groups-rings-fields

[coral shale]

on the topic of notation

[formal ermine]

my question isn't related to galois theory

this is field theory

[coral shale]

oh wait not nlab

i think chm

shared columbia thingy

notes

||Field theory is galois theory ||

[tribal niche]

The galois group of a galois extension K/L is the group of automorphisms of the extension which fixes L pointwise

[formal ermine]

🙄

I mean

as in

[coral shale]

jk

[formal ermine]

my course hasn't started galois theory yet

[coral shale]

well this is like galois stuff already

but the prerequisites to understanding

[formal ermine]

we did the primitive element theorem

that was the last thing we did today in lecture

[next obsidian]

https://stacks.math.columbia.edu/tag/09GZ

[tribal niche]

U guys are like 2 weeks behind my exact course lol 💀

[lavish gull]

galois theory depends on field theory i think, not the other way around

[tribal niche]

Holy shiz

[next obsidian]

I mean studying embeddings of stuff into the alg closure

[coral shale]

yushhh i remember this

[next obsidian]

This is already basically Galois theory

[formal ermine]

every algebra course on the world doesn't start at the same time

[coral shale]

ah yeah thats the word

embedding

[tribal niche]

Yeah of course lol it's just really cool that we are learning the topics at the same time

Since I'm 2 weeks behind my class 😭

[lavish gull]

you can look at the fundamental theorems, they are different

[tribal niche]

There are few things you can do with fields without utilizing galois theory as far as I know, though I could be very mistaken

[lavish gull]

yeh like the field axioms

[coral shale]

i feel like i get so confused when galois go to finite fields

idk

and infinite im happy with

[tribal niche]

Char 0 is definitely much simpler imo

[lavish gull]

except that you get the non-separable case, which doesn't happen in Char != 0 i think

[tribal niche]

I'm still a galois novice though I kinda suck I basically failed the galois theory exam in my course

[coral shale]

its probably precisely because char 0 => separable

[tribal niche]

Yeah, but we can ignore non separable extensions I hate them anyway

[lavish gull]

ignore extensions of Q, the rationals?

[coral shale]

x^n+y^n = (x+y)^n is based

tho

[coral shale]

?

[lavish gull]

afaik finite fields always have a cyclic galois group, so that case is easy too

[lavish gull]

i see, you mean the opposite. ignore non-separable extensions

[next obsidian]

You can prove it inductively

Handle the degree 2 case

[coral shale]

prove what?

[next obsidian]

Then use the theorem that if there’s one subgroup of every order dividing the order of the group, you’re cyclic

[coral shale]

oh ic

[next obsidian]

This relies on knowing the structure of finite fields tho

But that’s kinda easy

Actually you don’t even need induction

[coral shale]

oh whats this

not sure i hearda it

[next obsidian]

By induction, all smaller subgroups are cyclic

Now you basically do some screwy element counting

[lavish gull]

that isn't sufficient to be cyclic afaik

[next obsidian]

And arrive at a contradiction if it isn’t cyclic

[knotty frigate]

FLT is sad

[lavish gull]

why sad?

[coral shale]

it just got disproven

[next obsidian]

Chmowned

FLT stays losing

[lavish gull]

you mean Freshman's Dream, not FLT?

[coral shale]

flt in non-char 0

we have infinite solutions ez

[knotty frigate]

no i do mean fermats last theorem

[formal ermine]

is the splitting field of $(x^2 + 1)(x^2 + 3)$ (roots are $\pm i$ and $\pm i\sqrt{3}$) $\bQ(i, \sqrt{3})$ or $\bQ(i, i\sqrt{3})$

[cloud walrus]

rectangle cube

[next obsidian]

Those are the same

[coral shale]

indeed

[next obsidian]

Lmfao

[formal ermine]

yeah I was thinking that

but I had a brain fart

thanks

?

[coral shale]

ive found a c though lmao

[knotty frigate]

he was responding ot me

[lavish gull]

FLT is a^n + b^n = c^n, nothing to do with distributivity as far as i know

[coral shale]

c = a+b

given the right field

ok, was silly joke

[knotty frigate]

yeah it was

[lavish gull]

well there are a couple of active questions

[knotty frigate]

but whatever

[lavish gull]

what interests me is this

[coral shale]

according to chm, induction.

so the induction step is all the subgroups are cyclic (except the biggest one)

and from that hopefully u get there

[next obsidian]

https://math.stackexchange.com/questions/525643/at-most-one-subgroup-of-every-order-dividing-lvert-g-rvert-implies-g-cyclic

[coral shale]

at most

[lavish gull]

thanks!

[next obsidian]

Literally exactly the same strength tho lmfao

[coral shale]

bruh

[next obsidian]

If at most then is cyclic so actually there is one

[coral shale]

wait is the statement u said

not wrong

[next obsidian]

I said exactly 1 subgroup of each order dividing the group

[coral shale]

no u said 1

the way i read it it was 'there exists'

[next obsidian]

What

[coral shale]

rip

[lavish gull]

it is well known imo that, a cyclic group of order N has a subgroup of every order dividing N, but is the converse true?

[next obsidian]

Yes but it’s the same!

If you have at most one then you’re cyclic by this theorem

So actually there’s exactly one of every order didicing |G|

[coral shale]

yh i read yours as at least

[lavish gull]

are you saying this @next obsidian

[coral shale]

wait ur saying both are true?

[chilly ocean]

it's false then consider (C_2)²

[next obsidian]

No

[lavish gull]

seems to be the suggestion

[next obsidian]

You need exactly one

I never claimed it works if you have more

[coral shale]

wait mines kinda silly isnt it

[next obsidian]

But this is all you need to show finite fields Galois groups are cyclic

[coral shale]

yeah my interpretation is silly

given n divides n

[lavish gull]

ok then i don't undertand

[next obsidian]

Because there’s only one intermediate field of any possible order

[lavish gull]

this isn't a trivial thing imo

[coral shale]

mine is a tautology

hahaha

[lavish gull]

how would you show it?

[next obsidian]

Show what?

[lavish gull]

show the claim

[next obsidian]

Which claim?

[lavish gull]

the claim i replied to, where you say "show..."

[coral shale]

wait all finite field galois groups are cyclic?

