groups-rings-fields

[glad osprey]

for finite extensions, normal and splitting is the same

[frigid epoch]

Is the galois group of Q-bar over Q known

[wraith cargo]

I mean we understand a lot of it's structure

idk what exactly you mean by known

[frigid epoch]

Can it be decomposed as some product or free product of finite groups for example, and if so, which finite groups

Does it perhaps admit simplicial structure; are there rules for writing down how it's generated

[rocky cloak]

Every normal extension is the splitting field of some family of polynomials. If by splitting field you mean of a single polynomial, then any infinite normal extension will do

[rocky cloak]

I guess this is basically the field of algebraic number theory

[thorn jay]

how so? I can imagine a little but not familiar with algebraic number theory at all

[frigid epoch]

Hmmm

Z is not a subgroup of Gal(Q), right?

[rocky cloak]

Possibly helpful discussion
https://mathoverflow.net/q/2791/157483

[rocky cloak]

There's definitely elements of infinite order in Gal(Q-bar / Q)

[glad osprey]

I see, so we can consider Q-bar as the splitting field of the set of all polynomials?

[frigid epoch]

Oh huh

[rocky cloak]

Yes that's right. Any normal extension is the splitting field of the set of polynomials that split in it

[glad osprey]

In char 0 every algebraic extension is primitive, so any splitting field of finite degree is a splitting field of a single polynomial, right?

[rocky cloak]

Any finite algebraic extension is primitive yes

[glad osprey]

Thanks

[rocky cloak]

More generally, any finite seperable extension is primitive.

And a field is called perfect if any finite extension is seperable.

Any field of characteristic 0 is perfect, and any finite field is perfect

[glad osprey]

I see I don't believe in the existence of infinite fields of positive characteristic, so what you're saying is that every field is perfect?

[south patrol]

Bruh

Idk I find this amusing as someone who mostly works with positive characteristic fields

[glad osprey]

Infinite fields of positive characteristic is a scam invented by Big Algebra to sell you irreducible non-separable polynomials

[frigid epoch]

Can you do p-adic stuff in char q

It's past 11pm, I shouldn't be asking potentially difficult questions in fields I'm not familiar with

[thorn jay]

It'd be like someone telling me they dont believe in operations with an arity higher than three

[south patrol]

I don't rly know what this would mean

[tardy hedge]

Whats up Fellow Mathletes 👋

[rocky cloak]

The only arities I believe in are 0, 1, 2 and n

What is this "three" you speak of?

[thorn jay]

the arity found in the Mal'cev condition for permutable congruences

[chilly ocean]

I don’t believe in 3-categories

1? Fine. 2? Fine. $\infty$? Sure.

[cloud walrus]

NAT Enthusiast

[south patrol]

Actually a prof I know is specifically working w 3 categories aha

I believe it is basically cause you often study n-categories in conformal field theory in connection to n-manifolds

So yeah then specialise that

But the n here is more (oo, n)

[tardy hedge]

What is an n-category?

[minor fulcrum]

there are a lot of definitions

the better question is: what is an \infty-category

[tawdry venture]

Are left R-modules closed under scaling by elements of the ring?

Is that what this is saying?

[ripe crest]

Are you familiar with group actions?

Where the action is a homomorphism between a group G and the automorphism group of the object that G is acting on

The basic example is that of G-sets, where a group acts on a set

In the case of R-modules we have a ring acting on an abelian group. So there is more structure here, but the principle or idea of an action remains the same

[tawdry venture]

iirc my professor said something about modules are abelian groups in disguised or something

or abelian groups are modules in disguise

[ripe crest]

Yes abelian groups are Z-modules

See if you can see why

[tawdry venture]

I mean they're nice and they commute

Z-modules refer to vectors with Z entries right?

[ripe crest]

Vectors are elements of a vector space, Z-modules are not vector spaces because Z is not a field

[tawdry venture]

oh that's a problem (no good)

[ripe crest]

Sure. But let's walk through it. What is the actual action here?

[tawdry venture]

multiplication (or scaling) by elements of the ring

I'm assuming that's what they mean by R x M to M

Your input is R*M and output is M

[ripe crest]

Right, and if I have the group {e, a, b}, what would be an alternative way to express a*a*a*b*b using the language of a Z-module?

[tawdry venture]

any combination would work right? Since Z-module is an abelian group, I can commute the multiplication of the elements

so b*b*a*a*a would be one way

[ripe crest]

Yeah, but how could we express this using the language of modules?

[tawdry venture]

doesn't matter if I put the scalar on the left or right, it yields the same result?

[ripe crest]

Yeah

[tawdry venture]

wait so that means in Z-module

the left and right modules are equivalent

oh duh because Z is a commutative ring

[tawdry venture]

Are you implying that if my ring happens to be a field, then my modules are vector spaces?

I'm thinking contrapositive here

[ripe crest]

Yes, that is what a vector space is

[tawdry venture]

If Z is a field, then Z-modules are vector spaces

[ripe crest]

Modules are generalized vector spaces

[ripe crest]

Here, what I was pointing towards was 3a*2b

[tawdry venture]

ohh lol

[ripe crest]

We have the scalars from Z

[tawdry venture]

oh cause the vector entires would be 3a, 2b

[ripe crest]

And the action is just multiplication

[ripe crest]

Well, if this is a vector space, then a and b are vectors

So 3a and 2b are also vectors, just scaled

And 3a*2b is also a vector

(I should have used + instead of *, my bad)

[tawdry venture]

oh ok

Thank you!

I think I understand modules a little better now

Also in general

The ring that we are working with in modules

Are we taking elements from that ring to be our scalars?

[ripe crest]

That's right

[tawdry venture]

Sounds like it when you talk about Z-modules

ah ok

[ripe crest]

Try viewing a vector space, like R^2 as a field action. Where the scalars are from R, which is a field, and the vectors are from (R^2, +), an abelian group

[tawdry venture]

When we are taking scalars from R (I'm assuming you mean set of real numbers here), we don't take 0 right? Since 0 doesn't have an inverse

[ripe crest]

That might help in understanding, because all k-vector spaces are k-modules

[tawdry venture]

oh wait do you mean the rotational group?

[ripe crest]

We do take 0

[tawdry venture]

R^2?

[ripe crest]

The additive identity does not have a multiplicative inverse in a field

[tawdry venture]

If we're talking the set of real numbers and being a field, shouldn't we throw away 0?

or I'm confusing myself

[ripe crest]

No, fields have both 0 and 1

And 0 in a field does not have a multiplicative inverse

[tawdry venture]

oh im confusing about 0 being a unit

[ripe crest]

R (the set of real numbers), with addition and multiplication, is an example of a field

[tawdry venture]

I agree there

[ripe crest]

By R^2 I mean the Cartesian plane, analogous to R meaning the number line

[tawdry venture]

When you are saying vectors from (R^2, +), are you talking about vectors with 2 entries?

[ripe crest]

That's right

[tawdry venture]

oh ok

[ripe crest]

<1, 2> + <5, 4> = <6, 6>

Just basic stuff

[tawdry venture]

dang fields are nice

[ripe crest]

Yeah

[tawdry venture]

ok now I understand a little more about modules, I'm going to get back to my homework. Thank you again! @ripe crest

[ripe crest]

Take care!

no problem

[tawdry venture]

This is my work

I'm not sure if I'm doing this correctly. I can't help but feel I'm missing something. Do I need to check for linear independence?

[tough raven]

The idea is right but your calculations are off.

Firstly, each entry is not a multiple of k (although I think that's just a typo). Secondly, when you write ai = a + nk, the n has to depend on i, so it should be ai = a + ni k for some a, n_1, ..., n_n ∈ ℤ. Then your next equation becomes (a_1, ..., a_n) = a (1, ..., 1) + (n_1 k, ..., n_n k).

From there, maybe you should try to guess what the correct basis is.

a correct basis*

[south sinew]

Can anyone introduce a good book and video to learn modern algebra?

[frigid epoch]

There's a lecture on here! But I don't know how far it is advanced

[south sinew]

What do u mean a lecture 👀

[frigid epoch]

Check the events on this server

[velvet hull]

they're pretty far in, they were doing group actions last time I checked ~1 month ago

[tardy hedge]

Dummit and Foote is a great book for me but I remember it being kind of scary when I first looked at it , before i learned more algebra

[thorn jay]

Micheal Penn has a great video series on his second channel Math Major

You'll have to get exercises elsewhere (as he only gives a couple) but the content itself is good

[tawdry venture]

I see! Thank you! (and yes that was a typo lol)

[south sinew]

Okay, thanks for all the reply

[glad osprey]

How can I show that Q(sqrt(2) + sqrt(2)i) = Q(sqrt(2), i)? The <= inclusion is obvious, but I don't quite see the other direction

[wraith cargo]

Square it

[south patrol]

A common trick is to consider inverses.

[wraith cargo]

What I mean by square it is to square sqrt(2)+sqrt(2)i

[south patrol]

like here the inverse is 1/4 sqrt(2)( 1-i)

but also squaring works lol

[glad osprey]

hmm, so squaring gives 4i, which shows that i is in Q(sqrt(2) + sqrt(2)i). How do I show that sqrt(2) is in Q(sqrt(2) + sqrt(2)i)?

