Mathematics - 2017-12-05 (page 1 of 5)

Akiva Weinberger

00:01

> https://en.wiktionary.org/wiki/prætentious

lol

Ted Shifrin

hi DogAteMy, orbit

Julius

00:21

Hi. Any hints on how to find the asymptotic order of $P_K(1+K^{-\alpha})$ with $\alpha\in(0,2)$ if $P_n(x)$ denotes the Legendre polynomial? There are results when $n$ goes to infinity, but I couldn't find anything for my case, in which also $x>1$ rather than $x\in[-1,1]$.

user685252

00:46

winterbash2017.stackexchange.com

orbit-stabilizer

@TedShifrin hello ted

@Daminark ive got a good group theory question

Argon

01:35

Is there an easy way to show $|\text{Aut}(S_4 \oplus \mathbb Z_2)|=48$?

orbit-stabilizer

@Argon, what's the hard way?

Argon

@orbit-stabilizer The hard(est) way is to enumerate them all

orbit-stabilizer

@Argon ah yes

anon

@Argon write down a 2x2 matrix with homomorphisms in each entry

Argon

@anon How do you mean? Any homomorphism?

Ted Shifrin

01:45

heya anon :)

anon

not quite any

see the end of my answer here

hi ted

it's not terribly hard to prove the fact, and could be useful beyond just a single exercise

Argon

@anon Thanks, I'll take a look

anon

the rest of my answer covers H^2 which is different from HxK mind you

Simply Beautiful Art

Hi chat.

For anyone who's curious, the calculus group is officially over.

Ted Shifrin

May the stooopid idea of classes on MSE rest in peace.

Applause for those who gave it a try.

anon

01:49

also kind of reminiscent of goursat's lemma, which Tobias explains nicely here

Simply Beautiful Art

For anyone who's curious and insane, I wrote an explanation of SOAP, which makes really big numbers.

Ted Shifrin

@Kasmir: Hello, but it's way past your bedtime.

Simply Beautiful Art

Lol

Ah man.

I feel like I'm at the limit of the largest numbers I can write in a decently sized program. Spent so much time these past few weeks working on it.

At least I'm at the limit of what I'm comfortable with. That much is certain.

orbit-stabilizer

Why does a non-trivial homomorphism exist iff $q|p-1$? I know, given $H,K$, with orders $p,q$ respectively, and $q<p$, we have $Aut(H) \cong Z/(p-1)Z$. So for the homomorphism from $K $to$ Aut(H)$ to exist, q must divide $p-1$ since the image is a subgroup, but why are those sufficient conditions rather than just necessary conditions?

@SimplyBeautifulArt I think it would be useful to have a room going for people self-studying the same thing

Simply Beautiful Art

Ordinality?

Trying to understand extraordinarily large numbers.

orbit-stabilizer

02:04

Hah, not my cup of tea

Simply Beautiful Art

4

Q: How to disprove this fallacy that derivatives of $x^2$ and $x+x+x+\cdots\ (x\text{ times})$ are not same.

Possible Duplicate: Where is the flaw in this argument of a proof that 1=2? (Derivative of repeated addition) $$x^2=\underbrace{{x+x+x+\cdots+x}}_{x \text{ times}}$$ $$\therefore \frac {dy}{ dx}\\ =\frac{d}{dx} \left( \underbrace{{x+x+x+\cdots+x}}_{x \text{ times}} \right) \\ = \u...

^ More ways to disprove your classmates.

Ted Shifrin

Oy vey.

Simply Beautiful Art

@orbit-stabilizer Also, check my profile.

Needs an update though

orbit-stabilizer

@SimplyBeautifulArt I did, what was I supposed to find?

Simply Beautiful Art

@orbit-stabilizer I had thought your first comment to me was for a room for studying very large numbers.

Hence link to one chatroom as well as an off-stackexchange chat place for more.

orbit-stabilizer

02:06

"I think it would be useful to have a room going for people self-studying the same thing"

I just meant that in general. Like I would be interested in a room discussing Aluffi's Algebra

Simply Beautiful Art

Oh haha

I wonder

What are the best quotes of the year from this chatroom?

