groups-rings-fields

[tardy hedge]

What textbook are u using?

[white oxide]

oh cool thanks

how exactly does $F[X]/\bigl(p(X)\bigl) \simeq F[\alpha]$ and $F[\alpha]$ being an integral domain imply that $p(X)$ is irreducible?

[cloud walrus]

okeyokay

[white oxide]

oh sorry to interrupt my b

[coral spindle]

Well suppose it were reducible, and try and see why that would mean F[X]/(p(X)) would not be an integral domain

[white oxide]

oh

is it just that if $p(X) = h(X)g(X)$ then $\bigl(h(X) + p(X)\bigl)\bigl(g(x) + p(X)\bigl) = 0$

[cloud walrus]

okeyokay

[next frost]

yes

[hollow imp]

my condolences

[next frost]

r u at berkeley?

[hollow imp]

ya

[next frost]

halp me

I have a midterm in 1 hour

in fucking dwinelle

all the way across campus

[wraith oak]

Anyone know what l.o.t means?

[summer path]

lower order terms

?

[wraith oak]

That makes sense, thx

[torn warren]

For this definition, it means for example, $x^{360}-1=0$ is solvable, the primitive root is $a=e^{i 1^\circ}$ (I use degree to measure the angle), but what is $\cos(1^\circ)=?$ This is not constructible.

[cloud walrus]

WT

[torn warren]

wait, is this solvable?

[white oxide]

is there any non-annoying way or do i just have to solve a system of equations to get (\alpha - 1)^{-1} in the form

because i got to a system of equations but it's inconsistent

i was able to get the first expression as a linear combination of the basis elements

[torn warren]

just multiply and equals 1, then expand and compare coefficients

[white oxide]

wdym

[torn warren]

what is wdym?

[white oxide]

what do you mean

[torn warren]

(a alpha^2+b alpha+c)(alpha-1)=1

expand and compare coefficients

and reduce alpha^3 by using this

[delicate orchid]

have you managed to do the first part okeyokay?

[torn warren]

can you help me on my questoin?🥹

[delicate orchid]

quite literally no, I hate galois theory

[torn warren]

why

[delicate orchid]

and you wrote e^i1^\circ which is an abomination

[delicate orchid]

too many words for things that are the same

[white oxide]

yeah i meant that i was able to express (a^2 + a + 1)(a^2 + a) as a linear combination of the basis elements

bro's a mod

[delicate orchid]

yur

[sonic coral]

would the order of N_G(P cap Q) being 36 be a contradiction since i havnt used any sylow 2’s yet and already have 216 elements of order 7 in a group of 252

[white oxide]

oh smart

[torn warren]

but I am confused on the x^360-1=0, is this solvable?

[delicate orchid]

magma won't compute the galois group for me so am I actually going to have to think about cyclotomic polynomials chat

is that something I'm going to have to do

[white oxide]

wait i don't know why i didn't decide to multiply both sides by alpha - 1 oop

[coral spindle]

Too much is changing in this server

can't handle it smh

Gl with your cyclotomic polynomials, wew, and NOTHING else lmao

[torn warren]

yes, it is cyclotomic polynomials

the book gives this lemma, but it depends on the cyclotomic ploy is always solvable, right?

[delicate orchid]

They’re all I need

[torn warren]

what is smh?

[coral spindle]

Google it

[delicate orchid]

I don’t know enough Galois theory to comment

[topaz solar]

Shaking my head at this question

[torn warren]

oh no

[torn warren]

thankyou..

[delicate orchid]

sharps what's the minimal polynomial of the splitting field of x^a-1

[topaz solar]

I think these are the same statement?

[topaz solar]

Y-(x^a-1)

[torn warren]

it is (x-r1)...(x-rk), where r1, ..., rk are all the primitive roots

[delicate orchid]

makes sense

[topaz solar]

You should get a cyclic thing I think?

Which I mean, finite subgroup of field so

[delicate orchid]

computer says no

[topaz solar]

Insane

Math is a hoax

Outer automorphisms are fake

[torn warren]

computer? do you ask GPT?

[delicate orchid]

apparently Gal(x^8-1) (I don't know the notation nor do I care) is C_2^2

[delicate orchid]

no I use a CAS

fuckin gpt

[topaz solar]

Yeah I’d believe that

[delicate orchid]

it's the right order at least

oh yeah the groups gonna be order phi(a) obvs

[topaz solar]

Sounds about right for flopping da circle around or wtv

[coral spindle]

The galois group of the nth cyclotomic polynomial is (Z/nZ)*

[delicate orchid]

yeah I buy this as well

[coral spindle]

So duh doy, it's Abelian

[torn warren]

x^8-1 and x^4+1 has the same splitting field, because they have the same cyclotomic polynomials

[coral spindle]

Wym don't you trust me??? >:(

[topaz solar]

Yeee cyclic-y momento

[coral spindle]

You can see this super easily if you just consider where the generator lands

[delicate orchid]

sorry I only trust BOOytjie

[coral spindle]

I AM HE, HEATHEN

[topaz solar]

Wew has gon mad with power

He can no longer yell for mods however

[coral spindle]

Some primitive root of unity must land on a power of itself, and those powers are primitive exactly when the exponent is coprime to n

A shoddy explanation I know but it is enough to recover at least that it's a quotient of the group of units

[delicate orchid]

I may hear the siren's call but I will continue to sing

[topaz solar]

You are the siren

[coral spindle]

Yeah I hear the sirens.... 👮

[delicate orchid]

anyway back to the galois theory

[torn warren]

and the elements are relative primes to n, hence includes phi(n) elements, where phi(n) are Euler function.

[topaz solar]

:theuniversehasamostmelodioushum: or something

[coral spindle]

One might even call that the definition of phi(n)

In fact... many do!

[topaz solar]

I think I just had my third eye opener by realizing Galois groups are the things permuting realizations of types

Which is like, comically obvious but I just now thought about

[torn warren]

For this example, how to show the splitting field is contained in some radical extensions? how to choose alpha1,... alpha4?

[coral spindle]

Lol

I've heard there's some kind of generalised galois theory for models

[topaz solar]

“Yes it permutes polynomial solutions”

[delicate orchid]

nobody does this

[topaz solar]

You can definitely look at like, automorphisms f fixing some set X (pointwise or setwise as appropriate) and clearly tp(a/X) = tp(fa/X) or such

But in big enough models you can get converses and such

Idk how related stuff from Galois stuff works out, or structures on those associated aut groups, but I do know of this

But uh, types = “orbits of automorphisms” holds literally for “rich enough” models, which you can oftentimes see fixed as a universe

[torn warren]

If by definition the splitting field of $x^n-1$ on Q is Q(r1, ..., rn), where r1,.., rn are roots, then apparently $r_k^n=1\in Q$ for $k=1,2,...n$, so the radical-extension-field is just Q(r1, ..., rn)? Can I say this?

[cloud walrus]

WT

[tardy hedge]

Question i needed to do was compute factor group Z6xZ4/<(3,2)> and i noticed some things and was thinking yooo is this new group gonna be cyclic? And it was!

Then apparently it means that Z6xZ4 is called metacyclic

[delicate orchid]

indeed it does

[white oxide]

i'm a little bit confused here, what exactly is C? Lang says it's a class of extension fields, but then proceeds to say things like "k \subset FE is in C" so is C a class of inclusions or smt?

or is this just imposing some relations on the class of extension fields

[karmic moat]

an extension field is a pair of fields

so k \subset F is an extension field and F \subset E is an extension field

tbh i've seen it more in the notation of like k/F

[white oxide]

oh yeah i guess you have to refer to what it's an extension of

true

well this dude calls integral domains entire so...

WHERE MY CHARACTER THEORISTS AT

how does one show that a finite abelian group of order n has exactly n characters

at first i thought about the homomorphisms determined by the generators of the group

which is euler-phi(n)

but alas

no....

[coral spindle]

Well here's one way: Classification of finite Abelian groups + what you know about characters of cyclic groups.

[glossy crag]

Show that a cyclic grou of order n has exactly n characters, show that the character group of a product is the product of character groups, and use the classification theorem for finite abelian groups.

[coral spindle]

Conversely you can do it purely by counting: the irreducible characters of an Abelian group are all of degree 1, and the sum of the squares of the degrees of the irreps is the order of the group.

[white oxide]

hm okay i'll try to show this when i have time

reason i'm asking is because maclane is using it as an example and i'm unfamiliar with it

like for example what's the difference between DG and D(G)

or are they same thing and it was just a typo

cuz i thought that the character group was the set of all homomorphisms from G to the nonzero complex numbers under multiplication

[coral spindle]

Oh you're not talking about rep theory characters

These are slightly different in the way they're presented

[white oxide]

maybe i should skip this example then LOL

[coral spindle]

Linear characters of G are in bijection with Hom(G, R/Z) because the group of units of C is isomorphic to R/Z

But n.b., only linear characters

Characters of degree 1

[white oxide]

oh yeah i guess that makes sense

[coral spindle]

Frustratingly, some texts use 'characters' to mean linear characters into some field, possibly other than C. So that may be the confusion.

