groups-rings-fields

[latent anvil]

Wouldn't it be (a + bω)(a - b - bω)?

Since ω and -1 - ω are galois conjugates

[cerulean siren]

I don't know what a galois conjugate is

[latent anvil]

Yeah I'm talking to zoph, sorry

[mild laurel]

oh whoops yeah

I was thinking it was just -ω in my head

[latent anvil]

Yeah I made the same mistake

[mild laurel]

anyways, what does that simplify to

[cerulean siren]

oh you're talking about the norm

[latent anvil]

Yup

a^2 - ab - abω + abω - b^2 ω - b^2 ω^2 = a^2 - ab + b^2, right?

So the norm is N(a+bω) = a^2 - ab + b^2

@cerulean siren do you know about norms?

[cerulean siren]

In the context of euclidian domains

but not how to find one without having it given

[latent anvil]

Ah yeah this isn't a Euclidean norm

The general defintion for a number field (thing which you get by adjoining finitely many algebraic numbers to Q) involves Galois theory

[cerulean siren]

you can adjoin multiple things to a field?

[latent anvil]

Well what happens when you adjoin one thing?

You get another field, right?

[cerulean siren]

ohhh

sort of like F[x, y] is the same as F[x][y]

[latent anvil]

Yeah exactly

And it's really the same thing

Instead of polynomials in the one thing you adjoin, you look at polynomials in both

Anyways, we have a function N(a+bω) = a^2 + b^2 - ab on Z[ω] which is multiplicative (you can check this explicitly)

Oh sorry using ω for zeta_3 is pretty common

And N(7) = 49, so if 7 = (a+bω)(c+dω) then we'd have to have N(a+bω) = N(c+dω) = 7

So in addition to the equations above we get c^2 + d^2 - cd = a^2 + b^2 - ab = 7

[cerulean siren]

oh right you need 4 systems of equations to solve for 4 variables

now I have no idea how I was supposed to do this lol

[latent anvil]

ac - bd = 7
ad + bc - bd = 0
a^2 + b^2 - ab = 7
c^2 + d^2 - cd = 7

Just so we can keep track

[mild laurel]

Nah from here, you can just see that you can get an element with norm 7 since 3² + 2² - 6 = 7

[latent anvil]

Oh okay sure

[cerulean siren]

wat

[latent anvil]

So 3+2ω has norm 7

[cerulean siren]

oh I see you're using the last equation

[latent anvil]

And 2 + 3ω

[cerulean siren]

I'm confused where you got equation 2

and 3 and 4 I definitely couldn't have gotten on my own

[latent anvil]

Equation two comes from multiplying (a+bω) (c+dω) and setting the ω coefficient to 0

And yeah 3 and 4 are pretty magical unless you know more stuff

Anyways, (2+3ω)(2+3ω^2) = 4 + 6ω^2 + 6ω + 9 = 13 + 6(ω^2 + ω) = 13 - 6 = 7

Zoph was using more Galois theory to say it suffices to find any non-real element of norm 7, and this is the explicit example

[cerulean siren]

I still don't see how I could have done this on my own 🤷 but oh well hopefully the professor will have it in the solutions

[latent anvil]

Yeah, sorry

I'm not sure either

[cerulean siren]

nah not your fault

[latent anvil]

I might be missing something?

But it seems hard

[cerulean siren]

I spent like 2 hours on this so I'm glad to know it wasn't something silly lol

[latent anvil]

Definitely remember the comparing coefficients trick

Also I can teach you about norms if you want, they're pretty cool

[cerulean siren]

sure, why not

[latent anvil]

So suppose you have an irreducible polynomial f(x) with rational coefficients

And that it factors over C as (x-α1)...(x-αn)

Then we get a field F = Q(α1,...,αn) by adjoining all the αs

So the elements are polynomials in the roots of f

This is the smallest field containing Q over which f factors completely

Does that make sense so far?

[cerulean siren]

so F is a subset of C?

[latent anvil]

Yup

[cerulean siren]

ok I think I'm with you

[latent anvil]

Okay I just realized this is way too hard to explain except for n = 2

So I can explain norms for quadratic extensions

Does that make sense?

[cerulean siren]

quadratic extensions means 2 factors? sure

[latent anvil]

So we have an irreducible polynomial f with roots α, β

And actually to get F we only have to adjoin one of them

[cerulean siren]

why's that?

[latent anvil]

Since f(x) is divisible by (x-α) in Q(α)[x], meaning the other factor (x-β) is in Q(α)[x] as well, which means β is in Q(α)

I think you can do it with the quadratic formula too

[cerulean siren]

oh hmm since for it to be divisible you need a second factor

makes sense

[latent anvil]

Yeah, we're using the fact that some bigger ring where f factors completely, like C[x], has unique factorization

So because α and β are both roots of the same irreducible polynomial, we call them "Galois conjugates"

[cerulean siren]

roots of the same irreducible polynomial
🤦

but I'm following

[latent anvil]

?

[cerulean siren]

irreducible implies it has no roots, you're talking about two different fields at once

[latent anvil]

Oh, sure

So the fact that they're Galois conjugates implies that there's a bijection σ : Q(α) -> Q(α) such that σ(0) = 0 and σ(1) = 1 and σ(x+y) = σ(x) + σ(y) and σ(xy) = σ(x) σ(y) (this says that σ is a "field automorphism") and finally σ(α) = β

Explicitly, since any element of Q(α) just looks like p + q α, we must have σ(p+q α) = p + q β

[cerulean siren]

my professor was talking about how all transcendentals t, s form an isomorphism under Q[t] <=> Q[s], this sounds kind of similar

err I think that's what he said I have to reread my notes

[latent anvil]

That is true, but I'm not sure it's related

[cerulean siren]

well it's the same idea, you're forming an isomorphism between all the roots adjoin Q

[latent anvil]

Oh yeah that's a good way of looking at it

And actually there's a much nicer way to say that

Where f might not be quadratic

[cerulean siren]

p is an element of Q?

[latent anvil]

If α is a root of any irreducible polynomial g with Q coefficients, then Q(α) ≈ Q[x]/(g(x))

So if you have two different roots, you can compose the isomorphisms to get ones between Q(α) and Q(β) swapping the generators

Yeah, p is an element of Q

[cerulean siren]

ok you've lost me now, what does Q[x]/<g(x)> have to do with it?

[latent anvil]

Define a ring homorphism φ : Q[x] -> Q(α) by sending x to α. This is clearly surjective, right? What's the kernel?

This isn't what I was talking about, it's just a tangent based on the stuff you brought up

well it's the same idea, you're forming an isomorphism between all the roots adjoin Q
@cerulean siren

I'm saying this idea works even if f isn't quadratic

And so Q(α) and Q(β) might be different fields

[cerulean siren]

ok I see what you're trying to prove, I'm just a little confusd by the ideal you introduced

[latent anvil]

The ideal is the kernel of that map

Which is also (g(x))

[cerulean siren]

let's go back to galois theory 😆

[latent anvil]

Sorry

Okay the so the "norm" I mentioned above is the function N : Q(α) -> Q given by N(x) = x σ(x). Explicitly, N(p+q α) = (p+qα)(p+qβ) = p^2 + pq(α+β) + q^2αβ

So there's general Galois theoretic reasons that this lands in Q, but you can also see it explicitly

Can you see why? Try to prove it!

