Let $X$ be a Banach space over the complex field $\mathbb{C}$ and $\mathcal{B}(X)$ the algebra of all bounded linear operators on $X.$ I want to show that $\mathcal{B}(X)$ is a prime algebra.
My attempt: Let $A, B\in\mathcal{B}(X)$ such that $A \mathcal{B}(X)B=\{0\}.$ I want show that either $A=0$ or $B=0. $ On contrary assume that $A\neq 0$ and $B\neq 0. $ Since $\mathcal{B}(X)$ is unital algebra, we have $AB=0,$ that is, $A(B(x))=0$ for all $x\in X.$ Since $B\neq 0,$ there exists an element $x_0\in X$ such that $B(x_0)\neq 0.$ Now, we have $A(B(x_0))=0.$
From here, I want to get a contradiction, but don't how to procced further.
Remark: I have read somewhere that the statement ``$\mathcal{B}(X)$ is a prime algebra " is a consequence of Hahn-Banach Theorem.
If $B\neq 0$ there is $x_0$ such that $y_0:=Bx_0\neq 0.$ Fix $z\neq 0.$ There exists a bounded linear operator $C$ for which $Cy_0=z.$ Thus $$0=ACBx_0=Az$$ Since $z$ is arbitrary we get $A=0.$ The existence of the operator $C$ can be proved as follows. There exists a bounded linear functional $\varphi$ such that $\varphi(y_0)=1.$ Then $Cx=\varphi(x)z$ is a bounded linear operator satisfying $Cy_0=z.$ The existence of $\varphi$ follows from the Hahn-Banach theorem.