groups-rings-fields

[chilly ocean]

is there a lemma for this qq

[next obsidian]

For what

@chilly ocean

[chilly ocean]

i figured it out brute forcing

[next obsidian]

Anyway

I think you can figure out how to factor that poly

In a cheater way

Write x^2 = y

Turns into y^2 + y + 1

[chilly ocean]

showing x^4+x^2+1 resucible

[next obsidian]

Use quadratic formula to find the roots in terms of y

Translate to x^2

[chilly ocean]

ive done this

yeah you get roots

square roots

not in Z[x]

[next obsidian]

Yeah and sqrt(y) = x

I mean you want to factor it like TTerra did right?

[chilly ocean]

yeah i figured out

[next obsidian]

Ok

[chilly ocean]

but i wonder if there is better way

[next obsidian]

I mean that is a good way

[chilly ocean]

it for sure works

[next obsidian]

You just use quadratic formula and it kinda spits it out

[chilly ocean]

but feels messy

[next obsidian]

I think that’s clean

[chilly ocean]

i tried factoring like ttera

[next obsidian]

Yeah but you can get that factorization without brute forcing it

[chilly ocean]

didnt go smoothly because i got binomial that isnt in Z[x]

brb ill show

[next obsidian]

You ought to be able to get from here to the good factorization

I am trying to think how, I think it probably is clean

Must think

Oh

Brain work

Fuck so like

I think you need to break that up into products of linear a

Then swap

And you should get like

(x - z)(x - z-bar)(x - w)(x - w-bar)

Which turns into two real quadratics

Like basically you have to find

[chilly ocean]

o true

[next obsidian]

Sqrt((1-isqrt(3))/2)

But yeah this is kinda shite

[next obsidian]

Cuz you want to write this as like difference of squares

I think

And then it should just turn out that you get conjugate pairs

Mainly cuz like you have (1 +- isqrt(3))/2

And then like that gives you the conjugate-ness

That’s my chmonkeytuition

[chilly ocean]

mm

[next obsidian]

But yeah this kinda sucks deek

Anyway in general like

I guess you posit it it’s a product of quadratics

And write it out in general

Then use Lin alg to find the coefficients IG

[chilly ocean]

yeah

its just really measy lol

[next obsidian]

Eh

I think it’s not as bad as it could be

[chilly ocean]

because i have to check for or cases

it could be worse thats tru

[next obsidian]

I mean you can always assume it’s

(x^2 + ax + b)(x^2 + cx + d)

Like Gauss’s lemma says it’ll factor in Z if it factors in Q

And even w/o that like

You can always assume the polynomial it factors into are monic

By just like moving the leading coefficient over, you know the coefficients on x^2 have to be inverse

So just move them over

Then it’s just 4 unknowns and it is kinda shit sure but you get like 4 equations for x^3,x^2,x,and constant term

In 4 variable

So you can throw it into a matrix and do like rref to find a solution

Or maybe not, this maybe isn’t linear

Whatever

Chmonkey

[uncut girder]

Chmonk

[chilly ocean]

ye

it boils down to basic HS algebra

i ended up getting that a(d-b)=0

so a is 0 or d-b=0

a=0 runs into a contradiction

d=b works but then you need to consider if b=d=1 or -1

you have contradiction when they are negative

the reason its messy is because showing the contradiction is messy

i dont need to show all the details though

[regal arrow]

Question, what's up with F_1?

Idk much algebra but I'm getting that "field with one element" is touchy

Why so?

[scarlet estuary]

it formally doesnt exist

but one can define structures that "feel like" what it'd be if F_1 existed

in that you can put geometries on them that "agree with" what we'd expect geometry on F_1 to be

in practice, you dont need to worry about this unless youre one of like

10 people in the world

for everyone else, it suffices just to say F_1 doesn't exist since a field must have at least two distinct elements.

(if youre wondering why we dont just let {0} constitute F_1, it breaks all our theorems so we'd need to revise all the statements, and some definitions too)

(and doesnt really behave as we expect a field to)

[final pasture]

char 2 breaks half of the theorems too but we allow it tho

[scarlet estuary]

but fields of char 2 at least feel like fields

like gross, ill-behaved fields

but like fields

[scarlet estuary]

if youre interested in this, see https://arxiv.org/abs/0806.2401

just note that its, uh, pretty prereq heavy

[final pasture]

(I was joking)

[prisma thunder]

I'm still stuck on that problem :/

Ugh.

I thought I figured it out but I think I reached a dead-end. Hmm.

OOHHHHHHHHHH

Wait

[next obsidian]

^^Maff^^

[prisma thunder]

@next obsidian

[next obsidian]

Yes

[prisma thunder]

Suppose $L = F(\alpha)$ and $n$ denote the degree of the minimal polynomial of $\alpha \in L$ over $F$; since $L\backslash F$ is Galois with Galois group $G$, if we consider the field of extensions $F \subseteq F(\alpha) \subseteq L$, we have
$$|G| = [L:F] = [L:F(\alpha)][F(\alpha):F] = [F(\alpha):F] = n,$$
so then if $\sigma \in G$, there are $n$ distinct Galois conjugates $\sigma$ of $\alpha$.

[cloud walrus]

eM

[prisma thunder]

I figured out the reverse direction but I'm trying to figure out how to word the forward direction.

[next obsidian]

Ummm

This is like

If the extension is degree n

And there’s n Galois conjugates

Then L = F(alpha)?

[prisma thunder]

We want to show if L = F(alpha), we have to show the images of alpha under G are all distinct.

Which I guess

[next obsidian]

Okay so

G has size n = degree of L over F

Is degree of min poly

Yeah?

[prisma thunder]

Yes

[next obsidian]

Consider (x - beta_i) over all beta_i in the orbit of alpha under G

So this has degree = # of Galois conjugates

Notice this is fixed under the action of G

Cuz all you do is permute roots

So that this poly lives in F[x]

Thus alpha is a root of this poly

So it must have degree <= n, but also bounded by n cuz there’s at most |G| = n Galois conjugates of alpha, so it’s degree n

Aka orbit is size n

[unique berry]

I feel like I'm missing something obvious here but first off what ring is the principal ideal of x even an ideal of?

[next obsidian]

R[[x]] bruv

@prisma thunder idk if this is how you came up with the other direction but this is how I would do it

I think it’s pretty simple to state and stuff

Uh, it follows because Z[i] is a UFD

It’s just like a general fact about them that like gcds exist and they operate like dat

[unique berry]

I'm a bit iffy on the second part of the problem though

I'm guessing it has to do with how if the ideal wasn't maximal then there'd be nonunit elements

but idk how to really prove that

[next obsidian]

Just mod out by (x)

You get R

R[[x]] -> R defined by set x = 0

Kernel is (x)

Surjective

Apply 1st iso

[unique berry]

oh yeah

[unique berry]

I'm guessing this is a common trick for polynomial rings

[next obsidian]

I mean, yes

But this is a power series ring

The same idea works tho

The issue is just like, you can’t define a map R[[x]] -> R by sending x to like 1

Because infinite sum

[unique berry]

ok yeah

1st iso is so op

Just find some homomorphism and you already get an isomorphism for free

[next obsidian]

[chilly ocean]

ok

if i have to make a field with 144 elements

Does F_12[x]/(x^2+1) work?

where F_12 is field with 12 elements

wait bruh

idk if this exists lol

yeah nvm idk field with 12 elements

or 144

[upbeat juniper]

finite fields can only have prime power order

144 is not a prime power

[chilly ocean]

uh

ig

ik it has to do with inverses

but i forget exactly why

[upbeat juniper]

every finite field is a vector space over the subfield generated by 1, which is some F_p

[chilly ocean]

def doesnt clarify understanding

why are fields generated by 1 of prime power?

