groups-rings-fields

[amber burrow]

im using dummit and foote

because most of the exercises in some earlier chapters (even late exercises), I can kinda work out a solution in my head

but im worried that i wont gain the pattern recognition

to be able to solve a large problem

[chilly ocean]

book name please

[chilly ocean]

why dont you send one problem here and see people how they are doing it

[chilly ocean]

and yes dont expect any pattern some times you have to proov by contradiction but sometimes you have to proove by some trivial rizzics example argument

[woven trout]

This is from an old prelim exam

https://www.colorado.edu/math/graduate-program/graduate-resources/preliminary-exams

Spring 24

[chilly ocean]

i will send you the solution of that problem , because it is done in my class

[agile meteor]

is this correct statement?

[chilly ocean]

mention your base field

[agile meteor]

real numbers

[vagrant zinc]

crack

[subtle crypt]

What is a reducible form?

If you mean that it's factored into irreducible polynomials, then yes, that's a correct factorization into quadratic polynomials and the roots of those aren't real numbers, so you can't reduce anything to linear polynomials.

[chilly ocean]

onto map from noetherian module M to M is one one

[crystal vale]

Yes that result I really like, no reason

[tame sierra]

i think this is true when the field is Q

[rocky cloak]

Yes.

More generally it holds for rings such that Z -> R is an epimorphism.

Z/n should be clear. For (subrings of) Q:

f(1/b x) = b/b f(1/b x) = 1/b * b f(1/b x) = 1/b f(b/b x) = 1/b f(x)

[chilly ocean]

the rest part of the claim you can solve yourself, and here M need not be free

[chilly ocean]

why 0 ideal is maximal ideal of M_n(R) where n is biggere than equal to 2

[rocky cloak]

This is true when R is a field.

You can show that the ideals of Mn(R) are exactly Mn(J) where J is an ideal of R

You can show this pretty explicitly by just multiplying your ideal by elementary matrices

[chilly ocean]

professor said that it is true for any commutative ring with unity

he said 0 ideal is not left /right maximal ideal

[rocky cloak]

Well maybe they misspoke or you misunderstood.

But the maximal (twosided) ideals of Mn(R) corresponds exactly to maximal ideals of R. You don't even need R to be commutative

[chilly ocean]

okay

[chilly ocean]

i just saw my POD and you are right

[elfin wraith]

The proof is quite straightforward, it’s a good exercise, I enjoyed doing that last semester

[chilly ocean]

yes the idea is to multiply the matrix with E_(pq) which has entries 0 except pq entry

[acoustic igloo]

true or false?

Every finite subset of R^2 is an algebraic variety

i think {(n,0) | n an integer} won't be a variety

oh that's infinite duh

riiight right ok i guess it's true

although...

[next obsidian]

Lmfao

[acoustic igloo]

the solutions manual says it's true but i don't see why it must be true

[next obsidian]

Is a single point an algebraic variety

[acoustic igloo]

yeah

it's the intersection of two lines

[next obsidian]

Aren’t varieties closed under finite union

[acoustic igloo]

i think you can multiply the elements of the bases pairwise to create a new basis

whose variety is the union

[next obsidian]

Yes

[acoustic igloo]

nice

thanks 🙏

$V(F)\cup V(G)=V({fg\mid f\in F, g\in G})$

[cloud walrus]

Axe

[acoustic igloo]

like that, right?

[thorn jay]

Yee

[acoustic igloo]

sweet

[thorn jay]

The wonders of a field being an integral domain

[acoustic igloo]

hmm

[keen badge]

Let $G,H$ be groups and let $\phi:G \to H$ be a homomorphism.
Is it true that if $\psi:H\to G$ is a homomorphism such that $\psi \circ \phi \cong id_G$ and $\phi \circ \psi \cong id_H$ then $\phi$ and $\psi$ are isomorphisms?

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[keen badge]

(It's not a h.w. assignment, I'm just curious if the two identity conditions mean that both $\phi$ and $\psi$ are bijections)

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[acoustic igloo]

the condition is that psi phi = the identity map, right?

[wraith cargo]

idk what \cong means lmao

[acoustic igloo]

and phi psi = the identity map in H

yeah it should be = right?

then it should work

[thorn jay]

Isomorphic to the identity?

But yes

I believe so? Assuming that isomorphic to the identity means that they differ from the identity by an isomorphism

This is nothing group-theoretic by the way, its just set theory

[thorn jay]

Maybe they differ by an automorphism?

I did

[keen badge]

yes

I'm sorry

I am dealing with different notations, and got them all mixed up

I meant equal

[thorn jay]

Then its basically by definition of bijectivity

[acoustic igloo]

$\phi(a)=\phi(b)\implies\psi\phi(a)=\psi\phi(b)\implies a=b$, so one-to-one

[cloud walrus]

Axe

[thorn jay]

At least, its a fairly simple exercise to show the existence of an inverse of f iff f is both injective and surjective

[glad osprey]

This is basically the definition of isomorphism in any category, it just happens that in the category of groups a bijective homomorphism is an isomorphism

[acoustic igloo]

$\forall h\in H: \phi(\psi(h))=h$, so $\phi$ is onto as well

[cloud walrus]

Axe

[keen badge]

Yes, I just can’t help myself…
It’s like plugging 5 + 7 into a calculator, I guess.
I just needed the reassurance.

[thorn jay]

Well there is definitely a strong underlying reason for it. It has to do with the properties of the forgetful functor

Now I'm guessing its because of the monadicity rather than adjointness

(Monadicity basically means that the category of groups is a category of algebraic structures over Set)

[acoustic igloo]

is {(x,0) | x >= 0} an example of a subset of R^2 that is not an algebraic variety?

does this question belong in a different channel?

ah ok that makes sense

i'm not sure

maybe we can throw out any term containing y

the resulting polynomial in one indeterminate has infinitely many zeros

lol

[tardy hedge]

System of a down bio?

[acoustic igloo]

can i though?

i think i should take a break

thanks for the help

the next section is extension fields

[tardy hedge]

W

[vagrant zinc]

True

[wispy light]

Let g(x) = f(x,0). How many zeros does g(x) have (at least)? How many polynomials in R[x} do you know that have that many zeroes?

[proper jolt]

suppose that R' is an R algebra and a is an ideal then is it true that a \otimes_R R' = aR'?

[proper jolt]

this isn't a homework question, but I was thinking that I can use the fact that R' \otimes R[X] = R'[X]. and then replace the variables X with a. Does that work?

[next obsidian]

No this isn’t true

There’s a surjection from the left to the right but the failure of injectivity is determined by Tor^R_1(R’,R/a)

This works for R’ actually any A-module M

[chilly ocean]

How to show adj(AB)=adj(B)adj(a) over a commutative ring with unity

over a field it is easy

[harsh onyx]

I'm trying to tackle a question where I happen to be dealing with double cosets over the same subgroup, namely the set H\G/H = {HgH : g in G}, is there a special name for such a case?

[chilly ocean]

I DONT THINK

[harsh onyx]

?

Ah, perhaps this would've been better suited to #advanced-algebra

[primal beacon]

i guess write A as the image of a generic matrix (X_ij) and B as the image of a generic matrix (Y_ij), you then need to prove it over Z[X_ij, Y_ij] which is easy because it is a domain so you can go the fraction field. the map Z[X_ij,Y_ij]->R is well defined/behaved because of commutativity and unit and adj(AB), adj(A),adj(B) are indeed the image of adj((X_ij)(Y_ij)), adj((X_ij)),adj((Y_ij))

tell me if you think its wrong but i believe it works

[chilly ocean]

can you do littile work and leave it to me to fill the gaps

[primal beacon]

there are all elements there you can check if it works directly

the map is X_ij -> a_ij Y_ij->b_ij

minors have coeffs polynomials in a_ij and b_ij

[chilly ocean]

please write in latex it is confusing

[chilly ocean]

double cosets is a part of group theory and outlined in dummit & foote excersises but i saw them first time modular form (hecke algebra) in a book of modular forms many results about double cosets will be given

[rocky cloak]

How is it any different over a field than any other ring?

[chilly ocean]

there is a problem actually

[chilly ocean]

if we see the actual proof for field it uses a property of non zero divisiors

[primal beacon]

im on phone so its not easy but, take a commutative unitary ring R and two matrices $A,B\in M_n(R)$. write $(a_{ij}){ij}$ and $B=(b{ij}){ij})$. Then A and B have coefficients in $C=\mathbb Z[a{ij},b_{ij};i,j]\subset R$, now you can prove the theorem in C

[cloud walrus]

k[r,a,y,a,n,e]

[primal beacon]

to prove that consider the ring $D=\mathbb Z[X_{ij},Y_{ij};i,]$ with $X_{ij},Y_{ij}$ unknows for each $1\leq i,j\leq n$ and consider matrices $A_{gen}=(X_{ij}){ij}$ and $B{gen}=(Y_{ij})_{ij}$

[cloud walrus]

k[r,a,y,a,n,e]

[primal beacon]

The map $D\to C$ given by $X_{ij}\mapsto a_{ij}$ and $Y_{ij}\mapsto b_{ij}$ extends uniquely to a ring homomorphism

[cloud walrus]

k[r,a,y,a,n,e]

[primal beacon]

so that $adj(A_{gen}B_{gen})$ maps to $adj(AB)$, $adj(A_{gen})$ maps to $adj(A)$ and same for $B$

[cloud walrus]

k[r,a,y,a,n,e]

[primal beacon]

so that you just need to prove the theorem in D

now $D$ is a domain and the theorem is true in its field of fraction. Finally since the adjoints have coefficients polynomials in the coefficients of the initial matrices the theorem is true in D and you are finished

[cloud walrus]

k[r,a,y,a,n,e]

[tough raven]

Then let k be the field of rational functions in as many variables as there are entries in A and B together and let the entries of A and B be those variables. You get that the entries of the LHS and RHS are the same polynomials in the entries of A and B.

Shoot, I forgot to check if it's already answered. Sorry.

[chilly ocean]

now we can use the non zero divisior property for polynomials, and apply homomrphism on adjoint to get the desired result

[chilly ocean]

Can anyone tell how to show uniqueness of elementary divisors

[rocky cloak]

You can do it by proving the more general theorem, Krull Remack Schmidt:

If a module can be written as a finite direct sum of modules with local endomorphism rings, then it can be done so in a unique way.

[tardy hedge]

Jagr the 🐐

[rocky cloak]

Notice End_R(R/p^n) = R/p^n is a local ring

[tardy hedge]

Is that the same reason why End_R(R) = R?

[rocky cloak]

Yup

The main idea is say
M = Sum Mi = Sum Nj
Then consider the compositions
Mi -> M -> Nj -> M -> Mi

Then the sum of these for all j will be the identity on Mi, so if it has local isomorphism ring, one of them must be an isomorphism, which implies Mi is a direct summand of Nj

[chilly ocean]

suppose we dont know that result then?

