groups-rings-fields

[lunar coyote]

then there exist some u,v where they are not multiples of each other

tryin to show this basically

[rigid cave]

well I = nZ

[lunar coyote]

at the end i get r_n which must be equal to n right because it's the smallest positive integer

[rigid cave]

If we are talking about ideals in Z

[lunar coyote]

yeah

z

[rigid cave]

I would do this like this: Assume that it's false, ie I is not a subset of nZ. Then there exists an a in I that is not in nZ. But since a is not in nZ then a=zn+r where n>r>0 (division alg.). This means that r = a -zn. Since -zn in nZ then r need to be in I. This is a contradicion since now r is in I and r is also less than n

[lunar coyote]

i see

would the way i did work?

[rigid cave]

I am not entirely sure tbh. How do you know that r_n = n?

[lunar coyote]

because u,v are not multiples of each other

and so the gcd must be the smallest integer that exist in the subset I

if i run euclids algorithm i will eventually reach the gcd(u,v)

idk if it would work

[rigid cave]

Yeah that sounds good to be, but don't take my word for it. I am still a noob

[lunar coyote]

bro i thought you was sick at abstract algebra

😆

[rigid cave]

Dude I am still in highschool....

[lunar coyote]

damn so you a maths enthusiast

since you're doing abstract algebra at this age

[rigid cave]

Well I am about to go into university so I guess that I should be old enough to do this. Anyways, let's hope that someone else answers!

[next obsidian]

To answer the question in that photo?

[lunar coyote]

wa

[next obsidian]

It’s basically just applying Bezout’s theorem

[lunar coyote]

my photo?

[next obsidian]

If there was something which isn’t a multiple of n, call this m

Yes

Then the gcd(n,m) < n

[lunar coyote]

idk whats bezouts theorem

[next obsidian]

Err

Bezout’s lemma

You should look it up

It’s a basic number theory result that’s important to know

But the point is simply that gcd(n,m) < n

[lunar coyote]

oh

[next obsidian]

And you can show that gcd(n,m) is in I

[lunar coyote]

ax+by = gcd(a,b)

[next obsidian]

Yup

[lunar coyote]

yeah didn't know the name

[next obsidian]

So you’ll always have ax+by in an idea because like

Well... it’s just the sum of a lot of things in I

[lunar coyote]

this is what i did

[next obsidian]

So anyway, you can just immediately contradict the minimality of n

Yeah you could use the Euclidean algorithm

[lunar coyote]

could you verify if my statements are true

[next obsidian]

I don’t understand what you’re dividing by after the first step

What is u_n?

[lunar coyote]

thats just the euclids algorithm process

of finding the gcd

cuz u can get the gcd from euclids algorithm

[next obsidian]

Oh you’re just finding the gcd of u,v?

[lunar coyote]

yeagh

maybe i should make it clearer

[next obsidian]

I mean if that’s the case then sure

I mean honestly I do t know how to find the gcd using the Euclidean algorithm lmao

[lunar coyote]

waaat

[next obsidian]

Which probably isn’t good

[lunar coyote]

thats basic numer theory

jk

[next obsidian]

Idk I don’t know any number theory

[rigid cave]

same

[lunar coyote]

but it is number theory

[next obsidian]

I mean it is 😔

[lunar coyote]

it's tedious

[next obsidian]

I just picked up any NT I needed along the way

[lunar coyote]

but what you gonna do if you don't have a caclulator and you have to figure out pi to 100 digits

you probably don't don't use euclids algorithm for that lmao

[rigid cave]

I would kill myself

[lunar coyote]

same

[next obsidian]

I guess I’m a bit confused how you know r_n is in I

[lunar coyote]

because n is the smallest positive integer

[next obsidian]

That’s in I

I mean the smallest positive integer is 1 lol

But n could be anything (well actually it has to be prime)

Wait no it doesn’t

It can be literally anything

[rigid cave]

yeah, if it's prime than it's a field, right?

[next obsidian]

No...

The quotient is

[rigid cave]

Oh...

[next obsidian]

It’s still just an ideal if it’s prime

[lunar coyote]

n.z is an ideal

[next obsidian]

Yeah

[lunar coyote]

but it can't be empty

so it must have a positive or negative or just 0

my arguement is that

[next obsidian]

No I understand lol

[lunar coyote]

there must exist a smallest integer

[next obsidian]

But my point is you’d need to argue r_n is in I

Obviously I has a smallest positive integer, this is just well-ordering

Like... I actually don’t buy this anymore

[lunar coyote]

wouldn't it be in I because it's the gcd(u,v)

[next obsidian]

The gcd is gonna be a multiple of n

But not necessarily n

I mean sure, but given your steps it’s unclear how that’s true

Anyway, it can’t possibly be right, like consider 2Z

And the elements 4,8

Their gcd is 4

So you couldn’t possibly prove the gcd of two arbitrary elements in I is just n

[lunar coyote]

àà

so back to the drawing board then

😢

[lunar coyote]

boss you still here i fixed it maybe this may be right 😄

[maiden ocean]

is there a nice relationship between the complement of a union of prime ideals and divisors?

specifically im trying to show that if S is multiplicatively closed then if x is in the complement of the union of prime ideals not meeting S, x is a divisor of some element in S

my idea was that we have a maximal ideal m containing x, and since x isnt in any prime ideal not meeting S, we have y in m cap S

and then somehow i think we should be able to show that xy is in S

but im not sure how to do that

[severe kelp]

You can reduce to the case where S = {1} by going to the localization S^{-1}A. Over S^{-1}A, your statement is that x is invertible (that x "divides" 1) if x is not in the union of all prime ideals

Of course all the prime ideals in S^{-1}A looks like S^{-1}p for prime ideal p not intersecting S and its inverse image under the natural map A -> S^{-1}A is just p again (not gaining anymore unlike some non-prime ideals)

[severe kelp]

@maiden ocean You understanding all these? Do you know localization?

[maiden ocean]

Kind of? I've seen it but passing to and back from localization is not something I'm familiar with

I think I'll get it if I think on it for a bit though

thanks!

i have seen the prime ideals in S^{-1}A stuff

so i think thatll help

[severe kelp]

Yea good

[maiden ocean]

oh i think i see, x being a divisor of an element of S is equivalent to x/s = 1 in S^{-1}A for some s

or at least x/s = 1 is sufficient

actually nah we just need to show that x/1 is invertible which is easier because x is contained in no prime ideal of A that doesnt meet S then f(x) = x/1 is not contained in any prime ideal of S^{-1}A: if it was then x/1 = y/1 for some y in p prime not meeting S, then xu = yu in p and x notin p so u in p

then p meets S, a contradiction

so now x/1 must be a unit in S^{-1}A and we have x/1 cdot s/t = xs/t = 1/1, and therefore xsu = tu in S

and this also proves that the saturation of a set is the smallest saturated set containing S

[drowsy quest]

so as groups, SU(2)/{\pm 1} and SO(3) are isomorphic as we can construct a homomorphism with kernel {pm 1} and image SO(3)

and lie group homomorphisms induce lie algebra homomorphisms as far as i understand, by evaluating the differential at the identity

my question is how does this induce an isomorphism between su(2) and so(3)?

basically how/why do we no longer have to mod out by the kernel

or alternatively i see why it induces a homomorphism su(2) to so(3) but not an isomorphism

[hot lake]

you would quotient by the kernel if the identity wasn't isolated in the kernel

since the lie algebra only sees a neighbourhood of the identity

it doesn't see -1

so no need to do anything

[drowsy quest]

i see

[drowsy quest]

could this statement be qualified by saying su(2)/(pm 1) = su(2) or no

wait never mind my previous message doesnt make any sense lol

could it be qualified by saying Lie(SU(2)/(pm 1)) = su(2)

[hot lake]

lie algebra of SU(2) / tangent space of {pm 1} at 1 (which is just {1}) = lie algebra of SO(3)

[drowsy quest]

ah i see

thank you!

