[proofplan] We first record the algebraic structure on double centralizers and check that the prescribed operations preserve the defining identities. The central point is that a double centralizer acts as an adjointable operator on the standard Hilbert $A$-module $A$, with the second component encoding the adjoint; this identifies $M(A)$ with a unital $C^*$-algebra of adjointable operators. We then verify that $a\mapsto(L_a,R_a)$ is a $*$-homomorphism, prove injectivity using a contractive approximate identity, and prove that the image is a closed essential two-sided ideal by direct multiplication formulas. [/proofplan] [step:Verify that the operations preserve double centralizers and have an identity] Let $(L,R)$ and $(S,T)$ be double centralizers of $A$, and let $\lambda\in\mathbb C$. Componentwise linear combinations satisfy the three double centralizer identities by linearity of multiplication in $A$. For the product, define \begin{align*} (L,R)(S,T):=(L\circ S,T\circ R). \end{align*} For $x,y\in A$, the left centralizer identity follows from the corresponding identities for $S$ and $L$: \begin{align*} (L\circ S)(xy)=L(S(xy))=L(S(x)y)=L(S(x))y. \end{align*} The right centralizer identity follows from the identities for $R$ and $T$: \begin{align*} (T\circ R)(xy)=T(R(xy))=T(xR(y))=xT(R(y)). \end{align*} For the compatibility identity, use first the compatibility identity for $(L,R)$ and then for $(S,T)$: \begin{align*} x(L\circ S)(y)=xL(S(y))=R(x)S(y)=T(R(x))y=(T\circ R)(x)y. \end{align*} Thus the product is again a double centralizer. Define maps $L^\sharp,R^\sharp:A\to A$ by \begin{align*} L^\sharp(x):=R(x^*)^*,\qquad R^\sharp(x):=L(x^*)^*. \end{align*} These maps are bounded and linear because $L$ and $R$ are bounded linear maps and the involution on $A$ is conjugate-linear and isometric. For $x,y\in A$, \begin{align*} L^\sharp(xy)=R(y^*x^*)^*=(y^*R(x^*))^*=R(x^*)^*y=L^\sharp(x)y, \end{align*} where the second equality uses $R(uv)=uR(v)$ with $u=y^*$ and $v=x^*$. Similarly, \begin{align*} R^\sharp(xy)=L(y^*x^*)^*=(L(y^*)x^*)^*=xL(y^*)^*=xR^\sharp(y), \end{align*} where the second equality uses $L(uv)=L(u)v$ with $u=y^*$ and $v=x^*$. Finally, applying the compatibility identity to $y^*$ and $x^*$ gives $y^*L(x^*)=R(y^*)x^*$, and taking adjoints gives \begin{align*} R^\sharp(x)y=L(x^*)^*y=xR(y^*)^*=xL^\sharp(y). \end{align*} Hence $(L,R)^*:=(L^\sharp,R^\sharp)$ is again a double centralizer. The pair $(\operatorname{id}_A,\operatorname{id}_A)$ is a double centralizer and is the identity for the above product. Thus the displayed operations are well-defined on $M(A)$ and have a multiplicative identity. [/step] [step:Identify a double centralizer with an adjointable operator on the standard Hilbert $A$-module] Regard $A$ as the standard right Hilbert $A$-module over itself, with right action given by multiplication and $A$-valued [inner product](/page/Inner%20Product) \begin{align*} \langle x,y\rangle_A:=x^*y \end{align*} for $x,y\in A$. If $(L,R)\in M(A)$, then $L:A\to A$ is a bounded right $A$-module map because \begin{align*} L(xa)=L(x)a \end{align*} for all $x,a\in A$. Define $L^\sharp:A\to A$ by $L^\sharp(x)=R(x^*)^*$, as above. For $x,y\in A$, the adjoint of the compatibility identity $y^*L(x)=R(y^*)x$ gives \begin{align*} (Lx)^*y=x^*R(y^*)^*. \end{align*} Therefore \begin{align*} \langle Lx,y\rangle_A=\langle x,L^\sharp y\rangle_A. \end{align*} Thus $L$ is adjointable on the Hilbert $A$-module $A$, and its adjoint is $L^\sharp$. [guided] The reason for introducing the Hilbert $A$-module structure is that the $C^*$-identity for multipliers is most naturally inherited from adjointable operators. We use the standard right Hilbert $A$-module $A_A$, whose right action is multiplication and whose inner product is the map \begin{align*} \langle \cdot,\cdot\rangle_A:A\times A\to A \end{align*} defined by \begin{align*} \langle x,y\rangle_A=x^*y. \end{align*} This inner product is linear in the second variable and satisfies $\langle x,x\rangle_A=x^*x$. Let $(L,R)\in M(A)$. The identity $L(xa)=L(x)a$ for all $x,a\in A$ says exactly that $L$ respects the right $A$-module structure. Since $L$ is bounded by definition of double centralizer, $L$ is a bounded right $A$-module map. To see that $L$ is adjointable, we must exhibit a bounded right $A$-module map $L^\sharp:A\to A$ satisfying \begin{align*} \langle Lx,y\rangle_A=\langle x,L^\sharp y\rangle_A \end{align*} for all $x,y\in A$. Define \begin{align*} L^\sharp(y):=R(y^*)^* \end{align*} for $y\in A$. This is the candidate forced by the compatibility identity. Indeed, applying the compatibility identity to $y^*$ and $x$ gives \begin{align*} y^*L(x)=R(y^*)x. \end{align*} Taking adjoints yields \begin{align*} (Lx)^*y=x^*R(y^*)^*. \end{align*} By the definition of the Hilbert-module inner product, this is precisely \begin{align*} \langle Lx,y\rangle_A=\langle x,L^\sharp y\rangle_A. \end{align*} Thus $L$ is adjointable and $L^*=L^\sharp$ as an adjointable operator. Conversely, suppose $T:A\to A$ is an adjointable right Hilbert $A$-module operator. Define a map \begin{align*} S:A\to A \end{align*} by \begin{align*} S(x)=(T^*(x^*))^*. \end{align*} Since $T$ is a right $A$-module map, $T(xa)=T(x)a$ for all $x,a\in A$, which is the left centralizer identity for $T$. Since $T^*$ is also a right $A$-module map, for $x,a\in A$ we have \begin{align*} S(xa)=(T^*(a^*x^*))^*=(T^*(a^*)x^*)^*=x(T^*(a^*))^*=xS(a), \end{align*} which is the right centralizer identity for $S$. Finally, for $x,y\in A$, adjointability gives \begin{align*} (Ty)^*x^*=\langle Ty,x^*\rangle_A=\langle y,T^*x^*\rangle_A=y^*T^*(x^*). \end{align*} Taking adjoints gives \begin{align*} xT(y)=(T^*(x^*))^*y=S(x)y. \end{align*} Thus $(T,S)$ is a double centralizer. The two constructions are inverse to each other because a double centralizer $(L,R)$ is sent to $L$, and the reconstructed second component is $x\mapsto (L^\sharp(x^*))^*=R(x)$. Therefore $M(A)$ is identified bijectively with the adjointable operators on the standard Hilbert $A$-module $A$. [/guided] Conversely, if $T:A\to A$ is an adjointable right Hilbert $A$-module operator, define \begin{align*} S:A\to A \end{align*} by \begin{align*} S(x)=(T^*(x^*))^*. \end{align*} Then $T(xa)=T(x)a$ because $T$ is a right $A$-module map. Also $S(xa)=xS(a)$ follows from right $A$-linearity of $T^*$ after taking adjoints. Finally, for $x,y\in A$, \begin{align*} xT(y)=S(x)y \end{align*} is the adjoint form of $\langle Ty,x^*\rangle_A=\langle y,T^*x^*\rangle_A$. Hence $(T,S)$ is a double centralizer. Let $\mathcal L_A(A)$ denote the set of adjointable right