groups-rings-fields

[coral shale]

high level?

[winter shore]

it makes sense that groups act on sets since groups are group objects in Set but the category of abelian groups doesn't seem very deeply related to Rings

I mean, what's special about abelian groups that make it the right category for rings to act on

[dull ginkgo]

“Acts on” like a group action sense but it also distributes

[coral shale]

I feel like this way of thinking is backwards

[dull ginkgo]

I’m pretty sure there’s “group modules” where groups act on another groups in a same vain

[coral shale]

I would say rings are defined as such because...

[rocky cloak]

Rings are monoids in the category of abelian groups, so if that's your justification...

[winter shore]

I did not know that!

then yes, that'd be a pretty convincing justification for me

thank you

[rocky cloak]

But I think a better explanation is just that a ring has two operations. The multiplication becomes the composition of actions, but then addition should correspond to something

[dull ginkgo]

Currying moment

[dull ginkgo]

Didn’t know there’s a notion of a “monoid in a category”

[winter shore]

but even then at this level, if you take multiplication to be the composition of actions you're already working with monoid objects :p

[dull ginkgo]

Homie’s going to be mortified by C* algebras

[rocky cloak]

Really it just comes down to End(A) being a ring when A is an abelian group and a monoid when A is a set

[dull ginkgo]

I can’t even think of how many fucking layers of binary operators are in one of those fuckers

[rocky cloak]

Thus it makes sense to define actions of rings and monoids on abelian groups and sets respectively.

And group action is just a special case

[winter shore]

good point, though we would also like it to be universal somehow? I.e. that rings act on a category C such that End(A) is a ring for each A in C. Not sure what the universal property should be, I'll be thinking about that

[rocky cloak]

I'm not sure what kind of property your looking for. An X-action of R on A is just an X homomorphism R -> End(A). So this makes sense whenever End(A) has the necessary structure to define an X homomorphism.

I guess you can also define actions internally. In which case a ring action is just a monoid action in the category of abelian groups

Though I don't really see any universal property in sight

[crystal vale]

And there are only 3 maximal ideal in Z/2024Z ?

[south patrol]

Sure

It's a product of three rings, each of which has a unique maximal ideal

[winter shore]

Not quite what I was looking for but I got: There is a forgetful functor F: Grp -> Sets such that for every group G, there is a canonical action f: G -> End(|G|) such that any other action G -> End(X) factors through it. Similarly with the forgetful functor Ring -> Ab

this doesn't really characterize Sets or Ab though but eh good enough

the fact that the tensor product does not appear in my description above further indicates that I'm not using all the data required

[rocky cloak]

Guess it appears indirectly in that when you think of End(A) as an abelian group you're using the monoidal structure of Ab

[winter shore]

oh right mb

[winter shore]

yeah I think this is as good as it gets, with no universal property in sight

[delicate orchid]

Z[[x]]^3

[boreal inlet]

Is x^2 + 1 irreducible in F_49[x]?

[south patrol]

Mods...

[rocky cloak]

Triple the power, triple the ideals! Buy the new power series today!

[south patrol]

Does Z[[x]] even have a unique maximal ideal

[delicate orchid]

no

[south patrol]

don't (p,x) work

[delicate orchid]

it's only the power series over local fields that are local

[coral spindle]

When would it be irreducible?

[south patrol]

ok

[coral spindle]

Or rather, when would it be reducible? What would that mean, more concretely.

[winter shore]

this is wrong btw, got it the other way around. It should be: For every set X there is a canonical action S_|X| -> End(X)

[boreal inlet]

Never mind, it is.

As x^2 + 1 is a degree 2 polynomial, it is reducible iff it has a root in F_49. I couldn't find a root, so I thought I was tripping and ran a simple python code, and none of the elements in F_49 returns 0 by evaluation on x^2 + 1

and lmao funny story this was in a graded assignment and i basically somehow proved that for a finite field F with characteristic p > 2, x^2 + 1 has a root in F iff |F| = 1 mod 4

Only after 2 hours after deadline was passed i got this

[winter shore]

huh I'm confused, then what's the root?

[nova holly]

https://en.wikipedia.org/wiki/Quadratic_reciprocity#Supplements_to_Quadratic_Reciprocity you might be interested in that!

[rocky cloak]

Might be a bug in your code then

[rocky cloak]

One construction is F49 = F7[x]/(x^2 + 1), so then what the root is is pretty easy. With another construction it would be a bit more work to find it explicitly.

[crystal vale]

Okay thank you

[boreal inlet]

yeah.... Let me actually check real proper

.....ok anyways me idot. F_49 is not Z/49Z. It's Z/7Z x Z/7Z, with multiplicative group as Z/48Z.

[south patrol]

I would be careful saying "it's Z/7 x Z/7"

only as an abelian group

[boreal inlet]

Yes, as a module

[south patrol]

But yeah you can just check that Z/48 has a primitive fourth root of unity

[boreal inlet]

It does, it's 12 right

[south patrol]

yeah that's one of them

Anyway one nice way to do it which is basically what jagr is saying is like

[boreal inlet]

Then i is a root

i is adjoined in there

[south patrol]

Yeah what I was gonna say was that you can just consider the splitting field over F_7

Like it's a degree 2 extension i.e. F_49

[coral spindle]

It's a bit misleading to call it i. There are no complex numbers in this field. There is a primitive 4th root of unity but it's not the same one as in C. Perhaps pedantic but it's worth mentioning this, for instance if you try and find 'i' in a field of characteristic 2 you run into problems.

[boreal inlet]

I'm actually wondering how the root is actually like

[coral spindle]

Well construct F_49 as F_7[x]/(x^2+1) and it's just x + (x^2+1)

[boreal inlet]

lmao

[coral spindle]

Letting \bar F_7 be the algebraic closure of 7, it's a fact that the group of units of this field is isomorphic to (Q/Z)_{7'}, which are the 7-regular elements of Q/Z. This means that these are the elements of Q/Z whose order is coprime to 7.

So you can picture it in that way.

It's not easy

[boreal inlet]

holy shit.

Isn't Q/Z is like homeomorphic to [0,1) or something

[coral spindle]

No

[boreal inlet]

Oh no

Yeah no

[coral spindle]

Not even R/Z is homeomorphic to [0, 1)

It's a circle. Picture a circle.

[boreal inlet]

This one's a circle i think

yeah

What I get is Q/Z is like an infinite group with each element with finite order

Because (a/b)^b = 0

Where 0 is the identity

[coral spindle]

Yup

[boreal inlet]

What is this even 💀

[delicate orchid]

not being an integral domain is a problem
skill issue

[coral spindle]

Wew I am going to send you a bouquet of roses. Now tell me your home address

[coral shale]

:fakenews:

[sonic coral]

this works, right?

[delicate orchid]

yeah I buy it

[sonic coral]

i just didn’t know if it was more proper to say it was irreducible in Z_3 so it’s irreducible in Q

[delicate orchid]

I mean, equating it to the 8th cyclotomic polynomial is very extra but it still works

[sonic coral]

one day he asked us why the 5th one was irreducible without acknowledging it was the 5th one so i felt that i should show him i’m acknowledging them more

[coral shale]

Id state the reason.

or is this one of the ones with no name

[delicate orchid]

I know the irred over Z => irred over Q as Gauss' lemma but it's usually just a corollary of the big boy Gauss' lemma

[hidden wind]

i am very much enjoying how this book takes as a starting point, the fact that group theory originated in the study of equations (solvability of the quintic)

the introduction is a great, short read

[celest cairn]

For anyone wanting to check, does this look good? My tutor wanted me to find the order of both of those Galois groups, and I presume he wanted me to determine them using the Galois Correspondence.

He gave me a problem earlier where I had to find the Galois group of a primitive 3rd root of unity, which i’d already established, so I simply just said that G is isomorphic to S_3 for simplicity.

[sonic coral]

wait does having no roots in mod 3 even help since it’s a quartic, so roots doesn’t imply reducible?

[delicate bloom]

they're relying on x^4+1 being irreducible in Z/3Z which is not the same thing as having no roots mod 3. So technically maybe they should be proving that as well instead of assuming it.

[sonic coral]

hm so i should really be assuming it factors in some way and getting a contradiction

like with the coefficients not being in Q

[delicate bloom]

maybe, I'm thinking probably there's a trick using the fact that it's a quadratic in y=x^2

[olive granite]

Also x^4+1 should be reducible mod 3

It's (x^2+x-1)(x^2-x-1)

[rocky cloak]

It might be better to just show directly

(x2 + ax + b)(x2 + cx + d) =
x4 + (a+c)x3 + (ac+d+b)x2 + (ad+bc)x + bd

Means a = -c, b=d=±1, (ac+d+b) = -a^2 ±2 = 6. Which is impossible.

[sonic coral]

thank you. i knew i could take this approach but I was trying to get around a direct factorization like this.

[sonic coral]

thank you. i knew i could take this approach but I was trying to get around a direct factorization like this.