[white oxide]

hmm R/Z being isomorphic to C* is kinda weird tho

[coral spindle]

Whoops

I didn't mean the group of units

I meant the group of the unit circle

mb

[white oxide]

oh yea that makes more sense then

[white oxide]

oh yea

[torn warren]

can anyone help me?

[round hull]

french-pilled

[torn warren]

why this definition works? for example, x^7-1=0, how to use this definition to show the radical expression for the roots?

for the primitive root r=exp(2pi/7), but what is the radical expression for this?

[round hull]

r^7 = 1 which is in F

[torn warren]

yes, but how this works? I mean by definition r^7=1 in Q, so Q(r) is the radical extension of Q, but how does this help to show r can be expressed into a radical form?

and what is the radical form for r

[cobalt heath]

What do you mean by radical form?

[torn warren]

r=exp(2pi/7)=cos(2pi/7)+i sin(2pi/7), how to write cos(2pi/7) into radical form?

[cobalt heath]

I forgot if that was feasible for all n or not.

But this is not obvious

Like, I think in some cases such cosine value might not be written in radical form

[torn warren]

and in general, x^n-1=0 are solvable by this definition, so how to write cos(2pi/n) into radical form

[cobalt heath]

I think there would be an assumption about root of unity.

[torn warren]

what assumption?

[cobalt heath]

Usually there is an assumption like, the base field F contains roots of unity.

Oh, you do not need that

[torn warren]

what you said is the next step, which is this lemma, and they consider two cases, the base field F contain or not contain the root of unity.

[cobalt heath]

I guess you use Galois theory to show that cyclotomic extensions can be found by taking strictly radicals.

Yes

[torn warren]

there should be an algorithm to write cos(2pi/n) into radical forms?

[cobalt heath]

I guess there would be algorithm for that, but you do not need explicit algorithms.

Just the fact that cyclotomic extension is a abelian is enough.

[torn warren]

cyclotomic poly needs to know all primitive roots, if n is fixed, then the Euler function phi(n) is known, so we just need to know the form of exp(i 2pi/n) , then everything will be done, but my question is how to find this?

[cobalt heath]

Nah, you do not need to find it explicitly

The abelian Galois group simply gives the result.

[torn warren]

why?

[cobalt heath]

Consider when a field extension is of prime degree. What would be such a field extension?

[torn warren]

wait, it is not nessesarily be prime degree

[cobalt heath]

Ugh, I mean

[torn warren]

the degree equals phi(n)

[cobalt heath]

Of course cyclotomic extension is not of prime degree

But recall what Galois theory was about.

[torn warren]

ok, then the G(K/F) is cyclic

[cobalt heath]

Well yea, it is even cyclic

Then you have chains of subgroups, where the index is prime at each point.

Since all these are abelian, they should all be normal as well.

Now what happens?

[torn warren]

why it is even?

the degree is prime

[cobalt heath]

Nah, I meant cyclic

Huh

[torn warren]

I don't get it, if the order of Galois group is prime, then why there is a chain of subgroups?

[cobalt heath]

Nah

Galois group of cyclotomic extension is cyclic. You got this, right?

It is not of prime order.

But it has chain of subgroups where the index is prime.

[torn warren]

it is isomorphic to the multiplicative group of Zn where the elements are relative primes to n

[cobalt heath]

Yea

Which is cyclic, but you really just need abelian here.

[torn warren]

yes, so each index is prime, and all subgropus of cyclic group are cyclic

since G(K/F) is abelian, all subgroups are normal

[cobalt heath]

Yep

[torn warren]

so each level's extension are normail extension to the previous level' extention field

[cobalt heath]

Indeed

[torn warren]

then what?...

[torn warren]

what can we conclude?

how to move to the radical extension

[cobalt heath]

...well that's why I told you to consider what is the extension of prime degree(index).

[torn warren]

sorry i am lost, where is the starting point?

[cobalt heath]

Huh?

[torn warren]

my question is to ask how the definition of the radical extension helps to express cos(2pi/n) into radical form?

so I am expecting some algorithm (based on the definition of radical extension) to write this into radical form, you said there is no need.

[torn warren]

so i am lost...

[cobalt heath]

Wellll

I mean

Have you followed what I said so far?

[torn warren]

but I don't get how your said is to solve this question

[cobalt heath]

It is nearly done

[torn warren]

your starting point is to consider a prime order cycl-poly

[cobalt heath]

Basically, this approach gives a way without explicitly constructing the root of unity with roots

[cobalt heath]

Nah

[torn warren]

how is related to the n-th primitive root of unity?

I don't get that

[cobalt heath]

It was just about a field extension of prime index.

Do not confuse this prime index extension with cyclotomic extension - cyclotomics are not usually of prime index at all.

[torn warren]

ok, so if F=Q, and then we consider an extension with degree p

so we forget cycl-poly

[torn warren]

hello?

[bleak abyss]

What's going on?

[torn warren]

just now i saw someone post a solution, then take it back

[bleak abyss]

I'm not asking from a mod pov

I'm asking from a math pov

So I can help

[sonic coral]

Is this line of reasoning correct? If I have a finite group G and assume it is simple, there exists a map from G to S_k where k is the index of some subgroup in G

so if G doesn’t divide k! then it is not simple

[bleak abyss]

That sounds right

[sonic coral]

but also, since i am assuming G is not simple, it has no subgroup of index 2, so i can actually map it into A_k and get a better bound, like k!/2

[bleak abyss]

Otherwise it has a kernel

[torn warren]

this is my question

[sonic coral]

this is the part i’m less sure about i guess

[bleak abyss]

It can be done but you'd have to do some work. Are you familiar with Gauss sums?

[torn warren]

no i don't..

[bleak abyss]

I guess you're saying, let k be the index of some H. You have a transitive action of G on G/H, so this gives a homomorphism G->S_k

Compose this with the sign homomorphism S_k -> \pm 1. Unless |G| = 2, in which case this problem is kinda lol, this homomorphism has to be trivial. So yeah the image of G lives in A_k

[bleak abyss]

I'll do this for zeta_5

zeta_5 solves x^5 - 1, and in fact we can divide by x-1 to get the cyclotomic polynomial x^4 + x^3 + x^2 + x + 1

[cloud walrus]

Sloth King Daminark

[torn warren]

what's zeta5? is zeta5=exp(i 2pi/5) ?

[bleak abyss]

Yea

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[torn warren]

ok, I can do the next steps

but question is , we have four roots, how do you set which one is exp(2pi/5) ?

[bleak abyss]

Geometrically

[torn warren]

and how to show the general case exp(i2pi/n)

[bleak abyss]

The calculation I did above was agnostic to which root of unity, and you'll notice that we get a \pm, so you'd need to think of which one holds in the particular case

I think doing it explicitly is gonna be hard in general, but I think these types of objects are useful for that. Look up "Gauss sums"

[torn warren]

this definition is a recursive way, so I guess there is some algorithm to build the radical expression recursively?

[bleak abyss]

So what I did for zeta_5 gives you that Q(sqrt(5)) is an intermediate extension

I think Gauss sums will be the way to go to get one step of the way, but doing it explicitly will be hard

[torn warren]

and just now @cobalt heath said we don't need the explicit form, but I don't know what he had in mind...

[bleak abyss]

A field extension is solvable in radicals iff its Galois group is a "solvable group"

So if you're just looking for a proof of the fact that cyclotomic extensions are solvable, you say well the Galois group is abelian, and thus solvable

There is some work needed to prove this, called "Kummer Theory"

[torn warren]

my question is not that far, it is just the initial definition: when they define the radical extension, then why this definition works? so I just want to see an example, like x^n-1=0, this satisfies this definition, but how can we express the root into radical form

[bleak abyss]

Yeah I'm saying, using some theory we know it can be done. But doing this example in general is not easy

This is where I said to look up Gauss sums, they're what we did for n=5 and I think are gonna be pretty important in general

But it's not a napkin proof

[torn warren]

the definition didn't say how to choose those alpha1, alpha2...

my question is how to choose them

[bleak abyss]

The definition just says "The extension is solvable if there exists such alpha_i"

Now if someone hands you a solvable extension and tells you it's solvable

I don't know of a way to compute the alpha_i by hand. Even in cyclotomic function example, it's some hard work. A common way of proving solvability of an extension is using group theory, that you'll do soon enough

[torn warren]

do you mean this?

[bleak abyss]

Yup

This gives you a way to show that such a_i exist. It might not give them to you explicitly (tbh I forget how the proof of Kummer theory goes, maybe if you work really hard and unwrap it it does? But it doesn't give you an easy way)

[torn warren]

it proves one way, but not the reverse direction

[bleak abyss]

I mean the proof can be done in general, it just isn't done by this book

[torn warren]

thank you, and do you know a good material to cover this part? friendly to beginers

Fraleigh's book is very brief on this section

[bleak abyss]

I have a pin in #book-recommendations for algebra books

Nutshell is, Artin is probably good

[grizzled crane]

Is $\mathbb{Z}_3$ and subgroup group of $\mathbb{Z}_6$?

my answer is no,

because the operation is different even though they are subsets

[cloud walrus]

.doc

[grizzled crane]

please verify my answer

[coral spindle]

Sure, if you're saying Z_3 = {0, 1, 2} and Z_6 = {0, 1, 2, 3, 4, 5} then one is certainly not a subgroup of the other.

