groups-rings-fields

[astral galleon]

I never took topology or geometry so i'll have to dig in deeper

[chilly ocean]

it's not really that interesting

i wouldn't dig deeper

[astral galleon]

Ok after thinking about it a little i'm starting to see it we consider all subsets so S_3 as open sets so for every element in S_n you just look at the neighborhoods when we "zoom" we can indeed see they are locally euclidean. But what does it mean for the group operations to be "smooth" sorry for sounding dumb i'm new lie groups I did take abstract algebra tho 😓

[chilly ocean]

someone posted a definition of smooth in the differential geometry channel not too long ago

[astral galleon]

all right i'll check sorry for the newbish questions

[smoky cypress]

Is there any point to Cayley’s theorem? Like does it have any significant consequences?

[chilly ocean]

What's the point of algebras over non-commutative rings if we're going to map them to the center anyway

Doesn't it induce a map on R/[R, R] anyway

[hidden haven]

Mostly historically important because groups were originally defined to be subgroups of S_n

[hidden haven]

Similarly, what is the point of defining representations of groups when they are representations of the group algebra anyway

[chilly ocean]

Hmm... I guess if you tensor an R-module by an R-algebra S then you can get an S-module. And now when you iterate then having algebras over non-commutative rings makes it more efficient

Yeah we shouldn't worry why study objects but if they arise in nature

[strong yacht]

Ohhh ok I think I briefly heard about this before but never understood, thanks for the rundown. Do you do pure maths in these years? If so, this could be really beneficial, because in the UK many people have no idea what a maths degree entails since school maths is very methods/computational, so having a prep year to see what you like/able to do could be good for deciding which way to go

[wooden ember]

That’s fair, and yes you do quite a bit of pure maths during that time (though I suppose that might depend on the institution)

[broken stirrup]

We know that nilpotent groups are close to be abelian. Let G be a nilpotent group and let's consider all pairs (g,h). Then there are |G|*|G| pairs. Is there any lower bound such that if G is nilpotent then at least x% of the pairs commute

For example there are 36 pairs in S_3 and half of them commute and we know that S_3 is not nilpotent group so it makes sense since we interpret nilpotents as almost abelians or close to be abelian

perhaps this is a nonsense question

[north sand]

I don't know but it reminds me of this: https://johncarlosbaez.wordpress.com/2018/09/16/the-5-8-theorem/

[broken stirrup]

HOLY S

this is amazing

[gritty sparrow]

Here is a proof that none exist: if we let P(G) be probability that two elements commute, we see that P(G)= |{conjugacy classes}|/|G| and if you consider the group of 3x3 upper triangular matrices with all diagonals 1 and entries in Z/pZ , you can show that conjugacy classes=p^2+p-1. So P(G)<= 1/p. Taking larger and larger p we can make this as small as we want. (Also note that this group is nilpotent since it is a p-group)

[wraith ether]

We have a cyclic group $G$ with order of 20 and a generator $a$, and the following question "How many distinct cosets are there of $H = \langle a^7 \rangle$?".

Is it fair to say that this $H$ is not a subgroup of $G$? Since if we want to calculate the order of $H$ under the assumption that $H$ is a subgroup, we have $H = \langle a^\frac{n}{t} \rangle$, so taking $\frac{20}{t} = 7$ would have to hold up, but 20 is not divisible by 7

[cloud walrus]

S3BAS

[north sand]

per definition of the brackets $\langle, \rangle$, $H$ is a subgroup

[cloud walrus]

Denascite

[wraith ether]

Okay yeah, I was confused by the theorem "for each positive divisor $t$ of $n$ there is exactly one subgroup $H \leq G$ such that $|H| = t$, and $H = \langle g^\frac{n}{t} \rangle$ but I see where I went wrong with my thinking, I cannot use this formula in the first place since 7 does not divide 20

[cloud walrus]

S3BAS

[wraith ether]

Hm not sure how I'm supposed to calculate the amount of distinct cosets then, do you have a hint on what theorem/property I can use?

[north sand]

well the first step would be to calculate |H|

that is calculate the order of a^7

[wraith ether]

The formula for that would be $a^7 = 1$ right? But I'm unsure how I'm supposed to use this formula if I don't know the module H is under

[cloud walrus]

S3BAS

[wraith ether]

Or $|a^t| = \frac{n}{gcd(t, n}}$ but for this I need the order of H

[cloud walrus]

S3BAS
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[north sand]

don't just spoil the answer

[wraith ether]

That was my suspicion all along, but I'm trying to find the theorem/property on how to show that

[south patrol]

Oh sorry

[cloud walrus]

Denascite

Denascite

[wraith ether]

Oh right I was thinking the order of the element depends on the set it is in for a second

Looking back on it not sure why I was thinking that

Hm because of the theorem I feel like it was implied you apply that formula once you know the cyclic group the element it is in

[north sand]

well no, generally the other way around

[wraith ether]

So basically, if you know any cyclic group and the order of that cyclic group the element is in, you can calculate the order of the element

[south patrol]

I think for this one you can just list out a few elements from <a^7>

[north sand]

if by order of the formula you mean order of any element a^t, then yes

cyclic groups are very nice and simple

[south patrol]

a^7, a^14,a^21 = a

But ye

[wraith ether]

I guess it does hold up under the assumption that G and H are under the same modulo

Since that's not specified in the assignment

[north sand]

not sure what you mean with modulo here

[wraith ether]

This is the assignment

It doesn't say for example that both G and H are under $\mathbb{Z}_{21}$

[cloud walrus]

S3BAS

[north sand]

well it doesn't need to

[south patrol]

Wdym

[north sand]

G is a cyclic group of order 20. that's enough

[wraith ether]

Yeah, I'm not disagreeing, just trying to understand

[north sand]

G doesn't have to be Z_20

it's only isomorphic to it

(which is another nice thing about cyclic groups. all cyclic groups of the same order are isomorphic)

[wraith ether]

I mean more like, shouldn't it at least be specified G and H are under the same group? Imagine G was Z_20, and H was Z_21, then the order of a^5 would change right?

[north sand]

well H is per definition a subgroup of G

ok granted, that is assumed within the notation $\langle a^t \rangle$

[cloud walrus]

Denascite

[wraith ether]

I thought the bracket notations just mean "generated by the element of a^5", does it mean "generated by a^5 under G" instead?

[north sand]

yes

[south patrol]

wdym by "under G"

[north sand]

because what else is a if not an element of G

[south patrol]

Well what does "generated by" mean fully in your definition

It should be the intersection of all subgroups of G containing a^7 or equivalent

[cloud walrus]

Denascite

Denascite

[wraith ether]

I think I'm overthinking it, I'm thinking along the lines of: a^5 just any number that happens to be a generator of G, but if you'd use that number to generate under Z_21, surely that would not be a subgroup of H

[south patrol]

I think you seem slightly confused about the idea of cyclic groups - there needn't be any sense of "under" Z_21 (im not sure why you mention Z21 when it seems irrelevant to the question at hand)

Maybe we could discuss the topic more from the beginning?

[wraith ether]

If you use a generator of 2 under Z_10, the resulting set is different from using that generator 2 under Z_20 right?

[north sand]

well yes. but just because in both cases we use the symbol 2 to denote that element, that doesn't mean it's the same thing we are looking at

in Z_10 the symbol 2 has the meaning of the residue class 2 + 10Z. in Z_20 the symbol 2 has the meaning of the residue class 2+20Z. those are different things

[wraith ether]

Yeah, that's fair

[north sand]

and if we say that G is a cyclic group of order 20 generated by a, then the symbol a also has meaning and part of that meaning is that it only makes sense if we treat it as an element of G

[wraith ether]

Yeah okay

That makes sense, that's the missing piece, thank you

[iron vessel]

So I have that $\eta = \sum_{\sigma \in H} \sigma(\zeta_p)$. I need to show that if $H \subset \left(\mathbb{Z}/p^2\mathbb{Z}\right)^{\times}$, the subgroup of prime order $p$, then $\eta = 0$. My idea was:

If $H$ is of order p, then it is cyclic. Furthermore, by Galois correspondence, $H$ corresponds to $\mathbb{Q}(\zeta_p)$. Additionally, $\zeta_p = \zeta^p$; the $p^2th$ root of unity; hence, $Aut(\mathbb{Q}(\zeta_p)/\mathbb{Q}) \leq Aut(\mathbb{Q}(\zeta)/\mathbb{Q})$. Furthermore, $Aut(\mathbb{Q}(\zeta)/\mathbb{Q}(\zeta_p)$ consists of precisely of all $\sigma \in Aut(\mathbb{Q}(\zeta)/\mathbb{Q})$ such that $\sigma(\zeta) = \zeta^{p+n} = \zeta_p^n$ for $n \in {1,\cdots,p-1}$.