[cloud solar]

Can someone explain me something

I have the polynomial f=X^p - X^(p-1) + a_(p-2) X^(p-2) +...+ a_1 X +1 in Z[X]

With p prime and p divides a_1,a_2,...,a_(p-2)

And I want to show f irreductible in Q[X]

And i took the residue polynomial of f in Z_p[X]

Which is g=X^p - X^(p-1) +1

and the artin-scheier polynomyal h=X^p - X +1 is irreductible over Z_p[X]

And here I got stuck

And in the solution they say g=X^p h(1/X) is irreductible over Z_p[X]

Can we do this thing with 1/x ?

And can someone explain why we do this

Is there a theorem that says we can use this even if h(1/x) is not a polynomial

[boreal inlet]

If f is irreducible in Z[x], by Gauss lemma you should be done, as the polynomial is already primitive (that is, the gcd of coefficients is 1)

[cloud solar]

I know that we can use the Gauss lemma this is why I took the residue polynomial over Z_p

To prove f is irreductible over Z

But i dont understand the part with h(1/x)

[south patrol]

The thing is the X^p

Basically it is like

Just reordering the coefficients

(You'll see what I mean)

[cobalt heath]

Hmmmm

[south patrol]

And you can check that a polynomial is irreducible iff it is irreducible after reordering coefficients like that

[cobalt heath]

This is harder than I thought

[cloud solar]

It is so akward to write g as a product of two things and one thing is not a polynomial

[cobalt heath]

But yeah, it induces ring homomorphism (isomorphism?)

[south patrol]

Yeah it is a little

[boreal inlet]

Wait, what does that mean

[south patrol]

Wdym

[boreal inlet]

I don't exactly understand what you meant by reordering coefficients. In what particular way, or is it just any permutation

[south patrol]

I don't think so

[cobalt heath]

Oh?

[south patrol]

No I mean that specific way

[cobalt heath]

Ah, it is only monoid homomorphism

[south patrol]

like f(x) -> x^(deg f) * f(1/x)

[cobalt heath]

Wait, maybe not?

[boreal inlet]

Ooh

It's completely reversed

[cobalt heath]

Hmm, why?

[south patrol]

As written, it sends x to 1 and 1 to 1

I think

[cobalt heath]

Ah, so it cannot be ring homomorphism

[south patrol]

Well okay that shows not an auto

[cloud solar]

So this is just reversing the coefficients

[south patrol]

Yeah basically

[cobalt heath]

So we are stuck with monoid homomorphism..

[south patrol]

It's enough for the thing I said

I've not thought about whether it gives an actual auto

[cloud solar]

And how this proves the "reversed polynomial" is irreductible because h is irreductible?

[south patrol]

Well if you let f* be the reserved poly then like

if f = gh then f* = g* h*

and vice versa

And then you can check stuff with degrees etc

[boreal inlet]

I don't think the direction "reversed is irreducible so the original is also" is true

It only works for one direction

[cobalt heath]

Hmmm, x^n pose an issue

The problem here is due to the failure of bijection, I guess

So basically, it should go like:
If f is irreducible, then f* is irreducible.

[cloud solar]

Yeah but h(1/x) is not a polynomial.. ugh... so g was the polynomial we wanted to prove is irreductible we wrote g as X^p×h(1/x) and h is irreductible. But X^p×h(1/x) is g* so g=g* in this case. But how h irreductible means g*/g is irreductible... I understood this trick is to reverse the polynomial but I dont understand how h(x) irreductible implies g/g* irreductible

[cobalt heath]

What is g and g*?

[cloud solar]

g=X^p-X+1

And g* is g with reversed coeff

[cobalt heath]

Ahh

[cloud solar]

And in this case g=g*

[cobalt heath]

Then since g is irresucible, g* is irreducible

[cobalt heath]

Doubt that

[cloud solar]

and of course g=g*=X^p × h(1/X) where h(X)=X^p-X+1 which is irreductible but how this implies g* or g is irreductible becuase h(1/X) is not a polynomial if it was let's say something like g=X^p × h(x) I would say yes it is pretty clear why g is irreductible

[cloud solar]

How g is irreductible this is what I dont understand :)))...I feel so dumb

[cobalt heath]

This

So h = g

[cloud solar]

Sorry g=x^p-x^(p-1)+1

I wrote it wrong

[cobalt heath]

Ah, so h = g*

[cloud solar]

How

[cobalt heath]

Huh

[cloud solar]

h(x)=g(1/x)×(1/x^p)

And g=g*

In this case

[cobalt heath]

It is x^p * g(1/x)

Did you compute it?

[cloud solar]

Yes sorry

[cloud solar]

Ok and we know that is irreductible

How this implies g irreductible

[cobalt heath]

This

[cloud solar]

Yeah but h(1/x) is not a polynomial

[cloud solar]

K finite field with q elements show that if q = 1 (mod 4) then f=X^4 + 4 has 4 roots in K

Clearly q=p^n with p prime

4 divides q-1

K* cyclic so there is unique H subgroup of K* with 4 elements

p≠2 so if z is a root of f then -z is also a root of f with z≠-z

Any hint

[terse crystal]

I can help half of it…
You only need to consider cases when p=1 mod 4, K=Fp and p=3 mod 4, K=Fp^2=Fp[x]/(x^2+1)
(recall that Legendre symbol (-1/p)=1 when p=1 mod 4, =-1 when p=3 mod 4). So both cases you have i from K such that i^2=-1. So if you can find a root z in K as you said, you would have 4 roots z, iz, -z, -iz

But I don’t know how to find one root z either….

[cloud solar]

interesting

Um I have a question

If f=X^(p-1)+...X+1 is the cyclotomic polynomial with p prime we know this is irreductible in Q[X] but then f(X^n) is irreductible in Q[X] for every n?

[terse crystal]

1+x^2+x^4=(x^2+1+x)(x^2+1-x)

[cloud solar]

Yeah

[cloud solar]

Im still struggling with this

[rocky cloak]

It might help to think about what the complex roots of this polynomial is

[cloud solar]

So I should find an element z in K* s.t. z^2=-1

[rocky cloak]

Yeah, or at least determining if such an element exists. But do you see what the complex roots are?

[cloud solar]

-1-i, -1+i, 1-i, 1+i

[cloud solar]

It has something to do with legendre symbol?

[rocky cloak]

Maybe you can do something with the legendre symbol, but there is a very simple group theory solution.

[rocky cloak]

So then you see that x^4 + 4 has roots iff x^2 + 1 has roots

[cloud solar]

Yes

It has something to do with the unique subgroup of K* with 4 elements

[rocky cloak]

Indeed

[south patrol]

Real

[cloud solar]

I feel so dumb

We dont have injectivity

I would say the element must be a where a generates that subgroup

[rocky cloak]

That's right, if x has order 4, then x^2 has order 2, thus equals -1

[cloud solar]

But how this implies x^2 = -1

Oh

Im idiot

Yeah

The only element with order 2 is -1

[cloud solar]

How can i prove the polynomial x^8 - 16 has at least one root in K

If q the order of K is =0,1,2 (mod 4) we are done

If q=3(mod 4)

Hm

X^8 - 16 = (X^2-2)(X^2+2)(X^4 + 4)

So this has roots if x^2-2 has roots or x^2+2 has roots or x^2+1 has roots

[formal ermine]

whats K

[cloud solar]

Finite field with q elements

[formal ermine]

i cant think of the elementary solution rn

[crystal vale]

I have a doubt there is a question which states that if V is infinite dimensional, then there always exists at least one linear independent sequence V1,V2,......, Vn for all positive integers n.

My friend said that if we take contrapositive statement such that there exist an integer say m such that any sequence u1,u2,.....,um are linear dependent, then he said that since it is linear dependent hence there exist non-zero u_k then he showed that we can write v_k in terms of remaining ones, and this v_k is arbitrary because we take any sequence of length m.

Now my doubt is, how can I show that for any arbitrary vector v there always a linear dependent sequence of length m where coefficient of v is non -zero.

[formal ermine]

What are you doubting?

[nova holly]

What does v coefficient mean in a linear dependent sequence? Is v,v,...,v m times a linear dependent sequence of length m with non-zero v coefficient?

[cloud solar]

I need some element like sqrt(2)

And i dont think the cyclic subgroup of order 4k+2 has some particular magic properties

[formal ermine]

Hm lets see

do u know galois theory

[cloud solar]

Nope... I should start learning some galois theory I think it might help for my olympiad

[crystal vale]

Last paragraph

[crystal vale]

Means I want to show that for any arbitrary vector v there is list of m-1 vector which span v

[nova holly]

what does spanning a vector mean

and what about the second question in my first message

[formal ermine]

thats not a doubt, thats a question

[dull marsh]

Not necessarily, consider Z3

[crystal vale]

Means there is some finite vector V1,......,vt such that I can write v=a1v1+ a2v2 + .....+ atVt

For your second question we assume any m length sequence is linear dependent so I think yes

[crystal vale]

What is the difference between these?

