groups-rings-fields

[fickle brook]

2D

mD, in general

[plucky flicker]

but that is a subset of D

[mild laurel]

do you require your subrings to contain the multiplicative identity?

[plucky flicker]

not necessarily

[vestal snow]

I have a question about function fields that I posted on the number theory channel. If someone could take a look, that would be pretty great.

[chilly ocean]

can anyone tell me how to prove that a set is a vector space?

[chilly ocean]

A set of elements V is a vector space when it is equipped with two operations, namely vector addition (+) and scalar multiplication (.), such that V under (+) forms an abelian group

so you could already start by proving that (V, +) is an abelian group

[chilly ocean]

yeah i saw this

so i need to prove all these? 😐

Sadly, yes

It would have been shorter if you were asked to prove that a set is a subspace of a given vector space

yeah that ik

Otherwise, you have the fact that every subspace is a vector space in its own right

So if you can prove that your space is a subspace of a larger vector space, its a shorter proof and it implies the desired property of the set being a vector space

[chilly ocean]

oh okay will try that

thanks!

[marble whale]

I have groups Q, + and Z, +. And their factor group (Q/Z, dot). I'm supposed to prove that the multiplication circledot is not defined properly. I came to the conclusion that if b = 0 and a = 1/2, then if Z stands for integers, we'd have integers + halves = integers which is false. However I'm fairly uncertain about this exercise and I'd appreciate some feedback

[chilly ocean]

by factor group you mean the quotient group right?

because Z is a normal subgroup of Q

[marble whale]

yeah, the course material is not in english and it seems to use some words that I can't translate

[magic owl]

Factor group is fairly common in older english texts

i think dummit and foote uses it

[chilly ocean]

whats the law of composition on Q/Z?

[marble whale]

law of composition? A word for word translation of the exercise would be "Give an example which proves that among cosets Q/Z multiplication circledot can't be defined as follows" and that's all I'm given

[oblique river]

Multiplication by 1 will be weird

on the one hand, 1*r should be equal to r, but on the other hand, 1 is an element of Z, so multiplying by 1 would be the same as multiplying by 0 in the factor group

[chilly ocean]

how to do this?

you show that if $a,b,c\in\bR$ satisfy $$ae^x+b\sin x + c\cos x = 0$$ for all $x$ then $a=b=c=0$

[cloud walrus]

(T*Terra, dq^i \wedge dp_i)

[chilly ocean]

that's what linearly independent means.

[chilly ocean]

yeah i know that but im unable to prove it.

This is something new. I'll try that.

[cloud walrus]

slimvesus

slimvesus

[chilly ocean]

ohh okay

i'll try that way

thank you so much.

[chilly ocean]

or use the wronskian formula

[chilly ocean]

that is something new for me but the way slimvesus suggested worked.

[chilly ocean]

thanks tho!

I'm gonna send you the formula, it generalises to any set of functions

[chilly ocean]

okay appreciate it

if the wronskian is non zero for some x in the domain of the functions then they are lin indep

[chilly ocean]

thanks a lot for this information.

[mystic jungle]

so first i assume ab^-1 is in H then (ab^-1)b is in Hb so a is in Hb which implies Ha = Hb

then i go the other way around and say Ha = Hb so I let x be in H then xa is in Ha = Hb which implies that xab^-1 is in Hbb^-1 which implies xab^-1 is in H

how do i get from there to ab^-1 is in H

[rigid cave]

Wait wait... If a group G acts on a set X without non trivial fixed points and R contains exactly one element from each orbit of G, then O_x intersect with R is just R, right? Here O_x is the set g*x for each g in G

[rustic crown]

didn't you say R contains exactly one element from each orbit, in particular with O_x, so the intersection is a singleton?

[rigid cave]

Wait, why is it just a singleton?

[rustic crown]

R contains exactly one element from the orbit O_x

[rigid cave]

Oh, what if I take the union of G and R then?

[rustic crown]

R is a subset of X and G is something different from X... you sure union is something you wanna consider?

[rigid cave]

Oops, I meant O_x union R

then it is just O_x, right?

[rustic crown]

if there is only one orbit, then R is a singleton contained in O_x

else R also intersects with O_y with O_y being another orbit (disjoint) from O_x

[rigid cave]

Okay I think I got it, thank you! I was stuck on a very confusing proof involving this but I think that I get it now!

[rustic crown]

np!

[chilly ocean]

pls help

[scarlet estuary]

have you tried anything?

[chilly ocean]

yes what i could see was that this doesn't span the entire P3

two more elements have to be added in S ryt?

[scarlet estuary]

yes

what should those be, do you think? can you find elements that arent currently in its span?

[chilly ocean]

no im very new to abstract algebra

[paper flint]

Try to think of proto-typical polynomials of at most degree 3.

Can you build them out as linear combinations of elements already in S?

If not, adding what could possibly fix the situation?

[chilly ocean]

no polynomials having constant terms can not be build

[paper flint]

Yep!!

That gives you a lead. You need a constant as well.

[chilly ocean]

but then why do i need two more elements for this?

one should work too

[paper flint]

Okay, suppose you added the constant 1 to S. Can you build x+1?

[chilly ocean]

ohhhh

thanks man!

[paper flint]

No worries; goodluck!

[rigid cave]

Can someone explain what the author means by "inducting up"? How does he define the fancy A and how does he know that g(fancy A) are pairwise disjoint?

[mild laurel]

can you define what G-paradoxical means?

[rigid cave]

here is the definition given in the text/pdf

[mild laurel]

inducting up just means that the author starts by defining fancy A_1 and then fancy A_2 and so on

[rigid cave]

Okay, so the fancy A_1 is the big mess in the center when i = 1?

[mild laurel]

only if g(A_1) and the rest of the g(A_i) aren't disjoint

[rigid cave]

Oh yeah, that is true. Thank you!

[loud root]

Hi im a bit confused

So I'm told that if f=a * f_0 with a in R, and f_0 a primitive in R[x] then since f is irreducible either a or f_0 are units. If a is unit then f is primitive .. I don't fully follow this argument

[mild laurel]

what does it mean to be a primitive?

[loud root]

all coefficients are coprime

and R is a UFD

[mild laurel]

yeah

so if a is a unit and f_0 is primitive, do you see why a * f_0 is also primitive?

