You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: 'In this paper we investigate the Lie structure of the Lie superalgebra $K$ of skew elements of a semiprime associative superalgebra $A$ with superinvolution. We show that if $U$ is a Lie ideal of $K$, then either there exists an ideal $J$ of $A$ such that the Lie ideal $[J\cap K,K]$ is nonzero and contained in $U$, or $A$ is a subdirect sum of $A''$, $A''''$, where the image of $U$ in $A''$ is central, and $A''''$ is a subdirect product of orders in simple superalgebras, each at most 16-dimensional over its center.'
author:
- |
    Jesús Laliena [^1] and Sara Sacristán\
    [Departamento de Matemáticas y Computación]{}\
    [Universidad de La Rioja]{}\
    [26004, Logroño. Spain]{}\
    [<PRESIDIO_ANONYMIZED_EMAIL_ADDRESS>]{}
title: Lie structure in semiprime superalgebras with superinvolution
---

[= 4em *Keywords: associative superalgebras, semiprime superalgebras, superin-*]{}

[=10em *volutions, skewsymmetric elements, Lie structure.*]{}

Introduction.
=============

The study of the relationship between the structure of an associative algebra $A$ and that of the Lie algebra $A^-$ was started by I. N. Herstein (see [@H1], [@H2]) and W. E. Baxter (see [@B]). Afterwards, several authors have made different contributions and generalizations to the subject (see for instance [@Er], [@La], [@Ma] ).

Regarding superalgebras, this line of research was motivated by the classification of the finite dimensional simple Lie superalgebras given by V. Kac ([@Ka]), particularly the types given from simple associative superalgebras and from simple associative superalgebras with superinvolution. In [@Go-S], thinking in simple associative superalgebras with superinvolution, C. Gómez-Ambrosi and I. Shestakov investigated the Lie structure of the set of skew elements, $K$, of a simple associative superalgebra, $A$, with superinvolution over a field of characteristic not 2. These results were later extended to prime associative superalgebras with superinvolution ([@Go-L-S]). It was specifically proved that the Lie ideals of $K$ and $[K,K]$ are of the kind $[J\cap K, K]$ for a nonzero ideal $J$ of $A$, if $A$ is nontrivial, that is with a nonzero odd part, and if $A$ is not a central order in a Clifford superalgebra with at most 4 generators.

This paper is devoted to the description of the Lie ideals of $K$, the set of skew elements of a semiprime associative superalgebra, $A$, with superinvolution \* over a commutative unital ring $\phi$ of scalars with ${1\over 2 } \in \phi$.

We notice that the Lie structure of prime superalgebras and simple superalgebras without superinvolution was studied by F. Montaner (see [@M]) and S. Montgomery (see [@Mo]).

For a complete introduction to the basic definitions and examples of superalgebras, superinvolutions and prime and semiprime superalgebras, we refer the reader to [@Go-S] and [@M].

Throughout the paper, unless otherwise stated, $A$ will denote a nontrivial semi-prime associative superalgebra with superinvolution \* over a commutative unital ring $\phi$ of scalars with ${1\over
2}\in \phi$. By a nontrivial superalgebra we understand a superalgebra with nonzero odd part. $Z$ will denote the even part of the center of $A$, $H$ the Jordan superalgebra of symmetric elements of $A$, and $K$ the Lie superalgebra of skew elements of $A$. If $P$ is a subset of $A$, we will denote by $P_H=P\cap H$ and $P_K= P\cap K$. The following containments are straightforward to check, and they will be used throughout without explicit mention: $[K,K]\subseteq K, \quad [K,H]\subseteq H, \quad [H,H]\subseteq K, \quad H\circ H \subseteq H, \quad
H\circ K\subseteq K$ and $K\circ K\subseteq H$.

We recall that a superinvolution \* is said to be of the first kind if $Z_H=Z$, and of the second kind if $Z_H\not= Z$.

If $Z\not= 0$, one can consider the localization $Z^{-1}A=\{z^{-1}a : 0\not= z \in Z, a\in A\}$. If $A$ is prime, then $Z^{-1}A$ is a central prime associative superalgebra over the field $Z^{-1}Z$. We call this superalgebra the central closure of $A$. We also say that $A$ is a central order in $Z^{-1}A$. While this terminology is not the standard one, for which the definition involves the extended centroid, if $Z\not=0$ both notions coincide (for more specifications see 1.6 in [@M]).

Let $A$ be a prime superalgebra, and let $V=Z_H-\{0\}$ be the subset of regular symmetric elements. Note that if $Z\not= 0$, $Z_H\not=0$. Also $Z^{-1}A=V^{-1}A$, since for all $0\not=z \in Z, a\in A$ we have $z^{-1}a=(zz^*)^{-1}(z^*a)$. It will be more convenient for us, in order to extend the superinvolution in a natural way, to work with $V$ rather than with $Z$. We may consider $V^{-1}A$ as a superalgebra over the field $V^{-1}Z_H$. Then the superinvolution on $A$ is extended to a superinvolution of the same kind on $V^{-1}A$ over $V^{-1}Z_H$ via $(v^{-1}a)^*=v^{-1}a^*$. It is then easy to check that $H(V^{-1}A,*)=V^{-1}H$ and $K(V^{-1}A,*)=V^{-1}K$. Moreover, $Z(V^{-1}A)_0=V^{-1}Z$ and $V^{-1}Z\cap V^{-1}H=V^{-1}Z_H$. We will say that the superalgebra $V^{-1}A$ over the field $V^{-1}Z_H$ is the \*-central closure of $A$.

