Hm well there is an elementary proof using the norm if you note (2+2i)(2-2i) = 4, so 4 = 0, and from there it is clear enough
#groups-rings-fields
south patrol
balmy python
for a counterexample i should probably be looking at some non abelian groups right
delicate orchid
for a counterexample i should probably be looking at some non abelian groups rig...
Yes, theres a small non abelian group that works
balmy python
Thats absolutely colossal
😭
balmy python
The group in my head is order 8
order 8 and non abelian wtf
hmmm
is it smth ive prob come across
delicate orchid
Yeah
balmy python
is it a subgroup of S4
delicate orchid
It is indeed
karmic moat
dimension 4
colossal
delicate orchid
It’s uncountably infinite mate
karmic moat
yeah it was a bad joke
delicate orchid
for a counterexample i should probably be looking at some non abelian groups rig...
Morally, this result shouldn’t be true: imagine G was a really big non-abelian group and H was a normal abelian subgroup, it would be surprising if G normalised every subgroup
delicate orchid
ohhh is it D8
Yup!
balmy python
i'm not very used to dihedral and permutation groups
i shouldve been last year
but i didn't take groups serious
delicate orchid
It’s the smallest group that has a counter example so I thought it would be good to go for
delicate orchid
Ok just think of it as <r, t | t^4 = r^2 = 1, rtr = t^3>
balmy python
so like (12)(43) is in D8
delicate orchid
I’ll try and think of another example using a matrix group
balmy python
I’ll try and think of another example using a matrix group
don't spoil though
i would rather figure it out with hints
delicate orchid
Ok I’ve got one
balmy python
Ok just think of it as <r, t | t^4 = r^2 = 1, rtr = t^3>
how could i make sense of this through geometry
delicate orchid
why would you
balmy python
as in
delicate orchid
t is the rotation r is the reflection of the square
balmy python
isn't D2n meant to give the symmetries by an n gon
karmic moat
how could i make sense of this through geometry
best question i've ever seen
delicate orchid
best question i've ever seen
You gotta go back: #geometry-and-trigonometry
karmic moat
:(
balmy python
thats uncountable i hope you know that
yeah i was being stupid 😭
forgot that Sn exists
looking at cantor distributions in probability mightve fried my brain a bit
thorn jay
for a counterexample i should probably be looking at some non abelian groups rig...
decomposition series would be great if this were true
quiet pelican
for a counterexample i should probably be looking at some non abelian groups rig...
I am so tempted to overkill this
I’ll spoiler this but like
Come up with an actual example because it’s informative
||Pick a group with two non-conjugate isomorphic subgroups which are mapped to each other by an outer automorphism
Take the semidirect product of that group with Z, where the action is by said outer automorphism||
delicate orchid
You can take S_4 instead of D_8 if that helps you
balmy python
You can take S_4 instead of D_8 if that helps you
it'd help if i look at D_8 so i actually try to understand other groups 😭
i did take S_4 originally but
thinking about all the elements sounded off putting
quiet pelican
I am so tempted to overkill this
I’ll spoiler this but like
Come up with an actu...
In some sense this example is universal, in that every possible example either has a subgroup that looks like this, or the same thing with ||Z/nZ||
quiet pelican
You can take S_4 instead of D_8 if that helps you
Does S_4 have any normal non-characteristic subgroups?
thorn jay
In some sense this example is universal, in that every possible example either h...
how so
quiet pelican
Oh wait I’m dumb
thorn jay
lol
delicate orchid
Does S_4 have any normal non-characteristic subgroups?
