groups-rings-fields

[coral spindle]

Nice

[delicate orchid]

D_8 has 3 iso classes of fusion systems on it, Q_8 only has 2.

trolled once more

anyway this makes sense now

maybe.

[coral spindle]

Well all we get is

Well

I'll think of the normal subgroup iso to C_4

[delicate orchid]

I'm far too tired to induce things from an index 2 subgroup in my head. Do you think I'm some kind of braniac?

[coral spindle]

Trust me it's OK, just think about the degrees

We have four characters down there. Two of them have inertia group the entirety of the group

[delicate orchid]

yeah, trivial and the (1,-1) mf

[coral spindle]

And Gallagher tells us therefore (assuming they extend -- they do) that there will be four linear characters

[delicate orchid]

I don't think you need Gallagher for that

[coral spindle]

And the other one? Degree two, since it's (as a class function) the sum of two linear characters

[coral spindle]

Allow me my abstract nonsense

So you get 1 1 1 1 2

[delicate orchid]

ah ok. Then the rest is trivial

[coral spindle]

Those are the degrees, and you even know the representations since you've described them as induced from linear characters!

Sick huh? I love this shit

[delicate orchid]

are u telling me a p-group has monomial characters??!?!?!??!

[coral spindle]

Stunning ik

You can get a lot out of this in the McKay setting, as you're likely aware

Since we always have a normal subgroup when we're working with da normaliza

[delicate orchid]

erm.. what the deuce?

[coral spindle]

In my lovely finite groups of lie type, the Sylow subgroup is often cyclic too. So we get an awful lot indeed.

[delicate orchid]

buddy. U ever heard of "A_n"?

[coral spindle]

Yes... for nice primes.....

[delicate orchid]

so... n square free...

got it....

[coral spindle]

Lol

[delicate orchid]

ok now u got me thinking

what are the sylows of the finite groups of lie type

[coral spindle]

For my beloved Sp_4(q), all the Sylow ell-subgroups are cyclic... for ell =/= 2 or p

[coral spindle]

!!!!!!!! I'm so glad you asked

Firstly: I have no idea for the prime 2

[delicate orchid]

A_n is easy, B_n, C_n, are easy unless p =2

I forget how D_n works

and the exceptionals are a finite problem

[coral spindle]

Secondly: in almost all cases, they're in the normaliser of a torus

[delicate orchid]

oh it's THAT kind of bullshit is it

[coral spindle]

And there are effective ways of computing the order of the normaliser of the torus

Very effective

Basically

[delicate orchid]

actually of course it is they're reflection groups lmfao

[coral spindle]

Compute the determinant of a certain action on the character lattice

and also

mix in the # of fixed points of the Weyl group

It turns out to be very nice indeed

We can't compute them explicitly, usually, since this involves calculating the Lang map. But we can compute them up to isomorphism.

[delicate orchid]

SNOORRREEEE

[coral spindle]

🤓

[delicate orchid]

compute them explicitly I don't want any more excuses

[coral spindle]

It's actually so sick Wew

I love it so much

[delicate orchid]

ok now do it for the complex reflection groups

all 37 of them

well, 34+3*\infty

[coral spindle]

erm 🤓 the Weyl group is a real reflection group always

[delicate orchid]

does he know

[coral spindle]

For reductive groups at least

Is there something I don't know? Mayhaps

[delicate orchid]

groups
oh no no no he does not know

[coral spindle]

Oh no

No no not this 'fyoo zhon'

[delicate orchid]

https://link.springer.com/content/pdf/10.1007/BF01237357.pdf

[coral spindle]

Oh right Spetses

Fine, fine

[delicate orchid]

he knows!

[coral spindle]

Gunter Malle scares me

[delicate orchid]

he should

[hot quartz]

sounds like you guys know how to handle sylow groups. can someone explain to me how to prove that a group of order 24 is not simple using the sylow theorems?
I don't think I can use the same argument I've used for other orders so far and I'm wondering what I'm missing:
If it has three 2-Sylow groups and four 3-Sylow groups, where is the contradiction?

[south patrol]

Hm maybe an element count works but hm

may be possible to use smth more fancy ig

[rustic crown]

the group acts on the three sylow 2 subgroups by conjugation, and this action is transitive. That gives a homomorphism G --> S_3

kernel is non-trivial and normal.

[south patrol]

yeah that is what i mean by more fancy - considering actions on groupsi snice

oof

[rustic crown]

hi walter

[agile burrow]

hi det

hope you're well

[south patrol]

hi walter and det

[rustic crown]

hi potato

[agile burrow]

hi potato

[south patrol]

hello hehe

[rotund aurora]

Let $B\subseteq A$ be an integral extension of rings. Let $I$ be some ideal of $B$. Then, the induced map $B/IB\to A/IA$ is also integral. This is injective if and only if $AI\cap B=I$?

[chilly ocean]

how would element count work? i'm learning this as well

[cloud walrus]

croqueta3385

[south patrol]

hm i'm unsure here, i think acting on sylow is nicested

[rustic crown]

how would you count elements from the sylow 2 subgroups? sylow 3 are nice and intersect only at {e}. so ig that gives (3-1)*4 elements of order 3

[south patrol]

i'm not sure about the 2 yeah

[south patrol]

sure yeah

seeing B contained in A is hurting my brain though

But yeah B -> A -> A/IA has kernel IA cap B, which of course contains I

so it becomes injective iff that kernel is just I

[rotund aurora]

I don't understand this example

what is the injection B-->A here?

B/x=k[y]/y^2 and A/x=k[z]/z^2

[hot quartz]

I'm fairly certain counting is the way I'm supposed to solve this task since they did the same for order 12 but I can't figure it out either

[rotund aurora]

well lol if k is char 2 you can send x to x and y to y

ah nvm

that works

because (xz-y)^2=x^2z^2-2yxz+y^2=-2y^2+y^2=-y^2

but then y is sent to zero

[dull ginkgo]

Oh god oh fuck

Welp, anyway $N(a + b\sqrt{pq}) = (a + b\sqrt{pq})(a - b \sqrt{pq}) = a^2 - b^2pq$

[cloud walrus]

THE TUBE

[dull ginkgo]

Assume $u = a + b\sqrt{pq}$ is a unit, then $N(u) = \pm 1$. assume $N(u) = 1 = a^2 - b^2pq$

[cloud walrus]

THE TUBE

[dull ginkgo]

going to look into irreducibles

[rotund aurora]

||sqrt(pq)sqrt(pq)=pq||

[dull ginkgo]

Let w = sqrt(pq)

Assume p is reducible in Z[w], then we have a non-unital x = a + bw in Z[w]/Z such that x | p. Likewise x' = a - bw | p, implying (a + bw)(a - bw) = a^2 - b^2pq | p. p is prime so a^2 - b^2pq = s1 or sp where |s| = 1

Assume a^2 - b^2pq = sp, then a^2 = (s + qb^2)p implying p | a. let a = pc

(pc)^2 - b^2pq = sp implying pc^2 - b^2q = s

-qb^2 = s (mod p)

q has an inverse modulo p, call it k, then:

b^2 = q^-1 (mod p) or b^2 = -q^-1 (mod p)

First case cannot happen, q is a nonresidue mod p. By Euler's criterion, (-1)^{(p-1)/2} = (-1)^2n = 1, thus -1 is a q-residue mod p, thus the second case cannot happen. Proof by exhaustion, a^2 - b^2pq cannot equal plus/minus p

Therefore p is irreducible (prime) in Z[w], but pq = w^2

which implies that p | w^2 => p | w

[rotund aurora]

I think this is overcomplicated. If p=ab then p^2=N(p)=N(a)N(b). If a nor b are units then N(a)=+- p, so that you can find integers x,y with x^2-y^2 pq=+- p. This leads to the equation p x^2-q y^2= +-1. Reducing mod p gives that q is a square mod p

note: you should use that -1 is a square mod p

[dull ginkgo]

I am going off of what I did on paper and I just wrote out the legendre symbol sorry about that

[dull ginkgo]

Alrighty yeah that can allow me to chop out some of the extra work

thanks

i have a quick question croqueta if you don't mind

[rotund aurora]

to finish the proof you have to show that p and sqrt(pq) are not associate (i.e., one is not the multiple of the other), but this is obvious

[dull ginkgo]

not seeing it off the top of my head let me think

p | sqrt(pq) => p^2 | pq => p | q (contradiction)

[celest furnace]

Looks good

[dull ginkgo]

Is the reason we can put a norm on Z[a] due to quadratics being equivalent up to scaling and translation?