[upbeat juniper]

by 1 I mean the element 1

[chilly ocean]

i can see how that justifies powers

but i mean why only prime powered fields, the prime part

why cant subfield generated by 1 be F_2n

as example

[upbeat juniper]

this goes back to the idea of characteristic

[chilly ocean]

yeah

i remember

[upbeat juniper]

since it is a finite field, there is some sum of 1s that add up to zero

[chilly ocean]

1n=0 is characreristic

the n

[upbeat juniper]

in a finite field the characteristic must be a prime number

[chilly ocean]

?

i thought thats for perfect fields

oh lol

all finite fields are perfect

[upbeat juniper]

yes

[chilly ocean]

but i forget details

[upbeat juniper]

soz I can't continue talking cos I have an exam starting soon

[chilly ocean]

👀

[hidden haven]

Characteristic is prime for any integral domain

Because if it is mn then in the domain, mn = 0 so m = 0 or n = 0 but then mn = m or n

[chilly ocean]

oh ok

i can see how assume characteristic is prime power makes sense from this

like if characteristic is p^2 then pp=0 implies p=0 or p=0 lol

so nvm

characteristic is always prime for integral domains

but why can order be prime power :/

[hidden haven]

Any finite field is then a vector space over F_p of some finite dimension n so cardinality would be p^n

[rustic crown]

there is another nice proof, say char F = p, now if q is any prime that divides |F| then there is an element a with additive order q by cauchy/sylow whatever.
so q * a = 0, treat q as an element of the field as well! so since a is not 0, we get q = 0, i.e. p divides q.

so there is only one prime divisor of |F| which proves it's a prime power

[next obsidian]

Honestly

If ur going through all this

Just classify finite fields already

IMO

Like the moment you start thinking about them, you might as well just go ahead and classify them

¯_(ツ)_/¯

[rustic crown]

[robust pollen]

is this true?

[next obsidian]

I think this is like

Bass-Serre conjecture

And it got proven I think

I’m gonna google

Bass-Quillen

[next obsidian]

You can find some info in that ^^

[robust pollen]

thanks

[next obsidian]

Actually I forget if this is only for fg

[robust pollen]

but uh... that's not true, is it? Quillen-Suslin is just "fg projectives are free", and here again is the assertion that Quillen-Suslin is "all projectives are free"

[next obsidian]

But there’s a proof in Lang

[robust pollen]

from Lang, tf is a unimodular vector and the unimodular extension property

[next obsidian]

HMMMMMMMMMM

oh

It’s like

Defined in that section

That and the section before it

Combined proves it

[robust pollen]

Yeah, I see

[next obsidian]

But also yes it seems fg is needed

Yeah idk

I can’t find anything when you drop finite generation

I kinda feel like it’s either open or just false

[robust pollen]

thank, I'll put [citation needed] then

[regal carbon]

Hello@robust pollen

[robust pollen]

from Bass "Big projective modules are free". So now I see it for fields, but if R is a pid, why is it connected?

I mean, R pid, why is R[x_1, ..., x_n] connected?

Oh, it's an integral domain

[robust pollen]

Hello

I think I'm dumb.

Take any ring R. Is the answer to "is the dual of a finitely generated projective R-module again projective?" as simple as

[hidden haven]

Connected rings?

[robust pollen]

$R^n = Hom_R(R^n, R) = Hom_R(P \oplus M, R) = P^* \oplus M^*$, where $P$ is our fg projective?

[cloud walrus]

expectTheUnexpected

[robust pollen]

those in which the only idempotents are 1 and 0. Which apparentley means that the spectrum is connected in zariski top, but uhhh

[hidden haven]

oh ok not too bad

[robust pollen]

But moldi, is my logic above correct for the duals of fg projectives? seems 2 trivial 😬

[hidden haven]

I do not understand it

[rustic crown]

for commutative rings this is the condition that it's not a product... why are products disconnected, how to picture it?

[robust pollen]

what but you know everything!

[rustic crown]

woah pretty neat

[hidden haven]

Primes in product are whole ring x prime

[hidden haven]

woah pretty neat

[robust pollen]

is it correct? I don't believe it lel

[rustic crown]

i can't geometrically think about spec yet 😦

[hidden haven]

Disconnection is V((1,0)) and V((0,1))

I'm not thinking geometrically

[rustic crown]

yea, i remember it was pretty simple that projectives are precisely direct summands of free

[robust pollen]

yeah, I know, but then I don't get why this is an actual exercise and not just a minor remark

[hidden haven]

You're just too good for the book

I'd recommend switching to a harder book

[rustic crown]

[robust pollen]

No, it's actually a course I'm TAing fugg

Actually, I found the non-trivial part

[hidden haven]

I'd recommend switching to a harder course

[rustic crown]

wait right

why is dual of R^n = R^n?

[robust pollen]

They haven't learned that/how/when Hom commutes with direct sums

[rustic crown]

i remember dualing kills torsion

so situation is good if you're working with fields

maybe check this up once more

[robust pollen]

But, dual is defined as M^* = Hom_R(M, R), no?

[hidden haven]

Isn't this by uni prop coprod

[rustic crown]

right

[hidden haven]

Hom(R^n, R) ≈ Hom(R, R)^n

[rustic crown]

my bad

[robust pollen]

I like russian math books where End_R(R) = R as rings. Because they write the composition f then g as fg

[hidden haven]

End_R(R) is R anyway though isn't it

Like notation shouldn't affect that

[robust pollen]

If R is not commutative, End_R(R) = R^op

[rustic crown]

yea, Hom(R, M) = M

oh rip

[robust pollen]

no, as rings, not modules

[hidden haven]

Wait what

Oh

Rings bad

Too many structure

[rustic crown]

yea lol, spec R is too much

[robust pollen]

join the dark side moldi, learn rngs. They have less structure.

[hidden haven]

[robust pollen]

Actually, a unit is a property and not structure, right?

[rustic crown]

property and structures are same thing ig

[robust pollen]

no

[hidden haven]

It's part of the structure if you want your substructures to contain it and morphisms to respect it

[robust pollen]

wat

[rustic crown]

elements can be thought of as 0-ary operation

and now you just want this operation to be preserved

[hidden haven]

This is a model theory question

[robust pollen]

But, like, an algebra can admit multiple non-iso bialgebra structure. Thus, being a bialgebra is not a property of an algebra. But a bialgebra admits at most one Hopf algebra structure (i.e. an antipode), so being a Hopf algebra is a property of a bialgebra.