[next obsidian]

I have you blocked so idk what you’re replying to

[coral shale]

i completely missed this

[next obsidian]

No

[coral shale]

sry edited

[next obsidian]

Galois groups of finite fields

Yes

[tribal niche]

no. the galois groups of the extensions of finite fields

[next obsidian]

It follows from the thm I said

[lavish gull]

the actual theorem is that "all galois groups of extension of finite fields are cyclic"

[coral shale]

looks like some fun thinking

[next obsidian]

The Galois group’s subgroups are in bijection with the intermediary fields

There’s at most 1 of those of any fixed degree

So by the thm I was talking about for finite groups

[lavish gull]

why do you have me blocked?

[next obsidian]

You see the Galois group is cyclic

[lavish gull]

not true in general

[next obsidian]

Idk, I block ppl when they’re excessively annoying so you probably kept saying some Hurb shit one day

[lavish gull]

annoying?

[lavish gull]

you have various claims outstanding. you are the excessive one

[next obsidian]

Yeah, so this is the type of shit that made me block you

[tribal niche]

isn't it enough to just consider the splitting field of x^(p^n)-x or something i vaguely remember the argument professor made for finite fields

[next obsidian]

Omegalol

[lavish gull]

insert some shit to sidestep the proofs

about to call bullshit on this entire server because of @next obsidian who never read
a) the fundamental theorem of galois theory
b) the fundamental theorem of field theory

[next obsidian]

Idk probably

[chilly ocean]

#groups-rings-fields

[next obsidian]

Dude TTerra you reading this shit?

[tribal niche]

TTerra is always reading

[chilly ocean]

no but let me read it now

[next obsidian]

Idk the fundamental theorem of Galois theory bro

[chilly ocean]

LMAOOOOO

this is good

trying not to laugh out loud in the library

[tribal niche]

it's basically that the degree of field extensions are associated with the index of a subgroup

[chilly ocean]

i think chmonkey knows this...

[next obsidian]

I literally do tho

Is my point

[tribal niche]

how are we arguing about fucking GALOIS THEORY

[next obsidian]

The structure of intermediary subfields is exactly the structure for the subgroups of the Galois group

[tribal niche]

of all things

[next obsidian]

Which is why it suffices to know there’s one finite field of every order

[tribal niche]

one of y'all gonna die in a duel at 20

[chilly ocean]

at 20

[next obsidian]

To know the Galois group has one subgroup of every possible order

Which by the theorem I stated for groups

Says the Galois group is cyclic

Why this means I don’t know the fundamental theorem of fields or Galois theory idfk

[tribal niche]

everyone getting mad over math proofs 💀

[next obsidian]

I’m not even mad I’m just confused

[tribal niche]

dawg i don't even know what a trie is and i have a data structures final in 2 hours i'm just sitting here laughing at this bruh

[lavish nexus]

damn
trie is actually a word

[stoic rose]

By 6.1 (2) it suffices to prove that there is an irrep on which Z(P) acts faithfully. If Z(P) has order p^k with generator g, this is the same as showing that there is an irrep rho such that rho(g) is a root of unity of order exactly p^k. If rho(g) was of stricly smaller order for every irrep rho then you can get a contradiction using the regular representation.

[coral spindle]

Me too 🫂

[agile burrow]

I'm trying to compute the homology of Q_8 with trivial coefficients. From my understanding, all of the homology is concentrated in a 2-primary component, which will be isomorphic to the invariants in the homology of a 2-Sylow subgroup under the action of Q_8 by conjugation.

Say we fix the 2-Sylow subgroup to be <i> = C_4, which in particular is normal, so we just have an action of Q_8 / <i> = C_2 on the homology of C_4. I'm confused on how to compute this action. In any case, I've already made a mistake in this setup because if I look up the homology of Q_8, we get that H_1(Q_8) = Z/2 + Z/2, which isn't even a subgroup of Z/4 = H_1(C_4)

[coral spindle]

Ehm, Q_8 is of order 8 right?

So its Sylow 2-subgroup is just itself

[agile burrow]

oh

man i feel dumb

[coral spindle]

🫂

Happens to the best of us (as this rightly shows)

[agile burrow]

hmm so is a spectral sequence the way to go about this?

[coral spindle]

Walter I do not know group cohomology, I just happened to notice that

[agile burrow]

fair enough, thanks friend

[coral spindle]

I can ask someone who does, if you'd like

[coral shale]

chill, jesus.

[coral spindle]

but their answer would take a while

[agile burrow]

I'll give it a shot first, at least I can check my answer online

[coral shale]

lmao they're out

[agile burrow]

thanks again

[coral spindle]

Finally banned yay

No worries

[coral shale]

no they left, im fairly certain

[coral spindle]

Oh that's cool too

[chilly ocean]

Tfw abstract algebra attracts weirdos

[coral spindle]

Kinda a math problem tbh

[simple mulch]

Hello

Take a group of order 2p

p odd prime

cauchy tell us there's $x,y \in G$ with $|x| = p$ and $|y| = 2$

[cloud walrus]

mns

[simple mulch]

since $\gcd(|x|,|y|) = 1$, I claim $\langle x,y \mid x^p = y^2 = 1\rangle = G$

[cloud walrus]

mns

[lavish nexus]

might need some other relation

[coral spindle]

It does indeed

[lavish nexus]

in case you have a semidirect product

[coral spindle]

<x, y | x^p = y^2 = 1> is an infinite group

[simple mulch]

well yeah, we have the normality of $\langle x \rangle \triangleleft G$

[cloud walrus]

mns

[lavish nexus]

yes it has the smallest prime index

so it's normal

but <y> is not necessarily normal

[simple mulch]

oh really?