[loud root]

no, this is what i'm stuck on

[mild laurel]

so the coefficients of f are just a * the coefficients of f_0 right

[loud root]

yes

[mild laurel]

so you know that the gcd(a_0, a_1, ... , a_n) = 1

[loud root]

yeah

[mild laurel]

so you're looking at the gcd(a * a_0, a * a_1, ... , a * a_n)

[loud root]

yes

gcd = a?

[mild laurel]

yeah

but a is a unit

[loud root]

I think I'm missing a very important piece xD

but gcd =/= 1?

because a =/= 1 necessarily

[mild laurel]

yeah, so gcd's in general rings are a bit weirder

It's kind of like how

gcd(-3,6) = 3 technically

but it also makes sense to say that gcd(-3,6) = -3 right

[loud root]

i guess

[mild laurel]

like, there's no real reason to prefer the positive gcd over the negative one

this happens in general rings too

[loud root]

I see

just brought up the formal definition too

[mild laurel]

yeah

so you should prove that if gcd(x,y) = a where a is a unit, then you also have that gcd(x,y) = 1

[loud root]

thank you very much!

[loud root]

hello again

I'm just reading this and I fail to see why "a f = b g and f primitive polynomial implies g primitive polynomial"

[chilly ocean]

What does primitive mean?

[loud root]

this

[acoustic pine]

what is the best way to write elements in a potentially infinite direct product/sum of groups?

[scarlet estuary]

context dependent

like if you have an indexing set I, you might write elements as alpha_j for j in I or whatever

[acoustic pine]

Thats what I was thinking

[wild sapphire]

is it valid to say that (1,4,2,6)(2,5,7) from S8 has order 6? they aren't disjoint so I couldn't calculate the least common multiple, so I applied the composition to get (1,6,2,7,5,4), and this is a cycle of length 6 so it's of order 6.

Just wondering if my reasoning is correct.

[wraith obsidian]

Unless you made an error in the calculation, this is correct.

[wild sapphire]

yeah i was just looking to verify the logic, thanks!

[south temple]

does anyone here know some nice books which cover semidirect products?

we started doing them in my class but they aren't covered in our text

[solemn rain]

dummit and foote

[acoustic pine]

Dumb question. Does the following commute in a non-abelian group: $(a^ib^j)(ab)^{-k}$

[cloud walrus]

deathcode

[acoustic pine]

I'd assume not

Or is there any good way to simplify this

[mint seal]

I think any simplification of that would amount to there being a relation in the group, which can't be assumed

so I don't think so

[acoustic pine]

yeah I realized I'm using a bad candidate for an isomorphism. If I use a different candidate, its much cleaner

[lavish pike]

$f(x)=2x^3+x^2+2x+2$ i can just plu 0,1,2,3,4 in and if any produce 0 then it is reducible right?

[cloud walrus]

NocuousNick

[chilly ocean]

what ring are you trying to check (ir)reducibility over?

[oblique river]

are you working over Z/5Z?

oh hi tterra 👋

[lavish pike]

Z5 sorry

[oblique river]

then yes that's true, because the poly is degree 3

[lavish pike]

well now wait a second whats that trick lol

[oblique river]

what do you mean

this is the trick

for polys of degree 2 and 3, (reducible) is equivalent to (has a root)

[lavish pike]

mmmm interesting lol

[lavish pike]

so like $f(x)=x^3-9$ in $Z_{11}$ $f(4)=0$ which also means that -7 is also a factor since 4=-7 in mod 11. so then it is reducible right?

[cloud walrus]

NocuousNick

[carmine fossil]

Yes

[thorn delta]

Let F/K be a finite dimensional Galois extension and F1, F2 be intermediate fields which are galois over K. How would I show that F1 \cap F2 is galois over K?

[mild laurel]

What have you done?

[thorn delta]

Well, at first I thought I would try to prove it directly: for u not in K there is an automorphism in Gal(F1\cap F2/K) which does not fix u, but I don't think that is the right approach

the fundamental theorem says that an intermediate field E is galois over K if Gal(F/K)^E is normal in Gal(F/K). Maybe that's the correct way to proceed?

elements of Gal(F/K)^E being automorphisms of Gal(F/K) which fix E pointwise

[next obsidian]

Do you know how the Automorphisms which fix F1 and F2 relate to those which fix F1capF2

[thorn delta]

hmm im not sure

[vestal snow]

Does i work here?

I think this question is asking us to find a degree 2 intermediate field

I don't know what it means by being explicit

[next obsidian]

I think like write it down

With actual coefficients

[vestal snow]

Like a+bi?

a,b in the rationals

[next obsidian]

I think it’ll be in terms of zeta_10

[vestal snow]

Oh wait

[next obsidian]

Maybe something like that

But you need actual numbers

[vestal snow]

Yeah got it

[latent anvil]

Why is i in that field?

This feels wrong since 4 doesn't divide 10...

[vestal snow]

You're right

[latent anvil]

I think by explicit they just want you to find a generator instead of citing a theorem that states one exists

Ah I think I see the answer

[vestal snow]

No spoilers

Until I fail at it

Then you can spoil it for me

[latent anvil]

Yeah, I wasn't going to say

Okay yes I did some computations and I do know the answer

you are welcome

[vestal snow]

zeta+zeta^3+zeta^7+zeta^9?

No wait

Nevermind

zeta+zeta^7?

[latent anvil]

hmm I didn't write it in the basis, let me check

[vestal snow]

It gets moved by the generator of the galois group, but not the generator squared

[latent anvil]

I believe this is degree 4

Are you allowed to use wolfram alpha?

[vestal snow]

Haha no I don't think so

[latent anvil]

huh

That's weird

well, all I can say is that I computed it explicitly and got something that looked degree 4

Hmm I was going to tell you why it was degree 4

In way you could check on any calculator

But I realized it would tell you how to get a degree 2 extension

So I will stay silent

[next obsidian]

Oh no

[vestal snow]

Isn't the idea to look at the fixed field of a two element subgroup?

