groups-rings-fields

[left estuary]

Understand by doing problems

Ok first I’ve read the definition of nil potent elements but still am confused on what it means

Says for x^n=0 where x is a unit of R right?

[chilly ocean]

no

an element x of R is nilpotent if x^k = 0 for some positive integer k

a nilpotent element cannot be a unit

[left estuary]

Ok

[chilly ocean]

unless 1 = 0, but that's not an interesting ring in the first place

[left estuary]

Can you help me define unit in a way I can understand

[chilly ocean]

elements with inverses

[left estuary]

cause right now the way I see it is a unit is the same as an element

[chilly ocean]

a unit is an element with an inverse

a unit in a ring R is an element x of R such that xy = 1 for some element y of R

well, xy = yx = 1, but we're dealing with commutative rings so it's okay for me to just write xy = 1

[left estuary]

Right that was part of group theory

[chilly ocean]

it might resemble something you've seen in group theory, but there is no group theory involved in the mere definition of a unit in a ring

[left estuary]

So I get that a unit is an element but where is the inverse located

In the set?

[chilly ocean]

"for some element y of R"

everything stays in the ring R

[delicate orchid]

ok dude you very clearly don't have the pre-reqs for this, don't mean to be rude

it would be way more productive to go back and read way simpler stuff

[chilly ocean]

i'm trying lol

[left estuary]

No no it’s working I understand

[left estuary]

But it’s working?

[coral shale]

If you read the introduction to proofs first

It would speed up progress a lot

You're literally going backwards

You dont realise it

[left estuary]

But I just figured out what a unit is and nilpotent element

[chilly ocean]

but can you comfortably prove basic facts about them?

[coral shale]

You didnt know what set builder notation is. Thats a problem.

[left estuary]

That’s what the questions are for

[left estuary]

I do now

[chilly ocean]

i will give you a basic exercise, one that you should be able to do right from the definitions

[delicate orchid]

prove the product of two units is a unit

[chilly ocean]

wew did it for me!

thanks wew

[delicate orchid]

explicitly, given two units x and y, show that xy is a unit

[chilly ocean]

someone post the picture of the girl from steins gate at the whiteboard

[left estuary]

Ok I think I know how to do this

x in R, y in R right that’s the first part

x^-1 in R y^-1 in R

[delicate orchid]

so far so good

(x^2) is yes, (x)^2 is not

(x)^2 is the product of the ideal (x) with itself

[coral shale]

right?

[left estuary]

I’m trying to figure out how to prove x+y can exist in R and then it’s inverse that’s what I need to solve

[coral shale]

or am i oopsing

nvm

[delicate orchid]

take for example (x^2) in Z[x] vs (x)^2 in Z[x]

[coral shale]

wew r u oopsing

(x)(x) = (xx)

[delicate orchid]

doesn't matter, (x)^2 still contains (x)

x+0 is in (x)^2

[agile burrow]

wew i think you are trolling

[left estuary]

Doesn’t that change for negative values?

[coral shale]

u dont know what my notation means.

[chilly ocean]

wew...

[delicate orchid]

I would never

[chilly ocean]

(x)(x) consists of thing that looks like sum of x^2 factor x^2 out

[left estuary]

Could you say x+x in R?

[chilly ocean]

and if you have thing that looks like x^2 f(x) do x(xf(x))

[delicate orchid]

R is closed under addition yes

but I think you're going in the wrong direction

[chilly ocean]

what does x + y have to do with showing that xy is a unit?

[left estuary]

Cause if it exists and it’s inverse exists then that would answer the question would it not?

[chilly ocean]

too v ague

[left estuary]

An element and it’s inverse

[chilly ocean]

you are being asked for the inverse of xy

x + y is irrelevant

[coral shale]

hmm

[chilly ocean]

... you did read the definition of a ring, right?

[left estuary]

assuming xy=yx

[chilly ocean]

you're not confusing xy with x + y, right?

[coral shale]

to give a concrete example, Z is a ring

[agile burrow]

maybe even before trying to prove something, you could think about examples of units? what are the units in Z?

[delicate orchid]

we can assume xy = yx yes

[left estuary]

Ye just didn’t understand it

[coral shale]

your approach doesnt work for x = 1, y = -1

[left estuary]

Unit only applies to positive integers?

[coral shale]

i mightve given something away

[coral shale]

no...

where did we ever mention positive integers anywhere

in this conversation

[chilly ocean]

side question: how much group theory did you study?

[coral shale]

its 0

they tried from 0, 2-3 days ago

[chilly ocean]

i am going to give in and say that you should be focusing on the prerequisites

[left estuary]

Ye I started a bit ago

[left estuary]

What are the prerequisites

[coral shale]

Introduction to proofs

That aside, when you do a proof, you should be writing all the relevant definitions down

[chilly ocean]

basic set theory. learn how to write proofs and read mathematics. for ring theory, be familiar and comfortable with all of the above and with some basic group theory

[coral shale]

When you are learning. Because using the wrong defn will never help

[left estuary]

I wanted to do ring theory so I studied sets then group theory

[chilly ocean]

i feel like you haven't gotten comfortable enough with those topics

these questions about ring theory you're being asked should be straightforward if you've prepared enough for the subject

especially if you've done group theory before

you should go back and strengthen your prerequisite knowledge

[left estuary]

But I’m not that’s the thing I’m jumping head first into something I’ve never heard about before and trying to learn the same ways the one who invented it did

I’m aware of that but I still try

[chilly ocean]

mathematics is not invented in a single night

this shit took years to develop

[coral shale]

the ones who 'invented' it had much higher prereqs than you

and possibly most of us in this chat

[left estuary]

Ofc not that doesn’t mean I think it will only take a day or 2

[coral shale]

In terms of mathematical maturity, Id expect

[delicate orchid]

instead of trying to learn the same ways some random dude learnt - read a book

[chilly ocean]

you can do better things with your time than spend more than a day or two on a single ring theory exercise that almost everyone would consider basic

[delicate orchid]

that's the best piece of advice possible, just read a book

[coral shale]

just read the intro to proofs pdf i linked. Its very short

For starters

[left estuary]

I got this far by using this method

[coral shale]

Then you can start with group theory

[agile burrow]

no offense, but you haven't gotten particularly far

[chilly ocean]

nah

[chilly ocean]

Yes, you got the illusion of getting far, which is arguably a bad thing.

[coral shale]

ong catgirl bak 💙

[chilly ocean]

there is basically no nice way to say it

[agile burrow]

from your message in #algebraic-geometry a few days ago, it seems like you've learned the group theory that you claim to know over the past 4 days

[chilly ocean]

but you're not ready

that is not to say that you will never be ready, though

go learn the prerequisites properly

read a book

[left estuary]

But this is how I prefer to learn

[chilly ocean]

and it's clearly not working

[delicate orchid]

you're not learning

you're tricking yourself into thinking you understand

btw

x, y are units => xyy^-1x^-1 = 1 so xy is a unit

[chilly ocean]

tough love but if you were learning you wouldn't be struggling this much with such basic concepts

[coral shale]

[chilly ocean]

that's not to say your style of learning doesn't work in general, but that it's just not working for you

[coral shale]

Would it be circular to claim R^x is a group

[left estuary]

But that style of learning is how I learned pretty much everything that school hasn’t taught me yet

[agile burrow]

It might benefit you to work with explicit examples too before you start proving things. Do you have a few examples of groups/rings you can try checking definitions with and applying statements to before you prove them in more generality?

