Register to reply

Consider the principal ideal

by mnb96
Tags: ideal, principal
Share this thread:
Nov26-09, 02:04 PM
P: 626
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.
Phys.Org News Partner Science news on
Physical constant is constant even in strong gravitational fields
Montreal VR headset team turns to crowdfunding for Totem
Researchers study vital 'on/off switches' that control when bacteria turn deadly
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: 626
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?

Register to reply

Related Discussions
Semi-groups and monoids Linear & Abstract Algebra 4
More algebra...MONOIDS, etc Calculus & Beyond Homework 10
Couple questions involving monoids/isomorphisms Calculus & Beyond Homework 8
References on monoids and applications? General Math 0