Proposition 4

If A possesses a unit element, the spaces J(A) and $\widehat{A}$ are quasi-compact.

It is enough to give the proof for J(A). Let $(T_j)$ be a family of closed subsets of J(A) whose intersection is empty. If the sum $\sum_j\Upsilon (T_j)$ were not equal to A, this sum would be contained in a maximal two-sided ideal I. The ideal I would be primitive (Lemma 7); since the subset $T_j$ is closed, hence equal to $V(\Upsilon (T_j))$, one would have $I\in T_j$ for every $j$, which contradicts the hypothesis. Thus one has $\sum_j\Upsilon (T_j) = A$, and therefore one can write $1 =x_1+\cdots +x_n$ with $n\geqslant 1$ and $x_i\in \Upsilon (T_{j_i})$ for every $i$. This implies that $\Upsilon (T_{j_1})+\cdots +\Upsilon (T_{j_n}) = A$, whence $T_{j_1}\cap \cdots \cap T_{j_n}=\emptyset$.

Suppose that the algebra A is commutative and unital. The Jacobson topology on J(A) is the topology induced on J(A) by the Zariski topology of the prime spectrum of A (AC, II, Definition 4, p. 125).

Suppose that A is commutative and that K is a topological field. The canonical isomorphism of K onto $\mathscr{L}(K)$ makes it possible to identify an element of $\mathsf{X}(A)$ with a representation of A in the vector space K, which defines an injective mapping of $\mathsf{X}(A)$ into $\widehat{A}$. One can therefore identify $\mathsf{X}(A)$ with a subset of $\widehat{A}$.