[nova holly]

I believe the correct way of saying it would be "a sequence which span contains v" then

so your question is answered?

[crystal vale]

Can you please explain more ?

v is arbitrary and I want to show that there exists a finite list which span v

[nova holly]

some set of vectors span a subspace, so spanning a single vector doesn't make a lot of sense

[nova holly]

v is a finite list that spans v (using your definition)

[crystal vale]

No I want a spanning list for vector Space V

[daring nova]

"a sequence whose span contains v" is the grammatically correct way btw

[nova holly]

thanks, English is hard 😅

[nova holly]

ok this wasn't clear at all

then a base of V would work?

[crystal vale]

That means a sequence v,v,......,v m times is a list which spans v right?

[nova holly]

using your definitions, yes

[crystal vale]

Base means basis right, that is question how I create a spanning list

[nova holly]

yeah sorry I can't speak english apparently

[crystal vale]

So my friend's proof is correct?

[cloud solar]

Yeah

[nova holly]

well the fact that any linear dependent sequence contains a non zero element is false, 0 0 0 is linear dependent

[cloud solar]

Can someone find an elementary solution to the problem with the polynomial X^8-16 ?

[dire siren]

if char(K)=2, then 0 is a root
if char(K)=/=2, we have:
case i) if 2 or -2 is a square, then the polynomial has a root
case ii) if 2 and -2 are not squares, then -4 is a square (refer to the fact that K* is cyclic to see this)
so there is t such that t^2=-4; but then (t+2)^2=4t, so ((t+2)/2)^2=t ||a/2 simply means a*2^(-1) ||
so you have t=s^2; and then s is a root of the polynomial

[crystal vale]

But i didn't say that, I say non zero coefficient

[nova holly]

no you said u1 .. um are linear dependent, which means they are vectors, then you asid u_k is non-zero, which is false

[crystal vale]

Oh sorry I mean there is u_k which has non zero coefficient

[cloud solar]

Can you explain the part with t^2=-4. Why we get this from K* cyclic?

[dire siren]

K*={1,a,a^2,a^3,...,a^(q-1)}, for some a

is 2 is not a square, then 2=a^(odd)

the same for -2

so then -4=a^(odd)*a^(odd)=a^(even)

[cloud solar]

Right

Thank you

[velvet steeple]

In the group axioms, do we need a unique neutral element or would it also be fine if we knew every element had a different neutral element?

[delicate orchid]

Show that the neutral element has to be unique

Also if a so called “neutral” element didn’t fix everything it just doesn’t satisfy the definition of being neutral so idk what that second part even means

[velvet steeple]

Basically my question is if our neutral element is allowed to behave like an inverse, where everything in our group has a different neutral element

[velvet steeple]

Show that the neutral element has to be unique
Do we have to?

What if we know every element has a neutral element

But not necessarily the same one

Isn't that fine too?

[delicate orchid]

If x is neutral then by definition of being neutral ax = a for all a

[velvet steeple]

Ah, for all a..

Then it must be unique

[delicate orchid]

I think this might be able to happen in Boolean monoids

Since x^2 = x for all x there

So every element acts like it’s own “neutral” element

[dire siren]

@cloud solar another way to deal with the second case is to factor X^4+4 as (X^2+2X+2)(X^2-2X+2)
and since the discriminant of these quadratics is -4, a square, they are going to have solutions

the whole idea behind this problem is that at least one of -1,2,-2 must be a square

[cloud solar]

I understand

[cloud solar]

This is interesting

[astral fractal]

I saw something like you can show a map between two groups is homomorphism if you can show that the generators of both groups have the same relations, but I'm confused as to why this method shows the map itself is homomorphism?

since this method does not comment on the map itself

[delicate orchid]

like this?

that lowercase r should be an R that's a typo in these notes

wait, no that's right. It's just written weirdly

[astral fractal]

i think that might be it

or wait

[delicate orchid]

waiting

[astral fractal]

im not sure actually

[delicate orchid]

well if the generators of H satisfy the relators R then just any embedding of X into the generators of H will work

[astral fractal]

i see

[delicate orchid]

if the generating set of H is smaller than X then uhh

[astral fractal]

So as long as the generators of one set X and the generators of a set H satisfy all the same relations then any function from X to H will be a homomorphism?

[delicate orchid]

oh no that's not what you mean - you literally mean "the set of relators in both groups are the same"

[astral fractal]

i meant the generators are different but the relations are the same

[delicate orchid]

well then this is just relabelling your generators

[delicate orchid]

any function from X into the set of generators of H - annoyingly the theorem doesn't give this set a name lol

ok from the top

[astral fractal]

X being a set of generators?

of G

[delicate orchid]

ok nevermind I'm just restating the theorem I've already posted

[astral fractal]

i think i understand

my question was in regards to this problem https://gyazo.com/2fb930c16032c0fde2c85324f0eb7754

[delicate orchid]

oh for fucks sake that's so much simpler

just post your actual problem in the future

[astral fractal]

my bad

i was asking in a general case

[delicate orchid]

it's ok but this is a lot more approachable

and your method would work

[astral fractal]

i'll look up a proof of this method online

i didnt know it had a name

Substition Test

[delicate orchid]

anyway

[astral fractal]

yikes

[delicate orchid]

D_2n = <a, b | a^2 = b^n = 1, aba = b^-1>, set c = ab then D_2n = <a, c | a^2 = (c)^2 = (ca)^n = 1>

our group G is <x, y | x^2 = y^2 = (xy)^n = 1>

these are isomorphic with iso x -> a, y -> c

[winter shore]

My book says:

The free R-Algebra on a set A is isomorphic to the monoid ring over the free monoid on A,

this was in the context of a finite A, is this not the case too for infinite sets A?

[tough raven]

I think it's true for infinite A.

[dull marsh]

Is this real chat

Ah wait I missed the part where x and y generate G

I was flabbergasted for a moment

[hearty ledge]

what book is this one exactly i see that he has published multiple books

[coral spindle]

Algebra

Algebra, by Hungerford.

[tardy hedge]

yoo dudes whats up

remember me

algebra is better than analysis

im tryna study analysis rn

[summer brook]

I'm doing row and column operations on a matrix to get something which presents an isomorphic Z-module, but I'm stuck on the third step. Does anyone know what is happening for it? Why does the matrix change dimension?

[rotund aurora]

Is there a ring whose ideals look like $I_1\subset I_1\subset\cdots\subset I_n\subset\cdots\subset \frk p\subset(1)$ where $\frk p$ is the only prime?

[cloud walrus]

Croqueta

[summer brook]

I think I've figured it out but would be worth clarifying. These are the allowable actions to preserve module isomorphism:

• Multiply row/column by unit
• Add any multiple of row/column to another row/column
• Swap rows/columns with other rows/columns (composition of above two)
• Delete or add a column or row of zeroes
• If j-th column is e_i, delete row i and column j

Is that correct?

[crystal vale]

What do you mean p is only prime?

[rotund aurora]

it has no other primes

idk this is pretty stupid

everything would be nilpotent

[crystal turtle]

Sure, there are plenty of examples. A fairly simple example is k[x]/<x^2>, where k is a field. Here the only ideals are 0 <= <x> <= <1> by the correspondence theorem.
Such examples tend to be fairly useful actually, because they are simple and easy to understand

Note that not everything is nilpotent; only the nonunits are

Well idk precisely that this is super common to have only one prime, more common to only have one maximal ideal (i.e. local rings), but the principal is still the same: easy to understand and naturally arises from quotienting and localization

[rotund aurora]

sure

[next obsidian]

If you’re Noetherian this basically says you’re an “infinitesimal thickening” of a field

Geometrically, a point (field) with some fuzzuzz

[rotund aurora]

this doesn't work, you should have I1 subset I2 subset ... subset In susbet ... for all n in N, and then at the very top the prime p

[next obsidian]

You want those all proper?

[crystal turtle]

Ah I misunderstood

I thought you wanted a finite chain

[rotund aurora]

proper subsets? yes

[crystal turtle]

lol consider my example where all I_n = 0

hehehe

[rotund aurora]

ye the inclusions should be proper

[crystal turtle]

No. Such a ring would satisfy the descending chain condition but not the ascending chain condition. However, for rings, it is known that DCC ==> ACC (and dim = 1 0) on ideals (if these are to be all of the ideals)

Uhh for commutative rings i guess

idk and idc about noncommutative ones

[rotund aurora]

yeah I was just worrying about commutative ones

[crystal turtle]

(see for instance Atiyah-Macdonald Theorem 8.5)

[next obsidian]

Dim 0

Not dim 1

[crystal turtle]

Oh right

[next obsidian]

It’s okay silly

:3

[crystal turtle]

i was thinking PIDs had dim 1 i guess

[rotund aurora]

ye

I had read that theorem somewhere, but just forgot lol

[crystal turtle]

:3c

[south patrol]

PID

[crystal turtle]

PID.

