groups-rings-fields

[solar glacier]

canu define those terms

[white oxide]

polynomials of degree 1 or higher

[solar glacier]

ill dm u rn

[slim kayak]

I love when I write a proof and just divide all my deltas and epsilon by a fitting number after the fact.

[white oxide]

oh do you mean semigroup and monoid

[formal ermine]

fwiw it's ||roots of unity in Q(sqrt(2)) x Z||

[solar glacier]

looooool

[rustic crown]

yea looks like it

[white oxide]

ok sweet

[solar glacier]

now check it!

🙂

[white oxide]

nah i mean it's pretty easy

[solar glacier]

omfg duhhhhh wow i cannot believe i missed that

[formal ermine]

yeah duh

[solar glacier]

true standard definitions

[formal ermine]

in general like

@solar glacier do you know what the ring of integers of a number field is

[rustic crown]

is

[slim kayak]

show there is no identity element (impossible) [almost died]

[solar glacier]

havent heard of this term before no

i wanna say its analgous

to ring of fractions

but integers

[formal ermine]

a number field is a finite extension of Q

[white oxide]

LMAO

[formal ermine]

the ring of integers contains all elements that are the root of a monic polynomial with integer coefficients

e.g. for Q, the roi is Z

for Q(sqrt(2)) it's Z[sqrt(2)]

[solar glacier]

ahh makes sense ok

btw i just messaged you the file

enjoy : )

[coral spindle]

det nooo I was gonna say that

you're so mean

spoiling all my fun >:(

[rustic crown]

oh gomen >.<

[solar glacier]

lol i thought the same for rad any square free integer

[rustic crown]

[coral spindle]

daijoubuweeb

[slim kayak]

hi gauss

[formal ermine]

and now there's this funny theorem that the unit group of the ring of integers of a number field is something something, and in the case of a quadratic number field, i.e. K = Q(sqrt(d)) (assuming d is square free), it's mu(K) i.e. finite when d < 0, and mu(K) x Z for d > 0. mu(K) here are the roots of unity in K

[rustic crown]

do it for a cubic now

[coral spindle]

Now I get to say that the ring of integers of Q(sqrt(5)) is Z[(1 + sqrt(5))/2]

[coral spindle]

You can't make me!!!!

[formal ermine]

ty

[solar glacier]

anytime!

my masters asvisor wreote the text

i wrote up the solutions that you see

not all solutions are complete yet

[rustic crown]

(i actually read "masters" as "mastews" )

[elder wave]

maybe read regular analysis first

[solar glacier]

p-adic number suystems and p-adic (induced)topology

[elder wave]

[formal ermine]

I don't know what I don't know

except like a million sequence convergence criteria

[solar glacier]

u have a copy of baby rudin

[formal ermine]

I've skimmed through the p-adic analysis book and it doesn't require more than basic real analysis / complex analysis

[rustic crown]

(singular analysis )

[solar glacier]

thats the god of all undergrad analkysis (inequality theory) texts

[solar glacier]

you should try the one by Serre

[formal ermine]

my prof said we should base our presentations on this one

[south patrol]

Nullstellensatz

[slim kayak]

Der Satz über die Nullstellen

[formal ermine]

https://www.youtube.com/watch?v=t1CRjNriCR8

[south patrol]

Genau

[solar glacier]

my fav theorem

[rustic crown]

i hear this word very often

[solar glacier]

ahh gotcha

[long nebula]

hilbert zero theorem

[south patrol]

Nullstellensatz is just assuming all polynomials have roots then all polynomials have roots

[slim kayak]

💀

[solar glacier]

LMFAO hi potato

[south patrol]

Hi My Math Your Math

[solar glacier]

its functions that are zero on zero set of ideal some power of them is in that ideal

[south patrol]

Ye

I(V(J)) = sqrt(J)

[slim kayak]

*if all the loner single variable polynomials have roots then all the social polynomials do too

[solar glacier]

i had to STATE it on my alg geo final and not prove it

i had prepared for the proof smh

[solar glacier]

precisely

i know the proove like the back of my hand

lol

one direction is trivial

[south patrol]

I am take comm alg exam in 2 weeks

[solar glacier]

i just took my alg geo final

got second highest score in class

[south patrol]

Gg

[rustic crown]

good luck uwu

[south patrol]

Thanks for the flex

Jk

[solar glacier]

lol

[delicate orchid]

$\sqrt(J) = \pm J$

[cloud walrus]

Wew

[south patrol]

Yh

[rustic crown]

[slim kayak]

I mean, it's not wrong...

[delicate orchid]

it's VERY wrong

[solar glacier]

it is wrong lol

[formal ermine]

it's the only truth

don't believe what the GOVERNMENT has been trying to TELL you

[slim kayak]

You defined minus ideal as something already?

[solar glacier]

rad J is the a \in R such that there is a k >=1 so a^k \in J

[formal ermine]

cumbersome definition

[solar glacier]

LOL

[formal ermine]

giga chads prefer the intersection over prime ideals one

[slim kayak]

Oh wait, i read that very wrong

[solar glacier]

it really is but hey its true lol

[south patrol]

rad J is just like

Yeah

[solar glacier]

thats more of a proposition i feel like

[obsidian sleet]

rad j^2

[formal ermine]

DEFINITION

[south patrol]

Anyway mmym

What is ur preferred proof

Of nullstellensatz

[solar glacier]

Rabnowitch trick

intriducting an auxilallry ideal

[south patrol]

Yes fair

[formal ermine]

who needs 10 pages of sheaf theory

[solar glacier]

exactly lol

[south patrol]

The presentation I've seen I quite liked

Uhh

[rustic crown]

sheaves are cute tho >.<

[solar glacier]

the trick of Rabitnowitch is SO CLICK

SLICK

[formal ermine]

hartshorne was too hard for me

I switched to uh

geometry of schemes

is easier

but they do so much sheaf theory at the beginning

[solar glacier]

i jus tpurchased it

i have the pdf

[rustic crown]

hartshorne hard >.<

[solar glacier]

yes chapter 2

[elder wave]

try Bosch

[formal ermine]

why purchase it if you can just read the first book of the bible

[solar glacier]

is there any ties w p adic and alg geo

[formal ermine]

yes

[slim kayak]

invented by algebraists to sell more theorems (weak nullstellensatz)

[solar glacier]

i need a hard copy lol

[solar glacier]

YES LOL

[formal ermine]

idk I like this one

[solar glacier]

REALLY?!?!?!?!?!?!?