[rotund aurora]

mmh I'm not sure I understand the question

[dull ginkgo]

Well when we append the root of a quadratic integral polynomial to Z

we still get a norm it seems

is this exclusive to quadratics?

[rotund aurora]

the norm is a quite general construction coming from field extensions

[dull ginkgo]

is it just the constant coeff of the min poly?

[rotund aurora]

yes

[dull ginkgo]

I see

[rotund aurora]

well

here it is implicit that if $\alpha$ is some algebraic number then we are talking about the norm coming from the field extension $\Q(\alpha)/Q$

[cloud walrus]

croqueta3385

[dull ginkgo]

Let a be a root of a monic integral polynomial of degree n

Z[a] ~= Z[X]/(p(X))

it's an integral extension so Z[a] has the structure of a rank-n free module over Z

[rotund aurora]

in this case the norm will be the product of the roots of the minimal polynomial of \alpha, or the constant coefficient of the minimal polynomial times (-1)^n where n is the number of roots. Note that not all roots of the minimal polynomial of alpha need live in Q(alpha)

[dull ginkgo]

but the issue is getting the norm from that constant coefficient? Would it just be the product of each element's conjugates. But what would be the "minimal polynomial" of an arbitrary element of Z[a]

[rotund aurora]

In general, if $L/K$ is a finite field extension and $G$ is the Galois group (of the Galois closure) of $L$ over $K$ then given $\alpha\in L$ you define the norm $N(\alpha)=\prod_{\sigma\in G}\sigma\alpha$

[cloud walrus]

croqueta3385

[dull ginkgo]

i haven't gotten to too much galois stuff yet

[rotund aurora]

well you just adjoin to L all the roots of all the minimal polynomials over K of elements of L

this will give the normal closure, to be Galois you will need separability but just assume L and K are field extensions of Q, so everything is separable

[dull ginkgo]

i will come back to this after I learn more about galois extensions

[rotund aurora]

note that for example the Galois group of Q(i) consists of the identity and the automorphism bringing i to -i

so then the norm of a+bi will be (a+bi)(a-bi)=a^2+b^2

[dull ginkgo]

i wonder at some point is Galois theory has a connection to characters

though I suppose automorphisms are

themselves "additive" and automorphic characters from G^X to G if you ignore 0

[rotund aurora]

you can also compute the norm from linear algebra, by regarding L as a vector space over K and looking at the linear transformation x-->alpha x. Then the norm is the determinant

[dull ginkgo]

Interesting chain of ideas

Characters are linearly independent over an integral domain, in particular fields

automorphisms are a special subset of characters, still linearly independent

I am going to toy with this

[cobalt heath]

Automorphisms were linearly independent? hmm

[chilly ocean]

what is the kernel of this action? or how can i see its non-trivial?

[south sinew]

Since there's someone stduy algebra, there must be someone study cryptography
DMs me, please, i wanna some communication

[lusty marlin]

Do you know what the kernel of a group action is?

[chilly ocean]

Intersection of stabilizers?

[lusty marlin]

Yes

[chilly ocean]

I know what it is, I couldn’t find it in this particular example

[rocky cloak]

Hint: 24 is more than 3!

[chilly ocean]

In that case, the fact that the action is transitive is irrelevant?

[rocky cloak]

Well, you need the kernel to not be everything either

[chilly ocean]

Okay, thank you

[summer notch]

Problem VII.4.5 from Aluffi: Can't we just say like, take the roots of g in F, and consider the field generated by those roots, which is clearly the unique splitting field

i don't understand why there is work to be done in this question

here is Aluffi's definition of splitting field

[glad osprey]

I'm trying to understand a proof that any group of order p^2 is abelian, for prime p. They do this by showing that either G is cyclic, or G is isomorphic to <a>x<b> for two elements a, b of order p. But they claim that <a> is normal in G because of one of the Sylow theorems. I don't get why. <a> is not even a Sylow p-group, it's just a p-group, and it's not the only one in G

[loud merlin]

Only if you see those automorphisms as group automorphisms instead of K-akgebra's

For the "subset part", I think the dependence part is false

[tough raven]

Fun fact: this is not true for division rings.

[tough raven]

Maybe the fact that such a field exists and is unique, so that you're entitled to call it "the" splitting field?

[tough raven]

When you write that G is isomorphic to ‹a› × ‹b›, do you mean the direct product of the groups ‹a› and ‹b› on the RHS?
Because that's already necessarily abelian (direct product of abelian groups is abelian). (And so ‹a› is normal, if that's still important.)

[dull ginkgo]

Wouldn’t be surprised

Could probably find one set of linear dependent characters over the ring of rational quaternions

[cobalt heath]

What is linear independence of group automorphism?

Meh I should know this but

[dull ginkgo]

Not really the group automorphism itself but the character it induces

[cobalt heath]

Hmmm

phi : G -> G induces.. uh

[dull ginkgo]

I.e group automorphism F^X to F^X

[cobalt heath]

What is X?

[dull ginkgo]

It’s the times symbol

I am lazy

[cobalt heath]

Ah, so you mean automorphism of unit group for fields?

[glad osprey]

Yeah, but they need to show that <a> is normal first. Then G iso to <a> x <b> is eventually the conclusion, which shows that G is abelian as you say

[tough raven]

Ah, maybe it's because [G : ‹a›] = p then. There is a theorem in finite group theory that if the index of a subgroup of a finite group is the smallest prime factor of the order of the group, the subgroup is normal.

[cobalt heath]

Does some texts include this one among one of Sylow theorems?

[tough raven]

I think they were referring to a field automorphism.
An automorphism of a field F can be viewed as a function from F to F, and the set of all automorphisms is a linearly independent subset of F^F (F-vector space with pointwise operations).

[tough raven]

Never that I've seen.

[cobalt heath]

Hmm

[glad osprey]

I guess it's just a mistake in the proof

[cobalt heath]

Ouch, just found out that miz just said automorphism, not group automorphism

I want die me ashamed..

[tough raven]

?

[tough raven]

Can't you use this?

[glad osprey]

I mean, it's a step in the proof that is wrongly attributed to Sylow. But the conclusion is correct

[dull ginkgo]

Well a character is a map from a monoid M to an integral domain or field K so a field automorphism can be forg’d into a character from the field’s multiplicative group into itself, since it must fix 0

[tough raven]

It's OK, the word ‘character’ probably understandably primed you to think about groups.

[cobalt heath]

Yeah, field automorphism is a character

[dull ginkgo]

Therefore group automorphisms are linearly independent

[cobalt heath]

Maybe I should have studied character theory explicitly, seeing how this terminology confuses me

Eh, perhaps too late

[dull ginkgo]

I don’t know much about character theory either

Tldr it’s just a monoid map into a group

*field

[weary spade]

Prove that Zp is a field for p prime

[dull ginkgo]

Use p being coprime to every nonzero element, and use Bezout lemma

[weary spade]

So for the sake of illustration let's let p = 11

Then every element is coprime to 11

So then we can construct the inverse by ax + bp = gcd(x,p)

So like for x = 5 we have

11/5 = 2 r 1
5×2 + 1 = 11
11 - 5×2 = 1
So 5^(-1) = -2 = 9

So you can construct it.

[tardy hedge]

groups of order 6 are either cyclic or S3 right?

[weary spade]

What about A6

[tardy hedge]

what is A6 again?

