Nonlinear Right Bi-Skew Commuting Maps: Parameter Rigidity and Semiprime Classification

1. Introduction

Let R be a ring with involution *. In each setting below, $\eta$ is a scalar from the stated coefficient ring. Put

$$a{•}_{\eta }b=a{b}^{*}+\eta b{a}^{*}.$$

We study arbitrary maps $\Phi :R\to R$ satisfying

$$\Phi \left(a\right)b^{*}+\eta b\Phi {\left(a\right)}^{*}=a\Phi {\left(b\right)}^{*}+\eta \Phi \left(b\right)a^{*}(a,b\in R).$$

(1)

Neither additivity nor homogeneity of $\Phi$ is assumed; any automatic additivity below is derived from the identity. The values $\eta =1$ and $\eta =-1$ give the right bi-skew Jordan and Lie cases, respectively.

The closest ring-theoretic result is Ansari et al. [1] [Theorem 2.2], giving a symmetric right-multiplier conclusion for $\eta =-1$ under a high-degree hypothesis; in the noncommutative prime PI case, this excludes precisely central degree two. Related work treats bi-skew commuting maps and preservers on von Neumann algebras [2,3], nonlinear *-commuting maps [4], and mixed bi-skew Jordan and Lie n-derivations [5]; Brešar [6] gives the classical background on centralizing mappings in prime rings.

Two features control the answer: the scalar action of the involution and, for first-kind PI-rings, the central degree. We treat all scalar parameters in the unital second-kind setting. For first-kind involutions and $\eta =\pm 1$, we prove rigidity on non-PI prime rings and prime PI-rings of central degree at least three, and classify the orthogonal and symplectic degree-two exceptions. These prime results yield Theorem 5, a classification for arbitrary 2-torsion-free semiprime rings without assuming an identity. Solutions are additive at $\eta =1$, whereas at $\eta =-1$ nonadditivity can occur only through the commutative central summand. Examples 1 and 2 show that the degree restriction is genuine; the semiprime decomposition likewise reflects genuine behavior.

The proofs use parameter reduction, Beidar–Martindale functional identities, central closure and orthogonal completion. The semiprime classification is not merely a product of prime classifications, since the commutative output may depend on the whole input. Section 2 gives notation and parameter reduction; Section 3, Section 4 and Section 5 treat the second-kind, prime first-kind, and semiprime first-kind cases; Section 6 concludes.

2. Preliminaries and Notation

Throughout the paper, the notation introduced in this section is used with the same meaning unless explicitly stated otherwise. All rings are associative; unless stated otherwise, they need not be unital. For a ring R with involution *, write

$$Z\left(R\right)=\{z\in R:zr=rz\mathrm{for}\mathrm{all}r\in R\},[x,y]=xy-yx,$$

and let $\mathfrak{C}\left(R\right)$ denote the two-sided ideal generated by the commutators. For an ideal I, ${Ann}_{\ell }I$ and ${Ann}_{r}I$ denote its left and right annihilators; when they coincide we write $AnnI$. We also use

$$Sym\left(R\right)=\{x\in R:x^{*}=x\},Skew\left(R\right)=\{x\in R:x^{*}=-x\}.$$

When the ambient ring is denoted by S, we write $S_{+}=Sym\left(S\right)$ and $S_{-}=Skew\left(S\right)$.

For a prime or semiprime ring, C denotes the extended centroid and $R_C=RC$ the central closure. We write $Q_s\left(R\right)$ for the symmetric Martindale ring of quotients, $Q_{\mathrm{ms}}\left(R\right)$ for the maximal symmetric ring of quotients, and $Q_{ml}\left(R\right)$ and $Q_{mr}\left(R\right)$ for the maximal left and right rings of quotients. Whenever a prime or semiprime involution is said to be of the first kind, this means that its extension fixes C elementwise. In the scalar setting of Section 3, second kind means that the involution on the base field is nontrivial. If R is prime PI, its central degree is the degree of the central simple C-algebra $R_C$.

When C is a field and S is a C-algebra, set ${adeg}_C\left(x\right)$ equal to the least degree of a nonzero polynomial in $C\left[t\right]$ annihilating x, with value ∞ when no such polynomial exists, and for $X\subseteq S$ put

$${adeg}_C\left(X\right)=\underset{x\in X}{sup}{adeg}_C\left(x\right).$$

Thus ${adeg}_C\left(X\right)>d$ means that X is not algebraic of bounded degree at most d. For a central simple algebra, Trd and Nrd denote the reduced trace and reduced norm.

In the semiprime part, $C=Z\left(Q_{\mathrm{ms}}\left(R\right)\right)$ is commutative von Neumann regular and self-injective, and hence the Boolean algebra $\mathcal{B}$ of its central idempotents is complete. We write $\mathcal{O}\left(D\right)$ for the orthogonal completion of a C-algebra D in the relevant quotient ring; in particular $A=\mathcal{O}\left(R_C\right)$. For $x\in A$, $E\left(x\right)\in \mathcal{B}$ is its central support, that is, the least central idempotent satisfying $E\left(x\right)x=x$. For $X\subseteq A$, put

$$E\left(X\right)=\underset{x\in X}{\bigvee }E\left(x\right).$$

If M is a maximal ideal of $\mathcal{B}$, let $MA$ and $MC$ denote the ideals generated by M in A and C, respectively, and write $A_M=A/MA$ and $C_M=C/MC$. If $S=eA$ is a central summand, $S_M$ denotes its image in $A_M$, and induced maps are marked by the same subscript, such as $T_M$ or $J_M$. When working in $e\mathcal{B}$, its maximal ideals are identified with the maximal ideals M of $\mathcal{B}$ for which $e\notin M$.

2.1. The Elementary Parameter Reduction

For the remainder of this section let $\Lambda$ be a commutative ring with involution, let R be a $\Lambda$-algebra with compatible involution, so that ${\left(\alpha a\right)}^{*}={\alpha }^{*}{a}^{*}$ for all $\alpha \in \Lambda$ and $a\in R$, and fix $\eta \in \Lambda$. Let $\Phi :R\to R$ be arbitrary and put

$$F(a,b)=\Phi \left(a\right)b^{*}-a\Phi {\left(b\right)}^{*}.$$

Lemma 1. 

For every map Φ one has

$$F{(a,b)}^{*}=-F(b,a).$$

Moreover, (1) is equivalent to

$$F(a,b)=\eta F(b,a)(a,b\in R).$$

(2)

Proof. 

The first identity follows directly from

$${(\Phi \left(a\right)b^{*}-a\Phi {\left(b\right)}^{*})}^{*}=b\Phi {\left(a\right)}^{*}-\Phi \left(b\right)a^{*}=-F(b,a).$$

Equation (1) says $F(a,b)+\eta F{(a,b)}^{*}=0$, which is (2). □

Proposition 1. 

Every solution of (1) satisfies

$$(1-{\eta }^2)F(a,b)=0,(1-\eta {\eta }^{*})F(a,b)=0,(\eta -{\eta }^{*})F(a,b)=0.$$

Consequently, if any one of the three displayed scalars acts injectively on R, then

$$\Phi \left(a\right)b^{*}=a\Phi {\left(b\right)}^{*}(a,b\in R).$$

(3)

Proof. 

Interchanging $a,b$ in (2) gives $F(b,a)=\eta F(a,b)$, hence $F(a,b)={\eta }^{2}F(a,b)$. Applying * to (2) and using Lemma 1 gives $F(b,a)={\eta }^{*}F(a,b)$. Comparison with $F(b,a)=\eta F(a,b)$ yields the third relation, while substituting the ${\eta }^{*}$ relation back into (2) yields the second. □

Corollary 1. 

Let Λ be a field fixed pointwise by the involution and let R be a nonzero Λ-algebra. If $\eta \in \Lambda$ and ${\eta }^2\ne 1$, then every solution of (1) satisfies (3). Thus, in the first-kind scalar setting, the only parameter values requiring separate analysis are $\eta =\pm 1$.

Proof. 

Since $1-{\eta }^2$ is a nonzero scalar, it acts injectively on the $\Lambda$-vector space R. Proposition 1 applies. □

The complementary scalar situation is treated next. It includes the usual complex conjugate-linear case.

3. The Scalar Second-Kind Case

Throughout this section, $\Lambda$ is a field with nontrivial involution, $\mathcal{A}$ is a unital associative $\Lambda$-algebra with compatible involution, so that ${\left(\alpha x\right)}^{*}={\alpha }^{*}{x}^{*}$ for all $\alpha \in \Lambda$ and $x\in \mathcal{A}$. Choose $\lambda \in \Lambda$ with

$$\mu =\lambda -{\lambda }^{*}\ne 0.$$

Thus $\mu$ is invertible. Let I denote the identity of $\mathcal{A}$; the commutator and annihilator notation is that of Section 2.

Lemma 2. 

For $K\in \mathcal{A}$, the following are equivalent:

(i)

$K[A,B]=0$ for all $A,B\in \mathcal{A}$;

(ii)

$K\in {Ann}_{\ell }\mathfrak{C}\left(\mathcal{A}\right)$.

Proof. 

Only (i)⇒(ii) needs proof. Since $[A,XB]=[A,X]B+X[A,B]$, left multiplication by K gives $KX[A,B]=0$. Multiplication on the right by an arbitrary Y then gives $KX[A,B]Y=0$. □

Theorem 1. 

Let $\eta \in \Lambda$ and let $\Phi :\mathcal{A}\to \mathcal{A}$ be arbitrary. If $\eta \ne 1$, then (1) holds if and only if

$$\Phi \left(A\right)=AH(A\in \mathcal{A})$$

for a unique $H=H^{*}$. If $\eta =1$, then (1) holds if and only if

$$\Phi \left(A\right)=AH+K{A}^{*},H=H^{*},K\in {Ann}_{\ell }\mathfrak{C}\left(\mathcal{A}\right).$$

The parameters are recovered from

$$H={\mu }^{-1}\left(\Phi \left(\lambda I\right)-{\lambda }^{*}\Phi \left(I\right)\right),K={\mu }^{-1}\left(\lambda \Phi \left(I\right)-\Phi \left(\lambda I\right)\right).$$

(4)

Proof. 