It’s centreless and Aut(S_4) is S_4
thorn jay
No this is still right
its right if one of the subgroups were normal
delicate orchid
We’re just proving that normal doesn’t imply strongly closed in a generalised fusion system it’s not that deep
quiet pelican
its right if one of the subgroups were normal
Oh yeah that was implied 🙃
thorn jay
smh
delicate orchid
I just remembered what my username is I gotta change ts back
thorn jay
We’re just proving that normal doesn’t imply strongly closed in a generalised fu...
yeah we're just proving that groups do not in general have the congruence extension property smfh
thorn jay
I just remembered what my username is I gotta change ts back
probably a banger idea
balmy python
Ok just think of it as <r, t | t^4 = r^2 = 1, rtr = t^3>
where does rtr = t^3 come from?
karmic moat
no i dont think i used that phrase right
delicate orchid
where does rtr = t^3 come from?
Reflect rotate and reflect back yields the same rotation just in the other direction
But really this is the definition of D_8
The fact that it acts nicely on a polygon is coincidence
balmy python
Reflect rotate and reflect back yields the same rotation just in the other direc...
ahh okkk
balmy python
The fact that it acts nicely on a polygon is coincidence
oh 😭
thorn jay
end of an era, really
an era that lasted like
half a day
delicate orchid
The fact that it acts nicely on a polygon is coincidence
Im gonna be called unhinged for this take btw
And it is not historically accurate
balmy python
so the way i would go to generalise this is:
D_2n = <r,t | t^n = r^2 = 1, rtr = t^n-1>?
thorn jay
well yes because its a pattern
balmy python
so the way i would go to generalise this is:
D_2n = <r,t | t^n = r^2 = 1, rtr = ...
or does this differ with odd n
okay lessgooo
delicate orchid
It’s the same
thorn jay
that D_n acts nicely on the n-gon is not a coincidence because its a pattern 
balmy python
now time to look at the actual problem at hand
delicate orchid
that D_n acts nicely on the n-gon is not a coincidence because its a pattern <:K...
Mere coincidence that they admit that particular representation
south patrol
that D_n acts nicely on the n-gon is not a coincidence because its a pattern <:K...
Or a definition
thorn jay
Or a definition
then its a pattern that theyre coxeter groups
delicate orchid
Groups should be defined abstractly it’s just the way of the world
delicate orchid
then its a pattern that theyre coxeter groups
THIS is acceptable
thorn jay
hell you could also let D_n act in the obvious way on R^2 and define the n-gon as the polygon defined by the orbit of (0, 1) ig
delicate orchid
Now we’re talking
thorn jay
let me cook guys let me cook
karmic moat
https://tenor.com/view/writing-too-fire-fire-writing-writing-fire-friendcord-gif-7801118907023267493
delicate orchid
The point is to emphasise the symmetry’s structure rather than the object itself
thorn jay
focussing on single objects or structures often doesnt reveal their structure enough
karmic moat
the point is to emphasize maps around the object than the object itself 🤤
delicate orchid
Im the worlds leading expert on groups of order 8
quiet pelican
Groups should be defined abstractly it’s just the way of the world
Groups should be defined as isometry groups of metric spaces
delicate orchid
Someone ping mods immediately
quiet pelican
Fundamental groups of topological spaces also sometimes ok
thorn jay
Groups should be defined as isometry groups of metric spaces
THAT is an insane take
quiet pelican
Someone ping mods immediately
/ban Wew
thorn jay
Fundamental groups of topological spaces also sometimes ok
so
every group
quiet pelican
so
every group
Sometimes ok
Not always ok
delicate orchid
Find me a topological space with any of these as it’s fundamental group
thorn jay
Groups should be defined as isometry groups of metric spaces
i can remember the set of isometries on R^n carrying some quandle structure
delicate orchid
And do NOT give me that classifying space shit
quiet pelican
Find me a topological space with any of these as it’s fundamental group
They’re finite so they’re virtually trivial so we don’t study them
Go back to your #combinatorial-structures 🙃
delicate orchid
No frob enjoyers?
thorn jay
Find me a topological space with any of these as it’s fundamental group
every group is the quotient of some free group so wedge some possibly infinite amount of circles together and glue relations using discs
delicate orchid
And do NOT give me that classifying space shit
AHEM
karmic moat
the only frob i care about is the ring endomorphism
delicate orchid
Might as well be
thorn jay
idk man do i LOOK like i know any algtop
delicate orchid
Alls well that ends well
thorn jay
or whatever tf classifying spaces are from
south patrol
No frob enjoyers?
