There may be several ways to prove that in a commutative ring with unity, maximal Ideals are prime ideals. I attempted the following proof via contradiction.
Let us consider a commutative Ring $R$ with an unity and let $I$ be a maximal ideal. Suppose there are two elements $r_1,r_2\notin I$ such that $r_1.r_2=i_1\in I$ where "." is the product operation. Consider the two ideals $I_{1,2}=R.r_{1,2}+I=\{r.r_{1,2}+i , r\in R, i\in I\}$. Both $I_{1}$ and $I_{2}$ contain $I$ and $\neq I$. Now $I_{1}.I_{2}=I$, which means that both $I_{1}$ and $I_{2}$ cannot be $R$ itself. So either $I_{1}$ or $I_{2}$ is an ideal which is not $R$ and contains but not equal to $I$, which contradicts the fact that $I$ is a maximal ideal.
If anybody can point out if there are any loop holes or wrong statements above, it will be very helpful.