groups-rings-fields

[chilly ocean]

@stone fulcrum can i ask u? 😃

[stone fulcrum]

Feel free to post it

[chilly ocean]

so this is a new base

this is polynomial in standard base

standard base

or basis whatever, not sure how u say it hahah

[somber bramble]

basis

[chilly ocean]

this is relation

matrix

equation

so what i'm wondering

does T represent a matrix which maps v to e?

oh and x1'-x4' represents values in new basis

coefficents

like the matrix maps x1' to e0, etc..

is that correct? 😛

[stone fulcrum]

There's a few transformations going on here.
-There's a transformation that maps the standard basis to your polynomial.
-There's a transformation that maps the new basis to your polynomial.
-There's a transformation that maps your standard basis to your new basis.

As we just discussed, matrix multiplication preserves each matrix's action on a vector, so we can use that here.

(Map from new basis to old basis)×(map from old basis to polynomial) = (map from new basis to polynomial)

Think of it like a trip.
(Going from A to B) × (going from B to C) = (going from A to C)

[chilly ocean]

aye

so i'm just wondering about the last step

[stone fulcrum]

None of this answers your question exactly because I'm not sure what you might be looking for here

[chilly ocean]

i just gave the rest of info to put it into concept

but what i'm wondering

matrix T basically tells the vector x1'-x4' how to map itself into standard basis?

for example it maps v0 into e0

[stone fulcrum]

I don't know what x1' is

[chilly ocean]

v1 into e1

[x1',x2',x3',x4']

the vector in last picture

[stone fulcrum]

OH I see

Yes this matrix maps the new basis into the old basis

[chilly ocean]

aye

aight that's all i wanted to know

😃

thank you so much @stone fulcrum @somber bramble both of u 😃

[stone fulcrum]

Np. Feel free to ask if you have anything else!

[steep quest]

does anyone have any recommendations for resources on magmas?

[oblique river]

it probably depends on what you want to learn about them. as far as I know, magmas aren't something that people just study for the heck of it

[somber bramble]

consider a geology book

[raw moth]

go to that slovakian city

[sick acorn]

@chilly ocean which group do you want to show is not cyclic? (Z/8Z)* or (Z/nZ)* (in general)?

if former, you might recall something about the order of generators of cyclic groups

[gilded marsh]

how u type that word?

[somber bramble]

what word

[gilded marsh]

that wor

word*

how?

[chilly ocean]

[gilded marsh]

hmmm

the word that @somber bramble typed

[somber bramble]

what, “geology”?

[gilded marsh]

yes

how

[somber bramble]

g-e-o-l-o-g-y

[gilded marsh]

hmmm

tell me pls

i know only little thing only

like...

font 1

font 2

font 3

/only

[somber bramble]

oh

~~word~~

just say you want struckthrough

[gilded marsh]

word

word

ouh thxxx

[chilly canyon]

From Lang :

What's the sense of aE ?

Is it

$$\mathfrak{a} E = \lbrace ax, (a,x)\in \mathfrak{a}\times E \rbrace$$

[cloud walrus]

Matplotlib:

[chilly canyon]

?

Then I can't visualize his map so well

[oblique river]

yes that's correct

it's just the things in E that are a multiple of something in a

[somber bramble]

wouldn’t that just be a

since ideals are closed under multiplications with elements from the ring

[oblique river]

for example, let's say R = Z and E = Z x Z and a = (2), then aE = (2)(Z x Z) = {2y | y in E}

[somber bramble]

oh wait

[oblique river]

which is just the submodule {(2x, 2y)} of Z x Z

[somber bramble]

things in the module not the ring

sorry

[oblique river]

correct.

[chilly canyon]

Okay, thanks 😃

But then, isn't his map weirdly-defined ?

The first one, not the induced

Like, for a class a in E/a and x in E, what is a.x ?

Is this the set {b.x, b in a} ?

[oblique river]

the idea is like

pick a lift of a to R

then multiply by x, now you get something in E

but oh wait, our lift of a wasn't totally canonical

it's only defined up to an element of a

so "ax" is only defined up to an element of aE

another way to think about it is just like

the map is:

lift a to E

oops fuck mistake

lift a to R

lift a to R, multiply by x, and then look at the image under the quotient map E --> E/aE

another way to say it using your notation is

if a in E/a then let's call that coset a + a

then a.x = the coset (ax + aE) in E/aE

[daring wolf]

non zero elements of GF(4) in GF(16) in terms of powers of alpha where GF(4) is a subfield of GF(16)

anyone know what are the non-zero elements?

[covert vector]

wat

α?

[oblique river]

GF(4) and GF(16) are fields

all fields have a "0"

and everything else is, well, not zero

[covert vector]

is α a generator of GF(16)*

if so then should be α^5

to be a generator of GF(4)*

cuz |GF(16)*| = 3×5 and |GF(4)*| = 3

[daring wolf]

yea alpha is a generator

but that's all that exercise asked o.o

so is that all the non-zero elements?

alpha^5?

@covert vector

[covert vector]

and what it generates

[daring wolf]

sorry if this is silly >< but how do we find what it generates? @covert vector

[covert vector]

multiply by itself over and over

do you know what a group is

[daring wolf]

yea but no other info was given

if we multiply itself we might very well never get back to 1

[covert vector]

but it's finite

[daring wolf]

that's why I am confused

yea exactly

GF are finite

[covert vector]

and it's a field

meaning if you ever got into a cycle

you can multiply by inverse

to get 1

[daring wolf]

alpha^5 times alpha^5 is alpha^10

then 15

then 20... etc

[covert vector]

yeah α^15 = 1

this is Lagrange's theorem

[daring wolf]

hmmmmm....

oh wow

I see...

dang thanks

[covert vector]

np

[daring wolf]

totally forgotten about it

[cedar gate]

In k[x]/(x^n), is x^(n+1)=0 or =x? I cant convince myself either way

Actually x^(n+1) = x*x^n, so it's in the ideal generated by x^n, so I suppose it should be 0

[somber bramble]

sounds right

[timber bay]

i'm just gonna put this here while I work on my other problems. I'm not sure exactly how to prove that first part. Thinking itll be some field properties or something

hopefully this other stuff im doing will give me some insight

[oblique river]

do you want a hint?

[timber bay]

i suppose

is it easy?

[oblique river]

I think that talking about the minimal polynomial is kind of a red herring

instead, look at the extension of F generated by alpha, F(alpha)

what is its degree over F?

F(alpha^2) is clearly a subfield of F(alpha). What could the relationship between those two fields be?

