A unital commutative ring having a square-zero maximal ideal is local

Question

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.

Answer 1

By assumption every element of $M$ is nilpotent and hence* is contained in every prime ideal of $R$. This immediately implies that $M$ is equal to every prime ideal, and we see that $R$ has exactly one prime ideal (which, by the way, is much stronger than having exactly one maximal ideal).

*Here we only need the trivial observation that every prime ideal contains every nilpotent element ($x^n = 0 \in P \implies x \in P$).

Answer 2

If $x\notin M$, then exists $r\in R, m\in M$ s.t. $$rx+m=1.$$ Then $$0=m^2=(1-rx)^2=1-2rx+r^2x^2. $$ i.e, $1=x(2r-r^2x)$. Therefore $x$ is invertible. Every element in $R-M$ is invertible $\implies $ $M$ is the unique maximal ideal.

Answer 3

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)