{m}))=\lambda.g$, where $\lambda\in F-\\{0\\}$ and
$g\in\mathcal{B}$ is complete with respect to $T_{u_{i}}$;

:

If $(\lambda_{1}.1_{G},\cdots,\lambda_{m}.1_{G})\in G^{m}$ is a Scalar-$f$
sequence, then $\newline
dom(u_{i}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m}))=dom(u_{i}(A_{1},\cdots,A_{m}))$.

###### Proof.

First of all, put $A_{j}=\sum_{k=1}^{Deg_{x_{j}}u_{i}}a_{j,k}$.

According to assumption, there exists a Type-$u_{i}$ sequence $T_{u_{i}}$.
Therefore, by Lemma 6.8:

$dom(u_{i}(A_{1},\cdots,A_{m}))=dom(u_{i}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m})=\newline
\pm
2^{l}(\prod_{r=1}^{m}n_{r}!)(\prod\limits_{(j,k(j))\in\mathcal{X}}a_{jk(j)}),$

where

:

$2l=deg(\psi(u_{i}))$;

:

$n_{r}=Deg_{x_{r}}beg(LT(f))$;

:

$\mathcal{X}(j):=\\{(j,k(j))|k(j)=1,\cdots,Deg_{x_{j}}u_{i}\\}$, and
$\mathcal{X}:=\bigcup_{j=1}^{m}\mathcal{X}(j)$.

∎

Following word for word the proof of Lemma 6.10, it is possible to prove the
following lemma.

######  Lemma 6.11.

Let $f(x_{1},\cdots,x_{m})=f_{0}+\sum_{t=1}^{n}f_{t}u_{t}$ be a test
polynomial, where $deg_{Z}beg(u_{i})\geq 1$ for some
$u_{i}\in\\{u_{1},\cdots,u_{n}\\}$. If there exists an Almost Type-$u_{i}$
sequence $(A_{1},\cdots,A_{m})$, then:

:

$dom(u_{i}(A_{1},\ldots,A_{m}))=\lambda.g$, where $\lambda\in F-\\{0\\}$ and
$g\in\mathcal{B}$ is complete with respect to $AT_{u_{i}}$;

:

If $(\lambda_{1}.1_{G},\cdots,\lambda_{m}.1_{G})\in G^{m}$ is a Scalar-$f$
sequence, then $\newline
dom(u_{i}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m}))=dom(u_{i}(A_{1},\cdots,A_{m}))$.

##  7\. Leading terms

In this section, we prove the important results that are going to be used in
the next three sections.

######  Lemma 7.1.

Let $f(x_{1},\cdots,x_{m})=\sum_{t=1}^{n}u_{t}$ be a test polynomial and let
$\newline (\lambda_{1}1_{G},\cdots,\lambda_{m}1_{G})$ be a Scalar-$f$
sequence. If there exists a Type-$LT(f)$ sequence $(A_{1},\cdots,A_{m})$,
then:

:

$dom(f(\lambda_{1}1_{G}+A_{1},\cdots,\lambda_{m}1_{G}+A_{m}))=dom(LT(f)(A_{1},\cdots,A_{m}))$.

Moreover, if $n>1$ and $u_{i}<LT(f)$:

$wt(u_{i}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m}))<wt(LT(f)(A_{1},\cdots,A_{m}))$.

###### Proof.

If $n=1$, the proof is an immediate consequence of Lemma 6.10. Suppose that
$n>1$ and consider $u_{i}<LT(f)$. It is enough to prove that no summand of
$\newline dom(u_{i}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m}))$
is complete with respect to $(A_{1},\cdots,A_{m})$.

:

Case 1: $degu_{i}<degLT(f)$. In this situation, a variable $x_{j}$ that
appears more times in $LT(f)$ than $u_{i}$ exists. According to assumption,
$(A_{1},\cdots,A_{m})$ is a Type-$LT(f)$ sequence. So, if
$Deg_{x_{j}}beg(LT(f))>Deg_{x_{j}}beg(u_{i})$, then the summand (summands) of
$u_{i}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m})$ has (have), at
most, $|supp(A_{j})|-2$ elements of $supp(A_{j})$. If
$Deg_{x_{j}}(beg(LT(f)))=Deg_{x_{j}}(beg(u_{i}))$, then the summand (summands)
of $u_{i}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m})$ has (have),
at most, $|supp(A_{j})|-1$ elements of $supp(A_{j})$.

:

Case 2: $degu_{i}=degLT(f)$ and $beg(u_{i})<_{lex-rig}beg(LT(f))$. In this
situation, there exists a variable $x_{j}$ that appears more times in
$beg(LT(f))$ than $beg(u_{i})$. Consequently, the summand (summands) of
$u_{i}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m})$ has (have), at
most, $|supp(A_{j})|-2$ elements of $supp(A_{j})$.

:

Case 3: $degu_{i}=degLT(f)$, $beg(LT(f))=beg(u_{i})$, but $\psi(u_{i})<_{lex-
rig}\psi(LT(f))$. In this case, there exists a variable $x_{j}$ that appears
in $\psi(LT(f))$ but does not appear in $\psi(u_{i})$. Consequently, the
summand (summands) of
$u_{i}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m})$ has (have), at
most, $|supp(A_{j})|-1$ elements of $supp(A_{j})$. We are done.

∎

######  Lemma 7.2.

Let $f(x_{1},\cdots,x_{m})=\sum_{t=1}^{n}u_{t}$ be a test polynomial
($deg_{Z}beg(LT(f))>0$), and $(\lambda_{1}.1_{G},\cdots,\lambda_{m}.1_{G})$ a
Scalar-$f$ sequence. If $f$ does not admit a bad term and there exists an
Almost Type-$LT(f)$ sequence $(A_{1},\cdots,A_{m})$, then:

:

$dom(f(\lambda_{1}1_{G}+A_{1},\cdots,\lambda_{m}1_{G}+A_{m}))=dom(LT(f)(A_{1},\cdots,A_{m}))$.

Moreover, if $n>1$ and $u_{i}<LT(f)$, then:

$wt(u_{i}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m}))<wt(LT(f)(A_{1},\cdots,A_{m}))$.

###### Proof.

If $n=1$, the proof is obvious.

Notice that $Deg_{pr(z)(LT(f))}\psi(LT(f))=0$. Suppose that $n>1$ and consider
$u_{i}<LT(f)$. It is enough to prove that no summand
$dom(u_{i}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m}))$ is
complete with respect to $(A_{1},\cdots,A_{m})$.

:

Case 1: $deg(u_{i})<deg(LT(f))$. This is similar to Case 1 of Lemma 7.1.

:

Case 2: $deg(u_{i})=deg(LT(f))$, but $beg(u_{i})<_{lex-rig}beg(LT(f))$.

Subcase 2.1: There exists $x_{j}\in X-\\{pr(z)(LT(f))\\}$
such that $Deg_{x_{j}}beg(u_{i})<Deg_{x_{j}}beg(LT(f))$. In this situation, it
is not hard to see that no summand of
$u_{i}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m})$ contains more
than $|supp(A_{j})|-2$ elements of $supp(A_{j})$.

Subcase 2.2: $Deg_{pr(z)(LT(f))}beg(u_{i})<Deg_{pr(z)(LT(f))}LT(f)$ and
$\newline Deg_{x}beg(LT(f))\leq Deg_{x}beg(u_{i})$ for all $x\in
X-\\{pr(z)(LT(f))\\}$.

Notice that if $Deg_{pr(z)(LT(f))}beg(LT(f))-Deg_{pr(z)(LT(f))}beg(u_{i})=1$,
$u_{i}$ and $LT(f)$ have the same multi-degree, then there is $x\in Y$ such
that $Deg_{x}beg(u_{i})<Deg_{x}beg(LT(f))$, because $u_{i}$ is not a bad term.
Moreover, if $Deg_{pr(z)(LT(f))}beg(LT(f))-Deg_{pr(z)(LT(f))}beg(u_{i})=1$,
$u_{i}$ and $LT(f)$ do not have the same multi-degree
($Deg_{pr(z)(LT(f))}\psi(u_{i})=1$), there must exist $x_{i}\in
X-\\{pr(z)(LT(f))\\}$ such that $Deg_{x_{i}}u_{i}<Deg_{x_{i}}LT(f)$. The
remaining part of the proof is similar to Case 1 of Lemma 7.1.