[south patrol]

This is why I suggest doing 1/ lol

Because it shows you have sqrt(2)(1-i)

Then it is easy to get sqrt(2) and i

[glad osprey]

ah, I see nice, thanks

[frigid epoch]

Or just multiply a = sqrt2+i sqrt2 by i

Then a-ia = 2sqrt2 and everything's good

[empty perch]

Hey folks, just a quick reminder that the Abstract Algebra Lecture series is meeting regularly again ^_^
We will meet to discuss Adjoints and the Spectral Theorem in about 15 minutes over in the ⁠⁠#1055201711679082516. More information can be found in our thread: ⁠⁠#1317307081535000606

[glad osprey]

To find the splitting field of x^4 + 1 over Q, I can show that every root can be written in terms of i and sqrt(2), so the splitting field is Q(sqrt(2), i). To show that this has degree 4, do I need to show that x^4 + 1 is irreducible, or is there a quicker way?

[south patrol]

it's much easier than that tbf

You can just add a single primitive 8th root of unity c

[glad osprey]

roots of unity in every algebra book I'm reading they don't show up until way after the meat of the Galois theory, so I haven't gotten that far yet

[south patrol]

sure

tbf given what you have, it is quicker

you can apply the tower law

[glad osprey]

I see, just [Q(sqrt(2), i) : Q] = [Q(sqrt(2), i) : Q(sqrt(2))] [Q(sqrt(2)) : Q] = 2 \cdot 2. Thanks

[south patrol]

np

[midnight wolf]

how is a vector field considered a field

[thorn jay]

It's a field of vectors

Field in the sense of a field of grass

Not field in the sense of simple commutative ring

[midnight wolf]

so the math people just called it a field for shits and giggles basically

[frigid epoch]

I'd sooner apply that to the algebraic fields

[thorn jay]

Yeah i have no clue where that name comes from

[coral spindle]

Different people name things differently. It wasn't some kind of conspiracy.

In real life, the same word can refer to many different things. This is just normal.

[midnight wolf]

nah that was intentional their all a bunch of trolls

[kind temple]

serre forgot why he gave the property of flatness its name

see the thread here

[tough raven]

You do need to show that the splitting field contains i and sqrt(2) as well (which is not difficult once ||you realise that the roots are the primitive 8th roots of unity (±1±i)/sqrt(2)||).

Does any finitely generated field extension have a maximum subextension which possesses a separating transcendence basis?

[glad osprey]

I guess it's strongly related to the fact that the 8th cyclotomic polynomial is x^4 + 1? I think I'm gonna follow Aluffi's suggestion and spend some quality time learning about them

Aluffi actually does cyclotomic polynomials before Galois theory, as opposed to Artin, so I should probably just learn about them now

[acoustic igloo]

Z2 x Z2 x Z2 has 7 subgroups of order 2 and 7 of order 4 right?

thanks 🙏

[potent condor]

if F and L are extensions of a field K, B is a transcendence basis of F\cap L, and I extend it to bases A u B and B u C for F and L, then how can I show that A u B u C is algebraically independent over K?

[acoustic igloo]

Give an example illustrating that not every nontrivial abelian group is the internal direct product of two proper nontrivial subgroups.
huh?

do the elements of a group need to be ordered tuples in order for the group to be an internal direct product?

[potent condor]

No, but there must be subgroups H, K of your group G such that G\cong H x K. And the elements of H x K are tuples.

[acoustic igloo]

hmm

So like Z6 is the internal direct product of its cyclic subgroups <2> and <3>

but Z9 is not the internal direct product of two proper nontrivial subgroups

is that right?

[potent condor]

this is right

[potent condor]

this is not right

[acoustic igloo]

but Z3 x Z3 is not isomorphic to Z9

[potent condor]

oh mb ignore me. i didnt realize 3 is a common divisor of 3 and 3

[acoustic igloo]

oh ok

thanks though

[thorn jay]

Weird phrasing for "give an example of an abelian group that is not isomorphic to a nontrivial direct product"

Lol

[toxic zephyr]

say we have an R module M, and R is a PID (i.e. an integral domain) and an R mod homomorphism phi from M to R (considering R as an R module, obviously). since R is an ID, then it's obvious that Tor(M) is a subset of ker(phi) right? like i can prove it in two lines, but if I just say it, then is would it be enough obvious to a reader as to not require proof?

further, would the generalization of this fact be that if we have phi: M to N and N is torsion free, then Tor(M) is a subset of ker(phi)?

[tough raven]

Yes, that's correct. (If you do end up needing to give a proof, any module homomorphism maps the r-torsion {m : rm = 0} to the r-torsion for any r ∈ R, and for the codomain the r-torsion is 0 for all non-zero r ∈ R.)

[acoustic igloo]

an abelian group of order 72 = 2^3 * 3^2 has exactly one subgroup of order 8 and may have 1, 2, or 7 subgroups of order 4

no i'm wrong

[tough raven]

Do you know that this is true?

[acoustic igloo]

it may have 1,3, or 7 subgroups of order 4

[tough raven]

BTW, any abelian group of any order p1^a1 ... pk^ak has exactly one subgroup of order p1^a1.

[acoustic igloo]

that makes sense

[tough raven]

And the number of subgroups of order p1^(less than a1) is the same as the number of such subgroups of the subgroup of order p1^a1.

[tough raven]

(By Sylow's Second Theorem, all of them are conjugate, but the group is abelian, so there are no non-identity conjugations.)

[acoustic igloo]

nice

[crystal vale]

Let $G$ be a finite group such that if we take any two subgroups $H,K$ of $G$ then one of them $ H \subset K $ or $K\subset H$ holds.
Then $G$ is a cyclic group.

\vspace{1.5cm}
Proof: $G$ has a finite order, let $p,q$ be two distinct prime divisor of $ | G |$, then by Sylow there exists $p$-sylow subgroup $H$ and $q$-sylow subgroup $K$, since $H\cap K ={e}$ hence its violates the given condition.
\vspace{0.5cm}

\Therefore $| G | = p^n$ for some $n\in N$.
\vspace{0.5cm}
$\textbf{ Claim: If there exist a subgroup of order $p^k$ then it is unique}$.
\vspace{0.5cm}
\Let $H$ and $K$ have the same cardinality, then by given condition let $H\subset K$, since both have same cardinality therefore $H = K$.
\vspace{0.5cm}
\Now I will prove that $G$ is cyclic by induction.
\vspace{0.5cm}
\For $ n = 1$ it is true.
\vspace{0.5cm}
\Now let it is true fo all group which has order $< | G |$.
\vspace{0.5cm}
\Since $|G| = p^n$, so its $Z(G) \neq {e}$. So now let group $G' = G/Z(G)$.
\vspace{0.5cm}
\So now we can apply our induction hypothesis to $G'$, just we have to prove that same $\textbf{Claim}$ for $G'$.
\vspace{0.5cm}

\I think we can prove that by canonical mapping $G\rightarrow G/Z(G)$.
\vspace{0.5cm}
\So now we have $G'$ is cyclic therefore $G$ is abelian.

\vspace{0.5cm}
\Now use the fundamental theorem of Abelian group, if it is not cyclic then I can construct subgroup $H,K$ in its isomorphic group such that $H\cap K = {e}$, so it will violates the given condition.
\vspace{0.5cm}

\Therefore it must be cyclic. Hence G is cyclic.

[cloud walrus]

Notknow🙇
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[crystal vale]

Is it correct?

[lucid shadow]

How should I list all subgroups of GL(3,2)≌PSL(2,7) and order them in the form of a lattice of subgroups?

[ivory ore]

hint for this

[ivory ore]

delete the command

[ivory ore]

oh boy for long long time i wanted to prove this but i got no motivation
may i ask where did you get the motivation to do it ?

order 168 groups 💔

[wraith cargo]

What are H and K?

[ivory ore]

groups

after solving the next two problems i think i will have a hint, however

[wraith cargo]

Finite groups or any groups?

[halcyon gulch]

are you going to prove it for finite or infinite or what

[wraith cargo]

Idk I'm just thinking

[wraith cargo]

Okay well here's a hint

[ivory ore]

for infinite i think i got it

RxR

nevermind

[wraith cargo]

Might be overkill but you can use Goursats lemma

[ivory ore]

alr, i will try to learn

[wraith cargo]

Easier hint
Notice that since N,H and K are all normal in H x K with pairwise trivial intersection
The you just have to show N commutes with everything in H and everything in K

This is an easier result to get

And it holds more generally

If A and B are normal subgroups with trivial intersection, then they commute

[ivory ore]

next problem was this

but its for abelian groups

[ivory ore]

i fear that is not what has been asked

i got what you are saying tho

alr I am tired, will try later

[rocky cloak]

If N commutes with everything in H and everything in K, then it will be in the center of HxK, hence abelian

[crystal vale]

i don't get it

[ivory ore]

sure, i will try to get bak to it after finishing the set

[ivory ore]

nothing serious but the input command of the latex is length which is why i told you to conside it to delete, which doesnt affect the generated latex

[crystal vale]

okay

[delicate orchid]

Very annoyed. You can shorten this proof substantially if you assume |G| is odd but of course p = 2 ruins it all

"subgroup lattice is a totally ordered set <=> the lattice is a line <=> G is C_{p^k}". Most of the work is in showing that G has to be a p-group, then you just check the subgroup lattice of (C_p)^2, so your p-group has to be rank 1. For odd primes we get that G is cyclic immediately but for p = 2 we have to manually check the generalised quaternion groups - which can be done inductively by just checking Q_8 but it's still annoying

[crystal vale]

actually i found the shorter proof

let x \neq e, if G = <x> then we are done. If it is not then there is y in G but y \not\in <x>, then <x> \subset <y>. Continue the procedure, this will stop at some point because | G | is finite, Hence G = <z> for some z \in G.