Taking $B=I$ in (1) gives

$$\Phi \left(A\right)+\eta \Phi {\left(A\right)}^{*}=A\Phi {\left(I\right)}^{*}+\eta \Phi \left(I\right)A^{*}.$$

(5)

Taking $B=\lambda I$ gives

$${\lambda }^{*}\Phi \left(A\right)+\eta \lambda \Phi {\left(A\right)}^{*}=A\Phi {\left(\lambda I\right)}^{*}+\eta \Phi \left(\lambda I\right)A^{*}.$$

(6)

Subtracting $\lambda$ times (5) from (6) and dividing by $-\mu$ gives

$$\Phi \left(A\right)=A{H}^{*}+\eta K{A}^{*},$$

(7)

with $H,K$ as in (4). The same definitions give

$$\Phi \left(I\right)=H+K,\Phi \left(\lambda I\right)=\lambda H+{\lambda }^{*}K.$$

Evaluating (7) at I and $\lambda I$ therefore gives

$$H-H^{*}=(\eta -1)K,$$

and

$$\lambda (H-H^{*})+(1-\eta ){\lambda }^{*}K=0.$$

Substitution yields $\mu (\eta -1)K=0$. Hence $(\eta -1)K=0$ and $H=H^{*}$.

If $\eta \ne 1$, then $K=0$. If $\eta =1$, substitution of $\Phi \left(A\right)=AH+K{A}^{*}$ into (1) cancels the multiplier terms and gives

$$K[A^{*},B^{*}]+[B,A]K^{*}=0.$$

(8)

Replacing A by $\lambda A$ and subtracting $\lambda$ times (8) gives

$$({\lambda }^{*}-\lambda )K[A^{*},B^{*}]=0.$$

Thus $K[A^{*},B^{*}]=0$, and $K\in {Ann}_{\ell }\mathfrak{C}\left(\mathcal{A}\right)$ by Lemma 2. The converse follows by direct substitution, using $\mathfrak{C}{\left(\mathcal{A}\right)}^{*}=\mathfrak{C}\left(\mathcal{A}\right)$. Uniqueness follows from (4). □

Remark 1. 

For $\Lambda =\mathbb{C}$ with complex conjugation one may take $\lambda =i$, and (4) becomes the familiar pair of formulas involving $\Phi \left(I\right)$ and $\Phi \left(iI\right)$. The argument depends on the invertibility of $\lambda -{\lambda }^{*}$ and therefore has no first-kind analogue.

4. Prime Rings with an Involution of the First Kind

In this section R is a noncommutative 2-torsion-free prime ring with a first-kind involution and $\eta \in \{1,-1\}$. We use the notation fixed in Section 2.

4.1. Automatic Additivity and Central Closure

Lemma 3. 

If $d\in R$ satisfies

$$d{b}^{*}+\eta b{d}^{*}=0(b\in R),$$

then $d=0$.

Proof. 

Replace b by $cb$ and compare with the displayed identity multiplied on the right by $c^{*}$. Since $\eta =\pm 1$,

$$cb{d}^{*}=b{d}^{*}{c}^{*}(b,c\in R).$$

Thus every $w\in R{d}^{*}$ satisfies $cw=w{c}^{*}$. Applying this relation to a product in two ways gives $w[y^{*},x^{*}]=0$, hence $w[u,v]=0$ for all $u,v\in R$. Since $w[u,vr]=w[u,v]r+wv[u,r]$, we get $wR[u,r]=0$. The commutator ideal is a nonzero ideal because R is noncommutative, so primeness forces $w=0$. Hence $R{d}^{*}=0$ and $d=0$. □

Proposition 2. 

Every solution of (1) is additive. If $z\in Z\left(R\right)$, then

$$\Phi \left(za\right)=z\Phi \left(a\right)(a\in R).$$

Proof. 

For additivity subtract the identities at $(a+c,b)$, $(a,b)$, and $(c,b)$. The defect $d=\Phi (a+c)-\Phi \left(a\right)-\Phi \left(c\right)$ satisfies Lemma 3. For central linearity, subtract z times the identity at $(a,b)$ from that at $(za,b)$ and use $z^{*}=z$. □

Lemma 4. 

Every additive solution $\Phi :R\to R$ of (1) extends uniquely to a C-linear map

$${\Phi }_C:R_C⟶Q_s\left(R\right)$$

which satisfies (1) on $R_C$.

Proof. 

If $c_i\in C$, $a_i\in R$, and ${\sum }_i{c}_i{a}_i=0$, put $d={\sum }_i{c}_i\Phi \left(a_i\right)\in Q_s\left(R\right)$. Multiplying the identities for $(a_i,b)$ by $c_i$ and adding gives

$$d{b}^{*}+\eta b{d}^{*}=0(b\in R).$$

Replace b by $cb$ and compare with this identity multiplied on the right by $c^{*}$. For $w=b{d}^{*}$ one obtains $cw=w{c}^{*}$ for all $b,c\in R$. Applying this relation to a product in two ways gives $w[y^{*},x^{*}]=0$ for all $x,y\in R$. Indeed, after renaming the variables, $w[u,v]=0$ for all $u,v\in R$, and

$$0=w[u,vr]=w[u,v]r+wv[u,r]$$

shows that $wR[u,r]=0$. Thus w annihilates the two-sided ideal generated by the commutators. The commutator ideal is a nonzero essential ideal of the prime ring R, so the quotient-ring action gives $w=0$. Thus $R{d}^{*}=0$, and the essentiality of R in $Q_s\left(R\right)$ gives $d=0$.

It follows that ${\Phi }_C(\sum c_i{a}_i)=\sum c_i\Phi \left(a_i\right)$ is well defined. The identity extends C-bilinearly, and uniqueness is immediate. □

4.2. The PI Case

Lemma 5. 

Let F be a field of characteristic different from two, let $n\ge 3$, and let $J\in M_n\left(F\right)$ be invertible with $J^T=\epsilon J$, $\epsilon \in \{1,-1\}$. Equip $M_n\left(F\right)$ with

$$X^{*}=J^{-1}{X}^{T}J.$$

If an F-linear map $T:M_n\left(F\right)\to M_n\left(F\right)$ satisfies (1) for $\eta =\pm 1$, then

$$T\left(X\right)=XP(X\in M_n\left(F\right))$$

for a unique $P=P^{*}$.

Proof. 

Put $U\left(X\right)=T\left(X\right)J^{-1}$ and $\delta =\eta \epsilon$. Right multiplication of (1) by $J^{-1}$ gives

$$U\left(A\right)B^T+\delta BU{\left(A\right)}^T=\epsilon AU{\left(B\right)}^T+\eta U\left(B\right)A^T.$$

Take $A=u{v}^T$, $B=x{y}^T$, and set $p=U\left(u{v}^T\right)y$, $q=U\left(x{y}^T\right)v$. Then

$$p\otimes x+\delta x\otimes p=\epsilon \left(u\otimes q+\delta q\otimes u\right).$$

(9)

Fix $u,v,y$ and vary x. Passing in both tensor factors to $F^n/Fu$ gives $\overline{p}\otimes \overline{x}+\delta \overline{x}\otimes \overline{p}=0$ for every $\overline{x}$. Because $dim(F^n/Fu)\ge 2$, this forces $\overline{p}=0$. Hence $U\left(u{v}^T\right)y\in Fu$. Interchanging the two rank-one matrices gives $U\left(x{y}^T\right)v\in Fx$. Therefore

$$U\left(u{v}^T\right)y=h(v,y)u$$

for a bilinear form h (linearity in u makes the scalar independent of u), and (9) gives $h(v,y)=\epsilon h(y,v)$. Write $h(v,y)=v^{T}Hy$ with $H^T=\epsilon H$. Then $U\left(X\right)=XH$ and $T\left(X\right)=XP$ with $P=HJ$. Finally $P^{*}=J^{-1}{P}^{T}J=P$. □

Theorem 2. 

Let R be a noncommutative prime PI-ring with first-kind involution and central degree $n\ge 3$. Every solution of (1) has the form

$$\Phi \left(a\right)=aq,q=q^{*}\in Q_s\left(R\right),Rq\subseteq R.$$

Proof. 

By Posner’s theorem [7] the central closure $R_C$ is a central simple C-algebra of degree n. Since a prime PI-ring is Goldie, $Q_s\left(R\right)$ lies in its classical quotient ring, which is $R_C$; see also [8] [Chapter 2]. Thus Lemma 4 extends $\Phi$C-linearly to $R_C$; keep the same letter for the extension. After a faithfully flat splitting-field extension $L/C$, the induced involution is adjoint to a nondegenerate symmetric or alternating form and hence has the matrix form of Lemma 5 [9] [Chapter I]; that lemma gives ${\Phi }_L\left(x\right)=x{q}_L$ with $q_L=q_L^{*}$. Since $R_C$ is unital, $q_L={\Phi }_L\left(1\right)=\Phi \left(1\right)\otimes 1$, so the coefficient descends to $q=\Phi \left(1\right)\in R_C$ and $q=q^{*}$. Restricting to R gives $Rq\subseteq R$, and taking involutions gives $qR\subseteq R$; thus $q\in Q_s\left(R\right)$. □

4.3. The Non-PI Case

We use the algebraic-degree notation ${adeg}_C$ and the spaces $S_{\pm }$ from Section 2. The degree here is the one occurring in the functional-identity theorem of Beidar and Martindale [10]; in the non-PI case Montgomery’s theorem will rule out a uniform algebraic-degree bound on the symmetric elements.

Lemma 6. 

Let S be a noncommutative centrally closed prime ring with extended centroid C and a first-kind involution, where C is a field of characteristic different from two. If either ${dim}_C{S}_{+}\le 1$ or ${dim}_C{S}_{-}\le 1$, then every element of $S_{+}\cup S_{-}$ is algebraic over C of degree at most two.

Proof. 

Suppose first that $S_{-}=Ck$ with $0\ne k\in S_{-}$. (The case $S_{-}=0$ would make the involution the identity and hence make S commutative.) For $s\in S_{+}$ there is ${\alpha }_s\in C$ such that

$$sk+ks={\alpha }_{s}k.$$

In the maximal quotient put $t=s-{\displaystyle \frac{1}{2}}{\alpha }_{s}1$. Then $tk=-kt$. Consequently $k^2$ commutes with every symmetric element and, trivially, with $S_{-}$; hence $k^2\in C$. If $k^2=0$, the displayed relation gives $k{S}_{+}k=0$, while $k{S}_{-}k=0$ is immediate. Since $S=S_{+}\oplus S_{-}$ as a C-space, $kSk=0$, contradicting primeness. Thus $k^2\in C^{\times }$.

