TITLE: Two-sided ideals of $B(H)$ are hereditary
QUESTION [6 upvotes]: 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?

REPLY [6 votes]: 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 $x = a \sqrt{y} (\sqrt{y} a^{*}) = aya^{*} \in I$.