groups-rings-fields

[delicate orchid]

important thing is there's one orbit

[long oracle]

Quite possible that the word was translated from some other language

[rocky cloak]

Transitivus means to move across. The group action moves one element to another

[delicate orchid]

ah it comes direct from latin. Nice

[long oracle]

Yeah not that this has anything to do with actual math, just a curiosity

[viscid pewter]

'transit' means 'a movement' generally in english

[long oracle]

I think I should look when was this term introduced. Weyl introduced group actions, didn't he?

[delicate orchid]

I'm not sure you'll get a better explanation than jagr's

[long oracle]

Any free action would do that, wouldn't it?

Except the identity

Perhaps I'm splitting hairs here....

[delicate orchid]

no, a free action can definitely have more than one orbit (take <(12)(34)> acting on {1,2,3,4})

[rocky cloak]

A free action doesn't necessarily move across the whole set.

That's the special thing about transitive actions, it moves across the entire set.

[long oracle]

Ah now I get it

Thanks both of you

[stark helm]

then suppose that alpha is a root in Zp(t), so t=(q(t)/r(t))^p=alpha^p, and take the derivative on both sides such that r(t)=0

[stark helm]

are there any ideas about how to show alpha can not be element in Zp(t) by writing (x-alpha)^p=x^p-t?

[rocky cloak]

Well if
t = (q/r)^p, that means
t r^p = q^p

Left side has degree
pdeg(r) + 1 and right hand side has degree pdeg(q). So they can't be equal.

[stark helm]

one question: so you mean left side has degree p+1, right hand side has degree p? but why t has degree 1?(since t can not be anything)? and t is not similar to r, so why we can just add degree 1?

[rocky cloak]

When you multiply two polynomials you add their degrees, so the degree of
t * r(t)^p is 1 + the degree of r times p.

[stark helm]

so here t is a polynomial with degree one? and also must degree(r)=degree(q), since we only know that alpha should be of the form alpha=q/r where q and r are polynomials?

[rocky cloak]

We don't a priori know anything about the degree of r and q, but no matter what they are the degree of q^p is divisible by p, while the degree of t r^p is not

[stark helm]

since t=(q/r)^p, it seems that if difference of deg(q) and deg(r) is divisible by p, then t must have deg(t)=0 mod p.

[celest furnace]

You’re really trying to show that you don’t have equality in the polynomial ring F_p[t]

So remember what you know about polynomials and apply that here—F_p(t) is just Frac(F_p[t])

[stark helm]

probably i need explanation about why deg(t)=1?

[rocky cloak]

That's kinda just the definition of degree.

Like the degree of a polynomial is the largest n such that t^n apears

So 1 has degree 0, t has degree 1, t^2 has degree 2 and so on

[stark helm]

but it seems that t=(q/r)^p, so won't the degree of t related to the difference of deg(q) and deg(r)? Also what will t be, since you said t can not be anything before

[stark helm]

I think you want to express that F_p(t) contain elements of rational functions?

[celest furnace]

If I could offer some words of advice—q and r are polynomials in t (q = q(t) and r = r(t)) so what jagr has done is write r^p(t) t = q^p(t) and deduce a contradiction from there by arguing about degrees in the polynomial ring F_p[t]

[rocky cloak]

I think you need to take a few steps back here.

A polynomial is just a formal sum a0 + a1t + a2t^2 + ...

The t is just a formal symbol, it's not anything specific.

When we have a ring R, then we can define a ring of polynomials R[t]. And we can define the degree of a polynomial, and it behaves well.

You seem to want to make t into something, but it is just a formal symbol. and by itself it is a degree 1 polynomial.

[stark helm]

oK, I got it, really appreciate your response

[alpine plank]

has anybody actually used the 2nd isomorphism theorem in a problem lol? halfway through my algebra course and I don't remeber using it once

[low wyvern]

I just opened a thread to ask a rings question (also for future questions) would someone mind taking a look https://discord.com/channels/268882317391429632/1210670975054512188

[celest furnace]

Used it a bunch in Galois theory but not much for just groups in my first course except for one time

[south patrol]

it kinda comes up a lot in general ig

but uh i haven't seen it explicitly referred to much

i for one never remember which is 2nd and which is 3rd

[rocky cloak]

Different authors use different numberings, so that's understandable

[lethal dune]

very common in comm algebra, ex localisation and stuff

[celest furnace]

I believe hes referring to the $GH/H \cong G / G \cap H$

[cloud walrus]

rbmuk

[south patrol]

Yes that's what it means to me too

[stark helm]

Can someone explain what does sigma|X mean?

[rocky cloak]

Sigma restricted to X

[stark helm]

Then what will be elements of Sx? (ie. all elements from X?)

[stark helm]

it seems somewhat confused that automorphisms sigma can be mapped to elements from symmetric group

[rocky cloak]

Why do you say that?
Sigma permutes elements of X, hence gives you an element of SX

[stark helm]

does permute mean something like sigma(a)=b?but all sigma fixes elements from X and sigma(X).

[coral spindle]

a function sigma : X -> X 'permutes' X if it is a bijection.

That's all that means

[stark helm]

all sigma from Gal(E/F) seems to fix all elements from X.(ie. sigma(alpha1)=alpha1 and so on) so it seems to me that it all maps to identity?

[coral spindle]

Sigma does not fix all elements of E, it only fixes elements of F.

[rocky cloak]

Why do you think that?

[stark helm]

so we will have a map to elements of Sx now, since elements from X will not be fixed

[vocal fern]

can someone give me some pointers for parts b and c?

[formal ermine]

idk just go through the definition for b

[stark helm]

so we will have sigma(alpha_i)=alpha_j as automorphisms from E to E?

These. will be form of elements in Gal(E/F)?

[formal ermine]

for c u have two antihomorphisms, one of them being self invertible

and the composition of two antihomomorphisms is a homomorphism

[rocky cloak]

Sigma maps elements of X to other elements of X yes

[stark helm]

is there any argument in order to argue that sigma->sigma|X is an homomorphism?

[tender wharf]

not totally related but interesting fact for b: if you take psi to be the trivial homomorphism the equalizer would be the kernel of phi

I almost thought it was category theory for a second

[formal ermine]

so true

average derp moment

[tender wharf]

i'm not doing category theory tho

[formal ermine]

thats why i said average derp moment

sees "equalizer"
must be category theory

[tender wharf]

oh wait I am taking it as a course.

[rocky cloak]

Not much. It's just that composition is still composition when you restrict a function.

[stark helm]

I think it is hard to imagine an arugment here, it seems like intuition

[rocky cloak]

I mean you can easily turn that into a formal argument.

(st)|X(x) = (st)(x) = s(t(x)) = s|X(t|X(x)) = (s|X t|X)(x)

Hence (st)|X = s|X t|X

[stark helm]

OK, so since we have a injective homomorphism, we finally conclude that Gal(E/F) is isomorphic to the subgroup of Sx?

[rocky cloak]

Yes

[solemn kettle]

some pointers for 2 and 4??

[rocky cloak]

For 2, A/M is a field extension of K. So if A is finitely generated and K is algebraically closed, what can you conclude?

For 4, you could use Artin-Wedderburn. Or just assume it has d+1 maximal ideals and use CRT to get a contradiction.

[stark helm]

does two automorphism refer to map from a+bi-> a-bi and a-bi to a+bi respectively? And also can we modify it into map from z to z instead of z to conjugate(z)?

[rocky cloak]

You don't need K to be algebraically closed either, just that A/M is a finite extension.

[coral spindle]

Both of the maps you've written are equal

The identity map, sending a+bi to a+bi, is the map z |-> z that you describe

[chilly radish]

And the map sending a+bi to a-bi and that sending a-bi to a+bi is also the same map

[stark helm]

so the two maps are referring to z->z and z-> conjugate(z) right?

[astral fractal]

I've been working on this problem for a lil while now and am stuck https://gyazo.com/29db92a701bc682b8c9ec22990b49799

[tender wharf]

where are you stuck at

the hint is actually quite neat

[astral fractal]

im a bit stuck on how to proceed

So far I showed if X is the set of diagonals of the cube, and s_4 is the group action acting on X, then the stabilizer of some diagonal i corresponds to S_3

[tender wharf]

okay now you need to find some kind of homomorphism

[astral fractal]

between S_4 to S_3 right

[tender wharf]

yes

the stabilizer idea is good

[astral fractal]

im not sure how it'll help me in constructing a homomorphism

which is where im stuck

i guess it'll show by orbit stabilizer theorem that S4/H has same cardinality as S_3 which may help me later on

[tender wharf]

intuitively H is the rotations of some face right

[astral fractal]

yes

I said it corresponds to 1 Type 1 rotation and 2 Type 3 rotations of a cube

https://gyazo.com/5eaee9412324b1926a10962285b77864

if you label the diagonals 1,2,3,4

[tender wharf]

fix a diagonal and rotate it

what do you get

(visualize this as the cube staying fixed, but the diagonal rotates)

[astral fractal]

hmm

[tender wharf]

so for example the middle cube in your image

that red line

rotate the red line by 90 degrees

[astral fractal]

right

it would map to another diagonal

[tender wharf]

yep

do you have an idea now?

[astral fractal]

hmm not really

[tender wharf]

okay so you observe that to change the stabilizer from one diagonal to another we're simply rotating

[astral fractal]

change the stabilizer as in..

[tender wharf]

more formally, stab(2) = rotate by something * stab(1)

the left cosets

[astral fractal]

ohh

right

[tender wharf]

so what would a good surjection be

[astral fractal]

i'll think about it its not totally clear to me yet

[tender wharf]

just say one intuitively for now

think about the fact that S_4 is the disjoint union of left cosets of stab (i) for a fixed diagonal i yes

[astral fractal]

ohhh

and they're 4 disjoint cosets of this stab(i) right

[tender wharf]

yeah

[astral fractal]

lets see

[tender wharf]

so a surjection could perhaps be "throwing out" the rotation information

[astral fractal]

wdym by throwing out the rotation information

[tender wharf]

an element of S_4 is in one of these left cosets yes?

[astral fractal]

yes

oh

Does this have something to do with that S_3 permutations can be represented in S_4?

is that what you meant by throwing out rotation information

[tender wharf]

basically the intuition is that we can get rid of the information about which diagonal we are taking the stabilizer of

[astral fractal]

let me reread some of the past messages and see

[tender wharf]

i kind of have to go rn

[astral fractal]

all good

i think what you've given me is good help but ill have to think about it

thanks for the help

[spare isle]

Is the Z/4Z notation equally valid for the reals, i.e. R/4R? (My textbook extends what they call Z_4 to R_4, but my professor prefers the Z/4Z notation [since it conflicts with cyclic groups I think].)

[chilly ocean]

what's 4ℝ equal to

[cobalt heath]

Yeah you gotta need to look into how the extension works

[spare isle]

That's a good point

Would there be some equivalent notation other than R_4, or is it just completely nonstandard in the first place?

[prisma ibex]

yeah another standard notation is "0" lmao

[spare isle]

I presume that's a joke?

[prisma ibex]

no?

[spare isle]

So 0 means R_4...

[prisma ibex]

R/4R is the trivial group

[spare isle]

I'm referring to R_4, not R/4R

R/4R was my attempt at using some other notation for R_4

[prisma ibex]

what is R_4 lmao

[spare isle]

^

[prisma ibex]

can you actually define it

[spare isle]

set of [0,4) where when (for a,b in said set) a +_4 b > 4, a +_4 b = a + b - 4

[prisma ibex]

oh do you mean R/4Z

[spare isle]

well the issue is they also do 2pi

although I guess 2piZ is valid, just doesn't yield a set of integers?

[prisma ibex]

I mean sure you can talk about R/rZ for any r in R

[spare isle]

I see, thanks

[prisma ibex]

this is a standard way to write the circle group

[lime warren]

is there a particular reason why this channel isn't just called abstract algebra?

[chilly ocean]

it used to be called that and then some people decided having homological algebra and people just starting out with group theory in their first proof based math course was a bad idea

[south patrol]

Abstract algebra is a funny term to me

[chilly ocean]

you'd have questions on Tor and Ext and whatever followed up by people having trouble proving things are subgroups

[south patrol]

The abstract is pointless other than to say it isn't high school right lol

[lime warren]

yeah that's fair enough lmao

[cobalt heath]

May I have struggles in figuring out if specific Tor is a subgroup of another group

[spare isle]

I guess you can say that the terrm is abstract 😏

[spare isle]

Half the people I know think that it's just a normal algebra class

[peak root]

One thing is that this channel is named more specifically, unlike advanced algebra or algebraic geometry, as a result, every now and then someone will post some high school geometry or algebra questions.

There have been discussions on changing algebraic geometry channel names to more specific terms, like schemes-varieties-stacks or something.

[languid trellis]

This happens rarely and people get crucified for doing it so it's a non issue really

[solemn garden]

when did that happen?

[prisma ibex]

All the time lmao

[glad osprey]

I'm trying to come up with a proof that R/4Z is isomorphic to the [0, 4) group defined above (lets call it S). Can we define a homomorphism phi: R -> S, phi(x) = x mod 4, then notice that ker(phi) = 4Z, then use the first isomorphism theorem to conclude that R/4Z ~= S?

[void cosmos]

is R the real numbers?

not sure your map would make sense

if R was to be the real numbers

[glad osprey]

Yeah, R is the real numbers. I think mod can be defined for reals: x mod n is the (unique) r s.t. x = n * q + r where q in N and |r| < n

[rocky cloak]

That's usually the way to prove that something is isomorphic to a quotient yes. And indeed it works

[glad osprey]

So you rarely think about the cosets directly, you just find an appropriate homomorphism and use the isomorphism theorem?

[rocky cloak]

If the quotient group is isomorphic to one you're already familiar with, then yeah sure

[void cosmos]

wow i didnt know that lmfao

what would sqrt(2) mod 4 be

sqrt(2)?

[rocky cloak]

But in this case you're just considering cosets, like the definition of the group operation in a quotient group is just that you pick two representatives and add them together. Which is exactly how you defined addition in S

So you are just working directly with the cosets

[glad osprey]

I see, thanks! Final question: is R/pZ isomorphic to R/qZ for any positive integers p and q? So there is in a way only one circle group?

[dull ginkgo]

Yes!

For any real p and q

In essence you are “rescaling” the quotient cosets,

[void cosmos]

yeah

[glad osprey]

Thanks 🙏

[dull ginkgo]

If you want a challenge try to describe the isomorphism for real numbers a and b

not really a challenge

multiplication by a fixed real number is a group automorphism of R

[rocky cloak]

Level 2 challenge, prove that any continuous homomorphism R -> R is multiplication by a real number.

[glad osprey]

I think just phi: R/aZ -> R/bZ, phi(x) = (b/a)*x is an isomorphism

[warm wyvern]

how do I calculate <a,b>/N where N is the normal group generated by abab^-1?

and thereafter its abelianization?

oh nvm

its ablianization is rather simple

just ZxZ/<(2,2)>

which is ZxZ_2, I think

[warm wyvern]

maybe it's just $\bZ\star \bZ_2$

[cloud walrus]

DarQ!

[rocky cloak]

One way to calculate it is to notice that this is exactly the defining relation of a semidirect product

[warm wyvern]

I don't know what semi-direct products are

[delicate orchid]

<a,b| abab^-1 = 1>
It’s like Z ltimes Z upon

[fading field]

thewitnessing

[warm wyvern]

I should do some algebra sometime, it's kinda funny how little algebra i know at this point

ehhhh, it can wait a semester I bet

[fading field]

i don’t understand why this is related to semidirect products actually

[rocky cloak]

Hmm, so what kind of answer are you looking for when you say "calculate it"

[warm wyvern]

thank jagr tho!

[fading field]

i have seen like the construction of one via group actions, but is there a way to make some connection to this presentation?

[warm wyvern]

I'll look into them

[rocky cloak]

bab^-1 = a^-1

Means you have a semidirect product of Z with Z of the action a |-> a^-1

[fading field]

oh omg

ty

[warm wyvern]

I dunno, like, to some extent I think just giving the presentation of the group is enough of a calculation, right?

ig I wanted something more concrete, like Z\str Z_2 or smth

that doesn't really work tho, does it

[rocky cloak]

Well $\mathbb Z \ltimes \mathbb Z$ is pretty concrete, if you know that it means. Otherwise I guess you would just need to describe the elements and how they multiply.

[cloud walrus]

jagr2808

[warm wyvern]

that makes sense

[fading field]

if you’re trying to describe the elements use the fact that abab^-1 is a commutation relation, so every element has a form a^ib^j, then for describing multiplication you only need to say how elements of this form multiply

(you get that ba=a^-1 b, so just at every point where you have an a to the right of a b, use this relation)

[cobalt heath]

This is not isomorphic to Z * Z_2 right

[rocky cloak]

It doesn't have any elements of order 2

[cobalt heath]

Guessed it, but I cannot think of why right off my head.

[stark helm]

w means e^(2pi*i/3). And I want to know if the two automorphism I find is correct here

[warm wyvern]

chat

I'm about to prove snek lemma

@warm urchin

[dull ginkgo]

salamander lemma

[summer path]

Have fun diagram chasing

I remember having that as a homeowork problem and just being like I don't think I need these 10 points

[stark helm]

if there is a function f(x)=x^3-1 in Q[x], since f(x) is separable and f(x)=(x-1)*(x^2+x+1) , we know that E=Q(w). and [Q(w):Q]=|Gal(E/Q)|=2, so we have two automorphism from E to E. So what are the two automorphism: z->conjugate(z) and z->z?

[dull ginkgo]

I liked proving how diagram chasing works, I don’t like doing it :)

[summer path]

What is there to prove about how diagram chasing works, you just have a commutative diagram and just going around chasing elements

[rocky cloak]

There needs to be more diagram lemmas with fun names.

Like this one https://www.researchgate.net/publication/341029857_The_Anaconda_lemma

[dull ginkgo]

It’s fun to consider how there’s like an underlying homological lattice

[rocky cloak]

How do you mean?

[warm wyvern]

[warm urchin]

good thing you already seen it

[languid trellis]

I'm struggling with an exercise in Jacobson: Show that for n \geq 3, the centre of S_n is of order 1.

The centre is a subgroup, so that implies that the only element in the centre is the identity map.

My approach so far has been, take a and b to be maps in S_n. As they are not the identity, each permutes at least two elements of [n]. Say a permutes at least j and k, and b permutes at least k and m. Then a(b(k)) = a(m), whereas b(a(k)) = b(j), which may not be equal. I'm unsure about my idea of "permutes at least 2 elements", and "may not be equal", (this idea breaks for S_2 because there are only two elements to permute, thus our construction breaks).could anyone advise if (1) this approach works, (2) is there a nicer way to go about things, and (3) a hint if so. 🙂

pls ping btw

[rocky cloak]

If a is not the identity, then there is some i, j different such that a(i) = j. From there I would split into to cases either a(j) = i or a(j) is different from i.

In both cases try to construct a simple permutation that doesn't commute with a.

[viscid pewter]

you can always write a permutation (a b c ... z) as (1 a)(1 b)(1 c)...(1 z)

or the reverse depending on how it works specifically i forget

anyway that might help

or not

[languid trellis]

Case 1: a(j) = i. Define b(i) = 1, b(j) = i. Then ab(i) = a(1) \neq i (by bijectivity of a) whereas ba(i) =i.

Case 2: a(j) \neq i. Define b(j) = i, b(i) = j. Then ab(i) \neq i, whereas ba(i) = i

And this only works when n \geq 3 because then I have to make a third choice in Case 1.

thanks jagr

[rocky cloak]

Yeah, so you sort of implicitly assumed neither i or j was equal to 1. So in general you would just have to replace 1 with something different from i and j

[viscid pewter]

or you can say that the center is normal but there are only very few normal subgroups of S_n

lol

[languid trellis]

Make sense.

[languid trellis]

this feels very like presupposing my conclusion

actually

[viscid pewter]

it does a bit but i think it's fine

[languid trellis]

feels sus tho lol

[viscid pewter]

i thought about it for 5 secs and concluded it was fine and that's just the most reliable thing in existence

[languid trellis]

lol

[warm wyvern]

Like a year ago!

[winged void]

I have question regarding modulo

i do not really understand

what is the bar suppose to mean

can somebody help with that

[rose prism]

mod n

[winged void]

well then what is the diffrence between a bar and not a bar

because they both mean mod n

[rose prism]

one is not mod n

no they dont

[winged void]

can you give an example

[rose prism]

no

look at the screenshot you sent

they are defining k-bar as k mod n

idk how to give an example of this

there are examples already in the screenshot

3-bar = 1002-bar

etc.

but the integers 3 and 1002 are in fact not equal

[winged void]

becasue 3/3 = remainder(0) and 1002/3 = remainder 0

i see that but i mean

we write as well 11 = 8 mod 3

is that not the same

[rose prism]

youre asking whether “11=8” is the same phrase as “11 = 8 mod 3”?

[winged void]

im aksing if its the same as bar yes

[rose prism]

what

[winged void]

if its the same as saying 11 bar = 8 bar

[rose prism]

again…

yes

it is the same

[winged void]

sure then i get it eventhough i do not like the notation because it reminds me of conjugate of a complex number

[rose prism]

you can alternatively view the elements of Z/nZ as sets of integers with the same remainder when divided by n

but i suspect that isnt a particularly helpful perspective at this stage

[winged void]

i see

[vivid tiger]

The bar is the equivalence class

[formal ermine]

you will see this more than once: in general $\bar x$ for some $x \in R$ usually means the image of $x$ under $R \to R/I$ for some ideal $I$ of $R$

[formal ermine]

or the image under some other map, but usually under a quotient

[vivid tiger]

The 2bar mod n is the set of all integers that equal 2 mod n.

(I prefer the notation [x] for this)

[winged void]

what do you mean with second bar

[vivid tiger]

What

[winged void]

the notation [x] i understand that

you mean 2 bar mod n

[vivid tiger]

[2] = {2,2+n,2+2n,2+3n,...}

e.g. if n=2, [2]=[0]=evens

[formal ermine]

$\bar 2$

[vivid tiger]

thanks

I forgot about \bar

[winged void]

well this is then equal to [0] and [1]

am i right

[vivid tiger]

What is then equal to it?

Mod what?

[winged void]

mod n

[vivid tiger]

Mod 2, [0] ≠ [1].

[winged void]

\bar 2 = [0] and [1]

[vivid tiger]

Mod n≥2, [0] ≠ [1]

[vivid tiger]

Why is [0] equal to [1]?

[winged void]

no equal to each other

[vivid tiger]

What?

[2]=[0] using mod 2, because they're the same set.

[winged void]

but i mean that when i say 2\bar this mean actuall 2 mod n and this equal to the equivalen class [0] and [1] together

[vivid tiger]

[1] ≠[0] because they are different sets.

[winged void]

not equal to each other

[vivid tiger]

Oh.

This is incredibly unstandard.

"\bar{2}" means [2]

[formal ermine]

huh what

[vivid tiger]

ℤ/2ℤ = {[0], [1]}

If you use \bar{2} to mean ℤ/2ℤ, everyone will come after you with pitchforks.

[formal ermine]

oh

wth

Z/2, Z/2Z, Z_2 (dont use this), Z/(2) are all of the conventional ones

[vivid tiger]

Z/2 is bad. Me no like.

[formal ermine]

im a Z/2Z guy but if i have to write quick i do Z/2

[vivid tiger]

Z/2 would make me think "??? is this the coset 1/2 Z in something like Q?"

Z_2 is quick!

[winged void]

i see but i mean

that z/2 is the same notation as 2 bar

[formal ermine]

i think of 2 adics then

[winged void]

that is what i want to say

[vivid tiger]

fair

[formal ermine]

Eh

[vivid tiger]

I've literally never used the p-adics.

[formal ermine]

2 = prime ideal (2) in Z

problem solved

[formal ermine]

analyst spotted

[vivid tiger]

I've been doing more algebra recently.

[formal ermine]

Good.

[vivid tiger]

By that I mean learning algebra that I really should've known, like the isomorphism theorems beyond the first.

[formal ermine]

i forgot 50% of my alg 1 course already

i just look up stuff when i need it

[vivid tiger]

Learning what a modular lattice is helped me finally understand the lattice isomorphism theorem

[formal ermine]

the stuff ive been learning the past year doesnt need any group theory beyond like simple groups

and silly me can only remember so much

[vivid tiger]

imo Sylow's theorem should be replaced in introductory algebra classes with something else

[formal ermine]

my alg 1 course did too much

[winged void]

i get it now @vivid tiger

it just other notation

thanks

[] = bar

[south patrol]

p-adics are analysis

Jk

[formal ermine]

unfortunately...

[chilly ocean]

why unfortunately?

[chilly ocean]

And why jk? i think its fair to call p-adics analysis since they have a complete metric

[spare isle]

Would it be right to say that all groups with prime order have only one distinct form? (i.e. group table)

[chilly ocean]

Yes, this is called isomorphism type. There is an unique group of order p up to isomorphism

[spare isle]

Interesting, thanks!

Oh and I guess that would mean they would be cyclic

[coral spindle]

Indeed, the only groups of prime order are cyclic.

[chilly ocean]

Fun fact: the first product of two primes such that the only group of that order is cyclic is 15. And the first product of three primes is 255

(and the pattern stops there)

[sly crescent]

What is the next one?

[chilly ocean]

i forgot, i just remember its not of the form 2^n-1

[spare isle]

darn, I was hoping to find it in oeis

[sly crescent]

It’s 5865

https://oeis.org/A264907

[spare isle]

thanks!

[chilly ocean]

you're welcome

[spare isle]

Also I just realized that the pattern for number of groups for a given order is the first entry to oeis

https://oeis.org/A000001

[viscid pewter]

god bless

they have their priorities straight

hmm

is there a list of oeis sequences with just brief descriptions anywhere

or a next button i can click

[spare isle]

wdym by next

[chilly ocean]

you can just edit the link

[viscid pewter]

do i really have to edit the url every time

[spare isle]

if you are searching, you can scroll through the matches. If you just want to look at a random ones, you can modify the number in the url.

[chilly ocean]

you can make a script to automate it

[viscid pewter]

that's like six whole key presses

that's too many

[young hinge]

(5)

[spare isle]

just type in a random sequence and scroll through them on a single page then

[young hinge]

or one highlight and 2 keys

man look at this one https://oeis.org/A000004

[viscid pewter]

ctrl-l (2)
backspace (3)
number (4)
enter (5)

[spare isle]

https://oeis.org/search?q=1%2C2%2C3&language=english&go=Search

[spare isle]

actually you need to click ctrl-l then right then backspace 😉

[chilly ocean]

install qutebrowser and make a macro to automatically increment the url with a key press, its pretty easy

[young hinge]

yea they said 6 keys

[spare isle]

it's 7

[viscid pewter]

why would i need to click

[young hinge]

I consider ctrl + L one key since its done with both hands at the same time

[viscid pewter]

it's literally two keys, this is not up for debate

[spare isle]

lmao

point is, you forgot to click right

[viscid pewter]

why would i need to click??

[spare isle]

fine, hit

or punch, whatever you fancy

[viscid pewter]

what is 'hit'

[young hinge]

it

[viscid pewter]

what am i clicking on

[spare isle]

right key

ctrl-l highlights

[viscid pewter]

what?

ctrl-l goes to browser line

the url

oh!

yes

ok i see

[spare isle]

yeah and highlights (in 95% of browsers)

[young hinge]

lol

[spare isle]

I'm an advid ctrl-l haver, I know my stuff 😉

[viscid pewter]

i literally did not realise, clicking right arrow is literally unconscious

[spare isle]

fr

[young hinge]

told you

[viscid pewter]

anyway 7 is definitely too many

[young hinge]

5 + 2

[spare isle]

just create a user script, easy

[viscid pewter]

but that would be like

[young hinge]

thats a whole lot of overhead

[viscid pewter]

dozens of keypresses

do you see the problem here

[spare isle]

but think about all the seconds those minutes of work would save!

I'm starting to see where all my time goes

anyway, groups am I right?

[chilly ocean]

with qutebrowser you can make little macros for little stuff like this in like 10 keypresses

qutebrowser sucks tho

but so does every other browser

which is unfortunate

[spare isle]

yeah and I'm locked in due to my over-dependence on some plugins

like vimium-c

[chilly ocean]

life is unfair

well qutebrowser uses vim-like keybindings

[spare isle]

yeah but I just recall it not being that good back when I checked it out many years ago

[chilly ocean]

yeah

[spare isle]

and vimium c has a ton of functionality

[chilly ocean]

how come it doesnt have macros

[spare isle]

good question

[hidden wind]

it do be cute though, i use it sometimes when i need to do some simple web stuffs

[chilly ocean]

me too besti

[winged void]

is this a good proof

let $n$ be a positive integer
\newline
\textbf{(a) }If (a) and (b) are integers congruent modulo (n), so (a \equiv b \pmod{n}), shows that (\text{ggd}(a, n) = \text{ggd}(b, n)). Thus, (\text{ggd}(a, n)), for (a \in \mathbb{Z}/n\mathbb{Z}), is well defined.
\begin{proof}
$ggd(a,n) = ggd(b,n)$
\newline
we know that $a \equiv b \mod n$ that they have the same remainder or in other words the following holds:
\begin{align*}
n | a &- b \
a - b &= qn \
a &= b + qn\
b &= a +- qn
\end{align*}
we know that the $ggd(a,n)| a$ and $n$ and furthermore we know that a equals $a = b + qn$ so this means that the $ggd(a,n) | b + qn$ this subsequently means that $ggd(a,n) | b$ and that $ggd(a,n) | qn$. we also know that $ggd(b,n) | b$ and $ggd(b,n)| n$. this actually means that $ggd(b,n) | a+(-qn)$ or $ggd(b,n) | a $ and $ggd(b,n) | -qn$ and therefore $ggd(b,n) | qn$ this also means that:
\begin{equation*}
ggd(a,n) = ggd(b,n)
\end{equation*}
\end{proof}

ggd or gcd are the same its the highest common divisor

[spare isle]

Just curious, what does ggd stand for?

[winged void]

its in dutch

[spare isle]

ah

[winged void]

grootste gemeen deler

greatst common divisor

is my proof right ?

[winged void]

if you can help me with that (:

[spare isle]

I'll let someone else look at it in addition, but so far I notice that you don't define q and you use b x qn at some point instead of b + qn?

[winged void]

oh that was a mistake

but its +

[cloud walrus]

Mootje

[spare isle]

I'm not sure that gcd(a,n)|b+qn => gcd(a,n)|b, gcd(a,n)|qn generally?

although I guess you are utilizing that a === b and qn has divisor of n

still not convinced it holds like that though

[winged void]

you are right actually

[spare isle]

anyway I'll let someone else look at it more since I have to get going

good luck!

[winged void]

sure thanks

(:

[prime sundial]

can every derived series be written as the homology for some chain complex? as in H_n(C) = G^n. if there is such a construction, can anyone point me in the right direction

[prisma ibex]

yeah take the chain complex where all the differentials are 0 lmfao

[prime sundial]

okok nontrivial

but fair enough

[prisma ibex]

replace this with a quasi-isomorphic chain complex

idk this is silly

you have to ask a more refined question if you want a good answer

[cobalt heath]

This works?

Wow

[prime sundial]

isn't this what i am asking though? to find a quasi isomorphic chain complex to the trivial one

but yeah i'll try to refine the question and come back

[prisma ibex]

yes but you have so much freedom to do this replacement that you can't expect an interesting answer without asking something more specific

[prime sundial]

makes sense

[prime sundial]

maybe the point here is that derived series are not interesting as a homology then?

or well, i guess you could say that about any homology then

i'll try and make sense of what i want to ask

[vivid tiger]

nGroupoid's answer had nothing to do with the derived series.

Give me any sequence, I can find a chain complex with that as the homology.

[prime sundial]

that was the point of my second answer

well not answer

but my second message

this stemmed from trying to give an example of homology to someone who had seen derived series before, and to say that they had seen a specific example of a broader concept

[vivid tiger]

So, quasi isomorphism just means a morphism that induced iso on homology, right?

[prime sundial]

i believe so

[vivid tiger]

Are the derived series a homology group of something interesting?

lol that's your question

But hey, now I get it 🙂

[prime sundial]

well i kind of just assume they are interesting

it is more about, how can we express the derived series using homology in a meaningful way, but that is too broad a question because of what ngroupoid said

[vivid tiger]

Here's a paper that I don't understand. Sorry if it's useless.

https://arxiv.org/abs/math/0407203

[prime sundial]

i think it is also out of my depth but looks like it touches on what i was asking

thank you for the link!

[cobalt heath]

Ahhh

Right, you can take homology of the group

And do fun stuffs, I guess

It's quite interesting how you can consider a group as a discrete group, and work on it topologically.

[winged void]

can somone helo with the folloowin

\textbf{(a) }If (a) and (b) are integers congruent modulo (n), so (a \equiv b \pmod{n}), shows that (\text{ggd}(a, n) = \text{ggd}(b, n)). Thus, (\text{ggd}(a, n)), for (a \in \mathbb{Z}/n\mathbb{Z}), is well defined.
\begin{proof}
(a \equiv b \pmod{n}), so this implies that $n | a - b$ or this is also equal to:
\begin{equation*}
a - b = qn
\end{equation*}
so this means that:
\begin{equation*}
a = qn + b
\end{equation*}
we then want to prove that
\begin{equation*}
gg(a,n) = gcd(qn+b,n) = gcd(b,n)
\end{equation*}
suppose that $ggd(b,n) = d$ this implies that $d| b$ and $d|n$ this also implies that $d| qn$ so this means that $d | qn + b$ this means that d is the common factor of $qn+b$ and also of $n$ but we now want to show that $d$ is the gr

[cloud walrus]

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

Mootje

[winged void]

let $n$ be a positive integer
\newline
\textbf{(a) }If (a) and (b) are integers congruent modulo (n), so (a \equiv b \pmod{n}), shows that (\text {ggd}(a, n) = \text{ggd}(b, n)). Thus, (\text{ggd}(a, n)), for (a \in \mathbb{Z}/n\mathbb{Z}), is well defined.
\begin{proof}
(a \equiv b \pmod{n}), so this implies that $n | a - b$ or this is also equal to:
\begin{equation*}
a - b = qn
\end{equation*}
so this means that:
\begin{equation*}
a = qn + b
\end{equation*}
we then want to prove that
\begin{equation*}
gg(a,n) = gcd(qn+b,n) = gcd(b,n)
\end{equation*}
suppose that $ggd(b,n) = d$ this implies that $d| b$ and $d|n$ this also implies that $d| qn$ so this means that $d | qn + b$ this means that d is the common factor of $qn+b$ and also of $n$ but we now want to show that $d$ is the gr
\newline
\end{proof}
​

[cloud walrus]

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

[winged void]

im not really sure how to continue

[waxen hedge]

Look at #❓how-to-get-help

[delicate orchid]

Asking for help here is fine that’s the entire point of these channels

[vernal flume]

Now that you've shown that d is a common factor of both a and n, you need to show that d is the greatest common divisor by introducing a number "c" such that it's a common factor of a and n, then show c<=d after some manipulations. Ping me to elaborate or you can try doing it yourself.

[winged void]

i see

[winged void]

i proved in another way

but not really sure if its correct

this is what i did

@vernal flume im not really sure if this is correct

[chilly ocean]

i am facing difficulties with part i

i tried looking at the contrapositive

also as cosets are disjoint besides the first term in the union the rest of a_2H_2 to others must equal the union of rest of cosets of H_1

I don't understand pls elaborate

How does being the largest H_i work

If we don't know any of them are finite yet

[delicate orchid]

The problem is you could have [G:H_i] = \infty with each H_i also infinite if their argument is what I think it is

Hiding the evidence

Anyway this is a very strange question and it’s making my head hurt

[chilly ocean]

Same

[delicate orchid]

I think it’s because it’s a finite union of these cosets, that has to be where it would break

[delicate orchid]

But again, this

Its clear that at least one of the H_i must infinite or the union would be finite. If there are an infinite number of cosets for all of the H_i then since it’s a finite union of other cosets can we conclude via a disjoint argument that only a finite number of cosets of each H_i are contained in this union, giving a contradiction?

[delicate orchid]

Using this idea

[chilly ocean]

I was thinking of doing the same

I couldn't put it down rigorously

[delicate orchid]

Gee wilikers…

[chilly ocean]

First of all if that then we have a finite union on one side and infinite one on the other

[dire siren]

I don't think it is true that at least one H_i is infinite
Part (ii) seems to contradict this

[chilly ocean]

Huh

How

[delicate orchid]

Then we would have a finite union of finite sets, contradicting the fact that G is infinite

Once again I reiterate that an infinite subgroup can still have infinite index

Z as a subgroup of R

Or S^1 as a subgroup of C^\times if you want two of the same cardinality

[dire siren]

I see, but part (ii) explicitly asks to show that G is the union of the cosets of finite index

[chilly ocean]

It doesn't imply that each H_i is finite?

[south patrol]

I think I have a proof

But not totally sure

[delicate orchid]

Yeah we agree there

[dire siren]

no, but if you drop the infinite ones, you are left with a counterexample

[chilly ocean]

Index being infinite dosent guarentee uhh

[chilly ocean]

Tell

[delicate orchid]

I’m on da edge of my seat

[chilly ocean]

I'll leave the source here incase it helps
: Problems in Abstract Algebra by A R Wadsworth

From groups ch

[dull ginkgo]

I’m trying to intuitively understand this problem

[delicate orchid]

It’s saying we can write g = a_ih_i for all g in G and finitely many a_i on an intuitive level I guess

[dull ginkgo]

Assume the indicies are ALL infinite. With the index and a specified coset we can find an partition of G for each a_i H_i with that coset in it. I’m wondering if we can hijack the infinite partitions that each contain one of the finite union’s sets

choice jumpscare

[chilly ocean]

Uhh probably not a very important observation but if something say b is in a_2H_2 then it can be written as bH_2

We can keep getting new letters if we assume every index is infinite

So starting from a_1 we get there is some H_i with an infinite number of representatives of H_1 in it

Then that H_i can be written as those things multiplied with it

Again it's assumed of infinite index

[vernal flume]

If a|b and b|a then a=+b or a=-b not a=b

But yeah in this case since gcd is always a positive integer, a=+b

[winged void]

Sure I will write both prove because I really love it

[chilly ocean]

Ok this has been resolved

Thank you everyone

https://math.stackexchange.com/questions/2616574/neumann-theorem-about-finite-unions-of-cosets

The author is apparently very mean

[delicate orchid]

Let’s have a peep

Inclusion exclusion crossed my mind I must admit but I didn’t see how it would be useful

Cheeky little poset

[south patrol]

Okay I am unsure it worked lol

[dull ginkgo]

So it’s a finite union, but over sets which are each a part of infinite partitions of the first set

I’m wondering if you can inductively prove it by somehow combining cosets of two subgroups

[delicate orchid]

If we’re to go with the inclusion exclusion nonsense then |a1H1 U … U anHn| = |H1| + … + |Hn| - mobius function junk

[delicate orchid]

I’m still not seeing how this would show something to do with the index

Strange

[dull ginkgo]

Are you guys still stuck on (i) lol

[delicate orchid]

I want to see if I can come up with the argument used without just Googling it but I may need to relent

[dull ginkgo]

Assume G, infinite, is the finite Union of cosets with infinite order. Firstly each of these subgroups are proper otherwise the index would be 1

[dull ginkgo]

I have an idea

The intersection of two of the cosets is either empty or is a coset itself in the intersection of the two subgroups

However the index of this subgroup doesn’t necessarily need to be infinite

[low wyvern]

Quick question; does the theorem from group theory that says that Znm is isomorphic to Zn X Zm if gcd(n,m) = 1 hold if we replace the groups with rings?

[vale locust]

@warm urchin nvm, it's literally just long snek

[languid trellis]

I can't figure out this proof at all. I don't understand how we justify f_i > e_i for some i. Can someone help?

o(g) = |<g>| and exp G = the smallest positive integer e s.t. x^e = 1 for all x in G.

[dull ginkgo]

Chinese remainder theorem?

In general if two two-sided ideals $X$ and $Y$ of ring $R$ satisfy $X + Y = R$ then $\frac{R}{X \cap Y} \cong \frac{R}{X} \times \frac{R}{Y}$

[delicate orchid]

If h^og isn’t 1 it has to have a bigger order?

[cloud walrus]

Dyfunction Executive

[delicate orchid]

Hence one of the exponents in the prime factorisation has to be bigger

[languid trellis]

I thought this, but im not convinced the order has to be bigger

[low wyvern]

ok thanks

[delicate orchid]

Yeah ok fair enough

You can argue this by splitting h into p-parts for each prime but ill keep it basic. If h^o(g) isn’t 1 then clearly o(h) cannot divide o(g), so one of the prime exponents must be higher

If o(h) did divide o(g) then h^(o(g)) = h^(ko(h)) = 1^k = 1

[dull ginkgo]

It holds for Z/nZ and Z/mZ because Bezout’s lemma implies there exists a and b such that an + bm = 1, so it generates all of Z. Therefore the ideals <n> and <m> are comaximal, and their intersection is <mn> by divisibility. Thus Z/mnZ is iso to Z/mZ x Z/nZ

[languid trellis]

ahhhhhhh

[dull ginkgo]

The proof of sylow’s theorems are so fucking simple I’m surprised I didn’t figure it out

[delicate orchid]

They’re easy once you know them

[languid trellis]

i swear to god working with divisibility is the hardest thing known to man

the rest of the argument is nice tho

[rocky cloak]

I think I have an inductive argument on the number of distinct subgroups that apear:

||If there's just one subgroup that appears, then G is the finite union of cosets means it has finite index||

||If G is such a union with n distinct subgroups, let Hn be one of those groups. If Hn has finite index we're done. If Hn has infinite index, then not every coset apears, hence the is a coset of Hn contained in the union of the other groups. By left multiplication this means any coset of Hn can be written as a finite union of cosets of the other groups. In particular the cosets that apear in the decomposition can be. Thus G is the finite union of cosets of the other n-1 groups. So by induction one of those has finite index.||

[dull ginkgo]

Group action on groups of prime power order
HOW THE FUCK DID I NOT THINK OF THAT

[languid trellis]

um spoilers sweaty

[dull ginkgo]

Why spoilers if it’s literally nothing to do with the ongoing convo

[languid trellis]

spoilers for me in about 3 weeks time

u just spoiled that captain america dies

good job

[dull ginkgo]

Don’t worry, now it’s trivial

wait no I didn’t get to it yet in Jacobson lmao, I just started last night and finished the exercises in Chapter 0 lmao

[languid trellis]

are you just doing the exercises?

[dull ginkgo]

No,

I’m reading it and doing the exercises when I get to them

Well, I alternate between doing the even and odd exercises

[languid trellis]

gotcha

[dull ginkgo]

I read over all the problems, but if I don’t immediately know how to do a problem I do it

[languid trellis]

I need to review my lin alg but I really cba taking notes so I think I'll implement ur strat

until i get to stuff I'm looking at for the first time

[dull ginkgo]

I know a lot of the content in Jacobson already but it’s worded much nicer than D&F

[languid trellis]

i plan on using D&F as an exercise book

[dull ginkgo]

Exercises are much harder though. But because it’s not for a class I used abbreviations for my understanding

Overly formalizing it takes way too long often times if I understand the solution on my own

[delicate orchid]

the important thing is that you're able to come up with the ideas

[dull ginkgo]

That’s what I focus on

[languid trellis]

being able to write is probabyl important too

esp if ur a maths student

[dull ginkgo]

If a problem uses repetitive wording I use ditto marks lmao

[dull ginkgo]

Engineering lol

I am an engineering major

[languid trellis]

thats why i specified

[low wyvern]

how do u get good at this lol>

[delicate orchid]

practice and crying

lots of crying

[languid trellis]

1. dont do physics

[low wyvern]

Like once I start a proof I am generally ok, but I can not always start

[dull ginkgo]

The problem is trying to keep at the level that the textbook is at because I know more than the textbook has provided thus far

This one took a bit for me to process Lmao

[languid trellis]

i skipped this one

[dull ginkgo]

But I use abbreviations or English sometimes like for that one

[dull ginkgo]

Though once I slowed down and actually read it, it’s fuckin awesome that you can “generate” a equivalence relation

I really liked that exercise

I am excited to get to the Galoisy stuff

But my goal is to try to take an alternate approach to the fundamental theorem of Galois theory

[languid trellis]

meh im stopping after chap 3 ithink

i want to do reid commalg

[delicate orchid]

I read these as double cosets for like 15 seconds

yeah this is cool

[dull ginkgo]

I want to prove Artin’s Lemma by proving F is an **algebraicly finitely generated extension ** of F^G

For G Finite subgroup of Aut(G)

[delicate orchid]

reminds me of generating a simplical complex from a bunch of 1-simplices

[dull ginkgo]

and then invoke Neother normalization

To make it algebraicly generated as a module

Which the dimension being |G| comes from that

I have an idea

Let F be a field, G a finite subgroup of Aut(G) of order N, and F^G the fixed point subring.

Because G is Finite, F over F^G is an integral extension and thus algebraic. Any intermediate ring must be a field.

Fix some x not in F^G, and consider the valuation map from F^G[X_1…X_N] that sends the indeterminates to the orbit of x under G, necessarily of cardinality N.

I want to prove that the map must be epimorphic

I can assume there’s an intermediate field K that is the image, I wonder if I can prove that it can’t happen

[chilly ocean]

Are the modules $Hom(M\wedge M,N)$ and $Alt(M\times M,N)$ isomorphic? Where can I find a reference for this statement?

[cloud walrus]

username0000

[delicate orchid]

this should just be the universal property of the alternating product

which is in turn a combination of the universal props of the tensor product and quotient

[dull ginkgo]

WAIT A MOMENT

[delicate orchid]

waiting

[dull ginkgo]

The monoid-algebra F^G[G]

Which is a commutative ring

I don’t really know if it’s a field

FINITELY GENERATED MODULE WAIT A

[dull ginkgo]

Wait a second, I think F^G[G] is isomorphic to F

[rocky cloak]

As F^G[G]-modules, yes. But as rings no. Since F^G[G] has zerodivisors

[dull ginkgo]

Oh good point, what is a zero divisor you can think of?

[rocky cloak]

1 - g for example

[dull ginkgo]

what do you mean 1 - g

1e + 1g

[rocky cloak]

1e - 1g yeah

[dull ginkgo]

Got ya

Trying to see what I can do with that ring

It’s not hard to show it’s a quotient ring of R[X_1…X_|G|] via either universal property or a similar construction to the tensor product

[chilly ocean]

putting a backslash \ before your _ and * will prevent discord from turning them into italics and stuff

[dull ginkgo]

I am too lazy to do it on mobile except for when I want to escape asterisks for jokes

[dull ginkgo]

Do you think I can prove Artin’s lemma possibly by using this ring

[rocky cloak]

Idk, seems it would be easier to just use the usual linear algebra way

[dull ginkgo]

The usual linear algebra way seems like magic

I had a brain wave

Chinese remainder theorem

[tardy hedge]

what is combinatorial commutative algebra?

[raw badge]

What even are fields

[drifting plover]

Is there software that computes the square root of an element in a multivariate polynomial ring modulo a certain ideal?

E.g. I believe that -2x^6 - 40x^3 + 16 is a square in Z[x, y] modulo (3x^4 + 12x, -x^3 + y^2 - 1), but I don't know how to compute the square root

[tender wharf]

this question feels more suitable for #math-discussion tbh

otherwise look in your abstract algebra textbook for the definition

[cobalt heath]

One can instead ask, tho, like

What even is a field of one element

[indigo ridge]

im having a little trouble understanding the proof of this theorem, if im not mistaken an ideal must be a subring of a ring, so what this correspondence is doing is using the fact I think that F_q is isomorphic to F_p[x]/f where q = p^n and f is an irreducible poly thats just my guess and im pretty sure im wrong

obviously the whole in my logic is that F_q is isomorphic to F_p[x]/f not F_q[x]/f

where q = p^n

[cobalt heath]

I think this argument is not using what is q

That is, e.g. I do not see the mention of p, and that q = p^n.

[indigo ridge]

yeah but i believe my professor is skipping steps

[cobalt heath]

Why though?

[indigo ridge]

because his notes are full of him skipping steps lol

[cobalt heath]

Ah

I don't think it is using "F_q is isomorphic to F_p[x] / f".

Rather, it is defining a new ring. And an ideal on the ring

[indigo ridge]

the new ring being all the polynomials of the form c_0+c_1x + ... + c_(n-1)x^n-1?

[cobalt heath]

Yep!

Basically you can identify that with F_q[x] / (x^n - 1)

By like, regarding "x^n = 1"

[indigo ridge]

so if we regard x^n = 1 then it follows that this new ring is a subring

or subfield

[cobalt heath]

It's just a new ring

Not really a subring of F_q[x], because multiplication is different

Like, on F_q[x] there is no zero divisor, but F_q[x] / (x^n - 1) has an obvious zero divisor of (x-1).

[indigo ridge]

I meant a subring of F_q[x]/(x^n-1)

[cobalt heath]

Ah, do you mean C?

[indigo ridge]

well our new C the ones with the polynomial types

[cobalt heath]

Yeah C is an ideal to make/identify (whether C is a ring or not.. is debatable)

[indigo ridge]

so for the most part.... yes?

[cobalt heath]

Yeah except that C does not have identity ( 1 )

[indigo ridge]

so how does it follow though that part is not exactly clear to me

[indigo ridge]

.

[cobalt heath]

It is showing that C can be considered an ideal, not a subring.

Do you recall what is an ideal?

[indigo ridge]

yeah

[cobalt heath]

The line "Since a(x)c(x) = .." is showing that C is an ideal of the "polynomial" ring F_q[x] / (x^n - 1).

[indigo ridge]

yeah I got that part but what Im interested in is for C to be an ideal, if im not mistaken it must be a subgroup of the quotient group F_q[x]/x^n-1 first

this is also one of the conditions for C to be an ideal

or is that wrong?

[cobalt heath]

Yeah, it should be.

Indeed, there is an abuse of notation happening here, by identifying C as a subset of F_q[x] / (x^n - 1).

Which is basically the correspondence given above

[indigo ridge]

is it an isomorphism?

[cobalt heath]

Isomorphism as an additive abelian group, yeah.

I admit this kind of identifiation may look like "skipping steps", but this is quite common approach in algebra (and other math).

[indigo ridge]

I see, so from my understanding he is showing there is an isomorphism to this specific subset of F_q[x]/x^n-1, is that correct?

or saying rather

[cobalt heath]

The isomorphism is immediate

You can also notice that.. "(c_0, .., c_{n-1})" is not given an algebraic definition, so you can simply regard it as a polynomial.

[indigo ridge]

Ok so it's obvious but this is what he was saying essentially right?

then my second question would be, do we regard x^n = 1 because 1 is a root of x^n-1 or something its not really clear to me why this is the case tbh

[cobalt heath]

F_q[x] / (x^n - 1) happens to work out well in this case.

[indigo ridge]

?

[cobalt heath]

Since you want the ring to have dimension n over F_q, you want F_q[x]/(f) where deg(f) = n.

[indigo ridge]

yeah yeah ofc

so how does x^n = 1 happen?

[cobalt heath]

With the setting f = x^n - 1, multiplication by x basically gives "cycling" the code.

[indigo ridge]

yeah

[cobalt heath]

For instance, with F_q[x]/(x^3 - 1),
x (2x^2 + 3x + 1) = 2 + 3x^2 + x = 3x^2 + x + 2
(2, 3, 1) -> (3, 1, 2)

[indigo ridge]

mhm, yeah it gives this effect because x^n = 1, how does x^n = 1 be though?

[cobalt heath]

Wdym "how does x^n = 1 be"?

[indigo ridge]

how is it that x^n = 1

[cobalt heath]

Ah, the ring F_q[x] / (x^n - 1) is basically like

Polynomials, but x^n - 1 = 0. So x^n = 1.

If you recall modular arithmetics, you might understand this better

[indigo ridge]

oh right

yeah ok I see lmao

because we are modding by x^n -1

thanks @cobalt heath

[cobalt heath]

No problem!

[tribal niche]

if you give me an abelian group G, can you always find a ring R such that G (with the inherited product) is an ideal in R?

(i assume rings are commutative and have an identity)

i've found an example of an abelian group which can't be made into a ring, but i'm not sure if you can ever not make it into an ideal

[viscid pewter]

seems hard

not sure how to approach this in a general sense

you could probably do it for finite abelian groups, hah

[rotund aurora]

shouldn't that example work?

mmh if I understand you want G to be embedded in the ring where the operation on G is multiplication on the ring, right?

in that case G has an identity element, so the ideal would also be a ring

but maybe you meant G should be embedded with respect to addition in the ring

[rocky cloak]

You mean that the addition in the ring is the operation of G? If so you can make this into an ideal.

Consider the ring ZxG with pointwise addition, and multiplication (m, g)(n, h) = (mn, ng + mh). Then G is an ideal.

[chilly ocean]

Let $h: \mathbb{C}^{\times} \rightarrow \mathbb{C}^{\times}$ be defined by $h(z) = z^{4}$. Prove that $h$ is a homomorphism of multiplicative groups and find the kernel of $h$.

[cloud walrus]

gian

[chilly ocean]

I'm comfortable showing h is a homomorphism

But is ker(h) = {i} ?

[rocky cloak]

The kernel are those elements that map to the identity, so for which z do we have
z^4 = 1?

[chilly ocean]

got it

thanks

[rocky cloak]

(so the kernel should have 4 elements)

[jagged widget]

Let $G$ be a finitely generated group . Then show that for fixed positive integer $k$ there are only finite number of normal subgroup of $G$ of index $k$. How to do this can someone can give a hint ?

[cloud walrus]

spectrum

[vocal patrol]

how do i prove for 0 < x < y and 0 < u < v, xu < yv for x,y,u,v being real numbers

it seems rather trivial but I cannot figure out what axioms to use

[mighty kiln]

xu < xv < yv

[tribal niche]

relate subgroups to homomorphisms to S_n

[tribal niche]

thanks, that's clever

[vocal patrol]

Thanks got it

Concluding with transitivity then

[long obsidian]

I don't know where else I would ask this question. This might belong in linear algebra. I guess this is a question about modules

What is the basis of the complex numbers C as a complex vector space?

I think as a real vector space the basis is 1,i

But as a complex vector space it is (1,i)? I don't get that. Is that saying if I pick an a+ib I can always find an c+id so that (c+id)(1+i)=a+ib?

[rocky cloak]

The complex numbers is a 1 dimensional complex vector space. So any nonzero complex number will give you a basis. For example {1} is a basis, but also {i} would be a basis or {1-i} or whatever.

Not sure where you're getting (1, i) from...

[delicate orchid]

it should hopefully be clear that C is a 1 dimensional vector space over itself by that exact reasoning you just gave

[long obsidian]

Oh sorry for the confusion that was from some specific thing I was reading.

I guess if I have a z in C it's a basis cause if I have another w in C then (w/z)z=w. Yeah so anything works thanks

[delicate orchid]

(1,i) generates C as well it's just a wonky copy of C

[viscid pewter]

wait, what is ng in this context?

oh, is it just addition n times?

[cloud walrus]

HausdorffT1

[old hollow]

Can anyone give an overview of the proof for why (Z/pZ)* is cyclic?

[clear fiber]

The quotient ring R[x]/(x^2 + 1) defines the complex numbers.
The quotient ring R[x]/(x^2) defines the dual numbers
The quotient ring R[x]/(x^2 - 1) defines the split-complex numbers

This is blowing my mind. That’s so amazing to me that there is this single method to produce all these weird and exotic algebraic systems. Abstract algebra is an absurdly powerful tool!!

There’s literally infinite possibilities of quotient rings to just try out and see what they’re like

[fast sentinel]

Hey, I need help doing a proof if anyone is not busy.

The proof is Let $\sigma$ be a cycle. Prove that if $\sigma$ is even, then $\sigma^{-1}$ is even as well.

[cloud walrus]

Reactorge

[fast sentinel]

I started by saying that since $\sigma$ is even, it can be written as a composition of an even number of transpositions. I'm unsure of where to go from there

[cloud walrus]

Reactorge

[ivory trail]

what's the inverse of a bunch of transpositions

[fast sentinel]

the same transpositions right?

[ivory trail]

well what happens to the order

[fast sentinel]

they reverse

so like (1 2) (2 1)

[vivid tiger]

sgn: S_n -> Z/2Z is a group homomorphism.

looking at cycle or transposition decompositions will work just as well

[vivid tiger]

((12))^{-1} = (21) = (12)

((12) (23))^{-1} = (32) (21)

err

((1 2 5) (4 3) )^{-1} = ??

notice how the lengths and number of groupings are the same

but you can also try the abstract approach: sgn is a group homomorphism, and then you can figure it out

[fast sentinel]

I was eventually able to figure it out! The proof I did was

Let $\sigma$ be a cycle, and let $\sigma$ be even. This means $\sigma$ can be written as a composition of an even number of transpositions. Let $\sigma = (a_1 a_2 a_3 \dots a_{2n+1})$, where n is a positive integer. If we take the inverse of this, it becomes $\sigma^{-1} = (a_{2n+1} \dots a_3 a_2 a_1)$. Sincee they have the same elements and the same length, $\sigma^{-1}$ must be even as well.

[cloud walrus]

Reactorge

[vivid tiger]

This works

Well, I skimmed it,

actually I didn't look at it

maybe it doesn't work

[vivid tiger]

anyways I was gesturing towards the easiest way:

sgn(σ^{-1})=sgn(σ)^{-1}=1^{-1}=1 so even

(here, Z/2Z thought of as ({-1,1},×))

that is, it's a group homomorphism, so sign of inverse is inverse of sign is inverse of +1 is +1.

likewise the product of an even and an odd is odd

[fast sentinel]

We haven't done the sgn stuff so I have no clue what that means 😭

[fast sentinel]

Why doesn't it?

[chilly ocean]

I wonder if this is related to the classification of quadratic forms over R

[narrow wagon]

is ${\frac{1}{2}k : k \in \mathbb{Z} }$ the smallest subgroup of $\mathbb{Q}$ containg 1/2?

[cloud walrus]

Dubs

[coral spindle]

Yes

[narrow wagon]

what about the smallest subring?

[coral spindle]

What do you think?

[narrow wagon]

let me try

[coral spindle]

Just test the ring axioms

That's a good first step.

[narrow wagon]

sure

closure of multiplication

[coral spindle]

Indeed it is not a ring.

[narrow wagon]

(1/2)^{k} l?

[coral spindle]

Prove it

First show it's a ring, then argue why it's the smallest.

If you can't prove any part of it, try disproving that part.

[narrow wagon]

let me see

Is this proof of closure of + correct?

[coral spindle]

looks vaguely right

I'm not going to check in detail

[narrow wagon]

I picked a,b in S and assume k1<k2 to factor (1/2) outside

[coral spindle]

I assumed you had

[narrow wagon]

I proved S is an abelian subgroup of +

a+(-b) in S whenever a,b in S

finally it’s closed under multiplication

[pliant forge]

wolg* k1<k2

[narrow wagon]

yeah

since Q is a ring, i assume associativity and distributivity is gifted for it’s elements

@coral spindle

is that correct?

[coral spindle]

Of course

No assumption is necessary, this is trivial to prove

[narrow wagon]

makes sense

thank you

now i have to argue this is the smallest by inclusion

[coral spindle]

Hint: assume you have a subring containing 1/2 and construct some elements in it

[narrow wagon]

that’s how i started with

i generalised it to this above

I’m wondering something similar to this, to prove 4Z is contained in 2Z, i can say any element in 4Z is a= 2(2l), is some multiple of 2.

[coral spindle]

Yes, what's the confusion

[narrow wagon]

i mean can i argue something like that

[coral spindle]

You just did

[narrow wagon]

but with the above set S

(1/2)^k l

[coral spindle]

Just try the proof

[narrow wagon]

sure

[viral tiger]

No idea where to start

The hint is suggesting that i show that the group $H = \mathbb{Z}^_{p^m} \times \mathbb{Z}^_{q^n}$ has two or more elements of order 2 to contradict the fact that H is cyclic

[cloud walrus]

Noah Song

[viral tiger]

But I'm lost on how to show that

[rocky cloak]

Do you know any element of order 2 in (Z/p^n)^*? Or what about in (Z/3)^* or (Z/5)^*?

[viral tiger]

(2, 1) has order two in (Z/3)^* or (Z/5)^*

[rocky cloak]

Yeah, so 2 is an element of order 2 in (Z/3)^*

Do you also know an element of order 2 in (Z/5)^*? What about (Z/p)^* or (Z/p^n)^*?

[viral tiger]

4 has order to in Z/5 * , this leads me to guess that elements of order two in (Z/p)^* are p-1, but im not sure why that is. For (Z/p^n)^*, I dont any intuition for, except a random guess would be p^n-1

I'm going to think about why elements of order two in (Z/p)^* are p-1

[rocky cloak]

Well, try to verify your guess. What happens if you square it mod p^n

[viral tiger]

Ah wait

So $(p -1)^2 = p^2 - 2p + 1$, mod p is just one

[cloud walrus]

Noah Song

[viral tiger]

So the element p-1 done twice is 1 mod p

but for (Z/p^n)^* , p^(2n-2) mod p^n is p^2 * p^-2 mod p^n, which is just 1 mod p^n

Nice

So for $H = \mathbb{Z}^_{p^m} \times \mathbb{Z}^_{q^n}$, we can choose two elements of order two:
(a, b) where $a = p^{m-1} , b = 1$
(c, d) where $c = 1, d = q^{n-1}$

[cloud walrus]

Noah Song

[viral tiger]

Therefore, the group is not cyclic.

Thank you so much !!

How difficult is this problem? I don't know if I'm meant to go into math or not. Is needing help on something like this a bad sign

[rocky cloak]

I wouldn't classify it as a particularly hard problem, but I wouldn't say needing help is a bad sign.

How to tackle problems is a skill you build up, and can take time. Everyone struggles in the beginning.

[coral spindle]

Its difficult is based on how comfortable you are with groups. If you are very experienced, it's simple. If not, it is difficult precisely because it requires identifying several relevant facts.

[rocky cloak]

Also I didn't help very much, so that shouldn't take away from what you accomplished.

[rocky cloak]

Seems this is not quite right yet.

p^(2n-2) = p^(n-2)p^n, so is 0 mod p^n. You seem to have reduced the exponent modulo n, or something to that extent.

[long obsidian]

Hey I'm trying to show $\mathbb{Z}/ n \mathbb{Z}\otimes \mathbb{Z}/ m \mathbb{Z}\cong \mathbb{Z}/gcd(m,n)\mathbb{Z}$. I know there is a bilinear map $f:\mathbb{Z}/ n \mathbb{Z}\times \mathbb{Z}/ m \mathbb{Z}\to \mathbb{Z}/gcd(m,n)\mathbb{Z}$ defined by $f([x]_n,[y]m)=[xy]{gcd(m,n)}$. So by the tensor product universal property there is some $f':\mathbb{Z}/ n \mathbb{Z}\otimes \mathbb{Z}/ m \mathbb{Z}\to \mathbb{Z}/gcd(m,n)\mathbb{Z}$ such that $f'\circ \pi=f$. It's not clear to me why this map is injective. For instance I can consider $6([3]_4\otimes [1]_2)+5([2]_4\otimes [1]_2)\in \mathbb{Z}/4\mathbb{Z}\otimes \mathbb{Z}/2 \mathbb{Z}$ but $f'(6([3]_4\otimes [1]_2)+5([2]_4\otimes [1]_2))=6([1]_2)+5[0]_2=[6]_2=[0]_2$. Is this not the right tensor product?

[cloud walrus]

HausdorffT1

[delicate orchid]

6(3 \otimes 1) = (3 \otimes 6) = 3 \otimes 0 = 0 \otimes 0
5(2 \otimes 1) = 10 \otimes 1 = 0 \otimes 1 = 0 \otimes 0

of course 0 will map to 0

my hint for this is bezout's lemma

[long obsidian]

Ahh okay thank you!

[coral spindle]

[That's so wild](#advanced-algebra message)

Are you and TStepped in the same class or sth

[delicate orchid]

ip logger MODS

[coral spindle]

🔪 I NEED THE IPs

[delicate orchid]

oh you recommended bezout's as well lol

[coral spindle]

Extremely similar question, exactly the same hint

[viral tiger]

Haha yeah it should be p^n - 1 instead?

Thanks again

[long obsidian]

Wait what about $[1]_{12}\times [4]_8 \in \mathbb{Z}/12\mathbb{Z}\otimes \mathbb{Z}/8\mathbb{Z}$ is this actually zero 0?

Cause from that map $f'( [1]_{12}\times [4]_8)=[4\cdot 1]_4$ is this just another cause where it's actually zero

[cloud walrus]

HausdorffT1

[coral spindle]

BEZOUT!!!!!!!!!!!!!!!!!!!!!!!!

USE IT!!!!!!!!

4 = 12 - 8, plug and chug

[delicate orchid]

smokes crack and dies

at this point I'd just start working with things that look like a (x) b it makes my head hurt less

[dire siren]

I think 1 (x) 4 = 1 (x) 3*3^(-1)*4 = 12 (x) 3^(-1) = 0

anyway, 3^(-1) is just 3 in Z/8Z, but I wanted to emphasize that it doesn't have to be calculated

[rocky cloak]

Guess it's worth noting that this aproach uses that 3 (=12/gcd(8, 12)) is relatively prime to 8, thus wouldn't work if you wanted to show something like
1 (x) 6 being 0 in Z/12 (x) Z/18

[long obsidian]

I'm not really sure how to apply bezout. If I have something in the kernel $f'([a]_n\otimes [a']_m + [b]_n \otimes [b']m)=[a\cdot a'+b\cdot b']{gcd(n,m)}$ then all that tells me is that $a\cdot a'+b\cdot b'=k\cdot gcd(n,m)$. I don't see how bezout tells me anything or even how to apply it. Is it something like $l\cdot gcd(a,b)=k \gcd(n,m)$ then you have the division of integers $gcd(n,m)|gcd(a,b)$ then this is enough to say that the original $([a]_n\otimes [a']_m)+([b]_n\otimes [b']_m)$ is trivial since a and b being divisible by gcd(n,m) means each term cancels out? I genuinely don't know basic arithmetic or even gcds

[cloud walrus]

HausdorffT1

[coral spindle]

What is the statement of Bézout's lemma?

This is not a trick question. I am asking you to recall the statement.

[long obsidian]

That if you have n,m integers there exists other integers r,s so that rn+sm=k gcd(n,m). Does this even hold for linear combinations with more terms i.e. kgcd(n,m,p)=rn+sm+tp

[coral spindle]

Yes it does apply to more terms, but this is irrelevant.

This is not the correct statement.

k has been added here for no reason

[long obsidian]

Woops

[coral spindle]

The statement is that for all n,m there exist r, s such that rn+sm= gcd(n,m). That's it.

Now I will tell you how to apply it: substitute rn + sm everywhere.

See what that gets you.

[indigo ridge]

if a is in the field F_2^m

I mean $F_{2^m}$ FYI

[cloud walrus]

jayz

[indigo ridge]

then is $a$ an m-tuple with elements in F_2?

[cloud walrus]

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

[coral spindle]

Uh depending on the construction, I suppose so yes

but this is totally unhelpful

[indigo ridge]

what do you mean?

[coral spindle]

It's not helpful to think of them as m-tuples

This is like

A good example is the construction of the integers. Technically integers are equivalence classes of pairs of naturals. Is that a helpful view? No.

[indigo ridge]

this is why though they ought to be m-tuples since the matrix here is 2m x 2^m -1

[indigo ridge]

fair enough

[coral spindle]

OK yeah sure I see why you'd say this.

F_{2^m} is, in particular, an F_2 vector space

So choose a basis for it. Tadah, you have an m-tuple of elements of F_2 describing some element of F_{2^m}

This is what's happening implicitly, I reckon.

[indigo ridge]

OK I see makes sense

[crystal vale]

If I have a non-empty set X and G is a group then the set {f | f:X->G } is a group, right?

[tardy hedge]

Algebra is the most epic math

I think that is just a fact

[barren sierra]

what is your operation on this set?

if you claim it's a group, give me an operation

and show the group axioms are satisfied

[reef trench]

What is the significance of learning about PIDs, IDs, UFDs and EDs?

I can accept why we need to learn about groups, rings and fields, as they come up quite a bit in mathematics

but those domains? I am not sure. The only ones I can really come up with are either Z, or Z[X], or a subring Z(D) of a field extension Q(D)

[crystal vale]

f•g(x)= f(x)°g(x) where ° is a group operation of G.

Then if I want all x belong to X,
f•g(x)=f(x) then take g(x) be an identity element of G.( g maps every element of X to identity element of G).

Hence g is the identity element of the new Group.

Similarly, associativity comes from °, and inverse also.
For closure since ° is closure so new set will be closure under •.

Is it right?

[barren sierra]

looks fine to me yea

[viscid pewter]

can you explain in more detail how you get an inverse

[barren sierra]

but yea ^

explain how to get an inverse

[crystal vale]

I don't know a lot, but you can learn from that how factorisation works, how the irreducible element and prime element are related to each other, how you can write an element of ring in terms of irreducible elements.

[viscid pewter]

i don't think g(x) mapping everything to the identity of G is an identity element

actually no it is nvm

[crystal vale]

Since here my identity element is g which map every element to the identity element of X.

So let x belong to X for f(x) I take g(x) which is inverse of f(x) in group G so for all x belong to X I take inverse g(x) is inverse of f(x) in G. Hence, f•g is identity element of new group.

[barren sierra]

ok yea looks fine to me