[cerulean siren]

you would need alpha+beta and alpha*beta to both be in Q, right?

[latent anvil]

Yup

So what do we know about α and β?

[cerulean siren]

they're both roots of g

[latent anvil]

Yup

And what do we know about f?

[cerulean siren]

what was f?

[latent anvil]

Oh sorry g

I think I originally used f

And then used g for the tangent

About Q[x]/(g(x)) ≈ Q(α)

[cerulean siren]

ok ok we know that g is irreducible in Q

[latent anvil]

Yeah

And how does g factor?

[cerulean siren]

(x - alpha)(x - beta)

[latent anvil]

Right

[cerulean siren]

so that gives you back a function with coefficients in Q!

[latent anvil]

Yeah!!!

[cerulean siren]

so we know -alpha * -beta is in Q

[latent anvil]

Right

[cerulean siren]

and also that x - beta -x - alpha is in Q

which is the same as -beta - alpha is in Q

and you can just factor out -1

[latent anvil]

Yup

[cerulean siren]

oh that's so cool!

[latent anvil]

Yeah ikr

So don't be scared of all the scary words

Like Galois conjugates

[cerulean siren]

can you teach my class 😆

[latent anvil]

If α, β are the roots of an irreducible polynomial f(x) = x^2 - ax + b, then we have a function N : Q(α) -> Z given by N(p+qα) = p^2 + apq + bq^2,

Now if you believe that the σ above is a multiplicative, it's really easy to see N(x) = x σ(x) is multiplicative

You should also be able to check it directly I think

[cerulean siren]

wait is this the same N in both cases?

p^2 + apq + bq^2 = x * theta(x) ?

[latent anvil]

Yup

sigma, not theta

[cerulean siren]

oh right

because it's a permutation of the roots

[latent anvil]

Because the definition of σ is σ(p+qα) = p + qβ

And I just expanded (p+qα)(p+qβ) out to give the nice explicit definition

Does that make sense?

[cerulean siren]

shouldn't that say p^2 + alpha*pq + beta*q^2 then?

you wrote a and b

(also, how are you typing greek letters?)

[latent anvil]

So I said f(x) = x^2 - ax + b

Wait hang on, I think you made a mistake

It should be p^2 + (α+β)pq + αβq^2

Right?

[cerulean siren]

I just copied what you wrote 😛

[latent anvil]

Oh gosh

Sorry

Oh no you didn't

I'm being careful

So -a and b are the coefficients of f

But - (α+β) and αβ are also the coefficients of f

Right?

Since f(x) = (x-α)(x-β)

[cerulean siren]

wouldn't it be (-alpha - beta)/2 ?

wait hold on

I'm bad at multiplying lol

yeah no you're right

ok, so -(alpha + beta) = -a and alpha*beta = b

I'm with you

ahhh and that's how you normally factor over the integers anyway

[latent anvil]

So we have this nice norm function

Which can be expressed in terms of the coefficients of f

And it's multiplicative

[cerulean siren]

https://i.kym-cdn.com/photos/images/original/001/619/361/8a0.jpg

[latent anvil]

In the case of Q(i), which you've probably seen, this is the usual norm N(p+qi) = p^2 + q^2

Right?

Since the polynomial is f(x) = x^2 + 1

So a = 0 and b = 1

[cerulean siren]

ok, I'm with you

[latent anvil]

Alternatively, σ(p+iq) = p - iq, so N(p+iq) = (p+iq)(p-iq) = p^2 + q^2

[cerulean siren]

because our original formula was p^2 + apq + bq^2

[latent anvil]

Right

[cerulean siren]

oh and this is why multiplying by the complex conjugate always gives you something in R!

never understood that before

[latent anvil]

Yeah that's a nice observation

Another better way to see it is Galois theory

But you'll get to that

So there's other stuff about the "ring of integers" which I'm not going to explain, and which let's us look at a subring of Q(α) and say the norm of an element in that subring is an integer

And then you can do stuff about factorization

Like we did above with Z[ω]

Okay, I hope you found this interesting

I don't know though about this area

It's like the beginning of algebraic number theory

But it's on my list

[uncut girder]

What are you guys talking about

[latent anvil]

The norm of an element in a quadratic number field

[cerulean siren]

that's why you edited phi: Q(alpha) -> Z to say it was to Q before

[latent anvil]

Yeah lol

I was getting mixed up

As I said, I don't really know this area

[cerulean siren]

@uncut girder starts here: https://discordapp.com/channels/268882317391429632/496784958430380033/692900697473286195

and then goes for ... a while

wow it's been more than an hour lol

[latent anvil]

Wow

I've been making dinner

[cerulean siren]

I've been pretending to do spanish homework

oh you never did say how you were typing greek letters haha

[latent anvil]

Oh lol yeah

You can install different keyboards on your phone

[cerulean siren]

you've been doing this from your phone ???

[latent anvil]

Yeah

I've been writing math on my phone for several years

[cerulean siren]

I guess if I can write code on my phone other people can write math 😆

[latent anvil]

I did get a small aws server so I could do homework on my phone a couple years ago

Coding on your phone is a great skill

[cerulean siren]

I just set up a desktop at home that's always on

and called it a 'server'

[latent anvil]

Ooh nice

[cerulean siren]

plus it's much faster than anything I'd be willing to rent on aws

[latent anvil]

Yeah I got the lowest tier

Lightsail or something

[cerulean siren]

is there an #offtopic channel somewhere?

[vestal snow]

I got a question about modules

So there's an example in dummit and foote which says that given a vector space V over F, the F[x] modules of V and the linear transformations of V form a bijection

[stone fulcrum]

Is F a field here? Or the usual ring?

Oop, you said vector space. Mb.

[vestal snow]

The way I interpreted it, is that given any module module action • we can find a transformation T such that p(x) • v = p(x) © v

Where I'm using © to represent the action induced by T

Is the statement correct?

[mild laurel]

Yes, but you have to be careful

As in, what is the transformation (T^2 + T + 1) for example

[vestal snow]

T(T(v)) + T(v) + v

[chilly ocean]

Are all vector spaces products of fields?

[mild laurel]

Yeah, thats the right idea

Okay well first, the product of two fields is not a field anymore

[vestal snow]

I think my book sweeps something under the rug

By abusing notation

It's dummit and foote btw

[mild laurel]

The abuse of notation is pretty common for this

[vestal snow]

I'll write down what I'm trying to say one sec

[stone fulcrum]

In a vector space,

• The vectors form an abelian group
• The scalars form a field
• There's a scalar multiplication that has a few other properties
  @chilly ocean

[chilly ocean]

yeah zoph u right the converse wouldnt be true I was wondering about that

thanks

Kanyex is there something obvious from the axioms I'm missing you mean to say ;-;

[mild laurel]

If you're asking if all vector spaces are isomorphic to F^n over F for some field F

then the answer is yes, if your vector space is finite dimensional

[chilly ocean]

pog

[mild laurel]

you should be able to show this

if your vector space is over F and has dimension n, then it is isomorphic to F^n

[chilly ocean]

that would be the proof true

[stone fulcrum]

Oh you are actually asking about a product of fields. Sorry, I misunderstood

[latent anvil]

wait is every vector space a product of Fs? That sounds fake

if we include the infinite case

isn't the infinite countable direct product much bigger than the direct sum?