[elder wave]

I've been reading some scheme/sheaf stuff on the side for a casual reading group and bosch has been good

[formal ermine]

timo tell me what a noetherian scheme is

[elder wave]

No

find out yourself

[south patrol]

sorry yeah i mean uh

[slim kayak]

For example pyramid schemes, they only work for finitely many steps of geometric growth before they collapse as well and the founder stabilizes in wealth

[south patrol]

I quite like Noether Normalisation => Zariski's lemma and show that f.g. algebra over a field is Jacobson

[rustic crown]

i like to phrase it differently, liek show given f in I(V(a)) then you wanna show f is nilpotent in k[x1, ..., xn]/a, so instead show (k[x1, ..., xn]/a)[1/f] is the zero ring. then it doesn't feel like a trick at all

[south patrol]

because uh

theoyr of jacobson rings gives u another way to think about rabinowitch too

all very cool

[solar glacier]

the proof goes as follows

[south patrol]

which proof

[solar glacier]

let $I=(f_1,...,f_m)$ where $f_i \in k[x_1,...,x_n]$

[cloud walrus]

MyMathYourMath

[solar glacier]

for Nullstellen

[formal ermine]

the proof

[south patrol]

Ah okay

[cloud walrus]

MyMathYourMath

[solar glacier]

then consider the ideal

[cloud walrus]

MyMathYourMath

[solar glacier]

Then the zero set for this is empty

[cloud walrus]

MyMathYourMath

[south patrol]

okay tbf like what is ur favourite proof of weak nullstellensatz lol that's probs more interesting

cause weak => strong is fairly easy

[solar glacier]

let $y = \frac{1}{x_{n+1}}$

[cloud walrus]

MyMathYourMath

[south patrol]

lol

[solar glacier]

then for large enough N letting y = g we get g^N \in I so g \in rad I

[white oxide]

i'm a little bit confused about (iv), why are all the left cosets of the form a^-1H? or is it because the set of left cosets partition G so that each coset necessarily contains a^-1 for all a in G?

[past temple]

because taking the inverse of all the elements in G gives u G

[white oxide]

ah ok got it thanks!

[white oxide]

how is it possible for E to be contained in an algebraic closure Zp of Zp if the algebraic closure of Zp is less than equal to p and if n > 1? like how can a field of p^2 elements be contained in a field of p elements

or am i reading this wrong

[long nebula]

the algebraic closure of Zp is not p elements

Zp is p elements, the algebraic closure of Zp has infinitely many elements (countably many to be precise)

[white oxide]

oh wait the algebraic closure is just all elements in some extension that are zeros of polynomials with coefficients from Zp?

[long nebula]

Not with coefficients from Zp

It's the smallest algebraically closed field which contains Zp

So Zp-bar also contains all the zeros of polynomials with coefficients from Zp-bar (and thus any subfield of Zp-bar too)

So Zp-bar definitely contains p^n zeros of x^(p^n) - x

But it also contains the roots of polynomials with coefficients in, say, Z_p^2, Z_p^3, etc

Since those are all splitting fields of polynomials over Zp

[noble hedge]

Can someone explain why all p-subgroups of a finite group G are contained in a Sylow p-subgroup? I understand that Sylow p-subgroups are p-subgroups of maximal size, but why can't there just exist a smaller p-subgroup outside of the sylow subgroup?

Like if G is a group of order p^n(k), where p isn't a factor of k, then why can't there exist a p-subgroup of order p^2 (assuming n>2) that isn't contained in a sylow p-subgroup, which would have order p^n?

Is it just from counting the elements of order p, or am I missing something?

I found this online, but there's no explanation (that I can see) for the highlighted line?

[long nebula]

If you have Dummit–Foote, this is part 2 of Theorem 18 of 4.5 (Sylow's Theorem), lemme try to paste the relevant bits of the proof

So let P be a Sylow p-subgroup and {P1, P2, ..., Pr} be the set of all conjugates, let Q act on this set via conjugation and let O1, ..., Os be the distinct orbits

[noble hedge]

that's pretty impressive that you can just pull that citation out so quickly

[long nebula]

Lol I found my copy and looked through the index for Sylow's theorem

Sure enough this statement is part 2 of it in Dummit–Foote

Of course the proof itself is a pain...

[noble hedge]

I should really get my hands on a copy, I've heard its name tossed around so many times now

[long nebula]

I like it a lot

[agile burrow]

I don't like that proof

[noble hedge]

It's not digesting well with me lol

[agile burrow]

If you have a p-subgroup H of G, then you can show the index of H in the normalizer of H is equivalent to the index of H in G mod p

That implies p divides the order of the quotient N_G(H) / H so you can apply Cauchy again, as in the proof of Sylow I

[agile burrow]

To see this, let H act on the set of left cosets of H in G by multiplication. Then a coset gH is fixed if and only if g is in the normalizer of H. Then you apply the result about actions of p-groups on sets to say that the number of fixed points is congruent to the order of the set mod p

Group theory is so cool

[noble hedge]

If P is a sylow p-subgroup and Q is just a p-subgroup, does a conjugate of Q being a subgroup of P imply that Q is a subgroup of some conjugate of P?

Like does $g^{-1}Qg \subseteq P\implies Q\subseteq gPg^{-1}$ or does that require something more, like normality?

[cloud walrus]

(𒀭)

[mighty kiln]

Yes, multiply by g and g^-1 on both sides

G ⊆ H ⇒ aG ⊆ aH, etc

[noble hedge]

yeah okay cool thanks

It's 1am and I'm dead inside, so I just wanted wanted a second set of eyes lol

Is there a nice way to show that if G has a sylow p-subgroup, P, of order p^n, then P contains a p-subgroup of order p^n-1 without showing that P contains subgroups for all powers of p?

[mighty kiln]

Divide by center until abelian, then divide by order p subgroups until you're left with an order p group

And take the kernel of all that

Center is nontrivial by class equation and order p subgroups exist by Cauchy

[noble hedge]

By that point it's just easier to show that the sylow p-subgroup contains p-subgroups for all powers of p

[mighty kiln]

Both are not too hard

[noble hedge]

showing all is just fairly straightforward induction and like only one quotient

well I guess one quotient per step

which I'm now realizing could just be applied to the sylow p-subgroup to begin with

and then just ignore the induction outright

[south patrol]

P of order p^n (n>1) and take an element g of the centre of order p. Then <g> is a normal subgroup and P/<g> has subgroups of order p^k for each 0 <= k < n by induction, so that P has subgroups of order k for each 1 <= k <= n by the correspondence theorem

That's how I'd do it lol

[chilly ocean]

what is the analog of euler's criterion in an arbitrary finite field?

[lethal dune]

what Euler theorem

[chilly ocean]

Sorry, the euler criterion for determining whether an element of $Z_p$ is a quadratic residue: if $a^\frac{p-1}{2}=1$ then a is a quadratic residue, if $a^\frac{p-1}{2}=-1$ then a is a non-residue

[cloud walrus]

Cauchy

[lethal dune]

All finite field of char 2 has a square root

As for other characteristics, It’s just counting argument

[chilly ocean]

what do you mean by counting problem

[rustic crown]

just replace p with q, and you're

your same old proof should work verbatim

[chilly ocean]

thanks 🙂

[minor glen]

Question about this exercise from Dummit & Foote: In the question I have been asked to show that a certain map is a homomorphism, but to do that do I not need to know the operation which defines the group?

[coral spindle]

The group is a subgroup of the nonzero complex numbers under multiplication.

[south patrol]

Yeah tbf would've been clearer if they emphasised C^x rather than C

[minor glen]

Interesting thank you! How were you able to deduce this?

[coral spindle]

By my intuition

[south patrol]

This is a common group

[delicate orchid]

it's very obvious that set is not closed under addition

1 is in that set, 1+1 is not

[minor glen]

Very fair

[delicate orchid]

but yeah

this is the circle group

it's very common

[coral spindle]

This is not the circle group.