We notice that in every semiprime superalgebra $A$, the intersection of all the prime ideals $P$ of $A$ is zero. Consequently $A$ is a subdirect product of its prime images. If each prime image of $A$ is a central order in a simple superalgebra at most $n^2$ dimensional over its center, we say that $A$ verifies $S(n)$.

If $M$ is a subsupermodule of $A$, we denote by $\bar M$ the subalgebra of $A$ generated by $M$. We will say that $M$ is dense if $\bar M$ contains a nonzero ideal of $A$.

In this paper, we prove that if $K$ is the Lie superalgebra of skew elements of a semiprime associative superalgebra with superinvolution, $A$, and $U$ is a Lie ideal of $K$, then one of the following alternatives must hold: either $U$ must contain a nonzero Lie ideal $[J\cap K,K]$, for $J$ an ideal of $A$, or $A$ is a subdirect sum of $A^\prime$, $A^{\prime \prime}$, where the image of $U$ in $A^{\prime \prime}$ is central and $A^\prime $ satisfies $S(4)$.

The following results are instrumental for the paper:

([@H2], lemma 1.1.9) If $A$ is a semiprime algebra and $[a,[a,A]]=0$, then $a\in Z(A)$.

([@M], lemmata 1.2, 1.3) If $A=A_0 \oplus A_1 $ is a prime superalgebra, then $A$ and $A_0$ are semiprime and either $A$ is prime or $A_0$ is prime (as algebras).

([@M], lemma 1.8) Let $A=A_0 \oplus A_1$ be a prime superalgebra. Then

1.  If $x_1 \in A_1 $ centralizes a nonzero ideal $I$ of $A_0$, then $x_1 \in Z(A)$.

2.  If $x_1 ^2$ belongs to the center of a nonzero ideal $I$ of $A_0$, then $x_1 ^2 \in Z(A)$.

([@Go-L-S], Corollary 2) Let $A$ be a semiprime superalgebra and $L$ a Lie ideal of $A$. Then either $[L,L]=0$, or $L$ is dense in $A$.

([@Go-L-S], Theorem 2.1) Let $A$ be a prime nontrivial associative superalgebra. If $L$ is a Lie ideal of $A$, then either $L\subseteq Z$ or $L$ is dense in $A$, except if $A$ is a central order in a 4-dimensional Clifford superalgebra.

We remark that the bracket product in lemma 1.1 is the usual one: $[a,b]=ab-ba$, but the bracket product in lemmata 1.3,1.4,1.5 is the superbracket $[x_i,y_j]_s=x_iy_j-(-1)^{ij}y_jx_i$ for $x_i\in
A_i, y_j\in A_j$ homogenous elements. In fact, the superbracket product coincides with the usual bracket if one of the arguments belongs to the even part of $A$. In the following, to simplify the notation, we will denote both in the usual way $[\ , \ ]$ but we will understand that it is the superbracket if we are in a superalgebra.

Lie structure of K.
===================

Let $A$ be an associative superalgebra and $M,S$ be subgroups of $A$. Define $(M:S)=\{a\in A :
aS\subseteq M\}$.

Let $U$ be a Lie ideal of $K$. We recall (see lemma 4.1 in [@Go-S]) that $K^2$ is a Lie ideal of $A$.

If $A$ is semiprime, then either $U$ is dense in $A$ or $[u\circ v,w]=0$ for every $u,v,w\in U$.

We have $$[u\circ v, k]= u\circ [v,k] + (-1)^{{\bar k}{\bar v}} [u,k]\circ v\in \bar U$$ [for every $u,v \in U$ and $k\in K$. And also for any $u,v\in U$ and $h\in H$ we get]{} $$[u\circ v,h]=[u,v\circ h]+(-1)^{\bar u \bar v}[v,u\circ h]\in  U,$$ [because $K\circ H\subseteq K$. Since $A= H\oplus K$ it follows that $[u\circ v, A]\subseteq \bar U$ for any $u,v\in U$. But for any $a\in A$]{} $$[u\circ v, wa]=[u\circ v, w]a + (-1)^{({\bar u}+{\bar v}){\bar w}} w[u\circ v,a]$$ [and so $[u\circ v,w] A\subseteq \bar U$ for every $u,v,w \in U$, that is, $[u\circ v,w]\in
 (\bar U:A)$. We notice that from the above equations we can also deduce that $[u\circ v,w]a\in
\bar {K^2}$ and so $[u\circ v,w]\in(\bar {K^2}:A)$.]{}