[latent anvil]

Hello chm

[next obsidian]

Buncho and Shamrock have fused

[latent anvil]

lol

I'm doing galois theory

So am buncho

@vestal snow yeah, I think that would give it to you

Just write down a generator

[vestal snow]

That's what I tried

[next obsidian]

I honestly would have solved this by just trying to bash out some random crap and find something whose minimal polynomial is degree two

[latent anvil]

hmm

So

[vestal snow]

We know that the function \zeta goes to zeta^3 is a generator

[latent anvil]

φ(ζ) = ζ^k where k is invertible mod 10

Right

Agreed

[vestal snow]

and applying this to zeta + zeta^7 twice

Oh

Arithmetic error

[latent anvil]

r i p

[next obsidian]

What's the Galois group here lol

[latent anvil]

yeah

You get ζ^3 + ζ

[next obsidian]

I don't remember shit about cyclotomic extensions

[latent anvil]

(Z/10Z)^*

[vestal snow]

Z_4

[next obsidian]

Ah

[latent anvil]

Oh wait why is it cyclic

Ah nvm I see

The 2 part vanishes

becuase the totient of 2 is 1

[next obsidian]

Yeah dummy classification of finite groups of order 4

[latent anvil]

I was worried it might be klein four for a sec

[vestal snow]

cyclotomics have cyclic groups right?

[next obsidian]

I mean can't you just directly see that zeta -> zeta^3 is an automorphism or whatever

[vestal snow]

Like always

[latent anvil]

No banana

[next obsidian]

and has order 4

[latent anvil]

take ζ8

This should be klein four, iirc

(Z/nZ)^* is not always cyclic

Only when n is prime

[next obsidian]

well

isn't it actually based on like

[latent anvil]

Okay yes not only

Eg n = 10

[next obsidian]

classification of when a primitive root exists

and like... we know when that is right?

I think the German dude walked me through it once

rip not remembering the name

[latent anvil]

@vestal snow what is (Z/8Z)^*?

{1,3,5,7}

right?

3^2 = 9 = 1 mod 8

5^2 = 25 = 1 mod 8

[next obsidian]

3^2 = 8

cringe

[latent anvil]

7^2 = 49 = 1 mod 8

so it's Klein four

does that make sense?

[vestal snow]

Yeah

[next obsidian]

3^2 = 8

cringe

[latent anvil]

what?

oh

[next obsidian]

3^2 = 8

cringe

[latent anvil]

thank you for your valuable contribution chmonkey

[next obsidian]

you're welcome

[vestal snow]

Guess you and I both are not in our best arithmetic form

[next obsidian]

ugh

that sounds like a legit math thing

arithmetic form

even a best arithmetic form sounds like a thing that could be a real thing

[latent anvil]

anyways banana your method should work

but my solution is more geometric

Or at least trigonometric

[uncut girder]

Arithemtic form

[vestal snow]

mfw I never learned trig

[latent anvil]

woah

[vestal snow]

Lmao yeah never really needed to

Until I gave my ACT and lost a lot of points on it because of trig

[latent anvil]

is ACT the same as the American test?

[vestal snow]

yes

It's like the SAT

[latent anvil]

Yeah I took it

[next obsidian]

Wait Sham

is yours based off regular polygons

or whatever the connection of the cyclotomic extensions to the geometry angle thingies is

[latent anvil]

yup

[uncut girder]

Unbased

Do not invoke trigonometry for Algebra problems

[acoustic pine]

If $\nu$ is a discrete valuation, is it the case that $\nu(a)=\nu(-a)$?

[cloud walrus]

deathcode

[next obsidian]

only if v(a) = 0 right?

v(-a) always equals -v(a)

so you're just asking for 2v(a) = 0 which is just saying v(a) = 0

[acoustic pine]

Oh shit my bad that was a typo

Yes I meant -\nu(a)

[next obsidian]

or wait

am I dumb

No I'm mega dumb

[sturdy marsh]

this is true for the p-adic valuation

[next obsidian]

v(1/a) is -v(a)

lol

[acoustic pine]

The characterization is a big confusing. We're given that $\nu(ab)=\nu(a)+\nu(b),\nu$ is surjective, $\nu(a + b) \geq \min{\nu(a), \nu(b)}$

[cloud walrus]

deathcode

[next obsidian]

yeah

Forget what I said about v(-a) = -v(a) lol

The axiom about the sum tells you literally nothing

[acoustic pine]

So then $\nu(a + (-a)) = \nu(0)$ for all $a$, but that means that $\nu(0)$ must equal $\aleph_0$ right?

[cloud walrus]

deathcode

[next obsidian]

it just tells you that infinity >= min{v(a),v(-a)}

[acoustic pine]

Or $\infty$

[next obsidian]

which says literaly nothing

[cloud walrus]

deathcode

[acoustic pine]

Right ok

I'm trying to show its an additive subgroup, and my first thought was show that for $a,b \in {x\in K^*\mid \nu(x)\geq 0}$ for some field $K$ is an additive subgroup using the one step subgroup test

[cloud walrus]

deathcode

[next obsidian]

that what's an additive subgroup?

[acoustic pine]

I'll rephrase

I'm trying to show that $R={x\in K^* \mid \nu(x)\geq 0}$ is a subring

[cloud walrus]

deathcode

[acoustic pine]

Showing its closed under multiplication is trivial

[latent anvil]

The answer to your original question is yes btw

[acoustic pine]

Its the addition that I'm struggling with for some reason

[next obsidian]

oh lmfao

[sturdy marsh]

v(-1) = 0

[next obsidian]

they differ by a unit

lol

hurb

v(a/-a) = v(a) - v(-a) = v(-1) = 0

[acoustic pine]

And every ring with unity has a $-1$

[cloud walrus]

deathcode

[latent anvil]

yep

to you and chm

[acoustic pine]

Ok, thats the only fact I needed

[next obsidian]

Hurb moment

[latent anvil]

Nice!

[acoustic pine]

I assumed that, but I didn't know how to prove it

[latent anvil]

Good intuition :)

[next obsidian]

Says valuations are ez

[acoustic pine]

I didn't say that lol, I'm just trying here lol

[next obsidian]

fails absolute basic things with them 3 days later

oh no

this is me

[acoustic pine]

Ahhh

[next obsidian]

I am making fun of myself

[sturdy marsh]

valuations are hard

[latent anvil]

> factoring polynomials

[acoustic pine]

Self deprecation, an old friend to be sure

[next obsidian]

Factoring polynomials is ez

[acoustic pine]

[next obsidian]

does like 4 lines of crazy inequalities

tfw couldn't factor a quadratic

[sturdy marsh]

I think my algebra final last year

had a problem which went

[acoustic pine]

Its kinda funny bc for me the last algebra class I took was our undergrad algebra 1 course, and now I'm diving headfirst into graduate algebra 2 without taking grad algebra 1

[latent anvil]

Would someone like to help me solve a "factoring polynomials" problem

[acoustic pine]

Wish me luck

[latent anvil]

Why is (x, y)(z, t) = (x, y) intersect (z, t)

[next obsidian]

Sham I am almost certain my proof works now that I thought more

[latent anvil]

chm has a proof but it's too messy for me to trust

[latent anvil]

good luck!