[delicate orchid]

oh that's very neat

[crystal vale]

yes

[delicate orchid]

you'll have to say a bit more I think, like why <x> must be a subset of <y> (it's obvious but still needs stating)

[crystal vale]

because by property of G, <x> \subset <y> or <y> \subset <x>, but since y not in <x> therefore later will not hold

[delicate orchid]

very nice

full marks!

[thorn jay]

I wonder if you can somehow translate this into UA

"If the subalgebra lattice is a line then must be finitely generated" seems doable at least

[crystal vale]

UA?

[thorn jay]

Universal Algebra

[crystal vale]

i don't know

[delicate orchid]

I imagine something similar works immediately for Artinian rings with this property => local

sorry, noetherian

[thorn jay]

The only thing we really can use is that the lattice must be algebraic

Well, in the general case

[delicate orchid]

it's really just "if there's only one ascending chain that terminates, the maximal element is unique"

because... there's only one chain lol

[thorn jay]

Well, say your lattice is uncountable

Then there are plenty chains :P (indexed by N that is)

[delicate orchid]

all objects are finite

and that still isn't really true, I'm imagining an unbounded half closed interval, it still has a unique maximal element even though as a poset it's uncountable

[thorn jay]

Alright, for a totally ordered set (X, <) there is an algebraic structure A such that Sub A ≈ (X, <) iff every element in (X, <) is compact

[tardy hedge]

Yee so true

[thorn jay]

Well

That's.. vacuously true

Great

[tardy hedge]

man what are yall even talking about

[delicate orchid]

I got no idea what this guy is talking about

[thorn jay]

Then take the natural numbers + infinity ordered in the usual way, and consider the chain
0 < 1 < 2 < 3 < ...
The sup of this chain is \infty, but doesn't terminate.

Now by a theorem from Birkhoff there is an algebraic structure A such that it's subalgebra lattice is isomorphic to the above totally ordered set. Hence there is a nonterminating chain of subalgebras
A_0 < A_1 < A_2 < ...
Of which the direct limit is A. This means that A cannot be finitely generated, hence in general we do not have that
Sub A totally ordered => A finitely generated

Nice!

Well, not nice

But fun proof

Now the next question is: when is it true that Sub A totally ordered => A finitely generated

[thorn jay]

Ive also got no idea what im talking about dw

[sturdy spear]

am i going in right path?

[thorn jay]

Alright, this would immediately mean that A is noetherian

A sublattice of a totally order lattice is totally ordered, naturally

So this would mean that Sub A totally ordered => Sub A finite

As any infinite totally ordered lattice contains a non-terminating chain

[thorn jay]

I really do cuz i got it completely wrong

[thorn jay]

I forgot one crucial detail which is that this is completely wrong

Smh

[thorn jay]

And this too

[glad osprey]

in c), you write that a reflection fixes two vertices, but that's only true for odd n unless I'm mistaken. Other than that, it looks good 👍

[thorn jay]

Okay this example should still work as I'm like 80% sure the natural numbers + infinity form an algebraic poset

[lucid shadow]

i like the symmetry of the fano plane

[ivory ore]

right, looks like you are going through richard's playlist

maybe I'm wrong

[sturdy spear]

Oh yeah, (iirc if n is even then it fixes two vertices, consider square and triangle).
I will correct it, thank you so much
(Sorry for late reply got busy in some urgent work, but it's algebra time now )

[acoustic igloo]

Show that a finite abelian group is not cyclic if and only if it contains a subgroup isomorphic to Z_p X Z_p for some prime p

i feel like it's helpful to write notes showing which statements imply which other statements

then you can fit them together like a puzzle

it was helpful here

[south patrol]

Nice

It is also fun to prove the following: a finite abelian group G is cyclic iff for every divisor d of #G (the order of the group), there is exactly one cyclic subgroup of size d

[acoustic igloo]

oh i noticed cyclic groups seem to have only 1 subgroup of each order

[south patrol]

Yeah, exactly. This is clearest if you it as Z/nZ for example and think what the subgroup of order d is

[ivory ore]

what does exponent mean here ?

[next obsidian]

https://ncatlab.org/nlab/show/exponent+of+a+group

[ivory ore]

alr

[velvet hull]

oh, it's just a funny way of saying the annihilator.
im assuming this terminology comes from somewhere else?

[next obsidian]

Cuz it’s the smallest exponent that kills shit

(Idk maybe ¯_(ツ)_/¯)

[acoustic igloo]

Let G, H, and K be finitely generated abelian groups. Show that if GxK is isomorphic to HxK, then G is isomorphic to H
how do i do this?

[sturdy spear]

Can i ask, Is it Rotman?

[lusty marlin]

Does anyone know of a counterexample to this when either the finitely generated or abelian conditions are removed?

Sorry if it's obvious but I can't think of any off the top of my head

[lone niche]

Something like K =prod Z^i where i is some index in an infinite set, and G=Z and H=0 should work I think

[lone niche]

Is the structure theorem for finitely generated abelian groups available?

[lusty marlin]

Ah, nice

Thanks

[lusty marlin]

And a counterexample using finitely generated non-abelian groups?

[lone niche]

no clue

[lusty marlin]

Hmm ok

[rocky cloak]

It's tricky but possible
https://math.stackexchange.com/a/349960/306319

Appears to be a small typo as H is a semidirect product of C1023 with Z

[lusty marlin]

Thanks

[crystal vale]

one direction is easy but how can i show if Z^n/ im phi is finite then phi is injective?

i know how we can extend phi to Q-vector space Q^n to Q^n

[tough raven]

Let k be an algebraically closed field. Every subfield of k(X) is k(p(X)) for some rational function p. Are all of these distinct from each other? When is k(q(X)) ⊆ k(p(X))?

Every subextension of k(X)/k, rather.

OK, k(q) ⊆ k(p) iff q ∈ k(p) iff q = f(p) for some rational function f, obviously.

[violet spade]

Hi, I am stuck on the first part of this question. I know that any number in $F_p$ is a root of $X^p-X$, since, by Fermat's little theorem, we have $a^p=a \mod p$. So for any $a \in F_p^*$, it is clear to me that $X^p-X-a$ has no zeros. But then, how do we conclude that this polynomial is irreducible? This polynomial could have a more complicated factorisation than just a linear one, right? So what is the idea?

[cloud walrus]

joel

[violet spade]

The frobenius map is given by $F: \overline{F_p} \longrightarrow \overline{F_p}$, where $x\mapsto x^p$. So would $F(x)=x^p=x \in F_p$?

[cloud walrus]

joel

[rocky cloak]

The idea is as given in the hint.

How does the Frobenius automorphism act on a root of this polynomial. More specifically how many conjugates will a root have?

[violet spade]

I mean, if $x$ is a zero, then $x^p=x+a$, right?

[cloud walrus]

joel

[violet spade]

My TA said something like, if $x$ is a root, then $x+1$ is also a root. Because $(x+1)^p=x^p+1^p=x+1+a$

[cloud walrus]

joel

[violet spade]

So this means that if we know one root, then we know all of the $p$ roots, right? But still, how do we move on from here?

[cloud walrus]

joel

[rocky cloak]

So do you know that a polynomial has coefficients in Fp if and only if it is fixed by the Frobenius automorphism?

Or formulated another way, that x and x^p have the same minimal polynomial

[violet spade]

What is "it is fixed"? Like the roots of the polynomial are fixed by the Frob automorphism?

[rocky cloak]

The coefficients

But equivalently, that the automorphism just permutes the roots

[violet spade]

This makes me confused, like isn't $a^p=a$ for any $a\in F_p$?

[cloud walrus]

joel

[violet spade]

or am i saying something stupid

[rocky cloak]

Yes exactly, and this is if and only if a is in Fp

[violet spade]

Right, yes I've seen this

What about permuting the roots?

[rocky cloak]

So if you have a polynomial, say
(x - a)(x - b)
then applying the automorphism to the coefficients gives
(x - a^p)(x - b^p)

[violet spade]

Like, in this case, the roots would not have to be in $F_p$ right?

[cloud walrus]

joel

[rocky cloak]

The only way for this to be the same polynomial is if
{a^p, b^p} = {a, b}

[violet spade]

because if they would be in F_p, then the roots would just be fixed under F_p

[violet spade]

yes\

this makes sense

[rocky cloak]

I mean, the roots can be in Fp, but then it's trivial

[violet spade]

yes then the permutation would be just the identity right

nothing interesting

[rocky cloak]

Alright, so then back to the original problem.

If x is a root of your polynomial, then the minimal polynomial of x has coefficients in Fp.

How many roots could it have?

[violet spade]

"If x is a root of your polynomial, then the minimal polynomial of x has coefficients in Fp." Why is this the case

[rocky cloak]

That's just the definition of minimal polynomial

[violet spade]

Or you mean, the minimal polynomial over the field F_p? Because then yea sure

[violet spade]

Yes okay i thought there was something else going on... that's why i got confused

[rocky cloak]

My point is just we can apply the reasoning from before

[violet spade]

I'm thinking p?