[delicate orchid]

a subgroup of

same difference

[south patrol]

Lol

Set of nth roots

[delicate orchid]

yur

[grand cliff]

Anyone have some cool examples of wreath product acting on some interesting sets

[delicate orchid]

complex reflection groups are C_n \wr S_m

*most

[grand cliff]

C_p as in cyclic group with p elements and S_m symmetric group over m elements

[delicate orchid]

yes

[grand cliff]

What does the "complex" part have to do with things

is there a natural way elements of the group map to some sort of complex functions

[delicate orchid]

you need pth roots of unity

if p = 2 they're called real reflection groups cause, well

R has the 2nd root of unity

actualyl wait it doesn't even need to be a prime

sorry mb

[grand cliff]

So we have an m-tuple of roots of unity

and an element of S_m

[delicate orchid]

yus, which acts in the natural way - permuting the entries of the tuple

[grand cliff]

of course

This is the product action iirc

[delicate orchid]

yeah it's the uhhh

restricted....? wreath product

I get them confused

[grand cliff]

Yeah idk, my course is specifically on permutation groups and group actions

[delicate orchid]

actually no it doesn't matter, S_n is finite

[grand cliff]

he's only introduced 1 wreath product, but I've seen multiple definitions online

[delicate orchid]

yeah if all of your groups/sets involved are finite it doesn't matter

[grand cliff]

Yeah, cool

[delicate orchid]

restricted just means you take the direct sum instead of the direct product iirc

so they're iso for finite sums

[grand cliff]

right okay

Nevertheless. So we have the elements of our complex reflection group:

Each element is either a tuple of m roots of unity OR it's a permutation of the m indices

[delicate orchid]

no it's both at once

[grand cliff]

Yeah

[delicate orchid]

or it's one of the 34 exceptional cases

[grand cliff]

okay, you can write an element as being a tuple * a perm

soz

[delicate orchid]

yeah

[grand cliff]

not there yet

Now presumably this has some relevance to the real world or something

[delicate orchid]

(a_1, ... a_n|\sigma) (or just (f|\sigma) where f is a map f : {1,...,n} -> C_p, which is a neater way of working with it imo)

[grand cliff]

• real world being geometry or smth

[grand cliff]

yep, that's exactly what my prof does

[delicate orchid]

yeah, I can tell you that if you want

but that's less to do with the wreath product and more to do with the groups themselves

[grand cliff]

True, I guess I have just learnt the 2 natural actions of the wreath product

but was looking for maybe a connection/use outside of group theory

IMO the "imprimitve" action of the wreath product seems like it would naturally model many symmetries

Where it acts on tuples from (a, b) f,x = (a^f(b), b^x)

[delicate orchid]

that looks very semidirect product-y yeah

I don't know of any applications outside of algebra though

[grand cliff]

anyway, please tell me about this complex reflections

Or is it too complicated

[delicate orchid]

so a regular, real reflection, works by fixing a hyperplane (subspace of codimension 1 or whatever) and then mapping the normal vector through the plane to it's "negative" kinda

[grand cliff]

Okay, so this is a reflection on R^n what you're describing. And I guess the wreath product somehow represents reflections in C^n somehow

[delicate orchid]

it would be the wreath product C_2 \wr S_n

C_2 because if you do the reflection twice you don't do anything

n because there's n reflections

anyway a complex reflection group is just generated by n transformations of order m (instead of 2) that fix a hyperplane

[delicate orchid]

I should clarify, there are exceptional real reflection groups as well

like G_2, E_6, E_7, E_8 blah blah blah

[grand cliff]

Why do we need the additional permutation bit?

[delicate orchid]

fix it pointwise, btw - I have a habit of missing out important words

[grand cliff]

The element (1, 0, 1) representing: reflect x axis, reflect z axis

[delicate orchid]

these elements don't commute with each other, it's not a direct product

you must be familiar with how you can generate S_n with just 2-cycles

[grand cliff]

yeah okay, was literally typing that it's probably to do with the structure of reflections

[grand cliff]

yeah

[delicate orchid]

yeah it's exactly the same thing

and it makes S_n a coexter group but that's for a different time

[grand cliff]

Same thing in what sense sorry 😅

[delicate orchid]

2-cycles can be seen as reflections if you let S_n act on the basis vectors, and their negatives in R^n

[grand cliff]

huh

true

That's such a good intuition lmao

[delicate orchid]

I'm gonna check a book for the actual construction, this seems wrong

OH

ok this is how it actually works

if e_1, ..., e_n are the basis vectors of R^n

then the 2-cycle (ij) is the reflection that sends $e_i-e_j$ to $e_j-e_i$

[cloud walrus]

Wew

[delicate orchid]

so it's not quite as nice but meh it's not that bad

[grand cliff]

Right, so you're reflecting between basis vectors

Still a very "symmetrical" action tbh

[delicate orchid]

yeah you're taking the line that goes "between" the basis vectors and flipping them according to that

so for n=2 this would be the reflection around y = x

[grand cliff]

ya

[delicate orchid]

I read the book a bit further, the action I described before and said didn't sound right is the action of C_2 \wr S_n on R^n

[delicate orchid]

and this one is just S_n

and now the reason I started this whole discussion is to COMPLAIN

C_2 \wr S_n are called reflection groups of type B_n ok ok no complaints there

S_n is of type A_n.....

WHY

THAT'S ALREADY USED YOU FOOLS

[grand cliff]

lmao

damn, that is quite confusing. S_n has reflection type of its normal subgroup

lololol

Group theory is so vast

[delicate orchid]

yeah it's so cool

[grand cliff]

Literally feels like we are walking into the open ocean lol

[delicate orchid]

this just so happens to be one little corner I know something about

[grand cliff]

It's completely wild that people can literally study this stuff for a lifetime

[delicate orchid]

yeah there's very very deep connections here to stuff like representation theory of lie groups

[grand cliff]

All i know about rep theory is matrices

[delicate orchid]

which I do not know much about

[grand cliff]

groups elements become matrices or something

[delicate orchid]

you take the corrisponding Weyl group of the root system associated to the lie group and it's a real reflection group that's basically all I know

[grand cliff]

ah yes

of course

This course is supposed to culminate with the O'Nan Scott Theorem

[delicate orchid]

the who theorem

googling

[grand cliff]

If you know her

[delicate orchid]

oh it's about maximal subgroups of S_n, nifty

OH that's why people care about S_n \wr S_m so much

[grand cliff]

I think the sole reason we are learning this is that Prof Praeger used to work at our uni

and literally all her phd students now lecture

so we all learn permutation groups

[delicate orchid]

all you need

[grand cliff]

true

Group theory is useless without group actions

[delicate orchid]

very jealous, my undergrad uni was starved of algebra content

[grand cliff]

don't worry, our analysis has been hijacked

[delicate orchid]

as a rep theorist I must agree

[grand cliff]

no topology or differential geometry

we have finite geometry

[delicate orchid]

I don't think my uni had any geometry at all

[grand cliff]

crazy

did you have a physics dept

[delicate orchid]

yeah

maybe it did actually but it was all very applied

[grand cliff]

fair

[zenith hollow]

GF(4) has a Galois field with 4 elements. These are it's tables for + and *
a) what's the identity element for +
b) what's the inverse element for +
c) whats the identity element for *
d) what's the inverse element for *

can you guys teach me how to read these fields?

[delicate orchid]

the identity element is the one that doesn't change anything when applied to it

so, in the oplus table on the left, which one doesn't change anything

[long nebula]

Quick clarification, it says "Let GF(4) be the Galois field with 4 elements" not has a Galois field

[long nebula]

To find out what (c\oplus d) is (for example), you check the row corresponding to (c) and the column corresponding to (d), and the entry there reads (b), so (c\oplus d=b)

[cloud walrus]

boolean_satisfiERIC

[long nebula]

Hope this helps @zenith hollow

[zenith hollow]

ok

ii can see that a + anything is itself

and anything + itself is a

so a is the neutral element for +

[long nebula]

Yes

Yes

[zenith hollow]

and some itself is the inverse

ahh ok i didn't know it was this easy to read them

• is more complicated though ^^

i'll thell you when i get it

a* anything is a. b * anything is itself so i guess b is the neutral

[long nebula]

yes!

