groups-rings-fields

[teal vessel]

I know, it's more a semantic than logical preference

[delicate orchid]

it's a logical preference as well

conjugation is a group action

[crystal turtle]

The elements that x commutes with are the ones fixed by conjugation

[delicate orchid]

so you can think of these things as stablisers of a group aciton

[crystal turtle]

As a group action/automorphism

[delicate orchid]

you can extend this to normalisers as well

N_G(H) is the set-wise stabiliser and C_G(H) is the point-wise stabiliser

[solemn dew]

if i have a o(G) = 3*2^2, will my 2-Sylow group have 4 elements?

[wraith cargo]

yes

[distant summit]

Is the answer false, true, false?

For (a), the inverse of a matrix might not have an integer in the entry concerned.

For (c), 40 does not divide 100.

[unkempt stream]

seems about right.

[delicate orchid]

yeah it's false true false

[distant summit]

Epic

[solemn dew]

guys Z/Zp where p is a prime is a field

is Z/Zp_1 x ... x Z/Zp_n also a field?

[formal ermine]

only in ohio

[chilly ocean]

(1, 0)

(0, 1)

uh oh!

[solemn dew]

hahah

ok it must be a field

[chilly ocean]

multiply them

[formal ermine]

all finite fields have order p^n!

[solemn dew]

what

[chilly ocean]

ignore him

[delicate orchid]

dunno how you concluded the exact opposite thing to what TTerra was showing you lol

[formal ermine]

by crt yours would be p_1p_2...p_n

[chilly ocean]

multiply these two elements jonathan

[chilly ocean]

because i didn't actually say what to do with the example...

[chilly ocean]

what do you get?

[solemn dew]

(0,0)

[delicate orchid]

takes a massive hit

[chilly ocean]

the zero element of this ring

right

can you multiply two non-zero elements of a field to get zero?

[solemn dew]

oh

no u cant

[chilly ocean]

there you go

[solemn dew]

nice

[formal ermine]

seems pretty fieldy to me...

[solemn dew]

thanks brother

[chilly ocean]

np

[fleet pelican]

bc you have to map 1 to 1, and that automatically fixes Q

[coral shale]

1 times 1/infinity = 0

[topaz solar]

Ok GT

[toxic zephyr]

i feel bad that gt has become a bit of a joke, but they didn't exactly do anything to stop it

[glossy crag]

who's GT?

[glossy crag]

You're gatekeeping me!

[hollow mica]

GT did it better

[sonic coral]

id like to hear this story

[toxic zephyr]

there's a person named gtbot who keeps sending messages about a new number system they're developing in #math-discussion. but they keep presenting it and responding to legitimate criticism in the most annoying way possible. being extremely vague about definitions mostly, which makes for an extremely unproductive discussion.

[sonic coral]

that sounds entertaining

i hope he hasnt been bullied enough to stop

[toxic zephyr]

I don't think he'll ever stop lmfao

[sonic coral]

guess i need to read up in that channel

[dull marsh]

The 1/0 guy?

[ornate vessel]

Why Z[x]/(5, x^2+4) is isomorphic to Z5[x]/(x^2+4)

[ornate vessel]

Maybe f+I to f+J

[coral shale]

(a, b) = (a) + (b)

R/(I+J) === R/I/J

[ornate vessel]

😮

[coral shale]

probably.

2nd iso

idk how to write it properly

[coral shale]

In any case, you should be able to write down the explicit bijection for yourself and verify it

[ornate vessel]

[f] -> [f with reduced coefficients]

Isn't this isomorphism

Xd

[coral shale]

@quiet cave = (R/I)/(J/I) = R/J

3rd iso

[solemn dew]

guys why is the answer 20?

i'm thinking

(5 pick 3) = 10, and to make a group two of these have to be together

for example
(123) and (132) needs to be paired

so we get a subgroup {e, (123), (132)}, therefore there will be 5 subgroups

but answer is 20??

[coral shale]

so u hopefully identified it must be C3

and because its S5, the only way to do this is with single 3-cycles as elements

(if S6 you could have double cycles)

but hmm

maybe this line of reasoning is missing something

[solemn dew]

yeah...

[coral shale]

No clue

[coral shale]

where is 20 from

[solemn dew]

solution from teacher

[coral shale]

i find it sus

[solemn dew]

an old test

[dim widget]

It’s not 5 it’s just 5 choose 3

[coral shale]

oh i didnt read what u did with 5 choose 3...

5 choose 3 does not differentiate between (123) and (132)

[solemn dew]

then i think that you can make this argument instead:

5.4.3 = 60 elements

divide by 3 since (abc) = (cab) = (bca)

so we have 20 elements now

...but to make a subgroup we need to pair two, for example (abc) and (acb), thus 10 subgroups

still wrong

https://www.iosrjournals.org/iosr-jm/papers/Vol8-issue3/O0838793.pdf

found this paper and it says 10 subgroups

i guess the teacher is wrong

[rocky cloak]

There are 20 elements of order 3 though, maybe that's what your teacher was thinking.

Also from the Sylow theorems the number of subgroups must be 1 modulo 3. So that's another way to see that the answer isn't 20

[solemn dew]

1 mod 3?

[rocky cloak]

The number of p sylow subgroups must be 1 modulo p, and divide the order of the group.

This is the third sylow theorem

[south patrol]

ig 1 mod 3 and dividing 40 narrows it down quite nicely lol

[solemn dew]

Oh thanks

[south patrol]

In fact of course you can also reduce to A_5 to remove a factor of 2 but doesn't change much

[coral shale]

potato

[south patrol]

Helo!

[distant summit]

Would someone check my answers please?

2. (a) $\sigma=(14)(23), \tau=(142) \implies \tau^{-1}=(241), \sigma\tau^{-1} (14)(23)(241)=(23)$.

(b) $\sigma$ is composed of two 2-cycles, and $lcm(2,2)=2$. Therefore $ord(\sigma)=2$. Then $ord(\tau^{-1})=3$ and $ord(\sigma\tau^{-1})=2$.

[cloud walrus]

Douglas

[delicate orchid]

Everything looks fine except your computation of \sigma\tau^-1

[coral shale]

how did you manage to parse it...

[delicate orchid]

With my eyes

[distant summit]

I wrote out the matrices

[delicate orchid]

assuming you're going left to right?

ok we go left to right

1 -> 4 -> 1

[distant summit]

Oh wait I think I found tau^-1 sigma

[delicate orchid]

2 -> 3
3 -> 2 -> 4
4 -> 1 -> 2

[distant summit]

Cos I wrote sigma above tau^-1

[delicate orchid]

so it's (234)

[distant summit]

For this, would it be sufficient to show that $m/\gcd(m,n)=\lambda/n$ where $\lambda$ is the lcm of m and n. Then since $\lambda$ and $n$ are fixed, there is no smaller multiple of $m$, and therefore $m/\gcd(m,n)$ is the least power of $g^n$ that equals $e$, hence it is the order.

[cloud walrus]

Douglas

[toxic zephyr]

the quickest proof of this i think would be to suppose (g^n)^k=e and then use the fact that gcd(m/d,n/d)=1 (where d=gcd(m,n)). you want to show that (g^n)^k=e iff m/d divides k.

[chilly ocean]

Let a be a root of p(x) = x^3+9x+6. How would one go about finding the inverse of a+1 in Q(a) ?

I have tried deducing equalities from a^3+9a+6=0, but I am wondering if I am not making some observation about p(x) that will point me in the right direction.

[summer path]

try manipulating x^3+9x+6 to take out an a+1, using say euclidean algorithm

[indigo ridge]

im trying to look for some readings that go into depth on the topic of the The Circle Group, more specifically showing that R/Z is isomorphic to S^1, explaining the intuition and all

I already have 1 but im looking for some more things to read on

[chilly ocean]

why not write one yourself?

[summer path]

you should be able to write one out yourself

or try to and ask here for hints as needed

[delicate bloom]

I use the remainder theorem, plug in -1, then subtract that off and divide by a+1 to see what's left

[chilly ocean]

Ahhh OK, thanks. So this is an algorithm for finding any inverse in a Euclidean domain?

[chilly ocean]

I don't quite follow. What are you plugging -1 into and why?

[delicate bloom]

that step is the remainder theorem, are you familiar with that

you can use that other method too, maybe best to focus on one method at a time, I don't wanna distract you too much

[coral steeple]

I'm trying to figure out how this theorem lets us identify quotient groups. What I'm understanding is that in general we can take the map $\pi:G\rightarrow G/N$ given by $\pi(a)=aN$, restrict its codomain to its image to get a surjection $\pi'$, and conclude that $G/N\simeq \operatorname{im}\pi'$. How does that sound?