Hilbert $A$-module operators $A\to A$. Therefore the assignment \begin{align*} \Theta:M(A)\to \mathcal L_A(A) \end{align*} defined by \begin{align*} \Theta(L,R)=L \end{align*} is a bijection from $M(A)$ onto the $C^*$-algebra of adjointable operators on the standard Hilbert $A$-module $A$. [/step] [step:Transport the $C^*$-algebra structure from adjointable operators] The preceding step shows that $\Theta:M(A)\to\mathcal L_A(A)$ is a bijection. It respects products because \begin{align*} \Theta((L,R)(S,T))=L\circ S=\Theta(L,R)\Theta(S,T). \end{align*} It respects involution because the adjoint of $\Theta(L,R)=L$ is $L^\sharp$, and \begin{align*} \Theta((L,R)^*)=L^\sharp. \end{align*} It also preserves the norm by definition: \begin{align*} \|(L,R)\|_{M(A)}=\|L\|_{\mathcal L(A)}=\|\Theta(L,R)\|. \end{align*} The module $A_A$ is a Hilbert $C^*$-module: the right action is multiplication, the inner product is $\langle x,y\rangle_A=x^*y$, and the induced norm satisfies \begin{align*} \|x\|=\|x^*x\|_A^{1/2}. \end{align*} By the $C^*$-identity, \begin{align*} \|x^*x\|_A^{1/2}=\|x\|_A. \end{align*} Thus completeness follows from completeness of the $C^*$-algebra $A$. The standard theorem on adjointable operators on Hilbert $C^*$-modules applies to this Hilbert $A$-module and gives that $\mathcal L_A(A)$, with composition, adjoint, operator norm, and identity operator, is a unital $C^*$-algebra. Transporting this structure through the isometric $*$-isomorphism $\Theta$ proves that $M(A)$ is a unital $C^*$-algebra. [guided] We now check exactly why the adjointable-operator theorem applies. The object on which the operators act is the standard right Hilbert $A$-module $A_A$: the module action is multiplication in $A$, and the $A$-valued inner product is $\langle x,y\rangle_A=x^*y$. Its induced norm is \begin{align*} \|x\|=\|x^*x\|_A^{1/2}. \end{align*} The $C^*$-identity in $A$ gives \begin{align*} \|x^*x\|_A^{1/2}=\|x\|_A. \end{align*} Therefore the Hilbert-module norm is exactly the original norm on $A$, and completeness follows from the fact that $A$ is complete as a $C^*$-algebra. The standard theorem on adjointable operators on Hilbert $C^*$-modules says that, for any Hilbert $C^*$-module $E$, the adjointable operators on $E$ form a unital $C^*$-algebra under composition, adjoint, and the operator norm. We have verified the needed hypothesis with $E=A_A$, so $\mathcal L_A(A)$ is a unital $C^*$-algebra. The map $\Theta:M(A)\to\mathcal L_A(A)$ is a bijection by the preceding step. It preserves multiplication, involution, and norm by the computations above in this step. Hence $\Theta$ transports the unital $C^*$-algebra structure of $\mathcal L_A(A)$ to exactly the displayed operations and norm on $M(A)$. Thus $M(A)$ is a unital $C^*$-algebra. [/guided] [/step] [step:Show that the canonical embedding is an injective star homomorphism] For each $a\in A$, define bounded linear maps $L_a,R_a:A\to A$ by \begin{align*} L_a(b):=ab,\qquad R_a(b):=ba. \end{align*} Associativity of multiplication in $A$ gives, for all $x,y\in A$, \begin{align*} L_a(xy)=a(xy)=(ax)y=L_a(x)y, \end{align*} \begin{align*} R_a(xy)=(xy)a=x(ya)=xR_a(y), \end{align*} and \begin{align*} xL_a(y)=x(ay)=(xa)y=R_a(x)y. \end{align*} Thus $\iota(a):=(L_a,R_a)$ belongs to $M(A)$. For $a,b\in A$ and $\lambda\in\mathbb C$, linearity is immediate: \begin{align*} \iota(a+\lambda b)=\iota(a)+\lambda\iota(b). \end{align*} The product formula gives \begin{align*} \iota(a)\iota(b)=(L_a\circ L_b,R_b\circ R_a)=(L_{ab},R_{ab})=\iota(ab). \end{align*} The involution formula gives, for $x\in A$, \begin{align*} L_a^\sharp(x)=R_a(x^*)^*=(x^*a)^*=a^*x=L_{a^*}(x), \end{align*} and \begin{align*} R_a^\sharp(x)=L_a(x^*)^*=(ax^*)^*=xa^*=R_{a^*}(x). \end{align*} Hence $\iota(a)^*=\iota(a^*)$, so $\iota$ is a $*$-homomorphism. We prove injectivity for every $C^*$-algebra $A$. If $A=\{0\}$, then the only element of $A$ is $0$, so $\iota$ is injective. Suppose now that $A\neq\{0\}$ and $\iota(a)=0$. Then $L_a=0$, so $ab=0$ for every $b\in A$. By [citetheorem:8570], $A$ has a contractive approximate identity; let $(e_i)_{i\in I}$ be such a net. Since $ae_i\to a$ in the norm of $A$ and $ae_i=0$ for every $i\in I$, it follows that $a=0$. Thus $\iota$ is injective. [/step] [step:Prove that the embedded copy of $A$ is a closed two-sided ideal] Since $\iota:A\to M(A)$ is an injective $*$-homomorphism between $C^*$-algebras, [citetheorem:8547] implies that $\iota$ is contractive. To identify its norm exactly, let $a\in A$ and let $(e_i)_{i\in I}$ be a contractive approximate identity for $A$, whose existence is given by [citetheorem:8570]. Since $ae_i\to a$ in $A$ and $\|e_i\|_A\le 1$, \begin{align*} \|a\|_A=\lim_i\|ae_i\|_A\le \|L_a\|_{\mathcal L(A)}. \end{align*} The reverse inequality follows from submultiplicativity: \begin{align*} \|L_a(b)\|_A=\|ab\|_A\le \|a\|_A\|b\|_A \end{align*} for all $b\in A$, so $\|L_a\|_{\mathcal L(A)}\le \|a\|_A$. Hence \begin{align*} \|\iota(a)\|_{M(A)}=\|a\|_A. \end{align*} Thus $\iota(A)$ is complete in the norm inherited from $M(A)$ and is therefore closed in $M(A)$. Let $(L,R)\in M(A)$ and let $a\in A$. The product formulas give \begin{align*} (L,R)\iota(a)=(L\circ L_a,R_a\circ R). \end{align*} For $x\in A$, \begin{align*} (L\circ L_a)(x)=L(ax)=L(a)x=L_{L(a)}(x), \end{align*} and \begin{align*} (R_a\circ R)(x)=R(x)a=xL(a)=R_{L(a)}(x). \end{align*} Therefore \begin{align*} (L,R)\iota(a)=\iota(L(a)). \end{align*} Similarly, \begin{align*} \iota(a)(L,R)=(L_a\circ L,R\circ R_a)=\iota(R(a)). \end{align*} Hence $\iota(A)$ is a two-sided ideal of $M(A)$. [/step] [step:Prove that the embedded ideal is essential] We use the [annihilator criterion for essential ideals](/theorems/8574) in [citetheorem:8574]. Let $(L,R)\in M(A)$ and suppose \begin{align*} (L,R)\iota(A)=\{0\}. \end{align*} From the formula proved above, \begin{align*} (L,R)\iota(a)=\iota(L(a)) \end{align*} for every $a\in A$. Since $\iota$ is injective, $\iota(L(a))=0$ implies $L(a)=0$ for every $a\in A$. Thus $L=0$. For $x\in A$, the compatibility identity for $(L,R)$ gives \begin{align*} R(x)y=xL(y)=0 \end{align*} for every $y\in A$. Let $(e_i)_{i\in I}$ be a contractive approximate identity for $A$, again supplied by [citetheorem:8570]. Applying the preceding equality with $y=e_i$ gives $R(x)e_i=0$ for every $i\in I$. Since $R(x)e_i\to R(x)$ in $A$, we obtain $R(x)=0$. Hence $R=0$, so $(L,R)=0$. Therefore every multiplier that annihilates $\iota(A)$ on the right is zero. By [citetheorem:8574], $\iota(A)$ is an essential closed two-sided ideal of $M(A)$. This completes the proof. [/step]