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

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.

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)

• How does primality of $N$ yield $M\subseteq N$? Commented Dec 31, 2023 at 16:45
• Do you mean that primality works as subsets too? I didn't know that. If $AB\subseteq N$ then $A\subseteq N$ or $B\subseteq N$. Is that true? How do I prove it? The definition of primality I know is that if $ab\in N$ then $a\in N$ or $b\in N$. Commented Dec 31, 2023 at 16:46
• @Nothingspecial The general definition of prime ideal is "$P$ is a prime deal iff it is an ideal, $P\neq R$, and for any ideals $A$ and $B$, if $AB\subseteq P$, then either $A\subseteq P$ or $B\subseteq P$." This is equivalent to the elementwise definition for commutative rings, but it is weaker (and the correct notion) for non-commutative rings. It is easy that the elementwise version implies this one. For the converse in the commutative setting, use principal ideals. Follow the link I gave for some discussion. Commented Dec 31, 2023 at 16:51
• @Nothingspecial The product of $M$ with $M$ is contained in $N$, with $M$ an ideal, so either $M$ is contained in $N$ or $M$ is contained in $N$. i.e., $M$ is contained in $N$, using the definition of "prime ideal". Commented Dec 31, 2023 at 17:00
• I see that it's more general and sometimes more useful definition... If I knew this definition, this problem would have been trivial... :) But my course, being an introductory one, focuses mostly on commutative rings with identity. Commented Dec 31, 2023 at 18:17

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 ($$x^n = 0 \in P \implies x \in P$$).

• $M\subseteq N(R)\implies M\subseteq P$ for any prime ideal $P$ of $R$. By maximality of $M$, we get $M=P$. Thus, $M$ is the only prime ideal of $R$. Now we use that fact that in a commutative ring with identity, every maximal ideal is a prime ideal to conclude the proof statement. Amazing..! I am getting so many different approaches to the same problem, it's worth asking questions here... Commented Jan 1 at 12:41
• Yes. It's just that simple. Commented Jan 1 at 12:44

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.

• If I understood correctly, $Rx+M=R$ because $M$ is maximal. Thus, $1\in Rx+M$. But I am not sure why the last implication is true. What does all elements of $R\setminus M$ are invertible imply that $M$ is the unique maximal ideal? Commented Dec 31, 2023 at 16:34
• Oh I got it, after some thought! If $N$ is another maximal ideal, it must contain one of those invertible elements (say, $x$) which is not in $M$. Thus, $Rx$ is contained in $N$. Now, $1\in Rx\subseteq N$. Hence, $N=R$ (contradiction) Commented Dec 31, 2023 at 16:38
• Assume $R$ has another maximal ideal $M_0\ne M$. Since $M_0$ is maximal, it can not be contained strictly in $M$, and $M_0\ne M$ by hypothesis. Therefore $M_0$ is not a subset of $M$, i.e. $M_0-M\ne\varnothing$. But if $a\in M_0-M$, then $a$ must be invertible since $a\in R-M$, implying $M_0=R$, which is absurd. Generally speaking, if an ideal consists of all of the non-invertible elements, then it is the unique maximal ideal. Commented Dec 31, 2023 at 16:46
• Well, the set of zero divisors in $\mathbb{Z}$ is $\varnothing$ and is therefore contained in $\{0\}$. Maybe it is true if $R$ has non-zero zero divisors? Let me think of the question. Commented Dec 31, 2023 at 17:11
• @Nothingspecial Ok, I think I have come up with a counterexample. Take $R=\mathbb{Z}_4[x]$, the polynomial ring on $\mathbb{Z}_4$. Then the zero divisors (including $0$) is the ideal generated by $\bar{2}$. But $\langle\bar{2}\rangle$ is not maximal since it is strictly contained in $\langle\bar{2}, x\rangle$. The latter does not equal $\mathbb{Z}_4[x]$. Commented Dec 31, 2023 at 17:29