I love Frobeniï
quiet pelican
idk man do i LOOK like i know any algtop
We should algtopify you
thorn jay
NO I DONT WANT TO BE A TOP
balmy python
Yes, theres a small non abelian group that works
8 elements sounds small
delicate orchid
I love Frobeniï
Ts mf just reinvented the nerve of BG right in front of my eyes potato
balmy python
it's annoying asf to even get the permutations of D8
delicate orchid
it's annoying asf to even get the permutations of D8
Why are you doing that? I gave you a presentation
quiet pelican
NO I DONT WANT TO BE A TOP
Embrace the cute algtop
thorn jay
I love Frobeniï
bro used to the right latin conjugation
balmy python
Why are you doing that? I gave you a presentation
oh wait you use the presentation itself 😭
south patrol
Embrace the cute algtop
Misplaced full-stop/period
south patrol
delicate orchid
Perhaps S_4 was the better suggestion
balmy python
😭
thorn jay
Embrace the cute algtop
oh homology i can do
learned it by doing rack/quandle homology
south patrol
learned it by doing rack/quandle homology
I feel sorry for you
karmic moat
the classic
quiet pelican
Unfortunately I don’t have any homotopy notes that aren’t handwritten
delicate orchid
learned it by doing rack/quandle homology
Mods?
thorn jay
its okay I've got Bredon
south patrol
Unfortunately I don’t have any homotopy notes that aren’t handwritten
Eh handwritten and typed up are the same up to homotopy
quiet pelican
learned it by doing rack/quandle homology
south patrol
But the space of such homotopies is unfortunately non-contractible
quiet pelican
The best type of quandlestuffs
karmic moat
omg i have mine too at home
south patrol
Lol Wew can I tell u smth frightening
I need to actually think about topological spaces for something now
delicate orchid
Lol Wew can I tell u smth frightening
Prthsod
south patrol
As in rather than homotopy types
delicate orchid
I need to actually think about topological spaces for something now
Im gonna be sick
karmic moat
i think the first book i bought was hoffman-kunze
thorn jay
Ts mf just reinvented the nerve of BG right in front of my eyes potato
apparently in trying to construct a top space for some fundamental group i described essentially what the claasifying space of a group is
south patrol
i think the first book i bought was hoffman-kunze
Did ng force you
Oh wait I misunderstood
For linear algebra? Ok cool
karmic moat
they have another one?
karmic moat
ah okay
south patrol
Another book with two German-surnamed authors
karmic moat
many such cases
delicate orchid
apparently in trying to construct a top space for some fundamental group i descr...
