groups-rings-fields

[oblique river]

and that coset represents a single function on V

[river fern]

When I use the Orbit-Stabilizer formula |O| * |N_G(H)| = |G|, I'm getting that |N_G(H)| = |G| which seems clearly wrong:

(one sec)

[oblique river]

arvindr could you wait a bit?

we're having another discussion about a different question right now

(if neither boti nor matplotlib responds in the next 5 minutes -- go for it)

[plucky flicker]

okay, it's clear now 🙂 thanks a @oblique river

[oblique river]

great!

[oblique river]

(okay arvindr you can go now lol)

[river fern]

I'm getting an orbit of 1 (since H is normal), so the formula gives that the number of conjugates is equal to |G|

Which doesn't make sense

I feel the number of conjugates of a normal subgroup should be 1

[oblique river]

the orbit is the set of conjugates

those are the same thing

that's not anything to do with the orbit-stabilizer theorem

that's just hte definition of orbit in this case

[river fern]

Wait, so what are the orbit and stabilizer in this case?

ohwait

oh i'm clear now

yeah all the elements of G fix elements of H, so the stabilizer has size |G|

makes sense

[oblique river]

also the second question is independent of the first btw

you're not supposed to assume H is normal anymore

in the second bullet point

[river fern]

oh yeah mbad

[chilly canyon]

May I continue with the previous question please ? (and thanks again for the explanation !)

[oblique river]

yeah go for it

[chilly canyon]

Similarly to the fact that k[V] is viewed as the set of polynomials on V, is its field of fractions k(V) viewed as rational fractions defined on V ?

[oblique river]

yep

[chilly canyon]

Okay, thanks !

[oblique river]

it's also often called the "function field"

[chilly canyon]

Moreover, in the example you gave with V:X²+Y²=1, one may use the 1st isomorphism theorem to show that (if I'm correct) k[V]=k[X,Y]/(X²+Y²-1)≅k[a,a⁻¹]. How to interpret this isomorphism in terms of the analogy with polynomials on V ?

[oblique river]

I don't see where that isomorphism comes from

what is your map from k[x,y] to k[a,a^-1]?

[chilly canyon]

I read that k[X,Y]/(X²+Y²-1)≅k[X,Y]/(XY-1), I don't remember how to apply 1st iso to obtain this now ! :x

[oblique river]

hmm I think that's true over an algebraically closed field

but I can't find a proof that shows it's true over R, for example

[chilly canyon]

Perhaps they assumed k was algebraically closed ? I have no clue... My question makes no sense, sorry !

[oblique river]

haha it's all good

[vestal snow]

Why is h^-1 not a polynomial in the y_i?

[oblique river]

any polynomial in the y_i will have a denominator that is a product of the g_j

but the denominator of h^(-1) is prime to all of the g_j

[vestal snow]

Okay, so you get a product of g_j = h * some polynomial in x_1,..., x_r

which means that product of g_j is in (h)

[oblique river]

mhm

[vestal snow]

if the product is 1, then we get a contradiction as h is not a unit

if there is at least one g_k in the product, we get that 1 is a multiple of g_k

I guess we should require that all g_i are non-units in the polynomial ring

[oblique river]

units in a polynomial ring are just the nonzero constants

[vestal snow]

yeah

Okay, thanks

That made sense

[oblique river]

👍 :)

[plucky flicker]

can i have one more quick question?

so, we defined the coordinate ring as k[V] = k[X]/I(V), but it can be defined as k[V] = k[X]/J, where J is the radical ideal of the ideal I, as well because of the strong of the Nullstellensatz, right?

[oblique river]

I(V) is a radical ideal

[plucky flicker]

thanks!

[next obsidian]

@vestal snow the statement that units in a polynomial ring are the non-zero constants is only true for being over a field btw

[vestal snow]

Yeah I figured

[next obsidian]

Over a general ring once can show that a polynomial a_nx^n +... + a_1x + a_0 is a unit iff a_0 is a unit and all a_i for i > 0 are nilpotent

[vestal snow]

The general statement is the units in R are the units in R[x] right?

[next obsidian]

Nope

[vestal snow]

Ah my bad

I remember that exercise from Dummit and Foote

[next obsidian]

It gets really weird with nilpotence

Yeah

Altho tbh I have never once used this fact lol

But it was on an old quals at my school haha

So maybe it’ll end up on my quals eventually

[vestal snow]

Just out of curiosity, have you seen the computation of cos(2pi/5) using galois theory?

[next obsidian]

I don’t think so

I’ve seen sort of a different direction

That some angles aren’t constructible with compass and straightedge using Galois theory haha

[vestal snow]

It's a pretty interesting proof

My algebra professor put it on our final last semester

In general, you can compute cos(2pi/p) for all p prime and p = 2^(2^n) +1

[next obsidian]

Oh huh those are mersenne primes right?

[vestal snow]

Using galois theory and the quadratic formula

Yup

I think that's what they're called

[next obsidian]

Ohhhhh

[sly nexus]

the virgin taylor series vs the chad galois theory

[next obsidian]

I think this might be related to the fact that the uh

Regular n-gons that are constructible are exactly the ones for n a mersenne prime or something

Or uh

A product of mersenne primes

Or something like that

I don’t quite remember but mersenne primes are related to constructibility of the regular n-gon

[vestal snow]

Maybe you're right

I think galois theory and group theory have been my favorite subtopics in algebra

Yet

[stone fulcrum]

Group is bae

[next obsidian]

Groups are nice

I like commutative algebra now

After doing AG

There’s something I find satisfying and refreshing about it

After like... seeing hard stuff in AG just reduce to commutative algebra

[stone fulcrum]

Cannot concur

[sly nexus]

group actions >>

[next obsidian]

I’m not really one that cares about “motivation” per se, but I like applications

[vestal snow]

AG just seems

[next obsidian]

I don’t do well with like “oh commutative algebra is important for AG” I like to see how it’s important

[vestal snow]

Messy

[sly nexus]

applications

[next obsidian]

It is messy lol

Applications with respect to other math Fractal

[sly nexus]

I know

Im joking

I like those kind of applications

[next obsidian]

I’m actually actively turned off when people say it’s applicable to like... bio stats or some shit lmao

[stone fulcrum]

Not as messy as attaching a distance function 😫

[next obsidian]

Banana, are you doing A-M for ANT then?

[sly nexus]

Im sorry my numerical algorithms are applicable to anything bro

[vestal snow]

Instead of using Hartshorne, I'm planning on using Quing Lui's book

Yeah

Though I plan on studying AG too

[next obsidian]

I see. Rip, have to learn AG anyway

I mean to do ANT you learn ANT, and then you also learn AG lmaoo

[sly nexus]

Ill probably never have to learn AG

[next obsidian]

Don’t take a class on it then

[sly nexus]

but I have interest in AT

[next obsidian]

I didn’t think I’d want to do it, then took a class and schemes were too fun for me to quit

Varieties can schlob on my knob tho lol they suck

[sly nexus]

if you want schemes, go to CT

[vestal snow]

I'm considering taking classes in universal algebra as well

[next obsidian]

Lol does your school have one?