[chilly ocean]

again...

[left estuary]

From integrals to summations

To trig

[formal ermine]

that's like barely anything

[left estuary]

To derivatives

To partial derivatives

[delicate orchid]

trig is the study of two functions and the rest are very computational

[coral shale]

You were asked an integral by parts the day before and I didn't feel you were comfortable with it

[delicate orchid]

this is big boy stuff now

[chilly ocean]

i'm gonna ask a personal question

are you a teenager

[left estuary]

Sure go for it

Ye

17

[chilly ocean]

touch grass

[solar inlet]

And then eventually you find out they're the same function

[coral shale]

This wasnt covered in school?

[formal ermine]

I'm taking this personally.

[coral shale]

calculus

and trig

[left estuary]

Never took calculus they wouldn’t let me

[coral shale]

???

[left estuary]

They didn’t even let me into precal

[coral shale]

?????

This is a highschool yes?

[delicate orchid]

not even eventually, cos(x) = sin(x+pi/2) is pretty early

[left estuary]

Yes

[elder wave]

good

[coral shale]

Im a bit confused about a school not letting you take a class

[left estuary]

This is why I don’t like learning things the normal way because I wasn’t allowed to

[solar inlet]

Oh yeah lol

[coral shale]

Is it because math grades or something

[chilly ocean]

are you not allowed to open a book and read it?

is someone looming over your shoulder preventing you from literally just... opening a book

[coral shale]

The normal way isnt even whats taught in highschool anywYs

[left estuary]

Yeah but I def understand the info because I have perfect 100s in my math classes rn

[coral shale]

Then ask to be moved up

[left estuary]

I tried

Hopefully it works this time

[solar inlet]

Anyways this belongs in a discussion channel not #groups-rings-fields

[chilly ocean]

discord mods gif

[coral shale]

alas. Anyways the solution isnt to jump from easy mode to final boss insane difficulty

[left estuary]

Fair enough

[delicate orchid]

excuse me mr moderator I have a doubt...

[coral shale]

mr mod mr mod, ur not gonna mute us now r u

[left estuary]

But the thing about math is it’s all connected you can learn simple things by studying the complex stuff

[agile burrow]

bruh

[coral shale]

no no no

[prisma ibex]

reading the above chat, my man you are delusional

[delicate orchid]

what the fuck

[coral shale]

its like trying to play a game backwards from the final boss to the easy stages.

thats never going to work

[delicate orchid]

how could you possibly hope to understand more complicated objects (objects which are built upon simpler ones) without being intimately familiar with simpler ones

[chilly ocean]

please consult with a member of the preferred sex

[delicate orchid]

they did this in FNAF: Security breach

[chilly ocean]

Who is? I didn't read

[next obsidian]

You never do

[chilly ocean]

don't read

[solar inlet]

No but I might throw a temporary Slowmode up

[left estuary]

By breaking down the complicated stuff into simple parts

[delicate orchid]

fair

[coral shale]

As wew has said, this completely goes against how math works indeed

It's built from the bottom

up
The prerequisites are there to help you slowly build up a solid foundation

[delicate orchid]

yes! exactly! so go learn about the simple parts!

[left estuary]

But it’s not like I have access to that info and know exactly what to learn

[coral shale]

You do.

i pointed you to the pdf
#proofs-and-logic message

[left estuary]

So I study the complex to figure out what I need to learn

[left estuary]

Yes and thanks for that

I appreciate it

[coral shale]

There isn't a single thing in there that's useless, if that's what you're thinking

There's no wasted value in understanding it all
https://i.imgur.com/fgLkeFo.png

[delicate orchid]

the fuckin temptation to link sheafification is unreal

[prisma ibex]

my mother teaches Russian language, usually at the first and second year level. The students are usually able to read short, simple texts. Every so often she will have a student who attempts to read Crime and Punishment in Russian by reading about 1 page a day and looking up all of the words in the dictionary as they go through.

For whatever reason, no student has been successful in doing this in her almost 30 years of teaching. Why do you think this is?

[solar inlet]

Don't you dare

[left estuary]

This is also the same method I used to learn programming

[coral shale]

Yes doesn't work great with math.
Programming you can go lookup the docs or se on how to do something.

[left estuary]

Vocab is one thing but understanding how words change for the context is what really allows you to do that

[delicate orchid]

also with programming you know you're doing something wrong because it breaks

[woven obsidian]

Just do all your math in Lean

[left estuary]

You can always double check your work

[prisma ibex]

if you're doing things that can be checked numerically sure, but if you're doing proofs then "checking" is not so clear, this requires more mathematical maturity

[coral shale]

At a higher of math level, its less about 'double checking your work', and about understanding the stuff.

[delicate orchid]

gave up with the second e could not be asked

[elder wave]

remove wew

[prisma ibex]

But yeah I would worry about essentially handicapping yourself by using the method you are using. There is a chance this will work for you, but probably a much larger chance that you're going to waste a year or two of your life studying things with extreme inefficiency before recalibrating a bit

[left estuary]

Ok then I’ll try it your way after I read the proofs what would be the next step

If I want to build from the group up then I need to know what the building blocks are and how to find them

[delicate orchid]

honestly following a university syllabus for guidance wouldn't be a bad idea

proofs -> basic group theory would be possible but shoving lin alg in between wouldn't hurt

[coral shale]

Introduction to group theory and see how you feel, I'd say.

[left estuary]

I did that last night I understand it but obviously haven’t fully learned all of its ins and outs

[formal ermine]

you learnt group theory in one night?!

[chilly ocean]

Artin's book starts with both lin alg and group theory

[left estuary]

The basics like all the rules and requirements

[formal ermine]

wut

[delicate orchid]

they mean the group axioms

[coral shale]

I don't think so. If you can't show product of 2 units make a unit in a ring, I doubt you can show identities and inverses are unique

[formal ermine]

proof by my textbook saying so

[left estuary]

Yeah that

[coral shale]

Thats not the basics. Far far from it.

[formal ermine]

@left estuary do you know what a normal subgroup is

[left estuary]

A_n?

[coral shale]

^ these are literally the 1st 2 results proven in any intro to group

Also. You either know something or don't. What's the point in guessing.

[left estuary]

Because I don’t know anything I’m not an expert in anything nor do I have a degree at any given point anything and everything I say could be wrong

[solar inlet]

you should really know linear algebra before any actual algebra

[coral shale]

That's not how it works. If you're learning properly, you should be confident in what the things you've learnt are

[solar inlet]

as well as probably some sort of understanding of like elementary discrete stuff like sets and equivalence classes

[coral shale]

Even better - you should have some confidence in explaining the subject to someone else

[left estuary]

The finemen technique

Are you talking about linear algebra as in linear equations?

[solar inlet]

no

[formal ermine]

vector spaces, linear maps, all that sort of stuff

[solar inlet]

well technically yes but not in the way you mean

[left estuary]

So basically just stuff like this?