[lusty marlin]

A_6 doesn't have order 6

[weary spade]

Nvm

[tardy hedge]

A6 is alternating group or smth?

[lusty marlin]

The group of even permutations in S_6

[tardy hedge]

oh ok thanks

which is more than 6 elements

n!/2 elements

trying to remember why its that though

[lusty marlin]

Construct a bijection between the odd and even permutations

[dull ginkgo]

Let G be order 6 and not cyclic, then there’s 3 Sylow 2 subgroups and G acts on them by conjugation. That gives a morphism from G to S_3, and because the groups have the same order, it’s an isomorphism

[south sinew]

do u guys have some recommanded books about group, ring and fields

I just don't understand why these concepts is proposed or introduced

So abstract, and hard to learn and understand

but why in an abstract way

[lusty marlin]

Dummit and Foote

[south sinew]

i don't think it is convenient

It is hard to understand

[lusty marlin]

Convenient for what? First of all, nobody is forcing you to study about these things

[south sinew]

I am not taking about the energy i spent
I am takling about the generation of this subject to have a better understanding

for me, To figure this out can make this subject easy to learn

[loud merlin]

Do you prefer reproving everything about an object that looks like Z or just see it's a ring and apply the whole general theory

That's the thing

[south sinew]

kind of understand

It's like summarizing the general law of the existence of specific

At what stage do u study this subject
high school?

[loud merlin]

No

Undergrad, usually second or third year maybe

[south sinew]

okay
I learn this to do my research in cryptograhpy

[loud merlin]

Highly depends on how your country's educational system works etc etc but whatever

[loud merlin]

Oh yeah that indeed might be useful

[tardy hedge]

At my school abstract algebra gets introduced in third year

Of undergrad

[south sinew]

same

but second year

@loud merlin wait, why your pfp YeWenjie?
U Chinese?

I suddenly realized it

[loud merlin]

No, I don't need to be Chinese to read the books / watch the show lol

[south sinew]

Three body is classic

I am a Chinese

[loud merlin]

Oh okay

[untold crag]

(i) help

[rocky cloak]

Think about 2^3

[untold crag]

too hard gonna come back to it i guess

[long obsidian]

6R as in the ideal sum of the ring with itself 6 times?

[rocky cloak]

How about this
What do you get if you expand
(x + x)^3 ?

[untold crag]

6R = {a+a+a+a+a+a | a in R}

[long obsidian]

How does that help you get 8x^3 don't you?

[rocky cloak]

Alright, so 8x^3 = 2x

Then you can simplify that a little bit

[untold crag]

sorry ill do this tmrw

[dim widget]

it’s 8, I got it!

[long obsidian]

How does this statement even make sense. Isn't the sum of two ideals a sup set of each ideal?? So 6R contains R. It seems like you're saying the ring is trivial if a^3=a. Come to think of it I can't think of a ring that satisfies this. Is the ideal sum not bigger than each ideal?

[delicate bloom]

way I see it, as long as the ring contains 1 you can make the element 6=1+1+1+1+1+1 and then 6R would be the principal ideal generated by 6

[rocky cloak]

It just refers to the elements of the ring that are multiples of 6.

You wouldn't think 6Z = Z for example

Don't even really need 1 in the ring, you just define it exactly like gkn1 suggested

[delicate bloom]

their response I don't believe was meant to imply that it was the ideal sum of R with itself 6 times. For instance since R+R = {a+b | a, b in R} while 2R = {a+a | a in R} are not the same thing. I think that's what's confusing you there

[dull ginkgo]

OH NO OH NO

[untold crag]

wat

[delicate orchid]

questions about rings with a = a^3 tend to have a reputation

[dull ginkgo]

I’ve tried a problem like this and it was hell

[dull ginkgo]

Rings where for each x there is a n > 1 where x^n = x are commutative

[delicate orchid]

https://tenor.com/view/erm-what-the-freak-puppy-stare-cute-reaction-gif-15215675160072457679

[rotund aurora]

Classic problem at this point

[untold crag]

well i'm glad other people find it hard, earlier i was beating myself up over not being able to do it

i dont understand what this 'let x be a symbol' means

is f actually a function and x its parameter

[summer path]

x is just some indeterminate variable

[untold crag]

so f is a function right?

[delicate orchid]

nope

[summer path]

f is just a formal polynomial

[delicate orchid]

functions need a domain and codomain, I see nothing of the sort here

[untold crag]

so is this touching like

I just dont understand ive never seen this concept before

[delicate orchid]

it's just a symbol

it's actually simplier to understand than whatever the hell the a_is are

[untold crag]

why are the a_ hard to understand?

also why can we multiply a_ by x even though x isnt necessarily in R?

[delicate orchid]

they're not - it's just that the x is even easier

it's literally just a symbol

[untold crag]

that doesnt make sense

like is this some formal syntax thing

should i have seen this concept before?

[sinful blaze]

They have written it in a weird way. The natural way would make to make the a_is constants denoting elements of R. Instead they're saying x is a symbol and the a_is are elements of R, which feels like a type mismatch though technically there's no problem

[spice whale]

this seems false
consider Z⟨X,Y⟩/(X²+1,Y²+1)

hm

actually i suppose you don't have (XY)^n = XY

[delicate orchid]

can we conclude that X.. yeah exactly

[untold crag]

f represents a range of different values depending on what x is right?

[spice whale]

hrm

[delicate orchid]

no, f is not a function

[spice whale]

hm

[untold crag]

what is it then

[delicate orchid]

a polynomial. It says right there

[spice whale]

uh
consider the quaternions on F_p?

[delicate orchid]

babble alert! babble alert!

do you mean like

[delicate bloom]

you can think of it as just giving you the form of adding and multiplying polynomials without evaluating them

[untold crag]

ohhh okay I see

it literally is a new type of object to me

[delicate orchid]

F_p<i,j,k>/(i^2+1, j^2+1, k^2+1, ijk+1)?

[spice whale]

yes

[delicate orchid]

to demonstrate the difference, as functions F_p -> F_p, x^p and x \in F_p[x] are the same due to fermat's little theorem

but they're very different as polynomials

[untold crag]

i dont understand your notation

[spice whale]

Fermat's tiny itty bitty theorem

[delicate orchid]

what exponent are we using on elements of F_p

[spice whale]

wdym

[delicate orchid]

well we need x^n = x for all x in your ring

[spice whale]

yeah 5

[delicate orchid]

right so F_5

[spice whale]

they all have 5

well p

in my head F_p is F_5

[delicate orchid]

p = 1 mod 4

my beloved

[spice whale]

actually all the elements of this ring have exponent 5

i suppose

because frobeanius

1 5 10 10 5 1

[delicate orchid]

there might be something of higher expontent we're missing. I do not have the will to think about it

ok I thought about it

(i+j)^5 weirds me out

,w expand (x+y)^5

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by wewladstbh

[delicate orchid]

hmm... you win this time...

[spice whale]

frobeanius wins again

^5 is an endomorphism

[delicate orchid]

hmm... but what about (j+k)^5....

[spice whale]

it's actually the Identity

[dull ginkgo]

I had to use this crazy equivalence relation and split the group down lmao

[delicate orchid]

I cannot wait until I never have to think about any of this ever again

[dull ginkgo]

Then don’t

[spice whale]

and ultimately this is a false statement

[delicate orchid]

alas, I must

[dull ginkgo]

You sure?

[delicate orchid]

check this shit out @dull ginkgo

[dull ginkgo]

Oh ignore 0

[spice whale]

yeah i've literally just given a counterexample

[delicate orchid]

lets not be too hasty now

[delicate bloom]

what was the counterexample

[delicate orchid]

quaternions on F_5, apparently

[spice whale]

F_5 quaternions

[delicate bloom]

which element in there fails

[delicate orchid]

I just don't like it...

[spice whale]

none of them

[delicate orchid]

I'm suspicious........

[spice whale]

frobenius endomorphism

since it's char 5

so x⁵ = x for every element x

[delicate bloom]

this expansion assumes i, j commute if you're trying to look at (i+j)^5

[rocky cloak]

You haven't really proven that x^5 = x.