I'm not being helpful, sorry

[chilly ocean]

zoph only said finite d but I'd like to see the counterexample you have too

for the infinite case

actually this would be cool to know about in general because the only infinite-d vector spaces I know are isomorphic to infinite products of fields

so I don't have a lot of variety ;-;

[latent anvil]

let V be a countable direct sum of Q with itself, as a Q-vector space. I claim this is not a product. For dimension reasons it's not a finite product. If it were an infinite product, it would contain a subspace isomorphic to the countable product of Q with itself. Call this space W. Then dim W <= dim V. In particular, W has a countable spanning set {w0,w1,...}. For each n, let Wn be the subset spanned by {w0,...,wn}. Then W is the union over all n of Wn, by how bases work, and each Wn is countable, so W is countable. However this is false, since we have an injection [0, 1) -> W by sending a number to its decimal expansion or w/e

the problem is that direct sums and direct products don't coincide in the infinite case

[latent anvil]

the infinite direct sum of a bunch of stuff consists of all tuples where only finitely many are nonzero

[vestal snow]

To show that 1 = 2, I need that k°v = k•v for all k in F and v in V

[gentle pendant]

And a word about the natural followup question, is every vector space a direct sum of copies of the field, yes if you assume the axiom of choice, undecidable without!

This is just a statement about the existence of a basis for an arbitrary vector field.

[latent anvil]

I think I used choice above too

Well for one thing I used dimension at all

I feel like you should be able to prove that a countable direct sum of Qs isn't a direct product of Qs without choice

Idk

[chilly ocean]

thanks yall :)

[gentle pendant]

yes, the former is countable, the latter is uncountable, no?

no need to talk about dimension I think.

[latent anvil]

Oh lol that's exactly what I did above yeah

I was just getting worried because I overcomplicated things

[gentle pendant]

lol

[latent anvil]

I don't need to reduce to the case of a countable direct product

I mean, you'd need to show it's not a finite direct product but then you can use dimension

[gentle pendant]

yeah, and no need to even formalise dimension for infinite dim vsps either, just mention existence of a finite spanning set, none can exist for the countable direct sum for obvious reasons.

[latent anvil]

Yep

[vestal snow]

Can one of you guys look at the image I sent earlier?

It's about modules and showing that there is a bijection between F[x] modules of V and the linear transformations of V

[latent anvil]

Yeah the thing you want holds by definition of your two operators, right?

[vestal snow]

It does?

[latent anvil]

Oh sorry I see

So you need to assume that the F[x] structure is compatible with the vector space structure

So the map F[x] × V -> V should be F-bilinear

Or rather the restriction to F×V should be the existing scalar multiplication map

[vestal snow]

Are we supposed to assume this or does it follow from the axioms of modules?

[latent anvil]

No you have to assume this

I mean, otherwise you just start with a vector space and then throw out all the vector space structure

Right?

Like, why assume it's a vector space at all

You could have two nonisomorphic vector spaces whose underlying sets are in bijection

And they would give a counterexample to this theorem

[vestal snow]

But you do use the vector space structure

When defining the action induced by the transformation T, you use scalar product with the coefficients of the polynomial

[latent anvil]

Right, so consider the following proposition

"Given a vector space V over F, the set of F-linear transformations V -> V and the set of F[x] module structures on the underlying set of V are in bijection"

Is that what you're trying to prove?

[latent anvil]

in the scratch paper you showed above, it looked like you started with both an F[x]-module structure and a vector space structure on V

[vestal snow]

Ah I think I see what you're trying to say

[latent anvil]

but in this page on dummit and foote, you start with an F[x] module structure and then construct both the linear transformation T and the vector space structure

I think this is an awkward way to present it tbh

I think of it as starting with a vector space structure on V and looking at F[x] module structure which extend that

but they're equivalent

neither is what you're doing though

if I understand correctly

[vestal snow]

So is it something like this

Given a F[x] module V, we can define operations to make V a vector space over F and define T a transformation on said vector space which induces the module action that we began with

[latent anvil]

I would say something slightly different

"Given a F[x] module V, we can define operations to make V a vector space over F and define T a transformation on said vector space"

and then "Given a vector space V over F and a transformation T on V, we can define a F[x] module structure on V"

and then that these processes are inverse

instead of trying to do it all at once

[vestal snow]

But this does not imply that the module structure at the end is the same as the one we started with

[latent anvil]

that's what I mean by "and then that these processes are inverse"

[vestal snow]

Oh okay

[latent anvil]

I'm just saying define both separately first

and then establish that they define a bijection

[vestal snow]

Okay so the idea is that one can define F vector space operations using F[x] module operations and vice versa

[latent anvil]

yup

[vestal snow]

Okay

Does the uniqueness follow as well?

[latent anvil]

if you can show that these are inverses, yeah

[vestal snow]

Thanks for the help

[latent anvil]

np

[chilly ocean]

wanna see a gross but funny abuse of notation 👀

F^n = n^F (caution, just a meme don't hate)
If F is a field and n is a natural number, specifically a von Neumann ordinal, the n-th dimensional vector space over F is isomorphic to the vector space of functions from n={0,1,...,n-1} to F.

[latent anvil]

But the set of functions from A to B is written B^A?

[chilly ocean]

oh fuck 😂

yeah I was wondering why it was the opposite of the cardinality for finite sets

the reason is its not and I just remembered it wrong lmaoo

too bad the notation is indistinguishable then so I can't make a shitty meme

[tribal pasture]

Why is R^n/IR^n considered an n dimensional vectorspace over R/IR? Is it because R^n/IR^n ≊ +^n R/IR and if I is the maximal ideal then IR is also the maximal ideal of R thus R/IR is a field.
Since for a given ring R, R^n is a module over itself we have that +^n R/IR is an n-dimensional VS over R/IR.

[mild laurel]

what does +^n R/IR mean

[tribal pasture]

The direct sum over R/IR n times: \oplus

[latent anvil]

Do you understand why it's considered an R/I-module?

For arbitrary I

[tribal pasture]

What is considered an R/I-module?

[latent anvil]

R^n/IR^n

[tribal pasture]

Well, because for a given ring K, K^n is a module over K. Set K = R/I = R/IR
Then we have that (R/IR)^n is a module over R/IR
Since R/IR is a field, we have that the former is VS

[latent anvil]

I mean you're identifying R^n/IR^n with (R/I)^n there, but yeah

And F^n is an n-dimensional F vector space

[tribal pasture]

Is that not correct since IR = I for a maximal ideal I?

yes yes

[latent anvil]

No. (R/I)^n is the set of n-tuples of cosets of elements of R by I, whereas R^n/IR^n is the set of cosets of tuples of elements of R^n by IR^n

Anyways, I'm not sure what your question is

[tribal pasture]

Well I am supposed to show that R^n/IR^n is an n dimensional vectorspace where I is a maximal ideal in a ring R

[latent anvil]

Didn't you do that above?

[tribal pasture]

Yeah that's what I was asking whether my explanation makes sense

[latent anvil]

Then yeah, R^n/IR^n is isomorphic to (R/I)^n as R/I-modules, with the obvious R/I-module structure on each. Since I is maximal, R/I is a field, so modules over it are just vector spaces

[tribal pasture]

If I want to lay out the isomorphism it would be this right,
R/IR ≊ R/I
Then
R^n/IR^n ≊⊕^n R/IR ≊ ⊕^n R/I ≊ (R/I)^n

[latent anvil]

IR = I, so the first thing is just an equality

[tribal pasture]

Oh yeah yeah! Sorry

[latent anvil]

Really only the second isomorphism has any content

The others are just equalities

[tribal pasture]

Yes yes. Thanks!

