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Consider the principal ideal 
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#1
Nov2609, 02:04 PM

P: 626

Hello,
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 AK[/itex], now I consider the principal ideal of [itex]K[/itex] generated by [itex]x[/itex], that is [itex]xK=\{xkk\in K\}[/itex]. The question is: if I consider another element [itex]x'[/itex] such that [itex]x'\in AK[/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. 


#2
Nov2609, 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] [tex]x'=y^2[/itex] It's trivial to verify [itex]x,x' \in AK = \{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. 


#3
Nov2609, 05:24 PM

P: 626

You are right. You easily found a counterexample.
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 deltafunction) 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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