[quaint tree]

Dumb question, but if I have a finitely generated group $G=\langle a_{1},\dots,a_{n}\rangle$, is $\langle a_{1},\dots,a_{n-1}\rangle\cong G/\langle a_{n},\rangle$? I feel like this should follow almost trivially from the fundamental homomorphism theorem, but everything I can think of requires $G$ to be abelian.

[cloud walrus]

Bannanachair Monarch

[chilly ocean]

Hm, intuitively it sounds right, yeah

[quaint tree]

I was about to cite that fact as part of a larger proof (involving induction on the number of generators), but something in the back of my brain said to verify it, and for the life of me I can't figure out a way to verify it for non-abelian groups.

[rustic crown]

I don't think that works

G = <a, a> = <a>
Then you're asking whether <a> is isomorphic to G/<a> = {1}

you need some sort of independence condition

[chilly ocean]

Ah yeah, I assumed they are not generatable by the previous elements

[rustic crown]

one straightforward condition i can think of is, if f(a1,...,an) = 1 is a relation among a1,...,an. That is f(a1,...,an) is a finite product of elements from the set {a1,...,an,a1^-1,...,an^-1}.

if for every relation f(a1,...,an) = 1 . then f(a1,...,a_{n-1},1) = 1 is also a relation then the claim is true.

[rustic crown]

for example in this case i have the relation a1^-1 * a2 = 1 but i can't replace a2 by 1

[chilly ocean]

I think for sure you can say G/<an> = <presentation of G with the added relation an=1>, and my guess is that you can remove the relation an=1 and the generator an as long as this doesn't "fuck" with the other relations

[rustic crown]

wait this still doesn't work

[chilly ocean]

The first claim or the second claim?

[rustic crown]

maybe it does... i need to verify it

if G = <X | R> then G = F(X)/<<R>>, and then <X | R, a> = F(X)/<<R, a>>

just using <<R>> meaning the smallest normal group of F(X) containing R

so by the theorems we do get a map from <X | R, a> --> G/<a>

oh ig this follows from the correspondence theorem then

[viscid pewter]

wait, hang on. Q8 is generated by <i, j>, <i> = Z4, Q8/<j> is not Z4?

or am i missing something

[rustic crown]

yea so the problem is i and j **** with each other

[viscid pewter]

so i think you need something stronger

[rustic crown]

this is a simpler counter example

[viscid pewter]

no, but even in the case where neither of them generate each other

[rustic crown]

i don't see if this reduces to some nice condition

[night bison]

How would I go about finding a representative for an equivalence class in a quotient ring

The specific problem is finding a representative of $[x^5]$ in $\frac{\mathbb{Q}[x]}{\langle x^3-5x^2+7x-9\rangle}$

[cloud walrus]

SelahW

[night bison]

lol

[chilly ocean]

x⁵ is a representative for [x⁵]

maybe you want one whose degree is <= 2

you could divide x⁵ by the generator of the ideal

youll get x⁵ = gq + r where deg r <=2

where g is the generator im too lazy to type

gq = 0 in the new ring so...

[night bison]

Ah

I got it I think

[drowsy quest]

does inner product of the character of a representation with itself being 1 imply that the representation is irreducible, for finite-dimensional representations of compact groups?

(i know it is true for finite groups)

[compact needle]

According to page 58 of http://math.tkk.fi/opetus/harmanal/pruju/group08.pdf

It is true, at least for unitary representations

[untold sapphire]

Does anyone know about how to construct free resolutions?

Ideally minimal ones

[rustic crown]

find minimal generating set

Then you get a map from R^(A) --> M

find minimal generating set for the kernel

So you then get the map R^(B) --> R^(A) --> M

keep on doing this

[untold sapphire]

Yeah that gives me the presentation matrix right

For the first step

[rustic crown]

presentation matrix is the matrix corresponding to the linear map R^(B) --> R^(A)

[untold sapphire]

Ok cool

[rustic crown]

but you can go further if your ring wasn't like a PID

[untold sapphire]

Is there a good way to check if it's minimal or do I basically need to eyeball it

Like what if I find generators that I can't remove any of them but they aren't minimal

I think that can happen?

[rustic crown]

in the end it probably depends on what you want to do with the minimal generating set

most nice things would be true for any resolution you take

[untold sapphire]

I need betti numbers

Well, graded ones

I think this problem is too hard ha

My supervisor suggested I look at it

[rustic crown]

i haven't read much about this, so i can't really say

[untold sapphire]

Ok thanks anyway

Do you know if there is another math discord for PhD students

I feel like this one isn't very active

[rustic crown]

sorry i don't know 😦

maybe you can try asking in #math-discussion or similar cause they are far more active?

[drowsy quest]

thank you!

[next obsidian]

This one isn't active?????

this is like the most active math server

I'm pretty sure computing minimal free resolutions is like.... a hard problem

Pretty sure commutative algebraists are interested in them because they're useful but hard to find explicitly

[rustic crown]

this is like 100x more active than any other server i'm on

[chilly ocean]

I hear jacobians server is active

u got inv?

[carmine fossil]

I think they want a high math server

[chilly ocean]

Would you want it if you could have it?

But no, I don't have an invite

Joining the forbidden server

jaco dm inv

It's like joining voldemorts army. It is a powerful one, but it is full of evil

Yeah but this one's the voldemorts one.

[golden pasture]

[rustic crown]

do you know that finite fields of same size are isomorphic?

try to look at the polynomial x^(p^2)-x

do you also know that any two splitting field of a polynomial are isomorphic, or are you still reading that?

if you don't, then we can avoid that path of splitting fields.

once you have a single field of size p^n then with the knowledge you have you can prove the uniqueness.

but the standard way to construct one such is by looking at splitting field of x^p^n-x

We want to show Fp[x]/(f) and Fp[x]/(g) are isomorphic

What we'll show is x^p^2 - x factors completely in both fields

hence we'll get for free that f and g both divide x^p^2 - x

because in Fp[x]/(f) the minimal polynomial of (coset of) x is f

but then it will also satisfy the polynomial t^p^2 - t in (Fp[x]/(f))[t]

In particular f factors in Fp[x]/(g)

Then get the map Fp[x] --> Fp[x]/(g) by sending x to a root of f

so f would lie in the kernel

which will give you a map
Fp[x]/(f) --> Fp[x]/(g)

This is a map between fields, hence is automatically injective. And both sides have same size, hence automatically surjective!

yep

yea you're using lagrange's theorem here. look at the multiplicative group which has size p^2 - 1.
now any non-zero element in the field satisfies a^(p^2 - 1) = 1
and hence also satisfies a^p^2 = a

0 also satisfies the last equation, which shows every every elment of Fp[x]/(f) satisfies t^p^2 - t = 0

now this is a polynomial of degree p^2, but you already know p^2 distinct roots of it

if you have a group G of size n then a^n = e in the group

because <a> is a subgroup

so order of a, which is |<a>| will divide |G| = n

so a^n = e

issokie, don't say that >.<

[rustic crown]

i was too excited that i forgot to write this lol

nanana

[warped bane]

hey everyone, I would appreciate it if someone could clarify this:

[rustic crown]

if you look at the field as an additive group then you get (p^2) a = 0

[warped bane]

if f^n = fofofo.....of

what is f^0?