[blissful ice]

Hello!
I am wondering if anyone know the idea behind general tensor multiplication of modules

in specific

M and N are R-modules

and the _S is extension of scalars

so the first step is normal
write extension by definition
associativity of R tensor products

then fron 2nd to 3rd expression we again use definition of extension of scalars

to get

((M_S \otimes_S S)\otimes_R N)

[cloud walrus]

zavzav:

[blissful ice]

does anyone know a proof of why that is isomorphic to

,$ M_S\otimes_S(S\otimes_R N)

[cloud walrus]

zavzav:

[blissful ice]

I only know associativity for tensors over same space

[wind steeple]

take the bilinear product : (m (x) s, s' (x) n) -> ((m (x) s) (x) s') (x) n

[blissful ice]

hmmmm thank you I will think about it and come back when I write it all out

so using this bilinear and it generates a tensor product

[wind steeple]

it's probably a natural isomorphism, then check for natural morphism you can write

[uncut zenith]

I would like to understand these equations but there seems to be some abuse of notion I am not aware of.
I guess R is a subring of S and M and N are R-modules.
So then you have $M_S$ that is a S-module so $M_S \otimes_S S$ makes sense (by using S both for the ring and the S-module that it gives rise to). But then we should have $M_S \otimes_S S \simeq M_S$.
Then I cannot make sense of $(M_S \otimes_S S) \otimes_R N$
Since the left part is a S-module but $\otimes_R$ ask for a R-module.

[cloud walrus]

nicolas-blanco:

[oblique river]

if R is a subring of S, then any S module is canonically an R module by restricting scalars

[blissful ice]

M and N are R-modules by definition
M and N are S-modules by extension of scalars using phi: R->S

so the full question is

M extended to module over S tensor N extended to S
(M_S\otimes_S N_S)

[cloud walrus]

zavzav:

[blissful ice]

and you have to check if you get the same if you do M x N as modules over R first

and then extend scalars after

[uncut zenith]

Oh, of course! Thanks for the clarifications.

[blissful ice]

but R is not a subring of S here
just exists a map from R->S which we use

[oblique river]

oh ok

[blissful ice]

I found it in hungerford's algebra

[gilded trellis]

erm, so I'm a but confused about this equality here :

what exactly is the group $$ (Z^{X}_n, . ) $$

[cloud walrus]

Der Gegenstand ist einfach.:

[gilded trellis]

(everything else is clear btw)

[latent anvil]

It's the group they just defined, U(n)

[gilded trellis]

but what is Z^{x}_n ?

[latent anvil]

It's U(n)

They're two names for the same thing

[gilded trellis]

oh

I see

heh

I thought it's some other set that I'm supposed to know about

[latent anvil]

Maybe they had previously defined it?

[gilded trellis]

I don't know

[latent anvil]

What reference is this?

[gilded trellis]

mathonline.wikidot.com

http://mathonline.wikidot.com/

(Great reference by the way ! Clear and concise with plenty of examples)

(It is made up of someone's math notes from when they were studying maths... he compiled all his TeX notes into a website)

[latent anvil]

This is a weird site

Cool!

[gilded trellis]

[final gulch]

@gilded trellis the x denotes the group of units of a ring
so, it is the integers mod n that have a multiplicative inverse in Z_n

[latent anvil]

Does anybody have a good practice problem for irreducibility criterion stuff? I'm trying to write a problem set and accidentally reused "find all irreducible polynomials of degree <= 4 over F2"

Most eisenstein's criterion stuff is too easy

The one nontrivial example I know is the cyclotomic one, which is done in dummit and foote

[gilded trellis]

@final gulch heh, I haven't studied rings yet ... but I'll keep that in mind and go back to this when I do ! thanks 😊

[latent anvil]

This isn't helpful if they don't know what a ring is

[final gulch]

this is the definition from page 226 of Dummit & Foote

I did say what it is in the case of Z_n
I'm just explaining what the notation refers to and what keywords they could use to look it up

[gilded trellis]

oh, I was familiar with this notation : (G,x,+) (or smth like that)

(from what I've seen)

[cold flint]

@latent anvil there are some rather difficult irreducibility problems here: http://yufeizhao.com/olympiad/intpoly.pdf. You can also probably make some stuff up involving Hensel's lemma.

[latent anvil]

Oh I hadn't heard of hensel's lemma

Thanks!

[gilded trellis]

@final gulch I'll leave that for later ... heh

I'm still struggling with groups... erm

stuck at the coset section again

Artin's Corollary 6.15

here :

I got confused at 6.14

and I'm currently looking for another resource that proves 6.15 (and elucidates 6.14)

what should I google for ?

[latent anvil]

First isomorphism theorem

Well hang on

Do you know that the set of cosets forms a groups?

[gilded trellis]

yeah

[latent anvil]

Then yeah, first isomorphism theorem

[cold flint]

only when you take a normal subgroup!

[gilded trellis]

alright ! thanks again 😊 !

ah, 4c ... that might've slipped me

hehe

[cold flint]

However, the kernel of a homomorphism is always normal

[gilded trellis]

I get a lot of information overload whilst reading this so... It's hard to remember everything

[latent anvil]

Yeah, that's a good clarification

Are you doing exercises?

[gilded trellis]

but doing exercises would help me remember later when I finish reading

[latent anvil]

Usually when I feel like that it's because I'm rushing through without really digesting it

And forcing myself to do exercises helps

[gilded trellis]

I've only done some of the exercises for section 1 of chapter 2 : groups

(and almost all of the exercises for chapter 1 : matrices)

should I do the exercises per section ?

or per chapter ?

[latent anvil]

I'm not sure how your book is laid out

If there are exercises at the end of each section, do them before moving on to the next

[cold flint]

Is this Artin's algebra? Are you self-studying?

[gilded trellis]

there are exercises at the end of each chapter

Is this Artin's algebra? Are you self-studying?
@cold flint yeah ...

I asked for a resource on Group theory

and someone recommended Artin's algebra

so I went with it .. heh

(we were supposed to study it but schools are closed ...)

(and our maths teacher is basically sending us notes and exercises and nothing else..)

[latent anvil]

That's tough

[gilded trellis]

Although the exercises are split according to sections ...

[latent anvil]

Oh okay

Then do them

Before moving on

[gilded trellis]

:""""(

heh, I'll try to do that from now on

even though it is a pain and I feel more like reading on whilst I'm doing them ...

[cold flint]

You don't have to do all of them. I've used the book and I know there's a lot

[gilded trellis]

yep

[cold flint]

I think it's a very good book but very hard to self-study

[gilded trellis]

60+ on the section on matrices ...

actually, I've also been using harvard's lectures (old ones) because they use the same book

but I've gotten so confused by them starting from lecture 6 (and 5 a bit) that I just decided to drop them

[cold flint]

Joseph Gallian has a nice book which is probably a bit easier. Also, if you're interested in just reading a more casual, intuitive book on group theory, there's the book Visual Group Theory

[gilded trellis]

I thought about switching books

but I'm a bit worried about losing motivation and just dropping it all together ...

