The Wedderburn decomposition in finite-dimensional alternative superalgebras of

characteristic 3

M. C. López-Díaz, Departamento de Matemáticas, Universidade de Oviedo, Oviedo, España

Abstract The Wedderburn decomposition in finite dimensional alternative superalgebras of characteristic 3 is studied. It is proved that the decomposition exists provided the semisimple quotient superalgebra does not contain certain simple superalgebras. Counterexamples are given which show that the restrictions cannot be omitted. Together with the similar results by N.A.Pisarenko for the characteristic ≠ 2, 3 case, the results of the paper give a description of Wedderburn decomposition for alternative superalgebras of characteristic ≠ 2.

1 Introduction

The classical result by R.D.Schafer (see [2]) says that the principal Wedderburn theorem on the decomposition of a finite dimensional algebra into a direct sum of a semisimple subalgebra and the radical is true for alternative algebras. In [1], N.A.Pisarenko considered the question of whether this theorem could be extended to alternative superalgebras. He has proved that, for alternative superalgebras of characteristic ≠ 2, 3, the decomposition exists provided the semisimple quotient superalgebra does not contain simple superalgebras of two types: F [u] and M(1|1) (see the definitions and notations below). He also gave two counterexamples showing that the restrictions could not be omitted.

We study the problem for finite-dimensional alternative superalgebras of characteristic 3. In this case, due to [3], the new simple nontrivial superalgebras appear which are not associative: B(1, 2), B(4, 2), and B(E, D, γ). We prove that the decomposition exists provided the semisimple quotient superalgebra does not contain simple superalgebras of four types: F [u] and M(1|1) (Pisarenko’s restrictions); B(1, 2) and B(E, D, γ) (the new restrictions in characteristic 3). We give two counterexamples which show that the new restrictions neither can be omitted.

Joining our results with those by Pisarenko, we obtain finally that if A is an alternative superalgebra over a field F of characteristic ≠ 2, with the solvable radical N such that the semisimple quotient superalgebra A/N is separable and does not contain subsuperalgebras of types F [u], M(1|1), B(1, 2), and B(E, D, γ), then a Wedderburn decomposition holds in A.

2 Preliminaries

Let A = A_{0} + A_{1} be a superalgebra over a field F. Then, A is called an alternative superalgebra if the superidentities

(x_{i}, y_{j}, z_{k}) = (−1)^{ij+1}(y_{j}, x_{i}, z_{k}) = (−1)^{jk+1}(x_{i}, z_{k}, y_{j}),

where (x, y, z) = (xy)z − x(yz), i, j, k ∈ {0, 1} and x_{i} ∈ A_{i}, y_{j} ∈ A_{j}, z_{k} ∈ A_{k}, hold in A.

In this case, it is easy to prove that A_{0} is an alternative algebra and A_{1} is an alternative bimodule over A_{0}.

Lemma 1 [1]. Let A = A_{0} + A_{1} be a superalgebra over a field F with the unity 1; then 1 ∈ A_{0}.

Proposition 1 [1]. Let A = A_{0} + A_{1} be an alternative superalgebra over a field F, charF ≠ 2, with the unity 1. Let e_{i} ∈ A_{0}, i = 1, 2, ..., n, be mutually orthogonal idempotent elements such that e_{1} + ... + e_{n} = 1 and A_{ij} = (e_{i}A)e_{j}. Then

1) (e_{i}, x, e_{j}) = 0, ∀x ∈ A, i, j = 1, ..., n,

2) A = ⊕_{i,j=1}^{n}A_{ij} is a direct sum of vector spaces,

3) A_{ij}A_{jk} ⊆ A_{ik}, ∀i, j, k = 1, ..., n,

4) A_{ij}^{2} ⊆ A_{ji}, ∀i, j = 1, ..., n, i ≠ j,

5) A_{ij}A_{kl} = {0}, ∀i, j, k, l = 1, ..., n, k ≠ j, and (i, j) ≠ (k, l).

According to [3] and [6], simple finite dimensional alternative superalgebras B = B_{0} + B_{1} over an algebraically closed field F, charF ≠ 2, are of one of the following types:

1) B = B_{0} = C, a Cayley-Dickson algebra over F (see [2, 6]).

2) B = M_{n}(F)[u], where, for an algebra A, a superalgebra A[u] = A+Au is defined as an extension of A by an odd central element u such that u^{2} = 1. (In [1] this superalgebra was denoted by (M_{n}(F), M_{n}(F))).