[rustic crown]

but what i want is to look at the non-zero elements of the finite which forms a multiplicative group of size p^2-1. So we'll get a^(p^2-1) = 1

[rustic crown]

since you can compose the map with itself, it must have same source and target. And then there is also a notion of identity on that object!
f^0 = id
f^(n+1) = f^n o f

yep!

(pardon my english, i can't type correct grammar when i'm doing math >.<)

[warped bane]

is that by convention or rather a consequence?

[rustic crown]

more like a definition?

[warped bane]

i see

[rustic crown]

in the end you want this formula to hold f^(n+m) = f^n o f^m

[warped bane]

yeah sure

[rustic crown]

if f is also invertible, then you want it to hold for all integers

[warped bane]

that makes sense

thanks a lot!

[rustic crown]

[untold sapphire]

Yeah, there is a recent paper that finds a closed form minimal free resolution for monomial ideals that I'm trying to understand

And its built on the idea that you can compute multi graded betti numbers of monomial ideals using some simplicial homology

The second thing is the part that I'm stuck on at the moment - it's detailed in the book by Miller and Sturmfels but it's super terse

https://www.google.com/url?sa=t&source=web&rct=j&url=https://pages.uoregon.edu/njp/MS.pdf&ved=2ahUKEwjUjbO835bwAhUitHEKHaKmCpYQFjAWegQIJhAC&usg=AOvVaw1xDhPEW58NltFGjsxvl7Pa&cshid=1619262664930

The first chapter

And its really weird, like you build a bunch of simplicial complexes out of your ideal and then compute their usual simplicial homology and somehow that gives you what you want

[untold sapphire]

That sounds ominous

[vale idol]

how can we identify if a cartesian product is the same as a direct sum?

for instance i'm looking at abelian groups of order 6750, we have the product Z50xZ135
the lcm of 50 and 135 is 1350 so the elements of the group have order no greater than that
is it immediate that the group is isomorphic to Z5+Z1350 ?

i presume not, because if we have the product Z25xZ27xZ10 this also has lcm 1350 but isn't isomorphic to Z50xZ135... right ?

[hot lake]

what do you know about finite abelian groups ?

[vale idol]

they're finitely generated modules over Z

[hot lake]

do you know stuff like chinese remainder theorem ?

[vale idol]

yea

[hot lake]

or the theorem that classifies finite abelian groups

well then you can just use the chinese remainder theorem to split up everything as much as you can, then see if they are the same

[vale idol]

i'm not sure if it's the theorem you're referring to but i know how to find abelian groups of a given order in terms of elementary divisors and invariant factors

[hot lake]

yeah that

Z50 = Z2 * Z25, Z135 = Z27 * Z5, so Z50 * Z135 = Z2 * Z27 * Z5 * Z25

Z1350 = Z2 * Z27 * Z25, so Z5 * Z1350 = Z2 * Z27 * Z5 * Z25

so yeah those two are the same

Z25 * Z27 * Z10 = Z2 * Z27 * Z5 * Z25 too

[vale idol]

... huh, i didn't think about it like that, thanks!

[hot lake]

Z50 * Z135 = (Z2 * Z25) * (Z27 * Z5) too I guess

[unique juniper]

$f(x) = (x-1)\dots(x-n)-1$

anyone can help to show this is irreducible in Z[x] for all n?

[cloud walrus]

Yes

[hot lake]

sounds like something einsenstein criterion maybe can do ?

[unique juniper]

ive done some testing

eisenstiens criterion wont help

[hot lake]

or maybe not

[cursive temple]

just let a be a root of f(x), now $(a-1)(a-2)\ldots (a-n) = 1$

[cloud walrus]

Pappa

[cursive temple]

and now all of the factors of the LHS have to be integers, and their product is 1

[unique juniper]

i dont get it

how does that help

[hot lake]

that shows there is no factor of degree 1

[cursive temple]

oh rip it was to show irreducibility

mb

[unique juniper]

but uh why do we need that

[hot lake]

well it's a start I guess

[cursive temple]

okay i think ive got a soln now

[unique juniper]

nice

[cursive temple]

assume to the contrary, that $f(x) = g(x)h(x)$. Now $-1 = f(1) = g(1)h(1)$, and similarly for 2, 3, $\ldots$, n. So $g(x) = -h(x)$ for all $x \in {1, 2, \ldots, n}$

[cloud walrus]

Pappa

[cursive temple]

We get that g(x) = -h(x) because their product is -1, so one takes the value 1 and the other takes the value -1

now look at the polynomial $p(x) = g(x) +h(x)$

[cloud walrus]

Pappa

[cursive temple]

and how do i finish

[hot lake]

you say that it's either 0 or some degree >= n polynomial

[cursive temple]

yeah

[hot lake]

so it has to be 0 if the factorisation is nontrivial

[carmine fossil]

You could have just ended with f=-(h)^2 but f(n+2)>0

[cursive temple]

sure

i was thinking on using the fact that g and h monic

and so wlog g has leading coefficient 1 and so h has leading coefficient -1 so contradiction

[unique juniper]

nice

thanks

[carmine fossil]

You could have used Lagrange interpolation to get the same result in a slightly different way

[unique juniper]

im not familiar with that

actually i have a question

what do you guys mean by trivial factorization

[carmine fossil]

p=u*(vp) where u and v are units

[unique juniper]

oh right

so after considering p(x) = g + h(x)

p has degree of 0

dont think i understood how that finishes it

[carmine fossil]

p is zero at n points

[unique juniper]

yah

oh

ok'

so we assumed that f(x) = g(x)h(x), how does that contradict that? not too sure

so h(x) + g(x) = 0 for all x <= n

does that somehow mean that h(x)g(x) = 0 for all x <= n

[carmine fossil]

you can conclude p=q(x-1)(x-2)...(x-n) where q is a constant=g(x)+h(x)

Suppose q is not 0. Now,this forces one of them to have a degree which is atleast n. Let's say deg g=n then deg h=0 since deg(gh)=n. So,h has to be a constant polynomial and under the constraints is forced to be 1 or -1

Which means h is a unit

[unique juniper]

wiat

q has to be zero tho

?

[hot lake]

q is 0 at x=1...n

so either q=0 or q has degree >= n

[unique juniper]

yeah

[hot lake]

we have to study both cases

[unique juniper]

oh yes

ok isee

what about in the case that q = 0

[carmine fossil]

g(x)=-h(x)

f(x)=-h(x)^2

Consider f(n+2)

[unique juniper]

i see

sorry but i dont understand the case where q is not 0

[unique juniper]

how is this done

[hot lake]

if q is not 0 then it has degree at least n. Since both g and h have degree at most n, at least one of g or h must have degree n, and so the other one must have degree 0, so the factorisation is trivial

[carmine fossil]

If p(a)=0 then p(x)=(x-a)q(x) for some q of deg (deg p-1)

[unique juniper]

oh

[hot lake]

wait here I was thinking q was g(x)+h(x)

[unique juniper]

alright i think i got it

i think ive got a slightly better way to end this

$(x-1)\dots(x-n) = -g(x)g(x)$

[cloud walrus]

Yes

[unique juniper]

just compare the leading coefficient

[chilly ocean]

can somebody give me a hint as to how to prove that if a finite group is the direct product of its Sylow subgroups, then for each divisor d of its order there is a normal subgroup of order d?
I feel like it all comes down to proving that the non maximal p-subgroups are normal in their respective Sylow p-subgroup (I think I've shown that that is enough) but I can't do it.

maybe I'm just missing something obvious...

