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The annihilator records which ring elements kill a specified subset of a module.

For an exhaustive nonnegative increasing multiplicative filtration, each left ideal $I$ induces a homogeneous ideal $\mathrm{gr}I$ in the associated graded ring. If that graded ring is a left Noetherian ring, lift a finite homogeneous generating set of $\mathrm{gr}I$ to elements of $I$. For any element of $I$, subtract a combination of these lifts with the same leading symbol; the remainder has strictly lower degree. Induction on the nonnegative degree proves the lifts generate $I$. There is no completion hypothesis because the cancellation terminates in finitely many steps. The analogous result holds for right ideals.

A module is cyclic if one element $m$ generates it: every element is $rm$ for some scalar $r\in R$. The generating element need not give unique coefficients. The module homomorphism $R\to M$, $r\mapsto rm$, is surjective. Over a commutative ring, its kernel is the annihilator ${\mathrm{Ann}}_R(m)$, givingThus cyclic modules are exactly the quotient modules $R/I$ for ideals $I$ in $R$. For a left module over a noncommutative ring, use a left ideal and interpret the quotient as a module, with no claim that it is a quotient ring. See John A. Beachy's Abstract Algebra Online, section 10.1.

The module $\mathbb{Z}/6\mathbb{Z}$ is generated by $\overline{1}$, and also by $\overline{5}$, whose successive multiples are $\overline{0},\overline{5},\overline{4},\overline{3},\overline{2},\overline{1}$. The element $\overline{2}$ generates only $\{\overline{0},\overline{2},\overline{4}\}$, a submodule with three elements. A cyclic module can be infinite and free, as $\mathbb{Z}=\mathbb{Z}\cdot 1$ shows, or finite and torsion, as $\mathbb{Z}/6\mathbb{Z}$ shows. One generator counts directions of generation, not the number of elements and not freedom from relations.

For a state on a C-star algebra $x$ on a unital algebra $A$, put $⟨[a],[b]⟩=x(b^{\ast }a)$ on $A/N_x$, with the inner product linear in its first argument. The Cauchy–Schwarz inequality for positive C-star functionals makes the null vectors pair to zero with every vector. Moreover $c^{\ast }c\le \Vert c{\Vert }^{2}1$ gives $x((ca{)}^{\ast }(ca))\le \Vert c{\Vert }^{2}x(a^{\ast }a)$, so $N_x$ is a left ideal and $[a]\mapsto [ca]$ is well-defined and bounded by $\Vert c\Vert$. Completion gives a Hilbert space $H_x$ and a star-representation ${\pi }_x(c)[a]=[ca]$. The adjoint identity follows from $x(b^{\ast }ca)=⟨[a],{\pi }_x(c^{\ast })[b]⟩$. The unit vector ${\xi }_x=[1]$ is cyclic and recovers the state as $x(a)=⟨{\pi }_x(a){\xi }_x,{\xi }_x⟩$.

In a noncommutative domain that is a left Noetherian ring, two nonzero principal left ideals always intersect nontrivially. If $Ra\cap Rb=0$, a relation ${\sum }_{i=0}^m{u}_{i}a{b}^i=0$ first forces $u_{0}a\in Ra\cap Rb$, then $u_0=0$. Right cancellation of $b$ and repetition give all $u_i=0$. Hence ${\sum }_{i\ge 0}Ra{b}^i$ is an infinite direct sum of nonzero left ideals, whose finite partial sums violate the ascending chain condition. A nonzero member of $Rr\cap Rs$ supplies the equality $s^{\prime }r=r^{\prime }s$ with $s^{\prime }\ne 0$, proving the left Ore set condition for all nonzero elements.

The left regular module of a unital associative algebra $A$ is its underlying additive group with $A$ acting by left multiplication. Its submodules are exactly the left ideals of $A$. The associative algebra is a semisimple algebra precisely when this module is a semisimple module.

A maximal left ideal is a proper left ideal contained in no larger proper left ideal. In a unital Banach algebra it is norm closed: its closure is a left ideal, and if that closure were the entire algebra it would contain elements of $L$ within distance less than one of $1$, hence invertible by the Neumann series. A proper left ideal contains no invertible element. The quotient $A/L$ is consequently a quotient Banach space and a simple module.

