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The intersection of maximal left ideals is the intersection of annihilators of simple modules on the left, hence is a two-sided ideal. If $j$ belongs to it, $1-rj$ has a left inverse for every $r$. Its left inverse also differs from $1$ by a radical element and has a left inverse; this makes the original inverse two-sided. The identity $(1-ba{)}^{-1}=1+b(1-ab{)}^{-1}a$ then shows that $1-jr$ is invertible too. If $j$ were outside a maximal right ideal $I$, one would have $1-jr\in I$ for some $r$, a contradiction. Applying the same argument to the opposite ring proves equality.

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 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:

Put $K={\bigcap }_{L}L$, where $L$ ranges over all maximal left ideals. For each such $L$, the action on the simple module $A/L$ has a primitive ideal $P_L$ as its kernel. An element of that kernel annihilates $1+L$, so $P_L\subseteq L$. The intersection $J$ of all primitive ideals is consequently contained in every $L$, hence $J\subseteq K$.

Conversely, take $x\in K$ and any nonzero simple module $E$. For every nonzero $v\in E$, simplicity makes the evaluation map $A\to E$, $a\mapsto av$, surjective. Its kernel is a maximal left ideal, so it contains $x$. Thus $xv=0$ for every nonzero $v$, and also for zero. The element $x$ therefore belongs to the annihilator of every simple module, hence to every primitive ideal. This provesSince the primitive ideals are two-sided ideals, this argument also makes $J$ a two-sided ideal, despite the one-sided ideals in this alternative description.

A primitive ideal is the kernel of an algebraically irreducible unital representation, equivalently the annihilator of a module for a nonzero simple module. Irreducibility here excludes all proper nonzero invariant linear subspaces, not just closed ones. In a Banach algebra the simple modules can be represented as $A/L$ with $L$ a maximal left ideal. Such an ideal is closed: if its closure were all of $A$, it would contain an element close enough to $1$ to be invertible by the Neumann series, contradicting properness. Maximality then makes its closure itself. Thus the quotient carries a Banach-space norm and a bounded left action.

Let $P$ be a maximal proper two-sided ideal. The quotient ring $R=A/P$ is nonzero, unital and simple as a ring. By the Zorn lemma, choose a maximal left ideal $L$ of $R$. The simple module $R/L$ has an annihilator which is a proper two-sided ideal of $R$. Simplicity forces that annihilator to be zero. Its pullback to $A$ is exactly $P$, proving every maximal proper two-sided ideal is primitive. If one wants the Banach representation explicitly, realize the same module as the quotient of $A$ by the pullback maximal left ideal, which is closed by the preceding argument.

First, a maximal left ideal $L$ is closed. Its norm closure is again a left ideal, since multiplication is continuous. If that closure were all of $A$, some $\ell \in L$ would satisfy $\Vert 1-\ell \Vert <1$ and would be invertible by the Neumann series. But a proper left ideal cannot contain an invertible element: ${\ell }^{-1}\ell =1$ would put $1$ in $L$. Maximality therefore forces $\overline{L}=L$.

Consequently $X=A/L$ is a nonzero quotient Banach space. It is a simple module for left multiplication by $A$, because the inverse image of any submodule under $A\to A/L$ is a left ideal containing $L$, and hence is either $L$ or $A$. Since $La\subseteq L$, right multiplication defines a well-defined bounded module endomorphismIndeed, changing $u$ by an element of $L$ changes $ua$ by an element of $L$, and taking the infimum over representatives proves the quotient norm bound. Right multiplication commutes with the left action by associativity.

Choose $\lambda$ in the nonempty spectrum of an element of $T$ in the Banach algebra $\mathcal{B}(X)$. Both the kernel and range of $T-\lambda I$ are submodules. If this module endomorphism were nonzero, simplicity would force its kernel to be zero and its range to be all of $X$. It would be a bounded bijection of Banach spaces, with bounded inverse by the bounded inverse theorem, contradicting the choice of $\lambda$. Thus $T=\lambda I$. Applying this identity to $1+L$ yields $a-\lambda 1\in L$. If also $a-\mu 1\in L$, then $(\lambda -\mu )1\in L$; since $1\notin L$, $\lambda =\mu$. ThereforeThis is the scalar right action on a maximal-left-ideal quotient; it does not assume that $L$ is a two-sided ideal or that $A/L$ is an algebra.

Now consider the center of an associative algebra $Z$. For each $u\in A$, the map $z\mapsto zu-uz$ is a continuous linear map, so its kernel is closed. Their intersection $Z$ is therefore closed. The identity belongs to $Z$, as do sums and scalar multiples of its elements. If $z,w\in Z$, then for every $u\in A$,Thus $zw\in Z$. Also $zw=wz$, since $z$ commutes with every element of $A$, including $w$. Hence the center is a closed commutative unital subalgebra, and is a Banach algebra in the inherited norm.