[cloud walrus]

person2709505

[summer path]

this is probably a faster solution considering that you're looking for the inverse of a linear polynomial; the remainder theorem also follows pretty easily from euclidean algorithm

[delicate bloom]

yeah exactly what I was thinking, this is a distraction since it's for the linear case

[barren sierra]

What you said is a bit backwards to the statement. One way it's commonly used is if you want to show G/N is isomorphic to H, you construct a surjective map G -> H such that the kernel of the map is N. Then first iso tells you the desired isomorphism

[delicate bloom]

plus when I say divide by (a+1) i'm thinking of not doing literal synthetic division or w/e i'm thinking of just like playing with the polynomial which is a little tricky but easy and quick enough to do, but still more tricks

[summer path]

if they don't know remainder theorem, they could reasonably try to do a quick proof of it, then use it; though doing it the "default" way is probably better for figuring out what's going on

[coral steeple]

Ok so you might have to be a bit creative to actually construct the homomorphism?

[summer path]

i also forgot about synthetic division

[barren sierra]

Sometimes but a lot of times it's "obvious"

[coral steeple]

I see, thanks

[chilly ocean]

Yes, I am familiar with it. I will assume you mean that you mean to plug in -1 for the constant in the linear factor (x-r) in the theorem. This gives you p(x)=q(x)(x+1) - 4, pass to the quotient which gives that 0 = q(x)(x+1)+4, and so -4=q(x)(x+1). Then the inverse is (-1/4)q(x) ?

[delicate bloom]

yeah exactly

the polynomial q(x) you get is x^3+9x+10 so I'd then step through by trying to make factors of x+1, like x^3-x+10(x+1), then x^3+x^2-x^2-x+10(x+1) = ... etc

well basically just did the whole thing whoops but w/e

[toxic zephyr]

i was trying to find methods to find square/cube/nth roots in the field Q(sqrt2) and this gave me an idea. if alpha has minimal polynomial p(x), then the solution to p(x^n)=0 will be an nth root, right? curious if anyone else has any other methods for finding cube roots. for example, how could i find that 45+29sqrt2=(3+sqrt2)^3

solving a 6th degree polynomial isn't exactly ideal

one observation i made is that N(45+29sqrt2)=45-2(29^2)=343=7^3. so we can conclude that the cube root has N(x+ysqrt2)=7. but idk how i can leverage that very usefully

[frigid lark]

We have this theorem

But it might not be that useful as determining whether or not a in k^p is going to be hard

[toxic zephyr]

well the polynomials will always be of the form x^(2n)+px^n+q=0, so there is that

i guess we just need to divide a quadratic out of that

[celest furnace]

What is the 4 doing here?

[frigid lark]

Also this theorem will only tell you if the n'th root is not it Q(sqrt2)

[frigid lark]

2 is a little bitch, requires more work in these types of theorems

ah this screenshot doesn't give enough context

Consider the case when n = 4, and a = -4b^4

CF Lang Ch6 section 9

[frigid lark]

I think this theorem will say whether or not the root will be in k(sqrt(2)), so we consider polynomials with coeff's in Q(sqrt(2))

[frigid lark]

Maybe you could leverage whether the norm is in Q^3

I mean N(45 + 29sqrt2) in Q^3

[toxic zephyr]

we can always bring it into Z(sqrt2)

sorry brb

[frigid lark]

wdym bring it into Z(sqrt2)?

If a in Q(sqrt2), then N(a) in Q, then N(a^3) = N(a)^3 in Q^3. So if you have a b such that N(b) is not in Q^3, then the cube root of b is not in Q(sqrt2). I feel like I'm misinterpreting you cause that statement is obvious.

I think I've got something for cube roots

You could solve the 2 equations (substituting 2b^2 = s - a^2 to get what a and b are

But this doesn't scale too well

Ah sorry, N(a + sqrt2 b) should be a^2 - 2b^2

[toxic zephyr]

just multiply by the common denominator (or factor it out) shouldn't affect the cube root right?

[frigid lark]

Consider sqrt2, it's cube root is not of the form a + sqrt2 b

[toxic zephyr]

sqrt2 has a cube root?

[frigid lark]

Do you mean if a has a cube root in Q(sqrt2), there exists a k in Z such that ka has a cube root in Z(sqrt2)

[frigid lark]

Well in a bigger field

[toxic zephyr]

yeah that's what i mean

i just want to make sure that WLOG i can consider cube roots in Z(sqrt2)

[frigid lark]

Sure, but why?

[toxic zephyr]

easier to work with i think. since then can we say that a is a cube in Z(sqrt2) iff N(a) is a cube in Z?

actually that isn't trivial is it...

[frigid lark]

Well if a is in Z(sqrt2) then N(a) is in Z

[toxic zephyr]

well i guess tbh i was only really considering Z(sqrt2) originally so idk maybe we can just ignore Q for now and stick with Z

can something in Z(sqrt2) have a cube root in Q(sqrt2) but not Z(sqrt2)?

[frigid lark]

Maybe, I think if a = b^3 in Z(sqrt2), and if b is in Z(sqrt2), then N(b)^3 = N(a) in Z, thus N(b) is an integer

I think that's as strong as you can get

[frigid lark]

A counterexample to that is 4 + 2sqrt2

I think an approach you can do is to consider the norm and compare coefficients to get a polynomial of degree 3 in a or b, solve that to get a or b, get the other term, and cube your result

but that's only if a cube root exists

in Q(sqrt2)

And to solve the polynomial rational root theorem is enough since you only care about rational solutions

[toxic zephyr]

i see, that's disappointing. i wonder if it's true after dividing out any common factor, but imma guess not.

[toxic zephyr]

i still think that taking the minimal polynomial and dividing out a quadratic from p(x^3) should work. i think you can probably argue that you can split off a quadratic iff it has a cube root.

[chilly ocean]

If you quotient F_2[x] by an irreducible polynomial of degree 3, the powers of the root form a cyclic group of order 8. What's the general explanation behind this?

I understand in the quotient, any polynomial equivalence class can be represented by a polynomial of degree less than or equal to 2 by division with remainder, and that there are 2^3 = 8 of them.

...

I think I just answered my own question, but if you have any extra input, feel free to give it!

My last question was, how does one see that one always gets every possible polynomial of degree less than 2 over F_2 as a power of the root. But this can also be seen by division with remainder.

[frigid lark]

I guess my approach only requires you to find rational roots of a degree 3 polynomial

which is a bit easier on the computation

[coral steeple]

I'm looking for a homomorphism $\phi: G={\textrm{ invertible upper triangular real 2x2 matrices}}\rightarrow H$ whose kernel is the subset of $G$ which has all diagonal entries equal, but I'm not sure where to look

[cloud walrus]

person2709505

[formal ermine]

phi(x) = 0

kernel has the subset you're looking for

[coral steeple]

well isn't the kernel of that function too large? I don't want ker phi to just contain this set; I would like set equality

[wraith cargo]

What is H

[coral steeple]

Anything

I'm looking for G/{ equal diagonals }

Which would be isomorphic to im phi

(If I understand correctly)

[chilly ocean]

there's an obvious map from G to (R \ {0})^2 for which the pre-image of the diagonal ∆ is the subgroup you'd like to write as a kernel. compose the map at the start with the quotient map by the diagonal and you'll have what you wanted

[coral steeple]

What is the quotient map? Google returns definitions from topology

[chilly ocean]

map to the quotient group

[coral steeple]

Nevermind yep

[cloud walrus]

person2709505

[chilly ocean]

more explicitly, you have one map from G to (R \ {0})^2 given by [[a, b], [0, c]] -> (a, c). you also have the diagonal subgroup ∆ of (R \ {0})^2 consisting of all pairs of the form (r, r), and you can form the quotient group (R \ {0})^2 / ∆. you have a natural map to this quotient group. compose for a map from G to (R \ {0})^2 / ∆.

it is given by [[a, b], [0, c]] -> (a, c)∆. the kernel is the set of such matrices with (a, c)∆ = ∆, i.e. (a, c) in ∆, and this is the same thing as saying that a = c

this definitely isn't the only way to come up with such an H and phi, but it's one

[coral steeple]

I'll think about that a minute lol

[chilly ocean]

writing the subset where two things are equal as the pre-image of the diagonal under a certain map is a natural thing to do (and quite useful). since you wanted specifically a kernel and not just any pre&image, it was a just a matter of taking the quotient

of course, you can get much simpler than (R\0)^2/∆. this thing is isomorphic to R\0 anyways. i just wanted to outline a general way you could tackle this

[[a, b], [0, c]] -> ac^{-1} does the job just fine

[coral steeple]

Ok I get it now thanks!! Now I know at least one application of quotient groups too

[void cosmos]

yo

if R is a ring ( not necessairly having identity )

and I is a modular left ideal

then R(R/I) !=0

now is the proof of this is to assume this is zero , choose the element r(e+I) --> re is in I --> r is in I and since r was arbitrary I = R and a contradiction? (implying R is just zero ring )

im confused

anyone help?