[cobalt heath]

How did you define Z_3 and Z_6 ..

[grizzled crane]

Z3 is the the residue set under addition modulo 3

similarly Zn

[cobalt heath]

Residue set?

[grizzled crane]

remainders upon division by m

[cobalt heath]

Ah, so basically what Boyt said

[grizzled crane]

Yep

[cobalt heath]

Yea it all depends on how you identify these.

[grizzled crane]

How do we define alternatively?

I’m introductory

[cobalt heath]

By equivalence classes.

[grizzled crane]

quotient group?

[cobalt heath]

Like $\bar{1} = {1,4,7, …}$

[cloud walrus]

Absta

[grizzled crane]

This is also equivalance classes but partition of Z

[cobalt heath]

Yes your Z_3 is commonly referred to as the quotient group Z / 3Z.

[grizzled crane]

What’s the notation /3Z?

[dark elk]

3Z = multiply every integer by 3

So 3Z = multiples of 3

[grizzled crane]

subgroup of Z

yes, 3Z

[grizzled crane]

so we remove them

from Z3?

[dark elk]

Because it's a normal subgroup, you can quotient

[grizzled crane]

I haven’t reached there yet i guess

[dark elk]

Z/3Z = {3Z, 1 + 3Z, 2 + 3Z}

[grizzled crane]

I’ve pretty much only reached till permutation group

[cobalt heath]

Hm but yea, Z_3 itself is not subgroup of Z_6 without alternative identification, even in this way.

[dark elk]

Here 1 + 3Z = numbers that are 1 above a multiple of 3 and 2 + 3Z = numbers that are 2 above a multiple of 3, so we get back to residues

[chilly ocean]

Why is the field generated by F_p and some finite set of elements from the algebraic closure still finite?

[coral spindle]

An algebraic extension generated by a single element is finite-dimensional, so in this case the field remains finite. Induction solves the rest.

[chilly ocean]

I understand, thank you

[tardy hedge]

lol the section on cosets, normal subgroups and factor groups have taken me so long to understand at a level im happy with

I really liked this section though

after this we are done with groups and going to fields and polynomials

[uneven jackal]

how exactly is schur's lemma used here ? it feels like what's used here is that if L would be non-zero, it would contradict irreducibility of the reps

I mean I'm especially confused since Schur in the book is only concerned about self maps of representations

is there an easy way to prove there's no nonzero maps between inequivalent reps in the infinite dimensional case ?

[molten rivet]

I'm having trouble with a group classification problem, I'm trying to find all groups of order 174 (2*3*29). For the abelian case we clearly just have the cyclic group. For the nonabelian case, clearly there's a normal sylow 29-subgroup, and there's either 1 or 58 sylow 3-subgroups.

In the case with 1 sylow 3-subgroup, we take a product of the sylow 29 and sylow 3-subgroups to get a normal group of order 87, and then we can take a semidirect product with one of the sylow 2-subgroups to get up to 3 possible groups.

In the case with 58 sylow 3-subgroups, I got that there are either 1, 3, or 29 sylow 2-subgroups. If there is only 1, we take the product of that with the sylow 29-subgroup to get a normal group of order 58, and then semidirect that with a sylow 3-subgroup. But, since Aut(Z_58)=Z_28 we can't take a semidirect prodcut here. Now, the 3 and 29 cases are where I don't know what to do, we can't just take an easy semidirect product or anything obvious that I can see.

[tardy hedge]

It would be cool to go through linear algebra again, after having learned some abstract algebra

[dim widget]

I would do it this way: there is always a normal subgroup of order 29, so there is an exact sequence $H \to G \to X$ where $X$ is a group of order 6. If $X$ is $Z/6$ then you can show that the short exact sequence is split (think about why). If X is $S_3$ then there is a short exact sequence $K \to G \to Z/2$, which is obviously split, and you have to classify groups of order 3*29 and the actions of Z/2 on them.

[cloud walrus]

n-Coskeletal E-Girl

[dim widget]

then classifying groups of order 87 is easy

[sonic coral]

To show a group of order 252 is not simple, can we use the fact that we have 36 sylow-7 subgroups of order 7, and so 216 elements of order 7. We also have either 4,7,28 sylow-3 subgroups. We can’t have 4 by constructing a map into S_4, so we have atleasy 7 sylow-3 subgroups. But then 36(7-1)+7(9-3)> 252 which is a contradiction to the order of the group

like i feel like i assumed the best case scenario, the smallest number of sylow 3 and the biggest intersection and still had too many elements so i’m done?

[dim widget]

why cant you have 4?

your group is of size $497$

[cloud walrus]

n-Coskeletal E-Girl

[dim widget]

so there could be a nontrivial map to S_4 which acts transitively on the 3-sylows

ah but i see you are trying to prove that its not simple

okay

[dim widget]

this is good! this works

[tardy hedge]

To show if homomorphisms are injective we can just look at the kernel , but why cant we do this for some function u would study in calculus from R to R, or something like that?

Those functions cant be considered homomorphisms of R?

[lusty marlin]

Recall the definition of homomorphism

[tardy hedge]

f(ab)=f(a)f(b)

So most functions u look at in calculus dont have that property

[lusty marlin]

Right, so a function from ℝ to ℝ that doesn't satisfy the definition of a homomorphism isn't a homomorphism

Which part of this is confusing?

[tardy hedge]

Its not really

Already knew why, but was wondering if there was some more insight somewhere

Also i think maybe a bigger problem is that a set function cant even really be considered functions between groups normally. Like i suppose f: R -> R, f(x) = x^2 + 1 cant even be considered a function between groups because you are using two different operations within R for the output of f

[rocky cloak]

Part of Schurs lemma says that there is no nonzero map between inequivalent irreducible representations.

This is easy to prove, by just noting that the image and kernel of a homomorphism is a submodule.

[uneven jackal]

what if the linear map has dense image and zero kernel ?

I know the "full" version of schur's lemma for finite dimensional reps, but for infinite dimensions I don't quite see the argument

[dim widget]

the point is that the group is compact and all of the representations are unitary

so then this reduces to the proof of the finite dimensional case, if you take "image" to mean the image in topological vector spaces which should be the closure of the image

[uneven jackal]

I don't quite get what you mean sorry

I get the unitarization part

but how does it solve this problem ?

[dim widget]

unitary reps of a compact group are hilbert space sums of finite dimensional representations

perhaps you have not proved this yet?

[uneven jackal]

ohh right yes I (technically) did

[dim widget]

anyway and distinct finite dimensional irreducible reps have no intertwiners by the usual argument

(intertwiner meaning G-equivariant hom)

[uneven jackal]

right so in this case it's just that the irreps in question are finite dimensional

[dim widget]

yes basically, although schur's lemma is a very general principle

[uneven jackal]

thank you, missed that detail embarassing lol

[dim widget]

not embarassing! no worries

[rocky cloak]

I see, but aren't all irreducible representations finite dimensional anyway?

[uneven jackal]

it does make sense indeed, if G is compact, morally for a fixed vector u, pi(G)(u) should be compact and G-invariant

[delicate orchid]

this can easily be seen by realising that all compact groups are finite

[torn warren]

how to find the book? when I click it takes me here

[plain sedge]

[karmic moat]

#book-recommendations message

[torn warren]

oh, thank you! I found it.

Artin and Dummit, which one is good for the Galois theory part?

[molten rivet]

thanks for the help! that makes sense.

[white oxide]

how exactly sigma being injective imply that sigma(E') has the same dimension of E'? is it just because sigma just permutes the generators of E'

[coral spindle]

Rank-nullity

[white oxide]

oof don't remember that from lin alg

[coral spindle]

dimension of quotient space

Or you could just

say the image is isomorphic to the space

hence of the same dimension

[white oxide]

oh shit that's facts

[coral spindle]

You should burn rank-nullity into your mind

[white oxide]

i plan to go over lin alg over winter break

that class is a haze in my mind

[coral spindle]

f : V → W linear implies dim im f + dim ker f = dim V.

That's it

[white oxide]

oh

fundamental theorem of linear maps

bruh why is lang not using kronecker for showing a polynomial over a field has a root in some extension

my favorite theorem :(

[south patrol]

yeah i'd just say that

[wraith oak]

What does the symbol at the bottom mean?

[south patrol]

The field with 71 elements

[wraith oak]

Are those elements the numbers 0 to 70?

[barren sierra]

checks if 71 is prime

Yes those are the elements of F_71

[uncut girder]

71 is kind of an underrated prime

[rocky cloak]

How would you rate it?

[broken quartz]

quick question. https://sites.math.washington.edu/~jarod/math404A-spring21/hw7-solns.pdf

can someone look at problem 7.3 and tell me whether 0.1 and 0.2 are necessary or one or the other is fine

just tryna understand the proof

[topaz solar]

71

[broken quartz]

dragonslayer do u think u could tell me if 0.1 and 0.2 are needed in the proof or 0.1 or 0.2. cuz it says we will show two ways.

