2.1. Variety of all $K$-algebras

The homogeneous variety $\mathcal {A}_K$ of all $K$-algebras has the empty set of defining relations. Consider $K\{Y\}$, the free $K$-algebra on a set $Y$. It is well known that every non-associative word $w$ of degree at least two has unique representation in the form of a product of two non-associative words. Hence, one can introduce a total ordering in the set $V(Y)$ of non-associative words as follows. Order the words of length one (variables) arbitrarily. Assuming that the words of length $n$, $n\geq 1$, have been already ordered in such a way that words of smaller length precede words of greater length, then given two words $w_1,\,w_2\in V(Y)$ of length $n+1$ and represented as a product of two non-associative words of lesser length $w_1=u_1v_1$, $w_2=u_2v_2$ we define $w_1< w_2$ if, and only if, $u_1< u_2$ or $u_1=u_2$ and $v_1< v_2$. Observe that this total ordering of the non-associative words satisfies that if $w_1< w_2$ then $ww_1< ww_2$ for all $w\in V(Y)$.

Proof. Consider $\mathfrak {A}=\mathcal {A}_K$ and $\mathfrak {A}_F=\mathcal {A}_F$ in Setup 4.

Suppose $p_n$ is the $p_i$ of greatest degree $\geq 1$ and with the greatest word in the support among the $p_i$'s.

If $\mu \left (\sum \nolimits \limits _{i=1}^{n}{c_i\otimes p_i}\right )=\sum \nolimits \limits _{i=1}^{n} c_ip_i =0,$ then,

\begin{align*} 0 & = \left(\sum_{i=1}^{n}c_ip_i\right) p_n-p_n\left(\sum_{i=1}^{n} c_ip_i \right) \\ & = \sum_{i=1}^{n} c_i\left(p_i p_n - p_n p_i\right) = \sum_{i=1}^{n-1} c_i\left(p_i p_n - p_n p_i\right) \end{align*}

Hence,

\[ \mu\left(\sum_{i=1}^{n-1} c_i\otimes \Big(p_i p_n - p_n p_i\Big)\right)=0. \]

By the minimality of $n$, we have $0=p_i p_n-p_n p_i \textrm { for }i=1,\,\dotsc,\,n.$ This equality implies that $p_i$ and $p_n$ have the same maximal word in its support. Thus, there exists $\lambda _i\in K$ such that $p_i-\lambda _i p_n$ has a lesser maximal word in its support. Since

\[ (p_i-\lambda_i p_n)p_n -p_n(p_i-\lambda_i p_n)=0 \textrm{ for }i=1,\dotsc,n, \]

we obtain that $p_i=\lambda _ip_n$ for each $i=1,\,\dotsc,\,n.$ This contradicts the fact that the $p_i$'s are $K$-linear independent.

2.2. Variety of commutative $K$-algebras

The homogeneous variety $\mathcal {C}_K$ of commutative $K$-algebras has the defining relation $x_1x_2-x_2x_1$. The free commutative $K$-algebra on a set $Y$ will be denoted by $\mathcal {C}_K\{Y\}$. We will need a result and a definition from [Reference Shirshov8].

Let $Y$ be a non-empty set. Consider the words on $Y$. Words of length one will be called regular and ordered arbitrarily. Assuming that regular words of length less than $n$, $n>1$, have been already defined and ordered in such a way that words of smaller length precede words of greater length, a word $w$ of length $n$ will be called regular if

1. (1) $w=uv$ where $u$ and $v$ are regular words;

2. (2) $u\geq v$.

We order the regular words of length $n$ defined in this way, declaring that $w_1=u_1v_1< w_2=u_2v_2$ if either $u_1< u_2$ or $u_1=u_2$ and $v_1< v_2$. Then we declare the regular words of length $n$ to be greater than regular words of smaller length. By [Reference Shirshov8, Theorem 1], the collection of all regular words form a basis of $\mathcal {C}_K\{Y\}$.

Proof. Consider $\mathfrak {A}=\mathcal {C}_K$ and $\mathfrak {A}_F=\mathcal {C}_F$ in Setup 4.

Suppose $p_n$ is the $p_i$ of greatest degree $\geq 1$ and with the greatest word in the support among the $p_i$'s.