[rocky cloak]

I wasn't supposing you did...

[chilly ocean]

dont go to general case

[rocky cloak]

Okay, then just take my hint and replace Mi with R/(pi^ni)

[chilly ocean]

Mi -> M is inclusion?

[rocky cloak]

Yes

And M -> Mi the projection

Since M is a direct sum

[chilly ocean]

same for M-> Nj

[rocky cloak]

Yes

[chilly ocean]

but i dont know local isomorphism

[rocky cloak]

Then don't worry about it.

Just ask yourself what are the homomorphisms R/(p^n ) -> R/(p^n) and when are they isomorphisms

[chilly ocean]

identity one is isomorphism

[rocky cloak]

It's always a good idea to look at the simplest possible cases.

What are the homomorphisms Z/4 -> Z/4 for example?

[chilly ocean]

1 goes to 1 and 1 goes to 3 , are isomorphisms

[rocky cloak]

So what seperates the isomorphisms from the non-isomorphisms?

[chilly ocean]

1 goes to an element which is relatively prime to 4

[rocky cloak]

Yup.

So then if the sum of homomorphisms is an isomorphism, could it be that the terms are all non-isomorphisms?

[gusty thistle]

it follows from the fact that it holds over the reals

if a polynomial identity with integer coefficients holds over a field of characteristic 0 then it holds for all commutative rings

i see it was already answered but its one of my fav things i had to share it lol

[tardy hedge]

For E/F algebraic and considering some map E -> Fbar, can I say something about how that map maps inseparable elements of E over F?

[rocky cloak]

Purely inseperable elements will map to themselves (or to something unique at least), but in general the situation is the same as any algebraic element. It will map to one of its conjugates.

[tardy hedge]

purely inseparable elements will map to themselves because they dont have any other conjugates?

[rocky cloak]

Yup

[tardy hedge]

so hom(E, Fbar) is the same size as hom(Es, Fbar) because the elements in E\Es are the purely inseparable ones, so there is always only one spot for them to go in the maps anyway?

[rocky cloak]

Kind of yeah.

E/Es is purely inseperable. That doesn't mean the elements of E are necessarily purely inseperable over F, but where they are mapped are determined by where Es is.

[south patrol]

Indeed this is a way to define separable degree

You should have Hom_F(E, F bar) though, as you probably know lol

[rocky cloak]

To take an example, let
F = F2(t)
and let s be a root of
f(x) = x^4 + x^2 + t
and let E = F(s).

Then Es = F(s^2), s is purely inseperable over Es (being the root of x^2 - s^2), but
f(x) = (x-s)^2 (x - s+1)^2
so s and s+1 are conjugates.

[south patrol]

Actually maybe you don't for your purposes, but what I wrote is normally what you will be interested in anyway

[rocky cloak]

So a homomorphism may map s to either s or s+1, but if you know where s^2 is mapped, then s has to map to the unique square root of that

[elfin wraith]

That seems whacky, I was not aware of that

[gusty thistle]

cool right

i like it because at first it seems illegal and not true but its also trivial

[elfin wraith]

Why on earth is that true

[gusty thistle]

think about it -_0

you can do it

[gusty thistle]

the special case implies the general case

[rocky cloak]

I guess it's sort of hard to make polynomial identitets that only hold in some rings.

Besides like 2x = 0, what can you really do

[elfin wraith]

I guess yeah, but it seems very surprising at first that something which is true over such a nice ring would be true over them all

[gusty thistle]

yes !

rings are the best

rings and fields 😌

[rocky cloak]

Well I guess what's hidden in there is that a polynomial identity is an extremely constrained thing

[elfin wraith]

Yeah also true I suppose, I definitely see what you mean now about once you reason it out it is somewhat trivial, I guess torsion is pretty much the only concern

[rocky cloak]

Not quite a one liner, but
||If it holds in Q, then by continuity and density it holds in R. Then it holds for algebraically independent elements, hence holds in Z[x1, ..., xn]. By homomorphisms it holds in all commutative rings.||

[gusty thistle]

||If K is a field of char. 0 then Z[x_1,...,x_n] injects into K[x_1,...,x_n] which injects into the set of maps K^n -> K as K is infinite||

[white oxide]

Let k be an infinite field. If v is nonzero in k, is it true that {1, v, v^2, ..., v^d} form a linearly independent set?

[south patrol]

v = 1

[south patrol]

Linearly independent over what? Z?

[white oxide]

Over k I guess

[white oxide]

Oh yeah

Oops

[south patrol]

Well k is a k-vector space of dimension 1

So they are always dependent

[white oxide]

Oh yeah

[south patrol]

Over Z it is still false by considering like k = Q say

[white oxide]

This seems pretty easy to prove via something like linear independence but I've been stuck 😹

[barren sierra]

Here's an example. f(x, y) = x^2 + y^2 so f(λx, λy) = ???

then you can factor out something nonzero

And use the fact that k[x, y] is a domain

[rocky cloak]

You can think of this as a polynomial in lambda.

Then what you're trying to prove is that a polynomial that is 0 everywhere must be 0

[white oxide]

Are you getting that we can write $\sum_{k = 0}^d \lambda^k f_k(a_1, \dots, a_{n + 1}) = 0$?

[cloud walrus]

okeyokay

[barren sierra]

saying anything more would give it away

Where did u get a sum

NVM I can't read

[white oxide]

Lol nah ur goo

[barren sierra]

Ok my hint is slightly less useful but still applicable

I guess my hint gets you what you wrote

[white oxide]

Sorry, what do you mean by polynomial in lambda in this case?

[barren sierra]

And then apply what jagr wrote

[rocky cloak]

Treat lambda as a variable

[barren sierra]

Act as if you're in the ring k[a1, ..., an][λ]

[white oxide]

Ohh okay

[barren sierra]

You really do need that the field is infinite

[rocky cloak]

Like what is written here.

The fk(...) are just numbers, so this is a polynomial in lambda

[white oxide]

Is there a result saying that if p(a) = 0 for all a in k \ {0} where k is a field and p in D[x] where D is a domain, then p is actually the zero polynomial?

I guess if that's true then that makes sense yeah

[rocky cloak]

If k is infinite and sits in the center of D it should be true

I guess it's easier in the case D=k which is the case here though

[white oxide]

Sorry I'm a little confused, I get that we can view it as a polynomial with a variable in lambda, but are the coefficients in k or k[x_1, \dots, x_{n + 1}]? Since we're saying that \lambda^0 f_0(a_1, \dots, a_{n + 1}) = 0 where f_0(a_1, \dots, a_{n + 1}) are in k

Ah okay I guess that's what I just asked

[rocky cloak]

the coefficients are in k yeah

[white oxide]

Okay thanks, that solved things!

[rocky cloak]

I'm not sure the result has a name, but a degree n polynomial having at most n roots is well known

[white oxide]

Oh yeah I remember that one lol

[wintry sluice]

why do group representations need not be isomorphisms?

[rocky cloak]

Would be kinda boring if they were. Then we could only do group representations of GLn

[thorn jay]

It'd defeat the whole point of using group representations, lol

[coral spindle]

If you are actually intending to ask why group reps don't have to be injective homomorphisms, i.e. faithful, the answer is that we get a more interesting theory if we consider ones that aren't as well as ones that are.

The goal is often to understand the irreps, and irreps need not be faithful.

[wintry sluice]

ok thanks

[south patrol]

Also just like

Group act on stuff "in nature" all the time

And we study that

It isn't just a matter of wanting to study a group by embedding it in something else (though this is sometimes useful too!)

[barren sierra]

the only reps worth studying

[coral spindle]

We could only study ONE SINGLE representation of GLn

[rocky cloak]

Only one up to isomorphism maybe

[tardy hedge]

Boytjie Returns

[coral spindle]

Hmmm don't get used to it

[tardy hedge]

Noooooo

I like the bio update though

[tardy hedge]

That emoji lmao

Or emote i suppose

Its a funny one

[eager willow]

Not true, even up to isomorphism. Take n = 1. Then the weight -1 rep is not iso to the weight 1 rep. They are inequivalent, i.e. non conjugate isomorphisms GL1 -> GL1.

Likewise there

there's the inverse transpose representation of GLn for any n, and this is not isomorphic to the standard n-rep, but it is coming from an (outer) automorphism of GLn

[coral spindle]

Damn good point

[eager willow]

still a good reduction for why the maps G-> GLn really shouldn't be isos because such reps are simply the outer automorphism group of GLn, practically by definition, and rep theory needs to be built around studying more things because there are more things in nature.

[acoustic igloo]

i have a problem asking me to find the degree of sqrt(pi) over Q(pi)
i think it's 2 because sqrt(pi) is a zero of x^2-pi, but how do i know x^2-pi is irreducible?

[eager willow]

Perhaps what does it mean for any f(x) = x^2 - a to be irreducible in a field with some element a? Not in the form f(x) = g(x)h(x) for degree 1 polynomials g, h

[acoustic igloo]

if i can show it has no zeros in Q(pi) that would suffice

[eager willow]

sure, so can some rational function of pi generate pi's own square root?

[chilly ocean]

No one of them is isomorphism

[sturdy spear]

If G is an Abelian group and |a|=m, |b|=n are finite. Then what is the order of ab?
Claim: |ab|=mn.
Certainly (ab)^mn =1 so |ab|<=mn.
Assume d be any power such that (ab)^d = 1. Let d=(mm)q +r so (ab)^r=1 or a^rb^r=1

Idk how to proceed further

[eager willow]

consider a = 2 and b = 4 mod 8

[acoustic igloo]

sorry i have to go, i'll think more about this later

[sturdy spear]

|2| = 4 and |4|=2
But |8| = 1 oops

[eager willow]

wait wait I oopsed you

because mod 8 is written additively

so that's the wrong multiplication. The value of ab is actually 6

[sturdy spear]

Oh yes |2+4|=|6| = 4 i guess

[eager willow]

no 6 mod 8 has additive order 4. But still not |2||4|

[sturdy spear]

I edited it (thank you )

So |ab| seems lcm(m,n)

[eager willow]

|4 + 4| =/= lcm(2, 2)

[sturdy spear]

True

[eager willow]

basically I think you gotta go case by case in the structure theorem for finite abelian groups

[chilly ocean]

Now in your proof use definition of lcm

[sturdy spear]

I don't have much space to play

Just definition of order i guess

[sturdy spear]

Doesn't it fail?
See counterexample #groups-rings-fields message

[eager willow]

play? oh I see you don't know this theorem yet

[sturdy spear]

Yes

[eager willow]

well consider two cases.

1. a = b^l or b = a^l for some integer l.
2. Neither is true.

I'm fairly certain these are the only considerations necessary to arrive at what I know from the structure theorem.

basically, the lcm guess should hold if they are 'linearly independent' and it shouldn't hold when they're linearly dependent, but some other considerations can work.

