TITLE: Does every infinite-dimensional Banach algebra contain an infinite-dimensional subalgebra with second-countable primitive ideal space?
QUESTION [7 upvotes]: Let $A$ be an infinite dimensional Banach algebra.  Even if separable the primitive ideal space of $A$ need not be second-countable when endowed with the hull-kernel topology. Can we at least find an infinite-dimensional subalgebra with this property?
For C*-algebras, separability implies second-countable primitive ideal space.

REPLY [3 votes]: The answer is no.
Let $A=A(D)$ be the disk algebra. Every character on $A$ is a point evaluation at some $x\in D_1$, the closed unit disk of $\mathbb{C}$, i.e. the spectrum $\sigma(A)$ of $A$ is $D_1$. Let $D_r = \{z\in\mathbb{C}: |z|\leq r\}$ for $r<1$. Then, the restriction of the hull-kernel topology on $\sigma(A)$ to $D_r$ is cofinite topology. In fact, if $S\subseteq D_r$ is not finite, then $k(S) = \{0\}$, so $h(k(S)) = \sigma(A)$. Consequently, the hull-kernel topology on $\sigma(A)$ is not even first countable.
Second, let $B\subseteq A$ be an infinite-dimensional closed subalgebra of $A$. Pick an $r<1$ and define an equivalence relation on $D_r$ via
$$x\sim y \Leftrightarrow f(x) = f(y)\hspace{8mm} \forall f\in B.$$
By the basic properties of analytic functions, it is not difficult to show that the equivalence classes of $\sim$ are at most finite. Thus, $D_r/\sim$ is an uncountable set. Moreover, $D_r/\sim \subseteq\sigma(B)$. Proceeding the same as above, the restriction of the hull-kernel topology on $\sigma(B)$ to $D_r/\sim$ is the cofinite topology. We conclude that the hull-kernel topology on $\sigma(B)$ is not even first countable.
On the positive side, if a commutative Banach algebra $A$ contains a separable regular (some sources use "completely regular" rather than "regular") subalgebra $B$, then the hull-kernel topology on $\sigma(B)$ is second countable. The reason for this is the fact that the Gelfand topology and the hull-kernel topology coincide on a spectrum of a regular Banach algebra, e.g. Theorem 3.2.10 in [Palmer, "Banach Algebras and the General Theory of *-Algebras I"]. Clearly, the Gelfand topology of a separable Banach algebra is second countable.
Finally, I think the following related note might be useful although it is not asked in the question. Let $F=\{x_n:n\in\mathbb{N}\}$ be a countable subset of $D$ that has an accumulation point in $D$. Let $V=\{(f(x_n))_{n\in\mathbb{N}} : f\in A(D)\}$ with the norm adopted from $A(D)$. Then, the map $f\to(f(x_n))_{n\in\mathbb{N}}$ maps $A(D)$ isomorphically onto $V$, where $\sigma(V)$ is countable. Thus, the hull-kernel topology on $\sigma(V)$ is second countable. Paraphrasing, there exists countable subsets $F\subseteq\sigma{(A(D))}$, which are dense in $\sigma(A(D))$ in the hull-kernel topology of $\sigma(A(D))$, and for which the restriction of the hull-kernel topology to $F$ is second countable.