[cloud walrus]

Évariste Galois

[iron vessel]

Finally, $\eta = \sum_{\sigma \in H} \sigma(\zeta_p)$

[cloud walrus]

Évariste Galois

[iron vessel]

which is equal to $\sum_{\sigma \in Aut(\mathbb{Q}(\zeta)/\mathbb{Q}(\zeta_p)} \sigma(\zeta_p)) = \zeta_p^{p-1} + \cdots + 1 = 0$. As required.

[cloud walrus]

Évariste Galois

[iron vessel]

I'm wondering if this looks acceptable. Or if I have made a mistake

[iron vessel]

in here, $\zeta$ represents $\zeta_{p^2}$.

[cloud walrus]

Évariste Galois

[urban ice]

@tribal moss@strong yacht A follow-up on the problem I posted yesterday

A friend gave me the tip to look for linear functions

And sure enough, it worked

[cloud walrus]

Syst3ms

Syst3ms

Syst3ms

Syst3ms

[latent night]

I am really struggling to understand this proof. Can someone explain what they mean by {t_f} in bijection with the set of nonconstant monic polynomials f(x) \in k[x]?
My understanding is that they assign one indeterminate to every monic polynomial in k[x]. And then they create a new polynomial ring, in possibly infinitely many indeterminates k[fancyT]. Then they look at the ideal I generated by all the polynomials in only one indeterminate t_f in our new ring
Does this sound right?
Then they say that this has to be a proper ideal (contained in a maximal ideal M by zorns lemma), because if it was all of k[fancyT], then we could write 1 = \sum a_i f_i(alpha_i), which obviously is impossible since by V.5.7 we can create an extension where all the polynomials f_t has a root. (edited)
Here is prop V.5.7

[urban ice]

What's footnote 5?

[latent night]

Nothing. It just comments on how k[fancyT] might have infinitely many indeterminates, which hasn't been covered much in the book.

[gritty sparrow]

this is not exactly right, they are not looking at ALL polynomials in one variable, only the ones of the form f(t_f) note that the polynomial f and the subscript f of t are the same

so if we have a variable $t_{x^2+1}$ we only put $t_{x^2+1}^{2}+1$ in the ideal

[latent night]

Yeah, that makes sense.

[strong yacht]

whoa, that looks slick

[latent night]

wait

[cloud walrus]

chmonkeynumber1fan

[latent night]

No, yeah that makes sense.

Sorry its a little confusing to have polynomials of polynomials kind of.

[north sand]

it's not polynomials of polynomials tho. the subscripts just look like polynomials

[latent night]

Yeah I know

[urban ice]

manner of speaking

[north sand]

if you want you can also index every such polynomial and then do t_1, t_2 etc

[urban ice]

(not quite, the field might not be countable)

[north sand]

oops yeah true

well index it by some uncountable set I guess

[urban ice]

well the set itself serves as a good indexing set in that case

because cardinality can go far

[north sand]

which is why I said if you want, not that you should

[tribal moss]

Cool!

[latent night]

This proof is quite clever. Thanks for the help @urban ice @gritty sparrow

[gritty sparrow]

Np

[urban ice]

yw

[simple valley]

how to identify a group of order 96

it's centerless and has S_4 subgroup

[iron vessel]

Could use a look at my proof 🙂

[delicate orchid]

what are the sylow theorem results for groups of order p^m*q again

[cloud walrus]

mniip

[simple valley]

that's all I can say rn

[fervent helm]

Just started reading Serge Lange's book on Abstract Algebra, and while I'm pretty sure this isnt really abstract algebra yet (as its the first chapter, which I assume are prereqs) I wouldnt really know where else to put it

[next obsidian]

I don’t mean to be rude, honestly, but Lang’s algebra book is really hard. This is a super classical proof of the infinitude of primes, due to Euclid more than 2,000 years ago, so if you don’t see how to prove this I would strongly suggest you pick an easier book. Perhaps something like Hungerford’s book

[fervent helm]

I see, it was recommended to me by someone I know who helps me with math

[strong yacht]

Looks correct to me, maybe just slightly more detail on why the auts in Gal(Q(zeta_p^2)/Q(zeta)) are what you said they are

[simple valley]

it has elements of order 8 so according to GAP this narrows it down to SmallGroup(96, 64)

[chilly ocean]

I don't think this is in his abstract algebra book?

[strong yacht]

The way I like to see what you posted about: Q(zeta_p^2)/Q is degree p(p-1) and Q(zeta_p)/Q is degree p-1, so Q(zeta_p^2)/Q(zeta_p) is degree p(p-1)/(p-1) = p (tower law). Then as you observed (zeta_p^2)^p - zeta_p=0, so the min poly of zeta_p^2 over Q(zeta_p) is X^p-zeta_p, with coefficient of X^{p-1} being 0 (so the sum of roots are zero, and the set of roots is just the orbit of zeta_p^2 under H, similar to our discussion yesterday)

[chilly ocean]

or do I misremember something

[fervent helm]

it is there, in the first chapter on Integers

[chilly ocean]

no, the first chapter is on group theory

[next obsidian]

wait, is this his intro to abstract math book?

Basic Mathematics?

[fervent helm]

It's titled "Undergraduate Algebra" Third Edition

[next obsidian]

Oh, he has an undergraduate book too!

Never mind, this is appropriate

Sorry, the most famous Lang book on algebra is this very hard graduate level book

[chilly ocean]

oh. I never knew that

[next obsidian]

I apologize

[chilly ocean]

yeah this is really basic number theory

[next obsidian]

To address the actual question you asked, this number has to have a prime factor, but this is going to be one of those numbers in the large product. If you try to divide by any of the primes N, you should end up with something that looks like

an integer + 1/N

This isn’t an integer, which tells you that N doesn’t divide that number, but if that’s the case no prime divides this number, which is a contradiction

[chilly ocean]

Assume by contradiction that there's a finite amount of prime numbers and define N
N > 1 so it has some prime q dividing it
Since N-1 is a product of all the primes, it's of the form kq for some k. So q divides N-(N-1) = 1. But no prime divides 1

[astral galleon]

Dumb question still talking about that DRP talk would real valued functions under composition not form a lie algebra ?

[next obsidian]

No

You need to do something like [f,g] = f•g - g•f I think

[astral galleon]

figures 😦

[tribal niche]

Lang is a graduate level text. I'm studying it right now after taking an undergraduate level algebra class and it is still difficult. I would suggest starting with something a bit simpler like Fraleigh, especially if this is your first time studying algebra

Oh wait apparently you are studying his undergrad text

Ignore me and proceed lol

[astral galleon]

Thinking along those lines so would on ehave to show that [f,g] = f(g(x))-g(f(x)) does not equal [g,f] = g(f(x)) - f(g(x)) I think a basic example can be ported over to do this

Sorry for being newbish it took me a couple of days to get a vague idea of what a lie group even was 😢

[next obsidian]

I mean no, you need to show this satisfies the jacobi identity

[astral galleon]

ahhh kk makes sense

[next obsidian]

Which is formal

[astral galleon]

What would be something that dosen't form a lie algebra been at this for days and haven't come up with a good nonexample

[chilly ocean]

but [f, g] does equal -[g, f]

[chilly ocean]

what kind of example do you want? every vector space admits a lie bracket, so i guess you want something that satisfies only some of the lie bracket axioms?

[astral galleon]

i'm looking for an example that is indeed vector space but does not form a lie algebra

[chilly ocean]

let me elaborate: do you want a "bracket" that only satisfies some, but not all, of the lie algebra bracket axioms?

[astral galleon]

yeah not all of the lie algebra brakcet axioms

[chilly ocean]

here's a natural example from differential geometry: given a 2-form \omega on a manifold M, you can construct a "bracket" on the space of sections of TM x R -> M which satisfies the jacobi identity if and only if \omega is closed.

specifically, the sections of this bundle are pairs (X, f) of a vector field and a smooth function, and the bracket is given by [(X, f), (Y, g)] = ([X, Y], Xg - Yf - \omega(X, Y)). a short computation shows that the "jacobiator" is equal to \pm 1/2 d\omega(X, Y, Z). this example is important in lie algebroid theory (courant bracket with a modified function term)

there are certainly easier examples, but that's the first that comes to mind

[iron vessel]

Ohh this is a clean argument. Thanks for the detailed explanation!