Like, that's an upper bound

I don't think it's the lowest upper bound haha

[rocky cloak]

So we established that the Frobenius map would need to permute the roots

And x is one of the roots

So what must the other roots be at least

[violet spade]

Well, F(x), F^2(x), etc

[rocky cloak]

And how many is that?

Or what are those explicitly

[violet spade]

x^p, x^p^2, x^p^3 right?

[rocky cloak]

Well, sure

But what is x^p

[violet spade]

Well, I'm thinking x does not necessarily lie in F_p, so can we say that much about it?

[rocky cloak]

Well, we can say quite a lot about it.

In fact you already did earlier

[rocky cloak]

.

[violet spade]

right

i'm stupid haha

okay... so now?

[rocky cloak]

So the roots are
x, x+a, x+2a, ...

[violet spade]

I was thinking about permuting the roots and the frobenius automorphism

[violet spade]

yes

[rocky cloak]

Which is how many?

[violet spade]

Like our last root would be x+(p-1)a right? Since x+pa=x again

So it would be exactly p roots?

[rocky cloak]

Yup, so the minimal polynomial of x has degree p

[violet spade]

Since p definitely does not divide a

[violet spade]

... So X^p-X-a IS the minimal polynomial

thank you!!

[south patrol]

Artin Dchreier

[rocky cloak]

Big D energy

[late python]

Hi can someone explain the difference between the a cyclic group and cyclic subgroup I don't understand the difference since they are both groups that can be generated by some single element in them

[coral spindle]

A cyclic subgroup of a group is a subgroup which is cyclic

The confusion you have is mysterious to me

If you know what a subgroup is, there should be no confusion

[late python]

Ok I understand now

I was reading the definition incorrectly

thanks for clarifying

[violet spade]

@rocky cloak And for $F_q$, we have $q=p^n$ right? Which means that $x^{p^n}-x-na=0$ if $x$ is a root of $X^p-X-a$.

[cloud walrus]

joel

[tough raven]

No, no n.

[violet spade]

Is this even useful? I'm trying to find the irreducible factors. They must all have degree p?

[tough raven]

A zero x of X^{p^n}-X-a satisfies x^{p^n} - x - a = 0 by definition.

[violet spade]

Yes

[rocky cloak]

The irreducible factors will all have degree p yeah, by the same argument as before

[tough raven]

No, it's either irreducible or splits.

Over F_q.

[violet spade]

So $x^p=x+a$. So $x^{p^n}=(x^{p})^{p^{n-1}}=(x+a)^{p^{n-1}}=x+na$

[cloud walrus]

joel

[violet spade]

By induction

[tough raven]

a ∈ F_p?

[violet spade]

Is it possible to write a more explicit form?

[violet spade]

F_p*

So without 0

[tough raven]

OK, sorry. I think I have misunderstood what you are trying to do, so ignore my previous comments.

[tough raven]

It is true that this polynomial is either irreducible or splits into linear factors, over any field containing F_q, though.

[rocky cloak]

Why?

[violet spade]

I mean, isn't that just that it's either reducible or irreducible? Like, that's obvious?

Oo wait

linear factors

hmm

[rocky cloak]

Like x^q = x + a, so any root will have exactly p conjugates

So you split it into factors of degree p

[violet spade]

Why not q conjugates? Because we are working in $F_p$?

[cloud walrus]

joel

[rocky cloak]

pa = 0

[violet spade]

yes

[rocky cloak]

It doesn't really matter if you take a from Fp or Fq

[violet spade]

But like, would $X^q-X-a=(X^p-X-a)^n$?

[cloud walrus]

joel

[violet spade]

Feels like sophomores dream haha

[rocky cloak]

No, it has all distinct roots

[violet spade]

So $X^q-X-a = (X^p-X-a)(X^p-X-2a)\dots$?

[cloud walrus]

joel

[violet spade]

I think i'm stupid haha

[rocky cloak]

That doesn't seem right either no

[violet spade]

I'm very much thinking

haha

[rocky cloak]

One thing to notice if that you add any element of Fq you still get a root

[acoustic igloo]

Fraleigh

[acoustic igloo]

yes

[sturdy spear]

Cool

[rocky cloak]

So once you have one irreducible factor, that should give you all the other ones

[violet spade]

So like, $X^q = X+a$ implies $X^q = X+b$, for any $b\in F_q$?

[cloud walrus]

joel

[rocky cloak]

No, that would just be saying a=b

[violet spade]

This makes me feel like it's the product over all elements

So $(x+a)^q=x+a+a=x+2a$, meaning $x^q+a^q=x^q+a=x+2a$, so $x^q=x+a$. Thus $x+a$ is a root

[cloud walrus]

joel

[violet spade]

And x+a also has p conjugates

[loud osprey]

hey does anyone know if theres a formula to work out the conjugacy class size, like in this one for example

like is there a formula that would give me 6

[violet spade]

Isn't this just 1 times 3!

[rocky cloak]

Yeah, so in the case where n is relatively prime to p, then
X^p - X - a/n
should be one of the factors, like you said before.

Then the other factors would be
X^p - X + b^p - b + a/n
where b runs over representatives for Fq/Fp.

When n is a multiple of p you may have to do something else clever.

[loud osprey]

but what about for different cycle types like (1,2)(3,4)

[violet spade]

Wait, why is it a/n, not an?

[rocky cloak]

It's just thinking about the number of ways to make such a cycle.

There are 4! ways to assign values to
(a,b)(c,d)
swapping a, b gives the same result, same for c and d and same for swapping (a,b) with (c,d) so that's 4!/8 = 3

[violet spade]

Like, i did this?

[rocky cloak]

n * a/n = a

[tough raven]

Precisely: if K is any field containing F_q and p any polynomial over K such that all monomials in p have degree a power of p, then for any root x of p, x + F_q is the full set of roots of p. So if p has one root, it splits.

... which doesn't imply what I said.

[gleaming shell]

For a cyclic group G of order n, and some d that divides n, how many elements in G have order d? Im doing an undergraduate study, and this information will help!

[violet spade]

phi(d)

[gleaming shell]

ty

[violet spade]

i think

yes this is correct

[violet spade]

I'm still confused....

Like, how do we get the a/n??

[lone niche]

Well I would start by decomposing G,H,K into products of cyclic groups via the theorem, and then compare the factors of GxK and HxK. Kind of like if we have a * b * d = a * b * c numbers, we expect c and d to be equal.

also make sure you read carefully the statement of the theorem

[rocky cloak]

It's just like you wrote before.

If x^p = x + a/n
then
x^p^n = x + na/n = x + a

[acoustic igloo]

are you referring to the uniqueness of the factors?

if the factors of G and H were not the same, then the facors of GxK and HxK would not be the same, and GxK would not be isomorphic to HxK

maybe

[violet spade]

And the goal is to find other conditions such that x^p^n = x+a right? Because then we've found other irreducible factors. And for that, we use the Frob automorphism?

Like, would x+a work?

I think so, yes?

[rocky cloak]

Like, if you find an irreducible polynomial where the roots satisfy x^q = x + a, then that will divide x^q - x - a

[violet spade]

Yes

But aren't we trying to find all of the irr factors of this? Or is this not the question?

[lone niche]

yes

[lone niche]

yes what you wrote is true, but you essentially turned the statement into its contrapositive, but maybe you can prove it easier that way i.e. fill in the gaps easier?

[acoustic igloo]

i seem to have the habit of changing things to the contrapositive

uhm

yeah i'm not sure how to fill in the gaps without doing that

[violet spade]

So, you know that any abelian group can be written as $G\cong \mathbb{Z}/p_1^{n_1}\mathbb{Z} \times \dots \times \mathbb{Z}/p_k^{n_k}\mathbb{Z}$?

[acoustic igloo]

yes

[cloud walrus]

joel

[violet spade]

Clearly, if G,H,K are abelian, then the direct product are also abelian right?

[acoustic igloo]

yes

[violet spade]

Well, then $G\times K$ would be $\mathbb{Z}/q_1^{n_1}\mathbb{Z} \times \dots \times \mathbb{Z}/q_k^{m_k}\mathbb{Z}$ and $H \times K$ would be $\mathbb{Z}/r_1^{l_1}\mathbb{Z} \times \dots \times \mathbb{Z}/r_k^{l_k}\mathbb{Z}$

[cloud walrus]

joel

[violet spade]

If they are isomorphic, what would you be able to deduce?

@acoustic igloo

[lone niche]

doesn't change the proof, but before I go to sleep I should mention that there should be some Z^n there since the problem asks for finitely generated

[acoustic igloo]

the factors are the same?

yeah it should have Z^n

[violet spade]

Well, up to the chinese remainder theorem, yes

[acoustic igloo]

huh?

[rocky cloak]

Once you have one irreducible factor you can easily find all, because they're related by shifting the roots by elements of Fq

[violet spade]

Yes this is what i was thinking

[violet spade]

Like, for example, $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}/6\mathbb{Z}$

[cloud walrus]

joel

[violet spade]

what i'm saying is, that once you have this product, then you could write everything as "compact" as possible, so just use CRT

and then yes, you can conclude that their factors are the same. Because otherwise, they would not be isomorphic.

