groups-rings-fields

[frank cosmos]

where am i confused at

[shell agate]

Sorry you're talkin about the elements of (Z/3Z)/S

[frank cosmos]

correct

[shell agate]

Is thought you were saying that Z/3Z is {{0,1,2},{1,2,0},{2,0,1}}

Anyway

[frank cosmos]

oh lol

[shell agate]

I'm not sure this is relevant to their question

[frank cosmos]

? I was just showing an example of dividing a group into equivalence classes for a quotient group since it seems to be what they are confused about.

oh well

[shell agate]

As for this, are you asking why the last to lines say "subgroup" and not jut subset?

[frank cosmos]

im wondering why do I need the fact that subgroups of abelian groups are abelian when comuting that the inverses of abelian towers are abelian

I only use the isomorphism theorems

[shell agate]

But they are already having a hard time understanding Z/pZ, so I think quotienting Z/pZ would not help them

[frank cosmos]

ok

[shell agate]

Well, you have an abelian tower of G', and a morphism from G to G'

[frank cosmos]

yes

[shell agate]

Gi/Gi+1 is not isomorphic to G'i/G'i+1

[frank cosmos]

because its not surjective?

[shell agate]

yes

so it's just a subgroup

[frank cosmos]

not sure what you mean...

[shell agate]

rather, it's isomorphic to a subgroup of G'i/G'i+1

[frank cosmos]

ohh

thanks!!

[shell agate]

no problem

[frank cosmos]

also this

[sharp marsh]

@hot lake @rustic crown i did my best to prove

also this alternative using ring isomorphism

[novel parrot]

tips for this question?

i dont know where to start

[hot lake]

so uh which part is the proof that "pZ is maximal => p is prime" and which part is the proof that "p is prime => pZ is maximal", or are you just doing a chain of equivalences

[sharp marsh]

Chain of equivalence

In the starting i stated that if we can prove it is a finite commutative integral domain we can conclude that it's a field so pZ will be maximal

[hot lake]

ah so you already know theorems like "I is maximal <=> R/I is a field" and "Zp is a field <=> p is prime" ?

[sharp marsh]

Yes

[hot lake]

then yeah the alternative one is the easiest

[sharp marsh]

Also yeah I think i should have explicitly stated the theorems before using them

[hot lake]

are you saying Zp and Z/pZ are literally equal

idk about your course but I probably wouldn't bother proving that Zp and Z/pZ are the same thing

I'm not even sure where you are taking your ring operations from in your isomorphism proof

[sharp marsh]

Please elaborate

[hot lake]

well normally you need to show that f(a+b) = f(a)+f(b) where + and + are the corresponding ring operations and same with x

so here your proof is just stating that it is the case

and usually you would have to go and use the definitions of + and +

so that we know you aren't making things up

you have to show that the addition in the equivalence class world is the same addition as the one in the remainders world

[sharp marsh]

Since it's the set of integers i automatically used it without thinking. I'll define them from the next time

[hot lake]

however that last one is defined

if your argument is f(a+b) = a+b = f(a)+f(b) without explaining why the definitions of additions make those true you aren't really bringing much to make a proof

though in this case uh

like I said I wouldn't bother trying to explain why Z/pZ and Zp are isomorphic

because when you look at the definitions it's just the same thing

and yeah the whole exercise reduces to pZ is maximal <=> Z/pZ is a field <=> p is prime

and pointing out that those two equivalences are theorems from the course

presumably you defined Zp as an example of a ring before defining ideals and quotient rings

[sharp marsh]

Off topic, some seniors of mine has stated that i need to practice my proof writings more be it number theory or calc. But with only reading the book and teh internet i get a lot of things wrong. Can you recommend how do I know if my proofs are wrong or correct.

And processors are very very very unreliable

[hot lake]

your proofs are very unclear logically speaking

so you proved the isomorphism

and then said "therefore pZ is maximal only when Z/pZ is a field"

here you made no mention of the theorem form the course saying I is maximal <=> R/I is a field

and "only when" is not the same thing as an equivalence

you are just saying "if pZ is maximal then Z/pZ is a field"

when you could be saying "pZ is maximal <=> Z/pZ is a field"

which by the way comes straight from the theorem in the course and has nothing to do with the isomorphism with Zp

so using "therefore" as if you needed the isomorphism there is wrong

then you go to the next step

and you write "<=> Z/pZ is a field only when Zp is a field"

so you copied your last sentence and replaced Z/pZ with Zp because they are isomorphic

however putting an "<=>" to introduce that step is super wrong

you already know the first sentence is true

you can just say "since I know Zp and Z/pZ is isomorphic, we now have : insert statement"

you have no businees saying "the first statement is true if and only if the second is true, oh and does anyone remember if it was true in the first place ?"

[sharp marsh]

😓

[hot lake]

then you go to the next step

now you are introducing the fact that Zp is a field <=> p is prime

taken straight from your course

[novel parrot]

can someone help me

[hot lake]

and you are claiming that the statement in line 2 is equivalent to that fact from the course

and then after all that

you says "this implies that pZ is maximal <=> p is prime"

so yeah the collection of facts proves that

don't put <=> signs as if there were equivalences going on

[somber sleet]

may I ask you where these rings are contained?

[hot lake]

yes that's perfect

[sharp marsh]

oh I know what I did initially I literally wrote my whole thought process even though those only broke my proof

[hot lake]

you should try and make it clear what comes from a theorem what comes from a definition, what comes from a computation

and then writing the logical steps is uh.. from practice

the way you put "therefore" and "<=>" carries some meaning

[sharp marsh]

thank you so much for being patient with me! I will keep these tips in mind when solving my next problem

[novel parrot]

the ideals are in an arbitrary ring

[rustic crown]

say p is a prime in A. and show that A/p is a field

[novel parrot]

to need to show that descending chain condition holds in A/p ?

[rustic crown]

pick an element a outside p, and see what happens to its powers

yep, that's also a good idea.

[novel parrot]

but then what?

[rustic crown]

cause now you know that A/p has descending chain condition and is also an integral domain.

[novel parrot]

ok

[rustic crown]

if a is any non-zero element, can you find an inverse? (in A/p)

[novel parrot]

i can try

i dont know hwo to find inverse

[rustic crown]

you know something about descending chains

[novel parrot]

we need to construct an descending chain right

[rustic crown]

yep

what's a nice chain you can construct using the element a

[novel parrot]

no idea, maybe let I_1 = (a)

for the first ideal

[rustic crown]

okie :3

[novel parrot]

is it wrong

[rustic crown]

nope

[novel parrot]

kk but then what ideal to be I_2 ?