[chilly ocean]

you can probably come up with an easy "artificial" example (e.g. take an existing lie bracket and modify it to make one of the axioms fail)

you do not have a vector space

[astral galleon]

shoot 😩

can't come with anything

Maybe taking the lie bracket of a polynomial that isn't factorable since the Jacobi identity is equivalent to a system of polynomial equations would work since at least from what i've read there indeed has to exist a solution. But i'll have to come back to this topic when i've learned a lot more

[simple valley]

https://people.maths.bris.ac.uk/~matyd/GroupNames/73/C4^2sS3.html

[frank fiber]

does the minimal number of generators of a non abelian group is preserved under isoclinism?

[mighty spade]

just divide N by a prime, and you get an integer + fraction, and thus the prime does not divide N, contradiction

lang's algebra isn't that hard, i don't think. I covered part 1 while in a mental hospital, and I have really fond memories of it

i got to spend the majority of my day working on lang and munkres

and I actually found munkres harder

[fiery berry]

how do i prove the 4th claim here without transfinite induction? Is that possible?

[next obsidian]

I don’t think you need that lmfao

Take an arbitrary a in the union, it lies in some E_alpha, so it’s algebraic

@fiery berry

I totally misinterpreted this, you totally need transfinite induction

These aren’t just algebraic over the base

These aren’t just algebraic over the base field, it’s like extending E_i+1 algebraic over E_i transfinitely

[tough raven]

Is it true then?

Consider the chain
k < … < k(t^8) < k(t^4) < k(t^2) < k(t) < k(t^1/2) < k(t^1/4) < …

If it was a well-ordered chain it would make sense.

[next obsidian]

I mean it’s definitely true lol

[toxic zephyr]

im not sure how to prove that f(pq)=f(p)f(q) here

unless its just as simple as p(sqrt2)q(sqrt2)=(pq)(sqrt2)

and im overthinking it

[chilly ocean]

just write out pq as a polynomial, stick sqrt(2) in, and show that this is the same as p(sqrt 2)q(sqrt 2)

[chilly ocean]

what you wrote here is exactly what it means for f to be a homomorphism (you also need to show it respects addition and sends 1 to 1, but these should be obvious)

[odd flame]

This is a very silly question possibly but uh how do i do 2a

Any hints

Havent working with polynomials with more than 1 variable

[untold basin]

Why if 10^m is congru modulo nZ implies that 10 is inversible in Z/nZ ?

@delicate orchid Answering is more helpful pls :)

[delicate orchid]

your question makes absolutely no sense

I have no clue what you're asking

[untold basin]

What don't you understand ? I thought what I've said is understandable in English

[delicate orchid]

unfortunately it isn't, I'll try and help though

10^m is congruent to what?

[untold basin]

1 mod nZ

Which is the class of equivalence of 1

[delicate orchid]

ah ok, I can answer this now

[untold basin]

Wait I just didn't type it lol

[delicate orchid]

10^m = 1 => 10*10^(m-1) = 1
so 10^-1 = 10^(m-1)

I believe this shows it

[untold basin]

I have trouble to understand what does " 10^m is congruent to 1 mod nZ"

Because I have and integer in one side and a set on the other

[delicate orchid]

you can read "congruent" as "is equivalent"

[untold basin]

Is it implicit that 10^m is 10^m mod nZ ?

[delicate orchid]

so 10^m is in the equivalence class of 1 mod nZ

[delicate orchid]

10^m is in the equivalence class of 10^m yup

[untold basin]

Ok I have understood ty

[fiery berry]

ok then i guess i have to figure out what transfinite induction is :D but it makes sense that you need some stronger principle than induction to prove this

[marsh quarry]

I understand how to solve this problem by just visualizing the cases in my mind

but is there a way to solve it using group theory?

[chilly ocean]

Maybe you can use Burnside's lemma here

This should be number of orbits where we have G the group of rotations of a cube, I think

So you have different transformations of a cube and you want to count how many fixed points does a given transformation g give. This is |X^g|

You sum them all and divide by number of transformations |G|

Yep, this works

[sinful mirage]

how to show the map $\rho:V \to V \rho(w)=\int_{G} gv (gv, w) dg$ is $G$ equivariant?

[cloud walrus]

ProphetX

[jagged meadow]

What is gv and what does (., .) mean? Inner product?

@sinful mirage

[sinful mirage]

G v is action of The group on the vector v

Inner prod yes

[vast quiver]

this’ll probably be one of those nice change of variables things. Have you seen the proof Maschke’s theorem (for finite groups)? I think it’s gonna the same trick

er idk this isn’t really my ally, haven’t learned much rep theory

[prisma shuttle]

does anyone knwo any good papers on polynomial mappings over Z/p^nZ?

in other words paper studying the following question (sort of like a generalization fo lagrange interpolation)?

For what tuples (f(0), f(1), \cdots, f(n-1)) does there exist a polynomila in Z/p^nZ[x] that takes on the values in the tupe

[tribal moss]

It is clear that you need f(ap) and f(bp) to differ by a multiple of p, f(ap²) and f(bp²) to differ by a multiple of p², and so forth.

I suspect that's the only condition.

[tough raven]

Do you mean p|(a-b) => p|(f(a) - f(b))?
Or is it not required that f(1), f(p+1) differ by a multiple of p?

[gusty tangle]

good second book on abstract algebra? I would like if it shed some light onto linear algebra too

[prisma shuttle]

i think troposhere is basically just saying that if x and y differ by a multiple of p^n, then f(x) and f(y) have to differ by a mutliple of p^n

@tribal moss i suspected that too

but i am not too sure how to actually prove it

like show it is sufficient

by generating an actual polynoimla

[tough raven]

Yes, that's what I said.

[prisma shuttle]

like in lagrange interpolation

[prisma shuttle]

oh oops i misread

yes u are correct

[kindred jay]

anyone know how to solve this?

Determine all homomorphisms $f \colon \mathbb{Z} \oplus \mathbb{Z}_3 \to \mathbb{Z}_2 \oplus \mathbb{Z}_2$

[cloud walrus]

scorpio1147

[kindred jay]

i kinda know how to do it when it isnt external direct product but im sorta stuck here

[stark sigil]

I see 2x2 matrices where each entry is a homomorphism from one factor of the LHS to one factor of the RHS

[prisma shuttle]

wut do u mean by external direct product?

[stark sigil]

You can fill in the details

[prisma shuttle]

i think that distinction only matters when u have an infinite number of rings in the product?

[kindred jay]

oh these are group homomorphisms btw

this was on an old exam tho before we learned rings

[tough raven]

Wait upto n-1 or upto p^n-1?

[prisma shuttle]

oh that's a typo

yes p^n-1

[tough raven]

I was going to ask if you can find an interpolating polynomial for
(1,1,3,3)

mod 4

[prisma shuttle]

lemme see

[tough raven]

Are you familiar with the universal property of the direct product?

[prisma shuttle]

hmmmm i can

't seem to find one?

idk ima starting to feel like the condition isn't strong enough

[tough raven]

Same

[prisma shuttle]

but idk wut other conditions are needed

[tough raven]

X^2 - X + 1
(1,1,3,3)

I think

OK so my approach (a Hensel lifting-like argument) fails

Because X^2 - X vanishes mod 2

[prisma shuttle]

i found this paeper that seems to talk about this type of thing

but ima finding it really difficult to read

its here

https://arxiv.org/pdf/math/0103046.pdf

it talks about smth related to "lifting" (namely cycle lifting) not too sure if thats the same as your hensel lifting thing

[tough raven]

Hensel lifting is about taking a solution mod p^k and picking representatives for the coefficients mod p^k+1 such that you get a solution mod p^k+1

Only that's for solutions to a factorisation problem f=gh for g,h

But something similar seemed to be happening here

The problem is that you can't just adjust the choice of coefficients mod p^k+1

[tough raven]

Or you get stuck trying to pick 1/3 for 1 and 0/2 for 2 to turn the constant polynomial 1 into a solution

You also need to allow yourself to add polynomials like X^2 - X which vanish mod 2 but not 4

[prisma shuttle]

so i think the origianl claim that tropshere proposed is not sufficient right

theres def a lot ore condition

[tough raven]

No, their condition could be (and I'm guessing, is) correct. My approach to proving it was wrong.

[tough raven]

How many such tuples are there?

(p^k)^p × (p^(k-1))^(p^2-p) × … I think?