[carmine fossil]

Fix a p and let's say G=PxQx... Where P is the p Sylow subgroup. Let R be another p Sylow subgroup. Therefore an element in R can be written as a tuple (a_1,a_2,a_3...)

Now by lagrange a_2,a_3... all have to be identity(since x^a=1 and x^b=1 implies x^gcd(a,b)=1)

Now each element in {(x,1,1..)|x in P} can be identified with an element in P

Which means the element in R is in P

Which means R=P,i.e., There's exactly one p Sylow subgroup

@chilly ocean

[chilly ocean]

didn't we already know that? if the product is direct the Sylow subgroups are normal and therefore unique

[carmine fossil]

Yea in that case,Non maximal p-subgroups being normal is trivial

Because if g is in P
xgx^-1 is in xPx^-1=P for all x

[chilly ocean]

is P the maximal p-subgroup?

[carmine fossil]

Yes

All maximal p-subgroups are conjugate to each other

[chilly ocean]

But that doesn't tell me that the non maximal p-subgroups are normal in P right? that just tells me that if Q is a non maximal p subgroup and x is in P then x^-1 Q x lands inside P,am I missing something?

[carmine fossil]

x^-1Qx=Q is the normality condition

mb

[chilly ocean]

I know, but what you've told me doesn't prove x^-1Qx=Q

it only proves that it's in P

[carmine fossil]

That doesn't feel true

[chilly ocean]

I know

[carmine fossil]

I think the best approach is to induction

Assume every p-group upto order p^k has a normal non maximal subgroup of every order below it

prove that holds for p^k

Assume that holds true till |G|=p^k
Take G,such that |G|=p^{k+1}

Now |Z(G)|>1 by class formula

Which means there's a normal subgroup of order p

Call this subgroup N

Then G/N has a normal subgroup of order p^m for any m<k

[chilly ocean]

(I'm going afk for a bit, thank you for helping me)

[carmine fossil]

And for every normal subgroup P in G/N you have a normal subgroup in G with order |P||N|

and you are done

[limpid pilot]

Hi everyone.. could anyone answer me why linear algebra is useful in group theory?

I know there's a link between both called representation theory

[carmine fossil]

You know ring actions?

[limpid pilot]

Not really

[carmine fossil]

You know group actions?

[limpid pilot]

yes kinda

[carmine fossil]

So,You basically let a ring act on an abelian group to get a structure called module

And vector spaces are special cases of these structures when the rings are fields

[limpid pilot]

Hmm this is a bit too complex

[carmine fossil]

Ok, Let's say you make a group G act on another group H:
along with the usual group action properties you add a new property:
(h.a)(h.b)=h.(ab)

Under this property action of h on G can be completely defined by action of h on it's generators

In other words,The generators "form a basis" and action of h is like a "Linear Transform"

This is like a pretty important property in group theory

Because see semi direct products

@limpid pilot

[wraith obsidian]

My brief answer without technical details would be: When you represent a group by linear transformations, you have the advantage that this representation always „decomposes“ into complementary parts (unless it's „simple“). Groups don't have this; not every (normal) subgroup allows for a complement (e.g. take C2 in C4). This makes the structure of representations simpler to study.

For instance, we can represent the cyclic group of three elements C3 by linear maps ℝ³→ℝ³ by cycling the basis vectors, i.e. the generator x of C3 would be assigned the linear map Tx: e1↦e2↦e3, the neutral element e would be represented by the identity over ℝ³, and x² would be represented by (Tx)².
This representation of C3 is in some sense the „simplest conceivable“, but even that decomposes into two „subrepresentations“: All of these transformations leave the line through e1+e2+e3 invariant. And the plane orthogonal to it is also invariant, giving us actually an „even simpler“ representation where we just rotate a plane by thirds!

These „even simpler“ representations tell one a lot about the group. But please don't ask me what, lol.

[limpid pilot]

Well thank you so much @carmine fossil and @wraith obsidian for the great explanations. It helped me a lot and I'm one step closer. It's extremely appreciated!

[carmine fossil]

if you want,here's an example

define G=<h,k|h^3=1,k^2=1, khk^-1=h^2>

let k.h=khk^-1

this can be seen as Z_2 acting on Z_3 as k.(x)=2x,if you let 2x to mean x^2

[limpid pilot]

There’s a link between linear algebra and group theory called representation theory. It converts the world of groups into the territory of linear algebra. Linear algebra specifies the arithmetic behind different kinds of linear transformations. In the representation theory a matrix is assigned to each element in a group following certain rules and it also needs to respect the relationship between elements in the group. If those guidelines are respected, we can define it as a representation of a group. This is exactly why linear algebra is useful in group theory. Basically, the groups can be extremely complicated and this kind of representation of groups provide a simplified overview of these groups that resembles a simpler version in the format of linear transformations. With the linear transformations we have a simplified look at the groups and are able to understand the groups without having to get into the complexity of it.

What do you think of my explanation?

[carmine fossil]

I think so

Also, idk why but representation theory feels more like module theory,rather than linear algebra in particular

[chilly ocean]

hey @carmine fossil I just wanted to thank you for bearing with me and for helping me solve my problem

[wraith obsidian]

Well… a k-linear representation of G is a module, over the group ring kG
and the invariant subspaces / subrepresentations are just submodules

or are you aiming at something different

[carmine fossil]

idk any representation theory but yea I think that feels right

[strong valve]

not clear what you did

can you explain a bit more and concretely

[fossil lodge]

I think you should look more closely

[lunar coyote]

tried it again is it correct?

[rustic crown]

a few things...
where you do (=>)
This means you're assuming the thing on the left and you're going to prove the thing on the right. So you know I is not R, hence it doesn't make sense to start with "Assume I = R"
But i think you meant <= here. so that's okay

where you actually say <=, i don't understand how union got in the picture to begin with. And R^* contains 1 and I contains 0, so assuming their union is empty is just wrong... how does that prove anything?

[lunar coyote]

i was trying to do proof by contradiction

assuming the left is false

and attempt to do that with each side

[rustic crown]

the left thing is false when I = R

but when you're doing => then to do a proof by contradiction you'll assume the thing on the right is false and try to see where things go wrong

[lunar coyote]

o

well i fucked up

thought it was the otherway round

[lunar coyote]

would this be the correct way to start?

[rustic crown]

yea makes sense
but that implication is always true, the right contains 0 whether or not I=R.

[lunar coyote]

o

so is there no point in trying to prove that impliction then if it's always true

because i'm trying to prove by contradiction

[rustic crown]

you need to show intersection is empty in this case!!!

[maiden ocean]

im trying to show that if N is the nilradical then A/N absolutely flat implies that every prime ideal of A is maximal

i know already that A is absolutely flat iff A_m is a field for all maximal m

so if i take a prime ideal p of A then i can pass to A/N and then (A/N)_m (or the image of m in the quotient whatever) and p is a prime ideal in (A/N)_m which is a field so p = (0) in (A/N)_m

so then for x in p it passes to x + N

and then ux + N = 0 for some u not in m, right?

i.e ux nilpotent

im not really sure where to go from here

maybe im going about this all wrong xd

[golden pasture]

i believe fastest way is show that spec A is hausdorff

[hollow comet]

The part about the nilradical is irrelevant, since the primes of A are the primes of A/N

[golden pasture]

yea

[hollow comet]

So with that reduction in mind, I will just show that in an absolutely flat ring A every nonzero prime ideal is maximal. Since a quotient of an absolutely flat ring is again absolutely flat, we may reduce the problem to showing that A/P is a field for every prime ideal P. The claim is that any absolutely flat ring which is an integral domain is automatically a field.