And the original statement is true, so there can't be any counter examples

[delicate orchid]

I know I was just taking the piss by asking wolfram

[spice whale]

does it assume they commute

ah

[delicate orchid]

that's precisely why I'm still sus of it

[spice whale]

i see

we have the trivial ring i believe

[delicate orchid]

wait haven't we done this before

wait

[spice whale]

kind of

[delicate orchid]

[Just a burning memory gradually fades in]

[spice whale]

the commutative quaternions are the trivial ring

[dull ginkgo]

Burning memery

[spice whale]

but the one we've given now is nontrivial

it just doesn't have x⁵ = x

[dull ginkgo]

Do you mean the abelianization of the qaternions

[delicate orchid]

take a guess... brainaic...

[dull ginkgo]

consider any commutator of the Q_8 group lmao

[spice whale]

(i+j)^5 is something funny exterior algebra

in that you flip the signs or whatever

[delicate orchid]

uhhhhhh ok!

wait

if that's true it should be 0

[spice whale]

it might be

wait

no it's not

is it?

no because you get the funny i⁵ and j⁵ at the end

uh

what is it actually

[delicate orchid]

odd degree elements square to 0 in graded commutative rings

[spice whale]

yeah but it's not commutative innit

[delicate orchid]

did I say it was

[spice whale]

yes

[delicate orchid]

I said graded commutative not commutative graded

[spice whale]

graded commutative rings

what

[dull ginkgo]

fuck this

[delicate orchid]

$xy = yx(-1)^{\text{deg}(x)\text{deg}(y)}$ is graded commutative

[cloud walrus]

Wew Lads Tbh

[spice whale]

oh

[delicate orchid]

question is, is what we have graded commutative.... probably!

probably not

oh well!

too many loops spoil da broth...

[spice whale]

(i+j)⁵ = 2i+2j i think

[delicate orchid]

where have the k's gone

oh that's a 5

can u write it normally

[spice whale]

plus minus plus minus

or whatever

[delicate orchid]

I thought it was a 2

[rocky cloak]

(i + j)^2 = i^2 + ij + ji + j^2 = -2
So (i+j)^4 = 4, so
(i+j)^5 = 4(i+j)

[spice whale]

you thought what was a 2

[delicate orchid]

don't

oh

the 5

[spice whale]

i see

[delicate orchid]

what else looks like a 2

[spice whale]

1

kinda

7 does

[delicate orchid]

wow somebody was actually bothered to do the explict calculation

[spice whale]

if you look at it funny

[dull ginkgo]

Anyway the first step of that psycho problem for me

Is establishing (what feels like) torsion

Then i had to split it into commutative subrings

[rocky cloak]

Is it possible to do something clever, or is it just horrible calculations all the way down?

[dull ginkgo]

In particular, subrings of Z[x] for x in R

[dull ginkgo]

call the minimum n > 1 : x^n = x the rank of x, r(x)

it’s easy to show x^k = x for k >= r(x), k = 1 (mod r(x) - 1)

assume we have some n > 1

k = lcm(r(x) - 1, r(nx) - 1) + 1

(nx)^k = n^k x^k = nx = n^kx
(n^k - n)x = 0

[rocky cloak]

Oh, you're letting n depend on x as well

[dull ginkgo]

So each x has an additive order

[dull ginkgo]

Y e s

[rocky cloak]

Sounds rough

[dull ginkgo]

So every element has an additive order

Back

So ord(x) | (n^lcm(r(x) - 1, r(nx) - 1) - 1)n

For each n

[rotund aurora]

Mmh maybe try doing some concrete case like x^3=x for all x or that a finite noncommutative field is commutative

[dull ginkgo]

I basically considered Z*(x)

[dull ginkgo]

Anyway as a break

I need to show Z[sqrt(w)] is a euclidean domain where x^2 - w = 0 is irreducible over Q

don't tell me to latex it

i swear to

[next obsidian]

I’m pretty sure if you use the standard function on C

You will end up with what you desire

Idk where the irreducible assumption comes in but it will

Because I said so

[dull ginkgo]

i have to use the norm function

[next obsidian]

Yeh

[dull ginkgo]

x = a + bs {s = sqrt(w)} has norm N(a + bs) = a^2 - sb^2 which is nonzero for each x

[next obsidian]

Yeah

[dull ginkgo]

(if it is, then (a/b)^2 - w = 0 giving x^2 - w a linear factor which can't happen)

[next obsidian]

[dull ginkgo]

so then

Assume we have elements x = a + bs, y = c + ds \neq 0

We want to find elements q = u + vs, r = p + qs such that
yq + r = x and |N(r)| < |N(y)|

this is harder than I thought it would be

Basically we just need to find a q that satisfies |N(x - qy)| < |N(y)|

oh wait it's not a general statement

sqrt(-10) is not UFD

[long obsidian]

Say R a ring and b an ideal and x an element of R. Are xb and (x)b different sets?

[south patrol]

I guess the left side may fail to be an ideal for example

As an extreme example, take b = R

Then you are comparing xR with (x)

Then the two definitely needn't coincide

[barren sierra]

any hints on (ii) I'm being dumb

I feel like I need to use the fact that x in H itself

[terse crystal]

G is direct sum of A and B, where A is sum of n many cyclic groups of order powers of 2, B sum of cyclic groups of order prime powers. Then clearly H=(Z/2Z)^n. Try calculate sum of elements of (Z/2Z)^n

[barren sierra]

Ohhhhh

Use classification, I see

[terse crystal]

Yeah

[barren sierra]

Thanks

[terse crystal]

Np

[dull ginkgo]

good god

[south patrol]

lol

Oh writing g for an additive group hurt my brain slightly

I saw sum of g and assumed this was the like central idempotent of the group ring or whatever

up to scaling

[cobalt heath]

Oh, what is this?

[dull ginkgo]

(i) if g is not an involution (order 2 element), then g^-1 is distinct from g and is also in G, thus will cancel when all added up. The only elements that won't cancel like this are involutions. Therefore we can strip out G\H from the sum and still have an equality. This implies that x is in H since H is closed under addition.

(ii) Assume |H| > 2. H must contain an element other than x and 0. Let y be the sum of elements of H other than x and 0, which is not an empty sum. This implies x = y + x + 0 => y = 0. So the sum of elements other than x and 0 is still 0

assume for x and y in H, x + y = 0, then x = y because both are involutions

wait

this feels like a contradiction

Can |H| > 2?

[barren sierra]

Yea (i) I got the same proof

[cobalt heath]

(Z/2Z)^2

[barren sierra]

Consider (Z/2Z)^2

Yea you got to my failed proof attempt lol

[dull ginkgo]

let me ponder for a second

[terse crystal]

Hint: you ||pairing f and (1,…,1)+f, see yourself how many (1,…,1) you get||

[barren sierra]

Yea it's a counting thing

That's where you need > 2

[cobalt heath]

Wonder if one could do some kind of reduction to prime power

[dull ginkgo]

|H| must be a power of 2

Every element is by definition an involution (is of order 2)

by legrange's theorem, |<a>| = 2 | |H|

so no other prime can divide |H| as it'd have a larger cyclic group

[south patrol]

Do you know what group rings are

Well i guess i should say group algebras which is preferable

[dull ginkgo]

omfg

[dull ginkgo]

so uhhhh

it's not

Just a bit weird to work out the details

[barren sierra]

Oh

Maybe it was too weird for me

It felt really circular

[dull ginkgo]

2 | |H| because each element is an involution right

And we stripped out {0,x} so H\{0,x} has an even number of elements

if it's empty we have a contradiction, otherwise there has to be another element we can strip out the same as the first. We have induction that leads to a contradiction unless x = 0 i think

[cobalt heath]

Yes

[long obsidian]

Oh whoops I see. The difference between xR and (x)R is somewhat whether the ring R is commutative. Maybe I could find an example of these being different in a matrix algebra or something like that

[cobalt heath]

R[G]

[cobalt heath]

Ah so it is enough to analyze that.