Isn't that right?

[hidden haven]

[rustic crown]

lol so many words

[hidden haven]

Ooga booga but ok.

[robust pollen]

ok, uhhh... like, a semigroup

[hidden haven]

[robust pollen]

wait

uh

lel

[hidden haven]

[robust pollen]

Meh. I can't formulate this in Set, because every set has a unique comonoid structure.

[hidden haven]

[rustic crown]

woah

algebraic structures that are not glorified sets

[robust pollen]

quoi

[hidden haven]

quoi

[rustic crown]

quoi

[robust pollen]

(french for "what?!" btw)

[rustic crown]

(thanks for telling)

[hidden haven]

Question ignored

[robust pollen]

You're welcome, I know not everyone has read EGA

[rustic crown]

ohhh that was a question

[hidden haven]

Pretend

[rustic crown]

i thought this was nice, that people are interested in studying things that are not directly some added structure to set

[robust pollen]

Ah, if. But what I mean is that a monoid is unital in a unique way, right? So, of course, if you talk about the category of unital monoids, then the unit is part of the structure, but I don't even know what I want to say anymore.

[hidden haven]

No even if it's unique, if you don't make it part of the structure you still shouldn't call it a unital monoid

[robust pollen]

uh, I don't get what you mean.

[hidden haven]

Because morphisms from or to this will not need to respect the unit

[rustic crown]

then probably i didn't get what your statement meant lol

[hidden haven]

Imagine not learning what a comonoid is in your course

Your teacher must have been worse than mine

[robust pollen]

let me make it precise: In the cartesian monoidal category Set, every set carries a unique coalgebra structure.

actually, I just added the words "cartesian monoidal"

[rustic crown]

which course are you talking about lol

[robust pollen]

sure, i guess. Do we mean different things by "property" and "structure", then? I thought I was using it the way, but maybe I'm wrong

[hidden haven]

Category theory course

[rustic crown]

F

[hidden haven]

I am using structure to mean model theoretic structure

[robust pollen]

I'll have you know that I'm a model theory pro and totally know what you're talking about

[hidden haven]

Would expect nothing less from you 🥰

[rustic crown]

meanwhile me who thought that there was only one type of structure

[robust pollen]

what do you mean by "type"

is a structure an algebra over an operad

[rustic crown]

like one definition

[hidden haven]

I assume the model theoretic type

[rustic crown]

i can't think of a proper definition of structure

what's the model theory definition?

[hidden haven]

Did you not attend the tutorial where I defined it

[rustic crown]

you said something like topological spaces then are weird

[hidden haven]

I don't thing you attended lul

[rustic crown]

what did you define during this then?

[hidden haven]

During what

I'm talking about the logic tutorial that I did when topology tutorial was cancelled by hanny

Took advantage of free slot this was before cat theory tutorials started

[rustic crown]

you're once talking about why continuous maps aren't model theoretic structure maps

[hidden haven]

Pretty sure I didn't devote a tutorial to that

[rustic crown]

i think it was just on DMs

[hidden haven]

Ye

[cloud walrus]

expectTheUnexpected

[robust pollen]

Since the class of fg projectives is self-dual, and since the dual of a projective module over a field is projective, so this is in some sense the simplest example (since Z is a euclidean domain, i.e. closest thing to being a field, and that direct sum above is the simplest non-finitely generated module over it)

[neon warren]

Abc

[chilly ocean]

I was "manually" proving that distinct up to real multiples and irreducible implies coprime for elements in R[x] (R= reals here) of degree 1 and 1 and degree 1 and 2, but can't quite figure it out for degree 2 and 2, is this even true? Is there some high powered theorem that would be more obvious to use instead to show this?

well ok just googled it too 💀

makes sense, you can't get anything smaller containment wise from the sum than the whole ring cause we're assuming maximal (via irreducible generator) anyway

and assuming distinct up to real multiple and irreducible also take care of the case where you add two things in the same ideal for degree <= 2

so that will never happen
much better than my "manual" proof it was yucky yucky

[robust pollen]

well, what's your group?

[hidden haven]

If it's mod 3 then it's in Z[x]/3Z[x]

[frigid bough]

im having brain poop; if I have two groups G,H and a representation of both on a single vector space V; does V furnish some representation of some G semidirect product H?

[hidden haven]

But it doesn't live there Unless you mean ignore mod 3 in which case yes all linear monic polynomials are irreducible over integral domains

[cloud walrus]

Yeetus

[hidden haven]

The representations are maps from kG and kH to GL(V). These should induce a map kG ⊗ kH → GL(V), and kG ⊗ kH ≈ k[G × H]. You might want to check some of these things I'm not completely sure about this

[hidden haven]

This irreducibility check is happening in the quotient

So your question just with Z3[x]

And same answer applies

Because Z/3Z is an integral domain

Yes. Implicitly using the isomorphism Z[x]/3Z[x] ≈ (Z/3Z)[x] here

[plucky flicker]

Do you have any idea how to show that if for a character chi(g) = 0 for all g except for the identity, then chi(1) is divisible by the order of G

[frigid bough]

Im kinda rusty, what morphisms are these? by kG, ,.. I assume you mean group algebra; what type of morphism are you having from kG -> GL(V)? because GL(V) is not a vector space over k; so I'm confused where you are getting the induced map from the tensor product

[plucky flicker]

any hint would be appreciated

[robust pollen]

should be End(V) I think

[frigid bough]

oh sure

[hidden haven]

k-Algebra morphism

[frigid bough]

so they are algebra morphisms

[hidden haven]

Oh right should be end

mb

[frigid bough]

is the claim then that if you have a V that furnishes representations of G,H, then it automatically furnishes a representation of G x H? because I thought this was wrong but maybe im poopoo brain

[hidden haven]

Yeah I think that should be the case

Working it out concretely might help verify

Though I don't see any obvious ways of making (g,h) act on V 😵‍💫

[robust pollen]

doesn't it suffice that (chi, chi) = 1?

[plucky flicker]

why would it be 1? chi is not irred

not neceserally

[robust pollen]

oh.

[hidden haven]

But it should be an integer tho right?

[frigid bough]

wait this can't be right; what if I just took a representation of G semidirect H; then forgot about the semidirect structure, now I have a representation V of G,H separately; but somehow you end up with a representation of GxH?

[hidden haven]

Wait chi(1) is always dimension of rep

Ye it's probably wrong then lol

Is kG ⊗ kH ≈ k[G × H] true?

I think it is but 😵‍💫

[frigid bough]

I'm big confused

[hidden haven]

And I hope tensor product is coproduct in category of k-algebras

Lol 😵‍💫

[frigid bough]

wait thats not dimensionally correct? for finite groups, left hand side has dimension dim(G)dim(H) but right hand side is dim(G)+dim(H)?

[robust pollen]

No, for sets |S x T | = |s| |T|, no?