[lavish nexus]

I mean
D_3

[simple mulch]

was about to use this

[coral spindle]

<y> is typically not normal, yeah

[lavish nexus]

nah it's not always a direct product

[simple mulch]

fk

[coral spindle]

However, if you deal with the case where it is a direct product, you'll get the other case pretty easily via some Sylow theory

[lavish nexus]

plus that would mean all groups of order 2p is abelian

[coral spindle]

Or well, idk if you even need the Sylow theory

[lavish nexus]

if you have 2 | |Aut(Z_p)|

then it could be semidirect

but right Aut(Z_p) is like p-1 so

[simple mulch]

Well I am trying to show that the groups of order 2p are all isomorphic to either Z2p or D2p

and I started with cauchy

[lavish nexus]

yeah if it is semidirect you get D_p

[simple mulch]

got both x and y

and the normality of <x> in G

then I want to show their product is G

[lavish nexus]

their product has to be G

consider the order

just consider |HK|

[simple mulch]

so that I can manipulate the presentation of the group $\langle x,y \mid x^p = y^2 = 1, yxy^{-1} = x^i\rangle$ to show the cases for the cyclic and the dihedral case

[cloud walrus]

mns

[simple mulch]

|HK| = |H||K| = 2p

[lavish nexus]

yes so it's the whole thing

HK here is either H x K or H semidirect K because one of them is normal

then just work out how you map the nontrivial element y to Aut(X)
should be exactly the second one in the relation

[lethal lava]

Oh ok, I think I see. If rho(g) has order p^(k-1) or less for all irreps rho, then reg(g) would have order p^(k-1) (where reg is the regular representation), which is a contradiction because the regular representation is faithful. Is that correct?

[simple mulch]

hum

[lavish nexus]

because it has to have order 2
so there's only two choices for i

[simple mulch]

I don't see how it would yield the result, there must be some theorem that's missing

[lavish nexus]

one gives you the direct product

[simple mulch]

well yeah

[lavish nexus]

the other one the semi

[simple mulch]

yeah I know i either is 1 or p - 1

because it can be shown that $x^{i^2-1} = e$

[cloud walrus]

mns

[simple mulch]

so $p \mid (i+1)(i-1)$

[cloud walrus]

mns

[lavish nexus]

I think you're done at this point

[simple mulch]

Well yeah, I think the detail that $\langle x,y \rangle$ generates the whole group is missing

[cloud walrus]

mns

[lavish nexus]

uh they generate the two subgroups

[simple mulch]

well yeah

[lavish nexus]

then HK has order 2p

HK is contained in the group

and has the same order

so it is the group

[simple mulch]

and they have the same order

yeah indeed

that's what was missing

thanks

[lavish nexus]

yw

[stoic rose]

Yep

[lethal lava]

Great, thank you so much 🙏

[stoic rose]

That's a nice problem

[steep scroll]

how do i prove this?

[coral shale]

what have u tried

[steep scroll]

Nothing, i have no intuition

[coral shale]

Have you thought of say an explicit example to make sense of this

[steep scroll]

i proved the first part

[coral shale]

Thats what I would do

[steep scroll]

no thats a good idea

[coral shale]

So I would recommend

V is R3
W is some line

and X this line moved in a straight line

[steep scroll]

i confused why U contains W

[coral shale]

if this all makes sense

So say W is the x-axis

Then do u agree the set V/W is all the translations of this x-axis?

its W plus some vector (essentially)

{v + W} right

[steep scroll]

Yes, it'd be sets of of horizontal lines?

[coral shale]

yes

ok so now let X be a subspace

so you can see how V/W is 2D

let X be the subspace where u move the x-axis along the y-direction

W = {(x, 0, 0)}
X = {(0, y, 0) + W}

Now think what U must be

It needs to contain W and U/W needs to 'look like' X

[steep scroll]

U is a subspace of V, so um maybe like some set of points on the y axis

[coral shale]

hm well think about subspaces that contain the x-axis

first of all - what must be the dimension of U for all this to make sense

[steep scroll]

U must be a horizontal line?

[coral shale]

no

W is 1D
X is 1D

X = U/W

So U has to be 2D?

(more formally i should be writing dim(W) = 1, etc)

[steep scroll]

why does U have to be 2D?

[coral shale]

dimensions of the quotient

dim(A/B) = dim(A)-dim(B)

right?

not sure if uve seen this

[steep scroll]

ohhhhhhhh yeah

[coral shale]

So now u need to consider planes that include the x axis

In particular, what comes out when you quotient such planes by the x axis

And that will answer this example

=======
As for proving the general case...
You're going to have to work explicitly with the algebra, but I imagine the thought process wont be too far off

[steep scroll]

Oh that makes sense

[coral shale]

In particular the 'tough' part is figuring what U should be for the general case

But the example should clue you in

[steep scroll]

so if U is planes hat include hte X axis

then U/W is ??

[coral shale]

So by defn, U/W = {u + W}

the particular W being the x-axis

[coral shale]

and remember we want this to be equal to X

[steep scroll]

is U the set of all 3d planes?

[coral shale]

no

You need to figure out what U is

we established it had to be a 2 dimensional subspace of V

V = R^3
W = {(x, 0, 0)}
V/W = {v + W}
X = {(0, y, 0) + W}

W subspace of U (subspace of V)
X = U/W = {u + W}

---

to sum up everything so far

[steep scroll]

im sorry

im still lost at how to get U

[coral shale]

sure so did you follow

how Ive written X in 2 different ways

X = {(0, y, 0) + W}
X = {u + W}

Using this, you need to figure out what U should be

So first of all, U needs to contain the y axis or this fails

[steep scroll]

u is the set of lines with y component only?

[coral shale]

no u is members of U

sry i wrote it casually

$$X = {(0, y, 0) + W: y\in \bR}$$
$$X = {u + W:u\in U}$$

[coral shale]

So yes U needs to contain the y axis

but remember, U also needs to contain W

so it contains both x and y axis

[steep scroll]

U set of planes where z = 0?

[coral shale]

just the one plane

the xy plane

perhaps try to draw diagrams of everything in this example to make sense of it

[steep scroll]

oh that makes sense

for the xy plame

u could do it at z=1 too?

[coral shale]

thats not a subspace

[steep scroll]

oh bc it doesnt contain the zero vector

thank u so much

i think i understand it now

[chilly ocean]

https://en.wikipedia.org/wiki/Correspondence_theorem

You can take $U = \bigcup X$

[cloud walrus]

Blitz

[coral shale]

animal-noises-v1

[coral shale]

With finite field stuff, do ppl have some sort of visualization? With extensions of Q in C, I can at least kindof visualize in the complex plane

I never got comfortable with finite field stuff

[delicate bloom]

Something that helps is to know is the multiplicative group is a cyclic group.