[timber bay]

i have to go. be back in a few minutes

well we know that the extension is odd degree, right? i'm not sure how to actually find the degree.

[oblique river]

what is the relationship between degree of minimal polynomial and degree of the extension?

[timber bay]

theyre the same

[oblique river]

yes

so F(alpha)/F is odd

what could F(alpha)/F(alpha^2) be?

(in degree)

[timber bay]

would it be 2 @oblique river

is alpha^2 considered the min poly in this situation?

[oblique river]

well it better not be 2

because degree is multiplicative

since you have a tower F(alpha) / F(alpha^2) / F

the degree of F(alpha) / F(alpha^2) divides the degree of F(alpha)/F

can you write down a polynomial with coefficients in F(alpha^2) that alpha satisfies

[timber bay]

why do we say F(alpha) is over F(alpha^2)? theyre both just over F, not necessarily over each other, right?

[oblique river]

because F(alpha^2) is contained in F(alpha)

that's all it means to be a field extension

[timber bay]

oh okay so if we have an extension Q(beta), then Q(beta^2) is under Q(beta) right?

[oblique river]

always

[timber bay]

just to make sure i understand it right

these field extensions are really tricky for me to grasp

[oblique river]

as is Q(beta^2 + 3*beta + 5)

being a field extension just means one field is a subset of the other

[timber bay]

any combination

[oblique river]

that's it

[timber bay]

yeah

[oblique river]

beta^2 is contained in Q(beta)

and thus Q(beta^2) is contained in Q(beta)

in the same way that Q(sqrt(2)) is contained in Q(4th root of 2)

because sqrt(2) = (4th root of 2)^2

there's nothing happening "under the rug" here

[timber bay]

yeah that makes sense

so the degree of F(alpha^2) is the same as F(alpha)?

and [F(alpha):F(alpha^2)] = 1?

[oblique river]

that's what you're trying to prove

subfields usually don't have the same degree

Q(sqrt(2)) / Q does not have teh same degree as Q(4th root of 2) / Q

[timber bay]

arent they both 2?

[oblique river]

what is the minimal polynomial of 4th root of 2 over Q

[timber bay]

x^4-2?

[oblique river]

what is the degree

[timber bay]

oof

shh

lol

im just still thinking of degree as the num of elements in the basis. and i only see those two extensions as having 2 element baseis

so we know there is an f(x) in F such that f(alpha) =0 but its not right to say that F(alpha) = 0

[oblique river]

degree is the number of elements in a vector space basis

which is different than the number of field generators you need

Q(4th root of 2) requires 4 elements in a vector space basis; one possibility is {1, 2^(1/4), 2^(1/2), 2^(3/4)}

that should probably not be your first reflex when you see degree

[timber bay]

okay i get that F(alpha)/F(alpha^2) must be odd btw

[oblique river]

I think about field extensions all the time and I basically never ever need to use a basis

[timber bay]

hm

i guess because im more comfortable with the idea of a basis than anything else

[oblique river]

so part of what you have to do now is practice using other notions :)

like sometimes bases are useful

but these are fields, not just vector spaces

you need to be thinking in terms of properties of them as fields, not just properties as vector spaces

[timber bay]

yeah i suppose

thats good

[oblique river]

for example: thinking about this problem in terms of bases will get you nowhere

[timber bay]

thanks for helping me realize that

[oblique river]

because we don't even know what the degree of any of these things are!

just that F(alpha) / F i sodd

let's go back to F(alpha) / F(alpha^2)

can you find a polynomial with coefficients in the base field that is satisfied by alpha

that will give you a bound on the degree of F(a) / F(a^2)

(here by bse field I mean F(a^2))

[timber bay]

is this saying [F(alpha):F] = odd?

am i understanding that correctly

[oblique river]

yes

degree of extension F(a) / F is exactly equal to the degree of the minimal polynomial of a over F

[timber bay]

if thats true and we know F(a)/F = odd, and that F(a^2) is a subfield of F(a) then isnt that enough to say that F(a^2)/F is odd? since theyre multiplicative

[oblique river]

(that should be something you've proven already)

[timber bay]

and an odd times an even is an even

[oblique river]

what you want to do is prove that F(a^2) is equal to F(a)

just saying that the degree is odd isn't enough

[timber bay]

i think the notation is throwing me off a lot

yeah i just didnt realize we had already proven the first part of a

part a

[oblique river]

oh, yeah

[timber bay]

practically

[oblique river]

yes we have done that

because F(a^2) is contained in F(a) so by multiplicativity of degrees, F(a^2) / F has odd degree, and thus the min poly of a^2 over F has odd degree

[timber bay]

yeah exactly

[oblique river]

the second part is slightly more interesting

[timber bay]

anything to do with simple extensions?

i assume no since we dont assume deg is finite

[oblique river]

what? odd numbers are finite

the degree is finite

[timber bay]

it says in the question to not assume the field extensions are finite degree

that's why i figured it wouldnt have to do with simple extensions

[oblique river]

the question you linked me said nothing about that

also, it tells you alpha is algebraic, which automatically emans that F(alpha) / F is finite

maybe E / F isn't finite but who cares? we aren't dealing with E at all

[timber bay]

oh sorry i was looking at the other question i was working on

[oblique river]

in any case, you were right that simple extensions aren't relevant here haha

[timber bay]

lol

[oblique river]

let's go back again to F(a) / F(a^2)

can you find a polynomial with coefficietns in F(a^2) that has a as a root

[timber bay]

i struggle with finding polynomials with this sort of stuff

i think you helped me last week i think you understand lol

[oblique river]

the fact that this problem is abstract (i.e. no actual numbers) is helpful here. don't get confused.

we literally have nothing to use to solve this problem

except for alpha and oddness

[timber bay]

great!