Ur one quandle away from reinventing homotopy colimts
south patrol
You know a nice basic but sometimes troll question
What is the fundamental group of BS^1
thorn jay
Ur one quandle away from reinventing homotopy colimts
enpeace homotopy theory arc is something yall arent ready for
delicate orchid
What’s B even mean here what are we delooping
south patrol
Well S^1 is a group
delicate orchid
News to me
thorn jay
I need to actually think about topological spaces for something now
hey it could be worse
south patrol
hey it could be worse
Like what
thorn jay
Well S^1 is a group
whuh but S^1 isnt finite how can it be a group
south patrol
I mean okay at least these spaces are not far off being manifolds
south patrol
whuh but S^1 isnt finite how can it be a group
First year group theory be like
quiet pelican
whuh but S^1 isnt finite how can it be a group
Finite groups don’t exist
delicate orchid
No groups larger than 7^4 exist
south patrol
whuh but S^1 isnt finite how can it be a group
Actually S^1 is a finite space
quiet pelican
No groups larger than 7^4 exist
Behold, Z/1000Z
south patrol
for a suitable definition of finite space
delicate orchid
Behold, Z/1000Z
My theorem remains unchallenged
karmic moat
where does 7^4 come from
quiet pelican
My theorem remains unchallenged
Behold, Z/10000Z
south patrol
Idk I mean like these I am considering are at least locally compact Hausdorff
delicate orchid
What is the fundamental group of BS^1
That’s easy, pi_1(K(S^1, 1)) of course
south patrol
Could be horrible ig
thorn jay
idk anything worse than topological spaces
personally find them quite nice but i am used to working with algebraic closure operators so
south patrol
Q lol
quiet pelican
Behold, Z/10000Z
It micophobic that mico no was able to to calculate
south patrol
That’s easy, pi_1(K(S^1, 1)) of course
Loud incorrect buzzer
delicate orchid
Behold, Z/10000Z
WTF
quiet pelican
thats abelian, doesnt count
D_10000
thorn jay
damn youre good
south patrol
Metacyclic groups when amazoncyclic and googlecyclic groups turn up
quiet pelican
Metacyclic doesn’t count
S_10000
south patrol
I don’t care about optimal lower bounds
Valid
south patrol
S_10000
Has at least 10 elements
Consider the group of permutations of quarks in the universe
quiet pelican
I can give 9999 elements
karmic moat
Has at least 10 elements
oh yeah? name them
quiet pelican
I can give 9999 elements
(12)
(13)
(14)
…
thorn jay
No groups larger than 7^4 exist
mathieu groups?
quiet pelican
(1 9999)
delicate orchid
oh yeah? name them
Hydrogen helium lithium beryllium boron carbon nitrogen oxygen fluorine neon
delicate orchid
mathieu groups?
They’re sporadic groups not groups
south patrol
oh yeah? name them
(11),(12),(13),(14),(15),(16),(17),(18), (19) and (139 193 184 999)
thorn jay
They’re sporadic groups not groups
M11 × M11
balmy python
ummm @delicate orchid after quite a lot of distractions i think i got it
take the subgroup of all t in D8, this is obviously normal since rtr^-1 = t^n-1
the subgroup of all t is cyclic of order n for all t since t^n = 1 so the group itself must have n elements
wlog call it H = {e,t,t^2, t^3}
then, K= {e,t^2} is a normal subgroup of H but rt^2r^-1 = t^3 which is not in K so K is not a normal subgroup of D8
that was much easier than i thought after being given that definition of D2n 😭
delicate orchid
ummm <@233993116418441216> after quite a lot of distractions i think i got it
t...
You picked the only normal index 4 subgroup. t^2 is central
rt^2r = (rtr)^2 = t^6 = t^2
delicate orchid
Well no both are true
The idea is that r and rt^2 (or rt and rt^3) are conjugate in D_8 but generate an abelian subgroup
Anyway it’s midnight I gotta go to bed
thorn jay
goodnight
karmic moat
good night wew
thorn jay
maybe let the bedbugs bite.. this time...
chilly ocean
every group is the quotient of some free group so wedge some possibly infinite a...
I haven’t checked it out formally, but I think this can actually be extended to find a space with fundamental groupoid(restricted to the right basepoints) isomorphic to some arbitrary groupoid?
Take some groupoid G with some presentation. The 0-cells are the vertices of the graph in the presentation, the 1-cells the edges, and glue 2-cells along the relations as above. Then, the fundamental groupoid of that space, restricted to the 0-cells, is the original groupoid I think?
Haven’t been interested enough to actually try to prove it or google it though
Could very well be wrong
thorn jay
I haven’t checked it out formally, but I think this can actually be extended to ...
well the problem it seems is that the objects of the fundamental groupoid are taken to be the points in X
so if you have some fundamental groupoid on a finite number of objects youre gonna have some serious problems lol
chilly ocean
well the problem it seems is that the objects of the fundamental groupoid are ta...