Subcase 2.2.1: $Deg_{pr(z)(LT(f))}beg(LT(f))-Deg_{pr(z)(LT(f))}beg(u_{i})>1$.
The analysis and the conclusion of this subcase are similar to Subcase 2.1.

Subcase 2.2.2: $Deg_{pr(z)(LT(f))}beg(LT(f))-Deg_{pr(z)(LT(f))}beg(u_{i})=1$,
$u_{i}$ and $LT(f)$ do not have the same multi-degree. If
$Deg_{pr(z)(LT(f))}\psi(u_{i})=0$, the analysis is similar to Case 2 of Lemma
7.1.

:

Case 3: $degu_{i}=degLT(f)$, $beg(u_{i})=beg(LT(f))$, but $\psi(u_{i})<_{lex-
rig}\psi(LT(f))$. This is similar to Case 3 of Lemma 7.1.

∎

######  Lemma 7.3.

Let $f(x_{1},\cdots,x_{m})=\sum_{t=1}^{n}u_{t}$ be a test polynomial
($deg_{Z}beg(LT(f))>0$), and let
$(\lambda_{1}.1_{G},\cdots,\lambda_{m}.1_{G})$ be a Scalar-$f$ sequence. If
$f$ admits a bad term and there exists a Type-$LBT(f)$ sequence
$(A_{1},\cdots,A_{m})$, then:

:

$dom(f(\lambda_{1}1_{G}+A_{1},\cdots,\lambda_{m}1_{G}+A_{m}))=dom(LBT(f)(A_{1},\cdots,A_{m}))$.

Moreover, if $u_{i}\neq LBT(f)$:

$wt(u_{i}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m}))<wt(LBT(f)(A_{1},\cdots,A_{m}))$.

###### Proof.

It is enough to prove that if $(A_{1},\cdots,A_{m})$ is a Type-$LBT(f)$
sequence and $u_{i}\neq LBT(f)$, then no summand of
$dom(u_{i}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m}))$ is
complete with respect to $(A_{1},\cdots,A_{m})$.

:

Case 1: $u_{i}<LBT(f)$. It is enough to repeat word for word the proof of
Lemma 7.1.

:

Case 2: $LBT(f)<u_{i}\leq LT(f)$, but $u_{i}$ and $LBT(f)$ do not have the
same multi-degree. First, notice that $deg(LBT(f))=deg(u_{i})$. In this
situation, there must exist $x_{i}\in X$ such that
$Deg_{x_{i}}u_{i}<Deg_{x_{i}}LBT(f)$. The remaining part of the proof is
similar to Case 1 of Lemma 7.1.

:

Case 3: $LBT(f)<u_{i}\leq LT(f)$, $u_{i}$ and $LBT(f)$ have the same multi-
degree. By the definition of bad term: $deg(LT(f))=deg(u_{i})=deg(LBT(f))$,
and $\Pi_{1}(Z)(LT(f))=\Pi(Z)(LBT(f))$. Thus, $\Pi(Z)(LBT(f))\leq_{lex-
rig}\Pi(Z)(u_{i})\leq_{lex-rig}\Pi(Z)(LT(f))$.

Subcase 3.1: $\Pi(Z)(u_{i})=\Pi(Z)(LT(f))$. In this situation,
$\Pi(Y)(u_{i})\leq_{lex-rig}\Pi(Y)(LT(f))$, because $beg(u_{i})\leq_{lex-
rig}beg(LT(f))$. On the other hand, $\newline \Pi(Y)(LT(f))<_{lex-
rig}\Pi(Y)(LBT(f))$. So, there must exist $x_{i}\in Y$ such that
$Deg_{x_{i}}beg(u_{i})<Deg_{x_{i}}beg(LBT(f))$. The remaining part of the
proof is similar to Case 2 of Lemma 7.1.

Subcase 3.2: $\Pi(Z)(u_{i})<_{lex-rig}\Pi(Z)(LT(f))$. In this situation,
$\Pi(Z)(LBT(f))=\Pi(Z)(u_{i})$. By the definitions of leading term and bad
term, there must exist $x_{i}\in Y$ such that
$Deg_{x_{i}}beg(u_{i})<Deg_{x_{i}}beg(LBT(f))$. The remaining part of the
proof is similar to Case 2 of Lemma 7.1.

∎

######  Proposition 7.4.

Let $f(x_{1},\cdots,x_{m})=f_{0}+\sum_{t=1}^{n}f_{t}u_{t}$ be a test
polynomial.If there exists a Type-$LT(f)$ sequence
$T_{LT(f)}=(A_{1},\cdots,A_{m})$, then there is a Scalar-$f$ sequence
$(\lambda_{1}.1_{G},\cdots,\lambda_{m}.1_{G})$ such that:

$dom(f(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m}))=\lambda.g$,

where $\lambda\in F-\\{0\\}$ and $g\in\mathcal{B}$ is complete with respect to
$T_{LT(f)}$.

###### Proof.

Suppose without loss of generality that $LT(f)=u_{1}$. According to Corollary
4.6, there exists a Scalar-$u_{1}$ sequence
$(\lambda_{1}1_{G},\cdots,\lambda_{n}.1_{G})$ such that
$f_{1}(\lambda_{1}1_{G},\cdots,\lambda_{m}1_{G})=\alpha.1_{G}\neq 0$.

Consider the polynomial $g=\sum_{t=1}^{n}u_{t}$. According to assumption,
there exists a Type-$u_{1}$ sequence $(A_{1},\cdots,A_{m})$. Therefore, by
Lemmas 6.10 and 7.1:
$dom(g(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m}))=dom(u_{1}(A_{1},\cdots,A_{m}))=\lambda.g\neq
0$, where $g\in\mathcal{B}$ is complete with respect to $T_{u_{1}}$. Moreover,
if $n\geq 2$ and $u_{i}<u_{1}$:

$wt(u_{i}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m}))<wt(u_{1}(A_{1},\cdots,A_{m}))$.

On the other hand, by Lemma
6.10,$dom(u_{1}(A_{1},\cdots,A_{m}))=dom(u_{1}(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m}))$
and consequently:

$dom(f(\lambda_{1}1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m}))=f_{1}(\lambda_{1}.1_{G},\cdots,\lambda_{m}.1_{G}).dom(u_{1}(A_{1},\cdots,A_{m}))=\alpha.\lambda.g$.

∎

######  Proposition 7.5.

Let $f(x_{1},\cdots,x_{m})=f_{0}+\sum_{t=1}^{n}f_{t}u_{t}$ be a test
polynomial
($deg_{Z}beg(LT(f))>0$). If $f$ does not admit a bad term and there exists an
Almost Type-$LT(f)$ sequence $AT_{LT(f)}=(A_{1},\cdots,A_{m})$, then there is
a Scalar-$f$ sequence $(\lambda_{1}.1_{G},\cdots,\lambda_{m}.1_{G})$ such
that:

$dom(f(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m}))=\lambda.g$,

where $\lambda\in F-\\{0\\}$ and $g\in\mathcal{B}$ is complete with respect to
$AT_{LT(f)}$.

###### Proof.

It is enough to repeat word for word the proof of Proposition 7.4, with two
exceptions: instead of Lemma 6.10, we will apply Lemma 6.11; instead of Lemma
7.1, we will apply Lemma 7.2. ∎

######  Proposition 7.6.

Let $f(x_{1},\cdots,x_{m})=f_{0}+\sum_{t=1}^{n}f_{t}u_{t}$ be a test
polynomial
($deg_{Z}(beg(LT(f)))>0$). If $f$ admits a bad term and there exists a
Type-$LBT(f)$ sequence $T_{LBT(f)}=(A_{1},\cdots,A_{m})$, then there is a
Scalar-$f$ sequence $(\lambda_{1}.1_{G},\cdots,\lambda_{m}.1_{G})$ such that:

$dom(f(\lambda_{1}.1_{G}+A_{1},\cdots,\lambda_{m}.1_{G}+A_{m}))=\lambda.g$,

where $\lambda\in F-\\{0\\}$ and $g\in\mathcal{B}$ is complete with respect to
$T_{LBT(f)}$.

###### Proof.

It is enough to repeat word for word the proof of Proposition 7.4, with two
exceptions: instead of Lemma 7.1, we will apply Lemma 7.3; we assume that
$LBT(f)=u_{1}$. ∎

##  8\. Case 1: $G_{can}$ and $G_{\infty}$

In this section, we describe the $\mathbb{Z}_{2}$-graded identities of
$G_{can}$ and $G_{\infty}$.