For $u\in S_{+}$ we have $[t,u]\in S_{-}=Ck$, and therefore

$$[t^2,u]=t[t,u]+[t,u]t=0.$$

Also $t^{2}k=k{t}^2$. Thus $t^2$ commutes with S inside the maximal quotient. Since S is centrally closed, the commutant of S there is its extended centroid C, and hence $t^2\in C$. Thus every $s\in S_{+}$ satisfies

$${\left(s-\frac{1}{2}{\alpha }_{s}1\right)}^2\in C,$$

and every element of $S_{-}$ is a scalar multiple of k with $k^2\in C$. So the degree is at most two.

The symmetric space cannot be zero: otherwise every element is skew, so $x^2$ is both symmetric and zero for every x, whence $xSx=0$ and primeness forces $S=0$. Thus, if ${dim}_C{S}_{+}\le 1$, write $S_{+}=Cs$ with $0\ne s\in S_{+}$. Write $s^2=\alpha s$. If $\alpha =0$, then $s{S}_{+}s=0$. Since $[s,k]\in S_{+}$ for $k\in S_{-}$, we also have $sks=0$; hence $sSs=0$, again contradicting primeness. Thus $\alpha \ne 0$ and $e={\alpha }^{-1}s$ is a nonzero idempotent. For $k\in S_{-}$, the commutator $[e,k]$ is symmetric, so $[e,k]=\beta e$ for some $\beta \in C$. Multiplication by e on both sides gives $0=e[e,k]e=\beta e$, whence $[e,k]=0$. The element e therefore commutes with both $S_{+}$ and $S_{-}$, and hence with S. Primeness then forces $ex=x$ for every $x\in S$, so e is the identity of S. Thus $S_{+}=Ce$, while $k^2\in S_{+}=Ce$ for every $k\in S_{-}$. Again every symmetric or skew element has degree at most two. □

Lemma 7. 

Let S be as in Lemma 6 and assume

$${adeg}_C(S_{+}\cup S_{-})>2.$$

Let $\gamma ,\delta :S\to C$ be C-linear maps satisfying

$$\gamma \left(a\right)b^{*}+\eta \delta \left(a\right)b=\delta \left(b\right)a+\eta \gamma \left(b\right)a^{*}(a,b\in S),$$

(10)

where $\eta \in \{1,-1\}$. Then $\gamma =\delta =0$.

Proof. 

By Lemma 6, both $S_{+}$ and $S_{-}$ have C-dimension greater than one.

If $\eta =-1$, putting $a,b\in S_{+}$ in (10) gives $(\gamma -\delta )\left(a\right)b=-(\gamma -\delta )\left(b\right)a$, hence $\gamma =\delta$ on $S_{+}$. Putting $a,b\in S_{-}$ gives $(\gamma +\delta )\left(a\right)b=-(\gamma +\delta )\left(b\right)a$, hence $\gamma =-\delta$ on $S_{-}$. With $a\in S_{+}$ and $b\in S_{-}$ the residual identity then reduces to a scalar multiple of b equal to a scalar multiple of a. Since $S_{+}\cap S_{-}=0$, both scalars vanish. Thus $\gamma =\delta =0$ on $S_{+}\cup S_{-}$.

If $\eta =1$, the pairs from $S_{+}\times S_{+}$ and $S_{-}\times S_{-}$ give, respectively,

$$(\gamma +\delta )\left(a\right)b=(\gamma +\delta )\left(b\right)a,(\delta -\gamma )\left(a\right)b=(\delta -\gamma )\left(b\right)a.$$

Since both eigenspaces have dimension greater than one, $\gamma +\delta =0$ on $S_{+}$ and $\delta -\gamma =0$ on $S_{-}$. A mixed pair then gives

$$-\gamma \left(a\right)b=\gamma \left(b\right)a(a\in S_{+},b\in S_{-}),$$

up to the common nonzero factor 2. Again the two eigenspaces intersect trivially, so $\gamma$ vanishes on both, and then so does $\delta$. Finally $S=S_{+}\oplus S_{-}$ because 2 is invertible in C. □

Lemma 8 

(Beidar–Martindale specialization). Let S be a prime ring with involution, let C be its extended centroid, and let $Q_{ml}\left(S\right)$ be its maximal left ring of quotients. Assume $charC\ne 2$ and

$${adeg}_C\left(Sym\left(S\right)\cup Skew\left(S\right)\right)>4.$$

Suppose maps $E_1,E_2,H_1,H_2:S\to Q_{ml}\left(S\right)$ satisfy

$$E_1\left(y\right)x+E_2\left(x\right)y+x^{*}{H}_1\left(y\right)+y^{*}{H}_2\left(x\right)=0(x,y\in S).$$

(11)

Then there are $q_{12},q_{21}\in Q_{ml}\left(S\right)$ and maps ${\lambda }_i,{\mu }_i:S\to C$ such that

$$\begin{array}{cccc}\hfill E_1\left(y\right)& =y^{*}{q}_{12}+{\lambda }_1\left(y\right),\hfill & \hfill E_2\left(x\right)& =x^{*}{q}_{21}+{\lambda }_2\left(x\right),\hfill \end{array}$$

(12)

$$\begin{array}{cccc}\hfill H_1\left(y\right)& =-q_{21}y-{\mu }_1\left(y\right),\hfill & \hfill H_2\left(x\right)& =-q_{12}x-{\mu }_2\left(x\right).\hfill \end{array}$$

(13)

Proof. 

This is the part of [10] [Theorem 3.1] needed here. In the notation of that theorem take $m=2$,

$$I=L=\{1,2\},J=K=⌀.$$

The general functional identity then reduces exactly to (11). For this index pattern the degree bound in that theorem is

$$max\left\{\right|I|+|K|-1,|J|+|L\left|\right\}=2,$$

so our stronger hypothesis ${adeg}_C>4$ certainly applies. Specializing the standard-solution formulas to this index pattern gives (12)–(), after renaming the two constant quotient-ring coefficients. Only this two-variable specialization is needed below. □

Lemma 9. 

Let R be a noncommutative prime ring with a first-kind involution and extended centroid C, where $charC\ne 2$, and assume

$${adeg}_C\left(Sym\left(R_C\right)\cup Skew\left(R_C\right)\right)>4.$$

If an additive map $\Phi :R\to R$ satisfies (1) for $\eta \in \{1,-1\}$, then

$$\Phi \left(a\right)=aq(a\in R)$$

for a symmetric element $q\in Q_s\left(R\right)$.

Proof. 

Extend $\Phi$C-linearly to $R_C$ by Lemma 4, and keep the same letter for the extension. Put $\Theta \left(x\right)=\Phi \left(x^{*}\right)$. Then

$$\Theta \left(x\right)y-\eta \Theta \left(y\right)x-x^{*}\Theta {\left(y\right)}^{*}+\eta y^{*}\Theta {\left(x\right)}^{*}=0.$$

(14)

Equation (14) has the form (11) with

$$E_1\left(y\right)=-\eta \Theta \left(y\right),E_2\left(x\right)=\Theta \left(x\right),H_1\left(y\right)=-\Theta {\left(y\right)}^{*},H_2\left(x\right)=\eta \Theta {\left(x\right)}^{*}.$$

Lemma 8 therefore gives $q_{12},q_{21}\in Q_{ml}\left(R_C\right)$ and central maps ${\lambda }_i,{\mu }_i:R_C\to C$ satisfying

$$\begin{array}{cccc}\hfill -\eta \Theta \left(y\right)& =y^{*}{q}_{12}+{\lambda }_1\left(y\right),\hfill & \hfill \Theta \left(x\right)& =x^{*}{q}_{21}+{\lambda }_2\left(x\right),\hfill \\ \hfill -\Theta {\left(y\right)}^{*}& =-q_{21}y-{\mu }_1\left(y\right),\hfill & \hfill \eta \Theta {\left(x\right)}^{*}& =-q_{12}x-{\mu }_2\left(x\right).\hfill \end{array}$$

Because the left sides are C-linear, the four central maps are C-linear.

Comparing the first two formulas for the same variable gives

$$x^{*}(q_{21}+\eta q_{12})+{\lambda }_2\left(x\right)+\eta {\lambda }_1\left(x\right)=0.$$

We shall also use the following elementary observation. If $u\in Q_{ml}\left(R_C\right)$ and $\nu :R_C\to C$ satisfy $xu+\nu \left(x\right)=0$ for all x, then $u=0$ and $\nu =0$. Indeed, $R_{C}u\subseteq C$; if $su\ne 0$ for some s, then $\left(rs\right)u=r\left(su\right)\in C$ for every r, forcing $R_C\subseteq C$ and hence commutativity. Thus $R_{C}u=0$, and essentiality gives $u=0$. It follows, with $q=q_{21}$, that

$$q_{12}=-\eta q,{\lambda }_1=-\eta {\lambda }_2,{\mu }_2=-\eta {\mu }_1.$$

Writing $\lambda ={\lambda }_2$ and $\mu ={\mu }_1$, we obtain

$$\Theta \left(x\right)=x^{*}q+\lambda \left(x\right),\Theta {\left(x\right)}^{*}=qx+\mu \left(x\right).$$

(15)

For $a=x^{*}$ set $\gamma \left(a\right)=\lambda \left(a^{*}\right)$ and $\delta \left(a\right)=\mu \left(a^{*}\right)$. Substituting (15) into the original identity and canceling the terms containing q gives

$$\gamma \left(a\right)b^{*}+\eta \delta \left(a\right)b=\delta \left(b\right)a+\eta \gamma \left(b\right)a^{*}.$$

Lemma 7 therefore gives $\gamma =\delta =0$. Hence on $R_C$,

$$\Theta \left(x\right)=x^{*}q,\Theta {\left(x\right)}^{*}=qx.$$

Restricting to R yields $Rq\subseteq R$ and $qR\subseteq R$, so $q\in Q_s\left(R\right)$. Since $q\in Q_s\left(R\right)$, the involution is defined on q. Taking involutions in the first equality and comparing with the second gives $(q-q^{*})R=0$. Hence $q=q^{*}$, and therefore $\Phi \left(a\right)=aq$ for every $a\in R$. □

Theorem 3. 