[dull ginkgo]

wait nvm

[dull ginkgo]

y not equal to x or 0 in H implies (y + x) is in H and is also not equal to x or 0

I keep trying to find an easier route but it's just harder than showing it's a power of 2 and using contradiction lmao

Does this use the lemma that every rational number is within or a half distance of an integer?

the multiplicative part is easy (conjugate is also in D)

x = (a + b*sqrt(-3))/2 for a = b (mod 2)

(a^2 + 3b^2) = (a^2 + 3b^2) = 4a^2 = 0 (mod 4)
so N(x) = (a^2 + 3b^2)/4 must be an integer for each x

[barren sierra]

I solved this by doing (ii) first and then (i). How do you solve (i) without (ii)?

[dull ginkgo]

Honestly (ii) feels like the de facto way to prove (i)

but uh, maybe take [x_1,x_2]?

Or conjugation?

[barren sierra]

those just feel like (ii) in disguise

[barren sierra]

I think so

[dull ginkgo]

Showing it's euclidean seems to elude me

[dull ginkgo]

Because it kinda is?

[barren sierra]

yea lol

[dull ginkgo]

For finite G two subgroups having coprime orders is equivalent to having trivial intersection

[lusty marlin]

No, two subgroups can have trivial intersection even without having coprime orders

They're not equivalent

[dull ginkgo]

Oh shit really?

Oh wait

Justified one in my head

[lusty marlin]

Lol take any two distinct subgroups of order 2

[dull ginkgo]

N_1 and N_2 is normal, so both are fixed by conjugation by G

Thus g_1 g_2 g_1^-1 is in N_2

[barren sierra]

coprime orders implies trivial intersection but not the converse yea

[barren sierra]

this + whatever you were probably going to write further is basically my proof of (ii) lol

[dull ginkgo]

Nvm the route i was gonna try leads to trivial intersection anyway

[long obsidian]

If R is a commutative ring and A and B are subsets of R. Then the product of sets AB is a well defined set.

If A and B are ideals then the product is also an ideal AB.

Say in this case that A and B are ideals then it's the case that the ideal product AB is exactly the same thing as the subset product AB right? I think I basically asked the same question but I wanna be sure if this is obvious to anybody

[dull ginkgo]

For commutative rings yeah they are the same

For noncommutative rings it is sums over the subset product

aka the Abelian additive group generated by the subset product :3

[barren sierra]

yea the definition of the product of ideals is the sum of such products of elements (in a commutative or noncommutative ring)

[dull ginkgo]

Wait i am mistaken because i was thinking of the group theoretic def of ideals for a second

It’s sums over the product lmao

[agile burrow]

Yeah, they don't agree even for commutative rings

[dull ginkgo]

I was thinking of principal ideals for a second

For commutative rings it is just aR but for noncommutative it is sums over RaR

My brain is frying today Jesus

[dull ginkgo]

Gonna take a shot at this

I need some help at the euclidean domain part

We have x = x_r + x_i sqrt(-3) \neq 0, y = y_r + y_i sqrt(-3)

1/y = y*/N(y) where y* is conjugation here. Issue is that I cannot state much about if 2 divides N(y) or not

[dull ginkgo]

@long obsidian Consider (a)(b). a^n*b^m might be in the subset product, but a^n b^m + a^m b^n isn't

[agile burrow]

Sorry I'm trying to juggle rn

[long obsidian]

Thanks okay it seems like the difference is whether it's a single term.

[dull ginkgo]

Oh. D^x is isomorphic to Z/6Z

idk if that's useful for any proof

[dull ginkgo]

y'know what. fuck this shit

i am constructing the eisenstein integers and constructing an isomorphism

[delicate bloom]

I'm thinking if you were given an example earlier with the Gaussian integers, instead of the basis {1, i} you can use the basis {1, 1/2+sqrt(-3)/2} and do the same trick with parallelograms

[dull ginkgo]

I basically used the primitive 6th root instead of the primitive 3rd because it’s more obvious to conclude

But yeah

[delicate bloom]

that is what I put

;P

[dull ginkgo]

Did you do Jacobson

[delicate bloom]

I hardly know him...

[dull ginkgo]

The textbook crying emoji

[delicate bloom]

nope lol

is that an algebra book

[dull ginkgo]

That’s where the q is from

Questions tend to be more difficult than D & F from my experience though I stopped D&F at groups

[delicate bloom]

I've heard of it now that I think of it

[dull ginkgo]

This set took me like, an hour and a half when in D&F it took me like a quarter of the time per problem set

[delicate bloom]

but mostly I learned algebra out of Artin, D&F, and Lang and none from alluffi nor jacobson

[dull ginkgo]

There is a big ol string of problems about Gaussian primes

Surprisingly none about Eisenstein primes. Idk about those, might try to prove some stuff about em

[delicate bloom]

yeah try to see if you figure out which prime split, ramify, or remain inert

[dull ginkgo]

I think I know how to do all of them

[celest furnace]

D&F is peak 🔥🔥🔥 I’ve almost read the entire book! Which is so wild to me

[dull ginkgo]

Gonna try them tomorrow

[cobalt heath]

Wow, how did you find time to read entire book?

[celest furnace]

Not all the exercises but I took algebra for this year and we covered basically all of it

[cobalt heath]

Hmm, do you read the textbook when you study?

I’ve nearly never done that

[delicate bloom]

silly question but, what's the alternative

[cobalt heath]

Not a silly question at all! For me, I just review what I heard in the lecture.

[long obsidian]

Do you guys think atiyah and matasuma's commutative algebra pretty much have the same material in them?

[cobalt heath]

Actually I almost don’t, just try to learn most in the lecture

[next obsidian]

Fuck no

This isn’t a matter of opinion

If you look at the table of contents you see AM has way less

[long obsidian]

Okay I thought matasuma had more but I couldn't interpret the table of content exactly cause I don't know how much is left in the exercises. Thanks

[celest furnace]

Used to not but this quarter I do

[sonic coral]

i just finished a whole year of algebra and we also did practically the whole book minus a module theory

[celest furnace]

Feels great right

[sonic coral]

i feel like i’ve learned a lot but there are still some parts of galois theory i don’t feel completely comfortable with yet, yeah

[kind temple]

sorry, what is the first question even saying? it reads like gibberish to me

[crystal vale]

if H is a subgroup of G, let S be the set of all right cosets of H then let there is a homomorphism between G and A(S). if o(G) does not divide i(H)! then kernel of that mapping is larger than {e}, because if kernel is {e} then image set is a subgroup of A(S) but by Lagrange theorem o(G) divide i(H)! which is not possible by hypothesis, is it correct argument?

[kind temple]

what are A(S) and i(H)?

[crystal vale]

A(S) is a set of all automorphism of s and i(H) is a index of H, o(G)/|H|

[mighty kiln]

Do you know what a polynomial ring is

[kind temple]

i just reread it. i had a TIA reading this. the "for which a second polynomial" part tripped me up

[kind temple]

thanks

[rapid junco]

Is there specific name for polynomials raised to some kth root

And are these well studied

Looking for a source

[untold crag]

with commutative ring R, is R[x] = R[y]?

i dont get how we can use x like this

i thought it was just part of the notation for constructing a polynomial

[elfin wraith]

They’re certainly isomorphic. The indeterminate x or y is just notation.

So it’s things of the form $r_nx^n+r_{n-1} x^{n-1} + \cdots + r_1 x + r_0$ where $r_i$ are elements of the ring R and your $x$s are just indeterminates

[cloud walrus]

Nope

[untold crag]

i understand, thx

[runic lava]

Only knowing that G is a group and H a subgroup of the center of G, how would you describe/write G/H ?

I’m struggling to see what the group looks like

[kind temple]

how do you want to describe it?

[runic lava]

Just with {}

[kind temple]

{gH : g in G} isn't good enough?