It is well known that $[y_{1},y_{2}],[y_{1},z_{2}],z_{1}z_{2}+z_{2}z_{1}\in
T_{2}(G_{can})$. Moreover, due to the graded identity $z_{1}z_{2}+z_{2}z_{1}$,
we have $z_{1}^{k}\in T_{2}(G_{can})$ for $k>1$.

######  Theorem 8.1.

Let $F$ be a field of $charF=p>2$ and $|F|=q$. The $\mathbb{Z}_{2}$-graded
identities of $G$ (with grading induced by $\phi_{0}$) follow from:

$[y_{1},y_{2}],[y_{1},z_{2}],z_{1}z_{2}+z_{2}z_{1}$ and
$y_{1}^{pq}-y_{1}^{p}$.

###### Proof.

Let $I$ be the $T_{2}$-ideal generated by
$[y_{1},y_{2}],[y_{1},z_{2}],z_{1}z_{2}+z_{2}z_{1}$ and
$y_{1}^{pq}-y_{1}^{p}$. Suppose by contradiction that $I\subsetneq
T_{2}(G_{can})$. Thus, there would exist a polynomial
$f=\sum_{i=1}^{r}\lambda_{i}v_{i}\in T_{2}(G_{can})-I$, where
$v_{1},\cdots,v_{n}\in Pr(X)$ is (are) distinct. Note that $f$ is not a
$p$-polynomial. We may suppose that $f$ is a test polynomial of the form:

$f=f_{0}+\sum_{i=1}^{l}f_{i}y_{1}^{A_{i,1}}\cdots
y_{m}^{A_{i,m}}[z_{1},z_{2}]\cdots[z_{n-1},z_{n}]\ \ \mbox{if}\ \ n\ \mbox{is
even}\ \mbox{or}\newline f=f_{0}+\sum_{i=1}^{l}f_{i}y_{1}^{A_{i,1}}\cdots
y_{m}^{A_{i,m}}z_{1}\ \ \mbox{or}\newline
f=f_{0}+\sum_{i=1}^{l}f_{i}y_{1}^{A_{i,1}}\cdots
y_{m}^{A_{i,m}}z_{1}[z_{2},z_{3}]\cdots[z_{n-1},z_{n}]\ \ \mbox{if}\ \ n>1\
\mbox{is odd}$.

Without loss of generality, we suppose that $LT(f)=y_{1}^{A_{1,1}}\cdots
y_{m}^{A_{1,m}};A_{1,1},\cdots,\newline A_{1,m}>0$ and
$f=f(y_{1},\cdots,y_{m},z_{1},\cdots,z_{n})$. Notice that $f$ does not admit a
bad term.

We prove the theorem by the following cases.

:

For even $n$. In this situation,
$(e_{1}e_{2}+\cdots+e_{2(b_{1})-1}e_{2(b_{1})},\cdots,\newline
e_{2(\sum_{j=1}^{m-1}b_{j})+1}e_{2(\sum_{j=1}^{m-1}b_{j}+1)}+\cdots+e_{2(\sum_{j=1}^{m}b_{j})-1}e_{2(\sum_{j=1}^{m}b_{j})},e_{M+1},\cdots,e_{M+n})$,
where $b_{j}=A_{1,j}$ and $M=2(b_{1}+\cdots+b_{m})$, is a Type-$LT(f)$. On the
other hand, by Proposition 7.4, we have $f\notin T_{2}(G_{can})$, which is a
contradiction.

:

For odd $n$. In this situation, the sequence above is an Almost Type-$LT(f)$
sequence. On the other hand, by Proposition 7.5, we have $f\notin
T_{2}(G_{can})$, which is a contradiction.

∎

Now, we describe the $\mathbb{Z}_{2}$-graded identities of $G_{\infty}$.

######  Theorem 8.2.

Let $F$ be a field of $charF=p>2$ and $|F|=q$. Then the
$\mathbb{Z}_{2}$-graded identities of $G$ (with grading induced by
$\phi_{\infty}$) follow from:

$[x_{1},x_{2},x_{3}],z_{1}^{p}$ and $y_{1}^{pq}-y_{1}^{p}$.

###### Proof.

Let $I$ be the $T_{2}$-ideal generated by the three identities above. Suppose
by contradiction that $I\varsubsetneq T_{2}(G_{\infty})$. According to
Proposition 6.1, there exists a test polynomial
$f=f_{0}+\sum_{i=1}^{n}f_{i}u_{i}\in T_{2}(G_{\infty})-I$.

We may suppose without loss of generality that:

$LT(f)=y_{1}^{a_{1}}\cdots
y_{n_{1}}^{a_{n_{1}}}y_{n_{1}+1}^{a_{n_{1}+1}}\cdots
y_{n_{2}}^{a_{n_{2}}}z_{1}^{b_{1}}\cdots
z_{m_{1}}^{b_{m_{1}}}z_{m_{1}+1}^{b_{m_{1}+1}}\cdots
z_{m_{2}}^{b_{m_{2}}}\newline
[y_{n_{1}+1},y_{n_{1}+2}]\cdots[y_{n_{2}},y_{n_{2}+1}]\cdots[y_{l_{1}},z_{m_{1}+1}]\cdots[z_{m_{2}-1},z_{m_{2}}]\cdots[z_{l_{2}-1},z_{l_{2}}]$,

where $n_{1}<n_{2}<l_{1},m_{1}<m_{2}<l_{2}$;
$a_{1},\cdots,a_{n_{2}},b_{1},\cdots,b_{m_{2}}>0$.

Suppose that $f=f(y_{1},\cdots,y_{l_{1}},z_{1},\cdots,z_{l_{2}})$. Consider
the following map:

$\phi:\\{y_{1},\cdots,y_{l_{1}},z_{1},\cdots,z_{l_{2}}\\}\rightarrow G\newline
$

$y_{1}\mapsto\sum_{l=1}^{a_{1}}e_{4l-2}e_{4l}\newline \cdots\newline
y_{n_{1}}\mapsto\sum_{l=a_{1}+\cdots+a_{n_{1}-1}+1}^{a_{1}+\cdots+a_{n_{1}}}e_{4l-2}e_{4l}\newline
y_{n_{1}+1}\mapsto
e_{4(\sum_{l=1}^{n_{1}}a_{l})+2}+\sum_{l=1}^{a_{n_{1}+1}}e_{4(\sum_{l=1}^{n_{1}}a_{l})+4l}e_{4(\sum_{l=1}^{n_{1}}a_{l})+4l+2}\newline
\cdots\newline y_{n_{2}}\mapsto
e_{4(\sum_{i=1}^{n_{2}-1}a_{i})+2(n_{2}-n_{1})}+\newline
+\sum_{l=1}^{a_{n_{2}}}e_{4(\sum_{i=1}^{n_{2}-1}a_{i})+2(n_{2}-n_{1})+4l-2}e_{4(\sum_{i=1}^{n_{2}-1}a_{i})+2(n_{2}-n_{1})+4l}\newline
y_{n_{2}+1}\mapsto e_{4(\sum_{i=1}^{n_{2}}a_{i})+2(n_{2}-n_{1}+1)}\newline
\cdots\newline y_{l_{1}}\mapsto
e_{4(\sum_{i=1}^{n_{2}}a_{i})+2(l_{1}-n_{1})}\newline
z_{1}\mapsto\sum_{l=1}^{b_{1}}e_{2l-1}e_{M+2l}\newline \cdots\newline
z_{m_{1}}\mapsto\sum_{l=b_{1}+\cdots+b_{m_{1}-1}+1}^{b_{1}+\cdots+b_{m_{1}}}e_{2l-1}e_{M+2l}\newline
z_{m_{1}+1}\mapsto
e_{2(\sum_{l=1}^{m_{1}}b_{l})+1}+\sum_{l=b_{1}+\cdots+b_{m_{1}}+1}^{b_{1}+\cdots+b_{m_{1}+1}}e_{2l+1}e_{M+2l}\newline
\cdots\newline z_{m_{2}}\mapsto
e_{2(\sum_{l=1}^{m_{2}-1}b_{l})+2(m_{2}-m_{1})-1}+\sum_{l=b_{1}+\cdots+b_{m_{2}-1}+1}^{b_{1}+\cdots+b_{m_{2}}}e_{2(l+m_{2}-m_{1})-1}e_{M+2l}\newline
z_{m_{2}+1}\mapsto e_{2(\sum_{l=1}^{m_{2}}b_{l}+m_{2}-m_{1})+1}\newline
\cdots\newline z_{l_{2}}\mapsto
e_{2(\sum_{l=1}^{m_{2}}b_{l})+2(l_{2}-m_{1})-1}$,

where $M=4(\sum_{i=1}^{n_{2}}a_{i})+2(l_{1}-n_{1})$.