[rocky cloak]

Yeah, that's right, specifically if r=e you would get that e is in I

[void cosmos]

yea got it

ty

another bad question

or nvm i got it

[solemn dew]

any idea how to do b?

any element in our field can be written as {a + bx + f(x) }

that is how far i've come lol

have also proven f(x) is irreducible in F[x]

[delicate orchid]

yeah b) seems a bit annoying to do

[open sluice]

what does it mean to be an inverse of an element

[solemn dew]

i have no idea hahah

[delicate orchid]

concerning

[open sluice]

/r/elihafcui5m (explain like I have a final coming up in 5 minutes)

[open sluice]

what sets a field apart from an integral domain

[delicate orchid]

how are you defining a field without inverses

[glossy crag]

It's the same way you find inverses in Z/(n), you find the Bezout coefficients (polynomials f,g so that (3+4x)f+(3x^2+3x+1)g=1). You can do this backtracking through the Euclidean calculation of the gcd of 3+4x and 3x^2+3x+1 or using the extended Euclidean algorithm (which is the same thing, just optimised a little).

E.g. in Z/(7), to find the inverse of 5mod7
7=1*5+2
5=2*2+1,
therefore
1=5-2*2=5-2*(7-5)=3*5-2*7,
so 3mod7 is the inverse of 5mod7.

Do the same thing for polynomials.

[solemn dew]

alright i gotcha brotha

will try

[glossy crag]

By Bezout coefficients I mean Bezout's theorem: if R is a PID and d is the gcd of a and b, then there exist elements x and y such that ax+by=d. If d=1 (x and y are coprime), then y+(a) is the inverse of b+(a) in R/(a).

Both Z and F[x] are PIDs, so it works word for word the same way in both settings.

[teal vessel]

thought process involving Q8 and centralizers in it: for all elements a,b of Q8 that aren't 1 or -1, ab = -ba, which directly implies that for all members of Q8, the centralizer of that element in Q8 is either Q8 (for 1 and -1) or just {1,-1}.

does this track correctly or am I missing something here?

Q8 being the basis elements of the quaternions {±1, ±i, ±j, ±k}

[lusty marlin]

What you have said here is a nice way of 'looking at things', but you should keep in mind that the symbol '-' doesn't have meaning on its own in the context of the group Q_8 without any additional structure.

[delicate orchid]

this is why I prefer writing M for -1 or something when thinking about Q_8

and no, clearly "-i" is in the centraliser of i as is i itself so the centaliser cannot just be {1, -1}

[teal vessel]

oof, yeah, all elements self-commute, obvious things are obvious.

[lusty marlin]

Yes

[lusty marlin]

*Commute with themselves as well as their inverses

[delicate orchid]

or more generally, <x> is always a subgroup of C_G(x)

[teal vessel]

at least I've saved time by not having to check every element by hand.

[lusty marlin]

Let ρ: G → GL(V) be a representation of a group G on a complex vector space V. How do we prove that the matrix ρ(g) is diagonalizable for all g in G?

[coral spindle]

Have you perhaps seen the lemma that says that the irreps of Abelian groups are 1d?

[dim widget]

Is G finite?

[delicate orchid]

more over is G abelian

[lusty marlin]

Yes. I forgot to add this

[coral spindle]

G doesn't have to be Abelian

[solar vessel]

you ever heard of Weyl's unitary trick?

[lusty marlin]

Nope. New to representation theory

[coral spindle]

No, then?

[lusty marlin]

Yeah I haven't seen it yet

[coral spindle]

OK.

[dim widget]

If G is finite then each matrix \rho(g) has finite order. Thus the minimal polynomial of \rho(g) divides X^m - 1 for some m. The minimal polynomial of \rho(g) therefore has no repeated roots, so \rho(g) is diagonalizable

[lusty marlin]

Ah
Thanks!

[delicate orchid]

that's a very nice argument

[barren sierra]

I gotta learn rep theory one day

[delicate orchid]

same

[barren sierra]

Seems to keep coming up in things I'm interested in

Just finding the time is hard

[lusty marlin]

What sort of stuff are you interested in?

[delicate orchid]

oh yeah it's everywhere

turns out R-mod is a really nice category lol

[barren sierra]

algebraic combinatorics and computational group theory.

[lusty marlin]

Nice! Algebraic combinatorics is something I would really like to explore

[delicate orchid]

it's so useful for computational group theory it's unreal

[barren sierra]

I know very little about CGT

Just small bits of it came up in my REU

[delicate orchid]

would you struggle with trying to impliment a bunch of weirdo functions to model a group or just do a matrix multiplication

[lusty marlin]

Makes sense

[barren sierra]

Ye

I've been meaning to dive into a book about it but again, time

And then algebraic combo seems fun but like the stuff I'm trying to read about RN is so over my head

[delicate orchid]

I recommend representations and characters of groups by James & Liebeck when you have the time

[lusty marlin]

Is research on finite groups mostly computational nowadays?

[delicate orchid]

it's mostly not on finite groups at all it's using other structures to probe finite groups

[coral spindle]

great book

[delicate orchid]

it's the one I read to learn rep theory

[coral spindle]

They actually look at real Schur indices in that book, although ofc they don't call it that

[delicate orchid]

and then I attempted to read Fulton-Harris

[coral spindle]

And they even prove the formula for the Frobenius–Schur indicator

[lusty marlin]

That book is extremely intimidating

[delicate orchid]

it is

[lusty marlin]

I opened it like once before switching to Serre

[delicate orchid]

it's how I learnt about the funny tetris squares albeit

[lusty marlin]

lol what

[delicate orchid]

if you want an approach more focused on character theory (the based approach) then Isaac's book is good imo

GoldenPhoneix typing out a novel atm

[teal vessel]

sometimes I need to type up a proof and not hit send because the process of typing up the proof makes it clear if anything went wrong. Anyone else get that?

[delicate orchid]

this is why whenever I think I've proven a result I wait a few hours then rewrite it

[coral spindle]

That's called thinking lmao

[delicate orchid]

no I refuse to do that

the thinking goes into coming up with a viable proof in the first place

[teal vessel]

like, I have it sitting in my clipboard here, but tbh I don't really need it rubberstamped when I already know what's going on with it, especially when I'm being stubborn and using facts that might not be the most efficient lol

showing that for G=s3, A={1, (1 2 3), (1 3 2)}, the centralizer of A in G is A, and the normalizer of A in G is G

1. S3 ≅ D6, therefore showing the equivalent statement in D6 satisfies the same structure.

2. A ≅ {1, r, r²} (mutatis mutandis)

3. the centralizer of a set is equal to the intersection of the centralizers of its elements (not strictly necessary, but is the path I chose because I can)

4. The centralizer of identity is the total group

5. C(r)=C(r²)={rⁿ | n ϵ ℤ}
   C1. C(A) ≅ {1, r, r²} ≅ A

6. C(A) ≤ N(A)

7. srs = r² ϵ A which means s is a member of (the set isomorphic to) N(A)

8. by Lagrange's theorem, |N(A)| divides |G|, and |N(A)| ≥ 4, which implies that |N(A)| = |G|
   C2. Since N(A) ≤ G and |N(A)| = |G| < ∞, N(A)=G

I'm not being dumb again with forgetting something obvious, right? Some notation here is adjusted or omitted because it is unambiguous.

(final question being rhetorical, as I have since answered it by writing)

[delicate orchid]

yeah this is right but my god is this overcooked

just conjugate by (12), (13) and (23)

you'll see that none of them fix (123) or (132)

but they fix A

and what the FUCK does mutatis mutandis mean

bros got the grimore out summoning egregores

[crystal turtle]

[teal vessel]

latin for something to the effect of "making the appropriate changes in scope"

[delicate orchid]

or just do it in S_3?