[sturdy spear]

Ah i see. Let me try to figure it out

Thank you so much datorangeguy

[acoustic igloo]

sorry i was watching the white lotus finale, so good

but i think the answer is no, right?

this isn't really justified by any of the theorems in this chapter, as far as i can see

well it probably is and i just can't see it

[acoustic igloo]

if f(pi)/g(pi) were a zero of x^2-pi, with f,g in Q[x], we would have
f(pi)^2 - pi*g(pi)^2=0, and then pi would be a zero of
f(x)^2-xg(x)^2 and pi would be algebraic over Q

[eager willow]

right, so for any transcendental t in C we have Q(t) is isomorphic to Q(t^2)[x]/(x^2 - t^2). In particular [Q(t) : Q(t^2 ) ] = 2. Take t = sqrt(pi) and you're done.

[acoustic igloo]

[chilly ocean]

how you have seen the isomorphism

[eager willow]

first, there is a Q(t^2)-linear map of rings Q(t^2)[x] -> Q(t), taking x to t. This is automatically well defined since R = Q(t) is a ring and a vector space over the field K = Q(t^2). Any element y in such a ring R has a well defined K-linear map of rings K[x] -> R taking x to y.

Second, this map takes the polynomial x^2 - t^2 to 0, because it takes x to t and also it takes t to t, since it is Q(t)-linear

Therefore there is a well defined homomorphism even after quotienting by the ideal, and we have our homomorphism Q(t^2)[x]/(x^2 - t^2) -> Q(t).

To show it's an isomorphism, check that Q(t^2)[x] -> Q(t) is surjective and has actually the ideal (x^2 - t^2) is the entire kernel.

Isomorphism from the quotient ring is then the first isomorphism theorem.

We proved (x^2 -t^2) was a maximal ideal, it's generated principally by an irreducible. Therefore since the map from the quotient is a well defined homomorphism, (x^2 - t^2) is the entire kernel: the map has image including not just 0, so the kernel is a proper ideal containing (x^2 -t^2). Therefore the kernel is (x^2 - t^2).

The map is surjective because, after descending to the quotient, we see that t is in the image (it's the image of x), and so the image of the quotient is a subfield of C containing t.

But Q(t) is the smallest field containing t.

[chilly ocean]

You just shown that (√π) is transdental using π is transdental isn't it?

[chilly ocean]

M_i is direct summand of Nj how

[rocky cloak]

You have maps
Mi -> Nj -> Mi
that compose to the identity

[chilly ocean]

yes M_i is direct summand and the we can change the roles of M_i and N_j

[rocky cloak]

Yes, or use the fact that Nj is indecomposable

[chilly ocean]

oh thats also fine

[chilly ocean]

please send a reference ofthis proof

[late python]

How I find the left and right cosets of $D_4 \text{ in } S_4$? Is there an easier way than to compute all elements?

[cloud walrus]

rabbits_advocate

[late python]

since $S_4$ is 4!

[cloud walrus]

rabbits_advocate

[thorn jay]

D_4 meaning the dihedral group of 8 elements?

[late python]

yes

[thorn jay]

Also, it depends on the specific subgroup of S_4 isomorphic to D_4

I dont think theres an easier way besides computation for cosets, right? At least there are only 5 unique ones in total :P

[pearl fog]

The galois grp of x^4-6x^2+2 over Q should be D4 since galois ext galois grp etc, where roots are $\pm\sqrt{3\pm\sqrt{7}}=\pm\alpha, \pm\beta$, but when I do $(\alpha, -\beta, -\alpha, \beta)$ on $a\alpha+b\beta+c$ to find the fixed elements in splitting field where a, b, c in Q I get -a=b and b=a? Or does it rlly not fix linear combinations of roots but something like $\sqrt{2}$ or $\sqrt{7}$ instead?

[cloud walrus]

Cro

The galois grp of x^4-6x^2+2 over Q should be D4 since galois ext galois grp etc, where roots are $\pm\sqrt{3\pm\sqrt{7}}=\pm\alpha, \pm\beta$, but when I do $(\alpha, -\beta, -\alpha, \beta)$ on $a\alpha+b\beta+c$ to find the fixed elements in splitting field where a, b, c in Q I get -a=b and b=a? Or does it rlly not fix linear combinations of roots but something like $\sqrt{2}$ or $\sqrt{7}$ instead?
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[late python]

by lagrange's theorem there are only 3 distinct ones

but i suppose I have to go through all 24 elements in S_4 to find them?

then also do that again to find the left cosets scince the group is not abelian

[pearl fog]

Then how to find a basis of splitting field using the two roots instead?

[thorn jay]

Left and right cosets may be different

5 as you count the actual subgroup twice

[rocky cloak]

I don't really have a reference at hand, but I'm sure it's covered in most books on module theory

[thorn jay]

I guess multiplied by 3 also for the conjugates

[chilly ocean]

not in dummit foote

[south patrol]

Well ig usually when people count cosets a sign is chosen right

[rocky cloak]

It is proven on page 467, but I guess you're saying that they have a different proof

[chilly ocean]

yes

[rocky cloak]

I see the proof is given in theorem I.4.6 in Frobenius algebras by Skowronski and Yamagata.

They have the assumption that the modules are over a finite dimensional algebra, but that is only used to show it can be written as a direct sum of things with local endomorphism ring. The uniqueness step is unchanged

[chilly ocean]

i dont have have that book, please send link or screen shots of the proof

[thorn jay]

They were talking about left and right cosets of the abstract group D_4 in S_4

[thorn jay]

.

[south patrol]

Ih sure oop

[thorn jay]

:3

[south patrol]

Chungus

[chilly ocean]

thankyou majesty

[pearl fog]

Ok I worked it out but I cannot find any auto of order 4 that send roots to roots, all of them are order 2? Cuz by image once I send alpha1 to alpha2 then alpha2 sends to alpha1? I am very confused pls help

[tardy hedge]

Wassup Algebros

We Algebraing Today 🔥

Who Up Algebraing Rn ⁉️

[wispy light]

Let $f(x) = x^2 - 6x + 2 = (x - 3)^2 - 7$. We construct the splitting field of $f(x^2)$ in two steps:
\begin{itemize}
\item The splitting field of $f(x) \in \mathbb Q[x]$ is $F = \mathbb Q(b)$, where $b^2 = 7$.
\item The splitting field of $f(x^2) \in \mathbb Q[x]$ is $E = \mathbb Q(c,d)$, where $c^2 = 3 + b$ and $d^2 = 3 - b$.
\end{itemize}

Of course, the roots of $f(x^2)$ are ${ \pm c, \pm d }$. A straightforward calculation shows that
$$(cd)^2 = c^2 d^2 = (3 + b) (3 - b) = 9 - b^2 = 9 - 7 = 2 \implies cd \notin F.$$

Now we deduce that
\begin{itemize}
\item Since ${ c, d, cd }$ are all quadratic over $F$, we have that $\mathrm{Gal}(E/F)$ is isomorphic to the Klein group $V_4$.
\item Since $F/\mathbb Q$ is a quadratic extension, $\mathrm{Gal}(E/F)$ is a normal subgroup of $\mathrm{Gal}(E/\mathbb Q)$ of index $2$.
\end{itemize}

So far, we have constructed the elements of $\mathrm{Gal}(E/\mathbb Q)$ that fix $F$, hence fix $b$. Now we have to construct the elements that swap ${ b, -b }$.

[cloud walrus]

Eduardo León

[wispy light]

By Eisenstein's criterion, $f(x^2)$ is irreducible, so $\mathrm{Gal}(E/\mathbb Q)$ acts transitively on its roots. Hence there exists $\tau \in \mathrm{Gal}(E/\mathbb Q)$ such that $\tau(c) = d$. Notice that $\tau(d^2) = \tau(2 / c^2) = 2 / \tau(c^2) = 2 / d^2 = c^2$, so we have $\tau(d) = c$ or $\tau(d) - c$. And, at this point, I think I've given you enough generators to reconstruct the isomorphism type of $\mathrm{Gal}(E/\mathbb Q)$.

[cloud walrus]

Eduardo León

[wispy light]

Really, from $\tau(c) = \pm c \iff \tau(d) = \pm d$, you can see that $\mathrm{Gal}(E/\mathbb Q)$ respects the partition of ${ \pm c, \pm d }$ into ${ \pm c }$ and ${ \pm d }$. So you can think of ${ c, d, -c, -d }$ (in this order) as the vertices of a square, and $\mathrm{Gal}(E/\mathbb Q)$ will act on it by symmetries of the square. Therefore, $\mathrm{Gal}(E/\mathbb Q)$ is a subgroup of $D_8$ of order $8$ (by earlier calculations)... which has to be all of $D_8$.

[cloud walrus]

Eduardo León

[glad osprey]

Still grinding Galois theory 💪 I've started reading Galois Theory by David Cox now

[thorn jay]

Logic has snuck into my universal algebraic geometry

Isomorphism between the lattice of closed sets of some topology on a set of congruences (analogous to ideals/normal subgroups) and the lattice of infinite conjunctions of finite disjunctions of equational identities (ordered by implication) quotiented out out by equivalence in some class K of algebras

[thorn jay]

If you take only the formulas consisting if infinite conjunctions, then you get the complete lattice of regular closed sets, and can use some version of the Nullstellensatz to get a correspondence between kernels of maps into ISP(K) and equivalence classes of these infinite conjunctions in K

[wispy light]

I cleaned up the argument a bit.

Let $f(x) = x^2 - 6x + 2$. We compute the splitting field of $f(x^2)$ in two stages:
\begin{itemize}
\item The splitting field of $f(x)$ is $F = \mathbb Q(b)$, where $b^2 = 7$.
\item The splitting field of $f(x^2)$ is $E = \mathbb Q(c,d)$, where $c^2 = 3 + b$ and $d^2 = 3 - b$.
\end{itemize}

Since $F/\mathbb Q$ is a quadratic extension, $H = \mathrm{Gal}(E/F)$ is a normal subgroup of $G = \mathrm{Gal}(E/\mathbb Q)$ of index $2$. We shall compute $H$ first.

By construction, $c$ and $d$ are quadratic over $F$. And so is $cd$, because
$$(cd)^2 = c^2 d^2 = (3 + b) (3 - b) = 9 - b^2 = 9 - 7 = 2.$$

Therefore, $H$ is isomorphic to $V_4$, generated by the elements
$$
\alpha : \begin{cases} c \mapsto c, \ d \mapsto -d, \end{cases} \qquad
\beta : \begin{cases} c \mapsto -c, \ d \mapsto d. \end{cases}
$$

By Eisenstein's criterion, $f(x^2)$ is irreducible. Hence, there exists $\gamma \in G$ such that $\gamma(c) = d$. Moreover,
$$\gamma(d)^2 = \gamma(d^2) = \gamma(2 / c^2) = 2 / \gamma(c^2) = 2 / d^2 = c^2.$$

Therefore, either $\gamma(d) = c$ or $\gamma(d) = -c$. The latter is more convenient, because it has order $4$:
$$\gamma : \begin{cases} c \mapsto d, \ d \mapsto -c. \end{cases}$$

Notice that $\beta = \gamma \circ \alpha \circ \gamma^{-1}$. Therefore, $\alpha$ and $\gamma$ generate a subgroup of $G$ that properly contains $H$. But the only such subgroup is $G$ itself. And you can see that $\alpha$ and $\gamma$ satisfy the same relations as in the usual presentation of $D_8$.