[vestal snow]

They offer it as a topics course

[sly nexus]

what is universal algebra

[next obsidian]

Also come to think of it, I wonder what channel universal algebra best fits in for this server

[vestal snow]

I think this one

[next obsidian]

Abstract algebra and universal algebra seem so distant tho

[vestal snow]

Really?

[next obsidian]

Idk, I feel like universal algebra has a more categorical feel from my very very very limited knowledge of it

[vestal snow]

I haven't studied much of it, but someone said that universal algebra is similar to what the first week of a first course in group theory looked like

[sly nexus]

seeing the wiki definition, it seems way cooler than AA

but honestly, yes
this seems the right channel

[vestal snow]

Though you kind of have to restrict yourself with UA

Because you're not allowed to exclude specific elements from having a certain property

For instance, you can't study fields

Because all elements except 0 have inverses

Now that I think about it, perhaps it would fit in the model theory channel

[shy bluff]

Wait question

What is the difference between a prime ideal and a normal ideal? Like in my notes we've defined an ideal as such

But how does that differ from a prime ideal? Like this is our definitiono f a prime ideal

Are these not saying the same thing?

[sly nexus]

nothing says you cant get a,b not in I and get ab in I for an ideal

[shy bluff]

Hrmmm

Oh

So say for exmaple, the ideal 12Z, we have that 3 and 4 are'nt in it but 3 * 4 is?

[sly nexus]

ye

but it cant be a prime ideal

[shy bluff]

Ah

So can I say that any time that I have a non-prime ideal I, that there must exist some a, b such that ab in I but neither a nor b are in I?

[sly nexus]

yep

that is the direct negation

and ofc

a in I, b in R

implies ab in I

(also with the right)

Id tell you the intuition behind prime ideals but Id have to recall them

but it should be immediate any pZ for p prime is a prime ideal

[shy bluff]

Ah

Oke

Yea that's what I thought was but i think it's just kinda confusing for me becauseit's like not just one element but like an entire set?

[sly nexus]

@shy bluff wym

[shy bluff]

Idk I'm having a somewhat hard time grasping ideals/rings/quotient rings/etc

You know how when you learn something you go from like "I know nothing -> I know the words but not the meanings -> I know the meanings but not what they actually mean -> I knoww hat they actually mean but not how to use them -> I know how to use them and what they mean -> I truly understand this topic"

I'm somewhere between "I know the words but not the meanings" and "I know the meanings but not what they actually mean"

[sly nexus]

thats maths

[shy bluff]

Ok yes but this is harder than other maths

[sly nexus]

look at what the ultimate chad told us

[shy bluff]

Oh

I see

[sly nexus]

Im just kidding
try doing more examples

[shy bluff]

It's just kinda hard because uh this course is going really fast and I'm barely keeping up with lecctures which dont' have any examples

[sly nexus]

I assume this is your intro to proofs?
because you said it is harder than other maths

[shy bluff]

Group/ring theory

Err introduction to groups and rings

[sly nexus]

ye
but have you had "proof" subjects before

[shy bluff]

Yes

[sly nexus]

oh

[shy bluff]

I had uh one course on introduction to proofs and proof techniques, and a proof based linear algebra course ... which was fine

[sly nexus]

which book are they using

[shy bluff]

As I could wrap my head around the linear algebra course mostly?

Dunnit and Foote, but not really following it much

Prof teaches off of his own notes kinda

[sly nexus]

well
linear algebra can be extremely hard to ok
depends on how they give it

[shy bluff]

I guess

this is true

[sly nexus]

what part specirically are you having a hard time grasping?

[shy bluff]

I'm at the point where I don't know how to put into words what it is that I don't know

I think that I'm having a bit of trouble at least with ideals and quotients

[sly nexus]

how long has the subject been going for?

[shy bluff]

Uhhhh since May? I've asked the prof for help several times and like.. it helps but the material piles on very quickly and he only has office hours once a week, so I get a rough understanding of the previous week's material and then we move onto the next week and it's based on the previous week's material and so it just gets worse

my uh marks in this course are atrocious due to the fact that I'm basically guessing all the quizzes as they're due before his office hours happen

[sly nexus]

dont rush it

[shy bluff]

And I've like been spamming this channel kinda asking for help with homework questions because uh I'm just not understanding them? Part of it is just because it's online though

[sly nexus]

give yourself more time

you seem to understand the definition

just need to get used to it

[shy bluff]

Yea I think so

[sly nexus]

you immediatelly understood the nuance

youre being unfair to yourself
practice more and youll do well

[shy bluff]

Yea

I think that I need to retake this course when in person courses reopen though

[sly nexus]

I could help, but I rushed af through rings
so Im not sure I could teach

[shy bluff]

Well

I'm just learning as I do my homework

[sly nexus]

follow the book

may take a bit, but you can try

[thin flume]

I agree AA is hard

[shy bluff]

Is this a normal amount

[sly nexus]

true

[thin flume]

I'm retaking the class in the next semester too rip

[shy bluff]

I feel like I'm just googling everything

I guess

No I'm staring at the wikipedia definition of something or reading 10000 stackexchange explanations of something

[sly nexus]

cant say I havent stack exchanged a lot of functional analysis

[shy bluff]

Idk

I guess I'm just not used to this

Also like my marks are bad relative to the class

[sly nexus]

also, wiki can be rushed sometimes

[shy bluff]

So uh I feel like I'm doing something wrong

Oh ok

[sly nexus]

may be helpful to get an online book or smth

[shy bluff]

Like I really want to take more because I think that the topic is very interesting?

But at the same time I'm pulling high 50s low 60s on the weekly quizzes and taking 2-3 days to work on them when theyare supposed to take a couple hours

[sly nexus]

a lot of undergrad subjects were like that

that being said, you need to practice problems and get used to assimilating definitions

getting a definition and being avle to apply it is a slowly developed skill

and depending on the def, it may take very long

[shy bluff]

maybe this is the lack of mathematical maturity that my friend said I had

[sly nexus]

but dont let this discourage you

its just a matter of practice

[shy bluff]

Hrm

Ok well

Question

In order to show that something's a local ring I need to show that it has a unique maximal ideal right?

[timid hull]

Yes

[shy bluff]

And a localization of a ring S^{-1}R means that all of S is the denominators, while all of R is the numerators?

[timid hull]

I mean yeah but there is also a equivalence relation on S^{-1}R ie we say a/b=c/d (where a,b\in R ,c,d\in S) iff there is a s\in S such that s(ad-bc)=0

[shy bluff]

Oh

Wait what? Why the s in S such that s(ad - bc) = 0?

I get the first half

I don't know if I saw the 2nd half

[timid hull]

I mean not all rings we encounter are integral domains that's why we include the second condition ... Like how do we define rationals?

We define them as a equivalence class of a/b such that a/b=c/d iff ad-bc=0 where a,b are integers.

