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Let $R$ be a commutative ring without identity.

My question: is it true or false that every maximal ideal of $R$ is primary?

(An ideal I of R is said primary if is proper and

$\forall a,b\in R, ab\in I \Rightarrow a\in I \vee b\in\sqrt{I}$,

where $\sqrt{I} = \{r\in R| \exists n\in\mathbb{N}^+ : r^n \in I \} $. )

In a unitary commutative ring this is trivially true, because in such a ring is true:

$I$ maximal $\Rightarrow$ $I$ prime,

and in every ring holds:

$I$ prime $\Rightarrow$ $I$ primary.

I'm not sure about what happens in the non-unitary case. I've some results about the primary ideals, but mainly in unitary rings.

user26857
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Barry Allen
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  • There are many posts here that a maximal ideal need not be prime (e.g., here). We could check posts with "primary". – Dietrich Burde Feb 18 '20 at 09:28
  • I've done some research before to post, but i haven't found an answer; an eventual counterexample must be a maximal ideal that is not prime (else is primary), and i have check some of such examples, finding only primary ideals... so still i'm not sure if the statement is true or false. – Barry Allen Feb 18 '20 at 11:43

1 Answers1

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Definition: A simple ring (not necessarily commutative or with 1) is a non-zero ring $A$ that $A^2\neq0$ and it has no two-sided ideal besides the zero ideal and itself.

Suppose $R$ is a commutative ring and $I\subset R$ is a maximal ideal such that it is not primary. We want to get a contradiction. Then there are $a,b\in R$ such that $ab\in I$ but $a\not\in I$ and $\{b, b^2,\dots\}\cap I=\emptyset$ (obviusly, in this case, $a\neq0,b\neq0$ and $R^2\neq0$). Consider $R/I$; it is a simple (commutative) ring. It is not so difficult to prove that $R/I$ is a field (in fact, Theorem: Any commutative simple ring is a field). Since $a\not\in I$ and $b\not\in I$, we have $\overline{a}\neq0$ and $\overline{b}\neq0$ in $R/I$. Hence $0\neq\overline{a}\overline{b}=\overline{ab}$, i.e., $ab\not\in I$. This contradiction anwers your question.

freshman
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  • $(Z_2, +)$ with the zero multiplication is a commutative simple ring that isn't a field, so in the non-unitary case it seems that your theorem doesn't work. Am I doing something wrong? – Barry Allen Feb 19 '20 at 09:15
  • By definition, $R$ is simple ring if $R^2\neq0$ and its only ideals are $0$ and $R$. See HERSTEIN - Noncommutative rings or MATEJ BREŠAR - Introduction to noncommutative rings (page 6). I should put it in the answer, but sometimes things are clear in our mind. Sorry, I'll edit the answer. – freshman Feb 19 '20 at 13:19
  • But in this way, if $I$ maximal but not prime, is $R^2 \subseteq I$ (this is a standard result), so $\forall a\in R, a^2\in I$ and $a^2+I = I$, that implies $\left(\frac{R}{I}\right)^2 = 0$, and $\frac{R}{I}$ is not simple, thus the theorem can't be applied. Am I missing something? I appreciate your help, i'm a beginner. Thanks. – Barry Allen Feb 19 '20 at 16:01
  • Of course in this case the theorem can't be applied. I'm not applying the theorem for this assumption! My proof is based on "Suppose $I$ is maximal and is not primary" not on "Suppose $I$ maximal that isn't prime". Both are completely different and in my assumption I can use the theorem freely since $R^2\not\subset I$. I'm proving that maximal ideal is primary, not maximal ideal is prime. – freshman Feb 19 '20 at 17:04
  • Now i see my mistake. My objection was not on the assumption, but on the fact that the proof could't work in every case, but i forgot that we were reasoning by contradiction. I had doubts because i missed $b^2 + I \neq I$. Thanks for you help, i will consult the books you recommend too. – Barry Allen Feb 19 '20 at 18:23
  • My pleasure. You'll be a good mathematician. – freshman Feb 19 '20 at 18:39
  • Can you provide some reasoning for your theorem, that every commutative simple ring is a field? (In particular, why does there have to be a unity?) – math54321 Feb 21 '20 at 03:19
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    Sure I can! Let $R$ be a simple commutative ring. Take $x\neq0$, then $xR$ is a nonzero ideal in $R$ (because if $xR=0$ then $Ann(R):={a\in R : aR=0}$ is a nonzero ideal in $R$. Then $Ann(R)=R$. Hence $R^2=Ann(R)R=0$. But, by definition $R^2\neq0$). Okay, since $xR\neq0$ and $R$ is simple we have $xR=R$. Thus there exists $e\in R$ such that $xe=x$. For all $y\in R$, $y=xr$, hence $ey=e(xr)=(ex)r=xr=y$. Therefore, $e$ is the unity (let's denote it by $1$). Using the same reasoning earlier, there exists $z$ such that $xz=1$. Thus any nonzero element $x$ has an inverse. – freshman Feb 21 '20 at 04:38
  • Thanks for providing this proof - the definition of simple seems to vary depending on the source (e.g. some don't require $R^2 \ne 0$), so it's good to have this. +1 – math54321 Feb 21 '20 at 21:48