[rustic crown]

something inside (a)

so the generator should be a multiple of a

[novel parrot]

ohohoh

you mean powers of a

[rustic crown]

[novel parrot]

:d

[rustic crown]

yee so what does the hypothesis tell you now

[novel parrot]

so now (a^n) = (a^n + k)

k is any integer

n is large enough

so we are looking for inveres to a

[rustic crown]

lets take k=1

[novel parrot]

oh

so something like (a^n+1)(r) = a^n

[rustic crown]

[novel parrot]

nice

[rustic crown]

and finally integral-domainness

[novel parrot]

yep

thank you

do you think that is a common exam question?

or too easy/hard for exam?

[rustic crown]

i can only see this as a part of some larger question. don't think it would be asked independently

[novel parrot]

oh

[somber sleet]

how do you know if a polynomial is irreducible on a ring like F7

[formal ermine]

you can try using eisenstein

[somber sleet]

I think with x^3+x+1 I can't use it

[coral spindle]

Eisenstein is not going to be useful in a finite field

I don't know any clever tricks for checking if a particular polynomial over a finite field is irreducible, but the good thing is that you only need to check finitely many things

In particular, if your polynomial is of degree <= 3, then you can just check if it has a root, which involves just... substituting the finitely many field elements and checking lmao

[somber sleet]

This is exactly what I did

but is this rigorous

like how can I motivate it?

[novel parrot]

you can look into quotients of F7[x]

[coral spindle]

Wym? If a nonlinear polynomial has a root, then evidently it is reducible

this is due to the division algorithm

[somber sleet]

yes exactly, but how do I argue the opposite

[coral spindle]

The opposite being...?

[somber sleet]

can I say that by proving that no elemnts of Fn is a root of my polinomial, then it has to be irreducible

[coral spindle]

Well that's not going to be sufficient in general

Let me think of an example, quickly

If we're working in F_2, then the polynomial x^2 + x + 1 is irreducible, since it is a degree 2 polynomial with no roots

[formal ermine]

I was thinking of the same example

[coral spindle]

but (x^2 + x + 1)^2 is not irreducible, despite having no roots

[formal ermine]

you beat me to it

[coral spindle]

Haha great minds 🧠

[somber sleet]

like I see it

but I don't know how to argue ti

[coral spindle]

Argue what, then?

I just showed that the thing you said wasn't true

[somber sleet]

that it is irreducible

[coral spindle]

OK, what's the polynomial you're looking at

[somber sleet]

this one exactly

[coral spindle]

Right

OK so if it is reducible, how is it going to reduce?

Think about the degrees of its factors

[somber sleet]

oh contradiction

I see

in two polynomials fg where if I call my polynomial p then deg(p)=deg(fg)

[coral spindle]

that's right

[somber sleet]

this means that either f or g have to have degree 3

[coral spindle]

Well yes, but in particular, if p is reducible then either f or g has to have degree 1!

Which means...

[somber sleet]

either f or g has to be a linear polynomial

like ax+b

or am I trippin

[coral spindle]

And what does that tell you about the roots of p

[somber sleet]

that I have to have at least one in F7 right?

[coral spindle]

That's exactly right

[somber sleet]

otherwise you can't decompose it

[coral spindle]

So if p is reducible, then it has a root

Now we just take the contrapositive of this statement

:)

N.b. this only worked because p was degree 3

[somber sleet]

I get it now, thank you for helping me

[coral spindle]

No worries DG

[somber sleet]

does it always work if my degree is prime?

I'd say yes

by the same argument(?)

[coral spindle]

No it doesn't

remember deg(fg) = deg(f) + deg(g)

[somber sleet]

oh yes you are right

thank you

[coral spindle]

Anytime

[somber sleet]

Is it true that thst R[X] is a pid only if R is a field?

[oblique river]

Yeah

[formal ermine]

yeah we talked about this yesterday

.

[somber sleet]

is there a case where R[X] could be a field?

[elder wave]

no

you can show that x never has an inverse by degree reasons for example

[formal ermine]

over the field with one element it is

[elder wave]

okay explain what the field with one element is then

[formal ermine]

it's the zero ring but as a field

[elder wave]

[somber sleet]

1!=0 (?)

[formal ermine]

it's a joke

it doesn't exist

1 isn't prime

[chilly ocean]

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

[delicate orchid]

oh yeah? then why are sets F_1-vector spaces

GL_n(F_1) is iso to S_n... if it's not real then how am I spitting factoids

[formal ermine]

wait what how

[delicate orchid]

I wrote them in a misleading way, it's more that the first statement implies the 2nd

[delicate orchid]

this probably explains it better than I could

[chilly ocean]

basically, field with one element = funny meme

[delicate orchid]

there's some weird categorical construction that makes these ideas concrete but F_1 is definitely not a field

[formal ermine]

F1 IS A FIELD!!!!1!1!1!1 DON'T BELIEVE WHAT THE GOVERNMENT WANTS YOU TO THINK,,,,

[elder wave]

[coral shale]

whoa whoa whoa what the heck

What are the premises to this statement?

'Let R be a commutative ring with identity' ?

[formal ermine]

yes

[oblique river]

It maybe looks more surprising than it is. The implications
R field => R[x] euclidean => R[x] PID
are standard (the first is polynomial long division, the second is a general fact about any commutative ring). The final implication
R[x] PID => R field
Is maybe the most surprising but ultimately not that deep; consider the ideal (r, x) for some r in R.

[somber sleet]

can somebody tell me how do I find this isomorphism, I know chinese remainder theorem, but I don't get how to use it in this case

[chilly ocean]

I don't think this is about Chinese remainder theorem?

[somber sleet]

[chilly ocean]

180 = 2^2 * 3^2 * 5 for example

so you can decompose C_180 as product of C_2^2, C_3^2 and C_5

and I'd do the same to 150, 70 etc.

oh well there is probably like a faster way to see this instead of writing it in this form

[rustic crown]

is this called something else in the group case? i call it CRT here as well :p

[chilly ocean]

Idk. I refer to it only in ring case

[rustic crown]

yea that's a proper thing to do

but can't resist it if you use the Z/nZ notation instead of C_n

[chilly ocean]

like if you look at factors of 2, you have C_4 x C_2 x C_2 x C_4 x C_4 here so 3 copies of C_4 and 2 copies of C_2

so d) looks plausible

c) too

okay they all have 3 copies of C_4 and 2 copies of C_2

now maybe look at the prime 3

[rustic crown]

luckily you don't have to go beyond 7

[chilly ocean]

you have two C_9 and two C_3

so a) and b) still have this

but c) doesn't!

and d) doesn't either

so we know it must be a) or b)

and 1260 has a 7 in it so a) can't be good either

oh wait no

there's a factor of 7 in the original group anyway

you have 4 copies of C_5 in a)

but original group has two C_5 and one C_25

so they're not isomorphic

so its b)

[somber sleet]

[rustic crown]

dun cri

[tender wharf]

[somber sleet]

if I have Z[X]/(3,X^2+X+1) what is the best way to prove if this is a field or not? I thought about writing it as (Z/3Z[X])/(X^2+X+1) I mean, I know that Z/3Z[X] is a ring and my Ideal is contained there, so it would be enough to prove that my polynomial is irreducible right?

over Z/3Z

which is not I guess

[formal ermine]

(that's only true for pids fwiw)

(but yeah it's a pid)

[somber sleet]

what exactly?