We claim that $A(\bar {K^2}:A)\subseteq (\bar {K^2}:A)$. Indeed, for any $x\in (\bar {K^2}:A), a,b\in
A$ $$axb=(-1)^{(\bar x + \bar b)\bar a} (xb)a+[a,xb]\in \bar {K^2},$$ [because $[\bar {K^2},A]\subseteq \bar {K^2}$ (for any $t,s\in K^2$, $[ts,a]=t[s,a]+(-1)^{\bar s \bar a}[t,a]s\in \bar K^2$). Hence $A(\bar {K^2}:A)A\subseteq (\bar {K^2}:A)A\subseteq \bar {K^2}$. But $K(\bar U:A)\subseteq (\bar U:A)$ because for any $x\in (\bar U:A), k\in K, a\in A$]{} $$(kx)a=[k,xa]+(-1)^{(\bar x +\bar a)\bar k}(xa)k\in \bar U,$$ [because $[K,\bar U]\subseteq \bar U$, and so $\bar {K^2}(\bar U:A)\subseteq (\bar U:A)$. Therefore, we finally get]{} $$A(\bar {K^2}:A)A(\bar U:A)A\subseteq \bar {K^2}(\bar U:A)A\subseteq \bar U,$$ [and since $[u\circ v,w]\in (\bar U:A)$ and also $[u\circ v,w]\in (\bar {K^2}:A)$ it follows that $A[u\circ v,w]A[u\circ v,w]A \subseteq \bar U$. Thus, since $A$ is semiprime, either $[u\circ
v,w]=0$ for any $u,v,w\in U$ or $U$ is dense in $A$.]{}

We note that the ideal contained in $\bar U$ in the above Lemma, $J=A[u\circ v,w]A[u\circ v,w]A$, is also a $*$-ideal, that is, $J^*\subseteq J$.

Let $A$ be semiprime, and let $U$ be a Lie ideal of $K$ such that $[U\circ U,U]=0$. Then

1.  $u\circ v\in Z$ for every $u,v \in U_0$.

2.  $u\circ v=0$ for every $u,v\in U_1$.

Assertion (i) is proved as in Theorem 5.3 of [@Go-S], and (ii) as in Theorem 3.2 of [@Go-L-S].

Next we deal with the second case of lemma 2.1, that is, when $[u\circ v,w]=0$ for any $u,v,w\in U$ (and therefore when $u\circ v\in Z$ for every $u,v\in U_0,$ and $u\circ v=0$ for every $u,v\in U_1$), and we will study the prime images of $A$.

Let $P$ be a prime ideal of $A$. We will suppose first that $P^*\not= P$. In this case $(P^*+P)/P$ is a nonzero proper ideal of $A/P$ and we claim that $(P^*+P)/P\subseteq (K+P)/P$. Indeed, if $y\in P^*$ then $y+P=(y-y^*)+ y^*+P\in (K+P)/P$. Also if $U$ is a Lie ideal of $K$ we have that $(U+P)/P$ is an abelian subgroup of $A/P$ and satisfies $$[(U+P)/P,(P^*+P)/P]\subseteq ([U,K]+P)/P\subseteq (U+P)/P.$$

Therefore $(U+P)/P$ is a Lie ideal of $(P^*+P)/P$, and $(P^*+P)/P$ is an ideal in $A/P$, a prime superalgebra. Of course if $u\circ v\in Z$ for every $u,v \in U_0$ and $u \circ v =0$ for any $u,v\in
U_1$, then the same property is satisfied in $A/P$, that is, $(u+P) \circ (v+P) \in Z_0 (A/P)$ for every $u+P, v+P \in (U_0+P)/P$, and $(u+P)\circ (v+P)=0$ for any $u+P, v+P\in  (U_1+P)/P$. Let us analyze this situation. We notice that the assumption that $A/P$ has a superinvolution is not required. We state first a useful lemma.

Let $A$ be a prime superalgebra, $I$ a nonzero ideal of $A$ and $U$ a subset of $A$ such that $[U,I]=0$. Then $U\subseteq Z$.

For any $u_k\in U_k, a_i\in A_i, y_j\in I_j$, applying $[U,I]=0$ we get $$u_k(a_iy_j)= (-1)^{(i+j)k}
(a_iy_j)u_k=(-1)^{ik} a_i(u_ky_j).$$

[Since $A$ is prime it follows that $u_ka_i=(-1)^{ik}a_iu_k$. On the other hand, given $u_1 \in U_1$ we have $[u_1,I_0]=0$, and applying lemma 1.3 (i), $u_1\in
Z_1 (A)$. Hence for every $u_1 \in U_1, a_1 \in A_1$ we have $u_1 a_1 = a_1 u_1 = - a_1 u_1$, that is, $a_1 u_1 =0$, and, because $u_1 \in Z(A)$ and the primeness of $A$, $U_1 =0$ and $U\subseteq Z$.]{}

Let $A$ be a prime superalgebra, and let $I$ be a nonzero proper ideal of $A$. Suppose that $U$ is an abelian subgroup of $A$ such that $[U,I] \subseteq U$, $u\circ v \in Z$ for every $u,v \in U_0$, and $u\circ v=0$ for every $u,v\in U_1$. Then either $A$ is commutative, or $A$ is a central order in a 4-dimensional simple superalgebra, or $U\subseteq Z$.