Let R be a noncommutative 2-torsion-free prime ring with a first-kind involution. Assume either that R is non-PI, or that R is PI of central degree at least three. Then for $\eta \in \{1,-1\}$ every solution of (1) is automatically additive and has the form

$$\Phi \left(a\right)=aq,q=q^{*}\in Q_s\left(R\right),Rq\subseteq R.$$

(16)

Conversely every map in (16) is a solution.

Proof. 

Automatic additivity is Proposition 2. If R is PI, apply Theorem 2. Suppose that R is not PI. If the symmetric elements of the central closure $R_C$ were algebraic over C of uniformly bounded degree, then Montgomery’s theorem [11] [Corollary 2] would imply that $R_C$ is a PI-algebra. Since $R\subseteq R_C$, this would make R a PI-ring, a contradiction. Thus the symmetric elements, and hence $Sym\left(R_C\right)\cup Skew\left(R_C\right)$, have no uniform algebraic-degree bound. In particular their algebraic degree is greater than 4, so Lemma 9 applies. The converse follows by direct substitution. □

4.4. The Degree-Two Case

In the degree-two case R is prime PI. By Posner’s theorem $R_C$ is a central simple C-algebra of degree two, and, as in the proof of Theorem 2, $Q_s\left(R\right)$ lies in the classical quotient ring $R_C$. Hence Lemma 4 extends $\Phi$ to a map $R_C\to R_C$; the formulas below restrict to R exactly when they preserve R.

Proposition 3. 

Let A be a central simple algebra of degree two over a field F of characteristic different from two, equipped with a first-kind involution. Every solution $T:A\to A$ of (1) is additive and F-linear, and is as follows.

$$\begin{array}{ccc}\mathit{involution}& \eta & \mathit{solution}\hfill \\ \mathit{orthogonal}& 1& T\left(x\right)=xh,h=h^{*},\hfill \\ \mathit{orthogonal}& -1& T\left(x\right)=xh+s(Trd\left(x\right)1-x^{*}),h=h^{*},s=s^{*},\hfill \\ \mathit{symplectic}& -1& T\left(x\right)=\lambda x,\lambda \in F,\hfill \\ \mathit{symplectic}& 1& T\in {End}_F\left(A\right),B(Tx,y)=B(x,Ty),\hfill \\ \end{array}$$

where $B(x,y)=Nrd(x+y)-Nrd\left(x\right)-Nrd\left(y\right)$. The parameters h, s, and λ, whenever they occur, are uniquely determined.

Proof. 

Additivity and F-linearity follow from Proposition 2. All conditions are linear over F, so equality of the asserted solution space with the kernel of the defining linear system may be checked after a faithfully flat field extension. Choose a field extension $L/F$ large enough both to split A and to put the involution into its standard split form. Write $A_L=A{\otimes }_{F}L\simeq M_2\left(L\right)$ and denote the scalar extension of T again by T.

First suppose the involution is orthogonal, so that on $A_L=M_2\left(L\right)$ we have $x^{*}=x^T$. A direct comparison of the coefficients in (1) for the sixteen pairs $(E_{ij},E_{kl})$ gives, when $\eta =1$,

$$\begin{array}{cc}T\left(E_{11}\right)=\left(\begin{array}{cc}a& b\\ 0& 0\end{array}\right),\hfill & T\left(E_{12}\right)=\left(\begin{array}{cc}b& c\\ 0& 0\end{array}\right),\hfill \\ T\left(E_{21}\right)=\left(\begin{array}{cc}0& 0\\ a& b\end{array}\right),\hfill & T\left(E_{22}\right)=\left(\begin{array}{cc}0& 0\\ b& c\end{array}\right).\hfill \end{array}$$

Thus $T\left(x\right)=xh$ with $h=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)=h^T$. For $\eta =-1$, comparison of the same coefficients gives

$$\begin{array}{cc}T\left(E_{11}\right)=\left(\begin{array}{cc}a& b+q\\ 0& r\end{array}\right),\hfill & T\left(E_{12}\right)=\left(\begin{array}{cc}b-q& c\\ -r& 0\end{array}\right),\hfill \\ T\left(E_{21}\right)=\left(\begin{array}{cc}0& -p\\ a& b-q\end{array}\right),\hfill & T\left(E_{22}\right)=\left(\begin{array}{cc}p& 0\\ b+q& c\end{array}\right).\hfill \end{array}$$

Writing

$$h=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right),s=\left(\begin{array}{cc}p& q\\ q& r\end{array}\right),$$

the four displayed formulas may be written as

$$T\left(x\right)=xh+s\left(tr\left(x\right)1-x^T\right).$$

The six scalars $a,b,c,p,q,r$ are determined by the four images above. Consequently the symmetric matrices h and s are uniquely determined.

For the symplectic involution take $x^{*}=J^{-1}{x}^{T}J$ with $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. Then $x^{*}=tr\left(x\right)1-x$, and the polarized Cayley–Hamilton identity gives

$$x{y}^{*}+y{x}^{*}=B(x,y)1.$$

(17)

For $\eta =1$, the two sides of (1) become $B(Tx,y)1$ and $B(x,Ty)1$, respectively. Hence the required condition is $B(Tx,y)=B(x,Ty)$. At $\eta =-1$, substitution of the four matrix units gives

$$T\left(E_{ij}\right)=\lambda E_{ij}(1\le i,j\le 2),$$

for some scalar $\lambda \in L$, hence $T\left(x\right)=\lambda x$ on $A_L$.

Conversely, direct substitution verifies each of the four families listed in the proposition.

The defining system and the asserted solution spaces are defined over F. After scalar extension to L their kernels agree by the calculations above, so faithful flatness gives the same equality over F. The intrinsic formulas, and in the orthogonal case the unique parameters h and s, therefore descend to the original algebra. □

Example 1. 

For $R=M_2\left(F\right)$ with transpose and $\eta =-1$,

$$\Phi \left(x\right)=tr\left(x\right)1-x^T$$

is the special case $h=0$, $s=1$ of Proposition 3. It is not a right multiplier. The same formula restricts to the nonunital prime ring $M_2\left(3\mathbb{Z}\right)$: it sends $3E_{11}$ to $3E_{22}$, whereas $\left(3E_{11}\right)q$ has zero second row for every right-multiplier coefficient q.

Example 2. 

Let $R=M_2\left(F\right)$ carry the symplectic involution $x^{*}=tr\left(x\right)1-x$, and take $\eta =1$. By (17),

$$x{•}_{1}y=x{y}^{*}+y{x}^{*}=B(x,y)1.$$

Hence the defining identity is equivalent to

$$B(Tx,y)=B(x,Ty),$$

so every B-self-adjoint endomorphism of the four-dimensional space $M_2\left(F\right)$ is a solution. Since B is nondegenerate on this four-dimensional space, its self-adjoint endomorphisms form a space of dimension $4(4+1)/2=10$. By contrast, $Sym\left(M_2\left(F\right)\right)=F1$ for the symplectic involution, so the right-multiplier solutions $T\left(x\right)=xh$ with $h=h^{*}$ form a one-dimensional space. Together with Example 1, this shows that the central-degree hypothesis in Theorem 3 is necessary for both signs.

5. Semiprime Rings with an Involution of the First Kind

Now let R be a semiprime 2-torsion-free ring with first-kind involution; no identity is assumed. We use the notation of Section 2. The injectivity of multiplication by 2 makes $2R$ essential, so 2 becomes invertible in C. Fix $\eta \in \{1,-1\}$.

5.1. The Commutator Annihilator and Defects

For ideals of a semiprime ring the left and right annihilators coincide. With $I=\mathfrak{C}\left(R\right)$, put

$$N=AnnI.$$

Lemma 10. 

The ideal N satisfies

$$N\subseteq Z\left(R\right),N^{*}=N,R/N\mathrm{is}\mathrm{semiprime},Ann\mathfrak{C}(R/N)=0.$$

If the involution is of the first kind, then it fixes N elementwise.

Proof. 

Since I is a two-sided ideal, N is a two-sided annihilator ideal and $NI=IN=0$. If $x\in N\cap I$, then $Rx\subseteq I$ and hence $xRx=0$; semiprimeness gives $x=0$. For $n\in N$ and $r\in R$ the commutator $[n,r]$ belongs both to N and to I, so $[n,r]=0$. Thus $N\subseteq Z\left(R\right)$. Since $I^{*}=I$, the equality of the left and right annihilators also gives $N^{*}=N$.

Annihilator ideals of semiprime rings are semiprime ideals, and the corresponding quotient is semiprime; see, for example, [8] [Chapter 2]. Finally $\mathfrak{C}(R/N)=(I+N)/N$. If $x+N$ annihilates this ideal, then $xI\subseteq N\cap I=0$, and similarly $Ix=0$; hence $x\in N$ and the annihilator in the quotient is zero. For a first-kind involution the extension to the extended centroid is the identity. Since $Z\left(R\right)$ embeds in the extended centroid, every element of N is therefore symmetric. Consequently, if $n\in N$ and $b\in R$, then $bn\in N$ and $n{b}^{*}={\left(bn\right)}^{*}=bn=nb$. Thus $b\mapsto n{b}^{*}$ is already the symmetric right multiplier $b\mapsto bn$, explaining why the second-kind exceptional term has no separate analogue here. □

Proposition 4. 

Let $\Phi :R\to R$ be a solution of (1), and put $D(a,c)=\Phi (a+c)-\Phi \left(a\right)-\Phi \left(c\right)$.

(i)

For either sign, $D(a,c)\in N$.

(ii)

If $\eta =1$, then $D(a,c)=0$; hence every solution is additive.

(iii)

If $\eta =-1$, every map $\Gamma :R\to N$ is itself a solution of (1).

(iv)

For either sign, $\Phi \left(N\right)\subseteq N$, and the induced map on $R/N$ is additive.

Proof. 

Subtracting the identities at $(a+c,b)$, $(a,b)$, and $(c,b)$ gives

$$D{b}^{*}+\eta b{D}^{*}=0.$$

(18)

Replacing b by $cb$ in (18) and comparing with (18) multiplied on the right by $c^{*}$ gives

$$cb{D}^{*}=b{D}^{*}{c}^{*}(b,c\in R).$$

Thus every $w\in R{D}^{*}$ satisfies the twisted-central relation $cw=w{c}^{*}$. Applying it to a product in two ways shows that w annihilates every commutator, and replacing one factor by a product then gives $wI=0$. Hence $R{D}^{*}\subseteq N$. Equation (18) also gives $DR\subseteq N$. If $i\in I$, then $Di\in N\cap I=0$ by Lemma 10; therefore $DI=0$, and the equality of the two annihilators yields $D\in N$.