[acoustic igloo]

combining them helps us?

[violet spade]

Well, for the sake of argument, it is nice to do this. What we want to do is say $n_1=l_1,n_2=l_2,\dots, n_k = l_k$ and $q_1=r_1,\dots,q_k=r_k$

[cloud walrus]

joel

[violet spade]

We can say this, but not directly, because for example, if on one hand we have $G$ is $Z/2Z \times Z/3Z$, and $H$ is $Z/6Z$, well, then $k=2$ for $G$, and $k=1$ for $H$, you know?

[cloud walrus]

joel

[violet spade]

So because of the chinese remainder theorem, this is clear

Because of what was stated before

[acoustic igloo]

but we're not dealing with remainders

i don't get it

[violet spade]

Have you seen the chinese remainder theorem for groups?

[acoustic igloo]

no

[violet spade]

aha

Basically, if $gcd(x,y)=1$, then $\mathbb{Z}/x\mathbb{Z} \times \mathbb{Z}/y\mathbb{Z} \cong \mathbb{Z}/xy\mathbb{Z}$.

[cloud walrus]

joel

[acoustic igloo]

yes

[tardy hedge]

How can u argue that the frobenius homomorphism Fp^n -> Fp^n is an automorphism using some
Vector space argument ?

I think it was mentioned in lecture today but i missed it

[violet spade]

So, what i'm trying to say is, whenever any two prime's gcd is 1, then just put them together

[violet spade]

or prime power rather

[violet spade]

So, if $gcd(p^n,q^m)=1$, then rewrite this direct product according to the Chinese remainder theorem, and you'll get that all the primes with the powers are the same

[cloud walrus]

joel

[rocky cloak]

Well, it's injective, and Fp^n is finite

[tardy hedge]

An injective hom of vector spaces of same dimension must be surjective?

[rocky cloak]

Yes, that's true. But also any injective function on a finite set is a bijection

[violet spade]

f: A --> B, injective, and |A|=|B| is finite, then f is a bijection

[tardy hedge]

An automorphism of a field must always fix its prime subfield right

[coral spindle]

Well it must send 1 to 1, and 1 generates the prime subfield

[tardy hedge]

Yea

[tardy hedge]

Oh its injective becuz its a field hom yes lol

[ivory ore]

you can count it from the cycle and the fact that same cycle types are conjugates

[tardy hedge]

You cant have a field hom between finite fields F1->F2 where charF1 is bigger than charF2 because then it wont be injective

Can u have charF1 smaller than charF2

[tough raven]

You can't have any homomorphism between fields of different characteristic (why?).

[rocky cloak]

You can only have a ring homomorphism R1 -> R2 if the characteristic of R2 divides that of R1

[tardy hedge]

Yeah and that comes from the group hom properties of order of elements

[violet spade]

@rocky cloak So we'd have $X^q-X-a=(X^p-X-a/n)((X+1)^p-(X+1)-a/n)\dots$? This doesn't seem right to me, but on the other hand, these are all the roots right??

[cloud walrus]

joel

[violet spade]

Like, for X^p=X+a, then X+1 also a root

[rocky cloak]

(X+1)^p - (X+1) is just X^p - X

So you've just written the same factor twice

[violet spade]

Yes this is what i was thinking...

but then, how would we do it?

[rocky cloak]

.

[violet spade]

damnit why did i not read this properly

... 😦

These representatives could look like {1,p+1,2p+1,...}, right?

[tardy hedge]

This is pretty basic but the characteristic of a ring being n implies na = 0 for all a, but it doesnt mean that a has additive order n right

Just that the order of a divides n

[lusty marlin]

Yes

[rocky cloak]

I mean, 1 = p+1, so these are all the same.

Unless you're using p to mean something else

[tough raven]

Unless n is prime, in which case that does imply that a = 0 or the order is n.

[violet spade]

Then, what would these representatives look like?

[rocky cloak]

I mean they'll look like elements in Fq, so I guess it depends on your construction of that

[violet spade]

Suppose you have this definition

Then what would you get?

[rocky cloak]

Like if you have Fq = Fp(a), then Fq = (Z/p)^n spanned by 1, a, ..., a^n-1

Then a natural choice for representatives for Fq/Fp could be the span of a, a^2, ..., a^n-1

(just removing the basis v vector that dspans Fp)

[violet spade]

So you can extend F_p by one element and then that would generate F_q?

well, apparently

is the case a not coprime to n similar, or completely different?

I was thinking, maybe just choose gcd(a,n)=k and then k must show up in the answer

[rocky cloak]

There exists such an element yeah, in fact there exists an element that generates the multiplicative group of Fq even.

[violet spade]

How would you write the set of representatives for F_q/F_p?

[violet spade]

Since it is cyclic right?

[rocky cloak]

Yup

[violet spade]

Like, what notation would you use to denote this?

Or just, let A be a set of representatives for F_q/F_p, and then write X^q-X-a as the product over that

I would have never been able to come up with this by myself though...

this exercise is kinda crazy

I don't know how not being co-prime would be a a problem? @rocky cloak

Like, what is the difference between a,n being coprime or not coprime?

[mental silo]

do subrings require multiplicative identity?

im going through hungerford right now and there's some stuff about like "the ring 3Z of all multiples of 3" but to me that's not a ring cause theres no mulitplicative identity

im js bamboozled

[violet spade]

I'm sorry

Yes. Definitely

they do need a multiplicative identity

I thought you meant inverse

[mental silo]

acc?

oh ok

[violet spade]

But yes 1 has to be in there

[mental silo]

i mean i was thinking that you wouldnt use it so you could have non trivial ideals as subrings

so when all these texts talk about 3Z (not Z3) how is that even a ring if it doesnt have multiplicative identity

[chilly ocean]

Rings are sometimes defined without identity(sometimes these are called rngs, no identity)

If the book in question defines a ring explicitly, make sure to read that rq to check.

[mental silo]

yea it says ring has identity

i may be bugging idk ill ask my prof

all this notational disagreement is so annoying why cant we js come to a consensus

[violet spade]

I have a feeling Z/3Z is meant...

[rocky cloak]

If n=0 mod p you can't divide by it

[elfin wraith]

Got feedback on a homework today that the marker was confused by my nonstandard notation saying the two groups of order 6 and S_3 and C_6, they’ve never seen C_6 and I should have wrote Z/6Z

Like brother I agree they’re isomorphic but that’s a ring, how have you never seen C_n denote a cyclic group

[mental silo]

yea me too

[south patrol]

That is very silly lol

Z/6Z is also standard notation for the group but ye

[violet spade]

@elfin wraith :/

[hidden wind]

flashback to my first course in abstract algebra which wrote Zn for Z/nZ

[frigid epoch]

I mean it'd sure be nice to have that notation for the Very Common cyclic groups instead of the Very Niche p-adic integers

[thorn jay]

I'd use 3Z in referral to the ideal of Z generated by 3, which is a rng

[elfin wraith]

Yeah I know a lot of people use that, I wouldn’t question it if I saw it, but like C_n or Z_n (not black board) are equally as common lol

And to my mind better, one is by definition the cyclic group on n elements, the other is equivalence classes of integers

They’re clearly isomorphic, but I wouldn’t define it to be the later

[thorn jay]

I suppose it depends
When working with abelian groups/modules I'll tend to the Z_n or Z/nZ
but when working with groups I'll tend to use C_n

[south patrol]

Sure I mean that just cause you said Z/nZ is a ring lol

Though Z_n and Z/nZ are just different notations (though Z_n is occasionally given a hacky definition)

Z_n is worse anyway lol given that the p-adics exist

[coral spindle]

Dang I wish there was a way to notate the cyclic group of order n in such a way that makes the generator "1" naturally and doesn't clash with any other notation whatsoever !

[thorn jay]

Me when i use 1 as the identity

Or should i switch back to "e"

Micheal penn-pilled

[coral spindle]

Guy who has only seen michael penn: dang this is like michael penn

SORRY I had to

[rocky cloak]

Imagine talking about the group ring Z[Z_n]

[rocky cloak]

Then you can write amazing things like
$(ag)^n = a^nng$

[cloud walrus]

jagr2808

[coral spindle]

Now that's when you write C_n tbh

Hey speaking of notation

Q_8 has 8 elements

D_8 has 8 elements

and they're extensions of the same groups

beat that

[rocky cloak]

No, bad Boytjie

[coral spindle]

But for real I like that Q_4n and D_4n have the same # of elements and fit into SESs with the same start and end

So... take that!!!

I get to die on this shitty hill

[hidden wind]

i tried my best to explain to a fellow student the other day why he shouldn’t think of Z/nZ = {0, 1, …, n-1} as a subset of Z… i gave up after a while

[rocky cloak]

I'm down with renaming the Quaternion groups if that makes you feel better

[coral spindle]

[thorn jay]

No, good Boytjie

[coral spindle]

Wikipedia calls them the dicyclic groups

and has... nice notation for them

[hidden wind]

Dic

[coral spindle]

Unfortunately it gets the numbers all wrong! smh

[thorn jay]

I got called out by someone for learning algebra from Micheal Penn when they saw me using e as identity qwq

[rocky cloak]

Good Wikipedia

[coral spindle]

Idk what to say, using e as the identity in intro group theory courses is way older than penn

More than anything it is a mark of beginners

[rocky cloak]

I'm not sure I've seen an intro book that uses anything else

[hidden wind]

who is micheal penn

[thorn jay]

Mm, that makes sense

Maybe I'm misremembering

[rocky cloak]

YouTuber that solves tricky integrals or something

[thorn jay]

He's done a couple courses on stuff

On his second channel, Math Major mainly

[hidden wind]

i see

will he help me solve easy integrals

[thorn jay]

I learned group theory using his abstract algebra series

[hidden wind]

actually nvm i don’t need no integrals mwahaha

(…for the moment)

[thorn jay]

Integrals aren't too bad

But man am I glad they don't come up in what I'm interested in

[thorn jay]

This begs the question though, why in every intro book? Is it simply tradition?