3) B = M(p|q), B_{0} = \{( \begin{matrix}X & 0 \\ 0 & Y \end{matrix} ) ∈ M_{n}(F), | X ∈ M_{p}(F), Y ∈ M_{q}(F)\}, B_{1} =

\{( \begin{matrix}0 & A \\ B & 0 \end{matrix} ) ∈ M_{n}(F) | A ∈ M_{p×q}(F), B ∈ M_{q×p}(F)\}, where p + q = n and p ≤[\frac{n}{2}].

4) B = B(1, 2), where char F = 3, B_{0} = F · 1, B_{1} = F · x + F · y, with 1 being the unit of B and xy = −yx = 1, x^{2} = y^{2} = 0.

5) B = B(4, 2), where char F = 3, B_{0} = M_{2}(F), B_{1} = F ·m_{1}+F ·m_{2} is a 2-dimensional irreducible Cayley bimodule over B_{0}; that is, B_{0} acts on B_{1} by

e_{ij} · m_{k} = δ_{ik}m_{j}, i, j, k ∈ {1, 2},
m · a = \overline{a} · m,

where a ∈ B_{0}, m ∈ B_{1}, a → \overline{a} is the symplectic involution in B_{0} = M_{2}(F). The odd multiplication on B_{1} is defined by

m_{1}^{2} = −e_{21}, m_{2}^{2} = e_{12}, m_{1}m_{2} = e_{11}, m_{2}m_{1} = −e_{22}.

6) The twisted superalgebra of vector type B = B(E, D, γ). Let E = F [t_{1}, t_{2}, ..., t_{n} | t_{i}^{3} = 0] be the algebra of truncated polynomials over F (char F = 3), D be a nonzero derivation of E such that E is D-simple, and γ ∈ E. Observe that a derivation D for which E is D-simple always exists: one may take, for instance, D = ∂/∂t_{1} + t_{1}^{2}∂/∂t_{2} + ... + (t_{1}...t_{n−1})^{2}∂/∂t_{n}.

Denote by \overline{E} an isomorphic copy of the vector space E, with an isomorphism mapping a → \overline{a}. Consider the vector space direct sum B(E, D, γ) = E+\overline{E} and define multiplication on it by the rules

a · b = ab, a · \overline{b} = \overline{a} · b = \overline{ab}, \overline{a} · \overline{b} = γab + 2D(a)b + aD(b),

where a, b ∈ E and ab is the product in E. Define a Z_{2}-grading on B = B(E, D, γ) such that B_{0} = E and B_{1} = \overline{E}. Then B is a simple alternative superalgebra of dimension 2 · 3^{n} over F.

The structure of semisimple alternative superalgebras of characteristic ≠ 2 is given by

Proposition 2 [4]. Let A be an alternative superalgebra over an algebraically closed field F of characteristic ≠ 2, and let N be the solvable radical of A. Then, the quotient superalgebra S = A/N is a direct sum of simple superalgebras of types 1)-6).

3 The Wedderburn decomposition

A finite-dimensional alternative superalgebra A over a field F is called separable in case, for every extension K of F, the superalgebra A_{K} is a direct sum of simple ideals. It is equivalent to saying that A = B_{1} ⊕ B_{2} ⊕ … ⊕ B_{r} is semisimple with the even part of the center Z_{i} of each simple component B_{i} being a separable extension of F (i = 1, ..., r).

The following theorem and corollary generalize to alternative superalgebras of characteristic ≠ 2 the Wedderburn principal theorem for alternative algebras.

Theorem 1 Let A be an alternative superalgebra over a field F of characteristic ≠ 2, and let N be the solvable radical of A. Let A/N be separable such that A/N = B_{1} ⊕ ... ⊕ B_{k}, where B_{i} are simple subsuperalgebras. Assume that B is a subsuperalgebra of A/N of the form B = B_{i_{1}} ⊕ ... ⊕ B_{i_{r}} and that no superalgebras of types F [u], M(1|1), B(1, 2), and B(E, D, γ) appear among B_{i_{j}}. Then, there exists a subsuperalgebra S of A such that S ≅ B.

Corollary 1 Let A be an alternative superalgebra over a field F of characteristic ≠ 2, and let N be the solvable radical of A. Let A/N be separable. If no superalgebras of types F [u], M(1|1), B(1, 2), and B(E, D, γ) are found among simple direct summands of A/N, then a Wedderburn decomposition holds in A.