The first-kind involution fixes the central ideal N elementwise. If $n\in N$, then $nb\in N$ and $n{b}^{*}={\left(bn\right)}^{*}=bn=nb$. At $\eta =1$, (18) becomes $2bD=0$. Since R is 2-torsion-free, $bD=0$ for every b, and semiprimeness gives $D=0$.

At $\eta =-1$, if $\Gamma \left(a\right)\in N$, then $\Gamma \left(a\right)b^{*}-b\Gamma {\left(a\right)}^{*}=0$ and $a\Gamma {\left(b\right)}^{*}-\Gamma \left(b\right)a^{*}=0$, so every N-valued map is a solution.

Finally, for $n\in N$, the right-hand side of (1) with $a=n$ lies in N. Modulo N the image of $\Phi \left(n\right)$ therefore satisfies the homogeneous equation. Since $Ann\mathfrak{C}(R/N)=0$, its image is zero. Thus $\Phi \left(n\right)\in N$. Together with the defect statement, this makes the map induced on $R/N$ well defined and additive. □

5.2. Passage to the Central and Orthogonal Completion

We now use the semiprime notation fixed in Section 2. The involution extends uniquely to $Q_{\mathrm{ms}}\left(R\right)$; see [12] and [8] [Chapter 2]. Realize

$$A=\mathcal{O}\left(R_C\right)\subseteq Q_{mr}\left(R\right)$$

in the maximal right ring of quotients; see [8] [Proposition 3.1.14 and Corollary 3.1.15]. Every element of A is locally an orthogonal mixing of elements of $R_C$. On such mixings put ${\left({\sum }_i^{\perp }{e}_i{x}_i\right)}^{†}={\sum }_i^{\perp }{e}_i{x}_i^{*}$. Common refinements show that this is a well-defined involution of A. If $x\in A$, choose a dense right ideal J of R with $x^{†}J\subseteq R$. Then $J^{*}$ is a dense left ideal and $j^{*}x={\left(x^{†}j\right)}^{*}\in R$ for $j\in J$, so the defining characterization of the maximal symmetric quotient gives $x\in Q_{\mathrm{ms}}\left(R\right)$. Hence $A\subseteq Q_{\mathrm{ms}}\left(R\right)$, and † is the restriction of the extended involution; we again write it as *. We use the remaining standard orthogonal-completion facts from [8] [Chapter 3, pp. 97–128]; see also [13] [Appendix B, pp. 243–248]. The precise quotient and stalk facts needed below are collected in the next lemma.

Lemma 11. 

The ring A is an orthogonally complete centrally closed semiprime ring with extended centroid C. Moreover,

$$Q_{\mathrm{ms}}\left(A\right)=Q_{\mathrm{ms}}\left(R\right),Q_{\mathrm{ms}}\left(eA\right)=e{Q}_{\mathrm{ms}}\left(R\right)(e\in \mathcal{B}).$$

For every maximal ideal M of $\mathcal{B}$, the ring $A_M$ is centrally closed prime with extended centroid the field $C_M$, and

$$\bigcap _{M\in Max\mathcal{B}}MA=0.$$

(19)

Proof. 

Orthogonal completeness and semiprimeness follow from the construction. Apply [13] [Theorem A.4] first to $R\subseteq R_C\subseteq Q_{ml}\left(R\right)$ and then to $R\subseteq A\subseteq Q_{ml}\left(R\right)$. It gives

$$Q_{ml}\left(R_C\right)=Q_{ml}\left(A\right)=Q_{ml}\left(R\right).$$

The same argument for the opposite rings gives equality of the maximal right quotients. Since $R\subseteq A\subseteq Q_{\mathrm{ms}}\left(R\right)$, Lanning’s invariance theorem [12] [Theorem 2.5] now gives $Q_{\mathrm{ms}}\left(A\right)=Q_{\mathrm{ms}}\left(R\right)$ directly.

Since A is orthogonally complete, $eA\subseteq A$. Thus, in the notation of [13] [Lemma B.2], $L_e=A$, and that lemma, together with its right-hand analogue, gives $Q_{ml}\left(eA\right)=e{Q}_{ml}\left(A\right)$ and $Q_{mr}\left(eA\right)=e{Q}_{mr}\left(A\right)$.

We spell out the symmetric corner. Let $(f,g,I,J)$ be a compatible pair representing an element of $Q_{\mathrm{ms}}\left(eA\right)$, where I and J are dense left and right ideals of $eA$. Using $A=eA\oplus (1-e)A$, put

$$\widehat{I}=I\oplus (1-e)A,\widehat{J}=J\oplus (1-e)A$$

and extend f and g by zero on $(1-e)A$. Density is checked on the two summands. The extensions are A-module maps, and compatibility follows from that of $(f,g)$ because all cross-products vanish. Lanning’s compatible-pair description [12] [Proposition 2.1] therefore embeds $Q_{\mathrm{ms}}\left(eA\right)$ into $e{Q}_{\mathrm{ms}}\left(A\right)$. Conversely, if $u\in e{Q}_{\mathrm{ms}}\left(A\right)$, the two one-sided corner equalities above supply dense left and right $eA$-domains on which multiplication by u takes values in $eA$; those actions are compatible. The constructions are inverse. Indeed, if $(f,g,I,J)$ represents $u\in e{Q}_{\mathrm{ms}}\left(A\right)$, centrality of e gives $f\left(\right(1-e\left)I\right)=0$ and $g\left(J\right(1-e\left)\right)=0$. Thus the zero extensions of the restrictions to $eI$ and $Je$ agree with $(f,g)$ on the dense domains I and J; the reverse restriction is immediate. Hence

$$Q_{\mathrm{ms}}\left(eA\right)=e{Q}_{\mathrm{ms}}\left(A\right)=e{Q}_{\mathrm{ms}}\left(R\right).$$

The center of $Q_{ml}\left(A\right)$ is therefore C. Since A is closed under the action of C, it equals its central closure and is centrally closed. Primeness of $A_M$ follows from [13] [Theorem B.9], while the Pierce-stalk theorem [8] [Theorem 3.2.15(iv)] identifies its extended centroid with the image $C_M$ of C. Thus $C_M$ is a field, and since $A_M$ is already a $C_M$-algebra, it is centrally closed. Finally, by [13] [Lemma B.8], $x\in MA$ is equivalent to $E\left(x\right)\in M$. An idempotent belonging to every maximal ideal of the Boolean algebra is zero, and this proves (19). □

Since $2\in C^{\times }$, its image in every stalk field $C_M$ is invertible. Hence $char{C}_M\ne 2$ and every prime stalk $A_M$ is 2-torsion-free.

Identities valid in A pass to every $A_M$. Since the involution is of the first kind, it fixes $\mathcal{B}$ and therefore descends to each $A_M$. Conversely, an equality in A which holds after passage to every $A_M$ holds already in A by (19). We use these facts together with the orthogonal-mixing principles in the references above.

Lemma 12. 

Let $f:R\to A$ be additive and satisfy (1) in A. Let $e\in \mathcal{B}$ be central with $Ann\mathfrak{C}\left(eA\right)=0$. Then

$$ea⟼ef\left(a\right)$$

extends uniquely to a C-linear solution $e{R}_C\to eA$.

Proof. 

Suppose ${\sum }_i{c}_{i}e{a}_i=0$, with $c_i\in C$, and put $d={\sum }_i{c}_{i}ef\left(a_i\right)\in eA$. Multiplying the identities for $(a_i,b)$ by $c_i$ and adding gives

$$d{b}^{*}+\eta b{d}^{*}=0(b\in R).$$

Because the involution is of the first kind, the same relation holds for every $b\in e{R}_C$ by C-linearity. Now let $b\in eA$. Choose an orthogonal family $\left(e_j\right)$ with join e and elements $b_j\in e{R}_C$ such that $e_{j}b=e_j{b}_j$. Multiplying the preceding relation by $e_j$ and then mixing over j gives

$$d{b}^{*}+\eta b{d}^{*}=0(b\in eA).$$

The proof of Proposition 4(i) uses only this homogeneous relation and semiprimeness. Applied inside the semiprime ring $eA$, it gives $d\in Ann\mathfrak{C}\left(eA\right)$. By hypothesis this annihilator is zero, and therefore $d=0$. Thus

$$\sum _i{c}_{i}e{a}_i=0⟹\sum _i{c}_{i}ef\left(a_i\right)=0,$$

so the proposed C-linear extension is well defined. Substitution shows that it still satisfies (1); uniqueness follows because $e{R}_C$ generates its C-linear extension. □

Lemma 13. 

Let $S=eA$ be a *-invariant central summand with $Ann\mathfrak{C}\left(S\right)=0$. Let $D\subseteq S$ be a C-subalgebra with $\mathcal{O}\left(D\right)=S$. Every C-linear solution $\psi :D\to S$ has a unique $\mathcal{B}$-linear extension $\tilde{\psi }:S\to S$ satisfying (1).

Proof. 

Let $x\in S$. By orthogonal density there is an orthogonal family $\left(e_i\right)$ in $\mathcal{B}$ with join e and elements $x_i\in D$ such that $e_{i}x=e_i{x}_i$. Define

$$\tilde{\psi }\left(x\right)={\displaystyle \sum _i^{\perp }}{e}_i\psi \left(x_i\right).$$

To see that this is independent of the representation, take a second representation $f_{j}x=f_j{y}_j$. On the common summand $e_i{f}_j$ we have $e_i{f}_j(x_i-y_j)=0$. Since D is a C-subalgebra and $\psi$ is C-linear,

$$e_i{f}_j\left(\psi \left(x_i\right)-\psi \left(y_j\right)\right)=\psi \left(e_i{f}_j(x_i-y_j)\right)=0.$$

Thus the two orthogonal mixings coincide. The same calculation shows $\tilde{\psi }\left(cx\right)=c\tilde{\psi }\left(x\right)$ for $c\in \mathcal{B}$, so the extension is $\mathcal{B}$-linear.

Given $x,y\in S$, choose a common orthogonal refinement for their local representations. On each central piece, (1) is exactly the identity for $\psi$ on D; mixing the resulting equalities proves (1) for $\tilde{\psi }$. Finally any $\mathcal{B}$-linear extension must agree with these local values, which proves uniqueness. □

Lemma 14. 

