$R$ is a prime ring in which the zero ideal is the intersection of finitely many maximal left-ideals.
Show that $R$ is primitive.
It was suggested to me that the best method to use is proof by contradiction.
My proof has run like this so far:
Suppose $R$ is not primitive; then
$\exists A_{i} \triangleleft R$ non-zero with $A_{i} \subseteq M_{i}$,
where $M_{i}\cap M_{2} \cap ... M_{n} = <0_{R}> $.
Then, $A_{1} A_{2} ...A_{n} \subseteq A_{1}\cap A_{2}\cap ...A_{n}$, and $A_{1}\cap A_{2}\cap ...A_{n} \subseteq M_{i}\cap M_{2} \cap ... M_{n} =<0_{R}>$.
But if R is prime, then $A_{1}A_{2}...A_{n} \neq <0_{R}>$.
Is this contradiction valid?