[acoustic pine]

You are such a wholesome person jeez

I appreciate you lol

[sturdy marsh]

find a valuation Q[x,y] ---> Z with residue field Q

[latent anvil]

oh

Interesting

wait is the residue field the residue field of the valuation ring you get from the valuation?

[next obsidian]

I could find you one with residue field Q[alpha] for alpha an irrational number

[latent anvil]

just making sure I understand

[next obsidian]

Wait no I can't

[latent anvil]

yeah this seems p tricky

imo

wait is it tricky

hmm

Maybe not

[sturdy marsh]

yes

it is

[latent anvil]

maybe it is

Fair enough

[next obsidian]

You want a valuation with positive degree terms being like Z[rational numbers] yeah?

or I guess non-zero

[sturdy marsh]

it just needs to be a valuation

[next obsidian]

I mean the valuation ring associated to it has to be between Z and Q

I think

Actually I guess Z is the minimal ring here, nothing is smaller than it

so really you want every non-constant to have negative valuation

Okay I think I have an idea

based on this exercise

the exercise wants alpha irrational but maybe I can just take alpha = 1 it works but

[latent anvil]

I do not see why

[sturdy marsh]

?

[latent anvil]

Oops lol

[latent anvil]

Meant this

I don't see why it has to be within Q

[next obsidian]

Because if it doesn't

then its fraction field would not be Q

right?

[latent anvil]

I thought we wanted the residue field

[next obsidian]

like you'd have a polynomial in there

wait wut

O_O

[latent anvil]

residue field Q

[next obsidian]

oh

[sturdy marsh]

yup

[next obsidian]

fuck this

nvm

[latent anvil]

lmao

[sturdy marsh]

luckily this was a take home final lol

took me a very long time to solve it

at one point I was convinced the question was wrong

[latent anvil]

Yeah seems very hard

[next obsidian]

bruh moment

[sturdy marsh]

the solution is kinda messed up

[acoustic pine]

Sorry to interrupt, does this reasoning track?

[sturdy marsh]

we have a valuation on Q((t)) with residue field Q

[latent anvil]

stap

I wanted to think more about it

[sturdy marsh]

ok

[golden pasture]

ohwait do you ||do some funny injection into Q((t))||

oml

[next obsidian]

why does 2v(-1) = 0 imply v(-1) = 0? This is true, but you want to be careful you know why

[sturdy marsh]

took me several days to come up with that lol

[acoustic pine]

Noted! The professor is very pedantic from what I've heard, so I'll clarify

[next obsidian]

For a non-discrete valuation you need to use ordered group stuff

[sturdy marsh]

I was doing AG at that time

[next obsidian]

but for a discrete valuation it's quite a bit simpler lol

[sturdy marsh]

so I kept trying divisors in spec Q[X,Y]

none of them work

[golden pasture]

yea thats the most natural approach

[sturdy marsh]

you always get an extension

[golden pasture]

i guess you need to just think in the opposite direction

how do i get Q

[next obsidian]

I'm going to k-word myself

[golden pasture]

problems do be like that sometimes

[sturdy marsh]

that final had many problems

spent 8 full days working on it

Another one was:

[cloud walrus]

Brofibration

[sturdy marsh]

this one was hard

but not as hard as the valuation one

[golden pasture]

the valuation feels like some imo training problem lol

once you get the key idea it breaks immediately

but theres not much to throw at it until you get the idea

[sturdy marsh]

yeah the hard bit is figuring out that you need to look at Q((t))

and that it is uncountable

it fits into the geometric picture in a way

I think you can make it a divisor-like thing in the classical topology

[sturdy marsh]

||and this is how you get the funny injection||

[smoky cypress]

So p adic integers and rational numbers are not subsets of each other in the field of p adic numbers right?

[mild laurel]

yes

[chilly ocean]

can anyone have a quick look at this se question? https://math.stackexchange.com/questions/4083947/construction-of-c-ellv-q-for-a-quadratic-space-v-q-and-proving-an-inject

[amber flicker]

So this is kinda algebra

I need to find the SA of this and V

I want to know what to do with the 135 degrees, can I translate it to mm can I use it in a formula to give me a more accurate Area or volume calculation. My teacher said I can just ignore the angle but i don't want to. I want to do it to the best of my ability

I didn't know where to post this

[golden pasture]

maybe some #geometry-and-trigonometry may be more appropriate

scroll up in this channel

youll see it doesnt fit

[carmine fossil]

Also,Is this civil engineering or smt?

[untold sapphire]

Are people here graduate students? Just about to start my PhD and I have no idea how to prepare for my first meeting with my advisor

[carmine fossil]

Better ask that in #discussion or smt

[untold sapphire]

Ok

Thanks

[amber flicker]

@untold sapphire are you getting a PHD in MATHS?!

[untold sapphire]

Yeah

[amber flicker]

I admire you

[untold sapphire]

I’m in the UK

[amber flicker]

I'm a citizen of Australia and a perm resident in the UK

Just have to file some paper work and ehhhhhhh

I'll probably do it after covid

[chilly ocean]

fuck yeah! maths!

[amber flicker]

I use to be terrible at maths

In my first day of middle school I was given a calculator and d's went to A's

Turns out I'm just bad at mental math

[golden pasture]

just wing everything

[granite compass]

Given an exact sequence $1 \rightarrow G_0 \rightarrow \dots G_n \rightarrow 1$ of finite groups of sizes $g_n$ for $G_n$ I'm trying to show that $\prod_{j=1}^{n} g_{i}^{(-1)^i}=1$. I tried first induction by the sequence size but this didn't work since from a sequence I can't get a one shorter sequence. Then I'm thinking that maybe since $g_1 = g_0 * k$ since the first map is injective but I don't know how this helps. Any ideas?