[rotund aurora]

I was just trying to see if the most stupid non-noetherian ring with just 1 prime would be possible

[hollow mica]

This is chinese-remainder theorem (CRT)

Is my assesment right that the "elementary" form of CRT (the first one stated) is just the statement that the below map of rings is a bijection

So in fact the "ring isomorphism" statements tells you more (that ring structure is preserved)?

[next obsidian]

Basically, yes

But the elementary statement is basically the same thing because the map clearly is a homomorphism

[hollow mica]

Ok I see

[errant shadow]

X is a finite set, R is some ring. Does this look like an isomorphism map?

F(x,R) is the ring of functions from X to R

This is the question statement in case my shit handwriting is getting in the way.

[formal ermine]

i can't read your handwriting

can you tex it

[quiet pelican]

That is the right map

[crystal turtle]

Yes that's correct but uhh depending on how you define n-tuples it's a virtually tautological question

[crystal vale]

If H and K has a finite index in G then H intersection K has a finite index in G, any hint?

[viscid pewter]

probably consider cosets of H & K in H

[crystal vale]

Then I have a statement which states that the number of distinct Cosets of the form hK , for h an element of H, is the same as the number of distinct Cosets h(H intersection K) for h an element of H

And why do I consider the cosets of H intersection K in H , I want Coset of H intersection K in G ?

[viscid pewter]

|G : H| |H : H&K| = |G : H&K|

[crystal vale]

But I want also proof that for any H< K<G I have | G: H| = |G:k| |K:H|

[viscid pewter]

you can prove that

[crystal vale]

I am stuck here

[crystal vale]

Can you explain to me when G is infinite then how can I estimate|G: H|

[viscid pewter]

it's just a number

consider Z: {0, 1, -1, 2, -2, ...}

and then consider 2Z:

{0, 2, -2, 4, -4, ...}

then 2Z has index 2 in Z

sometimes it might be infinite of course

but when it isn't

[crystal vale]

Yes

[crystal vale]

Let g1H=g2H means g2^(-1)g1 an element of H so it will be an element of K so g1K=g2K then what is the next step if these are correct?

[viscid pewter]

consider the representatives of K in G, and the representatives of H in K, and try and connect these to the representatives of H in G

[crystal vale]

Do I show that first it has a finite index or I assume that?

And I think for any gH , and I know the cosets of k partition G so g I can write g= g_s•k1, where k1 is an element of K and also cosets of H partition K so I can write k1 = k_t•h1 .....is it correct way ?

[viscid pewter]

you can assume it

it's just that it's fairly trivial to see that if either of them have infinite index then the overall thing will have infinite index also

so you can just consider the finite one

[crystal vale]

Yes but in case of H intersection K I should first show that it has a finite index I think

[viscid pewter]

again, reasonably easy to show the infinite case

or whatever

[errant shadow]

I am kinda bad at it. But I'll try. So the map basically takes any function, $f_i={(x_i, r_i)| 1<i<n, r_i \in R}$ to $(r_1, r_2,....,r_n) \in R^n$

[dire siren]

take a coset of HnK, say a(HnK)
notice that a(HnK)=aHnaK
so essentially a coset of HnK is the intersection of a coset of H and a coset of K

[cloud walrus]

CHTRGPT

[errant shadow]

Eh? What way do you have in mind?

[crystal vale]

Thank you but now I want a hint for if h<k<G then |G:h|= |G:k| |k:H|

[crystal vale]

Nice point

[dire siren]

did you figure how to solve the original problem?

[crystal vale]

Yes

Coset of Hnk is intersection of coset of H and K so it will be finite , right

[crystal vale]

Any hint for this?

[viscid pewter]

consider the representatives of K in G, and the representatives of H in K, and try and connect these to the representatives of H in G

[crystal vale]

I tried this but didn't figure it out

[crystal vale]

Look at the second paragraph

[viscid pewter]

yeah idk what that means

try it though

[dire siren]

@crystal vale isn't this just a consequence of |G:H|=|G|/|H|?

oh, nvm, G is not necessarily finite

[crystal vale]

No if G is not finite

[tacit ginkgo]

im trying to find the group table for the permutation group s3

would <1,2> <2,3> = <2,1,3> be right?

[south patrol]

Are you using the convention that like

It's the opposite order to composition

[tacit ginkgo]

I think so yeah

[south patrol]

Then yes

You can just plug in 1 and see where it goes and stuff

Nice

[tacit ginkgo]

Would <1,2> <1,2,3> = <1,2,3> also be correct?

[dull marsh]

No

[tacit ginkgo]

Not even when doing it the opposite order to composition?

[dull marsh]

Yeah

Retry

1 shouldn't go to 2 in the product

[tacit ginkgo]

what should it be?

[dull marsh]

<1, 3>

[tacit ginkgo]

why

[dull marsh]

1 goes to 2 in <1, 2> and then 2 goes to 3 in <1, 2, 3>

So 1 goes to 3 in the product

2 goes to 1 in <1, 2> and 1 goes to 2 in <1, 2, 3>, so 2 is fixed in the product

[delicate orchid]

When the regular action is not free

[tacit ginkgo]

what if u do it the other way round

[dull marsh]

With composition read from right to left, the product would be <2, 3>

[crystal turtle]

I'm just saying one way of defining n-tuples in R is as a function {0, 1, ..., n-1} --> R lol. With this definition, it's essentially immediate that functions from an n element set are the same as functions from an n element set

[tacit ginkgo]

Ok thanks

[vale locust]

I dont' get this definition

A_k(V) are the alternating k-tensors on V

why can we stop the direct sum at a n?

[errant shadow]

Ah yea basically

Yea that makes sense

But explicitly proving that homomorphism was a bit eh for me ig.

My prof is very particular about notation.

Last problem session he made me hunt down every single bracket I missed on the board.

[terse crystal]

There has to be two same elements in the basis

When k>n

[granite topaz]

unless I'm mistaken, A_n has only one tensor, the determinant

[unique anchor]

In a field of characteristic $p>0$ why is it that one can write an inseparable polynomial $f$ as $f(x) = g(x^{p^n})$ for some irreducible polynomial $g$ ?

[cloud walrus]

piss_master49

[glossy crag]

1. A polynomial is separable iff (f,f')=1 (valid over any field).
2. Thus if f is irreducible, then f is inseparable iff f'=0.
3. In char p f'=0 iff f=a_0+a_px^p+a_2px^2p+..., i.e. iff f(x)=g(x^p) for some g.
4. Thus in char p an irreducible f is inseparable iff f(x)=g(x^p) for some g, which is necessarily irreducible (otherwise f would be reducible).
5. Now repeat this until you can't anymore (total induction on deg f, if you want to be formal).

[delicate orchid]

precomp.... with da frob.....

[crystal vale]

If H and K are normal in G and H<K( h be a subgroup under K) then K/H is normal in G/H ...... I showed it, but I am thinking about why we take H<K if we do not have this condition then what happens? Then K/H is not defined?

[delicate orchid]

yeah like, how are you supposed to quotient by a group that isn't a subgroup

the cosets of H have to partition K, so if any part of H isn't in K this is a contradiction

[crystal vale]

To show K/H is a subgroup in G/H we need it, right?

[delicate orchid]

for K/H to make sense at all you need it

[crystal vale]

Yes but I thought if K<H then it was just H right?

[delicate orchid]

in order to form a quotient object we need an equivalence relation that plays nicely with the operations associated with that object. It turns out for groups that these equivalence relations are in correspondence with normal subgroups. If we take a group that's not even a subgroup the quotient is ill defined

you could just blindly define for arbitary K/H to be the set of equivalence classes under the relation a ~ b <=> aH = bH but this is just a quotient set

[crystal vale]

Yes but I just take K/H as a set of Cosets of H in K not as a subgroup

[crystal vale]

And this?

Oh then this single H is not subgroup in G/H

It just an element

But in G what is K/H if K<H ?

[delicate orchid]

what

[delicate orchid]

if you really want to do this completely pointless construction, do this

and K/H doesn't live in G

[crystal vale]

If I want to make a set of Cosets of H by an element in K then ?

Then this set contain only one element H, right ?

[fickle brook]

ugh ok so im back on my bullshit with the polynomial calculator thing

currently tackling factorization

https://en.wikipedia.org/wiki/Factorization_of_polynomials_over_finite_fields#Factoring_algorithms

Many algorithms for factoring polynomials over finite fields include the following three stages:

1. Square-free factorization
2. Distinct-degree factorization
3. Equal-degree factorization

im kind of struggling to see how this translates into a programmatic organization of the full factorization function

im gonna try to write some pseudocode here

to try and see if maybe that works

def factorize(f):
  # list_sqf is a list of squarefrees whose product is f
  list_sqf = factorize_squarefree(f)
  
  # list_ddf is a list of polynomials whose product is f
  # and those polynomials are each the product of several
  # same-degree irreducibles
  list_ddf = []
  for g in list_sqf:
     list_ddf += factorize_distinct_degree(g)
  
  # list_edf is a list of polynomials whose product is f
  # and which are all themselves irreducible
  # obtained from each element of list_ddf 
  # with equal-degree factorization
  list_edf = []
  for h in list_ddf:
    list_edf += factorize_equal_degree(h)

  return list_edf

is this the correct structure?

it still feels like a beast to implement

it is making my head spin so bad.

ping me if anyone replies

[gloomy night]

Looks like you're missing a += in the list_ddf step, but otherwise the structure overall looks good to me.