[rocky cloak]

They are two independent solutions

Though they reuse the part about geometric series

[broken quartz]

so either suffice to explain the statement if a|b, F+b_a inclusion F_p^b part

[rocky cloak]

Yes

[broken quartz]

cool thanks

[molten rivet]

So I've tried approaching this problem in a couple different ways, but I haven't had any luck. My idea was to use induction on m, but after I finished the base case I didn't really know how to progress. I mean for the induction step, clearly I just need to show that N_1 is equal to the commutator group, but I couldn't see how to do that.

[broken quartz]

Proof. (Existence) For $q=p^n$, consider $x^q-x$ in $\mathbb{F}_p[x]$, and let $F$ be its splitting field over $\mathbb{F}_p$. Since its derivative is $q x^{q-1}-1=-1$ in $\mathbb{F}_p[x]$, it can have no common root with $x^q-x$ and so, by Theorem 3.15, $x^q-x$ has $q$ distinct roots in $F$. Let $S=\left{a \in F: a^q-a=0\right}$. Then $S$ is a subfield of $F$ since

• $S$ contains 0 ;
• $a, b \in S$ implies (by Freshmen's Exponentiation) that $(a-b)^q=a^q-b^q=a-b$, so $a-b \in S$;
• for $a, b \in S$ and $b \neq 0$ we have $\left(a b^{-1}\right)^q=a^q b^{-q}=a b^{-1}$,so $a b^{-1} \in S$.
  On the other hand, $x^q-x$ must split in $S$ since $S$ contains all its roots, i.e its splitting field $F$ is a subfield of $S$. Thus $F=S$ and, since $S$ has $q$ elements, $F$ is a finite field with $q=p^n$ elements. (Uniqueness) Let $F$ be a finite field with $q=p^n$ elements. Then $F$ has characteristic $p$ by Theorem 6.2 , and so contains $\mathbb{F}_p$ as a subfield. So, by Lemma $6.4, F$ is a splitting field of $x^q-x$. The result now follows from the uniqueness (up to isomorphism) of splitting fields, from Theorem 5.18.

[cloud walrus]

.

[broken quartz]

does my proof answer the question properly?

[molten rivet]

wait the fact that (a-b)^q=a^q-b^q is called Freshmen's Exponentiation is hilarious to me lmao (also I can't help you unfortunately sry)

[broken quartz]

lmao

dw

[dim widget]

what have you proved so far? you did the first part?

anyway it's crucial to use that the quotients have prime order, so the abelianizations are cyclic

[broken quartz]

can someone help me prove this?

[delicate orchid]

Do you have Fermat’s little theorem? Or Lagrange’s theorem

[broken quartz]

fml

*fermat's little theorem

not f my life lmao

[ivory trail]

can you imagine why it would be an issue if a nonzero element wasn't a root of unity?

[delicate orchid]

Perhaps that approach is better rather than invoking field extensions

[broken quartz]

i mean id prefer using field extensions tbh

[broken quartz]

um lemme think

[delicate orchid]

Well if you know about field extensions surely you have lagranges theorem

[ivory trail]

consider the subset {z, z^2, z^3, ...} consisting of powers of a nonzero element z, and picture how it fits into your finite field

[broken quartz]

oh sorry i read lagranges as lebesgue lol

[delicate orchid]

Then the group of units is a finite group, apply Lagrange

[delicate orchid]

I prefer this method though, it’s lower level

[broken quartz]

im still not understbaiundg

does ths imply z is a root ofg x^n - 1

[delicate orchid]

Yes, in any finite field F_q every element is a root of x^(q-1)-1

By Lagrange

Alternatively, following rho’s method

If z wasn’t a root of unity z^k would be different for each k in N, so our field would be infinite

[broken quartz]

i think i got it actually

Let a be a nonzero element in $\mathrm{F}$. We want to show that a is a root of unity, which means that there exists a positive integer $n$ such that $a^{\wedge} n=1$.

Now, let's consider the set $S=\left{a, a^{\wedge} 2, a^{\wedge} 3, a^{\wedge} 4, \ldots\right}$. Since $F$ is a field, it follows that the set $S$ is closed under multiplication, and it contains only nonzero elements.

Because $\mathrm{F}$ is a finite field, and $S$ contains infinitely many elements $\left(a, a^{\wedge} 2, a^{\wedge} 3, \ldots\right)$, by the Pigeonhole Principle, there must be at least one pair of distinct positive integers $i$ and $j$ such that $a^{\wedge} \mathrm{i}=\mathrm{a}^{\wedge} \mathrm{j}$ for $\mathrm{i}<\mathrm{j}$.

Let's consider the element $a^{\wedge}(j-i)$. We have:
$$
a^{\wedge}(j-i)=a^{\wedge} j / a^{\wedge} i=\left(a^{\wedge} j\right) /\left(a^{\wedge} i\right)=1
$$

So, we've found a positive integer $(j-i)$ such that $a^{\wedge}(j-i)=1$. This means that $a$ is indeed $a$ root of unity, with $n=j$ - $i$ being a positive integer. Therefore, every nonzero element of the finite field $\mathrm{F}$ is a root of unity.

does this make sense

[cloud walrus]

.

[broken quartz]

why tf did i write wedge instead of exp bru

[crystal turtle]

$a^n$

[cloud walrus]

n-connected Ryx

[ivory trail]

that is so funny

[broken quartz]

yes i has a ltex sheet right next to me idk why i did that lmao

does my proof look alr? i used ur hint rho

[ivory trail]

• S is closed under multiplication basically by definition, not because F is a field
• i don't think at this level you can just state that "S does not contain 0" follows from "F is a field." you should give a better/clearer reason
• S does not contain infinitely elements. to invoke the pigeonhole principle, you can say that you have a map from the positive naturals to S sending n to a^n, but S is finite.
• i am not sure what the purpose of the second equality is (where you parenthesize your numerator and denominator)

other than that, looks good

[delicate orchid]

Holy moly that is a long proof

Let z in F non-zero, then because F is finite we have to have z^m = z^k for some m, k - which in turn implies z^(m-k) = 1

[ivory trail]

*distinct

🤓

[broken quartz]

bro wtf howd u simplify it in one line

[ivory trail]

oh yeah you don't really need to state that S doesn't contain 0

[broken quartz]

the beauty of proofs is screaming its name right here

[ivory trail]

if you have a^i = a^j then just multiply both sides by a^(-i)

[broken quartz]

do i not need to clarify anything in this

this proof is just relying on the def of finite field right

[crystal turtle]

:theheavenlytuneofawellwrittenproofbringsjoyandblissupontheearth:

[ivory trail]

do you need to clarify?

i would write "for some positive k < m" instead but then it's fine

[broken quartz]

what does z in F non-zero mean? im p sure ik what hes saying but idk

[broken quartz]

yea thats what i meant by clarifying

[ivory trail]

z is an element of F and z is not zero

[broken quartz]

alr yea thats what i was thinking

[delicate orchid]

nah you can have m < k it works as well, but yeah write this unless you want to write the extra z^n = 1 <=> z^-n = 1 step

[broken quartz]

[sonic coral]

can i get a hint to show that if G is a group of order 36 which has more than one sylow-3 that is isomorphic to Z_3 x Z_3, then G is isomorphic to A_4 x Z_3

i have gotten to the center of G being either order 3 or order 6, but i’m not sure how to contradict order 6

[cobalt heath]

Maybe I forgot but, perhaps you can quotient by Z(G) to obtain a group of order 6

Then see how it goes.

[sonic coral]

i wasn’t sure where my contradiction would be there

[white oxide]

could i get a hint for this problem please?

i'm supposing for contradiction that it's reducible

[topaz solar]

Uhh, could maybe go reducible -> divisible type route?

[white oxide]

do you mind expanding on that

let's see

[topaz solar]

I’m thinking that’s nicer?

[white oxide]

i tried some shit by dividing g by f

[topaz solar]

Well, you won’t be just outright dividing g by f, but it’ll factor to like h•j+qf or something ye?

[white oxide]

ye, as in i wrote it in accordance with the division algorithim

g = fq + r

[topaz solar]

But well I’m not sure that’s the way to go?

[white oxide]

what about using the fact that F(\alpha) intersection F(\beta) = F

using the coprime condition

[topaz solar]

Ye that

Dimension of F(a) vs dimension F(b)

[white oxide]

stole that from someone i know as a hint but need to think about that for a sec lol

[topaz solar]

Compared to F(a, b)

Which is clearly divisible by both

[white oxide]

oh write dimension F(a) dimension F(b) are coprime duh

[topaz solar]

The whole F”/F’/F thing

Ye that

[white oxide]

yeah lemme try doing a dimension argument maybe?