It's some sort of generalisation of this idea

[sly nexus]

basically you want the same fractions to be the same fractions

[timid hull]

We could have just defined the relation for S^{-1}R just like we did for rationals ie a/b=c/d iff ad-bc=0 but the localisation S^{-1}R has the property that every element in the set S is invertible in S^{-1}R so we can "weaken" the condition " a/b=c/d iff ad-bc=0" to "there exist a s\in S such that s(ad-bc)=0" as we can "divide" by s in S^{-1}R this actually matches our usual intution for rationals ...

[shy bluff]

Oh

I think that it's because for us we assumed that R is an integral domain when we defined ring of fractions

wait another question, can I ... mulitply? a ring?

Like say for example I have S^{-1}R, where S = R - P for some prime ideal

can I multilpy S^{-1}R by P?

[sly nexus]

you can do it elementwise

[shy bluff]

oke

Thank you

[sly nexus]

but not sure if that will still be a ring

[shy bluff]

That's ok

I can just have it bel ike

If a/b is an element of S^{-1}R

Then PS^{-1}R is the set pa/b for some a in R, b in S^{-1}, p some element of P

Right?

[timid hull]

Yeah but to be more precise it should be (p/1).(a/b)

[shy bluff]

oh ok

[chilly ocean]

aww man i'm so dumb someone help me out

[thin flume]

@chilly ocean #❓how-to-get-help

That doesn't look AA at all too..

[shy bluff]

Aren't these tow statements contradictory?

Like don't we require that h(c) not equal 0 in order for c to be in D?

[shy bluff]

hrm

Like I'm not really sure what we're supposed to do

A friend of mine says that apparently I'm looking at it backwards?

[11:42 PM]
You don't start with g and h and see what f you get
[11:42 PM]
You start with f and see which g and h you get```

but uh I don't relaly get it still :x

and he gave me the example (x^2 - 1)/(x - 1) = x + 1

what do these words mean

I can't tel leither

I emailed the prof too

Yea

That's my question

Like can you even do that?

Ok but that wont' give you the same domain if g and h are different right?

Like g = x + 1, h = 1 versus g = x^2 - 1 and h = x -1 will give you the same f

Yea like those are two different rational functions

imo

But like they reduce to the same f?

And something about equivalence classes?

No

He says that g/h and f/1 are equivalence clasess??

Well from the notes we have the scerenshot above

And in the definition we've defined f = g/h

Err in the question

Or rather, by the notes, we havet hat f = g/h if and only if (f, 1) ~ (g, h)

Which is true if and only if fh = g1

Yea

Err

Oh wait yes it is

Oh

Yea

Wait lemme check the definition of K(x)

hrm

K[x] is the ring of polynomials over K

Yea

oh wait yea

By the equivalence class definition, x + 1 = (x^2 - 1)/(x - 1)

oh wait

Ok I see

ok yea so we have that (f, 1) is an equivalence class

And any (g, h) that satisfy (f, 1) ~ (g, h) is good, but that each pair (g, h) will create a different D(f)?

😦

wynaut

[cloud walrus]

slimvesus:

[shy bluff]

oh ok

But what if we talk about it as being (f, 1) ~ (g, h) in K(x) instead of K[x]?

dwbi

oh

Ok

oh

Wait I'm still confused, why is 1 still an element of D(g, h) = D(x + 1, 1)? Is it because we can construct a different polynomial such that h(1) not equal to 0?

oooooooooooooooooooooooooooooh

So D((g, h)) is really the set of "c such that there exists some (s, t) ~ (g, h) such that t(c) nonzero"?

....

Ok

I see

Oke

That's all clear now

Thank you!

[river fern]

What is |Aut(Zp x Zp)|

Is it ||2(p - 1)^2||?

||(the number of ways you can map (0, 1) and (1, 0) -- either map them to (0, a) and (b, 0) or map them to (a, 0) and (0, b) for any 1 <= a, b <= p - 1)||

Just want to make sure -- I was wondering if I maybe undercounted

[timid hull]

$(p^2-1)(p^2-p)$

[cloud walrus]

WhyamIsohot?:

[river fern]

Wait, what's the construction?

[timid hull]

Aut(Zp x Zp)=GL_2(Zp)

You can express any map from Zp x Zp to Zp x Zp by a multpilication by a matrix .....

[river fern]

Ah ok

[elder valley]

||(the number of ways you can map (0, 1) and (1, 0) -- either map them to (0, a) and (b, 0) or map them to (a, 0) and (0, b) for any 1 <= a, b <= p - 1)||
@river fern you can map (1,0) to any nonzero element, which there are p^2-1 choices to pick from. Then you can map (0,1) to any element not in the span of where (1,0) was sent, which there are p^2-p choices. In total that gives (p^2-1)(p^2-p) maps

[river fern]

So I guess this could be generalized

Like |Aut(Zp x... x Zp (n times))| would be (p^n - 1)(p^n - p)...(p^n - p^(n-1))

And then something similar for Zp x Zq where p and q are distinct primes

Is there a well-known form for any finite abelian group?

[elder valley]

Probably

You have Aut(G x H) = Aut(G) x Aut(H) when the order of G and H are relatively prime

[river fern]

oh rly

ok I think it's clear that Aut(G x H) contains Aut(G) x Aut(H) (since the same mappings that are done in G and H will be feasible in G x H)

[river fern]

You have Aut(G x H) = Aut(G) x Aut(H) when the order of G and H are relatively prime
@elder valley From the abstract of this (https://link.springer.com/article/10.1007/s00013-005-1547-z), it seems like finding the automorphisms of the direct product of two groups of relatively prime order is nontrival

but maybe it's simpler if G and H are abelian?

[elder valley]

yeah sorry i was assuming the finite abelian case still

although i'm not seeing why it's not true in general. if the orders are relatively prime then any element of G has to map back to G, and same for H. so it seems the automorphism would decompose into the "G part" and the "H part"

i think that paper may be using "direct factor" to mean something else, like some common subgroup or something

[next obsidian]

For relatively prime order it 100% is just the product of automorphism groups

[solemn rain]

why

[next obsidian]

I did this problem last year

And also did it this year

[river fern]

It's definitely true for abelian (lemma 2.1 of https://www.msri.org/people/members/chillar/files/autabeliangrps.pdf)

is it true for non-abelian?