Proof. According to [1] and [2], we can reduce the problem of the existence of a Wedderburn decomposition in A to the case when the field F is algebraically closed, N^{2} = 0, and A/N is a simple superalgebra. Now, for the case char F ≠ 2, 3, the problem has been already solved by Pisarenko [1]. Let us consider the characteristic 3 case.

Lemma 2 Let A be an alternative superalgebra over a field F of characteristic 3, such that N^{2} = 0 and A/N ≅ C, a Cayley-Dickson algebra, or A/N ≅ M(p|q), where p = 0 or p+q ≥ 3. Then, there exists a subsuperalgebra S of A such that S ≅ A/N and A = S +N.

Proof. The proof of these facts in [1], carried out for the case of characteristic ≠ 2, 3, is valid for characteristic 3 too. •

Unfortunately, we cannot use Pisarenko’s proof for the case M_{n}(F)[u] since it uses essentially the characteristic ≠ 3 restriction. So we give an independent proof for this case.

Lemma 3 Let A be an alternative superalgebra over a field F of characteristic 3, such that N^{2} = 0 and A/N ≅ M_{n}(F)[u] with n ≥ 2. Then, there exists a subsuperalgebra S of A such that S ≅ A/N and A = S + N.

Proof. By the Wedderburn theorem for algebras, there exists a subalgebra in A_{0} isomorphic to M_{n}(F). Let e_{ij} be matrix units in A_{0}; consider the mutually orthogonal idempotent elements e_{ii}. We have the corresponding Peirce decomposition in A. Consider an element u_{11} ∈ (A_{1})_{11} such that u_{11}^{2} = e_{11} + r_{11}, where r_{11} ∈ (N_{0})_{11}; this element u_{11} always exists. Indeed, let ϕ : A ⟶ A/N = M_{n} + M_{n} · u be the canonical homomorphism. Denote the matrix units in A_{0}/N_{0} by e′_{ij} = ϕ(e_{ij}). Take the element u′_{11} = e′_{11} · u, then u′_{11} ∈ (A_{1}/N_{1})_{11} and (u′_{11})^{2} = e′_{11}. Consider a preimage u′ of u′_{11} in A_{1} with respect to ϕ, and put u_{11} = e_{11}u′e_{11}; then evidently u_{11}^{2} = e_{11} + r_{11}.

For any a_{11} ∈ A_{11} we define a_{ij} = e_{i1}(a_{11}e_{1j}), i, j = 1, …, n. The parenthesis here can be removed since (e_{i1}, a_{11}, e_{1j}) = −(a_{11}, e_{i1}, e_{1j}) = −(e_{i1}, e_{1j}, a_{11}) = 0, by Proposition 1. Then the following equalities hold:

e_{ij}a_{ik} = a_{ij}e_{jk} = a_{ik},  (1)
a_{ij}b_{jk} = a_{i1}b_{1k},  (2)
u_{ij}u_{jk} = e_{ik} + r_{ik},  (3)

for any b_{11} ∈ A_{11} and b_{ij}, u_{ij}, r_{ij} defined as above. If j = 1, equalities (1) and (2) are evident. Let j > 1, then

e_{ij}a_{jk} = e_{ij}(e_{j1}a_{11}e_{1k}) = −(e_{ij}, e_{j1}, a_{1k}) + e_{i1}a_{11}e_{1k}.

If j ≠ k, then (e_{ij}, e_{j1}, a_{1k}) = 0. If j = k, there are two cases: i = 1 and i ≠ 1. In the second one, (e_{ik}, e_{k1}, a_{1k}) = −(e_{k1}, e_{ik}, a_{1k}) = 0 also by Proposition 1. Now if i = 1, e_{1k}a_{1k} = 0. Indeed,

e_{1k}a_{1k} = e_{1k}(a_{11}e_{1k}) = −(e_{1k}, a_{11}, e_{1k}) + (e_{1k}a_{11})e_{1k} = 0

by Proposition 1. Hence (e_{1k}, e_{k1}, a_{1k}) = −(e_{1k}, a_{1k}, e_{k1}) = 0 in this case too.

We have proved the first of equalities (1). In a similar way, one can prove the second one.

Moreover, for j > 1 we have

a_{ij}b_{jk} = (a_{i1}e_{1j})b_{jk} = (a_{i1}, e_{1j}, b_{jk}) + a_{i1}(e_{1j}b_{jk})
= −(e_{1j}, a_{i1}, b_{jk}) + a_{i1}b_{1k}.