[zenith hollow]

but dunno what the inverse of * would be

cause the a is so weird

for b inverse is b, for c inverse is d, for d it's c

[delicate orchid]

yeah a doesn't have an inverse

[long nebula]

Remember in a field, 0 doesn't have a multiplicative inverse

[zenith hollow]

ahh it's like [0]?

[long nebula]

Yes

[delicate orchid]

a is 0 yeah

[zenith hollow]

niiice

[long nebula]

yup

[zenith hollow]

i love this example

hope this will come to the test

[solemn dew]

Z_n is not a group, right?

Z_n = {-n, ..., -1, 0, 1 , ... , n}, where n>=2

n * n = n^2 which is not in Z_n

Or is this the modular group?

Z \ {nZ} ?

[chilly ocean]

Z/nZ

[bitter shard]

Does anyone know what we refer to as Schreier method? Googling it, it gives me Schreier–Sims algorithm or schreier's lemma or Reidemeister Schreier method but i guess it isnt one of the above

[agile burrow]

Can you provide some context? I've heard of the Reidemeister Schreier method shortened to just Schreier's method before

[bitter shard]

I saw it in the Artin's theorem about the presentation of group of braids . It says the proof uses the Schreier method. Maybe its the Reidemeister Schreier then ?

[slim kayak]

Depends on how you defined the addition here

[slim kayak]

same here

[bitter shard]

oh ok good to know, thanks

[slim kayak]

Why is any subset with the induced topology a noetherian topological space also noetherian?

[rustic crown]

pretty much a definition chase with the fact that countable intersections of closed are closed.

[slim kayak]

Countable intersections of closed sets are closed in noetherian spaces?

bruh, mixed up open and closed sets

[rustic crown]

that's in arbitrary top spaces

take a decreasing chain B_n of closed subsets in the subspace Y of X, you can write B_n = A_n intersect with Y for A_n closed in X. now sadly A_n is not really decreasing anymore, but that's super easy to fix... replace A_n with intersection of A_i for i <= n

[south patrol]

Hm is that rly true

[rustic crown]

arbitrary intersect of closed = closed

[slim kayak]

thats in the definition

[south patrol]

Oh wait

Yeah I also misread

Lol

Was weird why you emphasised countable

[rustic crown]

[south patrol]

Emrkamwmsfnd

Lol

[slim kayak]

Shame on you det for making your msg's so misreadable smh

[rustic crown]

hehe :p

[formal ermine]

hi det

[rustic crown]

hi illu

[agile burrow]

topology

[formal ermine]

I have to do even more classic algebraic geometry now

I thought my newton okounkov theory course would only do algebraic curves stuff

but now we introduced the coordinate ring

the homework is finally getting fun though

but my latest homework sheet is just "apply bezout a bunch of times" lol

[rustic crown]

me know very little classical ag

[formal ermine]

I only know like very very basic stuff

[slim kayak]

But yeah, that alr took care of itself. Currently dealing with the irreducible components of the intersection of affine variety Y with dim r and a hyperplane H (dim n-1) having dimension r-1 when Y isn't in H. Trying to pull towers of prime ideals around the coordinate rings rn

Which isn't really working out

[rustic crown]

now that you say it, i realize i only needed finite intersections

[south patrol]

Classical

[agile burrow]

I am doing group theory

I am computing cohomology of D_{2n} for n odd

[formal ermine]

idk if this should go in this channel or adv nt. my prof said "O_K^times = mu(K) x Z^(r + s - 1) as abelian groups, where mu(K) are the roots of unity in K, the torsion subgroup of O_K^times" what does torsion subgroup mean in this context? is it just elements of finite order?

[agile burrow]

Yes

[formal ermine]

where does the name come from

if we view it as a Z-mod?

then it's just torsion elements innit

[agile burrow]

Yes

[formal ermine]

thanks

[agile burrow]

it is periodic 😮

[south patrol]

I am do rep theory / non comm alg

Basically was uhhh

[formal ermine]

scary non comm alg

[south patrol]

Well this was actually coincidentally commutative lol

[agile burrow]

I never understood how people identified like matrix groups over finite fields with various groups arising from studying symmetries of polyhedra

[south patrol]

Like give an example of a finite dimensional C-algebra A such that dim Z(A) > r, where r is the number of isomorphism classes of simple A-modules

[formal ermine]

what does it mean for a group action to be "simple transitive"?

[south patrol]

hm

[formal ermine]

for context

[agile burrow]

I've seen it for free + transitive

[south patrol]

Equivalently means free + trans

yeah

[formal ermine]

"If K/Q is galois, then G_(K/Q) acts simple transitive on Hom(K, C)"

[south patrol]

What are some examples lol

Oh okay

[formal ermine]

what is a free group action

oh

[south patrol]

g.x = x => g = 1

ig

[agile burrow]

Action of group on itself lol

[formal ermine]

I keep forgetting group actions

[tribal furnace]

for some x

[south patrol]

yh i think it's like

bad terminology lol

well

Easily confused terminology

[formal ermine]

yeah also transitive group action

I always confuse that one with uh

that other group action

[south patrol]

transitive is the only one i can remember lol

well faithful makes sense but comes up in other contexts anyway

[formal ermine]

also like, why is this true

and uh

how does this help us classify Hom(K, C)

my prof continues by saying that

complex conjugation operates on H = Hom(K, C) by left composition

and that if sigma bar = conjugation circ sigma = sigma then sigma factors as K -> R

and in that case sigma is a real embedding, and otherwise a complex embedding

and therefore H = { f_1, ..., f_r, sigma_1, sigma bar_1, ..., sigma_s, sigma bar_s }

where the f_is are the real embeddings and the sigma_i, sigma bar_is are the complex embeddings

and so [K : Q] = r + 2s

how do we even know that Hom(K, C) is finite?

[rustic crown]

so in general, say F/k is an algebraic extension, you define [F:k]_s = |Hom_k(F, k bar)|.
you then show this behaves multiplicatively in towers and for simple extensions, [F:k]_s <= [F:k]

[rustic crown]

free-ness is immediate, then a counting argument

[rustic crown]

better to use the word "pre composition" as left/right are confusing.

[formal ermine]

how can we use this here? k needs to be a subset of F, but if e.g. F = Q then k bar won't be C

[rustic crown]

oh right, should have said that...

if L is any alg-closed thing containing k, then you have a map
Hom_k(F, k bar) --> Hom_k(F, L)

but if you assume that F/k is algebraic then this is an iso. the image of any f in Hom_k(F, L) would consist of stuff in L that are algebraic over k, so f factors through k bar.