[next obsidian]

Yes

[river fern]

Of course it is true that Aut(G x H) contains Aut(G) x Aut(H) but I'm unsure how to prove the other dir

[next obsidian]

I am absolutely positive

Let G’ be the copy G x {1}

And H’ be {1} x H

[elder valley]

if f is an automorphism, then f(g,h) = f(g,0)*f(0,h), and f(g,0) has to be in G x {0} for any g if the orders are relatively prime. same for H

[next obsidian]

By coprimality any automorphism of G x H has to fix G’ and H’

Yeah basically what Auvera is saying

You abuse the fact that the order of (g,h) is the lcm of orders of g and h

Combined with the fact that order of elements of a group need to divide the order of the group

[river fern]

ahh ok makes sense^^

[elder valley]

basically f restricted to G is an automorphism of G, same for H

and the value of f on G and H determine f completely

[next obsidian]

That’s a great way to put it

[river fern]

ok that makes sense, i'm unsure what this paper is about then: https://link.springer.com/article/10.1007/s00013-005-1547-z

since it uses Hom(H, Z(K)) and Hom (K, Z(H)) to express Aut(G) where G = H x K

[next obsidian]

Yo

Look at theorem 2.1 in that pdf yo

Lmao

[river fern]

oh LMAO

[next obsidian]

The original one

[river fern]

it's not even abelian

🙃

[elder valley]

@river fern so to get back to your original problem, if G is finite abelian, then decompose G into it's primary decomposition and collect terms to put G in the form of G_1 x ... x G_n with |G_i| distinct prime powers. then Aut(G) = Aut(G_1) x ... x Aut(G_n). so the problem is reduced to computing Aut(H) for H an abelian p-group

[river fern]

ok

seems a little bit nontrivial:

where d_k and c_k have this definition:

[elder valley]

yeah looks ugly

[river fern]

Is it easy to show that p divides |Aut(H_p)|?

wait that's false for Zp

I wonder if it's true for everything else

[elder valley]

well given that theorem you can find when p will be a factor

[next obsidian]

I’m pretty sure

Oh hmm

My friend found the order of aut(C_p^n)

But maybe he didn’t actually like, classify it

[river fern]

Oh the formula gives it away

Look at the rightmost product

If any of the ei's are > 1, then since we have that c_i <= n, (p^(ei - 1))^(n - ci + 1) is divisible by p

And obv it's true for a direct product of Zp's

I wonder if there is a nicer way to prove that p usually divides |Aut(Hp)|

[elder valley]

you could try to construct an automorphism of order p

actually as you said, Aut(G) x Aut(H) is a subgroup of Aut(G x H), so if the group contains a copy of Z/p^nZ then the automorphism group contains Aut(Z/p^nZ). now let n>1

[shy bluff]

Given a field K, we have that K(x) contains K[x] which in turn contains K right? and that they're subrings of each other?

[elder valley]

Yes

Also K doesn't need to be a field for that to be true

[little tree]

So the factorisation of the polynomial is given and it's immediately clear that we can form a relationship between the roots and coefficients. The roots of the polynomial are all roots of unity. We know that the coefficient a_n (which we dont know is zero or not) is going to be formed out of the sum of products of roots of unity taken n at a time. Now we must show this can or cannot equal zero.
We know the sum HAS to come out to something real as the coeffieicnts are real. Therefore all terms which are not 1 or -1 cancel out (since 1 or -1 are the only real roots of unity). So now I can write the remaining terms as:
(some big product) * a root + (some other big product) * some other root + ...
Where (some big product) is a root of unity, as multiplying roots of unity gives another root of unity. Every single one of these either comes out to 1 or -1. (ie, I'm only considering the products which are going to come out to something real)
I do not really know where to continue with this line of reasoning, could someone help? I feel like it's kind of a pile of BS

Was suggested to post this here

We now know my idea actually was BS

I'd appreciate any ideas on this

I was relatively sure I could use the roots of unity to solve this

But since there's an infinite amount of them

It's hard

[shy bluff]

Err yea

In this problem though k is a field but yea that makes sense for it ot be true in general

[shy bluff]

for any K[x], where K is a field, can I say that it must have an element of the form (x + a)(x + b), where a and b are elements of K?

[woven delta]

x^2+(a+b)x+ab is always in R[x] for R a ring, not just a field

The polynomial ring is a formal object

[shy bluff]

ok epic

[woven delta]

Like the elements of the polynomial ring happen to represent functions on K, but 2 elements of K[x] which represent the same function aren't equal

It happens to be they are in some nice cases

But the elements themselves are purely symbolic

[elder valley]

@little tree this isn't an algebraic approach so doesn't fit this channel but one idea that comes to mind is that a_n is the nth coefficient of the Taylor expansion of P around 0, so a_n =P^(n)(0) / n!, which you can probably compute easily from P. My analysis sucks so I'm not sure how convergence and passing derivative through the limit will work

[shy bluff]

hrm

ok

for part c here I know that I have to construct some surjective homomorphsim from S^{-1}K[x] to D(c), but I'm not sure how/which one? I thought identity at first but I don't think that that works?

[elder valley]

hmm. i don't see how that is true

[next obsidian]

I think

For localization

You want to show the other thing satisfies the universal property

In general

[elder valley]

seems to me the RHS is K and the LHS is K[x]

@shy bluff what is your definition for S^{-1} R? because usually S needs to be a multiplicatively closed subset of R

[shy bluff]

Yea that' sour definition

That S is multiplicatively closed

[next obsidian]

It is

It’s the complement of a prime ideal

[elder valley]

ooooh, i misread the definition of S, i thought it was quotient

[next obsidian]

So D(c) actually isn’t just K

For example take c = 1

Wait nvm

Wait what is going on hmmm

[shy bluff]

Uh

You fix c

and you pick any g and h such that (g, h) ~ (c, 1) is my understanding

[next obsidian]

Yeah, but D(c) is like the domain

So it’s all of K

[shy bluff]

yea idk

[next obsidian]

Because like D(c) means for the constant polynomial

Then for any x in K,

You have c= c/1

And 1 evaluated at x is not 0

I feel like maybe

[shy bluff]

yea

[next obsidian]

D(c) is supposed to read D(x - c)???

[shy bluff]

no like

you can have x + 1/x+ 1

And that would still equal to c

[next obsidian]

Yeah but

Read that D(f) is a subset of K

It’s the values for which f is represented as a quotient

Where the denominator is non-zero at that value

[elder valley]

D(c) and D(x-c) are both K though

D(f(x)) is K for any polynomial f

[next obsidian]

Oh yeah that’s true

[elder valley]

for part c here I know that I have to construct some surjective homomorphsim from S^{-1}K[x] to D(c), but I'm not sure how/which one? I thought identity at first but I don't think that that works?
@shy bluff maybe try the evaluation map f(x)/g(x) --> f(c)/g(c)

may not be injective actually

[shy bluff]

Yea I don'tt hink that's injective

[elder valley]

wait since D(c) is K, it's a field, but not everything in the LHS is invertible since the numerator can be divisible by x-c

also it's a localization so it's a local ring with maximal ideal corresponding to (x-c)

[next obsidian]

Yeah lmao

[woven delta]

Maybe D(c) means all the elements f of K(x) with c in D(f)?

Otherwise D(c) being a subring of K(x) doesn't make so much sense

[next obsidian]

Oh that makes sense Liquid

That actually makes a lot of sense

Cuz then you’ve inverted everything except multiples of (x - c)

[woven delta]

Yeah

[next obsidian]

In that localization

Geez, if they like, WROTE THAT

[woven delta]

This problem is just bad lol

Maybe there was a previous problem which made this clear?

[shy bluff]

nope

I have no idea

Uhhh last night I talked about this problem here too

As far as I can tell, D(f) means all elements c of of K such that there exist g, h such that g/h = f and h(c) != 0

Also like it can still be a subring of K(x) cant' it? like it's just constant polynomials?