(happened to me before)

[latent anvil]

I wouldn't suggest it tbh. If you're mostly self studying you need to be consistent

[gilded trellis]

exactly !

[latent anvil]

It's easy to get wrapped up in this kind of stuff and never get off the ground

If you have more specific questions about coset stuff, we can try to help

[gilded trellis]

awww, thanks !!!

I'll be sure to ask here whenever I feel lost 😊

[chilly ocean]

I thought about switching books

gallian's text is good if you want handholding (but the majority of exercises aren't really helpful imo)

[gilded trellis]

Hmm, I'll keep that in mind :3 (btw, you can just @mention me heh)

[latent ingot]

Hello, I am stuck on this problem. I is a nilpotent ideal in R and M and N are R-modules (not finitely generated!). The right implication is trivial, but I get stuck on the other one.

I have tried writing out some definitions, but all I can get is that for all (m\in M) exists (n\in N) s.t. (m - \phi(n) \in IM), and I'm not sure how to proceed from this (or if I even can)

[cloud walrus]

Muf:

[cold flint]

So you know all $m$ is in the set $\phi(N) + IM$. Show it is in fact in the set $\phi(N)+I^2M$. Then keep going and use nilpotency...

[cloud walrus]

4c:

[latent ingot]

I see, I will try, thank you 😄

[uncut girder]

I dont understand what this is asking

how are the roots of x^p - x closed under addition?

[golden pasture]

cuz (a+b)^p=a^p+b^p in Fp(since it has characteristic p)

[mild laurel]

The roots of x^p - x are all contained in \overline{F}_p

so you just want to show that they're all actually in F_p

[vestal snow]

Should 22 have that R be commutative?

[solemn rain]

if G = p^a for a>=1 and prime p then G must have a nontrivial center

proof:

|G| = |Z(G)| + sum(|G|:C_G(g_i)) from i = 1 to n

where g_1 , g_2 , g_3 .. are representatives of conjugacy classes

p^a = |Z(G)| + sum(p^a:G_C(g_i)) from i =1 to n

p must divide sum(p^a:G_C(g_i))

and also p must divide |Z(G)| hence Z(G) is non trivial

i dont understand this proof

why must p divide them both?

[mild laurel]

Is there a way you can rewrite [G: C_G(g_i)]?

[solemn rain]

idk umm

|G|/|C_G(g_i))?

[mild laurel]

yeah, and what is |G|

[solemn rain]

p^a

for somea

oh okay lmao

yea yea

bad

and it must divide Z(G) cuz Z(G) is a subgroup or what?

or its bcause p divides p^a

so it must divide both the terms?

[mild laurel]

p divides p^a

[solemn rain]

yea

[mild laurel]

and it also divides the sum

[solemn rain]

so it must also

divide both

?

okay okay

i think i got it

tysm

[solemn rain]

i struggle with finding things directly

like i struggle with finding the center of some group

or the conjugacy classes

how would you do this in exercsies

for small groups mostly

ugh i struggle with 90% of problems is it just practice?

i get the content in the explanation but i just cant do problems

[jolly crypt]

one of my homework problems asks me to find the complete factorization of x^27 - x over F_3[x]

In class we've already proven a nice theorem which implies that it's essentially x(x-1)(x-1)(all the degree 3 irreducible polynomials)

But how to find the (all degree 3 irreducible polynomials)? There are 81 potential degree 3 polynomials which are irreducible. And trying to directly factorize x^27 - x hasn't proven very fruitful (you can easily get a product of 2 degree 12 polynomials, but stuck there).

[gilded trellis]

(I'm still a newbie to abstract algebra but I've found a similar question to yours in maths stack exchange : https://math.stackexchange.com/questions/1160050/to-factorize-x27-x-over-mathbb-f-3 )

(I don't even know what irreducible polynomials are and what they have to do with divisibility but .. you know... heh)

[somber bramble]

@jolly crypt there’s actually not as many candidates. first you can restrict yourself to monic polynomials since otherwise you can just divide by the leading factor anyway. that leaves just 27 polynomials. And then to find all the irreducible ones you can just find all the reducible ones (which is easier cause they must have a linear term so it’s just the set of multiples of an irreducible linear poly times a quadratic poly). and then take the complement

it’s still tedious but far from undoable at that point

(i don’t know if there’s a nicer approach)

oh yea the answer in the stackexchange thread would be a bit faster to work through I reckon

[brisk granite]

what is the notation for splitting field of some polynomial over some field K

Is there a way to write that or do I have to say it like I just did everytime

[somber bramble]

I’m not aware of a standard notation but I’d just write $\mathrm{sf}_K(f)$

[cloud walrus]

Sascha Baer:

[somber bramble]

leaving off the subscript if it’s clear

[gilded trellis]

the answers to this are (a) , (b) and (d) right ?

[upper pivot]

yep

[gilded trellis]

I've stumbled upon this document: https://sites.math.washington.edu/~colling/Teaching/402a08HW1sol.pdf which has a correction of that question and in which it is stated that (e) is a subgroup

which really threw me in for a loop

[upper pivot]

those matrices arent even in GL_2 (R) tho lol

[gilded trellis]

yep lol

detA = 0 here

[upper pivot]

and empty set cannot be a group, cause no identity

[gilded trellis]

and the empty set doesn't have the identity

yep lol

thanks ! heh

[upper pivot]

np

[gilded trellis]

btw, do you listen to anything whilst doing maths ?

I've been listening to some music but I've tired of it recently

and audiobooks ... are difficult because it's hard to focus on them

[upper pivot]

depends on how im feeling, sometimes i like musik while math, sometimes i like the quiet

[gilded trellis]

hmm, I see

thanks !

[final gulch]

music while math? more like music is math https://www.youtube.com/watch?v=F7bKe_Zgk4o

[stone fulcrum]

https://www.youtube.com/watch?v=F7bKe_Zgk4o

[golden pasture]

debatable but not here

[brisk granite]

Are both these answers correct?

these question don't really seem to have too much alg but it was in my alg book. let me know if I should post somewhere else

[vestal snow]

Maybe the analysis channel?

[final gulch]

I remember someone asking the same question maybe a week ago

with the same page from the same book

maybe you can search and find their discussion

[vestal snow]

Given a left (unital) module A (which is also a ring w/ identity 1_A) such that r•(ab) = (r•a)b = a(r•b) for all r in R, and a,b in A, prove that f: R -> A defined by f(r) = r•1_A is a homorphism making A an R-algebra

Is the question missing the hypothesis that R is commutative?

Because I don't think one can derive commutativity from the given hypothesis and an R-algebra is only defined when R is commutative

[brisk granite]

@final gulch where?

[final gulch]

I found it, but it was you lol

it was 8 days ago in #advanced-number-theory

[brisk granite]

oh yea lol

I didn't have a sol back then but now I do

[gilded trellis]

quick question that might seem stupid but I need to be sure of something :
If H is a cyclic subgroup of G generated by two elements a,b... then is $$ H = { a^{m}b^{n} and b^{m}a^{n} ; m,n \in \mathbb{Z} } $$ ?

[cloud walrus]

Der Gegenstand ist einfach.:

[chilly ocean]

then is H what

[gilded trellis]

?

[chilly ocean]

oh you mean is that how H looks like?

