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Consider the principal ideal

by mnb96
Tags: ideal, principal
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Nov26-09, 02:04 PM
P: 625
I have an infinite monoid [itex]A[/itex] and a submonoid [itex]K[/itex].
let's assume I pick up an element [itex]x\in A-K[/itex],
now I consider the principal ideal of [itex]K[/itex] generated by [itex]x[/itex], that is [itex]xK=\{xk|k\in K\}[/itex].
The question is:
if I consider another element [itex]x'[/itex] such that [itex]x'\in A-K[/itex] and [itex]x'\notin xK[/itex], is it possible to prove that [itex]xK\cap x'K=0[/itex] ?

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

PS: I clicked too early and now I cannot change the title into something better.
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Nov26-09, 03:33 PM
P: 424
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 [itex]A[/itex] be the free monoid on a singleton {y} so [itex]A=\{1,y,y^2,\ldots\}[/itex]. Let,
[tex]K = \{1,y^3,y^4,y^5,\ldots\}[/tex]
[tex]x = y[/itex]
It's trivial to verify [itex]x,x' \in A-K = \{y,y^2\}[/itex], [itex]x' \notin xK = \{y,y^4,y^5,\ldots\}[/itex] and:
[tex]xK \cap x'K = \{y^5,y^6,y^7,\ldots\}[/tex]

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.
Nov26-09, 05:24 PM
P: 625
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 [itex]A[/itex] of functions [itex]f(x)[/itex] (plus the delta-function) with the operation of convolution [itex]\ast[/itex].
[itex](A,\ast)[/itex] should now be a monoid, and [itex]K[/itex] can be, for example, the submonoid of the gaussian distributions [itex]g(x)[/itex].
At this point if we assume that [itex]fK \cap f'K \neq 0[/itex] it means that there exists some gaussians [itex]g,g'\in K[/itex] such that [itex]f\ast g = f' \ast g'[/itex].

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

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