[elder valley]

Otherwise D(c) being a subring of K(x) doesn't make so much sense
@woven delta i don't understand what you mean by this

[woven delta]

I mean by the interpretation you gave it makes sense because K can be seen naturally as a subfield of K(x), but it's kind of boring

And it doesn't seem like the sort of thing we would bother talking about

[elder valley]

yeah i agree with that

seems there's a typo somewhere because from what i can tell (c) can't be true going off of everything that's stated there

[woven delta]

I think my interpretation which I wrote above was what they meant to write

[shy bluff]

Idk

I emailed the prof

He hasn't responded yet

[elder valley]

i think you're saying that in (b) and (c), "D(c)" should be replaced with "rational functions defined at c"

[next obsidian]

Yup

[woven delta]

Yeah

[next obsidian]

Then it’s kind of intuitively clear why that localization is D(c)

Since you’re inverting everything not in (x - c)

Which everything without a root at c

[shy bluff]

hrm

that makses esnse

[river fern]

Given a group G, is it true that all Sylow-p groups are isomorphic to each other?

(since they are conjugate to each other)

Oh wait i think that is true

Since the map x -> g^{-1}xg is an isomorphism

[elder valley]

yeah, the conjugating element gives the isomorphism

[solemn rain]

yes

[shy bluff]

But uhhh I still don't really see how to find a surjective homomorphism between them?

[elder valley]

well given the change we suggested above, it's kind of trivial. you can consider your S^{-1}K[x] to be a subring of K(x), and it's exactly the rational functions defined at c

so like you said at the start it's essentially the identity map

[shy bluff]

Oh

Ok

[elder valley]

you'd need to show that everything in S^{-1}K[x] is defined at c for it to be well-defined

[shy bluff]

hrm

[river fern]

I'm stuck in part (a) (b and c follow trivially):

I know that AB = G, Ba = aB and Ab = bA (for any a in A, b in B)

[woven delta]

||We want to show aba^{-1}b^{-1} is the identity. aba^{-1} is inside B so aba^{-1}b^{-1} is inside B. Similarly ba^{-1}b^{-1} is inside A so aba^{-1}b^{-1} is inside A||

[oblique river]

look at aba^(-1)b^(-1)

smh liquid spoiling everything

[woven delta]

Lol

Read if you want to be spoiled

[river fern]

lol

[woven delta]

I'm not sure where the finiteness comes in at all

[oblique river]

probably so that you can conclude that injective ==> surjective

[woven delta]

Oh sure

[river fern]

ok cool makes sense

[oblique river]

it's not true as stated for infinite groups

[elder valley]

probably the last part: AB=G

[oblique river]

and the exact issue is that the map might not be surjective

[elder valley]

comes from |G|=|A||B|

[oblique river]

I mean, it comes from the map in part b being surjective

for infinite groups, |G| can still equal |A||B| but you don't get AB = G

for example, take G = Z x Z x Z, A = Z x {0} x {0}, and B = {0} x Z x {0}

[woven delta]

Oh I didn't even notice we were talking about Cardinality

[oblique river]

how else would you make sense of "|G| = |A||B|" for infinite groups?

[woven delta]

I thought we were talking about the standard definition of the inner direct product

Like G = AB

[oblique river]

I thought you were asking why the "finite group" hypothesis was necessary

and so I gave an example of how this theorem fails without it

[woven delta]

Serves me right for not reading the question correctly

[river fern]

Unsure how to do (c):

[bleak abyss]

Ur illegible

(Liquid not you arvindr9)

[river fern]

nono, this is a different question lol

For (a), ||you can basically show that there is either exactly one Sylow-3 group or exactly one Sylow-5 group implying that one of them is normal, the product of one group from each category will give a group of order 15.||

[oblique river]

how do you know that those elements commute though?

like if there's a unique sylow 5-subgroup for example, and you take one of the sylow 3-subgroups

call them A and B

[river fern]

Their intersection is 1

AB = BA

[oblique river]

I don't see how that's enough to prove AB = BA

unless both A and B are normal

as in the previous problem we talked about

[shy bluff]

By "everything in S^{-1}K[x] is defined at c", do you mean just show that theree are no elements s of S such that s(c) = 0?

[oblique river]

maybe it's fine, I haven't spent too much time thinking about it

[river fern]

For each b in B, Ab = bA

[elder valley]

By "everything in S^{-1}K[x] is defined at c", do you mean just show that theree are no elements s of S such that s(c) = 0?
@shy bluff yeah that's essentially what it boils down to

[river fern]

And |AB| = 15

Thus, the Ab's are disjoint

[oblique river]

why is |AB| = 15

[river fern]

|AB| <= 15 since each |Ab| has size 5

[shy bluff]

can't I just be like "S is all polynomials of K[x] except for the ones that have c is a root, as if a polynomial would have c as a root then it would be in (x - c)"?

[river fern]

15 | |AB| since |A| = 5, |B| = 3

[oblique river]

why is AB a subgroup of G

if you write something like aba'b'

oh wait I think I see

wait no I don't

why is aba'b' in AB

oh I see, mayvbe it's okay because A is normal

[river fern]

aba'b' = aa''bb' or smth

[oblique river]

so ba' = a''b

ah okay, so you only need one of them to be normal -- which I think you said a while ago but I didn't believe at first

[river fern]

I think for (b), since ||groups of size 15 are cyclic, the candidates are {phi(15), 2phi(15)}||

[elder valley]

can't I just be like "S is all polynomials of K[x] except for the ones that have c is a root, as if a polynomial would have c as a root then it would be in (x - c)"?
@shy bluff yeah that's probably good enough. if you wanted to be more thorough you could use division algorithm to show explicitly that the value is nonzero at c

[shy bluff]

oke

And then I can just say that because it's all s(c) nonzero, the inclusion homomorphism is well defined, surjectiev, with kernel {0} and thus tehy'er isomorphic?

[oblique river]

can there be two different subgroups of order 15?

[river fern]

maybe

there can't be more than 2

since all groups of size 15 are cyclic, implying disjointness besides the identity

[oblique river]

how does that imply disjointness

like, if we call the generators a and b, why can't a^5 = b^5

for example consider the group G = Z/15Z x Z/5Z

the elements (1,0) and (1,1) both generate a cyclic subgroup of order 15

but those subgroups aren't disjoint becuase they both contain (5,0)

now, I know that example has size larger than 30, so maybe you can use the 30 to rule out somethign like this

in fact, the subgroups generated by (1,k) for k in {0,1,2,3,4} are all distinct

and all cyclic of order 15

[river fern]

hmm ok

So one thing I realized -- to solve (c), I think you mainly need the fact that there are >= phi(15) = 8 elements of order 15

There are a few cases:

10 Sylow-3's, 1 Sylow-5: The number of elements in G is at least 8 + 10 * 2 + 4 + 1 = 33

1 Sylow-3, 6 Sylow-5's: The number of elements in G is at least 8 + 2 + 6 * 4 + 1 = 35

Therefore, the group has one Sylow-3 and one Sylow-5

yay

[elder valley]

if R is a ring and S is the group of units of R, then S^{-1}R is essentially the field of fractions of R right? because you're adding inverses to all the noninvertible elements

so what i'm getting at is that even if R is not an integral domain, you can still form the field of fractions in this way

[oblique river]

S is the set of things you invert

so if S is just the units, then S^{-1}R = R

because inverting the units doesn't do anything

[elder valley]

doh

i feel dumb

does taking the compliment of the group of units work? together with 1

[oblique river]

inverting zero-divisors is always a bad idea

https://en.wikipedia.org/wiki/Total_ring_of_fractions#:~:text=In abstract algebra%2C the total,that may have zero divisors.

there's no reason to artificially exclude the units from S

adding them into S doesn't change anything

[elder valley]

oh true. didn't think about 0

[shy bluff]

Question

If (p) is an ideal of R, then (p) contains all powers of p

and all multiples of p

Right?