[cobalt heath]

Is H = Z(G) here

[runic lava]

I’m so bad at this

[kind temple]

a subgroup of Z(G)

[runic lava]

Z/3Z is {0 + 3Z, 1 + 3Z, 2+3Z}

[kind temple]

G/H = {gH : g in G} is defined for any group and subgroup as the collection of left cosets.
However, G/H is a group iff H is normal in G

[runic lava]

I just can’t wrap my head around G/H

[cobalt heath]

Ah, subgroup - then idk if there are specific ones

Effectively, gH is like g + H, but you use operation of the group instead of addition

[cobalt heath]

Btw this kind of repr is also a hack in some sense

[runic lava]

I see

[kind temple]

this one just breaks everything up via modulo classes. so all multiples of three are treated the same here, etc

[runic lava]

Okay i may still be a little confused

[kind temple]

how sway

[runic lava]

No no okay i’m getting there, thank you !

I need to prove that if G/H is cyclic, then G is abelian

[kind temple]

whatchya got

[runic lava]

I will try to do this now that i have a better representation of G/H

[solar dock]

Oh that. Is G per chance finite?

[runic lava]

No more than what i shared

[runic lava]

Not specified nope, just that H is a subgroup of Z(G)

[solar dock]

Darn. No sylow.

[cobalt heath]

Maybe good to begin with G/H = <a>

[solar dock]

I just realized I never did this for the infinite case.

[kind temple]

G/H = <g_0H>

for some g_0 in G

finite, infinite, the proof should go through the same

[cobalt heath]

Guess the proof is kinda anticlimatic

[solar dock]

;-; My proof for the finite case used Sylow (I did this a year ago) so that machinery is out of my toolkit now.

[untold crag]

isnt the ideal principal anyway

by definition

oh

i missed the second half of the sum xd

i thought it was an equality

[kind temple]

try looking at ||xH and yH||

[solar dock]

I'll try this.

[cobalt heath]

Alternatively, you can look at G -> G/H and uh
Commutator? Idk
Traditional way works better

[solar dock]

Also uh question: Say you got a family of subgroups {Ki} of G indexed by some index set I. Look at the cosets. Are all equivariant maps under left multiplication action of G between these cosets of the family characterised as the action of an element of G^I by right multiplication on Prod(Ki) and the maps taking Hi->Hj?

[runic lava]

@kind temple i’m still kinda stuck ngl haha

G a group, H a subset of Z(G) such as G/H is cyclic. Prove that G is abelian

I think we can say that there is a g in G such as G/H = <gH>

And if n is the cardinal of <gH>, G/H = {g^kH^k, k ranging from 0 to n-1} ?

[delicate orchid]

for a hint: ||G/Z(G) is cyclic means you can write every element of the group as x^az for some fixed x, and z central||

[runic lava]

Okay i got it, thank you !

[crystal vale]

let F be the free group on a set S = {a,b,......} , and let G be a group. Every map of sets f: S-> G extends in a unique way to a group homomorphism T: F -> G.
They defined a_1a_2......a_n -> f(a_1)f(a_2)....f(a_n), how can i show that this mapping is well defined?
Since ab is equivalent to acc^(-1)b but then how f(a)f(b)= f(a)f(c)f(c^(-1))f(b) ?

[terse crystal]

f(x^-1) is defined to be f(x)^-1

[stuck cosmos]

Quick sanity check: Given a finite dimensional real or complex vector space $(V, +, \cdot)$, the representation
\begin{align*}
\rho : \text{End}(V) &\longrightarrow \text{End}(V)\
A &\longmapsto \rho(A) := A
\end{align*}
is irreducible, right?

[cloud walrus]

SK2099

[rocky cloak]

That's correct, you don't need finite dimensional either.

[coral spindle]

So you're viewing V as an End(V)-module. Indeed it's a simple module, even if V is infinite dimensional. Any nonzero vector can be sent to any other one by some element of End(V)

[stuck cosmos]

Thank you; just wanted to make sure.

[runic lava]

Hello, for all g in a finite group G, card(g Z(G)) = card(Z(G)) ?

[delicate orchid]

multiply by a group element is invertible, hence it's a bijection

[runic lava]

That’s what i thought yes, but it sounds insane ?

The little intuition i have is telling me wtf

[delicate orchid]

how does that sound insane, take the map sending x to xy, then this map has a two sided inverse given by x -> xy^-1

[coral spindle]

Then adjust your intuition

If K is some subset of the integers, |K| = |K+n| for any n. This is the same.

[delicate orchid]

the one thing I will never find intuitive is the whole "topological group is generated by a neighbourhood of the identity". I know the proof off by heart and yet I do not like it.

[coral spindle]

Simply close your eyes and repeat it 50 times

[rapid junco]

Is the following a well studied property: what is it called when if I give you a set of functions and look at their sum. If their sum is zero, then all the functions must be zero

[crystal vale]

But there is no information about f

[terse crystal]

f is an arbitrary map from S, set of variables to G, when we extend f to F<S> to G

We require that f(s^-1) is defined to be f(s)^-1

[delicate orchid]

the map T sends a word a_1...a_n to f(a_1)...f(a_n), we want T to be a group homomorphism, so we require T(a_1a_1^-1) = T(1) = 1, hence we require exactly what cogwheels said

[terse crystal]

Or put it this way, we define g(s)=f(s), g(s^-1)=f(s)^-1, g(Πs^ε)=Πf(s)^ε

g(s^-1)=f(s)^-1 is a rule, we require so

It has nothing to do with the original f

[crystal vale]

Yes we require but f is an arbitrary, so we define f(s^(-1)) = f(s)^(-1)

[crystal vale]

Yes

[terse crystal]

we define g(s^-1)=f(s)^-1

[delicate orchid]

If you go through the whole construction of the free group from it's generating set X, you'll know that it's actually words in X U X^-, where X^- is the set of symbols x^-1 for each x in X (followed by quotienting by the relation xx^-1 ~ 1). So really, we should be thinking about an extenstion of f to a map g from a set S U S^- with the restriction that g(s)g(s^(-1)) = 1

[terse crystal]

s^-1 isn’t in the domain of f

We define g(s^-1)=f(s)^-1, I used a different letter g to make you less confused to understand that it’s a extended, new map

[delicate orchid]

I'll match this notation to make it clearer

[terse crystal]

There is no such thing as f(s^-1), originally I thought you understood the same letter f is actually a new map

So in summary, given any map f:S->G, we define g:F<S>->G, by g(Πs^ε)=Πf(s)^ε. You can directly verify that g is a homomorphism, we call g an extension of f because any s from S we indeed have g(s)=f(s). Sometimes we just use the same letter f instead of g

Now all clarified right

[crystal vale]

Now I got it but why don't they introduce g in the statement why they just tell us that {....,a^(-1),a,...} maps to corresponding image

[crystal vale]

Thank you a lot

[terse crystal]

Np

Since there is no ambiguity, everything is clear, so doesn’t matter how an author writes it you will eventually get the correct version

[untold crag]

is it seen as messy/ugly to abuse proofs by contradiction

[terse crystal]

Math has no emotion, it doesn’t care or even know whether a method is beautiful or ugly , it doesn’t have any preference

[untold crag]

yeah but theres a culture right

[terse crystal]

It depends on you, what kind of methods you prefer

The statement remains the same

[untold crag]

i saw ppl saying reaching for proofs by contradiction will make you learn less from the proof (if youre doing it as an exercise) usually

maybe that would havew been a better q

[coral spindle]

I have no idea why people would say that.

I have seen people new to proofs prove something by first assuming that it's not true, and then proving that it is true without using the assumption that it isn't true, which is to say that proof by contradiction was unnecessary. But this is just lack of awareness.

[rocky cloak]

I think it depends, but if you're for example proving the existence of something, then you are in a sense learning more from the construction of said thing than just proving that it must exist.