Ted Shifrin

Spend days sorting through the star board?

orbit-stabilizer

'my dad sees me reading a book on Galois Theory "Remember, it's just a theory." '

Simply Beautiful Art

Lol, yeah I'm gonna do that I guess

orbit-stabilizer

$\Bbb{ON~SOME~PAPERS~ON~SOME~MEASURE~THEORY~TEXTBOOKS \\ AND~THEIR~USE~BY~SOME~PROFESSORS~IN~GRADUATE-LEVEL~COURSES}$

Ted Shifrin

02:09

stop that, orbit

Simply Beautiful Art

yes plz no gigantic chatjax text.

orbit-stabilizer

okay

Simply Beautiful Art

@orbit-stabilizer If you want to be showing some interesting starboard stuff, just do the links.

orbit-stabilizer

I dont know how :/

Simply Beautiful Art

When you've got a chat message you wanna link, such as this one, click the gray arrow on the right
<-----
and copy the permalink.

Argon

02:13

@anon I can understand until about the commutators. Why does that guarantee the automorphism is the matrix?

anon

@Argon check that the matrix applied to an element equals the automorphism applied to an element

racks brain to remember answer written 2 years ago

Argon

Hehe

Isn't it that matrix by definition?

anon

assume $\phi$ starts out as an arbitrary automorphism, then associate a matrix to it, then check the matrix and $\phi$ evaluated at a tuple produce the same result

Simply Beautiful Art

._. so much drama

Answer Lee

Can I ask a question? Can I say a metric space (X,d) metrizable? Thank you!!

anon

02:18

yes

Simply Beautiful Art

One does not simply ask to ask a question and then ask away anyways before anyone has the chance to respond.

Ted Shifrin

hmm .. "metrizable" = there is a metric. Hmm ...

@SimplyBeautifulArt One shouldn't even ask to ask.

Simply Beautiful Art

Yeah, and I think you were the one that said that policy to me first :P

Answer Lee

But why my professor said it is unclear. Cry.

Ted Shifrin

Professors suck.

Simply Beautiful Art

02:22

Also, why are most of the highly starred chat messages about flags needing to stop? x.x

Ted Shifrin

because there have been some infiltrators doing crazy flags in here

Simply Beautiful Art

:(

orbit-stabilizer

:41545189

Oops didn't work

Ted Shifrin

nods

Eric Silva

sup duds

orbit-stabilizer

02:28

When you've got a chat message you wanna link, such as this one, click the gray arrow on the right
<-----
and copy the permalink.

Gahh

Simply Beautiful Art

lol

@DHMO Man, I do miss you bud. I really do lol

orbit-stabilizer

I click the grey arrow. I click permalink. What happens next?

Ted Shifrin

heya Eric

orbit-stabilizer

16 mins ago, by Simply Beautiful Art

When you've got a chat message you wanna link, such as this one, click the gray arrow on the right
<-----
and copy the permalink.

I did !

it

Ted Shifrin

the permalink stuff here is highly confuzling.

orbit-stabilizer

02:30

Now that's out of the way. How many elements of order 25 are there in $C_{5^{100}}$. How does one go about answering this?

Eric Silva

@Ted I managed to acquire a copy of Bryant's EDS

Simply Beautiful Art

On chrome:

Eric Silva

will thumb through over break probably

Simply Beautiful Art

Then you can copy/paste a link to a chat message.

orbit-stabilizer

Got it, thanks @Simply !

Ted Shifrin

02:31

Well, the real tôme is the Bryant/Chern/Griffiths/Goldschmidt EDS book, Eric.

Eric Silva

ya that one

i didnt wanna write out all the names

Ted Shifrin

ohhh ... great.

it takes more than thumbing

Eric Silva

yeah i figured

Ted Shifrin

it's one of the few books I didn't discard when I moved

but if you want to work through things, I'll be glad to participate

I'm off to cook dinner

Eric Silva

mmk

have a nice meal

Ted Shifrin

02:33

I'll do that

Simply Beautiful Art

Starred messages I personally noticed and found interesting:

["Math is hard, guys. Don't do it if you don't want to struggle to learn."](https://chat.stackexchange.com/transcript/36?m=39667233#39667233)

["A **tan** is good **cos** it's not a **sin**."](https://chat.stackexchange.com/transcript/36?m=39535988#39535988)

["We definitely need to encourage more women re math."](https://chat.stackexchange.com/transcript/36?m=38508407#38508407)

["mathematics is not about numbers, nor is painting about ink, nor is music about notes."](https://chat.stackexchange.com/transcript/36?m=35840237#35840…

Oh, thanks for not working too.

Anyways, good night all

orbit-stabilizer

02:50

night!

Argon

03:15

@anon What makes $\alpha(H)$ and $\beta(H)$ normal?

