groups-rings-fields

[forest tundra]

i see caley hamilton theorems from matrices which i studied few months ago

[bleak abyss]

Linear algebra is part of algebra, and Artin doesn't wanna assume linear algebra background

If you don't wanna review linear algebra, read Jacobson

Or D&F

[warped fable]

Or just skip to group theory

[bleak abyss]

Anyway time for me to conquer this channel a bit and review last week's conference

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

So this should be in GL3 as long as the top left entry is non-zero and the bottom right 2x2 bit is in GL2

[cloud walrus]

Sloth King Daminark

[bleak abyss]

So that's $\frac{(q^3 - 1)(q^3 - q)(q^3 - q^2)}{(q-1)(q^2 - 1)(q^2 - q)} = q^2(1 + q + q^2)$

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Actually no I should divide further by q^2

The number of such matrices is what I said times q^2 since I forgot you can have q choices for the (1,2) slot and q for the (1,3) slot

So yes number of lines in F_q^3 is 1 + q + q^2

Number of planes should be the same

[cloud walrus]

Sloth King Daminark

[bleak abyss]

And that has the same count as before

Each plane contains q+1 many lines, that's just the order of P^1(F_q)

And we wanna say that each line is contained in q+1 many planes, which likely holds by a kind of symmetry but

Wrong type of algebra

You want #prealg-and-algebra

[sudden drum]

thanks 🙂

[delicate orchid]

oh my golly gosh it is le wholesome parabolic subgroup

[bleak abyss]

truuuu

But yeah so let me attempt to make that symmetry argument

So given a L and a plane P containing it, I'll map the line L to {(x,0,0)} by some element of GL_3

Let's call that element A

And let's let P' = {(0,y,z)}

A(P) + P' = F_q^3

So dim(A(P) cap P') = 1

So what I do is, I'm gonna map the plane P containing L to the line in P' which is given by A(P) cap P'

Question is, does this depend on the choice of A?

If not then I should be able to say this is a bijection from planes containing L to lines in P'

I mean really the answer is orthogonal complements basically

Out with it det

[rustic crown]

too shy now >.<

was gonna say something about this >.<

okie i'll say it >.<

yea but i would say that's more so because in early algebra you're studying sets with extra stuff, like it makes sense to say "consider a function f : X --> G from a set to a group", which if you're being pedantic is a map f : X --> UG.

aluffi introduces functors right before doing some "harder" linear algebra like tensor/symmetric/wedge stuff and later for some homological algebra. i think that's a good place for it. Other than that, I think i really find it useful to keep "rights adjoints and hom functors are continuous" and all that in mind while doing algebra as they change your perspective on proving a statement like Hom(-, A) turns co-products into products, from "yea this is an easy check, which i can do but i'm too lazy to" to "ofc, continuous preserves limits".

.<

[bleak abyss]

That's a pretty fair point yeah. I guess what I was going for with my statement was less

Ugh why is Aluffi already doing functors

So much as, I think the lack of category theory in other books isn't a huge problem, and not enough to counter Aluffi being a bit too slow/not the best problems/examples

[rustic crown]

right, aluffi not exactly a problem book. it's exercises are usually just to see if you understood the definitions than "solve this thing right now, which hardly anyone can. later you'll see when we develop more material that this was not that bad"

[bleak abyss]

Anyway, to formalize the whole shtick with orthogonal complements instead of dicking around like I was earlier

A plane in F_q^3 is given by some equation ax + by + cz = 0

So given a line l = span(x_0, y_0, z_0)

We're counting the set of (a,b,c) such that ax_0 + by_0 + cz_0 = 0

But now we can just think of x_0, y_0, z_0 as coefficients and a,b,c as variables

So gg

[summer path]

Never seen dami in aa before

[rustic crown]

dami does scawy aa

[lethal dune]

aaaaa

[bleak abyss]

Let us continue

So, this isn't in the notes but connects to my combo class

A "finite projective plane" in general is an incidence structure, so I have a set P of points, a set L of lines

And an incidence relation I \subset P x L, telling me which points are on which lines

And I want it to be such that:

• Any two lines share exactly one point
• Any two points live in exactly one line
• (Non-degeneracy condition) There are 4 points, no 3 of which are on a line

In this case, I'm gonna let the points be lines in F_q^3, and the lines are planes in F_q^3, and incidence is inclusion

Well, given two points, aka two 1-d subspaces, there's exactly one line (2-d subspace) containing them both, namely their span

Given two lines (2-d subspaces) V and W, we have V+W = F_q^3

So dim(V cap W) = dim(V) + dim(W) - dim(V+W) = 1, so their intersection is the one point (1-d subspace) they both contain

As for non-degeneracy, probably just take {<e_1>, <e_2>, <e_3>, <e_1 + e_2 + e_3>}

Now, in general, the degree of any point in a projective plane = the degree of any line, call this quantity n+1, and say that n is the order of the plane

Then you have n^2 + n + 1 points, and n^2 + n + 1 lines

In the case of P^2(F_q), the order is q (sensibly), and the q^2 + q + 1 tracks with what was said earlier

The proof of the fact that all degrees are the same and of the count of points and lines? Prob could do it another time, and frankly I did in combo way back when, so I won't emphasize now

So, two types of ways I can draw these guys. One is, points as points and lines as lines. The other as a bipartite graph, points are blue vertices, lines are red vertices, and an edge connects a point to a line which it's on

In particular, consider P^2(F_2). Order 2, so degree of things is 3, and #points = #lines = 7

Black vertex = point with label, white vertex = line labeled as set of 3 opints

[delicate orchid]

oh my golly gosh it's le wholesome octonians

[bleak abyss]

Lol

Anyway let's even go so far as to assign the points to subspaces

To see apartments

[bleak abyss]

This might start to be a bit more painful given that these guys have their labeling already

[uncut girder]

Isn't that the heawood graph

[bleak abyss]

Yeah, a name I just learned 😛

[uncut girder]

Its the graph you get by embedding a complete graph on 7 verticies into the torus and then taking the dual

[bleak abyss]

Yeah I played a bit on a whiteboard and see it now

The idea is to take 3 vertices consecutively, go to the other side (take the only edge that's not on the "boundary", go 3 vertices in the other direction, come back

[spice whale]

what do quadratic forms over the p-adics look like

up to equivalence

posted here because i wasn't sure where else to put it

[dim widget]

The key case to think about is in 3 dimensions

[bleak abyss]

Isn't there something called the Hilbert symbol or whatever that tells you about that?

[dim widget]

All of them can be diagonalized and it comes down to classifying squares in Q_p, understanding which 3 dimensional forms are equivalent, then doing induction to classify all of them

[bleak abyss]

Serre Course in Arithmetic I think discusses the stuff

[dim widget]

Yes but these kinds of manipulations are where the Hilbert symbol comes from and it can be done by hand in this case

[spice whale]

and then can you always go up from 3d

[zenith hollow]

this is the full answer for that sheet

it's just braindead to put b) as a task in the exam

[summer path]

i mean

if you know what it's supposed to look like

you can skip some steps

or just write it in terms of indices

but i do agree that's not a very good exam problem

[molten silo]

Can anyone help me with this question C(i)

[zenith hollow]

oh damn before i start forgetting

i * i = -1?

[coral spindle]

This is an unnecessarily long solution to this question. It is not hard at all to prove what is written here using indices rather than writing out each one.

[zenith hollow]

or 1/i = -i ?

[formal ermine]

jesus christ

[zenith hollow]

with my current knowledge i will have to write that should it come to the exam

but i think it's unlikely since i am sure they learned from their mistakes and saw that students had trouble with this

[coral spindle]

Are you aware that $(AB){i,j} = \sum{k=1}^n A_{i,k}B_{k,j}$?

[cloud walrus]

Boytjie (never-to-be-glomed)

[coral spindle]

This simple formula makes what is written very easy to prove

[zenith hollow]

i am not that advanced

[coral spindle]

This is the definition of matrix multiplication

I wouldn't call this advanced

[delicate orchid]

If you’re CS and don’t know that formula i really do wonder how you’re implementing matrix multiplication

[formal ermine]

function composition then look where the basis vectors land

[dim widget]

@spice whale yeah it turns out every form of dimension > 3 is a trivial form in n-3 dimensions + a nontrivial form in dimension 3, thus can be proved by induction

*this

[delicate orchid]

The higher forms get “glomed” down

[uncut girder]

What

Oh quadratic form

[south patrol]

lol this is what i initially suggested

[coral spindle]

🤝 we can lead a horse to water

[molten silo]

Can anyone help with this?