[coral shale]

in what particular way?

it kindof doesn't "mesh" in a nice way with the additive one right

[delicate bloom]

Pretty much, but I'm sure there are a few examples in part where it does

Nothing great comes to mind really though

Hopefully someone has something more useful to offer than me here lol

[kind temple]

i remember drawing (F_p)^n as an n-dimensional grid for n = 1, 2, 3 and trying to identify subspaces of it if that helps. not sure if it would help too much with extensions

[coral shale]

thats what i have in my head.

or n p-gons

[chilly ocean]

Hi, i was reading this https://en.wikipedia.org/wiki/Frattini_subgroup and it says that the frattini subgroup is analogous to the Jacobson radical. But i don't get it, wouldn't the analogous thing be to consider the intersection of all maximal NORMAL subgroups?

[chilly ocean]

oh, I know Frattini subgroup

read about it in Bergman's Universal Algebra book

🍔man

[formal ermine]

berg means mountain in german

[chilly ocean]

interesting

i really don t get why wikipedia calls it "analogous to the Jacobson radical"

wikipedia is not always good

true

[spice whale]

i need some help ((a) specifically)

$IJ \subseteq I \cap J$

[cloud walrus]

lambda cube (alison)

[next obsidian]

This one follows near immediately yeah?

[spice whale]

so all i need to prove is the opposite

[next obsidian]

Right

Anytime you have an ideal equal to R

This is equivalent to 1 in that ideal

So I would write down 1 as an element of I + J

And then see if you can prove the reverse inclusion

[spice whale]

hm

ok

[coral shale]

IJ := {polynomials in i, j with coefficients in R} ?

[chilly ocean]

a = x+y, x in I, y in J. a in both
x = a-y in J, y = a-x in I

[next obsidian]

{Sum i_k j_k, i_k in I, j_k in J}

[coral shale]

ig coefficients in R dont matter; ideal defn

[next obsidian]

Ur message sucks cuz it just is the answer instead of providing a hint that they can use to then try and find the answer themselves

[chilly ocean]

I technically didn't answer it. What I said is not even useful

[solar inlet]

Then why bother saying it at all?

Just to flex your intelligence? You're not clever you just need to go touch some grass.

[lavish spoke]

why has this server been on defcon 1 past few days

[delicate orchid]

This is a rather pedestrian discord for pedestrian mathematicians it seems

[long nebula]

btw the reason that this def doesn't work is that for example IJ doesn't necessarily include J

[chilly ocean]

i wish people here spent more time talking about math and less time making jokes and arguing

I just made a mistake in what I'm trying to prove is all

the point was that we were both wrong

[lavish spoke]

quick someone drop a "no u"

[chilly ocean]

sounds like you have a problem

[lavish spoke]

no u

[rustic crown]

abstract-chill uwu

[chilly ocean]

i want to show a polynomial irreducible over Q(i) -- my suspicion is that it is enough to show that this polynomial is irreducible in Z[i] which is straightforward

[next obsidian]

This just works

Just use Gauss’s lemma

Z[i] is a UFD (it’s even a Euclidean domain)

[chilly ocean]

there is a more general version of gausses lemma? the only one ive seen is for Q and Z not Q(i) and Z[i]

[next obsidian]

It works for any UFD

[chilly ocean]

i see.. thank you!

over any UFD.. thats powerful

This makes me wonder, are there non-UFDs for which gauss's lemma works?

[next obsidian]

I would think so, but that’s an interesting question

There’s some “well-studied” weakenings of UFD

But idk who studies them lmfao, it feels like something that’s probably niche and just at some random unis

[rotund aurora]

isnt that a recurrent theme of algebraic number theory? lol

[chilly ocean]

I'd say in algebra

[rustic crown]

🙈

[rustic crown]

what happens with the example D = Z + t*Q[t], this is one of the nice example of a ring i know where gcd makes sense but it's not a UFD

the fraction field is just F = Q(t)

[spice whale]

I've ended up with
for some i in I, 1-i in J

[lavish spoke]

write i + j = 1 for i \in I, j \in J and consider z \in I \cap J as z times 1

[next obsidian]

I think this isn’t what you want to do

Start with x in I\cap J, then write 1 = i + j

Then we know x = x•1

So now try substituting

Oh, this is what George said kekw

[spice whale]

yeah i got it from george

[spice whale]

thank you

and thanks chmokey

[spice whale]

i tend to go down long roads and get nowhere sometimes

[chilly ocean]

This message is just a bookmark to remind me to check this again later when i have time.

[toxic zephyr]

for the left module of $2\times2$ matrices over itself, would ${I}$ and ${e_{11},e_{22}}$ both be valid bases? since
$$A=IA=e_{11}A+e_{22}A$$

[cloud walrus]

nilpotent nix

[south patrol]

Sounds good to me, though linear independence of e11 and e22 must be shown ofc

[next obsidian]

they aren't tho? am I dumb

[south patrol]

Wait

Yeah I didn't check

[next obsidian]

not a ring

wow

[south patrol]

Lol

[next obsidian]

gonna delete so I look silly?

[south patrol]

Sorry chmonkers

Wait also like silly q but

You said left module and then did e11A + e22A

I assume that is a mistake

[chilly ocean]

they also said that {I} is a basis

something's not right here

[south patrol]

Isn't I a basis?

[next obsidian]

I is surely a basis lol

[south patrol]

Since it is linearly independent and as a submodule it generates the whole ting

since this is like

A ring viewed as a module over itself

[chilly ocean]

well. Take 2x2 matrices over R

how can the identity matrix be a basis

[south patrol]

they are viewing it as a left M(2,R) module

[next obsidian]

because A = AI

[south patrol]

So like it is spanning due to that ^

And lin indep as if AI = 0 then A =0

[next obsidian]

it's being considered a left-module over M_2(R)

not as an R-module

[south patrol]

ye

Uhhh so are e11, e22 lin indep in this module meh im sleepy and just on phone lol

[next obsidian]

absolutely not

I think

[south patrol]

Wait yeah no

just like

[next obsidian]

uhhh

maybe you need right mult too

[south patrol]

Take e22e11 + e11 e22

= 0 +0

[next obsidian]

so true

[south patrol]

But e22 and e11 are not zero

lol

[next obsidian]

you a real one for that

[south patrol]

Wait wdym lol

Idk the tone lol

[chilly ocean]

[toxic zephyr]

oops yeah i think so. it would be AI and Ae11+Ae22 i think. but the principle is the same

[south patrol]

Ye

But yeah those r spanning but not lin indep

[toxic zephyr]

not linearly ind?