One thing I noticed when looking at the wiki article is that their first "equal degree factorization" algorithm (Cantor-Zassenhaus) needs the input h to be a product of at least two irreducible factors of the same degree. When you iterate over list_ddf and pass each element to the equal degree factorization method, you might pass on an irreducible polynomial. I wonder what happens in the algorithm in this case, I haven't checked.

Also, in the same stage, they say on the wiki that the finite field needs to have odd degree, which the other stages don't - so you might need special handling of the even case.

[fickle brook]

the finite field needs to have odd CHARACTERISTIC you mean

[gloomy night]

Yeah, that's what I meant.

[fickle brook]

you might pass on an irreducible polynomial. I wonder what happens in the algorithm in this case, I haven't checked.
i mean that's not really an issue, i can just insert an irreducibility check somewhere.

cause i already have that implemented.

[gloomy night]

Good, so only the even characteristic thing might need some special handling. Didn't look like they mentioned it on the wiki.

Except for telling us to read Knuth's book. 😬

[hidden wind]

i’ve only had a look at vol 1 but it’s pretty nice

[gloomy night]

Does it still hold up? When I took Algorithms, we used one of the modern books (CLRS).

[winter shore]

I'm getting confused with group morphisms f : Zm -> Zn. Clearly, any such morphism is determined by the image of 1. Since m * 1 = 0 on the left, then we should have that m * f (1) = 0 on the right. i.e. m*f(1) = 0 mod n. Conversely this last condition induces a group homomorphism. I'm failing to characterize nicely this last condition

are ther any formulas for solving mx=0 mod n?

[south patrol]

Do you know when a map Z -> Z/n factors through the quotient

[winter shore]

yeah

it's the same condition mx=0 mod n

where x is the image of 1

[south patrol]

Oh sorry yes

[vale locust]

what's the basis of A_k(V)?

[dire siren]

yes
take d=gcd(m,n)
then m=ad and n=bd for some integers a,b with gcd(a,b)=1
try to solve from here

[south patrol]

Ye

[vale locust]

oh I think that's discussed in the following subsection, guess I'll check that one out!

[winter shore]

thanks

[dull ginkgo]

Hello chat do you know any cool applications of Chinese Remainder Theorem

[coral spindle]

Proving that the totient function is multiplicative

[vale locust]

ohhh

wait

it's because A_k(V) is just the 0 k-tensor when k>n

[cloud solar]

If K finite field and f a polynomial in K[X] with deg >=1 show there is n>=1 s.t. f(x^n) is reductible

I think I need to pick such n that has something to do with char

[south patrol]

exactly

[hidden wind]

oh god i’m stumbling over very elementary facts… like, three points uniquely determine a circle

oop wrong channel sowwy

[cloud solar]

Hmm

[dull ginkgo]

This is technically a linear algebra problem lol

[coral spindle]

I would call it an algebraic geometry problem.

[hidden wind]

blessed including the three colinear points case

[dull ginkgo]

Not if you use the vector space of homogeneous polynomials of a specified degree :troll:

[coral spindle]

Right.

[stark helm]

For this question, is it correct to construct a field Z2[x]/[x^3+x+1] by aruging x^3+x+1 is irreducible. Then assuming u be the root of x^3+x+1. and u^3=u+1. Therefore, we have elements 0, 1, u, u^2, u^3=u+1, u^4=u^2+u, u^5=u^2+u+1, u^6=u^2+1, is it a complete statement for this one?

[stark helm]

and also for this one, I tried it by assuming that an+a_n-1x^+...a0x^n is reducible(is that a correct start and how to proceed if it is true?)

[dull ginkgo]

if p(x) is your original polynomial

Then p(1/x) x^n is your one with the coefficients “flipped”

That might help?

[stark helm]

it helps

[dull ginkgo]

I haven’t done that exercise before I just pulled it out of my ass, glad that helps, gonna use it if I get to it in Dummit & Foote

[cobalt heath]

Sorry for interruption, but is there a reason why (on a field F of ch. 0) xCk and x^k seems 'similar'? Like, they both spans the polynomial space F[x], and they have 'same asymptotic behavior' considering R and C case.

[dull ginkgo]

what is xCk?

Binomial coefficient?

[cobalt heath]

Yep. Idk, maybe there is no real reason or something?

[stark helm]

one question, if p(x) is reducible in F[x], p(1/x) will definitely be reducible over F[x] right?

I consider substitue 1/x as u

[cobalt heath]

p(1/x) does not lie on F[x]

[stark helm]

Then we construct p(1/x)*x^n for p(x), i am a little confused about how will it be valid here?

it seems that there is no better to flip the coefficient

[cobalt heath]

Yep, you basically need to flip the coefficients

Maybe it helps to sleep on the reducibility?

[stark helm]

what does that mean?

[cobalt heath]

Sorry, I mean think about when a function is reducible

And its relations to flipping the coefficients.

[stark helm]

When we can factor it into two polynomial with more than 1 degree, then p(x) is reducible

[dull ginkgo]

it explains that you’d flip the factors too

x^np(1/x) = x^n(a(1/x)b(1/x)) so your x^n would distribute between the degrees of a and b, “flipping” their coefficients too :)

[cobalt heath]

Ohh, that's a great insight

[dull ginkgo]

that’s how I interpreted it

[stark helm]

so when we have p(1/x)x^n, since we know p(x) is reducible, then p(x)=a(x)b(x), and p(1/x)x^n=a(1/x)b(1/x)*x^n. and we can see that x^n will distribute to a and b seperately, is it true?
x^n will definitely have enough degree to distribute here

[cobalt heath]

Hmm, which part is confusing you here?

[stark helm]

i think I should understand now, before that it seems somewhat counterintuitive that p(1/x)=a(1/x)*b(1/x) can be argue directly

[cobalt heath]

Ahh, def take time. Usually it takes a while for concepts to stick

[dull ginkgo]

let S(p) be the multiplicative operation that sends a polynomial p to the polynomial of same degree with the coefficients swapped.

You can show S(pq) = S(p)S(q) as shown

It’s also an involution, I.e S(S(p)) = p

Assume p is irreducible, but S(p) = ab.

S(S(p)) = S(a)S(b) = p

But p is irreducible, and S preserves degree. Contradiction

[cobalt heath]

I had bumped into some counterexamples for this

Compute S(x^2)

[dull ginkgo]

S(x^2) = 1

oh shit yeah

[cobalt heath]

Yeah, things messy

[dull ginkgo]

No wait he forced the constant coefficient to be nonzero

So that preserves degree

[cobalt heath]

Yeah, that's why this map works in this specific case

[dull ginkgo]

And the divisors thus must also have nonzero constant coefficients

sick

[cobalt heath]

I guess S is some kind of retraction lol

[dull ginkgo]

Perchance

[cloud solar]

.

K has p^t elements

and (f(a))^(p^t)=f(a^(p^t)) for every a in K

Hm

Any hints

@south patrol

[rocky cloak]

I mean, you have the answer right there.

f(x^n) = f(x)^n when n=p^t

[cloud solar]

Yes but can we apply the rule (a+b)^n=a^n +b^n even for the variable X?

[rocky cloak]

Try thinking about an example, if n=2, what is (x + 1)^2 ?

[cloud solar]

Yeah right

Thanks

[cloud solar]

Let K be a field with 3^n elements. Show that f=X^4+X-1 is reductible iff n even

If f reductible then f has a root in K* or is just a product of two polynomials with degree 2

[rocky cloak]

Both of those cases would make f reducible yes

[cloud solar]

For the case when n is even can we say that there is a subgroup of K* with 8 elements

[rocky cloak]

A possible hint could be that ||the same statement is true for any degree 4 polynomial that is irreducible over F3 (or any polynomial of degree 2^k really)||

[rocky cloak]

You can say that, sure

[cloud solar]

But idk if that helps

[rocky cloak]

Even more general: ||if F is a finite field, f is an irreducible polynomial of degree d, and K is a degree n field extension of F, then f is irreducible over K iff n and d are relatively prime||

[cloud solar]

Should I use the field extension of the polynomial f

It has order q^d where q is the order of K and d the degree of f

[rocky cloak]

Yeah, if adjoining a root of f gives you a degree d extension, then f most me irreducible.

[cloud solar]

Where can I find a proof of this

[cloud solar]

.

This is pretty nice

[rocky cloak]

Well if K(a) has degree d, then the minimal polynomial of a has degree d, hence equals f

[dense flare]

Hello, below i'm posting a proof that if G is a p-group and H is its proper subgroup then H is a proper subgroup of its normalizer. I don't get highlighted part, why existence of this fixed coset other than H implies that gHg-1 is a subgroup of H. Thanks for help!