[coral spindle]

Looks like something Chmonkey would know

[white oxide]

how does 33.10 imply that the subfield K of the algebraic closure of F consists of the zeros of x^p^rn - x? wouldn't it just consist of the zeros of x^p^n - x?

also what is n in this case? (if F has p^r elements and not p^n elements)

[south patrol]

Not sure what you mean

K is by construction just teh roots of x^{p^{rn}} - x

It's not the same K as in 33.10

You just change n (in 33.10) to rn

[white oxide]

oh yeah oops

thanks

[tiny jolt]

If an element x^n (n>0) is in a prime ideal, does it follow that x is in that ideal?

[coral spindle]

Well x . x^{n-1} is in the prime ideal, so either x is or x^{n-1} is (or both, ofc)

by induction we're done

[tiny jolt]

Ohh right, I forgot you can induct

[next obsidian]

You need to see if you can remove an ideal, so you want to compute the three pair-wise intersections, and try to verify if any of these are primary. If not, it’s minimal.

Alternatively, you can try and compute associated primes of J itself and then use primary decomposition nonsense to see if it’s minimal, but this is probably harder to be quite honest

[white oxide]

i'm confused as to what purpose the sentence "because the elements of E are zeros of x^p^n - x, we see that f(x) is a factor of x^p^n - x in Zp[x]" serves? couldn't we just use the uniqueness of the irreducible monic polynomial f(x) in Zp[x] and basically apply the first half of the proof to E' to show that they're both isomorphic to Zp[x]/<f(x)>?

[molten silo]

I think the answer is <x^2,y^2,z> becuase it contains <x^2,z> and the radical of <x^2+y^2-z> is equal to the radical of <x^2,y^2,z> i think. so we can elimate that one

@next obsidian

[next obsidian]

I don’t understand what you mean

You’re saying you think you can remove that ideal?

[molten silo]

yes

i assumed that they have the same associated prime

the first to generators

[next obsidian]

Wait I’m sort of confused even lmao <x^2,y^2,z>\cap <x^2,z> is just <x^2,z>

[white oxide]

i guess the main confusion that i'm having with this proof is understanding why f(x) is the irreducible polynomial in Zp[x] for both E and E', is it just due to uniqueness of E and E' both being simple extensions over Zp?

[next obsidian]

So you can remove that for sure

[molten silo]

yes

[next obsidian]

So you have the first and last ideal

[molten silo]

agreed

[next obsidian]

And these have different radicals

So you can’t reduce it any further

[molten silo]

yes, which have the same associated prime right?

oh

[next obsidian]

I don’t think so?

[molten silo]

i see

[next obsidian]

Let me look at them again

[molten silo]

i thought they were the same

[south patrol]

The terminology for this is that prime ideals are radical btw

[next obsidian]

The radical of the last one is <x,z>

I don’t see how that could possibly be the radical of the first ideal

[molten silo]

no

[next obsidian]

how does x^n land inside of the first ideal

[molten silo]

I was thinking about the first and second one

[next obsidian]

Right, but you removed the second one

Since when you intersected it with the last one, you just get the last one

[molten silo]

dawm

i thought you remove the other one

[next obsidian]

Well no

[molten silo]

i understand now

[next obsidian]

You’re just saying that like

I\cap J\cap K = I\cap (J\cap K)

And here J\cap K = K because K < J

Yeah?

[molten silo]

i cant read that

lol

[next obsidian]

Idk what’s unreadable about that

$I\cap J\cap K = I\cap (J\cap K)$

[cloud walrus]

Chmonkey (glomed)

[molten silo]

ha

[next obsidian]

$And here J\cap K = K because K < J$

[cloud walrus]

Chmonkey (glomed)

[molten silo]

you keep the smaller one.

I some times get confused

[next obsidian]

Okay

Well I hope it’s clear now

[molten silo]

it is

thank you

[white oxide]

can i get some help here pls?

[frigid lark]

They're trying to show that it's the same f(X) in both cases

Have you done splitting fields?

[white oxide]

yea i have

[frigid lark]

Yeah, then you can show that each field of size p^n = q, is isomorphic to the splitting field of x^q - x

cf Lang Chapter 5 section 5

[frigid lark]

Over Z/pZ[x] ofc

And you can actually prove the existence of a field of size q by proving that the roots of x^q - x in Z/pZ[x] form a field

And x^q - x in Z/pZ[x] has no multiple roots since its derivative = -1

[white oxide]

oh yeah i remember seeing that for f(x) in F[x] any two splitting fields for f(x) are isomorphic

[white oxide]

yea fraleigh did that earlier

is there any reason why some people do Z/pZ[x] instead of Zp[x]? obviously they're iso but i haven't come across a canonical choice

[frigid lark]

Z_p refers to the p-adics

Honestly at this point you could just write F_p

or GF(p^r)

Since you've proved uniqueness

[white oxide]

p-adics?

[frigid lark]

https://en.wikipedia.org/wiki/P-adic_number

[white oxide]

why is $\sigma^b\sigma^{ad}(x) = \sigma^b(x)$?

[cloud walrus]

okeyokay

[frigid lark]

$\sigma^d(x) = x \implies \sigma^{ad}(x) = \sigma^d(\sigma^d( ... (\sigma^d(x))...))$ a times is also $x$

[cloud walrus]

Parrot Tea

[frigid lark]

$\sigma^d(x) = x \implies \sigma^{ad}(x) = \sigma^d(\sigma^d( ... (\sigma^d(x))...))$ a times is also $x$

[cloud walrus]

Parrot Tea

[solemn garden]

By the sylow theorems, a group of order 60 must have 1 or 6 subgroups of order 5, and these subgroups are all conjugate. But not all 5-cycles are conjugate in A_5, so is this a contradiction? Where did I go wrong?

[frigid lark]

I don't think A_5 has 2 conjugacy classes of order 5 elements, you've got 4! = 24 elements of order 5, and each subgroup of A_5 has exactly 4 of those elements in them (along with the identity), moreso this allocation is in fact a partition. Thus by a counting argument you have 6 subgroups of A_5 each having order 5, and each of them conjugate to each other by Sylow

You can tell that there are 4! elements of order 5 since an element of order 5 is a 5 cycle, for uniqueness fix the first entry, say 1, then for the second entry you have 4 choices, third entry 3 choices, and so on.

[solemn garden]

but 24 doesn't divide 60

[frigid lark]

One could also that all 5 cycles are of the form (1 $\sigma(2) ; \sigma(3) ; \sigma(4) ; \sigma(5) )$ for $\sigma \in$ sym${2,3,4,5}$

[cloud walrus]

Parrot Tea

[frigid lark]

Does that matter?

[solemn garden]

And the size of the conjugacy class divides the order of the group

So there are two classes each with 12 elements

[frigid lark]

what do you mean by the size of the conjugacy class?

The number of conjugacy classes?

[solemn garden]

No, I mean the number of elements in the conjugacy class

[frigid lark]

Why?

[solemn garden]

because number of elements in the conjugacy class is index of centralizer of any one of the elements of the conjugacy class

[frigid lark]

That statement says that the conjugacy class is a subgroup

not an individual element

So in your case you would have 6 conjugacy classes

Corresponding to the 6 distinct subgroups of A_5 with order 5

[frigid lark]

Elements don't have normalizers, subgroups do

[solemn garden]

I thought normalizer N(a) of an element a is defined as {n in G | na = an}

[frigid lark]

That's a centralizer

[solemn garden]

Aren't they the same thing
edit: for a single element a

[frigid lark]

No

For some $A \leq G$, $N_G(A) = {g \in G; |; gA = Ag}$

[cloud walrus]

Parrot Tea

[frigid lark]

That just means that the coset gA is the same as the coset Ag

[frigid lark]

Recall the proof of this statement, let H be a subgroup of G. Then G acts on the conjugacy classes of H by conjugation. The stabiliser here will be all elements g in G, such that gHg^{-1} = H, i.e. N_G(H). This action is obviously transitive. Thus by the orbit stabiliser theorem, the number of conjugacy classes of H is equal to [G : N_G(H)]. If G is finite then the number of conjugacy classes of H is equal to |G|/|N_G(H)|

[solemn garden]

the proof I know involves a bijection between the left cosets of N(H) and the conjugate subgroups

[solemn garden]

So A_5 can’t have a conjugacy class with 24 elements since 24 doesn’t divide 60

[solemn garden]

But I’m still confused about why the sylow theorems don’t imply that all 5-cycles in A_5 are conjugate

[frigid lark]

It's not a conjugacy class of the elements, its a conjugacy class of the subgroups of order 5

Of which there are 6

and 6 | 60

The Sylow theorem's explicitly refer to subgroups of G

Not necessarily elements

if K is a 5 subgroup of A_5, the Sylow theorem's state that all other 5 groups of A_5 are conjugate to K

[solemn garden]

I see

How can I check if a 5-cycle is conjugate to its square (in A_5)?

[frigid lark]

is this for the same question?