Let $T=\{x\in A: [x,A]\subseteq [U,I]\}$. Since $$[[[U,I],[U,I]],A]\subseteq [[U,I],[[U,I],A]]\subseteq [[U,I],I]\subseteq [U,I],$$

[we have $[[U,I],[U,I]]\subseteq T$. We notice that $T$ is subring because for any $t,s\in
T$,]{} $$[ts,a]=[t,sa]+(-1)^{\bar t \bar s + \bar a \bar t}
[s, at]\in [U,I].$$ [Let $T^\prime$ be the subring generated by $[[U,I],[U,I]]$. Since]{} $$[[[U,I],[U,I]],I]  \subseteq [[U,I],[[U,I],I]] subseteq [[U,I],[U,I]]$$ [it follows that $[T^\prime,I]\subseteq T^\prime$. We consider now two cases: a) $[T^\prime,I]=0$, and b) $[T^\prime,I]\not=0$.]{}

a\) If $[T^\prime,I]=0$, then $[[[U,I],[U,I]],I]=0$. By lemma 2.3 we get $[[U,I],[U,I]]\subseteq Z$, and so $[[U,I],[U,I]]_1=0$.

We claim that in this situation either $U\subseteq Z$, or $A$ is commutative, or $A$ is a central order in a 4-dimensional simple superalgebra. We present the proof of this in 6 steps.

1\. $[U,I]_0\subseteq Z$. By hypothesis $u\circ v\in Z$ for any $u,v\in U_0$, so since $[U,I]\subseteq U$ it follows that $uv\in Z$ for any $u,v \in [U,I]_0$. Hence, for any $u,v \in
[U,I]_0$, we have $$[u,v][u,v]=[u,v[u,v]]-v[u,[u,v]]=[u,[vu,v]]-[u,[v,v]u]=0$$

because $[u,v], vu\in Z$. Therefore, from the primeness of $A$, $[u,v]=0$ for any $u,v\in
[U,I]_0$. So since $[[U,I],[U,I]]_1=0$, $[u,[u,I]]=0$ for any $u\in [U,I]_0$, and therefore, by lemma 1.2 and theorem 1 in [@H3], $u\in Z(I)$, that is $[U,I]_0\subseteq Z$ because $A$ is prime.

2\. $[U_0, I_0]=0$. By step 1 we have $[u_0,[u_0,I_0]]=0$ for any $u_0\in U_0$, and again by theorem 1 in [@H3] and lemma 1.2, we obtain that $[U_0,I_0]=0$.

3\. $U_1 U_1\subseteq Z$. Let $u_1\in U_1, y_1\in I_1$, since $[U_1,I_1]\subseteq
[U,I]_0\subseteq Z$ we get $$[u_1^2,y_1]=u_1[u_1,y_1]-[u_1,y_1]u_1=u_1[u_1,y_1]-u_1[u_1,y_1]=0.$$ Therefore, since $u_1\circ v_1=0$ for any $u_1, v_1\in U_1$, $0=[(u_1+v_1)^2,y_1]=[u_1v_1+v_1u_1,y_1]=
2[u_1v_1,y_1]$ for any $y_1\in I_1$. And, since $[u_1,v_1]=2u_1v_1\in U_0$ because $u_1\circ v_1=0$ for any $u_1,v_1\in U_1$, we have $[u_1v_1,I_0]=0$ for any $u_1,v_1\in U_1 $ by step 2. So $[u_1v_1,I]=0$ for any $u_1,v_1\in U_1$, and then $u_1v_1\in Z$, because of lemma 2.3.

4\. $I_1(U_1)^3\subseteq Z$. From the steps 1 and 3 for any $u_1,v_1,w_1\in U_1, y_1\in I_1$ we get $[u_1,y_1]v_1w_1\in Z$, but

$$\begin{aligned}
[u_1,y_1]v_1w_1&=&[u_1,y_1v_1]w_1+y_1[u_1,v_1]w_1\\&=&[u_1w_1,y_1v_1]-u_1[w_1,y_1v_1]+
y_1[u_1,v_1w_1]+y_1v_1[u_1,w_1]\end{aligned}$$

[and since $[u_1w_1,y_1v_1]=0$, $y_1[u_1,v_1w_1]=0$, by step 3 and $u_1[w_1,y_1v_1]\in U_1[U_1,I_0]\subseteq
U_1U_1\subseteq Z$, we obtain that $y_1v_1[u_1,w_1]\in Z$, that is $I_1(U_1)^3\subseteq Z$ because $u_1
\circ w_1=0$.]{}

5\. Either $U_1U_1=0$ or $A$ is commutative. By step 4 we have an ideal of $A_0$, $I_1u_1^3$, contained in $Z$, and so $[A_1,I_1u_1^3]=0$, and by lemma 1.3 either $A_1\subseteq
Z(A)_1$ or $I_1u_1^3=0$ for any $u_1\in U_1$.