A minimal left ideal is a nonzero left ideal containing no smaller nonzero left ideal. Equivalently, it is a simple module inside the left regular module. Every nonzero associative algebra of finite dimension has one: choose a nonzero left ideal of least positive dimension.

Noncommutative algebra studies rings whose multiplication need not commute and their one-sided ideals and modules.

A left module over the unital $K$-algebra $A$ is a $K$-vector space $V$ with a $K$-bilinear action $A\times V\to V$ satisfying $(ab)v=a(bv)$ and $1v=v$. For a vector $v$, the map $A\to V$, $a\mapsto av$, is an $A$-module homomorphism. Its kernel is $\mathrm{Ann}(v)$; explicitly this is a linear subspace and $b(av)=(ba)v=0$ for $a\in \mathrm{Ann}(v)$, so it is a left ideal.

If $V=Av$, this homomorphism is surjective, and the module isomorphism theorem gives $V\cong A/\mathrm{Ann}(v)$. Conversely the quotient by a left ideal is generated by $1+I$. Thus cyclic modules are exactly the quotients $A/I$ by left ideals.

For the first example set $e=\mathrm{diag}(1,1,0)$. Right multiplication by $e$ keeps the first two columns and zeroes the third. Every matrix with third column zero equals its product with $e$. Hence $V=Ae$ is cyclic, generated by $e$.

Coordinatewise multiplication gives $V=A(1,1,0)$, so the third module is cyclic.

A submodule of a cyclic module need not be cyclic. In the second example the first-row space is a left ideal and hence a submodule of the regular module $A$. The regular module is cyclic, generated by $1_A$, whereas its first-row submodule has just been shown not to be cyclic. Thus the proposed inheritance property fails even for a finite-dimensional algebra.

A composition series is a chain $0=V_0⊊V_1⊊\cdots ⊊V_r=V$ of submodules with every factor $V_i/V_{i-1}$ simple; its length is $r$. Suppose $0\subset S\subset V$ is one of length two, and put $T=V/S$.

For any nonzero proper submodule $U$, either $U\cap S=S$ or $U\cap S=0$, by simplicity of $S$. In the first case, simplicity of $V/S$ forces $U=S$. In the second, the quotient map embeds $U$ nontrivially into $T$, so $U\cong T$ and its image is all of $T$. Hence $V=S\oplus U$, and $V/U\cong S$. Thus every nonzero proper submodule is simple and has simple quotient; every composition series has length two.

The Jordan–Hölder theorem states that any two composition series of a finite-length module have equal length and the same multiset of simple factors, up to isomorphism and order. The preceding argument proves this for length two: either $U=S$, giving factors $S,T$, or $U\ne S$, giving factors $T,S$.

The map $A\to S$, $a\mapsto ax$, is a surjective module homomorphism with kernel $I={\mathrm{Ann}}_A(x)$. The first isomorphism theorem for modules gives $A/I\cong S$ as left modules; $I$ need only be a left ideal. Every simple module is cyclic, since any nonzero $x$ generates a nonzero submodule. As a quotient of the finite-dimensional regular module $A$, it is finite-dimensional. Extend $I$ to a composition series of $A$; its simple quotient $A/I$ is one of the factors. Jordan-Hölder shows that its isomorphism type belongs to the fixed finite list of factors of any one composition series of $A$. Hence there are finitely many simple-module isomorphism types.

For the final assertion let $W$ be the image of $\psi$. Its kernel is $\bigcap I_j$. We prove $W=\prod S_j$ by induction on $k$. It is true for $k=1$. Projection onto the first $k-1$ factors is onto by induction. Its kernel on $W$ is a submodule of $S_k$, so is zero or $S_k$. If zero, $W\cong {\prod }_{j<k}{S}_j$. But its projection onto $S_k$ is onto, making $S_k$ a composition factor of that product, contrary to Jordan-Hölder and the pairwise distinct types. Thus the kernel is all of $S_k$. Surjectivity on the first factors together with this kernel proves $W$ is the entire product. The first isomorphism theorem now givesas left $A$-modules, without asserting that the left ideals are two-sided.