[coral shale]

no what............... nowhere close.

[left estuary]

Basic equations learning how to isolate variables and stuff like that?

[coral shale]

Desmos is very limited in visualising stuff - you're not going to be able to go higher than 2 dimensions (ok maybe 3 with some projection but not more)

[delicate orchid]

you could've posted a carly rae jepsen youtube music video and been closer to the mark

[formal ermine]

vector spaces, linear maps, all that sort of stuff

[uneven folio]

This channel has nothing to do with what you're talking about. Please stop flooding it with random stuff or your advanced access will be revoked.

[coral shale]

#math-discussion would be more appropriate

[formal ermine]

I feel like you're (kind of) missing context ryc

[solar inlet]

no ryc is right

[delicate orchid]

yes free up chat for me to ask my doubts yes good...

[solar inlet]

this is not the place to pretend you know what you're talking about

[uneven folio]

I don't care, this isn't the place for this. It's interrupting the flow of the channel. Have this conversation in #math-discussion or something if someone is talking here while having no idea what the channel is for.

[chilly ocean]

i was wondering about what would be the coequalizer of the endomorphisms of $Z$ given by multiplying by 4 and by 2 respectively. I think it will be $C_2$. For given an $f:\mathbb Z \to X$ such that $f(2n)=f(4n)$ it follows that the kernel of f contains $2 \mathbb Z$. So for a morphism from $C_2$ to $X$ that makes stuff commute we are forced to send 1 to $f(1)$ and this shoould be an homomorphism because image of f is isomorphic to either $C_2$ or $C_1$. Is this correct?

[cloud walrus]

Obviously A Catgirl

[next obsidian]

In an abelian category, the coequalizer of two maps A -> B is the cokernel of the difference of the two maps

[chilly ocean]

sorry i didn't mean in Ab

i mean in Grp

[next obsidian]

[chilly ocean]

so its not an abelian cat

[next obsidian]

https://tenor.com/view/my-reaction-to-that-information-gif-25633617

Uhhh time to think

[chilly ocean]

i think it should still here work cuz image of abelian group is abelian

[next obsidian]

Yeah I think this works

The only thing of note is that f(2n) = f(4n) just says that f(1) has order 1 or 2

I’m trying to see if there’s a general categorical thing like if the forgetful functor Ab -> Grp preserves (finite) colimits but I don’t see a way to get this for free

Tfw wrong side of an adjoint

[chilly ocean]

well it already doesn't preserve coproducts doesn'tn't it

maybe it preserves coequalizers tho?

[next obsidian]

Oh

¯_(ツ)_/¯

[chilly ocean]

would be nice to have a reference saying that it preserves coequalizers tho, cuz i don't trust my own proofs ;-,

[spice whale]

currently struggling trying to find an element in PSL(2,7) with an order 21 conjugacy class

[next obsidian]

Bruh

[solar glacier]

Question is 3+3i reduciduble or irreducible in Z[3i]

I put that it’s irreducible since it’s norm is 18 which is either 3 times 6 but no Element of Z[3i] has norm 3 or 6 or it’s 9 times 2 but the norm of no element is 2 forcing one to be 18 and one to be 1

[next obsidian]

Yes

Now you have to show anything of norm 1 in Z[3i] is a unit

And you’d be done

[solar glacier]

I’ve shown that part

Thanks 🙏

[gray junco]

Can someone help me understand how to do this? I think I know some abelian groups of order 24, but how do I know which of them are isomorphic?

idk what maximal order means

[rotund aurora]

Lol I was gonna give the same hint its given in the exercise before reading it

The maximal order is the largest integer n such that there exists an element of your group of order n

[gray junco]

Ok that makes sense. But how does that relate to ismorphic groups?

[rotund aurora]

Think about it

The first question you would ask: what would happen if two groups had different maximal orders?

[gray junco]

I'd assume they'd have different orders? Because the only possible orders of elements are the integers that divide the order of the group

[rotund aurora]

No...

That neednt be the case

Idk what to say to not simply tell the solution

Suppose G and H are two groups such that G has an element g of order n but H has no element of order n

Can they be isomorphic?

So if you have two groups with different maximal orders, you have a situation similar to that one

[rotund aurora]

||Try contradiction: look at the hypothetical image of g in H||

[rotund aurora]

This is a big spoiler: || example: Z/2Z x Z/2Z and Z/4Z ||

[gray junco]

what do you mean by hypothetical image?

[rotund aurora]

Way of talking

So assume you have an isomorphism begween G and H, call it f

Now look at f(g)

[rotund aurora]

|| Look at its order ||

[knotty frigate]

||S_4 instantly comes to mind
Oh and Z_24 and then D_12||

Just my guess

[knotty frigate]

Oh

[gray junco]

I thought permutation groups weren't abelian?

[knotty frigate]

My god I'm stupid

Should have read it more carefully

[knotty frigate]

Z_2xZ_12

Would that be another one?

Sorry if I'm being stupid I've just started AA

[rotund aurora]

Trivially, Z_24 right

[knotty frigate]

Wdym?

[rotund aurora]

Thats an example

Cyclic group of order 24

[knotty frigate]

Oh

Yeah

[gray junco]

Is U(35) another one?

[knotty frigate]

That's of order 35 no?

[rotund aurora]

No

Its order phi(35) I think, which is 24

[rotund aurora]

What is the maximal order of that?

Im sleepy lol

Isnt that cyclic?

[gray junco]

yeah

[knotty frigate]

Oh ok

[rotund aurora]

Wait no I think no?

[rotund aurora]

Ok thats Z_4 x Z_6, so I think that works

You should justify it

[knotty frigate]

Oh

[rotund aurora]

Forgot Chinese Remainder Theorem existed

https://math.stackexchange.com/questions/373708/when-is-the-group-of-units-in-mathbbz-n-cyclic

But idk why you looked at that group, you would just use the product of groups to give an example

[gray junco]

I was just thinking of groups with order 24. (7-1)(5-1) = 24

35=7*5

[rotund aurora]

Yes but that seems more complicated

[gray junco]

so the direct product of groups are non-isomorphic?

[rotund aurora]

Just look at the divisors of 24, it has many

The point is, product of groups is a very simple construction and will provide many examples

And the orders multiply

[rotund aurora]

Thats the result in elementary NT that characterizes for which integers n there are primitive roots modulo n. Idk why I didnt relate it to that. It is not trivial tho

[viscid pewter]

it varies

specifically if you have Z2xZ2 and Z4 they will not be isomorphic

but if they have distinct factors, then they are isomorphic

Z2xZ3 is isomorphic to Z6

[rotund aurora]

Perhaps try solve the following: let n and m be integers. When id Z_n x Z_m cyclic? || it is cyclic when gcd(n,m)=1 ||

[viscid pewter]

and like you have Z2xZ2xZ2 and Z2xZ4 and Z8 all distinct

[chilly ocean]

Is there a name for an element a of a ring such that a^n = a for some n>1? Ik that if you require n=2 it's called idempotent, but what about for any n>1?

