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Is there any non-monoid ring which has no maximal ideal?

We know that every commutative ring has at least one maximal ideals -from Advanced Algebra 1 when we are study Modules that makes it as a very easy Theorem there.

We say a ring $R$ is monoid if it has an multiplicative identity element, that if we denote this element with $1_{R}$ we should have: $\forall r\in R;\: r.1_{R}=1_{R}.r=r$

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Every non-zero commutative ring has a maximal ideal. But what do you mean by a non-monoid ring? –  Zhen Lin Dec 13 '12 at 15:45
@ZhenLin I add my meaning from monoid in the question. –  AmirHosein SadeghiManesh Dec 13 '12 at 15:53
I thought that this was simply "ring" or "unital ring", whereas a ring without such identity is "rng". –  Asaf Karagila Dec 13 '12 at 15:55
@AsafKaragila I heard "unital ring" for monoid ring before, it was good to mentioned, but the expressions "rng" is new for me, and for the other only "ring" it is used for monoid rings in notes that at the start of them, authors make agreement to call monoid rings simply "ring" because they don't want to observe non-monoids one in those notes. –  AmirHosein SadeghiManesh Dec 13 '12 at 16:01
@Chao: But without an identity you cannot guarantee that an ideal is proper if and only if $1$ is not in the ideal. So the union of a chain of ideals might not be a proper ideal. –  Asaf Karagila Dec 13 '12 at 16:42

2 Answers 2

up vote 2 down vote accepted

If $D$ is a valuation domain with unique maximal ideal $M$, then there are some conditions where $M$ is an example of a commutative rng with no maximal ideals.

As I remember it, one can choose a domain with a value group within $\Bbb{R}$ such that the group has no least positive element. Then, one can argue that the maximal ideal of that valuation domain $\{r\mid \nu(r)>0 \textrm{ or } r=0\}$ is a rng without maximal ideals.

Scanning the web, I think this pdf contains an argument of that sort.

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@rshwieb , thank you. –  AmirHosein SadeghiManesh Dec 13 '12 at 18:18

Not every non-unital ring (or rng) has a maximal ideal. For example take $(\mathbb{Q},+)$ with trivial multiplication, i.e. $xy=0$ for all $x,y\in \mathbb{Q}$, then a maximal ideal is nothing more than a maximal subgroup. See this question why such a group does not exist.

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Your answer was nice, and as I can choose only one answer, so I selected for rshwieb , because $\mathbb{Q}$ is a field with characteristic zero that is mentioned in the pdf he brought below page 2, so his answer cover your answer too. But your thought was nice too. –  AmirHosein SadeghiManesh Dec 13 '12 at 18:17

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