[solemn garden]

Yes

[frigid lark]

One can show that for s in S_5, s(1 2 3 4 5)s^{-1} = (s(1) s(2) s(3) s(4) s(5))

yeah, yeah in A_5

[solemn garden]

Oh I meant in A_5

[frigid lark]

You just have to check that if such an s lives in A_5

That shows that this statement is trivially true in S_5

[solemn garden]

Oh, then if the s necessary for conjugating a 5-cycle to its square is odd, then it isn’t conjugate

[frigid lark]

Well you would actually have to check for 5 s's

To account for cyclic shifts

There might be an easier way

[solemn garden]

(12345)^2 = (13524) which needs a 4-cycle to conjugate

[frigid lark]

So in this case sigma must send 1 to 1, 2 to 3, 3 to 5, 4 to 2 and 5 to 4

But there exists another sigma that sends 1 to 3, 2 to 5, 3 to 2, 4 to 4, and 5 to 1 that is also sufficient

and so on, and so forth

[solemn garden]

So the answer to my original question is that not all elements of a subgroup are conjugate to each other, and the conjugacy classes are split evenly across all subgroups

[solemn garden]

I understand

[frigid lark]

I don't know

[solemn garden]

My confusion is cleared

Thanks!

[frigid lark]

Alright, good luck

[formal ermine]

@solemn garden your username must surely be true in the dyadics

[solemn garden]

are dyadics same as 2-adics

[formal ermine]

it's just another name for them

[solemn garden]

if the 2-adics have characteristic 2, then it’s true in them

[formal ermine]

yeah the p adic numbers have characteristic p

[solemn garden]

totally

[formal ermine]

now prove that that's wrong

[solemn garden]

there is a field homomorphism of Q into Q_p

[formal ermine]

every field homomorphism is injective

completion theorem my beloved

[summer path]

#1100503863586455632

leaking lol

[formal ermine]

this is algebra related tho lol

[frigid lark]

Consider the element (1, 1, 5, 5, 5, 5, 69, 69, 69,...)

[empty rose]

what's that an element of?

[solemn garden]

Would (1,0,0,0…) and (2,0,0,0…) work as a counterexample?

[formal ermine]

you guys are doing it wrong

fuck inverse limits, all my homies hate inverse limits

use decimal expansions 👍

[frigid lark]

Or the element (0, 0, 4, 4, 4, 36, 36, 164, 420, 420,...)

[frigid lark]

I have only created padics using inverse limits

[formal ermine]

silly you

[frigid lark]

Not elements of 2 adics

[formal ermine]

I had that mindset too in the beginning

but the p-adic analysis way of defining them is superior

[frigid lark]

Keep in mind that I know 0 maths

[solemn garden]

Why

the only things I know about p adic numbers come from wikipedia

[formal ermine]

🫂

I barely know anything either

[frigid lark]

This server needs a good hug emoji

[formal ermine]

my favorite p-adic number is $(p - 1)^\infty$

[solemn garden]

What is an inverse limit?

[frigid lark]

Read Lang

[formal ermine]

wait

read this:

[frigid lark]

Or the start of a homological algebra book

[formal ermine]

https://people.math.harvard.edu/~smarks/mod-forms-tutorial/mf-notes/galois-reps.pdf

1.1

example 1.5 (1) defines the p adics using an inverse limit

[frigid lark]

Oh, Lang just leaves it as excercise

Like basically half of his theorems

[formal ermine]

trivial

left as an exercise to the reader

[frigid lark]

I like this style

[lethal dune]

1+2²+2³+...=-1 now makes sense

[rustic crown]

[summer path]

[slim kayak]

(Inverse) limits are for real ones

[summer path]

inverse limits uwu

[chilly radish]

That's literally the same exact thing but ok

Like there is actually no difference

[grand cliff]

With this question, I proved that G = Z in a slightly different way:

I argued by contradiction:
If G != Z then: we can get the quotient G/Z which will also be a p-group of lower order,
Then G/Z must have a nontrivial center too, so we can consider (G/Z)/Z which is again a p-group of lower order,
we can repeat this process. It is guaranteed to terminate since G is finite.

The fact that this process terminates means that on the last step we had a p-group with a trivial center which is a contradiction so G = Z.

just wanted to fact check that my logic is sound

[next obsidian]

No your logic is not sound

This would prove every p-group is abelian

The issue was how you concluded, this process terminating means you had some iterated quotient thingy where Z(G) = G, that’s what terminates the process

Not when Z(G) = {1}

[grand cliff]

ahh yeah true

this class is so hard 😭

But makes sense

[grand cliff]

this complaint is unrelated to this particular question

[formal ermine]

shhh don't tell them

I'm trying to pursuade them into doing p adic analysis

[dim widget]

@grand cliff you’re not very far off from a proof though, just need to work a little harder

[grand cliff]

Sadly I poisoned my mind with the answer

Unless there's a way to recover with my approach

[teal vessel]

so I'm trying to prove that |GL2(Zp)|=p^4-p^3-p^2+p without using the fact that in general |GLn(F)| is the product from i=0 to n-1 of (|F|^n-|F|^i) for finite fields.

I've determined that it is clearly less than p^4, since that would be the total number of possible 2x2 matrices with coefficients in the given finite field, and I thought I knew that you could subtract out p^3 for all the matrices such that the bottom row is a multiple of the top row, and then p^2 for all the matrices such that the bottom row is a zero row, but now I'm realizing that I didn't technically restrict my p^3 case to be non-zero rows, so I may be overcounting here. In any case, I'm not sure why I would need to re-add a copy of p into the order, and I'm a little stuck.

[dim widget]

How about you try thinking about the possibilities for the first row, then the possibilities for the second row, etc.?

[grand cliff]

Isn't that essentially what you do for the general fact?

[teal vessel]

dunno yet, haven't gotten to part 3 of the textbook where they talk about fields and prove this. I'm still in baby mode, you know, like how you spend the first week of calculus only knowing the limit definition of the derivative before you realize you can skip it 90% of the time.

[next obsidian]

You should do this

Use the linear algebra fact that a matrix is invertible iff it has full rank

[waxen mural]

dummit and foote moment

[teal vessel]

literally

[waxen mural]

what edition are you reading

[teal vessel]

3rd

[waxen mural]

me too!!

[teal vessel]

hmm.... ok so a nonzero determinant only works if ad=/=bc, which breaks down into two cases depending on the coprimality of a and b

but I'm thinking too hard already lol

[next obsidian]

Don’t use that characterization

Did you learn about column space, row space, etc

[teal vessel]

I did, but it's been a hot minute. iirc they share the same rank

[next obsidian]

The column space is the vector space spanned by the columns

[teal vessel]

time to pull out the Lalgebra text I used to refresh lol

[next obsidian]

A matrix is invertible iff this is the entire vector space

So, to put it in more actionable terms

A matrix is invertible iff the columns are all linearly independent

(Same goes for all rows)

So, to find all invertible 2x2 matrices you have to count all pairs of linearly independent 2 x 1 vectors

Does that sound doable?

[teal vessel]

I mean, it does, but clearly something has gone wrong in the way I'm counting, as for pairwise dependence, one must be a scalar multiple of the other, which I've already been looking at.

[next obsidian]

Okay so let’s look at it a combinatorial way

You first pick a nonzero vector

Then pick a vector which isn’t in the space spanned by the first

How many choices do you have at step 1, and at step 2

[teal vessel]

ah, got it.

given a non-zero vector (a b) you have a total of P choices for either vector, resulting in p^2 vectors, but removing the option (0 0) gives you (p^2-1) options. Choosing the second vector, you must now find a linearly independent vector, which can be accomplished by taking your vector (a b) and adding to it some vector (x y) s.t. x=/=y. This means that you have a new vector (a+x b+y) where a and b are fixed according to the previous choice, and you have p options for one of the new variables, and p-1 options for the other (since they cannot both be the same value). Given that the choice for a and b is independent of the choice for x and y, you get (p^2-1)(p^2-p)=p^4-p^3-p^2+p

/box

[next obsidian]

Yes!