[formal ermine]

oh that makes sense

thanks det

[rustic crown]

[steel light]

What’s your favorite algebraic structure? Groups, rings, modules, varieties, algebras, whatevers

[warped fable]

A rack: https://en.m.wikipedia.org/wiki/Racks_and_quandles

[glad stirrup]

i was only exposed to groups and rings (and a little bit of fields and vector spaces)

[warped fable]

Because its ridiculous that theyd be useful

[glad stirrup]

but i prefer rings and fields over groups

[agile burrow]

Groups are cool

[steel light]

Wow I’ve never seen this before LOL

[glad stirrup]

me either

i thought it said candles when i first saw it

[warped fable]

They’re pretty funny

[steel light]

I like abelian goops

Simple and symmetry purtty

[sonic coral]

i prefer abelian grapes

[dim widget]

quivers

[agile burrow]

I wanted to read a bit on quiver reps but never really got around to it

[slate tide]

Quivers show up in topological data analysis. I tell everyone that fun fact without actually knowing what a quiver is though 😦

[frigid lark]

Is this a marketing ploy by Gwyneth Paltrow?

[lethal dune]

vegitables

[prisma bluff]

tambara module

it's what is behind your minecraft updates (no joke)

[vagrant zinc]

Groups rings, modules varieties and Galois

[steel light]

Wtf am I reading

[vagrant zinc]

[prisma bluff]

if you have a burrito, you can often replace the ingredients in it without damaging the wrap

the tambara module is a vast generalization of that simple concept

😉

[white oxide]

is the field of complex numbers an infinite extension of the rationals because there's no way to express an irrational number with any scalars involving rationals?

[lethal dune]

No it's not possible, C is an extension of Q but not an algebraic one. Think of Q(x) as a field, C is that but with uncountable variables (and some algebraic ones)

[white oxide]

oh right because pi is transcendental over Q

ah ok i'll think of that thx

[south patrol]

Yeah note that [Q(a): Q] is finite iff a is algebraic over Q

So you can just pck anything non-algebraic

Howeever even just appealing to transcendent numbers is kinda overkill like

For example Q(2^1/n)/Q is an extension of degree n

[grand cliff]

@delicate orchid An Bn Cn

soz for interruption everyone

[south patrol]

How much blur do you want? Yes

[grand cliff]

this is a screenshot at max resolution

my lecturer just doesn't know how to record lectures lol

[south patrol]

Lol

It's funny cause all I can read is like

Complex lie algebras classifed by Killing

[grand cliff]

simple complex lie algebras classified by killing/c[some other author] 1880s

then some graphs

[delicate orchid]

And there they are

[warped fable]

Cartan?

[grand cliff]

yep

[south patrol]

Oh okay

Killing/Cartan and then a date with a scribble

[grand cliff]

the scribble is an s

[delicate orchid]

Killing and Cartan on a date

[south patrol]

K-I-S-S-I-N-G

[warped fable]

K I L L I N G

form

whats the deal with these dynkin diagrams anyway

what does it even mean lol never understood

[delicate orchid]

Shows how reflections compose with each other

Encodes the structure of a root system basically

[warped fable]

vertices are a basis for the Weyl group?

what do the edges mean

[delicate orchid]

The edges tell you the order of the product of two basis elements

[warped fable]

ah

[delicate orchid]

No edge => the product is another reflection, and then you label the edge with either 3,4 or 6 to show the order of the product

If you remove the restriction on the possible edge labels you get a Coexter diagram which classifies general coexter groups rather than just root systems

[warped fable]

whats a painless way of seeing the polytope st the Weyl group is reflection about the faces of it? something something Weyl chamber but this I also never really understood

[delicate orchid]

Oh uhh

[warped fable]

It’s sitting in the Cartan subalgebra I assume

[delicate orchid]

Yeah I really can’t tell you more than “something something Weyl chamber” I don’t really think about them geometrically that much

[warped fable]

gotcha

np, this was helpful

These stupid diagrams appear everywhere thats why I’m intrigued

[delicate orchid]

Although I think B_n acts on an n-cell, just from memory

[warped fable]

For example, singularities of maps f : C^2 -> C

[warped fable]

interesting

[delicate orchid]

Wtf are they doing there

[warped fable]

ask Arnol’d lol

[delicate orchid]

I shall be having a word

[warped fable]

f(x, y, z) = x^2 + y^2 + z^n+1 is A_n, iirc?

x^2 + y^3 + z^5 is E_8

https://en.m.wikipedia.org/wiki/Du_Val_singularity

[delicate orchid]

Spooky

[slim kayak]

Bro how did you get access to my profs handwritten script?

[lethal dune]

plot twist, he is your prof

[molten silo]

Can anyone help me with this last question

This is nakayama lemms

[warped fable]

@molten silo Corollary of Nakayama (this is the version I remember): If (R, m) is a local ring, M is an fg R-module, a set {x1, …, xk} generates M as an R-module if their images generate M/mM as a R/m vector space

Apply to your case with the maximal ideal (x, y, z) of C[[x, y, z]]

[molten silo]

okay thank you

[grand cliff]

@delicate orchid wtf is even going on, this feels like pokemon except for weird creatures its groups

[delicate orchid]

yeah so those are the only graphs which are positive definite, meaning if you take the associated matrix which is some mess involving a cosine, it's positive definite

the proof I've seen involved just finding all postitive definite coexter graphs and then taking out the ones that aren't dynkin diagrams

which is just like

17 different observations where you eliminate cases

[formal ermine]

this seems really trivial but it feels like it's wrong

given an irreducible polynomial f in C[x, y, z], its partial derivative wrt x and itself do not share a common factor as its pd has degree < degree f and f would then no longer be irreducible

[next obsidian]

This is true

This is essentially at the heart of why every extension in char 0 is separable, you can’t share a root with your derivative

[formal ermine]

right

[next obsidian]

This is possible in char p only because d/dx(x^p) = px^p-1 = 0

[south patrol]

Yes, in fact this is true for derivatives in A[x] whenever A has characteristic 0

Oh bruh

[next obsidian]

Piece of shit………

[formal ermine]

wait

can differentiating introduce repeated factors

like if f has no repeated factors/is irreducible, will f' wlog wrt x also have no repeated factors?

[next obsidian]

In char p, x^p+1

[formal ermine]

and in C[x, y, z]?

[next obsidian]

Idk, try and prove it can’t or find an example

[molten silo]

Guys pls help

[lethal dune]

K[x²] generated by one element

[molten silo]

<x^2>?

[lethal dune]

when can you say R[s] is an integral extension over R?

[molten silo]

When its contained in R and k[x^2] is integral over k[x]

i just cant find a standard f

Cant i just use f(y) = y-f , where f=y and f is in R{X}?

[lethal dune]

what polynomial does x² satisfy over k[x]

think of a something say a(x²)t²+b(x²)t+c(x²) s.t. putting t=x gives you 0

there's a very trivial one

[molten silo]

is that not the wrong way round?

coefficients are in R{x}

[lethal dune]

which one is larger? k[x] or k[x²]? in the inclusion sense

[molten silo]

k[x]

[lethal dune]

so which one is the extension

[molten silo]

k[x]

[lethal dune]

you have to find a polynomial in k[x²][t] that is satisfied by x

monic*

[molten silo]

a=0, b=-c*x

nevermind

[lethal dune]

you are close

[molten silo]

a=-c*x^2

b=0

[lethal dune]

and c?

does replacing t=x² give you 0?

[molten silo]

c=1

[rustic crown]

(t = x i think)

[molten silo]

no

it doesnt give 0

[lethal dune]

oops*

[rustic crown]

i'll suggest to write X = x^2

[lethal dune]

have I been messing up the entire time?

