groups-rings-fields

[obsidian sleet]

maybe is there a way to come up with a counterexample

[viscid pewter]

i don't know normalisers of subgroups well enough lol

HK = KH, ie. h1k1 = k2h2
HKH = h1k1h3 = k2h2h3 = k2h4 = KH

upon reflection it now seems false

like if HKH = K that would be ridiculously strong

when you take products of subgroups the result should be bigger

to be clear this just means my proof method doesn't work i think?

the actual true-falseness idk still

[obsidian sleet]

hmm

i feel like there could be a neat counterexample which i can't think of

im sure there is one

cus if one normalizes the other it's sufficient to form a subgroup yes

ah yes okay the stack exchange with the hookup

im bad with symmetric groups

no wonder

[viscid pewter]

??

link pls

[obsidian sleet]

https://math.stackexchange.com/questions/308479/hk-a-subgroup-neither-is-included-in-the-others-normalizer

[chilly ocean]

[obsidian sleet]

indeed it is

False.

[chilly ocean]

group theory moment

[obsidian sleet]

thanks for bringing it up i never thought of the converse lmao

[viscid pewter]

ic

ok that's neat ty

[obsidian sleet]

np

[viscid pewter]

damn, that really is a neat proof

[obsidian sleet]

Sylow strong.

[sly crescent]

What is the automorphism group of the sedenion algebra?

[balmy compass]

for this, is there a better way than writing out the multiplication tables?

like we can associate g with the cycle permutation (1 2 3) = x and f with (1 2) = y and then write out a multiplication table for both, but is there something faster?

[robust pollen]

If you Google it you will find that's it's the product of S3 with the exceptional lie group G2. No idea why tho

[robust pollen]

Yeh, do you know presentations of groups?

[balmy compass]

haven't heard that phrase, mind explaining?

[robust pollen]

for example the group $S_2$ has presentation $\langle t | t^2 = 1 \rangle$. You read this is as follows: build the free group on the symbol $t$, and then impose the relation $t^2 = 1$.
The free group on the symbol $t$ is just the set of words built from the symbols $t, t^{-1}, e$, and the group product is given by concatenation, i.e. $ttt \cdot tt = ttttt$. Also, $t^{-1}$ is viewed as the inverse of $t$ and $e$ is the unit.

[cloud walrus]

expectTheUnexpected

[robust pollen]

So if you impose on that set the condition $t^2 = e$, then most words will collapse, until you are left with only the set ${e, t}$, which is precisely the symmetric group $S_2$

[cloud walrus]

expectTheUnexpected

[robust pollen]

It's not hard to see that groups with the same presentation are iso. So now you need a presentation for $S_3$, and this is well-known. You can take $\langle s, t | t^2 = 1, s^3 = 1, st = ts \rangle$. Then check that your functions $f$ and $g$ satsify the relations in this presentation, and you know that the group is iso to $S_3$

[cloud walrus]

expectTheUnexpected

[robust pollen]

btw the unit e in the free group is ofc the empty symbol

[rustic crown]

technically you would need to show that f and g satisfy no other relations other than the ones built from those 3. this is usually done by looking at sizes. You'll only get the map S3 --> 'that group' which is surjective. If you show sizes are same, it will also be injective.

there is another cute way btw, at least for this situation.

[robust pollen]

true

[rustic crown]

these fractional linear functions act on the Riemann sphere C u {infinity}

[cloud walrus]

det

[rustic crown]

these are your 3 'letters' which can be permuted in anyway

action is simply evaluation

so since that group acts on the set of 3 elements, you'll get a group hom to S3, shouldn't be too hard to conclude it's an iso

[rustic crown]

I was wondering if there is a coordinate free analogue of Cauchy-Binet's formula. So say $R$ is a commutative ring with $1$ and $M, N$ are free modules of finite rank, and we have linear maps
[M \xrightarrow[]{\alpha} N \xrightarrow[]{\beta} M]
can we say something about $\det$ of the composite without breaking into matrices?

I'm not sure how would one think about the corresponding analogue for minors in the coordinate free situation.

[cloud walrus]

det

[rustic crown]

ping me up if there is something cute we can do here

[robust pollen]

can you do something with the exterior algebra?

[rustic crown]

yea, that's what i want to relate it to

the definition of det via exterior algebra is very sweet.

[robust pollen]

Hm, to me even the formula of Cauchy-Binet kind of seems to make explicit use of direct summands, if I want to write it down for free modules over rings. And that seems equivalent to choosing a basis = not coordinate free. What I mean is that, as you said, minors probably don't really work coordinate free?

[quaint tree]

What's the name for something that obeys all the ring axioms, except that + just has to be a group, not necessarily an abelian group?

[hot lake]

a ring

there are two ways to distribute (1+x)*(1+y), and you get x+y = y+x from comparing them

[quaint tree]

Oh so it doesn't exist if you use a nonabelian group? Damn, I guess I should have tested for existence before playing around with it and seeing it was interesting.

[robust pollen]

dang, there was a name for this, the "x-y -trick", where x and y were the names of two mathematicians, but I can't remember

ah, no, not trick. It was the "Eckmann-Hilton argument"

It states: if you have a two unital magma strucutres on a set, and if these distribute over one another, then they are actually the same and even commutative.

That can be used e.g. to show that an algebra object in the category of groups (or algebras) is just an abelian group (or algebra)

[hidden haven]

is the interchange law the same as distributing over one another

because the statement I know is that they should satisfy the interchange law

and you also get associativity for free

I think

ye you do

[robust pollen]

I guess "distribute" is not quite the right word, yeah. By my imprecise "distribute over one another" i mean "(a.b) x (c.d) = (a x c) . (b x d)"

[hidden haven]

right yeah

[robust pollen]

dass nice mang

[hidden haven]

mang

[ocean steppe]

Hey guys, i come up with a proof of Projective => Torsion less

let L be a free module, and P proyective
then L = P+Q => P is submodule of L but L is torsion less => P is otrsion less

someone can check if this proof works? im not absolutly sure of it and i didnt see it in any book

they all use that proyective => flat => torsion less

[ocean steppe]

(there exist an L free module by the definition of P)

[hidden haven]

Yeah that seems good

[ocean steppe]

thanks!

[random pendant]

$H_1$ and $H_2$ are subgroups of G(finite)$\$
Prove TFAE$\$

1. Permutation representation associated to action of G on $G/H_1$ and $G/H_2$ are isomorphic.$\$
2. For any conjugacy class [g] in G, $\abs{[g]\cap H_1}=\abs{[g]\cap H_2}$

My prof gave this question as part of our representation theory course

But I am finding it difficult to see the connection between these 2 statements

Any ideas on how to approach?

[cloud walrus]

Averisera

[random pendant]

Hello? Anyone?

[hidden haven]

the latter

rather

they are homomorphisms into group of bijections of any set to itself

not just group automorphisms

In that case the set is also G, so any bijection is fine

[hidden haven]

For 1 implies 2, notice that in G/H, the stabiliser of H is exactly H itself. This, along with the fact that isomorphisms of representation conjugate the stabiliser should give 2. I am not sure how to prove that 2 implies 1.

@random pendant

For 2 implies 1, we probably want to somehow show that H_1 and H_2 are conjugates of each other, and deduce that the quotient group action is the same, but I am too sleepy to think rn

[chilly ocean]

Think about (Z_2)²

[simple mulch]

Hey
So I have $v_1 \otimes v_3$ and $v_1 \otimes v_2 \otimes v_3$.
Can I represent it by $(v_1,v_3)$ and $((v_1,v_2),v_3)$?

[cloud walrus]

mns

[simple mulch]

asking because having (v_1,v_3) and ((v_1,v_2),v_3) could I add them as (v_1 + (v_1,v_2),v_3) ?

[hidden haven]

Not in the tensor algebra

[simple mulch]

hum ok

[proud bear]

x^4+1=(x^2+sqrt(2)x+1)(x^2-sqrt(2)x+1) has no real roots

[obsidian sleet]

$\langle x^p\mid x\in M\rangle$

[chilly ocean]

no space before last dollar

[obsidian sleet]

fuck

[cloud walrus]

sylow green (Yung cofe)

[obsidian sleet]

ok

i have no clue what this is supposed to denote

but apparently it's a characteristic subgroup of M

p is prime i think

[chilly ocean]

subgroup generated by pth power of elements of M?

[obsidian sleet]

thats what i guess too

[little root]

is it common to be confused by the sylow theorems

welp time to drop out

[chilly ocean]

just admit you can't understand sylow's theorems

[little root]

does anyone have any tips for understanding the proof of II

it seems so

arbitrary idk just a lot of orbits

[lethal dune]

send me some resources where I can read about subgroup counting / element order in product group (i.e. the direct product)

[next obsidian]

Honestly man

Just read it until you get it and maybe memorize bits of it

Come back later

It didn’t make much sense no matter what I did the first pass, but on the second pass it was really clear

[little root]

so later as in much later after I've learned more algebra, or later as in next week?