[prisma shuttle]

how did u get that?

and what is k?

my idea is that you have to start from the most "specific classes" and then "geenralize"

so take care of the p^{n-1}

then the p^{n-2}

and so on

that way they sort of form an "inclusion" chain nicely

[tough raven]

n 😅
p^n choices for 0‚…,p-1; p^(n-1) choices for p,…,p^2-1; etc.

[tough raven]

= p^(p^n + p^(n-1) + … + p)

[prisma shuttle]

oh yea i think that works

so like p^n * p^{n-1} * p^{n-2} * \cdots?

[tough raven]

p^p^n × …

But yes

[prisma shuttle]

oh shoot i don't think this works

in this paper

https://www.sciencedirect.com/science/article/pii/0022314X74900316

they show that the number of polynomial functions mod m is $\prod_{k=0}^{n-1} \frac{m}{\gcd(k!,m)}$

[cloud walrus]

JustKeepRunning

[tough raven]

Don't think what works?

Our formula for the number of such functions?

[prisma shuttle]

yea

like polynomial functions

[tough raven]

Or that these are the polynomial functions?

[tough raven]

What's n here?

[prisma shuttle]

like the formula is not the number of polynomila functions

[prisma shuttle]

$n$ is the least integer such that $m|n!$

[cloud walrus]

JustKeepRunning

[tough raven]

Oh so n is morally infinity

Just that the rest of the factors of 1

[prisma shuttle]

what?

wait

no like

idk what you mean by morally infinity

but for example if m=6 then n=3

cuz 6|3!

but 6 does not divide 2!

[tough raven]

Let n' be anything larger than n

You get the same product upto n'

[prisma shuttle]

oh yea i see wut u are saying

[tough raven]

You could take the product upto infinity and get the same answer

[prisma shuttle]

yea thats true

so u can take it to fininty

[tough raven]

Because all the factors are 1 for k >= n

[prisma shuttle]

but i am not sure that helps?

i mean ig u can

if that makes it easier to think about

[tough raven]

No, it just reassures me n is not important

[tough raven]

Yeah

For me

[prisma shuttle]

yea its not really improtant

[tough raven]

Anyway

[prisma shuttle]

but like the point is

like then if we use the infinity notion

the product is $\prod_{k=0}^{\infty} \frac{p^n}{\gcd(k!, p^n)}$

[cloud walrus]

JustKeepRunning

[prisma shuttle]

but this is significantly less than $p^{p^n+p^{n-1}+\cdots +p+1}$ or whatever we had

[cloud walrus]

JustKeepRunning

[prisma shuttle]

so that means theres a bunch of tuples satsifying troopsheres condition that don't yield actual polynomila functions

actually ima pretty sure i know the number of polynoimla functions

wait no nvm

[prisma shuttle]

the main problem that i am tryign to solve is determing the zero divisors of the ring of polynomial functions, and i feel like knowing which of these types of tuples "qualify" as polynomila functions would drastically help wiht solving it

[tough raven]

I think it's true that the zero divisors in any polynomial ring are those polynomials whose every coefficient is a zero divisor.

Oh, in the ring of polynomial functions.

[prisma shuttle]

idt this is true

i mean basically u are saying any monic polynomila in any polynomila ring is not a zero divisor

[tough raven]

Maybe; I might be remembering wrong.

[tough raven]

That is true, and easy to see by looking at leading terms.

[prisma shuttle]

oh wait nvm

i was thinking about polynomila functions

[tough raven]

Yep, I misread.

[prisma shuttle]

like basically there's this nice theorem in this paper

basically its just saying that if we restrict the first $n$ terms of $f$ to be less than $p,$ then next $n$ terms to be less than $p^2$, and so on, all the way up to $p^n,$ then we can generate all polynomila functions using these polynomials

[cloud walrus]

JustKeepRunning

[prisma shuttle]

like this set of polynomila, if evaluated at every point, will give us all polynomila functions

but the problem is now I am trying to find when a(x)b(x)=0 from the COEFFICIENTS of a(x) and b(x) (likes the bounds that I have are all for the coefficients)

which is a lot harder to tell if the product vanishes because this notion of "multiplicativity" no longer holds

when I had tuples, it was true that (a(0), a(1), \cdots a(p^n-1))(b(0), b(1), \cdots, b(p^n-1))=(a(0)b(0), a(1)b(1), \cdots, a(p^n-1)b(p^n-1))

but for a product of polynomilas, the product of coefficients no longer has this nice sort of behavior

which makes it a lot harder to study

and since its polynomial functions, like, u can't just "match" the terms to 0 or to be zero divisors in Z[p^{n}] either cuz its not necessary for all the coeffs to be 0 in order to have a 0 funciton

[upbeat fulcrum]

can anyone help me with this? How am I suppose to prove the following?

[prisma shuttle]

it says its a corollary do u have like a theorem before it or smth?

[upbeat fulcrum]

The professor just give it away

but he mentioned something about hensels lemma would be a good approach in proving this but I just dont know how to start

[tribal moss]

In fact I retract what I said. I imagined one could find a Lagrange interpolation polynomial that vanishes at all but one of the elements coprime to p, but in the light of day that seems to be nonsense.

[sinful mirage]

1-page proof of Peter Weyl theorem second part with not high level tools, machinery

[prisma shuttle]

i mean the condition u gave is def necessary

but idt it is sufficient

[weary bane]

Hello what does this notation stand for $B^{W_G}$ where $W_G$ is a finite group of automorphismes of B, B is the algebra of regular functions on a on algebraic torus and G is a reductive group?

[cloud walrus]

kh

[tribal moss]

Do you have any context this notation is used in?

[thorny knoll]

Hi! So apparently, given commutative rings with unity R, S with multiplicative set D in R, we have that $D^{-1}R$ is, up to isomorphism ,determined by the universal property:

For every ring hom $\phi : R \to S, \phi(D) \subset S^*$, there is a unique ring homomorphism $\psi : D^{-1}R \to S$ such that $\phi = \psi \circ \iota$, where $\iota$ is the canonical embedding into the localization.

I don't see how this property can determine the localization, as even R itself fulfils this property, but is not generally isomorphic to a localization over R. What am I missing?

[cloud walrus]

AzarTheKas

[stoic rose]

You're missing the condition that the map $\iota$ itself satisifies $\iota(D)\subset S^\times$ where $S$ is the localization.

[cloud walrus]

radiateur-man

[stoic rose]

With this condition the universal property determine the localization uniquely up to canonical isomorphism

[thorny knoll]

Oh wow that condition is just missing from my exercise sheet

[stoic rose]

When we say that the localization map is universal among all maps $R \to S$ sending $D$ to units, that implies in particular that it the localization map sends $D$ to units 😛

[cloud walrus]

radiateur-man

[thorny knoll]

Could you please elaborate on that? I don't immediately see how this is implied 😅

[weary bane]

Ok, it can only be viewed as the following statetement. If B is à complexe algebra and W a finite group of automorphism on B then B is a B^W module finitely generated

[tribal moss]

Hmmm.

[gritty sparrow]

probably the fixed points

[stoic rose]

Well, in general when we say that "X is universal among all Y satisfying condition C", that mean something like "X satisfy condition C and there is a unique map from X to Y for each Y satisfying condition C"

[gritty sparrow]

As in, the elements of the algebra fixed by all automorphisms in W.

[simple mulch]

We associate basis to finite dimensional vector spaces or is there any better word? (instead of associate)

[stoic rose]

Context?

[simple mulch]

Since V (space of functions) has finite dimension we can associate a basis of delta functions indexed by the elements of the domain of f in V

[stoic rose]

V is the space of functions X -> k with X a finite set?

[thorny knoll]

Alright, I guess this makes sense. Pretty much the first universal property we have to prove and the actual meaning of universal hasn't yet been covered. Thanks a lot!

[stoic rose]

It's not really a formal meaning, but you'll see that pretty much everything we call a "universal property" has more or less this form.

[simple mulch]

yes

[stoic rose]

Your sentence is somewhat weird since it's more the fact that the delta functions form a finite basis which imply that V is finite dimensional than the other way around.

