Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 10 / 3

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.

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.

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:

If $x\in J$, part (ii) makes $1-x/\lambda$ invertible for every nonzero scalar $\lambda$. Thus $\sigma (x)\subseteq \{0\}$ and $r(x)=0$: it is a quasinilpotent element. Conversely, let $I$ be a two-sided ideal consisting of quasinilpotent elements. For $x\in I$ and $y\in A$, $yx\in I$ has spectrum $\{0\}$, so $1+yx$ is invertible. Part (ii) implies $x\in J$. This proves $J$ is the greatest ideal contained in the quasinilpotent set, the Jacobson radical as greatest quasinilpotent ideal characterization.

A semisimple Banach algebra is one with $J=\{0\}$. For $B=\mathcal{B}(X)$, its natural action on $X$ is faithful and algebraically irreducible. Indeed, given $v\ne 0$ and any $w\in X$, the Hahn-Banach theorem gives a bounded functional $g$ with $g(v)=1$. The rank-one operator $z\mapsto g(z)w$ then maps $v$ to $w$. Any invariant subspace containing $v$ consequently contains every $w$. The faithful representation has kernel zero, so zero is primitive and $B$ is semisimple.

Nevertheless, choose a nonzero bounded functional $g$ and $0\ne x_0\in \ker g$, possible when $\dim X>1$. The nonzero rank-one operator $Tz=g(z)x_0$ satisfies $T^{2}z=g(z)g(x_0)x_0=0$. Hence $\sigma (T)=\{0\}$ and $r(T)=0$, although $T\ne 0$. This proves the quasinilpotent set of $B$ is not $\{0\}$; it is not the same as the radical.

For infinite-dimensional $X$, the finite-rank operators form a nonzero proper two-sided ideal: products with bounded operators are finite-rank, and the identity has infinite rank. Thus zero is not maximal. Even if maximal ideals are required to be closed, the compact operators give a nonzero proper closed ideal containing the finite-rank operators. The identity is not compact on an infinite-dimensional Banach space, since the Riesz lemma supplies a bounded sequence with pairwise distances bounded below. Therefore zero is primitive but not maximal in infinite dimension.