[oblique river]

myh point is that there really aren't that many possible things we could write down

[timber bay]

yeah

[oblique river]

so i'm not sure what to say other than to just giv eyou the answer here

[timber bay]

wait a second

[oblique river]

ok

[timber bay]

kind of want to think about it

[oblique river]

ok

the only thing you should focus on right now is "can you find a polynomial with coefficietns in F(a^2) that has a as a root"

(for this problem at least)

that is how you are going to solve this

let me at least give you a roadmap for the rest of the problem: let's say that we got lucky and found a quadratic polynomial over F(a^2) that had a as a root

that would mean that the degree of the minpoly of a over F(a^2) was either 1 or 2

(cuz the minpoly would have to divide whatever polynomial we wrote down)

but we know it can't be 2, cuz 2 is even... so therefore it's 1, and thus F(a) = F(a^2)

[timber bay]

so by coeff in F(a^2) that means the coeff is an integer or a combination of alpha^2?

i assume something like x-a^2 wouldn't work

or x^2-a^2

i know x-a wont work because a is not in F(a^2)

[oblique river]

I just mean that the coefficients of the polynomial you write down are allowed to be any element in F(a^2)

[timber bay]

yeah

[oblique river]

you're write that x-a^2 doesn't work, because the only solution to that is x = a^2

[timber bay]

its hard for me to understand exactly what that means

[oblique river]

and a is not equal to a^2

F(a^2) is a set of things

F(a^2)[x] is the polynomial ring over that

your polynomial should be an element of that

also you already said the right answer

it's x^2 - a^2

that's a polynomial

of degree 2

its coefficients are 1, 0, and a^2

all of which belong to F(a^2)

[timber bay]

okay

[oblique river]

and a is a root of that polynomial

because, well, a^2 - a^2 = 0

[timber bay]

so rational numbers are in F(a^2) as well?

not that its relevant just trying to understand it

[oblique river]

not necessarily

maybe F = F_121, the field with 121 elements

[timber bay]

hmm

[oblique river]

but F certainly contains 0 and 1

I mean, any field contains every integer

it's jsut that some of them might be equal to each other... like p = 0 in F_p

I have to go in a minute, hopefully you can finish this off yourself

using the outline I gave

[timber bay]

yeah i think i can

thanks a lot

[oblique river]

np and gl

[timber bay]

i appreciate the help

[timber bay]

maybe a dumb question but is the algebraic closure of something algebraically closed?

[thorny slate]

yeah

cuz it's an algebraic extension

so if something is algebraic over it it's algebraic over the base field

[timber bay]

makes sense

im callin the cops

[timber bay]

need to find the basis and degree of this extension field.

I feel like the degree is 9 but im not sure about the basis

it appears that each element has a minpoly of degree 3, and each have different minpoly's so that's how I get the 9

[thorny slate]

you can give it a nicer form

since you suspect the degree is 9 I imagine you already started to do that

[timber bay]

but im not sure about the basis. I thought it was 1, 2^(1/3), 2^(2/3), 6^(1/3) etc

i imagine itd be the same as Q/(a degree 9 poly)

[thorny slate]

what do you mean by etc

can you write down the 9 elements?

[timber bay]

i only get 7

1, 2^(1/3), 2^(2/3), 6^(1/3), 6^(2/3), 24^(1/3),24^2/3

[thorny slate]

yeah that's not how you do it

if you do this properly you would get 27

because you need the products of them

but many of those are going to depend on each other

so there's redundancy

why did you think it'd be 9?

[timber bay]

because it seems as if the min poly for each of the elements has degree 3

[thorny slate]

so then it would be 27

333

[timber bay]

oh yeah

is that not right?

[thorny slate]

nope

consider this

the minimal polynomial of i is x^2+1, of degree 2

the minimal polynomial of -i is the same, degree 2

and yet R(i, -i) / R = 2

another case, think sqrt(2) and 6rt(2)

what's [Q(sqrt(2), 6rt(2)) : Q]?

you would think 12 because the minimal polynomials are degree 2 and 6

but sqrt(2) = 6rt(2)^3

[timber bay]

is it just 6>

ohh

[thorny slate]

so Q(sqrt(2), 6rt(2)) = Q(6rt(2))

and the degree is 6

[timber bay]

at first i thought the degree would be 3 but thered not be enough elements to form a basis

since there's 3 elements youd need 4 elements in the basis?

[thorny slate]

huh

you know the degree of https://media.discordapp.net/attachments/496784958430380033/540668796310913044/unknown.png
is divisible by 3

[timber bay]

yeah i was thinking the basis would look like (1, 2^1/3, 6^1/3, 24^1/3

[thorny slate]

because it includes an extension Q(3rt(2)):Q

of degree 3

maybe this would help

split it up

[timber bay]

makes sense

[thorny slate]

into multiple extensions

[timber bay]

i dont think any of them are extensions of each other, are they?

[thorny slate]

[Q(rt(2), rt(6), rt(24)):Q(rt(2), rt(6))] [Q(rt(2), rt(6)): Q(rt(2))] [Q(rt(2)):Q]

and calculate each one directly

if you do this then you can just multiply the degree of the minimal polynomials that you find

the subtlety is that it's the minimal polynomials over the intermediate extensions

so for example the minimal polynomial of rt(6) over Q(rt(2))

and not just over Q

[untold sapphire]

using the good old tower law

[thorny slate]

you should notice how it works when computing it like that

[timber bay]

wouldn't each extension have degree 3?

[thorny slate]

compute them

[timber bay]

or two maybe

im not sure how you would compute that exactly

weve never done that in class

[thorny slate]

you find the minimal polynomial

and its degree

each extension is of the form [F(a):F]

the degree is the degree of a

[untold sapphire]

each time you extend the base field, you express the polynomial as a polynomial over the extension

is the key part to the calculation

[timber bay]

i don't know how to find the minimal polynomial of something like Q(rt(2),rt(6):Q(rt(2))

unless its just 2

or 1?

[untold sapphire]

how about trying to figure out a polynomial that has the roots you want, and then trying to see if it's reducible

[timber bay]

i dont know how to find a cubic polynomial like that

x^3-x isnt the answer

[untold sapphire]

what about $x^6 - 1$

[cloud walrus]

bdobba:

[untold sapphire]

?

[timber bay]

no?

cuberoot(2)^6-1=/= 0

cuberoot(6)-1=/=0

right?

[untold sapphire]

oh wait sorry it's not 6th root

[cloud walrus]

bdobba:

[untold sapphire]

idk if it's right I'm just playing about with it

[timber bay]

is it like (x^3-6)(x^3-2)?

and in that case shouldnt the total basis be 9 elements?

[untold sapphire]

is rt cube root?

[timber bay]

yea

[untold sapphire]

oh

I'm not saying your wrong, if none of them are extensions of the other then that gives you 9 elements yeah

but you need to check that

[timber bay]

i dont think theyre extensions of the other

i think theyre independent

[untold sapphire]

ah

wait

$sqrt(2)sqrt(6) = 2sqrt(3)$

[cloud walrus]

bdobba:

[untold sapphire]

so they are in the same field

and it's a degree 3 extension

I think

[timber bay]

i looked at the factoring of (x^3-24)(x^3-6)(x^3-2) and none of them had factors that also had zeroes of 2^(1/3), 6^(1/3), 24^(1/3)

that's square root

[untold sapphire]

yes it is! sorry

[timber bay]

lol

[untold sapphire]

yeah I suppose it must be 6 then

[timber bay]

for the first two elements?

rt(2) and rt(6)?

