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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.

Nothing special
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3 Answers3

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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)

Arturo Magidin
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  • How does primality of $N$ yield $M\subseteq N$? – Nothing special Dec 31 '23 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 $b\in N$. – Nothing special Dec 31 '23 at 16:46
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    @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. – Arturo Magidin Dec 31 '23 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". – Arturo Magidin Dec 31 '23 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. – Nothing special Dec 31 '23 at 18:17
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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.

Asigan
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  • 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? – Nothing special Dec 31 '23 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) – Nothing special Dec 31 '23 at 16:38
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    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. – Asigan Dec 31 '23 at 16:46
  • @Nothingspecial Nice observation! You figured it out before I finish typing. But I still recommend you read my proof because it is more detailed, and you can check if you understood correctly. – Asigan Dec 31 '23 at 16:49
  • Your general result will help me later... If all zero divisors of $R$ are contained in an ideal $I\neq R$ then $I$ is the maximal ideal of $R$. – Nothing special Dec 31 '23 at 16:53
  • Well, I am a little doubtful on whether this is correct. What if $R=\mathbb{Z}$ and $I={0}$? – Asigan Dec 31 '23 at 17:04
  • But zero divisor, by the definition I am used to, is a non-zero element which "divides" $0$. – Nothing special Dec 31 '23 at 17:06
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    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. – Asigan Dec 31 '23 at 17:11
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    @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 $\mathbb{Z}_4[x]$. – Asigan Dec 31 '23 at 17:29
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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 ($x^n = 0 \in P \implies x \in P$).

Martin Brandenburg
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  • $M\subseteq N(R)\implies M\subseteq P$ 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... – Nothing special Jan 01 '24 at 12:41
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    Yes. It's just that simple. – Martin Brandenburg Jan 01 '24 at 12:44