If $A_1\subseteq Z_1(A)$ then $A_1^2\subseteq Z$, and since $A$ is prime and $A_1+A_1^2$ is a nonzero ideal contained in $Z(A)$, because $A$ is nontrivial, we deduce that $A$ is commutative.

If $I_1u_1^3=0$, since $0=I_1u_1^3=(I_1u_1)(u_1^2A)$ and $u_1^2A$ is an ideal of $A$ because $u_1^2\in Z$ by step 3, then from the primeness of $A$ either $I_1u_1=0$ or $u_1^2=0$ for any $u_1\in U_1$. But if $u_1^2=0$ for every $u_1\in U_1$ we get $U_1U_1=0$ because $u_1\circ v_1=0$ for any $u_1,v_1\in U_1$ and if $I_1u_1=0$ then $0=I_1(u_1v_1)$ for every $u_1,v_1\in U_1$. From step 3 and because $A$ is prime we obtain that either $U_1U_1=0$ or $I_1=0$. But $I_1=0$ contradicts that $A$ is prime because then $IA_1=0$ and so $I(A_1+A_1^2)=0$ with $A_1+A_1^2$ a nonzero ideal of $A$. Therefore $U_1U_1=0$ in any case, when $I_1u_1^3=0$.

6\. Either $U\subseteq Z$, or $A$ is commutative, or $A$ is a central order in a 4-dimensional simple superalgebra. We consider $[v_1,z_1]I$ with $v_1\in U_1, z_1\in I_1$. It is an ideal of $A$ by step 1. For any $u_0\in U_0, v_1
\in U_1$ and $y_1, z_1\in I_1$ we have $$[u_0,y_1][v_1,z_1]I=[u_0,y_1]v_1z_1I+[u_0,y_1]z_1v_1I$$ [with $[u_0,y_1]v_1z_1I=0$ by step 5 and]{} $$[u_0,y_1]z_1v_1I=-y_1[u_0,z_1]v_1I+[u_0,y_1z_1]v_1I=0$$ [by steps 2 and 5. Since $A$ is prime we obtain that either i) $[U_1,I_1]=0$ or ii) $[U_1,I_1]\not= 0$, and then $[U_0,I_1]=0$.]{}

[= 3em i) If $[U_1,I_1]=0$ then for any $u_1\in U_1$, $u_1I_1$ is a nilpotent ideal of $A_0$ because by step 5 $(u_1I_1)(u_1I_1)=u_1^2I_1=0$, and since $A_0$ is semiprime by lemma 1.2, we deduce that $u_1I_1=0$. But then $u_1I_0A_1\subseteq u_1I_1=0$ and also $u_1I_0A_1^2=0$, that is, $u_1I_0(A_1+A_1^2)=0$ with $ A_1+A_1^2$ a nonzero ideal of $A$. By the primeness of $A$, $u_1I_0=0$, and so $u_1I=0$ and $U_1=0$. Therefore $[U,I]=[U_0,I]=[U_0,I_0]$, and $[U_0,I_0]=0$ by step 2, so by lemma 2.3, $U_0\subseteq Z$.]{}

[= 3em ii) If $[U_1,I_1]\not=0$, then $[U_0,I_1]=0$ and so by step 2  $[U_0,I_0]=0$ and from lemma 2.3, $U_0\subseteq Z$. Also $Z\not=0$ and we may localize $A$ by $Z$ and consider in $Z^{-1}A$, the Lie subalgebra $Z^{-1}(ZU)$ and the ideal $Z^{-1}I$, which satisfy the hypothesis of the theorem. Now we have also that $0\not= Z^{-1}Z$ is a field. By step 1,   $[U_1,I_1]\subseteq Z$, and hence]{} $$0\not=[Z^{-1}(ZU)_1,Z^{-1}I_1]\subseteq Z^{-1}I_0\cap Z^{-1}Z.$$ [Therefore $Z^{-1}I$ has invertible elements and so $Z^{-1}I=Z^{-1}A$. But then $Z^{-1}(ZU)$ is a Lie ideal of $Z^{-1}A$. Since $[Z^{-1}(ZU),$ $Z^{-1}(ZU)]=0$ because $U_0\subseteq Z$ and because of step 5, it follows from theorem 3.2 and its proof in [@M] that either $Z^{-1}(ZU)\subseteq Z^{-1}Z$ or $A$ is a central order in the matrix algebra $M_{1,1}(Z^{-1}Z)$. In the last case $A$ is a central order in a 4-dimensional simple superalgebra, and in the first case $Z^{-1}(ZU)\subseteq Z^{-1}Z$ and we can deduce from the primeness of $A$ that $U\subseteq Z$.]{}