[hidden haven]

No |G × H| = |G| × |H|

[frigid bough]

oh right my bad

[robust pollen]

coproduct in Ring is tensor product over Z, algebras are just beefed up Rings, Q.E.D.

(lmao)

[hidden haven]

I am convinced

I don't quite get your question 😵‍💫

You want to prove that the subset containing 1 and all order 27 elements is a subgroup?

That seems false

Unless you have abelian or something

[robust pollen]

that only implies that the order of ab divides 27 though, right?

[hidden haven]

Ye you would need to

Ye but you might be able to somehow show that product of order 27 elements is order 27 I guess

2 and 27 are coprime

I am convinced

[robust pollen]

Wait, does this really work? Don't you need something like kH is a kG module algebra, and then a kH-module inside kG-mod is a module over the smash product kH # kG, which I guess would be the group algebra of the semidirect product?

[hidden haven]

I have no clue what a smash product is

You and your words

[robust pollen]

I T ' S J U S T

[hidden haven]

This just shows that a^27 is e

(ab)^27 I mean

Not that (ab)² is not e for example

Order 27 means no smaller power is e

Like x^27 = e but if also x³ = e then x doesn't have order 27

Because order is smallest positive exponent such that element becomes zero

[frigid bough]

is the consensus now that you get a representation of G semidirect H (from a rep of G,H)? 😢

the only thing I think I'm sure of is at the least, you get a representation of G freeproduct H.. right? lol

[robust pollen]

H a Hopf algebra (then H-mod is monoidal), and A a (unital) algebra in H-mod, then the category of A-modules internal to H-mod is isomorphic to the category of modules over a certain unital algebra A#H, which is called the smash product of A and H. The underlying vrector space is just A tensor H, but the multiplication is funny and I cna never remember. something like
a x h * b x g = a(h_1 b) x h_2 g I think. (sweedler notation used)

[hidden haven]

The consensus is my identity 🤡

[frigid bough]

I'm not really sure the semidirect makes that much sense to me either because it 'breaks the symmetry' between G and H

I also dont know basic algebra so I could be poopooing it up

[robust pollen]

So, your situation was just "G, H groups, and V a representation of G and H simultaneously", but without any compatibility conditions?

[frigid bough]

yep

[hidden haven]

Hopf algebra more like hopfully I can understand what you said within the next year 🤧

[robust pollen]

I breave dat shi*

but I don't know anything beside basic Hopf algebra and quasi-Hopf algebras, lmao. That's what you get for doing physics, and switching to math too late

[robust pollen]

without any compatibility conditions i cannot see anything 😦

[frigid bough]

well, if you're a physics person maybe I can say my actual question...
fermions live in some representation of lorentz group, but they also come equipped with actions of other symmetries (like chiral, C,P, blahblah..) depending on what you're doing, and for some reason I've never seen what actual group you get for the symmetries of a single fermion field (with say, just spin indices), if you include

• lorentz, C, P, chiral
  the C and P do funny things with the lorentz and I'm sure you do end up with somme kind of semidirect product, was curious about the general case

I have no idea where to look, but all I really want is a decomposition of operators into irreps of this lorentz (semi) C (semi) P (semi) chiral; (by that I mean a decomposition of the exterior algebra, something like that, into 2quark operators, 4 quark, etc)

[robust pollen]

"well, if you're a physics person" lmao, I'm neither a physics nor a math person, I dun totally goofed w/ my choices mang

[unkempt comet]

vibe

[frigid bough]

living in the inbetween I see lol

[robust pollen]

But aren't you asking here for like the gauge groups of some field theory, e.g. the entire standard model?

[frigid bough]

ah so I'm just dealing with QCD, so my fermions just come with SU(3) color as well; but that doesn't interact with the spin indices so it'll just be a direct product I'm thinking

[robust pollen]

oh man I wish I remembered/understood anything. aren't particles e.g. fermion something like section in the associated vector bundle of some principal bundle for the gauge group? Actually, that might be wrong, because it sounds too global.

[frigid bough]

this is something I wish I understood too lol; its funny because in all physics textbooks essentially its always assumed that you have a trivial bundle, and so you can write down fermion/gauge fields as just like real functions over space without ever talking about the bundle structure

but yeah, its something like gauge field lives in the associated bundle to the adjoint representation, and fermion in the vector (standard 3 dimensional rep) associated bundle, if you're doing SU(3) gauge theory

[young dew]

can someone explain how to find order of an element in a finite field. specifically im working with F_64 generated by x^6 + x + 1, and i need to find the order of x

[stark sigil]

You can always do it the direct way...

Multiply x by itself until you get 1

[young dew]

im not sure what that means

like into our generator poly?

[stark sigil]

The order of an element is what?

[young dew]

how do i make x^m = 1

ok yes sorry

i know the order is the amount of times needed to get back to 1

but idk how to understand the process in this context

[stark sigil]

That's because your definition isn't precise enough to use

[young dew]

say i have x^10 how do i even equate that to be a number let alone 1

[stark sigil]

Amount of times of what?

Oh, well x^10 is an element of F_64 so it is a linear combination of 1, x^2, x^3, x^4, and x^5

Every time you multiply something by x you can reduce it mod x^6+x+1

[young dew]

amount of times needed to add that element back to get 1? (or is it multiply

[stark sigil]

Hmmm indeed

add or muiltiply

[young dew]

i think its add

[stark sigil]

Better get that straight

[young dew]

yes

[stark sigil]

Okay, then the answer is 2 because 2x = 0

or actually

you needed to get 1 or 0?

[young dew]

need to get 1

in the field generated by F_2 / <x^6 + x + 1>

[stark sigil]

Well adding x to itself creates the sequence x, 0, x, 0, x, 0, ...

Doesn't it?

[young dew]

i dont understand that

two iterations it just 2x

so they all stay in that same position of the vector when adding x

[stark sigil]

What is 2 in F_64?

[young dew]

1+x?

if i had to guess

[stark sigil]

😵

What is 2 in F_2?

What is 2 in F_2[x]?

What is 2 in F_2[x] mod any polynomial?

[young dew]

im sorry i dont know. you mean looking for remainder 2 when dividing by a poly in F_2[x]?

[stark sigil]

noo

Answer the first question, then the second one

then I will clarify the third one

[young dew]

F_2 is just 0 and 1

so i guess 2 must be 0?

F_2[x] is all poly with 0 or 1 as coefficients

how to do name one of those elements 2

[stark sigil]

If 2 is 0 in F_2 why would 2 not be 0 in F_2[x]?

After all... F_2 embeds in F_2[x]

How does 2=0 in F_2 lead to 2=/=0 in F_2[x]?

[young dew]

are you confirming 2 is 0 in F_2

im not sure the pattern to relate to F_2[x]

[stark sigil]

ok so

2 is 1+1

1+1=2

Excellent equation

[young dew]

:nods:

[stark sigil]

We have confirmed that 1+1=0 in F_2

because F_2 is defined to be Z/2Z

[young dew]

okay that makes sense

[stark sigil]

Now in (Z/2Z)[x] how can 1+1 become something nonzero?