[rocky cloak]

You want to talk about examples both of additive groups and multiplicative groups.

If you for example introduce 1 as the symbol for identity you create confusion with what is the identity in (Z, +) for example.

So then it's nice to have a seperate symbol other than 0 or 1.

So e for Einheit(?) or idEntity (😉)

[thorn jay]

Eenheid ✨

Although dutch uses "identiteit" iirc

😔

[rocky cloak]

Yeah, I would assume they took it from German "Einheitselement"

[thorn jay]

German looks so silly to me

Im sure dutch looks silly to germans too

[rocky cloak]

I don't know about looks, but dutch sure sounds silly

It's like soft german, interleaved with someone gurgling in the back of their throat a bit

[thorn jay]

Fairly accurate

Except throw in some weed too

[rocky cloak]

Not as silly as danish though, of course

[thorn jay]

What is the silliest language

Is that even a total order?

[rocky cloak]

Maybe piglatin

[thorn jay]

What is piglatin

[rocky cloak]

https://en.m.wikipedia.org/wiki/Pig_Latin

[alpine plank]

bro has not read captain underpants

[thorn jay]

I have not

Indeed

[rocky cloak]

Damn, a lost childhood

[tardy hedge]

ya i think when u learn new math its easy to forget the minds of younger math students and the level of abstraction they think at

i.e why beginner books emphasize "e" for identity instead of just 1

[thorn jay]

It really is that simple of a reason, lol

[tardy hedge]

yea lol

[elfin wraith]

Well yes of course it’s also a group, but idk I assume the canonical structure on Z/nZ is a ring, of course it’s immediately obvious from context and whatever, but in either case, it’s a silly object to take as the definition and I’ll die on that pedantic hill

this is why you don’t make analysists mark a homework on characters and why I shouldn’t have taken analytic number theory

Could be doing Galois theory instead but no I’m pissing about with logarithms

[tardy hedge]

I honestly cant believe it, but the linear algebra is genuinely making me confused here

Especially 14.5

Im imagining the first equation starts as sum of aixi = 0 then u apply sigma 1 to it

[chilly ocean]

0 for Abelian groups ofc

[thorn jay]

Yeah but thats just cuz theyre modules

[tardy hedge]

Im just confused on what we are taking these equations over

What are the xi’s? In F? In K?

I dont know how to think about that system of equations

[chilly ocean]

Huh never thought about it that way

I thought + and 0 for Abelian groups and • and 1 for general groups was entirely arbitrary

That makes sense though

[coral spindle]

A German suggested to me that the typical name "s" for the reflection generating the dihedral group along with a rotation stands for "spiegelung," which is a cute idea. If only r stood for rotation in German (it does not).

[thorn jay]

"Spiegeling" and "rotatie" if we're gonna do a nice anglicisme

[tardy hedge]

This equation taken over F must mean xi’s are 0 becuz alpha i are lin ind

[elfin wraith]

I assumed it was r for rotation and s for being next to r

[thorn jay]

I suppose rotatie isnt really an anglicisme

[rocky cloak]

Speiling and rotasjon, clearly Norwegian

[tardy hedge]

Ok i think i get it now. We think of these system of equations as like something from the K-vector spaces K^(n+1) -> K^n … then it must have a solution because linear algebra matrix thing

But then we can say these solution must not all be from F becuz thats when it contradicts F-linear independence

[coral spindle]

Ha I like this

Clearly Abel came up with it

[rocky cloak]

I see no other possible explanation

[coral spindle]

This is probably why, but I suppose it's not really possible to get a definitive etymological confirmation

It was while I was giving a talk, I just casually mentioned that r is nice because it stands for rotation, but I lamented that s didn't really stand for reflection – that's when Spiegelung was mentioned

[hidden wind]

do people actually use «speiling» ? :o i’ve only heard «refleksjon»

[elfin wraith]

Just start giving your talks in denglisch, shoehorn it in

[coral spindle]

Oh ja hier we see the dihedralgrouppen

Can you tell I speak kein Deutch?

[hidden wind]

https://tenor.com/view/sad-sad-cat-cat-cry-cat-cry-gif-18340701370131332749

[rocky cloak]

https://matematikkradet.no/ordliste/

List "spelling" as the main translation

But the anglisism is strong in this country ofc

[hidden wind]

mhm

[elfin wraith]

And yet they chose to use the American spellings smh

If it makes you feel better, Norwegian left its mark on English, mostly in Aberdeen, lots of Doric words have a lot of Norwegian influence

[rocky cloak]

Depending on where you draw the distinction between Norwegian and old Norse, I'd say there is a lot of influence on English.

[glad osprey]

such a long proof 😭 that's the kind of proof I would skip on first reading (and maybe second and third reading)

[tardy hedge]

Yeah it was kinda ass

I get the gist of it now though 😢

[chilly ocean]

Honestly just do all your lectures in anglo-saxon.

[tardy hedge]

Why does it specify if f(x) is separable over F. I thought separability of polynomial didnt depend on base field but im prolly wrong

Doesnt separable polynomial means it splits into distinct factors in its splitting field

[rocky cloak]

Could be they define f to be separable over F if it is a product of irreducible seperable polynomials.

That would make it an if and only if at least

[tardy hedge]

It says a polynomial is separable over F if it has no repeated roots

(In its splitting field)

[rocky cloak]

Alright, then the "over F" part does nothing

[tardy hedge]

Yeah

[violet spade]

@rocky cloak I've been trying and trying to find an irreducible factorisation in the case n = 0 mod p. I have yet to come up with an expression... I feel kinda dumb haha

But anyway, thanks so much for your help!!!!!

I very much appreciate it!!!

🙂

[rocky cloak]

Yeah, I'm not so sure myself.

The coefficients won't be in Fp for any of them, so will be kinda annoying I guess

[violet spade]

I'm not sure you have to actually explicitly write these out for the exercise...

but why does X^q-X-a split into irreducible factors, all of degree p again?

@rocky cloak

Like, suppose we have a root of $X^p-X-a$, then we know all $p$ roots.

[cloud walrus]

joel

[violet spade]

but this in turn, is not a zero of X^q-X-a necessarily right??

So i'm confused....

[rocky cloak]

Because each root has p conjugates
x, x+a, x+2a, ...

[violet spade]

yes I've read it back haha

Since $\alpha^{p^n}-\alpha=0$

[cloud walrus]

joel

[violet spade]

like, again if $\alpha$ is a root, then $\alpha+a$, etc

[cloud walrus]

joel

[violet spade]

now it makes sense again

great

[acoustic igloo]

is there notation for something like,
in a finitely-generated abelian group G, the number of copies of each cyclic group of prime power order when you write G as a direct product of such groups?
like Betti number but including the finite factors?

[acoustic igloo]

i'm looking for a clean way to make this argument

[late python]

is the order of 12 in $\mathbb{Z}_{12}$ 0 or 1?

[wraith cargo]

why would it be 1?

[acoustic igloo]

the order of 12 in Z is infinite

[late python]

z_12

[wraith cargo]

it's 1

[acoustic igloo]

the class represented by 12 in Z_12 is the same as the class represented by 0

[late python]

i don't know how to type it in latex

[acoustic igloo]

0 has order 1 so 12 has order 1

[wraith cargo]

you need to put the _12 outside the mathbb
And put it as _{12} so it recognizes that both digits have to go in the subscript

[cloud walrus]

rabbits_advocate

[late python]

Thanks

[toxic zephyr]

is there a simple way to create a "standalone" module with a given structure theorem decomp (for a PID R)?
ex. M'=Z^2+Z/(2)+Z/(6)
like can we define some non-vector-like Z module M that is isomorphic to that? maybe like a polynomial type thingy?
how about F(x)^2+F(x)/(x-2)+F(x)/(x^2-4)?
(+ meaning oplus direct sum)

[wraith cargo]

wdym "non vector like"

[tough raven]

This is assuming p ∤ n, right?

[toxic zephyr]

like the structure theorem decomp is like a vector with entries for each component. can we do something where there are no entries

[tough raven]

I think this is hard to do unless you decide what does and doesn't count as a "vector-like" example to you.

[wraith cargo]

errrr for finitely generated modules over a PID we can't
Because the structure theorem guarantees such a decomposition

[toxic zephyr]

ex. like instead of R2, we can do R[x]/(x^2)
now they're just 2nd degree polynomials, no entries

[tough raven]

Fg modules over a PID do come up often in various places and those will probabl y not be "vector-like". But whether it can end up with any possible structure decomposition depends on the example.