[topaz solar]

If it’s reducible your dimension of F(a,b)/F(a) goes down

But it’s divisible by both…

||so what’s the lcm||

[white oxide]

oh wait is this something like $[F(\alpha, \beta): F] = [F(\alpha, \beta: F(\alpha)][F(\alpha): F]$

[cloud walrus]

okeyokay

[topaz solar]

Yes

Also works if you swap a & b

[white oxide]

oh the dimension goes down because if you can factor g(x) = r(x)s(x) then beta is a zero of r(x) or s(x) right

ye that makes sense

no zero divisors

[topaz solar]

Also one of them has degree 1 at least so

[white oxide]

hUh

it's evening and my brain is not working

oh

misread

ARRRGH

i'm so close holy

so say deg f = m and deg g = n

then F(a, b)/F = F(a, b)/F(a) times F(a)/F

if g = rs with deg r < g, deg s < g and say r(b) = 0

with deg r = k

then we have dim(F(a, b)/F) = nk....

but in particular m divides this....

uhhhhhhhh

so then m must divide k

[topaz solar]

Say we have degrees x, y

g is reducible over F(a), so deg min poly b/F(a) = z < y

But xz is divisible by y

Since it’s y * deg min poly a/F(b)

@white oxide

[white oxide]

yeah i'm here

lemme think

about that

[white oxide]

here x refers to deg f right

[topaz solar]

Yes

what about ||lcm(x,y) in relation to gcd(x,y)||

[white oxide]

okay yeah that makes sense

[topaz solar]

Or just ||z < y, so xz vs xy||

So then ||lcm < xy||

[white oxide]

hm yea rn i'm trying to convince myself that y divides xz

[topaz solar]

xz = [F ab/Fb][Fb/F]

= something*y

[white oxide]

shouldn't it be Fa?

[topaz solar]

Those are the same final product

[topaz solar]

It doesn’t matter whether we stop by Fa or Fb as the intermediary

But [Fb/F] = deg g

[white oxide]

oh

ok i think i got it

yeah that makes sense

cool thanks!

[topaz solar]

The square shape <> thing

You should remember a similar thing from like, group theory stuff

[white oxide]

are you talking about lattices of fields?

[topaz solar]

Or like

Subgroups even

Not just field stuff

[white oxide]

oh yea well i meant that

yea there were some diagrams in lang

[topaz solar]

You remember mf Lagrange

[topaz solar]

Ye those too though, the Galois stuff is related since ya know

Obviously

[white oxide]

😂

[cobalt heath]

Can I ask how do I obtain lasting knowledge regarding the e.g. groups? Like the Sylow group stuffs

How do I avoid forgetting the theorems and concepts?

[white oxide]

for me, just doing as many problems as i can without looking back at the theorems/text

[cobalt heath]

Hmm, I recall doing that

Still forgot it after 5 or so years

😭

[topaz solar]

Use a lot of group actions and rederive details as necessary

[cobalt heath]

I forgot like, how groups of order p^2 q could be classified.

Group actions?

Oh actually, can I derive these facts more easily once I learn representation theory?

[molten rivet]

Ok so I'm back with the same question, turns out I missed a key part of the question so I was able to get a little farther, so it's pretty clear that m=n, now I'm struggling to see why you can't reorder the composition factors to get a different composition series, in which case this would be false.

[rotund aurora]

I think (elementary) group theory is mostly about techniques and clever constructions. So given some problem in group theory, you might want to look for (in no particular order)

• A well-behaved set Omega to act on
• A nice action on Omega, where the objects you are dealing with are characterized in terms of the stabilizers/orbits/etc of the action
• The group G that acts on Omega

Usually, you won't have clear choices for all three, and even none. But if for example you find a plausible Omega, it may be clearer what the action should be and what G should be.

Given a very general statement one should always keep in mind "general" actions, namely, conjugation and multiplication (on elements of the group or cosets...). In this case for example the difficult part will be to find Omega and G

I think by focusing on the techniques and methods, rather than on the results, it will be easier to remember group theory

[molten rivet]

I ended up figuring it out finally lol

[cobalt heath]

I see, thanks for great writeup!

[sonic coral]

did I do this right? This one along with 252 have been the hardest orders less than 300

[rotund aurora]

why do they say "internal" ?

[delicate orchid]

because G need not literally be P_1 \times P_2 \times ... P_n

just isomorphic to it

hence the internal direct product rather than the external one

it follows pretty much immediately from every Sylow being normal and the fact that if P, Q are sylow at different primes then they must intersect trivially

[rotund aurora]

Internal means that the map $P_1\times\cdots\times P_k\to G$ defined by $(p_1,\dots, p_k)\mapsto p_1\cdots p_k$ is an isomorphism?

[cloud walrus]

Croqueta

[delicate orchid]

sure, although that's not how it's usually phrased

[tardy hedge]

https://www.math3ma.com/blog/whats-a-quotient-group-really-part-1 this was good but it didnt go into why you need the subgroup to be normal

[untold basalt]

Suppose G is a group of order 385. Show that one of the 7-Sylow subgroups is a subgrup of Z(G).

[tardy hedge]

For Q24 doesnt the factor group have 5 elementsv

[untold basalt]

385=11×5×7 and I find there is only one 7-Sylow. How do I proceed to show that it is a subgroup of Z(G)?

[tardy hedge]

G has 25 elements and N has 5 elements?

25/5 = 5

[delicate orchid]

G has 20 elements

d has to be non-zero

[tardy hedge]

Oops ur right thank you becuz its invertible matrices

[dim widget]

it's a silly notation that predates thinking about groups up to isomorphism

[rocky cloak]

Since C7 is a normal subgroup, conjugation defines a group homomorphism G -> Aut(C7). What can such a map be?

[mighty spade]

I've been stuck on this problem embarassingly long:

If $I$ and $S$ are a disjoint ideal and submonoid of a commutative ring $A$, then for all $r\in A$, either

[(I+Ar) \cap S = \varnothing ]
or
[\langle S,r \rangle \cap I = \varnothing ]

[cloud walrus]

sergeEmbedding

[mighty spade]

this is equivalent to krull's separation lemma, which i am trying to prove

[mighty spade]

I PROVED IT NOW???
I've been at this for days, and NOW???
as soon as I ask for help?

it's like 4 lines, too

krull's separation lemma says
if M is maximal in the set of ideals disjoint with a submonoid S, then M is prime

and if x,y not in M,
X=M+xA
Y=M+yA both properly contain M.

since all ideals disjoint with S are contained in M,
X and Y are not disjoint with S.
that allows us to find
s1 = (m+xa)
s2 = (m+ya)

and thus
s1s2=s3= (m+xya)
and therefore
M+xyA is not disjoint with S.
this means M+xyA is not equal to M

therefore, no element of xyA is in M.
therefore, xy is not in M
thus, M is prime

i can't believe i came up with this in like 5 min after failing to solve this problem for days

[untold basalt]

@rocky cloak for all g in G I have g^-1 C7g=C7

[rocky cloak]

Sure, that's just saying it's normal

[untold basalt]

yeah, I don't get what your hint says@rocky cloak

[rocky cloak]

So are you aware that x -> g^-1 x g defines an automorphism from C7 -> C7?

[untold basalt]

yes

[rocky cloak]

And do you know what the automorphism of C7 are?

[untold basalt]

no but I see where you are going, let me figure them out

thanks

[rocky cloak]

np

[untold basalt]

Any $\varphi \in Aut(C_n)$ has the form $\varphi(x)=x^k$ with k coprime with n

[cloud walrus]

Seagull

[rocky cloak]

Indeed, so then Aut(C7) is a group with 6 elements

Now, how many possible homomorphisms can exist from G to a group with 6 elements?

[winter shore]

If I have $aHa^{-1} \subseteq H$, does it follow that $aHa^{-1} = H$? When the assumption holds for all $a$, then it's true, but what if we fix one $a$?

[cloud walrus]

Shiranai

[winter shore]

I'd think so but haven't been able to show it

[rocky cloak]

No, it's not true in general

[winter shore]

do you have an example? I think you're right now that I think about it

[lusty marlin]

If H is finite, then the statement is true. If not, it need not be true.

[rocky cloak]

For a counterexample consider G^Z for some group G. Then Z acts on G^Z by shifting the index.

Consider the semidirect product of G^Z and Z with respect to this action. And let H be the subgroup G^N. Then conjugation by a the generator of Z maps H to G^N+1

(N meaning natural numbers and N+1 meaning numbers that are n+1 for a natural number)

[wraith oak]

I'm not too sure how to tackle this problem. Any advice?

[rocky cloak]

Have you tried checking some examples?

[wraith oak]

Actually no

I'll give that a go

[untold basalt]

Oh I got it, since the order of the image of an element divides the order of the element there is only one such homomorphism and it maps g to the identity automorphism.

Thanks

[white oxide]

why exactly will the polynomials g involve a finite number of variables?

[rocky cloak]

That's just the definition of a polynomial. A polynomial is a linear combination of monomials, thus a finite sum of things that only involve finitely many variables

[white oxide]

oh yeah that makes sense

thanks

here what would the root of fsigma be? would it just be sigma(x)?

[white oxide]

any gs?

[topaz solar]

What’s f sigma

[white oxide]

$\sigma f$

[cloud walrus]

okeyokay

[white oxide]

with sigma applied to the coefficients

[topaz solar]

So you see that sigma k is basically just k since field ye?