It's gonna stay zero...

[young dew]

so 2 is 0 in F_2[x] also?

[stark sigil]

Ya

F_2[x] is the ring of formal polynomials in the symbol x with coefficients in F_2

[young dew]

ok i see what the elements look like

[hidden haven]

It's not true actually, because if a has order 27 then a³ has order 9

[young dew]

so the case of F_64 generated by x^6 + x + 1, we have poly up to degree 5 with coefficients 0 or 1

[stark sigil]

Yep

[young dew]

if im looking for the order of x do i need to find some integer m larger than 6 such that x^m mod x^6 + x + 1 =1?

[stark sigil]

Yes

You're well on your way to solving this problem now

[young dew]

thank you

this seems very tedious to do a lot of long division for so many m cases

[stark sigil]

There's a better algorithm than long divisioning every time

Just multiply the last answer by x

then do the reduction

and repeat

[hidden haven]

Yes 31 being prime should make it work

[young dew]

can you explain that? so say i try x^10 mod x^6 + x + 1 and i get remainder x+1 (making this up)
then the remainder for x^11 would be the (x+1)*x now? each time the remainder goes up by x?

[hidden haven]

Yes because if a and b are order p, (ab)^p = e and any smaller exponent that makes it e has to be a factor of p which can only be of a and b are inverses

[robust pollen]

Is there an elementary (= without localizations) proof of the fact that Q is a flat Z-module?

[next obsidian]

Uhh

Doubt it

I mean

Z is a PID so torsion-free <==> flat

But that’s not elementary

Honestly I’d argue that if you’ve defined flat

Then localization is just more elementary than the notion of a flat module

So why do you want to do this?

[hidden haven]

Pretty much what you wrote. What I was saying is that if order of ab is 1, then ab is identity so a and b are inverses of each other

[robust pollen]

yeah, saw that as well, but the proof is also not appropriate.

[hidden haven]

Appropriate for?

[robust pollen]

Really?

[next obsidian]

Yes…

Localization is simpler than the tensor product

Pls

You just add fractions

[hidden haven]

bruh

[robust pollen]

For people without Comm Alg. They are probably just hearing about projective and flat modules for the first time, since we're going the abelian category route

[hidden haven]

[next obsidian]

[hidden haven]

They can be

ab is still in H

Everything works

[unique berry]

Got a 66 on my exam 😭

My confidence is done for (I felt incredibly confident before, during, and after)

[next obsidian]

What if the avg was like 40

[unique berry]

no

average was a 90

[stark sigil]

Look at points lost because of what reasons

[next obsidian]

[unique berry]

The test was legit easiest shit ever

I just suck ass

[sturdy marsh]

do well on the next one

[unique berry]

I said that last exam and did worse lmao

seems like the harder I study the worse I do

[sturdy marsh]

[stark sigil]

.

[chilly ocean]

maybe you were just having a bad day

exams are stupid anyways

[unique berry]

I was

I legit was sick during the exam

shat myself twice

[chilly ocean]

well

i don't think your poor performance had to do with you (supposedly) being bad at math

[unique berry]

I mean I've done well on exams while shitting before

I did my linear algebra final while throwing up

got a 96

could've been 100 if I didn't forget rank nullity

but yeah next exam is a final

well not really a final

but like it's everything from part 2 of D&F

[unique berry]

This ain't topology bro

that class was hard af

I feel like the only way to learn topology is just to do it twice

[unique berry]

Well I'm not bad at math but like I'm not as good as my peers

[chilly ocean]

i didnt mean to imply you were, it was poor phrasing

[unique berry]

And it's not even a peer thing it's more me comparing myself to what I think I should be at

Like I think I should've nailed that exam

last exam I was even more unsure about and I got above the median

[past temple]

ive been completely stuck on how to approach this problem for a few hours now

I want to prove that for $m,n$ coprime, $F_{p^m} \otimes F_{p^n} \cong F_{p^{mn}}$

[cloud walrus]

pdk

[past temple]

where p is prime

and the tensor product is taken over F_p

[deep nova]

hey so im stuck on this problem; so from what i gathered, R doesn't have an identity because it doesnt and cant have the identity matrix, but im not sure what subring would have the identity in the same matrix form as R

can i use a subring with a matrix different to the one for R and call it a subring?

or am i supposed to work with this only

[south patrol]

there is a subring isomorphic to a more famous ring, if that gives a hint

[south patrol]

and yeah it has to be a subset of R so i'm not sure what you mean by this

[deep nova]

i meant like S can be array 1: (a 0) array 2: (0 b) but realized that wont be a subset at this point

but i still dont get how i can possible form an identity matrix out of what im given

there's always going to be that zero at the bottom right

[deep nova]

i really dont know any famous rings lol

just assume i know nothing bc i have a hard time learning anything from my textbook or lecture notes

[south patrol]

I'm not sure how to give a hint without giving it away unfortunately

[deep nova]

so far, what i know i can work with is the definition that S should not be an empty set, elements r,s in S should also have r*s in S and r-s in S

to have a subring S of R if S is a subset of R

and so going off of that, I tried doing subtractions and multiplications with different matrices in hopes of forming an identity

am i on the right track at least?

[south patrol]

I think you're probably getting too hung up on the identity, take a step back

Can you see any "obvious" subrings of R?

(the most obvious one to me has an identity)

[deep nova]

im not sure if i have a gap in my knowledge here regarding matrices and subsets of matrices, because thinking about it, i cant really figure out any unique subsets that wont just be ([a b] [0 0])

just for reference

this is the only example or mention

of matrices in my textbook regarding subrings

am i supposed to look at general linear groups?

but those only have multiplication as binary operation

i really dont understand what im supposed to be doing here

i give up lol idek anymore

[south patrol]

Yeah probably over complicating it dw lol

Another major hint is that diagonal matrices are nice lol

[deep nova]

i did think of diagonal matrices, but i didnt want to use it since i wasnt sure if that would necessarily give me a subset of R

[south patrol]

well pick diagonal matrices w bottom right 0

[upper cape]

for a ring A is xA[x] isomorphic to A[x]?

[unreal portal]

I don't believe so, if you have a map $A[x]\to xA[x]$, then no element will map to the identity polynomial 1

[cloud walrus]

cgodfrey

[stark sigil]

They're isomorphic as A-modules, that's as much as you can say

[next obsidian]

It only makes sense to define this in terms of A-modules

[prisma thunder]

Find a splitting field and Galois group over $\mathbb{Q}(\omega)$, where $\omega$ is a primitive $6$th root of unity, i.e., a root of $x^2 - x + 1$, for $x^6+3$ and $x^6-5$.

[cloud walrus]

eM

[prisma thunder]

If I look at $x^6+3$ first, I find the splitting field $L$ over $\mathbb{Q}$ is $L = \mathbb{Q}(i\sqrt[6]{3},\omega\sqrt[6]{3}, iw\sqrt[6]{3})$, which is both normal and separable. If $\omega \in L$, then we have the tower of extensions $\mathbb{Q} \subseteq \mathbb{Q}(\omega) \subseteq L$. From earlier, we can determine $L$ is normal and separable over $\mathbb{Q}(\omega)$, so $|\mathrm{Gal}(L/\mathbb{Q}(\omega))| = [L:\mathbb{Q}(\omega)] = [L:\mathbb{Q}]/2$.