[viscid pewter]

finite order

wait, no, rings aren't multiplicative groups lol

[hollow mica]

Suppose that V is a vector space of dimension 2n over C, and B: V x V -> C is a non-degnerate, skew-symmetric bilinear form.
If L is a subspace that is killed by B (B(v, w) = 0 for all v, w \in L), then L is contained in a bigger subspace R of dimension n that is also killed by B

this is in similar vain to this problem

oops sorry for the ping

last time we were able to prove that there exists a k-dimensional subspace that is killed for each k <= n

[hollow mica]

oh yeah and the statement is sound because one can prove that all subspaces that are killed are of dimension <= n

wait do you get this for free by rank-nullity

because for the form f_v = B(-, v), by rank-nullity we have dim(ker(f_v)) = 2n - dim(im(f_v)) = 2n - 1 (B is non-degenerate).

so if {w_1, w_2, ..., w_k} is a basis for W, we just need to show that the intersection of ker(f_{w_i}) over all i (1 <= i <= k) has dimension of at least n

wait I know that the dimension of the intersection two distinct subspaces of dimension 2n - 1 is 2n - 2

but is it true that the intersection of three distinct subspaces of dimension 2n - 1 is 2n - 3

or r distinct subspaces of dimension 2n - 1 intersect to a subspace of dimension 2n - r ?

[viscid pewter]

no

take three planes through a common line in R^3

or three lines through a common point in R^2

[hollow mica]

oh I thought maybe over C it would be different

[viscid pewter]

nnno

[next obsidian]

Same exact thing holds

[hollow mica]

sad life

[next obsidian]

Let V be generated by x,y,z be 3-dim

[viscid pewter]

life would be nonsensically easy

[next obsidian]

<x,y>, <x,z>,<x,y+z>

They intersect at <x>

Doesn’t matter what field

Hell, this works for modules too

[tribal moss]

It's true for three "generic" distinct subspaces (in a sense that can be made precise).

[next obsidian]

Lmfao I mean

[hollow mica]

oh yeah one of my conditions was pairwise non-intersecting (besides identity ofc)

[next obsidian]

One definition is exactly that they don’t share subspace of the wrong dimension

[next obsidian]

This is impossible

Like, based on just numbers

[hollow mica]

oh

yeah good one

i'm dumb

[next obsidian]

Your subspaces have too high of a dimension

[hollow mica]

well can I say anything

about the intersection of r subspaces of dimension 2n - 1 in a v.s. of dimension 2n

[next obsidian]

Usually it has dimension 2n - r

Usually can be made precise as tropo said

[viscid pewter]

the intersection has dimension between 0 and 2n-1

[next obsidian]

You can say more lol

The lower bound is 2n - r

[hollow mica]

wait

that works for my problem

now how do i show that

[next obsidian]

It follows by rank nullity

Take V a VS and L,K subspaces of codimension n,m

Look at the map L (+) K -> V sending (l,k) to l-k

[viscid pewter]

codimension??

category theory has gone too far

[next obsidian]

The kernel of this is the pairs (l,k) where l = k

Which you can see is isomorphic to L\cap K

By sending x to (x,x)

Then applying rank nullity you see that the codimension of L\cap K is at most n+m

I haven’t actually ran the numbers, but this should work

Probably

Codimension is just dim V - dim L

It’s how small it is relative to the ambient space

[hollow mica]

let f: L + K -> V be your map
dim ker(f) + dim im(f) = dim(L + K)
above you showed that ker(f) is isomorphic to L \cap K

dim im(f) is probably at most a + b (where dim L = a and dim K = b)

[next obsidian]

Let dim V = n, dim L = n-l, dim K = n-k

from rank nullity, we have that
(n - l) + (n - k) = dim im(f) + dim(L\cap K)

dim im(f) <= n so we see that

dim(L\cap K) >= 2n - l - k - n = n - l - k

so that codim L\cap K <= l + k

[hollow mica]

why is dim(L + K) = (n - l) + (n - k)

[next obsidian]

Direct sum

[hollow mica]

that's only true when they are disjoint

[next obsidian]

you're doing that abstractly

not as the sum inside of V

hence why I wrote (+)

your map isn't well-defined if you do the sum of subspaces

I think, idk

actually I guess it is

but this doesn't give you what you want

[hollow mica]

I don't get it

is (+) something different from L + K = {v + w : v \in L, w \in K}

[next obsidian]

would it be better if I wrote x?

it's the direct sum

direct product

they're the same

[hollow mica]

oh

[next obsidian]

it's literally pairs

[hollow mica]

and the basis is (v_i, 0), (0, w_i) where v_i is a basis for L and w_i is a basis for K

[next obsidian]

sure

[hollow mica]

oh now I see what you mean here

well kind of

[next obsidian]

yes, it makes no reference to them as subspaces of V

[hollow mica]

someone explained to me the connection between direct sums and products here once

I think it was you

[next obsidian]

idk

[hollow mica]

most of it went over my head

[next obsidian]

maybe

[hollow mica]

but like

[next obsidian]

my point is I'm pretending they aren't embedded into V

so you just take pairs of them

so that the dimensions just add up

so it's good?

[hollow mica]

everything's good

but

[hollow mica]

even if they were

we can still take the direct sum right

it's dimension is just smaller

[next obsidian]

I mean, yes

that's what I did

no

that's not a direct sum then

[hollow mica]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

[next obsidian]

you're taking the join of the spaces inside of V

this isn't a direct sum

[hollow mica]

yea makes sense

[hollow mica]

Under the same conditions, let M be a subspace of dimension n that is killed by B. Then I want to prove that there exists another subspace N killed by B such that M + N = V

so the dimension of N must also be n

I was thinking

let {v_1, ..., v_n} be a basis for M

now extend this to a basis for V: {v_1, ..., v_{2n}}

I want to show that the subspace <v_{n + 1}, ..., v_{2n}> is killed by B

[formal ermine]

ok so I wanna show that $x^3 + 3x - \frac{8}{3}$ is irreducible over $\bQ[x]$. I already tried the following things: \

1. use eisenstein on it, this doesn't work as we're in $\bQ$ \
2. use reduction modulo $\frac{8}{3}$ to get $x^3 + \frac{1}{3}x - 1$ over $\bQ/(\frac{8}{3})[x]$. I don't think this gets me anywhere \
3. use gauß's lemma to obtain $3x^3 + 3x - 8$ over $\bZ[x]$. can't use eisenstein nor rmp on this though

[cloud walrus]

an illuwuminator3

[woven obsidian]

(8/3) is just all of Q

[formal ermine]

oh

so can we do rm 1/3

to get x

which is obviously irreducible

[woven obsidian]

No, Q is a field so ideals are either trivial or all of Q

It's not hard to prove that $f(x) = a_nx^n+...+a_0$ is irreducible iff $x^nf(1/x) = a_0x^n +...+a_n$ is.

[cloud walrus]

AoiKunie

[woven obsidian]

Multiply your polynomial by $3$ to get $3x^3+9x-8$

[cloud walrus]

AoiKunie

[woven obsidian]

The inverted polynomial is then $-8x^3+9x^2+3$, which is irreducible by Eisenstein

[cloud walrus]

AoiKunie

[formal ermine]

ohhhhhhh

ok thanks

[formal ermine]

I don't understand why this is wrong tho

[woven obsidian]

reduction modulo 1/3?