If $\mu \left (\sum \nolimits \limits _{i=1}^{n}{c_i\otimes p_i}\right )=\sum \nolimits \limits _{i=1}^{n} c_ip_i =0,$ then, for all regular words $w$,

\begin{align*} 0 & = \left(\left(\sum_{i=1}^{n}c_ip_i\right)w\right) p_n-(p_nw)\left(\sum_{i=1}^{n} c_ip_i \right) \\ & =\sum_{i=1}^{n} c_i\Big((p_iw) p_n - (p_nw) p_i\Big) = \sum_{i=1}^{n-1} c_i\Big((p_iw) p_n - (p_nw). p_i\Big) \end{align*}

Hence,

\[ \mu\left(\sum_{i=1}^{n-1} c_i\otimes \Big((p_iw) p_n - (p_nw) p_i\Big)\right)=0, \]

for all regular words $w$. By the minimality of $n$,

\[ 0=(p_iw) p_n-(p_nw) p_i \text{ for all regular words }w \text{ and for }i=1,\dotsc,n. \]

From the commutativity of the product and the definition of regular words, this equality implies that $p_i$ and $p_n$ have the same maximal regular word in its support. Thus, there exists $\lambda _i\in K$ such that $p_i-\lambda _i p_n$ has a lesser maximal regular word in its support. Since $((p_i-\lambda _i p_n)w)p_n -(p_nw)(p_i-\lambda _i p_n)=0$ for all regular words $w$ and $i=1,\,\dotsc,\,n$, we obtain that $p_i=\lambda _ip_n$ for each $i=1,\,\dotsc,\,n,$ a contradiction.

2.3. Schreier varieties satisfying $x^{2}=0$

Let $\mathfrak {M}$ be a variety of $K$-algebras. We say that $\mathfrak {M}$ is a Schreier variety if any subalgebra of a free algebra in $\mathfrak {M}$ is a free algebra in $\mathfrak {M}$. The main examples of homogeneous Schreier varieties are: the varieties of all algebras [Reference Kurosh2], commutative and anticommutative algebras [Reference Shirshov8], Lie algebras [Reference Shirshov7, Reference Witt10], and algebras with zero multiplication.

Let $A$ be a free $K$-algebra in $\mathfrak {M}$. Let $(x_1,\,\dotsc,\,x_n),\, (y_1,\,\dotsc,\,y_n)\in A^{n}$. Let $V$ and $W$ be the $K$-subspaces of $A$ generated by $(x_1,\,\dotsc,\,x_n)$ and $(y_1,\,\dotsc,\,y_n)$, respectively. We say that a transformation $\tau \colon (x_1,\,\dotsc,\,x_n)\mapsto (\tau (x_1)=y_1,\,\dotsc,\,\tau (x_n)=y_n)$ is an elementary transformation if either:

1. (1) $\tau$ induces a non-singular $K$-linear transformation between $V$ and $W$, or

2. (2) $y_1=x_1,\,y_2=x_2,\,\dotsc,\,y_{n-1}=x_{n-1}$ and $y_n=x_n+u_n$ where $u_n$ belongs to the $K$-subalgebra of $A$ generated by $x_1,\,\dotsc,\,x_{n-1}$.

Observe that inverses of elementary transformations are also elementary transformations.

It is known that homogeneous Schreier varieties $\mathfrak {M}$ are Nielsen, see, for example, [Reference Lewin3] or [Reference Mikhalev, Shpilrain and Yu5, Chapter 11]. In other words, if $A$ is a free algebra of $\mathfrak {M}$, one can transform any finite set of elements $a_1,\,\dotsc,\,a_n\in A$ to a free set of generators of the free subalgebra generated by $a_1,\,\dotsc,\,a_n$ by using a finite number of elementary transformations and cancelling possible zero elements.