[cloud walrus]

eM

[prisma thunder]

I am very certain $L$ is "smaller" than its current current form, which I'm trying to determine I can hopefully eyeball $\omega \in L$ and also be able to determine the degree of the splitting field to obtain information about the order of the group and decide what group it is.

[cloud walrus]

eM

[prisma thunder]

Hmm.

[cloud walrus]

det

det

[prisma thunder]

Yes

Yes it is.

[rustic crown]

oh lol, i forgot that i^6 = -1

okie what i want to say is that notice in this case that omega is actually redundant

[cloud walrus]

det

det

[rustic crown]

We should be able to use Kummer theory directly

[prisma thunder]

So the group is...

Z_3

[rustic crown]

right, but that's from order stuff

using that x^6+3 is irreducible over Q

in general if you're taking splitting field of irreducible x^n - a and your base field has zeta_n, then galois group is cyclic of order n

So for x^6 - 5, if you show this is irreducible, you'll be good for the second part

if you want to use this simple theorem for the first, we'll need to factorize the polynomial to get something irreducible, x^3 - (2omega -1) works

[prisma thunder]

Eisenstein this b*tch

[rustic crown]

that gives your irreducibility over Q, what about Q(omega)?

[prisma thunder]

Hmm.

🤔

[cloud walrus]

det

det

[rustic crown]

we can compute the top right easily because omega is not real!

[prisma thunder]

Yes

[rustic crown]

So the top left has degree 6 and we're done 😄

[rustic crown]

the proof of this is very simple if you have't seen Kummer theorem

the cooler part is that converse is also true!

if you have a cyclic extension of degree n and the base field contains zeta_n, then the extension is obtained from splitting field of irreducible x^n - a

[prisma thunder]

It's Z_6

That was the last problem

With that said, I am finally finished with this godforsaken long-a**, distasteful, yet fun HW

I like Galois Theory but it feels there's so many tools splattered together in this colorful, Jackson Pollock-esque manner, it's both ugly & abstract, yet pretty

Alright, I'm done and tired and want snacks. Till next time 😌

[trim grove]

How can i find the number of elements in $A_6$ of order 2

Any hint

[cloud walrus]

Algebra

[rustic crown]

only way to get order 2 is when its cycle type only has 2-cycles and 1-cycles

2+2+2
2+2+1+1
2+1+1+1+1

first and last are odd, so ignore them

you need to find number of permutations which have cycle type 2+2+1+1

now it's a combi exercise

[trim grove]

6!/2^2 * 2! 2!

[trim grove]

Looks like my campus wifi not allowing me to use discord 🤔

[rustic crown]

Yep!

[trim grove]

Thankyou @rustic crown

[rustic crown]

[trim grove]

I have to find the intersection of $\langle \sigma\rangle \cap \langle \tau \rangle$

Where $\sigma= (1234)$ cycle and

$\tau= (1324)$ cycle

Solution i tried : as i can see both are of order 4 so creat a cyclic group of order 4

And also group sigma will have all tau elemnets

So order of intersection will be 4?

[cloud walrus]

Algebra

[rustic crown]

enumerate all 8 elements

[trim grove]

Please elaborate little bit

[rustic crown]

why do you say tau in <sigma>?

[trim grove]

Like if i satrt taking square of tau the its generated set will have sigma

[rustic crown]

<tau> = {(1324), (12)(34), (1423), (1)(2)(3)(4)}
<sigma> = {(1234), (13)(24), (1432), (1)(2)(3)(4)}

i only see the identity common

[trim grove]

I was thinking both have same cyclic structure so they will form same group,

[rustic crown]

lol

[trim grove]

Btw thankyou for your response

[dense mulch]

is anyone around to help me get started with this problem?

"In any subgroup H of S_n, either all of the permutations in H are even, or else half of the permutations in H are even and half are odd"

I'm given that all cosets have the same size and a reminder of the sign homomorphism
sgn : S_n --> Z_2

[final pasture]

Well assume all permutations aren't even

So there's atleast one permutation sigma that's odd

Find some bijection between even permutations of H and odd permutations of H using sigma

[rustic crown]

Hello Shika

[final pasture]

Hii

[rustic crown]

also if you like playing with maps, notice that subgroup means that H --> S_n is a group homomorphism. so composing with the sign map gives you another homomorphism
H --> Z_2
kernel of this is even permutations in H. so either the kernel is all of H or like Shika said, there is one odd permutation which means that this map is surjective!! and in that case kernel is half of H.

last fact follows from the "first isomorphism theorem" H/ker =~ Z_2 so taking sizes, we get |H| = 2 * |ker|

[dense mulch]

omg you guys are so helpful, thank you!!

[final pasture]

Oh yeah this is a cute way of doing it

[rustic crown]

maps are cute

I realized, i basically reproved second isomorphism theorem lol

[cloud walrus]

det

[rustic crown]

depending of whether H contains all even permutations or at least 1 odd permutation we get HA_n = A_n or HA_n = S_n

[candid kiln]

[rustic crown]

[candid kiln]

what was an even permutation again?

[rustic crown]

many equivalent definitions, one is number of inversions is even
so number of i < j such that sigma(i) > sigma(j)

:NE_capreevee:

[candid kiln]

[candid kiln]

I should get to AA soon, this stuff seems so nice as well

[next obsidian]

I'm a chmonkey

[candid kiln]

yes u are chmonkey

[rustic crown]

how many chmokeys are there?

[robust pollen]

imagine defining the localization of a module without tensor-hom adjunction and pullback functors. Why would you... 🙈

[trim grove]

How can i find number of permutations in S5 that keep 1 to 1? Any hint

[robust pollen]

Do you know how to compute the number of elements of S5?

[trim grove]

Yes

[robust pollen]

Then explain to me how you do it. In the course of the explanation, you will hopefully also see the answer to your question (about the order of the stabilizer of 1)

[trim grove]

I think permutations matrix can help

Like fix a11 and then total ways will be 4! ,I think

[robust pollen]

Yeah, sort of. 4! is correct, but you don't need matrices

[trim grove]

Ohh , i have to fix one so just move 2,3,4,5,so total waysa 4!?

[robust pollen]

S5 is the group of all permutations of a set of 5 elements, call that [5]. So, you need to describe all bijective function from [5] to itself. To describe such a function, you need only to say where it sends each element. It can send the first element to 1,2,3,4,5, then it can send the second element to only 4 elements (namely [5] without the point where it sent the first element), the third can only possibly be sent to 3 elements, and so on. So the total number of ways is 5!
Now if you fix 1, you're basically asking for all permutations of {2,3,4,5}, so a four element set, hence 4!

[robust pollen]

yes!