[teal vessel]

no :chad:

[delicate orchid]

based albeit

if you must pass to D_3

then D_3 = C_3 \rtimes C_2

and then the result trivially follows

[coral spindle]

It literally – almost word for word – means "having changed what should be changed"

[teal vessel]

yeh. I'm not fluent in latin, I pick up phrases but don't keep their english counterparts because I already have the meaning, so why double up the memory space?

[barren sierra]

My REU was on classifying some finite groups and basically the process was
Step 0: Find a set of parameters that describe the groups we care about (this was already done for us)
Step 1: so a bunch of math to figure out some finite bounds on these parameters
Step 2: computer go brr on finite bounds

[delicate orchid]

it's just that nobody actually uses latin ever

unless you're a lawyer

[teal vessel]

or read too much phil

[delicate orchid]

yeah the hard part is step 0 lol

step 1 can be tricky too albeit

[barren sierra]

Thankfully some dude in the 60s figured out step 0

[delicate orchid]

I got a classification problem for ya

[barren sierra]

They just didn't have GAP

[teal vessel]

but yeh, I can do things in S3, but I prefer the notation of D6 better. my brain wraps around it better for some reason.

[delicate orchid]

it's because you have the presentation

[barren sierra]

I've mainly been looking at Issacs. I'll check this out

[lusty marlin]

Hmmm

[delicate orchid]

D_6 ( 🤢 ) = <r, t | r^2 = t^3 = 1, rtr = t^2>

so you can just load these relators into your head and work in D_....6

[barren sierra]

My REU mentor also did a similar thing, just the whole step 0 for his thing was 2 papers and then the other steps were each a paper lol

[sweet echo]

veni vidi vici

[delicate orchid]

yeah that sounds about right

[teal vessel]

learning from D&F, they choose the D_2n notation. I know it's also common to call it D3

[delicate orchid]

yeah I'm still gonna cry about it though

[teal vessel]

I'm a french horn player, I'm used to transposition in my head. Call it what you want, I can read.

[delicate orchid]

mfs really be working in SD_4 with a straight face

[teal vessel]

except when suffering from can't-read disease

[barren sierra]

See the thing about D_n vs D_2n is that it's like plugging in a USB. You're going to flip the notation depending on who you're talking to and still get it wrong anyways

[delicate orchid]

I do this to myself

I call the dihedral group of order 8,16,32 D_n in my head and then everything else is D_2n

[teal vessel]

best way I've found with notational variations like this is that when bringing up a subject, use notation you prefer, but when joining a subject, use the notation established by the previous comments. Makes life easier when everyone picks a notation and sticks with it.

so if you ever mention dihedral groups, I'll use the notation you bring up XD

[delicate orchid]

yur, I only ever work with dihedral groups that are 2-groups hence why my mindset switches

[teal vessel]

speaking of, I need to redemonstrate this question for dihedral groups of order 8 and 10 now. Probably not gonna write them up since I got a basic idea of what I'm doing.

[delicate orchid]

gl

[teal vessel]

ooh, A in the order 8 question is gonna exploit the half-turn symmetry. This should be fun

[solemn dew]

hints on solving d?

[glossy crag]

They probably mean x^125-x: if L/K is an extension of degree n and K is finite of size q, then L is automatically the splitting field of x^{q^n}-x (if you don't know this, I can give you a derivation of this fact). If you're given just some random extension of finite fields and you need a candidate for whose splitting field this extension is, that's the most natural example.

[rocky cloak]

Just p=f works, or any irreducible cubic. Or like ocean man says, the standard construction of the finite fields are as the splitting field of x^p^n - x

[glossy crag]

I think mine's easiest, because to know that any irreducible cubic works you need to know there's just 1 extension of degree n and it is normal.

[solemn dew]

thanks brothers

[rocky cloak]

Yeah, depends what has already been shown about finite fields I guess

[solemn dew]

can i get the derivation of this fact?

[rocky cloak]

Hint: Lagrange's theorem

[glossy crag]

You're fine with K having size q (a prime power) or do you need it to be F_p?

[solemn dew]

i don't understand both so..

[glossy crag]

OK you do know there's a field of prime size for any p, right

Z/(p)

[solemn dew]

yes

[glossy crag]

OK, so if K is an extension of degree n of Z/(p), then it has p^n elements -- is that clear?

[solemn dew]

yes sir

[glossy crag]

The subset of nonzero elements of K is a group of size p^n-1, because every element has a multiplicative inverse by the definition of a field.

In a group G of size m g^m=1 for any g (by Lagrange's theorem the order k of g divides m, so g^m=g^{bk}=(g^k)^b=1^b=1).

Therefore a^{p^m-1}=1 for any nonzero a in K.

Therefore a^p^m=a for any a in K (obvious if a=0, if a\neq0 apply previous step).

Therefore every element of K is a root of the polynomial f(x)=x^{p^m}-x\in Z/(p)[x].

This makes K a splitting field of f over Z/(p): by definition, this means f must split into linear factors over K (which it does) and K must be the smallest field over which it splits (if K' is a subfield, then it necessarily misses some roots, since ALL the elements of K are roots).

IDK if you know what "separable/only has simple roots" means, but it can be shown that f satisfies this condition too.

The above works word for word if you replace p by an arbitrary (prime power) q.

[solemn dew]

ur a genius man

i'll have to read this a few times to understand it

but it is really appreciated

thanks

[glossy crag]

Lol no I'm not, this appears word for word in pretty much every algebra textbook.

[solemn dew]

well

being able to memorize that is a super power imo hahah

[glossy crag]

But I understand what you mean, I feel exactly the same way when someone more knowledgeable than me answers my questions on this server.

[teal vessel]

don't memorize, divide into processes, and with a little practice you can usually rebuild textbook definitions out of the process you understand.

[solemn dew]

The subset of nonzero elements of K is a

[toxic zephyr]

sorry for the ping but i found a really interesting way to do it as a sort of generalization of de moivre's theorem. i don't have a proof for why it works but it just does when N(x)>0 lol. not sure how to adjust it so that it works for when N(x)<0 yet.

an example showing 20+14sqrt2=(1+2sqrt2)^3
https://www.desmos.com/calculator/zpepb2rron

[chilly ocean]

The minimal polynomial of √2+√3 over Q is x^4-10x^2+1. How would one go about proving that it's irreducible? Eisenstein's theorem does not apply, and it's degree 4 so the only strategy I see is to suppose it factors, set up a system of equations, and derive a contradiction.

Alternatively, I've solved for the roots over C by a change of variables and thus have factored it over C, but does that imply it's irreducible over Q and if so, why?

[dim widget]

Yes because none of the roots satisfy a quadratic polynomial over Q

4 = 3 + 1 or 2 + 2 or 4 + 0 so if no rational roots and you write down roots which are not contained in a quadratic extension of Q then it's irreducible

[dim widget]

Anyway an easier way to see this is that when you reduce mod 5 you get the polynomial x^4 + 1 which splits into two quadratic factors, but modulo 3 you get x^4 + 2x + 1 which has only one linear factor

So it can't possibly be reducible over Q

[dim widget]

I don't know if you know this but this is a cool polynomial because it's one of the ones that factors modulo every prime p

[primal beacon]

if you know any galois theory you can prove that $$\mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3}) $$ then just show that the second one has degree 4

[cloud walrus]

rayane

[primal beacon]

over$\mathbb{Q}$

[cloud walrus]

rayane

[primal beacon]

ho thats cool

[dim widget]

Ah it would be if it were true! But actually the user posted the wrong polynomial

the minimal polynomial of sqrt(2) + sqrt(3) is x^4 - 10x^2 + 1

x^4 - 10x + 1 is irreducible but it's reducible mod 17, so it must correspond to a C_4, A_4 or S_4 extension

[round hull]

when deriving the duplication formula for elliptic curves, does it matter which field we take the elliptic curve to be on

my guess is yes, something different happens for different characteristic

i don't know anything by the way

[primal beacon]

you can learn about « division polynomials » which gives a formula for nP for any n and any p. In char 2 weird things happen for multiplication by 2

in any other char everything works fine with the usual duplication formula

[round hull]

well damn that sucks

i'm tasked to derive the duplication formula of y^2 + y = x^3 + x^2 + x + 1 over F_{2^3}

[chilly ocean]

I'm having trouble understanding this. I see that none of the roots will satisfy a quadratic polynomial over Q because they are square roots of square roots. I don't understand why you're writing partitions of 4 and how this implies that the polynomial is irreducible (fixed the polynomial).