[untold sapphire]

I can't think of a single element that generates $2^(1/3), 6^(1/3)$

[timber bay]

lol

me neither

[untold sapphire]

$2{^(1/3)}, 6^{(1/3)}$

so if that's the case yeah it's 9

I forgot Galois theory was tough lol

sorry if I wasn't loads of help

[timber bay]

all of this stuff is really tough to me

extension fields do not click with me

[untold sapphire]

I got a really good mark in the exam but I don't know how tbh

[timber bay]

lol

[untold sapphire]

ah ok, what part doesn't click?

[timber bay]

like most of it

all the applications of extension fields

[untold sapphire]

because the conceptual stuff is worth learning even if some of the computations (like this one) are tough

ok so basically

we want to have roots to everything in our field

Generally that's where this problem first arises

So if we come across equations like $x^2 + 1 = 0$

[cloud walrus]

bdobba:

[untold sapphire]

We want to add elements to our field so we can solve it

[timber bay]

yes

i got that far

[untold sapphire]

so we literally just add add the elements (the roots) that has the properties we want

in that case we add two elements $r_1, r_2$ such that $r_1^{2} + 1 = 0$

[cloud walrus]

bdobba:

[untold sapphire]

and the same for $r_2$

[cloud walrus]

bdobba:

[untold sapphire]

and it turns out if we add those elements to our field, and they satisfy the relationship that r_1 and r_2 do, we get the complex numbers over the reals

[cloud walrus]

bdobba:

[timber bay]

yep

got it

[untold sapphire]

ok, fair enough

That's basically all a field extension is doing, but just with other roots

[timber bay]

yeah

its all the fairly complex applications that i dont get

this was my hw

i just finished it

i dont understand 3 that well but kind of do

but finished the others

[untold sapphire]

one thing to note with 3, if you suppose that the intermediate field E isn't an algebraic extension, then F can't be either

so they all sort of have to build on each other to be algebraic, because each field contains the roots of a subfield

Also if it makes you feel any better

[timber bay]

oh that makes sense

[untold sapphire]

We still don't understand

$\text{Gal}_{\mathbb{Q}}(\overline{\mathbb{Q}})$

[cloud walrus]

bdobba:

[timber bay]

im not familiar with that notation

[untold sapphire]

over $\text{Gal}_{\mathbb{Q}}({\mathbb{Q}})$

[cloud walrus]

bdobba:

[untold sapphire]

the bar means every single possible extension

[timber bay]

the closure i get

not the Gal

[untold sapphire]

It's just another way of writing The Extension

[timber bay]

wait is that not closure?

[untold sapphire]

yeah the bar is closure

The Gal just means it's a Galois extension, which is a field extension with some nice properties

[timber bay]

so we dont understand (Q-bar)_Q?

oh okay

[untold sapphire]

nope

[timber bay]

interesting

[untold sapphire]

so don't feel too bad 😃

[timber bay]

ill feel bad come test day

2 weeks from yesterday

[pale quarry]

Can anyone help translate this to an english statement, where all values are integers?

∃m, ∀n, ∃k, n=mk

[versed crescent]

there exists an integer m, such that for all integer n, there exists integer k, such that n = mk?

[timber bay]

@untold sapphire apparently 24^(1/3) = 2(3^(1/3)) and 3^(1/3)*2^(1/3) = 6^(1/3) so at most our minpoly has degree 6

[thorny slate]

yeah 24^(1/3) is irrelevant as you see

so we just have the other two

[untold sapphire]

Huh, fair enough

[signal tusk]

Hey does anyone have a book suggestion (that I can find online) for introductory group theory?

[thorny slate]

rotman's "a first course in abstract algebra"

[stone fulcrum]

I enjoyed Fraleigh's abstract algebra, it was fairly easy.

[wise grove]

Aluffi, just skip all the category theory

[solar wyvern]

@signal tusk milne

[random crag]

theres a pdf of rotman’s intro to group theory as well

https://superoles.files.wordpress.com/2015/08/graduate-texts-in-mathematics-148-joseph-j-rotman-auth-an-introduction-to-the-theory-of-groups-springer-verlag-new-york-1995.pdf

[solemn yew]

Hey. I've just started with group theory and I'm struggling with a question. I can't figure out how to show that integers modulo n are closed under modular multiplication

Where n is a prime

In particular, I'm stuck of showing closure under group operation

There are a couple of other hiccups, but this is the foremost

[solar wyvern]

@solemn yew so you're trying to show (Z/pZ)*is a group right?

even if n \neq p closure shouldn't be an issue

[somber bramble]

closure is pretty much by definition, since you take thr product and then modulo them back into the set

[solemn yew]

Is that the same as {0,1,2...,n-1}? @solar wyvern

sorry im new at this!

[solar wyvern]

do you...not know quotient groups yet

[somber bramble]

(or, if you've defined it via equivalence classes, then it'll always be in one of the equivalence classes)

[solemn yew]

not yet

[solar wyvern]

do you know cosets?

[solemn yew]

i see @somber bramble

[somber bramble]

you don't need that knowledge to talk about integers mod p

you can simply define it as a group by itself

[solar wyvern]

there's really no reason not to know that much tho

[somber bramble]

One can be literally at the beginning of the class

[solemn yew]

this is pretty much the first chapter of the book im using @solar wyvern

[somber bramble]

integers mod p is like the second example of groups

[solar wyvern]

do you know what a group is

[solemn yew]

cosets etc are far away for now

yes i do know what a group is

[solar wyvern]

what's a group

[solemn yew]

set, binary operation

associative, closed, inverse and identity

i hope that's it

[solar wyvern]

probably

[somber bramble]

it is

[solar wyvern]

so a coset of a group is just this:
choose a subset (usually a subgroup) H ⊂ G,
then its coset aH (fora ∈ G) is the set of elements {ah:h ∈Ｈ}

[somber bramble]

honestly I don't think explaining quotient groups is productive

you're overloading information

they didn't ask for it

[solemn yew]

but is knowledge of cosets and quotient groups necessary to show this set is a group under the operation?