Does it contain uh the sum of p with elements of R?

i.e, is a + p in (p) for some a in R?

[oblique river]

no

(p) = {ap | a in R}

[shy bluff]

oke

hrm

what is the definition of comaximal?

Oh it's that they equal to R

[elder valley]

the ideals sum to the whole ring

[shy bluff]

And the sum of two ideal sis always a ni deal?

[chilly canyon]

Try to prove it by yourself, it's a good exercice

[sonic current]

proof as an exercise is a very cool thing in algebra

I think it's probably the only subject where it makes me sort of happy

[elder valley]

why algebra specifically?

[sonic current]

I feel I'm way more at ease with the structure

[chilly canyon]

I had a teacher that always said "this is straightforward", meaning it's just a matter of re-writing things and using the definitions

Usually it's just a tiny bit of abstraction but not a lot of huge reasoning

[shy bluff]

wait question, can I say "d | p" for polynomials d, p in R[x]?

[sonic current]

does algebra have an equivalent to epsilon-proofs? ewww

[chilly canyon]

The case of I,J ideals => I+J ideal is the perfect example

[sonic current]

yes @shy bluff

[shy bluff]

That just means that remainder is 0 right

[chilly canyon]

Yes you can

Yes

[shy bluff]

It just feels weird to say that when neither is a number?

[chilly canyon]

P | Q if \exists V s.t. Q=P×V

In any ring

[sonic current]

just accept it @shy bluff

[shy bluff]

Working on it

[sonic current]

i read that as 'divides'

[chilly canyon]

Works for matrices, for holomorphic functions, for many things

[sonic current]

so it's pretty natural

straight up writing the polynomials feels weirder

[elder valley]

if the ring is noncommutative is there a "left divides" and "right divides" though?

[chilly canyon]

@shy bluff to accomodate to these notions, I advice you have a look at the ring of Gauss integers Z[i]

[shy bluff]

Yea

[chilly canyon]

There are some fun exercices with showing it is Euclidian and shenanigans

if the ring is noncommutative is there a "left divides" and "right divides" though?
yes

[shy bluff]

Uhhhh no time to do that or else I would; this course moves too fast + other courses means no time to play with a lot of things

On the other hand I did manage to prove half of this question without any help

So

[sonic current]

yeah fuck math like that

[shy bluff]

That's progress

[sonic current]

math has no schedule

[shy bluff]

Not really the prof's fault, it's the other courses

[sonic current]

do it at your own pace

[shy bluff]

scheduling conflicts

[sonic current]

I know

uni in general

[elder valley]

seems like "A divides B" when A and B are matrices is ambiguous

[shy bluff]

mmm

I learn better in a classroom setting so

[chilly canyon]

seems like "A divides B" when A and B are matrices is ambiguous
This is an example of left/right divisibility

You even have left/right/two-sided ideals

[elder valley]

yeah but your previous comment was "works for matrices"

[chilly canyon]

And left/right group actions

[sly nexus]

a polynomial dividing the other is just straight up them being a factor of the other

[woven delta]

It works up to the group of units obviously

[shy bluff]

question: what are qbar and a_i bar?

[elder valley]

the image of the elements in the quotient ring

[shy bluff]

Image of which elements?

[elder valley]

qbar = q + (p_i)

[shy bluff]

I dont' quite nuderstand what that has to do with alll the different ones

[elder valley]

i don't understand that comment xD

[shy bluff]

Err

I don't really understand what the question is asking me to show

So I'm given a set of coprime elements of R[x]

And a set of a_i... a_n in R[x]

What is the relationship between a_1... a_n?

[elder valley]

there's no dependence between the a_i's

you have to find a single q which when you map it through the projection homomorphism phi_i : R[x] -> R[x] / (p_i), you have that phi_i(q) = phi_i(a_i)

for all i

[shy bluff]

And q is some single fixed element of R[x]?

[elder valley]

yeah, you have to show it exists

[shy bluff]

Oh I have to show that q exists

And it's teh same q for all p_i?

[elder valley]

yep

one q to rule them all

[sly nexus]

true

[shy bluff]

Wait another question, any coprime pair of ideals (p_i) + (p_j) = R[x], doesn't it? Because if they'er coprime they're comaximal?

[elder valley]

that's what part (a) shows

[shy bluff]

Yea ok

That's what I showed 😎

Uhhh is there any reason that I cant' just let a_i = q = 1? THen we'd have that phi(a_i) = phi(q) = 1 regardless of the quotient ring?

[elder valley]

you don't get to choose a_i, they are given

[shy bluff]

oh I'm given some a_i

and we konw nothing about a_i? They're just arbitrary elements that are fixed?

I'm guessing that this has something to do with Chinese Remainder Theorem maybe?

Id ont' really nudertand how to construct it bercause we don't have anything for what a_i are?

[elder valley]

it's exactly the CRT for polynomial rings

[woven delta]

Sorry so divisibility makes sense in UFD's up to the group of units

Sorry I forgot to mention the UFD part

[shy bluff]

hrmmm

[elder valley]

it's kind of a tough proof to come up with on your own

[shy bluff]

So because p_1... .p_n are coprime, we can write R[x]/((p_1) cup .... (p_n))) = R[x]/(p_1) x .... R[x]/(p_n)?

[elder valley]

at least to me it's not intuitive

well yeah but that's essentially the statement of the CRT

[shy bluff]

hrm

Do you have any hints :x

[elder valley]

i think you use induction on n if i remember right

[shy bluff]

oh

Is this some sort of known proof

:x

[elder valley]

in the sense that it's the proof of the CRT yeah xD

[shy bluff]

:x

Is this qusetion basically aksing me to prove CRT?

[elder valley]

the version for polynomials rings yeah

try to do it for n=2

[shy bluff]

Yea idk what to do about a_i and q

like all that I got uh is that

p_1, p_2 are coprime, so (p_1) + (p_2) = (d) = R[x]

And as (d) = R[x], we require 1 in (d)

and that there must exist some s, t in R[x] such that sp_1 + tp_2 = 1

[elder valley]

yeah that's good so far. look at what happens when you mod that equation by p_i

[shy bluff]

The equation sp_1 + tp_2 = 1?