[dim widget]

Because if it was reducible it would either have a linear or quadratic factor

[round hull]

so what goes wrong with the chord and tangent method

[primal beacon]

is it rlly possible to have an irred pol that splits for every prime ??

[dim widget]

Yes, not only that every poly with galois group the Klein four group has this property

[primal beacon]

ohhh

I was thinking there must be a 4 cycle

my bad

[dim widget]

yeppp

For odd degree any A_n polynomial has this property.

So you can just check if the discriminant is a square

[south patrol]

Yeah these are very cool things

[primal beacon]

ohhh riight I didnt know about that

wait did I get it

[open sluice]

the only poly like this I know is x^4 + 1

[dim widget]

now you know two!

[primal beacon]

hm its a bit deeper, lets just say basically in char p if p doesnt divide n then the n torsion E[n] is isomorphic to Z/nZxZ/nZ (in an algebraic closurz) while E[p^k]=Z/p^kZ or O, now E[p]=ker([p]) where [p] denotes mult by p so you see it depends

[primal beacon]

I actually disnt answer : the chord tangent works, you just need to rewrite it from 0 it won’t be the usual formula I believe

[round hull]

wait i don't know what torsion points have to do with this

oh

[primal beacon]

can you explain to me I think I didnt get it 😭

[round hull]

or like i don't think i get the point of this

[dim widget]

What is the square root of the discriminant?

[dim widget]

There's not much point, but it gives a vague justification for why something should go wrong at 2

or at least be different

[primal beacon]

a generator of the 2 extension or smth ?

no ahem

[dim widget]

you can still use the tangent method, just don't divide by 2 anywhere, which would be a problem if your curve was in the usual weierstrass form

[round hull]

thank you both

[solar vessel]

yes thank you

[primal beacon]

oh okay from the root definition but still im not getting it

[void cosmos]

yo yo stupid question

is there a ring that has itself as its annihilator?

[dim widget]

Not with unity

[void cosmos]

can you give an example plz

[dim widget]

You could just take the rng {0, x} with x^2 = 2x = 0

[dim widget]

So how does the symmetric group act on the square root of the discriminant?

When it acts by permuting the roots

[primal beacon]

does An change its sign ?

[chilly ocean]

OK, the rational roots theorem will imply it has no linear factors, so it if it factors, it has to do so into quadratics. If it factored into two quadratics then ... I'm not understanding the part about roots in a quadratic extension.

[void cosmos]

okay so now with this is it correct to say that any simple ring is primitive?

or no

with identity

any simple ring with identity is primitive? is this correct

[dim widget]

No exactly the opposite

[primal beacon]

sht

[dim widget]

since A_n exchanges an even number of roots

[primal beacon]

okay yes right

[dim widget]

i.e. the galois group is contained in A_n if and only if it acts trivially on the square root of the discriminant

iff the square root of the discriminant is rational

for other transitive subgroups of S_n you can make lagrange resolvents which do a similar thing, but they're not as nice.

[primal beacon]

thats clearer now, thats rlly cool !

[primal beacon]

yes this is pretty pretty it evenfeels weird

thank you !

[dim widget]

yes if a ring is simple as a module over itself and has identity it is primitive

unfortunately simple means that it has no nonzero proper left ideals

so it's not a very interesting statement

[void cosmos]

wait what

do you mean no two-sided ideals?

wouldnt the proof be to consider R/I where I is a maximal left ideal

and show that this has trivial annihilator?

[chilly ocean]

Stupid question: if a polynomial p(x) factors as g(x)h(x) over F, how do you show that this factorization is the same in a field extension? I think it's by uniqueness of factorization. Since p(x)=g(x)f(x) in F, then that equation will also hold in field extension and so that's the factorization in the extension as well by uniqueness.

[void cosmos]

yea

[dim widget]

If $gf = p$ then $gf = p$ no matter what

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

It doesn't matter if you make the coefficients bigger

still $g*f = p$

[cloud walrus]

Topos_Theory_E-Girl

[chilly ocean]

I think the key theorem I'm missing here is that if a polynomial (over Q?) factors into quadratics then it's roots have to live in a degree 2 extension. Why is this true (if it is)?

[dim widget]

because the roots will be roots of its factors, which will live in a quadratic extension because the factors are quadratic

if f(x) = (x^2 + x + 1)*(x^2 + 2x + 2) then the roots of f lie in one of those two quadratic extensions

[void cosmos]

im stuck

ssuppose S has infinite dimension over D , {s1,s2,...} is an infinite basis

define I_i to be the annihilator of {s1,s2,...,s_i}

then this gives rise to a descending chain I_1 > I_2 > I_3...

this chain stablizies

now what

probably something something annihilator of I_n = annhilator of I_n+something

so then i get some relation between the s_is like say rs_i-k_sk = 0 or something which is contradicts linear independance?

thats all i can think of

[pale elbow]

Hey folks, is there software out there that helps you design algebraic objects, or at least makes it easier if you're a beginner?

[rocky cloak]

I guess you can just mod out by the annihilator of S and then apply Artin-Wedderburn. Maybe a boring solution...

[pulsar vault]

Hi I have a question about eliptic curves. Say I have 2 polynomials P(x) and Q(x) which have different orders, and for a specific value of x, P(x) = Q(x). Is there any relation between the two elliptic curves at that specific value of x? Thank you

[delicate bloom]

what's the order of a polynomial

are these elliptic curves when written as y^2=P(x) and y^2=Q(x) or something?

[void cosmos]

idk artin-wedderburn yet

is there another solution

[rocky cloak]

There might be, but I don't have one of the top of my head. Alternatively, you can try to prove Artin-Wedderburn.

Or I guess the thing to prove is that R modulo the annihilator of a simple module is a semisimple ring, when R is artinian.

From there you get that S = Hom(R, S) = Hom(R/I, S) = End(S)^n where n is the number of summands of S in R/I.

[rocky cloak]

Okay, so this is basically still using Artin-Wedderburn, but I think you can condense it to a pretty simple road map:

1. wlog replace R by R/I where I is the annihilator of S.
2. Show that every minimal ideal of R is given by Re for an idempotent e (only tricky part)
3. Show that R = Re \oplus R(1-e) and that by induction+Artinianness that R = Re1 + Re2 +... + Ren
4. Show that Rei is isomorphic to S and conclude S = Hom(R, S) = End(S)^n , i.e. S is n-dimensional ove D

[primal beacon]

No these are called hyperelleptic curves

[open sluice]

you cannot write it as r^k for some k

so x != r^k for every k

in D_2n it turns out there are only two kinds of elements, those of the form r^k and those of the form sr^k
so it just means “take the latter”

[crystal turtle]

Since it's not a power of r, we know that it must be s times a power of r.
Every element on D_2n can be written as either r^k or sr^k. In this case, we know it's the latter

[open sluice]

the result as stated would not hold for powers of r

unless you’re in D_2 or something

[solemn dew]

guys do you know how to find roots in 4c?

for g(x)

i thought f(alpha) = alpha^3+alpha+1 = 0

hence we want to find a
g([Element of the field]) = 0

where f(alpha) = 0

or is it wrong?

[open sluice]

what's 4d?

[solemn dew]

c*

[dim widget]

@solemn dew they just want you to brute force it

Plug everything in and use the relation f(alpha) = 0 to reduce the resulting expressions to degree <= 2

If you get 0 it’s a root

[solemn dew]

thanks ❤️

and alpha^3 = 1 right?

[dim widget]

No alpha^3 + 1 = -alpha

[solemn dew]

oh right sorry

i meant alpha^4

but then it has to be -alpha^2-alpha

[dim widget]

Alpha^4 = -alpha^2 -alpha

Just start plugging expressions a + b alpha + c alpha^2 into g and see what happens

[solemn dew]

yes i found one root so far

[fleet pelican]

sagemath, not very easy to use though, you need some programming skill

[rocky cloak]

Okay, you can extend this to a proof.

Let m1, m2, ... be annihilators of elements of S. Then because R is artinian there is a finite set such that the intersection is the annihilator of S.
Then by Chinese remainder theorem R/Ann(S) = R/m1 x R/m2 ... x R/mn = S^n

[rocky cloak]

^

[void cosmos]

im sorry i couldnt reply i was in a league game

apologies sir

Then because R is artinian there is a finite set such that the intersection is the annihilator of S.

sorry im new to like the artinian mechanics

the chain stablizies

then what?