@maiden ocean Do you think you can proceed with that or do you want more hints?

[golden pasture]

sounds like the full solution already ngl

[hollow comet]

I mean, you have to prove that ID + absolutely flat gives a field

Hahaha

But maybe they have that as a known theorem or something

[golden pasture]

i believe it was one of AM's exercises lul

[untold sapphire]

Does anyone have a good textbook recommendation for Hopf Algebras?

[wraith obsidian]

I think ℝG is a good one

[untold sapphire]

?

Lol thanks I guess

If anyone has any actual recommendations that would be cool

I'm finding Sweedler a bit hard to read

[maiden ocean]

its not, though local + absolutely flat -> field was

[golden pasture]

i cant rmb at this point

has been too long ago

[maiden ocean]

i think this makes sense though I also have to show that the quotient of an absolutely flat ring is absolutely flat :p

but that seems not that hard by using the principal ideals are idempotent characterization

[golden pasture]

i think can directly show spectrum is hausdorff?

[maiden ocean]

i could but the exercise is to show that 4 things are equivalent including that haha

[golden pasture]

lmfao

well 2,3,4 are kinda immediately equivalent lul

[golden pasture]

actually you can just localize everywhere and done

[maiden ocean]

oh i got it

[golden pasture]

nisu

[chilly ocean]

i am absolutely flat too

[past temple]

Given a finite field extension E/F such that E is the splitting field of some subset S of F[x]

how do i prove that E is also the splitting field of a finite subset of S

[rustic crown]

if E = F then you can take the subset to be empty
Else pick a polynomial f1 in S and its splitting field E1. if E1 = E then the subset is {f1}, else
pick a polynomial f2 in S that doesnt' factor over E1, say E2 is the splitting field over E1, if E2 = E, then {f1,f2} does the job else
....

this has to end as the degree [En:F] keeps increasing

[final pasture]

your pfp is scary

also flat is justice

[rustic crown]

please say that's not Yui

is that a yes or a not >.<

[next obsidian]

Wtf is absolutely flat

[maiden ocean]

every module over it is flat

[next obsidian]

Weird

[maiden ocean]

no u

[next obsidian]

Fuck you

[maiden ocean]

@mods please ban immediately

[next obsidian]

Fuck you

I won’t die bitch

[shut halo]

What is the rank of a Lie Algebra?

[golden pasture]

samengl

[ember fable]

hi

[old lava]

https://en.wikipedia.org/wiki/Von_Neumann_regular_ring this apparently

[delicate orchid]

woah that's cool

[broken vale]

is this true?:
If G is a group, A is a normal, abelian subgroup of G, then for all a in A and g,h in G, gag^-1 = hah^-1

[delicate orchid]

it's trivially true for g,h in A

[broken vale]

ye

[delicate orchid]

I'll see if I can prove it for all in G

[old lava]

I don't think it's true

[broken vale]

pls thank you so much

[junior moth]

it isnt true

[broken vale]

me neither

[delicate orchid]

I'm leaning towards it not being true as well

[old lava]

I think you can look at S_3

for a counterexample

[delicate orchid]

surely it doesn'

yup that's exactly what I was about to say

S_2 in S_3

wait S_2 isn't normal in S_3

epic fail

[old lava]

<(1 2 3)> is

[delicate orchid]

true

[old lava]

oh wait nvm

it's not abelian

[broken vale]

okay so then why does my professor tell me that this action is independent of choice of g bar?

[delicate orchid]

<(1 2 3)> is isomorphic to C3 no?

[old lava]

wait nvm, ya

it is abelian

what am I saying

S_3 is a counterexample

[delicate orchid]

<(1 2 3)> isn't normal either lmao

I think the only normal subgroup has to be A3

[old lava]

it is normal though?

it's the unique group of order 3 in S_3, it necessarily is normal

[delicate orchid]

lemme double check I'm rubbish at permutation mulitplicaiton

[old lava]

you don't have to check anything

it's very obviously normal

it's characteristic even

it's also of index 2, another reason it's normal

it's one of the most obviously normal subgroups of all time

[delicate orchid]

wait hold on lemme check tho

[broken vale]

okay yes it’s definitely normal and abelian but can u guys thing of a g and h in S3 for a counterexample?

[delicate orchid]

https://tenor.com/view/troll-face-creepy-smile-gif-18297390

uhhh

(12) is my first thought

[broken vale]

ya

ya that works

[old lava]

ya, it does work

any element of order 2 works

with identity

[broken vale]

identity and (1,2) works

[old lava]

as the other one

[broken vale]

okay so then what the fuck my prof talking about

[old lava]

better ask your prof

[broken vale]

can’t he sucks LOL

[old lava]

get L danced on

then

[broken vale]

like am i missing something?

[delicate orchid]

I see the word homology and I pipi my pampers bro

[broken vale]

dw about homologous

homology*

it’s just an exact sequence of groups

[delicate orchid]

well he's said it's ga = gag^{-1}

[old lava]

why is anarchy chess leaking into this server

[broken vale]

so A->E is injective E->G is surjective

and image of A->E = kernel of E->G

[delicate orchid]

yeah I know lol

[broken vale]

somehow that and A being abelian concludes that g • a is independent of choice of g bar

and i have no idea why

[delicate orchid]

hold on g bar is in E not G

[broken vale]

correct

[delicate orchid]

spooky

[broken vale]

also when he writes

$\bar{g}a\bar{g}^{-1}$

[cloud walrus]

danmarino900

[broken vale]

i believe he means the image of a in E

[delicate orchid]

yeah under the module structure he's defined

[broken vale]

but i just don’t understand why the independence

[delicate orchid]

wait so the mapping from A to E is injective and E is the image A under this mapping correct?

[broken vale]

If A was in the center of E it would make sense

no

A->E is not surjective it’s injective

[delicate orchid]

ah yeah of course

yeah I'm out of ideas soz boss!

[broken vale]

all good ty for trying

[sturdy marsh]

any two representatives differ by an element of A. A is abelian.

[broken vale]

? i don’t understand

[sturdy marsh]

assume g and g' in E represent the same element in G (i.e. map to the same element in G)

then g^{-1}g' is in A

[broken vale]

ye

[sturdy marsh]

so g' = ga for some a in A

[broken vale]

wait no

[sturdy marsh]

?

[broken vale]

how’d you arrive at that conclusion

[sturdy marsh]

the sequence is exact

so ker(E --> G) = im(A ---> E)

[broken vale]

oh you mean it’s in im(A->E)

[sturdy marsh]

sure

[broken vale]

okay gotcha i’m with you now

[sturdy marsh]

so g' = ga

for some a in im(A ---> E)

agreed?