[white oxide]

yea

[topaz solar]

But im presuming S is your variables made up by basically adding every possible root?

Or rather, to each polynomial you give a root

Namely: X_f

And sigma f(X_f) = 0

[white oxide]

oh i thought lang constructed S to be associating each polynomial f in k[x] with a letter X_f

does he mean that they're the roots or smt

[topaz solar]

what

Why do you think m contains all f(X_f)

[white oxide]

wdym what

[topaz solar]

They’re symbols added to ensure everything has at least one root

[white oxide]

how do you know that tho

[topaz solar]

Because f(X_f) -> 0

It’s a root of f

In the field k[S]/m

Or, sigma(X_f) a root of f^\sigma specifically

By definition

[mighty spade]

https://archive.seattletimes.com/archive/?date=19990411&slug=2954422

this old article on pokemon is interesting

"Are Your Children Nuts About Little Cartoon Creatures From Japan?
Don't Know Why?

The Diagnosis Is Simple: -- Pokemon Fever"

wait wrong channel

[topaz solar]

If you’d prefer the logic version, you accept that every specific polynomial f can have an extension where it has a root, and you can add finitely many at a time, we can invoke compactness to get a field to get a root for everything at once basically

Might not wanna take maximals since choice, instead only prime ideals, but you can still weasel basically the same construction into a field and by basically the same way get that it’s alg closed

[white oxide]

cool thanks i went to the beginning of the section and am rereading and verifying everything, but will look once i get there

you a g fr

so some people include the condition that ring homomorphisms send 1 to 1 right

but it's not necessary

i'm confused

[rocky cloak]

It's necessary in that you get something different if you don't require it

Typically people who consider rings to be defined to have 1, also require homomorphisms to map 1 to 1

[coral spindle]

Let me put it this way

If you don't consider homomorphisms of unital rings to map 1 to 1, you are the devil and deserve everlasting punishment

Let me put it another way

In order to form a variety of algebraic structures in the universal algebra sense, rings must include a unit, and they must be preserved for everything to be right in the world.

I phrased that badly. What I mean to say is that if you want your rings to be unital, you need to include it as part of the data for a ring.

If you don't know any universal algebra, dw about it.

[white oxide]

i see

i think that makes sense

thanks

i'm struggling to see what would the multiplicative inverse for an element S in E be

oh

would it just be itself?

wait

no

if $x \in S$ then $x^{-1} = \sigma^{-1}(\sigma^{-1}(x))$

can't use any homomorphism properties since x is not in k sadly

nvm

i've confused myself

ohhh

it's going to be $\sigma^{-1}\bigl((\sigma(x))^{-1})\bigl)$

[cloud walrus]

okeyokay

[white oxide]

where $(\sigma(x))^{-1}$ denotes the inverse of $\sigma(x)$ in $F$

[cloud walrus]

okeyokay

[topaz solar]

Model theoretically: put the 1 in as a constant, not an existential, because those aren’t preserved under substructures

eg Fields

[coral spindle]

[topaz solar]

Yes, since multiply that with x

Mfw groups being simple is invisible to the theory

[rotund aurora]

I do not understand why rings without 1 are bullied, given that they arise naturally

[topaz solar]

\mu(0) of an ultrapower of the reals

Which is also graded-ish, but certainly not N graded

Should be like something like Z^\mu/Z additively

[rotund aurora]

What

[topaz solar]

Or uh the positive side anyway and throwing out 0

Since nothing stays in the same slot

It’d just be like, N^\mu/N graded and unital if you did |x| <= 1 though

[rotund aurora]

I missed some Dragonslayer ultraproduct classes, so I'm lost

[topaz solar]

$\mu(0)$ being those elements $x$ of $\bR^\mu$ so that $|x|<\frac 1 n$ for every natural $n$.

[cloud walrus]

Dragonslayer Sharp

[rotund aurora]

Is mu just some infinite set?

[topaz solar]

No it’s exactly the things satisfying those inequalities

It’s also an ideal of the finite elements

(And a prime for obvious reasons)

[white oxide]

can someone smart, sexy, and strong explain to me why $p(\alpha) = 0$

[cloud walrus]

okeyokay

[topaz solar]

sigma is basically defined to be an isomorphism yes?

So sigma(x) = 0 iff x = 0

Algebraically manipulate and expand through p(a) since you know sigma(p(a)) is 0

Because it’s basically defined that way

[white oxide]

oh wait

i just take sigma inverse of sigmap = 0

[topaz solar]

Yeah basically

Or, ya know, derive that sigma p(a) = p^sigma (xi)

= 0

[wraith oak]

What would make two polynomials over a field F not associates of each other?

Does it have something to do with irreducibility?

[rotund aurora]

Aren't the units just F?

I think f and g are associate iff f=ug for some unit u, and the definition applies for any ring

[wraith oak]

What do you mean by that

[topaz solar]

The things you can take inverses of are the field your coefficients are from

That’s it

[wraith oak]

I'm confused but don't know why

[topaz solar]

When are elements associate

[white oxide]

wait so here sigma is just the canonical projjection right

[topaz solar]

It’s what’s written as sigma there

Mod m

[white oxide]

oh so yes

👍

[wraith oak]

Say f and g are polynomials over a field F, then f and g are associates when there exists some a in F such that f = a*g

I'm not sure if a is an element of F or F[x]?

[topaz solar]

F because it says a in F

It’s that (f) = (g)

[wraith oak]

Okay, say x^2+1 and x are two polynomials over R. Then these are not associates because there's no real number a such that x^2+1 = ax. Does that work?

[topaz solar]

Yes

[wraith oak]

So does this counterexample work for the problem I'm solving?

[topaz solar]

I think so

[tardy hedge]

I got Jim pr

80lbs dumbbells seated overhead press for 8reps

[topaz solar]

Are those subgroups?

They do need to have the identity element

[peak rain]

Recently had a sheet question to prove that $f(x,y)=y^2-x^3-x^2$ is irreducible in $\mathbb K[x,y]$.

I did this by following a suggestion from someone on another server and think I used a generalisation of the Eisenstein criterion where the requirement for $p$ to be a prime integer is replaced by the requirement for $p$ to itself be an irreducible polynomial in $\mathbb K[x,y]$ Is this generalisation a valid deduction and how would one prove it in general ?

[cloud walrus]

Ama Dablam

[rocky cloak]

Yes, Eisensteins criterion works over any integral domain.

When R is an integral domain, P is a prime ideal in R and f(x) is a polynomial in R[x] such that the leading coefficient is not in P, the other coefficients are in P and the constant coefficient is not in P^2 then f is irreducible

The proof is exactly the same as for when R is the integers

[peak rain]

Thank you very much 😌

[round hull]

what's a nice noetherian ring where the lengths of ideal chains are unbounded

these probably exist right

[coral spindle]

Um... Z?

There is no finite bound on the length of ideal chains in Z

I think that's the nicest possible example

[round hull]

sorry

i meant prime ideal chains

oh ok no wonder lmao

[round hull]

answered

[sonic coral]

can i get a hint to show that if G is a group of order 36 which has more than one sylow-3 that is isomorphic to Z_3 x Z_3, then G is isomorphic to A_4 x Z_3

i don’t know how to show A_4 is normal in G so i can apply the direct product theorem with Z_3 that was contained in the center

[rocky cloak]

There is a famous example due to Nagata. Though I'm not sure I would call it 'nice'.

[teal vessel]

so I'm working on showing that D_8 ≅ <(1 2), (1 3)(2 4)> ≤ S_4 and I've worked out that if I can demonstrate a proper mapping between the generators of this subgroup and some of the elements in D_8, as long as I have at least one element that can be represented as a product of s and a power of r, I should be able to generate all of D_8. I've been calling these generators x and y, respectively, to save some time, but somehow I've managed to get that x(xy)^2=yx, which implies that yxy=yx which seems contradictory. dunno where I went wrong lol.

[teal vessel]

there we go: A=<(1 2), (1 3)(2 4)> ≅ <s, sr> = <s, r | s^2=r^4=1, rsr=s>, therefore A is isomorphic to D_8

I'm sure it would also work by symmetry with <s, sr^3>, and probably also with the other 6 pairs of order 2 elements x,y s.t. |xy|=4, but I like my objects smol for arbitrary standards of aesthetic beauty.

[teal vessel]

so thoughts: I'm showing that S_4 is generated by two distinct elements of order 4. Since their cyclic subgroups only share the identity, and their order is 4, that means that the subgroup they generate together is at least of order 8. Multiplied together they create an element of order 3, which implies that the order of their generated subgroup is a multiple of 12 (4x3=12) and at least of order 8. This leaves me with the subgroup being of orders 12 or 24. I'd like to be able to avoid multiplying out all combinations, because that's a lot of multiplying to do. can't I leverage the intersections of these subgroups to avoid actually doing all the multiplication? It seems to me that the intersection of two cyclic groups is trivial (contains only identity) iff the generator of one is not generatable by the other.