[somber bramble]

theyre solving "prove this is a group" examples

no it is not

[solemn yew]

yes

[somber bramble]

give them time to work through stuff cleanly

these are important exercises

[solemn yew]

i can see that closure is trivial because of the way the operation is defined

[solar wyvern]

might want to show that explicitly nonetheless

[somber bramble]

it's like i'm asking about how to add fractions and you start talking about fraction fields

[solar wyvern]

how do u have fractions without fraction fields...unless you just straight localize

[somber bramble]

you don't but you don't need to know them do do it

[solemn yew]

i had a similar question

the book im using defines the set U(n): {a: a is an integer less than n, such that a and n are coprime}

[solar wyvern]

unusual notation, typical group

[solemn yew]

im confused in showing closure in this set too

operation is modular multiplication

i take 2 elements a and b that lie in U(n)

then what exactly do i have to show?

[somber bramble]

show that ab is in U(n) too

[solemn yew]

(ab)mod n is coprime to n and less than n?

[somber bramble]

where ab is the modular multiplicatopn

[solar wyvern]

ab mod n < n

[solemn yew]

the latter is obvious

[somber bramble]

so what you wrote

indeed it is

so that's all

[solemn yew]

how do i show the former?

[somber bramble]

no, with ab I means modular mult, so in this case ab mod n

[solemn yew]

yes i got that

[somber bramble]

oh but you also need to show coprimality ye

[solemn yew]

yes

that's what im having trouble rn

[solar wyvern]

what happens if they're not coprime @solemn yew

[solemn yew]

let me see

[solar wyvern]

yeah, good idea is to just explicitly do a bunch of examples with modular arithmetic if you're not familiar

did you show a,b always coprime for n=p?

[solemn yew]

one more thing: (ab) in this case is [(a mod n)*(b mod n)] mod n right?

[solar wyvern]

not really much to show I mean but do you know that

[somber bramble]

yea, but you can also just write ab mod n

the extra mods won’t do anything

[solemn yew]

sorry, I'm trying to compile all the info!

@solar wyvern what is p here?

in n=p

[solar wyvern]

that's how multiplication is defined on the group @solemn yew
$$(a \mod n)(b \mod n) = ab \mod n$$
so you're basically showing this is well defined

[cloud walrus]

flimflam:

[somber bramble]

p usually stands for prime number

[solemn yew]

ok

[somber bramble]

just useful shorthand, not a formal definition

[solar wyvern]

yeah with group stuff p often used to mean prime, q often used for some prime power p^n

[solemn yew]

to show closure, we need to prove that ab mod n and n and coprime

[solar wyvern]

also maybe you don't care about this right now, but the notation U(n) means units of Z/nZ, i.e., (Z/nZ)*, but if you don't know about rings don't worry about it now i guess.

[solemn yew]

i can't figure it out... any other hint? i can't see what the contradiction should be if I assume they aren't coprime

i think i see...

suppose ab mod n and n aren't coprime, let x be their prime factor (idk if smallest is necessary here) then ab mod n = ab-x(n) [by division algorithm]

x divides ab and x divides n

that means x divides a or x divides b [by euclid's lemma]

but since a,n are coprime and b,n are coprime, this is a contradiction

does this seem correct?

[somber bramble]

I don’t understand the line with the division algorithm, but the last step is definitely what you have to do

I personally would say that if x divides ab mod n, then x also divides ab

which I believe is true, right?

actually now I’m unsure

[solemn yew]

for me, it's not intuitively obvious how that is

[somber bramble]

oh yea since it’s a factor of n

[solemn yew]

so i had to write it out

by the division algorithm line, i meant: when we divide ab by n, we get a quotient and a remainder

the formalization is called the division algorithm i guess

[somber bramble]

yea so ab = (ab mod n) + mn, for some integer m
and since x|n and x|(ab mod n), we have x|ab

and then x|a or x|b, provided x is prime

this is probably the same thing?

[solemn yew]

i guess so

[somber bramble]

but yea

so that concludes that

[solemn yew]

thanks a lot for your help! @somber bramble (im not gonna tag Katzen, they might be busy!)

[somber bramble]

I might be busy too

!

[solemn yew]

yes...

[somber bramble]

and procrastinating

however, in fact

[solemn yew]

but you stuck around to help me!

[somber bramble]

I’m lying in bed

being as unbusy as I can possibly be

enjoying my two weeks of vacation

[solemn yew]

hope you have fun, thanks again

need to get back to work now

[somber bramble]

yea I’mma watch some classic doctor who now

finally getting around to watching those

only about 50 years behind

[chilly ocean]

This seems like it is self explanatory but I have to proof it

I mean we know that from the axioms of a group that e must exist

[somber bramble]

what is ≤ used for here?

subgroup?

[cedar gate]

Subgroup

[chilly ocean]

yeah

[cedar gate]

Just show that the set containing the identity satisfies the group axioms

[somber bramble]

yea I mena it’s a very easy theorem

[chilly ocean]

idk

it just feels weird

being so simple

[somber bramble]

literally just checking that {e} is a group and that it’s a subset of G

a lot of theorems are really simple

that doesn’t make them unimportant

[chilly ocean]

alright

[somber bramble]

I mean there’s a really important theorem in analysis that states you can’t fly from Los Angeles to Buenos Aires without crossing the equator

(it doesn’t literally state that but you know)

(intermediate value theorem, if it wasn’t clear)

[fringe nexus]

Just to make sure, for 54

https://cdn.discordapp.com/attachments/442470054978650122/541333040492838922/image0.jpg

The answer is phi(n) right?

[north tide]

No, the answer is gcd(m, n).

[chilly ocean]

To show any group with 3 elements myst be the same group table we would just show that any group wtih three elements in its group table must be cyclic (S^k| k in Z) right?

[chilly ocean]

do you guys have examples of infinite groups under multiplication?

[somber bramble]

the complex numbers of the form $e^{i\theta}$ form a group under multiplication

[cloud walrus]

Sascha Baer:

[somber bramble]

obviously the multiplicative group of any infinite field (such as reals without 0) is an infinite group

the set of all invertible matrices with coefficients in an infinite field (e.g. ℝ) would be a multiplicative and non-commutative group

similar to my first example, the unit quaternions also form a group under multiplication, but unlike the unit complex numbers this one is not commutative

[chilly ocean]

thanks my dude

[solemn yew]

So there's a question. I'm mostly having trouble with understanding the notation

In the group Z, find: (a) <8,14>; (b) <8,13>; (c)<6,15>; (d)<m,n>; (e)<12,18,45>. In each part, find an integer k such that the subgroup is <k>.

I'm on basics of subgroups right now

What does <8,13> mean?

one more thing: the book I'm using (Gallian's Contemporary Abstract Algebra) uses <a> to mean the cyclic group generated by a (not sure if this is standard notation, thought I should clarify in any case)

[chilly ocean]

re: infinite groups

are there infinite groups where there is an element with finite order?