[rocky cloak]

So m1 > m1\cap m2 > m1\cap m2\cap m3 > ... This chain stabilizes so there is a finite number of maximal ideals whose intersection is the annihilator

[void cosmos]

m1 is the annihlator of {u1} and so on?

[rocky cloak]

Yes

[void cosmos]

oh yeah i see it

this number of maximal ideals is the "n" where it stablizies

like the ideals m1,m2,...,m_n

as in m_n+1=...

right?

[rocky cloak]

Exactly

[void cosmos]

okay so the interesction has to be ann of the whole deal okay

R/m1 x R/m2 ... x R/mn = S^n

how is that 😦

wait

wait how ;D

[rocky cloak]

R/m equals S

[void cosmos]

yea lmao

[rocky cloak]

You have a map R to S sending 1 to s

The kernel is the annihilator of s

So R/m1 = Ru1 = S, etc

[void cosmos]

yea so u quotient by the annihlator and use CRT

[rocky cloak]

Indeed

[void cosmos]

wait where did u use artin wedden here?

[rocky cloak]

I didn't, but it's pretty similar to the proof of AW

[solemn dew]

does anyone know how to find Galois group?

(c) and (d)

or where i can read about this?

[void cosmos]

such a simple proof

bro ur a genius

tysm

[dim widget]

@solemn dew it’s more an art than a science. When the degree is small you can use the fact that there aren’t that many groups of small order

The galois group acts transitively on the roots of a polynomial which generates it

So you’re looking for a transitive subgroup of S_3 in your case

Fortunately there are not so many…

The galois group is Z/3 if the conjugates of one roots are polynomials in that root

Otherwise it’s S_3

[south patrol]

in this case it is a little boring in that it ain't bad to compute the degree directly

and combine that with the embedding in S_3

[dim widget]

If you know what E is it’s not that bad

But some people don’t know what E is a priori

So they have to do b.)

[solemn dew]

oh

what is a transitive subgroup ?

i'll search it on the internet

thanks though

[dim widget]

One that sends any number from 1,…,n to any other

Via some element

[eternal pine]

I am not seeking a solution I just want to know if I am doing something wrong in here

The element 1 here should have the orbit of size 2 right, since the orbits of 1 should have {1,4}, and the stabilizers of 1 should also be 2 of them since I can have either an identity map or a map (2,6)(3,5)?

[delicate orchid]

I have no idea how to interpret this graph into a group action

is this a subgroup of S_6?

[eternal pine]

the group actions will be permutations

yes

Sorry, I should have mentioned it

I want to find the size of Automorphism over this graph

[formal ermine]

all hail the almighty groups

[delicate orchid]

ah ok you want Aut(gamma)

[eternal pine]

yeah, that's why I wanted to find size of orbit and the stabilizers

I am bit unsure about stabilizers

[delicate orchid]

why can't we also have (2,6) and (3,5) as seperate maps in Stab(1)?

ah right no edges between 3 and 5/2 and 6

[eternal pine]

yes

[delicate orchid]

I think I agree with you then

[eternal pine]

Ah alright, thanks!

Do you guys too generally go like "yeah just try it all"? I am self studying through some lectures and lecturers didn't specify anything like that

[delicate orchid]

I just started swapping stuff around in my head - there will probably be some clever way of finding the automorphisms using some theorem

[eternal pine]

I see I see

[delicate orchid]

nevermind it's NP-hard

[eternal pine]

oh yeah

I forgot

Thanks

but there could be special cases tho

[delicate orchid]

yeah there will be classes of graphs where you can say a lot

[eternal pine]

but then there is burden of remembering those cases

[delicate orchid]

like, I'm sure you could come up with a theorem about the automorphisms of say

paths, perfect groups, etc.

*graphs

force of habit

[eternal pine]

there's probably one for this one

[delicate orchid]

da cube

[eternal pine]

I asked someone to confirm and he just threw n! * 2^n where n is dimension lol

here it's 3

so

idk what that guy used

but then I found it has 6 stabilizers and 8 orbits

so same thing

[delicate orchid]

so as a cube this is all rotations + a reflection and then some weirdo maps on top

[eternal pine]

yeah

[delicate orchid]

like (24)(68), (13)(57)

interesting

well the n!*2 part comes from thinking about reflection groups of type B_n acting on R^n that bit is clear

[eternal pine]

I think orbits being 2^n is easily visualizable for me, the n! stabilizer part is bit ugh

[delicate orchid]

I assume the other maps are exactly like this, order 2 swappings of opposing edges

nifty

[dim widget]

@eternal pine if you view the 8 vertices of the cube as being at (a, b, c) where a, b, c are either +/-1 you can verify that the group is the group of 3x3 signed permutation matrices

[delicate orchid]

I knew it

also let me change my name this is really obnoxious lmfao

[dim widget]

Yeah wew basically already explained where the signs come from

[eternal pine]

I see I see, that's a cool way to look at it too

[delicate orchid]

I didn't really, I just kind of stated it

[dim widget]

Actually wew already explained everything but you may not know what a B_n coxeter group looks like lol

[eternal pine]

yes I don't know about B_n

I think I will just note that down and think more on that later

I do remember talking about hypercubes in computer network where instead of +/-1 it's just 0/1

and I was thinking along those lines

but I could not reach permutation matrices

lol

[dim widget]

Yes +/-1 makes the maps linear, otherwise they are only affine transformations which are annoying to write down

0/1 is better for transistors though 🙂

[eternal pine]

lmao indeed

Thanks!

[rocky cloak]

Another cool way to think about this. If m1 is the annihilator of both s1 and s, then Rs1 -> R/m1 -> Rs is an isomorphism mapping s1 to s. So Hom(R/m1, S) is the span of s1. Similarly Hom(R/m1 x R/m2, S) is the span of s1 and s2, etc. So s is in the span of s1, ..., sn if the annihilator of s contains the intersection of the annihilators of s1, ..., sn

[delicate orchid]

do you know about S_n being generated by cycles of the form (k, k+1)?

[eternal pine]

I do think you can decompose them to multiple cycles of possibly different lengths and not just 2

otherwise nope

and I don't know proof or something to this fact

it's second week into this

lol

[delicate orchid]

ah no worries

you can think of it as "you can rearrange anything however you want by just repeatedly swapping things with their neighbours"

[eternal pine]

ah right

[delicate orchid]

like bubblesort, since you're a cs dude

[eternal pine]

Yeah makes sense

[delicate orchid]

yeah so if you have S_n act on your basis vectors of R^n, these (k, k+1)s fix e_k+e_{k+1} where e_k is the kth basis vector, this action I believe is what gives you the n! stabilisers?

could be wrong on that

[dim widget]

They negate that difference, they fix the sum!

[delicate orchid]

gulp

it's ok we're working over R = F_2

[dim widget]

But yeah wolog the vertices of the cube are three letter binary words. Then permuting the digits is the stabilizer of the word 000, so it has 3! stabilizer.

Any vertex has the same stabilizer because the automorphisms act transitively

[delicate orchid]

oh my golly gosh this immediately leads into the interpritation of B_n as a wreath product how wholesome chungus

[dim widget]

@delicate orchid it is actually much better to work over F_2 unironically lmao

[delicate orchid]

cringe

C_m \wr S_n my beloved

[dim widget]

Then the flippy guys are just adding a word of the form 010 or 100 or 001 to our string without carrying

[eternal pine]

Oh yeah right, and that permutation needs to be applied on all of them for it to make sense

[delicate orchid]

oh I see what you mean by "working over F_2"

[eternal pine]

That's very nice

[delicate orchid]

yes it is much nicer

[eternal pine]

Damn,

[delicate orchid]

I'm too group pilled I didn't make the connection between C_2^n and F_2^n

[eternal pine]

that's quite a good pattern to spot

[dim widget]

Adding without carrying is I guess bitwise XOR while we are at it :p

[eternal pine]

yeah

I am studying ECC too apparently and I don't really understand GF 2^n construction yet, but nice to see everything devolve into XOR

[delicate orchid]

you know mod?