Link to this question on sx: https://math.stackexchange.com/questions/168963/proper-subgroups-of-finite-p-groups-are-properly-contained-in-the-normalizer

[terse crystal]

Hgh=Hg for all h -> Hghg^-1=H for all h

You know Ha=H implies a is from H right

[dense flare]

okay, im just stupid, thank you

[chilly ocean]

What are some simple examples of R-modules that are not R-algebras? I am trying to understand the difference between them a bit more.

[next obsidian]

This is kinda not the right question to ask

[dull ginkgo]

It’s not really an “explicit” difference because one is sort of a “refinement” of the other

[next obsidian]

Something is an algebra when we choose to endow it with that structure

[dull ginkgo]

The complex numbers and R x R (as a ring) are isomorphic as modules, but NOT as algebras

[next obsidian]

Anytime you consider “an R-module” it isn’t coming with any sort of ring structure so it isn’t an algebra

You can ask something like “what is an example of an R-module which cannot have an R-algebra structure compatible with the module structure”

[glossy crag]

Take R non-commutative and take I a left ideal, then R/I is an R-module, but not an R-algebra (not in any way I know at least).

[next obsidian]

https://math.stackexchange.com/questions/93409/does-every-abelian-group-admit-a-ring-structure

This is one such example in the case of R = Z, because Z-modules are abelian groups and Z-algebras are commutative rings

[dull ginkgo]

like (x,y)(a,b) = (ax,by) for R x R
But (x,y)(a,b) = (ax - by, ay + bx) for C

They have the same structure as a module, which in essence is just an abelian group with your ring structure “acting on it”. However, there can be many algebras “on it”

the isomorphism itself is (x,y) <-> x + yi

[cloud solar]

Can you explain it more please?

[cobalt heath]

Wow, I've wondering about this for quite a while. Thanks!

[rocky cloak]

K(a) = K[x]/(p(x)) where p is the minimal polynomial of a. So if K(a):K = d, then the degree of p is d. Since a is a root of f, p divides f. But p has degree d, so p=f.

[cloud solar]

Thanks

[pliant forge]

I am struggling to see how a+b = 0 \in C when a+b \in A \cap B, for A, B submodules of C

[delicate orchid]

probably because that statement isn't true

A = <(0,0,1), (0,1,0)>, B = <(1,0,0), (0,1,0)> as sub Z-modules of Z^3

[pliant forge]

ic. then im overlooking something

[quiet pelican]

Where/how did you get this question?

[pliant forge]

So i want to show that for the canonical map $A \oplus B \to C$ the kernel is $A\cap B$

[cloud walrus]

Eso

[quiet pelican]

Is isomorphic to?

[pliant forge]

no

[quiet pelican]

A intersect B is not a sub module of A oplus B

[pliant forge]

this is the hint i was given

[delicate orchid]

the idea of even writing K (+) L when K, L have non-trivial intersection doesn't make sense to me

[quiet pelican]

well
||try a -> (a, -a)
Where does that map to in K + L?||

[delicate orchid]

unless we're just formally taking a coproduct

[pliant forge]

wouldnt it be like a cartesian product kinda

[delicate orchid]

well yeah R-mod has biproducts so they're isomorphic but I just don't understand what the construction is supposed to be here.

Micose seems to get it so listen to her

[pliant forge]

hmm i suppose it is isomorphic to K \cap L then rather than equal

shouldve looked at it like a single element rather than a sum

[wide brook]

Hello everyone I have a question and I can’t figure out the solution. Suppose we have two polynomials f(x),g(x) in Q[x] which don’t share any common complex roots, why does this imply that the gcd(f(x),g(x))=1? If I understand correctly since the polynomials are over a field to prove the statement we only need to show that the gcd is a constant and that’s what I don’t get. Does not sharing any complex roots mean that they don’t share any roots?

[viscid pewter]

yes

[wide brook]

Why? Because we think real roots as complex roots too?

[rocky cloak]

Real roots are indeed complex roots

[viscid pewter]

yes

R < C

[rocky cloak]

The idea is that if h is a common divisor of f and g, then any root of h would be a root of both f and g

Any non-constant polynomial has a root, so there you go.

[wide brook]

Great, I get it! Thanks! The thing I’m saying about the gcd being 1 if it’s a constant is true, right? Because every non zero number in a field is invertible….

[rocky cloak]

That's right

[sudden condor]

do icosahedrons and dodecahedrons have the same symmetry group or did i make a mistake somewhere when deriving one of them

[dull ginkgo]

Vertex or face?

[sudden condor]

vertex

[dull ginkgo]

Think about it

same order but different group

Rotating about a vertex has order 3 in one, 5 in the other :)

Wait

[sudden condor]

makes sense

i thought something was fishy

[dull ginkgo]

I am incorrect

[sudden condor]

no way

that's so cool

[dull ginkgo]

They do have the same symmetry group up to iso

There’s really “two actions” which I actually think are generators. Rotating about a face and rotating about a vertex

but rotating about a face is rotating about a vertex in the other

[sudden condor]

oh yea

[dull ginkgo]

So that provides an iso

[sudden condor]

wow i can't believe i got it right

thanks

[dull ginkgo]

I am a moron

You had it right first lol

[viscid pewter]

they do

this comes from the fact that they're dual to each other

if you take the 12 points at the centres of the faces of a dodecahedron and connect them iff the faces they came from shared an edge, you get an icosahedron

and the same for if you take the 20 points at the centres of the faces of an icosahedron

[dull ginkgo]

What’s the presentation of that group lol. I know there’s the vertex and face rotations lol

[sudden condor]

the way i was drawing it, it was basically a graph, and the group actions were like permutations of the graph that preserved adjacencies

[sudden condor]

and drawing it like that led me to notice you could draw it alternatively like this

but it seemed too crazy to be left unchecked

[viscid pewter]

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

[dull ginkgo]

Thank you chat

[stark helm]

Do you think 6 is correct? I wrote the root a, a^2, a^3=a^2+1, a^4=a^2+a+1, a^5=a+1, a^6=a^2+a.

[rocky cloak]

Think you're missing 1 and 0 there

[stark helm]

does F(a) mean find number of roots of f(x), so why 0 and 1 should be add?

probably I understand it wrong?

[rocky cloak]

F(a) is the field you get by taking F and adjoining a.

f can't possibly have more than 3 roots, it being degree 3 and all

For example f(a^3) = a^6 + a^3 + 1 = a^2 + a + a^2 + 1 + 1 = a shows that a^3 is not a root.

[stark helm]

Then how to express 1 and 0 in terms of a?

it seems a constant coefficient, so I am confused if we need to express it by appearing some a

[rocky cloak]

You have a^7 = 1, if you insist on expressing it in terms of a. 0 is obviously not a power of a

[stark helm]

For this one, I can quickly know that a and a+1 is a root of f(x), therefore, i factor f(x) into (x-a)(x-a-1)(x^2-x-a^2). another elements of E should be 0, 1, a^2, a^3, a+1, a^2+a, a^3+a^2, a^3+a+1, a^2+1, a^3+a, a^2+a+1, a^3+a^2+a, a^3+a^2+a+1, a^3+a^2+1, a^3+1, but no additional elements match roots of f(x). is that correct?

since if f(x) has a root a in E, then we must have (x-a)|f(x)

[hollow mica]

for prime ideals, are the notions of “principal” and “contains a prime element” equivalent?

[glossy crag]

Nope: x is prime in K[x,y], but the maximal (in particular prime) ideal (x,y) that contains it is not principal.

[rocky cloak]

Check f(a^2)

[glossy crag]

Do you study in Norge @rocky cloak?

[rocky cloak]

I do

Trondheim to be precise

[glossy crag]

PhD? Not trying to dox you, just curious about Scandi academia.

[rocky cloak]

I'm doing a PhD yeah. And you can dox me if you want

Don't think it's hard to connect my username to my real name with some google-fu anyway

[glossy crag]

How is it? Did you stay home for reasons of comfort or is it a good place to study?

[rocky cloak]

I certainly think it's a good place to study, but I guess I'm here more for my personal life yeah

What about your self, have you journeyed far from home?

[glossy crag]

Not really, kinda dreading it :/. I'm a very domestic person.

[rocky cloak]

So you're about to?

[stark helm]

f(a^2)=a^8+a^2+1, which means a^2(a^6+1)+1=a^2(a^3)=a^5=a^3+a^2+1?

[glossy crag]

Dunno yet, will see

[rocky cloak]

And do you remember what a^3 + a^2 + 1 equals?

[glossy crag]

What do you do btw, smth algebraic?

[rocky cloak]

Representation theory of algebras

Plus some homological algebra I guess

[glossy crag]

I finally learned semisimple theory about half a year ago, took me a while. I mean semisimplicity, Artin-Wedderburn, etc.