[delicate orchid]

if the subgroup they generate is order 12 then it has to be A_4, but 4-cycles are not even permutations and thus cannot be in A_4, so there are some elements outside of A_4 in the group they generate, hence it has to be the entire group

[teal vessel]

something here seems non-obvious. I'm not recalling what A_4 is, which might be part of the issue

[delicate orchid]

the only subgroup of S_4 of index 2 is A_4 maybe

[teal vessel]

index is not something I'm familiar with at the moment.

I've been chatting in this channel for several months now but autodidact timetables are quite a bit slower than classical academia 😅

[delicate orchid]

ok then, A_4 is the only subgroup of size 12

you definitely know what an index of a subgroup is if you're writing out presentations of D_8

[teal vessel]

how to put this.... I probably have understanding of the concept, but I have not as of yet been introduced to the formal notion. The presentations of D_8 above have been composed intuitively based on patterns I'm noticing in these groups which I can recognize but can't yet explain precisely and efficiently. I know that the presentations above work because one is the definition I was given originally for dihedral groups, and the other can produce both s and r in isolation, which therefore means they must have closure as part of the group generated by those generators (which entails that D_8 would be a subgroup of the group generated by that presentation, subgroup in both directions implies equality, etc).

[delicate orchid]

index of H in G is |G|/|H|

the number of cosets

[teal vessel]

by context, I assume a coset is a companion subset to a group? ain't got to them yet lol

[delicate orchid]

... ok what the fuck is your course

insane syllabus

in that case I do legitimately think you are just expected to multiply everything out

do you have lagrange at least so you know that when you reach 13 elements in the group you can stop

[teal vessel]

my course is "read through every section of Dummit and Foote in order"

yes I have LaGrange

[delicate orchid]

ok good

[teal vessel]

this is why I ask these kinds of questions, because I'm teaching myself without a prof, so I'm double checking that things I see that I think might be patterns in these groups and exercises are actually patterns, and not just a quirk of the particular examples that the text is choosing to use.

[delicate orchid]

if you're asking about your presentation of D_8, it is correct yes

[teal vessel]

I'm being a bit more general about it. I guess I'll come back once I find 13 unique elements in this group lol

good thing is I went back to look at symmetric groups and was reminded that the inverse of a permutation is just writing its permutation backwards (then shifting 1 to be in front), so that minimizes work ig

[delicate orchid]

you should already have 9 so not much to do

[teal vessel]

true

there we go, 13 found, thus the order must be 24 and the only subgroup of order 24 in S_4 is S_4 itself.

I love when they spam notation at me I learned like 3 months ago and have to look up lol

[tardy hedge]

@teal vessel you are self-studying abstract algebra?

[teal vessel]

yes.

what else would I be doing with my life?

[tardy hedge]

what is your procedure for doing so? my course rn goes up to some ring and field theory but stops before galois theory

i want to self study further afterwards

[tardy hedge]

facts dude

im wondering what your procedure for self study is

[teal vessel]

well, my procedure so far has been roughly as follows:

1. be curious about something
2. get recommendations for a textbook
3. read said textbook, performing all available exercises (or at least most of them) until you reach a concept that you realize you should have learned on its own earlier.
4. read what you should have previously learned.

etc. I have read through and studied the foundations of linear algebra, axiomatic set theory, and some of multivar calc this way (got bored, I'll come back to it eventually). Now I'm going through abstract algebra. It's really just always having a problem cooking on the backburner while life happens around you. This problem here will probably take me the weekend because I don't get a lot of time working only on my math

[tardy hedge]

yeah

[teal vessel]

so I let it stew until something clicks, then finish the problem.

[tardy hedge]

working problems truly is where the learning happens, thats true

[delicate orchid]

this is a good method

but these "generator" exercises are really tedious, and you clearly know what you're doing for the most part

so don't be afraid to skip some of them

[teal vessel]

They're varied enough to focus more on trying to exploit the particular group I'm working with. I also am only doing 1 or 2 at a time, so it's good for spaced repetition. Otherwise I might end up with this the same as my linear algebra memory: really spotty except for real basic stuff like loving systems.

[teal vessel]

Teach others. I teach grade school math for my job, and I regularly use results from my learning as part of my teaching, it really solidifies the knowledge. The proof of √2 outside Q is a favorite of mine.

[tardy hedge]

Oh yea? Thats cool

What grades do you teach

[teal vessel]

everything

k-12+

[rocky cloak]

Who falls into the '+'-category?

[crystal turtle]

Everyone

[teal vessel]

I get the occasional college algebra or calc student.

I'll teach you my opinion on string theory and type theory, too, but I wouldn't take those opinions seriously, they're outside my expertise

[teal vessel]

the additive inverses of all those falling in the '-' category

[wraith oak]

How do you figure out whether x^3+x+1 is irreducible over Q? I was able to show whether it was irreducible over R, F_1, F_2, F_3 and F_5 but Q is kinda weird

[south patrol]

it's irreducible over Q iff it's irreducible over Z by Gauss's lemma

and being irreducible over F_p for any p implies irreducible over Z

:)

[south patrol]

Try filling in the details of this

[quiet heath]

Hello
I have a theorem to prove which says if (G,*) is a group and $a \in G$ then:
$a^m * a^n = a^{m+n} \forall m,n \in G$.
To prove this I need to prove 5 cases

1. Where m=n=0 then m+n=0
2. Where m,n are greater than 0
3. Where m,n are less than 0
4. Where m is less than 0 and n is greater than 0 such that m+n is greater than 0
5. Where m is less than 0 and n is greater than 0 such that m+n is less than 0

So I proved the first 4 cases but I stuck with fifth one (actually I did prove it but I'm not sure if what I did is right or wrong).

[cloud walrus]

Alireza

[rocky cloak]

A cubic is irreducible iff it doesn't have a root. And you can use the rational root theorem to show that the only possible roots are 1 and -1, neither of which are roots

[teal vessel]

RRT is beautiful

[south patrol]

and actually you can combine our methods

go to mod 3, say, then check no roots to show irreducible mod 3

then use what I said to pass to Q

[wraith oak]

kk

thx!

[winter shore]

Is "normalizes" a symmetric relationship when restricted to subgroups? Pretty sure not but I don't want to spend an half an hour testing by hand for counterexamples

[rocky cloak]

Take one group to be not normal, and the other to be the entire group

(assuming H normalizes K means that hK = Kh for all h in H)

[winter shore]

oh nice! yes that works

[teal vessel]

question: what field is F_3?

the integers mod 3 since 3 is prime?

[white oxide]

yup

[teal vessel]

cool, that helps a lot

these three problems seem related. The behavior of these matrices seems suspiciously similar to those permutations I was working on.

[torn warren]

I don't understand this example, why beta^3 is in Q?

what is the relation between beta and alpha?

[rocky cloak]

They're saying F is not a simple radical extension. I.e. F is not equal to Q(beta) where beta^3 is in Q

Simple radical meaning you can get F by adjoing a single nth root

[torn warren]

but why they choose beta^3 is in Q, maybe beta^7 is in Q?

[rocky cloak]

F has degree 3

[summer path]

Your polynomial is of degree 3

If jagr sends answer right before I do, I can't get gaslit!

[torn warren]

but the definition didn't say those n1, n2 must be the order of the extension field?

it only says exist n1, n2

[rocky cloak]

The question then is what can the degree of S(a) be in those cases

[torn warren]

is this a typo? it should be $a^n\in S$

[cloud walrus]

WT

[rocky cloak]

Yes should be in S

[torn warren]

let's go back to the initial symbols, if F is a simple radical extension of Q, then exist some beta, such that, F=Q(beta), which means exist some n, such that beta^n is in Q

so far is good, right?

[rocky cloak]

Yes

[torn warren]

because [F:Q]=3, and alpha is root for x^3-3x+1=0, which means beta can be written as the combination of alpha^2, alpha, and 1

but I can't deduce beta^3 is in Q

[rocky cloak]

So you know Q(beta) has degree 3, so the minimal polynomial of beta must have degree 3

[torn warren]

yes

[rocky cloak]

So consider the minimal n for which beta^n is in Q

[torn warren]

but how to show the minimal n=3?

[rocky cloak]

Just assume n is not 3, then show that the degree of the minimal polynomial wont be 3

[torn warren]

so if I assume the minimal n such that beta^n is in Q, then this n will be the degree of F over Q, since it was shown the degree is 3, so n must be 3, right?

[rocky cloak]

There is some complications since x^n - beta^n doesn't have to be irreducible.

[torn warren]

so we can only get $n\ge 3$

[cloud walrus]

WT

[rocky cloak]

You can also freely assume prime, since if n/m = p, then beta^m would generate the same extension with (beta^m)^p in Q

[torn warren]

I am confused, so can we simply say it is due to F is of degree 3?

or it can be greater than 3?

like 7?

[rocky cloak]

Well, there are some details when you really spell it out as you can tell

But a root of x^p - a will either have degree p or p-1 depending on where a has a pth root or not

[torn warren]

then it says $\beta \omega$ is also in $F=Q(\beta)$, I know both $\beta$ and $\beta \omega$ are the roots of the same ploy, say $f(x)=x^n-b=0$, but it didn't say F is the split field of f(x), how can we tell $\beta \omega$ is also in F?

