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 $\mathcal{L}(E)$ be the set of all adjointable operators on Hilbert module $E$, if $\mathcal{L}(E)$ prime algebra?
($E$ is Hilbert module over $\mathcal{K}(H)$, such that $\mathcal{K}(H)$ is the set of all compact operators on Hilbert space $H$.)
Below is the answer to the original version of the question:
What do you assume on $E$? If nothing then the answer is trivially no.
Let $E=\mathbb{C}\oplus \mathbb{C}\oplus \mathbb{C}$ be the three-dimensional abelian C*-algebra regarded as a Hilbert module over itself. Then $\mathcal{L}(E)\cong E$ as the algebra of adjointable operators on a C*-algebra is nothing but the multiplier algebra. As $E$ is unital we do not get anything new. Certainly, for any $r\in E$ we have $$(1,0,0)\cdot (0,0,1)=(0,0,0) = (1,0,0)\cdot r \cdot (0,0,1).$$