Idempotents in a finite ring (or finite semigroup)

hhh

Feb 8, 2016 10:48

@TobiasKildetoft I am trying to reformulate the task

"Let R be a finite, but not necessarily commutative ring, and let x\in R. Show that there exists k\in \mathbb N_{>0} such that x^k=x^{2k}$."

in terms of ideals so that I could solve it "in general" by ideal criterion.

Is this wrong "in general" approach here?

Tobias Kildetoft

@hhh What do you mean in terms of ideals? I don't see any reason to involve ideals here

r9m

@I'manartist no clue .. most of what I have seen about that particular series are on MSE itself .. and I haven't seen a 'no integrals' solution on mse

hhh

@TobiasKildetoft my algebra book has that "normal subgroups in group theory corresponds to ideals in ring theory"

Tobias Kildetoft

@hhh Sure, but why would that be at all relevant here?

I'm an artist

@r9m I see. What I saw used a lot of polylogarithms, crazy, ugly manipulations that can be all avoided.

Tobias Kildetoft

Feb 8, 2016 10:51

Also, that is a very vague version of "correspond"

hhh

@TobiasKildetoft Because R is ring so addition and multiplication are defined in it with identity element 1 and neutral element 0.

Tobias Kildetoft

@hhh Yes, I am aware of what a ring is

r9m

@I'manartist well I'll try to see if I can manage anything .. :-)

I'm an artist

@r9m Great! :-)

r9m

@I'manartist btw did you make any progress with that $\int_0^1 \log (1-x+x^2)\log (1+x+x^2)\,dx$ integral I asked you about couple of months back? :)

I'm an artist

Feb 8, 2016 10:56

@r9m I didn't continue working on that. It seemed to me it has some closed form with special values (unknown) of the polygamma. Maybe I'm wrong.

r9m

@I'manartist okay .. I have been dead stuck :| could make any useful progress :(

I'm an artist

@r9m I didn't continue I felt it doesn't have a nice closed-form. If it has a nice closed form, that's amazing.

r9m

@I'manartist maybe .. the problem originally came from O. Furdui and I mailed him about it .. he said he posed it as an open problem :)

I'm an artist

@r9m Really? I didn't know that. Then maybe he doesn't have a solution.

r9m

@I'manartist that's what he told me ..

I'm an artist

Feb 8, 2016 11:00

@r9m I think it's true. Maybe he needed that result for another important problem.

hhh

@TobiasKildetoft For now I cannot think about any other approach to prove it in general. The variety V(x^k-x^{2k}), the ideal <x> with some k>0 -- perhaps, thinking.

Some polynomial ring criterion needed.

Tobias Kildetoft

@hhh Do you see that there there are some $n,m\geq 1$ with $x^m = x^{m+n}$?

r9m

@I'manartist probably .. well product of log integral has been baffling us for a long time with no known algorithm that covers all the cases :|

I'm an artist

@r9m I have some helpful stuff under research that could improve that, but I still have to work a lot on that.

@r9m Yeah, I know.

r9m

@I'manartist :D Awesome!!

hhh

Feb 8, 2016 11:03

@TobiasKildetoft sure

I'm an artist

@r9m :D

Tobias Kildetoft

@hhh Ok, can you then show that also $x^{k+n} = x^k$ for all $k\geq m$?

I'm an artist

@r9m I have to finish some research and do then some shopping. I'm back in a couple of hours.

r9m

@I'manartist 'kay :) bye ..

Balarka Sen

@r9m How is life?

hhh

Feb 8, 2016 11:17

@TobiasKildetoft yes

Tobias Kildetoft

@hhh Ok, so if $n\geq m$ what do you see about the square of $x^n$?

Cody

hey Rd. I'm slightly awake :)

Tobias Kildetoft

@BalarkaSen Hi

Balarka Sen

Hello!

@Cody iirc, you're Galactus, right?

Cody

Yep. that's me

Balarka Sen

Feb 8, 2016 11:19

Nice to see you here.

Cody

I have chatted much in this area. Nice to know it's here.

Balarka Sen

Interesting, I don't remember seeing you here.

Cody

I have NOT chatted much is what I meant to write. Sorry, but this is not a good time for me. I am getting ready for work and just got up. I may have to catch up with you all later. I was responding to r9m's 'hello'.

Balarka Sen

Ah, OK. Sure, have a good day.

Cody

I noticed several of you all are writing books, so to speak. Compiling problems. One on Euler sums is indeed a good idea.

You all have a good day and we will speak later this evening. OK

hhh

Feb 8, 2016 11:29

@TobiasKildetoft $x^{2n}\geq x^{k+n}=x^k \forall k\geq m, n\geq m$

Tobias Kildetoft

@hhh $\geq$?

Huy

Feb 8, 2016 11:48

@MikeMiller: do you have any idea how it makes sense to talk about the translation length of a parabolic isometry? if we take the infimum of $d(x, f(x))$, this will always be zero for parabolic isometries (but not realized), so what could be meant by it? fix some point and then compare? does such a comparison even make sense?

or maybe just use the Euclidean metric and then it makes sense?

hhh

@TobiasKildetoft $x^{2n}=x^{k+n}=x^k \forall k\geq m, n\geq m$

Tobias Kildetoft

@hhh well, with $k = n$

hhh

@TobiasKildetoft $x^{2k}=x^k \forall k\geq m$

Tobias Kildetoft

@hhh why?

hhh

@TobiasKildetoft
Consider $n,m\geq 1$ with $x^m=x^{m+n}$.
With $k\geq m$, $x^{k+n}=x^k \forall k\geq m$.
With $n\geq m$, $x^{2n}=x^{k+n}=x^k \forall k\geq m, n \geq m.$
With $k=n$,

$$x^{2k}=x^k \forall k\geq m.$$

Tobias Kildetoft

Feb 8, 2016 12:04

@hhh You can't say both $k=n$ and have a for all $k$ afterwards.

r9m

@BalarkaSen well I am alive :) How are you doing?