A state on a C-star algebra is a positive linear functional of norm one. In the unital case this is equivalent to positivity and $x(1)=1$. A unital star-homomorphism $\pi :A\to \mathcal{B}(K)$ is a cyclic representation of a C-star algebra if it has a vector $\xi$ with $\overline{\pi (A)\xi }=K$. For the representation associated with a state, the cyclic vector is taken to have norm one.

Use inner products linear in the first argument, as in the displayed formula for the projection in the source. In the Gelfand–Naimark–Segal construction, setThe Cauchy–Schwarz inequality for positive C-star functionals makes this inner product well-defined on $A/N_x$. Indeed a null vector pairs to zero with every vector. The set $N_x$ is a left ideal: $c^{\ast }c\le \Vert c{\Vert }^{2}1$ impliesso $a\in N_x$ implies $ca\in N_x$. The same estimate shows that left multiplication induces a bounded operatorComplete the quotient pre-Hilbert space to $H_x$, and extend these operators by continuity. The formula for the inner product gives ${\pi }_x(c{)}^{\ast }={\pi }_x(c^{\ast })$, while multiplication and the unit are preserved. Put ${\xi }_x=[1]$. Then $\Vert {\xi }_x\Vert =1$, the vectors ${\pi }_x(a){\xi }_x=[a]$ are dense in $H_x$, andThis gives the requested cyclic representation for an arbitrary state.

For the specified unit vector $\eta$, the functional $x_{\eta }$ is complex-linear and obeys $|x_{\eta }(a)|\le \Vert a\eta \Vert \Vert \eta \Vert \le \Vert a\Vert$. MoreoverThus it is positive and has norm exactly one, proving that $x_{\eta }$ is a vector state on a C-star algebra.

The rank-one projection satisfiesIf $a\eta =0$, this is zero for every $\zeta$, so $a{p}_{\eta }=0$. Conversely, evaluation at $\zeta =\eta$ gives $(a{p}_{\eta })\eta =a\eta$. HenceIn particular $N_{x_{\eta }}=\{a:a\eta =0\}=\{a:a{p}_{\eta }=0\}$.

The concrete GNS space is the completion of $A/N_{x_{\eta }}$ withDefine $U_0[a]=a\eta$. The description of the null ideal proves it is well-defined, and the displayed equality proves it is a linear isometry with range $A\eta$. It therefore extends to an isometry $U:H_{x_{\eta }}\to \overline{A\eta }$. The extended range is closed, since its domain is complete, and contains the dense set $A\eta$. Thus $U$ is onto the required closure.

In fact here $A=\mathcal{B}(H)$, so $A\eta =H$: given any $\zeta \in H$, the bounded rank-one operator $b(v)=⟨v,\eta ⟩\zeta$ sends $\eta$ to $\zeta$. Thus $U$ is a unitary map onto all of $H$, and $U{\xi }_{x_{\eta }}=\eta$.

Finally, on the dense quotient vectors,Both sides are bounded operators, so this identity extends to all of $H_{x_{\eta }}$. Multiplication by the inverse unitary map givesThis is the GNS representation of a vector state, with the vector state, null ideal, cyclic vector and intertwining isometry all made explicit.

For a bounded linear functional on a unital C*-algebra, the state criterion isEquivalently it is positive and takes the unit to one.

In the Gelfand–Naimark–Segal construction, set $N_x=\{b:x(b^{\ast }b)=0\}$. This is a left ideal. Give $A/N_x$ the inner product $⟨[b],[c]⟩=x(c^{\ast }b)$, linear in the first variable, and complete it to a Hilbert space $H_x$. Left multiplication gives the bounded *-representation ${\pi }_x(a)[b]=[ab]$. The vector $\xi =[1_A]$ is cyclic and satisfies $x(a)=⟨{\pi }_x(a)\xi ,\xi ⟩$.