[rustic crown]

prolly not

[lethal dune]

ok cool

[rustic crown]

so you're trying to find an monic polynomial over K[X] such that x = sqrt(X) is a root

(try looking at the situation Z --> Z[sqrt2] if that makes thinking easier )

[lethal dune]

ye

[formal ermine]

hi det

[molten silo]

so does what i said work. c=1 b =0 and a=-x^2

[lethal dune]

umm close

-x² (x) + 1 ≠0

try exchanging a and c maybe

[molten silo]

yeah

mistake

thanks you boys

[lethal dune]

cool

[molten silo]

i got confused what was the coefficient and which the variable

[lethal dune]

yeah happens, the trick det mentioned might help you

[lethal dune]

^

[molten silo]

yeah

will do

[lethal dune]

@molten silo here's a follow-up. Can you show for any polynomial p(x) in k[x], k[p(x)] ⊂ k[x] is an integral extension

[marsh marlin]

Hi, I study the Tic-Tac-Toe's grid and I wonder about the vocabulary to use...

For example, let us consider these four grids :

[coral spindle]

Are you wanting to know the terminology of group actions?

[marsh marlin]

We understand that these grays are equivalent and that they have a similar structure.

[marsh marlin]

Yes.

[coral spindle]

OK yeah, we would say those are in the same orbit

If a group G acts on a set X, then we say X is a "G-set"

[marsh marlin]

Is it correct to say that these grids are "isomorphic"? Or maybe "equivalent" is more accurate? Or just "similar"?

[coral spindle]

No

You could draw attention to the equivalence relation x ~ y iff x = g.y for some g in G, and then you could say equivalent more precisely

(they are literally equivalent in the sense of an equivalence relation)

You could simply define the terminology 'similar' for that case. It would be a bit of a faux pas to call them isomorphic

[marsh marlin]

Yes, I have a definition that looks like it!

[marsh marlin]

Ok I was also thinking, isomorphism is more for sets or algebraic structures, isn't it?

[coral spindle]

Yes

[marsh marlin]

Finally what I would like is that if A and B are equivalents then:

• There exists f such that A = f(B).
• There exists f^-1 such B = f^-1(A).
• If C is playable on A then f^-1(C) is playable on B.
• If C is playable on B then f(C) is playable on A.
  But (normally) it corresponds to the symmetries and rotations of the square.

A and B are grids and C a move.

[coral spindle]

The first two are by definition. The second two you will have to prove.

The group of symmetries and rotations of the square is the dihedral group of order 8, aka Dih(8) or D_8, or sometimes D_4 depending on who you ask. You may find it helpful to look that up.

[marsh marlin]

I really have no idea how to do this...

[marsh marlin]

By studying the noughts and crosses I made this truth table:

[coral spindle]

This is not a truth table lol

This is called a Cayley table

[marsh marlin]

Is it the symmetry group of the square?

[coral spindle]

Yes, this is Dih(8)

[marsh marlin]

Oh sorry, I used to say "table de vérité" but it is indeed confusing with tables in Boolean algebra...

[marsh marlin]

I think I stumbled upon it unintentionally 😅...

[lime badge]

Quick stupid question, suppose we have a field $k$, $f(x) \in k[x]$ and F is the splitting field for $f(x)$ over $k$.

If $k \subseteq K$ is an extension such that f(x) splits as a product of linear factors, then I need to show there is a homomorphism $F \rightarrow K$ extending the identity on $k$.

This is just using the theorem on the uniqueness of splitting fields to get an iso between $F = k(\alpha_1,...,\alpha_d)$ (where $\alpha_i$ are the roots of $f(x)$ in $F$) and $G = k(\beta_1,...,\beta_d)$ (where $\beta_i$ are the roots of $f(x)$ in $K$), that extends the identity on $k$. This iso then extends to a homomorphism between $F$ and $K$, the latter of which contains $G$ right?

[cloud walrus]

Eternal Way

[lethal dune]

read it wrong

looks fine

[lime badge]

Ah ok, thanks!

[white oxide]

can somebody give me a hint for this question other than the one they gave? LOL i'm using the fact that if N is a proper normal subgroup of G then {e} < N < G, and so there exists a composition series containing N and trying to find a contradiction but so far can't come up with anythign

like i don't see how i'm supposed to use exercise 23 if i don't know if Hi+1/Hi is finite

like am i missing something? are simple groups finite?

oh wait can i just let M be a maximal proper normal subgroup of G

and show that there's no composition series which contains it because it's maximal

bruh i'm lost

[agile burrow]

Suppose you have a composition series for an infinite abelian group. If each quotient were finite, then the order of G would be finite, contradicting the assumption. Thus, at least one of the quotients is infinite

[white oxide]

ah rip i see

then it wouldn't be simple

thanks

[agile burrow]

Happy to help

[south patrol]

Nice

[glossy crag]

I want to show L/K finite and E/K any extension => LE/E finite, does this work: take some finite generating set x_1,...,x_n s.t. L=K(x_1,...,x_n) (e.g. K-basis), then E(x_1,...,x_n)/E is finite (since the x_i are algebraic over K => over E) and E(x_1,...,x_n) contains E and L => LE/E finite.

I suppose the same way one could show L/K algebraic => LE/E algebraic.

[rustic crown]

yee

(just the thing that to talk about LE you need both L, E to be subfield of some large enough field)

[next obsidian]

I think up to iso you can embed them in an alg closure and it won’t change

[glossy crag]

ofc

[white oxide]

a in an extension field E being algebraic over a field F imply that [F(a) : F] is finite? all we know is that it's the zero of some polynomial with coefficients in F, correct? or are there no polynomials of minimal infinite degree (sounds stupid, but just a clarification)

[next obsidian]

A polynomial has finite degree

[white oxide]

ah ok thanks i suppose i must have missed that in the definition

[next obsidian]

How do you even make sense of evaluating a power series at an element of the field?

Try to make sense of this for an arbitrary field and you’ll see that it’s a nonsensical notion, there’s no way to interpret it in full generality

[white oxide]

so in general if a and b are in an extension E of F such that a and b are algebraic over F then F(a, b) is finite since the dimension of F(a, b) is at most max{deg(a, F), deg(b, F)}?

[next obsidian]

That isn’t the correct bound, but it’s still finite yes

[rustic crown]

I think it would. Like E = L = Q(cbrt2) and now [EL:E] would change depending on whether you embed them in the same way or not.

Relabel E as Q[x]/(x^3-2) if that looks weird.

[next obsidian]

_<

I think this isn’t right

I mean embeddings which preserve the base field

[rustic crown]

Base is Q tho

[next obsidian]

O

Thonking

_<

True

[white oxide]

what would the correct bound be

[next obsidian]

This is weird tho because

Shouldn’t they both be iso to the tensor product as Vector Spaces

[rustic crown]

Tensor prod need not be a field

[next obsidian]

Figure it out, it’s more instructive than me just telling you

Yeh I know

[rustic crown]

And can have multiple max ideals

[next obsidian]

I know I mean as a VS

Hmmmmm

No this clearly is fake

I dunno

¯_(ツ)_/¯

[rustic crown]

.<

[lethal dune]

E ⊗_k F is a field iff it's dimension = dimE * dimF

ok that doesn't sound right, let me check

[next obsidian]

That’s always true

Lmao

[lethal dune]

yeah

this is the right one

[next obsidian]

It’s some shit about transcendence degree and crap I think

[lethal dune]

field iff [EF: k] =[E:k][F:k]

*algebraic

*finite

[zenith hollow]

GF(2), F2, Z/2Z are all the same thing? The vectorial space of that is used for bits in computer science?