[simple mulch]

yeah I know 😦

[stoic rose]

You can just say something like : "V has a finite basis of delta functions indexed by the elements of X"

[simple mulch]

Ok, more direct

thanks

[chilly ocean]

can anyone explain the first equation

how is it the same as orthonormality of columns?

write out what "orthonormality of columns" means in terms of the entries

so the scalar product has to be zero, but why with the conjugate of the other vector's entries?

complex dot product conjugates the entries of one of the vectors

for $z, w \in \bC^n$, $$\langle z, w \rangle = \sum_{k=1}^n z_k \overline{w_k}$$

[cloud walrus]

TTerra

[chilly ocean]

thanks

[chilly radish]

Here is my solution, but I don't see where I usedd the assumption n>4 or if it's necessary

The aim of the second part of the exercise was constructing a polynomial of degree n with galois group Sn, so I suspect since we've classified galois groups up to degree 4 this is why they excluded it, but i'm not sure

[cloud walrus]

ShiN

[chilly radish]

g_2,..,g_r are also irreducible

[lilac trench]

What is the general element of F(a_1,a_2,...,a_n)

My professor tells me its a ratio of polynomials, but I dont see it

[lethal dune]

ratio of polynomials

[tight latch]

I'm struggling with how to calculate an endomorphism

For example, say youre calculating EndZ(Q)

I think I'm struggling conceptually with what it actually means

[mighty spade]

by this, do you mean the set of endomorphisms from Q to Z, or the set of endomorphisms from Z(Q) to itself?

[devout crow]

or even the set of endomorphisms of Q as a Z-module?

[tight latch]

I am not really sure

From the definition I thought its the set of o s.t. o:Q->Q is a hom

But I dont see where Z comes in

[devout crow]

I think it's just the set of group homomorphisms Q --> Q

do you know what a module is?

[tight latch]

yes

for endomorphisms we want module homs right?

[devout crow]

do you understand why Z-modules are the same thing as abelian groups?

[tight latch]

Hmm

let me think about it

when we say module, is this the same as KG-module?

[mighty spade]

he means a module over a ring

i'm not sure what a KG module is

[tight latch]

oh its different then

[devout crow]

anyway this is an important idea, and it means finding all the Z-linear maps from the Z-module Q to itself is the same as finding all the group homomorphisms from the abelian group Q to itself

[tight latch]

Just found the definition, so from the way we've defined a module, they are all abelian groups with an action RxM->M

[chilly ocean]

vector space over a ring

[tight latch]

We've only defined left R-modules

So to find a Endomorphism, we need to consider where the generators of the module go? pls ping if you reply tysm ❤️

[devout crow]

this would work for something like problem 1.15 (the one below), but it's tricky here because Q doesn't have a minimal set of generators

[chilly radish]

In general modules aren't what's called free (aka act like vector spaces) so a map defined on generators will not necessarily be well-defined

In fact sometimes you don't even have a basis

Actually, Q is not even free

[devout crow]

if you want a hint, ||suppose f: Q --> Q is an endomorphism; then f(n) = nf(1). what is f(p/q) in terms of f(1)?||

[chilly ocean]

Yeah. Q is not a free Z-module

[tight latch]

p/qf(1)

[chilly ocean]

Because any two points x, y of Q satisfy nx+my = 0 for some non-zero n, m in Z

[devout crow]

yes! which means every endomorphism Q --> Q is determined by the image of 1

(or equivalently, by the image of any other nonzero element)

[tight latch]

which is why in the second q we can use where the generators go?

[devout crow]

sorry I wasn't clear, this is only true in the setting Q --> Q

[tight latch]

also, is f(n)=nf(1) true for every endomorphism, as in this is the definition?

no worries!! tysm for being so patient

[devout crow]

yes that's true for any abelian group

[chilly ocean]

This is from definition of Z-homomorphism

[tight latch]

amazing ty

so overall for this question, can we write the answer as {f(1) st f(n)=nf(1)}?

[chilly ocean]

{f:Q to Q : f(x) = x•a for some a in Q}

[tight latch]

whats x in that?

[mighty spade]

the argument of the function

[tight latch]

ngl idk what that is

forgive me my first year at uni was covid year they just cancelled my classes

[chilly ocean]

{f:Q to Q : exist a in Q, for all x in Q, f(x) = x•a}

[devout crow]

f is a function Q --> Q right

so it takes a number in Q

[delicate orchid]

is this even uni-tier

[devout crow]

and spits out another number in Q

the thing it takes in is the argument

[tight latch]

no need to shame me i'm trying to learn

[devout crow]

argument of a function = input of a function

[delicate orchid]

not shaming I'm just surprised is all
anyway, don't let me interrupt

[tight latch]

oh

lol

[chilly ocean]

Do you get this one?

[tight latch]

I understand that x is the input but i meant is x in Q? And is x fixed?

[chilly ocean]

I wrote that x is in Q

And that it has to hold for all x

[tight latch]

oh aorry i didnt see your second one

Yeahhh ok i understand that thank you!

[chilly ocean]

Thank xdres

[tight latch]

Thank you both so much!

So generally I can follow the method see where numbers go in f(n)=nf(1) and then this gives the set?

[devout crow]

sorry if I patronised you a bit explaining what the input of a function is lol

[tight latch]

no worries haha i've just never heard it called the argument before I've always called it input ahah

thank you so much for your help and patience

this was really doing my head in 😂

i think i know how to do the second q

[devout crow]

yeah, so in this situation we must have f(p/q) = p/q f(1) (this comes as a consequence of the endomorphism condition f(n) = nf(1), as well as how fractions behave)

which means all endomorphisms are of the form "multiply by something"

[tight latch]

Great tysm!!

[devout crow]

np

[tight latch]

Ty too Blitz <33

[chilly ocean]

You're welcome

[chilly radish]

bump, I don't see what i'm missing

IF I am missing anything

[tardy yacht]

didn't you use the assumption on your third step, when you constructed a polynomial of degree n-3?

[chilly radish]

Well, if n=2, then i'm done since I can just take f_3=g_1, if n=3 I can take f_3= xg_1 and if n=4 I can take g_1 plus 2 degree 1 polynomials with distinct roots

So it doesn't really affect the final outcome

I did it like that because that was the assumption in the exercise, but it's not necessary for that part

I'm in F_5 so I have the freedom to choose 2 distinct roots

[tardy yacht]

I will read this more carefully but I think it should be fine (?)

[chilly radish]

Same, I just dislike being given extra assumptions without justification

Makes me feel like I did smth wrong when I didn't use them

[tardy yacht]

I think you have a typo in your next to last line, you mean f_3=g_1g_2x right?

[tardy yacht]

yeah I get you

[chilly radish]

Yes

Indices can be confusing

[tardy yacht]

Yeah, I don't think you need the assumption lol

Does someone know how is the state of Differential Galois Theory these days?

[wise igloo]

heyyyyy so would anyone want to guide me through how to prove that Aut(Z_n) is isomorphic to U(n) where U(n) is the multiplicative group of integers modulo n?

[chilly ocean]

look where the generators go

[wise igloo]

to generators

[chilly ocean]

hint: an automorphism of Z_n is determined by where it sends 1

[wise igloo]

yeah but like idk what to do with this information

im kinda stupid

[chilly ocean]

the other way around might be simpler

every element of U(n) gives an automorphism of Z_n, right?

(by multiplication)

[wise igloo]

yes

[chilly ocean]

so you get a map U(n) -> Aut(Z_n)

injective? surjective?

[wise igloo]

injective right

wait no

wait

yeah

injective

[dull roost]

a

[heavy dagger]

Sanity check: If k is a field a polynomial in k[x] with degree d can have at most d distinct roots since k[x] is a UFD?

[delicate orchid]

Is that not the fundamental theorem of algebra

[chilly ocean]

No

That's for C[x]

[delicate orchid]

Im thinking about the one that states that a polynomial of degree n (over a field) has at most n roots

[chilly ocean]

This is basicaly because k is an integral domain

[chilly ocean]

The same thing works in any integral domain

[delicate orchid]

Cool to know it’s that general

[heavy dagger]

Doesnt it have to be a UFD?

[chilly ocean]

No

It works in any ring R with no zero divisors

[south patrol]

I mean any integral domain embeds in its field of fractions so it immediately follows from the field case ig

[chilly ocean]

The proof for fields is the same as the one for integral domains as well as any ring with no zero divisors

So this is just complicating it

[tribal moss]

Hmm, the argument for fields I can imagine depends on polynomial division, which doesn't immediately work in an arbitrary integral domain.

We can form the field of fractions an do the proof there instead, though.

(As Potato said).

[chilly ocean]

You can divide by any monomial

[tribal moss]

So in Z[x], what is x²-1 divided by 2x-2?

[chilly ocean]

That's not a monomial

[tribal moss]

Oh, monomial.

How is dividing by monomials enough? The proof I know needs to divide by arbitrary degree-1 polynomials.

Hmm, but they can be chosen to be monic, which saves us.

Okay.

A "monomial" is a polynomial with only one nonzero term (but its coefficient can be arbitrary). A "monic polynomial" is a polynomial whose leading term has coefficient 1.