[cloud walrus]

PAHUS

[hot lake]

use that |G/ker f| = |im f|

[granite compass]

Ahhh that should help!

thanks a lot

[wraith obsidian]

stupid question, but does a free submodule in a free module necessarily leave a free quotient?

Ah, I forgot about the simplest of the cases, the doubling ℤ→ℤ

[latent anvil]

rip

in fact, in a PID anything is a quotient of a free module by a free submodule

[wraith obsidian]

But this begs the question… how does meyer-vietoris work when we don't take values in a field?
I did such calculations when doing de rham, i.e. H(-,ℝ), and there it was just „dimension counting“ along the whole train.
But with lets say ℤ we aren't even guaranteed to get free modules for the missing cohomology groups

so how does that work then

(may be #point-set-topology although it's still technically more of a homological algebra question)

[sturdy marsh]

if you're asking, "do nontrivial extensions cause trouble", the answer is yes

[wraith obsidian]

Sort of, yes

So I should just find my emergency $\otimes \mathbb Q$ kit and try to put out the flames, right?

[cloud walrus]

lux

[sturdy marsh]

maybe

or in some cases you know the maps very well

so you can check if generators map to generators

[thorn delta]

how would i start on this problem?

[haughty shale]

Hi. What does a simple zero mean?

[next obsidian]

zero of order 1

[haughty shale]

Okay thanks! So a zero with multiplicity 1?

Thanks!

[fathom delta]

Can anyone tell me if I am insane? I got x = (1 2) when solving this permutation equation (1 3 2)x = (2 3) in S_3, but my professor marked it wrong. I redid it multiple times, and even checked it with a Cayley table. Where am I going wrong?

[rigid cave]

Looks good to me but do not take my word for it, I am still a noob

[fathom delta]

Yeah I'm pretty sure it's right, I am going to try to politely email him about it

[rigid cave]

Your solution definitely works tho, I do not see a problem with it

[fathom delta]

yeah I thought it's pretty simple, just multiply by (1 3 2)^-1 on the left for both sides

and (1 3 2)^-1 = (1 2 3)

Then (1 2 3)(2 3) = (1 2)

[chilly ocean]

I agree, probably your prof is being [redacted]

[rigid cave]

Your prof might think that your hand drawn two looks like alpha. Happened to me once lmao

[fathom delta]

eh maybe but I kinda doubt it. my 2's are pretty different from alphas

[rigid cave]

Yeah I know, it was just a bad joke

[fathom delta]

lmao

[viscid pewter]

i've got a bunch of questions

does anyone have any intuition for {(123...n), (12)} generating Sn

like you conjugate (12) by (123...n) repeatedly to get 2-cycles?

[latent anvil]

yeah so conjugation like this will give you all adjacent transpositions

Right?

[viscid pewter]

so i know it does but i don't get why

someone referred to it as 'changing bases'? i think

[latent anvil]

Do you know how conjugation works in S_n?

[viscid pewter]

or basis

uhhh

probably not

[latent anvil]

Yeah that's a great intuition actually

[viscid pewter]

it will be once i understand it

[latent anvil]

Ah sorry I am about to lose connection

[viscid pewter]

aaaaa

[latent anvil]

But basically, conjugating (a1 a2… an) by σ applied σ to each ai

and keeps the order of the cycle the same

Try to prove this

[wraith obsidian]

long shot, but what might be happening here is that you're using different conventions for cycle composition
There is the covariant one and the contravariant one and I'm not gonna pretend like I know which one is the „usual“ one everybody uses

[latent anvil]

So say you have a conjugation like σ^-1 τ σ kai

[viscid pewter]

i mean conjugation always keeps the order of the element the same

so that makes sense

[latent anvil]

This will take something like σ(a), then compute τ(a), then replace it with σ(τ(a))

It's like you're applying tau "under" the sigma

Sorry, disconnected

[rigid cave]

So if $\sigma = (1, 2, \ldots, p)$ and $\tau = (1, 2)$ then $\sigma^{-1} \tau \sigma = (23)$, but then $\sigma^{-1}(23)\sigma = (34)$ so the group contains all elements of the form $(m, m+1)$, but then $(12)(23)(12) = (13)$ in in the group etc. so the group contains every $(1m)$, but then $(1m)(1r)(1m) = (mr)$ is in the group and $S_n$ is generated by transpositions

[viscid pewter]

oh i thought you were just gonna disappear for 14 hours lol

that was nothing

[latent anvil]

thanks sister!

[cloud walrus]

Older Sister

[viscid pewter]

yeah i understand all of that, i just don't understand why conjugating by that should have that effect, or the general effect that shamrock said

[latent anvil]

Sure, so let's think about the general setup of σ^-1 τ σ

In fact, let's specialize to the case of a cycle τ = (a1 ... an)

what will this conjugated element do to a generic thing in 1,2,...,m?

[viscid pewter]

so s-1 (a1 ... an) s = s-1 (s(a1) s(a2) ... s(an))? is that right?

[latent anvil]

nah, no s-1 on the second side

[viscid pewter]

oh right

that's what we've ended up with

so can't be that as an intermediate

[latent anvil]

Hm, I'm not sure what you mean

[viscid pewter]

so i'm just trying to understand why we end up with s-1 (a1 ... an) s = (s(a1) s(a2) ... s(an))

[latent anvil]

Yup, no worries

[viscid pewter]

so obviously my first stab at breaking it down didn't work

[latent anvil]

So when are two permutations equal?

They're just functions, right?

[viscid pewter]

yeah

so f(a) = g(a) for all a

[latent anvil]

Right

So there's two cases here

cycles do one thing to stuff inside the cycle and one thing to stuff outside of it

[rigid cave]

Perhaps this picture help?

So conjugation can be thought of as just permuting the order of the elements that are about to get swapped.

[viscid pewter]

i'm sorry i do not understand that picture at all

[rigid cave]

This picture tries to illustrate that conjugation can be thought of as a permutation of the domain

[viscid pewter]

well

isn't composition of permutations just generally a permutation of the domain??

[latent anvil]

sorry kaisheng, maybe we should do an example

[chilly ocean]

This is like whether Sn acts as a left multiplication or right multiplication on 1,2,..., n right?