Notice that
$(\phi(y_{1}),\cdots,\phi(y_{l_{1}}),\phi(z_{1}),\cdots,\phi(z_{l_{2}}))$ is a
Type-$LT(f)$ sequence.

On the other hand, by Proposition 7.4, $f\notin T_{2}(G_{\infty})$ that is a
contradiction. ∎

##  9\. Case 2: $G_{k^{*}}$

In this section, we describe the identities of $G_{k^{*}}$.

It is easy to see that $z_{1}\cdots z_{k+1}\in T_{2}(G_{k^{*}})$ (see for
instance, Lemma 4.2 of [3]). Moreover $z_{1}^{p}$ is consequence of
$z_{1}\cdots z_{k+1}$ when $k<p$.

######  Definition 9.1.

An element $a\in SS$ is labeled as Type-1 (or $u\in SS1$) when:

:

$deg_{Z}(a)\leq k$.

Due to identity $z_{1}\cdots z_{k+1}\in T_{2}(G_{k^{*}})$, we have a version
of Proposition 6.1 for $G_{k^{*}}$:

######  Proposition 9.2.

Let $F$ be a field of $charF=p>2$ and $|F|=q$. Let
$f=\sum_{i=1}^{n}\lambda_{i}v_{i}$ be a linear combination from $Pr(X)$. For
modulo $\langle[x_{1},x_{2},x_{3}],z_{1}^{p},y_{1}^{pq}-y_{1}^{p},z_{1}\cdots
z_{k+1}\rangle$, $f$ can be written as:

$\sum_{i=1}^{m}f_{i}u_{i}$,

where $f_{1},\cdots,f_{m}$ are $p$-polynomials and $u_{1},\cdots,u_{m}\in SS1$
is (are) distinct.

######  Theorem 9.3.

Let $F$ be a field of $charF=p>2$ and $|F|=q$. The $\mathbb{Z}_{2}$-graded
identities of $G$ (with grading induced by $\phi_{k^{*}}$) follow from:

$[x_{1},x_{2},x_{3}],z_{1}\cdots z_{k+1},z_{1}^{p}$ and
$y_{1}^{pq}-y_{1}^{p}$.

###### Proof.

Let $I$ be the $T_{2}$-ideal generated by the four identities above. Suppose
by contradiction that $I\varsubsetneq T_{2}(G_{k^{*}})$. According to
Proposition 9.2, there exists a test polynomial
$f=f_{0}+\sum_{i=1}^{n}f_{i}u_{i}\in T_{2}(G_{k^{*}})-I$, where
$u_{1},\cdots,u_{n}\in SS1-\\{1\\}$.

We may suppose without loss of generality that:

$LT(f)=y_{1}^{a_{1}}\cdots
y_{n_{1}}^{a_{n_{1}}}y_{n_{1}+1}^{a_{n_{1}+1}}\cdots
y_{n_{2}}^{a_{n_{2}}}z_{1}^{b_{1}}\cdots
z_{m_{1}}^{b_{m_{1}}}z_{m_{1}+1}^{b_{m_{1}+1}}\cdots
z_{m_{2}}^{b_{m_{2}}}\newline
[y_{n_{1}+1},y_{n_{1}+2}]\cdots[y_{n_{2}},y_{n_{2}+1}]\cdots[y_{l_{1}},z_{m_{1}+1}]\cdots[z_{m_{2}-1},z_{m_{2}}]\cdots[z_{l_{2}-1},z_{l_{2}}]$,

where
$n_{1}<n_{2}<l_{1},m_{1}<m_{2}<l_{2};a_{1},\cdots,a_{n_{2}},b_{1},\cdots,b_{m_{2}}>0$.

Suppose that $f=f(y_{1},\cdots,y_{l_{1}},z_{1},\cdots,z_{l_{2}})$. Consider
the following map:

$\phi:\\{y_{1},\cdots,y_{l_{1}},z_{1},\cdots,z_{l_{2}}\\}\rightarrow G\newline
$ $y_{1}\mapsto\sum_{l=1}^{a_{1}}e_{k+2l-1}e_{k+2l}\newline \cdots\newline
y_{n_{1}}\mapsto\sum_{l=a_{1}+\cdots+a_{n_{1}-1}+1}^{a_{1}+\cdots+a_{n_{1}}}e_{k+2l-1}e_{k+2l}\newline
y_{n_{1}+1}\mapsto
e_{k+2(a_{1}+\cdots+a_{n_{1}})+1}+\sum_{l=1}^{a_{n_{1}+1}}e_{k+2(a_{1}+\cdots+a_{n_{1}})+2l}e_{k+2(a_{1}+\cdots+a_{n_{1}})+2l+1}\newline
\cdots\newline y_{n_{2}}\mapsto
e_{k+2(a_{1}+\cdots+a_{n_{2}-1})+(n_{2}-n_{1})}+\newline
\sum_{l=1}^{a_{n_{2}}}e_{k+2(a_{1}+\cdots+a_{n_{2}-1})+(n_{2}-n_{1})+2l-1}e_{k+2(a_{1}+\cdots+a_{n_{2}-1})+(n_{2}-n_{1})+2l}\newline
y_{n_{2}+1}\mapsto
e_{k+2(a_{1}+\cdots+a_{n_{2}-1}+a_{n_{2}})+(n_{2}-n_{1})+1}\newline
\cdots\newline y_{l_{1}}\mapsto
e_{k+2(a_{1}+\cdots+a_{n_{2}-1}+a_{n_{2}})+(l_{1}-n_{1})}\newline
z_{1}\mapsto\sum_{l=1}^{b_{1}}e_{l}e_{Q+l}\newline \cdots\newline
z_{m_{1}}\mapsto\sum_{l=b_{1}+\cdots+b_{m_{1}-1}+1}^{b_{1}+\cdots+b_{m_{1}}}e_{l}e_{Q+l}\newline
z_{m_{1}+1}\mapsto
e_{b_{1}+\cdots+b_{m_{1}}+1}+\sum_{l=b_{1}+\cdots+b_{m_{1}}+1}^{b_{1}+\cdots+b_{m_{1}+1}}e_{l+1}e_{Q+l}\newline
\cdots\newline z_{m_{2}}\mapsto
e_{b_{1}+\cdots+b_{m_{2}-1}+(m_{2}-m_{1})}+\sum_{l=1}^{b_{m_{2}}}e_{l+b_{1}+\cdots+b_{m_{2}-1}+(m_{2}-m_{1})}e_{Q+b_{1}+\cdots+b_{m_{2}-1}+l}\newline
z_{m_{2}+1}\mapsto e_{T+1}\newline \cdots\newline z_{l_{2}}\mapsto
e_{T+l_{2}-m_{2}}$,

where $Q=k+2(a_{1}+\cdots+a_{n_{2}-1}+a_{n_{2}})+(l_{1}-n_{1})$ and
$T=b_{1}+\cdots+b_{m_{2}-1}+b_{m_{2}}+(m_{2}-m_{1})$.

Notice that
$(\phi(y_{1}),\cdots,\phi(y_{l_{1}}),\phi(z_{1}),\cdots,\phi(z_{l_{2}}))$ is a
Type-$LT(f)$ sequence.

On the other hand, by Proposition 7.4, $f\notin T_{2}(G_{k^{*}})$ that is a
contradiction. ∎

##  10\. Case 3: $G_{k}$

In this section, we describe the $\mathbb{Z}_{2}$-graded identities for
$G_{k}$. Unlike papers [3] and [6], we will not use representation theory
methodology.

######  Lemma 10.1.

The following polynomials are graded identities for $G_{k}$:

:

a) $h_{1}(y_{1},\cdots,y_{k+1})=[y_{1},y_{2}]\cdots[y_{k},y_{k+1}]$ (if $k$ is
odd) (1);

:

b)
$h_{2}(y_{1},\cdots,y_{k+1},x)=[y_{1},y_{2}]\cdots[y_{k-1},y_{k}][y_{k+1},x]$
(if $k$ is even; $x\in X-\\{y_{1},\cdots,y_{k+1}\\})$ (2).