[formal ermine]

yes

[woven obsidian]

That's the same as reduction modulo all of Q so you're just left with 0

[formal ermine]

ohhhhhh

right

[formal ermine]

would you prove this by contradiction?

[woven obsidian]

Think you can do it directly

You only need to prove one direction by symmetry

[hollow mica]

why does the character table of the dihedral group not contain any imaginary entries

is there some sort of geometric reason behind this

[woven obsidian]

Ah no directly is bad. I'd do a proof by contraposition

[formal ermine]

lemme google that

ah

ok

thanks. lemme try

[hollow mica]

If ρ: G -> GL(V) is a finite-dimensional representation whose character is χ, how do I show that

ker(ρ) = {g \in G : χ(g) = χ(e)}

apparently the solution has to do with the only way k roots of unity sum to k is for them to all be 1

though I don't see how roots of unity are relevant here

I guess I'm asking why the kernel of a representation is the same as the kernel of its respective character

[woven obsidian]

$\rho(g)$ is the identity iff all eigenvalues equal $1$, as it is always diagonalizable

[cloud walrus]

AoiKunie

[woven obsidian]

On the other hand $g$ is in the kernel of $\chi$ precisely when the sum of all eigenvalues is equal to the dimension of $V$

[cloud walrus]

AoiKunie

[woven obsidian]

Your remark about roots of unity finishes the argument, as all eigenvalues are roots of unity

[hollow mica]

how do we know this?

[woven obsidian]

If $n$ is the order of the group, then $\rho(g)^n = \rho(g^n) = id$

[cloud walrus]

AoiKunie

[hollow mica]

this just says that $\rho(g)$ is nilpotent?

[cloud walrus]

mrean

[woven obsidian]

Well, if $v$ is an eigenvector with eigenvalue $\lambda$ then that calculations shows $\rho(g)^n v= \lambda^n v = v$

[cloud walrus]

AoiKunie

[hollow mica]

oh so we must have $\lambda^n = 1$ I see

[cloud walrus]

mrean

[hollow mica]

If $N$ is a normal subgroup of $G$, then there exists a finite collection of irreducible representations $\rho_i: G \to GL(V_i)$ such that $N = \bigcap \text{ ker}(\rho_i)$

[cloud walrus]

mrean

[hollow mica]

apparently to solve this we can consider the trivial representation of N

(the action of G on the set of cosets G/N)

[woven obsidian]

Yeah, and use the fact that the only group element that can be in the kernel of all character is the identity

Then apply this to the group G/N

[hollow mica]

I don't see how you extract the \rho_i just from this though

[woven obsidian]

All representations of G/N correspond to representations of G whose kernel contains N

[hollow mica]

well yeah because N is the identity in G/N and representations take the identity of a group to the identity map

what does that say though

oh I see what you're saying

but how do we know there exists representations that are irreducible

[hollow mica]

and how do we know that the intersection doesn't contain other stuff too

[woven obsidian]

Just use this and you get the identity in G/N i.e. N

[hollow mica]

oh

[woven obsidian]

If you intersect the kernels of all irreps in G/N

[hollow mica]

wait so does this imply all irreps are injective

[woven obsidian]

Not sure what you mean

[hollow mica]

oh wait i'm not sure either lol

I guess "intersection of kernel of irreps is trivial" and "the kernels of the individual irreps are trivial" are two drastically different things

I thought there would be a connection there

[simple mulch]

$\big\langle x,y | x^p=y^2=1, yxy^{-1} = x\big\rangle.$

[cloud walrus]

mns

[simple mulch]

So I have this presentation

Notice the last relactor tell us that the group is abelian

I want to somehow get the cyclic group of order 2p from this

[chilly ocean]

Is p coprime with 2?

[simple mulch]

yes

I think I can just use the fundamental theorem of finite abelian groups

the group is abelian and finite thus is the product of two cyclic groups of order 2 and p

[chilly ocean]

yeah if 2 is coprime with p then xy should generate the group

[next obsidian]

I almost let the intrusive thoughts win and got myself banned

[simple mulch]

indeed

[chilly ocean]

what?

[next obsidian]

Idk

I almost sent a message that would’ve been inappropriate for literally no reason

But I stopped myself becuz chmonkey’s have willpower

[chilly ocean]

Uh congratulations then

[coral spindle]

👏 proud of u monke

[next obsidian]

Thx

🙏

[formal ermine]

howonorable

[elder wave]

[chilly ocean]

do NOT try to abbreviate finite abelian group

[next obsidian]

Finabgrp

[barren sierra]

GFA (Group that's Finite that happens to be Abelian)

[rustic crown]

Ab^f

[formal ermine]

fab

more like

dab on those non abelian groups

[rustic crown]

ah, but fgab will be sad

Ab^{fg}

[formal ermine]

is this really an iff?

[delicate bloom]

it has to be, because it's an involution

[formal ermine]

what's an involution

I thought like

let f' = x^n f(1/x) as notation

[delicate bloom]

yeah I'm being opaque, specifically the operation you use to go from f to f' is the same as the operation to go from f' to f

[formal ermine]

let g(x) be an irreducible polynomial with nonzero constant term

then f(x) = xg(x) is reducible

but f(x)' is irreducible

no?

[delicate bloom]

how are you getting that?

n is the degree of f, right?

[formal ermine]

yes

[formal ermine]

because we define it as xg(x)

doesn't that immediately mean that it's reducible?

[delicate bloom]

but that has zero constant term

xh(x) is not irreducible

[delicate bloom]

ok back up, why did you introduce g(x) here when your next two lines don't use it?

[woven obsidian]

Ah yeah, you will need to add a condition that a_0 is not 0

[delicate bloom]

show me an example

[woven obsidian]

But then reducibility is trivial anyway

[formal ermine]

oh I confused h and g

sorry

[delicate bloom]

if a_0 is 0, it's reducible, so you don't have to add the condition yeah

[formal ermine]

ah

[woven obsidian]

E.g. x^2+x maps to x+1

But it will be an involution for nonzero a_0, so theb you have iff

[formal ermine]

thanks

this works in any (commutative) ring?

I would assume so

[woven obsidian]

Yeah the proof doesn't use any property of the ring

[formal ermine]

epic

thank you!

[woven obsidian]

Np

[delicate bloom]

iff means f(x) irreducible implies g(x)=x^nf(1/x) is irreducible and g(x) irreducible implies f(x) is irreducible

[woven obsidian]

Yeah

Clearly doesn't hold for the pair x^2+x and x+1

[delicate bloom]

right, because x^2+x is not irreducible

we don't need to make the claim that a_0 is nonzero

[woven obsidian]

But x+1 is irreducible

So there's no equivalence

[delicate bloom]

yeah you're right, I'm thinking about it backwards as if evaluating an entire truth table for "iff" which comes up to be false

[foggy merlin]

We know that every symmetric quadratic form is "diagonalizable" i.e. the matrix corresponding to that quadratic form has a spectral decomposition. I was wondering if this also holds for higher order forms, if there is some type of spectral decomposition for symmetric tensors.

