Commutative Monoid and Invertible Elements I am looking for interesting (naturally occurring?) examples of commutative monoids with "lots" of invertible elements and "lots" of non-invertible elements. An easy way to get examples is use the direct product of a commutative monoid with no invertible element other than the identity and a group. For example, $\mathbb{N}_0\oplus\mathbb{R}$ with the usual operations. Two questions:


*

*Are all commutative monoids isomorphic to a direct product of a commutative monoid with no invertible element other than the identity and a group? (Fallacious "proof": The set of invertible elements is a group and $M\equiv M/I\oplus I$.)

*If question 1 is true, then there probably aren't too many interesting examples. If it is false, what are some interesting examples (in addition to the counterexample for question 1)?


Thanks.
 A: There are many commutative rings which would suit you for examples.
For one, you can use the power series ring $R=\Bbb R[[x]]$. An element of R is a unit iff it has nonzero constant term. Secondly, a monoid consisting of no invertible elements would have to lie in the ideal generated by (x), but this ideal contains no idempotents, so the identity of the monoid of no invertible elements we seek can't exist.
This demonstrates pretty simply that the answer to 1) is "no."
You can immediately generalize to say that any commutative local ring's multiplicative monoid does this, as long as you are satisfied with the cardinalities of the sets of invertible and noninvertible elements.
A: The answer to Question 1 is NO, and the answer to Question 2 is there are many interesting examples. Let G be a finite rank torsion-free Abelian group, let M be the set of isomorphism classes [A] of direct summands A of G, and define an operation + on M by [A] + [B] = [A $\oplus$ B].
Then M is a commutative monoid with only one invertible element $[0]$ which may have, depending on your choice of G, weird properties like
$[A] + [B] = [A] + [C]$ but $[B]\ne [C]$
$[A]+[A]=[B]+[B]$ but $[A]\ne [B]$.