[solemn yew]

yes

[covert vector]

means group generated by 8 and 13

all integer-linear combination of those two

[solemn yew]

2 x 2 matrices with non zero determinants, (01,10)

order of this element is 2

i hope this is correct

thank you very much! @covert vector

[covert vector]

np

[chilly ocean]

example of infinite group with ele of finite order?

[solemn yew]

is my example incorrect?

i've just studied this

[chilly ocean]

oh i didnt notice sorry

[solemn yew]

operation is matrix multiplication

should've specified

[chilly ocean]

thanks my dude

[solemn yew]

also identity element will have order 1 for any group, so that's a trivial example too

np

[simple agate]

when you're proving that, say G, is a group in an exam, do you explicitly show the existence of an identity or should closure and inverses suffice?

[covert vector]

@fresh lodge wrong channel, use #prealg-and-algebra

@simple agate yes

and associativity

of the operation, if it is not obvious

[simple agate]

ok, thanks. I typically leave it out when writing notes and didn't want to cost myself marks on an exam

[covert vector]

obvious as in like, it's a known operation like + or *

but for example i could give an example

[simple agate]

associativity is usually easy since we generally work with matrices so I can claim it's inherited and don't have to prove it lol

[covert vector]

let X be a finite set, P(X) its power set

the collection of all subsets

let (P(X), Δ) be the structure in question

AΔB = (A\B) U (B\A)

show this is a group

here it's not obvious but yes Δ, symmetric difference of sets, is an associative operation

[simple agate]

so is the operation you defined the disjoint union?

[covert vector]

no

if it were disjoint union, there wouldn't be inverses

cuz things could only get bigger

[simple agate]

oops my bad I understood it wrong

[covert vector]

try to find the inverses, closure should be obvious

and maybe I shouldn't have given the identity

[simple agate]

really puts the abstract in abstract algebra. I'm thankful my assignment questions are generally matrices or working with real numbers

[covert vector]

:P

it's a decent exercise!

except for associativity

that's arguably very painful

but if you want practice try it out

[simple agate]

I'll give it a shot if I finish my homework for this week 😄 thanks

[covert vector]

np!

also

you don't actually need X to be finite

X can be any set, even infinite

and it checks out, the proof

this gives examples of infinite groups that are reasonable to grasp

[simple agate]

sorry I also had another quick question - do you know whether there's a specific name for defining a group as say, "G is generated by a, b with a^4 = b^2 = (ab)^2 = e"

my lecturer gave no real explanation to this yet it's in our assignment. I tried looking online but can't find much on it

[covert vector]

that's called a presentation of a group

[somber bramble]

(that’s D8 isn’t it?)

[simple agate]

D_4 I think

[somber bramble]

same thing

[simple agate]

which is order 8

[somber bramble]

that notation differs

^^

[covert vector]

D_n the superior notation

for dihedral of order 2n

[simple agate]

can I generate the full cayley diagram just from those equations?

[covert vector]

yes

[somber bramble]

I wasn’t quite sure if it was Dwhichever or Z4×Z2

[simple agate]

ok thanks, just wanted to know that before I start trying to write it out. it's a lot of work lol

[somber bramble]

ah but Z4×Z2 wouldn’t have (ab)² = e

that would be a²

[covert vector]

ya

[simple agate]

is D_4 isomorphic to S_4 which is isomorphic to Z2 x Z2?

maybe I'm thinking of the klein four-group idk

[somber bramble]

no, no, no

[simple agate]

I'm terrible at remembering this stuff lol

[somber bramble]

wait that was only two questions

S_4 has order 24

so that’s right out

and Z2xZ2 is abelian

[simple agate]

isomorphic to a subset of S_4 maybe is the right way to say

[somber bramble]

and of order 4

that it should be because S_4 can be seen as the rotations of a cube and D_4 as the symmetries of a square, and you can represent those on a cube

by rotating the cube and flipping it over

(isomorphic to a subset makes no sense, btw)

[simple agate]

which can be represented by 2 elements of S_4 right

[somber bramble]

(isomorphic to a sub__group__)

[simple agate]

sorry that's what I meant

I think you can prove that D_n is isomorphic to a subgroup of S_n for all n right?

[somber bramble]

dunno but I wouldn’t be surprised, afaik all finite groups are subgroups of some S_n

[simple agate]

ooh ok. do you have any suggestions on a book for like a second exposure to group theory?

[somber bramble]

I’m still only just through my first so no ^^

and it wasn’t even a dedicated group theory class

[simple agate]

I aced my 1st year abstract algebra class which says a lot about the quality of the class lol

[stone fulcrum]

What did they teach?

[simple agate]

basics of group homomorphisms, cyclic groups, cosets, lagrange's theorem, ring homomorphisms and ending with its application to the chinese remainder theorem

[somber bramble]

that seems like awfully little, that’s about half as much as we did in ours

[simple agate]

we never studied any general groups or structures though. it was all super basic and easy examples

excuse the terrible screenshot https://i.imgur.com/4kCqZBU.png

[somber bramble]

ours was roughly in three thirds:
-first we did a very tiny intro to category theory and a few other fundamental things like zorn’s lemma; then a bunch of stuff on ring theory starting with the integers and then general rings. e.g. polynomial rings, fraction fields, PIDs and UFDs, first isomorphism theorem
-then we had a large chunk on groups, going through all the definitions and a lot of examples, quotient groups, group actions, isomorphism theorems
-finally we did a bit on field extensions, and on modules, but both were super shallow

in the end I’m left a bit dissatisfied with everything, I don’t think I learned anything well

[simple agate]

that would be a graduate class here 👀 who was that aimed towards?

[covert vector]

lol @simple agate every finite group is isomorphic to a subgroup of S_n for some n

[somber bramble]

3rd semester undergrad

[covert vector]

wow nice

hmm

[somber bramble]

I mean teh very basic definitions of rings, groups and fields were already handled in linalg and analysis

[covert vector]

we didn't do any zorns

[simple agate]

I guess that makes sense lol thanks woog

[somber bramble]

but basically no interesting examples beyond the real numbers (fields) and matrices & polynomials (rings)

but bsically, we had all seen the definitions and worked with them before

but we hadn’t had any of the theory

like, linalg was all over arbitrary fields

complete with “this one thing only works if the field is not of characteristic 2”

I forgot where that was relevant. there was one proof I remember where it was “let V be a vector space over a field 𝕂 where 0≠2, then”

and analysis of course started with the axioms of the reals

which are a superset of field axioms

[simple agate]

my analysis class was shockingly easy and it feels wrong

[somber bramble]

mine was literally purgatory

I wonder if I’ll ever have an exam like that again. I hope not

[simple agate]

this covers all of analysis for my undergrad programme https://i.imgur.com/p3qku8r.png