[formal ermine]

let p,q be two polynomials. let f(p) = x^(deg p) p(1/x). this is a multiplicative function because it's the product of two multiplicative functions, namely p(x) -> 1/x and p(x) -> x^(deg p)
how does this show that for a polynomial f with deg > 1 and a_0 \neq 0 that f is irreducible iff x^(deg f) f(1/x) is irreducible

[junior stump]

Think of the function as reversing the coefficients of a polynomial with non-zero constant term. Then g is an involution so it suffices to show that if g(f) is irreducible then f is irreducible. If f = pq then since a_0!=0, p(0)!=0 and q(0)!=0. g(f) = g(p)g(q), so wlog g(p) is constant and so of degree 0, since reversing coefficents preserves degree p is also of degree 0 so constant, so a unit and f is irreducible

[formal ermine]

ah gotcha

thanks

[junior stump]

Key idea is degree is preserved, and degree characterizes units with polynomials

[stoic rose]

Why is $\alpha^{\vee}$ uniquely determined by $\Phi$ and $\alpha$?

[cloud walrus]

Antoine Labelle

[stoic rose]

(There is a typo, $s_\alpha$ should be $x \mapsto x - \langle x, \alpha^{\vee}\rangle \alpha$)

[cloud walrus]

Antoine Labelle

[next obsidian]

Phi spans V should be why

like, alpha^ should be determined by how it acts under the pairing <x,alpha^> for x in V

but this is determined just by what happens to <beta,alpha^> for beta in Phi

[stoic rose]

Hmm I agree that the reflection is determined by how it acts on Phi but how do we know exactly how it acts on Phi?

[next obsidian]

I was assuming that part of "Phi" meant how alpha^ acts on Phi

but maybe it's from the reflection part

[stoic rose]

Hmm the way I interpret that is that there is a unique alpha^v satisfying the properties given before.

It's not clear to me why these properties determine the action on Phi

[next obsidian]

yeah i dunno

Me bad with Lie algebras

[stoic rose]

Found the proof

[next obsidian]

ChmonkaS

[lethal lava]

Could anyone help me with exercise 18.2.15 in Dummit and Foote?

the hints are quite a scavenger hunt but I'll try to get all of them

pretty sure exercise 18 gives the forward direction immediately, but I don't know where to start with the backwards direction

[slender hamlet]

Hello

[chilly ocean]

say i have a polynomial in Z[x] of which i find a root p/q in Q, does my polynomial necessarily factor with the linear term qx-p in Z[x]?

i feel like it should but i guess there isnt guarenteed polynomial division in Z[x] by non-monic polynomials

[next obsidian]

You definitely want to take p,q coprime

Or else like, take 4/8 and 2x -1

[glacial narwhal]

how to solve 7.32 ?
I want ask a question.
the algebra is finite dimensional , so it has a finite basic ,say $b_1,b_2,\cdots , b_n$ .Any element of the algebra can be written as a linear combination of these basic elements?
IT is right? if no. please tell me how solve 7.32.

[cloud walrus]

傅佳灿

[tribal niche]

Let $n$ be an integer. Show that there is a Galois extension containing $\mathbb{Q}(\sqrt{n})$ with Galois group $C_4$, if we have $n=a^2+b^2$ for some $a,b.$

[cloud walrus]

monkeman

[lavish gull]

$\alpha = \sqrt{n} \implies \alpha^2 - n = 0$

[cloud walrus]

coderizer

[lavish gull]

so the minimal polynomial is quadratic for the case of a square root. the galois group has order equal to the degree of the minimal polynomial
how could it have a subgroup C_4?

[lavish gull]

do you mean any extension of Q containing sqrt(n), not necessarily the smallest?

[glacial narwhal]

my teacher write this， that is say： every nonzero commutative ring have maximal ideal ? it is right?

By the way, this proved by Zorn's lemma.

[elder wave]

Are you asking if the statement is true?

[south patrol]

That isn't quite what the statement is saying

It is saying smth stronger

Though it follows from what u said

[left estuary]

Still trying to learn ring theory lol

[left estuary]

Number 2 seemed easier so I tried that and I think I got it

[formal ermine]

sure

list them

[left estuary]

Alr well because it’s a cyclic group of integers with the modular 20 it means that every multiple of 20 would be 0 and 0^n=0 {n≠0} so all the multiples of 20 would be a nilpotent element

Now the only thing I’m confused on is the range

Is it [0,19] or infinite?

[formal ermine]

0^n = 0 even when n = 0

by definition of g^n

[elder wave]

every multiple of 20 is already zero in Z_20

[left estuary]

Ye

[elder wave]

what you just said is that 0 is nilpotent

[formal ermine]

^

you're missing some

[left estuary]

Yeah is there something wrong with that

[formal ermine]

it's not the only one

[left estuary]

20 is nilpotent

[elder wave]

0 is always trivially nilpotent

[formal ermine]

20 = 0

[elder wave]

20 is equivalent to 0 in Z_20

[left estuary]

Yeah cause the mod

Trying to figure out what else I’m missing

0^0 is 1 I thought when defined in the domain of ℝ

Alr

So what am I missing cause I don’t see how the other numbers could be 0

[formal ermine]

just go through all numbers

and list their multiples

[elder wave]

have you worked out how the operations on Z_nZ work

[left estuary]

Ye

[formal ermine]

then see when they are a multiple of 20 i.e. go to 0

[elder wave]

shush stop spoiling illum

[left estuary]

OH

ok I understand now

Well one is just 1*20 so would that be included?

[shell agate]

that's 0 again

[left estuary]

Yeah so 1 is included

[shell agate]

how so?

[left estuary]

Going off of this

[shell agate]

x is nilpotent if for some n, x^n=0

can 1^n=0?

[left estuary]

Yes

Ah ok

That multiple

Had a feeling that was a possibility for error

Ok i get it

[formal ermine]

(once you're done with this, try expressing the general solution to the nilpotent elements of Z/nZ in set builder notation)

[left estuary]

Alright I’ll give it a shot

[shell agate]

(maybe try to understand 20 before generalizing)

[left estuary]

10

[shell agate]

yes!

well done

[left estuary]

10^2 = 100 which under mod 20 goes to 0

[shell agate]

exactly

[left estuary]

10 and 2 are also factors of 20 which I find interesting maybe that means something

[left estuary]

So this would be under multiplication so x^1 x^2…

[formal ermine]

no

don't do the general case

do the n=20 case first

[shell agate]

they did find 10 already

[left estuary]

So this is ℤ/20ℤ that would list the remainders for x mod 20

[shell agate]

yes

[formal ermine]

fwiw x is nilpotent in Z/abcZ for a,b,c primes iff it's nilpotent in Z/aZ, Z/bZ, and Z/cZ

[south patrol]

Chinese remainder moment

[formal ermine]

in this case a = 2, b = 2, c = 5

(btw this is true for any amount of primes, I'm just on mowobile and cba writing latex)

[left estuary]

wait how do you get a b and c?