[eternal pine]

yeah ofc

[delicate orchid]

like, mod 2

yeah ok

it's just the integers mod 2

[eternal pine]

No I don't mean that

I mean how the irreducible polynomials are used to construct the whole field and thing

my textbook went into quite some depth about it, to construct those polynomials

[delicate orchid]

oh right like LITERALLY GF(2)

[eternal pine]

GF(2^n) I guess they call it

my bad

: P

[delicate orchid]

yeah you can get finite fields of order p^n for any prime p and any n

[eternal pine]

Yeah

Actually it asked the reader to prove themselves why finite fields should always have order p^n but I guess I will tackle that later

I moved on to study block codes

[delicate orchid]

there is a very elegant proof of that fact

[delicate orchid]

if you don't care about the specifics you can just pick any irreducible polynomial f of degree n, and then do F_{p^n} = F[x]/(f)

[eternal pine]

let's see, I will have to look more into it

[distant summit]

Maybe not best placed to ask, but if you have $3x \equiv 6 \mod 12$, does the solution have to be given $\mod 12$?

I.e. my initial approach was to guess $x=2$, so $x\equiv 2 \pmod 12$ are the solutions, but apparently $x\equiv 2 \pmod 4$ are the solutions

[cloud walrus]

Douglas

[solemn dew]

how do you know if it is Z_4 or Z_2xZ_2?

in b?

only thing i can think of is that Z_4 is not a field and Z_2xZ_2 is.

and since it is a field extension it has to be a field? i am not sure?

[delicate orchid]

Z_2xZ_2 is not a field

it's a group, and if you mean F_2xF_2 then that isn't even an integral domain let alone a field

the key difference between Z_4 and Z_2xZ_2 is that Z_4 has elements of order 4

so are there any automorphisms of order 4

[solemn dew]

oh right my bad

Z_m x Z_n is a field if gcd(m,n) = 1

[delicate orchid]

there are a lot of things wrong with that

one, no - the product of two (unital) rings is never an integral domain, let alone a field

secondly those are groups

Z_m x Z_n is cyclic if gcd(m,n) = 1

[dim widget]

@solemn dew you have to look at subextensions and use the fundamental theorem of Galois theory

A C4 extension has a unique quadratic subextension (do you see why?)

[solemn dew]

i don't

i'll look into it

thanks guys

!!

[delicate orchid]

I see!! miss!! miss I see!!! can I have a gold star!!!??!?!?!?!

[distant summit]

.

[solemn dew]

wait how is F_5 x F_5 not a field

[delicate orchid]

x = 2 mod 4 is just short hand for x = 4k+2, which are indeed the solutions to 3x = 6 mod 12

but I agree with you, that's a stupid way to phrase it

x = 2, 6, 10 mod 12 is far clearer

or hell, x = 4k+2 mod 12

[wicked patio]

What's the inverse of (1,0)

[delicate orchid]

or even more to the point, (1,0)*(0,1) = (0,0)

so there's zero divisors

this holds for any two rings

[delicate orchid]

hence

[wicked patio]

I contest that that's more to the point

[distant summit]

so $x\equiv 2 \pmod {12}$ would only give some solutions, but not all?

[cloud walrus]

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

[delicate orchid]

yeah

this is because Z/12Z is a yucky ring

and 3 is a zero divisor

[elder wave]

She wasn’t a kind stranger

[distant summit]

Would the Chinese Remainder Theorem would here? I tried it but something weird is happening.

The method I've been taught for computing these is to write a table with (left to right) a remainder colum $r_i$, a column for $N_i = N/n_i$ where $N=n_1 n_2 \cdots n_k$ (each $n_i$ is the modulus), for $x_i$ which is the inversion of $N_i$, and then a column for $r_i N_i x_i$, i.e. the product of the row.

Here I have $N_1=44, N_2=33, N_3=12$. Then $x_1=2$ because $44x_1 \equiv 1 \pmod 3$ is solved by $x_1\equiv 2 \pmod 3$; $x_2=1\pmod 4$ because of $33x_2\equiv 1\pmod 4$; and $x_3 \equiv 1 \pmod{11}$ because of $12x_3\equiv 1\pmod{11}$.

So this gives the last column as 352, 33 and 84.

Since $N=3\cdot 4 \cdot 11=132$, the solution is $x\equiv (352+33+84) \pmod{132} \equiv 73 \pmod{132}$. The problem is that while this appears to work for the second and third congruences, it does not work for the first, which is $x \equiv 2 \pmod 3$.

Where have I gone wrong?

[cloud walrus]

Douglas

[barren sierra]

3, 4, 11 are all pairwise coprime

so it should work

[delicate orchid]

the chinese remainder theorem should work

yeah

[distant summit]

Wolfram says 29 (so not 73)

[barren sierra]

agh it's been a while since I did this by hand

[distant summit]

Wait I found it

No clue how I got 352

That's completely wrong

It should be 176

Ok for some reason I doubled iot

Idk why tho

Yeah I get 29 now

Top tip: be careful with arithmetic

[glossy crag]

@distant summit

[glossy crag]

d is gcd(a,m) here

Your initial question was about linear congruences mod 12, right? Thought you could use this info.

[distant summit]

Yes

I was doing several practice questions for CRT

[pale elbow]

Hey folks, I wanted to do some review. What are some examples of posets and monoids in the wild?

[crystal turtle]

There are tons of examples for each, but here are a few "classical" ones
Posets: any power set (ordered by inclusion); Z, Q, R with their usual orders; positive integers ordered by divisibility (n ≤ m if n divides m)

Monoids: any group, of course, or the multiplicative structure of a ring with identity; the naturals with addition (including 0); the integers or naturals with multiplication

[tawny dune]

could someone explain this paragraph in dummit and foote to me? i dont really understand the reasoning in the underlined red sentences

[coral shale]

transpose it. . .

[tawny dune]

What do you mean transpose it

[coral shale]

nothing

[coral shale]

$\varphi : G/N \to H$

[coral shale]

$\varphi(gN)$ needs to be well defined

[coral shale]

There are multiple g's that make gN the same set.

to be precise gn for any n in N will work

gnN = gN

So if we define our output of the function in g

well, it needs to be always the same

That's the point being raised about independence

$\varphi(gN) \coloneqq f(g) = f(gn) = f(ng)$

[coral shale]

I suppose

[coral shale]

and you can probably show this f needs to be a homomorphism

hence f(gn) = f(g)f(n) hence f(n) = e

[tawny dune]

oh ok i get it i think

thanks

[coral shale]

oh im high,

this f is exactly that capital phi thing

[warm gate]

you can partially order the elements of a finite group by g<h when ord(g) divides ord(h) right, is that anything?

[coral spindle]

That doesn't make sense as written. You would conclude that equal elements are unequal there

g = g has ord(g) = ord(g) so this would mean g < g, but g = g so this cannot be.

[coral shale]

whats a superorder called

[coral spindle]

Do you mean preorder

[coral shale]

why not ord(g) < ord(h)

well tbh this isnt super in the way i meant unless =<

[coral shale]

I mean an order that contains another

the relations are a superset

[coral spindle]

A refined order

[coral shale]

ah ic

[coral shale]

ig this has problem in that elements with same order are equal as well oops

[coral spindle]

Idk it just seems like an order that you could impose but I don't see why

With things fixed as need to be, ofc

It doesn't play very nicely with the group structure afaict

[coral shale]

What could make sense for a chosen generating set which is independent

Is if one element divides another maybe?

but not quite with inverses hmm

oh if any generators are finite order it dont make sense

[coral spindle]

It does make sense, it just doesn't respect the group operations

Oh wait you were talking about a different order

as you were lmao

[warm gate]

meant less or equal

as in g≤h when ord g | ord h

[coral spindle]

Then typically you will have a <= b and b <= a when a =/= b, so this is still not a partial order

[warm gate]

right ic

idk anything at all abt preorders so that's no fun either

thx for sorting this out!

[coral spindle]

Yes as you say this would be only a preorder

The thing you can do with a preorder is observe that the condition a <= b and b <= a is an equivalence relation, so you can take equivalence classes. Then the preorder forms a true-blue partial order on the equivalence classes. Btw, this construction is a functor, adjoint to the forgetful functor from partial orders to preorders.

Anyway, the thing I've always seen people do with preorders is look at this induced partial order. But in this case, this partial order will just be (isomorphic to a) suborder of the divisibility order on naturals, so it's not terribly interesting.

[low wyvern]

Ay Im taking a course in abstract algebra this year and I would like some book recomendations if thats ok. I need them to be able to take me from zero knowledge to quite a high level

[formal ermine]

artin's algebra

usually your prof will recommend a couple of books

[low wyvern]

Yes but this course mostly only covers the group theory part and the recommendations seem a little basic. I would like to go to a more advanced level as I believe I am capable of it and also it will allow me to learn other areas of maths like alg top etc ...