[broken vale]

yes

[sturdy marsh]

for any x in im(A ---> E)

[broken vale]

alright i think i see it

[sturdy marsh]

g' x g'^{-1} = ga x (ga)^{-1} = ga x a^{-1}g^{-1} = g aa^{-1}x g^{-1} = gag^{-1}

[broken vale]

gotcha. okay thank you so much!

p(g^{-1}g’)=1 is what my brain was missing lol

p:E->G btw

[sturdy marsh]

right

a possibly simpler way to put it is

E acts on A by conjugation (as A is normal)

the restriction of the action to A is trivial (as A is abelian)

so the action factors through E/A (which is G)

[broken vale]

ugh

the term “factors through” reduces a year off my lifespan whenever i hear it

😂

so it’s like the elements of the fiber of any g (which kind of represent the same element in E/A) in G acting on any a in A coincide

it’s kind of like an action of E/A on A that’s well-defined since the sequence is exact

right?

[sturdy marsh]

it is well defined as A is abelian

we have E/A = G due to exactness

[broken vale]

and a quotient by a normal abelian subgroup is special since that allows you to “factor” the action of E on A to E/A on A?

[sturdy marsh]

no

the reason why it factors through A is because A acts trivially on A

because it is abelian

[broken vale]

are you saying this?: an action E on A factors to E/A on A if and only if the action of E on A, restricted to A on A, is trivial

[sturdy marsh]

yes

this is the first isomorphism theorem

[broken vale]

oh shit

[delicate orchid]

I thought it was when I saw E/A = G

[broken vale]

😮

how is that the FIT?

[sturdy marsh]

this is also the first isomorphism theorem but not what I meant

[delicate orchid]

hey I'll take that as a win considering how bad I am at sequence stuff

[sturdy marsh]

you have a map E ---> S_{A} given by the action

A \subset Kernel of the map

so the map factors through E/A

okay, it's not quite the first isomorphism theorem

but this is how you map out of quotients

[broken vale]

alright lol that makes sense though

[broken vale]

we just so happen to have this because the action of E on A in this case is conjugation and A is abelian and normal

[sturdy marsh]

yes

[broken vale]

dang

life makes sense now lol thank you

[thorn delta]

Its true that a polynomial in a finite field of char p splits in a splitting field of x^{p^k} - x for some k, right?

I have a theorem that says that if F/K is a splitting field of a polynomial of degree n then [F:K] <= n! and another theorem that says F is a field with p^n elements iff it is the splitting field of x^{p^k}-x for some k.

Combining these, if you were to take the splitting field of an arbitrary polynomial f of degree n in a finite field, then its splitting field is finite by the first theorem i listed, and therefore has a power of p elements. Then the second theorem shows that the splitting field must also be a splitting field of x^{p^k}-1 for some k

[oblique river]

@thorn delta that's right

Extensions of finite fields are unique -- For each prime p and each positive integer k, there is a unique field of order p^k

and it's the one you described: The splitting field over F_p of x^{p^k} - x

[thorn delta]

nice, thanks!

[chilly ocean]

Is there something an ideal can't be? Or can it be any subset within a ring?

erm... not every subset of a ring is an ideal, if that's what you mean

Why not? As long as we can "describe" it?

,w ideal of a ring

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by 88ddda#7872

[chilly ocean]

[hidden haven]

[chilly ocean]

i think you know, an ideal is a subset of a ring that is closed under addition and "absorbs" multiplication, in the sense if r is an element of R and i is an element of I, then r*i is in I

and is non-empty

[thorn delta]

im reading this on wikipedia. How does an automorphism in Gal(Q(sqrt a)/Q) induce an automorphism in Gal(C/Q)?

i mean, I have only seen extensions up to a splitting field. I'm not quite sure how to go about some even larger extension

[chilly ocean]

maybe every automorphism of Q(sqrt(a))/Q can be extended to C/Q by zorn's lemma?

this feels kind of weird though, i feel like there should be an automorphism that extends in a natural way, like fix as many elements as possible

i dont think so cuz needs axiom of choice to show theres automorpisms of C other than conjugation

ah hm

https://math.stackexchange.com/questions/412010/wild-automorphisms-of-the-complex-numbers

interesting

[thorn delta]

possibly a dumb question, but how does this part work that I've underlined?

[rustic crown]

That's because C/Q(T) is algebraic and C is algebraically closed, so by definition C is the algebraic closure of Q(T)

if F/k is algebraic and you have a map k --> L with L algebraically closed, then you can extend this map with zorn to F --> L

Since you have the map
Q(T) --> Q(T) --> C
You can extend it to
C --> C

[thorn delta]

oh yeah... 🤦‍♂️

[rustic crown]

This had to be an isomorphism else you get a non trivial algebraic extension of C

Wild Automorphisms are Wild.

-- Yui Hirasawa :p

[chilly ocean]

yui hirasawa?

[rustic crown]

She's from the anime K-On!

https://tenor.com/view/kon-fun-things-are-fun-anime-cute-yui-gif-16787421

Oh I became very active

[rustic crown]

thinking about it we can have an an automorphism of C that fixes the algebraic closure of Q other than the identity

[chilly ocean]

yui chan

[strong valve]

Was she not on season 2 and 3?

[rustic crown]

season 3

only 2 seasons

[strong valve]

Sad

[chilly ocean]

[rustic crown]

[delicate hawk]

is a submodule just a subgroup of a module?

[rustic crown]

and closed under multiplication by elements of the ring

[delicate hawk]

aah

[old lava]

I believe all you have to check is that a subset is nonempty and x - r y is in the subset for every x,y in the subset and any r in the ring

[final pasture]

I could watch this gif a whole hour

[rustic crown]

sameeee loool

:YuiGoggles:

[latent ingot]

[strong valve]

noooooooooooo

[quaint tree]

How many non-abelian groups are there of order $900=2^{2}\cdot3^{2}\cdot5^{2}$?

[cloud walrus]

Bannanachair Monarch

[next obsidian]

Idk, probably like at least 3

[rustic crown]

(what will you do with that information lol?)

[quaint tree]

I don't know, I was just bored and thought of the problem and couldn't figure it out

I'm wondering if it's a known result or if it's actually really hard

[final pasture]

fun fact:
for any n <= 63, there is less than 50 non abelians group of order n (at most 15 except for 2 numbers that I forgot)
and then, for n = 64, there is exactly 256 non abelian groups

[quaint tree]

Prime numbers have no nonabelian groups, as do squares of primes. If p is prime, there is exactly one of order 2p, which is the dihedral group.

n=64 is just an absolutely bonkers case from what I've heard.

[chilly ocean]

t

[unique juniper]

so

i dont know why p(theta) = 0 ?

like

whats the point of using theta like this

[delicate bloom]

probably helps to see an example like p(x)=x^2+1 is pretty popular with p(theta)=0

then you can represent complex numbers as a+b*theta

[unique juniper]

how is theta chosen?

[viscid pewter]

so suppose the generating elements are like, a, b, c

then you just get all elements of the form (a^m)(b^n)(a^p)(c^q)...

and collapse them in accordance with the relations

sure

sometimes it's quite difficult to tell if you've found them all

do you have a specific problem or

hmmm

i wonder if you can collapse this somehow

zx = xyz = xzy = yxy, nothing suggests itself

oh, yes

i mean if they're all distinct then it's really nice

and if they're not then it's still easier

so that's basically says that all you ever need is (x^i)(y^j)(z^k)

there's no need for nonsense like xyzyzyzyzxyzxyzxyzy that can't be collapsed

[delicate orchid]

With a group that small it’s probably feasible to write out the Cayley table

[viscid pewter]

ew

it's still a 12x12

if nothing collapses

[delicate orchid]

It’s gross but still

Would get the job done

[viscid pewter]

i think i'd just do it the straightforward way in this case

just write out all the (x^i)(y^j)(z^k) and multiply it by itself and twiddle it down with the relations

hmmm

cayley table might not be such a bad idea

eh, no, still overkill

well maybe

that looks about right

[delicate orchid]

That looks right to me

[viscid pewter]

no

just need to find the thing for all of them

tf

i count 12

in that set you gave

it might be, but it's not necessarily

for example in dihedral of the triangle thingy you have r^3 = s^2 = 1 with rs = sr^-1; then (rs)(rs) = (rs)(sr^-1) = 1, so order of rs is 2

in fact it's a reflection

[delicate orchid]

Yeah thats an example of a relation that causes a collapse

[viscid pewter]

you would need quite a convenient group for the hcf to do the thing, i mean the lcd would work for (finite) abelian groups i think? am i smart

[delicate orchid]

For order of the elements?