[formal ermine]

prime factorization of 20

[left estuary]

Oh

I’ll have to make a table of that later to better understand why it works

[formal ermine]

"why it works" is a bit more complicated

requires some ring theory

[left estuary]

Well yeah but if you write it out in the form of a set you can get a better understanding

At least for me

[left estuary]

I have a request

If it’s not too much effort

[formal ermine]

what request

[left estuary]

Could you give me a list of beginner friendly questions for ring theory

If I start with basic questions I can work my way up

[formal ermine]

no

ring theory requires group theory

[left estuary]

Alright

[formal ermine]

do that first

[left estuary]

Well then I’ll need to find a list of group theory questions

[formal ermine]

just read a book

it doesn't work like that

read a book man

[left estuary]

I do better with questions

[formal ermine]

books contain exercises and questions at the end of each section/chapter

either read a book or you'll no longer receive help from me

[left estuary]

Where can I find the books

[formal ermine]

#books message

[left estuary]

I’ll look into it in a bit

[formal ermine]

thank you and good luck!

[left estuary]

Thanks for all your help

I appreciate it and everyone else’s

[left estuary]

Would these be some good problems to start off with https://www2.math.upenn.edu/~qze/math314s18/algebraex.pdf I am doing some research on group theory books and practice problems and I found this

[formal ermine]

those are problems that you would find on the first homework sheet of an aa course

[left estuary]

https://www.jmilne.org/math/CourseNotes/GT.pdf this is a document I found for group theory

[left estuary]

Well then I think I’ll start there after I do bit of reading on this doc I found ^

[formal ermine]

yes that seems okay

[left estuary]

Alr cool I’ll do that now then since I have today off

[cunning dust]

You should probably read a book about proofs before doing any advanced math

[left estuary]

I have

[left estuary]

I read this one

[cunning dust]

Hmm you may want to familiarize yourself more with basic set theory, but that may be enough

[left estuary]

We will see how the group theory problems go if I still feel uncomfortable with them I’ll go do more basic proofs

[tribal niche]

Yeah so any extension where Q(sqrt(n)) is a subfield.

[unique valve]

fourth root of n

[tribal niche]

Extension is forced to be degree 4 anyway so it's also a degree 2 extension of Q(sqrt(n))

Well it has to have galois group C_4.

Fourth root of n could potentially have galois group Klein 4

[unique valve]

Doesn’t fourth root always have C4

[tribal niche]

No, just check n=3

It has Klein 4 I'm pretty sure

I know that the condition fails for n=3

And n=-1

[unique valve]

n=a^2+b^2

[hot lake]

An extension of Q(sqrt n) with galois group C4 over Q would be a quadratic extension of Q(sqrt n) so you could look at what possible galois groups those or their galois closure can have and when

[tribal niche]

I thought that all degree 2 extensions had to have extension Z/2Z because there is only one order 2 group up to isomorphism

[hot lake]

Over Q(sqrt n) yeah

[tribal niche]

The issue with n=3 is that there are 2 degree 2 elements in the galois group which send sqrt(3) to -sqrt(3) and fourth root of 3 to its negative @unique valve

I think that the extension has to be adjoining some complex z and some conjugate stuff because you notice modulus of z squared is a^2+b^2

[hot lake]

q(4th root of 3) isn't a Galois extension of Q

[tribal niche]

Oh

Wait yeah

Good point

Because it isn't the splitting field of any polynomial in Q

We can also take an arbitrary degree 4 extension Q(sqrt(a+bsqrt(n))) and proceed from there?

Also what does L* mean or Q* what is the star for fields? I know for groups it is referring to the multiplicative group but I'm not sure what it means for fields.

[hot lake]

it's also their multiplicative group

[tribal niche]

https://math.stackexchange.com/questions/2037782/how-do-i-shows-the-galois-group-of-a-field-extension-is-a-cyclic-group-of-order I found this but I don't really understand it what do they mean when they take Q*^2 and stuff

[hot lake]

K*² is the subgroup of the squares of elements of K*

[tribal niche]

I see

Thanks

[coral shale]

i dont have recommendations myself, but you might wanna try asking for some problem sets to test yourself on it

maybe ask in #math-discussion or #proofs-and-logic or #book-recommendations

[left estuary]

Preferably something like questions with the answers and work shown at the bottom of the pdf that way if I get confused or don’t understand how to solve a problem then I can go look and see how it works

[coral shale]

well if there are no answers you can try asking here sometimes

better to ask for a hint than check the answers if ur stuck anyways

[left estuary]

I’m more interested in the work than the answer itself

[coral shale]

ik - but theres a lot to be gained from figuring it out yourself

even with some hints

[left estuary]

Yeah I agree

But I also think having work shown in its entirety is important because I can see every little step and fill in the gaps in my head kinda like peices to a puzzle yknow

Just requires some self control to actually try and understand the work and not just copy and paste the answer which I believe I am capable of

Though both are good methods

So I’ll try both and see which works best for me

[chilly ocean]

READ A BOOK

[coral shale]

this only gets you so far. depends on context

Like depends on teaching style

Sometimes in a 'methods' kinda class - u get shown the methods

Like u say, and u apply those to the problems

But generally for the purer subjects this is less the case

And you would benefit a lot from getting used to figuring things out for yourself from some hints

[left estuary]

You can also apply the same things to books

[coral shale]

Otherwise down the line ur still gonna be unable to do questions because you can't think of new ideas

Like - there's constantly new ideas / innovations in abstract algebra. To me at least
There's no 'set of methods' to learn how to solve stuff. Learning the defns are one thing, but getting 'better' at solving problems is also importnat

[left estuary]

Since your reading someone else’s understanding of a subject however to truly understand something you need to be able to define it in your own terms

That’s why I like doing problems it allows me to do that

[coral shale]

yeah sure

thats the reading part

in the book. Read and make notes is a good way

But I'm referring to solving problems - you want to throw yourself in the deep end for those

[left estuary]

Yes same here

Let’s give your hint system a go rn I already read a bit of that document and question 1 looks simple enough to try out on that set of problems

If you have the time that is

[untold basin]

Hello,
I have to determine all the subfields of Q(zeta_12)/Q
I know that its Galois group is isomorphic to (Z/2Z)². Do I need to determine explicitly all the subgroups of (Z/2Z)² and then apply correspondance of Galois ?

[coral shale]

remindme what zeta_12 is. 12th root of unity?

[untold basin]

Yes it is e^{2 . pi . i/ 12}

[coral shale]

(thinking it through, but no guarantee I'll be able to help)

[woven obsidian]

There's not a lot of subgroups of a group of order 4

[unique valve]

Yeah there’s 3 I think

Which should correspond to 3 subfields

[coral shale]

i feel like theres 4 . Counting trivial and non-proper

[unique valve]

Counting those theres 5

[coral shale]

oh yh

[untold basin]

There is the group itself, {neutral element} and the others are, by Lagrange, of order 2

[unique valve]

I can think of two of the three relevant subfields

[untold basin]

But my question was, do I need to find them explicitly

[coral shale]

Yes if it asks you to 'determine'

[woven obsidian]

I mean you really only need to find how many there are

And then find that amount of subfields

But probably easier to find them explicitly and then look at fixed fields

[untold basin]

Actually, the prof told us
Show that this "treillis" (like the tree of subfields) is actually the right one

[unique valve]

Adjoining i gives you one, also a third root of unity

[untold basin]

That's what I want to do

[unique valve]

Or a 6th root of unity

[woven obsidian]

Looks like the fields have been found

[untold basin]

Okay but it doesn't prove anything
Maybe there is one missing

I need the fact that there is only 4 subgroups

[woven obsidian]

5

[untold basin]

How did we find them

[woven obsidian]

Just find the elements of order 2 in (Z/2Z)^2 and you're done

There's as many nontrivial subgroups as there are such elts

[untold basin]

Does it prove that there is exactly 5 subgroups ? I thought that this gives at least 5

[woven obsidian]

Any nontrivial subgroup have order 2

[untold basin]

Yes

[woven obsidian]

So they're cyclic and thus generated by an elt of order 2

[untold basin]

SO GOOD

2 is prime

Thanks to all of you

[woven obsidian]

Yeah the worst prime of them all lol

[untold basin]

XD

[woven obsidian]

Actually the field where you adjoin the third root of unity is the same as the field where you adjoin a sixth root of unity

[next obsidian]

Wat

?????

