Show that N is a maximal ideal in a ring R if and only if R/N is a simple ring, that is, it is nontrivial and has no proper nontrivial ideals. (Compare with Theorem 15.18.)

Data from Theorem 15.18

M is a maximal normal subgroup of G if and only if G / M is simple. 

Proof Let M be a maximal normal subgroup of G. Consider the canonical homomorphism γ : G → G/M given by Theorem 14.9. Now γ^-1 of any nontrivial proper normal subgroup of G/M is a proper normal subgroup of G properly containing M. But M is maximal, so this can not happen. Thus G/M is simple. 

Conversely, Theorem 15.16 shows that if N is a normal subgroup of G properly containing M, then γ[N] is normal in G/M. If also N ≠ G, then γ[N] ≠ G/M and γ[N] ≠ {M}. Thus, if G/M is simple so that no such γ[N] can exist, no such N can exist, and M is maximal.