Yeah that's why you restrict the fundamnetal groupoid to the 0-cells.
thorn jay
ah you said restrict to the 0-cells
uhh yeah looks about right
i mean i cant see any reason why that wouldnt be the case
chilly ocean
yeah it sounds right to me but I haven't actually tried proving it haha
waxen marsh
just wanna make sure I got this all right. A ring is a group closed under addition and multiplication, with associativity and the distributive property holds. A field is a ring with an additive identity and multiplicative identity? so when is a ring not a field?
that seems to follow my professors slides, but maybe im missing something?
if rings have to be abelian, then they have to be commutative, no?
tardy hedge
if rings have to be abelian, then they have to be commutative, no?
The addition + is commutative but multiplication doesnt have to be
waxen marsh
oh i totally did not mean to say subtraction in my first message. meant multiplication
thorn jay
just wanna make sure I got this all right. A ring is a group closed under additi...
a ring is not just a group, it has two operations (formally it has 5 but we often dont explicitly write negation, 0, or 1), one of which we call addition: this forms an abelian group, and the other we call _multiplication: this forms a monoid (it is associative and has an identity element, just like how in the real numbers multiplication is associative and we have the identity element 1). Furthermore, multiplication distributes over addition.
oh ew you guys dont require identity
thorn jay
that seems to follow my professors slides, but maybe im missing something?
whats it with math courses and using the most horrible notation for operations + not requiring unity
ts pmo
thorn jay
that seems to follow my professors slides, but maybe im missing something?
okay these notes really pmo man
pissing me off
didnt feel like going to sleep tbh
kind temple
why does it look like a word doc
thorn jay
why does it use oplus and otimes
kind temple
fr
kind temple
ig they are trying to emphasize that like
thorn jay
if it walks and talks like addition then it is addition
wait if these are slides...
is this made in powerpoint...
noble nexus
why does it use oplus and otimes
based burnside ring enjoyer
kind temple
ye
thorn jay
based burnside ring enjoyer
im sure that was the thought behind using oplus and otimes
lol
oplus is a noninvertible operation in my eyes idk
noble nexus
or I guess the representation ring
thorn jay
and it looks kinda ass and cluttered ehen its that small
noble nexus
yeah
thorn jay
when
kind temple
now
noble nexus
oplus is a noninvertible operation in my eyes idk
what about invertible modules....
thorn jay
what about invertible modules....
was talking more like "not in general invertible
istg if i make any more typos im gonna hurl my phone into a wall
noble nexus
was talking more like "not in general invertible
yeah i just playin
I will say burnside rings/representation rings are really cool rings imo
although they're basically just polynomial quotients but still
thorn jay
Serre's book on reps has a chapter on it iirc
noble nexus
any time where u take isomorphism classes of some object and make it into a group/ring/etc
its cool shit
thorn jay
its related to K0 right
noble nexus
picard groups or k theory or etc
idk if directly related there are a bunch of times in math where u take iso classes of smth and make it a group/ring
thorn jay
well its the grothendieck construction for a group out of a commutative monoid right
noble nexus
yeah
tbh I feel like that construction is too simple to be given the name "grothendieck group"
it always feels like the name sounds more complicated than what it actually is
vapid vale
tbh I feel like that construction is too simple to be given the name "grothendie...
that’s the fun of it though
what is math if not a speedrun to work with an object named after grothendieck
grothendieck group is like the credits warp
thorn jay
SNES mario wrong warp ahh
"I was just fucking around with monoids WHY IS GROTHENDIECK HERE"
eh utrechts alright I guess
luckily its not Wageningen or smt
or Amsterdam
idk man dont really go there
nah
Delft campus looks like a prison
+its applied which makes it a minus 1000000 immediately
karmic moat
everything is grothendieck
twilit wraith
is there a non-brutal way this can be done
ive shown that these are all the subgroups of that form as there are p+1 of them and they each contain p-1 elements that arent found in any other given subgroup
but that doesnt prove that there arent any other subgroups
thorn jay
notice that E is a Z/pZ-vector space, so every nontrivial subgroup is principal
so we can also treat is as the set of ax + by with a, b in Z/pZ
twilit wraith
im really not sure we can use vector spaces
thorn jay
so we can also treat is as the set of ax + by with a, b in Z/pZ
the notice that <ax + by> = <x + b/a y> for a≠0
twilit wraith
thats the problem
thorn jay
im really not sure we can use vector spaces
if youre doing group theory then no way you havent had linear algebra
twilit wraith
yes but we havent talked about vector spaces in group theory all semester
thorn jay
dont see why you wouldnt be able to use it, LA is general math knowledge every math student should have
twilit wraith
so im not sure if its like currently out of scope
twilit wraith
dont see why you wouldnt be able to use it, LA is general math knowledge every m...