Over Q?

[wooden ember]

Chmonkey is shocked

[woven obsidian]

If $\omega$ is a cube root, $-\omega$ is a sixth root

[cloud walrus]

AoiKunie

[lavish gull]

root of unity?

[wooden ember]

Yeh

[woven obsidian]

yes

[woven obsidian]

$\phi(2) = 1$ goes brrrr

[cloud walrus]

AoiKunie

[lavish gull]

what is the proof?

if $\omega^3 = 1$, then also $(\omega^3)^2 = 1^2 = 1$
so $\omega$ is a root of $x^6$ too
since $(\omega^3)^2 = (\omega^2)^3$, substituting $\omega$ with $-\omega$ is also true since
$\omega^2 = (-\omega)^2$

[cloud walrus]

coderizer

[lavish gull]

is this correct or obvious?
is it too much detail if it is obvious?

[woven obsidian]

You need to show it's not 2nd or 3rd root of unity also

[lavish gull]

ok, well you didn't say that

[coral shale]

i mean thats not hard to show (or visualize)

[lavish gull]

can you do it without using the particular roots of 2 or 3?

[woven obsidian]

Obviously I'm talking about primitive roots of unity

[lavish gull]

are you now making a distinction between primitive and non-primitive roots of unity?

[woven obsidian]

$(-\omega)^2 = \omega^2 \neq 1$ is $\omega$ is a primitive third root of unity

[cloud walrus]

AoiKunie

[woven obsidian]

And $(-\omega)^3 = -\omega = -1$

[cloud walrus]

AoiKunie

[lavish gull]

thanks, i understand. so you are excluding 1?

[woven obsidian]

No special values used

Yeah, there's no point in considering non-primitive roots

[lavish gull]

except that extra terminology for "primitive" is needed

primitive already has a lot of meanings

you could instead ask for the roots of $\frac{x^n - 1}{x - 1}$

[cloud walrus]

coderizer

[coral shale]

this also includes non-primitive roots??

primitive roots; ie. roots of the cyclotomic polynomial

[lavish gull]

how are you using primitive now? i thought @woven obsidian used it to mean roots, except for x=1

[coral shale]

no this is the definition

usual one afaik

[woven obsidian]

Yeah for example its meaning in "pimitive roots of unity"

[woven obsidian]

No I did not

https://mathworld.wolfram.com/PrimitiveRootofUnity.html

I said that the primitive third roots of unity was the third roots of unity that were not 1

[lavish gull]

ok, now i understand. it is the generating element of the cyclic group

but there is only one primitive root for every n

[woven obsidian]

no

There are phi(n) primitive nth roots of unity, where phi is the euler phi function

[lavish gull]

yes you are right. my mistake.

[woven obsidian]

In C they all have the form $e^{2\pi k /n} $ for some $k$ coprime to $n$

[cloud walrus]

AoiKunie

[lavish gull]

why is this needed?
is it a problem with my answer or is it part of your question??

[woven obsidian]

I had no question

[lavish gull]

ok, i read the message history and see that you were answering another question

[woven obsidian]

Yeah

[lavish gull]

i wish it was easy to factor over Q. it makes it hard to study field extensions because of this.

[formal ermine]

Let $f(x) \in \bZ[x]$ and $f(\frac{a}{b}) = 0$ for $a,b \in \bQ$ and $\gcd(a,b) = 1$. Show that $a \mid a_0$ and $b \mid a_n$.

[cloud walrus]

rectangle cube

[formal ermine]

I have no idea how I would start with this

does anyone have a hint for me?

[lavish gull]

this is saying that a/b is a fraction in it's lowest common form. is that obvious?

[formal ermine]

yes

[lavish gull]

what isn't obvious to me is the meaning of $a_0$ and $a_n$

[cloud walrus]

coderizer

[formal ermine]

$f(x) = a_nx^n + \ldots + a_1x + a_0$

[cloud walrus]

rectangle cube

[lavish gull]

i think this is the rational root theorem?

[next obsidian]

Isn’t this literally the rational root test?

[elder wave]

I think this might be the rational root theorem

[formal ermine]

the what?

[next obsidian]

I think you can get this from vieta’s formulas too

[formal ermine]

what's vieta's formula

[coral shale]

computing coefficients in terms of roots

[formal ermine]

ah

never knew that has a name

ok thanks guys

is $\bK[X][Y] = \bK[X,Y]$

[cloud walrus]

rectangle cube

[woven obsidian]

They're at least isomorphic

[formal ermine]

oki

[woven obsidian]

Equality depends on how you precisely define polynomial rings I guess

[next obsidian]

The right is defined as the left in most books

[formal ermine]

interesting

why do we include the algebraic closure of the base field in the definition of the galois set

does it have a category theory background? I'd assume so

[coral shale]

never heard of a galois set 👀

[formal ermine]

could be an error in my translation

[coral shale]

galois group???

define it for me 🙏

or show screenshot

[formal ermine]

Let L/K be an algebraic extension. We call ... the galois set of L/K and [L : K]_s = ... the separability degree of L/K.

menge = set in german

[next obsidian]

Oh this is the same as the separable degree

[coral shale]

im slightly lost

so whats the 'galois set'

in this defn

is it Hom_k(...)?

[formal ermine]

yes

[next obsidian]

For an alg extension L/K

[formal ermine]

yes

[coral shale]

oh ok thats a group isnt it

[next obsidian]

The number of K-embedding of L into the alg closure

[formal ermine]

idk

[coral shale]

yet to be proven i suppose

[formal ermine]

yeah

[next obsidian]

This is the same as the separable degree of L/K

So it’s equal to the degree iff the extension is separable

Which is half of being Galois

[formal ermine]

why do we write Hom_K(L, K bar) though

does this have something to do with cat theory

[tribal niche]

Galois group is the automorphisms of the extension that's essentially what the Hom is saying

[next obsidian]

…

You’re just asking how many ways you can embed L into the alg closure of K

Which fix K

[coral shale]

Hom is all the homomorphisms from L to K bar

[tribal niche]

Yeah automorphisms which fix K, my bad

[coral shale]

which fix K oops

not auto

[formal ermine]

ah ok thanks

[coral shale]

theres some link on nlab

i remember being shared by moldi or chm

[coral shale]

which might shed some more light