I have the following question in my homework.
Let $R$ be a commutative ring with identity and $M$ be a maximal ideal of $R$ such that $M^2=\{0\}$. Show that $M$ is the unique maximal ideal of $R$.
My attempt:
I see that it's trivially true if $M=\{0\}$ in which case, $R$ becomes a field. However, in general,
$M^2=\{0\}\implies m_1m_2=0 \ \forall \ m_1,m_2\in M$.
This means all elements of $M$ are zero-divisors (and nilpotent too).
What are other implications of $M$ being an ideal which is both square-zero and maximal?
I am not sure how to proceed. I am doing an introductory ring theory course so do not use advanced theorems/tools to guide me.
Best Answer
Let $N$ be a maximal ideal. We prove that $M=N$. Since $N$ is maximal in a ring with unity, $N$ is prime. (See also here)
Since $MM=M^2=\{0\}\subseteq N$, the primality of $N$ yields $M\subseteq N$. Since $M$ is maximal and $N\neq R$, we conclude that $M=N$.
(Note that this also works for noncommutative rings with unity, and more generally noncommutative rings with $R^2$ not contained in any maximal ideal)