[viscid pewter]

this

i think it would have to be, what

just (Z_n)^m or something

ok i am just wildly conjecturing

whatever

[delicate orchid]

No that sounds right to me

Sounds being the operative word lol I’ll have to think about it more

It’s def true for cyclic groups

[viscid pewter]

yeah the non-cyclic groups are the hard bit lol

maybe there is a shortcut i completely missed

[rustic crown]

like you noticed you can write every element in the form x^i y^j z^k
so there are at most 12 elements in the group. Lets consider the case when all these are different, so we have a group of order 12.
zx=xyz is same as saying [x', z] = y which means this group isn't abelian. now y and z commute so {1, y, z, yz} is a subgroup of order 4, since x'yx = z this subgroup is normal but not contained in the center.

if you recall there are only 3 non abelian groups of order 12, the dicyclic, dihedral and alternating. dicyclic doesn't have a normal subgruop of order 4, and in dihedral that subgroup is the center.

so if these elements were distinct, then the group would be A4. So it suffices to show that A4 admits that presentation.

if not, the group is smaller and we need to look further.

yep

[viscid pewter]

no

[delicate orchid]

I knew yz=zy would simplify loads somehow

And no |C5| doesn’t divide 12

[viscid pewter]

if there's Cn there's something with order n i believe

[delicate orchid]

And ^

There’s no elements of order 4

So C4 cannot be a subgroup

[viscid pewter]

lol, proof by knowing all the small groups

bloody hell

[rustic crown]

if you find elements x, y, z in A4 satisfying those conditions then you'd directly prove the isomorphism (so forget how you got the idea to look at A4 😛 )

word problems are hard

[viscid pewter]

there could be one isomorphic to S3

idk, lemme see

[rustic crown]

A4 doesn't have any subgroup of order 6

[viscid pewter]

probably not

yeah

looking at the presentation it just doesn't feel like there's a subgroup with either only x, y or x, z

i'm not sure what this has to do with anything, just because 6 divides 12 doesn't mean it's in there

[rustic crown]

sage: A                                                                         
Alternating group of order 4!/2 as a permutation group
sage: (x, y, z)                                                                 
((1,2,3), (1,2)(3,4), (1,4)(2,3))
sage: (y*x, x*z)                                                                
((1,3,4), (1,3,4))
sage: (z*x, x*y*z)                                                              
((1,4,2), (1,4,2))
sage: (y*z, z*y)                                                                
((1,3)(2,4), (1,3)(2,4))

so the proof now is very simple
G = <x,y,z | x^3=y^2=z^2=e, yx=xz, zx=xyz, yz=zy>

since x, y, z satisfy the relations, we get a group hom
G --> A4 which is surjective

but like we argued |G| <= 12

this means G must be iso to A4

[viscid pewter]

lovely! just one flaw

i'm not certain they know what A4 is yet 😄

[maiden ocean]

[rustic crown]

@strong valve stop me from being here

[strong valve]

I am going to write to modmail

[maiden ocean]

if A is an integral domain and K is its field of fractions i want to show that K is the direct limit of the submodules Ax for x in K. id appreciate a hint about what directed set is supposed to be indexing the Ax

AMs defn of direct limit would require homomorphisms from every Ax to Ay with Ax < Ay as well

so i cant just use like, Ax -> Ax^2 by sending x to x^2 or whatever

[viscid pewter]

and i return

since no one else pointed it out

it's only isomorphic to A4

it has non-trivial subgroups C2, C3, C2xC2

but it's not isomorphic to them

isomorphic means that they are essentially the same group

among other things, they have to be the same size

but C2 is gonna be smaller than the group

yes

erm

there just isn't

it's like, why would there be

well in D3 or D6 or whatever there's also three elements with order 2

so firstly why S3 specifically

and secondly just because those elements exist doesn't mean they make a subgroup

because it's just like

i mean if you take all the elements of order 2 in Sn generally, they generate the entire group

so it's wildly inconsistent

idk, it just isn't a thing

there's no reason why it should be true? idk

[unique juniper]

$x^3 +3x + 3$ is irreducible over Q and u is a root

[cloud walrus]

Yes

[unique juniper]

how do i find u's inverse?

[rustic crown]

u^3+3u+3 = 0

[unique juniper]

i tried some euclidean domain properties

[rustic crown]

multiply by u^-1

[strong valve]

😠

[unique juniper]

:o

i didnt realise lmao

thank you

[strong valve]

.

study complex ana

for me too

[rustic crown]

but i don't wanna

i don't like the prof

[maiden ocean]

nvm im a brainlet just order by inclusion lmao

[old lava]

true

and make the homomorphisms just the inclusions

it's that easy

[cloud walrus]

slimvesus

slimvesus

[next obsidian]

this is basically just saying that K is the union of submodules of that form

For "is the direct limit of sub things..."

you're looking basically at union

so the directed-ness of it will come from subset inclusion, you look at the system of
Ax < Ay (so Ax -> Ay via inclusion when Ax < Ay as sets)

Note: This could be wrong since i haven't done this exact problem but this is my chmonkeytuition

I think this is directed because if you write x = a/b, y = c/d, then I think both Ax and Ay are contained in e.g. A(1/bd) because x and y are in there

[vestal snow]

Does anyone know where I can find this book?

https://bookstore.ams.org/cdn-1615015548587/smfdm-18/

It's not on libgen

There are notes of the lectures the book is based off of, but they are not easy to read through

[sturdy marsh]

the ams bookstore

[unique juniper]

anyone know why its linearly dependent ?

[cursive temple]

you have n+1 elements and the degree of your extension is n

so they have to be dependent

[unique juniper]

oh

yes

mb

thanks

[frank haven]

Does anyone know if (x+1)^6 is considered reducible over Z_7

I would think it is because it's already given as a product of nonzero polynomials , but I don't know if the fact that it is factored already matters

[unique juniper]

yeah think so

[hidden haven]

Yeah it's reducible

[frank haven]

I'm trying to show that a monic polynomial P is irreducible in Q, but all I'm given is that P(-1)=7 and the reduction of P mod 7 is (x+1)^6

I think the problem is set up so that I should be able to use the modular irreducibility test but (x+1)^6 is not irreducible mod 7

So that doesn't really work

Now I'm thinking maybe I can use the fact that the roots of the polynomial in Z[x] must be divisors of 8 or any multiple of 8, but that doesn't seem too helpful as I'm not even given P, only the reduction of it

[rustic crown]

do you know eisenstein's criterion?

[frank haven]

Yeah

[rustic crown]

and have you seen how sometimes its convenient to apply it to shifts?

take f(x) = P(x-1)

so we know f(x) = x^6 (mod 7)

and constant term = f(0) = P(-1) = 7

what do these 2 conditions say about coefficients of f?