[cloud walrus]

Eduardo León

[pearl fog]

Deadline passed still thank you very much! Was thinking of the 2 step thing and figured out earlier

[long obsidian]

Can someone help me understand adjoining roots to to a polynomial ring? For example, with the real f(x) = x**2+x+10 the root r is not in R[x]. How do I think about R[x][r] as an R algebra?

[wispy light]

In commutative-land, an $R$-algebra is simply a ring $S$ with a ring homomorphism $f : R \to S$. To see how $S$ is an $R$-module, just define scalar multiplication as $r \cdot s = f(r) \cdot s$.

[cloud walrus]

Eduardo León

[long obsidian]

Maybe I can rephrase my question, Is it true that (R[x][y])/<f(y)> = R[x][r]? as R algebras

[wispy light]

This really needs to be made precise. Let $R \subset S$ be an inclusion of commutative rings, and let $s \in S$ be an arbitrary element. Let $f : R[x] \to S$ be the $R$-algebra homomorphism that sends $x \mapsto s$. Denote by $R[s] \subset S$ the smallest $R$-subalgebra of $S$ containing $s$. Then $R[s] \cong R[x] / \ker(\varphi)$.

[cloud walrus]

Eduardo León

[long obsidian]

Thank you, that is helpful and seems correct

[steady field]

Is this "2d rubik's cube" a group? If so which one is it?

[velvet hull]

yeah it's most definitely a group, actually computing it looks like a pain in the ass though

[steady field]

thanks

[velvet hull]

do you allow odd centers of rotation?

[steady field]

what's the best way to tell? I know for example 3x3x3 rubik's cubes are a group but 4x4x4 rubik's cubes are not.

[steady field]

No idea, this is just a video I found

[velvet hull]

...they should always be a group

[steady field]

4x4x4 is not

[velvet hull]

I am very sure is a group

let me double check

[steady field]

it's because of some parity iirc

https://math.stackexchange.com/questions/144424/rubik-s-cube-not-a-group

[velvet hull]

hmmm, it seems like there's some associatve thing going on

okay

that's interesting, I'll look into that more later

my original statement may not be true then, I'll retract it for now

[steady field]

yeah it has to do with permutations of the center pieces

ok

[glad osprey]

The 0x0x0, 1x1x1 and 3x3x3 rubik's cubes are the only ones that admits a group structure, because those are the only ones with trivial tangent bundle

[thorn jay]

You can formalise it as some quotient of the free group with four generators acting on the set of all states of that grid

But actually computing that quotient sounds like something I'd rather not do

[glad osprey]

I think it makes more sense to talk about the group acting on that set though, right? Someone writes in the comments of that MSE post that

What you are saying is that the action of the group of moves on the set of achievable positions of a cube with edges of length 4 or more has nontrivial stabilizers. Statements like "Rubik's Cube is a group" are so unclear that that it is pointless to discuss whether they are true or not!

[toxic zephyr]

i wrote this general logic to try to prove that if u in F has a minimal polynomial of odd degree over K then K(u)=K(u^2). but it seems to work as long as the minimal polynomial has any odd degree terms (and even, but if there were odd but no even it wouldnt be minimal).

any polynomial p(x)=o(x)+e(x) (o odd e even)
p(x)=x(e'(x))+e(x) e' also even
e(x)=E(x^2)
p(x)=xE'(x^2)+E(x^2)
if min poly f of u is f(u)=uE'(u^2)+E(u^2)=0 implies u=-E(u^2)E'(u^2)^(-1) in K(u^2)
thus, u in K(u^2) implies K(u)=K(u^2) (since K(u^2) in K(u) is trivial)

[steady field]

yeah I saw that

[toxic zephyr]

can we say that if u has a minimal polynomial with even and odd degree terms over K then K(u)=K(u^2)?

[thorn jay]

It just so happens that some group actions have the property that you can take some element of you set X as the "origin" and any other element in X corresponds to a unique element of the group acting on it

This is, canonically, the case with the left/right multiplication group action of a group on itself

[thorn jay]

This has to do with heaps, I'm pretty sure

[glad osprey]

Interesting are those the group actions with trivial stabilizers? Or is that just a necessary condition?

[thorn jay]

Yes, these are the group actions where the stabiliser of every element is trivial

No, if it were a necessary condition, then group actions would be boring tbh

Besides, nontrivial stabilisers can be dealt with using the orbit stabiliser theorem

[glad osprey]

wait, do you mean it's not a necessary condition, but just a sufficient one?

[thorn jay]

What?

Oh i thought you meant trivial stabiliser being necessary for being a group action

Im tired

[glad osprey]

ah, lol

[thorn jay]

Group actions are goated ngl

[glad osprey]

for real 🔥

I'm thinking about reading Actions of Groups by John McCleary actually https://www.cambridge.org/highereducation/books/actions-of-groups/9E2CF39D99D882E664FFAB8547B39469

[glad osprey]

I think you just need to use multiplicativity. We have K < K(u^2) < K(u), and [K(u) : K] = [K(u) : K(u^2)] [K(u^2) : K]. Now if [K(u) : K] is odd, what values can [K(u) : K(u^2)] take?

[toxic zephyr]

im guessing that's the intended proof. but like does my logic also work and prove a more general result?

[balmy wraith]

Hello everyone

I am a little confused on how we go from gtheta = g'theta to see that (gtheta)((g')^(-1)theta) = e_H

I know that we can show two cosets xH and yH are equal iff (xy^(-1) is in H

I think that's what is being applied here, but I'm just confusing on the flow of the proof in applying that theorem

[glad osprey]

hmm, your proof actually works for any polynomial such that f(u) = 0, not just the minimal polynomial. It seems too strong, but your proof looks correct

[velvet hull]

theta(g) = theta(g')
-> theta(g) [theta(g')]^-1 = e
-> theta(g) theta(g'^-1) = e
the key here is seeing that if theta is any group homomorphism, [theta(g)]^-1 = theta(g^-1)

[toxic zephyr]

nice, thanks!

[glad osprey]

if this was a peer review I would give it a thumbs up, but hopefully some other more knowledgeable peers can weigh in too

[balmy wraith]

and so from the last step, can we then say that
-> theta(g) theta(g'^-1) = e
-> theta(g * g'^-1) = e
-> g * g'^-1 in ker(theta)
-> gker(theta) = g'ker(theta)

thank you for pointing out that key property :D

[tough raven]

Reading the answer, I think this is what's going on:

There is a group G generated by the possible rotation moves on a n⨯n⨯n cube, and a set Y of possible states of the cube, with a "basepoint" x_0 = the "solved" state. G\ acts on Y, and we can define a set X to be the orbit of x_0 ("solvable states"). Now whenever a group G acts on a set, we can try to define a group structure on the orbit of an element x by (g⋅x)(h⋅x) := gh⋅x. If the stabiliser of x is a subgroup H, then the orbit of x is isomorphic (as a G-set to be precise) to the set G/H of left cosets of H in G, and this is the same as trying to define a multiplication on G/H. In particular, it's well-known from group theory that this will work iff H is normal (in fact, this is pretty much the canonical motivation for normal subgroups).

Apparently for n = 3, the stabiliser of x_0 is a normal subgroup of G, but for n = 4 it is not. But the natural structure of the problem - a group of moves acting on a set of cube states - is well-defined for any n. I think the group structure on the set of states is somewhat unnatural because it depends on a choice of "identity element" or "solved" state.

There is some ambiguity about when exactly two states are considered the same - should we just record the colours of all squares, or track and distinguish different squares of the same colour (so that if there's a move sequence starting from a particular state whose net effect is to swap some pairs of small cubes such that all colours are the same, do we choose to treat the initial and final states as the same or not?), or track even the orientations of centre squares (i.e. in the previous scenario, if all cubes are in the same position but some of the centre cubes got rotated, is that different or not?). In all of these cases, you can describe things by a group of moves acting on a set of states. I don't know for which of them the normal stabiliser subgroup phenomenon happens.

[tough raven]

Also, just going to note that the argument there suggests the normal stabiliser will fail for any n ≥ 4, in case anyone reading this thinks "parity" means it works for odd n.

[tough raven]

Yes, it is true that K(u) = K(u^2) whenever the minimal polynomial of u has any odd-degree terms. (Note: if it has only odd-degree terms then by irreducibility it has to be x, so u = 0 ⇒ K(u) = K(u^2) = K. So we can assume there is always at least one even-degree term.)

[tough raven]

It requires the fact that E'(u^2) ≠ 0. This would not be true if e.g. f had only even-degree terms (because E' = 0). And if the minimal polynomial has only even-degree terms, you can show that for any f, f(u) = 0 ⇒ f_even(u) = f_odd(u) = 0.

[fickle pagoda]

let F=Q(pi^3). a basis of F(pi) over F should be {1,pi,pi^2}.

but i dont know how we know this spans the whole thing. certainly it spans over F[pi] but im not sure how we are still guaranteed with denominators

[acoustic igloo]

i'm just learning this stuff, but isn't pi algebraic over Q(pi^3)? which means you don't need denominators to form F(pi), right?

oh god i'm so confused

[fickle pagoda]

i think im not sure why that fact is true

[steady field]

so to be clear if we were to distinguish between different middle pieces of the same color it would be a group right?

I think what I was remembering with parity is that there is a parity of the middle pieces which allows us to distinguish some of them from each other but not all four.

[fickle pagoda]

oh yeah there's a result that if f(x) in F[x] is irreducible with a root k, then a basis is {1, k, k^2, ..., k^[(deg f(x))-1]}

just completely forgot about it and was trying to derive from scratch

[tough raven]

I'm not sure. What you want, I think, is to distinguish enough information that any group element fixing a state has to preserve the position and orientation of every cube.

[tough raven]

You need {1,pi,pi^2} to span F(pi) = ℚ(pi) as a vector space over F = ℚ(pi^3). So can you write any rational function in pi, f(pi), as a linear combination whose coefficients are rational functions in pi^3, i.e., as p(pi^3) + q(pi^3) pi + r(pi^3) pi^2 where p, q, r are rational functions?

[cloud walrus]

Luke

[rocky cloak]

What have you tried so far?

[surreal kelp]

Algebra: chapter 0 is introductionary abstract algebra?