[lethal dune]

Same

Rather isomorphic

[empty rose]

i don't think i'd really call the field with 2 elements any more related to bits than... anything else with 2 elements

the important property of bits is just that there are two of them

[slim kayak]

Someone familiar with very elementary knot theory here? I am supposed to show that two out of three reidemeister moves leave the group of the knot diagram unchanged up to isomorphism. I worked it out with for the first and third now but showing it for the second kind got me stuck.

I'd obtain the relation x2 = x2-bar, so by slight abuse of notation I know that short of this new a and the relation x1 a x1-inverse = x2 the group is unchanged. But I have to somehow get rid of this new generator. In the special case I'd have some crossing which gives me the relation x2 = x1 xk x1-inverse in which case a is just equal to some existing generator and can be removed, but I don't see why such crossings must always exist or why the relation would emerge algebraically out of the other relations.

[lament bough]

sorry if this is an obvious question, so if a R-module G is just a homomorphism R -> End_Ab(G), is it true that R-Mod is just a (appropriately restricted) coslice category of arrows under R?

[next obsidian]

I don’t think this category has the same morphisms

[lethal dune]

maybe ask in topology

[oblique river]

Or #combinatorial-structures

[warped fable]

Just write the relation the other way around, a = x1^-1 x2 x1. In the Wirtinger presentation you’re adding a to the list of generators, along with the above relation which expresses a in terms of already existing generators.

So the group is unchanged, upto isomorphism. Its a redundant relation

Theres no need for a to be literally equal to another generator for it to be removable from the set of generators.

[slim kayak]

I am only adding that relation (x1^-1 x2 x1) if my knot diagram has a crossing involving x2 going under x1 while having the right orientation tho?

Nvm I misunderstood you, yeah okay.

That's literally a tietze transformation, 1=a(x1^-1 x2 x1)^-1 at least

[warped fable]

Indeed

[slim kayak]

Thanks

[formal ermine]

not sure if this is the correct channel

but what does conv mean in the context of newton polygons?

more specifically

$$\on{conv}\left\lbrace \bigcup_{(i, j) \in \on{supp}(f)} (i,j)\text{-quadrant} \right\rbrace$$

[formal ermine]

ah

convex hull?

[slim kayak]

Hey, does the dimension of the double dual of a vector space differ from the original vector space differ when the dimension is infinite?

[coral spindle]

Yes

The dimension of the dual of a vector space with countable basis will be uncountable

(if I'm not terribly mistaken lol)

[slim kayak]

Came up with the double dual functor being a naturally isomorphic to the identity functor only when the vector spaces are finite, but vector spaces of the same dimension are isomorphic.

[oblique river]

this is true

dim(V*) = dim(V) iff dim(V) is finite

[slim kayak]

Yeah, why though?

[rotund aurora]

the dual is F^kappa where F is the base field and kappa the dimension of V. If it had dimension kappa, then |F|*kappa=|F|^kappa as cardinals

[oblique river]

to be clear, that's a direct product

assuming the axiom of choice, any vector space V over F is a direct sum of some number of copies of F; let kappa be that cardinal. then you can prove (it's not too hard) that the dual is the direct product of kappa-many copies of F

so what's left is to do some cardinal arithmetic to show that if kappa is infinite then the direct product is larger than the direct sum

[south patrol]

Also clear that F^kappa embeds in it using the dual consrtuction right

(So at least the bound in cardinality is easier I mean)

[rotund aurora]

F^kappa is functions kappa-->F. FIx a basis B (of size kappa) for V. Then the dual is of course the set of functions B-->F, which is the same as kappa-->F (tho you have to choose a basis, so maybe its not so natural)

[oblique river]

is this still true even without the axiom of choice?

[lethal dune]

Hom( \oplus k,k)= \prod Hom(k,k)= \prod k

[oblique river]

like let's say that we had a vector space V over F which doesn't have a basis; in particular it's not finite-dimensional. what can we say about V*

[rotund aurora]

interesting

[oblique river]

i guess what i'm asking is if there's a constructive proof that "V = V* implies dim(V) is finite" without assuming that V has a basis

[rotund aurora]

wow I bet theres none

do we even know if V^* is nontrivial

[oblique river]

hmm

so apparently the statement "every subspace has a complementary subspace" in fact implies axiom of choice

if you could get an example where no 1-dimensional subspace has a complement

(a complement of W in V is another subspace U with V = W + U)

then the dual would be trivial

okay so apparently here is an example

there is a model of ZF where the dual of R as a Q-vector space is trivial

https://math.stackexchange.com/questions/3059508/existence-of-complement-of-a-subspace-without-zorns-lemma?noredirect=1&lq=1

[rotund aurora]

yeah

thats what I was thinking

but no idea how you would prove it

the Q-vector space R is my favourite vector space by far, so cool

[oblique river]

i think the idea is that you can set up a model which basically forces everything to be continuous, but there's no continuous projections from R to Q

or something idk

[slim kayak]

The Q-vector space C is my favourite too

[oblique river]

those are the same vector space

[rotund aurora]

xdd

[oblique river]

also, some more weird stuff here https://mathoverflow.net/questions/142609/linear-algebra-without-choice?rq=1

[formal ermine]

wtf I just learned that C is the only algebraically closed topological field that is both Hausdorff and locally compact

[rotund aurora]

yeah, had read that in the past

[formal ermine]

shouldn't this be a geq i, b geq j?

[oblique river]

yes

presumably

[formal ermine]

yeah seems like a typo

thanks

in the definition of the newton polygon of a polynomial: "the 'compact part' of the edge of the convex hull of the union of all quadrants" what exactly does "compact part" mean? do we just remove the x and y axis? (cuz then it's closed and bounded so by heine borel it's compact)

[white oxide]

can somebody give me a hint to show that the finite direct product of solvable groups is solvable?

[agile burrow]

Can you think of how to construct a composition series for the direct product of two groups

[white oxide]

yes i was thinking about it but i'm not so sure it's right

gimme a sec i'll upload my attempt at a constrution

construction

like this feels a bit hand-wavy idk

[slim kayak]

Make sure you got a good handle on the structure of (normal) subgroups of direct products and the quotient group of direct products.

[white oxide]

ah yea i kinda had to prove that the direct product of normal subgroups of two groups is a normal subgroup on the fly

[slim kayak]

I guess work out the other fact about the quotient groups of direct products, then once you have done that you just apply those facts and make live a lot easier

[white oxide]

well i could use this i think right

because i could just consider the additive group structure

[slim kayak]

Only the factor groups are required to be abelian and in the general case you won't be able to apply it.

Actually, what's your definition of solveable here? Subnormal with abelian facto group or the derived one?

[white oxide]

subnormal with abelian factor groups

[coral spindle]

No, a similar fact does not hold for product groups.

[white oxide]

ah rip

[slim kayak]

Cool cool, but yeah. The next term in the series are just normal subgroups of the previous term, so you can't use the ring version

[coral spindle]

The problem here is similar to the subring counterexample that I mentioned around when I posted the proposition about the ideals of the product ring

[white oxide]

maybe i can proceed by induction of some sort? the base case is clearly true, since H1 and K1 are abelian (H1/{0} iso to H1 and is abelian and similarly for K1) and then consider the two cases for when i = j and when i \neq j

because n does not have to equal m

[white oxide]

ok let me look back

[white oxide]

and the direct product of abelian groups is abelian

[slim kayak]

You mean the subnormal series of the groups have various lengths?