[rigid cave]

So this would kind of explain $\sigma \tau \sigma^{-1} = (\sigma(a_1), \sigma(a_2), \ldots)$

[cloud walrus]

ɹǝʇsᴉS ɹǝplO

[chilly ocean]

Generally, I thought everyone thinks it is a left multiplication, I have never seen it the other way

[viscid pewter]

hrnnn

[latent anvil]

I'm working with right multiplication here

[viscid pewter]

this thing we're talking about rn should work either way round as i understand it

[latent anvil]

No it won't kaisheng

The order would flip

[viscid pewter]

no but like generally

[latent anvil]

If you use left multiplication it'd be σ τ σ^-1

[viscid pewter]

the conjugation

oh wait

yeah ok

[latent anvil]

Yeah, left vs right multiplication reverses the order of multiplication in Sn

[viscid pewter]

it'd have the inverse effect

[latent anvil]

So let's try and example

σ = (134), τ = (152)

I'm claiming that σ^-1 τ σ = (352)

right?

[viscid pewter]

um

yes

right

[latent anvil]

So let's look at k.σ^-1 τ σ vs k.(352) for each k in {1,2,3,4,5}

what happens when k = 1?

(the dot is for the action of Sn on an element)

[viscid pewter]

yeah

it's fixed by (352)

[latent anvil]

Right

so then for the other one, we want to apply σ^-1, then τ, then σ

right?

[viscid pewter]

-> 4 by s-1, unchanged by t, -> 1 again by s

[latent anvil]

Right

[viscid pewter]

yeah so they have the same effect, as we expected

[latent anvil]

Right

What about 2?

[viscid pewter]

-> 3

wait

-> 2 -> 1 -> 3

yeah so it works again

[latent anvil]

Right

And something interesting happened here

It's like we applied τ to 2 and then translated back inside σ

I think 3 will be the most useful example

so what happens with 3?

[viscid pewter]

so you're saying it's like s(t(2)), right?

[latent anvil]

Right

and this is because 2 = s(t)

Anyways

Let's try 3

[viscid pewter]

3 -> 1 -> 5 -> 5

[latent anvil]

Right

So 3 = σ(1)

Then we apply τ inside

To get σ(τ(1))

So this is what the conjugated thing does

[viscid pewter]

no, yeah, i'm convinced that conjugation has the effect you're describing, i just don't understand why it should have that effect

[latent anvil]

I'm confused

[viscid pewter]

why should $\sigma^{-1}(a_1a_2...a_n)\sigma = (\sigma(a_1)\sigma(a_2)...\sigma(a_n))$

[cloud walrus]

12ƃuǝɥsᴉɐʞ

[viscid pewter]

i can see empirically it's true

but not intuitively why

[latent anvil]

Sure

So it's this thing I'm saying about applying τ underneath the σ

If you start with σ(a_k)

You'll end up with σ(τ(a_k)) = σ(a_{k+1})

The conjugation first undoes the σ to get a_k

Then applies τ

Then applies σ again

[viscid pewter]

hmm

[rigid cave]

Look at the picture I sent before. Sigma literally permutes the domain like in the picture I just sent

[viscid pewter]

one sec sorry

[rigid cave]

So sigma permutes the a_1, a_2 etc

[viscid pewter]

hmmmmm

i'm sorry, i'm just holding off on understanding the meaning of 'permutes the domain' in favour of understanding the doing/undoing thing

[rigid cave]

Hold on, let me send another picture

[latent anvil]

I think that's a good idea kai

Focus on one thing at a time

σ^-1 τ σ has the effect of undoing σ, applying τ, and then redoing σ

[viscid pewter]

yes

i can agree with that lol

ok so i understand like the local argument given the thing

[viscid pewter]

not sure about why we're starting with s(a_k) though

oh, wait

[latent anvil]

Because I'm claiming that the conjugation equals this relabelled cycle

So we need to check that it has the same behavior as the relabelled cycle

[viscid pewter]

oh, right

that's just the claim, yeah, it's sigma on each a_k

[latent anvil]

Yup

So on a thing in the cycle, it does the right thing

σ(ak).σ^(-1)τσ = σ(ak).(σ(a1)...σ(an))

[viscid pewter]

wait

this is so dumb lol but σ(ak).σ^(-1)τσ, is this not you conjugating the thing that's already been affected

no

oh wait is this a relabelling

[latent anvil]

Ukmm

[viscid pewter]

so like sigma(ak) is some al

[latent anvil]

I'm confused sorry

So τ = (a1…an)

σ is some opaque permutation

[latent anvil]

Rn I'm just claiming this

[rigid cave]

Maybe this helps?

Notice that the a_1, a_2 etc have been swapped by sigma

Maybe I am just disturbing you guys. Just focus on what buncho says

[latent anvil]

Sorry sister, it's just tough to try and understand two explanations at once

I think this is a good way to think about it but probably confusing to talk about rn

[rigid cave]

Yup, definitely

[viscid pewter]

ok so basically you're saying this is true:
σ^(-1)τσ = (σ(a1)...σ(an))

so for all al:
al.σ^(-1)τσ = al.(σ(a1)...σ(an))

so for for each al, there's a convenient ak such that σ(ak) = al
and then you're saying σ(ak).σ^(-1)τσ = σ(ak).(σ(a1)...σ(an)) is true

but σ(ak).σ^(-1)τσ = ak.τσ = ak.(σ(a1)σ(a2)...σ(an)) ??

ok so that was a looong wait and i didn't really do anything except probably mangle the notation

lemme try again

t = (a1a2...an)
so t(ak) = ak+1
then s(t(s-1(s(ak)))) = s(t(ak)) = s(ak+1)

so s(t(s-1(s(a1)))) = s(a2), s(a2) -> s(a3), etc.

so s(t(s-1))) = (s(a1)s(a2)...s(an))

oh, it works out

ok so

huh

ok well at some point during the last hour i reached the point of completely understanding it

and i don't know when the transition occurred

which is weird af

but now it's really obvious lol

there's a trick here somewhere

ok i'll reflect on that in my own time

@latent anvil @rigid cave thank you both for trying for most of an hour, your efforts bore fruit at some point

[latent anvil]

Thanks!