###### Proof.

Suppose without loss of generality that $k$ is an even positive integer. Since
$h_{2}$ is a multilinear polynomial, it is sufficient to evaluate it for
$\mathcal{B}$.

Let $S=\\{a_{1},\cdots,a_{k+1}\\}\subset(G_{k})_{0}\cap\mathcal{B}$ be a
multiset and let
$S^{\prime}=\\{b_{1}\\}\subset(G_{k})_{\alpha(x)}\cap\mathcal{B}$. If $S$
contains an element of even support-length, it is clear that
$H_{2}(a_{1},\cdots,a_{k+1},b_{1})=0$. If every element of $S$ has even
support-length, there are $a_{i},a_{j}\in S$ such that $supp(a_{i})\cap
supp(a_{j})\neq\\{\\}$. So $h_{2}(a_{1},\cdots,a_{k+1},b_{1})=0$ and we are
done. ∎

Let $T^{\prime}=(i_{1},\cdots,i_{l})$ and $T=(j_{1},\cdots,j_{t})$ be two
strictly ordered sequences of positive integers such that $t$ is even,
$l+t=m$, and $\\{1,\cdots,m\\}=\\{i_{1},\cdots,i_{l},j_{1},\cdots,j_{t}\\}$.
Let us next define:

$f_{T}(z_{1},\cdots,z_{m})=z_{i_{1}}\cdots
z_{i_{l}}[z_{j_{1}},z_{j_{2}}]\cdots[z_{j_{t-1}},z_{j_{t}}]$.

In the same way, let $T^{\prime}=(i_{1},\cdots,i_{l})$ and
$T=(j_{1},\cdots,j_{t})$ be two strictly ordered sequences of positive
integers such that $t$ is odd, $l+t=m$, and
$\\{1,\cdots,m\\}=\\{i_{1},\cdots,i_{l},j_{1},\cdots,j_{t}\\}$. Let us then
define:

$r_{T}(y_{1},z_{1},\cdots,z_{m})=z_{i_{1}}\cdots
z_{i_{l}}[y_{1},z_{j_{1}}]\cdots[z_{j_{t-1}},z_{j_{t}}]$.

######  Definition 10.2.

Let $m\geq 2$. Let:

$g_{m}(z_{1},\cdots,z_{m})=\sum\limits_{|T|\ \mbox{is
even}}(-2)^{\frac{-|T|}{2}}f_{T}(z_{1},\cdots,z_{m})$.

Moreover: $g_{1}(z)=z$.

######  Lemma 10.3.

The polynomial $g_{k+2}(z_{1},\ldots,z_{k+2})$ is a graded identity for
$G_{k}$.

###### Proof.

It is enough to repeat word for word the proof for Proposition 18 by [6]. ∎

######  Corollary 10.4.

The multilinear polynomials $[g_{k+1},y]$ and $g_{k+1}[z_{k+2},y]$ are graded
identities for $G_{k}$.

###### Proof.

First, note that $[z_{1},y],z_{2}[z_{1},y]\in T_{2}(G_{0})$.

According to Lemma 10.3, $g_{k+1}\in T_{2}(G_{k-1})$ when $k\geq 1$.

Let $G^{\prime}$ be the (unitary) Grassmann algebra generated by
$\\{e_{1},\cdots,\widehat{e_{i}},\cdots,e_{n},\cdots\\}$ ($\widehat{e_{i}}$
means that $e_{i}\notin V$).

Notice that $G^{\prime}=((G_{k})_{0}\cap G^{\prime})\oplus((G_{k})_{1}\cap
G^{\prime})$, and $G^{\prime}\cong G_{k-1}$ (as super-algebras). Thus
$g_{k-1}\in T_{2}(G^{\prime})$. So by doing basic calculations, we can
conclude that $\newline [g_{k+1},y],g_{k+1}[z_{k+2},y]\in T_{2}(G_{k})$ as
required. ∎

######  Corollary 10.5.

If $p>k+1$, then $z^{p}$ is a consequence of $g_{k+2}$.

###### Proof.

Note that
$g_{k+2}(z_{1},\cdots,z_{1},z_{2})=z_{1}^{k+1}z_{2}+(k+1)(-2)^{-1}z_{1}^{k}[z_{1},z_{2}]$.

So $g_{k+2}(z_{1},\cdots,z_{1})=z_{1}^{k+2}$, and we are done. ∎

######  Corollary 10.6.

The following polynomials are graded identities for $G_{k}$:

:

$h_{3}=g_{k-l+2}(z_{1},\cdots,z_{k-l+2})[y_{1},y_{2}]\cdots[y_{l-1},y_{l}]$
(if $l\leq k$ and $l$ is even) (3);

:

$h_{4}=g_{k-l+2}(z_{1},\cdots,z_{k-l+2})[z_{k-l+3},y_{1}][y_{2},y_{3}]\cdots[y_{l-1},y_{l}]$
(if $l\leq k$ and $l$ is odd) (4);

:

$h_{5}=[g_{k-l+2}(z_{1},\cdots,z_{k-l+2}),y_{1}]\cdots[y_{l-1},y_{l}]$ (if
$l\leq k$ and $l$ is odd) (5).

###### Proof.

Notice that there are two cases to consider: either $l$ is even or $l$ is odd.

:

Case 1: $l$ is odd. According to Corollary 10.4,
$g_{k-l+2}[z_{k-l+3},y],[g_{k-l+2},y]\in T_{2}(G_{k-l+1})$. On the other hand,
we can see that $h_{4}\in T_{2}(G_{k})$ ( and respectively $h_{5}\in
T_{2}(G_{k})$) if, and only if, $g_{k-l+2}[z_{k-l+3},y]\in T_{2}(G_{k-l+1})$ (
and respectively $[g_{k-l+2},y]\in T_{2}(G_{k-l+1})$).

:

Case 2: $l$ is even. According to Lemma 10.3, $g_{k-l+2}\in T_{2}(G_{k-l})$.
On the other hand, we can see that $g_{k-l+2}\in T_{2}(G_{k-l})$ if, and only
if, $h_{3}\in T_{2}(G_{k})$.

∎

######  Corollary 10.7.

Let $I$ be the $T_{2}$-ideal generated by the graded identities of type $(3)$.
In the free super-algebra $F\langle X\rangle$, we have:

$z_{1}z_{2}\cdots z_{k-l+2}[y_{1},y_{2}]\cdots[y_{l-1},y_{l}]\equiv\ a.b\ \
mod\ \ I$,

where $l\leq k$, $l$ is even, and

:

$a(z_{1},\cdots,z_{k-l+2})=(\sum_{|T|\mbox{is even and non-
empty}}-(-2)^{-\frac{|T|}{2}}f_{T}(z_{1},\cdots,z_{k-l+2}))$;

:

$b(y_{1},\cdots,y_{l})=[y_{1},y_{2}]\cdots[y_{l-1},y_{l}]$.

######  Corollary 10.8.

Let $I$ be the $T_{2}$-ideal generated by the graded identities of type $(4)$.
In the free super-algebra $F\langle X\rangle$, we have:

$z_{1}z_{2}\cdots z_{k-l+2}[z_{k-l+3},y_{1}]\cdots[y_{l-1},y_{l}]\equiv a.b\ \
mod\ \ I$,

where $l\leq k$, $l$ is odd, and

:

$a(z_{1},\cdots,z_{k-l+2})=(\sum_{|T|\mbox{is even and non-
empty}}-(-2)^{-\frac{|T|}{2}}f_{T}(z_{1},\cdots,z_{k-l+2}))$;

:

$b(z_{k-l+3},y_{1},\cdots,y_{l})=[z_{k-l+3},y_{1}]\cdots[y_{l-1},y_{l}]$.

######  Definition 10.9.

An element $a\in SS$ is labeled as Type-2 (or $u\in SS2$) when:

:

$deg_{Y}(\psi(a))\leq k$ and $deg_{Z}(beg(a))+deg_{Y}(\psi(a))\leq k+1$.

######  Lemma 10.10.

Let $u$ be an element of $SS$ with the following property:

$deg_{Z}(beg(u))+deg_{Y}(\psi(u))\geq k+2$ or $deg_{Y}(\psi(u))=k+1$.