If i ≠ j or i = j and k ≠ 1, then (e_{1j}, a_{i1}, b_{jk}) = 0. In case i = j and k = 1, we have

a_{i1}b_{i1} = (e_{i1}a_{11})b_{i1} = (e_{i1}, a_{11}, b_{i1}) + e_{i1}(a_{11}b_{i1}) = −(a_{11}, e_{i1}, b_{i1}) = 0,

since e_{i1}b_{i1} = e_{i1} · e_{i1}(b_{11}e_{11}) = 0, and so (e_{1i}, a_{i1}, b_{i1}) = 0 too. Hence, identities (2) are proved.

Furthermore,

u_{i1}u_{1k} = (e_{i1}u_{11})u_{1k} = (e_{i1}, u_{11}, u_{1k}) + e_{i1}(u_{11}u_{1k}) = e_{i1}(u_{11}u_{1k}).

Similarly, u_{11}u_{1k} = u_{11}^{2}e_{1k}. This yields

u_{ij}u_{jk} = e_{i1}(u_{11}^{2}e_{1k}) = (e_{i1}u_{11}^{2})e_{1k} = e_{i1}u_{11}^{2}e_{1k}
= e_{i1}(e_{11} + r_{11})e_{1k} = e_{ik} + r_{ik},

which proves (3).

Consider the element d = (u_{11}, u_{12}, u_{21}) = (u_{12}, u_{11}, u_{21}) = 0. If we extend d, then by (3) and (1) we get

d = (e_{12} + r_{12})u_{21} − u_{11}(e_{11} + r_{11}) = u_{11} + r_{12}u_{21} − u_{11} − u_{11}r_{11} = r_{12}u_{21} − u_{11}r_{11}.

By (2), we have r_{12}u_{21} = r_{11}u_{11}, and so r_{11}u_{11} = u_{11}r_{11}.

Denote U = (r_{11}u_{11})u_{11}, U′ = u_{11}(u_{11}r_{11}), and consider the element e = (u_{11}, r_{11}, u_{11}). Since r_{11} ∈ A_{0}, we have

e = −(r_{11}, u_{11}, u_{11}) = −U + r_{11}(e_{11} + r_{11}) = −U + r_{11},
e = −(u_{11}, u_{11}, r_{11}) = −(e_{11} + r_{11})r_{11} + U′ = −r_{11} + U′,

which shows that 2r_{11} = U + U′.

Now, let u′_{11} = u_{11} + r_{11}u_{11} ∈ (A_{1})_{11}. We have

(u′_{11})^{2} = u_{11}^{2} + U + U′ = e_{11} + r_{11} + 2r_{11} = e_{11}.

Starting with the element u′_{11}, construct as before the system {u′_{ij} | i, j = 1, …, n}. Then it follows from (1), (3), (and from the fact that r′_{ij} = 0), that together with the matrix units e_{ij} these elements form a base of the subsuperalgebra S isomorphic to M_{n}[u]. (The element u in this case is u = u′_{11} + ⋯ + u′_{nn}; it is easy to see that it lies in the center of S and u′_{ij} = e_{ij} · u.) •

Lemma 4 Let A be an alternative superalgebra over a field F of characteristic 3, such that N^{2} = 0 and A/N ≅ B(4, 2). Then, there exists a subsuperalgebra S of A such that S ≅ A/N and A = S + N.

Proof. By the Wedderburn decomposition in alternative algebras (see [2]), there exists a copy of M_{2}(F) in A_{0}. Let e_{ij} be the matrix units in A_{0} and consider the mutually orthogonal idempotent elements e_{11} and e_{22}, such that e_{11} +e_{22} = I_{2}. We have the corresponding Peirce decomposition in A. Let ϕ : A ⟶ A/N be the canonical homomorphism. Consider the elements u_{11}, u_{22} ∈ A/N such that ϕ(e_{11}) = u_{11} and ϕ(e_{22}) = u_{22}. It is clear that u_{11} and u_{22} are mutually orthogonal idempotent elements and u_{11} + u_{22} = I_{2}. Moreover, if we consider the Peirce decomposition in A/N, the image of A_{ij} under ϕ is included in (A/N)_{ij}.