The Hermitian matrix $\rho$ has eigenvalues ${\lambda }_{\pm }=1/2\pm |\alpha |$ and trace one. In an orthonormal eigenbasis $v_{+},v_{-}$,If $|\alpha |\le 1/2$, both weights are nonnegative and sum to one, so this functional is positive and normalized, hence a state. If $|\alpha |>1/2$, evaluating on the positive rank-one projection onto the negative-eigenvalue eigenspace gives a negative value, so it is not a state. ThereforeTo obtain the representation dimension, cyclicity of trace givesThus $A/N_{x_{\rho }}$ identifies isometrically with the matrix space $\{b{\rho }^{1/2}:b\in M_2(\mathbb{C})\}$, with left multiplication as the representation. If $r=\mathrm{rank}\rho$, this space consists, after a unitary change of basis, of arbitrary first $r$ columns and zero remaining columns. Its complex dimension is $2r$. It is already complete because it is finite-dimensional. Rank is one at $|\alpha |=1/2$ and two at $|\alpha |<1/2$, proving

Since $A$ is a nonzero associative algebra of finite dimension, it has a nonzero left ideal $I$ of least positive dimension. Such a minimal left ideal is a simple module under left multiplication.

For $a\in A$, right multiplication defines the module homomorphismIts kernel is a submodule of $I$, so its image $Ia$ is either zero or a simple module isomorphic to $I$. Each $Ia$ is a left ideal, since $c(xa)=(cx)a$.

ConsiderIt is a left ideal. It is also stable under right multiplication, because $(Ia)b=I(ab)$, so it is a two-sided ideal. Moreover $I=I1\subseteq J$, giving $J\ne 0$. The assumption that $A$ is a simple ring therefore forces $J=A$.

Only finitely many summands are needed: express each member of a finite basis of $A$ as a finite sum of elements of the $Ia$, and collect the finitely many indices used. After discarding zero summands, $A$ is a finite sum of simple modules. Part (a)(ii) makes its left regular module a semisimple module. This is exactly the definition of a semisimple algebra, proving that a finite-dimensional simple algebra is semisimple. The finite dimension assumption is essential to this assertion about a simple ring.

Put $x=I+E_{12}$, $y=I+E_{23}$ and $z=I+E_{13}$. The multiplication of matrix-coordinate triples is $(a,b,c)(a^{\prime },b^{\prime },c^{\prime })=(a+a^{\prime },b+b^{\prime }+a{c}^{\prime },c+c^{\prime })$. In particular $z$ is central, $xy=zyx$, and every element has the unique normal form $y^c{z}^b{x}^a$. ConsequentlyThe distinct normal-form monomials are the group basis, so this is an isomorphism, not merely a surjection from a presented algebra. This is the integral Heisenberg group algebra as a skew Laurent ring.

Here is the needed skew Hilbert basis theorem, including its proof. If $A$ is a left Noetherian ring and $\sigma$ an automorphism, let $J$ be a left ideal of $A[t;\sigma ]$, where $ta=\sigma (a)t$. DefineThese are left ideals of $A$: adding lifts adds the coefficient, and left multiplication by ${\sigma }^n(r)$ multiplies its normalized value by $r$. Multiplication of a lift by $t$ proves $I_n\subseteq I_{n+1}$. The chain stabilizes at some $N$. Choose finite generators $u_{nj}$ for each $I_n$, $0\le n\le N$, and lifts $f_{nj}\in J$ with $t^n$ coefficient ${\sigma }^n(u_{nj})$.

For $f\in J$ of degree $m\ge N$, write its normalized leading coefficient as ${\sigma }^{-m}(\mathrm{lc}f)={\sum }_j{r}_j{u}_{Nj}$. Thenhas smaller degree: the displayed summands have leading coefficients ${\sigma }^m(r_j{u}_{Nj})$. For $m<N$ use the generators for $I_m$ without the shift. Induction on degree expresses every element of $J$ in terms of the finitely many chosen lifts. Thus $A[t;\sigma ]$ is a left Noetherian ring. Passing to the opposite ring replaces $\sigma$ by ${\sigma }^{-1}$ and proves the corresponding right-sided result.