I think Aluffi explains this well

sorry I had to go, other people were messaging me about my actual classes 😅

[viscid pewter]

you gotta do what you gotta do

is this, like, a standard thing

[latent anvil]

yup

[viscid pewter]

i'm gonna be so mad if i spent ages trying to understand the intuition behind this independently

[latent anvil]

it's something I would expect someone to learn in a first group theory course

[viscid pewter]

and it's just all in the textbook a few chapters later

[latent anvil]

lmao

[viscid pewter]

ok but yeah ty

[latent anvil]

Can someone confirm that this is true for me

Let $A = k[x, y, z, t]/(xz, xt, yz, yt)$

[cloud walrus]

🐊 sɔoɹɔɯɐɥs

[latent anvil]

Then $A/(xy-1) \cong k[t]_t$

[cloud walrus]

🐊 sɔoɹɔɯɐɥs

[latent anvil]

So $(xy-1, x+1)$ is a regular sequence for $A$

[cloud walrus]

🐊 sɔoɹɔɯɐɥs

[next obsidian]

Maybe call it k[x]_x

So it’s more clear what x + 1 does on it

[thorn delta]

alright so im kind of lost on what the best way is to work with the compositum of subfields. For part a, i give what im pretty sure is an incorrect proof. F1 and F2 are the span of the alpha beta, but this only shows that F1 + F2 (sum of vector spaces) is in the span, not that F1F2 is in the span.

I've thought about trying to use the hint, but I'm not quite sure how thinking of the elements of F1 and F2 as rational functions makes it any easier to find a spanning set for F1F2. Anyone have any tips?

[mild laurel]

i think its close to being right

depending on how you defined the compositum, you could use the fact that the compositum is the smallest field that contains both F_1 and F_2

[thorn delta]

use the fact that the compositum is the smallest field that contains both F_1 and F_2
that's the only definition/characterization i have. My concern is that span{alpha_i beta_j} isn't a field tho

[mild laurel]

mhm, that's not that hard to show though

all you really need to show is that the product of two such elements is also in the span

[thorn delta]

ah, right, okay that makes sense

[latent anvil]

reposting this

[maiden ocean]

im pretty confused as to whats going on here

[latent anvil]

haha yeah

:/

it is a very confusing argument to see for the first time

[oblique river]

the proof or the statement or both?

[maiden ocean]

uh kind of both

is phi^n here supposed to be phi(x)^n

or like. phi iterated

[oblique river]

have you seen the cayley-hamilton theorem for matrices before?

yeah it's phi iterated. in a module you don't have multiplication of elements so you can't write like, phi(x) * phi(x)

[maiden ocean]

ah okay right that makes more sense

i thought it was that at first but then 2.5 confused me

[oblique river]

but the ring of endomorphisms on M is, well, a ring, and the multiplication in that ring is composition

in 2.5, with phi = id, then phi^n is just id^n, or id

[maiden ocean]

right

but i dont get where x is coming from?

[oblique river]

in 2.5, AM is defining x to be that sum, where the a_i are from 2.4

[maiden ocean]

oh

wait hahaha

ok i was being silly thank u

[oblique river]

np!

[maiden ocean]

but shouldnt that just be 0?

[oblique river]

careful about where things are living

in 2.4, that thing which is = 0

[maiden ocean]

ohhhh i see

right

[oblique river]

is 0 as a transformation on M

but in 2.5 that is an element of A

which, by 2.4, acts as the 0 transformation on M

[maiden ocean]

yeah in the ring of endomorphisms

[oblique river]

yep! the 0 in 2.4 is "0 in the ring of endomorphisms"

and there is a map from A into that ring defined by x --> "multiplication by x"

and the x in 2.5 is not 0 in the former but is 0 in the latter

[maiden ocean]

ya and similarly 1 here is the identity in the ring of endomorphisms right?

under our map from A into E(M)

[oblique river]

yep

this is also important if you look at 2.4, that big expression phi^n + ... = 0, is supposed to be an endomorphism on M. but the last term is a_n which is an element of A.

and this is how they're interpreting that as an endomorphism on M

left multiplication by a_n

[maiden ocean]

yeah okay i think i get it now

[oblique river]

have you seen the cayley hamilton theorem before?

like in lin alg

basically "if you plug a matrix into its char poly, you get 0"

that's what this is, but the modules version, although instead of defining the "char poly" of a module endomorphism, 2.4 is just stated as "there exists some polynomial satisfied by phi"

[maiden ocean]

i think i might have?

yeah looking this up i have i believe and it also makes the proof make sense lol

ok ty :^)

[oblique river]

yeah np -- i'm honestly surprised that AM doesn't even mention that but I guess the book is all about being as efficient as possible haha

no wasted words

[maiden ocean]

_<

its nice if ur proving everything urself which ive been trying to do

[oblique river]

ah that's nice

[next obsidian]

ayyyyyyyyyy determinant trick

based tbh

I honestly forget how you prove Cayley-Hamilton using the adjugate matrix or whatever

determinant trick is my friend

[latent anvil]

?

alex

you prove cayley hamilton in the obvious way

[next obsidian]

eh?

[latent anvil]

wlog you're algebraiclly closed

then diagonalizable matrices are zariski dense

[next obsidian]

Wtf?

[latent anvil]

the condition in cayley hamilton is zariski cloesd

and it's trivial for diagonalizable matrices

[next obsidian]

????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

wat

[latent anvil]

bruh just follow your nose

[next obsidian]

🤔

smh

[small bison]

it's so natural

[oblique river]

yeah that's actually such a slick proof haha

[latent anvil]

yeah lol

I think it's really cool

but I am ofc shitposting

[next obsidian]

I mean

I'll take that proof if I can understand it

lmao

Time to think

[oblique river]

"satisfying the char poly" is trivial for diagonal matrices

[next obsidian]

yeah

[oblique river]

and therefore for diagonalizable matrices

[small bison]

i like to just view V as a K[T] module

[oblique river]

since nothing changes under conjugation

[next obsidian]

Yeah

[oblique river]

diagonalizable matrices are zariski dense in the set of all matrices

[next obsidian]

but then what about these being Zariski dense?

In what sense can you view this as being dense

[oblique river]

(in fact, also dense in the usual complex topology)

[latent anvil]

over an algebraiclly closed field!!!

[oblique river]

oh yes sorry that is very important

[latent anvil]

the zariski topology is discrete for finite fields

[oblique river]

I was just thinking over C

[latent anvil]

np

[next obsidian]

Like

you're talking about the like

shitty Zariksi topology

like with...

points

[latent anvil]

yes lmao

[next obsidian]

🤮

[latent anvil]

i mean the actual zariski topology

not scheme nonsense

[next obsidian]

I see

okay

[oblique river]

but also like, in the complex topology

on C^(n x n)

[next obsidian]

tfw everything is C

[latent anvil]

yeah it's probably worth talking about that first

[next obsidian]

Nah

[oblique river]

the point is, "satisfying the char poly" is a closed condition

[next obsidian]

yeah

[oblique river]

both zariski closed and topologically closed

[next obsidian]

I mean you're literally looking at some polynomial right?