Let $X\subseteq A$ be orthogonally complete and contain 0. Then some $z\in X$ satisfies

$$E\left(z\right)=\underset{x\in X}{\bigvee }E\left(x\right).$$

Proof. 

Choose a maximal family of nonzero orthogonal idempotents $\left(f_{\alpha }\right)$ and elements $x_{\alpha }\in X$ with $f_{\alpha }\le E\left(x_{\alpha }\right)$. Mix $f_{\alpha }{x}_{\alpha }$ on these pieces and 0 on their complement; orthogonal completeness gives $z\in X$ with $E\left(z\right)={\bigvee }_{\alpha }{f}_{\alpha }$. If some $E\left(x\right)$ were not below $E\left(z\right)$, the nonzero idempotent $E\left(x\right)(1-E(z\left)\right)\le E\left(x\right)$ would enlarge the family, contradicting maximality. □

The next lemma is the local-to-global step used in the classification. Its first part obtains a global coefficient directly from the induced functional identity.

Lemma 15. 

Let $S=eA$, where $e\in \mathcal{B}$, and let $T:S\to S$ be an additive $\mathcal{B}$-linear map.

(i)

Suppose that, for every maximal ideal M of $\mathcal{B}$ with $S_M\ne 0$ (equivalently, $e\notin M$), the induced map $T_M:S_M\to S_M$ is multiplication on the right by a symmetric element of $Q_s\left(S_M\right)$. Then there is a unique

$$q=q^{*}\in Q_{\mathrm{ms}}\left(S\right)=e{Q}_{\mathrm{ms}}\left(R\right)$$

such that $T\left(x\right)=xq$ for all $x\in S$. Moreover, $Sq\subseteq S$.

(ii)

Suppose that every nonzero stalk $S_M$ is a central simple algebra of degree two with orthogonal involution and that $J:S\to S$ is the $\mathcal{B}$-linear map whose stalk is

$$J_M\left(x\right)={Trd}_M\left(x\right)1-x^{*},$$

where ${Trd}_M$ denotes the reduced trace on the central simple algebra $S_M$. If, in every nonzero stalk, there is a unique symmetric pair $q_M,s_M\in S_M$ such that

$$T_M\left(x\right)=x{q}_M+s_M{J}_M\left(x\right),$$

then there are unique symmetric $q,s\in S$ satisfying

$$T\left(x\right)=xq+sJ\left(x\right)(x\in S).$$

Proof. 

For part (i), the local multiplier formula gives

$$T_M\left(x\right)y^{*}=x{T}_M{\left(y\right)}^{*}(x,y\in S_M).$$

The intersection formula (19) therefore gives

$$T\left(x\right)y^{*}=xT{\left(y\right)}^{*}(x,y\in S).$$

(20)

Put $F\left(x\right)=T\left(x\right)$ and $G\left(y\right)=T{\left(y^{*}\right)}^{*}$. Replacing y by $y^{*}$ in (20) gives

$$F\left(x\right)y=xG\left(y\right)(x,y\in S).$$

(21)

Since a semiprime ring has zero left and right annihilator, (21) also implies

$$F\left(ax\right)=aF\left(x\right),G\left(ya\right)=G\left(y\right)a(a,x,y\in S).$$

Thus $(F,G)$ is a compatible double centralizer on the dense left and right ideal S. By the defining universal property of the maximal symmetric ring of quotients, there is a unique $q\in Q_{\mathrm{ms}}\left(S\right)$ such that

$$F\left(x\right)=xq,G\left(y\right)=qy.$$

Furthermore,

$$x{q}^{*}={\left(q{x}^{*}\right)}^{*}=G{\left(x^{*}\right)}^{*}=T\left(x\right)=xq(x\in S),$$

so density gives $q^{*}=q$. The inclusion $Sq\subseteq S$ follows from $T\left(S\right)\subseteq S$.

For part (ii), fix a maximal ideal M for which $S_M\ne 0$. Since $S_M$ is central simple of degree two, both local coefficients lie in $S_M$. Choose lifts $u_M,v_M\in S$. As 2 is invertible in C and the local coefficients are symmetric, replacing these lifts by $(u_M+u_M^{*})/2$ and $(v_M+v_M^{*})/2$ makes them symmetric without changing their images in $S_M$.

Consider the error set

$${\mathcal{E}}_M=\{T\left(x\right)-x{u}_M-v_{M}J\left(x\right):x\in S\}\subseteq MS.$$

It is orthogonally complete: this follows from the $\mathcal{B}$-linearity of T and J. Lemma 14 gives $z_M\in {\mathcal{E}}_M$ such that

$$E\left(z_M\right)=\underset{z\in {\mathcal{E}}_M}{\bigvee }E\left(z\right).$$

The support criterion [13] [Lemma B.8] for $MS$ gives $E\left(z_M\right)\in M$. Hence $f_M=e-E\left(z_M\right)\notin M$, and $f_M{\mathcal{E}}_M=0$. Thus one pair of lifts works simultaneously for every $x\in S$ on the Boolean neighborhood $f_M$.

The basic clopen sets $U\left(f_M\right)=\{M^{\prime }:f_M\notin M^{\prime }\}$ cover the compact clopen set $U\left(e\right)$ in the Stone space of $\mathcal{B}$. Choose a finite subcover $U\left(f_{M_1}\right),\dots ,U\left(f_{M_t}\right)$ and put

$$g_1=e{f}_{M_1},g_j=e{f}_{M_j}\prod _{i<j}(1-f_{M_i})(2\le j\le t).$$

Then the $g_j$ are pairwise orthogonal, $g_j\le f_{M_j}$, and ${\sum }_j{g}_j=e$. Set

$$q={\displaystyle \sum _{j=1}^t}{g}_j{u}_{M_j},s={\displaystyle \sum _{j=1}^t}{g}_j{v}_{M_j}.$$

Then $q,s\in S$ are symmetric and the required formula holds on every $g_{j}S$, hence on S. If two global pairs existed, their images in every stalk would agree by the uniqueness in the degree-two prime classification; (19) would then make the pairs equal. □

5.3. The Central Decomposition

We now describe the four central supports used in the classification. The construction is standard, but we record the point which connects it to the prime stalks. Let

$$h=\underset{x,y\in A}{\bigvee }E\left([x,y]\right),e_c=1-h.$$

Lemma 16. 

With $N=Ann\mathfrak{C}\left(R\right)$ one has

$$Ann\mathfrak{C}\left(A\right)=e_{c}A,N=R\cap e_{c}A,$$

(22)

and $E\left(N\right)=e_c$. The ring $e_{c}A$ is the largest commutative central summand, and

$$A_M\mathrm{is}\mathrm{commutative}⟺e_c\notin M.$$

(23)

Moreover, the orthogonal completion of the central closure of N is $e_{c}A$.

Proof. 

First note that

$$uAv=0⟺E\left(u\right)E\left(v\right)=0(u,v\in A).$$

Indeed, otherwise a prime stalk would contain nonzero images of both u and v, contradicting $u_M{A}_M{v}_M=0$; the converse is immediate. If $z\in Ann\mathfrak{C}\left(A\right)$, then $zA[x,y]=0$ for all $x,y$, and hence $E\left(z\right)E\left(\right[x,y\left]\right)=0$. Thus $E\left(z\right)h=0$, so $z=e_{c}z$. Conversely, $e_{c}E\left([x,y]\right)=0$ for every commutator, whence $e_{c}A$ annihilates $\mathfrak{C}\left(A\right)$. Therefore $Ann\mathfrak{C}\left(A\right)=e_{c}A$. Lemma 10, applied to A, makes this annihilator central, so $e_{c}A$ is commutative. If $fA$ is any commutative central summand, then $f[A,A]=0$, whence $f\le e_c$; this proves maximality.

If $r\in N$, it annihilates the commutators of $R_C$ by C-linearity and then those of A by orthogonal mixing. Thus $r\in e_{c}A$. The reverse inclusion in (22) is immediate, proving the second equality. For every nonzero $f\le e_c$, Lemma B.2 of [13] supplies an essential ideal $L_f=\{r\in R:fr\in R\}$, and $f{L}_f\ne 0$. Hence some nonzero $fr$ belongs to $R\cap fA\subseteq N$. Therefore $E\left(N\right)=e_c$. Certainly $NC\subseteq e_{c}A$, so $\mathcal{O}\left(NC\right)\subseteq e_{c}A$. Conversely, choose a maximal orthogonal family $\left(f_{\alpha }\right)$ with join $e_c$ such that $0\ne f_{\alpha }\le E\left(n_{\alpha }\right)$ for some $n_{\alpha }\in N$; the equality $E\left(N\right)=e_c$ makes the join equal to $e_c$. In the commutative regular ring C, viewing $N\subseteq Z\left(R\right)\subseteq C$, one has $n_{\alpha }C=E\left(n_{\alpha }\right)C$. If $x\in e_{c}A$, represent x locally by elements $a_i\in R_C$ on an orthogonal cover $\left(g_i\right)$, and refine with the $f_{\alpha }$. On $f_{\alpha }{g}_i$ we have

$$f_{\alpha }{g}_{i}x=f_{\alpha }{g}_i{a}_i=n_{\alpha }{c}_{\alpha i}{a}_i\in NC$$

for a suitable $c_{\alpha i}\in C$, because N is an ideal of R. Thus x is locally in $NC$, and $e_{c}A\subseteq \mathcal{O}\left(NC\right)$.