This proof includes the ordinary Hilbert basis theorem when $\sigma$ is the identity. Starting with the field $k$, it makes $k[y,z]$ Noetherian; localization at powers of $y,z$ makes $B$ Noetherian as well. Finally, for a left ideal $L$ in $B[x,x^{-1};\sigma ]$, its intersection $J$ with $B[x;\sigma ]$ has finitely many generators. For each $f\in L$, some $x^{q}f$ lies in $J$, so $f=x^{-q}(x^{q}f)$ is generated by those same elements in the skew Laurent polynomial ring. Right multiplication clears powers for a right ideal. Therefore $kG$ is a left Noetherian ring and a right Noetherian ring.

Write the negative filtration as $F_{0}R=R\supseteq F_{-1}R\supseteq F_{-2}R\supseteq \cdots$, with $F_{i}R{F}_{j}R\subseteq F_{i+j}R$. Completeness is used in the separated sense $R\cong \underleftarrow{\lim }R/F_{-n}R$; in particular the intersection of all filtration terms is zero.

For any left ideal $J$, its induced associated graded ideal is a left ideal of $\mathrm{gr}R$. Choose finite homogeneous generators and lift them to $j_1,\dots ,j_r\in J$, with orders $n_i\ge 0$, meaning $j_i\in F_{-n_i}R\setminus F_{-n_i-1}R$. Let $j\in J$. At each order $s$, the leading symbol of the current residual is a sum of the generator symbols with homogeneous coefficients. Lift these coefficients to $c_{i,s}\in F_{-(s-n_i)}R$ (zero when $n_i>s$) and subtract ${\sum }_i{c}_{i,s}{j}_i$. The new residual lies in $F_{-(s+1)}R$.

Iterating expresses $j$ as partial sums of ${\sum }_i{c}_i{j}_i$, with residuals tending to zero. For fixed $i$ the successive coefficient increments have orders tending to infinity. Separated completeness therefore supplies $c_i={\sum }_s{c}_{i,s}\in R$. Multiplication is continuous by the multiplicative filtration, soThe sum on the right is finite, despite the convergent process used to obtain its coefficients. It belongs to the actual left ideal generated by the lifts; no assumption that $J$ is closed was made. Every left ideal is consequently finitely generated, and $R$ is a left Noetherian ring. This is the left-handed complete negative filtered-graded transfer of Noetherianity; the separated-completeness hypothesis is essential to the limiting argument.

A representation of a Banach algebra is a complex-linear multiplicative map $\pi :A\to {\mathrm{End}}_{\mathbb{C}}(X)$. A nonzero irreducible action of a unital algebra automatically has $\pi (1)=I$: the image of the idempotent $\pi (1)$ is a nonzero invariant subspace and hence is all of $X$. An algebraically irreducible representation of a Banach algebra is a nonzero action if it has no nonzero proper invariant submodule, whether closed or not. A normed representation of a Banach algebra has $\pi (a)$ bounded for each $a$ on a preassigned normed vector space $X$; continuity of $\pi$ as an operator-norm-valued map is a further conclusion. This must not be confused with topological irreducibility, which excludes only closed invariant subspaces. Unitizing the algebra extends a nonunital action by $\pi (a+\lambda 1)=\pi (a)+\lambda I$, so it is enough to discuss the unital case.

For $\xi \ne 0$, irreducibility gives $A\xi =X$. The annihilator $L_{\xi }=\{a:\pi (a)\xi =0\}$ is a maximal left ideal: an intermediate left ideal gives an intermediate submodule of $X$. It is closed by the Neumann-series argument for maximal left ideals. Consequently $A/L_{\xi }$ transports a complete quotient norm to $X$, and left multiplication is contractive in this norm. Every algebraically irreducible representation of a Banach algebra can therefore be realized continuously on a Banach space. This alone does not prove continuity for its preassigned norm.

The kernel of $\pi$ is the intersection of the closed left ideals $L_{\xi }$ and is a closed primitive ideal. The Jacobson radical is the intersection of all primitive ideals, equivalently the intersection of maximal left ideals. Thus these quotient representations detect semisimplicity. Equivalent algebra representations are related by a bijective module homomorphism.