[south patrol]

Eh I mean like

One thing is that if you do a proof without contradiction then the proof can have clear, useful byproducts

Whereas if you use a contradiction a lot then that doesn't happen

[coral spindle]

So you mean non-constructive proofs, really

[south patrol]

Personally I prefer to give a direct proof if one is possible, but sometimes doing stuff by contradiction can be quicker cause you can make more assumptions on your objects

[south patrol]

Not really no

What I mean is like

If I assume A and prove B1,...,Bn on the way to C, then I know A => B_i

Whereas if I assume a contradiction to begin with, I'd have to be more careful to extract the consequences

[dim widget]

you would have to add ¬ to everything

the horror

[south patrol]

What lol

[dim widget]

¬ is the symbol in formal logic for not

[south patrol]

I am well aware

I mean that your proof may essentially consist of a bunch of stuff which follows from a contradiction

[south patrol]

For a proof by contraposition sure

But yeah you could rephrase it but more time/work needed

[dim widget]

yeah I agree that some things would only be a result of A and not C which is useless

anyway I think the argument for not using proof with contradiction if you can avoid it is just that it clarifies what is going on

[untold crag]

how do you express proofs by contradiction formally

what language do you have to use

[coral spindle]

This is getting a bit off-topic.

You can use any logic that has LEM. Even constructive logics like many type theories support proof by contradiction to prove things are false.

[untold crag]

sorry I forgot this is #groups-rings-fields

[crystal vale]

Let G be a group of order n, and let F be any field. Prove that G is isomorphic to a subgroup of GL_n(F). Any hint?

[delicate orchid]

Consider the group action of G on itself. Keyword: “permutation matrices”

[elfin prairie]

Hello

I found problem sets which are assigned from Rotman's UG Algebra textbook

altho I don't know what sections I should read before attempting the psets

https://www.math.utah.edu/~schwede/math435-spring2011/

can anyone help me out?

[crystal vale]

G is isomorphic to the subgroup of S_n and S_n is isomorphic to the subgroup of GL_n(F)

[delicate orchid]

Sure, that’s a good way to think about it

[crystal vale]

The subgroup of S_n is isomorphic to subgroup of GL_n(F), hence G is isomorphic to subgroup of GL_n(F)

[spice whale]

K. Lee

[celest furnace]

Agreed

For example any definition like connectedness must be done by contradiction (since it has a negation in its definition)

[left portal]

Hi everybody I'm Luis from Mexico, my pronouns are her/him. I really love math!

I have a question: is group theory fundamentally about the simmetry of an object? whatever object means?

[coral spindle]

Group theory's fundamental insight is that it doesn't matter what the object is.

And I suppose it is about symmetries of things, insofar as groups are sets of symmetries.

But in moment-to-moment group theory, we don't really think of it that way very often.

If you're wondering what group theory is 'about' I would suggest just reading an intro.

[celest furnace]

Historically group theory was just about looking about subgroups of Sn (symmetry groups) and that was im sure a lot of motivation

[spice whale]

ultimately it's about more than just symmetries of objects

[left portal]

of course is about more than that but everything I have read so far, they always come back to the idea of simetry

[coral spindle]

OK

Are the integers symmetries to you? Like the numbers -1, 0, 1, 2, 3 etc

Does this describe some symmetry?

[left portal]

As numbers, no. As objects in a group they do

[coral spindle]

No but yes. Well I have nothing else to say I suppose.

I think you can see how groups describe symmetries in an abstract sense, rather than a literal one.

[left portal]

for instance, in this(https://math.stackexchange.com/questions/2394650/mathbbz-is-the-symmetry-group-of-what) question they define the integers as a group of symetries of this object.

And that's the thing, can we create and object (let's say geometric object) that has as a group of symetrys a given group?

[coral spindle]

Yes

The coloured Cayley graph is an example

[solar dock]

Groups exist only to act on things some might say

[coral spindle]

There are lots of theorems that say that, given a group, we can construct some object which has that group associated with them in some way or another.

• Cayley's theorem says that groups can always be seen as acting on a set.
• There is a theorem saying that every finite group is the automorphism group of some graph.
• Classifying spaces are spaces whose fundamental group is a chosen group. A related space is the Eilenberg–MacLane space.

[left portal]

Ok, thanks!I need to think a little more about this

[untold crag]

y is this true

[wraith cargo]

aD \cap bD is some ideal in D
So since D is a PID it's generated by one element called c

[untold crag]

thx

[runic lava]

Hey, does anyone know why |SL(2,F_3)| = (3^2-1)(3^2-3)/(3-1) ?

The first part is counting the number of possible basis i got that

But why divide by 3–1 ?

[delicate orchid]

SL is the subgroup of matrices of GL with determinant 1

there are 3-1 possible values of the determinant in GL(2,3)

if you want to use first iso, then SL is the kernel of the determinant map GL -> F*, and so it's index in GL is equal to the number of units in your field

[runic lava]

Why -1 ?

[delicate orchid]

0 isn't a unit

[runic lava]

Ofc

[delicate orchid]

matrices in GL are invertible so their determinants have to be units

[runic lava]

Sorry i don’t quite get why it’s index is equal to this ?

[delicate orchid]

the map det: GL -> F* is surjective. By first iso we then have GL/SL \cong F*

[runic lava]

Thank you

Nvm i’m dying inside i can’t deal with this subject haha

[delicate orchid]

let M be a matrix in GL with determinant d, then you can write M uniquely as d*S for some matrix S in SL

[delicate orchid]

that's all that this means

[spice whale]

determinant? is that perchance a reference to yonder tail?

younger rail

[delicate orchid]

lets just say it's the immanent because your end is imminent

[spice whale]

end? is that mayhaps a reference to minecraft???

[delicate orchid]

shite game

[solar dock]

If I wasn't about to vomit I would make a joke about the discriminant

I am not joking

I am in severe distress and suspect food poisoning.

[coral spindle]

Get well soon!

[solar dock]

Never let yourself act on the set of bad food or you will find the toilet bowl is equivariant under vomit action

[delicate orchid]

https://tenor.com/view/itsover-wojack-gif-4367840179675491690

good luck out there, cadet

[coral spindle]

Wew you're too based for this world

[solar dock]

I love minecraft tho

Especially modded

With stomach medicine mod

[delicate orchid]

I've played it for 3k hours I'm the only one allowed to call it shite

lmk when u hit that 10 win streak in bed waurz then u can chat shit ok?

[coral spindle]

please refer to my previous message

[solar dock]

Self reference

https://tenor.com/view/jojos-bizzare-adventure-jojos-reference-cringe-gif-20882517

Text unimportant

[tribal moss]

Memes go in #chill.

[noble lynx]

how would I construct a non abelian group where every non identity element has order 3?

[delicate orchid]

The group of strictly upper triangular 3x3 matrices on F_3 has this property

and is in fact the smallest group with this property

how would I go about finding this? Well I'd note that any group with exponent 3 would have to be a 3-group, as if |G| had any other prime divisors then there would be a q-sylow subgroup with q \neq 3, contradicting the fact that every element has order 3. From there I would just try different 3-groups - it's clear that we have to start from at least order 27 because everything smaller is abelian

[noble lynx]

how did you know every group of order 9 is abelian?

[delicate orchid]

every group of order p^2 is abelian

new exercise, prove it

[coral spindle]

this is a standard theorem that one learns in a first course in group theory

Might be worth giving a hint. May I, Wew?

[delicate orchid]

if it's "p-groups have non-trivial centre" then no. Giving such a hint would make it far too easy!

[coral spindle]

Gasp can you imagine such a thing

[delicate orchid]

and I forbid you from telling asdf that p-groups have a non trivial centre!

they must never know!

[coral spindle]

And considering the quotient G/Z(G)? Even worse frankly

[loud merlin]

Just check it for every such group smh

[tribal moss]

"Strictly upper triangular" means there are ones on the diagonal?

[delicate orchid]

yus

it's the Heisenberg group on F_3

[tribal moss]

I would have understood that as zeroes on the diagonal, except that doesn't even make a group.