Balarka Sen

@r9m Good to hear. Well, good. Tomorrow's the last of madhyamik.

Hi @Huy.

r9m

@BalarkaSen oh! Awesome! :D Best of luck!!

hhh

@TobiasKildetoft With $k=n$,

$$x^{2k}=x^k \exist k\geq m.$$

Balarka Sen

@r9m Thanks.

Tobias Kildetoft

Feb 8, 2016 12:07

@hhh No. We are assuming that $n\geq m$.

Martin Sleziak

I see that you started with rings, but is this not just slight variation of the proof that every finite semigroup has idempotent. math.stackexchange.com/questions/353028/…

Or maybe I should have linked to the proof at ProofWiki, which seems similar to what you were doing above: proofwiki.org/wiki/…

hhh

@TobiasKildetoft With $k=n$,

$$x^{2k}=x^k \forall n\geq m.$$

Tobias Kildetoft

@hhh No, $n$ is fixed

You really need to brush up on how quantifiers work for this

hhh

@TobiasKildetoft With $k=n$,

$$x^{2k}=x^k$$

Tobias Kildetoft

@hhh Yes, but why call it $k$ then?

hhh

Feb 8, 2016 12:14

@TobiasKildetoft With $k=n$,

$$x^{2n}=x^n$$

Tobias Kildetoft

@hhh Why are you introducing a $k$ at all?

hhh

@TobiasKildetoft ?

Tobias Kildetoft

@hhh There is no reason to mention a $k$ at all here.

hhh

@TobiasKildetoft

Consider $n,m\geq 1$ with $x^m=x^{m+n}$.
With $k\geq m$, $\forall k\geq m \quad x^{k+n}=x^k$.
With $n\geq m$, $\forall n \geq m \exists k\geq m \quad x^{2n}=x^{k+n}=x^k.$
With $k=n$, $\forall n \geq m \exists k\geq m \exists k=n \quad x^{2n}=x^n$ so

$$\forall n \geq m \exists k\geq m \exists k=n \quad x^{2n}=x^n$$

which is equivalent to

$$\exists k\in\mathbb N_{>0} \quad x^k=x^{2k}$$

with a finite ring R and $x\in R$.

Tobias Kildetoft

@hhh Well, you now have it when $n\geq m$. So you need to show that this is sufficient

Martin Sleziak

Feb 8, 2016 12:34

@hhh What is it that you are trying to prove? Existence of idempotent in a finite ring?

Tobias Kildetoft

@MartinSleziak That any element in a finite ring has a power which is idempotent

Martin Sleziak

I see. So this follows from the fact that any finite semigroup has idempotent. (See the links I posted in my previous message.)

Since the set $\{x^k; k=1,2,3,\dots\}$ forms a semigroup with multiplication.

Tobias Kildetoft

@MartinSleziak Right (and in fact, one reduces to this case inside the semigroup for that proof anyway)

Martin Sleziak

The proof which does not need any prerequisites is the same as hhh is trying to write down: Show that there is a cycle and play with it until you find an idempotent.

hhh

$\forall n \geq m \exists k\geq m \exists k=n \quad x^{2n}=x^n$ is equivalent to

$$\forall k \geq m \quad x^{2k}=x^k,$$

I cannot yet see where the sufficiency, @TobiasKildetoft.

Martin Sleziak

Feb 8, 2016 12:40

I recall that I found the way this proof is written down in the book by Hindman and Strauss quite nice. And also some answers in the post on main contain quite clever arguments.

Tobias Kildetoft

@hhh There is no for all $n$ here. We have a fixed pair $(n,m)$ with a certain property (namely that $x^{m+n} = x^m$), and we need to show that we can also find such a pair with the added property that $n\geq m$.

Martin Sleziak

Now, I see where Tobias is going with this. That seems to be rather natural continuation of the direction in which hhh started the proof.

Tobias Kildetoft

@MartinSleziak I was also the one to start the direction actually

Martin Sleziak

I see. I was too lazy to go look too far back in the transcript.

hhh

$\forall n \geq m \exists k\geq m \exists k=n \quad x^{2n}=x^n$ is equivalent to $x^{2k}=x^k$ in some finite ring R $x\in R$ where the sufficiency has not been proved yet.

Tobias Kildetoft

Feb 8, 2016 12:49

@hhh Stop putting quantifiers at all.

macurie

Is it correct to pronounce $a\ge b$ as "a is greater than b" and $a>b$ as "a is strictly greater than b"?

It's the first one I'm worried about

hhh

@MartinSleziak but the proofs are about semigroups, not about rings, tea pause :)

Martin Sleziak

Feb 8, 2016 13:06

@hhh Well, if $R$ is a ring then $(R,\cdot)$ is a semigroup.

So if you have some result which is true for any semigroup/finite semigroup; then the same result holds for $(R,\cdot)$ if $R$ is any ring/finite ring.

Idempotents in a finite ring (or finite semigroup)

$Mathematics$ Mathematics

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