Proof of the uniqueness of maximal ideal Let $R$ be a commutative ring with $1$. Let $M$ be a maximal ideal of $R$ such that $M^2 = 0$. Prove that $M$ is the only maximal ideal of $R$.
 A: A maximal ideal is prime. Let $\mathfrak{m}$ be any maximal ideal and let $x\in M$; since $x^2=0\in\mathfrak{m}$, you get $x\in\mathfrak{m}$. Therefore $M\subseteq\mathfrak{m}$.
Note that the same holds if $M$ is a nil ideal.
A: Let $M'$ be a maximal ideal. Then $F=R/M'$ is a field and $M$ maps to an ideal $\overline M$ in this field with $\overline M\cdot \overline M=0$. Since $F\cdot F\ne 0$, we conclude $\overline M=0$, i.e. $M\subseteq M'$.
A: The other solutions have the best elementary solutions, so I'll try to give a distinct way.
The Jacobson radical $J(R)$ of $R$ contains all nilpotent ideals and is contained in all maximal ideals.
By these conditions, $M\subseteq J(R)\subseteq M$. Thus, the Jacobson radical is a maximal ideal. Thus, there cannot be any different maximal ideals, since there are no ideals between $M$ and $R$.
A: The first thing I thought of was the fact that if $M'$ is a maximal ideal of $R$ distinct from $M$, then $M$ and $M'$ are comaximal, i.e., $M + M' = R$. But then we can write $1=am+bm'$, where $m$ and $m'$ belong to $M$ and $M'$, respectively, and $a,b$ belong to $R$. Now, let $x$ be any element of $M$. Then $x=x\cdot1=axm+bxm'=bxm'\in M'$, since $xm\in M^2$ and $M^2=0$. Since $x$ is arbitrary, $M\subseteq M'$, which is impossible unless $M'=M$.