[gilded trellis]

yeah

in set builder

[chilly ocean]

a^m b^n yeah

thats fine

no need to add that 'and sth'

its already in the first one

I mean

[gilded trellis]

if m or n is negative ? or ..?

[chilly ocean]

if its cyclic its generated by one

cyclic means its generated by 1 element

so cant say cyclic generated by 2 elements I think

[gilded trellis]

okay, so just to avoid the XY Problem

erm

here is the original question :

wait,

oh

I phoqued up

😐

[chilly ocean]

😔

[gilded trellis]

the question didn't mention cyclic anywhere... :3

[chilly ocean]

yep thoght so

[gilded trellis]

it said, generated

so I immediately associated that with cyclic

... heh

[chilly ocean]

ok, then its set of all elements a^m b^n for m,n in Z

[gilded trellis]

(wt phoque is wrong with me)

ah, alright !

thanks 😊 !

[chilly ocean]

uwu

[gilded trellis]

uwuwuwu

(PS: the cat in your pfp is cute af)

[chilly ocean]

I agree

[gilded trellis]

hh

@chilly ocean hmm: this maths stack exchange thread's answer defines H as follows :

[chilly ocean]

oh yeah I mean its all combinations

[gilded trellis]

https://math.stackexchange.com/questions/1184981/if-g-is-generated-by-a-b-and-ab-ba-then-prove-g-is-abelian

yep it makes more sense that way

since we haven't proven the group is abelian (yet)

[chilly ocean]

I mean

[gilded trellis]

anyways, just wanted to let you know 😊

[chilly ocean]

yeah right

[gilded trellis]

?

just go on

heh

[chilly ocean]

no ure right

what I wrote kinda assumed its abelian

its just combiantions

[gilded trellis]

ah

(It's just that I sent the message as soon as you said "I mean" so I'm sorry for interrupting you. Gomennashai )

[chilly ocean]

lol its ok we cool mate

[gilded trellis]

kay uwu

[solemn rain]

Show that if the center of a group G is of index n in G, then every conjugacy class of G has
at most n elements.

proof : the number of elements in a conjugacy class is |G|/|C_G(g_i)) where g_i is a representative of the conjugacy class ( an element in it )

|G|/|Z(G)| = n

Z(G) is a subset of C_G(g_i) ( smaller in size ig lmao )

hence |G|/|C_G(g_i)) <= n

is that right?

( i made a mistake i thoguht Z(G) is bigger than C_G(g_i) )

[raw moth]

okay took me a little bit to see why the conjugacy class has size |G|/|C_G(g_i)| (it is probably a standard argument, ||conjugation action of G on itself||)

but yes that argument works

to show what you wanted

[vestal snow]

Let R be a commutative ring. Prove Hom(R,R) (over the ring R) is isomorphic to R as a ring

Are we supposed to assume that the left module action on R by R is just the ring multiplication?

[upper pivot]

thats what is usually assumed

[vestal snow]

Alright. Just wanted to clear that up

Is it possible to make R a module over R by defining the action as something else

I guess negative of the product might work

[latent anvil]

What kind of structure on R do you want to preserve?

[upper pivot]

yeah

R is an R-algebra

so any endomorphism really

[latent anvil]

Like, take any other module whose underlying set had the same cardinality

You can transport the structure of that

[vestal snow]

Yeah that makes sense

Yeah no that I think about it second question was pretty dumb

[latent anvil]

lol I didn't want to be rude

Scalar multiplication gives a map R×R -> R, which distributes over everything in the right way

But you already have such a map

[vestal snow]

Guess the lack of sleep is starting to become evident

[latent anvil]

Did you figure out your earlier questions about algebras?

[vestal snow]

Not really

I just assumed that the book forgot to mention that R is commutative

Though I would be really grateful if you could look at it and see if I'm right

[gilded trellis]

I'm not sure about how to proceed with my proof here :

Let V and H be two cyclic subgroups of G. |G| = n and |V| = |H| = r.
thus V = < a^{v} > and H = < a^{h} > for some v, h in Z and a in G. (The subgroups of a cyclic group are also cyclic).
we have :

$$
(a^{v})^{r} = 1 = (a^{h})^{r} \Rightarrow a^{r(v-h)} = 1
\Rightarrow v \equiv h \mod n
$$

[cloud walrus]

Der Gegenstand ist einfach.:

[gilded trellis]

Can I just conclude because we're working with a cyclic group that V and H are equal ?

(here is my proof for there being at least a cyclic group of order r :
let G = <a> be a cyclic group of order n
|G| = n
thus a^{n} = e the identity element of G.
we have r/n => there exists k of Z such that n = rk.
a^{n} = 1 => (a^{k})^{r} = 1. as such we can conclude that there exists H a cyclic subgroup of G of order r;
H = <a^{k}>
)

(currently doing exercises for the subgroup section so I can't use LaGrange's theorem or morphisms)

[chilly ocean]

@gilded trellis do you not have to prove that the subgroups are cyclic?

[gilded trellis]

no

that was already proven in the previous question

so I can just use that result

[chilly ocean]

Okay then do you not have to prove that if a has finite order n, then so a^s has order n/gcd(n,s)

[gilded trellis]

I've proven that above ... heh (I think that's a good proof for it)

(here is my proof for there being at least a cyclic group of order r :
let G = <a> be a cyclic group of order n
|G| = n
thus a^{n} = e the identity element of G.
we have r/n => there exists k of Z such that n = rk.
a^{n} = 1 => (a^{k})^{r} = 1. as such we can conclude that there exists H a cyclic subgroup of G of order r;
H = <a^{k}>
)
this..

[chilly ocean]

Can I just conclude because we're working with a cyclic group that V and H are equal ?
Why can you conclude that from the congruence relation?

I mean it is true, but something is missing in this proof

Because you have shown that a subgroup exists, though not this

Okay then do you not have to prove that if a has finite order n, then so a^s has order n/gcd(n,s)

That every element has this order

So that you can conclude this from the congruence

[gilded trellis]

ah, okay, thanks ! I'll look into proving that statement and I'll get back to you. (I have something else to do, so I'll get to it once I'm done with that heh...)

[tribal pasture]

Can somebody justify the “natural “ consideration

Secondly what do they mean by distinguishing N from the coefficients

[cold flint]

it just means you don't take coefficients in S for elements of N yet

What do you mean by the natural question

[tribal pasture]

I mean they say it's natural to consider the free Z-module, so , the collection of all finite commuting sums. I guess we need commuting to ensure the abelian structure in the image. Why the finiteness though? Is it so the sum remains well-defined (don't have to bother with convergence bs of the series)?
Also how does that distinguish N? I thought it had to do with the fact that writing sn as a formal product still leads me to have a copy of N inside S x N. But I wasn't sure

[cold flint]

you always impose finiteness; e.g. in a vector space. For instance Z has no infinite elements...

I think you're overthinking the word distinguish; it's just meant to restate the "no relations" phrase

[tribal pasture]

Oh and to get that part right as well, when they say no relations, they just mean there doesn't exist an s and n, such that sn = s_1n_1 +s_2n_2 right?

[cold flint]

not exactly. It's just literally S x N. If you know what the direct product is then just consider it as an abelian group (i.e. Z-module) and you can forget whatever's confusing you.

[tribal pasture]

Oh okay. I understand now. Thanks.
One more thing: if an author says the module S ⊗ N is generated as an S-module by elements of the form 1 ⊗n, what they mean is that for any element s ⊗n, we have s ⊗n =s(1 ⊗ n). Correct?

[cold flint]

That is indeed true but not exactly equivalent to that statement

[toxic zephyr]

I'm having a bit of trouble understanding normal subgroups. Does anyone have any examples of them that might help me understand their purpose/use?

If I understand correctly, every element in a normal subgroup can be written as a conjugate by an element in the parent group?

[mild laurel]

Another alternative definition is that normal subgroups are exactly the kernels of group homomorphisms

[ripe crest]

And normal subgroups are the only subgroups you can mod out by

[cold flint]

If you conjugate a normal subgroup (element-wise) by any element, you end up getting the same group. Conjugate subgroups can be considered "the same" in some sense; from this perspective normal subgroups are unique subgroups.

[toxic zephyr]

@cold flint So if I have something in a normal subgroup, it can be written as conjugate by an element in the parent group and stay in the subgroup. If I conjugate the element, it can still be written as a conjugate by an element in the parent group additionally and stay in the subgroup. So essentially I can conjugate elements in a normal subgroup by an element in the parent group as many times as I please and it stays in the subgroup?

@mild laurel So for any homomorphism, the kernel is a normal subgroup? As in, elements which get mapped to the identity can always be written as a conjugate by an element in the original group?

@ripe crest That's the next section in my text, so I assume it'll make more sense after I finish.

[solemn rain]

a normal subgroup is the kernel of some homomorphism

to make quotient groups you need normality

you need normality so that the operation defined on the quotient space is well defined

quotient groups are helpful to study groups ig

[cold flint]

@toxic zephyr exactly. And saying the kernel is normal means if you take all those elements mapping to zero and conjugate them, they will still map to 0.

This is pretty easy to see, since if $f(g)=1$, then $f(h^{-1}gh)=f(h^{-1})\cdot 1 \cdot f(h)=f(1)=1$ (writing identity = 1).

[cloud walrus]

4c:

[toxic zephyr]

Oh of course! That makes so much sense. Thank you!

[cold flint]

Yep! This also shows why you can only mod out by a normal subgroup if you want to get a quotient group, because if $G/H$ is a group, then $H$ is the kernel of the natural projection $G\rightarrow G/H$.

[cloud walrus]

4c:

[cold flint]

Or, if you want a more concrete argument, if $H$ isn't normal, i.e. $gHg^{-1}$ contains $a\not\in H$, then if you want to define a group action on cosets as $g_1H \cdot g_2H = g_1g_2H$, then you will fail as $gH \cdot g^{-1}H$ contains $aH$ and thus $a$, but it's supposed to just be $H$ which doesn't include $a$.

[cloud walrus]

4c:

[solemn rain]

if i have 2 permutations in S_n for some n ( x and y )

and both are conjugate

x = zyz^-1 for some permutation z

then both have same order?

and ig if both are conjugate

then they both generate the same orbit

am i right?

me new sorry

[latent anvil]

Any two conjugates in any group have the same order

Conjugation by a fixed element is an automorphism and automorphisms preserve order

It's not true that they'll have the same orbit. Let x = (12) and y = (23). Then (132)x(132)^(-1) = (132)(12)(123) = (23) = y, but x and y definitely don't have the same orbit

[solemn rain]

idk what automorphisms yet

okay

i got the second paragraph

i am abit fuzzy with classes and group actions in general

okay well

what wanted to get at

was pctually proving htis

Any two conjugates in any group have the same order

and my argument was like

using a cool theorem that i learnt ( and never saw its use )

that every group is isomorphic to a subgroup of S_n

[fading wagon]

@solemn rain actually, you don't need that, let them be a, b with orders n, m respectively, WLOG $n\leq m$. Then $a=cbc^{-1}$, so $e=a^n=cb^nc^{-1}$, so it follows that $b^n=e$, hence $n=m$ and we are done.

[cloud walrus]

Element118:

[solemn rain]

oh

(xbx^-1)^n = xb^nx^-1

u shoudl ahve said that cuz im slow XD

but here is the important part

if cl(a) = cl(b)

where cl(a) = {gag^-1 | for all g is in G }

why is it that a and b are conjugate

[fading wagon]

if a, b are in the same conjugacy class, then they are conjugate

@solemn rain what's your definition of conjugate?

[solemn rain]

a and b are called conjugate if there exists a g in G such that a=gbg^-1

[fading wagon]

a is in cl(a) (by let g=e), so a is in cl(b), so...

@solemn rain

[solemn rain]

why is a in cl(b)

lmal

oh cuz cl(a) = cl(b)

yea got it

sorry im very slow

[thick geyser]

Hey guys, you can only partition non-empty sets right?

[sharp sonnet]

the empty set is a partition of the empty set

[thick geyser]

Is it the only partition

[chilly ocean]

yes

[thick geyser]

I thought partitions were only defined on non-empty sets

[scarlet estuary]

depedns on definition i suppose, but i've always seen it defined such that the only partition of {} is {}

wikipedia agrees with me, and doesnt mention any alternative notion

[thick geyser]

What about an equivalence relation

Can you define an equivalencr relation on the empty set

You have to right?

Because every partition defines an equivalence relation

[sharp sonnet]

yes, the empty relation

(notice that the empty relation is vacuously transitive and symmetric, but only reflexive if the underlying set is the empty set)

[vital bluff]

I'm trying to prove using PMI that sigma from i=1 to n of (3i-2) equals (n/2)(3n-1) for all natural numbers n.
In the example my professor gave there was a series written out (1 + 3 + ... + (2n-1) = n^2)
And maybe it's just the notation that is throwing me off, but I'm really stumped with what to do

[sharp sonnet]

this is more #proofs-and-logic or maybe #discrete-math

[vital bluff]

Oh sorry, i'm in intro to abstraction i thought this was the right channel

i'll move

[tribal pasture]

Oh okay. I understand now. Thanks.
One more thing: if an author says the module S ⊗ N is generated as an S-module by elements of the form 1 ⊗n, what they mean is that for any element s ⊗n, we have s ⊗n =s(1 ⊗ n). Correct?

That is indeed true but not exactly equivalent to that statement
@cold flint Can you expand on what it would be equivalent to ?

Or saying generates here just is equivalent to saying that a module is finitely generated over the ring S?

[hot lake]

it also mean that every element of $S \otimes N$ is a finite sum of elements of the form $s \otimes n$

[cloud walrus]

Zef Klop 🍃 🌿 🌻:

[tribal pasture]

You mean 1 ⊗ n, correct?

[hot lake]

no I mean $s \otimes n$, because as you noted, $s.(1 \otimes n) = s \otimes n$

[cloud walrus]

Zef Klop 🍃 🌿 🌻:

[tribal pasture]

Oh it also mean that, gotcha

[leaden finch]

can someone help me with this one

part c

i did this but idk if its right

[chilly ocean]

sure, but -3=2 in Z_5 so that would make it easier

[leaden finch]

ohh so do f (2)

[oblique river]

what does the remainder theorem say

[oblique river]

no, I already know the answer

I want you to tell me what it says

not just copy paste from the book

[leaden finch]

ohh, hmm i have trouble explaining it in my words D:

[oblique river]

in your solution, what are you testing for

like, what about the answer to that calculation would tell you "yes" or "no"

for the question

[maiden ocean]

hmm

I'm not really sure how id go about this proof

"show that if an element of a ring has more than one right inverse then it has infinitely many"

i get that im probably supposed to do like, ua = ub = 1 or something, and then try to construct a new one out of a and b maybe?

but im not rly sure how id go about that

[tribal pasture]

Maybe note that u(aub) = (ua)(ub) = 1
And keep generating infinitely many inverses this way

In other words, if x,y are right inverses, then so is xuy

[maiden ocean]

aub = a though

thats not distinct

[tribal pasture]

Hmm what about u(a+a-b)? Which exists since (R,+) is an abelian group?

[maiden ocean]

i mean, sure it exists but

how do we know its an inverse

:c

or

wait

ua + ua - ub

[tribal pasture]

yes

[maiden ocean]

ok that works

h m m

ok so ig from here u just need to generalize a bit

[tribal pasture]

so na - (n-1) b

[maiden ocean]

ye

ok thanks

that works

[tribal pasture]

np

[maiden ocean]

oh wait

@tribal pasture what happens if your group is finite though

then theres a finite # of n for which na and (n-1)b are distinct

great does this mean i have to prove that if u is in a ring w/ two distinct right inverses the ring is infinite ._.

[tribal pasture]

I don't see any reason why would it hold true in that case, because the right inverse must exist in your group which is finite. So you already have an upper bound

ye probs

Goodluck

[maiden ocean]

yike

[tribal pasture]

I think the contrapositive would be easier. Ring is finite imply that [ ua = ub =1 => a=b]

[maiden ocean]

[peak jewel]

If my axes are (x=w, y=dB), and I have a point at (w=195,dB=19.4) how can I figure out when a line with a slope of +20dB/decade will cross y-axis at (w=1)?

[scarlet estuary]

this is not abstract algebra

see #prealg-and-algebra or a generic questions channel

[peak jewel]

Got you, thanks

[ripe crest]

all algebra is abstract

[solemn rain]

idk if i messed up the algebra please help

let G be a group

if phi is in Aut(G) and f_g is conjugation by g prove phi(f_g)(phi)^-1 = f_phi(g)

proof:

f_g = gxg^-1 for x in G

phi(gxg^-1)phi^-1 = phi(g)phi(x)phi(g)^-1phi = phi(g)phi(x)phi^-1phi(g) = phi(g)phi(x)g = phi(g)phi(x)phi(g)^-1 = f_phi(g) for some phi(x) in G ( surjective )

is that right?

iirc phi should map an element to its inverse

[hot lake]

phi(f_g) ?

[solemn rain]

what

[hot lake]

if phi is in Aut(G) and f_g is conjugation by g prove phi(f_g)(phi)^-1 = f_phi(g)

wtf is phi(f_g) ?

[solemn rain]

times ig

but ig times is composition right?

@hot lake

exercise 1

[hot lake]

ah yes

[solemn rain]

correct?

[hot lake]

f_g = gxg^-1 for x in G

no

[solemn rain]

f_g : a--> gag^-11?

[hot lake]

yes

[solemn rain]

okayy

so f_g(phi^-1) = gphi^-1g^-1

phi(f_g)(phi^-1) = phi(g)gphi(g^-1)

[hot lake]

no

just

[solemn rain]

oh my god

[hot lake]

just no

[solemn rain]

wtf is wrong

[stone fulcrum]

phi is an element
f_g is a function

[hot lake]

please never write fg(something) or phi(something) where something is not an element of G

phi is not an element, it's an automorphism of G

[stone fulcrum]

Aight I'mma head out

Oop

[hot lake]

fg is also an automorphism of G

[solemn rain]

okay

[hot lake]

you can compose them

usually by writing a ° or a \circ

between them

but you can't apply one to the other

that makes no sense

[solemn rain]

okay so

i wanan prove phi f_g phi^-1 is f_phi(g)

[hot lake]

you want to prove phi ° f_g ° phi^-1 is f_phi(g)

[solemn rain]

yea

now f_g sends an element a in G to gag^-1

[hot lake]

okay ?

[solemn rain]

so phi ° f_g ° phi^-1 = phi(gag^-1)°phi^-1

right?

[hot lake]

no ??

[solemn rain]

xd

why

[hot lake]

because phi ° f_g ° phi^-1 is independant of a

and phi(gag^-1)°phi^-1 depends on a

[solemn rain]

okay

how can i write f_g

then

[hot lake]

so the thing on the left is one automorphism

and the thing on the irght is god knows what

well

if you want to prove that two automorphisms F1 and F2 are equal

the only thing you need to do

is to show that forall g in G, F1(g) = F2(g)

[solemn rain]

f_g is an automorphism on G

right?

[hot lake]

okay bad letter choice from me

is to show that forall a in G, F1(a) = F2(a)

g is already taken

you want to prove that phi ° f_g ° phi^-1 is f_phi(g)

so you have to show that forall a in G, ( phi ° f_g ° phi^-1 ) (a) = ( f_{phi(g)}) (a)

[solemn rain]

yea okay

i always have problem with this

knowing what this means

when there is literally no operation

[hot lake]

well you are told what f_g does

and composition is composition

[solemn rain]

is f_g ° phi^-1 f_g(phi^-1)?

phi^-1 isnt in G

so i cant say that

how am i going to do itt then XDDD

[hot lake]

(f ° g) (a) is f(g(a)), that's the definition of composition

[solemn rain]

yeae

[hot lake]

so you have to show that forall a in G, phi ( f_g ( phi^-1 (a))) = f_{phi(g)} (a)

now you use the definition of f_g and f_{phi(g)} since it's the only thing you can do

[solemn rain]

f_g(phi^-1(a)) = gphi^-1(a)g^-1

phi(g)aphi(g^-1) = f_phi(g)

got it

tysm

[hot lake]

yes

[solemn rain]

so

fgh

is f(g(h(x)))

?

[hot lake]

no

(f ° g ° h) (x) is f(g(h(x)))

[solemn rain]

whats fgh

like in the exercise

its just written phif_gphi^-1

what does this mean then

[hot lake]

they omit the °

because it's clear from context

that the operation there is composition

[solemn rain]

yea thats what i did?

'is f(g(h(x)))'?

[stone fulcrum]

I may not know the notation. What does φ_σ(g) represent?

[solemn rain]

where is the input then

conjugation

by phi(g)

@stone fulcrum

in G

[hot lake]

the input is x

[stone fulcrum]

Oh I see

[solemn rain]

okay then why

no
@hot lake

[hot lake]

I probably said no because one side was an automorphism of G with no input given and the other side was an element of G

fgh is an automorphism of G with no input given

[solemn rain]

i alwaays hve this fuckign prob;em

when dealign with permutaion groups too

it just says phi XD

no input nothing

and i get confused

[hot lake]

f(g(h(x))) is the element of G that you obtain when you take x, apply h to it, take the result, apply g to it, take the result, and finally apply f to it

[solemn rain]

okay

okay

ty

another problem just wanan check

if G is an abelian group of order pq,where p and q are distinct primes, then G is cyclic

proof: suppose G is an abelian group of order pq where p and q are distinct primes

p | |G| and q | |G|

by cauchy there exists elements x and y such that x^p = y^q = 1

consier (xy)^pq

(xy)^pq = (x^pq)(y^pq) ( abelian )

= 1

hence G = <xy>

correct?

[hot lake]

no o..o

[solemn rain]

ph my god