[somber bramble]

that barely covers my analysis I

and we had two semesters

[simple agate]

you clearly go to a way better school than I do lol

[somber bramble]

statistically speaking, yes I do

unless you’re at one of the top universities in the US or the UK

[simple agate]

I go to KCL in London. it used to have a good reputation, maybe not so much anymore

[somber bramble]

afaik only oxbridge has higher ratings than ETH Zürich

(among UK schools)

but yes, I am very happy with my programme so far

[simple agate]

oh wow that's impressive. I wasn't aware that there were schools that good in continental europe (including switzerland)

[somber bramble]

In the 2019 edition of the QS World University Rankings ETH Zurich is ranked 7th in the world (3rd in Europe after Oxbridge)[5], and is also ranked 10th in the world by the Times Higher Education World Rankings 2018 (4th in Europe after Oxbridge and Imperial College London)[6].

it’s a good school

and, more importantly for practical purposes: it’s piss easy to get into it, and not expensive either

(note: living in zürich on the other hand, is very expensive)

[simple agate]

it sounds incredibly intimidating compared to the standard of work I'm used to lol

[somber bramble]

For Swiss students, ETH is not selective in its undergraduate admission procedures. Like every public university in Switzerland, ETH is obliged to grant admission to every Swiss resident who took the Matura.

oh I hear a lot of “you’re insane” when I tell them I study there

(Matura being what you get if you go through the whole 12 years of education [the last 3 of which are optional])

[simple agate]

the whole UK/US university monopoly never made sense to me when you're paying them (not to mention greatly overpaying)

[somber bramble]

ETH has a lot of foreign students for good reasons I reckon

but you have to be able to afford living in switzerland

[simple agate]

admission should be standardised and not have random/biased factors

[somber bramble]

well, there’s stipends

but still

they state on their website that while there’s practically no mandatory costs to study at ETH (tuition is like 600/semester? but it got raised recently idk how much it is now. still, nothing compared to other places), you need around 16k to 26k to survive (that’s swiss fracns, which is more or less equivalent to US dollars)

and I think those are reasonable numbers

[simple agate]

wow that's expensive

[somber bramble]

Zürich is one of the most expensive cities in the world, and it honestly doesn’t get much better evenif you commute from farther away

I managed to get a really cheap room (500/month) rather close to the school, but it’s in a tiny flat without a living room and only a tiny kitchen and bathroom

the room itself is nice tho

sharing the flat with two others

[simple agate]

london is similar. I live ~30 miles outside of central london and commute in, however when you factor public transport costs, it's really no cheaper due to the rising costs around london

[somber bramble]

usually you’re looking more at 700-800

and like 1000+ for studios

and ETH has next to no student housing

as in, it does have a bunch, but not nearly enough to keep up with demand

so it’s constantly full

and getting a room there is basically black market business

[simple agate]

do you have a campus or is it more of a collection of buildings in the city?

[somber bramble]

both

there’s two campuses

one is a proper campus

and the other’s one proper uni building + a scattering of houses

near the city center

like actually within 300m or so of central station

haupbahnhof being central station

[simple agate]

I'm jealous of those huge US campuses with appropriate student housing onsite

[somber bramble]

anything red belongs to ETH, dark red is the main building

dark grey is stuff belonging to a different university

I like the central location, it’s nice

[simple agate]

very similar to London. reminds me of LSE a bit, they have a bunch of buildings densely scattered

[somber bramble]

Höngg is a proper campus tho

[simple agate]

that's pretty

[somber bramble]

there’s a shuttle bus (~15-20 mins) between the two

Höngg is mostly physics, chem, bio, architecture and exams

^^

[simple agate]

what language is generally used for your university?

[somber bramble]

undergrad is about 50/50 German English, master’s courses are all english

first year was all german, this year it was about 50/50

[simple agate]

ooh that's interesting that they use both

[quiet cave]

I had to follow an abstract algebra course in German...

[sacred rapids]

Am I being dumb or misunderstanding this

[cloud walrus]

D1richlet:

[somber bramble]

yes (though I don’t knwo what group U(26) is so that’s all I can say)

unless it’s the group of unitary matrices in 26 dimensions

but that seems unlikely

[cloud walrus]

D1richlet:

[sacred rapids]

U(26) is the set of numbers relatively prime to 26

[somber bramble]

⟨5⟩ isn’t 2

is it?

well, actually I don’t know

come to think of it ^^

but how can there only be 12 numbers relatively prime to 26 when there are infinitely many primes?

oh is it like, the numbers relatively prime to 26 as a subgroup of Z/26Z?

[sacred rapids]

Yes

Sorry, I thought that was standard notation

[somber bramble]

in which case, ⟨5⟩ would be (5, 25, 21, 1)

oh I just don’t know the notation

it may well be

so that‘s not 2, but 4

you can’t just stop at 21, to get closure you have to get going until you get to 1

[sacred rapids]

Shit you’re right

Thanks!

Follow up

Does this hold?

[cloud walrus]

D1richlet:

[sacred rapids]

Because 3.12 asks for an element of infinite order, so the group cannot be finite, but I can’t figure out why

[somber bramble]

oof uh… I’d be surprised if it did not hold but I’d have to think through why it holds

oh yea hm

wait hold on, no, I don’t think it’s the same

hm lemme think through what that group there is

so the think you’re quotienting out is elements of the form (n6,n9)

to get Z6 × Z9 you’d have to mod out elements of the form (n6,m9)

for any two m,n

[sacred rapids]

Well suppose g has infinite order and is of the form (a,b)

Then 54(a,b)=(54a,54b) is not the identity

[somber bramble]

well, (a,b)+⟨6,9⟩

I like writing that stuff explicitly

keeps me from being confused

^^

so hang on, in the quotient, (a,b) = (c,d) if there exists an n, such that a+6n = c and b+9n = d

[sacred rapids]

Yeah

[somber bramble]

so to find an element of infinite order, you’ll have to find one such that as you add it to itself, you’ll never get that identity to hold

[sacred rapids]

Ohhhhh I think I got it

Because we’re modding out (6,9), (12,18), etc

then anything coprime to 2 or 3 should have infinite order

but I’d need to show that

Like (1,5)

[somber bramble]

I would actually assume all that’s needed is for the second element to be sufficiently larger than the first

so that it grows faster than 9n when the first grows as 6n

then you’d only have to check the first few elements before it’s clear you won’t get the identity

right?

like by the time you add (1,5) to itself six times, it’s (7, 35), which is already way beyond reach

so you’ll never be able to get it to the form (1+6n, 5+9n)

precisely because the two n have to be the same

[sacred rapids]

Yep

Thanks

Sometimes when I start explaining something it clicks

[chilly ocean]

bruh this IBL bs I swear. Why can't they just use a book instead of a packet

[bleak finch]

If |G| = 44 then G is not simple.
Proof. By Sylow's 3rd n_11 = 1 mod 11 and n \ 4. So n_11 = 1. By Sylow's 2nd the unique subgroup S_11 must be normal.
This seems too simple. What did I forget?