[next obsidian]

For me it was like… 6 months lol

I think it’s just a hard proof

And one of the first ones you see in algebra if you started with group theory and went in a standard order of things

[little root]

ok haha thank you for the advice

ya i feel like im just not quite enough comfortable with all these concepts atm

[random pendant]

Thanks! Will try along these lines

[rustic crown]

the first proof of sylow 2 i read felt very arbitrary, but one of the latter one was much much nicer.
the main idea is that if a p-group G acts on a set X, there |X| = |X^G| (mod p). that is, size of the set and size of the fixed points under the action are congruent. this is used to get the existence of non-trivial fixed points. if you carefully choose nice G and X in the statement, all the sylow theorems will follow very naturally!

like for sylow 2, say G is a finite group, H is a p-subgroup, and P a sylow p-subgroup. we can let H act on the set G/P by left multiplication, where G/P is the set of left cosets of P. we're doing so because |G/P| isn't divisible by p, so this will give us a nontrivial fixed point gP which is fixed by every h in H.
so hgP = gP for every h in H,
this means g^-1 h g in P for every H
so g^-1Hg is contained in P. or equivalently H is contained in gPg^-1.

that was pretty much the proof of sylow 2. notice how simple minded it was, there was hardly an calculation. it shows every p-subgroup is contained in some conjugate of a given sylow p-subgroup. therefore if you actually take H to be another sylow p-subgroup, say Q, then we'll get Q is contained in gPg^-1 which by size constraints, shows Q = gPg^-1.

[little root]

omg thank you--this is way nicer than what i learned!

my prof's proof was similar to that of dummit & foote. It used the same fixed point result but it had H acting by conjugation and it involved a lot more steps

[rustic crown]

you can give such short proofs for all sylow by choosing the actions cleverly

group actions are op

[hidden haven]

Neat

I have seen a proof of sylow

What actions do you choose for the other 2?

[rustic crown]

so the proof flow is
(1) prove cauchy, act by Z/pZ on a necklace with p-beads and color it by elements of G such that the product of all beads is 1. probably the only arbitrary action i've seen.
(2) if H is a p-subgroup, then act by H on G/H, by the exact same calculation, this will show [G:H] = [N(H):H] (mod p).
(3) if H is not a sylow p-subgroup, then [G:H] is divisible by p, so size of the group N(H)/H is divisible by p, so we can find a subgroup of order p, it looks like K/H with |K| = p * |H|

in this step we started with a p-subgroup H and got a bigger p-subgroup K, and more over H is normal in K.
(4) the sylow 2 as above,
(5) if H is any subgroup, then G acts on set of conjugates of H by conjugation. so number of conjugates is size of orbit of H. since the stablizer is N(H), that will be [G:N(H)]
(6) finally if P is a sylow p-subgroup, then by 2, [G:P] = [N(P):P] (mod p), this gives [G:N(P)] = 1 (mod p) which shows number of conjugates of P is 1 modulo p.

part 3 shows something stronger, not only a sylow-p subgroup exists, we can in fact find normal series to a sylow subgroup. more over, every p-subgroup will occur in some series because, by staying in the p-subgroup H, find a series up to H, then look at this series in G, and complete it to a sylow p-subgroup.

[hidden haven]

In 1, are you proving Cauchy? The proof I remember used induction

I didn't get your argument

[rustic crown]

oh that's a very famous proof, so i avoided the details

[hidden haven]

oh I'll look it up

In 2, what do you mean act by H?

[rustic crown]

these are like the prototypical types of action, act by a p group on some quotient by left multiplication

h * (gH) = (hg) * H

[hidden haven]

But H on G/H would be trivial then

[rustic crown]

why tho

[hidden haven]

Oh

[rustic crown]

H is a fixed point i agree

[hidden haven]

Right I see

[rustic crown]

i've haven't played a lot with G-sets, but i think this shows up automatically. like at least i have seen that transitive actions up to G-equivariance look like G acting on some G/H by left multiplication

[hidden haven]

Yeah I told you about that one

Or at least it came up during the free action discussion

[obsidian sleet]

good morning friends

i've solved pretty much the majority of this problem

i just don't know what the notation means

in the very last part

what the hell is that last subgroup

M' is the commutator subgroup

(the last subgroup would tell me that the p-group is elementary i guess)

[hidden haven]

Everything that's a pth power

[obsidian sleet]

i showed up to the part where the minimal subgroup is abelian and is a p group

oh

so its just taking everything in the group and raising it to the p^th power

and finite products of these things

[hidden haven]

Yeah, usually it's not a subgroup but solvability probably makes it so

[obsidian sleet]

hm hm

[hidden haven]

Oh right

They're taking subgroup generated by it

[obsidian sleet]

ye

[hidden haven]

Very cool

[obsidian sleet]

this subgroup is characteristic because any automorphism is gonna preserve the exponentiation to the p^th power right

so yea

that's fine i think

[hidden haven]

Ye

[obsidian sleet]

so then it's a char subgrp of the minimal normal subgroup and so there are a few cases. It shouldn't be proper because then M would not be minimal

so either the whole minimal group or trivial

one of these should lead to a contradiction.

[obsidian sleet]

ah so by cauchy theorem there is a subgroup of order p in this minimal subgroup

so then raising to the p power means that this char group is proper

so it has to be 1

hmmmm

what does this mean tho lol

means that raising everything to the p power in the minimal subgp gives identity

uhh

surely this doesn't just mean that the minimal subgroup was a cyclic group?

what

something is wrong here.

oops no it means that the form of elemnts is Z_p x ... x Z_p

since the lcm of orders of elemnts in this group is p

sure this is fine

[hidden haven]

does everything having order p imply this directly?

I don't see how

[next obsidian]

I think so

Oh hmm

No

I was gonna say that p-groups have no trivial centers

So you need the center to be Z_p

Then you can mod out

Center is Z_p

…

But you can’t sew it back together from that

[hidden haven]

yeah I have seen that they are always solvable

I don't recall abelian

[next obsidian]

I mean

Idk, I don’t know what happens when you need everything to be order p

I think for even n = 3 the result is true

You can classify p^3 groups

And it mimicked what happens when p = 2

[hidden haven]

right

[next obsidian]

But I’m unsure about higher order ones

¯_(ツ)_/¯

[chilly ocean]

classify groups of order d^n

[next obsidian]

Classify groups of order deez nuts

🥜

^^ deez nuts

[ancient rivet]

well they are small, trivial and you wont stop trying to send people to its image

[chilly ocean]

this is a serious channel

classify groups

classify rings

no thanks

[rustic crown]

classify sets of finite order

[hidden haven]

upto equality

[chilly ocean]

something that always bothered me is like, this kind of question, to "classify" something. it's not a rigorous question. i wish it could be turned into a rigorous question

[wooden ember]

i think you can by induction on n if the order is p^n

[chilly ocean]

why clown reaction, i think its a very valid and serious concern

[wooden ember]

it is a rigorous question though

[chilly ocean]

how?

[wooden ember]

you are finding all isomorphism classes

[chilly ocean]

what does it mean to find them all? at extreme case you could just define the class of all isomorphism classes very easily, but that wouldn't be considered useful

there are infinitely many so you can't write them all

[prisma ibex]

In a fixed order there are finitely many

So obviously you can classify them all in a given order

[chilly ocean]

so a good answer would be an algorithm that given a order n classifies all groups of that order?

[prisma ibex]

Sure

[chilly ocean]

that still has a trivial answer tho: brute force all possible group structures of that order and check which ones are simple

so perhaps the question should be to find an algorithm as efficient as we can to do this

[prisma ibex]

Yea it’s hard to do this quickly

Usually the better strategy is to list all the ways a finite group can be described as an iterated extension of finite simple groups and then compute the relevant ext groups

But that’s really hard too

[wooden ember]

hmm no this is harder than i thought

there might be something to do with the frattini subgroup but im not sure

[prisma ibex]

Yea describing finite groups of prime power order is really hard

[hidden haven]

Carla has become the proover

[wooden ember]

i mean we want to get explicit answers: you could do this sure but at the end of the day you wouldnt have a simple answer to everything because youd still be iterating through infinitely many orders and brute forcing everything

while at least for finite simple groups we have a list of everything, even if some families are infinite

[prisma ibex]

In a given order this is obviously a finite computation

The problem is also like

Okay say you list all the possible multiplication tables for groups of order n and then sort them into isomorphism classes

This tells you very little interesting information

[chilly ocean]

i'm aware we want explicit answers, it just bothers me that there seems to be no simple way to formalize what it means for an answer to be "interesting"

[wooden ember]

im still not quite sure what your problem is: your version of classifying say finite simple groups just wouldnt get you a complete explicit answer

it's not even a question of the answer being interesting, you just dont get an answer at all

[chilly ocean]

how do you define "explicit answer"?

