Ideal is maximal iff the zero ideal is maximal in the quotient ring Let $R$ a commutative ring with unit. Without using the ideal correspondence theorem for quotient rings, show that
\begin{align}
\mathscr{m} \subset R \, \, \, \text{maximal ideal} \Longleftrightarrow 0 \subset R/\mathscr{m} \, \, \, \text{maximal ideal}
\end{align}
I have to know the proof of this statement for the exam, but I don't have any idea how to prove it.
Any suggestion? Thanks in advance!
 A: Let $\pi: R \rightarrow R/\mathfrak{m}$ be the canonical projection.  The gist of the proof is looking at preimages/images of ideals under $\pi$, and employing a few elementary properties of ring homomorphisms.  Note that $\pi$ is surjective and has kernel $\mathfrak{m}$.
For a ring homomorphism $\phi: R \rightarrow S$, we recall that:
(1) If $J \subset S$ is an ideal, then $\phi^{-1}(J)$ is an ideal.
(2) If $I \subset R$ is an ideal, then $\phi(I)$ is an ideal when $\phi$ is surjective.
(3) $\phi\big(\phi^{-1}(J)\big) = J$ when $\phi$ is surjective (but always have $\subset$)
(4) $\phi^{-1}\big(\phi(I)\big) = I$ when $\ker(\phi) \subset I$ (but always have $\supset$)
If you haven't seen these before, proving them is a good place to start (more good exam material!)  I'll  just show the $\subset$ containment in (4).  Indeed we have $x \in \phi^{-1}\big(\phi(I)\big)$ iff $\phi(x) \in \phi(I)$ iff there exists $y \in I$ with $\phi(x) = \phi(y)$, equivalently $\phi(x-y) = 0$.  By assumption that $\ker(\phi) \subset I$, we get $x \in I$.  
Note that these properties taken together give us the correspondence between ideals of $R$ and ideals of $R/\mathfrak{m}$. In practice, the point is that we can use surjective homomorphisms to transport ideals back and forth between rings without losing any information, a very powerful tool.  
Proof that $\mathfrak{m} \subset R$ is maximal $\iff$ $(0) \subset R/\mathfrak{m}$ is maximal:
Suppose that $\mathfrak{m}$ is a maximal ideal of $R$, and that $0 \subset J$ is an ideal of $R/\mathfrak{m}$.  Then $\pi^{-1}(J)$ is an ideal of $R$ (1) and since $\mathfrak{m}=\pi^{-1}(0) \subset \pi^{-1}(J)$, we have that $\mathfrak{m} = \pi^{-1}(J)$ or $R = \pi^{-1}(J)$ by maximality of $\mathfrak{m}$.  In the first case this gives $0 = \pi(\mathfrak{m}) = \pi\big(\pi^{-1}(J)\big) = J$ (last equality by (3)), and similarly in the second that $R/\mathfrak{m} = J$.  Hence $(0)$ is maximal in $R / \mathfrak{m}$.
For the other direction, assume that $(0)$ is maximal in $R /\mathfrak{m}$.  If $\mathfrak{m} \subset I$ then $\pi(I)$ is an ideal (2) so it must be either $0$ or the whole ring $R/\mathfrak{m}$. We need to show that this implies $I$ is either equal to $\mathfrak{m}$ or the whole ring $R$.  If $\pi(I) = 0$ then we have $\mathfrak{m} \subset I \subset \pi^{-1}\big(\pi(I)\big) = \pi^{-1}(0) = \mathfrak{m}$, and we see $\mathfrak{m} = I$.  If $\pi(I) = R / \mathfrak{m}$, then $I = \pi^{-1}\big(\pi(I)\big) = \pi^{-1}(R/\mathfrak{m})) = R$ (first equality by (4))
