fierce perch
Im not sure hilberts syzygy-treatment classifies as modern resolutions?
#groups-rings-fields
tacit saffron
im really confusion
i have very loose understandin of fields
but i be trynna do this proof: prove that for all primes p, there exists a field of order p^2
but have no idea where that ^2 boy comes from
maiden ocean
hi
its like
F is a field u do
F[x]/(f)
actually
fuck i have to do work i cant help rn sorry 
tacit saffron
lmk whos giving you work i can swiftly ✨ remove them
maiden ocean
stone fulcrum
I can but I'm concerned you won't know some of that stuff oop
F[x]/(f) is a field if f is irreducible?
tacit saffron
lemme go back into my notes i might just be a little monkey brain
brb
yesss we did irreducibility stuff
stone fulcrum
Right! Let's use the field of order 2 as an example. That is, the field with just 0 and 1. Note that 1 + 1 = 0 for that to make sense
tacit saffron
ok yes
stone fulcrum
I've locked myself into a bad example nvm. Consider the field of order 3 instead rofl
As a polynomial over it,
x² - 2
is irreducible. If it was reducible, then one of its roots would belong to Z3. And ±√2 is not in Z3
tacit saffron
this might be very stupid but how is Z3 diff from Z
stone fulcrum
Z3 is sometimes referred to as the field of order 3. Really, I should be using GF(3) to refer to this field
tacit saffron
oki
stone fulcrum
Letting F be that field
F[x]/(x² - 2)
is a new field of order 3² = 9
The elements look like ax + b
Where x acts like √2. That is, x² = 2
tacit saffron
why squared
stone fulcrum
Basically, modding out x² - 2 is a way to fix the algebraic rule that x² = 2
So now you have ax + b because all of the higher order terms go back to lower order terms
Like, in this new field, x³ = x(x²) = x(2) = 2x
So you can't have x³ in this field
I mean you could, but it would just be 2x haha
stone fulcrum
So all elements look like ax + b
There's 3 ways to set a (0,1,2)
There's 3 ways to set b (0,1,2)
That's 3² = 9 possible choices
0
x
2x
1
x + 1
x + 2
2x
2x + 1
2x + 2
That's all 9 lol
And this is a field! Which is nice.
stone fulcrum
Nah I'm cool I like this construction
Now for the genius part. This works for 3² because there was a number p such that x² - p was irreducible.
In the case of GF(3), the number 2 can't be reached by squaring a different number
Is there a way to guarantee this will always work?
tacit saffron
p bein prime?
stone fulcrum
Nah just p being any number
But x² - p being irreducible
Ergo p is not the square of any element in the field
Take any field GF(n)
Take the function:
f(x) = x², GF(n) → GF(n)
It is enough to show that this function is not surjective!
Since the input and output have the same amount of elements, this is easy to show by one of the outputs repeating:
f(1) = 1
f(n - 1) = n² - 2n + 1 = 1
Ergo for n ≥ 3, there exists a number w that isn't the square of some element in the field and x² - w is an irreducible polynomial that can be used to construct the field of order p²
dim escarp
Hm strange that while I'm introduced to the fields in abstract algebra class yet prof did not mention GF yet
scarlet estuary
GF(n) is just the finite field of order n
it stands for "galois field", for reasons that will become clear when you cover galois theory
but people often just say "finite field" instead
since there's precisely one galois field of every finite order of the form p^k for a prime p [up to iso]
(and none for natural numbers that are NOT of the form p^k)
latent anvil
me if I didn't know the theorem about existence of finite field of prime order: tfw no GF
chilly ocean
but you're gay
chilly ocean
demonstrated by my gt midterm 
maiden ocean
thorn delta
engineer?
maiden ocean
(pejorative)
latent anvil
homo (pejorative)
latent anvil
faggot (complimentary)
chilly ocean
do my rg for me
latent anvil
ughh
chilly ocean
maiden ocean
i wonder what straight ppl think when they see gay ppl communicating
latent anvil
"wow, fucking queers"
carmine fossil
homo sapiens
maiden ocean
i asked what heteros think not what i think shamrock
chilly ocean
alright so like there's a second derivative on the left side so curvature should magically pop out on the right yeah
latent anvil
Yeah ttera I'm going to ignore that
chilly ocean
😌
maiden ocean
psa
latent anvil
It looks like shit :+1:
maiden ocean
twink is not a slur but if i ever called you it i meant it as one
chilly ocean
make a few twitter posts and you can turn it into a slur
latent anvil
lol
maiden ocean
shut up twink
maiden ocean
irrelevant
latent anvil
ttera twink confirmed??
latent anvil
ttera furry??