[stark helm]

a, a^2, a^3, a+1, a^2+a, a^3+a^2, a^3+a+1, a^2+1, a^3+a, a^2+a+1, a^3+a^2+a, a^3+a^2+a+1, a^3+a^2+1, a^3+1

[rocky cloak]

Yeah, on the homological side semisimple algebras are completely uninteresting 😛

[glossy crag]

Does this intersect with cryptography in any way? It feels combinatorial/computational enough for there to possibly be something, maybe.

[rocky cloak]

There's lots of relations with combinatorics, don't know about crypto, but maybe

[stark helm]

I made mistakes, but it will be a^8+a^2+1=a^2*(a^6+1)+1=a^2*(a^3+a^2)+1=a^5+a^4+1=a^2

[stark helm]

it should be the correct order

(x-a)(x-a-1)(x^2-x-a^2) this is what I do for factorizatoin

[celest furnace]

Is there a constructive way to do this? I said you can find F_p^n as the splitting field of X^p^n - X over F_p. This extension is separable since the minimal polynomial of each alpha in F_p^n would have to divide X^p^n-X, and that polynomial has distinct roots (so any divisor would also have distinct roots). Thus, there is a primitive elment alpha so that F_p^n = F_p(alpha). Since |F_p^n : F_p| = n, the minimal polynomial over alpha has degree n

[glossy crag]

@rocky cloak be honest, how cringe do you consider the current (past 15 years or so) pop-culture hysteria around vikings and Norse myth

[rocky cloak]

Is there a current pop culture hysteria?

[glossy crag]

Well it's dying down, but for a while it was pretty strong, shit was everywhere.

[rocky cloak]

a^8 + a^2 + 1 =
a^4(a + 1) + a^2 + 1 =
a^5 + a + 1 + a^2 + 1 =
a(a+1) + a + a^2 = 0

[rocky cloak]

Guess I didn't really think much of it then

So when would this dreaded moving of yours potentially come @glossy crag ?

[glossy crag]

Dunno, possibly in April, if I can get some extraneous shit done by then. I likely won't be able to, so this would get moved back to the next semester.

[rocky cloak]

Are you in a PhD now?

[glossy crag]

Lol no, I'm finishing my bachelor's. Fukken wish, still a long ways to go.

[rocky cloak]

Makes sense, and then you'd want to do a master somewhere else?

[glossy crag]

Yep.

The itinerant academic lifestyle is something I dread overall.

[stark helm]

OK, then since we have factorization (x-a)(x-a-1)(x^2-x-a^2), is it complete enough?

[stark helm]

I can not figure out how to express something like x-a^2

[rocky cloak]

Cuz you can't study what you want there, or good with some variety or

[rocky cloak]

What do you mean express?

[glossy crag]

Bit of both. Our faculty has (what I feel like) is a pretty big name in Lie groups, but I'm not really interested in that, otherwise I would stay and scramble about his feet 100%.

[rocky cloak]

So what are you interested in?

[glossy crag]

Not Lie theory (which is a shame, because he's a great lecturer and enormously knowledgeable). Just algebraic stuff in general, want to get a taste of "serious" number theory and our guys don't offer that.

[stark helm]

i mean since we have root a^2, is it ok to try factorize x^2-x-a^2?

[rocky cloak]

Yeah, but you might have a mistake in the factorization you have so far.

I think it should be
(x - a)(x - a-1)(x^2 - x - a^2 - a - 1)

[stark helm]

that seems far more reasonable

[rocky cloak]

So where do they do the serious number theory? Do you have some places in mind?

[glossy crag]

Bonn prolly.

[rocky cloak]

Yeah, that sounds about right

Bist du Deutsch?

[glossy crag]

Nope, just study here (in German).

[rocky cloak]

So then you have already journeyed far from home

[glossy crag]

Well no, because my family is here with me, lol. I am lucky in this respect.

[rocky cloak]

I guess traveling between big cities on the continent can be pretty short trips anyway.

Like my family lives 8 hours away, even though they're also in Norway

[glossy crag]

Woof.

What, do you travel up and down "the spine"?

[rocky cloak]

The spine?

[glossy crag]

I mean the northern coastline.

[rocky cloak]

Ah no, they live in the south

[glossy crag]

like the "outermost" bit

Because I can't imagine how else you might get distances of 8 hours (unless you're walking home).

[south patrol]

my 8 hours lives families away.

[glossy crag]

Is it really 8 hours down to e.g. Oslo?

[rocky cloak]

Trondheim-Oslo is about 7 hours by either train or car

[glossy crag]

REALLY? Either I have wildly underestimated the scale of Norway or you guys travel by oxcarts.

[rocky cloak]

If the weather is bad it takes a while longer by car

[south patrol]

how tf is that 8 hours or is norway bigger than i realised

lol

[rocky cloak]

It's 500km

[south patrol]

okay yeah

bigger than i realised

[glossy crag]

Maybe he travels up to Tromso and takes a plane.

[south patrol]

actually lol

I just looked in comparison to UK

yeah fair enough

lol

[glossy crag]

Damn you guys are pretty big.

Just the southern bit is like a 3rd of Germany.

[rocky cloak]

Norway and Germany have about the same area I think

[south patrol]

nice

[rotund aurora]

You can cross Norway in a straight line

walking

[glossy crag]

Is it possible to define the characteristic polynomial coordinate-free 🤔 ? I expect there is some exterior algebra definition as for det, but I don't know it.

[rotund aurora]

it's in the wikipedia page

[glossy crag]

Figures.

[glossy crag]

Ours is a shaming culture.

[rotund aurora]

but it makes sense since you can define the determinant and the trace coordinate-free, and these are two coefficients of the characteristic polynomial

the other coefficients are defined in a similar manner

[rotund aurora]

maybe there is a (-1)^k missing there

[glossy crag]

Could be, the coefficients are this times the sums of principal k-row minors, which I expect what the trace of \Wedge^k f is

[stark helm]

do you think there are four roots instead of three?

[rotund aurora]

so for k=1 you get trace(f), and this should be multiplied by -1

so yes a (-1)^k is missing

[stark helm]

so since x^2-x-a^2-a-1, does x-a^2 must be one of the factor here?

[rocky cloak]

A degree 4 polynomial you should expect to have 4 roots yes

[stark helm]

Then I get a, a+1, a^2 and a^2+1

so we factor it as (x-a)(x-a-1)(x-a^2)*(x-a^2-1)

[crystal vale]

9 question, I think there is no eigenvaule , right?

[barren sierra]

Nah there's an eigenvalue

wait uh

@crystal vale P(R) is polynomials over R right?

or is it space of differentiable functions?

[crystal vale]

It is polynomial over R

[barren sierra]

yea ok there is exactly one eigenvalue

I'll leave you to figure out what it is, but there is one

[crystal vale]

For non-zero vector ?

[barren sierra]

uh

does it have to be non-zero?

I thought eigenvalues could be 0

[crystal vale]

Yeah get it

For non-zero constant polynomial, right?

[crystal vale]

Thank you

[celest furnace]

Does anyone have a small hint for this one

Tried a bunch of things none of which worked

[rocky cloak]

A small hint could be to consider the minimal polynomial of alpha over F

A slightly bigger hint could be that ||F is uniquely determined by this minimal polynomial||

[chilly ocean]

Can anyone look over this proof (p is a prime ideal)

Let $R$ be a commutative ring. Let $I$ and $J$ be ideals of $R$ and let $\mathfrak{p} \in \text{Spec}(R)$ with $I \cap J \subset \mathfrak{p}$. Prove that either $I$ or $J$ is contained in $\mathfrak{p}$.
\begin{proof}
Let $i \in I$ and $j \in J$. Then $ij \in I$ and $ij \in J$, i.e. $ij \in I \cap J \subset \mathfrak{p}$. Therefore $i \in \mathfrak{p}$ or $j \in \mathfrak{p}$. Hence $I \subset \mathfrak{p}$ or $J \subset \mathfrak{p}$.
\end{proof}

[cloud walrus]

gian

[dire siren]

@chilly ocean the last conclusion is wrong, but it can be fixed

[chilly ocean]

How?

[dire siren]

the problem is that as you let i and j run through I and J, you indeed obtain that i is in p or j is in p, but you don't necessarily have the same case every time, so you can't conclude that one of those inclusions hold

[dire siren]

assume that neither I nor J are contained in p

so then you can find i in I-p and j in J-p

and the rest is just the same

[chilly ocean]

i will give that a try

[dire siren]

you basically solved it already

just have to add that line

[chilly ocean]

I'm confused by what this means

do you mean like I/p

The question lets us assume that $I \cap J \subset \mathfrak{p}$. But $I-\mathfrak{p} \cap J - \mathfrak{p}$ is not necessarily a subset of $\mathfrak{p}$?

[cloud walrus]

gian

[chilly ocean]

I'm confused by your notation

[dire siren]

@chilly ocean I mean \setminus

[chilly ocean]

ok right

but then how can we use the fact that I intersect J is a subset of p?

