Can Principal Ideals in Infinite Monoids Have Non-Empty Intersections?

[mnb96]

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.

 

[rasmhop]

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]<br /> x&amp;#039;=y^2[/itex]&lt;br /&gt; It&amp;#039;s trivial to verify x,x&amp;amp;#039; \in A-K = \{y,y^2\}, x&amp;amp;#039; \notin xK = \{y,y^4,y^5,\ldots\} and:&lt;br /&gt; xK \cap x&amp;amp;#039;K = \{y^5,y^6,y^7,\ldots\}&lt;br /&gt; &lt;br /&gt; I don&amp;#039;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.

 

[mnb96]

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&#039;K \neq 0 it means that there exists some gaussians g,g&#039;\in K such that f\ast g = f&#039; \ast g&#039;.

I haven't proved it yet, but intuitively it sounds strange that one could pick up an f&#039;\notin fK and get something equal to f \ast g by just convolving. But maybe I am wrong?