[white oxide]

ye because they're not the same group

if they were the same group they would have the same length right because all composition series of a group G are iso

[slim kayak]

That's fine, just extend their series with trivial groups until they match

[white oxide]

wait so does my construction work? i'm assuming that m < n and i'm just fixing m when the m groups "run out" (then i just count up the His until I get to Hn)

like for example H0 x K0 < H1 x K1 < H2 x K2 < ... < Hm-1 x Km - 1 < Hm < Km < Hm+1 < Km <Hm+2 <Km < ... <Hn-1 x Km < Hn x Km

[slim kayak]

Looks fine to me

[coral spindle]

There is a much simpler composition series

not that it matters for this proof

[slim kayak]

Simpler in what sense? Gluing together the individual series together seems pretty simple

[coral spindle]

H_0 x K_0 < H_0 x K_1 < ... < H_0 x K_n < H_1 x K_n < ... < H_m x K_n is a composition series

it's worth noting that this is a composition series, not simply a subnormal series

if you wanted only to prove that H x K is solvable, this would be the simplest way

[slim kayak]

I think that's what they wrote, just with different indexing convention

[coral spindle]

No that's not what they wrote at all

[slim kayak]

Ah, true that

[coral spindle]

In general (H1 x K1) / (H0 x K0) \cong (H1/H0) x (K1/K0) will not be a simple group

But yeah again this is immaterial for this proof

[white oxide]

ah so you counted with the Ks first and then the Hs

wait this looks similar to zassenhaus

but zassenhaus involved intersections

[coral spindle]

You can use this to show that the derived length of H x K is the maximum of the derived lengths of H and K

('this' being the construction given in the image, not mine)

[white oxide]

derived length?

wait i feel like i could possibly use the third isomorphism theorem here?

[slim kayak]

For what? For the quotients of a direct product?

[white oxide]

yea

[coral spindle]

The derived length is the smallest number n for which G^(n) is the trivial group. Here, G^(0) is defined as G and G^(x+1) = [G^(x), G^(x)]. A group is solvable if and only if such a number n exists

[white oxide]

idk i'm looking at it now and maybe it could apply??

[slim kayak]

First isomorphism theorem will be more helpful actually

[white oxide]

because i'm using induction rn and assuming that Hi x Ki / Hi-1 x Ki-1 is abelian for all i, j <= m - 1 where i = j and trying to show that Hi+1 x Ki+1/Hi x Ki is abelian, and i know that Hi-1 x Ki-1 is a normal subgroup of Hi x Ki which is a normal subgroup of Hi+1 x Ki+1 by construction

[white oxide]

hm o

k

[white oxide]

oh interesting ok

maybe it's time for good ol' contradiction

i'll prolly come back to this problem it's kicking my ass but thanks for the help guys

if K is an extension of a field F and E is a subfield of K that is the algebraic closure of F, how do we know that any f(x) in F[x] splits in E? because E doesn't necessarily have to contain all of the zeros of f(x), because it's just the set of elements that are algebraic over F in the extension K (or is the extension K assumed to contain all of the elements algebraic over F)

[coral spindle]

What's your definition of 'algebraic'?

Depending on the definition it could be trivial

[white oxide]

a is algebraic over F if it's the zero of some f(x) in F[x]

[slim kayak]

And your definition of algebraic closure?

[coral spindle]

So it's the zero of some irreducible, monic f(x) in F[x], no?

And your other polynomial, let's now call it g(x), factors into some element of F times irreducible polynomials in F[x]

[white oxide]

the subfield of all elements in an extension E of F that are algebraic over F

[white oxide]

oh right i forgot about existence

[coral spindle]

me too I often forget about reality

[white oxide]

lol

[white oxide]

i think this works

[snow talon]

Hey folks. I'm trying to teach myself some Abstract Algebra and as such I'm working through the Dummit and Foote textbook. I found some answer keys online which I use to check my work, but I'm not understanding this step:

In one of those answer keys

I recognize I'm probably being a little slow here, but if anyone could clarify I'd appreciate

[formal ermine]

s has order 2

s^2 = e so s = s^-1

[snow talon]

Isn't that what I'm aiming to prove in the first place though? Contextually, the question is:

3. Use the generators and relations above to show that every element of D_2n which is not a power of r has order 2. Deduce that D_2n is generated by the two elements s and sr, both of which have order 2.

With the presentation being:

D_2n = <r, s | r^n = s^2 = 1, rs = sr^-1>

[formal ermine]

look at the presentation

s^2 = 1

[snow talon]

I'm being silly

You're right

I just presumed I couldn't use that fact

Because I'm being asked to show that the elements have order 2

Ty

[hollow shore]

hey guys how do i solve this?

[formal ermine]

chinese remainder theorem

[hollow shore]

also is there a way to tell whether it is odd or even?

[wraith cargo]

is 4k+1 odd or even

for some natural number k

[hollow shore]

that is odd

i see

[wraith cargo]

ok cool

and since x is congruent to 1 mod 4 that means it's...

[hollow shore]

why is it that when one is odd and one is even

we have no solution?

[hollow shore]

odd yes i understand 🙂

thanks

[empty rose]

well the same number can't be both odd and even

[wraith cargo]

doesn't this imply that our n's have to be coprime

and 4 and 10 def aren't coprime lol

[empty rose]

although if it's something like x = 2 mod 5 and x = 1 mod 4 then that does have a solution because 5k+2 isn't necessarily even just because 2 is even

[hollow shore]

i see

how do u apply chinese remainder theorem for that system of congruence

i know how to do it for 3

like this

[empty rose]

well yes the chinese remainder theorem does require a set of numbers that are coprime, but you don't necessarily have to put in 4 and 10

[hollow shore]

but for part a

i dont think it met the condition for chinese remainder theorem

[formal ermine]

can divide 10 by 2 to get a solution in the least residue system

lcm(5, 4) = 20

then by crt you get x = 13 mod 20

which is the correct solution

[formal ermine]

this one actually meets the condition for directly applying crt

7 9 and 11 are all coprime

[empty rose]

they said "for part a" though

[hollow shore]

walk me through it pls?

[empty rose]

which is this one

[hollow shore]

part a yes

[formal ermine]

ah

I'm blind

I should sleep

[hollow shore]

part b was designed for chinese remainder theorem

in lecture i was taught to use tabular method

which im so confused about

[formal ermine]

you can like

take the lcm

of the residues

then see which numbers you can like reduce to get to that lcm by multiplying and which gets you to coprimeness

then apply crt

in this case the lcm is 20

so you divide 10 by 2

to get 5

5 and 4 are coprime

so you do crt with those residues

[wraith cargo]

how does this work

I'm horrible at like basic NT so I'm actually curious

[empty rose]

(...wait why are we in the abstract algebra channel for this?)

[empty rose]

i think the idea is like
we know x = 3 mod 10, which implies x = 3 mod 5
and then "x = 1 mod 4, x = 3 mod 5" is something you can apply crt to

(and the existence of x = 1 mod 4 in there means we haven't accidentally lost the information that x is supposed to be odd)

[hollow shore]

i dont understand how he got v and u

[hollow shore]

someone halppp

[misty meadow]

I'm very tired however

Could you use x-1 | 2

And x - 3 | 5 ?

@hollow shore