# Consider the principal ideal

Hello,
I have an infinite monoid $A$ and a submonoid $K$.
let's assume I pick up an element $x\in A-K$,
now I consider the principal ideal of $K$ generated by $x$, that is $xK=\{xk|k\in K\}$.
The question is:
if I consider another element $x'$ such that $x'\in A-K$ and $x'\notin xK$, is it possible to prove that $xK\cap x'K=0$ ?

If that statement is not generally true, is there an additional hypothesis that I could make to force $xK\cap x'K=0$ hold?

PS: I clicked too early and now I cannot change the title into something better.

I must admit that I have never heard of ideals in monoid theory, but just accepting your definition of xK I would say no.

Let $A$ be the free monoid on a singleton {y} so $A=\{1,y,y^2,\ldots\}$. Let,
$$K = \{1,y^3,y^4,y^5,\ldots\}$$
$$x = y[/itex] [tex]x'=y^2[/itex] It's trivial to verify $x,x' \in A-K = \{y,y^2\}$, $x' \notin xK = \{y,y^4,y^5,\ldots\}$ and: [tex]xK \cap x'K = \{y^5,y^6,y^7,\ldots\}$$

I don't immediately see an obvious property on A that would make it hold for arbitrary K except requiring A to be a group, or actually requiring exactly what you want.

You are right. You easily found a counter-example.
I will now focus my interest in finding a property that satisfies that.

I don't know if the following is a valid example, but it is an attempt.
I was thinking about the set $A$ of functions $f(x)$ (plus the delta-function) with the operation of convolution $\ast$.
$(A,\ast)$ should now be a monoid, and $K$ can be, for example, the submonoid of the gaussian distributions $g(x)$.
At this point if we assume that $fK \cap f'K \neq 0$ it means that there exists some gaussians $g,g'\in K$ such that $f\ast g = f' \ast g'$.

I haven't proved it yet, but intuitively it sounds strange that one could pick up an $f'\notin fK$ and get something equal to $f \ast g$ by just convolving. But maybe I am wrong?