maiden ocean
a metaphysical thing
latent anvil
lol
latent anvil
ttera furry??
cub
chilly ocean
not in that sense
maiden ocean
i cant be called a twink because i am still a minor
chilly ocean
maiden ocean
get cucked
latent anvil
lol
sturdy marsh
destroyed
latent anvil
what thing??
chilly ocean
latent anvil
LMAO
chilly ocean
maiden ocean
THATS SO GOOD
latent anvil
I have no idea
chilly ocean
its so good
carmine fossil
Ok, Let's do a test
latent anvil
Is it homophobia,if I say twinks should fight?
maiden ocean
screencapped
latent anvil
carmine fossil
Ok,That gets deleted
latent anvil
Interesting
carmine fossil
So,The pin is just a link,not a copy of that message
maiden ocean
yea
maiden ocean
i like that me sham and tterra have just seized #point-set-topology and now #groups-rings-fields as our shitposting grounds
latent anvil
But early!!
maiden ocean
but hom alg cool : (
latent anvil
you make a good point
maiden ocean
please? 
latent anvil
So like
I have taken multiple classes before 10am
And every time I end up like, not attending past week 6
Because I'm alseep
sturdy marsh
im also doing a hom alg class next quarter lol
latent anvil
nice!
maiden ocean
i wish i could choose classes like that
i wish i could just skip school if i didnt feel like going
latent anvil
Maybr this will get me back into Hartshorne
Lol, i started doing that when I was your age
In junior year of hs I stopped attending courses for the most part
maiden ocean
based
latent anvil
I show up...sometimes
maiden ocean
unfortunately truancy laws are a thing
latent anvil
Cringe
maiden ocean
yes
latent anvil
Lectures are just so fucking boring
chilly ocean
laws
latent anvil
90% of the time
maiden ocean
cringe indeed
hs just fucking sucks
i am not having fun
admittedly life in general sucks rn but hs specifically
carmine fossil
Why is hs graduation like a big deal in the us
chilly ocean
hs graduation was one of my favorite moments
because i was fucking done hs
highschool was atrocious
hated every second
maiden ocean
instead i am attending hs while my family is traumatized in the background monkaS
maiden ocean
it is unironically
maiden ocean
2 days ago i learned that my sister is an alcoholic so i just read hatcher all day instead of processing my emotions
chilly ocean
prof talked about
maiden ocean
and it worked
chilly ocean
was it fun
maiden ocean
yes
chilly ocean
did u learn anything neat
sturdy marsh
noice
maiden ocean
i am rereading now to make sure i understand
chilly ocean
one of the proofs brought up this thing called like
orientable double cover
i had never seen it before
maiden ocean
oh thats cool
sturdy marsh
yup, literally
maiden ocean
sturdy marsh
well connected manifolds are path connected so........
still no canonical choice
chilly ocean
i cant moth i have 4 classes 
maiden ocean
sad
sturdy marsh
Moth if you like algebra you might like May
maiden ocean
what is 10/10 cope sham
latent anvil
chilly ocean
im behind in two of them because
- i spent the entire weekend preparing for the alg midterm
- i spend all my time on RG cause i like it the most
maiden ocean
mhm
im planning on running thru AT again via tom dieck at some point @sturdy marsh
cuz i hear he does a lot
sturdy marsh
tom dieck is also a nice book
maiden ocean
i want to make sure i understand this stuff very well b4 i move on to more advanecd things
sturdy marsh
probably a good idea
maiden ocean
shamrock the problem is that thats not a good cope at all when ur coping with ur sisters substance abuse
its an anti-cope
latent anvil
lol that is a good point
maiden ocean
because it just ties back into what u are attempting to cope
the dialectic does not sublate
latent anvil
This also happens with math tho
like when I am angry at my math class
I cannot fix that by doing math
latent anvil
I would probably have felt the same if I knew as much math as you do
maiden ocean
hurb
sturdy marsh
I hated math in hs, but I didnt know much either lmao
latent anvil
this time 4 years ago I was doing calc 2, not Hatcher lol
latent anvil
tfw
maiden ocean
thats what i did most of middle school
latent anvil
I mean I say 4 years because I am 4 years older than you
maiden ocean
yes.