[chilly ocean]

I'm sorry, I've mixed the terminology up

Monic polynomial of course

[tribal moss]

Right, so over any integral domain, any nonzero polynomial is a product of monic linear factors with a leftover factor that has no roots.

Every root corresponds to one of the linear factors.

[chilly ocean]

Yes. And any ring with no zero divisors as well, since commutativity is not needed

[chilly ocean]

Conversely, if R has zero divisors, ab = 0, then (x-a)(x-b) has at least three roots, a, b and 0

or a = b, 2a = 0 and a^2 = 0 I suppose

So that (x-a)^2 = x^2

Then (x-a)(x-2) is as needed, or a = 2 (or 2 = 0)

So we're looking at a ring with 4 = 0

I wonder if you can push this further and show the converse

If 2 =/= 0 then (x-1)(x-2)(x-3) has roots 0, 1, 2, 3 which are all different

so we're looking at a ring "of characteristic 2", that is in which 2 = 0

I just realized it doesn't really make sense to speak of roots of a non-commutative polynomial

since if we have something like w(x) = ax, then x is supposed to commute with a, so w(x) = ax = xa

so roots could really only be defined for the center of our ring

so let's just assume it commutes

So we have a commutative ring with 2 = 0. Question is, if there are polynomials with more roots than their degree

[north sand]

something like x(x-1) over the ring of diagonal matrices for some big enough n over F_2 ? I didn't follow the whole discussion. not sure what you are currently trying to do

[stoic rose]

One context in which it would make sense is if the noncommutative polynomial has coefficients in k for some commutative ring k (could be a field for example). Then we can talk about roots of the polynomial in any k-algebra since by definition of a k-algebra the action of k commute with the multiplication in the algebra.

[chilly ocean]

oh, sorry, I said something different than I meant. I meant if w(x) is a polynomial with coefficients in a non-commutative ring

[stoic rose]

Sure, though in most contexts coefficients are always from a commutative ring

[tribal moss]

Yeah, polynomial with non-commuting coefficients don't really work in the first place. They're not closed under multiplication.

[stoic rose]

Otherwise you get weird stuff going on, like how do you multiply the polynomials aX and bX? Is it abX^2, but to do that you have to commute b and X which isn't really desirable

[tribal moss]

I suppose it'll work as long as we only ever evaluate them on central elements as values for X.

[chilly ocean]

2 = 0 and there is some non-zero a with a^2 = 0

[tribal moss]

Z[x]/<2,x²> does that.

[tribal moss]

In Z[X]/<2, X²+X>, all four elements satisfy a²-a=0.

[devout crow]

this follows from the factor theorem and induction on degree right?

[stoic rose]

Yes

[chilly ocean]

hmm... what's an example of such polynomial there?

[tribal moss]

I mean it satisfies 2=0 and x²=0.

[chilly ocean]

oh, no, I meant it has to have all 3 properties

[stoic rose]

The polynomial T^2-T has more roots than its degree

[chilly ocean]

in Z[x]/<2, x^2> ?

[stoic rose]

No in the Z[x]/<2,x^2-x>

Is has roots 0,1,x

[tribal moss]

As well as x+1.

[chilly ocean]

but does it have a non-zero element a with a^2 = 0

[tribal moss]

Then look at Z[x,y]/<2, x^2-x, y²>.

[chilly ocean]

wait, I got confused

I am trying to check if 2 = 0 and non-zero element a such that a^2 = 0 imply that there is a polynomial with more roots than its degree

[chilly ocean]

so I should look at this and check if it works

ax has 0 and a as roots this was easy

okay, this was actually really easy

if there are two zero divisors a, b then I can just consider polynomial ax

I was overthinking it a lot

but maybe it can be salvaged if we change the hypothesis to monic polynomials

which is a little stronger

so if there exist zero divisors, does there always exist a monic polynomial with more roots than its degree

and I've reduced it to case when 2 = 0 and a^2 = 0 for some non-zero a

[tribal moss]

Take the product of (x-b) for all nonzero b in the ring. Since the ring has zero divisors, 0 is also a root.

Hmm, no, that presumes the ring is finite.

Why not just (x-a)(x-b) for ab=0 with a and b nonzero?

[chilly ocean]

Because we can have a = b

[tribal moss]

And all multiples of a are either a or 0.

I see.

If there's a monic polynomial with too many roots, then there's one that's a product of monic linear factors.

[tough raven]

Polynomial division by monic polynomials is always possible; in particular, by (X-a), which gives you the Factor Theorem.
Then you can do the usual argument
p = (X-a1)p1 = (X-a1)(X-a2)p2 = …
because (a2-a1)p1(a2) = 0 and a2-a1 is cancellative.

[tough raven]

Oh, this was already posted. Sorry 😅

[patent ocean]

I need help with 5i). From 4) I know that the symmetries of the tetrahedron are isomorphic to S4. So what I have done is paired two points with each edge and try to find what the symmetries do with these pairs of points. I thought I was going in the right direction but so far my answer doesn't match with the answer given

[patent ocean]

actually I think I've found the answer but i don't fully understand how it is right

I applied the elements A4 onto the colourings

then try to find the no.of orbits from there

but i'm unsure why I cannot include the transpositions

[west violet]

well how many elements are in a4

and how many rotations are possible with a regular tetrahedron

[patent ocean]

oh so you can use reflections with edges?

can't

[west violet]

nope

@patent ocean i'm assuming you got the answer via burnside lemma?

@delicate orchid did i make a mistake?

[delicate orchid]

hold on lemme make sure I'm visualising this right

yeah I'm pretty sure the reflections of the cube aren't rotations

[patent ocean]

yeah burnside lemma/ orbit counting formula

I think I get it because it isnt possible to get a transposition when applying symmetries to edges

you either get a 3-cycle or double transposition

[west violet]

ok i might have generalized so i'll take back that statement. What i was trying to say is that when you have a line passing through 2 non-intersecting edges in a tetrahedron, you get rotations of 180 degrees

[delicate orchid]

yes that is very much true

[west violet]

thanks for the catch though

[delicate orchid]

no worries

[patent ocean]

through the middle of the edges?

[west violet]

yeah

midpoint of one edge to midpoint of another,non-intersecting

[patent ocean]

ah yeah

does my reasoning make sense though

when applying the symmetries on the tetrahderon to S4

relating*

[west violet]

haha yeah and these are the elements of a4

[patent ocean]

thanks

could you give me a clue for ii) because I don't think I'm approaching it right

I'm trying to find the no. of colourings that have 0 monochromatic triangles then subtracting from the total

but my calculations are just wrong

[west violet]

uh

you use PIE but i think they already told you that

[patent ocean]

total-(AnBnCnD) right?

[west violet]

what are A,B,C,D here?

are you saying PIE in general?

[patent ocean]

the four faces being not monochromatic

A=face 1 not being monochromatic

B= face 2

etc

[west violet]

oh there's an easier way to think about it

how many ways can you color the edges in a tetrahedron?

with n colors

[patent ocean]

(n^6+3n^4+8n^2)/12 ''essentially'' different ways

[west violet]

oh just in general

the total ways you can color a tetrahedron(distinct coloring doesn't matter for now)

[patent ocean]

edges or verticies

[west violet]

edges

[patent ocean]

n^6

[west violet]

correct

so now find the numbers of ways to color the tetrahedron so that you have 1 monochromatic triangle

this is our application of PIE

[patent ocean]

umm

(n^3)*[(n-1)^2]n?

[west violet]

no lets step back a bit

you have 6 edges

3 edges make up a triangle

Thus in order to have only 1 monochromatic triangle, two of those edges must be the same color as another edge

Thus there are n^4 ways to do this

(n)(n)(n)(n)(1)(1)

[patent ocean]

every edge is connected to 4 other edges

i think

[west violet]

yeah

but do you understand the reasoning above?

or should i draw a pic

[patent ocean]

i think i kinda understand

given two edges that are connected together and the same colour

actually nah i'm lost

[west violet]

hopefully the numbering makes sense

[patent ocean]

yeah

[west violet]

so 5 and 6 are not highlighted

[patent ocean]

right

[west violet]

in order to form a monochromatic triangle, 3 must be the same color as 5 and 6

[patent ocean]

yes

so those 3 edges must be same colour n

[west violet]

yup

[patent ocean]

then the other three can be any other colour?