[formal ermine]

hauptidealring = principal ideal domain = pid

[somber sleet]

so it's not a field right?

[rustic crown]

haupt = principal?

[somber sleet]

jaa

[rustic crown]

wait what do you then call principal ideal ring

[somber sleet]

yeah I asked my self the same

[rustic crown]

yee your reasoning is all good because Z/3Z is a field.

another way to see that is by noticing directly that
(3, x^2 + x + 1) = (3, x^2 - 2x + 1) = (3, (x-1)^2)

:p

but ofc what you did is nicer

[somber sleet]

dumb question incoming, be ready

[rustic crown]

[somber sleet]

I can do that because it's generated by 3 as well right?

so -3x is also contained in my ideal

damn

this is crazy

[rustic crown]

(a, b) = (a - b*q, b)

it's like doing the gcd algo

[somber sleet]

yes for sure, I mean it's also just by definition

like it should be obvious haha

[rustic crown]

it was a nice question

https://media.discordapp.net/attachments/796976320067272704/818150000482582528/612933479730249728.gif

[somber sleet]

If I pass this exam, it's all because of you guys

[formal ermine]

eh

hauptidealbereich = pid

hauptidealring = pir

technically speaking

but usually you just say hauptidealring for pid

[rustic crown]

oh so like the main train-station is hauptbahnhof

[formal ermine]

yeee

[rustic crown]

principal station

[formal ermine]

bahn = train

[somber sleet]

bist ja gar nicht so schlecht

[formal ermine]

hof = station

[rustic crown]

[somber sleet]

it's also an orbit in algebra

[formal ermine]

die bahn fährt auf der bahn

bahn can also mean track

[tender wharf]

ok this is intriguing, could you explain

[formal ermine]

.

[elder wave]

if a prof allows PIRs we call PID Hauptidealbereich and PIR Hauptidealring

[elder wave]

bad

[formal ermine]

cope harder

[rustic crown]

[elder wave]

but there's a terminology clash

[somber sleet]

It's just that bahn has different meanings as illu said, it's like a path(?)

[rustic crown]

det is learning more german words

[somber sleet]

do you know what an hof also ist

[rustic crown]

illu said it's station

so i would guess something close to station

[somber sleet]

it has like 10 different meanings

[rustic crown]

[somber sleet]

it's also like courtyard, schoolyard,court, farm and so on

but it's like the collection of something

you say bahnhof because it's where all bahnen meet(?)

it's so stupid somehow ahha

[rustic crown]

i found it strange that hospital in german was krankenhaus

like you expect some words to be more or less similar in all languages right

[somber sleet]

I mean, you can say Spital

[rustic crown]

ooh

[tender wharf]

mfw the only german word I know is zahlen

[somber sleet]

but I don't know if it's just swissgerman

[rustic crown]

i know that too :3

[tender wharf]

is zentrum a german word?

good old \bZ

[somber sleet]

yes

[tender wharf]

reminds me of Z(G)

[rustic crown]

oh is that why we use Z(G) and not C(G)?

[tender wharf]

yeah

I think

zentrum of G

[somber sleet]

It exactly is what we call it haha

[rustic crown]

[tender wharf]

also

eigenvector

and eigenvalue

[rustic crown]

i used to always call it zenter while learning group theory lol

[tender wharf]

zenter

im stealing that

[elder wave]

i have never heard someone call it that

[tender wharf]

let Z(G) be the zenter of G

[coral shale]

i propose we use this notation

[elder wave]

[tender wharf]

wow

but what if

[rustic crown]

[tender wharf]

we had H

[somber sleet]

I think it's a Suisse thingy

[coral shale]

u overlay the same 4 arrows

[tender wharf]

$Z(\Xi)$

[tender wharf]

ha

[elder wave]

ah

i see

[tender wharf]

how would you notate the centraliser then

[rustic crown]

you mean zentralizer

[somber sleet]

Zentralisator

[rustic crown]

this is giving me tranquilizing a dinosaur vibes

[prisma thunder]

Is every Dedekind domain D finitely generated as a Z-algebra (D ~ Z[x_1, ... , x_n]/I for some ideal I in Z[x_1, ... , x_n]) ring isomorphic to a number field K? I've been thinking about this and I'm not sure if such a counterexample exists.

[somber sleet]

is this only the identity?

[chilly ocean]

conjugations just swap elements {3, 5, 2, 4} in that cycle

so there is as much elements conjugate to a as there is 4-cycles formed from 3, 5, 2, 4

[hot lake]

you can get a 1 too

[chilly ocean]

ah yes

I'm sorry

[hot lake]

and if you try to conjugate the identity you get the identity

so you can't conjugate something nontrivial and get the identity

if that was your question

[rustic crown]

is k[[x]] f.g. as a Z-alg?

[chilly ocean]

so all 4-cycles in S_5

[somber sleet]

is it true that conjugate means a*σ= σ where sigma is in S5

the a should be up

like aσa^-1=σ

[chilly ocean]

aσa^-1 this is a conjugate of σ

[somber sleet]

this thing confuses me

like this is the conjugate

[chilly ocean]

when you conjugate cycles by permutation, the formula is nice

[somber sleet]

I agree on this

[chilly ocean]

$\tau(a_1...a_n)\tau^{-1} = (\tau(a_1)...\tau(a_n))$

[cloud walrus]

Blitz

[somber sleet]

yes you can just operate the permutation on the single element

exactly

so in this case

when are two elements conjugate?

[chilly ocean]

and thats the point - we can get every 4-cycle from a

[somber sleet]

okay

[chilly ocean]

in S_n?

[somber sleet]

yes, like you just have to operate sigma on it?

[chilly ocean]

when they have the same type of decomposition as disjoint cycles

[somber sleet]

OH WAIT

I get it

[chilly ocean]

the the amount of m-cycles are the same in each decomposition for each m

[somber sleet]

what do you get here?

[chilly ocean]

all possible 4-cycles

[somber sleet]

for the first question

[chilly ocean]

this is given by some Stirling number

[chilly ocean]

oh apply this formula

[somber sleet]

by the way it's formulated meand that tau is (2 3) right?

[chilly ocean]

(2534)

[somber sleet]

okay yes thank you

[chilly ocean]

yes

(23) is a function which swaps 2 and 3

so conjugating a by (23) just swaps 2 and 3

[somber sleet]

yes I know that

I was confused by the formulation of the question

[prisma thunder]

I’m not parsing that correctly. What is that notation?