There’s an easier way to establish the second number though

Which generalizes for more than just 2x2 matrices

You have p^2 vectors total

And have to avoid the 1-dimensional space spanned by the first vector

Which has p things

So p^2 - p

This works for say n x n better because at step k you have p^n vectors, and have to avoid the (k-1)-dimensional space spanned by your previous choices

So you have p^n - p^{k-1} choices

[teal vessel]

'tis true, but that's a generalization I'd probably come to by doing a couple constructive assessments. teaching elementary arithmetic and geometry has made me prefer concrete constructions (that and the fact that I taught myself linear algebra several months ago and haven't used it since)

[next obsidian]

I think this is concrete

And it’s a more instructive way to view it I think

You’re really just “avoiding bad vectors”

So it makes sense to count how many bad choices there are

This is really common in combinatorial arguments

Anyway, you got the right answer, I just wanted to highlight a different way to think about the count

[teal vessel]

never did combinatorics, to be fair. set complements make sense, but I'm still rusty on my space manipulations lol

"never did" as if I haven't been doing naive combinatorics since 5th grade

lol

[next obsidian]

Sure sure, the point is just sometimes it’s easier to count he complement

If you know the size of the ambient set

This sort of stuff is really common in all math where it’s like, you can dualize things and examine the opposite thing

And figure out stuff about the thing you actually care about

I just bring it up because it’s good to have in the back of your head

If you do topology and it asks you to show a set is closed, often times it’s much more obvious the complement is open

Etc etc

[teal vessel]

used to have it, lost amongst the more "important" stuff teaching elementary geometry and constructing everything out of triangles for a month straight. We love how the brain prioritizes facts and methods

"show that GLn(F) is non abelian for any n>1 and any field F"
uh... matrix multiplication is non-commutative, and all properties of matrix operations in R hold for arbitrary fields F (according to the blurb before the exercises, to be proven later). next?

sometimes exercizes seem to be quite funny in their ordering

[next obsidian]

This isn’t good enough

Take the subgroup of matrices which are just some fixed number x along the diagonal and 0 everywhere else

This is abelian

You can have abelian subgroups of nonabelian groups (in fact any subgroup generated by one thing satisfies this)

To do this you have to explicitly exhibit non commuting matrices in GL_n

[coral spindle]

I'm looking for some standard notation/conventions on this particular concept of "nearly semidirect products" that is occuring a lot for me. Any references are appreciated :)

[cloud walrus]

Boytjie (never-to-be-glomed)

Boytjie (never-to-be-glomed)

[coral spindle]

My particular question is just if anyone knows any standard conventions on this stuff

(Also I guess my second definition is 'inner' rather than 'outer' lol)

[teal vessel]

maybe not for a college course, but it's almost painfully trivial tbh. It's one of those things like saying "show that log(a+b)=/=log(a)+log(b)" because there exists a special case like "log(1+2+3)=log(1)+log(2)+log(3)" my autodidact brain: gestures wildly at definition of matrix multiplication as composition

[next obsidian]

No it isn’t sufficient at all

You need to know that you can have non commuting automorphisms which is not a priori obvious

You have to know the structure of vector spaces and swap some basis elements or something

[coral spindle]

It is not painfully trivial at all

Try to find two matrices in GL_2 that don't commute. You may be surprised that you cannot do so as easily as you seem to think

As it happens, if you can show that GL_2 is non-Abelian then you can use this to show that GL_n is non-Abelian for all n > 2. You may treat this an an exercise if it's not already clear.

[coral spindle]

Extension: find an example that works for any field F (maybe even of positive characteristic)

[teal vessel]

don't have sufficient knowledge of fields to do so

they're not until a couple sections later in the book

at least re: positive characteristic

[teal vessel]

just take two arbitrary matrices with entries a,b,c,d and w,x,y,z respectively. aw+by=/=wa+xc generally. It would be harder to demonstrate for some finite field, but I'm not restricted to finite fields, so I can freely use whatever values I like so long as the determinants are nonzero. (0 2 / 4 6) and (1 3 / 5 7) work just fine in the 2 case as a concrete example, and any extension to larger matrices will necessarily contain the combinations of these elements the same way (albeit with extra terms added). I'm not sure why that's non-obvious

[teal vessel]

btw @merry plaza needs to lurk on this so I can talk to him later lol.

[forest tundra]

why is every group sub group of itself??

[coral spindle]

For the same reason that every set is a subset of itself

[forest tundra]

aight

[forest tundra]

can it be related to the isomorphism property shown by groups?

[coral spindle]

Not in any interesting way

[forest tundra]

👍

[formal ermine]

take the identity, it has kernel 0 and is surjective so the group is isomorphic is itself. hope that helps 👍

[forest tundra]

sure , thanks i was thinking the same thing

also but what do you mean by kernal 0 there?

[formal ermine]

its kernel is trivial

[forest tundra]

sub group?

you mean trivial sub group?

[formal ermine]

its kernel is { 0 }

[forest tundra]

{0} is the trivial sub group of set of all integers with addition binary operator if that what you mean ? becaues it is an identity element

[lusty marlin]

They mean to say the subgroup consisting only of the identity element of the group under consideration.

[forest tundra]

oh all right i was confused by the naming convention i'm quite new to the gorup theory

[merry plaza]

i think i got an example that generalizes to all n>1 and all nontrivial fields if you wanna hear it, it only uses the assumption that 1 =/= 0 and doesn't mess with field characteristic

[teal vessel]

I have my own. I'll share it sunday

[simple valley]

trying to figure out representations of symmetric groups. Fulton-Harris constructs given a Young tableau a pair of subgroups $P, Q \le S_n$ that preserve the rows and columns respectively. Then we set $$a = \sum_{g \in P} e_g \in \bC[S_n]$$ $$b = \sum_{g \in Q} <sign>(g) e_g$$

[cloud walrus]

m...

[simple valley]

but I can't think of anything I can really do to these objects?

the first exercise after asks to prove $\bC[S_n]ab \cong \bC[S_n]ba$ and I'm stumped

[cloud walrus]

m...

[simple valley]

hmm I guess conjugates of P and Q by g are preserving rows/columns in tableau that have been acted on by g

[simple valley]

Am I supposed to use the fact that ab is idempotent? I don't think that was proven yet

x mapsto xa and y mapsto yb are two halves of an iso by idempotency of ab

[dim widget]

You can easily prove it yourself! In general any projection/averaging map like this will be idempotent for very formal reasons.

[simple valley]

It's a composition of two noncommuting maps though

[dim widget]

But b commutes with a on the image of a, since on the image of a, a is just the identiy as it is idempotent.

[simple valley]

Does b not escape the image of a?

a1b = ab1 but does ab1 = aba

[dim widget]

No it doesn't, but I suppose proving this is as hard as proving the whole statement. You could also try just proving it formally by multiplying a_\lambda by b_\lambda and collecting terms

[simple valley]

Sounds horrifying

[dim widget]

It's very easy!

The only relevant fact is that you are averaging over subsets of S_n which are closed under inversion

and the sgn(g) = sgn(g^{-1})

*that

[simple valley]

Ok that sounds less horrifying

[white oxide]

ah that makes sense, thanks!

[simple valley]

Reminds me of the covariant measure projection trick

[dim widget]

Hmmm I don't know enough about statistics to know if there's a more than formal similarity.

[simple valley]

oh no you also have an averaging map $X \mapsto \int_{U \in G} UXU^{-1}$

[cloud walrus]

m...

[simple valley]

this turns out to be an idempotent onto the commutant of G

and the projection is orthogonal in the trace inner product

[slim kayak]

Can someone explain to me why every non-trivial free product A x B, as long either A or B have order larger than 2, contains a free subgroup of rank 2?

[coral spindle]

Let's say $A$ contains distinct non-identity elements $a_1$ and $a_2$ and that $b \in B$ is not the identity. Then if we let $x = a_1 b$ and $y = a_2 b$, $\gen{x, y}$ is a free subgroup of rank 2.

Typo!

[cloud walrus]

Boytjie (never-to-be-glomed)

[slim kayak]

Oh. For the proof use ping-pong lemma with, I guess, the sets of reduced words starting with x or y?

[coral spindle]

I expect that would work 🤷

[slim kayak]

What did you have in mind?

[white oxide]

why does $|\sigma_i|$ have to divide $m$? I thought $m$ was the least such positive integer?

[cloud walrus]

okeyokay

[coral spindle]

Nothing but my intuition. I thought about writing a proof via the universal property though

[coral spindle]

I don't understand your confusion. If $m$ is some integer such that $\sigma^m = (1)$ then $|\sigma_i| \mid m$, it's really as simple as that.

[cloud walrus]

Boytjie (never-to-be-glomed)

[coral spindle]

We then later identify the smallest such integer m.

Note that $\sigma^m = (1)$ does \textbf{not} mean that $m$ is the order of $\sigma$!

[slim kayak]

Huh, so like send x and y to their formal bethreen in F(x,y) and determine the kernel?

[cloud walrus]

Boytjie (never-to-be-glomed)

[coral spindle]

No that's not the universal property, but you could do that too I suppose

[echo gull]

Yeah that is a relatively straightforward proof, intuitively we have have k-cycles as order k, disjointness (as composition is thus commutative) allows us to distribute powers in the way we anticipate, so you have that at the lcm it is surely the identity and an easy contradiction below the lcm.

[slim kayak]

What other universal property besides that of the free group/product you wanna use here?