[prisma ibex]

No you get a completely explicit answer just not a particularly meaningful one

[chilly ocean]

yea, there not being a definition of what it means for an answer to be intersting is what bothers me

[wooden ember]

How so though: to get an explicit answer of a complete list of the finite simple groups by brute force you need to perform an infinite process

So you can never get a complete answer surely

[prisma ibex]

In all orders sure

In a given order this is obviously a finite computation

[wooden ember]

Yeah exactly

Oh sure

[prisma ibex]

Of course there’s no definition that’s a subjective thing

[wooden ember]

In the context of a given order I do get the point of the question then ig

But yeah it’s rather subjective

I mean mathematicians choose to care about certain things: they don’t care about certain theorems because “logic dictates they are better”

[chilly ocean]

most agree that some kinds of results are more interesting than others tho

[wooden ember]

Sure

Because a result may in a sense be more powerful/beautiful/generalizable

But I don’t think these are concepts that can be made rigorous

[chilly ocean]

that kinda returns to the original issue. "Classifying finite simple groups" is not a rigorous question then, in the sense of getting an "interesting" answer

[prisma ibex]

In the sense of expressing the answer in an interesting way sure

That’s fairly subjective

Ignoring this, the actual classification problem is obviously a rigorous question

[long obsidian]

What separates a field from a euclidean domain?

[hidden haven]

Every non zero element of a field has a multiplicative inverse

[chilly ocean]

for example Z is an euclidean domain but not a field cuz some elements of Z have no multiplicative inverse

[random pendant]

how does isomorphism of representations imply stabilizers are conjugate?

[hidden haven]

H_2 ∈ G/H_2 is the image of some aH_1 ∈ G/H_1, and since this is an isomorphism, stab(H_2) = stab(aH_1). Try to work it out from here (you want to write the write side in terms of stab(H_1))

[random pendant]

we do not know that G/H_1 and G/H_2 are isomorphic right? We just know that F[G/H_1] and F[G/H_2] are isomorphic. How does this imply G/H_1 and G/H_2 are isomorphic?

[hidden haven]

Hold on this is not familiar to me then

What is F[G/H]?

F a field, action is on free vector space by permuting basis?

[random pendant]

um the set of the form \sum a(s)s where s\in G/H_1

basically the vector space

[hidden haven]

And a(s) is in F?

[random pendant]

a(s)\in F

[hidden haven]

ok damn

I'll have to think about this but don't expect much because rep theory is pretty new to me

Could you repost the question? The old chat is not loading for me

Others could also look at it then lol

[random pendant]

$H_1$ and $H_2$ are subgroups of G(finite)
Prove TFAE

1. Permutation representation associated to action of G on $G/H_1$ and $G/H_2$ are isomorphic.
2. For any conjugacy class [g] in G, $\abs{[g]\cap H_1}=\abs{[g]\cap H_2}$

what happened to Texit?

[hidden haven]

I am guessing the $ at the end of each line are messing with it? Not sure

[random pendant]

oh its offline

[languid meteor]

can anyone tell me what this notation N^(\sigma) is?

is it the image of N under \sigma?

[lethal dune]

no

$h^g$ means $ghg^{-1}$, but if you use that definition, N is not a characteristic :/ or is it?

[hidden haven]

Looking at the context, yeah

[hidden haven]

This is not standard

[languid meteor]

yeah I have no idea why he uses that notation but it works I suppose

[plucky flicker]

Does ring homomorphism preserve ordering? Like if f : R -> S is a ring homomorphism and I and J are ideals in R such that I is a subset of J then f(I) is a subset of f(J)?

[long obsidian]

Is there an easy example of a principal ideal domain that is not a euclidean domain?

[chilly ocean]

dummit foote I think shows Z[ (1 + sqrt(-19))/2] is PID but not euclidean

yea

[chilly ocean]

It seems kinda obviously true

like idk f(I) = f(J cap I) \subset f(J)

f(I) = {x in S such that f(a)=x for some a in I}

[chilly ocean]

Also iirc in Alaca/Williams intro alg nt they show more examples, but not 100% sure

some a in I always in J

[plucky flicker]

true

thanks

[lethal dune]

guess so, only saw it in one book, though it makes the conjugation thing act nice, like $g^h\cdot k^h = (gk)^h$

[chilly ocean]

Yeah I thought this is a standard notation, seen it used in few sources, also in my course

[long obsidian]

Thank you

[long obsidian]

What on earth is the difference between a multiplicative subset and an ideal. They look like exactly the same things

[chilly ocean]

nope

multiplicative subset contains 1 and is closed under multiplication of just its elements. ideal is an additive subgroup, not necessarily containing 1, closed under multiplication by any element of the ring and an element of the ideal

just put the definitions side by side

[woven delta]

Try modding out a ring by a multiplicative subset and you'll see the difference lol

[waxen hedge]

A multiplicative subset is just a submonoid

[long obsidian]

It's just confusing because they both seem to do quotient stuff lol. Maybe 'quid is right I'm gonna thinking more about what happens when modding

[dapper nebula]

submonoid under what? your mom?

[inland otter]

Hey,
Looking at some introductory notes about field extensions and there is a notation i dont quite get
So we write K for a field, we write $K[t_1,..., t_n]$ for the ring of polynomials in the n $t_1,\ldots, t_n$ variables on the field.
Now given a field extension L//K, and some elements t1, ..., tn in $L \setminus K$, we write$ K(t_1,..., t_n)$ to be the field generated by $K \cup {t_1, \ldots}$.
Now in my notes it is written that $ K[t1,\ldots, t_n] \subset K(t_1, \ldots, t_n)$ and the inclusion is proper in general.
How can this be? for me this is always an equality...

[cloud walrus]

dadaurs

[dapper nebula]

t is not invertible

[inland otter]

But L is a field, so t should be invertible right?

ah wait it t^-1 isnt contained in K

I see thanks, that makes a lot of sense

[ancient rivet]

lol... at work another developer was tasked with 'load balancing' incoming requests in this piece of software. Well I guess he thought he was being clever by taking the message id (which is 6 or 7 digit number) modulus the number devices to distribute the requests amongst, and route to the device based on the result. This might have worked okayish if the number of devices to distribute the traffic amongst was 7, but as it happened, there were 8 devices. Needless to say, the load wasn't very balanced. Heck, half of all requests would have been distributed amongst device zero and device 4.
I proposed instead finding a generator of the multiplicative group of integers modulo a prime p, raising the primitive root to the request's index, mod p, doing exponentiation by squaring and always taking the result of each step mod p to avoid overflowing the integer, but there was really a much simpler way of handing the routing that we opted for instead. I shouldn't require a coworker to know what a multiplicative group of integers mod p is in order to understand and make changes to my code at a later date, despite how deliciously evil that sounds...

It was the first time I was able to utilize my knowledge of abstract algebra at work, lol. And really, I still didn't USE it, but it was still fun to flex on the other devs..

Imagine taking random numbers modulus 8 and thinking the result was going to have some semblance of being evenly distributed, lol.

[mental plover]

We had the exact same problem. 😬

Algebraic structures come up a lot in software.

[ancient rivet]

Perhaps. Extract, transform, load type of business software not so much.

Thats okay, I make heavy use of mathematics in my hobby/side projects

[mental plover]

If the business software contains a cache of some fold over a mutable set (that supports random inserts/deletes) then it better be based on an Abelian group.

[ancient rivet]

I might, but that kind of stuff is already handled by the framework (C#). There is mutable collections, immutable collections and cache classes already written for you to take advantage of. But your right, if you had to write your own implementation of certain classes because the framework version wasnt cutting it (because your collection is too large, say) then you just might need to make use of such structures. For example, the Dictionary<TKey, TValue> class comes with a default comparer which is okay, but will yield too many collisions if your key is a string and you try and stuff more than 10,000 key/value pairs into the dictionary. In that case, you want to write your own IEqualityComparer for the Dictionary to use, and I have found great success using a multiplicative group of integers mod p to evenly distribute objects' hashcodes amongst all the possible values of a 32 bit integer. This dramatically improved the asymptotic performance of the dictionary to the point of hitting the limit for the ammount of memory the framework allows you to allocate for a single variable (2^31 bytes, or approx. 2 gigabytes).

just find the primitive root (a generator, call it g) for a prime p that is larger than 2^32, convert the string key into a numeric value by treating the string of characters as a base 256 digit (or use base 64 if the string is restricted to alpha-numeric characters), e.g. char[0] * 256^0 + char[1] * 256^1 + char[2] * 256^2 and so on, call that sum s. Then g^s mod p is the hashcode to return for that string. You obviously will want to make use of the BigInteger class. If you use BigInteger.ModPow method, muliplication of very large numbers is avoided and so the performance of the code is not an issue as one might suspect when looking at the size of the integers involved.

Pretty sure BigInteger.ModPow uses exponentiation by squaring, and takes the intermediate value modulus the mod for every iteration, as necessary, so raising a number g to a very large power k mod some integer m only takes $log_2(k)$ multiplications and up to the same number ($log_2(k)$) modular divisions.