[west violet]

and 1,2,4 can be any other color

[patent ocean]

right yeah im dumb

[west violet]

nah its hard to understand unless you have a good visual

3 can be any other color as well but 5and 6 must match 3

so (n)(n)(n)(n)(1)(1)

since 5 and 6 can only be the same color as 3

[patent ocean]

yeah that makes sense

[west violet]

however we only focused on one edge (which was 3)

we have four possible triangles in a tetrahedron but we only focused on one

thus we have 4n^4 possible colorings with 1 monochromatic triangle

[patent ocean]

oh ok

n^4 for each triangle

[west violet]

yup

Then you can try to find the number of colroings with 2 monochromatic tirangles

This one is a bit trickier

[patent ocean]

oh so then you do it for 2,3 and 4?

[west violet]

yup

in this way ,our final answer(using PIE) will be the number of colorings that don't have a monochromatic triangle

and then that quantity you subtracted from n^6 will be the number of colorings such that there is atleast 1 monochromatic triangle

[patent ocean]

wait how

i thought we were adding 1+2+3+4 monochromatic triangles

so how does that give no. of colouring that don't have a monochromatic triangle

my bad for being slow

[west violet]

4n^4 overcounts the number of colorings with 1 monochomatic triangles

if a coloring has 4 monochromatic triangles, it gets overcounted if we consider the number of colorings with 1 triangle, 2 triangles and 3 triangles

thus by PIE, we can guarantee that we don't overcount since we want different colorings

[patent ocean]

oh ok

[west violet]

does that make sense?

https://math.stackexchange.com/questions/688019/what-is-the-inclusion-exclusion-principle-for-4-sets

we're essentially doing pie on 4 sets

[patent ocean]

yeah 2,3,4 is overcounted in 1

[west violet]

yup

[patent ocean]

3,4 overcounted in 2

etc

yeah that formula is long af

It looks tedious to compute as well

[west violet]

well its a bit easier here since our sets are just the triangles themselves

[patent ocean]

for 2 triangles do I start with edges connected together or disjoint edges

[west violet]

so for two triangles, we can fix 4 edges

and 2 remaining edges can have n colors

so here, we're looking for pairs of triangles that are both monochromatic

[patent ocean]

oh ok

is it just 2 pairs of 2 edges that are joint

1 fixes 4 and 5 but 2 fixes 6 and 4

but there are also other possibilities

[west violet]

that's the only possibility for 2 triangles

remember an edge is connected to 4 other edges

thre are n^2 possible colorings for 2 triangles in a tetrahedron

since 4 are fixed and 2 can each have n possible colors

but there are (4 choose 2) pairs of triangles in a tetrahedrn

[patent ocean]

so 6n^2 in total?

i think I visualise it a bit differently maybe

I see that 5 sides must have the same colour and so there are n choice for these 5 sides and then another n choices for the remaining side that isn't within the monochromatic triangles

for 3 all edges but be same colour

so n possible colourings for 3 triangles

4n triples of triangles in tetrahedron

and 4 is just n

[west violet]

yup

and then im sure you've figured it out

4n^2- 6n^2 + 4n-n =

$$4n^2-6n^2+3n$$

[cloud walrus]

suck2015

[untold basin]

First one says "Find the order of G"
but I have trouble with second one

"For which integers n ≥ 1 is there an injective morphism ?"

Btw for first one I found that order of G is 6

I found that if f is injective, then n<=order(G)

so n is in {1,2,3,4,5,6}

By Lagrange, |Im(f)| divides |G| so |Im(f)| is in {1,2,3,6}. And Z/nZ is isomorph to Im(f) since f is injective so n is in {1,2,3,6}

If n = 6, Im(f) = G so Z/nZ isomorph to G and since Z/nZ is abelian so G too. Contradiction so n is in {1,2,3}

And I'm stuck here

Hope someone can help c:

[tardy yacht]

For n=2, you must send 1 to an element x such that x^2=Id (this is possible)

It is also possible with n=3, (Hint: look at tau)

[stark sigil]

A flashy way would be to notice that the minimal polynomial of $\sqrt{-5}$ is $x^2+5$ which is not irreducible over $\bF_3$

[cloud walrus]

Icy001

[stark sigil]

You can even reverse-engineer this to produce a non-invertible element of $\bF_3[\sqrt{-5}]$

[cloud walrus]

Icy001

[smoky cypress]

Well maybe a more intuitive way is to just go through the process of finding an inverse for an arbitrary element of this ring and see which step will fail which will lead to a way to construct a nonunit. So you start by noticing that each element can be written as $a+b\sqrt{-5}$ where $a,b\in\bF_3$. If this has an inverse then $(a+b\sqrt{-5})(c+d\sqrt{-5})=1$ for some $c,d\in\bF_3$, so multiplying it out gives $ac+(ad+cb)\sqrt{-5}+bd(-5)=ac-5bd+(ad+cb)\sqrt{-5}$, so $ac-5bd=1$ and $ad+cb=0$. Let's just assume $a\neq0$, then $ad+cb=0$ implies $d=-cb/a$, so $1=ac-5bd=ac+5bcb/a=c(a+5b^2/a)$. Now if you want to solve for $c$ you need $a+5b^2/a$ to be not $0$, but if you think about it there is some choice of $a,b$ that makes $a+5b^2/a=0$, and such choice of $a,b$ will mean that $a+b\sqrt{-5}$ is not invertible.

[cloud walrus]

Whoever

[smoky cypress]

@chilly ocean

[tribal moss]

Try multiplying 1+sqrt(-5) and 1-sqrt(-5).

[smoky cypress]

No problem and if you go through that whole thing you'll actually end up with 1+sqrt(-5) xD

[tribal moss]

Have you seen the concept of "zero divisors"?

A ring that has zero divisors can never be a field.

Namely, if pq=0 and p is nonzero, then q cannot have an inverse. If r were an inverse to q we would have
0 = 0r = (pq)r = p(qr) = p1 = p

[charred flume]

Find a polynomial $p(x)$ in $\mathbb{Q}[x]$ such that $\mathbb{Q}(\sqrt{1+\sqrt{5}})$ is ring-isomorphic to $\mathbb{Q}[x] /\langle p(x)\rangle$.

[cloud walrus]

Ji

[charred flume]

is p(x) the irreducible polynomial with degree n?

or is it something else?

[tribal moss]

p(x) is an irreducible polynomial. You have not defined n, but p certainly has a degree.

[charred flume]

ah yeah sorry

[tribal moss]

In fancier terms you're being asked for the minimal polynomial of sqrt(1+sqrt(5)).

[charred flume]

n is the deg($(\sqrt{1+\sqrt{5}},\mathbb{Q})$

[cloud walrus]

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

[charred flume]

ohh alright

thankss

[chilly ocean]

can u post the full problem

[cloud walrus]

CuriousTalon

[charred flume]

use mathpix

[cloud walrus]

CuriousTalon

[proud bear]

are you sure they weren't $Z_5[x]/(x^2-3)$ and $Z_5[x]/(x^2-4)$? the first one you wrote doesn't make sense. the second one is like a localization

[proud bear]

oh ok

[patent ocean]

Yeah thanks again for being patient with me

[wooden ember]

how might i go about showing (or disproving) that an extension by separable elements is separable

wait i think i have a counterexample maybe

[hidden haven]

I'm pretty sure it's true

One way is to prove lemma: A finite extension E/F is separable iff there are exactly [E:F] many F-embeddings of E into cl(F)

And then show that if you extend by separable elements, then this bound is met

[wooden ember]

yeah nvm i just dont know how to cube things lmao

[hidden haven]

You always have no of embeddings ≤ deg of extension

[wooden ember]

cl(F)?

[hidden haven]

Algebraic closure

[wooden ember]

we havent covered it yet but i guess i could try

[hidden haven]

Oh then idk how you'd prove this lol

[wooden ember]

huh i dont think we've shown this

[wooden ember]

oh i mean it's not an exercise i was just wondering

[hidden haven]

My prof for field theory did this one big lemma on like day 2 of the course

About when extensions of homomorphisms exist

And how to count them

And then the entire rest of the course is just spamming this same lemma over and over again

Here we are just counting the extensions of the inclusion of F into cl(F)

So ye this is pretty fundamental in my opinion lmao but idk some books seem to do field theory very differently

[wooden ember]

so what's the lemma precisely?