[simple valley]

sorry I don't see it

I only get sums over conjugates of P and Q

that don't match with sums over P and Q obvs

[lavish sigil]

We can view a complex number $a+bi$ as a real $2\times 2$ matrix $\begin{pmatrix}a&-b\b&a\end{pmatrix}$. Similarly, we can view a complex $2\times 2$ matrix as a real $4\times 4$ matrix by expanding each complex entry into a real $2\times 2$ block. In this sense, the group $\mathrm{SU}(2)$ is identified as a subgroup of $\mathrm{SO}(4)$. The latter has a natural action on the $3$-sphere $S^3\subseteq\mathbb{R}^4$, given by matrix multiplication. If we now have a finite subgroup $\Gamma\subseteq\mathrm{SO}(4)$, and we assume that the action of $\Gamma$ on $S^3$ is free, will $\Gamma$ be conjugate (in $\mathrm{SO}(4)$) to a subgroup of $\mathrm{SU}(2)$?

[cloud walrus]

gustavn64

[ripe basalt]

i think so actually

[lavish sigil]

Why?

[simple valley]

this may be an unsatisfying answer but SO(4) cap Z is a group of order 384 that is not cyclic and not dicyclic (I think?) and not one of the 3 exceptional subgroups of SU(2)

er, of order 192, but same reasoning applies

[lavish sigil]

But this doesn't act freely on S^3.

[simple valley]

how so?

it's the rotation group of the hypercube

[lavish sigil]

Maybe I misunderstood something. By SO(4) cap Z, do you mean those elements of SO(4) where all entries are integers?

[ripe basalt]

yeah i also did not understand

[simple valley]

yes exactly that

[lavish sigil]

The matrix diag(1,1,-1,-1) is a non-trivial element in it which has fixed points when acting on S^3, for instance

[ripe basalt]

nice

[simple valley]

oh wait I mixed up with faithful action

[lavish sigil]

Ah

[ripe basalt]

can't you just find a subgroup of SO(4) that acts freely that is larger than a subgroup of SU(2)?

wouldnt that be the thing to chase for a counterexample

[lavish sigil]

What do you mean "larger"? Containing one as a subgroup?

[simple valley]

SO(4) contains a copy of SO(2)xSO(2) (block diagonal)

so you have a family of products of 2 cyclic groups, that should act freely?

ah wait again no

[ripe basalt]

I think I just meant larger in the sense that the conjugation isn't an isomorphism

only an injection

[ripe basalt]

where did you see this problem

[spice whale]

are you sure about this

what about like

-a²-b²-c²-d² over the reals

[grand cliff]

For this exercise, I'm pretty sure I've got most of the steps, but I just don't know how to fill in the orange arrow area

Basically I need to show that if the normaliser of a stabiliser is equal to the group, then G will be a fixed point free permutation group (aka a semiregular group)

[teal vessel]

btw, thanks guys for actually holding me accountable for doing these exercises in this book, lol. I really don't feel like finding a set of generators and relations for Q_8, but it's nice to be able to bounce things off of others and see where I'm going. (the generators are i and j, and the relations are at least i^2=j^2=-1, ij=k but checking the completeness of the generating relations is tedious)

maybe I need "ijk=-1" in there but I'm not entirely sure if it's directly implied already

[frigid lark]

I don't think ij=k is a relation, if the generators are i and j

[spice whale]

i would do
i² = j²
i⁴ = 1
iji = j

i think that covers it

[teal vessel]

oh, yeah, because the only elements you want to reference in relations are generators and identity, oof

iji looks funny in this font lol

[frigid lark]

Oh, also in that category is i^2 = j^2 = -1

Cause -1

[spice whale]

yeah -1 isn't a thing a priori

[spice whale]

the last one is from
-ji = ij
i²ji = ij
iji = j

[frigid lark]

Just imagine everything is a unit group of some field

*ring

[teal vessel]

wow, sudden memory flash: relations can be thought of as "commutation rules" (paraphrasing loosely). dunno why I forgot that that's sort of the purpose of the relations, to be able to reduce unusual combinations to canonical forms.

[teal vessel]

haven't gotten to rings yet ;-;

the hard part of starting from basically nothing in an area of study lol

[spice whale]

can you always do this?

hm

my first thought is the group ring

[cloud walrus]

CoffeeMan

[frigid lark]

I skimmed stack exchange and finding the unit group of a group ring seems non trivial

[frigid lark]

I have no idea

[teal vessel]

always remember the distinction between "think about it like" and "it is." The former is an intuitive aid, while the latter is a rigorous notion.

[frigid lark]

I think a counter example may lie in groups of odd size (maybe ones that aren't equal to 2^n - 1)

[teal vessel]

in other news, I was supposed to be calculating the order of elements of the Heisenberg group over Z_2 earlier, and I realized that since the H-group is a set of diagonal matrices with only three degrees of freedom, I could express all elements of the group as a 3-dimensional vector and then calculate a matrix (or two) that squares or cubes the vector (in Z_2 all elements have order 3 or less). I'm proud of myself, even if the exponentiation matrices required using a variable dependent on an entry in the vector. Made it easier to keep track of than doing repeated matrix multiplication the whole time.

[frigid lark]

If we replace ring with field this statement is trivially false (a group of size 5 for example). Would there be any way to reduce a ring case down to a field case?

[fossil shore]

Oh, think I've found a simpler argument. The set of units in the field of fractions is finite while it is infinite in the other from what I can see.

[frigid lark]

Wait is $k[x]_{(x)}$ how you denote the field of fractions of $k[x]$?

[cloud walrus]

Parrot Tea

[spice whale]

field of fractions should just be k(x)

[fossil shore]

wops, sorry, meant ring of fractions

[spice whale]

ring of fractions is a field

[frigid lark]

Does the initial ring has to be an ID for that?

[fossil shore]

ye think so too

[frigid lark]

No, the set of units in $k(x)$ is definitely not finite. But you can simply argue that one of these is a field, and one of them is not

[cloud walrus]

Parrot Tea

[frigid lark]

You can inject Z into the unit group (1 maps to x)

[waxen mural]

Does sylow's theorem just state that if you prime factorise a finite groups order $|G| = p_1^{a_1}p_2^{a_2}\dots$ there are subgroups $H_n$ of G where $|H_n| = p_n^{a_n}$

[cloud walrus]

Jamarcus

[coral spindle]

That is one of the Sylow theorems, yes

[frigid lark]

I'm being a bit pedantic here, but you should probably say $|G| = p_1^{a_1}p_2^{a_2}...p_d^{a_d}$ cause it has finite size

[cloud walrus]

Parrot Tea

[fossil shore]

Is the ring of fractions a field here? How would you invert x/1+x?

[frigid lark]

(1+x)/x

[fossil shore]

But x is in (x)?

[coral spindle]

The field of fractions of any integral domain is a field, no matter what.

[fossil shore]

Yes, but we're not talking about the field of fractions

[frigid lark]

Wait, do you mean the ring of fractions of $S^{-1}K[x]$ where $S = (x)$?

[cloud walrus]

Parrot Tea

[fossil shore]

Yes

That's what I wrote

[frigid lark]

sorry, I haven't touched this in a while

[coral spindle]

Are you sure you don’t mean the localisation

[fossil shore]

Dunno about terminology. Using what Dummit and Foote use

[coral spindle]

In which case you would set S to be the complement of (x)

[fossil shore]

Yup, that then

[frigid lark]

arghgh, this is what doing 5 regression models a day gets you

[fossil shore]

lol

[cloud walrus]

CoffeeMan

[fossil shore]

I think the unit argument still works when k is finite

Dang, that still leaves the case of k infinite with finite characteristic

[frigid lark]

What are the maximal ideals of k[[x]]?

[fossil shore]

I think (x) only right?

Isn't it local?

As the valuation ring of k((x))?

[frigid lark]

I was just thinking of things that could work

Wait, if such an isomorphism, $f: k[x]_{(x)} \to K[[x]]$ existed, would it have to send x to x?

[cloud walrus]

Parrot Tea

[oblique river]

Those are not isomorphic

[frigid lark]

yeah striving for a contradiction

[oblique river]

Oh

Do you know that k(x) and k((x)) are not isomorphic

[frigid lark]

why?

[oblique river]

Because that would show that your two rings arent isomorphic

Because they have different quotient fields

[fossil shore]

What is k(x)?

[frigid lark]

quotient field of k[x]

[oblique river]

Wait who originally asked this question

[frigid lark]

CoffeeMan

[oblique river]

Oh

[frigid lark]

I'm just a person failing at doing maths at midnight for no reason

[oblique river]

Are they asking if the obvious map k[x]_(x) -> k[[x]] is a bijection

[fossil shore]

And I'm a person failing at doing maths

[oblique river]

Or just if there is an abstract isomorphicm

[fossil shore]

just any isomorphism

[frigid lark]

^

[oblique river]

Ty

Wait… cant you just do linear algebra

k[x]_(x) is a countable-dimensional vector space over k, but k[[x]] is uncountable-dimensional

[fossil shore]

I mean, ngl, wouldn't know how to argue that, but think I follow. But why is k[[x]] uncountable-dimensional?

Is N^N uncountable?

