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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: $\forall r\in R;\: r.1_{R}=1_{R}.r=r$

Julian Kuelshammer
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AmirHosein Sadeghimanesh
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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
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    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
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    Sometimes people write rng for a ring with no identity. – the L Dec 13 '12 at 15:57
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    @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
  • I think maybe we can use the same argument as in monoid case, which is extending an ideal again and again, repeat many times, possibly uncountable times, until we get a maximal one. This proof requires Zorn's lemma and it also works on the non-monoid rings? I didn't see any difference made by the existence of identity. – Chao Chen Dec 13 '12 at 16:13
  • @ChaoChen so you are saing that all rings should have a maximal ideal? Your words don't proof it because you didn't mention to anything that gauranties to earning maximal ideal when you were saing "extending an ideal again and again, repeat many times, possibly uncountable times, until we get a maximal one"! – AmirHosein Sadeghimanesh Dec 13 '12 at 16:27
  • Sorry I didn't make it very precise. Consider the set of all ideals, they form a partial order under the '$\subseteq$', and every chain of them is bounded (union of all ideals in a chain is also an ideal), so according to Zorn's lemma, there is maximal element among them, which is a maximal ideal. I didn't see 'identity' in this proof, but I am not sure if I missed something. – Chao Chen Dec 13 '12 at 16:39
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    @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
  • I got it! Sorry I definitely missed something, thank you! – Chao Chen Dec 13 '12 at 16:45
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    FYI, when I hear the term "monoid ring", I think of the concept related to "group ring" i.e. the ring $\mathbb{Z}[M]$ consisting of formal sums of monoid elements with the product is given by the monoid operation. e.g. $\mathbb{Z}[\mathbb{N}] \cong \mathbb{Z}[x]$. (the isomorphism sends a formal natural number $[n]$ to the monomial $x^n$) –  Dec 13 '12 at 17:53

2 Answers2

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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.

Community
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Julian Kuelshammer
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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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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.

rschwieb
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