Proof. Since $\mathfrak {M}$ is a Schreier variety, the subalgebra $B$ of $K_\mathfrak {M}\{Y\}$ generated by $p,\,q$ is a free subalgebra of $K_\mathfrak {M}\{Y\}$. There exist elementary transformations

\[ (p,q)\rightarrow \dotsb \rightarrow (x,y), \]

such that $x,\,y$ (and $y$ may be zero) are free generators of $B$. If the rank of $B$ is two, then there exist elementary transformations (the inverses of the previous ones)

\[ (x,y)\rightarrow \dotsb \rightarrow (p,q). \]

Since $B$ is free on $x,\,y$, the elementary transformations induce automorphisms of $K$-algebras. Hence, there exists an isomorphism of $B$ that sends $x\mapsto p$, $y\mapsto q$. Hence $p,\,q$ are free generators of $B$. Since $pq=0$, and $p^{2}=q^{2}=0$, it implies that the free algebra in the variety $\mathfrak {M}$ on two free generators has zero product. Therefore, all the algebras in $\mathfrak {M}$ satisfy the identity $x_1x_2=0$, a contradiction. It implies that $B$ is a free algebra of rank one. Since $x^{2}=0$, the free subalgebra generated by one non-zero element is of dimension $1$. Now $B$ is generated by $q$, as desired.

Between the homogeneous Schreier varieties that satisfy the conditions of Lemma 4, we highlight the following:

• • The variety of Lie $K$-algebras $\mathcal {L}_K$. Its defining relations are: $x^{2}_1$ and $(x_1x_2)x_3+(x_2x_3)x_1+(x_3x_1)x_2$.

• • The variety of anticommutative $K$-algebras $\mathcal {AC}_K$. It has the defining relation: $x_1x_2+x_2x_1$.

Proof. Consider $\mathfrak {A}=\mathfrak {M}$ and $\mathfrak {A}_F=\mathfrak {M}_F$ in Setup 4. Then

\[ \mu\left(\sum_{i=1}^{n-1} c_i\otimes p_ip_n\right)=\mu\left(\sum_{i=1}^{n} c_i\otimes p_ip_n\right)= \sum_{i=1}^{n} c_i p_ip_n=\left(\sum_{i=1}^{n}c_ip_i \right)p_n=0\cdot p_n=0. \]

By the minimality of $n$, $p_ip_n=0$ for all $i=1,\,\dotsc,\,n$. By Lemma 4, and the fact that $p_n\neq 0$, we get that $p_i=\lambda _i p_n$, where $\lambda _i\in K$ for $i=1,\,\dotsc,\,n$, a contradiction.

2.4. Variety generated by special Jordan algebras

Let $K$ be a field. If $A$ is an associative $K$-algebra, we define on the $K$-vector space $A$ a new multiplication $\circ$, which is connected with the associative multiplication by the formula $x\circ y=\frac {1}{2}(xy+yx)$. In this way, a new $K$-algebra is obtained and it is denoted by $A^{(+)}$. The $K$-algebra $A^{(+)}$ is a Jordan algebra (that is, satisfies $x_1x_2=x_2x_1$ and $(x_1^{2}x_2)x_1=x_1^{2}(x_2x_1)$). If $J$ is a $K$-subspace of $A$ which is closed with respect to the operation $\circ$, then $J$ together with $\circ$ is a subalgebra of $A^{(+)}$, and consequently a Jordan algebra. Such a Jordan algebra is called special Jordan algebra. The variety generated by all special Jordan $K$-algebras will be denoted by $\mathcal {SJ}_K$ and it consists of all $K$-algebras which can be obtained as homomorphic images of special Jordan $K$-algebras.

Let $Y$ be a set. Consider $K\langle Y\rangle$, the free-associative $K$-algebra on the set $Y$. The $K$-subalgebra $\mathcal {SJ}_K(Y)$ of the algebra $K\langle Y\rangle ^{(+)}$ generated by the set $Y$ is the free special Jordan $K$-algebra on $Y$ and it is the free $K$-algebra on the set $Y$ in the variety $\mathcal {SJ}_K$.

Proof. Consider $\mathfrak {A}=\mathcal {SJ}_K$ and $\mathfrak {A}_F=\mathcal {SJ}_F$ in Setup 4.

Suppose $p_n$ is the $p_i$ of the greatest degree. If it is of degree zero, then all $p_i$ are of degree zero and the result follows from the case $n=1$ because there exist $a_i\in K$ such that

\[ \mu\left(\sum_{i=1}^{n}{c_i\otimes p_i}\right)=\mu\left(\sum_{i=1}^{n} c_ia_i\otimes 1\right)=0 \]

From now on, we suppose that $p_n$ is of positive degree.

Claim 1: $p_i$ and $p_n$ commute as elements of $K\langle Y\rangle$.