[south patrol]

Yeah

[toxic zephyr]

oh i see

hmm would it be possible to construct a 2d basis for this module?

[agile burrow]

I'm not sure, but I believe you'd want R to be non noetherian, or else the matrix ring is noetherian and it satisfies the invariant basis property

[south patrol]

Also uh random but I've seen online that sometimes u can find irreducible characters of a group by considering exterior/symmetric powers; is there much of a method to decomposing these though or is it just smth that occasionally works

Since usually I'd only decompose a character by already knowing the character table lol

[spice whale]

I've also gotten a bit stuck on c here

I've tried a bunch of different approaches and none have clicked together

Like I'm not sure what the isomorphism actually is

this being true isn't intuitive to me

[south patrol]

are I,J coprime/comaximal?

[spice whale]

I+J = R

[south patrol]

Yeah comaximal, cool

Hint: chinese remainder theorem!

[spice whale]

yeah I've tried that

[chilly ocean]

An isomorphism should obviously be $$r\mapsto (r+I, r+J)$$

[south patrol]

It should be immediate from that

[cloud walrus]

Blitz

[spice whale]

ah

[south patrol]

R/IJ \cong R/I x R/J right

[spice whale]

this isn't obvious idt

anyway

[south patrol]

Oh to me that just is the chinese remainder theorem

well I guess usually would have intersection but here the two coincide

[chilly ocean]

Look @south patrol

[south patrol]

at least assuming R is commutative lol

[chilly ocean]

It was different here

[south patrol]

Oh yeah lol okay they have a different form of CRT

but mine follows from that ig

[chilly ocean]

I'd just probably prove it directly lol

[spice whale]

thanks i got it from this

[chilly ocean]

Hi, can anyone explain me intuitively that what does it mean when I apply a relation on a particular set ?

[chilly ocean]

Pls tag me while answering

Like what does a relation on a particular set do ?

[formal ermine]

it's a condition for two elements to be equal under that relation

[chilly ocean]

A relation can be defined on only one set ?

[formal ermine]

I think an example would help:

[chilly ocean]

Like if I have set A = {1,2,3} and I define a relation on Set A then -
R = { (1,1) (2,2) (3,3) }

Am I right ?

[formal ermine]

that would be the definition of your relation, not the implication

it would imply that 1 is equivalent to 1 under R

[chilly ocean]

Oh, Then what type of questions can made on implication of relation ?

Can you give me any one example ?

[formal ermine]

if our set is $\bZ$ and our relation is $$R = \Set{(x, y) | \text{$x$ and $y$ have the same parity}}$$

[cloud walrus]

rectangle cube

[elder wave]

#proofs-and-logic

[formal ermine]

then we'd have $[3]_R = \Set{\ldots, 1, 3, 5, 7, \ldots}$ because they all have the same parity

[cloud walrus]

rectangle cube

[formal ermine]

[3]_R means "what elements in the set are equivalent to 3 under the relation R"

but yeah, go to #proofs-and-logic

[chilly ocean]

What is parity ?

[formal ermine]

odd/evenness

[chilly ocean]

Oh, but i think relations and functions are a part of abstract algebra ?

[formal ermine]

arithmetic is also part of abstract algebra, as you have to do addition sometimes

but that doesn't mean it's well-fit for this channel lol

[chilly ocean]

Sorry bro 😅

[coral spindle]

Relations and functions are not specific to abstract algebra. #discrete-math or #proofs-and-logic are more appropriate

[chilly ocean]

I didn't knew that, sorry

[chilly ocean]

This ring is really cool! It seems to be what happens if you take the ring of integers Z and freely adjoin a new element t that is divisible by every non-zero integer! (i didn't check this so might not be exactly that but still very cool)

[rustic crown]

also it's a bezout domain, so not just you can talk about gcds, you can also write them as linear combinations of your elements

in other words, every finitely generated ideal is principal

[next obsidian]

[rustic crown]

[rustic crown]

the proof i know is just a case bash, so not really enlightening

[coral spindle]

Perhaps it's best analysed as a graded ring? There are some obvious generalisations, right

Because the ring Z + tZ + t^2Z + ... + t^nZ + t^{n+1}Q[t] has similar properties

(At least I think!)

[rustic crown]

[wooden ember]

yo this is the first rings and modules exercise sheet i complete without asking for any help on this discord

proud moment

[formal ermine]

https://tenor.com/view/he-is-the-messiah-monty-gif-19570851

[chilly ocean]

can we perchance see the sheet?

[pastel cliff]

congrats

ive asked for help on most if not all of my hw's for the past yeaer

[languid hare]

[rustic crown]

me has also troubled chmonkey every now and then

[solar glacier]

question

how do I show 4x+2 is irreducible over Z[x] but Reducible over Q[x]

on my exam i did the silliest thing and modded out by the ideal generated by 4x+2 and said that if 4x+2 goes to 0 then -1/2 is not in Z

which i know is not the correct argument

[agile burrow]

Is it irreducible over Z[x]?

[solar glacier]

i wanna say so

though you can write it as 2 times 2x+1

[agile burrow]

You might have it the other way around, it's reducible over Z[x] but irreducible over Q[x]

[solar glacier]

oh no

shux

[agile burrow]

Sorry to be the bearer of bad news

[elder wave]

How could you

[agile burrow]

But yeah, the idea is that 2 is a prime element in Z, but it's a unit when you pass to Q

[rustic crown]

hi walter

[agile burrow]

hi det

[rustic crown]

[agile burrow]

Hope you're doing well around this holiday season

[rustic crown]

i saw a frozen lake today

it was so cool :p

[agile burrow]

what a lovely sight

[rustic crown]

[solar glacier]

so since 2(2x+1) in Q , 2 is a unit 4x+2 is irreducible?

[chilly ocean]

altho that is true, that is not why 4x+2 is irreducible

[agile burrow]

well, you'd have to show that every factorization of 4x + 2 over Q[x] would have a unit

[solar glacier]

how do you do that

every possible factorization?

