maximal ideal is prime

Theorem. In a commutative ring with non-zero unity, any maximal ideal is a prime ideal.

Proof.  Let $𝔪$ be a maximal ideal of such a ring $R$ and let the ring product $rs$ belong to $𝔪$ but e.g.  $r\notin 𝔪$. The maximality of $𝔪$ implies that $𝔪+(r)=R=(1)$.  Thus there exists an element  $m\in 𝔪$  and an element  $x\in R$  such that  $m+xr=1$.  Now $m$ and $rs$ belong to $𝔪$, whence

$$s=1s=(m+xr)s=sm+x(rs)\in 𝔪.$$

So we can say that along with $rs$, at least one of its factors (http://planetmath.org/Product) belongs to $𝔪$, and therefore $𝔪$ is a prime ideal of $R$.