[frank haven]

The leading coefficient is 1 and constant coefficient is 7?

[rustic crown]

leading could also be something like 8 right

but when you reduce mod 7, it becomes 1

oh nvm, we're given its monic

[frank haven]

wait, so first off, (x-1)(x+1)^6=x^6?

Mod 7

[rustic crown]

nope degree of left is 7 right is 6. they can't be equal.

[frank haven]

That's what I thought you wrote

f(x)=p(x-1) and f(x)=x^6

Sorry if I'm being really dumb

[rustic crown]

i wrote f(x) = x^6 (mod 7)

[frank haven]

But the thing that is giving me trouble is we don't know what p(x) is only it's reduction

[rustic crown]

that the condition we're given. p(x) reduces to (x+1)^6 and so p(x-1) reduces to ((x-1)+1)^6

[frank haven]

Oooh

Sorry I thought you were multiplying for some reason

My bad

[rustic crown]

oh lol

[frank haven]

So eisenstein says that if x^6 is irreducible then out original is too?

[rustic crown]

x^6 isn't irreducible 😶

[frank haven]

Yeah I know

[rustic crown]

so this is right, what about the other coefficients?

[frank haven]

They must also be multiples of 7

[rustic crown]

yep!

[frank haven]

So I guess p=7 satisfies eisenstein

[rustic crown]

yeo!

[frank haven]

For f(x)

But then how do that prove it for p

[rustic crown]

the proof is pretty elementary say you have an integer polynomial f and a prime p such that p doesn't divide the leading term, but divides all other terms. hence f = a*x^n (mod p)
lastly we're given that p^2 doesn't divide the constant term. in this case we could conclude the irreducibility.

because if f= g*h is a non-trivial way to factor, then going modulo p, we get a non-trivial factorization of ax^n, so looks like bx^s * cx^t
this is non-trivial so s,t >= 1

but this says that constant term of g and of h are both divisible by p, so constant term of g*h is divisible by p^2. this is a contradiction

[frank haven]

Ooh that makes sense

Wow what an awesome use of eisenstein

I think my prof was trying to jebait us into using the modular irreducibility

I think i got it now, thanks for the help

I guess I'll go rigorize this now

[rustic crown]

okie

[chilly ocean]

my dream is to one day have a question be answered by det so in the end they put that eevee emoji

[carmine fossil]

give det the eevee role

[rustic crown]

do we already have such a role, other than the one SashaMomo has?

[carmine fossil]

Yea,That role

[rustic crown]

.<

don't lewd eevee

,av

[cloud walrus]

det#7067's Avatar

Click here to view the image.

[rustic crown]

i was thinking of making a life-sized eevee papercraft

[cloud walrus]

Godel

[bleak crystal]

could someone help me finish out a proof?

i have a test in like 20 minutes and i need to get this done before then, the pace of the questions chat is just too slow rn

here is the question at hand

here is my work so far

[fading wagon]

looks fine so far I guess

yeah then from property (ii) we have J=I or J=R

we can break into cases there

and note if J=R then 1_R in J

[bleak crystal]

and we know that I is a subset of J how?

okay forget it I'll just skip over that part

[fading wagon]

yeah J=(a)+I

[bleak crystal]

how does this make a field?

[fading wagon]

we need to find inverses

if J=R which is the typical case, then we know that 1 is in (a)+I

so 1=ra+i for some r and i

then r is the inverse

[bleak crystal]

okay good enough

i have another question but I'll have to come back later because i have a test

i still don't really understand but i don't have time to understand

[untold sapphire]

What are the prerequisites for learning about linear algebraic groups? I've done a course in classical Algebraic Geometry and in finite dimensional Representation Theory, is that enough?

I don't really know anything about Lie Groups or Algebras beyond the definitions

[delicate orchid]

linear groups are just matrix groups right?

*groups that are isomorphic to matrix groups

[untold sapphire]

I believe they also carry the additonal structure of being Zariski closed subgroups

(over the real or complex numbers anyway, which is what I'm interested in)

[delicate orchid]

could've sworn all you needed was a homomorphism from the group to GLn

either way I think you'll be fine but reading up more on lie groups won't hurt

[untold sapphire]

I think you also need that the subgroup of n x n matrices of GLn is defined by polynomial equations

I'm not an expert though obviously, so you might be right

[delicate orchid]

I'm not an expert either so you might also be right lmao

[untold sapphire]

lol, fair enough

[chilly ocean]

yo I'm trying to find the smallest field L such that Q(3^(1/3)) \subset L and L is normal over Q. So that field is going to be Q(3^(1/3), epsilon_3) where epsilon_3 is a complex root of unity of order 3, since it's a splitting field for x^3 -3. Is there anything more to say here though, should I argue more why this is the solution?

That seems intuitively obvious but I'm just asking how to make it more rigorous I guess

[next obsidian]

I’m currently using Humphrey’s book. I think those are roughly all you need, combined with some time, and willingness to look up some results

Just a word of warning, if you use Humphrey’s book just know that there’s some results he will just state, but not prove.

An example is that a constructive set H contains a set U which is open and dense in the closure of H. I ended up looking at Borel’s book which had a proof

There’s 3 books just called “Linear Algebraic Groups” which are the standard text, by Humphreys, Springer (an author named Springe, not the company), and Borel

I’d just pick one to use as a main text, but look at the others as you need

[bleak crystal]

i am back

Let R be an integral domain
Show that R is a field IFF every ideal of R[x] is a principal ideal
and here's what i have thus far

does this seem like shit

[next obsidian]

Why do you need to assume R[x]/(x) is iso to R, it simply is

[bleak crystal]

idk he wrote it down in hints and we technically haven't proven it in class yet

[rustic crown]

you need to know the fact that finite normal extensions are precisely splitting field of some polynomial. if L is a normal extension then the irreducible polynomial x^3-3 should factor, so Q(crbt(3), omega) is contained in L. But as Q(crbt(3), omega) is a splitting field, its normal, but as L was the smallest normal extension with crbt(3) you get L containted in Q(crbt(3), omega).

Hence L = Q(crbt(3), omega)

[hidden haven]

You'd have to show that splitting field extensions are normal, depending on your definition of normal and what you are allowed to assume

[rustic crown]

🤝

[chilly ocean]

yeah we showed that no0rmal iff its a splitting field

[hidden haven]

🤝

[next obsidian]

It’s really simple, just take the map R[x] -> R given by sending x to 0, and sending the constants to themselves, and that proves it

[hidden haven]

Cool then you're done

[bleak crystal]

okay there you go then

it doesn't matter much anyway, since the question doesn't really ask about that

[next obsidian]

I mean if you’re going to use that result, it does

[delicate orchid]

Or the map R[x]/(x) -> R
Where r+(x) -> r

[chilly ocean]

thanks, needed confidence cause Im not sure about anything nowadays

[next obsidian]

You can’t just assume something unless you’re doing contradiction

Anyway, it’s unclear what the second part of your proof seeks to do

[bleak crystal]

he verbatim said we can

assume, i mean

[chilly ocean]

I dare you homie you react with the mfkin apple one more time

[next obsidian]

Then don’t write assume I guess, just say “as...”

[hidden haven]

[bleak crystal]

"You may assume this result"

[next obsidian]

Anyway, I still don’t see what the latter part means

[bleak crystal]

I'm just peeling it directly from the assignment

[untold sapphire]

Cool thanks a lot!

[bleak crystal]

idk I'm just trying to follow the hints