In particular, if you act with S_n on [n] in the natural way, then the stabilizer in S_n of a single element is isomorphic to S_{n-1}

this is so dang nice

[thorn delta]

what book is this?

[robust pollen]

my notes I'm writing right now

[thorn delta]

ah okay epic

[regal carbon]

Hello as a physics major I'm preparing for abstract algebra but i also have linear algebra. That that gets easy if i prepare for abstract algebra?

[robust pollen]

And actually, if f : R -> S is a ring extension, the pullback functor will always have S \tensor_R - as a left adjoint.
So I now state that fact as a lemma, and then define the localization functor simply as the left adjoint to the ring extension R -> S^-1 R. I like that a lot better than e.g. Rotman does it

[regal carbon]

........

[robust pollen]

Some parts in abstract algebra will be very well possible without linear algebra, but for some parts in the abstract you will also need a good grasp of linear things. What are your topics more precisely?

[regal carbon]

Topics for linear or abstract algebra?

I haven't looked at linear algebra portion but I'm pissed off weather that's gonna be easy or hard

[robust pollen]

Both, I guess. For example, if you only do basic group theory in abstract algebra, then that would be pretty disjoint from linear algebra (assuming basic group theory = no representation theory)

[regal carbon]

So linear algebra is completely different from abstract?

[robust pollen]

no

But, okay, what are you doing in abstract?

[regal carbon]

Groups

And there are also rings and fields

[trim grove]

I want to show that <x^4+4> is not prime in Q[x], ?

Here Q[x] is pid so every prime ideal will be maximal so irreducible, i f i shomehow showed that x^4+4 is irreducible them i am done? Is this ok?

[robust pollen]

In your first sentence, you want to show "not prime", and in the second sentence you want to show "prime"?

[trim grove]

I want to say that if i showed it is reducible , soory for that.

[robust pollen]

ok. and how do you show it's reducible?

[trim grove]

Just write that e^2x - 1 in from e^x( e^x- e^-x) and try to solve

[trim grove]

This is what i am trying to do, here I can't use root method becuse polynomial is of degree 4, can't use eienstine criteria.

[waxen merlin]

i don't get that sorry

[trim grove]

After some research i come accrose sophie germain. Formula

[next obsidian]

Here’s a really, really meme way to do this

Let y = x^2 then this becomes
y^2 + 4

Factor this as
(y + 2i)(y - 2i)

So we turn into
(x^2 + 2i)(x^2 - 2i)

Write 2i as a square, so let w be a primitive 8th root of unity, like e^ipi/4

we can now factor this as

(x - sqrt(2)iw)(x + sqrt(2)iw)(x - sqrt(2)w)(x + sqrt(2)w)

Err we should be able to like

[trim grove]

Btw Thankyou @next obsidian

[next obsidian]

Turn this into (x - alpha)(x + alpha-bar)

Or something like this

Gah!

Time to do a think

Okay so

I think you do

(x - sqrt(2)iw)(x + sqrt(2)w)(x - sqrt(2)w)(x + sqrt(2)iw)

So let’s expand the first two

We get

x^2 + sqrt(2)(w - iw)x -2iw^2

Note iw^2 = i^2 = -1

So the constant is 2

And I feel like

w -iw should be 0 or some shit

:kek:

But it isn’t

[robust pollen]

wot

[next obsidian]

Maybe I’m supposed to pair the negatives up

In which case the constant becomes -2

and we have w + iw

Or no

Negative

This is e^pi/4 - e^3pi/4

So the imaginary part is the same

So this is purely real

And is equal to like uhhhh fuggin…

cos(pi/4) - cos(3pi/4)

Which is…

Sqrt(2)?

Is that correct???

Yes!

It is

[trim grove]

Hope so

[next obsidian]

So this turns into x^2 + 2x + 2

Then do something similar for the other one

It’s probably the same thing with some negatives somewhere

I think

x^2 - 2x + 2

Yes!

So this is a meme way to factor it lol

I just think it’s fun to like

Do bullshit like this

Factor it linearly in C

Rearrange

[robust pollen]

that was pretty sweet man

[next obsidian]

Get real coefficients

[trim grove]

I appreciate your efforts

[next obsidian]

[trim grove]

How can i show x^4-x^3+x^2-x+1 is irreducible in Q,

I know about cyclotomic polynomial is irreducible in Q , and this given polynomial looks like ( little bit) cyclotomic polynomial, but I don't get which approach should i use to show that this is irreducible, any hint?

[hidden haven]

Take the x ↦ -x automorphism of Q[x]

[trim grove]

How to use this?

[hidden haven]

Irreducibles map to irreducibles under ring isomorphisms

That is the idea behind proving the irreducibility of the pth cyclotomic polynomial too

x ↦ x+1 is the automorphism you use there

And then Eisenstein

[trim grove]

Thankyou @hidden haven

[woeful flint]

So I have $\mathfrak{a}_1,\cdots,\mathfrak{a}_n\triangleleft R$ where $R$ is a ring. Also, $M$ is an $R$-module. I've defined a homomorphism $\phi:M\to\frac{M}{\mathfrak{a}_1M}\times\cdots\times\frac{M}{\mathfrak{a}_nM}$ by $\phi(m) = (m + \mathfrak{a}_1M,\cdots,m + \mathfrak{a}_nM)$. I've shown that $\phi$ is surjective and now I'm trying to prove that $\text{Ker}(\phi) = (\mathfrak{a}_1\cap\cdots\cap\mathfrak{a}_n)M$. The inclusion of the RHS is trivial but I'm having trouble showing the other direction. Does anyone have any hints?

[cloud walrus]

Kraft Macaroni

[woeful flint]

I should probably mention that $\mathfrak{a}_1,\cdots,\mathfrak{a}_n$ are pairwise coprime

[cloud walrus]

Kraft Macaroni

[woeful flint]

nevermind i got it

[potent briar]

@hidden haven i think i understand it now. In R^3 If your linear functional has let's say 1 variable then the linear functional only evals to 0 on the 0 vector. if it has 2 variables then it evals to 0 on a whole plane, if it has 3 variables it evals to 0 on a line

by "having x variables" I mean like f(v) = x "has 1 variable" (with x being a component of the vector v)
f(v) = x - y "has two variables" f(v) = x - y + z "has 3 variables"

and well the only vector that makes all linear functionals go to 0 is the 0 vector

[hidden haven]

The number of linear functionals matters, not the number of variables. If you have 1 variable, then the set of things that map to 0 is not just the 0 vector, it's all vectors with 0 in the coordinate corresponding to that variable which is an (n-1) dimensional subspace @potent briar

The zero set will always be (n-1) dimensional or n dimensional

This is by the rank nullity theorem if you have seen linear maps by now

[lethal dune]

12?

[hot lake]

smells like eisenstein criterion

[lethal dune]

it does

[hot lake]

is 101 a prime number ?

[prisma thunder]

sniff sniff

[lethal dune]

it is

what about this? "Every non-commutative ring has atleast 10 elements"

[hot lake]

what is your favourite example of noncommutative ring

actually hmmm

[lethal dune]

u wanted me to say Q8?

