Why the set of all bounded operators on Hilbert space is a prime?

Question

In abstract algebra, a nonzero ring $R$ is a prime ring if for any two elements $a$ and $b$ of $R$, $arb=0$ for all $r$ in $R$ implies that either $a=0$ or $b=0$. Or for any two ideals $A$ and $B$ of $R$, $AB={0}$ implies $A={0}$ or $B={0}$.

Let $\mathcal{B}(\mathcal{H})$ be the set of all bounded operators on Hilbert space $\mathcal{H}$ , Why $\mathcal{B}(\mathcal{H})$ is a prime algebra?

Answer 1

Assume $a,b \in B(H)$ satisfy $a \, B(H) \, b = 0$, and $b \neq 0$. We want to prove $a=0$.

Since $b \neq 0$, there is some $x \in H$ with $b(x) \neq 0$. For any $y \in H$ there is some $r \in B(H)$ with $r(b(x))=y$ (why?). Conclude $a(y)=0$.

Answer 2

Suppose $a,b \in B(H)$ are such that $a,b \ne 0$. Let $x \in H$ with $bx \ne 0$. By Hahn-Banach there is an $\ell \in H^*$ such that $\ell(bx) \ne 0$. Let $y \in H$ with $ay \ne 0$. Define $r \in B(H)$ by $rz = \ell(z)y$, all $z \in H$. Then $$ arb(x) = a(y)\ell(bx) \ne 0$$ Hence $arb \ne 0$.