[rustic crown]

pick a factorization

[solar glacier]

2, 2x+1

[rustic crown]

an aribtrary one >.<

4x+2 = f * g

[solar glacier]

ah

[rustic crown]

can you see why one of them is forced to be a constant?

(non-zero constants are units in the ring Q[x])

[solar glacier]

since the product is of nonzero elememts ?

or the polynomial is nonzero elements

[rustic crown]

(oh so i was saying we wanna show exactly one of them is a unit, so it suffices to show it is a constant)

so do you see why when you multiply to polynomials and you get a linear polynomial, then exactly one of the factors was a constant :p

lol idk what else to say without just telling the answer

[solar glacier]

this has to do with degree of the polnynomial being 1?

[chilly ocean]

yes

[solar glacier]

and degree of product is sum of degrees

[formal ermine]

what is an example of $[\bK[a] : \bK]$ not being finite

[cloud walrus]

rectangle cube

[formal ermine]

for a field extension $\bL/\bK$ and an $a \in \bL$

[cloud walrus]

rectangle cube

[chilly ocean]

have you tried finding one?

[formal ermine]

yes

can't think of one

[chilly ocean]

think harder :^)

have you heard of 🏳️‍⚧️cendental elements

[elder wave]

Lol

[formal ermine]

yes

why doesn't $\bQ[\pi]$ have $\Set{1, \pi}$ as a $\bQ$-basis though

[cloud walrus]

rectangle cube

[sharp sonnet]

how to write pi^2 as a Q linear combination

[elder wave]

pi^2

[formal ermine]

ah

got it

thanks

[elder wave]

Fuck

[sharp sonnet]

also you have to be careful

you mean Q(pi)?

[formal ermine]

no

[sharp sonnet]

is Q[pi] a field?

[formal ermine]

no

[chilly ocean]

[formal ermine]

because [Q[pi] : Q] isn't finite

[sharp sonnet]

[chilly ocean]

uhh

[elder wave]

[formal ermine]

in my lecture we had that K[a] is a field <=> [K[a] : K] < infty

[chilly ocean]

[sharp sonnet]

ok sure

but you talked about field extensions

Q(pi) is a field

[formal ermine]

yeah

[sharp sonnet]

and the degree is still not finite

and you asked about field extensions

so i just wanted to mention that

[formal ermine]

now I'm confused

[sharp sonnet]

yes?

[chilly ocean]

why are you confused

[formal ermine]

what does Q(pi) have to do with my question

[chilly ocean]

it has to do with this one

oh i see that s sneaky

[sharp sonnet]

Q(pi) is a field extension of Q, unlike Q[pi]

[formal ermine]

I'm talking about R/Q tho

[chilly ocean]

you should make stuff like that clear at the start

[formal ermine]

idk I thought it was clear cuz I was talking about a general field extension L/K and then oac talked about an example so I thought of pi which is transcendental in R/Q. mb, should've specified it next time

[sharp sonnet]

oh ok, i misread

but Q[pi] works, right, so all good?

[formal ermine]

yes

all gucci

[sharp sonnet]

the confusion (on my end) is here i think; its uncommon to use [L:K] when L is not a field

[formal ermine]

my prof uses it a lot lol

[sharp sonnet]

calling them rn

[spice whale]

does Fermat's little theorem apply to any element of a finite field of characteristic p

[coral spindle]

So, FLT should be seen as a specific case of Euler's theorem

Let me just state that for you in case you don't recall

$\varphi(m)$ is defined to be $|Z_m^\times|$, i.e., the size of the multiplicative group of the ring $\mathbb Z_m$. Euler's theorem states that for any $x \in \mathbb Z_m^\times$, we have $x^{\varphi(m)} \equiv 1 \mod m$.

[cloud walrus]

Boytjie

[coral spindle]

Now ofc, this is just the theorem of Lagrange in a specific case.

So the answer should become a lot clearer now

It's a no. However, x^{p^n} = x in the finite field F_{p^n}

[spice whale]

i see

[coral spindle]

This is just a consequence of the Theorem of Lagrange again

One more thing

Maybe you know this already, but this theorem is very nice

[formal ermine]

what's the name of this font

[coral spindle]

Let $F$ be any field, and let $G \leq F^\times$ be a finite subgroup of the multiplicative group. Then $G$ is cyclic.

[cloud walrus]

Boytjie

[coral spindle]

It's a clone of Zapf's masterpiece, Palatino. I think I used the newpxfont and newpxmath packages.

[formal ermine]

interesting

[spice whale]

trying to prove all finite integral domains are fields
i think I've got it now

[agile burrow]

I'm curious to hear how you prove it

[formal ermine]

||s -> rs is bijective||

[agile burrow]

I suppose I meant I'm curious how Ally proves it

[formal ermine]

right lol

[spice whale]

by pigeonhole forall a, a^n = a^m for n > m

and you can cancel since R is a domain

so forall a^(n-m) = 1

qed

[coral spindle]

Good proof

[agile burrow]

That's clever, yeah

[formal ermine]

I wouldn't have expected pigeonhole here

[spice whale]

i was gonna do some monoid stuff in the middle

[formal ermine]

me likes

[spice whale]

but figured out the cancellation part skips it

[coral spindle]

The same proof gives you the fact that any cancellative semigroup (i.e. one in which ab = ac implies b=c) is a group

[chilly ocean]

This proof is just an applied pigeonhole principle

[coral spindle]

(As long as it's nonempty!)

[chilly ocean]

Uh?

