Let $1\leq p<\infty.$ Denote $S_p(\ell_2)$ be the set of all compact operator $x$ on $\ell_2$ such that $Tr(x^p)<\infty.$ Define $\x\_{S_p(\ell_2)}:=Tr(x^p)^{\frac{1}{p}}.$ This makes $S_p(\ell_2)$ a Banach space. What is the largest closed twosided ideal in the Banach algebra of set of all bounded linear maps on $S_p(\ell_2)$?
Suppose $X$ is a Banach space that has the following property: (A) If $T$ and $S$ are in the space $L(X)$ of bounded linear operators on $X$ and the identity $I_X$ on $X$ factors through $T+S$, then either $I_X$ factors through $T$ or $I_X$ factors through $S$. It is more or less obvious that if $X$ satisfies (A) then $M_X:= $ all $U$ in $L(X)$ s.t. $I_X$ does not factor through $U$ is the largest ideal in $L(X)$. Probably $S_p$ satisfies (A); maybe it is even in some paper (maybe by Arazi and Lindenstrauss?).

1$\begingroup$ Link to an online copy of the ArazyLindenstrauss paper, for any readers following this up: numdam.org/item/CM_1975__30_1_81_0 $\endgroup$ Jun 19 '19 at 17:06

$\begingroup$ @Tomek. Why do you think that the fact that $S_1$ is the projective tensor product with $\ell_2$ with itself will be helpful? $\endgroup$ Jun 28 '19 at 5:01
This is not a complete answer but maybe it will shed some light. Btw. your question is slightly illposed as we do not know whether $B(S_p)$ has a unique maximal ideal, unless $p=2$.
Bill in his answer refers to the set $\mathscr{M}_X$ defined here. And indeed, it is very likely that the space $S_1$ has the property that $\mathscr{M}_{S_1}$ is closed under addition being the projective tensor product of $\ell_2$ with itself. The space $S_1$ is primary (by a result of Arias and Farmer) and is also isomorphic to the $\ell_1$sum $\ell_1(S_1)$.
If I am not mistaken, all known examples of primary spaces $X$ isomorphic to their infinite sums have the property that $\mathscr{M}_X$ is closed under addition, in which case, their algebras of operators have unique maximal ideals. However, there is no general result that would say it is a theorem, I am afraid.
As for $p=2$, the situation is trivial because in that case $S_2$ is a separable Hilbert space.


$\begingroup$ @ Tomek. Why do you think that the fact $S$ $\endgroup$ Jun 28 '19 at 5:01

1$\begingroup$ @SamyaKumarRay please read the third paragraph of my answer. Thanks. $\endgroup$ Jun 28 '19 at 7:16