[wraith cargo]

R[x]/(x^2) is isomorphic to R (+) R as an R-module

[toxic zephyr]

yeah exactly so that's an example of what I'm sort of talking about

[wraith cargo]

but that is vector like....

[toxic zephyr]

but how do you mix in some torsion with that?

[wraith cargo]

well over a field there's never any torsion

[tough raven]

Hmm, I see. How about R[X]/(X^n, a_n, a_{n-1} X, ..., a_1 X^{n-1}), where your module is M = R/(a_n) (+) R/(a_{n-1}) (+) ... (+) R/(a_1), with a_1 ∣ ... ∣ a_n (we allow a_n = 0)?

[wraith cargo]

but if you look at something like Z, then you can have something like Z^r (+) torsion

[toxic zephyr]

so for the example I gave, do something like
Z[x]/(x^2,6,2x)? or did I mess up the construction

[tough raven]

For example, ℤ[X]/(X^4, 0, 0 X^1, 6 X^2, 2 X^3).

[toxic zephyr]

oh I see

[tough raven]

(The 0's are just to illustrate how it goes in general.)

[toxic zephyr]

so the
0 corresponds to Z
0x corresponds to Z
(that ends the free part, now onto the torsion)
6x^2 for Z/(6)
2x^3 for Z/(2)
and then Z^4 because the module has 4 components?

then we have
a0+a1x+a2x^2+a3x^3

[tough raven]

Another way to say it would be M = ℤ/0ℤ (+) ℤ/0ℤ (+) ℤ/6ℤ (+) ℤ/2ℤ (+) ℤ/1ℤ (+) ℤ/1ℤ (+) ...

And then you read off 0X^0, 0X, 6X^2, 2X^3, 1X^4, 1X^5, 1X^6, ... as the generators

and remove redundancies.

That's what you said.

[toxic zephyr]

oops

no nvm I see. multiplying by 6 would eliminate both torsion components but not 2, that would only eliminate one (the x^3 part).

okay that makes sense

but what about the F[x] example? that seems a bit more complicated

maybe like F[x1,...,x4]? or something? not sure

would it be
$$F(x)^2\oplus F(x)/(x-2)\oplus F(x)/(x^2-4)$$
$$\cong
F[x,y]/(y^4,0,0y,(x^2-4)y^2,(x-2)y^3)$$
?

[tough raven]

Which one?

My example does work for any PID.

[tough raven]

F[x, y]/that, I guess.

Ah, no.

You need the factors in descending order by divisibility for this to work.

[toxic zephyr]

ohh right like how we had 6x^2,2x^3

[tough raven]

Try to figure out why it has to go that way.

[cloud walrus]

eigentaylor (STfFGMOaPID)

[toxic zephyr]

okay yeah I understand this better now. this is what I got when I tried to generalize it
so like I get what each a(i)X^(n-i) means. but like... does the X^n have some connection to the module besides just "make sure there are only n degrees of freedom"? like what goes wrong if we don't include it?

or I guess since we're defining the isomorphism via the unit for the ith component mapping to x^(i-1), then it's to make the map surjective?

[tough raven]

Yes, it's just to limit to n factors.

[toxic zephyr]

okay thank you! you were very helpful

[rare cipher]

hi i just have a general question. my teacher was talking about big problems in math

"Does there exist a lattice order on the field of complex numbers ℂ that is compatible with the field operations?"

does any1 know the answer to this? im pretty sure this was how my professor phrased it

[next obsidian]

I know at the very least you cannot have an order which has every two elements comparable, as if so if i < 0 you have i^2 > 0 but this is -1 which is < 0, and if i > 0 then i^2 > 0 but again a contradiction

I’m not sure if you can somehow adapt this to say something about a lattice

[crystal vale]

why do we taking finitely generated projective module, isn't every finitely generated module over k is vector space hnece it is free?

[errant wedge]

Sanity check, it's not true in general that if a group G has subgroups H and K, and if N is a normal subgroup of K, then HN is a normal subgroup of HK right? We need H to be normal too?

[tough raven]

It's a free k-module, but not necessarily a free (or even projective) k[x...]-module.

[tough raven]

No. For example, if K = G and N ⊆ H this says that if N ⊆ H ⊆ G with N normal in G, then H is normal in G.

[errant wedge]

fantastic thanks

[lone niche]

I'll give you the proof I had in mind, but with integers. I think it should generalize to the group problem. Main thing is that I will be avoiding using division because we don't have that in the decomposition of groups.

Let a,b,d be positive integers. We are also given that ad=ab. Consider a factorization of ab and ad and a,b,d each into prime powers. Since the factorizations are essentially unique, it must be that the product of the factorizations of a and d must equal to the factorization of ad (might need to combine some prime factors). Similarly for a and b. By assumptions, we have that ad=ab, so the factorizations of the products of a and b and the factorizations of the products of a and d must be equal. And after accounting for the prime factors of a, it must be that b=d (need to elaborate more on this part, and you should do it by writing out general general decompositions, but doing some examples might help, say a=2^3, b= 2^2 * 3^3 =c).

[crystal vale]

i got it, but if every finitely generated module M over k has FR, right? Then why we don't do induction ?

[tough raven]

They do do induction.

[tardy hedge]

hehe dodo

he said do do

[tough raven]

https://tenor.com/view/melon-watermelon-dodo-gif-4922037

[crystal vale]

but they are taking projective module

i am saying if i show that every finitely generated module over K[x1,..,xn] has FR then we are done for projective too

[acoustic igloo]

thanks

i wrote an answer earlier today

[lone niche]

you can ping me on reply btw and i actually prefer that

[acoustic igloo]

okay

[tough raven]

Because they want to conclude that it's stably free.

I don't know how they show projective + FR => stably free, but it's certainly not true for any module with FR.

[crystal vale]

i got it, thank you

[ivory ore]

hint?

well, what does commuting with a transposition mean

[acoustic igloo]

a non-identity element sigma of S_n must move some element a to b
if n >= 3 there must be a third element c
we can check if sigma commutes with (b, c)

[ivory ore]

elements inside the transposition have to be necessarily fixed by the candidate ?

oh the candidate can interchange (do or undo) the transposition elements too ig

[acoustic igloo]

sigma commuting with a transposition tau means
sigma tau = tau sigma

[ivory ore]

oh yeah

let's say we are trying to find sigma's for the centre

[acoustic igloo]

yeah they could fix or interchange the elements of the transposition

[ivory ore]

hmm then if we try to turn any big map into transposition it can take form of (12)(23)

and if two such transpotion shows up

we are done

nothing can commute with them

except the trivial shit

does the argument sound

idk what to do for A4

think of the polynomial \to \pm1 ?

[acoustic igloo]

what if it's (12)(23)(34)(56)
then it will commute with (56)

actually we don't need to show that nothing commutes with it, only that some permutation doesn't commute with it

[ivory ore]

oh yeah

my wording was wrong, good find

well, no kinda confused now

[restive birch]

clearly the rubiks cube group can be embedded into S48. can it be embedded into any smaller symmetric group? i cant find actually anything about this online

[ivory ore]

is it okay ?

hope this is fixed now

[woven trout]

for this one, is he asking for a single monic polynomial of degree 3 in Q whose roots are the ones he gave, or does he want a different one for each root he gave?

I can figure out a degree 3 monic polynomial in Q for each root individually, but not one with both of those values as roots

[somber sleet]

Guys do you have in mind and R-module, where r(x)m = 0 but both r and m are not 0

[delicate orchid]

Z (x) Z/2Z, r = 2, m = 1 but this is a little silly

just pick anything with R-torsion

[somber sleet]

This is exactly the easy example I wanslooking for

[rocky cloak]

Could also do
Z/3 (x) Z/2
if you want everything to be 0

[delicate orchid]

I thought that one was TOO silly

[spare nymph]

Lol

[crystal vale]

I have to find two elements in A_5 such that they are not conjugate in A5 but are conjugate in S5.

Is there any way other than computation ?

[frigid epoch]

You could try conjugating random things in A5 with elements in S5\A5

[tough raven]

This may not be much better but: A5 is a union of conjugacy classes of S5. Consider one such conjugacy class C = orbit(x), S5 acts on it (by conjugation, of course). If you restrict this action to A5, you get the conjugation action of A5. So the orbits of C under the action of A5 give the splitting of the single S5-conjugacy class C into A5-conjugacy classes. Now, if G_x = Stab(x) = {y ∈ S5 : y commutes with x} is the centraliser of x, then it is not too difficult to see that A5 ⋅ x = A5 G_x ⋅ x is equal to S5 iff A5 G_x = S5. (In general the A5-orbits of C are in bijection with the double coset space A5 \ S5 / G_x.)

So for each conjugacy class of S5 contained in A5, you pick some element x and need to check whether A5 Z = S5 where Z = centraliser(x); by cardinality considerations, this is equivalent to [Z : A5 ∩ Z] = [S5 : A5] i.e. 2. (The only other possibility is [Z : A5 ∩ Z] = 1 ⇔ Z ⊆ A5; so the conjugacy class splits into two iff Z ⊆ A5.)

This analysis works for the conjugacy classes of any normal subgroup of any group.

[mossy slate]

I just realized something really interesting

We can think of incrementation as the most basic operation. Addition can be thought of as repeated incrementation. Multiplication can be thought of as repeated addition. Exponentiation can be thought of as repeated multiplication.