[coral spindle]

I forgot to say finite

whoops

[spice whale]

this one is cool as well
though i think you need to do a contradiction to prove it's injective

shouldn't be too hard

[formal ermine]

nah

[coral spindle]

Showing surjectivity is the harder part

[formal ermine]

rs = rs' <=> r(s - s') = 0 <=> s = s'

[chilly ocean]

Yeah

[spice whale]

that's just finiteness

[coral spindle]

mmm yes

but prove it

[chilly ocean]

you also forgot to say non-empty

[elder wave]

☝️

[coral spindle]

nah actually I didn't

[chilly ocean]

apologies then

[spice whale]

this is just pigeonhole again i think

[chilly ocean]

Those semigroups generated by one element are called monogenic

[coral spindle]

Yes

[spice whale]

i was gonna say that

but i kept second guessing

[chilly ocean]

They turn out to be groups when elements are periodic, I think it's called

Which is not the case for infinite ones as then they turn out to be natural numbers

Hence the difference

[south patrol]

good theorem

[coral spindle]

One of the greats tbh

[chilly ocean]

Assuming cancellative here of course

[south patrol]

well i guess you can be more general lol

if G is any finite group with <= 1 subgroup of order n for each n, then it's cyclic ig

[spice whale]

not necessarily

[coral spindle]

Well, either 0 or phi(n)

No need to delete I wasn't roasting you

[south patrol]

lol ye

i forgot the exact statement

since i've only basically proven this directly lol

[solar glacier]

question wheenever this ones established

[coral spindle]

Maybe it is possible to reduce that to |{elements of order n}| <= phi(n)

But ofc we can probably say |{elements of order dividing n}| <= n

[south patrol]

I guess is the standard way to show $\sum_{d\mid n} \phi(d) = n$ just to consider elements of $\mathbb Z/n\mathbb Z$?

[cloud walrus]

potato

[south patrol]

there is a proof in terms of like

1/n, ..., n/n and reduce

but i imagine that is basically the same lol

[chilly ocean]

Ya

[coral spindle]

I think so, if we use (Z/nZ)/(dZ/nZ) \cong Z/dZ, where dZ/nZ is isomorphic to Z/(n/d)Z I think

[spice whale]

trolled

||none of them||

[south patrol]

uh okay yeah maybe i'm just thinking about it a different way lol

Just like φ(d) is the number of elements of order d right

Ah okay i see what you mean, sure

the subgroup of order n/d is dZ/nZ

[coral spindle]

I'm not sure phi(d) is the number of elements of order d

Like, additive order d?

Oh right yes. I see.

[solar glacier]

question

in general to determine wether or not an element in the gaussian integers

you you suppose it equals two elements then take norms

[south patrol]

Whether an element of what

like idk what the question is

[solar glacier]

say for instance in Z[3i]

[south patrol]

Oh checking irreducibility?

[solar glacier]

checking irreducibility of say 2+3i

well that has norm 13

which is

[formal ermine]

prime

[solar glacier]

yes

thus one needs to hve norm 12

1

**

norm 1

thus a unit

[formal ermine]

yes

[solar glacier]

and lets say for instance 3+3i in Z[3i]

thee norm is 18

which has possiblilites 3 and 6

which no element os Z[3i] has nor 3 nor 6

other option would be 2 times 9

but no element has norm 2

so only option is 18 and 1

so for eexample is 3+4i reeducible

and is my reasoning for why 3+3i is irreduc correct

[formal ermine]

1 + 1i has norm 2

ah wait we're in Z[3i]

[solar glacier]

yes

[formal ermine]

I thought Z[i]

oops

[solar glacier]

lol

so is my reasoning for why 3+3i is irredu correct

[formal ermine]

looks correct

[solar glacier]

im trying to find an element thats reducible

turns out 3+4i is irreducible as well

[formal ermine]

take two non-units then multiply them

[solar glacier]

ahh

[spice whale]

18

[solar glacier]

youre right

6 also

[spice whale]

is reducible

true

[formal ermine]

hi ally

[spice whale]

hi

Z[3i] is not a UFD despair

[formal ermine]

lmaooo

I fucking hate this shit

time to rant

my homework keeps introducing new notation without explaining it

[spice whale]

18 = 2•3² = (3+3i)(3-3i)

[elder wave]

is it just lin alg notation that you don't know again

[formal ermine]

no

Gal(L/Q)

[spice whale]

this is normal notation

[formal ermine]

yes

but we haven't even started with galois theory in the lecture yet

[spice whale]

automorphism group of a Galois extension

[formal ermine]

what is a galois extension

[spice whale]

a field extension with some specific properties

[formal ermine]

what properties

lemme just google

ah

normal and separable

[elder wave]

this is all on wikipedia

[formal ermine]

yes

[chilly ocean]

hi ally

[solar glacier]

is there a way using degrees of polynomials to determine irreducibility of polynomials in polynomial rings over either Z or Q

[coral spindle]

Not really

[solar glacier]

and can someone give me a small polynomial say linear which is irr in Q not in Z or vis versa and have me solve it out

[coral spindle]

There are some cases in which it's easier

e.g:

1. any polynomial of degree 1 is irreducible.
2. a polynomial of degree 2 is reducible iff it has a root
   This holds for any field*.

[solar glacier]

over Z and Q?!

[spice whale]

hello

[coral spindle]

Use Gauss' lemma for Z.

[solar glacier]

i havent been taught gauss' lemma

[formal ermine]

f is irreducible over R iff f is irreducible over Q(R)

[solar glacier]

ohj that i have been taught

[chilly ocean]

Thank you. I was getting increasingly sadder

[spice whale]

i was gone

[solar glacier]

so how do you determine wether or not a given polynomial is irr or red in Z[x] or Q [x]

if its degree 1

[rotund aurora]

its difficult lol

[solar glacier]

always red if youre over a field

but how do you show this

[solar glacier]

lol

[rotund aurora]

tbh, I have never cared about "factorizations" of polynomials which are just scalar multiples

and I think that's what you are talking about

[chilly ocean]

I just wanted to make a joke because rectangle cube did out of the blue but it got real

[spice whale]

[coral spindle]

Hey how do you illuminate a large area? Fire works! Hahahahahahaha

[chilly ocean]

Let's drink?

[elder wave]

🍻

[spice whale]

Carla!!!

hi!

not really I've talked to you on tau

but hello regardless

yes algebra is cool

[coral spindle]

Now we drink

[south patrol]

Now I want a drink

[spice whale]

where did carla go

oh ban evasion

[chilly ocean]

What's a cube anyway I omly know cube diagrams from category theory

[spice whale]

[formal ermine]

a cube is like a triangle but more polygony

[chilly ocean]

Commuting one?