(Sorry for the trivial questions, this is new to me)

Daminark

How can one show that the symmetries of the cube gives $S_4$ without doing the business about actions on opposite vertices?

Like I'm having an extremely bad time processing that picture, is there a sort of combinatorial way to execute this?

anon

for your purposes you want to write $\alpha(H)$ and $\beta(K)$. because they are images of normal subgroups under automorphism / surjection

Argon

Oh of course thanks

anon

@Daminark first figure out all the rotations of S_4 then see what the cycle types are of the corresponding actions on space diagonals

which doesn't avoid the diagonals like you want, but the diagonals are the best way to see it

like, the 90 degree face rotation has cycle type (abcd)

the 180 degree face rotations have cycle type (ab)(cd)

now do rotations around axes through vertices and edges

after explaining cycle types and drawing a nice picture I gave this exercise to a nursing student and she succeeded writing them all down

Daminark

Wait I think we can embed the tetrahedron into the cube, and I've been convinced that rotations of that are A_4

anon

03:21

you're tackling the problem like an air bender, you need to be an earth bender and just do it :P

Daminark

And I'm completely incompetent in visual stuff

So I'm trying to dodge that

anon

invest in some piper cleaners

Eric Silva

oh shit that sounds like a fun way to visualize stuff

anon

I want a dry erase sphere

and a knobbly wobbly

that reminds me of a problem that came to mind thinking about S4 acting on the cube. Can a proper subgroup of a finite group contain a representative of every conjugacy class?

orbit-stabilizer

@anon that's a good question

anon

03:27

googled and MO says no apparently

orbit-stabilizer

Yes, makes sense

wait

Yes, it does.

Eric Silva

my guess is counting

anon

yeah

orbit-stabilizer

Suppose there existed such a group, then let the elements of that group act on themselves by conjugation.

Mhmm.... no.

Eric Silva

cause if you had such a subgroup $H \leq G$ youd have $\bigcup_{g \in G} gHg^{-1}$

Argon

03:29

I'm getting confused around the section regarding "free products, semidirect products, and wreath products." It appears to me that in my case only diagonal matrices can be the automorphisms (?). Can I just then enumerate the number of automorphisms of $H$ and $K$ and find the number of matrices that can exist?

Daminark

Oh yeah for a proper subgroup H that union can never be everything

anon

@Argon I give the general form of matrices in ${\rm Aut}(H\times K)$, and whether or not they are all diagonal depends on if there are any group homomorphisms $H\to Z(K)$ or $K\to Z(H)$

if there are such homomorphisms, then the corresponding nondiagonal matrices still define automorphisms of $H\times K$

Eric Silva

yeah counting gives you an inequality $|G| - 1 \leq (|H| - 1)[G:H]$

which implies $|G| = |H|$

if i ever taught people group theory id assign this as a prob

i rate it 7.5/10

Daminark

It's p good, I know Frank assigned his class the stuff but Jack didn't

anon

now how to generalize to the lie setting appropriately

Argon

03:33

Urg this stuff makes no sense

orbit-stabilizer

^ I feel that

anon

@Argon given $\phi$, define $\alpha,\beta,\gamma,\delta$ appropriately. prove $\alpha,\beta$ must be invertible, and $\beta,\gamma$ land in the centers of their codomains. prove $\phi$ evaluated at $(h,k)$ matches the result of evaluating the matrix at $(h,k)$. conversely, prove every such matrix defines an automorphism.

consider it an exercise to try on your own instead of someone else's writing to read.

(after defining what $\alpha,\beta,\gamma.\delta$ are anyway)

Eric Silva

@anon what's a good place to read about rep theory of compact groups

I've got a final coming up and Farb never gave us references for the compact groups stuff

anon

I wouldn't say I'm a good place to ask, as I just read pdfs and play fast and loose with convergence of integrals/sums

Eric Silva

hmmmk

Daminark

03:53

Oh here is a fun problem

Prove the average degree of a graph is $< \sqrt{2m}$

anon

what's m?