[hot lake]

no that's a group

I wanted to see "a ring of matrices"

M2(Z/2Z) isn't quite small enough but it has a noncommutative subring

[lethal dune]

oh

IK the statement is wrong but don't have any example in mind

[robust pollen]

But now you do, right?

[lethal dune]

no

[robust pollen]

@hot lake gave you a good hint. How many elements does the ring of 2x2 matrices over F_2 have?

Do you know a subalgebra with half as many elements? Is it commutative or not?

[unique berry]

matricies

[lethal dune]

nvm got it now, didn't realise that was a hint

[runic hemlock]

fun fact: you can replace 101 with any number n, and it's still irreducible. Though the proof gets a bit harder when n isn't prime.

[prisma thunder]

Pretty sure this was a Putnam/Olympiad problem

Or some variant

[wraith obsidian]

Does anybody have a good reference which talks about monoids (and perhaps monoid actions / representations)?
I want to understand how they differ from groups; from what I've understood so far the existence of inverses is crucial because it allows for the orbits to form a partition and everything attached to it

[lethal dune]

you have proof for that?

[runic hemlock]

Yes, I know a proof.

[lethal dune]

would like to see or I won't be able to sleep tonight

[chilly ocean]

agreed

[civic tapir]

https://kconrad.math.uconn.edu/blurbs/gradnumthy/schurtheorem.pdf

El proof

[lethal dune]

[barren sierra]

In $\mathbb{Z}_7[x]$ all irreducible polynomials have degree less or equal to 2

[cloud walrus]

Spamakin🎷

[barren sierra]

it's a true or false question. Not sure what theorem or ideas to use to determine the answer

My guess is false

but I can't confidently prove either way

[lethal dune]

what about x⁶+x⁵+x⁴+x³+x²+x+1?

[chilly ocean]

1 is the root

[lethal dune]

oh we had to find one for Z⁷ and not Z

[chilly ocean]

x^3-2 is irreducible I think

[hidden haven]

Do you have access to the construction of splitting fields

[barren sierra]

no

[hidden haven]

F

[barren sierra]

idk what that is

[barren sierra]

I guess I'm more asking "hm how do I do this in general if I can't just come up with an example"

[chilly ocean]

just come up with an example

[rustic crown]

there are irreducibles of each degree

over finite fields

[next obsidian]

Z_7 is a field

[chilly ocean]

I guess maybe you can just do some mod stuff

[next obsidian]

So the claim implies any cubic has a root

Find a cubic without a root

This is my chmonkey strategy

[barren sierra]

hm ok

[hidden haven]

is there an elementary proof of det's statement

[rustic crown]

idts

[barren sierra]

or a link to said proof

hm

[hidden haven]

I can only think of take splitting field of x^p^n - x then primitive element

[rustic crown]

that basically gives you that there are fields of any size p^n

[hidden haven]

right

[chilly ocean]

what I did was just check what x^3 equals to mod 7

[rustic crown]

or equivalently what it doesn't equal by noticing (a^3)^2 = 1 mod 7

[lethal dune]

are there other methods of checking irreducibility other than roots, eisestein, reduction?

[next obsidian]

Nope those are the only ways in the world

[lethal dune]

anything practical

[chilly ocean]

wolfram alpha

[next obsidian]

I feel like determining irreducibility is like

Verifiably “hard”

As in I bet it falls into some sort of complexity class or like

[lethal dune]

hard fr integers

[next obsidian]

Insert some other meme that constructivist use

[lethal dune]

they are even 1 stage higher

[barren sierra]

yea Eisenstein and Gauss are all I know

[cursive temple]

There are some other memey criteria like perrons and cohns but they arent very useful

[lilac trench]

Can someone help? Im trying to build some geometric intuition behind normalizers, centralizers, and the center of a group

[chilly ocean]

Uhh what do you mean by geometric? I don't think that works for any group in general

[lilac trench]

like when we talk about the possible symmetries on say, a square, where any given symmetry can commute with

r^2 for instance

so that r^2 is in the set of centralizer

[shell brook]

can someone just sanity check my idea for this question. I just show the quotient is equal to {(e, h)(G x {e}) : h \in H}, and so the isomorphism is obvious?

[chilly ocean]

seems okay

you could also use the first isomorphism theorem

come up with a surjective homomorphism G x H -> H with kernel G x {e}

though you'll probably end up getting the same isomorphism as in your idea

[shell brook]

i was thinking using first iso but wasn't sure how

cuz showing quotient is what it is feels sorta janky for some reason

ill try rewriting with first iso

thx pptea

[chilly ocean]

pptea

😋

yea quotients can be weird to work with directly. isomorphism theorems are good tools for dealing with em

[barren sierra]

Struggling to show that if $G$ is abelian, $|G| = 9$, and there are no elements of order $9$, then we have that $G \simeq \mathbb{Z}_3 \times \mathbb{Z}_3$

[cloud walrus]

Spamakin🎷

[barren sierra]

I know order of elements must divide order of the group

and only the identity can have order 1, we have no elements of order 9

so we must have elements of order 3

😬 not sure where to go

My idea is that I pick elements g, h in G such that <g> intersect <h> = {e} and |<g>| = |<h>| = 3

and then every element in G is a combination of powers of g and h

and get a natural isomorphism from there

but idk how to do that

basically show that if <g> intersect <h> = {e}, <g>, <h> normal (free from def of abelian) and <g><h> = G we have isomorphism

it's that last part and existence of that first part that is tripping me up

[formal bough]

For the existence, can't you just pick some g of order 3, remove <g>, and pick some other h

[barren sierra]

Can we?

That was my thought

But wasn't sure if that was correct

Ok then I need to show <g><h> = G

Clearly ⊆ is trivial

Other direction is not

[chilly ocean]

Ig need to show g^ih^j are different for different i and j?

gh^2=/=g^2h=/=gh=/=g^2h^2

[thorn delta]

think about |<g><h>|.

[barren sierra]

Well I'm trying but idk lol

I need to show that something like gh^2 =/= e, g, g^2

For example

[thorn delta]

trying considering the map H x K --> HK
given by (h,k) mapsto hk. Compute its kernel, and compare orders and stuff

[next obsidian]

Any group of order p^2 is abelian

Cuz it either is Z/p^2Z or (Z/pZ)^2

Maybe try to prove this, the idea is if you have an element of order p^2 you’re done

If not, you have the identity and then everything else has order p

Fix a g which is order p, then consider H = <g>, this is a copy of Z/pZ

Then H is normal cuz index p, or find another way to prove it

G/H is order p so also Z/pZ

Then you can find a splitting map

Or you can manually show it, like look at k not in H

k also is order p

Then G breaks up as

H, kH,…,k^p-1H

[barren sierra]

What is splitting

[next obsidian]

A map going backwards

A umm

Left inverse?

[barren sierra]

Yea I'm trying this but idk how to show it

[next obsidian]

Well

[barren sierra]

That's exactly what I want

I think

[next obsidian]

They’re all different right?

You know cosets partition right?

So suppose k^nH = k^mH for different n,m