Does a Banach algebra version of "the sum of a closed subspace and a finite dimensional subspace is always closed" exist?

Question

In the setting of Banach spaces, it is well know that if $M$ is a closed subspace of a Banach space $X$ and $F$ is a finite dimensional subspace of $X$, then $M+F$ is closed.

Does a Banach algebra version of the aforementioned result exist? That is, if $M$ is a closed right ideal of a Banach algebra $\mathcal{A}$, $F$ is a minimal right ideal of $\mathcal{A}$, whether or not $M+F$ is closed too?

In particular, let $a \in \mathcal{A}$, $R(a):=\{x\in \mathcal{A}: ax=0\}$. If $a\mathcal{A}$ is closed and $R(a)$ is of finite order (i.e. a finite direct sum of minimal right ideals), whether or not $a\mathcal{A}+R(a)$ is closed?

Background: In operator theory, an operator $T$ acting on a Banach space $X$ is called upper semi-Fredholm if its kernel $ker(T)$ is finite dimensional and its range $ran(T)$ is closed. It is well known that $T$ is upper semi-Fredholm if and only if $T$ maps a bounded but not totally bounded set to a bounded but not totally bounded set. As a corollary, we know that if $T$ is upper semi-Fredholm, then $T^{n}$ is also upper semi-Fredholm.

At present, our aim is to give a suitable definition of semi-Fredholm elements in Banach algebra: $a \in \mathcal{A}$ is upper semi-Fredholm if $R(a)$ is of finite order and $aA$ is closed.

Answer 1

Here is a counterexample using similar ideas as my comments:

Let $E$ be a Banach space, $M,F \subset E$ be two disjoint closed subspaces s.t. $M+F$ is not closed. Let $B = E \oplus_{\ell^1} \mathbb{C}$ equipped with the multiplication,

$$(x,\alpha)(y,\beta) = (\alpha y+\beta x, \alpha\beta)$$

i.e., $(0,1)$ acts as the identity and the multiplication, when restricted to $E$, is trivial. Since $B$ has a unity of norm $1$, we have,

$$B \hookrightarrow \mathbb{B}(B), b \mapsto L_b$$

where $L_b: B \to B$ is given by left-multiplication by $b$, is an isometric algebra embedding. We thus regard $B \subset \mathbb{B}(B)$. Now, for each pair of unit vectors $x, y \in F$. Let $\varphi_x \in B^\ast$ be s.t. $\varphi_x(M) = 0$ and $\varphi_x(x) = 1$. Let $s_{x,y} \in \mathbb{B}(B)$ be given by $s_{x,y}(b) = \varphi_x(b)y$. Now, let $A$ be the Banach subalgebra of $\mathbb{B}(B)$ generated by $B$ and all such $s_{x,y}$. We observe that, for any $e,f \in E$,

$$s_{x,y}L_e(f, \alpha) = s_{x,y}(\alpha e) = \alpha\varphi_x(e)y = L_{\varphi_x(e)y}(f, \alpha)$$

Thus, by definition of $\varphi_x$, it is easy to verify that $M,F \subset A$ are left ideals and $F$ is minimal, but $M+F$ is not closed. Reversing the order of multiplication on $A$ provides a counterexample for right ideals.