[chilly ocean]

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

[rustic crown]

oh it's the ring of power series over the field k. you wanted a dedekind domain, so i started by thinking about PIDs first

[somber sleet]

ohhh, I didnt have this yes in the lecture

[rustic crown]

combi is so hard

[chilly ocean]

well, I'm just mentioning that there is probably some way of doing it more using Stirling numbers

I like to mention a result or two like that sometimes when talking about related thing

[somber sleet]

I'm gonna read it, sounds interesting

[somber sleet]

It's better for me to understand things, so thank you!!

[chilly ocean]

first focus on your exams and stuff

[prisma thunder]

Ahhhhh okay I figured it was the ring of power series 😂 alright we’re on the same page now

[chilly ocean]

so first you choose 4 elements from 5 so you get 5 possibilities
and then you get how many 4-cycles you can get from 1, 2, 3 and 4

[rustic crown]

so won't a finite type Z-alg be countable?

[chilly ocean]

which should be 3! = 6 (we can assume the cycle looks like (1abc) then 3! choices for a, b, c

[rustic crown]

so to get a counter example. find an uncountable PID

[prisma thunder]

Yes

[prisma thunder]

Oh hmmmm I didn’t think of that!

[somber sleet]

It's the best way to show that an ideal is maximal, showing that the factorring is a field?

[tired horizon]

can anyone give me a good book or any source to understand field extensions?

in english or french

[chilly ocean]

It probably depends on the context, there's probably some tricks as with everything too... but for me, yeah ig

Or directly from definition... probably even easier

[pliant forge]

could i get a root, a, where F(a) = F(a^2) but minimal polynomial of a is even

[chilly ocean]

root a of what?

[pliant forge]

of a minimal polynomial

irreducible over F

[chilly ocean]

ah so F is a field

[pliant forge]

yeah just any example

[coral shale]

jesus, I thought F was a function

[pliant forge]

I mean an example where its the case that F(a) = F(a^2) for some root a, but [F(a):F] = 2n for some n \in N

[chilly ocean]

so this is equivalent to a = p(a^2) for some polynomial p(x)

[coral shale]

why specify even - whats your odd example?

oh nvm i see it

[pliant forge]

the converse is true for whenever [F(a):F] = odd

[chilly ocean]

oh I guess I'm assuming a has finite order or whatever its called

[coral shale]

does picking a field where a = a^2 work

[chilly ocean]

a = a^2 always has two solutions, 0 and 1

which are in every field

[coral shale]

mod 2 then

[chilly ocean]

hm?

[coral shale]

but i dont think u get even min poly do u

nvm me silly

[chilly ocean]

oh you mean Z/2Z

[coral shale]

x^2 + x + 1
Roots of this are the 2 cyclotomic cube roots of unity? Done?

[pliant forge]

i have no idea what you just said

[coral shale]

cube roots of unity

their min poly is xx+x+1

right

[pliant forge]

whats cube root of unity

[coral shale]

x^3 = 1

[pliant forge]

o

[coral shale]

the solutions to this

ignoring 1

,w expand (x^2+x+1)(x-1)

yh so i think this fits what u want

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Shuri2060#9525

[coral shale]

x and x^2 are complex conjugates

[formal ermine]

my intuition tells me this is true

ok so

we need gcd(|a|, 2) = 1

5th cyclotomic polynomial works, doesn't it

yeah even 3rd one works

@coral shale is right

[coral shale]

very nice edit right there

totally didnt see anything

[formal ermine]

shhhhhh

[delicate bloom]

each prime p you can take that cyclotomic polynomial too to get infinitely many examples

is it possible to get all n, not just when n=(p-1)/2

[formal ermine]

any cyclotomic for n > 2 works (assuming gcd is 1)

because phi is even for any n > 2 (or 3? I forgor)

[delicate bloom]

yeah true I guess since the kth cyclotomic polynomial has degree phi(k) then I should really be asking when we have [F(a):F]=2n != phi(k)

[pliant forge]

thx btw

[delicate bloom]

first several cases to start looking at if I didn't do it wrong are: 14, 26, 34, 38, 50, 62, 68, 74, 76, 86, 90, 94, 98

[formal ermine]

but for even numbers isn't a^2 not a generator?

[delicate bloom]

wdym, in the x^2+x+1 example both a and a^2 generate the same extension of Q

[formal ermine]

yeah but that's 3

[delicate bloom]

I don't follow, it's degree 2

[formal ermine]

and gcd(3,2) = 1

it's the third cyclotomic polynomial

[delicate bloom]

yeah I see what you mean now

so only the odd cyclotomic polynomials

[formal ermine]

yip

and n > 2

and all n that satisfy those conditions work cuz phi is always even for n > 2

[delicate bloom]

I guess constructing a degree 14 irreducible polynomial where the roots a and a^2 both generate the extension is tough

[formal ermine]

yeah cyclotomics won't work

14 is a nontotient

[delicate bloom]

the obvious first thing to try of x^{2n} - p won't work

probably that's what motivated OP's question in the first place looking at eisenstein polynomials

what other irreducibility tests are out there to try to use

[strong moth]

any particular reason why these two diagrams are necessarily equal?

[formal ermine]

reduction mod p maybe, but I'm not sure that would help here

[formal ermine]

what's the context

[strong moth]

zx calculus doesn’t state why

[formal ermine]

?

[strong moth]

i’ll show the page

hold on crossing a street

[delicate orchid]

I mean, it's easy to see they're isomorphic as graphs

[strong moth]

how are they isomorphic as graphs

[delicate orchid]

you just move the heart box left

oh and flip the other one over, I suppose

unless there's some difference between a left and a right connection

[strong moth]

i don’t see it they have different connections on the boxes

and different labels for the boxes

even if you move them around

[delicate orchid]

who cares about labels? that's the point of an isomorphism

but no the connections are the same, that part is important

[delicate bloom]

I think they're trying to say if you have 2x=6 then 2x and 6 don't look the same but you can do "algebra" on them to solve for x basically by adding boxes etc I guess

[delicate orchid]

again, I'm assuming these are graphs - if you're treating them as braids or something then idk

[strong moth]

that makes sense

i’ll try to color code what i see on mobile and walking so be a sec

[delicate orchid]

if I draw them like this is it more obvious why they're the same?

[delicate bloom]

to me these aren't the same, but since the labels on the vertices are different, then it's possible that you can have the same net effect on two inputs with one output

[formal ermine]

they commute so they are the same to me

like

commute in the same way

yk

[delicate bloom]

i dont

[delicate orchid]

yes, they're iso as directed graphs illuminator - but if labels matter then I agree with merosity

[delicate orchid]

and as I said before

[delicate bloom]

can you link it

[strong moth]

i see the blue lining up

the red lining up

but the green and purple

are on different sides of the box

[delicate orchid]

are all of the operators involved commutative?

[strong moth]

i dunno because this is first few pages of the book doesn’t give any machinery of what’s valid

just says they’re equal

so wondered if there is some underlying set of moves that would make them equal if we assumed certain things are valid

it looks valid to me if you can move the wire from green position to purple position and vice versa

and make the respective move for the black wire?