[elder valley]

yeah

[shy bluff]

It just becomes uh tp_2 + 1 right? because the p_1 portion goes to 0?

[elder valley]

t p_2 = 1 you mean

[shy bluff]

Err yea

Tha tshould've been "tp_2 - 1 = 0"

Which is the same thing

[elder valley]

so instead of getting 1, we want to get a_i

so you can modify the expression "s p_1 + t p_2" so that you get a_i mod p_i, given that you know s p_1 mod p_2 = 1 and vise versa

[shy bluff]

Hrm

So then wouldn't that be like... q_1p_1 + q_2p_2 + .... + q_np_n = a_1a_2....a_n?

[elder valley]

what are the q_i?

[shy bluff]

Well I'm thinking that they're all relatively prime right

So we should be able to go and find some q_1, q_2... so on

such that they sum to 1? But instead make the sum a_1... a_n?

as they're relatively prime they have to be able to generate the entire ring

So there must be some combination of q_1... q_n in R[x] such that the sum of them equals to any element in R[x]

It's like if they're relatively prime then they span the space of R[x] right?

[timid hull]

Wait first try the case n=2 .... You will get the idea of what to do from that .....

[elder valley]

yeah i agree

[shy bluff]

well from the case n = 2 we had sp_1 + tp_2, if we go and take quotient of phi_1, we get tp_2 = 1

Right?

And I want to somehow modify this so that I get a_i mod p_i?

[timid hull]

Yeah

[elder valley]

yeah

[shy bluff]

so what I want is that I want phi_1(sp_1 + tp_2) = a_i mod p_i

[elder valley]

phi_i

[shy bluff]

Err

$\phi_1(sp_1 + tp_2) = tp_2 = a_1 mod p_1$

[cloud walrus]

Liria ^(;,;)^:

[shy bluff]

Is this what I want?

[elder valley]

yeah. in the i=1 case the sp_1 term is killed so effectively you can ignore it when trying to make it come out to a_1 instead

so basically, modify "t p_2" so that you get a_1 instead of 1

[shy bluff]

can't you just make it a_1t?

[elder valley]

yeah

[timid hull]

not quite but close

[elder valley]

guess it depends what "it" was referencing lol

[shy bluff]

tp_2 becomes a_1tp_2?

[elder valley]

yes

[timid hull]

yup

[shy bluff]

or rather just mulitply everything by a_1

bceause it doesnt' matter what you multiply sp_1 by, when you quotient it the term poofs

[elder valley]

it doesnt matter for the p_1 quotient

but it will effect the p_2 quotient

[shy bluff]

Yea it on ly affects the p_2

Am I ... trying to shift a_i around such that we have something of the form a_ip_jq_j, where we would have q_1p_1 + ... q_np_n = 1, and wehave i not equal to j?

[elder valley]

the indices will get mixed a bit. but i wouldn't really worry about the general case if you use induction. finish out the n=2 case and the induction step won't be that different

induction also gives a constructive proof for how to find q

assuming you use extended euclidean algorithm to find the bezout coefficients

[shy bluff]

D: I can't just say that they exist because they're coprime?

[elder valley]

in this proof you can

but there's a constructive way to compute them also

[shy bluff]

Ok so we have

[elder valley]

"a_2 s p_1 + t p_2 = a_2" is not true

but you don't need that

[shy bluff]

I see

[elder valley]

you're almost there

you are applying phi_i to different elements though. you need a single element q

[shy bluff]

oh

Oh wait what if I just multiply the entire thing by a_1 or the entire thingby a_2? that would givee us what we're looking for?

[timid hull]

no

[shy bluff]

Wait question, is this bit of the CRT on wikipedia relevant

oh wait

oh

This is actually what the question is asknig for

[timid hull]

Yes lol

[shy bluff]

q bar is my x

ok

[elder valley]

yeah that's for integers but the proof is pretty much identical

their x is your q

[shy bluff]

oh ok

Now I see wha tit's asking for

[elder valley]

i'm pretty sure the induction proof won't use the fact that your ring is K[x]. i think you just need commutative with 1

[harsh steeple]

I was wondering if someone can help explain to me the Lie vector space isomorphism between the tangent space vectors at identity and L(g) (the vector field of left invariant vector fields over the Reals)? Im confused as to how the isomorphism is made.

[hot lake]

you take f in L(g) and you map it to f(identity)

[harsh steeple]

then how would you go in reverse?

Because if it is a vector space couldnt the vector be written as a composition of two different actions?

[hot lake]

.... what ?

you go in reverse

by taking a vector x in the tangent space of the identity

and mapping it to the field

that takes a point g

and attaches to it the vector T_id(g)(x)

where T_id(g) is the differential at the identity of the multiplication by g map

maybe I shouldn't denote that multiplication by g map as 'g'

well, if g=e

the multiplication by e map is the identity

and so the differential at the identity is the identity of the tangent space at the identity

so you attach x to e

my god that's an awful sentence

[harsh steeple]

I guess what im confused about is, how do you know a L(g) field exists that includes that specific vector at identity and how do you know it is unique?

[hot lake]

write down what it means to be a left invariant vector field

[harsh steeple]

That for all g within G it takes whatever the point gets mapped to and attaches the tangent vectors it originally had?

[hot lake]

can i have something much much more precise

[harsh steeple]

Sure X as a field is left invariant if for all g within G, Lg.(X(h) = X(lg(h))

Where Lg. is the pushforward

[hot lake]

I think you forgot to quantify h

[harsh steeple]

Where h is a point on the smooth manifold that G sits on?

[hot lake]

and lg is left multiplication by g ?

[harsh steeple]

yes

[hot lake]

I don't usually make a difference between G and the smooth manifold

well then if you have decided on a value for X(e)

this constraint " forall g,h in G, Lg.(X(h)) = X(gh) " tells you, by plugging h = e, that forall g in G, X(g) = Lg.(X(e))

so you have no choice for X(g) once you have decided a value for X(e)

[harsh steeple]

So then what makes the isomorphism injective? How can we map a vector at identity to a distinct vector in L(g)?

[hot lake]

what

[harsh steeple]

Any tangent vector at identity, a element of L(g) will map it to itself correct?

[hot lake]

an element of L(g) is a left invariant vector field

[harsh steeple]

yes

[hot lake]

if it's a map it's a map from G to the reunion of the tangent spaces of G

I'm not sure I'm making sense of what you're saying

[harsh steeple]

You said that forall g,h in G, Lg.(X(h)) = X(gh) right? So that means for any g which is an element of L(g), Lg.(x(h)) = X(g(h)) right?

Which means ALL elements of L(g) map the same vector to itself correct?

[hot lake]

what the

why would you call g an element of L(g)

[harsh steeple]

g as the left invariant vector field which is an element of the set of all left invariant vector fields

[hot lake]

if you give the letter g to the lie group, to elements of the lie group, and to vector fields on the lie group then you're going to be confused

[harsh steeple]

Ok

So then g as an element of L(G)

[hot lake]

so g is a vector field on G ?

[harsh steeple]

yes

[hot lake]

can you call it X?

instead of g ?