[rustic crown]

yesh

[oblique river]

Yes

[fossil shore]

ah, ok

[rustic crown]

|N^N| >= 2^|N|

[fossil shore]

Ok, my set theory sucks, sry

[rustic crown]

it okie >.<

https://media.discordapp.net/attachments/796976320067272704/818150000482582528/612933479730249728.gif

[frigid lark]

Cantors diagonalization's argument?

Can't let my irrational hatred of set theory get me banned

[oblique river]

Isnt this a standard linear algebra fact? That $\bigoplus_{\mathbb{N}} k$ and $\prod_{\mathbb{N}} k$ are not isomorphic?

[cloud walrus]

Buncho Bananas

[fossil shore]

Mb, my uni decided to cut the linalg course into 2 parts and place the second part on graduate level, so...

ye, should prolly finish the linalg book this summer instead so I won't have problems like this

[frigid lark]

at least you did your uni's linalg course

[rustic crown]

~~parrottea didn't? ~~

[frigid lark]

parrot tea got a weird prohibition

[oblique river]

Hmm so maybe its not “standard” and does involve some minor cardinal arithmetic

[daring nova]

Because it doesn't work when it's infinite ?

Yeah ok

The infinite case is not taught in undergrad

[oblique river]

Idk what youre saying, that statement is true for all fields

[frigid lark]

F_1

[oblique river]

Idk what you mean by “the infinite case” but i did learn this in an undergrad class :/

[daring nova]

Because you indexed it with N

I only know of finite direct sums

[frigid lark]

Finite $\cong$ Boring $\cong$ not making my head hurt

[cloud walrus]

Parrot Tea

[rustic crown]

finite = combi = hard

[daring nova]

I like combinatorics

[frigid lark]

Isn't this an algebra chat, shouldn't we be good at counting?

[daring nova]

It doesn't make my head hurt

[rustic crown]

i can count to 3 >.<

1, 2, 3....? ... n?

[daring nova]

1, \dots, n, \dots

[frigid lark]

1 => 2 => 3 therefore 3 => n

Proof by seduction

Wait

[daring nova]

Indukcion

[frigid lark]

Duck coin?

Can I eat it's egg?

[coral shale]

[frigid lark]

I'll trade

[rustic crown]

you're a bird!!!!

.<

[daring nova]

He doesn't exist

[rustic crown]

no eating babies

[frigid lark]

Not a baby

No partner

[rustic crown]

[lethal dune]

[rustic crown]

ryu

[frigid lark]

Gn

[rustic crown]

oyasumi

[frigid lark]

Sayonara

[sweet echo]

I've come across the phrase "Morita progenerator", in Assem's rep theory book with no explanation. Should I take it to be the P that shows up here in Morita's theorem? Perhaps someone who knows can confirm, but seems the most likely

[dim widget]

Over the reals forms are just determined by their signature! But you were asking about p-adic forms I had thought.

[spice whale]

oh this is a p-adic thing

[spice whale]

I'd be interested in the proof of that 3d reduction thing then

[lethal dune]

how do I show this, D is a domain (not necessary commutative) and M_n(D) is semisimple then D is a division ring?

[chilly radish]

For some n?

[oblique river]

i think this is in the first half of serre's "course in arithmetic"

[echo gull]

"course in arithmetic" will never not be the funniest title to me

[lethal dune]

yeah no n is specified

[white oxide]

if we replace E and Zp with GF(p^n) this statement still holds right

in other words the elements of GF(p^n) are exactly the zeros of the polynomial x^{p^n} - x in GF(p^n)[x]

[lethal dune]

correction: polynomial x^{p^n}-1 in GF(p)[x]

the statement you wrote is not technically wrong but it's not the direct translation of the theorem

[white oxide]

huh okay i'll have to think about that thanks

[white oxide]

i'm a bit confused about this question, is it asking me to show that x^p^n - x is the product of all the monic irreducible polynomials in Zp[x] of degree d dividing n or that the only factorization of x^p^n - x into monic irreducible polynomials are those of degree d dividing n

[rustic crown]

former

[white oxide]

uhhh LOL could i get a hint im kind of stuck, i could upload what i wrote but my handwriting is so fucking messy

don't know if i'm on the right track or not negl

second pic is me fucking around and trying to figure shit out

[teal vessel]

while we're at it: I'm trying to prove that for some isomorphism M: G -> H, |M(x)|=|x| for all x in G. Assuming towards contradiction that such is not the case, I've determined that there exists some x such that one of the following must be true: M(x)^n=1 and x^n=/=1, or the reverse case. Something about the reverse case bothers me, though. Am I right in saying that it cannot exist because homomorphisms map identities together?

I suppose that's part of fully exploring the contradiction for the proof, but I'm not fully sure if that is accurate or not.

[white oxide]

<@&286206848099549185>

[echo gull]

is |x| denoting order here?

[teal vessel]

yes

[echo gull]

you don't need contradiction

[teal vessel]

probably not, it's just a method I like exploring to start with

there are many proofs I like using contradiction for that don't really need it

[white oxide]

<@&286206848099549185>

[echo gull]

$Let \theta: G \rightarrow H$ be an isomorphism mapping. Assume that the order of $x$ is precisely $n$. Then it follows $\theta(x)^{n}$ = \theta(x^{n}) = \theta(e) = e.$

[cloud walrus]

kodiak
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[echo gull]

I'll leave you to do the minimality aspect.

[teal vessel]

that's the part that I'm working through. I know that the exponent commutes with the homomorphism, but I'm using contradiction for minimality

[echo gull]

you just use injection

[teal vessel]

quoting it at me kinda defeats the purpose of creating the proof tbh.

[echo gull]

I should've marked it as a spoiler, my bad

[white oxide]

<@&286206848099549185>

sorry i can't help it

my fault

[teal vessel]

it's fine, I can just not read it until after.

[echo gull]

but my general advice would be to get away from the habit of contradiction, I am also a habitual lover of contradiction but it can cloud direct arguments like this

[teal vessel]

in either case, I'd be employing my sanity check that homomorphisms map identity to identity

[echo gull]

sure, but there are practical and philosophical reasons to avoid contradiction unless it is necessary

[teal vessel]

I'm also working a proof by contradiction because the second exercise is to show that this doesn't work for general homomorphisms, which would be shown much more quickly in a branch of the proof which allows for minimum counterexample

[echo gull]

if you are interested in a counterexample: ||just consider a mapping between a cyclic group and a product of cyclic groups of non-relatively prime order like $\mathbb{Z}{4} \rightarrow \mathbb{Z}{2} \times \mathbb{Z}_{2}$. There are easily homomorphisms on these groups but they have different order structures||

[teal vessel]

it's a good thing I already finished both branches lol

[lavish sigil]

I didn't really see it as a problem formulated in this form anywhere, but the motivation comes from Riemannian geometry. You can look at the manifold $S^3/\Gamma$, where $\Gamma$ is a subgroup of $\mathrm{SO}(4)$ acting freely on $S^3$. This is a Riemannian manifold. The question is essentially asking whether it would suffice (up to isometry) to look at the subgroups of $\mathrm{SU}(2)$.

[cloud walrus]

gustavn64

[west sinew]

Quick question: What's the dimension (as a vector space, of course) of a free Lie algebra on a set X?

Surely infinite if X is infinite, but what of when X is finite? Still infinite?

[ripe basalt]

No? I mean take a 1 element set and have a finite dim lie algebra act on it

as in the free action on the set gets you back to the lie algebra itself

[west sinew]

And if X has cardinality >1?

I mean I'm trying to see if [x, y], [x, [x, y]] and so on and so forth will just always be linearly independent (in which case it's infinite dimensional) or if skew-symmetry/Jacobi identity will produce enough identities to make the dimension finite. Doesn't seem clear on the face of it

[white oxide]

can i get a hint for this question pls

[formal ermine]

think about Fp(a)

[white oxide]

is that Zp(a) where a is a zero of x^{p^n} - x?

well i guess all fields of prime order p are isomorphic so it doesnt matter

[formal ermine]

yes

[formal ermine]

all finite fields of some order are unique up to iso

[white oxide]

yee

[white oxide]

sorry i'm really confused as to where to go with this, i know that F_p(a) = GF(p^n)

[formal ermine]

nu

a need not be a primitive element

[white oxide]

what? how does 31.4 show that each alpha has degree dividing n? it only mentions two fields here, namely F and Zp

[white oxide]

<@&286206848099549185>

[tiny jolt]

You could write [F:Z_p] = [F:Z_p(alpha)] . [Z_p(alpha):Z_p]. The second term is the degree of alpha over Z_p

[white oxide]

ah yeah that makes sense thanks

[umbral temple]

Elements of K can be characterised as being solutions to the equation t^p=t

Now if L isn’t K, then a isn’t 0 and so f should be irreducible

[rustic crown]

K can be an extension of F_p though

also not having roots doesn't automatically tell you it is irreducible

(unless i misinterpret whut you said >.<)

how did you show L = K(alpha)? it should be a simple check if you realized that

yep exactly! so if g(t) was a factor of f(t) what would it look like?