[delicate orchid]

I agree with you yeah

[strong moth]

k maybe i’ll just skip this part of the book since it doesn’t give any valid operations

[strong moth]

and maybe it’s what @delicate bloom said here

[delicate orchid]

it is indeed

[strong moth]

k

it looks like it’s 2 inputs 1 output for one box and 1 box with just an output and no inputs ?

for both diagrams

[hot lake]

they're saying that you can manipulate equations a bit like the usual algebraic equations

that is IF the first equation between diagrams is correct

then you can add boxes and stuff to get other equations

[strong moth]

ok that makes sense

tyall

yeah this is from something called zx calculus. i think they later define some sort of algebra on the string diagrams

[tulip cedar]

maybe this is the new standard for string diagrams

[sonic coral]

my professor showed me his atlas of finite groups today, and i can tell you that i’ve never seen a bigger book in my life

[formal ermine]

[untold basin]

Hello, how can I start proving that the algebraic closure of Q is countable ?

[coral spindle]

Read about Skolem–Löwenheim 🤷

[formal ermine]

in general the algebraic closure of a non finite field has the same cardinality as the field

[runic hemlock]

the set of polynomials in rational coefficients is countable, and each one has finitely many roots

[spice whale]

are these like
category theory string diagrams

[strong moth]

yes i believe so

[spice whale]

cool

[untold basin]

Okay thanks guys 👍

[formal ermine]

what?

every element in a field is invertible

[meager jolt]

lol

I am incredibly dumb

[formal ermine]

lol don't worry

I make these silly mistakes all the time too

[coral shale]

let me introduce you to 0

[meager jolt]

You'd think I would've internalised this by now after taking Galois theory

[formal ermine]

"but what about the empty set"

[coral shale]

[coral shale]

allow me to correct - all units in rings are invertible

[formal ermine]

shuri, they asked if some number is a unit in some field

[coral shale]

idc

[south patrol]

Yes, some number is always in a unit in some field

[white oxide]

yeah someone told me

anyways is there a way

[prisma shuttle]

can someone explain why k[x_1, x_2, ,\cdots, x_n] is a graded algebra

and wut ppl mean when they say its "graded by degree"

ima learning about algebras for the first time so its kinda hard ofr me to wrap my head aroudn

[rustic crown]

algebra = vector space + ring

you say it's graded if you can write it as a direct sum ⊕_{n in Z} A_n such that A_n * A_m lies inside A_{n+m}

you say k[x1, ..., xn] is graded by degree since you can decompose it like above where A_d = homogeneous polynomials of degree d

because you can write L = K(alpha)^{⊕n} where n = [L:K(alpha)].

[chilly radish]

And the characteristic polynomial of m_a on K(a) is exactly the minimal polynomial

Did I steal your line det

[rustic crown]

yee pretty much >.<

[chilly radish]

Sorry!

[rustic crown]

it's uwu

Eso >.<

[pliant forge]

someone send me the paramedic

[rustic crown]

[chilly radish]

Cardano

[rustic crown]

ig if you really wanna avoid calculations using roots, you can use the formula for discriminant in terms of the resultant, which then can be computed via det of a 5x5 sylvester matrix

[next obsidian]

WTF why is det in a 5x5 Sylvester matrix????

[pliant forge]

LMFAO

[next obsidian]

I thought det was in 🇩🇪

[elder wave]

🇩🇪

[rustic crown]

(you'll have to explain that to det >.<)

[solar glacier]

question

if F is a field if F[x,y]/(y^2-x) iso to just F[y]

[rustic crown]

yep

[solar glacier]

another question

wait lemme double check my calculation

is x^4 - 2x + 2 irreducible over Z

[next obsidian]

Yes

[delicate bloom]

what was your calculation

like, high-level not nitty-gritty details

[solar glacier]

showing x^4+4x^3+6x^2+2x+1 is irreudicible over Z

im substituting in x-1

[next obsidian]

Okay nice

[solar glacier]

which reduced down to x^4-2x+2

[next obsidian]

It becomes this?

[delicate bloom]

chmonkey knows why I asked haha

[next obsidian]

Cool

[solar glacier]

i have the calculations

here look

[next obsidian]

Mero was asking how you showed this is irr

[chilly radish]

this combined with what det said

so first of all, to establish that this is tru

true

[chilly radish]

do you agree that $\alpha$ satisfies the \emph{characteristic} polynomial of $m_\alpha$

[cloud walrus]

Not ShiN

[chilly radish]

I mistyped my bad

[solar glacier]

so how do we show x^4-2x+2 is irreducible

[chilly radish]

hint: cayley-hamilton

right

[delicate bloom]

Eisenstein's criteria

[chilly radish]

now what happens if you plug in 1 to this equation

how does m_a act on the element 1

[solar glacier]

ahh

[chilly radish]

yea

[solar glacier]

its field of fractions right

[chilly radish]

so you get that alpha satisfies the char poly

now in the case K(a)/K, what is the degree of the char poly

right

and what's the degree of the min poly

[solar glacier]

so does this work

to showing x^4-2x+2 is irreducible

by einsteins criteria 2 divides both -2 and 2

but 4 does not divide 2

so its irr over Q

and by Gauss lemma

its irr over Z

[chilly radish]

exactly!

[next obsidian]

Bro why are you passing to Q

Eisenstein just works over Z

[chilly radish]

ok so we established that for K(a)/K, the min poly and the char poly coincide

[solar glacier]

does einstein crieteria work over z too

[next obsidian]

And if it’s irr over Q it’s irr over Z, you don’t need Gauss’s lemma for that

[solar glacier]

i didnt know that

[next obsidian]

Gauss’s lemma says irr over Z gets it for you over Q as well

[chilly radish]

now, do you see why L/K can be written as [L:K(a)] disjoint copies of K(a)?

[next obsidian]

Eisenstein holds for polynomials over any UFD

[solar glacier]

oh thats right

[chilly radish]

and primitive

[solar glacier]

sorry new to this portion of alg

[chilly radish]

ok, so this basically immediately follows from how you prove the tower law for degrees

take some basis $\beta_1,\ldots,\beta_n$ of $K(\alpha)/K$, and take some basis $\gamma_1,\ldots,\gamma_k$ of $L/K(\alpha)$. You surely know that ${\beta_i\gamma_j}_{i,j}$ is a basis of $L/K$. Now, note that $\operatorname{span}(\beta_1\gamma_j,\ldots,\beta_n\gamma_j)\cong K(\alpha)$ as $K$-Vector spaces for all $ 1\leq j \leq k$. Does this make sense so far?