[timber bay]

Am I wrong in my understanding when I say that you can't have an ideal generated by a constant without it being the whole ring?

for a

[somber bramble]

no, as in you’re not wrong

[timber bay]

okay good

[somber bramble]

if you have a consant in it, then you have 1 in it

becaue F is a field

[timber bay]

which is still trivial

[somber bramble]

and an ideal containing 1 contains the whole ring

[timber bay]

so if f(x) = 0 then the ideal is only the zero poly, and when f(x) = c, its the whole ring

[somber bramble]

(note that this would not be the case if F wasn’t a field, e.g. in ℤ[x], the ideal generated by 2 isn’t all of ℤ[x])

[timber bay]

yeah okay cool

[timber bay]

I'm not exactly sure how to prove part a. I think I understand b pretty well. I'm feeling itll be using the properties of an isomorphism. But there's likely a more nuanced approach.

oh it just wants us to test the elements doesnt it

oof

you know any well defined proofs out there? i never really understood how to prove something is well defined, and we only have done it like once

not exactly sure what that means

you have to show that phi(xy) = phi(x) phi(y)

that's simple, is that all i would have to do for it to be well def?

so it just follows from field properties

well nevermind

how does it being well defined imply its an iso?

im not familiar with that terminology

sorry

hmm

i need to think about that some more

but what youre saying makes sense logically

thanks

[rapid tusk]

same tbh

[timber bay]

what

[outer solar]

Could someone help me study? Im honestly so completely lost in my class rn. Im a bit behind and just absolutely confused by everything all the time.

[scenic harness]

@outer solar help you how?

[outer solar]

I'm reviewing the online notes and trying to do homework but honestly abstract algebra is so abstract that I am having trouble even getting base concepts. I guess id appreciate going over the notes with someone along with the hw and just reviewing it together and seeing if it can be reworded more simply or something?
(Also in terms of going over the hw I in no way intend to get too much help. The kind of questions I would ask would be so general that it just tells me the vague direction to head for)

[untold sapphire]

You're better off breaking it down and posting the specific parts you don't understand here

You're more likely to get help that way

[tulip barn]

Or ask about the specific concepts you're having trouble with.

[outer solar]

Okay thank you. I guess I'll see best how to word them and post them in a bit. Thank you!

[random crag]

@timber bay well defined just means that it basically is what you think it is. So if you’re calling it a function, then it is indeed a function

do you still need help with the other questions as well?

[outer solar]

I'm very confused. It seems by definition this is already proved?

[scenic harness]

Well, if you assume it to be a subgroup then of course it's a subgroup.

Try to prove that the set defined in Definition 41 is a subgroup instead.

[outer solar]

AH! I gotcha. We know by the problem and the definition that the centralizer is the equation. We have to prove that such an equation can only result in a subgroup of G, given the relationship g (element of) G, right?

[scenic harness]

Not sure I understand correctly but yes, I hope.

For example: "the centralizer is the equation."
The centralizer isn't the equation. The centralizer is all elements of G that obey the equation.

[outer solar]

yes that is what i meant more properly worded. the centralizer is defined by the RHS of the equation."

[somber bramble]

I mean basically you just need to show that whenever two elements fulfill that equation, their product does too, and same for the inverse

[outer solar]

okay i think i get it. Thanks!

[timber bay]

@random crag Just saw your message. If part c is clear to you, some pointers would be nice

[covert vector]

@timber bay if ξ is a field isomorphism, then ξ(xy)=ξ(x)ξ(y)

pick x=y=something nice

[timber bay]

what is this supposed to show me @covert vector

what am i supposed to show with picking a nice x and y

[covert vector]

if x=y=√2, then ξ(√2)²=ξ(2) = 2, since 2 is rational

[timber bay]

ohhhh

okay

so you know ξ(2) = 2 because ξ(1) = 1 and you know this because ξ is an isomorphism

and then you can just bring out a power of 2 on both sides and get that ξ(sq(2)) = +/- sqrt(2)

[stone fulcrum]

No, we are looking for ξ that preserve Q. That is, ξ(Q) = Q for any member of Q

√2 is not a member of Q, so we don't know what ξ(√2) is

[timber bay]

is my reasoning not correct?

i just used rules from isomorphisms

[covert vector]

yes it's correct

idk what kaynex is on about

well like, I do but we're past that

[stone fulcrum]

@timber bay
Your reasoning is correct! I forgot that ξ(2) = 2 naturally. I suppose this isn't something to think about when the base field is Q

[timber bay]

lol

thanks

[stone fulcrum]

But you are looking for automorphisms that preserve the base field. A better example might be the extension from Q[√2] to Q[√2, √3]. In this, ξ(√2) = √2

[whole basalt]

Hey y'all.

Do you think this is clear enough?

Specifically, is it clear enough that I'm not claiming that a single example like this proves that 3 is prime in Z 😛

[stone fulcrum]

Is this a video?

[whole basalt]

Nah, it's a still image

For mathsubgre Twitter

[stone fulcrum]

Yeah it's very clear to me

... I mean I already know the concept though so EHHhhh

But I like the examples, makes it concrete

[whole basalt]

Cool

I'm trying to make sure that for just about every tip I publish, I have either a diagram or an example

[stone fulcrum]

There's a possible confusion. 12 = 3×4, and 4 divides 4

Ergo 4 is prime

[whole basalt]

Perhaps a note of "Since we've found a counterexample."

Here's another one for an analysis concept. (Yeah I know this is the algebra chat but I don't want to just go to another channel just to paste it :P)

[solar wyvern]

@whole basalt post group version of cauchy stuff

actually I'll post it later, @spark plank remind pls

you might like actually

[whole basalt]

...there's a group version of Cauchy stuff?

[somber bramble]

let’s say G is some group, any group, and we’ll interpret its operation as addition.
does there exist a field F such that G is a vector space over F?

(I will not consider F_un a valid answer)

[solar wyvern]

you mean like a group ring F[G]? @somber bramble