in this site maximal ideal is prime
Proof. Let $ \mathfrak{M}$ be a maximal ideal of such a ring $ R$ and let the ring product $ rs$ belong to $ \mathfrak{M}$ but e.g. $ r \notin \mathfrak{M}$. The maximality of $ \mathfrak{M}$ implies that $ \mathfrak{M}\!+\!(r) = R = (1)$. Thus there exists an element $ m \in \mathfrak{M}$ and an element $ x \in R$ such that $ m\!+\!xr = 1$. Now $ m$ and $ rs$ belong to $ \mathfrak{M}$, whence $\displaystyle s = 1s = (m\!+\!xr)s = sm\!+\!x(rs) \in \mathfrak{m}.$ So we can say that along with $ rs$, at least one of its factors belongs to $ \mathfrak{M}$, and therefore $ \mathfrak{M}$ is a prime ideal of $ R$.
I dont understand this part: s=1s=(m+xr)s=sm+x(rs)$\in m$. Why is x(rs)$\in m$? isnt it supposed to x(rs)$\in (m)$? and why sm $\in m$ since we dont know if s $\in m$ or not, yet.