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Let $R$ be a commutative ring and $m\subseteq R$ be a maximal ideal. Can you describe the set of prime ideals of the $R/m^2$. Are they all maximal ?

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2 Answers 2

up vote 10 down vote accepted

$m/m^2$ is a maximal ideal of $R/m^2$, and it is nilpotent; it is therefore the nilradical, and being prime, is the only prime ideal of $R/m^2$.

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Prime $\rm\ P = I + m^2\supset m^2\ \Rightarrow\ P\supset m\ \Rightarrow\ P = m\ $ since $\rm\ m\ $ is maximal.

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