leaden finch
can someone help me with this one
i checked for these things
i was struggling on showing injectivity and surjectivity
golden pasture
theres a more general problem in elementary number theory:
x = a_ mod n_i with n_i coprime, find x mod prod(n_i)
||chinese remainder theorem||
dim escarp
I wonder if group theory is like over 2 millenia old
scarlet estuary
well, the modern definition of a group was only given in the early 1900s
chilly ocean
fucking idiots in the 1800s didn't even know what a group was
scarlet estuary
but i'd imagine people have intuitively noticed that the dihedral group of the triangle is kind of like the permutation group on 3 elements for a while
@chilly ocean well they kinda did
they had galois shit
they just didnt have the modern conception of a group
chilly ocean
im joking
scarlet estuary
in fairness they didnt have the modern conception of a function either
so give them a break
chilly ocean
i'm sure in the 2100s they'll be working on some shit that's way beyond us now
what's gonna be the 2100s equivalent of a group or a function
dim escarp
Lol
maiden ocean
oh ok
@marsh fractal um ok basically the key here is that like
since the division algorithm still holds
we can pretty easily show that Z/nZ is a PID
marsh fractal
ok
ocean ermine
hey, what can i know about the sum of a finite generating set of finite abelian group G?
light tusk
Or for any composite number modulus
sturdy marsh
yeah so in particular, not a PID
ocean ermine
oh i was wrong in my question
i meant - what can i know about the sum of the orders of the generating of G
maiden ocean
agh yeah sorry i just defaulted to PID its just a principal ideal ring
or principal ring
whatever you want to call it
light tusk
@ocean ermine The simplest case is a cyclic group. a generator would have order of the group. However, there can be more than one generating set - if you already have a generating set, and you include some other superfluous element, then you get another generating set
golden pasture
since the division algorithm still holds
more generally cuz Z/nZ is a quotient of Z and all ideals in Z is principal, all ideals in Z/nZ are principal
maiden ocean
mhm
ocean ermine
<@!436270854578503680> The simplest case is a cyclic group. a generator would h...
hey 🙂 what i asked is, if i know that in G (finite abelian) i have <g1,g2,...gk> as generators, what would be the sum of O(gi) ?
if k =1 it is O(gi) = O(g1) = |G|
maiden ocean
i dont even remember the proof for that but im assuming it follows from the correspondence thingy or whatever its called
of ideals of the quotient ring
maiden ocean
yea
light tusk
@ocean ermine so if ${1,a,b,c}$ is a cyclic group of order 4, then any subset except for ${1}$ is also a generating set
cloud walrusBOT
Apopheniac:
light tusk
So the possible sums here are: 4, 4+4, 4+4+4
ocean ermine
So the possible sums here are: 4, 4+4, 4+4+4
but i am asking about the orders not the value
light tusk
all of those 4's i typed are orders of elements
ocean ermine
if it is a cyclic group there is only one generator. it can be a b or c, but it is 1
calling <a,b,c> generator set is abuse of my notation
light tusk
that is if the set of generators is chosen to be a minimal set of generators
ocean ermine
yes, i want the minimal set
and i also change my question to be the multipication of the orders, not the sum
is Multi(O(gi)) = |G| ?
light tusk
Hmmm, well have you gotten to classification of finite abelian groups?
ocean ermine
i did, long time ago and i remember not a lot, sadly
i know that every abelian group can be represented as multipication of cyclic groups
and those "g1,g2,g3" of the minimal generators are the generators of the respected cyclic groups
light tusk
Yeah, so I think that would help. You have the cyclic case, now see what happens with the direct product of two cyclic groups
light tusk
yes
I kept having to delete stuff realizing you had already said what I was about to say, lol
light tusk
yay!
carmine fossil
If (G/A) $\cong$ B ,can I always say
G $\cong A \cross B$?
cloud walrusBOT
DrunkenDrake:
carmine fossil
G,A and B are groups
carmine fossil
Example?
maiden ocean
do u know about short exact sequences
carmine fossil
Somewhat
maiden ocean
so like nice property of SES is if u have 0 -> A -> B -> C -> 0 then u get C iso B/im(f) where f is the map from A to B
so theres a specific kind of short exact sequence called a split exact sequence where u get smth even better:
B = C oplus A
see the splitting lemma
carmine fossil
ok
maiden ocean
for a counterexample try like um
S_3/A where A is the alternating subgroup
this is isomorphic to Z2
we know this doesnt work bc the alternating subgroup for S_3 is abelian as is Z2 but S_3 isnt
so Z2 oplus A cant be iso to S3
golden pasture
look up group extensions
glossy yoke
Splitting is a little bit delicate for groups... the usual criterion for modules doesn't analogize perfectly. If you have a SES of groups like 1 -> A -> B -> C -> 1 and you have a splitting map B -> A, then B = A x C. But if you have a splitting map C -> B, you don't always get a direct product of A and C. Sometimes you get a semi-direct product.
next obsidian
The issue is the image of C under that splitting map might not be normal
Also, separate question,
In A = k[x_0,...,x_n] let I be the ideal generated by all elements of total degree r and higher. Why is A/I Artinian?
Hmm maybe I can consider it as a module over itself and show it has finite length or some shit?
golden pasture
Also, separate question,
In A = k[x_0,...,x_n] let I be the ideal generated by a...
A/I is a f.g. k algebra here if im reading correctly
next obsidian
Yup
It’s also noetherian
I don’t know of a nice criterion for a finite type Noetherian k algebra to be Artinian though 😥
Wait a second...
F
Ari I think you meant it’s module finite... right?
@golden pasture
Since it’s generated by all the monomials of degree < r
golden pasture
yes
golden pasture
literally same qn in ags then XD
next obsidian
Okay dope that makes sense LOL
I read that but didn’t get what it meant by finite dimensional over k
LOL
golden pasture
XD
next obsidian
Like as a vector space?
I think that follows from classification of Artinian rings right?
Blah blah sum of local Artinian yadda yadda
golden pasture
a quotient of k[x_1, x_2,..., x_n]
the if f.g. as k algebra then
artinian <-> f.g. as module
comes from like nullstellensatz ish
(ok a bit sloppy lol
f.g. k algebra: quotient of k[x1,x2...,xn]
finite k algebra: finite (as module)
)
next obsidian
At some point me and shamrock classified finite type artinian k algebras
next obsidian
Yeah
We phrased it in terms of what the ideal you quotient by satisfies IIRC
It was kinda messy and not too useful sounding LOL
Anyway thanks that should’ve been more obvious lol
I needed this to show the saturation of a homogeneous ideal in that ring is equal to the normal ideal in high enough degrees
golden pasture
icic
next obsidian
The trick is basically
Since I^sat is f.g. You can find an r so that S_rI^sat is a subset of I
Then it holds for all k >= r
So we can consider I^sat/I as a module over S/S_>=r
And it’s finite
So if S/S_>=r is Artinian so is I^sat/I
But an Artinian graded module has degree 0 for high enough stuff
By just looking at the submodule of stuff generated by degree >= 0,1,2,3,4,... in turn
But then I^sat = I in high enough degree
And then you pog out
maiden ocean
chmonkey quick question
next obsidian
Yea
next obsidian
With B 0????
maiden ocean
mhm
next obsidian
Is this a trick question I feel like that Tor should always be 0
maiden ocean
yea makes sense to me
maiden ocean
right
oh maybe i messed up computing homology bc then that would get that this homology is all zero
oh jeez
hm
next obsidian
Rip lol
maiden ocean
oh i think i just got a super cursed case where kunneth theorem tells us literally nothing bc everything around the center homology is zero
maiden ocean
this makes no sense how is the homology all zero
i have to have fucked up somewhere
@next obsidian wait what about the case where B = Z for Tor
next obsidian
yup
maiden ocean
FUCK YEAH
next obsidian
It's naturally iso to identity
so to the strongest degree possible it does nothing
golden pasture
was thinking Tor_1^R lol and read the chat and was super confused
golden pasture
proposition: Z=R for every ring R
maiden ocean
golden pasture
mmmmm
golden pasture
what a chad
golden pasture
when are you publishing this result
next obsidian
write R as a decimal
maiden ocean
Yes.
next obsidian
when are you publishing this result
Right after my paper on astounding properties of bounded entire functions gets accepted
maiden ocean
currently i am bored in linear
maiden ocean
that sounds even more boring
next obsidian
You need it
maiden ocean
hurb for what
next obsidian
Nope
next obsidian
You shouldn't be allowed to do AT
next obsidian
without being able to do the multivariable chain rule
maiden ocean
next obsidian
You need to be punished
golden pasture
did sloth come on discord and like
next obsidian
with computations
golden pasture
wow at ppl
next obsidian
Max
next obsidian
it's Max
golden pasture
i blame max tbh
maiden ocean
its not max
golden pasture
maiden ocean
i did linear algebra
next obsidian
You just think that Moth
next obsidian
It was Max
maiden ocean
it was actually dami
next obsidian
@bleak abyss is this true?
next obsidian
Or is Sloth simply ing your name
next obsidian
"he made me read a calculus book before I did multivariable"
maiden ocean
he made me do H&K too
which is why i became an algebra memer
well he made me read H&K and then made me read jacobson
and THATS why i became an algebra memer
golden pasture
hartshrone have caused me to become a algebra memer
i blame hartshrone
for making me learn com alg
next obsidian
I'm an algebra memer and I'm doing topology and did Complex anal and will do reals next year
maiden ocean
what would i even use multi for
next obsidian
For not being a dum dum?
maiden ocean
why would i care about that
golden pasture
to learn about manifolds?
next obsidian
Imagine if your prof
maiden ocean
i just want to compute cohomology
next obsidian
is trying to talk to you about derived whatevers
golden pasture
i just want to compute cohomology
truely wasted
maiden ocean
it wont even matter im going to learn it before i go to uni
i have to learn it at my hs
so
golden pasture
:o
next obsidian
Hurb
maiden ocean
cuz i ran out of other math classes
golden pasture
teaches actual multivar
maiden ocean
if u mean like multivariate analysis obv im going to learn that eventually
maiden ocean
why would i need to know that for AT i dont need any analysis memes
i told u im doing rudin soon
golden pasture
i actually was surprised you havent done analysis lol
maiden ocean
hurb
maiden ocean
based
im not good at analysis though i think i need to actually read
but i think ch2 will still be boring because its just topology on R
maiden ocean
yea thats where i got bored
cuz i was like "wow special case of point set"
"great"
golden pasture
then skip the damn chapter
maiden ocean
hurb
maiden ocean
i dont know the notation
but i will read it soon!!
stop bullying me im getting there
next obsidian
SAD!
maiden ocean
ur sad
golden pasture
ill bully you for not knowing complex anal next
maiden ocean
i hope u and shamrock take ur stupid 9 am homology class and suffer together
next obsidian
I am okay with this
golden pasture
wow toxic
maiden ocean
punished shamrock
bleak abyss
Multi is for nerds
bleak abyss
I'm just gonna have Hegel do smooth manifold theory and backfill
maiden ocean
yes
this sounds good
: )
i have to learn it in school anyone im not going to learn how to compute gradients or whatever on my own smh
it is :uninteresting:
bleak abyss
Tru
next obsidian
🥱
bleak abyss
When it's time you can read Spivak Calc on Manifolds
next obsidian
Dami you're a bad influence
maiden ocean
i read 3.1 so far it seemed pretty cool!
next obsidian
Hey Dami
maiden ocean
im rereading rn going to do the problems after
universal coeff thm was cool
derived functors go brrr
bleak abyss
I feel like Hatcher's treatment of cohomology just did not stick for some reason I mainly just remembered how he was like
Hey if you want a product
Well you do this and then that
But uh
rip
next obsidian
did you know that if S is a graded ring with S_0 a field and S finitely generated over S_0 by S_1 then the saturation of an ideal is equal to the original in high enough degrees
bleak abyss
But wait if you're contravariant
maiden ocean
sadness
bleak abyss
Diagonal map gg
maiden ocean
i think it is kind of sticking with me?
next obsidian
This follows as I^sat/I is f.g. over S/S_>=r for some r and S/S_>=r is artinian
bleak abyss
Chmonkey: interesting
next obsidian
so I^sat/I is artinian but an artinian graded module is 0 in high enough degrees
maiden ocean
i spent like a day on the first 4 pages trying to make sure i got the intuition lol
next obsidian
This is very important I am sure
so you should remember this result for the rest of your life
Oh sorry, was I interrupting I didn't realize
bleak abyss
Played
maiden ocean
sad
also dami im going to read tom dieck at some point after
i am very determined to have a super solid grasp of AT fundamentals
golden pasture
be a degen like me! cant imagine things
bleak abyss
Ah nice yeah that one from glancing at it seems more formal than Hatcher for sure
golden pasture
way better
bleak abyss
And yeah same lol
golden pasture
do much better
maiden ocean
but visualizing is fun
golden pasture
hatcher lost me at pictures
bleak abyss
Visualization is for nerds
next obsidian
All I need to visualize is draw circles
and smaller circles in those
and that's worked for me so far
bleak abyss
See Hegel?
maiden ocean
: (
vital quail
how do you visualize T^2 - {x} retracting onto S1 v S1 @next obsidian 🤔
oblique river
just draw circles and then draw smaller circles smh
golden pasture
how do you visualize T^2 - {x} retracting onto S1 v S1 <@!146419489083228160> 🤔
t2 is a b2 with proper identification
b2 - {x} is s1 with proper identification
chilly ocean
This is going to reveal how stupid I am. But i'll do it anyway.
This is the solution to "Prove Z(G) is a subgroup of G".
Can someone explain to me what the "ax = xa | ∀ x exists in G"? Where does that come from, and how do you get there?
nvm misread
yeah
I get that, but like... how do you come up with that definition? and what does it mean?
😢
stone fulcrum
Z(G) is the set of all elements a in G such that ax = xa for all x in G
chilly ocean
abelian group nice, you want to study abelian thing, but not all group abelian ,so you look at abelian part
stone fulcrum
That's just what Z(G) is haha. Z(G) happens to have some nice properties is all
chilly ocean
G / Z(G) cyclic implies G abelian 😌
ax = xa just means its commutative, *?
i cant lol
ok, maybe i should rephrase it
"Whats the significance of ax = xa?"
viscid pewter
ax = xa means a commutes with x
chilly ocean
ok, last question (pwomise): Does this solution look correct? lmao. Does it prove Z(G) is a subgroup of G?
Sometimes solutions are wrong
viscid pewter
looks good
chilly ocean
vital quail
G / Z(G) cyclic implies G abelian 😌
prove this in 25 characters or less
chilly ocean
<xZ(G)>=Z(G)
trivial
chilly ocean
dim escarp
Lol
latent anvil
@chilly ocean related problem
Let G be a p group
Then G/[G, G] cyclic implies G abelian
it's a good problem
dim escarp
What is [G, G]? I'm sure I've learned it and I'll need it in the future as well but I can't recall
thorn delta
commutator subgroup
i.e. the subgroup generated by all of the commutators [g,h] = ghg-1h^-1
dim escarp
Oh I see
chilly ocean
Oh, what was the Wikipedia page?
chilly ocean
Well, it seems certainly true that semidirect products cannot be classified with the order of groups fixed eg Z3 x Z2 (semidirect product here) is always abelian since only autormophism of Z2 is the identity, but Z2 xZ3 can be non abelian
I guess I would read the wiki page as saying if the order of K and H are already fixed then you only need to look at homomorphisms H to Aut(K)
chrome hinge
Anyone here?
thorn delta
no
chrome hinge
I got a doubt about polynomial rings
Lmao
If K is a field, the set of unities of K[x] is exactly K right?
I mean the constant polynomials
next obsidian
Yes
chrome hinge
Nice
Now heres my question
In the ring K[[x]] of formal series, all elements with non zero constant term are invertible
next obsidian
Is that true? I don’t think so
thorn delta
i think it is true
chrome hinge
Theres an explicit inverse formula
next obsidian
Oh wait that’s for polynomials
Nvm haha
Yeah, things are better for the power series
chrome hinge
Seems like you can help me then
next obsidian
Hurb
chrome hinge
I dont understand why the same formula doesnt apply for polynomials
next obsidian
The idea is
So I proved this like
1.5 years ago
If I remember correctly essentially when you do this for power series
You do something in degree n
But it adds stuff in degree > n
Then you just fix stuff there
But it messes things up in higher degrees
The idea is you just sort of
“Throw the remainder to the end”
And since the thinfs infinite it just kind of goes into the void
Is how I thought about it
chrome hinge
Hmm kinda weird tbh
next obsidian
I don’t remember the formula off the top of my head but I did make it
And if you look at it you should maybe be able to see how you’d arrive to it
And it’s basically what I said
I think. Let me look at the formula again
chrome hinge
I actually replaced the inverse formula in the product definition and it just vanished every non constant term
chrome hinge
So i thought: well the same would happen in polynomials
next obsidian
The issue is that b_n might never become 0 for n sufficiently high enough
chrome hinge
Ya its a recursive formula
next obsidian
So for a power series inverse this isn’t an issue
But to check if it’s an inverse
You can simply multiply the two, then check the degree 1 stuff
Oh hey look I chose b_1 so that like this is fine
Look at b_2 oh hey I chose it so this is fine...
And for any n you chose b_n so the degree n stuff is fine
chrome hinge
Exactly, and every term is just vanishing every time
next obsidian
But a power series only has terms of degree n over n in N
So you basically just keep putting off “finally fixing it for real”
Into infinity lol
Like you could try the same thing for a polynomial inverse but you have to cut it off at some point
And if you cut it off at n then you can’t deal with the > n terms
chrome hinge
But for a polynomial there arent such terms...
next obsidian
No like
chrome hinge
Theyre zero
next obsidian
You aren’t fixing f(x) when you choose your next b_n
You’re building up an inverse g
Step by step
So like in order to fix the bottom term you make a g_0
So that f(x)g_0 has 1 has a constant
Now f(x)g_0 has a degree 1 term
So you take g_0 and make it degree 1 by adding a b_1x term
In order to kill the degree 1 term of f(x)g_0
By doing this now you have f(x)g_1(x) which has a constant term 1
And a linear term 0
BUT you’ve increased the degree hy 1
Then you do the same with g_2...
So as you’re fixing the lower degree terms the degree of the polynomial keeps getting bigger
When g is a power series this doesn’t matter you can just do it forever
Even if you increase the degree it doesn’t matter
chrome hinge
Ooh im starting to see it
next obsidian
You eventually get to it
chrome hinge
The thing is the inverse would be apower series and not a polynomial?
next obsidian
Yup
chrome hinge
Oooh
next obsidian
You finish off with g_infinity
But for any finite n g_n has arbitrarily high terms you need to deal with
Err
chrome hinge
I see i see
chrome hinge
Honestly i have to think more of it
next obsidian
So here’s my advice
chrome hinge
But i kinda got the idea
next obsidian
Don’t look at the formula
Rederive it
Then you’ll see why it won’t work with a polynomial
next obsidian
Np
chrome hinge
For your time and help
oak grove
im confused about this one
let $p$ is prime. show that an element has order $p$ in $S_n$ iff it's cycle decomposition is a product of commuting $p$-cycles.
dummit and foote 1.3.14
cloud walrusBOT
jan Niku:
oak grove
they aren't saying literally a single element has order p right
the first part means "for each element, it has order p"
or i think i might be completely misunderstanding the topic
chilly ocean
they're saying that given any element of S_n, it has order p if and only if its cycle decomposition is of that form
oak grove
they say then
give a case where it doesnt work if p is not prime
and they give the example (1 2)(3 4 5)
this does not seem to mesh with the question
the elements have dissimilar order
it actually seems like
chilly ocean
well the problem statement is an if and only if
oak grove
i mean i know its the case but it seems like if each element in S_n is order n, then the entire thing is a product of n-cycles
like theres nothing to prove
that would just have to be the case
and this counterexample doesnt satisfy the conditions of the problem
thorn delta
you know that an "element" of Sn is not necessarily a cycle, right jan?
thorn delta
i mean, its a product of disjoint cycles, but that's all you can say
oak grove
? what element would not be a cycle
i dont think i saw any groups that werent composed of cycles in this chapter but im a little dense
thorn delta
for example, (1 2)(3 4) is not a cycle.
oak grove
i guess im confused, isnt each of those an element?
maybe i missed the point of this whole chapter 
hmm i think im gonna restart this chapter
i did fine on all the questions up to this one
thorn delta
so, an element of Sn is just a bijection from {1,2, ..., n} to itself. Permutations don't have to be cycles, but the point is that you can split up permutations into the composition of disjoint cycles
oak grove
i have dumb questions
if a permutation can be written as a product of disjoint cycles there should be no way to write the permutation as a single cycle right
i do think i need to redo it tho
i think i missed the point
S_n contains multiple permutations right
its just the collection of all those bijections
thorn delta
in the case when you have a permutation that's already a cycle, cycle decomposition is just itself, a single cycle.
S_n is the group of all bijections from an n element set to itself.
oak grove
okay
thorn delta
so yea, n! elements in general
oak grove
then elements of S_n are not necessarily cycles, but are permutations/functions, then (1 2)(3 4 5) has order lcm(2,3) but is not a composition of disjoint 6-cycles
😄
since p has factors (p,1) then you must have |x|=p for all x in S_n
got it, thanks
thorn delta
|x|=p for all x in S_n
wait what does this mean?
not every element of S_n has order p
thorn delta
ah ok
fair shard
so a normal subgroup is a subgroup that is closed with conjugation, and it's easy to prove from that that G/H is a group. this "closed under conjugates has a few equivalent forms such as aH=Ha for all a in G, and I think there is something else about how if xH=yH, then xy^-1 is in H but that one might just be for any average subgroup. Can anyone explain an intuitive link between those 2 facts? Like what being closed under taking conjugates makes G/H so groupy?
thorn delta
you need the operation $(xH)(yH) = xyH$ to be well defined.
cloud walrusBOT
kxrider:
thorn delta
so, for any subgroup H, you get the equivalence relation x ~ y iff xy^{-1} \in H automatically, but it doesn't necessarily respect taking group products unless you have normality
idk if that answers your question
thorn delta
npnp
fair shard
and maybe the other thing is if xH=Hy idk
or maybe it has to do with any combination of x and y where one is inverted will be in H
something like that
the inversion thing is how I actually tried to find intuition for it
carmine fossil
If H is normal you get xH=Hy implies xH=Hx
Because x=hy for some h implying Hy=Hx
fair shard
cause I was thinking of a differentiation operator as like a homomorphism and the kernel will be constant functions, so constant functions should form a quotient group and that made perfect sense since if 2 functions have the same derivative them their difference is a constant function, meaning in this quotient group language that if they are mapped to the same place with differentiation then their difference is in the kernel and I like abstracted that to general groups. Don't know if that is a good way to think about it
but that's an abelian group so I guess that's a completely different question
but homomorphisms did help me understand quotient groups more
carmine fossil
I think that's pretty accurate
fair shard
conjugation has some magical properties I think I just haven't submitted myself to its power yet
thorn delta
quotients in general are pretty magical
carmine fossil
Conjugation is also nice with permutations
carmine fossil
Are you familiar with the cycle notation?
fair shard
yes
ohhhh
I think I know what you're saying
the conjugation of the permutation is the same cycles but each element is replaced with it's image by the permutation
carmine fossil
Yes
fair shard
is that what you were gonna say?
nice
I love that
I also love the normal subgroups you can find
cause I always interpret it as like
when you do a homomorphism from G to G/H there is like some aspect of the permutation that you are choosing to keep
such as whether or not they fix 3 or whether they are odd or even stuff like that'l
I'm new to group theory so this stuff still impresses me
chilly ocean
ethan since you're doing LA you should compare the stuff you're learning in gropu theory to the things you know from linalg
upbeat juniper
@fair shard I feel the same way as you haha
it was kinda mind blowing when I realised that homomorphisms are essentially projections onto quotients
fair shard
yes!! so cool
near yarrow
functions from a finite set A to itself are surjective if and only ifthey are injective. but does it apply for countably infinite sets? like the set of natural numbers?
vital quail
no, like for N you can just map 1 and 2 to 1, then n to n-1 from 3 and up
golden pasture
^ (dedekind infinite sets, set can be put in bijection to a proper subset of itself)
near yarrow
is there a way to prove $S_{3}$ that is the set of all permutations of a group of 3 elements has the smallest order of any non-abelian group?
cloud walrusBOT
jayanthrg:
carmine fossil
Try constructing non abelian groups of order less than 6
Order 1 is trivially abelian
Order 2 is cyclic
Order 3 is cyclic
Order 4 is either Z_4 or Z_2 x Z_2
Order 5 is cyclic
near yarrow
got it
maiden ocean
yea
really only order 4 needs to be verified bc groups of prime order are always cyclic
chilly ocean
isn't that exactly what is written above
maiden ocean
yes 
i am providing a more general fact that they might not already know
south storm
Is it always true that (up to isomorphisms) any group of order n is a cross product between cyclic groups of which the product of their order is n?
carmine fossil
No
chilly ocean
that is not true
carmine fossil
D8
south storm
Oh alright
carmine fossil
If you meant semi direct product,still no
south storm
No I meant Cartesian product
upbeat juniper
that's true for abelian groups
near yarrow
really only order 4 needs to be verified bc groups of prime order are always cyc...
is this some theorem which is proved?
well i am just starting group theory. still in the early stages.
carmine fossil
Just write down all the possible group tables
near yarrow
i'll try that
languid meteor
Hey all, just wanted to ask what the 'absolute value' symbols are in the context of rings
like what would |R| be
carmine fossil
Number of elements in the ring
languid meteor
ah ha, gotcha
near yarrow
what happens to the nth dihedral group $D_n$ as we let n tend to infinity? does it reduce to a trivial group?
cloud walrusBOT
jayanthrg:
carmine fossil
Why would it reduce to a trivial group
near yarrow
the rotated circle wont be any different right?
carmine fossil
It's clearly not the same
next obsidian
Taking a limit in this sense is a bit more finicky than in calculus
I’m not sure how you could make a system you could take the (presumably) colimit of
I think you have inclusions based on divisibilitu?
And that might make an inverse / directed system
I mean I think D_n embeds into D_2^kn
I’m not sure if it holds for any divisibilitu
And presumably the inclusion is compatible so you can make some sort of system
I think it would be a directed system so you get a colimit
I just have no idea what it would be
Ah okay so instead of injections we get surjections
I only vaguely know what it means
Pro finite group is some like topological group stuff
Not really
It’s important for number theory?
And like Galois whatever shit
They’re complicated I think
Not really
You can make inverse limits like
Physically
I think you take the like
Product of all the groups
And then mod out by relations
Sure, but category theory doesn’t really make it any more tangible haha
Maybe if you learn category theory you get more comfortable with intangible objects
But I think in general you show a specific group IS a pro finite group
By showing it satisfies the universal property and then boom you have something to deal with
You should learn inverse / direct limits
They’re useful in general
Nope
It’s in Lang
And even Atuyah MacDonald
The value is that direct limit is exact
Or like left exact I forget
And any module is a direct limit of its finitely generated submodules
So for example you know how something is flat iff tensor with it is exact?
By doing a direct limit you can check only finitely generated modules
Ah okay
Well it’s useful when you need to know like ___ is exact for any module
Because you can check on only finitely generated ones
By doing a direct limit
As long as that direct limit commutes with ___
Yeah this probably
Doesn’t matter until you do more homological things
golden pasture
just do am