[cloud walrus]

Marduk

[long obsidian]

Im asked to show $\hbox{Hom}_{R}(A,B)$ the set of R-module homomorphism is an abelian group under addition. But it's not even clear to me that it is closed under addition. The summation of two homomorphism might have a larger codomain than B. What is the addition here?

[cloud walrus]

fajitas

[chilly ocean]

what do you mean larger codomain

[frank fiber]

let f,g:A:->B R-module homomorphism then f+g:A->B is defined by f+g(x)=f(x)+g(x)

[frank fiber]

here f(x)+g(x) is also in B because B is a module

[long obsidian]

You're right 😎

[fossil shuttle]

nice!

does the language you work in have any standard hashing functions in the library

[marsh fossil]

idk if this is the right place but can someone explain what the identity is for the group of points of an elliptic curve mod p?

[golden pasture]

the point at infinity if you're using the usual weiestrass presentation

[marsh fossil]

okay, so if i do the group operation with two points and get an intersection line that has a vertical slope, then they're inverses of eachother and the result is the infinity point i basically union to the group?

[golden pasture]

i guess so?

[lavish nexus]

the invariant factors are x^2 and x^3
so any matrix commuting preserves both of the summands of Q[x]/(x^2)+Q[x]/(x^3)

not sure how to proceed next

[lavish nexus]

OK I got the basis vectors for the invariant subspaces and then reduced it to commuting a 2*2 and 3 *3 block

got something like this

is this how to do it?

[rustic crown]

why do you say that the commuting matrix will preserve the summands?

[lavish nexus]

If two matrices commute they should have the same invariant subspaces?

[rustic crown]

the zero matrix commutes with everything

if we call M the Q[x]-module given by that matrix on the vector space Q^5, a commuting matrix in this language corresponds to a Q[x]-linear map. that's all.

we're saying that M is isomorphic to Q[x]/(x^2) + Q[x]/(x^3) so just need to find Q[x]-linear endomorphisms of this.

[waxen hedge]

Of (A{0}, x), assuming it doesn't contain 0

[dapper nebula]

just a joke my friend....

[waxen hedge]

Just to make sure it was clear

[chilly ocean]

[next obsidian]

Yummy

When I have to talk to other ppl about Associated primes I just straight up say Ass M

Also Catfan you never DM’d me for cool prize :(

[dapper nebula]

hey

[chilly ocean]

what prize

[next obsidian]

Can’t know until dm

:3

[chilly ocean]

im scared

[next obsidian]

No scared

:D

[nova plank]

Can I dm for a prize? 👀

[candid kiln]

👀

[sour plume]

@robust pollen So I've been sick all week, so I've not had a lot of brainpower to think about this yet in detail, but do you think the statement "If $M$ is a free module over a Hopf algebra $H$, then so is $M \otimes_k M$" that you proved can also be extended to freeness of $\Lambda^2 M$? The proof for $\otimes_k$ that you gave is a little unsymmetric in both arguments, so I'm a little wary. The action at least makes sense if $H$ is cocommutative.

[cloud walrus]

Lartomato

[sour plume]

Sorry for the lazy question! But perhaps you have an insight which makes this super trivial from our previous discussion, since you seem pretty well-versed with Hopf algebra stuff

[robust pollen]

Hah, coincidentally I was also sick for a few days last week ✋ sickbros

But so I think my proof actually also shows if M and N are free, then M tensor N is free

[sour plume]

Ye exactly

[robust pollen]

What was unsymmetric about the proof?

[sour plume]

I guess the first identification $H \otimes M \cong H \otimes \tilde M$ relied on having a copy of H in one of the arguments, and you can't necessarily make create that situation with the exterior algebra, since the exterior algebra of two different vector spaces does not make sense

[cloud walrus]

Lartomato

[sour plume]

But perhaps that's silly, maybe I just need to track the final isomorphism $H^{(I)} \otimes H^{(I)} \cong H^{(I) \times (I)}$ and see if that one still makes sense when $\otimes$ is replaced by $\wedge$

[cloud walrus]

Lartomato

[sour plume]

I've not gone through the details yet, but I think you plug the antipode in one of the arguments somewhere along the proof, but not in the other argument, so that made me skeptical

awww shit I really need to go grocery shopping; If anything comes to mind which makes the question trivial or so then let me know, but if you think all I need to do is write the same argument out again, then you don't need to babysit me, I've just been too lazy to try 😛

But in either case thanks for your time!

[robust pollen]

Uhh... So, you similarly have W tensor H = W~ tensor H (at least in the finite dimensional case I think, because if I'm not mistaken this requires a bijective antipode. I don't remember: does the UEA of a lie algebrahave a bijective antipode?)

[robust pollen]

Grocery shopping on a Sunday

[sour plume]

The Netherlands are a godless place and I love it

But yeah, antipode for U(g) is just S(x) = -x for x \in g I believe

[robust pollen]

Anyway, I'll also go for a walk soon, then maybe think about the question

Oh, we're (almost) neighbors 😬

[sour plume]

Oh! Germany or Belgium?

[iron vessel]

Netherlandian here too

[robust pollen]

Not doxxing myself brah

[sour plume]

Lmao oki

[plucky flicker]

Hi! So I have the following problem: shot that a module left R-module M is injective iff every f:I->M has a lifting to f_bar:R->M whenever I is an essential left ideal of R. I have the following solution for the => implication:

,rotate

[plucky flicker]

is this okay, or am I missing something?

why would I need that I is essential?

I used the injective test lemma

I guess the proof for the <= direction goes the same way, but idk why would we need that I is essential

I is essential iff whenever J is a nonzero proper ideal of R then I intersected with J is nonzero

[hidden haven]

yes forward implication is trivial because the second statement is a special case of the first one

I am not sure what the injective test lemma is

[hidden haven]

makes sense then

also in the problem statement I don't think it is right to use the word lift there

you are talking about an extension

[plucky flicker]

i was thinking about that too

[hidden haven]

wait in your diagram

the second one

the map at the top should still be the inclusion right

[plucky flicker]

well from the first diagram I think yes, the second diagram shows what we want, we want such a g that lifts f to f-bar

and the inclusion map is a good choice by injectivity, right?

[hidden haven]

every f:I->M has a lifting to f_bar:R->M
This means that you already have a map from I to M and you want to extend it to a map from R to M, and this extension has to be along the inclusion I to R

So the inclusion map is fixed

you have to show the existence of the map on the right

[plucky flicker]

wait, we are discussing the <= direction right now?

[hidden haven]

no

forward

[plucky flicker]

so suppose M is injective, then we have the first diagram

[hidden haven]

what you want to show is that whenever you have this situation
I → R
↓
M
then you can complete this to a commutative triangle by R → M

[plucky flicker]

exactly

[hidden haven]

but you put a "there exists g" on the map I → R

the "there exists" should go on the map R → M

[plucky flicker]

ahh right

but there exists such a g from R to M, by injectivity

or am I missing something?

[hidden haven]

yes that is all

forward direction is trivial

[plucky flicker]

but why would I by essential lel?

we need it only for the <= direction?

forward direction is true for all left ideals

[hidden haven]

The point of the theorem is that if you want to check injectivity, it suffices to check the special case of only essential ideals

[plucky flicker]

then it is also true for all essential left ideals

[hidden haven]

rather than all ideals

[plucky flicker]

ohh wow

so what would be the idea for the <= direction?

[hidden haven]

I found a lemma about essential submodules that seems helpful

For any proper left ideal J, there is an ideal J' so that J ∩ J' = 0 and their internal direct sum is essential

Pretty sure this solves it

[hidden haven]

Should be doable from zorn's lemma, not sure if we need any additional hypotheses

[dull root]

I have the following question: Let G be a finite group, and phi some \C-representation of G, and chi the associated character of that representation. What is the ker(chi) even a kernel of(like what map?) It is obvious ker(chi) is normal in G, and that ker(phi) \leq ker(chi), but is there supposed to be equality?

[upper pivot]

in general no

as long as an element maps to a trace 0 matrix, it will be in ker(chi)

so like, consider the trivial rep of GL(2,C) or smth, and you get that matrix with 1,-1 on diagonal is ker(chi) but not in ker(phi)

[dull root]

So then I do not see what the kernel in ker(chi) refers to. What homomorphism is it a kernel of?