Each of these operations has an inverse. They are, respectively, decrementation, subtraction, division, and taking roots.

The set of positive integers is closed under incrementation. But it is not closed under the inverse-- under decrementation. In order to create a set that is also closed under decrementation, we have to expand the set to include 0.

This new set is closed under addition, but again, it is not closed under the inverse operation-- subtraction. In order to make it closed under subtraction, we have to include the negative integers.

This new set is closed under multiplication, but not under inverse multiplication, i.e., division. To make it closed under division (excepting division by 0), we must expand the set again to include all rational numbers.

This new set is closed under exponentiation, but not under the inverse. To make it closed under the root operation, we have to expand it to include all imaginary (or complex?) rational and radical numbers.

Each new operation has to expand the set in order for its inverse to also be universally applicable within the set.

[thorn jay]

Slight gripe that decrementing 0 yields -1 so closure under decrementing is already closure under additive inverses

But yeah, true

You have this tower of algebraic completions
semiring without zero -> semiring -> ring -> field -> algebraically closed field
Called algebraic because every step is extending your structure to meet some algebraic requirement

[frigid epoch]

Why is logarithmic closure not discussed

[thorn jay]

Then, if you want to get to the real numbers, you notice that at the step "field" (rational numbers) you have a notion of distance, namely the absolute value, giving a metric space. As this works nicely with addition and multiplication and whatnot, you can complete this metric space (filling all the holes) and get another field out of it, yielding the real numbers

And then you can do the same algebraic closure to get the complex numbers

[thorn jay]

Logarithmic closure is fairly messy

I believe

[frigid epoch]

Well any real >1 can be taken as the base for a logarithm which then generates all others(usually e)

So you only need to iterate log2's (say) on the rationals, then do the field closure, hence repeat etc.

[glad osprey]

speaking of fields, I've heard that the algebraic closure of the p-adics is C, ie. the same as for the reals, but are the p-adics actually isomorphic to the reals as a field? I know the metric is very different, but what field theoretic difference is there between the reals and the p-adics?

[coral spindle]

They are very far from being isomorphic to the reals

For p odd, the p-adics contain (p-1)st roots of unity for example

[mossy slate]

Yeah I realized shortly after I sent it that I made a big error in my wording 💀

[coral spindle]

What's that react to say, sheddow?

If you need clarification, I am talking about primitive such roots of course

[glad osprey]

I don't mean to convey confusion, it's just interesting

[mossy slate]

But yeah basically, each time you add the inverse, the set of possible outputs of the operation is a superset of the possible inputs

[glad osprey]

what about the 3-adics vs the reals? They both contain 2nd roots of unity, right?

[coral spindle]

And I didn't mention the 2-adics :P

I cannot, off the top of my head, give you an example of a strictly non-real algebraic element in the 2- and 3-adics

But you can construct these things via Hensel's lemma

[glad osprey]

I see

[thorn jay]

Upwards closure

My beloved

[glad osprey]

I'm guessing p-adics are mostly used in analytic number theory? I haven't really seen them used elsewhere so far

[coral spindle]

Algebraic number theory, really

And they're important in various bits of rep theory

They're surprisingly broad, I'd say

[glad osprey]

Interesting

[thorn jay]

How are they used there?

[coral spindle]

ANT gives us tools that relate local fields, their number rings, and finite fields of characteristic p

So it gives us a way to pass between fields of char 0 and fields of char p, really

This was used by Brauer to produce a character theory of finite groups in characteristic p

This was the first real instance of this

In some sense it is more natural to see representations of groups as lying in the algebraic closure of the p-adics than over Q, and certainly not the reals or complex numbers (ew)

See also: Navarro's conjecture

Aka the Galois–McKay conjecture, whatever you want to call it

[thorn jay]

My favourite disagreement on naming is "Burnside's lemma" and "The lemma that is not Burnside's"

[thorn jay]

Is there an underlying reason for this? Or is it really "it works because it works"

[coral spindle]

This remains unclear

[thorn jay]

Because i know that linear reps of (finite) groups in complex vector spaces are nice, because of the algebraic closure and char 0 of the complex numbers

I suppose the alg closure of the p-adics also have these nice properties but simply more of them?

[coral spindle]

Well ok let's be clear here: the niceness is just because of algebraic closure and being char 0 -- there are plenty of other such things

But when we're actually constructing the characters, right

We see them as being in some algebraic extension of Q

Right? For example it is usual to take some symbol x in the character table and give its minimal polynomial over Q

When we're constructing these Brauer characters – whatever these are – it's natural then to see them as living in an algebraic extension of Q_p

Now there are... complicated relationships between these Brauer characters and ordinary characters

So in some sense we would like to be able to see them as living over the same field

This is broad and vague, I know

[thorn jay]

I think I get it

[tough raven]

Explicitly, for p = 2, X^2 + X + 1 has a simple zero in ℤ/2ℤ hence in ℤ_2 ⊆ ℚ_2 but no zero in ℝ. For p odd, take any negative representative a of a non-zero quadratic residue modulo p - for example, a = 1-p - then X^2 - a has a simple zero in ℤ/pℤ hence in ℤ_p ⊆ ℚ_p but no zero in ℝ.

[coral spindle]

Nice, thanks Raghuram

[tough raven]

I wonder if this can be extended - say, to find a monic polynomial irreducible over ℚ which has a root in finitely many given completions and is irreducible in finitely many other given completions.

[tough raven]

In fairness, you only need to adjoin all roots of unity to ℚ, neither the full algebraic closure nor any transcendental elements.

I wonder if this is connected to Kronecker-Weber.

[thorn jay]

I noticed that you can approach representation theory of groups fully using universal algebra, as a representation and morphisms of representations can be fully faithfully described not unlike modules.
And in general for a variety of algebras V of type T you can get the variety of V-representations of G simply by adjoining unary operations f_g to the type which act as T-endomorphisms

Can anyon point to resources about this? I'm sure it's been done using category theory at least lol

hell, this can be done for any algebraic structure which has an underlying group where the group operations act as endomorphisms on the reduct

[gilded fulcrum]

do finite fields have odd # of elements?

[violet spade]

Z/2Z

@gilded fulcrum

[gilded fulcrum]

other than that one

[violet spade]

Finite fields always have p^n elements, where n is a positive integer, and p is prime. So the only case when a finite field has an even number of elements, is when p=2.

[gilded fulcrum]

Alright thanks

[coral spindle]

There are infinitely many (non-isomorphic) finite fields with an even number of elements.

[thorn jay]

exactly countably many of them in fact!

[coral spindle]

Indeed

But this is not so impressive, because there are only countably many isoclasses of finite rings anyway.

And you know very well this is a general statement

[thorn jay]

:P

I am indeed one who should know that yes

it being true for any variety of algebraic structures

a far more interesting question being "what finite orders actually occur in this variety"

the only nontrivial ones (either the full set of integers or just 1) I can think of are vector spaces over finite fields and boolean algebras

tbh

[coral spindle]

And of course one of those examples encapsulates the other

[thorn jay]

in a sense I guess

[gilded fulcrum]

if we take the product of all nonzero elements in a finite field, is it equal to 1?

[thorn jay]

boolean algebras <-> boolean rings -> F_2-vector spaces

[thorn jay]

yes, because the additive group is abelian

wait, no that only works in even ordered groups

[coral spindle]

Not sure how that follows, enpeace

[thorn jay]

nvm

odd*

odd*

not even

❌ even

[coral spindle]

Think about this. I can pair up all the elements with inverses, so that:
$$ \prod_{x \in F\setminus{0}} x = \prod_{x \in F\setminus{0},\ x = x^{-1}} x $$

[cloud walrus]

$\mathbf{Boytjie}$

[coral spindle]

Therefore you can reduce your question somewhat, right?

You may want to know that the group F\{0} is cyclic for all finite fields F, and you should have your answer.

[gilded fulcrum]

ok

[coral spindle]

This response does not fill me with hope

[gilded fulcrum]

your response fills me with hope

[coral spindle]

Great

[coral spindle]

Hold on. What ordered groups are finite?

[thorn jay]

even-ordered groups

groups of even order

[coral spindle]

Great

[thorn jay]

but then I, quickly corrected myself

to groups of odd order

I do not have my mind in the game today

[spare nymph]

Let's see what the discuss is about 👀

[tardy hedge]

Lol

[glad osprey]

The idea is that ||we only need to look at the product of all elements of order 2, right? Which in an integral domain is 1 and -1, so the product of all non-zero elements is just -1?||

[rocky cloak]

Indeed. This is also known as Wilson's theorem.

Well, maybe not exactly the same as Wilson's theorem, but the proof is the same

[glad osprey]

ah, I'm familiar with that from number theory, but never seen it in the context of rings before

[thorn jay]

I've seen it in the context of complete mappings and mutually orthogonal latin squares

[glad osprey]

that doesn't count, you see everything in the context of latin squares

[thorn jay]

at least, the notion of multiplying every element

[thorn jay]

😭

everything is connected..

I read in "the" Latin square book

not kidding

it's very dense, so mainly good for references

[coral spindle]

Yes and this generalises to all finite fields, of course, since the group of units is cyclic ||there is exactly one element of order 2, if there are any at all.||

[thorn jay]

yeah okay, this proves that an abelian group has an orthogonal mate iff it has not exactly one element of order 2