i took it with the same prof last semester
and we only went over R-vector spaces
barely C-vector spaces i guess
thorn jay
everything works the same so it really doesnt matter
ig besides the existence of a norm
twilit wraith
i guess i could say like "there is an obvious isomorphism from E to Zp x Zp"
which is basically the same idea
thorn jay
yeah
and then let ax + by, cx + dy be linearly independent and G their span
=> (ad - bc)x in G
and theyre linearly independent so ad - bc ≠ 0
=> x in G
same for y, so G = E.
this would be a sketch of a proof that every nontrivial subgroup is principal
twilit wraith
yeah
kind temple
everything is grothendieck
always has been
cloud walrusBOT
rocky cloak
but that doesnt prove that there arent any other subgroups
Every element has order p, so each nontrivial cyclic subgroup is Z/p. Z/p has p-1 non-trivial elements.
That's (p^2 - 1) / (p-1) = p+1 subgroups
twilit wraith
Every element has order p, so each nontrivial cyclic subgroup is Z/p. Z/p has p-...
I see
I ended up having to generalize it to p^n
But i finally got it solved I think
Sort group elements into equivalence classes based on if they generate the same subgroup
By construction these classes are subgroups
Each class has p-1 elements
So there are p^n - 1 / p - 1 subgroups
crystal vale
So if R = C[x1,..,xN] then are we considering R[y]/(gy-1) ?
rocky cloak
So if R = C[x1,..,xN] then are we considering R[y]/(gy-1) ?
Yes, which is the same as C[x1, ..., xN, y]
crystal vale
How is this the same as [x1,..,xN, y] ?
rocky cloak
I mean R[y] is
crystal vale
Okay
So here substitute y = 1/g means? Are we mapping this equation to R[y]/(gy-1)?
rocky cloak
So here substitute y = 1/g means? Are we mapping this equation to R[y]/(gy-1)?
You might be able to think about it that way, but I think it might be better to just think of it as a polynomial identity that holds whenever g is nonzero.
The final formula clearly also holds when g is 0, so then just holds on the nose
supple pecan
OH right that makes sense yeah thanks so much
crystal vale
So it is exactly the substitution of R[a,b] ring of polynomial over R in two variables to a = sin x and b = cos x ?
rapid cave
Yeah, all polynomials in sinx, cosx
crystal vale
So it is isomorphic to R[x,y]
crystal vale
I mean isomorphic to R[a,b]?
rapid cave
Thats R[a,b]
crystal vale
What if I send a to sin x and b to cos x ?
Oh there is relation between sin x and cos x
rapid cave
Oh there is relation between sin x and cos x
This might be important
crystal vale
Yes
If I have to show it is an integral domain, is there any better way than computation?
Can I see this ring isomorphic to some familiar ring?
rapid cave
I think yes
rapid cave
Oh there is relation between sin x and cos x
Using this
supple pecan
hi im confused on the notation used here. We've learned that S_X is the set of all bijections from X to X. what does S_2 mean and why does its multiplication table have matrices? how does function composition work w matrices
crystal vale
I think yes
How?
rapid cave
How?