(hint: ||what's the coefficient of t^(deg g - 1) in g||)

nah, i mean more like... you know all the roots of f, so you also can factorize g fully! can this g actually be a polynomial over K though?

pretty much!

because otherwise you'll have like some j * alpha + something where 0 < j < p

this is similar to how you show that x^q-a is irreducible when a is not a p^th power.

tubuwu

[summer path]

detuwu

[rustic crown]

did you eat food today?

[umbral temple]

Doesn’t really change anything. I don’t take p to be prime but a power of a prime

Still all elements of K must satisfy t^p=t

[rustic crown]

what about an infinite extension?

[umbral temple]

No it’s given that K has degree p>0

[rustic crown]

also p is the char of K, so it is prime

[umbral temple]

p isn’t infinite

[umbral temple]

Fields can have degrees that are powers of primes

p doesn’t have to be prime

[rustic crown]

there is no degree for a field usually, degree is defined for an extension no?

and degree of [F_p^n:F_p] = n

[umbral temple]

I might have messed up terminology, I meant order as in number of elements in the field

Sorry!

[rustic crown]

it okie >.<

https://media.discordapp.net/attachments/796976320067272704/818150000482582528/612933479730249728.gif

[umbral temple]

But otherwise, I think my argument works since f evaluates to a

?

[rustic crown]

not sure

so we only know that x^q-x-a has no roots, as every element of F_q satisfies x^q-x

[umbral temple]

Oh oh oh I see. Question says K has characteristic p and not order p

That breaks my argument

I really messed up my terminology

[rustic crown]

.<

[umbral temple]

🤦‍♂️

[summer path]

[hidden kite]

Det kawaii as always ><

[spice whale]

i found a paper and it's saying that they're all 3d or 4d + n copies of the hyperbolic plane

not 3d + trivial

[coral shale]

if a function and its derivative share roots, that tells you something more about f

It relates to the quadratic case when disc = 0, and I take it this generalises ig...

[frigid lark]

I got it down to $D = u_1 p^{40} + u_2 q^{24} + u_3 p^8 q + u_4 p^4 q^2 + u_5 p^2 q^4 + u_6 p q^8$

[cloud walrus]

Parrot Tea

[frigid lark]

For some unknowns

I assumed that D is a homogeneous symmetric polynomial of degree 120

I may have messed up my counting

[dim widget]

@spice whale I’m pretty confident in what I said, however since some forms are equivalent it’s possible to describe these things in many different ways.

What is the “hyperbolic plane” as a quadratic form? Is it x^2 - y^2

[frigid lark]

Yeah D is the evaluation of a homogeneous symmetric polynomial g(X_1,...,X_5) of degree 8 + 6 + 4 + 2 = 20, sending X_i to a distinct root of f

then $D = u_1q^4 + u_2p^5q$

[cloud walrus]

Parrot Tea

[frigid lark]

Cause since G is symmetric we write it with the basic symmetric polynomials

and then evaluate noting that that the evaluation of the first, second, and fourth elementary symmetric polynomial is 0, and the third one is p, the fifth one is q

Then we just have to find two suitable polynomials to get some simultaneous equations and solve for u_1 and u_2

[spice whale]

yes

[grand cliff]

Why is the condition Stab(G, α) ≤ H ≤ G important for orb(α, H) to be a block of imprimitivity?

in permutation groups

The proof is helpful, but not super enlightening since it's very algebraic

[dim widget]

@spice whale then I think this is wrong, perhaps you’re looking at a classification of hermetian forms? For instance if p is 1 mod 4 then -1 is a square so the “hyperbolic plane” is just the usual Euclidean quadratic form.

[formal ermine]

just partially learnt this. fancy stuff

[hardy tundra]

know any cool way to solve this? :

Let G a group , Z the center of G , and f\in Aut(G). Prove that f(Z) is a subgroup of Z.

[lethal dune]

Z(G) char G

I'm not sure about the result I just wrote and too lazy to check atm. Try to show f(Z(G)) ⊂ Z(G)

remember g= f ( f^{-1}(g))

[formal ermine]

this is true but just a harder version of the problem they're trying to solve lol

[chilly radish]

It's not really harder

It's basically equivalent

||just apply the inverse||

[lethal dune]

Yeah

[formal ermine]

oh I forgot the definition of a characteristic subgroup for a second lmao

[chilly radish]

You can use this, but you don't really need this, you just need surjectivity

[lethal dune]

Sure

[next obsidian]

It’s intrinsically defined

[spice whale]

yeah it is
the paper is very explicitly about quadratic forms so like

idk

chapter 4 and 5

they prove the u-index of Q_p is 4 so my statement is true

[dim widget]

@spice whale I don’t know what the u-index is… but the statements you’ve been making don’t seem true/seem taken out of context. If I were you I would try to sit down and prove these things myself at this point. It’s very easy to see what’s going on which is why I don’t really have any doubt about what I’m saying about this

The three relevant things are: every quadratic form can be diagonalized, then diagonal quadratic forms are classified by their discriminant in K/K^2, their rank, and the product of the hilbert symbols of the diagonal entries, and taking direct sum with the trivial form only changes the rank

I would at this point give your 4th recommendation to read Serre

[south patrol]

Well it is possible to compute the discriminant of f = x^n + ax + b in general by using the relation with the norm of the derivative evaluated at a root and rearranging cleverly

[coral shale]

repeated factors

right

[chilly ocean]

just take a = x

[south patrol]

I misread lol

My apologies

[south patrol]

You can use a) with L=K and use the linear independence applied to a suitable vector space to conclude

[lethal dune]

@south patrol want a Galois problem?

[chilly ocean]

duel over woman

[celest furnace]

But win

[chilly ocean]

yes, Ultimate Chad

[south patrol]

Yes pls

I have exam soonish

[lethal dune]

Show for an irrational $\alpha \in \overline{\mathbb Q}$, there exists an extension $E/\mathbb Q}$ not containing $\alpha$ s.t. any finite extension $F \subset \overline{\mathbb Q}$, $E/F$ is cyclic

basically this one

[south patrol]

Hm

Okay I will have a go :)

Thank u

[south patrol]

Btw, I'm assuming this is a past paper?

So I'm wondering how you managed to prove that # Galois group = [K:k] without using this as part of that

We did it a slightly different way but the standard way basically uses the method of this question

Ah okay sure nice

[spice whale]

i downloaded it

[south patrol]

any hints for this btw?

I am aware this is a special case of smth more general with cycle types tbh so trying to show that uh

If f in Z[x] is monic, separable and irreducible in F_p[x] then the Frobenius in a splitting field(which just cycles through the roots) corresponds to an n-cycle in the Galois group over Q

But I see no obvious way to relate the two things

[zenith hollow]

yo guys

if we look at the gorup (G, *)

and want to define the inverse element of *

does it feature all a^-1 element of G that are inverese of a

but i have to throw an exception here?

\{0}?

[next obsidian]

What the hell are you saying

??

[zenith hollow]

the heck

[south patrol]

* isn't an element of G, for starters

[next obsidian]

What does throwing an exception mean? This isn’t a program

[next obsidian]

And you don’t need to define an inverse

[zenith hollow]

dude

i am talking about the inverse wth

[south patrol]

??

[zenith hollow]

how is it called in english

[south patrol]

"the inverse element of *"

[elder wave]

i understand the screenshot

[formal ermine]

das hat nichts zu tun mit dem was du geschrieben hast lol

[south patrol]

implies * is an element

[elder wave]

and still don't know what you're saying

[next obsidian]

It’s a property of being a group that for any g in G, an element which we call g^-1 exists such that gg^-1 = g^-1g = e

[south patrol]

Same here lol

[next obsidian]

One can then show such a g^-1 is unique

[zenith hollow]

all a^-1 element of G

[formal ermine]

KRANKENWAGEN

[zenith hollow]

i said it ehre

[next obsidian]

What does that sentence mean?

[south patrol]

Klaus maybe just say it auf Deutsch lol given there are three other german speakers active rn lol

[formal ermine]

cereal bars are delicious

[zenith hollow]

alter ich will nur wissen ob es in Gruppen ein inverses bezüglich * für alle elemente gibt, oder ob die 0 davon ausgeschlossen ist

[formal ermine]

für alle elemente

[zenith hollow]

so schlecht kann mein englisch nicht sein

[formal ermine]

per definition

[zenith hollow]

das wundert mich

[formal ermine]

hättest "with respect to" sagen müssen

heißt bezüglich

[formal ermine]

0 * 0 = 0, wieder die identität

[zenith hollow]

aso wieso gilt das bei körpern nicht?