[harsh steeple]

Sure

[hot lake]

so we can have "forall g,h in G, Lg.(X(h)) = X(gh)" ?

[harsh steeple]

Just to clarify g and h are group elements of the lie group G?

[hot lake]

yes

[harsh steeple]

We defined that for a left invariant field all vectors in the field are mapped to themselves correct?

[hot lake]

no

a vector field on G is a map X from G to the reunion of the tangent spaces, such that forall g in G, X(g) is an element of the tangent space of G at g

forall g, the multiplication by g map is differentiable, so forall g and h, you can differentiate it at h to obtain a linear map from the tangent space of G at h to the tangent space of G at gh

and the field is left-invariant if when you apply it to X(h) you obtain X(gh)

for any g and any h

[harsh steeple]

So wherever a point gets mapped to it is guaranteed to have its same tangent space (the map may move points around but the reunion of the tangent spaces is identical to their original points)?

[hot lake]

I have no idea what you are saying

a point gets mapped - by a vector field - to a tangent vector

a point gets mapped - by the multiplication by g map - to another point

a tangent vector gets mapped - by the differential of blablabla - to a tangent vector from another tangent space

so since there are many maps around, please clarify which ones you are talking about

[harsh steeple]

Sure, sorry I was just introduced to the topic today

A left invariant vector field is such that what a point is multiplied by g map it is sent to another new point.

But

Its tangent vector at that new point is identical to what it was prior to the multiplication of g.

[hot lake]

no

a left invariant vector field is a map X from G to the reunion of the tangent spaces such that it satisfies (differential of (left multiplication by g map) at h) applied to X(h) = X(gh), forall g and h in G

[harsh steeple]

So a tangent vector at a point before and after g map transformation is identical?

[hot lake]

not sure what that means

[harsh steeple]

At each point on G a tangent space is given with a vector given by the left invariant vector field

[hot lake]

usually when you transform somethings it is no longer identical to what it was before

well a vector field is the data of a tangent vector for each point yes

[harsh steeple]

Ok so if you transform G by the map g each point will be transformed correct?

[hot lake]

the "left-multiplication by g" map is a map from G to G, it maps h to gh

[harsh steeple]

ok so every point h gets sent to gh then?

[hot lake]

I suppose ?

[harsh steeple]

and whatever tangent vector was at h gets sent to point gh?

As per the definition of left invariant?

[hot lake]

no

the differential of that multiplication by g map

at h

is a linear map from the tangent space of G at h to the tangent space of G at gh

the field is left invariant if forall g,h in G, Lg.(X(h)) = X(gh)

I really don't like your notation Lg

[harsh steeple]

use whatever youd like my dude, im just appreciative that your helping me out!

[hot lake]

maybe it should be T(mu_g)_h or something

mu_g : G -> G

defined by mu_g(h) = gh

T(mu_g)_h : the differential of mu_g at h

is a linear map from T_h to T_gh

if X is a vector field

X(h) is in T_h

you can apply T(mu_g)_h to it

and obtain T(mu_g)_h (X(h)), which is in T_(gh)

you can also consider X(gh) which is in T_(gh)

if it somehow happens that X(gh) = T(mu_g)_h (X(h)) for every g and h, then you say X is a left invariant vector field

(also T_g is my notation for the tangent space of G at g)

[harsh steeple]

Ok

So if X(h) = X(gh) after the pushforward it is left invariant?

[hot lake]

if X(gh) = T(mu_g)_h (X(h)) it is left invariant

[harsh steeple]

The analogy im thinking is like an eigenvector with eigenvalue 1, that once the transformation is applied it remains the same

[hot lake]

absolutely not

[harsh steeple]

So what is wrong with the analogy?

I think that might be what im stuck on

[hot lake]

it's a bad one

well

hmm

you can push the whole vector field

say, (push_g(X)) is the vector field Y defined by
Y(h) = T(mu_g)_(g-1h) (X(g-1h))

then you can say X is left invariant if push_g(X)=X forall g in G

I think

something like that

lemme check if that makes sense

yeah

and like that you can define an action push of G on the set of vector fields

(i don't remember if it's a left action or a right action)

and then left-invariant vector fields would be vector fields that are fixed by that action

[harsh steeple]

So then for left invariant vector fields, every vector when pushed forward stays the same? As in the vector doesnt change from T_g to T_gh

[hot lake]

you can't compare vectors from Tg with vectors from Tgh

it makes no sense

you can't say they are equal nor that they are different

well it's more like they're different because they're from entirely different vector spaces

[cloud walrus]

Have a Banana:

Have a Banana:

Have a Banana:

[next obsidian]

Suppose all r_i < n_i then their sum would be less than m

Like m is defined to be the minimal number such that this proof works

Because each r_i <= n_i - 1

But m = sum(n_i - 1) + 1

[vestal snow]

does m = sum(n_i) not work?

[next obsidian]

It will

Any m larger than theirs does

But for like elegance they went for the minimal m to make the proof work

[vestal snow]

I still don't see how their choice of m works

[next obsidian]

So suppose all r_i < n_i

[vestal snow]

like assume r_i < n_i for all i

[next obsidian]

Then r_i <= n_i - 1

We required that the sum of r_i = m

[vestal snow]

Ah got it

[next obsidian]

But m = sum(n_i - 1) + 1

[vestal snow]

What I was doing was that I was summing it up before converting it to <=

[next obsidian]

Oh lmao

But from there

You can see why this is the minimal m

Since you get over <= by exactly 1

To push it into <

If that... makes sense lol

[vestal snow]

It does

Thanks!

Also, this is the problem I was talking about @next obsidian

[next obsidian]

Aiyaiyai

[vestal snow]

This final was super fun

Here's another cool problem from it

[olive mirage]

Finite Field stuff is just so cool

[olive mirage]

if only we could all be so lucky

[next obsidian]

I hate finite fields

I would literally rather work with a general field

Even though then it applies to finite fields too

Only because finite fields are so nice that you can say more concrete stuff about them

But I just hate working with them

[oblique river]

lmao

[olive mirage]

I love finite group theory, so the fact that I get to throw all that around is fun

[next obsidian]

I like finite group theory

[olive mirage]

I vividly remember one of my professors saying "Finite group theory is nothing but combinatorics"

and at the time I did not believe it.

[oblique river]

everyone knows that combinatorics is just algebraic geometry over F_1

[olive mirage]

hahaha

I think I would mean "is" just as in, literally that's what combinatorics is. I think my combinatorics course is a lot of "how much can I have a semester long group theory course without ever saying the word 'group'?"

[oblique river]

haha yeha

what I said was just a meme

which maybe I would understand better if my brain were big enough

[olive mirage]

n! / (k! * (n-k)! ) is just the orbit stabalizer theorem applied to three symmetric groups

[oblique river]

I mean, I guess it kind of makes sense, since finite sets and group actions and stuff are vector spaces over F_1

[olive mirage]

gross!

[next obsidian]

F_1

Oh brother

😓

[woven delta]

Abelian groups are just modules over the trivial ring

[oblique river]

abelian groups are Z-modules