I think you can build an isomorphism to R[x,y]/(x^2+y^2-1)
And x^2+y^2-1 is irreducible
rapid cave
hi im confused on the notation used here. We've learned that S_X is the set of a...
These are not matrices.
You write a permutation of 1,....,n as where it sends each element
And the notation is that in the top row you have the inputs
1 2 3 .... n
And in the bottom row you have the outputs
f(1) f(2) .... f(n)
crystal vale
I think you can build an isomorphism to R[x,y]/(x^2+y^2-1)
I see
crystal vale
So if R = C[x1,..,xN] then are we considering R[y]/(gy-1) ?
Sorry to ping @rocky cloak, can't we just directly take a variable y such that gy = 1, so then y is a function in C[x_1,...,x_n] but what if g is zero somewhere then how do I define y?
delicate orchid
hi im confused on the notation used here. We've learned that S_X is the set of a...
they're not matrices, they're permutations written in two column notation
the WORST way of representing them 
crystal vale
Any hint for e?I know some of kernel elements are x^2-y, xy-z, x^3-z, y^3-z^2
thorn jay
the WORST way of representing them <:reallyMad:1399553225408905216>
imagine not using cycle notation
thorn jay
I have to use one-line notation 💔
how can you not like cycle notation
delicate orchid
I have to use one-line notation 💔
congrats on graduating middle school
karmic moat
I like cycle notation, one line is diff
One-line is u write ur matrix in two lines then throw away the top one
delicate orchid
just use cycle notation and put a lil 🖕 at the end
karmic moat
congrats on graduating middle school
Thanks bro cant wait for high school… I’m a bit nervous, are seniors really bullies?
thorn jay
yes theyre all psychopaths really
karmic moat
just use cycle notation and put a lil 🖕 at the end
At the board should I write 🖕or do the gesture or both
thorn jay
One-line is u write ur matrix in two lines then throw away the top one
i love how this is actively harmful notation
karmic moat
It’s the standard in flag varieties apparently
My prof said “one line is used in flag varieties and cycle notation everywhere else”
Or maybe he said alg combinatorics idr
It’s a headache
delicate orchid
red flag varieties perhaps
karmic moat
W
I keep making errors because i start writing one-line then accidentally switch to cycle for one step
Then everything falls apart
thorn jay
ehy cant you just use cycle notation
karmic moat
I gotta drink the kool aid
tulip otter
the aura is on another level 
crystal vale
You might be able to think about it that way, but I think it might be better to ...
Can you please explain this?
tardy hedge
Thanks bro cant wait for high school… I’m a bit nervous, are seniors really bull...
Ya theyre gonna shove u in lockers
And theyre also gonna say give me ur lunch money nerd
karmic moat
Shoving me in a box call that filling in a young tableau
thorn jay
And theyre also gonna say give me ur lunch money nerd
they broke ass
tardy hedge
WE are mathematicians WE are broke 🥀
thorn jay
preach
rocky cloak
Sorry to ping <@581774454091415553>, can't we just directly take a variable y su...
I'm not sure what you're asking exactly.
When g is nonzero you can pick the value of y to be 1/g. When g is 0 there's obviously no such y.
In the beginning you have an equation that holds for all values of x1, ..., xn, y.
In particular it holds for the choices of x1, ..., xn such that g is nonzero and y = 1/g.
Thus substituting in y=1/g you get an equation that holds for all x1, ..., xn such that g is nonzero.
crystal vale
Uniqueness:
I am trying for uniqueness, if T and T' are two objects which satisfy the given condition.