Daminark

Number of edges

anon

or is figuring out m part of the exercise

k

Eric Silva

I feel like average degree should be 2m/n

where n is number of vertices

anon

it is

orbit-stabilizer

04:07

anon r u in hs too?

anon

ha no

so I guess $2m/n<\sqrt{2m}\iff 2m<n^2$ means we look at the latter now

or $m<n^2/2$, but that follows from $m\le\binom{n}{2}$

so, done

Eric Silva

im satisfied

im procrastinating work by doing less important work

what has become of my life

anon

a rooted tree of procrastinations within procrastinations

Daminark

@anon okay that's another way to do it

What I had in mind was using the adjacency matrix, which is why I liked it so much

Eric Silva

easyboi counting mode

Daminark

04:14

The sum of the entries of the adjacency matrix is gonna be $2m$, but it has to be less than $n^2$

Because this is a 0-1 matrix, and the diagonal has to consist of zeroes

Eric Silva

i give it a 6/10

Daminark

no u

Eric Silva

6/10 aint bad

Daminark

Lol perhaps

Eric Silva

Ds get degrees

Daminark

04:17

But yeah I dunno why but I was just cracking up so hard at this problem, like I put it in my "favorite problems ever" list

More for being cute than being hard, but still

Eric Silva

i dont get excited by counting graph things really

Daminark

Lol yeah, I've gotten the vibe that we have... pretty different taste

Eric Silva

i just kinda get bored doing it

i get excited if it shows up in connection with something else

like turning something into combinatorics gets me intrigued

orbit-stabilizer

I've got my group theory exam tmr morning...

gah

Daminark

04:32

@orbit good luck

@Eric yeah I haven't thought about connections with other things so much

I just sorta found that the stuff in algebra that I've enjoyed most was the Sylow counting, and mostly because I just liked the exercises themselves

orbit-stabilizer

@Daminark Why is this true?

Why must a group of order 182 have a subgroup of order 91?

Daminark

Okay so if you have a group of order 182

You know by Sylow that you must have a normal subgroup of order 7, right?

orbit-stabilizer

yah

Eric Silva

i think about most math in terms of connections with other things

Daminark

Call that $K$. Now you also know by Sylow that you have /some/ group of order 13, which may not be normal, yeah?

orbit-stabilizer

04:35

Yeah, that's true.

Either 1 of them, or 14

Daminark

Call that $H$. Then because $K$ is normal, you have $HK \le G$

orbit-stabilizer

Okay. I don't see why it has to be a subgroup, lemme prove it

gimmie a sec

Daminark

Can you prove that if $H$ and $K$ are subgroups of $G$, that $HK \le G \iff HK = KH$?

orbit-stabilizer

Oh boy this does not bode well.

Gimmie a few more seconds

One direction is easy. Suppose $HK = KH$. Then $HK \leq G$ because $e \in HK$ obvs. And since $hk = k'h'$, we have that $HK$ is closed under multiplication. And ofc inverses exist because $(hk)^-1 = k^{-1}h^{-1}$.

Daminark

Yup

That direction suffices for this problem

Because $K$ is normal so ofc $HK = KH$

orbit-stabilizer

04:50

Okay, how do you do the other direction?

Daminark

Well, assume $HK \le G$

Then for $hk \in HK$, you know $k^{-1}h^{-1} \in HK$

So $k^{-1}h^{-1} = h'k'$ for some $h'\in H$ and $k'\in K$

But then $hk = k'^{-1}h'^{-1} \in KH$

orbit-stabilizer

Ah

Daminark

So that does it. So since in our above question, $K$ was normal by virtue of $n_7 = 1$, we know that $HK \le G$

orbit-stabilizer

Okay, so $HK \leq G$. Let's continue.

Daminark

Well, we know that the size of $HK$ has to be $7\times 13 = 91$

(This is because $H\cap K = \{e\}$)

orbit-stabilizer

04:53

Yep.

Daminark

So that's why $G$ has a subgroup of order 91, meaning index 2

So that's normal

BUT

If you have a group of order 91, just abstractly, let's look at it

orbit-stabilizer

Ah, and we don't have to consider any semidirect product to find what group of order 91 there is because it's cyclic.

Daminark

We know $n_7 = 1$ and $n_{13} = 1$ easily

So then we have a normal, cyclic subgroup of order 91

orbit-stabilizer

Wait.

Right. Got it.

Daminark

Here is where you have to use the semidirect product

Let's call $HK = N$

orbit-stabilizer

04:55

You mean G?

$HK = G$

Daminark

No... $G$ is the group of order 182, $HK$ is the subgroup of order 91

orbit-stabilizer

Oh I see, I thought we were relabelling K to be the group of order 91. Sorry

Daminark

Oh no it's the same as before

orbit-stabilizer

Okay, let's call $HK = N$.