If $\mu \left (\sum \nolimits \limits _{i=1}^{n}{c_i\otimes p_i}\right )=\sum \nolimits \limits _{i=1}^{n} c_ip_i =0,$ then, for all $z\in \mathcal {SJ}_K(Y)$

\begin{align*} 0 & = \left(\left(\sum_{i=1}^{n}c_ip_i\right)\circ z\right)\circ p_n-\left(\sum_{i=1}^{n} c_ip_i \right)\circ (z\circ p_n) \\ & = \sum_{i=1}^{n} c_i\left((p_i\circ z)\circ p_n - (p_i\circ (z\circ p_n))\right) = \sum_{i=1}^{n-1} c_i\left((p_i\circ z)\circ p_n - (p_i\circ (z\circ p_n))\right). \end{align*}

Hence,

\[ \mu\left(\sum_{i=1}^{n-1} c_i\otimes \Big((p_i\circ z)\circ p_n - (p_i\circ (z\circ p_n))\Big)\right)=0. \]

By the minimality of $n$, $0=(p_i\circ z)\circ p_n-p_i\circ (z\circ p_n)$ for $i=1,\,\dotsc,\,n$. Using that $\mathcal {SJ}_K(Y)$ is a subalgebra of $K\langle Y\rangle ^{(+)}$, we get

\begin{align*} 0 & = (p_i\circ z)\circ p_n-p_i\circ (z\circ p_n)\\ & = \frac{1}{2}((zp_i+p_iz)\circ p_n-p_i\circ(zp_n+p_nz))\\ & = \frac{1}{4}( zp_ip_n+p_izp_n+p_nzp_i+p_np_iz- (p_izp_n+p_ip_nz+zp_np_i+p_nzp_i)) \\ & = \frac{1}{4}[z(p_ip_n-p_np_i)-(p_ip_n-p_np_i)z]. \end{align*}

This implies that $p_ip_n-p_np_i$ commutes with any variable $z\in Y\subseteq \mathcal {SJ}_K(Y)$. Hence $p_ip_n-p_np_i$ is in the centre of $K\langle Y\rangle$, that is, $p_ip_n-p_np_i\in K$. But this is only possible if $p_ip_n-p_np_i=0$, because the term of zero degree in $p_ip_n-p_np_i$ is zero. Therefore, $p_i$ and $p_n$ commute in $K\langle Y\rangle$ and the claim is proved.

Claim 2: There exists $\lambda_i \in K$ such that $p_i=\lambda _i p_n$ for $i=1,\,\dotsc,\,n$. A contradiction with the fact that the $p_i$'s are $K$-linear independent.

By [Reference Cohn1, Corollary 6.7.7] and Claim 1, there exists a polynomial $u\in K\langle Y\rangle$ of degree at least one such that $p_i\in k[u]$, $i=1,\,\dotsc,\,n$.

From (2.1), $\sum \nolimits \limits _{i=1}^{n} c_ip_i=0$. Thus, for all $z\in \mathcal {SJ}_K(Y)$,

\begin{align*} 0& =\left(\sum_{i=1}^{n} c_ip_i\right)\circ (z\circ p_n^{2})-\left(\left(\sum_{i=1}^{n} c_ip_i\right)\circ p_n \right)\circ(z\circ p_n)\\ & = \sum_{i=1}^{n} c_i\Big( (p_i\circ (z\circ p_n^{2}))-(p_i\circ p_n)\circ (z\circ p_n) \Big) \\ & = \sum_{i=1}^{n-1} c_i \Big( (p_i\circ (z\circ p_n^{2}))-(p_i\circ p_n)\circ (z\circ p_n) \Big), \end{align*}

where we have used that $(z\circ p_n^{2})\circ p_n=p_n^{2}\circ (z\circ p_n)$ in the last equality.