Looking at the product in B(4, 2), we have m_{1} ∈ (A/N)_{12} and m_{2} ∈ (A/N)_{21}. Choose x_{1} ∈ (A_{1})_{12} such that x_{1}^{2} = −e_{21} − n_{21} where n_{21} ∈ (N_{0})_{21}, and let x_{2} = e_{12}x_{1} ∈ (A_{1})_{21}. Note that ϕ(x_{1}) = m_{1} and ϕ(x_{2}) = m_{2}.

Consider

x_{1}x_{2} = x_{1}(e_{12}x_{1}) = −(x_{1}, e_{12}, x_{1}) + (x_{1}e_{12})x_{1}
= (e_{12}, x_{1}, x_{1}) + (x_{1}e_{12})x_{1} = (e_{12}x_{1})x_{1} − e_{12}x_{1}^{2} + (x_{1}e_{12})x_{1}
= −e_{12}(−e_{21} − n_{21}) + (e_{12}x_{1} + x_{1}e_{12})x_{1}
= e_{11} + e_{12}n_{21} + (e_{12}x_{1} + x_{1}e_{12})x_{1}.

Now, by Proposition 1, x_{1}x_{2} ∈ (A_{1})_{12}(A_{1})_{21} = (A_{0})_{11} and (e_{12}x_{1} + x_{1}e_{12})x_{1} ∈ (A_{0})_{22}. So, (e_{12}x_{1} + x_{1}e_{12})x_{1} = 0 and x_{1}x_{2} = e_{11} + e_{12}n_{21}.

Similarly, we obtain

x_{1}^{2} = −e_{21} − n_{21},
x_{2}^{2} = e_{12} + e_{12}(n_{21}e_{12}),
x_{1}x_{2} = e_{11} + e_{12}n_{21},
x_{2}x_{1} = −e_{22} − n_{21}e_{12}.

Put n_{12} = e_{12}(n_{21}e_{12}) ∈ (N_{0})_{12}, n_{11} = e_{12}n_{21} ∈ (N_{0})_{11}, and n_{22} = n_{21}e_{12} ∈ (N_{0})_{22}. Consider the products e_{ij}n_{kr}, with i, j, k, r ∈ {1, 2}. By Proposition 1, we have

e_{11}n_{11} = n_{11} = n_{11}e_{11},
e_{22}n_{11} = 0 = n_{11}e_{11},
e_{12}n_{11} = e_{12}(e_{12}n_{21}) = (e_{12})^{2}n_{21}) = 0,
n_{11}e_{12} = (e_{12}n_{21})e_{12} = e_{12}(n_{21}e_{12}) = n_{12},
e_{21}n_{11} = e_{21}(e_{12}n_{21}) = −(e_{21}, e_{12}, n_{21}) + e_{22}n_{21}
= (e_{12}, e_{21}, n_{21}) + n_{21} = 0 + n_{21} = n_{21},
n_{11}e_{21} = (e_{12}n_{21})e_{21} = (e_{12}, n_{21}, e_{21}) + e_{12}(n_{21}e_{21})
= −(e_{12}, e_{21}, n_{21}) + e_{12}(n_{21}e_{21}) = e_{12}(n_{21}e_{21}).

But n_{22}e_{21} = (n_{21}e_{12})e_{21} = (n_{21}, e_{12}, e_{21}) + n_{21}e_{11} = −(n_{21}, e_{21}, e_{12}) + n_{21} = n_{21}. So, n_{21}e_{21} = (n_{22}e_{21})e_{21} = (n_{22}, e_{21}, e_{21}) + 0 = 0, and hence n_{11}e_{21} = 0.

In a similar way, we obtain e_{ij}n_{kr} = δ_{jk}n_{ir}, with i, j, k, r ∈ {1, 2}.

Now, consider ε = n_{11} + n_{22}. It is clear e_{ij} · ε = n_{ij} = ε · e_{ij} and (ε, e_{ij}, e_{kr}) = 0. Let x̃_{1} = x_{1} + x_{1}ε and x̃_{2} = x_{2} + x_{2}ε. A straightforward calculation gives x̃_{1}^{2} = −e_{21}, x̃_{2}^{2} = e_{12}, x̃_{1}x̃_{2} = e_{11}, x̃_{2}x̃_{1} = −e_{22}, e_{ij}x̃_{k} = δ_{ik}x̃_{j}, and x̃_{k}e_{ij} = \overline{e_{ij}}x̃_{k}, with a → \overline{a} the symplectic involution in M_{2}(F).

