Vtyjkyuk

Let $R$ a commutative ring with unit. Without using the ideal correspondence theorem for quotient rings, show that
beginalign
mathscrm subset R , , , textmaximal ideal Longleftrightarrow 0 subset R/mathscrm , , , textmaximal ideal
endalign

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!

Let $pi: R rightarrow R/mathfrakm$ 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 $mathfrakm$.

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) $phibig(phi^-1(J)big) = J$ when $phi$ is surjective (but always have $subset$)

(4) $phi^-1big(phi(I)big) = I$ 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 $x in phi^-1big(phi(I)big)$ 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 $R/mathfrakm$. 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 $mathfrakm subset R$ is maximal $iff$ $(0) subset R/mathfrakm$ is maximal:

Suppose that $mathfrakm$ is a maximal ideal of $R$, and that $0 subset J$ is an ideal of $R/mathfrakm$. Then $pi^-1(J)$ is an ideal of $R$ (1) and since $mathfrakm=pi^-1(0) subset pi^-1(J)$, we have that $mathfrakm = pi^-1(J)$ or $R = pi^-1(J)$ by maximality of $mathfrakm$. In the first case this gives $0 = pi(mathfrakm) = pibig(pi^-1(J)big) = J$ (last equality by (3)), and similarly in the second that $R/mathfrakm = J$. Hence $(0)$ is maximal in $R / mathfrakm$.

For the other direction, assume that $(0)$ is maximal in $R /mathfrakm$. If $mathfrakm subset I$ then $pi(I)$ is an ideal (2) so it must be either $0$ or the whole ring $R/mathfrakm$. We need to show that this implies $I$ is either equal to $mathfrakm$ or the whole ring $R$. If $pi(I) = 0$ then we have $mathfrakm subset I subset pi^-1big(pi(I)big) = pi^-1(0) = mathfrakm$, and we see $mathfrakm = I$. If $pi(I) = R / mathfrakm$, then $I = pi^-1big(pi(I)big) = pi^-1(R/mathfrakm)) = R$ (first equality by (4))