We know that if $R$ and $I$ an ideal of $R$, then there is a bijection between the prime ideals of $R$ containing $I$ and the prime ideals of $R/I$. It is given by $P\mapsto P/I$. Is it true that this map gives a bijection between the maximal ideals of $R$ containing $I$ and the maximal ideals of $R/I$ ?
2026-03-31 08:02:10.1774944130
Bijection beteween maximal ideals
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The converse is indeed true. If $I \subset M \subset R$ is a maximal ideal of $R$, then $R/I/M/I \cong R/M$ by the third isomorphism theorem for rings, so $M/I$ is a maximal ideal of $R/I$. Note that the same argument shows that prime ideals in $R$ containing $I$ project to prime ideals in the quotient $R/I$.