Daminark

So yeah we now know that there is some subgroup $T$ of order 2

orbit-stabilizer

04:57

By sylow right

Daminark

Yeah

We know that $TN = G$

orbit-stabilizer

And $N \cap T = \{e\}$

Daminark

So this is the internal semidirect product of $T$ and $N$

orbit-stabilizer

And $|NT| = |N||T| =|G|$ which means $NT = G$.

Right.

Huh, that NT looks off

Daminark

This means that $G$ is isomorphic to some external semidirect product of $N = C_2$ and $T = C_{91}$

orbit-stabilizer

04:58

Yes.

So now we need to classify the automorphisms?

Daminark

Yeah, first we find all the automorphisms, and then we see which ones give isomorphic groups

Or...

We want to find all homomorphisms from $C_2$ to $\text{Aut}(C_{91}) \cong C_6\times C_{12}$

orbit-stabilizer

Why can't we look at automorphisms from C_91 to C_2?

Daminark

$C_{91}$ is the normal subgroup

orbit-stabilizer

Right. my bad.

Daminark

So you have to take a map from $C_2$ to the automorphism group of $C_{91}$

orbit-stabilizer

05:01

Now, why is $Aut(C_{91}) \cong C_6 \times C_{12}$.

Daminark

But yeah so you can either take the trivial automorphism, in which case the group is $C_{182}$, or you can map to a subgroup of order 2 of $C_6\times C_{12}$

Well, you know $C_{91}\cong C_7\times C_{13}$

orbit-stabilizer

Yeap

Daminark

And we know there's no homomorphism from $C_7$ to $C_{13}$

orbit-stabilizer

You have 6 degrees of freedom in the first one and 12 in the second and so it works?

Daminark

So $\text{Aut}(C_7\times C_{13}) \cong \text{Aut}(C_7) \times \text{Aut}(C_{13})$

orbit-stabilizer

05:03

Oh I see. We disregard the trivial homomorphism?

Daminark

Yeah

I mean the other way to do it is to just find $C_{91}^{\times}$

orbit-stabilizer

Right, which is just $\phi(91)$.

Daminark

Well, the order is $\phi(91)$

orbit-stabilizer

Ah yes.

@Daminark Okay so we want to a homomorphism from C_2 to C_6 x C_12

Daminark

But it's a general fact that if $G$ and $H$ have coprime orders, then $\text{Aut}(G\times H) \cong \text{Aut}(G) \times \text{Aut}(H)$, which is easy to do when you note you don't have homomorphisms between them, so I'll move on

orbit-stabilizer

05:07

That's good to know

Daminark

So either the homomorphism is trivial or it sticks you in a subgroup of order 2, and I think there are just 3 of those in $C_6\times C_{12}$

$(3,0)$, $(0,6)$ and $(3,6)$ should be it

So now we have at most 4 groups of order 182, and we can find which are isomorphic

orbit-stabilizer

How did you come up with the subgroups of order 2 in C_6 x C_12?

Daminark

A subgroup of order 2 is an element of order 2

But you have to be order dividing 2 in each coordinate

So that only gives 3 options

orbit-stabilizer

Oh, that's right

So, 4 homomorphisms. 1 being trivial.

Daminark

So now I recommend just bashing them out to see which are isomorphic. You'll get a $D_{182}$ in there, wouldn't surprise me if there weren't any others tbh

Anyway, I'm dead tired and so imma slep

orbit-stabilizer

05:11

I see, and by bashing them out, you mean computing them by applying the autmorphism?

Alright, thank you! Goodnight!

This was extremely helpful!

Daminark

Yeah, like see how the automorphism acts on $C_{91}$ and find if any of them do the same thing. And yeah no problem, good night!

orbit-stabilizer

lol, welcome back

Daminark

Lol I came to log out from my phone :P

See you!

Leaky Nun

08:42

$\Bbb Q(x,y) \cong \Bbb Q(x+y,x-y)$?

Tobias Kildetoft

@LeakyNun Yes

Leaky Nun

as fields?

long time no see @TobiasKildetoft

Tobias Kildetoft

@LeakyNun Yes, as fields. In fact, one can argue that they are equal

Leaky Nun

indeed, one can

$\overline{\Bbb C(x)} \cong \Bbb C$?

Secret

09:03

So to continue where I left off yesterday...