The algebra of module endomorphisms is exactly $\mathbb{C}I$. Indeed, if a module endomorphism $D$ sends a fixed cyclic vector $\xi$ to $a\xi$, then $D(b\xi )=ba\xi$. It is induced by bounded right multiplication by $a$ on $A/L_{\xi }$, so is bounded for the quotient norm on an irreducible Banach module. Every nonzero module endomorphism is injective and surjective by irreducibility, and its inverse is bounded by the same argument. This commutant of an operator algebra is a closed division subalgebra of the bounded operators on this Banach module; the Gelfand-Mazur theorem makes it $\mathbb{C}I$.

The Jacobson density theorem now says that for finitely many linearly independent ${\xi }_1,\dots ,{\xi }_n$ and arbitrary ${\eta }_1,\dots ,{\eta }_n$, one element of $A$ sends all ${\xi }_j$ to ${\eta }_j$. Here is the induction behind it. The one-vector assertion is $A{\xi }_1=X$. Assuming the first $n-1$ coordinates can be prescribed, set $L=\{a:a{\xi }_j=0(j<n)\}$. Its image $L{\xi }_n$ is an invariant subspace. If it is $X$, it adjusts the last image without changing the first ones. If it is zero, the rule $(a{\xi }_1,\dots ,a{\xi }_{n-1})\mapsto a{\xi }_n$ is a well-defined module map $X^{n-1}\to X$. Each coordinate map is scalar by the module endomorphism result, forcing ${\xi }_n$ to be a linear combination of the earlier vectors, a contradiction. This proves the assertion.

We can now give Johnson's continuity argument with its crucial construction. If $X$ is finite dimensional, the closed kernel has finite codimension and $A/\ker \pi$ is finite dimensional, so $\pi$ is continuous. Suppose instead that $X$ is infinite dimensional. LetThis is an invariant submodule, because the orbit map of $\pi (b)\xi$ is the orbit map of $\xi$ composed with the bounded map $a\mapsto ab$. Thus either $Y=X$ or $Y=0$.

To rule out $Y=0$, choose linearly independent unit vectors ${\xi }_0,{\xi }_1,\dots$. There are elements $a_n\in A$ such that, with $c_n=a_n\cdots a_1$,For completeness, construct them in the canonical quotient-norm Banach module. At stage $n$ the current tail vectors are independent. In the closed left annihilator of its first vector, density permits arbitrary images of any finite list of the remaining vectors. The condition that finitely many such images are independent is open in this Banach subspace and dense: perturb any element by $t$ times an element prescribing independent images; a suitable determinant is a nonzero polynomial in $t$, with only finitely many exceptional values. The Baire category theorem intersects these open dense conditions for all initial finite tail lists. The resulting $a_n$ kills the first current vector and retains independence of the entire remaining tail. This is the countable annihilation sequence in an irreducible Banach module.

The gliding-hump continuity principle supplies the other ingredient: if $E,F$ are Banach spaces, $S:E\to F$ is linear, $T_j:E\to E$ and $U_n:F\to F_n$ are bounded, and $U_{n}S{T}_1\cdots T_m$ is continuous whenever $m>n$, then $U_{n}S{T}_1\cdots T_n$ is continuous eventually. The product form of this principle is recorded in Proposition 1.3 of Thomas's paper. Here is the contradiction mechanism. Rescale so $\Vert T_j\Vert ,\Vert U_n\Vert \le 1$ and put $P_m=T_1\cdots T_m$. If infinitely many diagonal maps are discontinuous, choose increasing $n_j$ and successively tiny $x_j$ soAlso choose $\Vert x_j\Vert \le 2^{-j}/(1+{\max }_{i<j}\Vert U_{n_i}S{P}_{n_i+1}\Vert )$. The convergent sum $x={\sum }_j{P}_{n_j}{x}_j$ has, after its $j$th term, a tail factoring through $P_{n_j+1}$. Applying that continuous composite bounds the tail contribution by one. The display forces $\Vert U_{n_j}Sx\Vert >j$, contradicting $\Vert U_{n_j}Sx\Vert \le \Vert Sx\Vert$. Only finite decompositions and the continuous tail composites are used; no unjustified interchange of a discontinuous map with an infinite sum occurs.