[rocky cloak]

It is, but it's not a conventional textbook. It tries to teach algebra with the language of category theory.

[delicate orchid]

prayer

[amber burrow]

why isn't $1 \trianglelefteq {1,-1} \trianglelefteq Q_8$ a composition series of $Q_8$?

[cloud walrus]

donut123

[coral spindle]

The quotient Q_8/{1, -1} is of order 4 and hence is not simple

You need to refine your series just a bit

[amber burrow]

oh i forgot thats a requirement

[coral spindle]

Knowing the definition of a composition series does help lol

[amber burrow]

lmfao yeah

[thorn jay]

I guess you could say you need to refine your understanding of them

[tough raven]

If the composite A → B → C is a localisation of commutative rings, must A → B be an epimorphism? (Of course, B → C must.)

[rocky cloak]

I mean,
Z -> Z[x] -> Z[1/2]
is a localization

Or
R -> R x R[S^-1] -> R[S^-1]

[acoustic igloo]

#help-8 message
there's a problem with my proof that sqrt(pi) is transcendental

i didn't show that g(x) is nonzero

i'm working on another problem which asks me to show that a rational function of a transcendental is another transcendental. i'm hitting the same problem where i need to show that a certain polynomial is nonzero

[tough raven]

Show that if g = 0, then either f_1 = f_2 = 0 or x is a square of a rational function (namely f_1/f_2). Then show that the latter cannot happen.

The abstract way to do that question BTW is to prove that the product of algebraic numbers is algebraic. Then sqrt(pi) is algebraic ⇒ sqrt(pi)⋅sqrt(pi) is algebraic.

[acoustic igloo]

maybe that will be easier

[south patrol]

Now show that pi is not algebraic

Lol

[acoustic igloo]

riiight

i don't think it will be easier to prove the product of algebraic numbers is algebraic

[coral spindle]

ACtually it's quite easy to prove that

[acoustic igloo]

oh

i will think about it a little more

[coral spindle]

You are aware that a complex number y is algebraic iff Q[y] is a finite-dimensional Q-algebra?

[acoustic igloo]

no

[coral spindle]

Do you know what a field extension is?

[acoustic igloo]

that's the chapter i'm doing now

[coral spindle]

Great

If I have a field extension F <= E, do you know what I mean if I write [E : F]?

[acoustic igloo]

not exactly

[coral spindle]

OK what do you mean by that

[acoustic igloo]

i just know it's called the index

[coral spindle]

Do you mean you do know but you can't remember

Great yes this is the dimension of E as a vector space over F

A finite extension E >= F is one where [E : F] is a finite number.

Fact: if x is algebraic over F then the extension F[x] is finite

You should hopefully be able to prove this fairly directly

Together with the tower law and this observation, the result that the product of algebraic numbers is algebraic comes out

[acoustic igloo]

that's stuff the book hasn't covered yet

[coral spindle]

Well great! You'll cover it soon.

[acoustic igloo]

they're not actually asking me to prove sqrt(pi) is transcendental, so maybe i don't have to do that

but they are asking me to prove that if alpha is transcendental over F, every element of F(alpha) that is not in F is transcendental

any way to do that before i get to vector spaces?

[coral spindle]

Yes it is very straightforward

Hint: Elements of F(alpha) are quotients of polynomials in in alpha with coefficients in F

Let me rephrase that

[acoustic igloo]

the problem is showing h(x) is constant

or wait, what do i need to show...

h(x) must be the zero polynomial. i need to show that this implies f(alpha) and g(alpha) are constant, i guess

[coral spindle]

Hint: multiply

Oh wait nvm you've got there

Where has h(x) come from?

[acoustic igloo]

multiplying by the denominators

[coral spindle]

But 0 times g(alpha)^n is 0

Why do you need to put the constant a_0 on the other side?

You're done!

[acoustic igloo]

hm i'm not so sure

[coral spindle]

Well ok, take your time

[acoustic igloo]

see i didn't use the premise that f(alpha)/g(alpha) is not in F yet

this should make it more clear how i got h(x)

i'm gonna eat dinner, i'll be back later

[tough raven]

I think it needs transitivity of algebraicity in towers though.

Or not, nvm.

[acoustic igloo]

How do I show that x is not the square of a rational function?

For the proof about sqrt(pi)

[acoustic igloo]

For this one I can check the solutions manual

i think i can handle the case when the degrees of f and g differ

wait the solution manual doesn't worry about showing h(x) is nonzero

this is what it says

am i overthinking this?

[tough raven]

Treat it analogously to showing that 2 is not a square of a rational number.

[acoustic igloo]

i think i got it, we have xg(x)^2=f(x)^2 and g(x) is nonzero
F[x] is an integral domain, so xg(x)^2 is nonzero and so is f(x)^2, and thus f(x) is nonzero
this means the left side has odd degree and the right side has even degree; a contradiction

thanks for the help 🙏

[finite turtle]

can anyone explain how to obtain the penultimate statement about [a,T] x [ab] generating the semi direct product, and how to show isomorphism to S_3 x Z_2

[velvet hull]

the notation is kind of confusing, but I believe that <a,T> x <ab> does not generate T xx V, but rather tells you what the isomorphism is

[velvet hull]

namely $\phi: T \rtimes_{\varphi_1} V \xrightarrow{\sim} S_3 \times \mathbb Z_2$,

where $\phi( 1 \rtimes_{\varphi_1} \langle ab \rangle) = 1 \times \mathbb Z_2$ and $\phi(\langle a, T \rangle \rtimes_{\varphi_1} 1) = S_3 \times 1$

[cloud walrus]

HChan

[velvet hull]

(the "<a,T>" is abusive-ish notation on the author's part, it's supposed to represent the subgroup of the semidirect product generated by T xx 1 and 1 xx a)

[rocky cloak]

It's not really abuse of notation.
T is a subgroup of G and a is an element of G and <a, T> is the subgroup generated by a and T.

[velvet hull]

not properly though, it should be Tx1 and 1xa

[rocky cloak]

G is not defined as an exterior semidirect product.

T and V are sylow subgroups

[velvet hull]

oh, right

[rocky cloak]

But yeah the point <a, T> and <ab> are normal subgroups that don't intersect and generate G, so G is the direct product of them

[finite turtle]

how does one see that we have an isomorphism?

[velvet hull]

to write down the isomorphism explicity:

$\phi(ab) = (e,1)$ (I'm using $e$ here to denote the identity because $1$ is not the identity of $\mathbb Z_2$)

$\phi(a) = ((12),e)$

$\phi(t) = ((123),e)$ where $t$ is the generator for $T$.

[cloud walrus]

HChan

[velvet hull]

you can check for yourself that this extends to a well-defined isomorphism

(you can take the presentation of S_3 to be {x^3,y^2,xyxy})

I'm pretty sure the latter isomorphism between <a,T> and S_3 is not unique, but this is the one I found

[finite turtle]

thanks. the way the solution i posted is written i thought there might be a more direct way to show isomorphism

[velvet hull]

well, there is an intuitive argument as to why they are isomorphic (and that's how I found phi), but I don't immediately see a simple way of showing the isomorphism that doesn't involve some kind of construction:
The order of a is 2 and the order of t is 3. So the obvious isomorphism is to just send a to (12) and t to (123). And S_3 is generated by those two permutations, so all we have to do is check this is well-defined

[vivid kestrel]

is it reasonable to say that the study of group representations is the study of ways a group can act linearly on a given vector space

[proper jolt]

hello, is this equation correct?

where V(I) is defined as follows

the equation is from a homework problem that asks me to prove it, but i don't think it's correct

[rocky cloak]

Should be unions.

Or possibly
V(I1 \cap I2 \cap .... Ik)

[proper jolt]

alright nice thks

[thorn jay]

It would have been \cup then inside the V

[rocky cloak]

\cap innside, \cup outside

[thorn jay]

Oh thats what you meant, i was slightly confused the "or possibly"

[tardy hedge]

For (c), are we keeping in mind the fact that Aut(F27) acts on the roots of x^27-x which just happen to be all elements of F27

Im not sure if im thinking of it properly

[glad osprey]

I don't have an answer to your question, but for a) and b) is the answer phi(26) = 12 in both cases? I'm not sure I understand the difference between generating the field vs the multiplicative group

[tardy hedge]

(b) was 12, the answer to (a) was 24

The solution for (a) just said the only subfield of F27 is F3. So there are 27-3 = 24 generators.

[glad osprey]

ah, that makes sense

For the first part of c) the answer is 3 right? Since F_27 is a splitting field we have |Aut(F_27)| = |Gal(F_27 /F_3)| = [F_27 : F_3] = 3, correct?

[rocky cloak]

In general the number of automorphisms is always less than or equal to the degree.

So just showing that the extension is Galois, or just explicitly giving the three automorphisms will do

As for the action I guess it's nice to keep in mind that the orbits of the Galois group are the conjugates of an element

So it just comes down to knowing the degree of the minimal polynomial of these elements

[tardy hedge]

Orbits of the Galois group G mean you take an element x and consider the set gx for g in G?

With G acting on F27 for the first part

[thorn jay]

Thats what orbits are so I'd assume so :P

[tardy hedge]

Sir Yes Sir 🫡

[amber burrow]

so i did this exercise that says that if G/Z(G) is cyclic, then G is abelian

but isnt G abelian if and only if it equals Z(G)?

[rocky cloak]

yup

[tardy hedge]

Is the degree of min poly of these elements all the same? So the size of all orbits are the same so number elements / size of orbit = # orbits?

[rocky cloak]

So another way to put it is that G/Z(G) is cyclic iff it is trivial

[amber burrow]

i was trying to do this exercise where |G| = pq, and if p does not divide q-1, G is abelian

you case work on order of Z(G)

if |Z(G)| = p

Then |G/Z(G)| = q

which is prime

so cyclic

so abelian

[amber burrow]

oh

ok that makes sense

"Hence we may assume Z(G) = 1.
Sec. 4.4 Automorphisms 135
If every nonidentity element of G has order p, then the centralizer of every nonidentity
element has index q,"

(both p and q are prime)

why is the centralizer of every nonidentity element have index q?

[rocky cloak]

Well, the centralizer of g is not everything (that would put g in the center), so can't have index 1.

And it contains the subgroup generated by g which has p elements, so has index at most pq/p

[amber burrow]

i got that, by why is it not just p?

[rocky cloak]

Because p doesn't divide pq/p = q

Interesting exercise though. I would have thought one already knew Cauchy's theorem or the sylow theorems before this, making this step redundant

[amber burrow]

they actually did have cauchy's theorem before this

but they use it for a different variation on this

to prove that its cyclic

that kinda had me confused too, cuz i thought you just go like it has to have a group of order p, and a group of order q

[crystal vale]

there is only one trivial homomorphism Z/nZ to C, because C has no non-identity element which has finite order, right?

but now i have to find the homomorphism Z/nZ to C\ {0}

i meant group homomorphism

[crystal vale]

so in this case image of 1 has finite order which divides n

so total number of homorphism is n? Because it sum of all euler function of divisor of n?

[glad osprey]

If n=p_1 ... p_k for primes p_i (not necessarily distinct), is the composition series of Z/nZ always of length k+1? Since at each step you remove exactly one prime in order for G_i/G_{i+1} to be simple

[thorn jay]

k, as your series is
H_0 = 1 < Z_p1 < Z_p1p2 < ... < Z_p1...pk = H_k

[tough raven]

Yes. Alternatively, the number of n^th roots of unity in ℂ is n.

[glad osprey]

There are k+1 groups in that chain, but I don't know how we define the length of a composition series, does that mean it has length k?

[tough raven]

The length is usually defined to be the number of adjacent pairs.

Because e.g. that is the number of factors G_i/G_{i+1}, length(G ⨯ H) = length(G) + length(H), etc.

[glad osprey]

I see, so the trivial series 1 < G has length 1?

[tough raven]

Yes (unless G = 1 maybe).

[thorn jay]

Oh yeah thats a good reason for it

[thorn jay]

Yee

[hidden wind]

girls

i totally didn’t spent almost two weeks finding this snake-proof of the four-lemmas

[tardy hedge]

Hey, in question (c), we are considering Aut(F27) as a group action on the set from (a), and then as a group action on the set from (b). How do we know that elements acting on those sets result in elements still in that set?

[rocky cloak]

Well, carefully examine the definition of those sets and the definition of a field automorphism I suppose

[tardy hedge]

A generator a for the field F means you take the smallest field containing a, and that should be F?

What is a generator of a field?

[velvet hull]

if you assume char 0 for instance, 1 generates Q

[rocky cloak]

It is just what you said. An element such that the only subfield containing it is everything

[tardy hedge]

Ok cool just wanted to make sure

[rocky cloak]

Also the empty set does

[velvet hull]

It doesn’t sit right with me very much in the same vein that Z[x] is generated by x as a comm unital ring

[thorn jay]

But thats not an element

Usually

[rocky cloak]

It's a set of elements

[tardy hedge]

oh, ok cause automorphisms should take generators to generators and units to units ...

[rocky cloak]

Automorphism preserves most stuff really

[tardy hedge]

yeah

[thorn jay]

That would be a generating set, not a generator :P

[elfin wraith]

That’s pretty sick, I hadn’t seen that before

[delicate orchid]

just do it with a spectral sequence next time it's much quicker

[hidden wind]

i dunno what a spectral sequence is

[tardy hedge]

My prof argued something in (c) saying Aut(F27) acts on those sets with no stabilizer. This would mean that (for the set in (a)) the size of each orbit is 3 right, so there would be 24/3 = 8 orbits. Aut(F27) acts with no stabilizer is because: the stabilizer of an element is a subgroup of G, and the only subgroup of Aut(F27) is {1} and Aut(F27), but the only elements fixed by Aut(F27) are those in F3 which are not in the set from (a) or (b)?

[tardy hedge]

Wow only comment since 5 hours

[tough raven]

I betrayed the channel for #advanced-number-theory

[tardy hedge]

😢

I like that sad bread emoji

[green fox]

Consider the standard action of $GL_2(\mathbb{Z})$ on $\mathbb{Z}^2$ by matrix multiplication.
Determine all the stabilisers of this action.

I am struggling to find the stabiliser of a general vector $v \in \mathbb{Z}^2$. I have tried just solving the equation $Av = v$, but have a few cases and am not sure if I have the correct answer/ if this is the correct approach.

[cloud walrus]

Luke

[crystal vale]

Yes

[woven dune]

Consider the action of A in GL2Z on an integral basis for Z^2

[green fox]

What does this mean?

[woven dune]

See what happens just for v=(1,0) and w=(0,1).

Then, if a (nonzero) point of Z2 is stabilized, what is a condition required for the determinant of A?

The way you’re doing it originally should work I think though

[green fox]

Is it true that the only element of the stabiliser of a general $(x,y)$ not in the integral basis is $I$?

[cloud walrus]

Luke

[green fox]

I used the condition that $det(A - I) = 0$

[cloud walrus]

Luke

[green fox]

And $det(A) = \pm 1$

[cloud walrus]

Luke

[woven dune]

Yeah this is basically it. DetA has to be 1. If it’s negative it reverses the vector and therefore doesn’t stabilize it

So what I would do is find matrices which have determinant 1 and preserve the vectors (1, 0) and (0,1). You can write the vector (x,y) as x(1,0) + y(0,1) for x,y integers. Can you take it from there?

[green fox]

Yes, but are you sure this is an $\iff$ scenario?

[cloud walrus]

Luke

[green fox]

Just because a matrix fixes the basis vectors thus fixes the vector, the converse is not obviously true

[woven dune]

Should be

[green fox]

this doesn't seem right, what if a matrix has two distinct eigan values, it is a clear contradiction

[woven dune]

While I was driving home I realized I spread misinformation mb dawg

Give me a sec I need paper for this

[crystal vale]

if i want to determine all group endomorphism of R to R, how can i do that? I know it gives f(q) = qf(1) for all q in Q, and if f is continuous then f(x) = xf(1) for all x in R. But when f is not continuous, how can i do that? Is there exist non-continuous function R to R such that f(x+y) = f(x) + f(y) ?

[velvet hull]

my guess it's likely going to be some kind of huge set determined by the free choice of the image of uncountably infinite many real numbers (that form a minimal generating set of R over Q), because you're essentially trying to understand the equivalence classes of real numbers that are algebraically dependent to each other over Q

[tough raven]

Yep, plenty. The keyword here is Cauchy's functional equation. But any even slightly well-behaved solution usually has to be scalar multiplication.

E.g., for any other solution, its graph is dense in ℝ^2 (in particular, it is discontinuous everywhere, is not monotone on any interval, etc.).

[lusty marlin]

All group endomorphisms ℝ → ℝ are given by linear maps from ℝ to ℝ when it is considered as a vector space over ℚ

[crystal vale]

Yes

[lusty marlin]

The first proof I saw for the density of the graph of such a function is among my favourite proofs of all time

[crystal vale]

Can I get the result, I want to prove that?

[rocky cloak]

One trick you can use. If you scale v down so that its entries are relatively prime. Then you can make a basis containing it.

But even without this it should be possible to just set up the equations

[crystal vale]

I have to show that Z cannot be Q-vector space.

If it is Q vector space then there exists ring homomorphism Q to End(Z) but since End(Z) is isomorphic to Z, as ring isomorphism.

And Q is field, so Q must be subring of Z, but then Q must be cyclic as additive structure but it is not, so it is not possible.

Is it correct?

End Z is ring isomorphic to Z right?

[velvet hull]

I would appreciate it if someone could validate my intuition

[crystal vale]

I have to find the order for the matrix, let A be that matrix then A^n = [ cos nx -sin nx, sin nx cos nx], but how can I choose n such that nx = 2kπ ?

[delicate orchid]

well you usually can't - most rotations don't have a finite order

[crystal vale]

Yes exactly

[crystal vale]

Can anyone verify it?

[delicate orchid]

so.... what's the problem?

[crystal vale]

But the author asked to determine the order so here the order is not finite

[delicate orchid]

are your rings unital? If not then no - Z is inital in Ring so there can only be one ring homomorphism Z -> Z

[crystal vale]

End Z is a set of all group homomorphism of Z

[delicate orchid]

then yeah it's Z

[crystal vale]

So is it correct?

[delicate orchid]

so there's two cases here, one where your angle isn't a rational multiple of 2*pi, and one where it is - the former case is what we've discussed

[crystal vale]

Yes

[delicate orchid]

??? then what

what were you asking?

[crystal vale]

No I got it, I was confused

[delicate orchid]

no worries

[twilit creek]

Hi, how can I improve this stupid proof ? do you have some suggestion to make it elegant ?

[delicate orchid]

yeah it's fine

[crystal vale]

Okay thank you

[elfin wraith]

As a general rule of thumb you don’t want to start a sentence with maths, you should have a sentence which leads into it

Also your second sentence is redundant, that’s what linear independence is

I think the proof is about as simple as it could be, but it could probably be written better. Think of your proof as a conversation you’re having with someone, trying to convince them it’s true. It should read like that

[twilit creek]

yepp I note

ok i'll improve, thanks for your advices

[crystal vale]

I want to understand the group of rotation of the cube, or say some rigid body, how can I understand it?

[twilit creek]

Do you have an idea how to show rigourously that the degree of the field extension Q(sqrt(2), sqrt(7)) over Q(sqrt(2)) is two ?

[delicate orchid]

well it's either degree 1 or 2, so show that sqrt(7) isn't in Q(sqrt(2))

or show that the minimal polynomial of sqrt(7) over Q(sqrt(2)) is quadratic, which is essentially equivalent

[barren sierra]

All subrings (not necessarily with unit) of the rationals are given by localizations of Z at some multiplicatively closed subset of Z right? So to characterize all subrings I must characterize all such subsets?

Oh I guess also nZ for some n as well whatever

[coral spindle]

You mean subrungs of the rationals, right? Localisations of Z are not typically going to consist of integers

But yeah I think these will look like localisations of nZ for some n. For example 3Z[1/2] I think is an example?

[thorn jay]

the only invertible elements of Z are 1 and -1
localisation changes that

so they cant be subrings of the integers

[barren sierra]

Yea I meant subrings of rationals whoops

Ok so I think that identifies all unital subrings of the rationals

Without unit though seems trickier

[thorn jay]

I don't think it changes much

maybe identify the nonzero positive integer n closest to 1 contained in your subrng, and then try to prove if its a localisation of nZ

[coral spindle]

Ah but 3Z[1/2] is not unital, is it?

What I mean is the rung generated by 3Z and 1/2, so the 'localisation' of 3Z at {2, 4, 8, ...} whatever that means

[barren sierra]

But (2) is not a multiplicatively closed subset of 3Z

So I guess it's not a localization

[barren sierra]

That's sort of my hangup

[glad osprey]

This is really obvious, but just to be sure: Z/4 and Z/2 x Z/2 have isomorphic composition factors, right?

[next obsidian]

Yeah

[barren sierra]

I guess I could just describe it as nZ[1/p : p in S] for some S subset of the primes

And forget the word localization exists

[thorn jay]

oh how nice the world would be if isomorphic composition factors implied isomorphism

[coral spindle]

Semisimple rings say hello

[coral spindle]

Here's a thought. This is a localisation of the Z-module 3Z. So there is some sense in which it is a localisation. I don't know how it would work in rung-land, but it's an idea

[glad osprey]

Do isomorphic composition factors imply isomorphic vibes atleast?

[coral spindle]

There are some properties that rely only on the composition series, sure

For example solvability

[thorn jay]

they imply equal cardinality in the finite case

:P

[coral spindle]

It's hard to pin down which properties two groups with the same composition series will share, I think

I think it's quite subtle

[thorn jay]

sometimes can make all the difference though

[coral spindle]

Yup

[thorn jay]

I will introduce a new property of groups

a group G satisfies the shoinky property if and only if the free product of its composition factors is isomorphic to the automorphism group of some boolean algebra

[thorn jay]

I've got a property that is implied by isomorphic composition factors

✨

[tough raven]

Grothendieck group moment

[coral spindle]

I don't really know much about (algebraic) K-theory but it is arguably the study of exactly this, isn't it lol

[worthy solar]

$\mathbb{R} \oplus \mathbb{R} \text{ and } \mathbb{R} \times \mathbb{R}$

and if the former makes sense or not
(in the context of algebra stuff like group rings and fields
because i was looking at the ring of integers and the quotients of it and the book uses the "direct product" but doing stuff like Z/4Z x Z/5Z but I was taught in group theory to use the Z/4Z oplus Z/5z
and i dont know if it is because the latter was the additive structure
and now i am trying to talk about it as an entire ring

what is the diff between
$\oplus \text{ and } \times$

[cloud walrus]

Brandon7716

Brandon7716

[coral spindle]

As a set you can consider them the same thing

[worthy solar]

like i dont konw if the symnols are specific to specific to looking at it as a group vs a ring

or group under addition vs group under multiplication

😖

[coral spindle]

Any nontrivial ring will never be a group under multiplication

You are right, we talk about R x S when we are talking about products of rings

When we are talking about Abelian groups (or modules) we want to talk about \oplus, the direct sum

[worthy solar]

okay

[coral spindle]

I am thinking of a good way to describe why we want it to be this way

[tough raven]

Well, me neither. But the Grothendieck group at least is precisely about ignoring the extension problem.

[coral spindle]

Here are a few reasons:

• In the sense of category theory, the (finite) product of rings is a categorical product, but not a categorical sum. Whereas if we look at Abelian groups, it is both the product and a sum.
• In the case that we are taking the product of infinitely many rings, when we look at the underlying group structure, this is different from the infinite sum of the underlying rings.

[thorn jay]

I suppose we use \oplus for modules because it is then easier to see the summands (components you sum) as submodules, which is often easier

[worthy solar]

i guess i dont even have a concrete defintion of what these mean

[thorn jay]

also has to do with the subtle (but important) difference of infinite products and infinite direct sums, and I've personally seen infinite direct sums more in module theory than products so I guess thats why they prefer it

[worthy solar]

like this is pretty much how the book defines it

[coral spindle]

Yup

[thorn jay]

(also something something classification of finitely generated modules over PID?)

[worthy solar]

so my impression is it is just to note that we are not just looking at the componetwise additive structure but now we have also this notion of componentwise multiplication

and so the x is used

🤷‍♂️

[thorn jay]

basically

[coral spindle]

It is fair to say that it is mostly convention :)

[thorn jay]

I think the oplus looks nicer too

not as nice as otimes

[worthy solar]

not sure how the details pan out if you take non finite or amount of these products or sums

or if that even makes sense

[next obsidian]

I use oplus when I am using its coproduct property and times when I use its product property so true

[coral spindle]

Let's not worry about that for now. You'll come across it later. I just wanted to explain that there is a genuine difference in the infinite case.

[worthy solar]

🙏 got it

[worthy solar]

I dont think i will

but thanks for believing in me 🥹

[thorn jay]

what if you have to use both in the same proof

go go gadget change notation without telling?

[next obsidian]

I just switch

[coral spindle]

Lmao

[next obsidian]

I actually just haphazardly pick one tbh

[tough raven]

I think I do this actually

I think of it as signalling that the proof uses the identification between them.

[next obsidian]

To me

oplus means both

Cuz there’s also a coproduct

Symbol

[thorn jay]

I simply dont work with structures which allow a nice \oplus 😎

[next obsidian]

Unless it’s infinite in which case…

[thorn jay]

dont have to choose when you can't even define "finitely many nonzero values"

[tardy hedge]

Holy Moly the font!

[next obsidian]

Jesus….

I regret clicking that

[thorn jay]

not even math mode for "Z/4Z oplus Z/5Z"

qwq

[coral spindle]

Maybe I could summarise by saying:

When you write R (+) R, I think "that's an Abelian group!" whereas when you write R x R I think "that's a ring!"

[next obsidian]

This is true af

[worthy solar]

thats about what i got from it

well

[thorn jay]

omw to view R x R as the underlying multiplicative monoid

[worthy solar]

it is quite nice

[thorn jay]

comic sans??

[coral spindle]

Comic Neue

[worthy solar]

just think, someone approved my preamble

and now we all get to enjoy the result from it

[thorn jay]

they need to be fired

[tardy hedge]

i aint seen a font like that since elementary school

[thorn jay]

immediately

[coral spindle]

It's genuinely a well-made font.

And the choice is obviously to rile you up ya dinguses

[thorn jay]

imagine an academic paper in this font

special comic sans math symbols too

[tardy hedge]

yes comic sans hate is overdone sure

its still funny though

[tough raven]

True.

[coral spindle]

Comic sans is less well-made than comic neue but yes, it still stands

[tardy hedge]

boytjie im listening to aphex twin rn

[thorn jay]

whats the difference between them?

[thorn jay]

based

[coral spindle]

They are different typefaces made by different people. But here is a side-by-side comparison of one of the fonts:
https://upload.wikimedia.org/wikipedia/commons/thumb/2/20/Comic_Neue_and_Comic_Sans_comparison_(DYK).svg/1920px-Comic_Neue_and_Comic_Sans_comparison_(DYK).svg.png

[thorn jay]

I would too but I'm too busy listening to dnb artists maybe 2 people in the netherlands have heard of

[thorn jay]

its way more bearable

at least

[tough raven]

What is .svg/.svg.png?

[coral spindle]

There are just lumpy bits in Comic Sans that have been removed in Neue. The kerning is better. The proportions are better imo. There are also several other fonts in the Comic Neue family

[thorn jay]

its the future

[tardy hedge]

oh yea what were some good dnb artists

i forgot about ur suggestions

[thorn jay]

Noisia

Sleepnet

[tardy hedge]

ok

[thorn jay]

IMANU
Buunshin
(I'm seeing them live in a month THEYRE COMING TO MY CITY EXCUSE ME)

[thorn jay]

(the artist I was referring to at that moment was Circumference)

[elfin wraith]

Mozey, Chase and Status, 4amKru, Bou, Gray, Phibes

Nia archives too of course, but this isn’t too algebraic

[thorn jay]

Chase and Status, Bou are peak

[thorn jay]

Yes it is, Noisia's logo is 4 circles

Which is like algebraic geometry

So close enough

[glad osprey]

I'm guessing you do it subtly though, like:
$R \oplus S$ blah blah $R \rotatebox[origin=c]{-12}{$\oplus$} S$ blah blah $R \rotatebox[origin=c]{-24}{$\oplus$} S$ blah blah $R \rotatebox[origin=c]{-45}{$\oplus$} S$

[cloud walrus]

sheddow

[next obsidian]

Nononono otimes and oplus are way different

[thorn jay]

Coproduct in CRing vs coproduct in R'-Mod where R and S are naturally R'-modules from context

[next obsidian]

The tensor product would be a coproduct in R’-alg

I mean at least I would implicitly be taking the tensor product over R’

[thorn jay]

Defeated by my own words

What if you have different R' though

[glad osprey]

Is a composition series of G basically just a path through the tree (or lattice I guess) of its normal subgroups?

[delicate orchid]

such that the quotient of each successive pair of subgroups is simple, yes

you described a subnormal series

[thorn jay]

No

Its in the lattice of subnormal subgroups

(Such that each quotient is simple)

[delicate orchid]

good point. That's a subtle but important detail

[thorn jay]

Take D_8 for example

Its composition series is
1 < 2Z_4 < Z_4 < D_8

[delicate orchid]

although honestly we might as well just write $1 \triangleleft H_1 \triangleleft ... \triangleleft H_n \triangleleft G$ at this poin t it's just the easier way of thinking about i t

[cloud walrus]

☻ Wew Lads Tbh ☻

[thorn jay]

But evidently 2Z_4 is not a normal subgroup of D_8

[delicate orchid]

le 1 < Z(D_8) < C_4 < D_8 has arrived

[thorn jay]

Wait really?

Nvm my statement then

Oh yeah ofc

Dumb

[glad osprey]

Ah yes, I just realized: if H is normal in G, there may be a subgroup N < H that is normal in H but not in G, right?

[thorn jay]

Yee

Subnormal series are very powerful tho

Ive seen them used in automorphism tower stuff

Very.. weird proof

[delicate orchid]

I'm trying to think of an explicit one where there must be at least one subgroup that's subnormal. S_4 doesn't work

[glad osprey]

But if we're looking at paths in the lattice of subnormal subgroups we can skip this qualification right? Because a path by definition can't "skip" any nodes, so each subgroup must be maximal

[thorn jay]

No

Take the subnormal series
1 < G

Lol

[delicate orchid]

ah got it, C_3 \wr C_3 should work

[thorn jay]

Freaky ahh product of groups

[thorn jay]

You misunderstand "path" haha

I tend to call it a chain

[delicate orchid]

C_2 wr C_3 probs doesn't work cause Q_8 is Hamiltonian

[thorn jay]

(I.e. a chain in a poset (P, <) is a sequence of elements
a_0 < a_1 < a_2 < ... < a_n
Not necessarily distinct)

[glad osprey]

You mean in the case G is not simple? I'm thinking about the full lattice of subnormal subgroups of G, so there would be a maximal subgroup between 1 and G. Like, take a group G, draw a tree of all its subnormal subgroups, then you can draw a path from the root G to a leaf, which would necessarily be the trivial group (I'm talking about paths as in computer science, dunno if there is a better term for lattices)

maybe this only makes sense for finite groups

[thorn jay]

There is not a tree of all subnornal subgroups

Its a lattice, im pretty sure

[toxic zephyr]

so in the remark is K0=K'' and then AutK0(F)=K0'=K'''=K'? is that what it's saying?

[thorn jay]

Also "chain of subnormal subgroups wouldnt even be correct lol"

Dumb me

A subnormal series is a chain of subgroups
1 = H_0 < H_1 < ... < H_n = G
Where each is a normal subgroup of the next

[glad osprey]

You can make it a tree by just not merging the nodes though, right? Either way, I think it doesn't matter, since you can draw a path from the root to a leaf anyways

[thorn jay]

But thats very counterproductive

Just use a lattice

Its way easier to visualise

[glad osprey]

yeah, I guess

[thorn jay]

A tree would just convolute things

[glad osprey]

but anyways, my idea is that if I want to find a composition series (or actually all composition series) of a finite group, I can just draw the lattice of subnormal subgroups, then each path from the root of the lattice correspond to a composition series, is that correct?