First to recap the definition of a measure:

Let $\mu : X \to \Bbb{R} \cup \{\pm \infty\}$ be a measure. Then

1. $\mu (\varnothing) = 0$

2. $\mu (X) \geq 0$

3. $\mu \left( \bigcup_{i \in I} \mu (X_i)\right) \leq \sum_{i \in I} \mu (X_i)$

Alessandro Codenotti

You need $I$ to be countable and the $X_i$ taken from some fixed $\sigma$-algebra of subsets of $X$ (which is why I find it more intuitive to begin with an outer measure and construct a measure from it)

Secret

09:18

is the trouble of dealing with an uncountable sum the only reason why we cannot define a measure with uncountable subadditivity, or is there more fundamental reasons to that?

Alessandro Codenotti

If you want points to have measure 0 and some set to have positive measure you can't have uncountable additivity (anyway an uncountable sum can be finite iff all but countably many factors are 0)

Leaky Nun

@AlessandroCodenotti is it false when $I$ is uncountable?

wait, yes it is

Secret

I see, since all elements in a set are points and points have zero measure thus allowing uncountable additivity will result in basically a sum of zeros which is always zero in the end, thus making all sets have measure zero

ok, now to revise on outer measure...

(also typo in the very first line, I should be saying the positive extended real ray, otherwise I will have negative values in the measure)

Alessandro Codenotti

Points don't have to have measure zero, but it's nice when they do

Leaky Nun

@Secret for example here's an uncountably subadditive measure: enumerate the rationals $f:\Bbb N \to \Bbb Q$. Define $\mu(A) = \displaystyle \sum_{n=0}^\infty \begin{cases} 2^{-n} & f(n) \in A \\ 0 & f(n) \notin A \end{cases}$

it's even defined everywhere

it fails translation-invariance

Secret

09:31

@LeakyNun hmm..., since $f$ is an enumeration, does it mean it is well ordering the rationals and thus under this measure, $\mu (A) = \frac{1}{2^n}$ if only the nth rational is in $A$?

Leaky Nun

yes, and it does not follow

Secret

I mean, if only the nth rational (and no other rationals) are in A, then by that formula, $\mu (A)$ should only have $\frac{1}{2^n}$ contributed to it

---
Outer measure: Let $m : 2^X \to \Bbb{R}_{\geq 0} \cup \{\infty\}$ be an outer measure. Then:
1. $m (\varnothing) = 0$
2. $A \subseteq B \implies m(A) \leq m(B)$
3. $m\left( \bigcup_{i=1}^{\aleph_0}A_i \leq \sum_{i=1}^{\aleph_0}m(A_i)\right)$

Given some $E \subseteq X$. $E$ is measurable under $m$ if $\forall A \in 2^X$:

$m(A) = m(E \cap A) + m(E^c \cap A)$

Therefore the Lebesgue outer measure is the infimum of the covers of $E$ formed by closed intervals

Let $\lambda^*$ be the Lebesgue outer measure.
Now to figure out how to prove $\lambda^*[a,b] = b-a$

Mary Star

10:32

Hello!!

A patient is sitting in the waiting room of a doctor's office. We assume that its waiting time in minutes is exponentially distributed with parameter $\lambda = 0.2$. Within what time will the patient be treated with probability $0.9$?
The patient waited $5$ minutes without being called. How long does he have to wait with probability $0.9$?

I have done the following:
$$P(X\leq x)=1-e^{-\lambda x}\Rightarrow 0.9=1-e^{-0.2x} \Rightarrow e^{-0.2x} =0.1 \Rightarrow x \approx 11.513$$
Within the first $11.5$ minutes the patient will be treated with probability $0.9$.

mick

12:39

0

Q: Abelian monoids and integer polynomial composition?

Consider distinct integer polynomials of distinct degree. Let $*$ denote composition. Consider when $a,b,c,d$ forms an abelian monoid. Many questions arise naturally. For instance 1) can we classify them ? 2) Is every finitely generated abelian monoid a submonoid of a nonfinitely generated a...

calculus polynomials commutative-algebra function-composition monoid

Any ideas ?

Tobias Kildetoft

@mick Most of that seems to not make any sense to me

mick

Why not ??

Tobias Kildetoft

what do you mean by "Consider when a,b,c,d forms an abelian monoid"?

mick

abc's are polynomial , They form abelian monoid by composition

Tobias Kildetoft

but no set of four integer polynomials will form a monoid under composition.

at least not when they have distinct degrees

I think there is only one way to have four polynomials form a monoid under composition, and that is by choosing $x$, $-x$, $-x+1$ and $-x -1$

Ordinality?

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$Mathematics$ Mathematics