For every $z\in Z$, the inclusion $Lz=zL\subseteq L$ allows the first argument to apply. Let $\chi (z)$ be the unique scalar for which $z-\chi (z)1\in L$. Equivalently, the induced right multiplication on $A/L$ is $T_z=\chi (z)I$. Sincethe map $\chi :Z\to \mathbb{C}$ is a complex-linear unital character of an algebra. It is surjective because $\chi (\alpha 1)=\alpha$, and its kernel is precisely $L\cap Z$. ConsequentlyIndeed any ideal properly containing the kernel would have a nonzero image in the field $\mathbb{C}$ and thus would be all of $Z$. The algebra character is also continuous: $|\chi (z)|=\Vert \chi (z)I\Vert =\Vert T_z\Vert \le \Vert z\Vert$, although continuity is not needed for maximality.

Take rings and modules to be unital. Let $J_{\ell }$ denote the intersection of the maximal left ideals. For each nonzero vector $v$ of a simple module $M$ on the left, the map $R\to M$, $r\mapsto rv$, is onto and its kernel is a maximal left ideal. Consequently $J_{\ell }$ annihilates every simple module on the left. Conversely, annihilating $R/L$ forces an element to belong to the maximal left ideal $L$. Thuswhich shows that $J_{\ell }$ is a two-sided ideal.

For $j\in J_{\ell }$ and $r\in R$, the left ideal $R(1-rj)$ cannot be proper: a containing maximal left ideal would also contain $rj$ and hence $1$. Therefore $u(1-rj)=1$ for some $u\in R$. Here $u=1+urj$ also differs from $1$ by an element of $J_{\ell }$, so it too has a left inverse $v$. Multiplying the first equality by $v$ gives $1-rj=v$; hence $(1-rj)u=1$. We have proved that $1-rj$ is a unit, not just left invertible.

The identityis verified by multiplying on either side, using $(1-ab{)}^{-1}(1-ab)=(1-ab)(1-ab{)}^{-1}=1$. It implies that $1-jr$ is a unit as well. If $j$ were outside a maximal right ideal $I$, then $I+jR=R$, so $1-jr\in I$ for some $r$. A proper right ideal cannot contain a unit, giving a contradiction. Thus $J_{\ell }$ belongs to every maximal right ideal. Applying the same argument to the opposite ring proves the reverse inclusion. This proves the Jacobson radical is independent of handedness:For the zero ring the intersections are interpreted as the whole ring, so the equality remains valid.

If a two-sided ideal $N$ satisfies $N^m=0$, then for $x\in N$ and $r\in R$, $(rx{)}^m=0$. The finite geometric series ${\sum }_{i=0}^{m-1}(rx{)}^i$ inverts $1-rx$. The unit criterion for the Jacobson radical, also proved above by the maximal-ideal argument, gives $N\subseteq J(R)$. This assertion concerns a nilpotent ideal; arbitrary nilpotent elements of a noncommutative ring need not lie in its Jacobson radical.

For the formal power series ring over the p-adic integers, a series $f={\sum }_{i\ge 0}{a}_i{t}^i$ is a unit exactly when $a_0$ is a unit of ${\mathbb{Z}}_p$. Necessity follows by comparing constant coefficients. If $a_0$ is invertible, construct the inverse recursively by $b_0=a_0^{-1}$ and $b_m=-a_0^{-1}{\sum }_{i=1}^m{a}_i{b}_{m-i}$. Therefore the nonunits are exactly the series whose constant coefficient lies in $p{\mathbb{Z}}_p$. They form the unique maximal ideal $(p,t)$, and the quotient is ${\mathbb{F}}_p$. Hence

A primitive ideal of a unital associative algebra is the annihilator of a module for a nonzero simple module. Every maximal proper two-sided ideal is primitive: its quotient is a simple ring, and any simple module over that quotient has zero annihilator there. In a unital Banach algebra, a simple module can be realized as $A/L$ for a maximal left ideal; this ideal is closed, making the representation a bounded algebraically irreducible action on a Banach space.

For a unital irreducible action and $\xi \ne 0$, the annihilator $L_{\xi }=\{a:\pi (a)\xi =0\}$ is a maximal left ideal and is closed. Thus $A/L_{\xi }$ supplies the displayed complete norm on the module, with contractive action. Every module endomorphism is bounded for this quotient norm: its value at the cyclic vector is $a\xi$, so it is induced by bounded right multiplication by $a$ on the quotient. The Gelfand-Mazur theorem then makes the module's commutant consist of scalar operators. This constructs one compatible norm; it does not by itself establish continuity in a different preassigned norm.