Finally, the set $\left\{\right[x,y]:x,y\in A\}$ is orthogonally complete. Lemma 14 gives a commutator z with $E\left(z\right)=h$. Now $A_M$ is commutative exactly when every commutator lies in $MA$, which, by [13] [Lemma B.8], is equivalent to $h\in M$. Since M is a maximal Boolean ideal and $e_c=1-h$, this is equivalent to $e_c\notin M$. □

On $(1-e_c)A$, put $S_4(x_1,\dots ,x_4)={\sum }_{\sigma \in {\mathfrak{S}}_4}sgn\left(\sigma \right)x_{\sigma \left(1\right)}\dots x_{\sigma \left(4\right)}$. For a central summand $fA$ and either $p=S_4$ or $p(x,y)=[x{x}^{*},y]$, let $X_p$ be its value set. Each $X_p$ contains zero and is orthogonally complete: by multilinearity for $S_4$; for the second, orthogonal mixing gives $[x{x}^{*},y]={\sum }_i^{\perp }{g}_i[x_i{x}_i^{*},y_i]$, since $g_i^{*}=g_i$. By Lemma 14, choose $z_p\in X_p$ with $E\left(z_p\right)={\bigvee }_{u\in X_p}E\left(u\right)$. For $g\in \mathcal{B}$, $g\le f$, central mixing gives $gA\vDash p=0⟺gE\left(z_p\right)=0$; hence $e\left(p\right)=f-E\left(z_p\right)$ is the largest such idempotent. If $f\notin M$, then

$${\left(fA\right)}_M\vDash p=0⟺X_p\subseteq MA⟺E\left(z_p\right)\in M⟺e\left(p\right)\notin M.$$

Indeed, $z_p\in X_p$, $E\left(u\right)\le E\left(z_p\right)$ for all $u\in X_p$, and $x\in MA\iff E\left(x\right)\in M$ by [13] [Lemma B.8]; the final equivalence uses maximality of M. Set $e_2=e\left(S_4\right)$ for $f=1-e_c$ and then $e_s=e\left([x{x}^{*},y]\right)$ for $f=e_2$; put $e_o=e_2-e_s$ and $e_h=1-e_c-e_o-e_s$. These are pairwise orthogonal central idempotents

$$e_c,e_o,e_s,e_h\in \mathcal{B},e_c+e_o+e_s+e_h=1.$$

(24)

The equivalence above, together with (23), Posner’s theorem and Amitsur–Levitzki [7,14], and the degree-two first-kind dichotomy [9] [Chapter I], gives the following local meaning:

• $e_{c}A$ is commutative;
• every prime localization of $e_{o}A$ is degree two with orthogonal involution;
• every prime localization of $e_{s}A$ is degree two with symplectic involution;
• every prime localization of $e_{h}A$ is either non-PI or PI of central degree at least three.

In particular, each of the four supports is determined intrinsically by identities of the ring with involution, and the prime results of Section 4 apply on exactly the corresponding localizations.

The degree-two operations needed below can be constructed without assuming that arbitrary elements of a stalk quotient lift to a quotient ring. The next lemma globalizes the reduced trace Trd and the polarized reduced norm B from Proposition 3 to the corresponding degree-two central summands.

Lemma 17. 

The ring $e_{2}A$ is unital, with identity $e_2$. On $e_{o}A$ there is a unique C-linear map

$$Trd:e_{o}A⟶e_{o}C$$

whose induced map on every stalk is the reduced trace. On $e_{s}A$ there is a unique symmetric C-bilinear form $B_s:e_{s}A\times e_{s}A\to e_{s}C$ satisfying

$$B_s(x,y)e_s=x{y}^{*}+y{x}^{*}.$$

(25)

On every stalk its induced form is the polarized reduced norm, and $B_s$ is nondegenerate.

Proof. 

Put $S=e_{2}A$ and work in the Boolean algebra $e_2\mathcal{B}$. For a maximal ideal M of $e_2\mathcal{B}$, choose $u_M\in S$ lifting the identity of $S_M$. The two sets

$$\{u_{M}x-x:x\in S\},\{x{u}_M-x:x\in S\}$$

are orthogonally complete and lie in $MS$. Applying Lemma 14 to each set and then using [13] [Lemma B.8], we find that the support joins of both sets belong to M. Thus there is $f_M\in e_2\mathcal{B}\setminus M$ on which both sets vanish. Applying the finite Stone-space refinement used in Lemma 15 and mixing the finitely many resulting local identities gives an element $u\in S$ which is a two-sided identity of S. In $Q_{ml}\left(S\right)=e_2{Q}_{ml}\left(A\right)$ this identity is the central idempotent $e_2$; hence $u=e_2\in S$.

It remains to construct the trace on the orthogonal part. Fix a nonzero stalk ${\left(e_{o}A\right)}_M$. Its skew subspace is one-dimensional, and each nonzero skew element $\overline{k}$ is invertible: after a splitting extension the involution has the form $X^{*}=H^{-1}{X}^{T}H$ with $H^T=H$, and $H\overline{k}$ is a nonzero $2\times 2$ alternating matrix. Choose lifts $k,l\in e_{o}A$ of $\overline{k}$ and ${\overline{k}}^{-1}$, replacing k by $(k-k^{*})/2$. After restricting to a Boolean neighborhood $f\notin M$, we have

$$k^{*}=-k,kl=lk=f.$$

For $x\in f{e}_{o}A$ set

$${\tau }_f\left(x\right)=(xk+k{x}^{*})l.$$

(26)

In a degree-two central simple algebra with orthogonal involution,

$$xk+k{x}^{*}=Trd\left(x\right)k$$

for every nonzero skew k; this follows after a splitting-field extension and a congruence from the case $X^{*}=X^T$ and $k=\lambda \left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, where it is a direct $2\times 2$ computation. Faithful flatness then descends the identity [9] [Chapter I]. Thus (26) is the reduced trace on every stalk over the central piece f. Its values commute with $fA$ in every stalk, hence globally, and so lie in $Z\left(fA\right)=fC$. The formulas obtained on two such neighborhoods agree in every stalk of their overlap, hence agree by (19). The finite Stone-space refinement used in Lemma 15 therefore patches them over a finite orthogonal cover to the asserted C-linear map Trd. The same stalk argument proves uniqueness.

Finally, on each symplectic degree-two stalk the element $x{y}^{*}+y{x}^{*}$ is the polarized reduced norm of x and y, multiplied by the identity. Hence $x{y}^{*}+y{x}^{*}$ is central in every stalk and therefore globally. Since $e_{s}A$ is unital with center $e_{s}C$, there is a unique symmetric C-bilinear form $B_s:e_{s}A\times e_{s}A\to e_{s}C$ satisfying (25), and its induced form on every stalk is the polarized reduced norm. If $B_s(x,y)=0$ for every y, then the image of x is zero in every stalk by local nondegeneracy; hence $x=0$ by (19). □

We therefore write

$$J_o\left(x\right)=Trd\left(x\right)e_o-x^{*}(x\in e_{o}A)$$

and use $B_s$ in the intrinsic form (25).

The only part of the extension argument not covered by Lemmas 12 and 13 is the commutative support at $\eta =1$. The following compatible-pair argument handles it without an identity.

Lemma 18. 

Assume $\eta =1$ and let $\Phi :R\to R$ satisfy (1). Then there is a unique symmetric element $q_c$ in the maximal quotient of the commutative support, identified with $e_c{Q}_{\mathrm{ms}}\left(R\right)$, such that

$$e_c\Phi \left(a\right)=e_{c}a{q}_c(a\in R).$$

(27)

Moreover, this formula extends C-linearly to $e_c{R}_C$ and then to $e_{c}A$.

Proof. 

If $N=0$, then $e_c=0$ and the assertion is empty; take $q_c=0$. Assume $N\ne 0$. By Proposition 4, $\Phi$ is additive, $\Phi \left(N\right)\subseteq N$, and the first-kind involution fixes N elementwise. Put ${T=\Phi |}_N:N\to N$. For $m,n\in N$, the identity at $(n,m)$ is

$$2mT\left(n\right)=2nT\left(m\right).$$

Since R is 2-torsion-free,

$$mT\left(n\right)=nT\left(m\right)(m,n\in N).$$

(28)

Put $K={Ann}_{R}N$. Then $N\cap K=0$, and $I=N\oplus K$ is an essential ideal of R: if an ideal J misses N, then $JN=0$, so $J\subseteq K$. For $r\in R$ and $m,n\in N$, relation (28) gives

$$m\left\{T\right(rn)-rT(n\left)\right\}=\left(rn\right)T\left(m\right)-rmT\left(n\right)=0.$$

The expression in braces belongs to N, whose annihilator in itself is zero; hence $T\left(rn\right)=rT\left(n\right)$. Since $N\subseteq Z\left(R\right)$, this is also the corresponding right-module identity.

Define $\widehat{T}:I\to R$ by $\widehat{T}(n+k)=T\left(n\right)$. It is an R-bimodule map, and (28), together with $NK=0$, says

$$\widehat{T}\left(x\right)y=x\widehat{T}\left(y\right)(x,y\in I).$$

An essential ideal of a semiprime ring is dense, so $(\widehat{T},\widehat{T})$ is a compatible pair on a dense ideal and represents $q_c\in Q_{\mathrm{ms}}\left(R\right)$ with $T\left(n\right)=n{q}_c=q_{c}n$. The ideal I is *-invariant and $\widehat{T}\left(x^{*}\right)=\widehat{T}{\left(x\right)}^{*}$ for $x\in I$; hence $q_c^{*}=q_c$. Also $e_c\widehat{T}=\widehat{T}$, so $e_c{q}_c$ and $q_c$ have the same pair on I; density gives $q_c=e_c{q}_c\in e_c{Q}_{\mathrm{ms}}\left(R\right)$.

We next show that the same element $q_c$ gives the $e_c$-component of $\Phi \left(a\right)$ for every $a\in R$. Fix $n\in N$. Since N is a central *-ideal fixed elementwise, both $n\Phi \left(a\right)$ and $a\Phi \left(n\right)$ lie in N and are symmetric. The identity at $(a,n)$ therefore reduces to

$$2n\Phi \left(a\right)=2a\Phi \left(n\right)=2an{q}_c.$$

Canceling 2 gives

$$n\left(\Phi \left(a\right)-a{q}_c\right)=0(n\in N).$$

Lemma 16 gives $E\left(N\right)=e_c$. If $u\in e_c{Q}_{\mathrm{ms}}\left(R\right)$ and $Nu=0$, regularity of C gives $E\left(n\right)u=0$ for every $n\in N$; taking the join of these supports gives $u=0$. Applying this to the last relation proves (27). It also proves uniqueness of $q_c$. The formula defines the required C-linear extension on $e_c{R}_C$, and orthogonal mixing extends it to $e_{c}A$. □

5.4. The Orthogonally Complete Case

Theorem 4. 

Assume $R=A=\mathcal{O}\left(R_C\right)$ and let $e_c,e_o,e_s,e_h$ be the central idempotents in (24).