Let $I$ be the $T_{2}$-ideal generated by $[x_{1},x_{2},x_{3}]$, the graded
identities $(1),(2)$, and the graded identities of types $(3)$ and $(4)$. For
modulo $I$, $u$ can be written as a linear combination of $SS2$.

###### Proof.

First, note that if $deg_{Y}(\psi(u))>k$, then $u$ is a consequence of $(1)$
or $(2)$. In this way, we may assume that $deg_{Y}\psi(u)\leq k$.

We may suppose without loss of generality that:

$u=y_{1}^{a_{1}}\cdots y_{n_{1}}^{a_{n_{1}}}y_{n_{1}+1}^{a_{n_{1}+1}}\cdots
y_{n_{2}}^{a_{n_{2}}}z_{1}^{b_{1}}\cdots
z_{m_{1}}^{b_{n_{1}}}z_{m_{1}+1}^{b_{m_{1}+1}}\cdots
z_{m_{2}}^{b_{m_{2}}}[y_{n_{1}+1},y_{n_{1}+2}]\newline
\cdots[y_{n_{1}+l},z_{m_{1}+1}]\cdots[z_{m_{2}-1},z_{m_{2}}][z_{m_{2}+1},z_{m_{2}+2}]\cdots[z_{l_{2}-1},z_{l_{2}}]$,

where
$m_{1}<m_{2}<l_{2},n_{1}<n_{2}<n_{1}+l;b_{1},\cdots,b_{m_{2}},a_{1},\cdots,a_{n_{2}}>0$
and $deg_{Z}(beg(u))+deg_{Y}(\psi(u))=k+2$.

Thus, according to Corollary 10.8:

$u\equiv a.b.c\ \ mod\ \ I$,

where

:

$a(y_{1},\cdots,y_{n_{2}})=y_{1}^{a_{1}}\cdots
y_{n_{1}}^{a_{n_{1}}}y_{n_{1}+1}^{a_{n_{1}+1}}\cdots y_{n_{2}}^{a_{n_{2}}}$;

:

$b(z_{1},\cdots,z_{m_{2}})=(\sum_{|T|\mbox{is even and non-
empty}}-(-2)^{-\frac{|T|}{2}}f_{T}(z_{1},\cdots,z_{m_{2}}))$;

:

$c(y_{n_{1}+1},\cdots,y_{n_{1}+l},z_{m_{1}+1},\cdots,z_{l_{2}})=[y_{n_{1}+1},y_{n_{1}+2}]\cdots[y_{n_{1}+l},z_{m_{1}+1}]\cdots\newline
[z_{m_{2}-1},z_{m_{2}}][z_{m_{2}+1},z_{m_{2}+2}]\cdots[z_{l_{2}-1},z_{l_{2}}]$.

Then, after applying the graded identity $[x_{1},x_{2},x_{3}]$ to $b.c$, we
are done.

When $deg_{Z}(beg(u))+deg_{Y}(\psi(u))>k+2$, the proof is similar by inductive
arguments. To arrive at this situation, we must replace $a$ by:

$y_{1}^{a_{1}}\cdots y_{n_{1}}^{a_{n_{1}}}y_{n_{1}+1}^{a_{n_{1}+1}}\cdots
y_{n_{2}}^{a_{n_{2}}}z_{1}^{b_{1}-c}\cdots z_{k_{1}}^{b_{k_{1}}}$,

where $k_{1}\leq m_{2},b_{k_{1}}-c\geq 0$, and
$b_{1}+\cdots+b_{k_{1}}-c=deg_{Z}beg(u)-(k-l+2)$. ∎

######  Definition 10.11.

An element $u_{i}\in SS$ is labeled as Type-$3$ ($u_{i}\in SS3$) when the
following conditions hold:

:

$u_{i}\in SS2$;

:

If $deg_{Z}beg(u_{i})+deg_{Y}\psi(u_{i})=k+1$, then
$Deg_{pr(z)(u_{i})}\psi(u_{i})=0$.

For lemmas 10.12 and 10.14, we use some of the arguments of Lemma 20-b by [6].

######  Lemma 10.12.

Let $I$ be the $T_{2}$-ideal generated by $[x_{1},x_{2},x_{3}]$ and the graded
identities of type $(3)$. In the free super-algebra $F\langle X\rangle$, we
have:

$z_{2}\cdots
z_{k-l+2}[z_{1},z_{k-l+3}][y_{1},y_{2}]\cdots[y_{l-1},y_{l}]\equiv(\sum_{J}\beta_{J}f_{J})[y_{1},y_{2}]\cdots[y_{l-1},y_{l}]\
mod\ \ I$ (if $l\leq k$ and $l$ is even)

for some $\beta_{J}\in F$, $J\subseteq\\{1,\cdots,k-l+3\\}$. Moreover, if
$|J|=2$, then $1\notin J$ and $\beta_{J}=-1$.

###### Proof.

First, note that
$[g_{k-l+2}(z_{1},\cdots,z_{k-l+2})[y_{1},y_{2}]\cdots[y_{l-1},y_{l}],z_{k-l+3}]$
is a graded identity for $G_{k}$, because
$g_{k-l+2}(z_{1},\cdots,z_{k-l+2})[y_{1},y_{2}]\cdots[y_{l-1},y_{l}]\in
T_{2}(G_{k})$. For modulo $I$:

$[z_{1}\cdots
z_{k-l+2}[y_{1},y_{2}]\cdots[y_{l-1},y_{l}],z_{k-l+3}]+[a[y_{1},y_{2}]\cdots[y_{l-1},y_{l}],z_{k-l+3}]\equiv
0$,

where $a=\sum_{|T|\mbox{is even and non-
empty}}(-2)^{-\frac{|T|}{2}}f_{T}(z_{1},\cdots,z_{k-l+2})$.

It is well known that $[uv,w]=u[v,w]+[u,w]v$ for $u,v,w\in F\langle X\rangle$.
Therefore, we conclude that:

$[z_{1}\cdots z_{k-l+2}[y_{1},y_{2}]\cdots[y_{l-1},y_{l}],z_{k-l+3}]\equiv
z_{1}[z_{2}\cdots
z_{k-l+2}[y_{1},y_{2}]\cdots[y_{l-1},y_{l}],z_{k-l+3}]+z_{2}\cdots
z_{k-l+2}[y_{1},y_{2}]\cdots[y_{l-1},y_{l}][z_{1},z_{k-l+3}]\ \ mod\ \ I$.

Thus:

$z_{2}\cdots
z_{k-l+2}[z_{1},z_{k-l+3}][y_{1},y_{2}]\cdots[y_{l-1},y_{l}]\equiv-
z_{1}[z_{2}\cdots
z_{k-l+2}[y_{1},y_{2}]\cdots[y_{l-1},y_{l}],z_{k-l+3}]-\newline
[a[y_{1},y_{2}]\cdots[y_{l-1},y_{l}],z_{k-l+3}]\ \ mod\ \ I$.

Applying successively the graded identity $[x_{1},x_{2},x_{3}]$ and the
expression $[uv,w]=u[v,w]+[u,w]v$, we are done. ∎

######  Corollary 10.13.

Let $I$ be the $T_{2}$-ideal generated by $[x_{1},x_{2},x_{3}],z_{1}^{p}$ and
the graded identities of type $(3)$.

If $v\in SS2,deg_{Z}beg(v)+deg_{Y}\psi(v)=k+1,2\mid
deg_{Y}\psi(v),\mbox{and}Deg_{pr(z)v}\psi(v)=1$, then:

$v\equiv\sum_{i=1}^{n}\lambda_{i}v_{i}\ \ mod\ \ I$,

where $v-\sum_{i=1}^{n}\lambda_{i}v_{i}$ is a multi-homogeneous polynomial,
and $v_{1},\cdots,v_{n}\in SS3$.

###### Proof.

Let $v=z_{1}^{a_{1}}\cdots
z_{n}^{a_{n}}[z_{1},z_{n+1}][y_{1},y_{2}]\cdots[y_{l-1},y_{l}]$ such that
$a_{1}+\cdots+a_{n}=k-l+1;a_{1},\cdots,a_{n}>0$.

Choose a convenient graded endomorphism $\phi$ such that
$\phi(z_{1})=z_{1},\cdots,\newline
\phi(z_{k-l+2})=z_{n},\phi(z_{k-l+3})=z_{n+1}$.

For modulo $I$:

$z_{1}[\phi(z_{1}\cdots
z_{k-l+2})[y_{1},y_{2}]\cdots[y_{l-1},y_{l}],\phi(z_{k-l+3})]+\newline
z_{1}[a[y_{1},y_{2}]\cdots[y_{l-1},y_{l}],\phi(z_{k-l+3})]\equiv 0$,

where $a=\sum_{|T|\mbox{is even and non-
empty}}(-2)^{-\frac{|T|}{2}}\phi(f_{T}(z_{1},\cdots,z_{k-l+2}))$.

Following word for word the proof of Lemma 10.12, we conclude that:

$(a_{1})v\equiv
z_{1}(\sum_{J}\beta_{J}\phi(f_{J}))[y_{1},y_{2}]\cdots[y_{l-1},y_{l}]\ mod\ \
I$,
for some $\beta_{J}\in F$, $J\subseteq\\{1,\cdots,k-l+3\\}$. Moreover if
$|J|=2$, then $Deg_{z_{1}}(\psi(\phi(f_{J})))=0$.

Generally,if $v\in SS2,deg_{Z}beg(v)+deg_{Y}\psi(v)=k+1,2\mid
deg_{Y}\psi(v),\mbox{and}Deg_{pr(z)v}\psi(v)=1$, then (by algebraic
manipulations):

$v\equiv\sum_{i=1}^{n}\lambda_{i}v_{i}\ \ mod\ \ I$,

where $v-\sum_{i=1}^{n}v_{i}$ is a multi-homogeneous polynomial and
$v_{1},\cdots,v_{n}\in SS3$. ∎

######  Lemma 10.14.

Let $I$ be the $T_{2}$-ideal generated by $[x_{1},x_{2},x_{3}]$ and the graded
identities of type $(5)$. In the free super-algebra $F\langle X\rangle$, we
have:

$z_{2}\cdots
z_{k-l+2}[z_{1},y_{1}][y_{2},y_{3}]\cdots[y_{l-1},y_{l}]\equiv(\sum_{J}\beta_{J}r_{J}(z_{1},\ldots,z_{k-l+2},y_{1}))[y_{2},y_{3}]\cdots[y_{l-1},y_{l}]\
mod\ \ I$ (if $l\leq k$ and $l$ is odd)

for some $\beta_{J}\in F$, $J\subseteq\\{1,\cdots,k-l+2\\}$. Moreover, if
$|J|=1$, then $1\notin J$ and $\beta_{J}=1$.

###### Proof.

The proof is similar to that demonstrated in Lemma 10.12. In this case, note
that due to the graded identities of type (5), we have:

$[z_{1}\cdots z_{k-l+2},y_{1}]\cdots[y_{l-1},y_{l}]\equiv a.b\ mod\ I$, where

:

$a=[(\sum_{|T|\mbox{is even and non-
empty}}-(-2)^{-\frac{|T|}{2}}f_{T}(z_{1},\cdots,z_{k-l+2})),y_{1}]$;

:

$b=[y_{2},y_{3}]\cdots[y_{l-1},y_{l}]$ (if $l\geq 3$), or $b=1$ (if $l=1$).

∎

######  Corollary 10.15.

Let $I$ be the $T_{2}$-ideal generated by $[x_{1},x_{2},x_{3}],z_{1}^{p}$ and
the graded identities of type $(5)$. If $v\in
SS2,deg_{Z}beg(v)+deg_{Y}\psi(v)=k+1,2\nmid deg_{Y}\psi(v),\newline
\mbox{and}\ Deg_{pr(z)v}\psi(v)=1$, then:

$v\equiv\sum_{i=1}^{n}\lambda_{i}v_{i}\ \ mod\ \ I$,

where $v-\sum_{i=1}^{n}\lambda_{i}v_{i}$ is a multi-homogeneous polynomial,
and $v_{1},\cdots,v_{n}\in SS3$.

As a consequence of Lemmas 10.10, 10.12 and 10.14, we have the following
proposition.

######  Proposition 10.16.

Let $F$ be a field of $charF=p>2$ and $|F|=q$. Let $I$ be the $T_{2}$-ideal
generated by $[x_{1},x_{2},x_{3}],z_{1}^{p},y_{1}^{pq}-y_{1}^{p}$, the graded
identities $(1),(2)$, and the graded identities of types $(3),(4)$, and $(5)$.
Let $f=\sum_{i=1}^{n}\lambda_{i}v_{i}$ be a linear combination from $Pr(X)$.
For modulo $I$, $f$ can be written as:

$\sum_{i=1}^{m}f_{i}u_{i}$,

where $f_{1},\cdots,f_{m}$ are $p$-polynomials and $u_{1},\cdots,u_{m}\in SS3$
is (are) distinct.

Now, we describe the $\mathbb{Z}_{2}$-graded identities of $G_{k}$.

###  10.1. Main result

######  Theorem 10.17.

Let $F$ be a field of $charF=p>2$ and $|F|=q$. The $\mathbb{Z}_{2}$-graded
identities of $G$ (with grading induced by $\phi_{k}$) follow from:

  * •

$[y_{1},y_{2}]\cdots[y_{k},y_{k+1}]$ (if $k$ is odd) (1);

  * •

$[y_{1},y_{2}]\cdots[y_{k-1},y_{k}][y_{k+1},x]$ (if $k$ is even and $x\in
X-\\{y_{1},\cdots,y_{k+1}\\})$ (2);

  * •

$[x_{1},x_{2},x_{3}]\ \ (3)$;

  * •

$g_{k-l+2}(z_{1},\cdots,z_{k-l+2})[y_{1},y_{2}]\cdots[y_{l-1},y_{l}]$ (if
$l\leq k$ and $l$ is even) (4);

  * •

$g_{k-l+2}(z_{1},\cdots,z_{k-l+2})[z_{k-l+3},y_{1}][y_{2},y_{3}]\cdots[y_{l-1},y_{l}]$
(if $l\leq k$ and $l$ is odd) (5);

  * •

$[g_{k-l+2}(z_{1},\cdots,z_{k-l+2}),y_{1}]\cdots[y_{l-1},y_{l}]$ (if $l\leq k$
and $l$ is odd) (6);

  * •

$z_{1}^{p}$ (7);

  * •

$y_{1}^{pq}-y_{1}^{p}$ (8).

###### Proof.

Let $I$ be the $T_{2}$-ideal generated by the eight identities reported above.
Suppose by contradiction that $I\varsubsetneq T_{2}(G_{K})$. According to
Proposition 10.16, there exists a test polynomial
$f=f_{0}+\sum_{i=1}^{n}f_{i}u_{i}$, where $u_{1},\cdots,u_{n}\in SS3-\\{1\\}$.

One of the three cases listed below can occur:

:

Case 1: $deg_{Z}(beg(LT(f)))+deg_{Y}(\psi(LT(f)))\leq k$;

:

Case 2: $deg_{Z}(beg(LT(f)))+deg_{Y}(\psi(LT(f)))=k+1$ and $f$ does not admit
a bad term;

:

Case 3: $deg_{Z}(beg(LT(f)))+deg_{Y}(\psi(LT(f)))=k+1$ and $f$ admits a bad
term.

Case 1. Suppose without loss of generality that:

$LT(f)=y_{1}^{a_{1}}\cdots
y_{n_{1}}^{a_{n_{1}}}y_{n_{1}+1}^{a_{n_{1}+1}}\cdots
y_{n_{2}}^{a_{n_{2}}}z_{1}^{b_{1}}\cdots
z_{m_{1}}^{b_{m_{1}}}z_{m_{1}+1}^{b_{m_{1}+1}}\cdots
z_{m_{2}}^{b_{m_{2}}}\newline
[y_{n_{1}+1},y_{n_{1}+2}]\cdots[y_{n_{2}},y_{n_{2}+1}]\cdots[y_{l_{1}},z_{m_{1}+1}]\cdots[z_{m_{2}-1},z_{m_{2}}]\cdots[z_{l_{2}-1},z_{l_{2}}]$,

where
$n_{1}<n_{2}<l_{1},m_{1}<m_{2}<l_{2};a_{1},\cdots,a_{n_{2}},b_{1},\cdots,b_{m_{2}}>0$.

Suppose that $f=f(y_{1},\cdots,y_{l_{1}},z_{1},\cdots,z_{l_{2}})$. Consider
the following map :

$\phi:\\{y_{1},\cdots,y_{l_{1}},z_{1},\cdots,z_{l_{2}}\\}\rightarrow G\newline
$

$y_{1}\mapsto\sum_{l=1}^{a_{1}}e_{k+2l-1}e_{k+2l}\newline \cdots\newline
y_{n_{1}}\mapsto\sum_{l=a_{1}+\cdots+a_{n_{1}-1}+1}^{a_{1}+\cdots+a_{n_{1}}}e_{k+2l-1}e_{k+2l}\newline
y_{n_{1}+1}\mapsto
e_{1}+\sum_{l=a_{1}+\cdots+a_{n_{1}}+1}^{a_{1}+\cdots+a_{n_{1}+1}}e_{k+2l-1}e_{k+2l}\newline
\cdots\newline y_{n_{2}}\mapsto
e_{n_{2}-n_{1}}+\sum_{l=a_{1}+\cdots+a_{n_{2}-1}+1}^{a_{1}+\cdots+a_{n_{2}}}e_{k+2l-1}e_{k+2l}\newline
y_{n_{2}+1}\mapsto e_{n_{2}-n_{1}+1}\newline \cdots\newline y_{l_{1}}\mapsto
e_{l_{1}-n_{1}}\newline
z_{1}\mapsto\sum_{l=1}^{b_{1}}e_{R+l}e_{l_{1}-n_{1}+l}\newline \cdots\newline
z_{m_{1}}\mapsto\sum_{l=b_{1}+\cdots+b_{m_{1}-1}+1}^{b_{1}+\cdots+b_{m_{1}}}e_{R+l}e_{l_{1}-n_{1}+l}\newline
z_{m_{1}+1}\mapsto
e_{R+b_{1}+\cdots+b_{m_{1}}+1}+\sum_{l=b_{1}+\cdots+b_{m_{1}}+1}^{b_{1}+\cdots+b_{m_{1}+1}}e_{R+l+1}e_{l_{1}-n_{1}+l}\newline
\cdots\newline z_{m_{2}}\mapsto
e_{R+b_{1}+\cdots+b_{m_{2}-1}+m_{2}-m_{1}}+\sum_{l=b_{1}+\cdots+b_{m_{2}-1}+1}^{b_{1}+\cdots+b_{m_{2}}}e_{R+m_{2}-m_{1}+l}e_{l_{1}-n_{1}+l}\newline
z_{m_{2}+1}\mapsto e_{S+1}\newline \cdots\newline z_{l_{2}}\mapsto
e_{S+(l_{2}-m_{2})}$,

where $R=k+2(a_{1}+\cdots+a_{n_{2}})$ and
$S=R+(b_{1}+\cdots+b_{m_{2}})+m_{2}-m_{1}$.

It is not difficult to see that
$(\phi(y_{1}),\cdots,\phi(y_{l_{1}}),\phi(z_{1}),\cdots,\phi(z_{l_{2}}))$ is a
Type-$LT(f)$ sequence.

On the other hand, by Proposition 7.4, $f\notin T_{2}(G_{k})$ which is a
contradiction.

Case 2. Consider the following map (in map defined below, we are going to
agree on a little abuse of language: $\sum_{l=1}^{b_{1}-1}e_{k+l+1}e_{l}=0$,
in the situation that $b_{1}=1$):

$\phi:\\{y_{1},\cdots,y_{l_{1}},z_{1},\cdots,z_{l_{2}}\\}\rightarrow G\newline
$ $z_{1}\mapsto e_{k+1}+\sum_{l=1}^{b_{1}-1}e_{k+l+1}e_{l}\newline
\cdots\newline
z_{m_{1}}\mapsto\sum_{l=b_{1}+\cdots+b_{m_{1}-1}}^{b_{1}+\cdots+b_{m_{1}}-1}e_{k+l+1}e_{l}\newline
z_{m_{1}+1}\mapsto
e_{k+b_{1}+\cdots+b_{m_{1}}+1}+\sum_{l=1}^{b_{m_{1}+1}}e_{k+b_{1}+\cdots+b_{m_{1}}+l+1}e_{l+b_{1}+\cdots+b_{m_{1}}-1}\newline
\cdots\newline z_{m_{2}}\mapsto
e_{k+b_{1}+\cdots+b_{m_{2}-1}+(m_{2}-m_{1})}+\sum_{l=1}^{b_{m_{2}}}e_{k+b_{1}+\cdots+b_{m_{2}-1}+(m_{2}-m_{1})+l}e_{l+b_{1}+\cdots+b_{m_{2}-1}-1}\newline
z_{m_{2}+1}\mapsto
e_{k+b_{1}+\cdots+b_{m_{2}}+m_{2}-m_{1}+1},\cdots,z_{l_{2}}\mapsto
e_{k+b_{1}+\cdots+b_{m_{2}}+l_{2}-m_{1}}\newline
y_{1}\mapsto\sum_{l=1}^{a_{1}}e_{M+2l-1}e_{M+2l}\newline
y_{n_{1}}\mapsto\sum_{l=a_{1}+\cdots+a_{n_{1}-1}+1}^{a_{1}+\cdots+a_{n_{1}}}e_{M+2l-1}e_{M+2l}\newline
y_{n_{1}+1}\mapsto
e_{b_{1}+\cdots+b_{m_{2}}}+\sum_{l=a_{1}+\cdots+a_{n_{1}}+1}^{a_{1}+\cdots+a_{n_{1}+1}}e_{M+2l-1}e_{M+2l}\newline
y_{n_{2}}\mapsto
e_{b_{1}+\cdots+b_{m_{2}}+(n_{2}-n_{1}-1)}+\sum_{l=a_{1}+\cdots+a_{n_{2}-1}+1}^{a_{1}+\cdots+a_{n_{2}}}e_{M+2l-1}e_{M+2l}\newline
y_{n_{2}+1}\mapsto e_{b_{1}+\cdots+b_{m_{2}}+n_{2}-n_{1}}\newline
\cdots\newline y_{l_{1}}\mapsto e_{b_{1}+\cdots+b_{m_{2}}+l_{1}-n_{1}-1}$,

where $M=k+b_{1}+\cdots+b_{m_{2}}+l_{2}-m_{1}$.

Notice that the sequence
$(\phi(y_{1}),\cdots,\phi(y_{l_{1}}),\phi(z_{1}),\cdots,\phi(z_{l_{2}}))$ is
an Almost Type-$LT(f)$ sequence. However, by Proposition 7.5, $f\notin
T_{2}(G_{k})$.

Case 3. In this situation, notice that
$deg_{Z}(beg(LBT(f)))+deg_{Y}(\psi(LBT(f)))\leq k$. Suppose without loss of
generality that:

$LBT(f)=y_{1}^{a_{1}}\cdots
y_{n_{1}}^{a_{n_{1}}}y_{n_{1}+1}^{a_{n_{1}+1}}\cdots
y_{n_{2}}^{a_{n_{2}}}z_{1}^{b_{1}}\cdots
z_{m_{1}}^{b_{m_{1}}}z_{m_{1}+1}^{b_{m_{1}+1}}\cdots
z_{m_{2}}^{b_{m_{2}}}\newline
[y_{n_{1}+1},y_{n_{1}+2}]\cdots[y_{n_{2}},y_{n_{2}+1}]\cdots[y_{l_{1}},z_{m_{1}+1}]\cdots[z_{m_{2}-1},z_{m_{2}}]\cdots[z_{l_{2}-1},z_{l_{2}}]$,

where $n_{1}<n_{2}<l_{1},m_{1}<m_{2}<l_{2}$.

Consider $\phi:\\{y_{1},\cdots,y_{l_{1}},z_{1},\cdots,z_{l_{2}}\\}\rightarrow
G$ as in the Case 1.

It is clear that
$(\phi(y_{1}),\cdots,\phi(y_{l_{1}}),\phi(z_{1}),\cdots,\phi(z_{l_{2}}))$ is a
Type-$LBT(f)$ sequence.

On the other hand, by Proposition 7.6, we have that $f\notin T_{2}(G_{k})$. It
is a contradiction and we are done.

By checking these three cases, we have $I=T_{2}(E)$ as required. ∎

##  11\. Acknowledgements

My sincerely thanks go to my entire doctorate board for their useful comments,
remarks and suggestions. Thanks specially to V.R.T. da Silva, a member of my
doctorate board, for spotting a flaw in the original proof of Theorem 8.1.

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