[upper pivot]

i mean its just asking when chi=0, althought it should corrospond to the ker of some homomorphism as it is normal

let me think about if it corrosponds to some nice homo or if its useful

ok sorry chi=0 is the wrong thing to say

now that i have thought about it

its chi= chi(1)

[dull root]

chi=chi(1) = deg of rep, yep

[upper pivot]

yeah i cant really think of any nice homomorphism this is like, the kernel of

i dont think thats important tho

we are just calling it ker so that it alligns with the ker(phi) ig

[dull root]

Let G be finite abelian group and chi some character on G. I want to show $\sum_{g \in G} \lvert \chi(g)\rvert^2 \geq \lvert G \rvert \chi(1)$. I'm having trouble starting this... If this was an irreducible character, we would have something easier, namely $\sum_{g \in G} \lvert \chi(g)\rvert^2 \geq \lvert G \rvert$, and this is true by orthogonality relations. The more general case also looks like some sort of inner product structure, but I'm not seeing how to use it, especially this abelian part has to come in somehow in this general case. The abelian part was used to show \chi(1) = 1 in the irreducible case

[cloud walrus]

MasakaBakana

[torpid mesa]

What is the meaning of the notation $\mathbb{C}_0$?

[cloud walrus]

Gunther

[weak oriole]

The irreducible case is simple right?
Can you see that each $\chi_(g)$ should be some root of unity?

[cloud walrus]

Noob666

[dull root]

Yes, I figured out that the irred case is trivial. The hard part is the non irreducible case.

[weak oriole]

Ah, cool
Hint : || |z|^2=zz* , the conjugate of z ||

Hint 2: || the inner product of two characters is the sum $\chi_1(g) \overline{\chi_2(g)}$ ||

Thanks

[dull root]

I was thinking of talking the trivial character that maps everything to \chi(1) and inner producting that with chi

[weak oriole]

Or even inner product of chi with itself works

[dull root]

I don't see where the $\chi(1)$ comes in though. $1 = \langle \chi, \chi \rangle = \frac{1}{\lvert G\rvert} \sum_{g \in G} \lvert \chi(g) \rvert ^2$

[cloud walrus]

MasakaBakana

[weak oriole]

Why the "1="?

[dull root]

wait, nvm the 1= is only true if its irreduc

[weak oriole]

Ya

[dull root]

Why is the inner product defined as \chi(1)?

[weak oriole]

it is not, it is not

break down chi as sum of orthogonal irreducibles
Then <\chi, \chi> is just sum of squares of stuff

[dull root]

Since G is finite, every rep can be broken into a sum of irreduc rep, so characters can be broken into a sum on irred chars?

[weak oriole]

Yes, and if $\chi_i$'s are the irreducibles, and $m_i = \langle \chi, \chi_i \rangle$, what's $\langle \chi, \chi \rangle$

[cloud walrus]

Noob666

[dull root]

its just the sum of the m_i's right

[weak oriole]

almost there

its $\sum m_i^2$

[cloud walrus]

Noob666

[dull root]

So $\lvert G \rvert \sum m_i^2 = \sum_{g \in G} \lvert \chi(g) \rvert^2$

[cloud walrus]

MasakaBakana

[weak oriole]

Yep, finish by figuring out what's $\chi(1)$ in terms of $m_i$'s

[cloud walrus]

Noob666

[dull root]

I don't see it. We have $m_i = \langle x, x_i \rangle = \sum_{g \in G} \chi(g)\overline{\chi_i(g)}$ where $chi_i$ is irred. I don't see how this tells us something about $\chi(1)$. Is there something about $G$ abliean being used here, since this has not used that at all

[cloud walrus]

MasakaBakana

[weak oriole]

So we have
$\chi = \sum m_i \chi_i$ right

[cloud walrus]

Noob666

[weak oriole]

So, $\chi(1)=\sum m_i \chi_i(1)$
Now we use abelianness
$\chi(1)=\sum m_i$

[cloud walrus]

Noob666

[dull root]

Why is the m_i in the sum. We broke chi intro irred char, so $\chi = \sum chi_i$ no?

[weak oriole]

So think of it this way

You have a vector space, (with a inner product) and a orthonormal basis, say $e_i$'s.

Then any element $v$ in that vector space is
$v= \sum \langle v,e_i \rangle e_i$

[cloud walrus]

Noob666

[weak oriole]

It has $m_i$ copies of $\chi_i$ in itself

[cloud walrus]

Noob666

[dull root]

I know that irreducible characters form a orthanormal basis for class functions of G

But here we are using that any character is a vector in the space of class functions?

[weak oriole]

its not hard to show that any character is a class function

its just trace of some matrix at the end of the day

[dull root]

Yes, the trace of a matrix is the same under conjugation

[weak oriole]

If this stuff seems blurry, maybe revisit the section on orthogonality of characters and all

[dull root]

@weak oriole So I understand the idea, but there are a couple details that make me hesitant. We seem to have equality, but I think the best we can guarantee is the following: $\lvert G \rvert \chi(e) \leq \sum_{g \in G} \lvert \chi(g) \rvert^2$, so I'm not sure why we had equality everywhere. We had $\sum_{i=1}^k \langle \chi, \chi_i \rangle = \langle \chi, \chi \rangle = \frac{1}{\lvert G\rvert} \sum_{g \in G}\lvert \chi(g) \rvert^2$. This decomposition of $\chi$ into irred characters does not use all basis elements, whereas $\chi(1) = \sum_{i=1}^N \langle \chi, \chi_i \rangle$ uses all the basis elements, and throws out the ones which are orthagonal. I'm not seeing why we don't have straight up equality based on this arguement

[cloud walrus]

MasakaBakana

[weak oriole]

We have $|G| \sum \lvert \chi(g)\rvert^2 = \sum m_i^2$

[cloud walrus]

Noob666

[weak oriole]

And $\chi(1)=\sum m_i$

[cloud walrus]

Noob666

[dull root]

Yes and the $m_i = \langle \chi, \chi_i \rangle$ right?

[cloud walrus]

MasakaBakana

[weak oriole]

Yep, I don't see the issue

[dull root]

I only get $m_i$ and not $m_i^2$. We know $\frac{1}{\lvert G \rvert} \sum \lvert \chi(g) \rvert ^2 = \sum \langle \chi, \chi_i \rangle

[cloud walrus]

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

[dull root]

Where that comes directly from decomposing chi into irred characters, and using the linearity of inner product

[weak oriole]

How did you get that last inequality?

$\langle \chi, \chi \rangle = \sum \langle \chi, \chi_i \rangle$

[cloud walrus]

Noob666

[weak oriole]

that's not true

[dull root]

Since $\chi = \sum \chi_i$, we have $\langle \chi, \chi \rangle = \langle \chi, \sum \chi_i \rangle$

[cloud walrus]

MasakaBakana

[weak oriole]

Ya, that's what I'm saying

Say you're $V_i$ is the ith irreducible representation

Then given any other representation $V$ we have

$V= \oplus V_i^{m_i}$

[cloud walrus]

Noob666

[weak oriole]

For integers m_i

Are you reading any particular book

[dull root]

We loosely follow Martin Issac's book

[weak oriole]

You get this right?

[dull root]

Yes

[weak oriole]

So a general character looks like $\chi = \sum m_i \chi_i$

[cloud walrus]

Noob666

[dull root]

Yes, I understand that part since chi_i form a orthnormal basis of class fucntions of G

so $\langle \chi, \chi \rangle = \sum m_i \langle \chi, \chi_i \rangle = \sum m_i^2$

[cloud walrus]

MasakaBakana

[dull root]

And since $\chi(1) = \sum m_i$, we have $\langle chi, chi \rangle \geq \chi(1)$, so I see where the inequality comes from then

[cloud walrus]

MasakaBakana

[weak oriole]

Cool!

[dull root]

Also, I have another question which I think has a easy answer, but I can't see it. Is it possible to have a irreducible character of a finite group that vanishes everywhere except at the identity?

[weak oriole]

Prove that for every irred char other than the trivial one,
$\sum \chi(g)=0$

[dull root]

Yes, that result shows it cannot happen, since we would not have a 0 sum.

I see

[cloud walrus]

Noob666

[dull root]

I think I can prove that yea

[hot lake]

uuuuh

that is not true

[cloud walrus]

MasakaBakana

[lavish nexus]

thank you

[regal arrow]

Yo I'm trying to prove that a field K has only two ideals, and also find those ideals.

My buddy and I instantly found that these ideals are the trivial ideal, and the entire field.

To prove it we supposed that J is a nontrivial ideal of K which is not K itself. Since J is nontrivial it contains a nonzero element x and thus it contains all elements of K which have the form rx, and the same for the inverse of x.

This just gave me a little room to work with, so now I assumed that k was not a part of J

this means that we cannot write rx=k, for any rx in J

I finished the proof by stating that this implies that k-rx is not contained in K, which is a contradiction under the field axioms.

[next obsidian]

Just show I = the whole ring iff I contains a unit

[regal arrow]

Did I have a massive brain fart or is this okay

[next obsidian]

if it's the whole ring it contains 1 so you're gg ez

if it contains a unit, say u, then it contains uu^-1 = 1 by absorption

then it contains x1 = x for any x in R

so it's R

I think this is basically what you did but I think it's easier to just prove this general statement then apply it

[regal arrow]

Sure thing. Thanks gamer

[oblique leaf]

how do i show that $x^4+9x+3$ is irreducible in $\mathbb{Q}(\sqrt{-2})[x]$? I have tried using the contrapositive of Gauss lemma to work in $\mathbb{Z}(\sqrt{-2})[x]$ and using Eisenstein's Criterion. But 3 doesn't seem to be a prime there.