Therefore in case a) we have obtained that either $U\subseteq Z$, or $A$ is commutative, or $A$ is a central order in a 4-dimensional simple superalgebra

b\) We suppose now that $[T^\prime, I]\not=0$. We recall that $[T^\prime, I]\subseteq T^\prime$. Consider $[[T^\prime,I],T^\prime]$. We claim that $I[[T^\prime,I],T^\prime]I\subseteq T^\prime$. Indeed, let $x\in T^\prime, y\in [T^\prime,I]$ and $a\in I$. Since $[T^\prime, I]\subseteq T^\prime$ and $T^\prime$ is a subring, $$[x,y]a=[x,ya]-(-1)^{\bar x \bar y}y[x,a]\in T^\prime.$$

[ Now, let $b\in I$; we get]{} $$\begin{aligned}
b[x,y]a &=&[b,[x,y]]a + (-1)^{(\bar x+ \bar y) \bar b}[x,y]ba \\ &=& -(-1)^{\bar y ( \bar b +
\ \bar x)}[y, [b,x]]a-(-1)^{\bar b \bar x+ \bar b \bar y}[x,[y,b]]a\\&\ &+(-1)^{(\bar x+ \bar y)\bar
b}[x,y]ba\in T^\prime.\end{aligned}$$

[Therefore, by the primeness of $A$, $T^\prime$ is dense if $[[T^\prime,I],T^\prime]\not=0$.]{}

If $[[T^\prime,I],T^\prime]=0$, then $$[[[U,I],[U,I]]_0,[[[U,I],[U,I]]_0,I]]=0$$

[so by theorem 1 in [@H3], $[[[U,I],[U,I ]]_0,I]=0$, and applying now lemma 2.3 we have $[[U,I],$ $[U,I]]_0 \subseteq Z$. We denote $V=[[U,I],[U,I]]$ and we have that $V$ satisfies the same conditions as $U$, that is, $V$ is an abelian subgroup of $A$ such that $[V,I]\subseteq V$, $u\circ v \in Z$ for every $u,v\in V_0$, and $u\circ v=0$ for every $u,v\in V_1$, because $V\subseteq U$. Since $V_0\subseteq Z$ we observe that $V$ has, like $U$ in case a) steps 1 and 2, the following properties: $[V,I]_0\subseteq V_0\subseteq Z$ and $[V_0,I_0]=0$. From this we can prove steps 3, 4, 5 and 6 in a) exactly in the same way but now taking $V$ instead of $U$. So we obtain that either $A$ is commutative, or $A$ is a central order in a 4-dimensional simple superalgebra, or $V\subseteq Z$. But if $V\subseteq Z$ we can apply case a) and we obtain that either $U\subseteq Z$, or $A$ is commutative, or $A$ is a central order in a 4-dimensional simple superalgebra.]{}

It remains to consider the case when $T^\prime$ is dense in $A$. We denote by $J=I[[T^\prime,I],T^\prime]I$ and so $J\subseteq T^\prime$. From the definition of $T$ and because $T^\prime \subseteq T$ we know that $[T^\prime, A]\subseteq [U,I]$, and therefore $[J,A]\subseteq [U,I]\subseteq U$. By hypothesis $u\circ v \in Z$ for any $u,v\in U_0$, so $u\circ v\in Z$ for any $u, v\in [J,A]_0$.

We assume first that $u\circ v=0$ for any $u,v\in  [J,A]_0$. Then $1/2 (u\circ u)=u^2=0$ for any $u\in [J,A]_0$ and since $A_0$ is semiprime by lemma 1.2, we can apply lemma 1 in [@La-M] and we have $[J,A]_0=0$. Therefore $[J,A]=[J,A]_1$ and then $[J,A]$ is a Lie ideal of $A$ such that $[[J,A],[J,A]]=0$. From theorem 3.2 and its proof in [@M] it follows that either $[J,A]\subseteq
Z$ or $A$ is a central order in a 4-dimensional matrix superalgebra. If $[J,A]\subseteq Z$, since $[J,A]=[J,A]_1$, we get $[J,A]=0$ and now by lemma 2.3, $J\subseteq Z$, and so $A$ is commutative.

Suppose now that there exist $u,v\in [J,A]_0$ such that $u\circ v\not= 0$. Then $Z\not= 0$, and we may form the localization $Z^{-1}A$. Since $[J,A]\subseteq [U,I]\subseteq U$ we have $[Z^{-1}J,Z^{-1}A]\subseteq [Z^{-1}(ZU), Z^{-1}I]\subseteq Z^{-1} (ZU)$, and so from the hypothesis of the theorem for any $u,v\in [Z^{-1}J, Z^{-1}A]_0$ we get $u\circ v\in Z^{-1} Z \cap Z^{-1}J$. But $Z^{-1}Z$ is a field and so $Z^{-1}J$ has some invertible element forcing $Z^{-1}J=Z^{-1}A$. Therefore $[Z^{-1}J, Z^{-1}A] = [Z^{-1}A, Z^{-1}A]\subseteq Z^{-1}(ZU)$ and again by the hypothesis of the theorem it follows that $[Z^{-1} A, Z^{-1} A]_1 \circ [Z^{-1} A, Z^{-1} A]_1=0$. We apply now lemma 2.6 in [@M] and we obtain that $Z^{-1}A$ is commutative (superalgebras of the type (b) and (c) in the lemma do not satisfy the condition $[Z^{-1} A, Z^{-1} A]_1 \circ [Z^{-1} A, Z^{-1}
A]_1=0$), and so $A$ is commutative. This finishes the proof.