Hence

\[ \mu\left(\sum_{i=1}^{n-1} c_i\otimes \left[(p_i\circ (z\circ p_n^{2}))-(p_i\circ p_n)\circ (z\circ p_n)\right]\right)=0. \]

By the minimality of $n$, and using Step 1 and $\mathcal {SJ}_K(Y)\subseteq K\langle Y\rangle ^{(+)}$,

\begin{align*} 0 & = p_i\circ (z\circ p_n^{2}) - (p_i\circ p_n) \circ (z\circ p_n) \\ & = \frac{1}{2}p_i\circ (zp_n^{2}+p_n^{2}z)-\frac{1}{4}(p_ip_n+p_np_i)\circ (zp_n+p_nz) \\ & = \frac{1}{4}(p_izp_n^{2}+p_ip_n^{2}z+zp_n^{2}p_i+p_n^{2}zp_i)\\ & -\frac{1}{4}(p_ip_nzp_n+p_ip_n^{2}z+zp_np_ip_n+p_nzp_ip_n) \\ & = \frac{1}{4}(p_izp_n^{2}+p_n^{2}zp_i-p_ip_nzp_n-p_nzp_ip_n)\\ & = \frac{1}{4}(p_i(zp_n-p_nz)p_n-p_n(p_nz-zp_n)p_i). \end{align*}

Hence, $p_i(zp_n-p_nz)p_n=p_n(zp_n-p_nz)p_i$ for all $z\in \mathcal {SJ}_K(Y)\subseteq K\langle Y\rangle ^{(+)}$ and $i=1,\,\dotsc,\,n$.

Let now $x\in \mathcal {SJ}_K(Y)$ such that $x$ does not commute with $u$. By [Reference Cohn1, Corollary 6.7.4], $u$ and $x$ form a free set over $K$. Then, rewriting the last equality, we obtain $p_i(u)(xp_n(u)-p_n(u)x)p_n(u)=p_n(u)(xp_n(u)-p_n(u)x)p_i(u)$.

Suppose that the degree on $u$ of $p_i(u)$ is smaller than the degree of $p_n(u)$. Considering the lexicographic order in $K\langle x,\,u\rangle$ with $x< u$, we obtain that the greatest monomial on the right-hand side of the equation is obtained from $-p_n(u)^{2}xp_i(u)$, which cannot be obtained from any other monomial on the left-hand side of the equality. A contradiction.

Hence $p_i$ and $p_n$ are polynomials on $u$ of the same degree. Thus, there exists $\lambda \in K$ such that $p_i(u)-\lambda p_n(u)$ has degree smaller than the degree of $p_n$ and

\[ (p_i(u)-\lambda p_n(u))(xp_n(u)-p_n(u)x)p_n(u)=p_n(u)(xp_n(u)-p_n(u)x)(p_i(u)-\lambda p_n(u)). \]

By the foregoing, the only possibility is that $p_i=\lambda _i p_n$ for some $\lambda _i\in K$.

2.5. Variety of non-commutative poisson $K$-algebras

In this subsection, we deviate from our notation and consider a variety of algebras endowed with more than one product. More precisely, the variety $\mathcal {NCP}_K$ of all non-commutative Poisson $K$-algebras.

A vector space $P$ over a field $K$ endowed with two bilinear operations $x\cdot y$ (a multiplication) and $\{x,\, y\}$ (a Poisson bracket) is called a non-commutative Poisson algebra if $P$ is an associative algebra under $x\cdot y$, $P$ is a Lie algebra under $\{x,\, y\}$, and $P$ satisfies the Leibniz identity: $\{x \cdot y,\, z\} = \{x,\, z\} \cdot y + x \cdot \{y,\, z\}$ for all $x,\,y,\,z\in P$.

Let $Y$ be a set. The free non-commutative Poisson $K$-algebra $\mathcal {P}_K(Y)$ is constructed as follows. Let $\mathcal {L}_K(Y)$ be the free Lie $K$-algebra on $Y$ and suppose that $X=\{x_1,\,x_2,\,\dotsc \}$ is a $K$-basis of $\mathcal {L}_K(Y)$. Then $\mathcal {P}_K(Y)$ is the free associative algebra on the set of free generators $X$. Using the Leibniz identity one can uniquely extend the Lie bracket $\{x,\, y\}$ of $\mathcal {L}_K(Y)$ to a Poisson bracket $\{x,\, y\}$ on $\mathcal {P}_K(Y)$, and $\mathcal {P}_K(Y)$ becomes a Poisson algebra.

We would like to finish noting that one can mimic the proof of Theorem 3 to show that the variety $\mathcal {NCP}_K$ is Reichstein when $K$ is an uncountable field.