[dire siren]

just like you did before

so you get that i is in p or j is in p, which contradicts our assumption

[chilly ocean]

oh i get it

Assume towards contradiction that $I \not\subset \mathfrak{p}$ and $J \not\subset \mathfrak{p}$. Let $i \in I$ and $j \in J$ such that $i,j \notin \mathfrak{p}$. Then $ij \in I$ and $ij \in J$, i.e. $ij \in I \cap J \subset \mathfrak{p}$. Therefore $i \in \mathfrak{p}$ or $j \in \mathfrak{p}$. But this is a contradiction, hence $I \subset \mathfrak{p}$ or $J \subset \mathfrak{p}$.

[cloud walrus]

gian

[chilly ocean]

Thank you!!!!! @dire siren

[cloud solar]

Can someone explain me this problem: G group with 4n+2 elements show there is a normal subgroup with index 2

I know I should use Cayley theorem

There is f isomorphism f:G->S where S is a subgroup of S_(4n+2)

And if I pick an element of order 2 from G let's say a then f(a) is gonna be a permutation with order 2

So f(a) is a product of disjoint 2-cycles

But why there are 2n+1 disjoint 2-cycles in this product?

And the 2-cycles are also from S? Or just somewhere in S_(4n+2)

[earnest hare]

can I think of a normal subgroup S as a subgroup where it's normalizers is the entire group G

[rustic crown]

yeep

[crystal turtle]

meep

[earnest hare]

thanks

[coral spindle]

Well yes such things exist lol

You can just take any subgroup with nontrivial normaliser, then just look at the normaliser

The normaliser in the normaliser is the whole normaliser!

So at the very least, you have a strategy for finding an example

[coral spindle]

Clarifying this slightly poetically cryptic statement:

If $H \leq G$ then setting $K = N_G(H)$ we have firstly that $H \unlhd K$, and secondly that $K = N_K(H)$.

To write this even more directly: $N_{N_G(H)}(H) = N_G(H)$.

[cloud walrus]

Boytjie

[coral spindle]

Wait lmao I misunderstood the question

[dull ginkgo]

I haven’t really used Normalizer outside of one question lmao

[coral spindle]

OK

[dull ginkgo]

Yeah lol

[coral spindle]

It's not "is it possible to think of an example of...", it's "is it correct to think of..."

Mb

[dull ginkgo]

I’m too lazy to mathematically work this out symbolically right now but is the normalizer of a subgroup S, N, the largest subgroup of supergroup G such that S is normal in the subgroup

[coral spindle]

Yes.

[dull ginkgo]

cool, was making sure my old interpretation was right lmao

Which was absurd, but I considered the normalizer N(S) for subset S of G to be the largest subgroup of G such that <S> is normal in it lol

[coral spindle]

That is also fine.

[dull ginkgo]

Just a bit insane

[coral spindle]

Not really.

[earnest hare]

🐊

[dull ginkgo]

After a while I started getting annoyed with pure symbolic descriptions

[cloud solar]

.

[crystal turtle]

Why would that be absurd or insane?

[dull ginkgo]

Monomorphism I think

[dull ginkgo]

Needless overcomplication

[crystal turtle]

No?

[cloud solar]

But cayley theorem says isomorphism

[crystal turtle]

That's s completely valid way to understand the normalizer. If anything, it highlights the point of the normalizer

Mfw having more than one way to understand something makes it insane

[coral spindle]

(writes down reasonable extension of definition) those fuckin algebraists are off their rockers

[dull ginkgo]

Cayley’s theorem states that it’s isomorphic to a SUBGROUP of S_|G|

[crystal turtle]

Yeah that's what they fucking wrote in the first place

[dull ginkgo]

I thought they said isomorphic to

I may have misread it

I misread it lol

[crystal turtle]

"G is isomorphic to a subgroup of S_{4n+2}"

[dull ginkgo]

yeah

[cloud solar]

.

S subgroup

[dull ginkgo]

I misread it sorry homie

[dire siren]

@cloud solar I think you have to use the actual construction given in the proof of the Cayley theorem rather than just Cayley theorem

[cloud solar]

Action from G to G just the regular multiplication right?

[dire siren]

yes, g is sent to phi_g, where phi_g is multiplication by g to the left/right

and then notice that phi_g doesn't have fixed points

[dull ginkgo]

I was more focused on the subgroup inclusion lattice esque structure for a while and someone told me to not do that lol

[cloud solar]

Unless g is e

[dire siren]

yes

in your case, g is an element of order 2, so that's enough to guarantee that there are no fixed points, and so the number of 2-cycles must be 2n+1

[cloud solar]

Ohh

I understand

We need to have the maximum number of 2-cycles otherwise the permuation has a fixed point but thats false because a has order 2 (≠e).

And the sign of the permutation phi_a must be odd (the sign is (-1)^(2n+1)

So for every order 2 element in G we get an odd permutation

And G iso to Imf where Imf subgroup of S_(4n+2)

And now we can take function pi:Imf->{-1,1} pi(sigma)=sgn(sigma). This map is surjective. This is a morphism so by first isomorphism theorem Imf/ker(pi) isomorphic with {-1,1} which is isomorphic with Z/2Z

So ker(pi) has index 2 in Imf.

But since G is isomorphic Imf so ker(pi(f)) has index 2 in G

So our normal subgroup of index two is the kernel of the morphism pi(f) right?

[dire siren]

@cloud solar I think it is f^(-1)(ker(pi))

but yes, it's right

[dull ginkgo]

I am still sorta obsessed with the ideal structure of R[X] based on R

[delicate orchid]

google "algebraic geometry" for more information

[dull ginkgo]

It can be shown prime ideals in R[X] way “lay over” prime ideals in R, and Noetherian-ness is inherited from R, so is Jacobson-ness

[dull ginkgo]

Too scary

[dull ginkgo]

The main proofs I’ve seen are using the leading coefficient map (which I don’t like because it’s not a ring homomorphism lmao) or using the prime laying over property but that requires choice for showing that every ideal is contained in a prime ideal (via the maximal ideal) so that needs strong choice

[delicate orchid]

these are such nothing objections

[dull ginkgo]

:(

[tough raven]

Ooh fun!

[tough raven]

Proofs of?

Hilbert basis theorem?

[dull ginkgo]

Yes

[stark helm]

I actually feel wondered about if K={e,} it seems that {e }is also a normal subgroup of H, can someone explain why this is wrong?

The most confusing stuff is that I can prove it, but also can figure out an counterexample( that seems true)

[rustic crown]

if you're picking K = {e} then H/K = H needs to be abelian.

[plain bloom]

Is there a way im supposed to know the group operation they're considering the homomorphisms under

cuz they don't specify

and idk whether we're considering the kernel to map to the additive or the multiplicative identity

[coral spindle]

Yes, it should be clear

[delicate orchid]

it's immensely obvious from context

[plain bloom]

ok

[plain bloom]

true

[delicate orchid]

Z/8Z isn't a group under multiplication, for instance

[chilly ocean]

Let $A = \begin{bmatrix} 1 & 1 \ 0 & 1
\end{bmatrix}$ over $V = \mathbb{F}_2$. Then $\rho : \frac{\mathbb Z}{2 \mathbb Z} \to GL(V)$ given by $\rho(0) = I$ and $\rho(1) = A$ is an irreducible representation right? I'm confused because $A$ is a root $X^2-1 = (X+1)(X+1)$ and I was told that when this happens then the matrix is diagonalizable. But it doesn't seem to be the case as the only eigenvectors of $A$ are multiples of $\begin{bmatrix} 1\0 \end{bmatrix}$.

[cloud walrus]

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

[south patrol]

When the eigenvalues are all distinct, yes it is diagonalisable. But that isn't the case here!

[chilly ocean]

oh yeah ofc

[south patrol]

Indeed uh

A matrix is diagonalisable iff its minimal polynomial is a product of distinct linear factors

Here the minimal poly is seen to be (x-1)^2

indeed it cannot be x-1 since x is not the identity

[chilly ocean]

yeah

I guess its because I was first taught over perfect fields (char 0 even)

so minimal poly product of more than 1 linear factors => distinct factors

[south patrol]

Well F_2 is a perfect field

[chilly ocean]

oops

[south patrol]

(any finite field is)

[chilly ocean]

yeah uh

[south patrol]

I'm not sure what you mean by this

[chilly ocean]

i guess the algebra GL(F_2^2) isn't perfect in some sense then?

[south patrol]

I'm not sure what that'd mean either hm

I think it's just that you didn't realise x^2 - 1 doesn't split into distinct factors over F_2, which is fair enough right

[chilly ocean]

Can the minimal poly of a matrix over C be a product of more than 1 equal linear factors?

[south patrol]

Also, I assume you mean V = F_2^2

[south patrol]

Yes

Any non-diagonalisable matrix

For example, non-zero nilpotent matrices exist

and their minimal poly is x^n for some n > 1

[chilly ocean]

right, of course

[south patrol]

(This is true over any field since you can always do
[0 1 ]
[0 0])

[chilly ocean]

Anyway, over char not 2, there are only two irreducible representations of Z/2Z right?

up to isomorphism

the trivial one and the one given by \rho(1)=-1

both 1-dimensional