If $R$ is an integral domain then $R-\{0\}$ equipped with the original multiplication can be recognized as a commutative and cancellative monoid. The inversible elements form a subgroup $R^*$ and it is wellknown that each finite subgroup of this group is cyclic. My question is:
Can any monoid with these properties be recognized as $R-\{0\}$ where $R$ is some integral domain?
Maybe you know some easy counterexample. Thanks in advance.
Any commutative and cancellative monoid $M$ that is finite and non-cyclic, or finite and for which $|M|+1$ is not a prime power, provides a counterexample.
Take for example $M=(\Bbb{Z}/2\Bbb{Z})^2$ to be the Klein four-group. If $M$ is isomorphic to $R-\{0\}$ for some integral domain $R$, then $R$ is a finite integral domain and hence a field. But then $R^*=R-\{0\}$ is cyclic, a contradiction.
Or take for example $M=\Bbb{Z}/5\Bbb{Z}$. If $M$ is isomorphic to $R-\{0\}$ for some integral domain $R$, then $R$ is a finite integral domain and hence a field. But there is no field of six elements, a contradiction.