# The intersection of all ideals of a monoid

While reading about sandpile groups, I read the (rather important) assertion that the intersection of all ideals of a commutative monoid is a group , but was unable to find a proof.

Unless the monoid is already a group, the identity element of the monoid will clearly not be in this intersection (as a commutative monoid with no proper ideals is a group, and any ideal containing the identity is improper), and finding an element that acts as an identity on this intersection is the only difficult part of the proof: if we have such an element, then the intersection of the ideals is itself a monoid, but has no proper ideals, and so is a group.

Given the complexity of the identity elements of sandpile groups, I suspect that finding such an element could be difficult, however.

Does anyone know how to show the existence of an element acting as the identity on the intersection of all of the ideals in a commutative monoid?

• By chance, is the given monoid commutative? Oct 18, 2013 at 23:03

Even if $M$ is commutative, the intersection of all ideals might be empty. Take $(\mathbb{N}, +)$: the ideal generated by $n$ is the set $I_n = \{m \in \mathbb{N} \mid m \geqslant n \}$ and the intersection of all these sets is empty.

Now, if $M$ is a finite commutative monoid, then the intersection $I$ of all (nonempty) ideals is a commutative group. Indeed $I$ is nonempty since it contains the product of all the elements of $M$. Being a finite semigroup, it contains an idempotent $e$. Let $x \in I$. Then the ideal $xM$ is contained in $I$ and hence is equal to $I$. In particular $xM = eM = I$. Thus $x = ey$ and $e = xz$ for some $y, z \in M$, whence $ex = eey = ey = x$ and $x(ze) = (xz)e = ee = e$. Thus $I$ is a monoid with identity $e$ and since $ze \in I$, it is an inverse of $x$. Thus $I$ is a group.

The result is also true if $M$ is a compact topological monoid.

It is not true. There are (even finite) semigroups without proper ideals which are not groups (so called completely simple semigroups). If you join an extra identity to such a semigroup you get a monoid with one ideal which is not a group.

Addendum: Let $S$ is a semigroup (not necessary a monoid), $M$ is intersection of all its ideals. Suppose $M\ne\emptyset$. Then $M^2=M$ since $M^2$ is an ideal. Let $I$ is an ideal of $M$, $a\in I, x\in S$. Since $a=bc$ for some $b,c\in M$ then $xa=(xb)c\in I$. So $M$ is a simple semigroup. If it is commutative then it must be a group (Clifford, Preston, Algebraic Theory of Semigroups, the end of Section 1.1).

• Could you give an example of a monoid without proper ideals that is not a group? What I gather from here is that the only thing precluding such a monoid from being a group would be noncommutativity of the idempotents. Oct 18, 2013 at 19:12
• @Karl Kronenfeld: Such an example is the bicyclic monoid (en.wikipedia.org/wiki/Bicyclic_semigroup). Nevertheless thank you very much for your remark. Owing to you I found a mistake in my answer and corrected it. Oct 18, 2013 at 19:44
• The op probably means commutative. Oct 19, 2013 at 3:18
• I do in fact mean commutative. I will edit the question to reflect this. Oct 19, 2013 at 3:25