Next we consider the cases when $P^*= P$ and the involution on $A/P$ is of the second kind or of the first kind.

Let $A$ be a prime superalgebra with a superinvolution $*$ of the second kind. Let $U$ be a Lie ideal of $K$ such that $u\circ v \in Z$ for every $u,v \in U_0$, and $u\circ v=0$ for every $u,v\in U_1$. Then either $U\subseteq Z$ or $A$ satisfies $S(2)$.

If $*$ is of the second kind we know that $Z_H=\{x\in Z : x^*=x\}\not= Z$. We may localize $A$ by $V$ and replace $U$ by $V^{-1}(Z_HU)$ and $A$ by $V^{-1}A$. The hypothesis remains unchanged, so we keep for this superalgebra the same notation $A$, and now $Z$ is a field. Let $0\not= t\in Z_K$. Then $H=tK$ and $A=tK+K$. It follows that $[ZU,A]\subseteq ZU$, $u\circ v \in Z$ for every $u,v \in ZU_0$, and $u\circ v=0$ for every $u,v\in ZU_1$. By theorem 2.4, either $ZU\subseteq Z$, which implies that $U\subseteq Z$, or $A$ satisfies $S(2)$.

Let $A$ be a prime superalgebra with a superinvolution $*$ of the first kind. Let $U$ be a Lie ideal of $K$ such that $u\circ v \in Z$ for every $u,v \in U_0$, and $u\circ v=0$ for every $u,v\in U_1$. Then either $U\subseteq Z$ or $A$ satisfies $S(4)$.

If $u^2=0$ for every $u\in U_0$, applying theorem 3.3 in [@Go-L-S] we obtain that $U=0$. Suppose then that $u^2\not= 0$ for some $u\in U_0$. By theorem 3.4 in [@Go-L-S] we get that either $U\subseteq Z$ or $A$ is a central order in a Clifford algebra with either 2 or 4 generators.

Combining the above results we obtain

Let $A$ be a semiprime superalgebra and $U$ a Lie ideal of $K$ with $u\circ v \in Z$ for every $u,v
\in U_0$, and $u\circ v=0$ for every $u,v\in U_1$. Then $A$ is the subdirect sum of two semiprime homomorphic images $A^{\prime}$, $A^{\prime \prime}$, such that $A^\prime$ satisfies $S(4)$ and the image of $U$ in $A^{\prime \prime}$ is central.

Let $T^\prime=\{ P: P$ is a prime ideal of $A$ such that $A/P$ satisfies $S(4)\}$ and let $T^{\prime \prime}= \{ P : P$ is a prime ideal of $A$ such that the image of $U$ in $A/P$ is central$\}$.

If we consider $P$ a prime ideal of $A$ such that $P^*\not= P$ we know from theorem 2.8 that either $A/P$ is a central order in a simple superalgebra at most 4-dimensional over its center, or $(U+P)/P$ is central. If we consider $P$ a prime ideal of $A$ such that $P^*=P$, it follows from lemmata 2.5, 2.6 that either $A/P$ is a central order in a simple superalgebra at most 16-dimensional over its center, or the image of $U$ in $A/P$ is central.

So every prime ideal of $A$ belongs either $T^\prime$ or $T^{\prime \prime}$. Then $A^\prime$ is obtained by taking the quotient of $A$ by the intersection of all the prime ideals in $T^\prime$, and $A^{\prime \prime}$ is obtained by taking the quotient of $A$ by the intersection of all the prime ideals in $T^{\prime \prime}$. This proves the theorem.

We finally arrive at the main theorem on the Lie structure of $K$.

Let $A$ be a semiprime superalgebra with superinvolution $*$, and let $U$ be a Lie ideal of $K$. Then either $A$ is a subdirect sum of two semiprime homomorphic images $A^\prime$, $A^{\prime \prime}$, with $A^\prime$ satisfying $S(4)$ and the image of $U$ in $A^{\prime \prime}$ being central, or $U\supseteq
[J\cap K, K]\not= 0$ for some ideal $J$ of $A$.

From lemmata 2.1 and 2.2 we know that either $U$ is dense in $A$, and so there exist a nonzero ideal $J$ such that $J\subseteq \bar U$, or $u\circ v \in Z$ for every $u,v \in U_0$, and $u\circ v=0$ for every $u,v\in U_1$. In the second case we obtain by theorem 2.7 the first part of the theorem. So suppose that $J\subseteq \bar U$.