[oblique river]

therefore, boom

yeah

the char poly is polynomial in the entries

[next obsidian]

Yeah

,w factor 42069

[oblique river]

now to prove it for all fields just embed a field in its algebraic closure. that's injective and so if you get 0 over the bigger field you get it over the smaller field too

[next obsidian]

yeah I mean

that bit was obvious haha

[oblique river]

i thought all of it was obvious

[next obsidian]

but doesn't this only give it to you for vector spaces?

or was Cayley-Hamilton only true for vector spaces lmao

Okay you know what

[oblique river]

this whole conversation started from moth trying to prove the modules version haha

[next obsidian]

oh

yeah

I mean can you not prove the module version in some trash adjugate matrix way

[sturdy marsh]

you can

[next obsidian]

whatever the normal proof is

Okay

but F that

determinant trick

[small bison]

just plug in your matrix

[next obsidian]

Oh like

[sturdy marsh]

I think my favorite proof for the vector space case is

modules over PID

[next obsidian]

det(M -tI) plug in M and get

det(M - M) = det(0)

[oblique river]

yes chm that is the meme

[next obsidian]

swag

I wrote that down on a final

because it was a bonus problem

and I didn't know what the fuck to do

I got 0

[latent anvil]

?

[next obsidian]

Abstract linear algebra

[latent anvil]

you really didn't think to prove this is a zariski closed condition?

[next obsidian]

340

[small bison]

you got 0 as you should

[next obsidian]

yeah

sorry Sham

[latent anvil]

and that diagonalizable matrices are zariski dense?

[small bison]

since the matrix satisfies the cayley hamilton theorem

[next obsidian]

no

[latent anvil]

lmfao okay lol and you thihnk you can do AG now

[next obsidian]

😭

I didn't even know what a zero divisor was at that time

[sturdy marsh]

cayley-hamilton is a pretty unreasonable problem to put on an exam

[next obsidian]

It was just a bonus problem but idk how he expected anyone to prove it

tbh it was for vector spaces

and we had gone over canonical forms

I just was asleep the last 3 weeks

[sturdy marsh]

lol

[next obsidian]

Guess who still doesn't understand canonical forms

dab

learned em 3 times baybee

forgot em 3 times

[chilly ocean]

the module decomp theorem is a black box and then canonical f orms follow from that

[next obsidian]

what module decomp theorem

the one that lets you like do shit about submodules?

[chilly ocean]

the one that decomposes modules, 42069

[next obsidian]

something something a generating set from a subset of the basis for the free module

thanks

:)

[chilly ocean]

what's the acronym

ftfgmpid

[latent anvil]

the stopvspoakmod alex

[next obsidian]

heh

eh?

setup to

[latent anvil]

Structure theorem lol

For finitely generated models over a PID

[next obsidian]

but I can't understand your acronym

structure theorem of principal vector spaces principal over a kmodule

:)

[latent anvil]

lll

[next obsidian]

III

[chilly ocean]

all these number names make me feel like i'm having a fever dream every time i open discord

[next obsidian]

I went to twitter and saw a real name

and wondered why it wasn't a number

[acoustic pine]

Oh my god

Why

[next obsidian]

April fools

[acoustic pine]

[next obsidian]

we now operate on Russia time

[chilly ocean]

overthrow mniip

,ban mniip

[cloud walrus]

This may only be done by a moderator!

Apr 1, 2021, 1:10 AM

[chilly ocean]

,kick mniip

[cloud walrus]

This may only be done by a moderator!

Apr 1, 2021, 1:10 AM

[next obsidian]

,ban mniip

[cloud walrus]

This may only be done by a moderator!

Apr 1, 2021, 1:10 AM

[next obsidian]

,kick mniip

[cloud walrus]

This may only be done by a moderator!

Apr 1, 2021, 1:10 AM

[next obsidian]

Woog kick mniip

,woog kick mniip

[chilly ocean]

,fuck mniip

[next obsidian]

,mniip mniip

[latent anvil]

Same 8. 71749

[latent anvil]

.pin

[latent anvil]

Wait what the fuck

LMAO

THE PIN BOT

IS A NUMBER

[small bison]

now this is meta

soon the server will be but a number as well

[barren sierra]

Wait can I be a number?

Oh fuck I already am

[vestal snow]

Any idea how to do this?

Here's my attempt

I am fairly certain my proof is wrong

Because I used nothing special about f_n during the second part

[chilly ocean]

I didn't read the whole argument (I got lost in some parts), but are you counting the roots in (-infty, b1), or just checking that a root exists? Eg how do you know there is only 1 root?

Overall, this feels like a real asshole calculus problem, I wish/hope there is a nicer solution

Er, wait sorry, nvm

I think it looks good, but I did not 100% carefully read it

Intuitively, isolated non double roots sound like they should be "not fucked up" by scaling and translating up and down

[vestal snow]

Yeah this problem sucked

Thanks for verifying though

[vestal snow]

How do I do c)?

[golden pasture]

the key is that either $\sqrt p\in\mathbb Q\left(\zeta_p\right)$ or $\sqrt{-p}\in\mathbb Q\left(\zeta_p\right)$ for a odd prime $p$

[cloud walrus]

142824

[vestal snow]

Is this a known fact?

Like is this something that most books present as theorems?

[scarlet estuary]

corollary of kronecker-weber

[golden pasture]

uhh

idk

but it isnt too hard to prove

not rlly

[scarlet estuary]

but if you dont want to prove it in full generality

[golden pasture]

kronecker weber doesnt bound what N

[scarlet estuary]

you can do it specifically yeah

not too bad

[golden pasture]

it tells you one exist

[scarlet estuary]

ah

good point

[vestal snow]

I'm kinda confused how we were supposed to come up with a proof without knowing this

[vestal snow]

Isn't Kronecker-Weber a theorem from class field theory?

[mild laurel]

yes technically

but you can prove it without class field theory techniques

[vestal snow]

Interesting