[cloud walrus]

Not ShiN

[chilly radish]

alright cool, so this shows us that $L$ as a $K$-vector space is a direct sum of these spans, each of which are isomorphic to $K(\alpha)$, and there are $[L:K(\alpha)]$ copies in total. Now, how does $m_\alpha$ act on each of these subspaces? (In terms of its action on $K(\alpha)$)

[cloud walrus]

Not ShiN

[chilly radish]

to make it a bit more explicit, we've written $\ L=K(\alpha)\gamma_1\oplus\cdots\oplus K(\alpha)\gamma_k$ and we are now examining the action of $m_\alpha$ on $L$ by examining the action on each of these subspaces

[cloud walrus]

Not ShiN

[chilly radish]

yes

every element in $K(\alpha)\gamma_j$ is just something of the form $\xi\gamma_j$ for $\xi\in K(\alpha)$ right, so $m_\alpha(\xi\gamma_j) = \alpha\xi\gamma_j = m_\alpha(\xi)\gamma_j$. The point is that $m_\alpha$ acts on each of these subspaces exactly the same way

[cloud walrus]

Not ShiN

[chilly radish]

That is, if we were to write $M=[m_\alpha]{\beta}$, the matrix representation of $m\alpha:K(\alpha)\to K(\alpha)$ according to the basis $\beta = (\beta_1,\ldots,\beta_n)$, then the matrix representation of $m_\alpha$ on $L$ would just be a diagonal of these $M$s

[cloud walrus]

Not ShiN

[chilly radish]

no problem!

yea from here it's just using the properties of the determinant of a block-diagonal matrix and our initial result that we established

it's definitely not trivial though imo

like idk if 'recall' here is like "figure it out" or if you were actually supposed to know this

ahh ok

fair enough

lmao wow

[south patrol]

Very based

[sharp peak]

What are some examples of a commutator ideal of a lie algebra?

[tender wharf]

any work?

!show

[flat tree]

Show your work, and if possible, explain where you are stuck.

[tender wharf]

oh god thats mathfrak

hint: ||think Z_5 x Z_5||

[lethal dune]

#prealg-and-algebra

[tender wharf]

ok maybe at least #proofs-and-logic

[lusty marlin]

nah not really

[chilly ocean]

i think that #proofs-and-logic is appropriate

[summer path]

If $A$ is a $k$-algebra with generators $x_1,\dots,x_r$, why is $A$ infinitely generated? am I misunderstanding something?

[cloud walrus]

Tubular Cat

[lethal dune]

Infinity generated as modules

[summer path]

is it correct to say that A is finitely generated as a k-algebra (by {x_1,\dots,x_r}) but infinitely generated as a module (by a bunch of various degree monomials in x_1,..,x_r)?

[next obsidian]

I mean it’s still possible that it’s finitely generated as a module by those elements

So I dislike how they’ve worded it

It’s just possible that A is not finitely generated as a module

Take k[x] over k as an example

But then k[x]/(x^2) is an example where it’s still finitely generated as a module

[summer path]

i think i am confused by the wording lol

[next obsidian]

This is correct but the thing is you might not need every monomial

[summer path]

yeah ok

[next obsidian]

Like in the k[x]/(x^2) example

You don’t need anything beyond x

Because the crap just ends up as 0

In general you can say that if you’re generated as an algebra by finitely many elements x1,…,xn all of which satisfy monic polynomials over k, then the thing is finite as a k-module

[summer path]

i guess i understand the intention of writing the fact that A can be infinitely generated, but it need not be

[next obsidian]

Yes

[summer path]

thanks!

[next obsidian]

[somber sleet]

If G is a group of order 8 I know that there are elements of g with the order 1,2,4,8. Let now g in G be the element with order 4, then I define N:= <g>, and I know that N is normal in G. If I take an x in G\N why is x^2N=N? I somehow don't see the direct connection, cause I'd have (xN)(xN). Is xNx=N because x is the inverse of x?

[quiet pelican]

There isn’t necessarily an element of order 4
And if there is x^2N = N is just lagrange’s theorem applied to G/N

(Counterexample to the first: C2 x C2 x C2)

[somber sleet]

I know that there might be one

[quiet pelican]

Then see the second part

[somber sleet]

I'm actually trying to find groups of order 8 up to isomoprhism

[somber sleet]

I somehow don't see the argumentation

[quiet pelican]

G/N has magnitude 8/4 = 2
So by Lagrange, (xN)^2 = eN so x^2 N = eN

[somber sleet]

pff okay that was easy hahahaha

sorry, somehow didn't see it

thank youuu

[formal ermine]

CAYLEYTABLESCAYLEYTABLESCAYLEYTABLESCAYLEYTABLESCAYLEYTABLESCAYLEYTABLESCAYLEYTABLES

[somber sleet]

we didn't do it yet 💔

[formal ermine]

cayley tables are a very basic thing lol

also I was just shitposting

[quiet pelican]

That sounds like pain for order 8

[formal ermine]

now do it for order 40

[somber sleet]

my prof doesn't like it I guess

[formal ermine]

yeah my prof didn't talk about them either

it's just like

a multiplication table

[quiet pelican]

Sylow deals with this quite quickly ||every one is the semi direct product of some G of order 8 with C5||

[formal ermine]

woosh

[quiet pelican]

I recognised the joke
It just wasn’t funny

[somber sleet]

I think this is it haha

nobody describes a group with its table

[formal ermine]

yeah gruppentafel is cayley table

[somber sleet]

how would you write Z[X]/(X^2+X+1) as a set

is this like the set of polynomial of degree one?

and also is it true that a prime ideal in R is maximal if R is a pid?

[quiet pelican]

Yes

[quiet pelican]

And yes

[somber sleet]

I don't see why though, is there like a proposition or a theorem(?)

[somber sleet]

does this only hold in pid's?

[quiet pelican]

The way I would think of it is that you can “get rid” of any terms of degree > 1 using x^2 = -x - 1

[quiet pelican]

I wouldn’t think so

[somber sleet]

so if my ideal (...) is a polynomial of degree n then Z/(...)[X] is always the set of polynomials of degrees n-1(?)

[quiet pelican]

Yes

[somber sleet]

I'm asking this, because I was trying to see if (3,x^2+x+1) is maximal or prime in Z[X]

I think that it is not maximal

but it still could be prime right?

Z[X] is not a pid

[quiet pelican]

It’s not prime

But theoretically, if it were a different ideal, yes

[somber sleet]

I don't see why though

[quiet pelican]

x^2 + x - 2 = (x-1)(x + 2) is in the ideal

[somber sleet]

so x^2+x+1 is not prime in Z/3Z right?

so it's not a prime ideal

[quiet pelican]

Yes

[somber sleet]

I got it, thank you for helping me

[somber sleet]

were we wrong(?)

like this is reducible mod 3 or am I tripping(?)

[elder wave]

„Maximal aber nicht prim“

[quiet pelican]

I mean, this is a factorisation with neither factor in the ideal

Anything in the ideal of degree less than 2 is divisible by 3

[somber sleet]

in my opinion the answer above is wrong

[quiet pelican]

Actually maybe not

Actually it will be

I was thinking if we could cancel down terms using a multiple of 3 and of the poly, but we can’t (just get a multiple of 3)

[somber sleet]

how do you know if this is irreducible or not?