[thorn jay]

No, the problem is that its not just a chain of subnormal subgroups

Its a chain of subgroups such that each is normal in the next

[thorn jay]

.

So i guess there your tree idea is good?

But still, what you usually do is take a subnormal series and refine it until its maximal right

[glad osprey]

I might be formulating myself poorly, but how about this: I start with G, then find all maximal normal subgroups N_1, N_2, ... Then for each N_i I find all maximal subgroups that are normal in N_i, say N_i_1, N_i_2, ... until I eventually hit {1}. Then for example G > N_1 > N_1_2 > ... {1} is a composition series, and all composition series can be found this way. Does that make sense?

[thorn jay]

Yeah

[vestal sapphire]

Is there some easy way to show that this is just sgn(pi)? (sigma is a permutation aswell)

[toxic zephyr]

if someone could confirm my understanding that'd be appreciated

[south patrol]

How are you defining sgn?

[vestal sapphire]

by definiton its almost exactly the same as on the picture, but we replace sigma(i) with i and sigma(j) = j. This thing seems obvious to me but for some reason I cant think of a good argument why. So far Ive just written that its because sigma is a bijection

[south patrol]

Actually that argument does work if you are careful lol

Just justify why it's okay considering that sigma may change the order

[vestal sapphire]

Hm thats weird, I got this product by trying to prove that sgn(pi sigma) = sgn(pi) sgn(sigma) and so far I got to sgn(pi sigma) = sgn(sigma) * the thing u see above

oh okay so it works

[vestal sapphire]

yeah but thats the part i have trouble with

[south patrol]

Probably the best way is to notice: what happens to each term when you switch iand j?

[vestal sapphire]

well nothing really happens

[south patrol]

Exactly

[vestal sapphire]

Hm so maybe a good argument would be:
sigma is a bijection so every pair (i, j) is achieved by some (sigma(k), sigma(l)) but it might be in the wrong order (i>j), but that doesnt change the product.

because what Id like to do is just write this product and remove the sigmas

[south patrol]

Yes exactly

Another equivalent thing would be to notice you can just index the product by unordered pairs of distinct things

So that the ordering goes away entirely lol

By the way, the reason I asked which definition you used is that there many different equivalent definitions - probably the best is that the sgn function is the unique group homomorphism S_n -> {-1, 1} which sends transpositions to -1

[vestal sapphire]

Yes, Ive seen some other definitions. Its probably defined this way because we were doing permutations prior to homomorphisms, so we didnt officially now what a homomorphism is. Personally I always with of sgn as of (-1)^number of moves to sort the permutation

anyway, thanks a lot for help

[south patrol]

Npp

[vast tangle]

Does this proof look good?

the second to last line should read "... implies a has finite order"

[coral spindle]

I think you're overcomplicating this a little bit. For example, you don't at all need to deal with the case that H=G seperately.

But yeah the logic is fine

[vast tangle]

Is that the only part that's a bit extra fluff?

[coral spindle]

It's a bit wordy I guess

But yes, there is no other extraneous logic

[thorn jay]

I guess you could show it directly that finite order => in H => identity

Would be a little cleaner, without the contradiction

[south patrol]

This is what I'd do too

[thorn jay]

But for the rest I agree with Boytjie, nothing wrong with it :3

[vast tangle]

alright, i'll write up that approach separately for completeness sake, thank you all

[thorn jay]

Dedication

[glad osprey]

Hmm, yeah, I think so... But I'm not sure thinking about the fixed group of the fixed field of the fixed group is the easiest way btw, what book is this? Is using apostrophe for the fixed field/group common?

[toxic zephyr]

hungerford and I'm not sure but it's what we do in class

but in class we went over that

• L'''=L' for all subfields
• H'''=H' for all subgroups of G
• F is Galois iff K''=K

this K0 remark seemed similar and I wanted to make sure I was understanding correctly

but like yeah basically it's Galois if K is the only thing fixed by G (or, alternatively, if anything outside of K can be moved by something in G), right?

like that's the basic idea of a Galois extension?

something like F=Q(2^1/3) isn't Galois because any root of x^3-2 in F (i.e. 2^1/3) must be mapped to another root. but the only root in F is 2^1/3. so the only map in G is the identity, which fixes all of F.

[velvet hull]

my intuition for a Galois extensions is that it is a looser form of algebraic completeness

[toxic zephyr]

am I on the right track?

[velvet hull]

so the extension contains everything that is "algebraically indistinguishable" from things in the extension

[tough raven]

Yes. It also says that F/K is Galois iff K'' = K (and so F/K0 is Galois for a general algebraic extension).

[glad osprey]

Yeah, that sounds correct 👍 The apostrophe is a map between subgroups and intermediate fields, which is called σ and τ here. For any field extension, Galois or not, this is what's called a Galois connection, which is basically a weak form of isomorphism between preorders. When the extension is Galois, this connection is upgraded to an actual isomorphism, ie. E = τ(σ(E)) and G = σ(τ(G)), and the lattices of intermediate fields and subgroups are isomorphic (but reversed)

[tough raven]

No, but it often helps to locally introduce notation for something you're going to talk about a lot and I think that's what it is.

[glad osprey]

I'm not sure I like that the apostrophe is used for the maps in either direction, but I guess with context it's not that big a deal

[thorn jay]

Galois connectionnn

Peak

[glad osprey]

Also this theorem is pretty nice, it gives some context to why it's called a normal extension

[tough raven]

Actually, since Galois theory played a role in motivating group theory, I suspect it's the other way round.

Which raises the question of where whichever one came first got the word 'normal' from.

[glad osprey]

Yeah, good point. I suspect they consulted the list of overused terms in mathematics, or maybe that was actually its first entry

[crystal vale]

is it true that there is no injective homomorphism S_n - A_(n + 1) for any n? I think yes it is true for every n > 1

it is easy to check that when n is even it is not possible, so we have to consider the case when n is odd

Now n = 3 is easy, for n >= 5, for this if there exist a subgroup of order n! in A_( n+ 1) then there will be non -trivial homomorphism A_( n + 1) -> S_( n + 1 )/2, and it cannot be injective homomorphism so it implies A_(n+1) has non-trivial normal subgroup, which is not possible

[ivory ore]

finite p group with more than one maximal subgroup

how to think about the number of least(max) number of such maximal subgroup?

[vagrant zinc]

Name of book?

[proper jolt]

does the first equation work if I_1,..., I_k were just sets

or must they be ideals?

[proper jolt]

forgot to put definiton of V(I)

[acoustic igloo]

what if you consider the sets {x,x-1} and {x,x-2}
then V({x,x-1}∩{x,x-2})=V({x})={0}
but V({x,x-1})∪V({x,x-2})=∅

[proper jolt]

oh yeh that's true thanks

[sturdy spear]

if $|a|=m$ and $|b|=n$ in an Abelian group, then $|ab|=mn$ if gcd$(m,n)=1$ otherwise idk if there is formula for $ab$\
like in $Z/4Z$, $|2|=2$ but $|2+2|=|4|=1$ idk how to relate these results.

[cloud walrus]

Abstract Afzal

[rocky cloak]

NORMAL SUBGROUP. According to Kramer (p. 388), Galois used the adjective "invariant" referring to a normal subgroup.

According to The Genesis of the Abstract Group Concept (1984) by Hans Wussing, "The German Normalteiler (normal subgroup) goes back to H Weber , Lehrbuch der Algebra, vol. 1, Braunschweig, 1895. p.511 and is possibly linked to Dedekind's term Teiler (divisor), which was employed in ideal theory" [Dirk Schlimm].

https://mathshistory.st-andrews.ac.uk/Miller/mathword/n/

I don't think Galois really gave names to types of extensions, as most was framed purely in terms of roots of polynomials

[rocky cloak]

It will divide the lcm at least

If d is the gcd, then
a^d and b^d has relatively prime orders, hence there product has order lcm/gcd. So the order should be a multiple of lcm/gcd

[ivory ore]

is this the gist ?

are there other uses of frattini ? that i should be aware of!

[velvet hull]

G is nilpotent iff Phi(G) is nilpotent

[chilly ocean]

why every rng is an ideal of some ring

[rocky cloak]

A set generates G iff its image generates G/Phi(G).

In particular all minimal generating sets has the same number of elements, equal to the dimension of G/Phi(G), and an element of the Frattini subgroup is always redundant in a generating set

[rocky cloak]

Freely adjoining a unit to your rng gives a ring which has your rng as an ideal.

In fancy abstract nonsense words the inclusion of the category of rings into the category of rngs has an adjoint.

And there is an equivalence of categories between Rng and the slice category Ring/Z

[ivory ore]

ahh nilpotent, was thinking of that

haha, i think its after this section of sylow

i vaguely can recall this: nilpo iff all sylow subgroups normal

[ivory ore]

now the "analogy" to being analogous makes more sense, for whatever that means

[chilly ocean]

what is the ring containg all continuous function with compact support as an ideal

i dont know any category theory yet and may i will not

[rocky cloak]

Like you're asking what you get from this construction? Otherwise it's an ideal in the ring of all continuous function

This construction would just add in the constant functions with integer values (and then any sums of those with the original functions)

[sturdy spear]

how did you come to know this?

sorry i am blind enough to see this

[rocky cloak]

if l = lcm
Then (ab)^l = a^l b^l
which equals...

[sturdy spear]

oh i see

[proper jolt]

im confused with this proof how can we choose the value a so that it has the prime decomposition we want it to have? in this case a = I times the product of the Q_i^{e_i}. Also A is a dedekind domain

[chilly ocean]

movie scene

[tough raven]

Oh yeah, I think I've seen that in Bourbaki.

[glad osprey]

(it's Aluffi if you weren't sure btw)

[thorn jay]

Based

[vestal sapphire]

Assuming that we know that if sgn(pi) = 1 then pi can be written as a product of cycles (1 i j) how do we show that if sgn(pi) = 1 then pi is a product of cycles (1 2 i)? We dont have to use the first part, I jusy mentioned it because it might be useful

[rocky cloak]

What is (1 2 j) (1 2 i) (1 2 i)?

[vestal sapphire]

(i j 1) = (1 i j)

holy crap

how did you come up with it so fast

like really ive been thinking about it for at least two hours, but I focused on a different approach

[vestal sapphire]

Is there a similar trick if we want pi to be a product of (i (i+1) (i+2)) cycles? Im really sorry about those silly questions, but Im really bad at this stuff sometimes.

[rocky cloak]

Well, if (1 2 i) generates it, that means (1 i j) is a product of such. So just try to multiply some of them together and there you go.

I gues more or less the same should work for your other example, though you may need to do some induction

[vestal sapphire]

ookay thanks a lot

[rocky cloak]

A useful general concept is that if s and t are permutations, then
s t s^-1
is the same as t except with the symbols reshufled according to s