Two-sided ideals of $B(H)$ are hereditary

Question

I seem to recall that (not necessarily closed) two-sided ideals of $B(H)$ are hereditary. Is that true?

If it is, can anyone post a proof/reference?

Answer 1

Yes, this is true and it's actually due to Calkin himself. See Theorem 1.6 in

J. W. Calkin, Two-Sided Ideals and Congruences in the Ring of Bounded Operators in Hilbert Space, Annals of Mathematics Second Series, 42, No. 4 (Oct., 1941), 839-873.

Calkin proved that ideals of $B(H)$ are in one-to-one correspondence with certain subsets of $c_0$ that he termed ideals sets. Now Theorem 1.6 tells you that the spectral set of a two-sided ideal is an ideal set which, by definition, is hereditary.

It is maybe worthwhile to add that this is not true for ideals of $K(H)$, however I don't have a counter-example off the top of my head.

Answer 2

Maybe it is in order to add that the same holds for arbitrary von Neumann algebras. Indeed, let $I \lhd M$ be a two-sided ideal in a von Neumann algebra $M$. We want to show that $0\leqslant x \leqslant y \in I$ implies that $x \in I$. It follows that $\sqrt{x} = a \sqrt{y}$, where $a \in M$ is a contraction that vanishes on $(Ran(\sqrt{y}))^{\perp}$ (this is the generalised polar decomposition); the fact that $a$ can be taken to be an element of $M$ follows from the bicommutant theorem. Now, using self-adjointness of $\sqrt{x}$, we can write $x = a \sqrt{y} (\sqrt{y} a^{*}) = aya^{*} \in I$.