Apply this principle with $E=A$, $F=\mathcal{B}(\overline{X})$, where $\overline{X}$ is the completion in the original norm, and $S(a)$ the bounded extension of $\pi (a)$. Set $T_j(a)=a{a}_j$ and $U_n(R)=R{\xi }_n$. ThenFor $m>n$ this is zero, hence continuous. For $m=n$ it is the orbit map of a nonzero vector, hence discontinuous if $Y=0$. This contradicts the principle. Therefore $Y=X$.

Finally, apply the Uniform boundedness principle on the Banach space $A$ to the continuous maps $a\mapsto \pi (a)\xi$ indexed by $\Vert \xi \Vert \le 1$. They are pointwise bounded because each individual $\pi (a)$ is bounded in the original norm. It givesThis proves Johnson's continuity theorem for irreducible normed representations, including when the original normed vector space is incomplete. The quotient norm, countable density construction, and gliding hump explain the substantive steps rather than assuming the desired continuity in advance.

It suffices first to treat unital algebras and a unital homomorphism. A surjective homomorphism between unital algebras automatically preserves the identity. If $B=0$, continuity is immediate. In the nonunital case, extend to the forced unitizations by $T^{+}(a+\lambda 1)=T(a)+\lambda 1$. This is still surjective and the unitization of a semisimple algebra is semisimple: its radical projects to zero in its scalar quotient, and its remaining intersection with the original algebra is its original radical. Continuity of the extension implies continuity of the original map.

An algebra homomorphism decreases the spectral radius, even without any continuity assumption, because it takes an inverse of $\lambda 1-a$ to an inverse of $\lambda 1-T(a)$. Suppose $a_n\to 0$ in $A$ and $T(a_n)\to b$ in $B$. Fix any $c\in B$ and choose $u,v\in A$ with $T(u)=b$, $T(v)=c$. PutConsider the polynomial $q_n(z)=T((1-z)w_n+zw)=(1-z)d_n+zd$. Its value at one is exactly $d$. On the outer circle $|z|=R$,On the inner circle $|z|=1/R$, spectral-radius decrease givesApplying part (i) in $B$ and then letting $n\to \infty$, with $R$ fixed, givesThis holds for every $R>1$, so $r_B(cb)=0$ for every $c\in B$.

This forces $b$ into the Jacobson radical. To see the criterion directly, if $b$ were outside a maximal left ideal $L$, then $L+Bb=B$, so $1=l+cb$ for some $l\in L$, $c\in B$. The equality $r_B(cb)=0$ makes $1-cb$ invertible, but this element equals $l\in L$, impossible in a proper left ideal. Hence $b$ lies in every maximal left ideal. Since $B$ is semisimple, $b=0$.

We have shown that the separating space of $T$ is zero. If $a_n\to a$ and $T(a_n)\to y$, apply the result to $a_n-a$ to obtain $y=T(a)$. Its graph is closed, and the closed graph theorem now provesThe crucial estimate uses part (i) with an exactly fixed central value; it does not assume the false general assertion that the spectral radius is continuous under norm limits.

Suppose $x\in J$ and put $u=1+yx$. If the left ideal $Au$ were proper, it would lie in a maximal left ideal $L$. But $yx\in J\subseteq L$, so $u\in L$ would imply $1\in L$. Thus $Au=A$, and $u$ has a left inverse $v$ with $vu=1$.

A left inverse alone is insufficient in a general operator algebra. Here $v=1-vyx$ differs from the identity by an element of $J$. The same argument supplies a left inverse $w$ of $v$. Then $w=w(vu)=(wv)u=u$, hence $uv=wv=1$. So $u$ is genuinely invertible.

Conversely, if $x\notin J$, part (i) gives a maximal left ideal $L$ with $x\notin L$. Maximality implies $L+Ax=A$, so $1=l+ax$ for some $l\in L,a\in A$. Therefore $1-ax=l\in L$ cannot be invertible, since a proper left ideal contains no invertible element. Taking $y=-a$ disproves the required universal invertibility. We have proved the unit criterion for the Jacobson radical in the requested form: