groups-rings-fields

[tacit hemlock]

so then im double counting

wait so, i can simply find ANY generator of Cn and proceed?

not necessarily 1?

but 1 is always the most obvious one for Z_n, so people pick it out of ease?

but, say, for Z_7

I could also use 3?

since gcd(3,7)=1

[rocky cloak]

Yeah, you could if you wanted to. But usually picking 1 is the easiest

[toxic zephyr]

oh yeah duh. 2am brain. strictly nonnegative doesn't even really mean anything lol

[vapid stone]

Hi! I want to show that if given a ring $\mathcal{Z}$ has the property that for any ring $R$, there is a unique homomorphism $\mathcal{Z} \to R$, then $\mathcal{Z} \cong \mathbb{Z}$. I have already shown that the property holds for $\mathbb{Z}$, and that the unique morphism $\mathbb{Z} \to \mathcal{Z}$ is injective, but I'm stuck afterwards. I think this question is a primer for category theory ($\mathbb{Z}$ would be the unique initial object in the category of unital rings), but once again I'm unsure of how to proceed. Any ideas would be appreciated, thank you!

[cloud walrus]

LF

[next obsidian]

What happens when you compose the map from both Z’s

And can you think of another map with the same domain and codomain

[vapid stone]

Ohh right, the composition would be the identity map?

[next obsidian]

Swag

You now proved that objects satisfying universal properties are isomorphic via unique isomorphism

Now you’re ready to glue things over affine open covers of schemes

🫡

[vapid stone]

thats so cool, thank you!

[chilly ocean]

How is that related to universal properties?

[next obsidian]

You need to prove a cocycle condition which basically says when you compose two maps it gives a third one. All of these maps are coming from a universal property and thus you get the equality for free

The way it goes is “you want a thing that solves some universal property over X”

You show it exists over an affine scheme

And that when you restrict to open subschemes, that solves the universal property

So you cover X with affines, take the object over each affine. On the intersection you have two things solving the same property so isomorphic

On triple intersections, you have the compatibility of the isomorphisms from the uniqueness part

Thus you can glue the things

[summer path]

what does chmuwu imagine topological glue to taste like

[next obsidian]

Yum yum

Like mmmmm

Maybe like

Sugary syrup

[summer path]

tubu imagines topological glue to be like gingerbread frosting

[rustic crown]

det thought glue tasted somewhat salty

[summer path]

det tasted glue?

[rustic crown]

dun remember

det used to do a lot of papercraft when det was tiny, so might have

[fading field]

bruh

[dull ginkgo]

hi smay

[prime sundial]

is the subspace lattice of an arbitrary vector space (can be uncountable) complete?
i can see that there are arbitrary meets since arbitrary intersections will be subspaces, but does the idea of the join of W and V being W + V extend to an uncountable collection of subspaces?

[rustic crown]

yee

given {V_i} you can define ∑ V_i as the union of all finite sums of V_i

[prime sundial]

alright i see. thank you

[prime sundial]

with regards to that, it is evident that T(V + W) = T(V) + T(W). i assume the same argument about the uncountable sum being made up of finite sums from the V_i can be used here to show
T(∑ V_i) = ∑ T(V_i)

T being a linear trasnformation, should have clarified

my goal is to show that an injective linear map induces a homomorphism of the subspace lattices for the domain/codomain

[rustic crown]

yep it works

equivalently you can define ∑ V_i as the image of the map ⨁V_i --> V

[prime sundial]

great, thank you det

[vivid tiger]

really you should ask what it's like to huff topological glue

[vestal marten]

For problems such as this, is it reasonable to use a calculator to obtain the prime factorization / modulo?

Maybe I'm missing something, but 6545 was not factoring nicely in my head

[barren sierra]

sure but like idk divide by 5 and move on with your life

the factorization is "obvious" if you go in a step by step manner

[chilly ocean]

6500 + 45

[dusk whale]

Im learning about composition series and I'm a bit confused why the condition that the composite factors are simple is necessary

[tawdry plover]

,rotate

[tawdry plover]

For the 8th bit a linear functional of such kind is completely determined by what it maps the countable basis of V to

So basically this question asks for a basis of sequence space of Q right?

Which shouldn't be countable because it has uncountable cardinality and if it had a countable basis then we can union the vectors as countable unions?

[rocky cloak]

Well the condition isn't "necessary", that's just part of the definition.

If you don't require the factors to be simple, it's usually called a filtration instead.

The main reason to care about composition series as opposed to arbitrary filtrations, would be the Jordan--Hölder theorem. Saying that the composition factors are unique.

[fleet citrus]

why you can union the vectors as countable unions?

[tawdry plover]

Every vector is a finite sum linear combination of basis elements

Oh wait youre right

I can't guarentee that by fixing an N for size of the sum

[tawdry plover]

How are set of all rational polynomials countable

I can just mimic the proof right @fleet citrus

[hidden wind]

why is the kernel (\Gamma_7) interesting here? Wouldn't that just be all (I +7\mathbb Z I) ?

[cloud walrus]

rødbet

[agile burrow]

Not quite. For instance, you'd have matrices like [1 7 \ 14 1] in the kernel

Part of why the kernel is interesting is because SL(2, Z) doesn't act freely on the upper half plane, but Gamma_7 does.

[crystal vale]

Hint, let G be a group, how can I show that S= {xyx^(-1)y^(-1) | x and y belong to G } this set is closed under operation?

Let a= x_1y_1x_1^(-1)y_1^(-1) be an element of S and b= x_2y_2x_2^(-1)y_2^(-1) then for ab what is my new x and y such that ab=xyx^(-1)y^(-1) ?

[untold garnet]

It is not closed, though…

[crystal vale]

But it makes a subgroup of G , so it must be closed

[untold garnet]

S is not a subgroup of G (in general).

[crystal vale]

Counter example?

[crystal vale]

Isn't the commutator subgroup of G?

[untold garnet]

Look up the definition of the commutator subgroup.

[untold garnet]

I don’t know one. I think it’s a bit messy.

[crystal vale]

41

I think S will be the commutator subgroup of G

[untold garnet]

Pay close attention to the notation.

It’s not a set they’re using.

[hidden wind]

woops right, and thank you i was not aware of the free action thing 🙏

[crystal vale]

x^(-1)y^(-1)xy and xyx^(-1)y^(-1) are not identical ? Means they generate the same set, right?

[untold garnet]

No, this is unimportant. The point is that N is not the set containing the x^{-1}y^{-1}xy.

[crystal vale]

Oh means S is not a subgroup, N generated by set S

[tawdry plover]

Generated by and set is not the same thing

[summer path]

the smallest group with commutator set not a subgroup has order 96

hi det

[rustic crown]

hi tubu

[delicate orchid]

this is an insane fact to know, do you know the SmallGroup number?

[summer path]

i only know it because there was a week where people wouldnt stop talking about it

[dull ginkgo]

Group proofs suck balls

[delicate orchid]

I must've missed that week

definitely gonna remember it now though, that's cool

[summer path]

(irl)

[delicate orchid]

I still missed it!

[rocky cloak]

Apearantly there's two

They have GAP ids (96,3) and (96,203) respectively.

[delicate orchid]

:thecosmoshumswithatunemostsweet: thank you jagr...

[summer path]

what's a SmallGroup number

[delicate orchid]

there's a database of all groups of order less than uhhh

2000...??? without 1024?

I forget

and basically all computer algebra systems have it inbuilt so u can access small groups really easily

[proud spindle]

2000 iirc

[delicate orchid]

found the full list

craaazzyy

[summer path]

except 1024

[delicate orchid]

1024 is most of them tbf

[chilly ocean]

2-groups are crazy

[summer path]

two many of them

[delicate orchid]

there's similar exponential growth for all p-groups it's just that 2^10 is WAY smaller than 3^10

[proud spindle]

idk why powers of 2 have so many group structures (up to isomorphism)

[rocky cloak]

1024 had to be skipped, as there are additional 49 487 367 289 nonisomorphic 2-groups of order 1024

[proud spindle]

WHAT

[delicate orchid]

like, if we ever classify the p-groups of order p^8 it'll be approaching the sheer computational effort of the classification for finite simple groups

[delicate orchid]

yeah I wasn't kidding, 1024 is almost all of them

[chilly ocean]

Have the p^7 ones been classified?

[delicate orchid]

they have, remarkably

[proud spindle]

i remember trying to check how many groups there were (up to isomorphism ofc) of order 2^8 and i didn't know about p-groups being the way p-groups are so i was astonished by there being around 56k of them

[delicate orchid]

impressive computation

[chilly ocean]

What about p^n q ones?

For how high n have those been clasisfied

[proud spindle]

i didn't compute a thing i just found it on a wiki lel

same number shows up on OEIS

[delicate orchid]

not sure, but their behaviour is well known

so I imagine it's not hard (thought wise, it'll be a fuck ton of computation) to do it for p^7 using the existing classification

[proud spindle]

iirc the highest power of 2 that's been calculated was 2^10 (for group structures)

[coral spindle]

Yes

[delicate orchid]

this is very easy to verify

in fact, you verified it on the first line

[crystal vale]

Then for the question, Let G be a group. Let H be the subgroup of G generated by {x^2 | x be element of G}. Let K be the commutator subgroup of G then K is subset of H.

So if I can write any xyx^(-1)y^(-1) as a sequence of a^2b^2c^2 so any element of K can be written as a sequence of g_k^(2), so K is a subset of H.

Is it correct?

[delicate orchid]

wait hold on I don't buy it

$xyx^{-1}xyx^{-1}xy^{-1}xy^{-1} = xyx^{-1}x^2y^{-1}$

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

then u cancel again

ok now I buy it

[coral spindle]

Yes. Where is the confusion?

[crystal vale]

I want confirmation

[coral spindle]

OK.

[crystal vale]

Thank you

[delicate orchid]

u should have more confidence in yourself

[summer path]

i have no confidence in myself either

[delicate orchid]

u should have more confidence in yourself

[chilly ocean]

how can i have confidence in myself when i often make stupid mistakes

[next obsidian]

By going

[chilly ocean]

but when i was a kid in school people said my smile was ugly so i stopped laughing forever

[dull ginkgo]

[next obsidian]

Damn

[night lantern]

Hey ! Do you know any free access book/website online which could be a good starting point to learn about symmetry groups ?

[coral spindle]

Do you mean the symmetric group?

I’m not sure what exactly you mean by a symmetry group, but if you just want to learn about groups, Artin is one standard choice amongst many textbooks.

[vivid tiger]

a

(so, a significant part of mathematics - get enough textbooks about subjects and you'll see a "symmetry group" somewhere)

[coral spindle]

I don't like the implication that I don't know what the group of symmetries of an object is...

But it seemed to me that this was simply to generic a thing to study, and perhaps the asker simply wanted to learn group theory :) I'm sure you agree this is a reasonable thing to infer.

Remember of course that arguably, every group is a group of symmetries of some object :)

[vivid tiger]

sorry, wasn't trying to imply that

[coral spindle]

Understood, I'm glad

[rain grove]

Hello I need some help with this problem:

Let φ : R → R endomorphism of an additive group of real numbers. Let's say φ is continuous. Prove that there exists such a ∈ R, so that φ(x) = ax for all x ∈ R.

I dont even know where to start

[coral spindle]

Here's a start: try proving it for a map phi : Q → R, and then try to extend the result to R → R by thinking about continuity. Note, I wrote Q → R, not Q → Q!

[vivid tiger]

an additive group of real numbers (as in, a subgroup of R,+) or the additive group of real numbers?

e.g.,
f: R->R arbitrary such that f(0)=0 is technically an endomorphism of the subgroup {0} after restriction.

you probably meant an endomorphism of the additive group of real numbers

but yes, as Boytjie says, a good trick is to identify \N with 0,1,1+1,1+1+1..., do things with \N, extend your result to \Z, extend it to \Q, and then use continuity to extend it to \R.

[rain grove]

I still don't know how to start

[coral spindle]

Have you given my hint any thought?

[rain grove]

i think i understand your idea,
something like this:
φ(n) = a * n
and then φ(n/m) = n/m * a ... ?

but dont i have to know that φ(1) = a?

[coral spindle]

φ(x) = ax

let x = 1

What do you get?

[rocky cloak]

I mean. Surely phi(1) is something. Might as well call that number a

[rain grove]

ohh that makes sense

how do i get from φ(n) = n * a to φ(1/n) = 1/n * a

[coral spindle]

What's phi(1/n) + phi(1/n) + ... + phi(1/n), n times?

[sly rain]

I read a chapter on free groups, which went over the characterisation via universal property and the explicit construction.

Can someone put the essence of the characterization via universal property into words?

The way i understand it, it gives us that up to isomorphism there is only one free group on sets of the same cardinality.
And gives us easy group homs out of the free group.
What else?

[mint root]

group homs out of the free group is kind of the whole point: If you take a free group on 2 generators, maps out of it are precisely determined by where you send those 2 generators

[coral spindle]

Take a map out of some group. The image of that map is a quotient of the group, which you can think of as making more things equal in the group: you add more relations between the elements.

The free group on n elements is therefore the group that has exactly the relations common to every group, which is to say only those relations guaranteed by the axioms of group theory and no more.

I guess I haven't addressed a different part of the universal property but it's close enough.

[sly rain]

Thanks, this is a nice way of thinking about it and i didnt make this connection.
Also thanks 23zeta

[sonic coral]

it’s example 8 on page 180 of dummit and foote

[vivid tiger]

...wait what?

huh?

oh, "commutator set"

[cobalt heath]

Hmm, what is a commutator set?

[vivid tiger]

set of commutators

commutator subgroup is subgroup generated by set of commutators

[cobalt heath]

Ah, I thought set of commutators always form a group

[vivid tiger]

me too, actually, but, apparently not!

[cobalt heath]

Yeah, quite surprising

[dull ginkgo]

Product of commutators need not be a commutator lol

$[a,b][c,d] = aba^{-1}b^{-1}cdc^{-1}d^{-1}$ noticeably doesn’t necessarily have to be a commutator

[cloud walrus]

Mizalign #1 simp

[vivid tiger]

(XY-YX)(WZ-ZW)=XYWZ+YXZW-YXWZ-XYZW is also not a commutator necessarily!

[dusk whale]

Could use a hint for this problem

[chilly radish]

Choose a composition series containing H (How do you always find one).

and remember the composition factors are abelian

i'm not sure about the 'normal in G' part, that seems wrong

I think maybe it's supposed to be normal in H

[rocky cloak]

Some hints:

• ||H is also solvable, and there's a very natural choice of subgroups, H > H1 > H2 > ... Hn > 1, such that Hi/Hi+1 is abelian||
• ||This choice is the derived series, and Hn is an abelian subgroup of H||
• ||The derived series consist of characteristic subgroups, fixed by any automorphism||
• ||Conjugation by elements in G is an automorphism on H||

[rocky cloak]

I think it should be normal in G. Doesn't really seem much harder to find one normal in G, than to find one normal in H anyway.

[chilly radish]

oh fair enough you're right

I forgot about that

I haven't done finite group theory in a very long time

[true ingot]

Can anyone give any hint for the following? I've been trying for a while but can't seem to solve it..

Let H be a normal subgroup of G and let $f:G \to G/H$ be the quotient projection. Suppose that, for some $g \in G$ the order $m$ of $f(g)$ is coprime to $|H|$. Show that then $f^{-1}(f(g))$ contains an element of order $m$.

[cloud walrus]

Faputa

[rocky cloak]

A hint could be that if m and |H| are coprime. Then there exists an n such that n|H| is congruent to 1 modulo m

[true ingot]

This follows from Bezout identity right? Thank you this looks like something direction I didnt try yet... I will try it

[true ingot]

Hm Im not sure how exactly I should use this information. So I know this implies that g^(n|H|) is also in f^(-1)f(g), however I am not sure how to get an element with an order m using that information 🙁

Perhaps I am too much blinded by my previous attempt, where Ive noticed that g^(m+1) is inside f^(-1)f(g) and have attempted to construct an element of order m using this information..

[rocky cloak]

Well, what's the order of g^n|H|?

[true ingot]

Some multiple of m, right?

[rocky cloak]

Yes, but which multiple?

[true ingot]

I didnt know we can get more information than what I said. Let me see

[fringe heath]

could i get some help with this please? i'm unsure how to proceed

[chilly radish]

First isomorphism theorem

[true ingot]

I guess the answer is probably one but how do we see that?

[rocky cloak]

Well, try to compute (g^n|H|)^m

Remember how you can manipulate exponents, and Lagrange's theorem

[fringe heath]

oh
is it just |A_5| * |ker \phi|

since it's surjective so image is A_5

[true ingot]

Ahh ok. I forgot that a consequence of Lagrange's theorem is that an element of a group raised to the group order is the identity 😳.
So g^(n|H|)^m = (g^m)^(n|H|) = e (since g^m is from H).

[crystal vale]

Here, they define vector space, so a unique element statement means, scalar multiplication is well defined operation, right?

[coral spindle]

In particular it will mean that, yes.

But you are not looking at the whole statement.

"... such that the following conditions hold." means that there is more you need to read!

The uniqueness is referring to the fact that ax will be uniquely defined by the properties listed below.

[delicate orchid]

I ain't readin no thing....

[coral spindle]

Read books du horensohn

[delicate orchid]

ich werde nie ein Buch lesen....

[crystal vale]

Okay, thank you

Let F be a field and p be a polynomial over F with degree n.
Then if p(x)=0 for all x belongs to F, then how can I show that all p=0.

If F is an infinite field then maybe I use that if p is non-zero polynomial then p has at most n roots but it contradicts it. Hence p=0.

But if |F| = n ? I think there is another easier way

[delicate orchid]

if |F| = p then the polynomial x^p-x is a counter example to this

there must be some other condition

your approach for infinite fields works

[crystal vale]

Then maybe they are not considered a finite field

[coral spindle]

Consideration doesn't play a part

[crystal vale]

Or I misunderstood the statement

In a field F if I take V consists of all sequences { a_n} in F makes a group under + (same additive operation of F) ? I think yes because of the field property it follows groups property

[crystal vale]

I think it works even when I take rings, right?

[coral spindle]

You can prove this by hand

Just try proving it

[delicate orchid]

sequences are just elements of the countable product of the ring with itself, and products of rings are rings. But prove it explicitly yourself as an exercise

[crystal vale]

I just wanted confirmation because many times I made silly mistakes, thank you

[crystal vale]

Yes

[devout kernel]

Hey all, can someone please help me get a better understanding of the proof here? In my case I need to show that $F[a] = F[a^2]$ which I think is the same?
https://math.stackexchange.com/questions/1393510/prove-that-e-f-alpha2

So I think $a^2\in F[a]$ since F is a field, therefore we have $F[a^2] \subseteq F[a]$\

Now the next part is ok but then $x^2-a^2 \in F[a^2][x]$ because this set contains all the polynomials which $a^2$ is root of? therefore we get that the extension is indeed smaller equal 2 (degree of that polynomial) and thus it must be odd so 1.

is it right? wrong?

[cloud walrus]

mtr123

[uneven bobcat]

You're correct that $a^2$ is in $F[a]$ since $F[a]$ is a field, and thus $F[a^2]\subseteq F[a]$.

$F[a^2][x]$ is the set of all polynomials whose coefficients are in the field $F[a^2]$. Thus it contains all polynomials of which $a^2$ is a root, but also many other polynomials. But yes you're correct that $x^2-a^2\in F[a^2][x]$.

[cloud walrus]

kr1staps

[devout kernel]

Thank you very much, it’s more of a complex question but if you could help me understand the second part here too I’ll appreciate

[uneven bobcat]

This result depends on two very important theorems in basic field theory. If you're unfamiliar with these and/or their proofs, I recommend first finding them in whatever text you're using, and understanding them.

1. Given any tower of field $K/E/F$, we have that $[K:F] = [K: E] [E : F]$, and

2. Suppose that $E/F$ is an extension, and that $a\in E$ is a root of $f(x)\in F[x]$. Then $[F(a) : F] \leq \deg f(x)$.

Returning to your particular problem, since we have a tower $F[a]/F[a^2]/F$, we know that $[F[a] : F] = [F[a] : F[a^2]] [F[a^2] : F]$. Observe that
$F[a^2][a] = F[a]$. Since $(x^2-a^2)\in F[a^2][x]$, and has $a$ as a root, by applying 2) we see that
$[F[a] : F[a^2]] = [(F[a^2])[a] : F[a^2]] \leq \deg (x^2 -a^2) = 2$.

Suppose for the sake of contradiction that $[F[a] : F[a^2]] =2$, then the formula we obtained from 1) becomes $[F[a] : F] = 2 [F[a^2] : F]$, but this would make $[F[a] : F]$ even, contrary to our assumption. Therefore $[F[a] : F[a^2]] = 1$ which is only possible if $F[a] = F[a^2]$.

[cloud walrus]

kr1staps

[devout kernel]

Damn I forgot to send the new link 😅

https://math.stackexchange.com/questions/3108395/degree-of-splitting-field-for-fx-xp-2-over-mathbbq-p-prime

Sorry

[rocky cloak]

Are you asking for help understanding the solution here? If so what parts do you understand and which parts don't you understand?

[delicate orchid]

I don't see how u can get more detail than what kr1staps has already said

consider the minimal polynomial of a over F(a^2), then contradict using the tower law or whatever

[rocky cloak]

Well, the other link they sent is a completely different problem

[delicate orchid]

oh

oh well!

[devout kernel]

The p-1 part, all of how you get it is what confuses me

[devout kernel]

With that special polynomial

x^p-1/x-1

[delicate orchid]

yeah that's the minimal polynomial for any pth root of unity for p a prime

special case of cyclotomic polynomials and whatnot

so what part are you confused about, is it the application of Eisensteins?

[devout kernel]

This

And the application of the second Einstein

[delicate orchid]

ok since you're refusing to be more specific than "the entire solution" I'll explain the Eisensteins to you.

[devout kernel]

I honestly don’t understand this part, why they chose to even use this polynomial, and how it help us get the rest of the answer

[delicate orchid]

If you expand out that ((x+1)^p-1)/x you get x^{p-1}+...+px, eisenstein gives us that this polynomial is irreducible immediately

now we have that f(x) is irreducible iff f(x+a) is irreducible for any a cause we're working over a field

gimme a minute to remember how this proof goes

ah yeah u show that the shifting is an automorphism

let f_a be the map sending a polynomial p(x) to p(x+a), then it's clear that f_a(1) = 1, and f_a(p(x)+q(x)) = f_a(p(x))+f_a(q(x)) and likewise for multiplication. So this is a ring automorphism with inverse given by just shifting back (the map f_-a). Hence p(x) is irreducible if and only if p(x+a) is irreducible.

which is why they're able to use the fact that phi_p(x+1) is irreducible to conclude that phi_p(x) is

[south patrol]

Cyclotomics hot

[delicate orchid]

this explaination is lame so I'll give you an intuitive one

[dire siren]

or it's just straightforward to prove instead f reducible <=> f(x+a) reducible

[south patrol]

Yes

Well you mean irreducible occ

[delicate orchid]

f(x+a) is shifting "the graph" of f(x) horizontally. So if f(x) doesn't intersect the x-axis it certainly ain't gonna do it after you move it side ways

[delicate orchid]

I felt like this was pretty straight forward

[south patrol]

I should also say that like

These cyclotomics and this trick are very well known

It's not like this person pulled it out of nowhere

So yeah like the trick is pretty cool

[devout kernel]

This way you prove x^(p-1)+x^(p-2)+…+1 is indeed irreducible right? Also why did we get this polynomial in the first place?

Thanks for helping all of you 🙏🙏

[delicate orchid]

indeed you do. And we get it by noticing that a pth root of unity "x" clearly satisfies x^p-1, that's what a root of unity is, but x^p-1 isn't reducible

so you factor it as (x-1)(x^(p-1)+....+1) and then show that the thing on the right is irreducible

[dire siren]

I meant reducible; so proving non p <=> non q

and in fact we only need the =>

which is a one-liner

f(x)=g(x)h(x) implies f(x+a)=g(x+a)h(x+a), q.e.d.

[south patrol]

Ye

[delicate orchid]

raagghh etc. etc.

[hollow mica]

why is schurs lemma for modules sooooo much easier to understand than the one for groups

[delicate orchid]

modules form an abelian category so every suboject is "Normal"

I haven't thought about schur's lemma for groups since I learnt about it but the fact that simple groups still have subgroups is very annoying

[hollow mica]

I see

[chilly ocean]

yeah, groups form only a semiabelian category so its only semi as easy to understand

[delicate orchid]

https://tenor.com/view/true-feel-that-thats-true-gif-7410050725709839510

no cokernels

[chilly ocean]

wut? groups homomorphisms do have cokernels

[delicate orchid]

u tellin me

wait

no you're right

wait no I don't buy it again

img(f) need not be normal

[chilly ocean]

the cokernel is the normal closure of the image iirc

[devout kernel]

I understand, but I’m still not fully understanding why we want that (x^p-1)/(x-1) in the first place?

[delicate orchid]

any excuse!!!

[chilly ocean]

it satisfies the universal property so its a cokernel

[delicate orchid]

x^p-1 = (x-1)(x^p-1+...+1)
so (x^p-1+...+1) = (x^p-1)/(x-1)

[delicate orchid]

yeah I'm just annoyed I didn't think of that

and thus will be making fun of you to make myself feel better

[chilly ocean]

it is pointless for i am a stoic

[delicate orchid]

even better, it's a win-draw rather than a win-lose situation!

[devout kernel]

Yeah but he didn’t say why we got that polynomial in the first place or have I missed it? I understand the computation there but not why we got there in the first place

[delicate orchid]

what is a pth root of unity

not a trick question, what is one

[devout kernel]

cis(2pi/p)?

[delicate orchid]

we're not in the complex numbers

it's something that satisfies x^p = 1

or x^p-1 = 0

hence if we wish to adjoin a pth root of unity to a field K, we should try K[x]/(x^p-1) as our first guess. But this doesn't work cause x^p-1 isn't irreducible

[delicate orchid]

ergo we factor it and find the largest irreducible factor

[devout kernel]

Feeling really silly now but why we wanted the pth root of unity? 😅

[delicate orchid]

idk dawg??? it's ur problem!!!

oh right we're doing tower law

[devout kernel]

But how did we get from Q[2^1/p, root]:Q[2^1/p] to this?

[delicate orchid]

we want the degrees of [Q(2^1/p) : Q] and [Q(root) : Q]

although at this point just go re-read the stack exchange post

[devout kernel]

I think I got it, thank you very much!

@delicate orchid

[sonic coral]

is this the correct idea or should I be using the frobenius map?

[south patrol]

Yes that is exactly right

I would add "over K" at the end of the first sentence for emphasis

But yes

[sonic coral]

cool thank you

[south patrol]

And also that t^1/p generates K over F

But yes

Method and idea are completely correct

[sonic coral]

i was also thinking you could argue that there was one root by the injectivity of the inverse of the frobenius map, would that also work

[south patrol]

The Frobenius doesn't have any inverse though

Not surjective on either K or F

[sonic coral]

maybe i am explaining myself wrong, but i was thinking like x^p = t = y^p and then saying that x=y so we only have one root

[south patrol]

Yeah though then you have to show F_p(t) has no primitive pth roots of unity

Which is sort of clearly the case but additional work when you can just do your current argument

Like you can just say uh

[sonic coral]

where should this be stated?

[south patrol]

oh sorry

You mean

x^p - y^p = (x-y)^p

But then that is more or less the same argument

[south patrol]

Idk I'd just say at the start

[sonic coral]

yeah i see that, okay

[languid trellis]

I don't see how we get from gA \cap A = \emptyset to G acts imprimitively on S

I can see why gA = A implies G is imprimitive, because if we take A = g-orbit of a certain x in S, gA = A and the g-orbits partition S

pls ping

[delicate orchid]

a group action on X is primitive iff it's isomorphic as a G-set to G/H for some maximal subgroup H. Cosets are disjoint

I'm not pinging you hahaha

mainly because this is a non-answer

[languid trellis]

i menat to ask why gA \cap A = \emptyset implies G is not primitive. I feel the theorem is what you just said, because the book already proved that the actions of G on S and G on G/H are isomorphic as g-sets

[delicate orchid]

we'll have to somehow construct a block of the action using that fact

if gA = A for all g it's obvious

both statements together

we get uhh

disjoint union $\bigsqcup_{g \in G} gA \subseteq S$

but how do we know it isn't S itself?

"block of S" doesn't make sense, ignore that

[cloud walrus]

Wew Lads Tbh

[languid trellis]

fyi defn of primitive here is that the only two partitions of S stable under the group action are {S} and each element {x_i} of S

[delicate orchid]

yeah, sorry "block" might not be standard terminology

oh wait

[delicate orchid]

this is a partition of S because the action is transitive

hence the action preserves this partition by our assumptions on A

[languid trellis]

does gA \cap A = \emptyset imply that any two gA are different?

[delicate orchid]

yur

[languid trellis]

cus we can left multiply

[delicate orchid]

yus

[languid trellis]

yah ay ah yah

ok will get back to that in 30

doing analysis now

[delicate orchid]

SNORE

[languid trellis]

thank you mr wew lads

for the help

[delicate orchid]

I shall continue to not ping you

[languid trellis]

fuck you

[white wraith]

Every time I get confused on something like this it ends up being a simple answer

But how do you know that group has p elements?

Apologies if it's obvious, I've been doing maths for 14 hours straight, not thinking properly

[delicate orchid]

take a break?

[white wraith]

I might, don't think it will improve my understanding when I get back

[delicate orchid]

what does p_e mean

[white wraith]

p^e for some e >= 2

They didn't mean to subscript it

Mistake

[delicate orchid]

ah ok

it's only p-elements because p^e = 0 lol

so once you've reached x = p you're back at x = 0, so the elements look like {1, 1+p^{e-1}, 1+2p^{e-1}, ...}

[white wraith]

Oh

[white wraith]

I return to my original statement above the image

[delicate orchid]

well tbf when they're trying to teach elementary group theory by doing this nonsense they tend to end up butchering it

like what's your definition of Z_n

[white wraith]

(They're not, this is for Number Theory)

[delicate orchid]

yeah, I should've guessed

[white wraith]

But it was a question about groups so I put it here

[white wraith]

It's also not teaching, it's something I found when trying to figure out a proof for a research paper lmao

Although it is teaching for some university, I feel bad for whoever is taught by it

[lilac mango]

Any hints to prove that given any field F, then F[x] has infinitely many ireducibles?

Ofc the case where F is infinite is easy since x-a is ireducible for any a in F

[rocky cloak]

Do you know Euclids proof that there are infinitely many prime numbers?

Because the proof is the same

[void cosmos]

https://tenor.com/bTf7F.gif

this is literally how jagr types guys

[south patrol]

Proof by explicitly counting for finite fields

Actually I guess that kinda would work lol. Like if there are only finitely many irreducibles over some F_{p^k} then there'd not be arbitrarily large finite fields of char p

But knowing that all finite fields are of that form is more work

In any case I'm joking as your argument generalises much better

[rocky cloak]

I guess proving something like
x^q^n - a is irreducible in a field with q elements might be doable.

[lilac mango]

Oh lmao

[south patrol]

Yeah

Though like ig the point is it doesn't split right

[rocky cloak]

Euclids proof is much easier anyway

[south patrol]

Yeah lol

I forget lol, what conditions do we need to apply that argument

Like for a general ring R

Infinite UFD or smth

Maybe a condition on units to avoid trivialities like an infinite field

[rocky cloak]

I guess you need some condition to make sure 1 + the product isn't a unit

[south patrol]

Yeah so finitely many units suffices

Cause you can take like

1 + (p_1 ... p_k)^N for different N

Until you don't get a unit

[rocky cloak]

But finitely many units seems very strong

[south patrol]

Using the ufd property I mean

That is true

Though it does hold for F[x] and F finite at least lol

But hm

[lilac mango]

You could also change the 1 instead of changing the powers of N

[south patrol]

True yeah lol

[lilac mango]

Which I think generalizes more?

[south patrol]

u + ... for all units u

nice

Well

What if u + units is another unit

Idk I don't see how it works fully

I guess you'd need to show you don't just cycle around all units

But I don't see why that should be the case

[rocky cloak]

Hmm, so let x be the product of all the primes. Then for every y
1 - yx
is not a multiple of any primes, hence is a unit. So x is in the radical. But the radical is 0, contradiction.

So a UFD is either a field or has infinitely many primes

[south patrol]

Nice

Though is the jacobson radical necessarily 0

I haven't thought about it for ufds

Uhhh

Like can't you have local ufds or am I smoking

Yeah I'm p sure lots of examples exist from AG

[rocky cloak]

Yeah, you're right. I'm being silly thinking of the nilradical

[south patrol]

Dw.

[rocky cloak]

And as you say k[[x]] is a UFD with only one prime

[south patrol]

Though this would work for jacobson ufds then lol

Now I'm confused

Found someone saying a ufd which is not a field has infinitely many primes on stack exchange

[rocky cloak]

Link?

[south patrol]

https://math.stackexchange.com/a/4210656/932674

[rocky cloak]

They seem to be counting primes not up to associates

So 2 and -2 are two different primes

[south patrol]

Oh yeah okay lol well that is boring

[little shadow]

If G (possibly infinite) is p-divisible is G/pG trivial?

It should be but im not sure how to prove it for myself, you can have A/B be non-trivial even if B is isomorphic to A

[south patrol]

To me this is basically the definition of p-divisible

p: G -> G is surjective so pG = G right

[little shadow]

Yes that was my thinking. Its a subtle difference between = and "is isomorphic"

[south patrol]

But this is an actual equality

Though qhen you say p-divisible, do you mean p:G -> G surjective

[white oxide]

big wagner fan?

[little shadow]

yes

[little shadow]

of course iam

[south patrol]

As p-divisible group usually means something else

Ah okay

[white oxide]

🤯

[little shadow]

i thought, for all intended purposes, = and is isomorphic are the same, its just a matter of labeling the elements

[south patrol]

Not when it comes to things like quotients here

for example Z/nZ is nonzero even though Z and nZ are isomorphic groups

[little shadow]

yes precisely

although i guess Z isnt P divisible for any p

[south patrol]

Sure though that wasn't the point here

The important thing is like

Two inclusions give the same quotient if they are isomorphic in a manner compatible with the inclusion maps

[little shadow]

😵‍💫

[celest furnace]

There’s notions of stronger isomorphisms—like induced ones

[little shadow]

I've never really heard of that

[south patrol]

Hm I guess more like natural ones

[little shadow]

Is the p map an automorphism for p-divisible groups?

surjective endomorphism sounds like it should be injective aswell, atleast for "nice" groups

[celest furnace]

Surjective endomorphism is redundant, right?

[icy totem]

No

[rocky cloak]

Not necessarily. Consider for example Q/Z. Multiplication by p (or any integer) is surjective, but not injective.

Don't know what qualifies as "nice" to you though. It holds for finitely generated abelian groups.

[little shadow]

I see, thank you

[celest furnace]

What is an endomorphism?

Ah I was thinking of epimorphism

[icy totem]

Ahh ok

Yes that would be redundant then ofc

[rocky cloak]

A map from something to itself

[vivid tiger]

the Sylow subgroups are isomorphic.

(because they are conjugate - thus these are even inner isomorphic)

yet it's obviously useful to think of them as being different, so that you can ask questions like "how many?" and "if i take an element in one and multiply by an element in the other, what happens?"

[teal wind]

Could someone check my work on these problems:

[vagrant zinc]

Name of book?

[summer path]

That looks like a pset

Considering it says homework 5 in the middle

[crystal vale]

I don't know if this is the correct channel for this question or not,

Let W_1 and W_2 be subspaces of a finite dimensional vector space V. I need to find the necessary and sufficient conditions on W_1 and W_2 so that the dimension of the intersection of W_1 and W_2 are the same as the dimension of W_1.

I think sufficient conditions will be W_1 must be a subset of W_2 but I am not sure for the necessary condition

[chilly ocean]

#linear-algebra would be the best channel for it

[chilly ocean]

What happens if W_1 is not a subset of W_2?

[crystal vale]

I thought as if W_1 is a subset of W_2 then intersection will be W_1 so it will be true so maybe I consider it as sufficient condition

[chilly ocean]

Yes

But what if it isnt a subset?

[crystal vale]

Intersection will be the proper subset of W_1, right?

[chilly ocean]

It will be a proper subset, yes

[crystal vale]

So I think the dimension of the proper subset cannot be equal to the dimension of that set

[chilly ocean]

Why do you think that?

Also dimension is defined for subspaces not subsets

In this case it is a subspace tho

[crystal vale]

Here the intersection is subspace, right?

[chilly ocean]

Yes, intersections of subspaces are subspaces

[crystal vale]

If they have the same dimension, then the basis of a proper subset can generate the set W_1 , it contradicts that intersection set is proper set of W_1

[chilly ocean]

Why cant the basis of a proper subspace generate W_1?

[crystal vale]

Let S be a proper subspace of W_1 then if basis of S generate W_1 that means every element of W_1 belongs to S, so S will not be proper subset of W_1

[chilly ocean]

Good, also why does S having the same dimension as W_1 imply the basis of S generates W_1?

[crystal vale]

Because, say dimension of W_1 will be n, then S has n linear independent vectors, so I know that if the cardinality of set of linear independent is the same as dimension then that set span vector space

[chilly ocean]

Yes, any linearly independent set of cardinality n will generate W_1

If you know that, what you want to prove follows

[crystal vale]

So the necessary conditions and sufficient conditions are the same

[chilly ocean]

if you mean that

[chilly ocean]

This is a necessary and sufficient condition

It is

[crystal vale]

Okay, thank you

[kind wave]

Are there any books or papers related to constructions with a marked ruler and a compass which define this notion properly ? I've studied Wantzel's theorem and I'm now trying to show that if you add marks to the ruler, constructible numbers by this means are still algebraic over Q, but I'm lacking a rigorous definition

[vagrant zinc]

W_1=V

If I am not wrong, you should ask in linear algebra, they should have mastered that exercise.

[rotund aurora]

mark pi on the ruler?

[kind temple]

rigorous definition of what?

[crystal vale]

But I think the necessary condition is that W_1 must be the subset of W_2. If it is not then dimensions can not be equal.

[vagrant zinc]

I'm sorry I couldn't help you much, but you make me think that your thought makes sense, it's more likely that it's not a subspace of its own but the same, maybe, I don't know.

[hidden wind]

ok that took way too long

(admittedly i kept putting it down instead of just taking the time i needed, which turned out to be roughly an hour split over a week)

[rare monolith]

Hi, I would like to show that $Z\oplus Z/Z\cong Z$ (Z being the group of all integers under addition) using the equivalence relations of the quotient group. So far I wrote down what's enclosed in the picture. However, I don't see how something of the form ( , ) could be in Z?

I genuinely hope this is the right channel to ask this in, I apolegise in advance if this is a mistake.

[cloud walrus]

schadowpop

[south patrol]

Dw correct channel

[rocky cloak]

In order for Z(+)Z / Z to make sense you have to realize Z as a submodule of Z(+)Z somehow.

A natural way might be to consider the elements (n, n)

[vagrant zinc]

Que agradeble sujeto

[rare monolith]

The only things I'm acquainted with is equivalence classes or the fundamentel theorem of homomorphisms. I think I can show it via the second but I'd rather show it via the first which has become more intuitive to me.

For context i'm a physics master that does a mathematical physics course. We're kinda thrown in the dark

[rocky cloak]

Yes, that's fine. But in order to define G/H, you need H to be a subgroup of G.

So you have to be a little precise what you mean by Z(+)Z / Z.

There are many subgroups of Z(+)Z isomorphic to Z, and it isn't clear which one you mean

[rare monolith]

I see you're correct

[rocky cloak]

Like the statement "(a, b) is in Z" doesn't make sense, since (a, b) isn't an integer.

[rare monolith]

Yes that's why I was confused

[delicate orchid]

now you've got me thinking how many there are

ah there's one for each rational number between 0 and pi

[rocky cloak]

Very fancy way of saying 'countably many' you got there

[delicate orchid]

that corrispondence gets you a generator though

so there is actually extra information there

[rare monolith]

The reason I'm asking this question is because I'm trying to follow this example of Nakahara where he computes the 0-th Homology group of the simplicial complex in the picture

Now the 0-cycle group is just ${n p_0+m p_1|n,m\in Z}$

[cloud walrus]

schadowpop

[rare monolith]

and the 0-boundary is just ${np_1-np_0)}$

sorry wait let me correct that

[rocky cloak]

Then the 0-boundaries are given by the image of the boundary map, which maps (p0p1) to p1 - p0

So then you're looking at the subgroup of elements of the form (n, -n)

[cloud walrus]

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

[rare monolith]

Yes I suppose so

I thought this would be Z because it's just one number

that's the way it was explained in class I think

[rocky cloak]

Yes, it's isomorphic to Z

But there are many subgroups isomorphic to Z

[rare monolith]

but then I'd have to formally define an isomorphism of (n,-n) to Z

f:(n,-n)-> n perhaps

[rocky cloak]

Not really, because the thing you're interested in is the subgroup (n, -n). So it's not really that important if it's isomorphic to Z or not

[rocky cloak]

But this is correct yes

[rare monolith]

I see, the reason I'm doing that is because in class we usually wrote things in terms of Z

So wehn we had expressioins like $Z\oplus Z\oplus Z/(Z\oplus Z)$ we'd just call it Z

[cloud walrus]

schadowpop

[south patrol]

honestly i think this is bad notation, lol

[rare monolith]

rip

[south patrol]

I would personally never write it like that

[rocky cloak]

Yeah, so often one can infer from context what is meant. But the notation is leaving out a lot of relevant information

[rare monolith]

I hate how non-rigorous math physics lectures can be sorry for my negativity haha. Please, let's look at it properly then

[south patrol]

In context it seems this should be {(a,b,0)} or {(a,0,b)} or (etc) and those do all give the same quotient

but i don't think it is good notation

[rare monolith]

I need to compute the 0-th homology group by expressing the quotient (n,m)/(n,-n) group for n,m integer

[rocky cloak]

So the way I usually find the easiest is to use the first isomorphism theorem.

But you can also construct the cosets explicitly like you were trying to before

[rare monolith]

this one?

[rocky cloak]

Yeah, that one

[south patrol]

though it should be emphasised that the important bit is not that there is just an iso, but how it is constructed

[rare monolith]

hmmm

then I need to find some homeomorphism f s.t. ker f = H with H subgroup G consisting of elements (n,-n)

I can have f:(m,n) -> m+n then

[south patrol]

(homomorphism)

[rare monolith]

we have f:(n,-n)->0 s.t. ker f = H

im f = all integers

done?

(for physicist rigor I'm quite content but I probably skipped a lot of required math rigor making you frown currently haha)

[rocky cloak]

Guess you should probably also check that no other elements except (n, -n) is mapped to 0. But yeah, good job!

[rare monolith]

hmm well n+m=0 iff n=-m. This implies elements like (n,-n) and (-m,m) are in the kernel of f

But these two elements are the same right, can both be written as (n,-n)? So I think there are no other elements except (n,-n) in the kernel of f

If this line of reasoning is fine, I want to thank you for your help. It was very helpful (& guiding me in the right direction made me come up with the answer myself!)

Wait this is no good (edit: nvm got confused by reading homeomorphism as homomorphism life is good)

[rocky cloak]

They did choose annoyingly similar words for those two

[hidden wind]

i love both homomorphisms and homeomorphisms but especially homeomorphisms

sorry homomorphisms

[rare monolith]

as a physicist I love eulers number

always fun to work with

[celest heath]

I love diffeomorphisms

[south patrol]

How about homeomorphisms which are homomorphisms

[coral spindle]

As a person who eats food, I love spoons

[celest heath]

Forks>

[delicate orchid]

As a person who drinks, I love spoons

[rare monolith]

Okay but now I have a somewhat similar question on the simplicial complex $K={p_0,p_1}$. I have that $Z_0(K) = {np_0+mp_0|n,m\in Z}$ and $B_0(K)= {}$ (empty set). Now I'd like to show that the quotient group $Z/B$ is $Z\oplus Z$ but I'm having a hard time here finding the $0$-th homology group using a homomorphism

[cloud walrus]

schadowpop

[rare monolith]

I have also considered all the equivalence classes but I'm not sure, because for two elements c and g in Z0 we need c-g in B0. But given B0 the empty set, there is no such g (c being the representative)

Should B0 actually be a set containing zero?

my textbook tells me I should define $B_0$ to be 0 but I'm not quite sure what that means, would that be the set containing on element (being $0$)?

[cloud walrus]

schadowpop

[rare monolith]

In that case the proof is trivial I think

[coral spindle]

0 is the zero group; the unique (up to isomorphism) group with only one element.

We call it 0 because we are working with Abelian groups.

[hidden wind]

gives me full bodily reactions

i haven’t actually looked properly into topological groups yet

[kind wave]

just the unit length is marked

[kind wave]

constructible using marked ruler and compass

one could probably say a point A is so iff there exists B such that AB = 1 and (AB) is constructible

but I'm not sure if this works

[rotund aurora]

... ok are you given any points?

Usually the classical set up is 2 distinct points + ruler + compass, you dont need to mark the unit length because its given by the two points

[kind wave]

yes, I'm given at least two

yes but if you allow the unit length to be "moved" you can trisect the angle and duplicate the cube

[rare monolith]

Ah this is helpful; I see in my textbook they define 0 as the unit element of Abelian groups. This is indeed a group under the addition operation, trivially.

[kind wave]

i'm trying to show you still cannot square the circle

[rotund aurora]

I told you already, let pi be marked then you can square the circle

[dull ginkgo]

You’re asserting a priori that \pi is transcendent over \mathbb{Q} right

[kind wave]

what is marked is the length between the two 2 initial points

[kind wave]

yes

[dull ginkgo]

And you’re trying to construct a square of area \pi right

[kind wave]

i'm trying to show it's impossible even by adding two marks on the ruler

[rotund aurora]

No dude, you are contradicting yourself

🤦‍♂️

[dull ginkgo]

Marking two points on a ruler allows you to “store” a length

[kind wave]

the only stored length allowed is 1

[dull ginkgo]

Marking points on a compass is like defining a variable in the field

The only polynomials you can solve using a compass and a ruler are quadratics

[kind wave]

well I guess I am not being clear enough and this is just really poorly documented

[dull ginkgo]

Mainly by vertue of most of these constructions being the intersection of a quadratic (the circle) and a line

You can’t really solve cubics or beyond, especially transcendentials, with your tools

[kind wave]

I know, I've studied Wantzel's proof

[dull ginkgo]

I’m pulling this out of my ass via intuition

[kind wave]

you can solve cubics if you allow segments of length 1 to be placed anywhere

[rotund aurora]

It's really well documented. You are just not making any sense

[kind wave]

what is documented is constructions using a ruler with no marks and compass

[dull ginkgo]

OH you can place them anywhere

[rotund aurora]

Sure dude... you know more than us

[dull ginkgo]

This is extremely up to interpretation here homie

[rotund aurora]

Why would you even ask if it's not documented? Lol

[dull ginkgo]

what do you mean that you can place “lines of length 1” anywhere

Like you’d have to “know where” to place em first

[kind wave]

well I'm trying to figure out a rigorous definition for this

[rotund aurora]

You can solve everything if you can place segments of length 1 anywhere

[dull ginkgo]

Wait

@kind wave let me try to parse. Given a point (constructed), you can place a line segment of length 1 there at any rational multiple of 2pi?

[kind wave]

https://fr.wikipedia.org/wiki/Nombre_constructible#Constructibilité_à_la_règle_graduée_et_au_compas

[rotund aurora]

Actually, you obtain all real numbers

[kind wave]

can't find the English equivalent for this

but no rigorous definition is given

[rotund aurora]

I told you the definition of constructible number already. And pretty sure there is a wikipedia article

[kind wave]

I know it

[dull ginkgo]

wait this is interesting

[kind wave]

I'm trying to extend it

[dull ginkgo]

You can construct all of the closure positive rational closure of Q with this technique I assume

[rotund aurora]

It's not, you can construct the point (cos x, sin x) then project onto the x axis to obtain cos x

For all x

[kind wave]

what I'd like to show is just the inclusion

[rare monolith]

I think I spoke to soon about the proof being trivial. Consider a simplicial complex $K={p_0,p_1}$. This has $0$-chain group $Z_0={np_0+mp_1|n,m\in Z}$ and $0$-boundary group $B_0=0$. Then trouble occur when I try to compute the 0-th homology grouop.

How is the unit element even a subgroup of $Z_0$?

[cloud walrus]

schadowpop

[dull ginkgo]

I came up with a weird alternate idea of a way to construct shit and now I’m parsing what it’s closure is

Kinda got distracted

I have an idea of how to interpret that

[kind wave]

French Wikipedia reads "Combined with the compass, the graduated ruler is more powerful than the conic intersection methods used by the ancient Greeks to solve so-called solid problems. The two instruments can be used to construct all the points in the complex plane obtained from the rationals by iterating the resolution of equations of the of the second or third degree, and to solve certain problems involving equations of the fifth or sixth degree."

[kind wave]

^^

[dull ginkgo]

If you have two points, one the center and one the vertex, you can construct a circumscribed regular polygon of any side count with the vertex around the center

[kind wave]

hmm

[dull ginkgo]

That space generated should be closed under taking “rational powers”

Now that I’ve overcomplicated the problem, have fun!

[kind wave]

well technically you could construct any angle

[dull ginkgo]

As a limit

[rotund aurora]

This is like adding all the roots of unity

[dull ginkgo]

Yeah

[tawny magnet]

https://fr.wikipedia.org/wiki/Neusis is this what you are looking for

[rotund aurora]

Or cos(2pi/n), sin(2pi/n) if you wanna keep it real

[dull ginkgo]

you can also get the Chebyschev poly roots by projecting onto a line

[rotund aurora]

Then if you close that with ruler and compass its just the splitting field of all polynomials of degree <=2 with coefficients nth roots of unity or rational numbers

[kind wave]

not sure if this precisely is the same thing

[dull ginkgo]

Yeah

Which really only works up to 4

Because of Abel-Ruffini

You can’t construct the whole “real” closure of Q

[tawny magnet]

in the english version it says this tho

[kind wave]

oh alright

anyway

gonna try to sort this out

sorry if my poor math vocabulary was confusing

thanks for the help

[glad osprey]

I often feel like the textbook solutions to exercises are needlessly complex. Here is the solution to an exercise asking to prove that a division ring has exactly two idempotent elements:

If $a \neq 0$, why not just multiply both sides of $a^2 = a$ by $a^{-1}$, leading to $a=1$ ?

[cloud walrus]

sheddow

[delicate orchid]

yeah you have a good point lol

[glad osprey]

It's annoying, because sometimes there's a good reason why they do it in a cumbersome way, but I can't always tell when

[fleet citrus]

That's so relatable 🥲

[rocky cloak]

I guess a reason here is that they're solution extends better to integral domains.

Still a little bit convoluted, but the fact that a(a-1) = 0 is somewhat useful to mention.

[glad osprey]

Yeah, I suppose thats true. Still, if I have a(a-1) = 0 in an integral domain, I would just immediately conclude a = 0 or a-1 = 0, instead of the weird a-1 = (a^(-1)a)(a-1) thing they did

[cloud solar]

Let p prime and n>=p. Find the number of degree n polynomials in Z/pZ[X] with an injective polynomial function. What i did: let f be a polynomial with the above property. Since the polynomial function of f is injective and because Z/pZ is finite we know the function is a bijection. We have an unique element r in Z/pZ with f(r)=0. So the polynomial f has only one root in Z/pZ. I somehow want ți attack the coefficients using this bijection. I tried looking at the product of f(x) when x is not r = product when x≠0 of x = -1. If p is not 2 we have the same thing but for sums and they equal 0. If the degree of f was p-2 i could get a property about of the coefficients looking at the rank of the cyclic matrix only with the coefficients. Some hints?

Another thing we can consider the polynomial f be monic and from Bezout we would have f=(x-r) g where g is a polynomial with degree n-1 and g doesnt have any roots in Z/pZ

[rocky cloak]

Note that any function can be realized as a polynomial. So you can reduce the question to how many polynomials define the same function.

[chilly ocean]

Is this question harder for Z/nZ where n is composite?

[rocky cloak]

I think it's probably a little harder yeah

[dire siren]

you got that f is a bijection; that's right
so for each permutation (a_0,a_1,...,a_(p-1)) of (0,1,...,p-1), you want to count the number of polynomials f such that f(0)=a_0, f(1)=a_1, ..., f(p-1)=a_(p-1) ......... (1)

||use Euclidean division: f=(X^p-X)q+r, for some polynomials q and r with deg(r)<p
[btw, I didn't assume f to be monic]||

||now prove that (1) is equivalent to r(0)=a_0, r(1)=a_1, ..., r(p-1)=a_(p-1) and deg(q)=n-p||

||the first condition uniquely determines r, and the number of polynomials q with deg(q)=n-p is easy to compute||

[rocky cloak]

||there are p! Injective functions on Z/p||
||Each is given by a polynomial of degree at most p-1||
||To get a polynomial of degree n add f(x)(x^p - x) where f has degree n-p||
||There are (p-1)p^(n-p) many such polynomials, so in total p!(p-1)p^(n-p)||

[cloud solar]

Thanks, guys!

[chilly ocean]

Youre welcomr

[coarse shore]

Quick question from someone kinda new to group theory: is a similarity transformation just a more general way of saying symmetry transformation?

[coarse shore]

nevermind i was completely wrong

[vivid tiger]

similarity with matrices is conjugation

[coarse shore]

thank you

[languid trellis]

How would i even begin to show that there are two groups of order p^2 up to iso? I already showed that every such group is abelian, and I'm pretty sure that these groups should be C_p^2 and C_p × C_p. I would appreciate any ideas, thanks all!

Please ping 🙂

[still dew]

Not allowed fundamental theorem of abelian groups?

In anycase wouldn't say if your group has no p^2 order element then pick a non identity element which by Lagrange has order p

Then it's normal

As the group is abelian

Then G ~= <a> x G/<a> ?

[languid trellis]

How do we know the non identity has order p? Is it not possible for it to have order p^2 if G = C_p^2 ?

Oh wait

[still dew]

Yes then it's cyclic

[languid trellis]

I can't read

[still dew]

I assumed it not to be

[languid trellis]

Yeah lol

[still dew]

Other than this just pick another <b> b not in a which should also have order p

Then uhhh

they don't have any non identity element in common

And abelian group so they commute and form a subgroup

G is just <a> x <b>

[still dew]

By counting argument this subgroup is G itself

[languid trellis]

How did we get here?

[still dew]

Okay that's taken more from the proof of fundamental theorem (don't need to do that)

[still dew]

Instead this should be fine @languid trellis

[still dew]

Easily identified as <a><b>

[languid trellis]

Right we have an iso <a>x <b> to a^kb^j which has order p^2 because p Is coprime to itself

[still dew]

p is coprime to itself ??

[languid trellis]

Wait

Gcd(p,p) = p

Right

Ofc

[languid trellis]

And this is not an iso

the element ab has order p not p^2

[still dew]

No element is supposed to have order p^2

Our group is not cyclic remember

And that is an iso

[languid trellis]

Lemme try recap where we've gotten so far

If our group has an element of order p^2, then we pick this element. Then our group is clearly iso to C_p^2

If our group has no element of order p^2, then by lagrange, it has an element of order p, call this a.

<a> is an abelian group, so it is normal in G.

[still dew]

No need for this part

[languid trellis]

Now, choose another element with order p, disjoint from <a>, call it b.

[still dew]

just pick another b not in <a>

Yep

[languid trellis]

Claim: <a> × <b> is iso to G

[still dew]

Identify <a>x<b> as <a><b>

[languid trellis]

Wdym by <a><b>?

[still dew]

Just product of two subgroups

[languid trellis]

Which one

[still dew]

It's also a subgroup

[languid trellis]

Just group op?

[still dew]

op?

[languid trellis]

As in the product by the group operation

[still dew]

Yes

[languid trellis]

Hm

It's not entirely clear that this should define another subgroup

[still dew]

In anycase this map works as it's surjective

And as G is finite

It automatically becomes Injective

[languid trellis]

I see

[still dew]

HK is subgroup iff HK=KH

Check this

[languid trellis]

I thought there'd be another condition

[still dew]

H and K are subgroups obviously

[languid trellis]

Then we have hkh'k'= hh'kk'

So stuff works

And (hk)^-1 = k^-1h^-1 = h^-1 k^-1

OK I'm convinced now

Wait

We have to replace elements

We don't have direct commutativity

[still dew]

yes

[languid trellis]

We have H, K normal then HK normal

[still dew]

But k^-1h^-1 is some h'k'

[languid trellis]

Yeah exactly

My bad

[still dew]

Do u see why this is a surjection

[languid trellis]

So <a><b> Is a normal subgroup of G, with order larger than p, so it has order p^2 and is thus iso to G

[still dew]

Yes it having order p^2 gives the surjection

[still dew]

We haven't checked that this is Homomorphism yet

But that is also easy to check

[languid trellis]

Mhm

[still dew]

At the end we have a bijective map

[languid trellis]

Right I see

[still dew]

Which is homo

[languid trellis]

Recap: G has element p^2, then we have an iso C_p^2 to G. If not, then we pick 2 elements of order p and consider the direct product, which is iso to the product group, which has order larger than p so is equal to G. Moreover, there is a clear iso <a> x <b> to C_p x C_p

And G only has elements of order 1, p or p^2

So we are done

Just details left to fill out

[still dew]

👍

Now you could do this with abelian groups of order higher powers of primes

Or any number in general

[languid trellis]

Thanks so much for the help

[still dew]

Np 🙂

[still dew]

You'll encounter this along with fundamental theorem of finitely generated abelian groups

[languid trellis]

That's in the modules chapter. Only like 120pages away lol

[still dew]

Huh

Which book

[languid trellis]

Jacobson: Basic Algebra I

[dull ginkgo]

No Sylow theorem is the next section lol

[signal dome]

what is a functor

[chilly ocean]

#category-theory

What are the applications of numerical analysis to algebra in particular group theory?

[crystal turtle]

What's with the 3x cross post asking about applications of numberical analysis to several different fields lol

Like what's the context

[summer path]

This looks very spammy, you might as well have just asked in adv lounge about applications to other areas

Also I'm pretty sure they cross posted in almost every channel lol

[chilly ocean]

i'm trying to get motivation for numerical analysis

[chilly ocean]

hmm you're right i guess i could. thanks for feedback

[south patrol]

3x is understatement

[crystal turtle]

Oh I have most of them muted lmao, make that 7 times

[delicate orchid]

Google “all of engineering”

[crystal turtle]

What the fuck is that

[delicate orchid]

It’s when people build things for actual people in reality. Like rocket ships and false hair implants

[chilly ocean]

idc about nonmath applications

[languid trellis]

So it's a 40 year old man playing kerbal space program

[dull ginkgo]

or a wew

[rocky cloak]

Well, computational algebra / numerical algebra is very useful to get your hands on explicit examples, so that's an indirect application. But I'm not sure that would fall under numerical analysis...

[mighty kiln]

Are you be @postgraduate

[chilly ocean]

what do you mean?

[mighty kiln]

Oh you be @postgraduate

[chilly ocean]

What?

[pliant forge]

your role

[chilly ocean]

What about it Eso?

[pliant forge]

arki asked if you were a postgraduate, and then realized you had the role indicating you probably were, i believe

[chilly ocean]

why did he ask that?

[pliant forge]

I suppose because of your interest in mathematical applications of numerical analysis, rather than engineering

[dull ginkgo]

Does a lot of group theory come as a consequence of the independence of characters lemma?

[delicate orchid]

some of it does

[dull ginkgo]

Actually I mainly mean the bijective-ness of the main correspondance for finite field extensions

[delicate orchid]

idk wtf ur talkin bout!

[vivid tiger]

hey, some of engineering isn't numerical stuff! e.g. coding theory, cryptography, apparently some telecom stuff but I'm not sure I believe them.

[vivid tiger]

explain?

I understand independence of characters

[delicate orchid]

telecom is fourier which is like................. like ok u get a pass but only by a technicality of FFT existing

[vivid tiger]

And the whole orthogonal basis for class functions

[vivid tiger]

no not that stuff

that's definitely numerical

other telecom stuff

[delicate orchid]

ohhh you mean scamming people by charging way more than you need to

riggghttt

[delicate orchid]

idk it's just one of those things that comes up occasionally

I recall someone using it to prove some analysisy thing

oh yeah

the functions {e^\lambda t, \lambda real} are linearly independent over R

you can prove this using linear independence of characters

[vivid tiger]

...please elaborate

also methinks one needs the compact group integral thing to do this

[delicate orchid]

yeah

I don't remember how it goes

[wooden fulcrum]

Hello

Why do I need a ring to contain the identity to prove that each proper ideal is contained in a maximal one?

If I have a chain C in the set of all ideals containing I, then since they're all proper and totally ordered by set inclusion, there is bound to be an element that is missing from them all hence U C would be proper also no?

No!

I see it lol

[delicate orchid]

when you union of the chain there is no gua- yeah you get it now

[wooden fulcrum]

Helpex writing it out

Yeah xddd

Wait or do I...

[south patrol]

Had never realised that was the case lol so thanks for mentioning

[delicate orchid]

concluding the union is not the entire ring really needs the fact that we can't have any units in any of the ideals in C

[wooden fulcrum]

Coule I not use Zorn's lemma to prove that there exists and element not contained in any of the members of C?

If C' is the set of x in R, x not in c for c in C then that is a chain right

Reverse order

[delicate orchid]

it's a set

but I presume you want to put the obvious filtration on it

C'_m = {x \in R : x not in C_n, n \leq m}

[wooden fulcrum]

Yes

Ah but I can't show hmm

[delicate orchid]

ok and why is this always non-empty?

[wooden fulcrum]

C is a set of proper ideals so theres always something missing from each C_n

And C is ordered by set inclusion so if something is in a C_n then it's also in all C_m for m≥n right

[delicate orchid]

true

[south patrol]

Why would that mean there's an element not in any of the things

for example R is the union of the sets [-n,n]

[wooden fulcrum]

Ah is an infinite union the set inclusion supremum?

[south patrol]

Yeah

Exactly

[delicate orchid]

in a sense, yes

[south patrol]

Union is the supremum of the sets you union over

[wooden fulcrum]

Ah then that makes sense

Is there no way to do it otherwise, it feels weird that you can union sets none of which contain x but the union does? Nvm i see this isnt the caee now

Wait or

Ah yeha

Infinite moment

[south patrol]

In fact lol

You can turn subsets of a given set into a ring

and i imagine if you play around you can make it non-unital and stuff to make it have this issue

but idk

[wooden fulcrum]

So uh

I forget how the example i wanted to ask about went

[rocky cloak]

It might be nice to look at an example. If the product of any two elements in the ring is 0, then any subgroup is an ideal. So you just need to find an abelian group with no maximal subgroup.

Consider the 2-Prüfer group
Z/2^infty = {a/2^n}/Z < Q/Z

The proper subgroups are exactly the ones generated by 1/2^n and you have
(1/2) < (1/4) < (1/8) < ...
so there is no maximal one.

You might also notice that the union of all the proper subgroups is not proper anymore

[spice whale]

i think these two things are equivalent for groups

since Grp is cocomplete

[rocky cloak]

Which two things?

[spice whale]

no maximal proper subgroup and the union of all proper subgroups being the whole group

[rocky cloak]

Well, you need some chain condition. Like the union of proper subgroups of Z/2 x Z/2 is everything

[spice whale]

hm
yeah actually I'm wrong

i wasn't actually thinking of maximal

[celest furnace]

Won't union of all proper subgroups always be the whole group unless the group is cyclic?

[rocky cloak]

Yeah, if an element isn't contained in any proper subgroup then it's a generator

[spice whale]

I'm actually completely wrong actually

the union of two subgroups is not the colimit of the inclusions

[delicate orchid]

no maximal proper subgroup and the union of all proper subgroups being the whole group
The C_p^2 in question

[spice whale]

wait

[delicate orchid]

and yes, unions are not coproducts

[rocky cloak]

If you include all intersections it is

[spice whale]

yeah you should just ignore everything i said

[spice whale]

consider <a> and <b> in F_{a,b}

[rocky cloak]

Right, groups can be nonabelian as well

Forgot about that

[delicate orchid]

do you have a single r/math post to back that claim up

[rocky cloak]

Wait, let me quickly post to Reddit

[spice whale]

this is not the reason
consider <a> and <b> in F^Ab_{a,b}

[delicate orchid]

this is what I do but with the journal of algebra instead of reddit

[spice whale]

ignore my terrible notation

[delicate orchid]

what are we doing?

coproducts of groups?

[rocky cloak]

Ah right, union doesn't mean sum either

[spice whale]

the colimit of the inclusions of these subgroups into the whole group is not their union

[rocky cloak]

Direct limit equals union though

Fun fact

[delicate orchid]

unions of groups aren't groups so why should I be surprised by this

[spice whale]

exactly

[delicate orchid]

????????

[spice whale]

this is one of the reasons why i was so wrong

[delicate orchid]

u said urself that group was cocomplete what's going on

CHAT I'M SCARED

anyway I'm over it now

[spice whale]

Grp is cocomplete yes

[delicate orchid]

the actual coproduct is way cooler

[spice whale]

free product

[delicate orchid]

really nice example of when biproducts don't exist

[rocky cloak]

Now I'm wondering. Is there any simple characterisation of (abelian) groups with a maximal subgroup. Or perhaps better, where every proper subgroup is contained in a maximal one?

Finitely generated is sufficient, but probably not necessary

[delicate orchid]

once more I need to remember that infinite groups exist

so like a "local" group kind of thing

a single maximal subgroup?

or just A maximal one?

because I really. REALLY. Don't want to think about Zorn's right now

[delicate orchid]

this is just cyclic of prime power order right?

[coral spindle]

I think so, yes

[delicate orchid]

yeah, otherwise if M is your maximal proper subgroup, there will be some element x not in M that generates a proper subgroup - unless you're cyclic of prime power order lol

[coral spindle]

Doesn’t Noetherianity give us what we want? As in, Noetherian Z-modules. The maximal subgroup isn’t unique but it guarantees existence, and the converse is obviously false

[wooden fulcrum]

Thank you that's a great example

a*b=0 forall a,b just from defining it to be as such right

[delicate orchid]

ahh that's smart

[coral spindle]

An Abelian group has a maximal subgroup iff it has an Noetherian quotient

Whoops

Meant Noetherian

[delicate orchid]

drop the abelian

[coral spindle]

Saying so little

[delicate orchid]

a group has a maximal subgroup iff it has a quotient

one of the stupidest things I've said in the server

[sly crescent]

(Is nontrivial)

Wait no

The trivial group has itself as a quotient

[rocky cloak]

Yeah Noetherian equals finitely generated for groups I guess.

But finitely generated is all you need, even for modules over non-Noetherian rings

[coral spindle]

Nice

[rocky cloak]

So the question is whether there's a characterization that catches infinitely generated ones

[coral spindle]

Iff it has a finitely generated quotient

[rocky cloak]

Might as well reach for simple quotient at that point

[coral spindle]

Idk this seems relatively impossible