Finally, consider the subsuperalgebra S of A with S_{0} = M_{2}(F) and S_{1} = F x̃_{1} + F x̃_{2}. Then, S ≅ B(4, 2), and the proposition is proved. •

Now, Theorem 1 follows easily from Pisarenko’s results [1] and the lemmas proved in this section. •

In the next section we will show that in the remaining cases we can not guarantee that a Wedderburn decomposition exists.

4 Counterexamples in characteristic 3

Let A be an alternative superalgebra over a field F of characteristic 3, with the solvable radical N such that N^{2} = 0 and A/N is a simple alternative superalgebra of one of the four exceptional types. We will show that in this case we can not guarantee that a Wedderburn decomposition exists.

Observe first that the counterexamples constructed in [1] by Pisarenko for the cases A/N ≅ F [u] and A/N ≅ M(1|1)(F) are valid for the case of characteristic 3 too. So, we have to consider only the cases B(1, 2) and B(E, D, γ).

We remind that for an A-superbimodule M, the opposite superbimodule M^{op} = M_{0}^{op} + M_{1}^{op} is defined by the conditions M_{0}^{op} = M_{1}, M_{1}^{op} = M_{0}, and the following action of A: a · m = (−1)\overline{a}am, m · a = ma, for any a ∈ A_{0} ∪ A_{1}, m ∈ M^{op}. It is straightforward to check that if M is alternative than so is M^{op}. A regular superbimodule, Reg A, is defined on the vector superspace A by an action of A coinciding with the multiplication in A.

Example 1 (A/N ≅ B(1, 2)). Let N = (Reg B(1, 2))^{op}, and consider the split null extension C = B(1, 2) + N of the superalgebra B(1, 2) by the superbimodule N. We have C_{0} = F · 1 + F · x^{o} + F · y^{o}, C_{1} = F · x + F · y + F · 1^{o}, where m^{o} ∈ M^{op} denotes the image of an element m ∈ M under the odd identical isomorphism id : M → M^{op}. Now, we change the multiplication in C, by setting x^{2} = y^{o}, y^{2} = x^{o}, and keeping all the remaining products the same. Denote the new superalgebra by A; then it is easy to see that A is still alternative and N is an ideal of A with A/N ≅ B(1, 2). Let a ∈ A_{1}, a = αx + βy + δ1^{o}; then a^{2} = α^{2}y^{o} + β^{2}x^{o}. Consequently, for any a ∈ A_{1} we have a^{2} ≠ 0 if a ∉ N, which proves that A does not contain a subsuperalgebra isomorphic to B(1, 2).

Example 2 (A/N ≅ B(E, D, γ). Let F be a field of characteristic 3, E = F [x |x^{3} = 0], D = ∂/∂x, γ = 0, B = B(E, D, 0), and N = Reg B. Denote by φ the identical isomorphism of superbimodulos φ : B ⟶ N. Consider the split null extension C = B+N and change the product in it by setting (¯1)^{2} = φ(x^{2}) and keeping the rest of the products the same. One can easily check that the resulting superalgebra A is still alternative, with the radical N such that A/N ≅ B. Below ab stands for the product in A and a ·_{B} b for that in B.

Let a ∈ A_{1}, a = a′ + φ(a′′) for some a′, a′′ ∈ B_{1}. We have a^{2} = a^{′2} + φ(a′ ·_{B} a′′ + a′′ ·_{B} a′) = a^{′2} = a′ ·_{B} a′ + α^{2}(¯1)^{2} = α^{2}φ(x^{2}), where a′ = \overline{α1 + a′_{0}}, a′_{0} ∈ F x + F x^{2}. In particular, if a^{2} = 0 then a = \overline{a′_{0}} + φ(a′′). Let also b ∈ A_{1}, b^{2} = 0, then we have again b = \overline{b′_{0}}+φ(b′′), where b′_{0} ∈ F x+F x^{2}, b′′ ∈ B_{1}. Consider ab = \overline{a′_{0}}·\overline{b′_{0}}+n, n ∈ N. Evidently, \overline{a′_{0}} · \overline{b′_{0}} = \overline{a′_{0}} ·_{B} \overline{b′_{0}} ∈ F (\overline{x} ·_{B} \overline{x^{2}}) = F x^{2}. So, there are no two elements in A_{1} with zero square such that their product were 1. Since we have ¯1 ·_{B} x̄ = 1, this proves that there is no a subsuperalgebra in A isomorphic to B.

References

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