[cloud walrus]

WT

[rocky cloak]

F is Galois, thus normal

[torn warren]

oh, right! it mentions in the begining, so it got omega is in F, but the minimal irr-poly for omega is of degree 2, but Q(omega) doesn't divide F, hence it is a contradiction, right?

Can this example be a part of some theorem? it seems like "if the Galois group is not the full symmetric group, then it must not be a simple Radical extension?"

[rocky cloak]

The full symmetric group is definitely too strong. But the splitting field of x^p - a should contain the pth roots of unity. So the Galois group will either be (Z/p)^* or the semidirect product of Z/p with (Z/p)^*

[torn warren]

why it is either? there are p-1 automorphisms, hence the Galois groups is ismorphic to the multiplicative group Zp^* only?

[rocky cloak]

So the splitting field of x^p - a is F(a^1/p, w) where w is a primitive pth root of unity.

Then either F(a^1/p) = F(w) or they are different. If they're equal, then the Galois group is (Z/p)^*.

If they are different then x^p - a, must be irreducible and F(w) is a normal extension with galois group (Z/p)^*. So the Galois group is an extension

Z/p -> G -> (Z/p)^*

And if you just write out the Galois automorphisms explicitly you see that this is a semidirect product

[torn warren]

a^1/p stands for the real pth root?

[rocky cloak]

Just any root really

[torn warren]

could you give an example for F(a^1/p) = F(w) ?

[rocky cloak]

Well it's equivalent to x^p - a having a root. So will be something like a^1/p = rw for some rational number r

Easiest example being a=1 I guess

[torn warren]

x^3-1 ?

[rocky cloak]

Sure

[torn warren]

and for the case #2, the example is x^3-2, hence the intermediate field is F(w), where w^p=1, there are p-1 automorphism on F(w), and the splitting field K=F(a^1/p, w) has degree [K : F(w) ]= p , so [K: F]=p(p-1) , right?

[rocky cloak]

Yes

[torn warren]

can we generalize it into composite number n? then the case #1 will be the multiplicative group with order $\phi(n)$, and the case #2 will be the order $\phi(n)(\phi(n)+1)$ ?

[cloud walrus]

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

[torn warren]

This is incorrect in general? For example, if $w^3=1$, and let $F=Q(w)$, so $F$ is a radical extension of $Q$, but $[F:Q]=2$, since $w^2+w+1=0$

hence for the simple radical extension $F=Q(a)$, if $n$ is the minimal integer such that $a^n\in Q$, then we can only guarantee $n\ge [F: Q]$

[cloud walrus]

WT

[rocky cloak]

So F(w) will have degree phi(n) and (Z/n)^* will be a quotient of the Galois group, but the kernel might depend on a I think.

Like the Galois group of x^6 - 4 and x^6 - 2 should be different I think

[cloud walrus]

WT

[rocky cloak]

Because x^6 - 2 is irreducible, while x^6 - 4 = (x^3 - 2)(x^3 + 2)

[torn warren]

thank you, I will work on this example!

[chilly ocean]

Hi, a serious question ... is Lam: A First Course in Noncommutative Rings actually meant for a one-semester course on noncommutative algebra? It seems like an insane amount of content for just one semester???

[rotund aurora]

what is the multiplicative structure of the algebraic closure of F_p

[cloud walrus]

jayz

[rocky cloak]

Should be

{a/b in Q | b relatively prime to p} / Z

I think. Because Fp-bar is the direct limit of all finite subextensions, whose multiplicative group is cyclic. So it's the direct limit of Z/(p^n - 1). And for any b relatively prime to p, there is an n such that b divides p^n - 1.

[rotund aurora]

ye, so basically, all complex (p^n-1)-roots of unity (taking all possible n), right

but ig that's a nicer characterization

[cedar vault]

Are there any theorems similar to warning's or chevalley's theorem but the degree of your polynomial is greater than or equal to the number of variables?

I was asked to post this here as well

[cedar vault]

Or any theorems that make a statement about the roots when you have lesser variables than degree, not looking for an exact analogue here.

[glossy crag]

That's clearly wrong (1/x^4+1 is definitely not equal to 2/x^2+x+2), the numerators of the fractions should be written linearly (you don't automatically know they're constant), i.e. Ax+B and Cx+D (proceed then as you did before).

[white oxide]

if $L$ is an extension of a field $k$, $L$ being algebraically closed doesn't imply that $L/k$ is algebraic right?

[cloud walrus]

okeyokay

[glossy crag]

Not at all, take C/Q for example (an algebraic extension of Q is countable, but C is uncountable).

[rocky cloak]

If L/k is algebraic and L is algebraically closed, then that is exactly saying that L is the algebraic closure of k.

If L is bigger than the algebraic closure then the extension won't be algebraic.

[white oxide]

i see

thanks

why does f have a root alpha in E?

[glossy crag]

E is algebraically closed and f can be considered as a polynomial over E (since k<=E), so there's a root of f in E, which then necessarily belongs to k^a, since this element is algebraic.

[white oxide]

ohhh right

makes sense

thanks

and is alpha in k^a because it's algebraic over k^a, which implies that it's algebraic over k and by construction it must be in k^a?

[dim widget]

yes

even better: your group maps to the group in question in a natural way, by reducing mod p

[untold basalt]

The only group of order 6 is the cyclic one right? If G has order 6 I can choose g in G and either <g>=G, and we are done, or wlog we can suppose <g> has order 2, but this means I can express G as a direct product <g><h> for some h in G of order 3.

[crystal turtle]

No

That does not mean that G is a direct product

Consider S_3

[untold basalt]

oh I have to chek they are normal

shiit that's true

lol

brain fart

[cloud walrus]

Seagull

[untold basalt]

Is this ok?

[white oxide]

by $(\sigma\beta_1, \dots, \sigma\beta_n)$ differing by a permutation from $(\alpha_1, \dots, \alpha_n)$, do they mean that for a suitable permutation $\tau$, $\alpha_i = \sigma\beta_{\tau(i)}$ for $i = 1, \dots, n$?

[cloud walrus]

okeyokay

[rocky cloak]

Yes, that's what that means

[crude pivot]

Can I get some help with the second part of b ii please :))

[torn warren]

here, "the field" refers to F or F(a^1/n) ?

I think it refers to F(a^1/n) ?

[white oxide]

if $[E: F] = 2$ and $p(x) \in F[x]$ such that $p(\alpha) = 0$ for some $\alpha \in E$, must we have $\text{deg } p \leq 2$?

[cloud walrus]

okeyokay

[torn warren]

false

p=x^3

[white oxide]

ah facts

if $[E: F] = 2$ can we assert that $E = F(\alpha)$ where $\alpha$ is a zero of a irreducibe polynomial $p(x) \in F[x]$ of degree 2?

[cloud walrus]

okeyokay

[white oxide]

oh yes

every finite extension is finitely generated

[torn warren]

If E is a separable extension, then E:F=2 means it is finite, hence it is simple extension, but if E is not a separable extension, maybe this is false.

[white oxide]

hmm yeah i'm just trying to show that if [E: F] = 2 then it's a normal extension

i kinda forgot what separable means tho

[torn warren]

to show it is normal, you need to show it is separable and splitting field

[white oxide]

well i'm going off of Lang's definition, he defines separable extensions in the next chapter

i do remember that's how fraleigh defined it

[torn warren]

how does he define normal?

[white oxide]

any extension field satisfying these conditions is normal

which is equivalent to fraleigh's definition

at least i hope

anyways, i'm trying to show Nor 1 - if the assertion that [E: F] = 2 => E = F(a) for some a in E, then any automorphism is determined by where a is sent, but i mean there's only two choices, a and -a right so it has to be a automorphism

idk

[torn warren]

all these are equivalent for the definition of splitting field, it didn't say anything about separable or not

[white oxide]

yes, but what i'm saying is that his and fraleigh's definition are most likely equivalent if they are to follow mathematical convention

[torn warren]

where does he say this is the definition for "normal"?

[white oxide]

on the next page

oh is E normal literally because if E = F(a) then E is the splitting field for irr(a, F)

[torn warren]

can you send the photo for that page?

If normal = splitting , then why waste one word?

I don't believe it

[torn warren]

is what?

[white oxide]

bro's skeptical

[torn warren]

what does he mean "distinguished"?

[rocky cloak]

You don't have to be seperable to be normal.

[chilly radish]

Not closed under towers most likely

Unlike separability, being algebraic etc

[rocky cloak]

If alpha is an element of E, then the degree of its minimal polynomial divides the degree of E. So if [E:F] is prime, then E = F(alpha) for any alpha in E \ F

[torn warren]

this book defines normal=spliting+separable

[rocky cloak]

That is quite unusual I think. Usually you call an extension that is normal and seperable a Galois extension

[rocky cloak]

These characterizations are equivalent to the standard definition (no seperability involved)

[torn warren]

this book avoids the use of Galois extension, this word didn't appear in the book