[north sand]

could check for zeros

[somber sleet]

yes

2 is a zero

[north sand]

probably the best thing to check first if you are given a polynomial of stupidly high degree

[somber sleet]

isn't there a better way

[elder wave]

is this an Altklausur from 2016 by any chance

[somber sleet]

it was taken in January 2016 yes haha

[quiet pelican]

Probably not here

[coral shale]

brute force is rapid?

its almost always brute force for small char

[elder wave]

profs at my uni do that as well

include the year in some exponent or something

[somber sleet]

my prof is gonna do it again this year

for sure

it's 2023 right

[elder wave]

indeed

[chilly ocean]

Are you one of those ppl that need to be reminded after new year because they write it wrong on postcards etc

[median pawn]

where does this cool equation come from? f is in R[X]

[oblique river]

Looks like it might come from considering forward differences

[rustic crown]

i think that's the (n+1)th derivative is 0. where by derivative i mean (Δf)(x) = (f(x) - f(x-1))/(x - (x-1)) = f(x) - f(x-1)

[median pawn]

interesting, why should this be true anyway?

also where does this discrete notion of derivative show up?

[rustic crown]

i've only seen it in some combi stuff

[oblique river]

In your problem, apparently :p

[rustic crown]

because Δ(x^n) = poly of degree n-1, so iterating it will kill the whole thing. Δ satisfies all the nice properties you want it to :3

[chilly ocean]

Δ = Id - shift might be also a good thing to notice, together with binomial theorem

[median pawn]

that's pretty smart

[median pawn]

which shift hmm

[chilly ocean]

shift(f)(x) = f(x-1)

is what I mean

this formula they wrote is basically using binomial theorem for Δ^(n+1) = (Id-shift)^(n+1)

[rustic crown]

that's quite nice Blitz

[chilly ocean]

and that Δ^(n+1) f(x) = 0 for polynomials of degree <= n for degree reasons as det already been saying

(i.e. deg(Δf) = deg(f) - 1 when deg(f) > 0, Δ is degree decreasing)

[median pawn]

this seems believable, i.e., that putting triangle^{n+1}(f(x)) = 0 would produce the equation

[median pawn]

idk what u mean by binomial theorem here

[rustic crown]

you have operators on the ring R[x]

consider the two operators Id and shift

Id(f) = f, shift(f) is the polynomial which satisfies shift(f)(x) = f(x-1)

[chilly ocean]

Binomial theorem is a pretty general formula

it holds in any commutative ring tbh

here you have... a ring of operators on polynomials

multiplication being composition, and addition being pointwise addition

I mean

sure this might not be a commutative ring but whatever

Id and shift commute is important part

so you have

[cloud walrus]

Blitz

[median pawn]

that's too good

[cloud walrus]

Blitz

[median pawn]

my mind is blown today

never used binomial theorem on operators except when i was really doing functional analysis lol

thanks a lot!!!

[pliant forge]

Let 0 \to X \to Y \to Z \to 0 be a SES where X and Z are noetherian R-modules. How can we deduce that Y must also be noetherian?

[elder wave]

You can use that quotients of noetherian modules are noetherian again and the fact that the SES and Y being noetherian implies that X and Z are noeth.

Take a bunch of quotients and construct new sequences

Maybe there’s a better way

[pliant forge]

Since X, Z noetherian any set of submodules will have a maximal element. So let Y_0 \subset Y_1 \subset \dots be an ascending chain of submodules of Y.
Then we have the sets S_x := {X \cap Y_i | Y_i in ascending chain of submodules in Y} and S_y := {p(Y_i) | p map from Y to Z in SES}
uhh..

[rustic crown]

we need to show middle is noetherian right

[elder wave]

Yes I mean that you use that fact in the new sequences you construct

Ah wait

I might be talking BS one sec

[pliant forge]

S_x would have a maximal element so any other Y_i for some large i would still be the maximal right

similarly for S_y

[elder wave]

ignore what i said i misread and mistook it for a different exercise

[pliant forge]

then we might induce SES from this such that 0 \to Y_{i+1}/Y_i \to 0

so Y_i+1 = Y_i

[rustic crown]

yea the idea is correct

[formal ermine]

common timo L

[feral agate]

Can someone help me to prove the underlined statement?

If I argue by contradiction no idea how to make progress

I guess I don't see why they necessarily need to be unique and you cannot have something like (4 2 3 3 4), when 4 is your a_1

[coral shale]

think its bijection idea

[vivid tiger]

so suppose sigma^n=sigma^k and wlog suppose n ≥ m
then sigma^(n-k)=x

[rotund aurora]

Let R be a ring, p a maximal prime ideal of R and M a finitely generated R-module. If pM=M does it follow that M=0 ?

[vivid tiger]

but oh no's, n-k < m, now done

[feral agate]

But that only guarantees the a_1 = a_m right and not the uniqueness for the elements in between

[vivid tiger]

i'm assuming sigma is invertible

[feral agate]

oh nvm I think I get it

[vivid tiger]

the argument i gave shows that the sigma^i (x) must be distinct. i don't actually have any clue what context this is in, nor what the sigmas are because i haven't read most of that

[feral agate]

Thanks for the inspiration

[vivid tiger]

this reminds me of the proof stuff used in e.g. Lagrange's theorem or other order stuffs

in that i used the exact same reasoning

[glossy crag]

Quick check: this should work verbatim for infinite splitting fields as well, right?

[chilly radish]

Baumslag-Solitar Group

[chilly radish]

How can one have an infinite-degree splitting field?

Perhaps I'm misunderstanding

Splitting fields are obtained by adjoining finitely many algebraic elements

[rotund aurora]

consider infinitely many polynomials

[glossy crag]

It's not about infinite degrees, it's about infinite families of polynomials. For the relevant definition see e.g. Lang.

[chilly radish]

Ah ok

[flint crater]

Bit confused. There is a dutch and english version of a book for class and one defines a representation of G of K (G a finite group and K a field) as a group homomorphism from G to GL(n, K) for n >= 0 and the other as a group homomorphism from G to Aut_K(G). Are these the same thing or is there a difference?

[formal ermine]

the first one is what you would typically call a matrix representation

a (more "general") rep is a group hom from the group to invertible operators on Kn

a rep can be made into a matrix rep by picking a basis

[chilly radish]

That should Aut_K(V) for some finite dimension vs tho

[flint crater]

Ohhh okay I think I might get it

[torpid bronze]

Is there someone who can help me with basic quaternion multiplication

[elder wave]

[torpid bronze]

this emote xd

Before starting, am I right to say that multiplying a vector from a n-dimension to a vector from a (n-1)-dimension will give a vector from the n-dimension

[coral spindle]

There is no notion of vector multiplication that I know of which gives you something like that

[torpid bronze]

Ok then, how is this guy multiplying a non-pure quaternion with a 3D vector and having as a result a 3D vector

[coral spindle]

The quarternion describes a rotation of 3d space. The 'multiplication' is best thought of more like an application of this rotation to a vector.