[delicate orchid]

mb, I've only seen it as 1s on the diagonal

[rocky cloak]

I would call them upper triangular unipotent matrices

[tribal moss]

That does make sense when the context is already matrix groups in particular.
My train of thought was something like, what would the "strict" variant of "$a_{ij} = 0$ unless $i\le j$" be?

[cloud walrus]

Troposphere

[tribal moss]

(Both of these are useful concepts. The ones with 0 on the diagonal make up the Lie algebra of the ones with unit diagonal).

[coral spindle]

Unipotent radical of the Borel

I would like to imagine there's some way of discovering this abstractly

Like, ok. Such a group has to be a 3-group and assume it has derived length two. So let's make the assumption its centre is isomorphic to Z_3, but then Aut(Z_3) is a 2-group and there's no action of a 3-group on a 2-group and we can't form a semidirect product. Idk, I'm rambling, but there's a way to discover this in that way.

Aut(Z_3 x Z_3) has how many elements... (9-1)x(9-3) = 48? So there is an action of Z_3 in there. So there could be a semidirect product Z_3^2 \rtimes Z_3 that works, and from then it's just a calculation

[barren sierra]

any hints on these? Never encountered PGL outside of this problem so I'm very stuck

[tribal moss]

It's equally true for GL(2), and the arguments would be the same.

The P just means that you quotient out the difference between matrices that are scalar multiples of each other.

[coral spindle]

OK now I'm curious. What's this automorphism of Z_3^2 of order 3? (x, y) |-> (x+y, y). Nice. So it's the matrix [ 1 1 \ 0 1]. That's really cool.

[rotund aurora]

||(iii) is just linear algebra I think. (iv) is just saying that you can bring p1,p2,p3 to 0,1,infty by a unique transformation, and the fourth coordinate is something else different from 0,1,infty||

[long obsidian]

Say $\mathfrak{a} \triangleleft A$, $f:A\to B$ is a ring hom and $\mathfrak{a}^e$ denotes the extension of ideals by f and say $r(\mathfrak{a})$ denotes the radical of the ideal.

I wanna show that $r(\mathfrak{a})^e \subset r(\mathfrak{a}^e). $ say for simplicity $c_1 f(x_1)+c_2 f(x_2) + c_3 f(x_3) \in r(\mathfrak{a})^e $ and say $n_j $ such that $x_j^{n_j}\in \mathfrak{a}$ then is it as simple as using an exponent like $2(n_1+n_2+n_3)$ so that $(c_1 f(x_1)+c_2 f(x_2) + c_3 f(x_3) )^{2(n_1+n_2+n_3)}\in r(\mathfrak{a}^e)$?

[cloud walrus]

HausdorffT1

[delicate orchid]

yeah the 0 diagonal ones seem to come up everywhere in lie theory

[dull ginkgo]

Assume if x | a and x | b implies x | d for x in D

[south patrol]

I think one way to do this is just to use the characterisation in terms of ideals for a PID

obviously E needn't be one but we don't need that

[dull ginkgo]

I was about to hyper shortcut it using Bezout statement of GCD

e.g divisor-minimal element of (a) + (b) and use PID

[south patrol]

yeah

so (a,b)=(d)in D

[dull ginkgo]

GOD I HATE THAT NOTA

[south patrol]

Now in E, note this remains true

[dull ginkgo]

oh wait

just ideal

jacobson uses (a,b) for gcd which like, implodes my brain when also working with ideals like, in the same goddamn sentence

[south patrol]

oop

i'm fine with it for Z cause there's almost no ambiguity

But outside that, ouch

[dull ginkgo]

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

attrocities

[cobalt heath]

(a, b) = ((a, b)) 👀

[dull ginkgo]

sniped :3

[south patrol]

but yeah think about this

[dull ginkgo]

is this by extending the ideals?

[south patrol]

ye

[dull ginkgo]

ic

[south patrol]

you can explicitly just write like lol

(a,b) = Da + Db

etc

but also kinda clear from viewing it as "smallest ideal containing a,b"

[dull ginkgo]

I write it as (a)_D because I'm a psycho

[south patrol]

I like just Da as it is very literal too

and you can do funny things

[dull ginkgo]

wait pid's are commutative right.

right.

he just says domain.

[cobalt heath]

Usually domain is when it is commutative

[south patrol]

yeah

[south patrol]

yes

[dull ginkgo]

i thought commutative domains are integral domains

[dull ginkgo]

H[X]

[south patrol]

maybe

[cobalt heath]

Hmm

[south patrol]

i don't think it matters for this question though

mayybe idk

[dull ginkgo]

I think H[X] where H is the quaternions

[south patrol]

but worrying about gcds in a non-commutative setting like

sounds HORRIBLE

[dull ginkgo]

this is also Jacobson

[south patrol]

like you'd have to distinguish between left and right divisors right

lol

[dull ginkgo]

bi-divisors are weird

and rare

so who cares

it's commutative, we are abelist here

[cobalt heath]

How's Jacobson going, miz

[dull ginkgo]

commutative geometers are abelists

i want to do this

so badly

but I need to finish

[weary spade]

what book is this

[dull ginkgo]

exercises are either really fun or awful and there is 0 in between

law of excluded middle holds here

[cobalt heath]

Woah

[dull ginkgo]

i really don't want to do these

[cobalt heath]

Skip this >.>

[dull ginkgo]

no I must do it

5 and 6 are intuitive

[cobalt heath]

Wow, dedication

[dull ginkgo]

like i debate writting them down

[cobalt heath]

Polynomials lol

[dull ginkgo]

not that they are inherently confusing it's that I want to not have carpal tunnel

Wait so

[cobalt heath]

Same

[dull ginkgo]

my brain just fried and wanted to stop working

[cobalt heath]

Da + Db notation difficulty?

[dull ginkgo]

idk how to formally prove this

like that this implies it holds in E

[cobalt heath]

Ah

You can do it by showing Ea + Eb = Ed, I guess

And obv d cannot be 'smaller'

[dull ginkgo]

how does Ed = Ex cause issues if x is strictly in E\D

[cobalt heath]

I think that would be fine, because in this case, d = x * unit

[dull ginkgo]

for non associated x

[cobalt heath]

Ed = Ex means d = x * unit, no?

[dull ginkgo]

oh shit yeah

[cobalt heath]

Ah wait sorry, this is not obvious; Still true for domains

[dull ginkgo]

if d | x but not symmetrically then xE is a subset of dE

but Ea + Eb = Ed

[south patrol]

Ea + Eb = E(Da + Db) = EDd = Ed

[dull ginkgo]

trying to prove how that implies d is gcd(a,b) again

[south patrol]

Ah sure

[dull ginkgo]

assume d | x | a and b, but NOT x | d

Ea + Eb = Ed, there's an equality so for d = xy
ua + vb = yx = d

[south patrol]

Then c | a and b implies c| d

From that last bit

[dull ginkgo]

oh wait

[south patrol]

Like we didn't even have to use that fact that Ea+Eb = Ed

All you need is that ax + by = d and d|a,b

That is true in D and remains true in E

[south patrol]

Actually this is also equivalent to (d) = (a,b) so it provides another proof that this is ture in E

[dull ginkgo]

How does this lift to E

still confused by that

[south patrol]

This is equivalent to saying like

Well okay so I mean it's still ture in E

So d is in the ideal generated by a and b (in D or E)

And since d | a and b, both a and b are in the ideal generated by d (i D or E)

So (a,b) = (d) in either

But that is not necessary for this proof tbf

[dull ginkgo]

what is pid used here for

[south patrol]

From the argument earlier

[dull ginkgo]

wouldn't this hold in GCD domains then

[south patrol]

Well

I used the fact rhat (a,b) = (d) in a PID

[cobalt heath]

I think so

[south patrol]

Idk if that is always true

[cobalt heath]

Assumng Da + Db = D(a, b) in GCD domains

[dull ginkgo]

oh no

[south patrol]

Ig you need Bezout domains

[dull ginkgo]

yeah that's a Bezout domain

[south patrol]

Tautologically