[cloud walrus]

alyosha

[oblique leaf]

<@&286206848099549185>

[lethal dune]

3 isn't but 1+sqrt(-2) is

@oblique leaf

[summer path]

I want to explicitly write out the group generated by (1,2,3) in S_3. I think I'm doing something stupid and I can't figure out what.
(1,2,3) --> (1,3,2) -> (1)(2,3) -> (1,2)(3) -> (1,3)(2), but (1,2,3)(1,3) = (1)(2,3), which is not the identity.

[rustic crown]

what did you do to go from (132) --> (1)(23)?

[summer path]

be stupid

thanks

[rustic crown]

🙈

[wooden ember]

General rule of thumb, if you’re coming up with a subgroup bigger than the order of the generator something fishy is going on

In this case 3 so if you start with the identity, then (123) and then (132) you should expect there to be nothing else subsequently

[rustic crown]

lol i can kinda relate with your status... i once wrote momomorphism for monomorphism

okie so given a vector v in R^n, we need to write it as sum of something in U and something in the orthogonal complement... so what would be a good start? Is there a nice choice of the "something in U"?

ah sure... so we need to show that U + U^{perp} = R^n. so if we pick v in R^n, we need to find u and u' such that v = u + u' and u is in U and u' in U^{perp}

okie let's go to help-9

[iron vessel]

Hey guys, how do you factorize f(x) = 6x^6 +3xy^7 + 3xy^5 - 18 in Z[x,y]?

I attempted this by using the ring isomorphism : Z[y]/(3,y^2 + 1) isomorphic to (Z/3Z)/(y^2 + 1)

Then using a map from Z[y] to Z[i] which is a homomorphism with kernel y^2 + 1

not sure if this is the right way to go, and would maybe like a short pseudoproof 🙂

[rustic crown]

so a factor of 3, is clear and next it would be cool if we can show
2x^6 + xy^5(y^2 + 1) - 6 is irreducible.

so killing that y^2+1 looks like a good idea

but maybe don't kill 3

[iron vessel]

Indeed, that was my logic too

[rustic crown]

because we know that x^6 - 3 is irreducible over Z[i] by eisenstein

[iron vessel]

yes

Thats what I had once I applied the map to f(x)

and I also used Eisensteins

But from x^6 - 3 being irreducible in Z[i], meaning that 2x^6 + xy^5(y^2 + 1) -6 is irreducible in Z[x][i], can I just conclude that f(x) = 3 * (2x^6 + xy^5(y^2 + 1) -6) is irreducible in Z[x,y]?

actually I had 2(x^6 -3) once I applied the map, but 2 and x^6 -3 are both irreducible

[rustic crown]

ah, right we would need some more work

okie won't it be better to directly kill y?

then we won't need to deal with the complicated ring Z[i]

and worry how 2 factors

[iron vessel]

what do you mean kill y?

[rustic crown]

use the map Z[x, y] --> Z[x] sending y to 0

so (2x^6 + xy^5(y^2 + 1) -6) is sent to 2(x^6 - 3)

[iron vessel]

yeah but that is achieved with the map Z[x][y]->Z[x][i]

because y^2 + 1 is in the kernel of the map

[iron vessel]

@rustic crown is this enough to show irreducibility in Z[x][y]?

[rustic crown]

idts... i don't want to brute force it and otherwise it doesn't really want to be irreducible

[iron vessel]

ok thanks anyways

[winter thorn]

https://youtube.com/playlist?list=PLelIK3uylPMGzHBuR3hLMHrYfMqWWsmx5

Abstract algebra lectures from a Harvard guy

[plucky flicker]

How can one describe injective objects in the category of finite groups? Maybe we can use Cayley's thm, but I can't write it down precisely

[smoky hedge]

You mean writing every finite group as the symmetry group of some object?

[plucky flicker]

yeah

[smoky hedge]

Yeah let me give you a link

https://m.youtube.com/watch?v=AZUDhtnz-Do&list=PL8yHsr3EFj51pjBvvCPipgAT3SYpIiIsJ&index=2

The ending describes how one does that iirc

[rotund shoal]

Hey can someone help me with this problem from dummit and foote? it's 14.2.30. Here's the question:

Tau in question is just the automorphism k(t) -> k(t) defined by t(f(t)) = f(1/t)

so to start out, if L is the fixed field, clearly k(t+1/t) is gonna be a subset of L, but I'm having trouble showing that the opposite inclusion holds - which is making me wonder if there's a better way to do this

(Also, in this case the subgroup generated by tau is simply {tau,1}, where 1 denotes the identity)

I feel like this is a super simple problem, but I just need a little bit of help to get rolling

[rustic crown]

what's [k(t) : k(t + 1/t)]?

so can you think of a polynomial that t satisfies, but over the field k(t + 1/t)

[rotund shoal]

I'm having some difficulties working on this field in general tbh. I'm used to fields of the form Q(alpha) for some alpha lol. So when you say a polynomial that t satisfies, what do you mean?

[rustic crown]

i see, so the general thing to know is that if G is a finite group of automorphisms of a field F, then the extension F/F^G is extremely nice

it will satisfy all the cool properties you define over a course of field theory

okie so let's make a substitution, i'll call t + 1/t = s
what i'm asking is can you write down a polynomial in k(s)[x] which has t as a root?

i.e. if you replace x with t, that will equal 0

[rotund shoal]

Ah so I'm basically trying to find the degree of the minimal polynomial for t when viewing the field in this way?

[rustic crown]

yep. that's how you calculate degrees of simple extensions!

if F is a field, the degree [F(alpha) : F] = deg of min poly of alpha over F

in our case, F = k(s)

F(t) = k(s, t) = k(t)

[rotund shoal]

Ah ok! I'm looking for one now!

[rustic crown]

okie, let me know if you find a nice one

then we'll try to use some theory and see how one can get to that polynomial without really going into the wild and finding it

[rotund shoal]

Great thank you so much for the help!

Wait so potentially stupid question: since t is arbitrary as well, wouldn't the only such mapping just be the zero map?

[rustic crown]

not really, we have the s to play around with

the coefficients can involve s

recall s = t + 1/t

[rotund shoal]

ah ok

so what about something like s^2*t - t^2 - 1?

[rustic crown]

s^2?

[rotund shoal]

oh oops I meant st - t^2 - 1

[rustic crown]

yep!

so the polynomial is x^2 - sx + 1

is this the minimal polynomial of t? [over the field k(s)]

[rotund shoal]

So my gut is telling me no, but I don't quite see a better one haha

[rustic crown]

lol it is actually the minimal polynomial

[viscid pewter]

baited

[rustic crown]

it's degree 2, if there was a smaller degree poly, then it would have to be linear and only possibility is x - t

[rotund shoal]

oh because k(t+1/t) clearly isn't k(t) so the degree is at least 2?

[rustic crown]

yep, that's also a perfectly fine reason

[rotund shoal]

Ah nice!

[rustic crown]

so yea, you can conclude [k(t) : k(s)] = 2, so what can L really be?

[rotund shoal]

I don't quite see the next step tbh

[rustic crown]

yea, so as you've already noted k(s) is clearly a subfield of L

[rotund shoal]

yeah, but couldn't L be of pretty much any degree?

[rustic crown]

well 2 is a pretty smol number

and L is no way going to be equal to k(t)

as the element t isn't fixed by the automorphism

[rotund shoal]

Oh yeah! L must have degree 2

so then by the standard tower theorem, L has to be k(s)!

[rustic crown]

L on it's own cannot have a degree

[rotund shoal]

I meant over k(t)

[rustic crown]

it's a subfield of k(t)

but yea, i get what you mean

[rotund shoal]

oh shit yeah

wait

now I'm confused lmao

[rustic crown]

oh oops

we have [k(t) : L] * [L : k(s)] = 2

[rotund shoal]

so either k(t) = L or L = k(s)

and thus since tau doesn't fix t

[rustic crown]

yep!

[rotund shoal]

L must equal k(s)!

[rustic crown]

in general if F is a field and G is a group of automorphism, you can show [F : F^G] = |G|

[rotund shoal]

Sick!

[rustic crown]

that proves one direction of the galois correspondence!

[rustic crown]

okie, so here this is what i meant.
say you were working with an extension F/F^G and wanted to figure out the minimal polynomial of an element alpha in F

so one thing you can say pretty easily is that, if f is a polynomial with coefficients in F^G and has alpha as one of its roots, then the polynomial f fixed by the group G! which means under an automorphism of G, alpha will map to another root of f

[rotund shoal]

Yep I follow!

[rustic crown]

so you wanted to find a polynomial with t as a root, so look at all the places t can go to

here we only had 1 non-trivial automorphism

which sends t to 1/t

so t can go to one of two things, {t, 1/t}

so look at the polynomial with those as roots!