[hidden haven]

Let me try and write it down on some paper

It's long

[wooden ember]

oh dont waste your time if it's that long haha

[hidden haven]

Don't have my notes from then with me

Nah I won't include proof

But will I remember all the cases he covered

[wooden ember]

yeah looking back at my notes i think the only cases of extending homomorphisms we've looked at is extending an isomorphism into the splitting fields of a related polynomials or into simple extensions

[hidden haven]

Damn

Very sad

[wooden ember]

sadge

[wooden ember]

so just to be sure by F-embedding here you mean an injective F-algebra homomorphism right

[hidden haven]

Injectivity is redundant

I mean homomorphism that maps the copy of F inside E to the copy of F in cl(F)

Ye I think that's exactly an F-algebra homomorphism

[wooden ember]

that's not necessarily injective though right? or am i being stupid

[hidden haven]

In particular it is a field homomorphism

Fields only have 2 ideals

1 has to map to 1 so can't be in the kernel

So the kernel is the 0 ideal

[lethal dune]

cool lemma

[wooden ember]

omg yeah im dumb

[lethal dune]

if that can be used to prove entire galois theory

[hidden haven]

This is sort of a paraphrasing, there were like 4 cases that we studied specifically in this second part

[wooden ember]

oh you mean this is just counting the number of ways to extend to a particular field

[hidden haven]

But you can probably get the idea of how counting roots and counting extensions are similar, and this is why when discussing F-automorphisms (extensions of the identity) the roots and separability/normality results come in

[wooden ember]

yeah we didnt prove that but it's pretty intuitive

[hidden haven]

Ye

But you can write down specific cases

[wooden ember]

i thought you meant how many field extensions existed or smth

[hidden haven]

And there's value to having the lemma prewritten

[wooden ember]

like how many different E's for an F

[hidden haven]

Oh lol

No of homomorphism extensions not field extensions

So ye the earlier claim follows from this stuff rather quickly

[hidden haven]

This one

[wooden ember]

right that makes sense

[hidden haven]

That's the best case because Ω has all the roots and the polynomials are all separable

[wooden ember]

you're just looking at all the roots of min poly and checking that they each contribute to an embedding

[hidden haven]

Yes

[wooden ember]

okay makes sense

[hidden haven]

But in a weird way

Because this happens inductively

If you try the induction you'll see that you want minimal polynomial over F[a,b,..,something]

Like you've already extended a bit and have to extend more

[wooden ember]

right

[hidden haven]

So minimal polynomial is not over F anymore but things still work out lol

[wooden ember]

i see

and extending by one separable element, i have [E:F] = # of roots of min poly = number of F-embeddings so that shows extensions by separable elements are separable right

[hidden haven]

Ye

[wooden ember]

coolio

thanks a bunch

field theory is whack

[hidden haven]

[wooden ember]

we agree that in part 3 we need alpha to be a root of f still right?

it's not very clear that it keeps the hypotheses of part 1

here its saying in 3) that if L is the splitting field of f then the minimal poly of alpha splits in L

and part 1 tells us that the galois group permutes roots of f

in principle alpha is an arbitrary element of L an arbitrary extension, its only assumed it's a root of f in part 1

seems to me necessary for part 3 too though

[hidden haven]

What does corps mean

[cursive temple]

field or fields

[hidden haven]

I see

Ye you need f to have some root in L

Wait no

I mean you probably need alpha to be a root of f

[wooden ember]

yeah exactly thought so

this was confirmed in the proof, where at one point they say "because alpha \in L..."

[strong moth]

someone asked me if every ring is a monoid.

i told them i wasn’t sure if that question made sense?

because you have to pick under which operation either a ring would be a monoid under: + or *, and in the case of + it’s a monoid since under + a ring is a commutative group and under * it’s a monoid since rings are monoids under * with left / right distr.

so without specifying the operation of consideration it doesn’t make sense to ask if “every ring is a monoid”

is my response correct?

[hidden haven]

Yes

[strong moth]

thanks!

[north sand]

sidenote, to prevent italics, you can use \*

[tough raven]

You prove this by induction one simple extension at a time, right?
If p_min(α) is q(X^p^k) where q is irreducible separable, then the number of extensions of embeddings of F to F(α) should be deg(q).
So is this inequality actually a divisibility relation?

With ratio of degree to number of extensions a power of the characteristic

[hidden haven]

I think divisibility will mess up in the inductive step

Because depending on what root you map the first generator to, you will get a different number of choices for the next one

[wooden ember]

[L:L^p] isnt necessairly finite right

[wooden ember]

i need a gut check, completed an exercise where i ended up showing midway that K is field-isomorphic to K^p for K a field of characteristic p>0

is this correct?

[hidden haven]

Ye frobenius map is an isomorphism

[wooden ember]

coolio

[hidden haven]

don't think so

You can have F[{x_i}]/F[{x_i^p}]

[wooden ember]

should be able to take something like F_p(t_1,t_2,....)

[hidden haven]

ye

i ranges over any infinite set

[wooden ember]

yeah

[delicate bloom]

just cause it's one line, since it's clearly surjective, just show injectivity by x^p=y^p means 0=x^p-y^p = (x-y)^p so x-y=0, I think that should do it

[wooden ember]

yeah i mean i did prove it it just didnt sound realy right with my intuition

[delicate bloom]

I guess I figured since you were unsure and didn't post your proof that meant your proof was really long or something weird was going on with it

better safe than sorry

[wooden ember]

yeh thanks

[wooden ember]

i hadnt proved it that way because i only really proved it as a biproduct of another result

i was uniquely extending automorphisms from K^p to K

[hidden haven]

In general field homomorphisms are injective so that isn't needed

[delicate bloom]

lmao true

[tough raven]

I worried about that, but it depends only on the minimal polynomial of β over F(α), which doesn't depend on the choice of embedding of F(α)
I think

Is the number of extensions 1 for a purely inseparable extension?

In that case, IIRC there is some separable/purely inseparable parts for any extension and the number of embeddings should be equal to the degree of the separable part, hence dividing the overall degree.

[hidden haven]

yes

[hidden haven]

The number of embeddings of Q[sqrt2, 4th rt2] into R should be a counterexample

[tough raven]

What

Any extension of Q is separable anyway, right?

Wait

[hidden haven]

I mean the number of choices available for 4rt2 depend on choice of sqrt2

[tough raven]

Isn't it supposed to be embeddings into an algebraic closure?

[hidden haven]

oh then you're good in any case

[tough raven]

Right

But does it divide the degree

[hidden haven]

ye

2 embeddings

but the reasoning don't work

[wooden ember]

okay im gonna sound like an idiot but ive literally been staring at the polynomial x^p-a for about 1h trying to determine how to show it's irreducible when K is of characteristic p and a is in K\K^p

a hint is all i need

[hidden haven]

In its splitting field, it splits as (x-b)^p

[wooden ember]

sure yeah i did that

[hidden haven]

So the splitting is (x-b)^k (x-b)^(p-k)

[wooden ember]

yeah

[hidden haven]

So b^k is in K

for k strictly between 1 and p

[wooden ember]

why can you say this

[hidden haven]

constant term of one of the factors

[wooden ember]

god dang

ok stop there ill try and complete

and then go die in a ditch somewhere

[hidden haven]

I remember struggling with the same problem once lmao

Solved it together with saketh in the end

[wooden ember]

aight well that makes me feel less bad lol

[wooden ember]

is there any particular reason we're choosing b^k instead of b? I mean we have k*(-b) in K since that's the degree k-1 coefficient of (x-b)^k so we would straight away get that b is in K since k<p no?

[hidden haven]

I guess that works better ye

[wise igloo]

so like could someone help me with this

pls

:>

[hidden haven]

The image of 1 under the automorphism determines the automorphism uniquely

Check which images of 1 give you automorphisms

[wise igloo]

it sends 1 to some a in U(n) depending on the automorphism

this is what I have written:
$$\text{Aut}(\mbb Z_n)={\sigma_a:a\in U(n)}\cong U(n)$$ where $\sigma_a:\mbb Z_n\to\mbb Z_n, \sigma_a(1)=a$

[cloud walrus]

gmod

[wise igloo]

oh ok actually I saw a proof of it

makes sense

[simple mulch]

So, I have 12 characters from a group. But there's only 11 conjugacy classes. How could I find the one that's not from an irreducible representation?

[delicate orchid]

inner product with itself

if it's 1 it's irreducible

[simple mulch]

yeah I did it

and its one

but shouldn't I have 11 characters? One for each conjugacy class?

[delicate orchid]

what field are you over

[simple mulch]

Or am I missing something? We know there's the same number of irreducible representations as conjugacy classes. So, since every irreducible representation is completely described by its character. And I have 12 characters?

[delicate orchid]

cause unless it's something bizarre then yes you should have 11

unless there's actually 12 conjugacy classes

[simple mulch]

its 11, the field is C

[delicate orchid]

super odd