[dim widget]

You don't need that much beyond group theory to learn basic alg top. On the other hand Dummit and Foote or Lang's algebra are good references. Lang's algebra is quite difficult but has every basic topic you could possibly want in it.

[low wyvern]

Want my maths knowledge to be spread evenly across analysis and algebra

[dim widget]

should take about a year to learn physics yeah

sounds good!

[sly rain]

What would be an easy/nice way to show that Z[X]/(X^5+2X-6) is not a field? The polynomial is irreducible so we know the quotient is a integral domain.
Now for example (X^5+2X-6) is truly contained in (X^5+2x-6,2) which is a proper ideal, thus not maximal thus quotient not a field.

If I now want to show that (X^5+2X-6,2) is proper, we want to prove that there are no polynomials such that 1=2b+g(X^5+2X-6).
I think if you reduce mod 2, you have 1=g(X^5) and you can argue by degree that its not possible.

What I dont fully understand is: Why does it mean that its not possible in Z[X] aswell?

[delicate orchid]

if two things aren't equal mod 2 then they can't possibly be equal in Z

an odd number can never be equal to an even number - in essence

[coral shale]

0 is even and odd...
jk

[rocky cloak]

2 is a pretty odd prime

[coral shale]

0 is even, odd, negative, positive, what else...

[formal ermine]

when the fuck did 0 become odd

oh wait nvm 0 = 2 * -1/2 + 1

right...

[open sluice]

0 is real, 0 is imaginary

[coral shale]

good.

[solemn dew]

i don't think 0 is positive nor negative

if x > 0 then it is positive, if x<0 then it is negativ

e

i think that is how positivity and negativity is defined

then 0 should be neither

[open sluice]

i think every number is positive because all numbers are good

[solemn dew]

isn't the answer wrong since any orbit is a subset of X, so then it is impossible for the orbit to be 72?

[chilly ocean]

That seems right to me

[solemn dew]

so my reasoning is wrong then?

[chilly ocean]

No, what you said seems right to me

[solemn dew]

oh okay

looks like the inverse to me

[chilly ocean]

I think that's inversion

[open sluice]

cursed notation

[solemn dew]

wdym

[coral shale]

its not cursed in plaintext

sotrue

whos gonna type out g^-1 = h^-1 blah blah in plaintext

in latex pretty cringe tho

[delicate orchid]

it's cursed irregardless

[coral shale]

no u

(gh)' = h'g' now what u gonna do

worried about g^-2? just do g'g'

[delicate orchid]

g''

[coral shale]

no.

[vague granite]

It’s a well known fact that a binary operation on a group is commutative iff the inversion map is a homomorphism

This gives you a sleek way of showing that rings are commutative groups as negation is a form of multiplication, which by the definition of a ring is a homomorphism

[chilly ocean]

what is the addition in a ring if not defined to be abelian

[coral shale]

well known fact

what even is the inversion map

[chilly ocean]

g -> g^{-1}

[coral shale]

oh thats like

uh

antihomomorphic

normally

[formal ermine]

a group is commutative iff the inversion map is a homomorphism

[coral shale]

what.

[chilly ocean]

thanks illuminator

[formal ermine]

no problem. however I think we need to mention that a group is commutative iff the inversion map is a homomorphism

[coral shale]

sod off.

[formal ermine]

now I have to look up yet another word

[coral shale]

i dont know how negation is defined to be homomorphic

in the ring axioms

[chilly ocean]

multiplication by -1

[coral shale]

or whichever ring axioms u wanna go by

Yeah

But the axioms I know are rings are abelian groups blah blah

[solemn dew]

any idea how to show b?

[formal ermine]

abelian

[coral shale]

[formal ermine]

maybe we need to mention that a group is commutative iff the inversion map is a homomorphism one more time

[agile burrow]

One way might be to consider that a normal subgroup is a union of conjugacy classes

[chilly ocean]

So then a normal subgroup of A^5 has to be some sum of the sizes of the conjugacy, and you rule out all sums except 1 and 60 by divisibility considerations?

[agile burrow]

That's right

[rapid junco]

does there exist an origin symmetric polytope in R3 that has an even amount of reflectional symmetries?
in even dimensions this is easy to find even amount of reflectional symmetries... take the hypercube, but I think in odd dimensions this is false.

[solemn garden]

What is an antihomomorphism

[chilly ocean]

f(gh) = f(h)f(g)

[rocky cloak]

Doesn't the tetrahedron have 6 reflectional symmetries?

I would think all the platonic solids work, but I haven't checked.

[woeful sage]

wait they changed the algebra channels

I see split up #abstract-algebra into undergrad/grad channels?

[ashen heron]

#advanced-lounge message

[vale locust]

is there a way to view the free product of two groups, say, G and H as a quotient of the free generated by the elements of both G and H

[solar vessel]

ofc

just map every element to itself

[delicate orchid]

yeah, you quotient out by the relators of G and the relators of H

but as daybroken said, since there's nice maps into the free product and the free group generated by the elements of G and H from G U H, there is a unique map induced by the universal property from the free group into the free product

[rustic crown]

equivalently, take free products of counits F(G) --> G and F(H) --> H to get F(G u H) = F(G) * F(H) --> G * H

[delicate orchid]

actual nerd ^

[spice whale]

that's not a ring axiom though

like

-x = -1x is such a weird axiom compared to x+y = y+x

-1x + x = (-1 + 1)x = 0x = 0

hm

0x + x = (0+1)x = x
0x = 0

hm i suppose you don't need a commutativity axiom

you do for semirings though

[vale locust]

what're the relators of G/H?

[delicate orchid]

I highly suggest you learn about presentations if you're at all interested in free groups

[vale locust]

I do know a bit about presentations

but I'm only interested in free groups as far as van kampen is concerned lol

[wraith cargo]

van kampen is all about presentations

[vale locust]

interesting

[topaz solar]

<@&268886789983436800> we got some spam

[open sluice]

does an amazon delivery team count as a group

[chilly ocean]

mods are asleep

[chilly radish]

Good morning

[elder wave]

sniped

[lethal dune]

[summer path]

,ti 102705166754549760

[cloud walrus]

The current time for shinmitsu is 08:00 PM (IDT) on Thu, 10/08/2023.
tubularcat is 7 hours behind, at 01:00 PM (EDT) on Thu, 10/08/2023.

[summer path]

[glossy crag]

I'm in the middle of learning about crossed products, have you got a preferred source?

[delicate orchid]

le semidirect product has arrived

[chilly radish]

I don't, sorry

I've used them in pretty specific circumstances so I don't have a good general resource

[rustic crown]

.<

[coral shale]

Didn't u go through that whole pdf or whatever introducing presentations. Dont say a bit, sully.

[delicate orchid]

word salad

[solemn dew]

any idea how to describe the elements of F(alpha) and F(alpha^2)?

[rocky cloak]

You don't need to describe the elements to solve the exercise

[solemn dew]

any hints then?

[rocky cloak]

F(alpha^2) is a subextension of F(alpha)

What do you know about the degrees of subextension

[solemn dew]

is it degree 2?

[dim widget]

No

[rocky cloak]

Could you maybe express [F(alpha):F] in therms of the other

[solemn dew]

if it is a subextension

(i don't know how you concluded that)

then [F(alpha):F] = [F(alpha):F(alpha^2)][F(alpha^2):F]

[dim widget]

Yass

[rocky cloak]

alpha^2 is contained in F(alpha)

Hence the subextension

[solemn dew]

right, because alpha * alpha is in F(alpha)

[dim widget]

And can you bound the first term on the RHS?

[solemn dew]

i don't know how to find [F(alpha):F(alpha^2)]

is it the degree of a polynominal with root alpha^2?

[dim widget]

It’s the degree of the smallest polynomial with coefficients in F(alpha^2) and root alpha

[rocky cloak]

A hint is that there is a priori two possible things it could be

[solemn dew]

x^2-alpha^2, one solution is alpha other is -alpha? two possibilities?

so then the degree is 2

hence [F(alpha):F(alpha^2)]=2

oh okay i think i get it

the degree is stated to be odd, but now it is even so we get a contradiction

so they must be equal to each other

u guys are good thanks 😄

[rocky cloak]

Yeah, so the point is that the degree is at most 2. It could be less, if that polynomial isn't irreducible