(x - t)(x - 1/t)

x^2 - (t+1/t) x + 1

in general, if alpha can be mapped to {alpha_1, ..., alpha_r}

then we can look at (x - alpha_1) ... (x - alpha_r)

the coefficients of this polynomial are symmetric in all alpha

[rotund shoal]

Oh dang!

[rustic crown]

so if you apply an automorphism now, those coefficients remain fixed. another way to see is that after applying the automorphism, the factors (x - alpha_i) are permuted

but since we're working over F/F^G

anything which stays fixed is by definition in F^G

so that polynomial is over the fixed field

[rotund shoal]

gotcha!

[rustic crown]

you'll probably see this theory in sometime

[rotund shoal]

It's weird since I'm honestly pretty comfortable with a lot of this stuff over extensions of a field like Q, but then my brain just shuts off when we get to fields like k(t) lmao

probably just haven't done enough practice problems!

[rustic crown]

yee, you'll get good soon

[rotund shoal]

mind if I run my answers to the other parts in this question by you when I solve them?

[rustic crown]

sure!

[rotund shoal]

Thanks so much!

So for the second one, much like the first one, tau (sigma)^2 has order 2, so it suffices to just find the fixed field of tau (sigma)^2. So then I defined s = t(1-t), and noticed the extension field will satisfy the polynomial x^2 - x + s, but is this the minimal polynomial?

I think it's irreducible over the field in question since s doesn't have a square root in the field, but I'm not positive

[rustic crown]

you can paste in the same reason as before

[rotund shoal]

got ya!

so the whole second part is basically the exact same reasoning

[rustic crown]

yep!

[rotund shoal]

wait in general if I'm trying to find the fixed field for some <theta>, does it suffice to show that theta fixes it?

Intuitively that makes sense, but I want to make sure that's the case before I potentially waste too much time haha

[rustic crown]

theta fixes what exactly?

[rotund shoal]

sorry I wasn't clear: if we have a potential fixed field L, does it suffice to show that theta fixes L?

[rustic crown]

nope

[rotund shoal]

Ah ok haha

[rustic crown]

why wasn't the answer just k in the first two parts

[rotund shoal]

Oh true lmao

I meant does it suffice to show that L is the fixed field of theta?

I guess the way I said it 0 would always be the fixed field of literally every subgroup lmao

[rustic crown]

0 isn't a field

[rotund shoal]

Alright switch that to "the prime field" haha

[rustic crown]

right

[rotund shoal]

but it does suffice to show that L is the fixed field of theta, right?

[rustic crown]

didn't get you lol.... depends on how you choose the "potential fixed field L"

if you make the wrong choice, it doesn't suffice

[rotund shoal]

So we'd use the same technique to find it as before

[rustic crown]

not sure if i understand it correctly... the way we're finding it, it's guaranteed to be fixed by <theta>

[rotund shoal]

gotcha. I guess I just gotta get looking haha

dang I'm really struggling to find this fixed field ngl

so I also know from a previous problem that <tau,sigma> is isomorphic to s3, so I tried to find some relation in there to get a decent description of tau sigma, but I had 0 luck

[rustic crown]

okie so notice that this polynomial is the minimal polynomial of alpha = alpha_1

[rustic crown]

that's what we really started with

so if alpha_1 is a root, then since the orbit of alpha under G is {alpha_1, ..., alpha_r} all of these are forced to be roots!

[rotund shoal]

so in this case our alpha would be t?

[rustic crown]

yep

you can look at the field k(e_1, e_2, ..., e_r) where e_i are the coefficients of f = (x-alpha_1) .... (x-alpha_r)

it's clear like last time that k(e1, ..., er) is contained in F^G

but we're not sure if it equals F^G

[rotund shoal]

so alpha_1 is t and the others are functions of t?

[rustic crown]

yep

[rotund shoal]

gotcha

[rustic crown]

since f is irreducible over F^G, it has to be irreducible over k(e1, ..., er)

that's a smaller field, if it breaks here, then it breaks over F^G

F = F^G(t) over F^G has degree equal to r

but we can say the same for F = k(e1, ..., er)(t)

if this is confusing, maybe try to do it concretely like the first and second part

[rotund shoal]

The thing I still don't get is how we use this to figure out what our fixed field is. I'm really confused how we find the generator

[rustic crown]

oh, we get that the fixed field is generated by the coefficients of the polynomial f

in the first case it was (x - t)(x- 1/t)

the coefficients are t+1/t and 1

in the second case it was (x - t)(x - (1-t))

the coefficients are 1 and t(1-t)

since 1 is already in k, we can ignore it as a generator

[rotund shoal]

oh so in this case will it just be (x-t)(x-sigma(tau(t)))?

(and we can find the tau(sigma(t)) pretty easily, I just haven't yet

[rustic crown]

what's sigma * tau? does it have order 2?

[rotund shoal]

I meant the composition tau sigma

which is the automorphism we want the fixed field of in part 3 of the question

[rustic crown]

could you tell me once more what sigma is?

ik tau(t) = 1/t

[rotund shoal]

sigma(t) = 1/(1-t)

[rustic crown]

oh cool

[rotund shoal]

so the minimum polynomial will be (x-t)(x-(1/(1-1/t)))?

[rustic crown]

so under tau*sigma, t is mapped to t/(1-t)

[rustic crown]

nah

wait

need to ocnfirm the order of tau*sigma

[rotund shoal]

tau sigma haha

I'm at least going with it as tau(sigma(t))

[rustic crown]

oh it does have order 2

yep, then that's right

[rotund shoal]

so then the generator will be found by the coefficients of that min poly?

[rustic crown]

yep

t + t/(t-1) and t * t/(t-1)

which both happen to be same

[rotund shoal]

so then the answer will just be k(t^2/(t-1))!

[rustic crown]

yup

[rotund shoal]

and all that's left to do is this same exact thing but for sigma!

[rustic crown]

yee, but this time order of sigma is 3 i think

[rotund shoal]

so there will be an extra alpha?

which will just be sigma^3(t)?

[rustic crown]

nah, sigma^3 is identity

[rotund shoal]

sorry I meant sigma^2(t)

[rustic crown]

under sigma, t is sent to 1/(1-t)

and that is further sent to 1 - 1/t

so need to look at the three coeeficients

[rotund shoal]

I really appreciate you helping me out with all of this!

[thorny fiber]

Hey hey, I have a single question :
If p is a prime number, are the subsets of Z/pZ all distinct from Z/pZ itself ?

I still feel so insecure about this theme so it would be nice that someone ensures it for good

[next obsidian]

What do you mean by this?

Subsets of Z/pZ are distinct from all of Z/pZ unless it’s the subset which is everything

But I don’t think that’s what you’re asking

[thorny fiber]

AH yes, there is only the two trivial subsets here

[next obsidian]

Eh?

No there’s many many subsets

2^p many

Do you mean subgroup?

[thorny fiber]

OW yes

God dammit

It was supposed to be for Z/(p^n)Z originally

[next obsidian]

Yes, then indeed there’s only the trivial one and everything

[thorny fiber]

n>=1

[next obsidian]

And in Z/p^nZ you have one subgroup of every order p^k with k <= n

Which is a copy of Z/p^kZ

[thorny fiber]

Isn't it written kZ/p^nZ with k/p^n then ?

[next obsidian]

Uh

You could write it that way

[thorny fiber]

But like p^n is a prime factorisation, so k is only factorisable by powers of p ?

[next obsidian]

That sort of shows what elements it is inside of Z/p^nZ

But I think it’s more common to just say it’s a copy of Z/p^kZ inside of Z/p^nZ

But you’re right that in terms of concrete objects, it will look like that

What I mean here is that if you want to think of Z/p^nZ as like

[m]

Equivalence classes

Then yeah it’ll look like the subgroup formed by {[km]}

[thorny fiber]

I am writing it like that because it helps me relate with the subgroups of Z being the kZ

[next obsidian]

Sure

[thorny fiber]

Ice-E

By the way, since I'm talking about distinct groups here

I was looking through properties of equivalence classes

Which would make me have to know about what the inverse of an equivalence class is

But I know that not every equivalence class have an inverse

Would there always be an inverse in Z/p^nZ because p is a prime number ?

[next obsidian]

Yes

This is the content of Bezout’s lemma

Oh no

Not in Z/p^nZ no

For example, [p] will never have an inverse

[thorny fiber]

Only in Z/pZ then

[next obsidian]

The things with an inverse in Z/nZ are those coprime to n

This is because of Bezout’s lemma

In Z/pZ everything is coprime to p, cuz p is prime

[thorny fiber]

Maybe it would be better to discuss it in vc honestly

[next obsidian]

I actually have to leave soon

[thorny fiber]

Aaaa

[next obsidian]

Do you know the statement of Bezout’s lemma?

