7
$\begingroup$

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

  • 0
    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 $x^n$)2012-12-13

2 Answers 2

3

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.

  • 0
    @rshwieb , thank you.2012-12-13
11

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.

  • 2
    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.2012-12-13