[formal ermine]

weil körper zwei binäroperationen haben

nicht nur 1

[elder wave]

Unter addition ist es ne abelsche gruppe

[formal ermine]

und wenn du ein inverses zu 0 hast dann geht alles kaputt

[elder wave]

da gibts auch überall ein Inverses

[south patrol]

Aber jedes Element besitzt ein additives Inverses

Lol

RIF

[formal ermine]

vllt verwirrt es dich dass wir die identität in einer gruppe mit 0 bezeichnen

[elder wave]

aber nicht unbedingt ein multiplikatives (bzw nur 0 hat keins)

[formal ermine]

könnten es ja auch 1 oder e nennen

[elder wave]

jo

[zenith hollow]

warte was

wieso muss dasselbe nicht in Gruppen gelten?

in der Gruppe (G, *)

[south patrol]

Was wäre 0 in einer arbiträrer Gruppe? Es gibt in einem Körper zwei Verknüpfungen und 0 ist das additives Neutralelement und hat also nichts (a priori) mit der Multiplikation einer Gruppe zu tun

[elder wave]

da steht's doch

K ohne 0 ist eine gruppe mit multiplikation

[zenith hollow]

aso

das hab ich übersehen

[elder wave]

passiert

[zenith hollow]

Aber sowas (R, *) wäre zum Beispiel keine Gruppe wenn die 0 dabei ist?

[formal ermine]

ja weil 0 kein inverses hat

[zenith hollow]

ah ok passt danke

[south patrol]

de nada

[south patrol]

Du passierst

[formal ermine]

ok fuck I'm having a brainfart here

[south patrol]

isch ausch brudi

Been stuck with the same Galois problem for ages

[formal ermine]

I'm stuck on a definition

[south patrol]

It follows trivially from the definition

but nah okay fair oof

[formal ermine]

ah

got it

[south patrol]

noic

[formal ermine]

why does this book have so many typos

[south patrol]

which book oop

[formal ermine]

katok

found another typo

[long nebula]

Lol wir haben zu viel Deutsche

[south patrol]

( @velvet elk do you happen to have any hints for the q I did above btw? Idk if your notes covered stuff about cycle types and reduction mod p that I'm missing lol)

Or is the proof that no n cycles => always reducible mod p actually just easier lol

[south patrol]

This q lol

[prime sundial]

trying to find the homology modules for $C_n = \mathbb{Z}/8\mathbb{Z}$, and boundary maps $x \mapsto 4x \pmod 8$\
i found $H_n \simeq \mathbb{Z}/2\mathbb{Z}$. i don't think it's a heavy computation if anyone would be able to confirm this for me

[cloud walrus]

maximo

[south patrol]

Yes, each kernel is 2Z/8Z and each image is 4Z/8Z so we get Z/2

[prime sundial]

perfect, that's what i got. thank you potato

[prime sundial]

sorry potato for hijacking the channel. i have another question
what is meant by u mapping Hn(C) to Hn(D)?

i've shown boundaries -> boundaries and cycles -> cycles. i feel like im missing something fundamental about extending a homomorphism to act on a quotient module

ah is this saying u(Zn(C)) = Zn(D) and u(Bn(C)) = Bn(D)

[south patrol]

Not quite, it means u(Z_n(C)) \subseteq Z_n(D) etc

Hopefully it's clear why that leads to the map on homology

[prime sundial]

right this is clear to me

sorry for ping

oh i think it's coming back to me
if u is the chain complex map, then the induced map u_* : Hn(C) -> Hn(D) is just given by u_*[x] = [u(x)] right

if so it's clear to me yeah

hm i think im confused. is the question asking me to define that induced map, or is that understood already. there is no mention of it in the text

ok i found something on mse that is a little more general. i'll try to go off of that

thank you again potato

[prime sundial]

yeah ok it seems like we just know Hn(u)[x] = [u(x)] for any [x] in Hn(C), it just wasn't mentioned

[smoky ivy]

Let $K$ be a number field and $\alpha_1, \ldots \alpha_n \in \mathcal{O}K$ a $\mathbb{Q}$-basis of $K$ such that $\langle \alpha_1, \ldots, \alpha_n \rangle{\mathbb{Z}} \subsetneqq \mathcal{O}_K$. The initial problem I wanna solve is to prove that there is some type of algorithm to find an integral basis. Now, I'm proving this right now, and I'm struggling to show that $\mathcal{O}K/\langle \alpha_1, \ldots, \alpha_n \rangle{\mathbb{Z}}$ has an element of order $p$ (prime). Any ideas?

[cloud walrus]

lewis

[primal beacon]

well for a general p just use cauchy’s theorem once you know the index is finite but you can have strong restrictions on p do you know how ?

[rustic crown]

sounds pretty alg-nt, i feel the following works, but not super sure. me never studied alg-nt properly

if K is the splitting field and pO_K = (P_1 ... P_r)^e , then i think the assumptions imply that P_i have residue field F_p^n. If D is a decomposition group, i.e. subgroup of Gal(K/Q) fixing one of the primes say P, then it will give you an injective map D --> Gal(F_p^n/F_p) = Z/nZ. but order of D would be [K:Q]/r = ne and it so this is also surjective and e = 1. so you get to pullback the frobenius and get yourself a nice n-cycle in D.

[smoky ivy]

not sure how, how would you do that

[hardy tundra]

Thanks! @lethal dune @chilly radish @formal ermine @next obsidian

[primal beacon]

you have two Z- modules of rank n one is a submodule of the other, finiteness of the index comes from usual results on modules over pid’s, for the other part well p will divide the discriminant of $K$over $\mathbb{Q}$ and that’s because if you have a sub $\mathbb{Z}-module$ of the free $\mathcal{O}K$ you gain an « adapted base » I forgot the exact name of your $\mathbb{Z}-module$ and a matrix sending the base of $\mathcal{O}K$ to the base of the $\mathbb{Z}-module$ the determinant of which is exactly the initial index now use this matrix and its determinant to find a formula to link both the discriminant $D{K/\mathbb{Q}}=D{\mathcal{O}K/\mathbb{Z}$ and $D{\mathbb{Z}[\alpha_i,i]/\mathbb{Z}$

[cloud walrus]

rayane
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[primal beacon]

if you don’t see it I can give you the answer, btw in fact in general you just have to worry about non-totally ramified primes, also this is number theory so this isn’t the right place to talk about that

[south patrol]

Ye using the decomp group is definitely how to do it with alg nt

[primal beacon]

even more $p^2$ divides the discriminant

[cloud walrus]

rayane

[zenith hollow]

@south patrolpotato are you speciliased in abstract algebra only?

[smoky ivy]

just didn't know if this would be considered advanced number theory, but how is the determinante exactly the initial index, that's the part that i don't quite see

the rest i can figure that out

[primal beacon]

if $(e_i)_i$ is a $\mathbb{Z}$-base of $\mathcal{O}_K$ there are some $a_i$ such that $(a_ie_i)_i$ is a $\mathbb{Z}$-base of $\mathbb{Z}[\alpha_i,i]$ now first the morphism sending the $e_i$to the $a_ie_i$ has determinant the product of the $ a_i$’s immediatly (the matrix is diagonal) and $\mathcal{O}_K/\mathbb{Z}$ is isomorphic to $\bigoplus_i\frac{\mathbb{Z}}{a_i\mathbb{Z}}$ which shows the determinant and the index coincide

[cloud walrus]

rayane

[primal beacon]

oh yeah I didn’t mention the matrix going from the $\alpha_i$ to the $a_ie_i$ is invertible in $\mathbb{Z}$ so has determinant +-1 which justifies that the two index are the same

[dim widget]

If f mod p was irreducible then Frobenius acts transitively on the roots!

[cloud walrus]

rayane

[south patrol]

No

[south patrol]

Yes I meant like relating it to the Galois group over Q

But yeah seems you need a big-ish theorem relating Galois group over Q to mod p stuffs

[zenith hollow]

i could need a German in proofs and logic to proofread some of my stuff -_-

[formal ermine]

a lot of people here speak german

[smoky ivy]

^

[formal ermine]

KRANKENWAGEN

[next obsidian]

What is Krankenwagen

[formal ermine]

ambulance

[next obsidian]

Or should I say KRANKENWAGEN?

Why do you yell it

[formal ermine]

funnily enough you can also translate ambulance as ambulanz in german

[formal ermine]

for the dramatic effect

[next obsidian]

Of what?

[formal ermine]

it

[next obsidian]

Proving you speak German?

[formal ermine]

it's a meme

you're not cultured enough to understand it as it seems

[next obsidian]

[frigid lark]

Ich kann nicht Deutsch sprechen

[dim widget]

You don't need a big theorem, you just need to know that automorphisms mod p lift to the p-adics, you can show more generally that this is true for any henselian local ring/complete hausdorff local ring.