The identity $$[xy,z]=[x,yz]+(-1)^{\bar x \bar y+\bar x \bar z}[y,zx]$$

[can be used to show that $[\bar U,A]=[U,A]$. Hence $[J\cap K, K]\subseteq [\bar U,
A]=[U,A]=[U,H]+[U,K]$. But $[U,H]\subseteq H$, and $[U,K]\subseteq K$, so $[J\cap K, K]\subseteq [U,K]
\subseteq U$.]{}

Finally, suppose that $[J\cap K, K]=0$, then $[u\circ v,w]=0$ for any $u,v,w \in J\cap K$ because $[uv,w]=u[v,w]+(-1)^{\bar v
\bar w}[u,w]v=0$. So by lemmata 2.1, 2.2 and theorem 2.7 it follows that for each prime image, $A/P$, of $A$ either its center contains $((J\cap K)+P)/P$, or $A/P$ is a central order in a simple superalgebra at most 16-dimensional over its center.

We claim that if the image of $J \cap K$ in $A/P$ for some prime ideal $P$ of $A$ is central, then $A$ is as described in the first part of the conclusion of the theorem.

Let $P$ be a prime ideal such that $P^*\not= P$. If $(J+P)/P\not=0$, then since $A/P$ is a prime superalgebra we get $((J\cap P^*) +P)/P \not=0$, and so we have $((J\cap
P^*) +P)/P\subseteq ((J\cap K)+P)/P \subseteq Z_0(A/P)$, that is, $A/P$ is commutative. So $A/P$ is commutative unless $J\subseteq P$. And if $J\subseteq P$, then by the proof of lemma 2.1 we know that $A[u\circ v, w]A[u\circ v,w]A\subseteq P$ for any $u, v, w \in U$, and because $P$ is a prime ideal we deduce that $[u\circ v,w]\in P$ for any $u, v, w \in U$. But now by Lemma 2.2 and since $[u\circ v,w]+P=0$ for any $u,v,w \in U$, it follows that $A/P$ satisfies the conditions $u\circ v\in Z$ for any $u,v\in ((U+P)/P)_0$ and $u\circ v=0$ for any $u,v \in ((U+P)/P)_1$. By theorem 2.4 we obtain that either $(U+P)/P\subseteq Z_0(A/P)$, or $A/P$ satisfies $S(4)$.

And if $P$ is a prime ideal such that $P^*=P$ then $A/P$ has a superinvolution induced by \* and $K(A/P)=(K+P)/P$. In this case if $((J\cap K) +P)/P=0$ we get $(J+P)/P\subseteq (H+P)/P=H(A/P)$, and therefore $(J+P)/P$ is supercommutative. But then for any $a,b \in A/P$ and $y,z \in (J+P)/P$ it follows that

$$\begin{aligned}
yabz&=&(-1)^{(\bar b + \bar z)(\bar y + \bar a)}(bz)(ya)=(-1)^{\bar b (\bar y +
\bar a)} b(ya)z\\
&=& (-1)^{\bar b \bar y + \bar b \bar a+ (\bar a +\bar z)\bar
y} b(az)y= (-1)^{\bar b \bar a} ybaz,\end{aligned}$$

[and since $A/P$ is prime $ab = (-1)^{\bar a \bar b} ba$, that is, $A/P$ is supercommutative. Now from lemma 1.9 in [@M], $A/P$ is a central order in a simple superalgebra at most $4$-dimensional over its center. And if $((J\cap K)+P)/P\not=0$ then $Z_0(A/P)\not=0$, so by localizing at $V=(Z_0(A/P)\cap H(A/P))-\{0\}$ we can suppose that $Z_0(A/P)$ is a field, which we denote by $Z$. We will replace $V^{-1}(A/P)$ by $A/P$ and $V^{-1}((J+P)/P)$ by $(J+P)/P$. Then if $0\not= t\in ((J\cap K)+P)/P$ we have $tH=K$ with $H=H(A/P), K=K(A/P)$, so $K=tH\subseteq K\cap J\subseteq Z$, and also $tH=K\subseteq Z$ and $H\subseteq t^{-1}Z\subseteq Z$. Therefore $A/P$ is a field.]{}

Finally we have

Let $A$ be a semiprime superalgebra with superinvolution $*$, and let $U$ be a Lie ideal of $K$. Then either $[J\cap K,K]\subseteq U$ where $J$ is a nonzero ideal of $A$ or there exists a semiprime ideal $T$ of $A$ such that $A/Ann T$ satisfies $S(4)$ and $(U+T)/T\subseteq Z_0(A/T)$.

By theorem 2.8 we have that either the first conclusion holds, or, for each prime ideal $P$ of $A$, either $A/P$ satisfies $S(4)$ or $(U+P)/P \subseteq Z_0(A/P)$. Let $T$ be the intersection of the prime ideals $P$ of $A$ such that $(U+P)/P\subseteq Z_0(A/P)$. Then $Ann T$ contains the intersection of those prime ideals $P$ such that $A/P$ satisfies $S(4)$. So we get that $A/Ann T$ satisfies $S(4)$, and this proves the result.

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[^1]: The first author has been supported by the Spanish Ministerio de Educación y Ciencia (MTM 2004-08115-CO4-02) and both by the Comunidad Autónoma de La Rioja (ANGI 2005/05).