[thorny fiber]

x and y are coprimes, iff there exists p and q relative integers such that px+yq=1 ?

[next obsidian]

Yes

It actually goes the other way

If those p,q exist, then x and y are coprime

Note that if we move yq to the right

[thorny fiber]

I rewrote it

Lol

[next obsidian]

This says px = 1 mod y iff x and y are coprime

[thorny fiber]

Aaaaah nice

[next obsidian]

But px = 1 mod y means that x is invertible in Z/yZ

So this is how you get the statement about inverses

Anyway, I must go

[waxen hedge]

What's very confusing about Z/nZ is when you mix up additive order and multiplicative order in (Z/nZ)*
You have an easy criteria for being a generator in the additive group (with Bezout's theorem) but you know nothing about the multiplicative order when you have an additive generator

[thorny fiber]

So when we are working in a ring you say ?

[waxen hedge]

Yes

[thorny fiber]

Have a good day !

Luckily enough, I am only working with groups

For now

[waxen hedge]

Actually some cryptography is based on this (the discrete logarithmic problem)

[thorny fiber]

Ah sorry I asked the question because I was having a hard time with a big homework during the whole week-end and I couldn't unlock myself from so much problems in it

[waxen hedge]

Don't feel sorry ! I was just rambling don't worry

It's good to ask questions

[thorny fiber]

Well I would have too much to ask and I would probably monopolise the whole channel, which is not a good think imo

Knowing $p$ and $q$ are prime numbers, and
[
G_1=\mathbb{Z}/p^n\mathbb{Z} ; G_2=\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/q\mathbb{Z} ; G_3=\mathbb{Z}/p^2\mathbb{Z} \times \mathbb{Z}/q\mathbb{Z}
]
I have found all of their respective subgroups, as listed up above in the previous conversation.
I now need to find all of their maximal distinct subgroups.
The real problem I am facing is for the "distinct" part.
What would be a distinct subgroup in Z/nZ in general ?

[cloud walrus]

Epsilia aka Epe

[waxen hedge]

I think that it probably means : find all subgroups H in G such than

• H ≠ G
• if H' is another subgroup of G such than H is contained in H', then either H'=H or H'=G

[thorny fiber]

Exactly, it's just that it's never H'=H but H'=G

[waxen hedge]

So that means you found one

[thorny fiber]

I mean $p^{n-1}\mathbb{Z}/p^n\mathbb{Z}$ would be one ?

[cloud walrus]

Epsilia aka Epe

[waxen hedge]

Yes exactly ! But {0} is not

[thorny fiber]

Yes, because it doesn't have any inverse

[waxen hedge]

Euh wait it's probably the other way
(p) ≥ (p^2) ≥ ... (p^{n-1})

(if a number is divisible by p^2 it's divisible by p but not the other way)

[thorny fiber]

Yeah sorry

The one set that would contain the most amount of number of $G_1$, while being a subgroup of $G_1$, without being $G_1$ itself

[cloud walrus]

Epsilia aka Epe

[thorny fiber]

Could only be $p\mathbb{Z}/p^n\mathbb{Z}$ then

[cloud walrus]

Epsilia aka Epe

[waxen hedge]

Yes
Here with Z/p^nZ is not hard because it's a cyclic group, with cardinal a prime power

[thorny fiber]

The problem now is that it is not distinct from $G_1$ though

[cloud walrus]

Epsilia aka Epe

[waxen hedge]

It's a bit harder with, let's say, Z/pqZ

[thorny fiber]

There would be 2 different ones then

[waxen hedge]

Yes

[waxen hedge]

Why ?

[thorny fiber]

One with p at the "numerator" and an other one with q

[thorny fiber]

Because [p] wouldn't have an inverse, wouldn't it ?

[waxen hedge]

Is 1 contained in pZ/p^nZ ?

[thorny fiber]

Let me think

Z/p^nZ is just the set {[0], [1], ..., [p^n - 1]}, right ?

[waxen hedge]

(I have to go; for this one, remember that if [p] has an inverse in pZ/p^nZ then [p] also has an inverse in Z/p^nZ)

[waxen hedge]

Yes

Try to write pZ/p^nZ

[thorny fiber]

I'm trying for the past 5 minutes lol

[waxen hedge]

It's made of multiples of p, modulo p^n

[thorny fiber]

So it would be what we had in Z/p^nZ, but multiplied with p

In terms of equivalence classes

So we wouldn't have [1]

[waxen hedge]

So ||0, p, 2p, 3p, ... p^{n-1}||

[thorny fiber]

Yes

[waxen hedge]

(it's the answer)

[thorny fiber]

Good I got it !

Just didn't dare to say it

[waxen hedge]

Great !

You have to make mistakes in order to make progress

[thorny fiber]

Yeah, not daring is a big problem of mine lol

Anyway, have a good day, thanks a lot for the help !

[thorny fiber]

Wait it doesn't make sense

Wasn't it 1, p, p², ..., p^{n-1}

?

Because yes if the groups are with the addition

Then we would have 0, p, 2p, ..., (n-1)p

Then the notaton p^n would be defined as np

The same way that p^-1 would be defined as -p

[prisma thunder]

Trying to figure out if 3x^2+2 has any solutions in F_p for p>3.

[chilly ocean]

yes

F_5 3

[prisma thunder]

Oop, should be 3x^2+2, not the - one.

[chilly ocean]

yes

F_5 1

[prisma thunder]

Schmoovin’

[prisma thunder]

Maybe not wait

Then since when is it the case (p-2)/3 is not a square then? Feels like I’ve seen this somewhere before

[chilly ocean]

who knows geometric algebra here?

[lethal dune]

[chilly ocean]

I wanna know how to represent a geometric algebra as a tuple of non-empty sets and operations

you know how you'd represent a field as (X, +, *) with a bunch of axioms

how would you do that for a geometric algebra?

[lethal dune]

#algebraic-geometry

[hidden haven]

Is geometric algebra algebraic geometry?

[lethal dune]

That's literally what I was asking myself while typing that

[chilly ocean]

[hidden haven]

[kind temple]

#serious-discussion message

[chilly ocean]

oh sorry i forgot that this is a math server

clifford algebra is commonly called "geometric algebra" by physicists (and even by william king clifford himself)

[chilly ocean]

you might be familiar with clifford algebra I'm not sure what the math people learn lol anyway basically in it you can add vectors and scalars and instead of the cross product there's the wedge product, if you wedge two vectors together you get a bivector, if you wedge a vector and a bivector you get a trivector, if you wedge k vectors together you get k-vectors

[hidden haven]

oh I have seen a youtube video on this

also my question was rhetorical lol

[chilly ocean]

that's where I first saw it too lol

oh lol

I needed to know how you can represent a clifford algebra as an n-tuple of sets and operations

[lethal dune]

we all started from there

[random pendant]

$H_1$ and $H_2$ are subgroups of G(finite)$\$
Prove TFAE$\$

1. Permutation representation associated to action of G on $G/H_1$ and $G/H_2$ are isomorphic.$\$
2. For any conjugacy class [g] in G, $\abs{[g]\cap H_1}=\abs{[g]\cap H_2}$

[cloud walrus]

Averisera

[random pendant]

Putting it here again

Still can't do it

For 1-->2, if I can somehow show that H_1 is conjugate to H_2, I am done and for that I need G/H_1 is isomorphic to G/H_2 as G-sets

Any ideas on how to get that?

[winter thorn]

Can someone explain to me what an isomorphism is?

[serene radish]

Do you know what a homomorphism is?

[weak oriole]

Sounds like the exterior algebra

[frank fiber]

Is an equivalence between two objects

[winter thorn]

Okay, thanks.

[hidden haven]

[viscid pewter]

i fear your power

[chilly ocean]

an invertible homomorphism

a homomorphism is mapping between two algebraic structures(that are the same type like groups, rings, vector spaces etc.) that preserve their operations

like for example a linear transformation between vector spaces is a homomorphism because it preserves the operations of vector spaces (addition and scalar multiplication)

and if that mapping is invertible

it's an isomorphism

someone please correct me if I'm wrong

[hidden haven]

It is usually better to say invertible homomorphism than bijective homomorphism because there are structures in which bijective homomorphisms may not be invertible (eg partially ordered sets)

invertible means that the inverse is also a homomorphism

[chilly ocean]

I see

thanks

[viscid pewter]

??

[tawny pine]

@chilly ocean this is the formal def. one should also include intuition of identification of 'same' structures

[hidden haven]

???

[viscid pewter]

could i get an example pls

that's wild

[hidden haven]

a < b,c, where b and c are non comparable, and a<b<c

these are 2 orders on {a,b,c}

identity map is a one way homomorphism

[viscid pewter]

aaaaah

ok i see

[hidden haven]

in general, relational structures will have this property, because "preserving relations" is defined as R(a_1,...,a_n) → R(f(a_1),...,f(a_n)), instead of an iff

[viscid pewter]

urgh

[chilly ocean]

wdym?