[coral spindle]

A cube is a bunch of 3-simplices if you glue them up good-like

[spice whale]

a cube is ${x \in \bR^n : |x|_\infty = 1}$

[cloud walrus]

lambda cube (alison)

[rotund aurora]

(+-1, +-1, +-1)

[formal ermine]

only in ohio

[rotund aurora]

guys

the parametrization of the vertices of the icosahedron and dodecahedron is very easy

[coral spindle]

In algebraic group theory, a "torus" is a direct product of k^\times. Which I think is a bit of a funny name

[rotund aurora]

if you learn them, they might save your life if you have to face some question about them, for real

[south patrol]

Maximal torus of my face

[coral spindle]

🅱️orel 🅱️ubgroup

[south patrol]

Very true

[coral spindle]

we all agree

[coral spindle]

go on then I'm waiting! I want to see this!

[rotund aurora]

wikipedia

[coral spindle]

Neat!

[rotund aurora]

where $\varphi=\frac{1+\sqrt 5}{2}$

[cloud walrus]

Croqueta

[coral spindle]

ofc ofc

[chilly ocean]

But is it a commutative icosahedron?

[coral spindle]

It takes the train

[south patrol]

I used icosahedrons yesterday lol

Well char table of A5

Phi comes up

[coral spindle]

And ofc phi does make an appearan—jinx

[rotund aurora]

the icosahedron is actually cool. I should study more geometry, and read Klein

[south patrol]

Not sure if there is a decent way to do it without ico

[coral spindle]

Ehm let me think

[south patrol]

Well at least, it seems most efficient to use it

Lol

[coral spindle]

Iirc you can use some orthogonality bullshit to get it

but it's not easy

[south patrol]

might have another go but eh

[coral spindle]

I think using the icos is equally hard tbh

[south patrol]

Think is like there are only two obvious characters and you need 5

Well with the icos you like

Trace of a rotation is easy

[coral spindle]

Yees indeed. I don't remember how I got it.

[south patrol]

which narrows it down a lot

Ye

[chilly ocean]

Do people study categorical geometry or something. Is that a thing

[coral spindle]

Does categorical rep theory count?

[chilly ocean]

Forget I asked. Obviously they do

[formal ermine]

@deft ferry if e and e' are both neutral elements then e' = e' e = e

[deft ferry]

ty 🙂

how to prove it?

[tribal moss]

Check the definition directly. It's not a "be clever" exercise, it's a "be sure you remember the definitions" exercise.

[solar glacier]

question

if youre irreducible over Q[x] are you irreducible over Z[x]

[formal ermine]

yes

gauß's lemma

[next obsidian]

Not true

You require the polynomial to be primitive, meaning the coefficients have gcd 1

Take for example 2x, this is irreducible over Q[x], but not over Z[x] since 2x = 2•x

[formal ermine]

(I forgor this condition)

[solar glacier]

but 4x+2 is irreducible over Q but reducible over Z

[celest cairn]

Hi

I was trying to find the minimal polynomial of $\mathbb{Q}(\sqrt[6]{2}) / \mathbb{Q}$ using Galois theory. For example, we can compute the minimal polynomial of $\mathbb{Q}(\sqrt{2}) / \mathbb{Q}$. The minimal polynomial of $\mathbb{Q}(\sqrt{2}) / \mathbb{Q}$ can be computed using the formula $\prod_{\sigma \in Gal(L/K)}(x-\sigma(\alpha)) = (x-\sqrt{2})(x+\sqrt{2}) = x^{2}-2.$ How would I do this with $\mathbb{Q}(\sqrt[6]{2}) / \mathbb{Q}$?
I tried computing this but kept getting different roots and wasn’t getting anywhere.

[cloud walrus]

Sapphire Gaming

[rotund aurora]

Its easy to see what are the roots in that case. Hint: ||sixth roots of unity||. But it seems what you want to do is find the Galois group of the splitting field of the polynomial x^6-2

And that is simply || Q(z, 2^(1/6)) where z is a sixth root of unity ||. Then you use the fact that the degree of the extension is the same as the order of the group and see that the possibilities for the automorphisms are limited

[solar glacier]

ist the other way around irr in Z gives irr in Q but not vis versa as 4x+2 is irr in Q but red in Z

[south patrol]

That is why the hypothesis of being primitive is important

[solar glacier]

is there anyone who wouldnt mind checking 30 problems of algebra problems for me lollllll

actually only like half of them

i have them in TeX in a pdf

only 12-27

theres only 27 questions

sorry

[high cypress]

Omg I’m trying to read it

But what does $H \le G$ mean?

[cloud walrus]

fifty_two

[high cypress]

I forgot

[winter yew]

Subgroup or same group

[solar glacier]

subgrou

[high cypress]

Just asking, why can’t they use $H \subseteq G$ ?

[cloud walrus]

fifty_two

[solar glacier]

thats subset

[high cypress]

I’m embarrassed to ask

But what is the difference?

[winter yew]

group is set + binary operation, not just a set

[solar glacier]

subgroup

has the property that

for all a,b \in H ab^-1 \in H

[high cypress]

Is that it tho?

[solar glacier]

yes and nonempty

thats all it means for subset to be subgroup

[high cypress]

So is $1 \in \mathbb{Z}$ a subgroup?

[solar glacier]

the singleton 1

[high cypress]

Even if it’s only one element?

[solar glacier]

in \z?

i wanna say yes vacusouly

but its not a subgroup as it doesnt contain the additive identity, am i right?

[cloud walrus]

fifty_two

[void cosmos]

no! (edited)

the trivial subgroup

[solar glacier]

vacuosuly right

[winter yew]

under multiplication I believe it is subgroup. Ander + - no

[solar glacier]

but z isnt a group under times

[void cosmos]

no

[solar glacier]

it doesnt have mult inverses

[void cosmos]

then

[high cypress]

Wait wait wait

[void cosmos]

yea

[winter yew]

ah, yes. true

[high cypress]

Why does a subgroup need an inverse?

[void cosmos]

in general if G is a group and e is the identity

then {e} is a subgroup

[solar glacier]

^

[void cosmos]

in this case no as Z isnt eveen a group under multiplication

{0} would be the trivial subgroup of Z

[solar glacier]

exactly what i was thinking

yes

0

sincee its the additive identity

[high cypress]

Wait

[solar glacier]

would u mind checming my work lol

[high cypress]

What about it’s inverse?

[void cosmos]

no i wouldnt

post

[void cosmos]

inverse of 0 is 0

[solar glacier]

0+-0=0

or just 0+0=0