N.b. you are not multiplying together two vectors, you are multiplying a quarternion and a vector. If someone goes "um ackshually quarternions form an R-vector space" I will become the joker

[elder wave]

Mr Boytjie

☝️

[torpid bronze]

Ok well how should I process

[elder wave]

Have you considered that quaternions form an R vector space

[coral spindle]

🤡 hi guys do u like my new makeup

[torpid bronze]

bruuuuuuuuuuuuuuh

[rustic crown]

you're cute uwu

[glossy crag]

If D is a Dedekind domain and F its field of fractions, what could Jacobson mean by F(D)? Do you think it's a typo and he means E(D) (although that doesn't really change things)?

[oblique river]

I think he is defining two things

See the “subfield(subring)” as well

[rustic crown]

oh

[oblique river]

F^(1/q) is the subfield of Fbar of elements v such that v^q in F.

Also

D^(1/q) is the subring of Fbar of elements v such that v^q in D.

[rustic crown]

should have written "(respectively)" somewhere... thats really confusing

[oblique river]

@glossy crag

I agree, det

[torpid bronze]

Finally, it worked

https://tenor.com/view/happy-dance-way-to-go-gif-24438686

[glossy crag]

Oh crap, I hadn't realised that's what he meant!

Thanks @oblique river and @rustic crown.

[delicate bloom]

might be * is overloaded and it's implemented as something like q * p = qpp^-1

[elder wave]

Good catch

[glossy crag]

I'm blanking, why I'^{q-1}J'^{q}\subset J'?

[warm shoal]

Could anyone help me w this?

[south patrol]

It should just be a direct computation

like see how πaπ^-1 acts on π(a_i) and stuff not in the image of π

[warm shoal]

i dont understand that tho

like

pi is a permutation

like (1, 2, 3, 4) (top row)

(4, 3, 2, 1) (bottom row)

[vagrant zinc]

How can I get the subgroups out of the Cayley table?

My problem is
Z_3×Z_3
I have to take out all the subgroups of these

[warm shoal]

and a is some cycle

like (xyz)

what does pi(ai) even mean

like that matrix (....)(x) ?

@south patrol

[tender wharf]

I mean we can use lagrange here

no

[tender wharf]

pi(a_i) is what happens when you use pi on a_i

like say pi = (123)

and a_i = 2

then it just sends 2 to 3

[warm shoal]

so pi(2) =3

?

@tender wharf ❤️

[vagrant zinc]

I have not read about it

[tender wharf]

yeah pi(2)=3

[vagrant zinc]

I hardly go in subgroups

[opal osprey]

Let $G$ be a finite group of order $2^{6} \cdot 3^{3}$ and suppose by way of contradiction that its number of Sylow $2$-subgroups is equal to $9$. Denote by $S$ its set of Sylow $2$-subgroups. Then the second Sylow theorem guarantess us that the action:
$$
\Psi : G \rightarrow \text{Perm}(S)
$$
given by $g \mapsto \Psi_{g}$ where:
$$
\psi_{g}(H) = gHg^{-1}
$$
Is well defined and transitive. Notice that a fixed point of this action corresponds precisely to a normal Sylow $2$-subgroup of $G$.
\bigbreak
We assure that this action has at least one fixed point. Indeed, if it didn't, then $\Psi$ would correspond to a free and transitive action of $G$ on $S$, thus implying that $2^{6} \cdot 3^{3} \neq G| = |S| = 9 $, which is a contradiction. This then implies the existence of a fixed point of the action $\Psi$, thus of a normal Sylow $2$-subgroup of $G$. This is again a contradiction, since this would imply that $G$ has a unique Sylow $2$-Subgroup, since maximal $2$-subgroups are unique, contradicting our hypothesis.
\bigbreak
We conclude then that $G$ can't have exactly $9$ Sylow $2$-subgroups.

[cloud walrus]

MisterSystem

[opal osprey]

Could someone verify if this particular argument is valid???

I am pretty sure there's a mistake somewhere

But I can't find it

oh

I fucked up the definition of free action

that's the mistake

[lethal dune]

ye

you should get |G/N(H)| =|S|

also like your argument works for all group then all syllow groups are normal

[smoky cypress]

So C[G] is isomorphic to the product of irreducible representations right

Is there a similar statement for R[G]

Where R is the reals

[somber sleet]

Is Z[X] a PID?

[next obsidian]

What do u think

[somber sleet]

I mean Z is Euclidean, so I'd say yes

Where the Euclidean norm is the absolute value

[next obsidian]

What facts do you know about PIDs

[somber sleet]

Or am I wrong?

Every PID is euclidean(?)

[next obsidian]

That’s not true

[somber sleet]

No wait

I wanted to say the contrary

[next obsidian]

Okay but that doesn’t help here

If you’re trying to ascertain if Z[x] is a PID (unless you want to try and prove it’s Euclidean)

[somber sleet]

I know the definition of PID, which is that every ideal is proper

[next obsidian]

I don’t think that’s what you meant to type

[somber sleet]

What do you mean?

[next obsidian]

Well, no ring has the property that every ideal is proper

Because the ring itself is an ideal

[somber sleet]

Oh wait you say principal in English

[next obsidian]

Yeah, proper means “is not the entire set”

[somber sleet]

So a PID is an integral domain where every ideal is principal

[next obsidian]

Okay, but do you know any other properties? You can disprove that Z[x] is a PID easier if you know some other properties of PIDs, but you can also just prove it directly

[somber sleet]

Do you have a hint?

You don't have to tell me the whole answer

[next obsidian]

Not in any fashion that doesn’t just solve the problem

Just find an ideal that can’t possible be principal

I say “just” to mean that there’s nothing fancy you have to do

[somber sleet]

(2,x)

Was it this one?

[next obsidian]

Now prove it can’t be principal

And you’re done

[somber sleet]

Yes I got it, thank you 🤭

[next obsidian]

[somber sleet]

[chilly ocean]

this has absolutely nothing to do with abstract algebra whatsoever

[stone notch]

@chilly ocean Thank you for pointing me in the right direction

[chilly ocean]

You could use that R[x] is pid iff R is a field if you know where it comes from

[jovial flint]

where can I find a proof of the cardinality result mentioned here? $V$ is a finite subset of $\mathbb{Z}^d$ and $\Delta$ is the lattice laplacian

[cloud walrus]

Bilboswaggins

[zinc heath]

Let $R$ be a commutative ring and $I$, $J$, $K$ be three ideals in $R$. Then show that if $I \subseteq J\cup K$, then $I \subseteq J$ or $I \subseteq K$.