(a) If $\eta =1$,every solution is additive and has the unique form

$$\Phi \left(x\right)=(1-e_s)xq+T_s\left(e_{s}x\right),$$

(29)

where

$$q=(1-e_s)q=q^{*}\in (1-e_s)Q_{\mathrm{ms}}\left(R\right),(1-e_s)Aq\subseteq (1-e_s)A$$

and $T_s:e_{s}A\to e_{s}A$ is additive, $\mathcal{B}$-linear, and self-adjoint for $B_s$:

$$B_s(T_{s}x,y)=B_s(x,T_{s}y).$$

(b) If $\eta =-1$,every solution has the unique form

$$\Phi \left(x\right)=\Gamma \left(x\right)+(1-e_c)xq+e_{o}s{J}_o\left(e_{o}x\right),$$

(30)

where

$$\Gamma :A\to e_{c}A\mathrm{is}\mathrm{arbitrary},q=(1-e_c)q=q^{*}\in (1-e_c)Q_{\mathrm{ms}}\left(R\right),$$

$$(1-e_c)Aq\subseteq (1-e_c)A,s=e_{o}s=s^{*}\in e_{o}A.$$

Conversely every map in (29) or (30) is a solution.

Proof. 

We use the Boolean localizations described above. Equalities obtained in every $A_M$ may be pulled back to A by (19).

For $f\in \mathcal{B}$ and $x\in A$, subtracting f times the identity at $(x,b)$ from the identity at $(fx,b)$ shows that $d=\Phi \left(fx\right)-f\Phi \left(x\right)$ satisfies $d{b}^{*}+\eta b{d}^{*}=0$ for every $b\in A$. The argument of Proposition 4(i), now inside A, gives $d\in Ann\mathfrak{C}\left(A\right)=e_{c}A$. Consequently the complementary map $T=(1-e_c)\Phi$ satisfies

$$T\left(fx\right)=fT\left(x\right).$$

(31)

Assume first that $\eta =1$. Proposition 4 gives additivity. On $e_{c}A$, Lemma 18 gives a unique symmetric multiplier $q_c$. On the complementary summand the commutator annihilator is zero. By (31), applying Lemma 12 to $T=(1-e_c)\Phi$ with $R=A$ and $e=1-e_c$ returns T itself (because $A_C=A$), and therefore makes it C-linear on $(1-e_c)A$. It is already $\mathcal{B}$-linear by (31), so it preserves the central summands $e_{o}A,e_{s}A,e_{h}A$.

The $\mathcal{B}$-linearity also shows that $MA$ is invariant under the extended map for every maximal ideal $M\subseteq \mathcal{B}$; hence the map induces a solution on each stalk $A_M=A/MA$.

On each prime localization of $e_{o}A$ or $e_{h}A$, Proposition 3 or Theorem 3 gives a unique symmetric right-multiplier coefficient. Hence (20) holds in every stalk. Applying Lemma 15(i) on the two supports gives symmetric elements $q_o\in e_o{Q}_{\mathrm{ms}}\left(R\right)$ and $q_h\in e_h{Q}_{\mathrm{ms}}\left(R\right)$ such that

$$e_o\Phi \left(x\right)=e_{o}x{q}_o,e_h\Phi \left(x\right)=e_{h}x{q}_h(x\in A).$$

The range of $\Phi$ gives $e_{o}A{q}_o\subseteq e_{o}A$ and $e_{h}A{q}_h\subseteq e_{h}A$.

On $e_{s}A$ put $T_s\left(x\right)=e_s\Phi \left(x\right)$. Every symplectic degree-two localization satisfies $B_s(T_{s}x,y)=B_s(x,T_{s}y)$ by Proposition 3; therefore the same identity holds globally. With $q=q_c+q_o+q_h$ we obtain (29). Uniqueness follows from the uniqueness in Lemmas 18 and 15, together with the definition of $T_s$.

Now let $\eta =-1$. Proposition 4 shows that the $e_c$-valued part is unrestricted; set $\Gamma \left(x\right)=e_c\Phi \left(x\right)$. The map on $(1-e_c)A$ is additive and, by Lemma 12 and (31), is C-linear and componentwise. On every prime localization of $e_{s}A$ or $e_{h}A$, the local classification gives a unique symmetric right-multiplier coefficient. Lemma 15(i) gives symmetric $q_s\in e_s{Q}_{\mathrm{ms}}\left(R\right)$ and $q_h\in e_h{Q}_{\mathrm{ms}}\left(R\right)$ with the required range properties.

On $e_{o}A$, Proposition 3 gives in every prime localization a unique pair $(q_M,s_M)$ with

$$T_M\left(x\right)=x{q}_M+s_M{J}_o\left(x\right).$$

Lemma 15(ii) produces symmetric $q_o,s\in e_{o}A$ satisfying the displayed formula on all of $e_{o}A$. Combining $q_o,q_s,q_h$ gives the unique element q in (30). These patching steps use the explicit support and finite-refinement argument of Lemma 15; no identity element of A is used. The $e_c$-part determines $\Gamma$, and the uniqueness clauses of that lemma determine q and s.

Conversely, each displayed formula satisfies (1) on the four central summands. For $\eta =-1$ the commutative summand contributes zero to both sides for every $e_{c}A$-valued map. A symmetric right multiplier satisfies the identity by direct substitution. Nondegeneracy of $B_s$ shows that an additive self-adjoint $T_s$ is automatically C-linear (apply $B_s$ to $T_s\left(cx\right)-c{T}_s\left(x\right)$), so it induces maps on the stalks. The $s{J}_o$ and $T_s$ terms satisfy the identity in every degree-two stalk by Proposition 3, and hence globally by (19). Orthogonality of the summands then gives the identity on A. □

5.5. Restriction to the Original Ring

Theorem 5. 

Let R be an arbitrary semiprime 2-torsion-free ring with first-kind involution, not necessarily unital, and put $A=\mathcal{O}\left(R_C\right)$.

(i)

At $\eta =1$, every solution is the restriction to R of a unique map of the form (29). Conversely, parameters $(q,T_s)$ define a solution on R if and only if

$$(1-e_s)aq+T_s\left(e_{s}a\right)\in R(a\in R).$$

(32)

(ii)

At $\eta =-1$, every solution has in A the form

$$\Phi \left(a\right)=\Gamma \left(a\right)+(1-e_c)aq+e_{o}s{J}_o\left(e_{o}a\right),$$

(33)

where $q,s$ have the properties of Theorem 4 and $\Gamma :R\to e_{c}A$ is arbitrary. Conversely, these data define a solution on R if and only if

$$\Gamma \left(a\right)+(1-e_c)aq+e_{o}s{J}_o\left(e_{o}a\right)\in R(a\in R).$$

(34)

Proof. 

Suppose first that $\eta =1$. Proposition 4 gives additivity on R. Lemma 18 extends the $e_c$-component to the commutative support. On $1-e_c$ the commutator annihilator is zero, so Lemma 12 first extends the map C-linearly to $(1-e_c)R_C$, and Lemma 13 then extends it uniquely to $(1-e_c)A$. The two extensions are orthogonal and therefore combine to a solution on A. Theorem 4 gives (29). Since the extension agrees with the original map on R, condition (32) is necessary. Conversely, (32) says exactly that the restriction of the solution on A takes its values in R.

For $\eta =-1$, set $\Gamma \left(a\right)=e_c\Phi \left(a\right)$. Proposition 4 shows that the complementary map $(1-e_c)\Phi$ is additive. The same two extension lemmas give a unique solution on $(1-e_c)A$, to which Theorem 4 applies. Reattaching the arbitrary $e_c$-valued map gives (33). Again the only additional condition needed for a formula in A to define a map $R\to R$ is that its value lie in R for every $a\in R$, which is precisely (34). □

Corollary 2. 

If R is a finite direct product of commutative rings and prime first-kind rings, the range conditions in Theorem 5 decouple factorwise. On centrally closed factors they are automatic when the parameters belong to the corresponding factors. Only the arbitrary commutative output at $\eta =-1$ may depend on all input components.

Example 3. 

Let F be a field of characteristic different from two and consider

$$R=F\times M_2{\left(F\right)}_{\mathrm{o}}\times M_2{\left(F\right)}_{\mathrm{s}}\times M_3\left(F\right),$$

with the componentwise involution, where the second factor has transpose, the third has the standard symplectic involution, and the last again has transpose. Then the four idempotents in (24) are simply the four coordinate idempotents. Thus $e_c$ selects F, $e_o$ the orthogonal $M_2\left(F\right)$ factor, $e_s$ the symplectic $M_2\left(F\right)$ factor, and $e_h$ the $M_3\left(F\right)$ factor.

At $\eta =1$, Theorem 4 says that the first, second and fourth coordinates are symmetric right multipliers, whereas the third coordinate may be any endomorphism self-adjoint for the polarized reduced norm. At $\eta =-1$, the first coordinate may be an arbitrary function of the whole input, the third and fourth coordinates are right multipliers, and the second coordinate has the additional term $s(Trd\left(x\right)1-x^{*})$. This finite product is the simplest model of the four supports occurring in the general semiprime theorem.

6. Concluding Remarks

The results obtained above concern two different scalar behaviors of the involution. Over a field with nontrivial involution, Theorem 1 determines all solutions for every scalar parameter $\eta$; the complex conjugate-linear case is a special case. The value $\eta =1$ is the only case in which the term involving the annihilator of the commutator ideal may occur.

For a semiprime ring with a first-kind involution, the cases $\eta =1$ and $\eta =-1$ are described by Theorem 5. On the non-PI components, and on the PI components of central degree at least three, both cases reduce to a symmetric right multiplier. The degree-two exceptions are those of Proposition 3. For $\eta =-1$, besides the arbitrary commutative output, the orthogonal degree-two summand has the $s{J}_o$ term; for $\eta =1$ the additional maps occur on the symplectic degree-two summand.

Remark 2. 

The semiprime result is not simply a direct product of the prime results. For $\eta =-1$, the component with values in the commutative summand may depend on the whole input element. This is the reason for passing to the orthogonal completion and for retaining the range conditions in Theorem 5.

Remark 3. 

For $\eta =-1$ and ${dim}_C{R}_C>4$, the prime multiplier conclusion is already contained in [1] [Theorem 2.2]. Beyond it, we treat the $\eta =1$ first-kind prime case, the complete degree-two analysis for both first-kind involution types, and arbitrary 2-torsion-free semiprime first-kind rings. Characteristic two is a natural further direction, since $Sym\left(R\right)=Skew\left(R\right)$ and this decomposition collapses.