So by universal property, there exists unique j such that j : T -> T' such that j(g) = g', similarly there exists unique j': T' -> T such that j'(g') = g.
h = j(j') : T' -> T' such that h(g') = g'. Now is it comes from the universal property that h = id, because universal property says there exists unique mapping k:T' -> T' such that k(g') = g' and identity can be the choice for k, and it is unique so k = identity
so h = identity
alpine plank
Any hint for e?I know some of kernel elements are x^2-y, xy-z, x^3-z, y^3-z^2
Thats a good guess for the generators, but notice that if x^2 - y = 0 and x^3 - z = 0 it already implies xy - z = 0 and y^3 - z^2 = 0. To show that they generate the kernel try showing every polynomial f(x,y,z) in C[x,y,z] can be written as f = g(x) + (y - x^2)h(x,y,z) + (z - x^3)k(x,y,z)
balmy python
in an isomorphism, do same order elements map to each other?
thorn jay
yes
elfin wraith
in an isomorphism, do same order elements map to each other?
As a general rule to guide your intuition, you should think of an isomorphism as equality for algebraic objects. That is, if you only look at the algebraic properties of your object, they should be completely indistinguishable. That being said, you should be able to prove this result pretty quickly from the definition of a homomorphism
valid fox
If $H \cong K$ with $H$ and $K$ normal subgroups of $G$, then is $G/H \cong G/K$
cloud walrusBOT
Tiessie
thorn jay
no
velvet hull
If $H \cong K$ with $H$ and $K$ normal subgroups of $G$, then is $G/H \cong G/K$
unforunately no, consider Z/4Z x Z/2Z with normal subgroups Z/2Z x {1} and {1} x Z/2Z
valid fox
Yeah no
thorn jay
G/H depends on the specific inclusion i : H → G rather than the group structure of H
valid fox
That is crazy
tardy hedge
Basic question but if K1/N = K2/N can you conclude K1 = K2?
elfin wraith
Yeah this kinda flies in the face of my general intuition advice given above, but no its not true unfortunaltly lol
thorn jay
Basic question but if K1/N = K2/N can you conclude K1 = K2?
= as groups?
valid fox
Everything is preserved under isomorphism they said
thorn jay
Everything is preserved under isomorphism they said
algebraic properties of the group itself are preserved
velvet hull
Everything is preserved under isomorphism they said
well yes, that is true! so the correct interpretation would be that G/H ~= G/K implies H~=K
thorn jay
the isomorphism class of G/H is not an algebraic property of the group H
i have doubts believing that
tbh
tardy hedge
Which 1
thorn jay
~~well yes, that is true! so the correct interpretation would be that G/H ~= G/K...
this
tardy hedge
I guess my question is like the reverse lol
velvet hull
i have doubts believing that
yeah I'm actually having second thoughts as well
elfin wraith
I think basically the same counter example as above works
Take like Z_2(+)Z_4 and the subgroups Z_2(+)<(0,2)> and 0(+)Z_4, those should both give you Z_2
thorn jay
Basic question but if K1/N = K2/N can you conclude K1 = K2?
this is fairly poorly worded because if K1/N = K2/N then yes of course K1 = K2 but thats like trivial. It is not the case that K1/N ≈ K2/N => K1 ≈ K2
elfin wraith
I guess a trivial example could be rings R\subsetneq S and then take N to be the ideal generated by a unit in R
Edit) im just a fucking idiot
I think that works, both should just give you zero
tardy hedge
this is fairly poorly worded because if K1/N = K2/N then yes of course K1 = K2 b...
It’s not trivial to me, something about it confused me
thorn jay
well K1 is the union of all elements of K1/N and K2 is the union of all elements of K2/N
so if K1/N = K2/N, then K1 = K2
i just dont think its useful to think on the level of sets and elements lol
elfin wraith
~~I guess a trivial example could be rings R\subsetneq S and then take N to be t...
Lets pretend I didnt talk about ideals over a field for a minute
thorn jay
i was about to ask "ideal generated by a unit"?
elfin wraith
Im doing galois theory rn im a bit field coded
But yes I think my example does now work lmao
Well no I guess it still doesnt, im a fool
But theres something out there, ill come back when I decide to have a brain wave
elfin wraith
Yeah thats much better lol, idk why I didnt just keep running with that as an example like I did for the question before
thorn jay
~~well yes, that is true! so the correct interpretation would be that G/H ~= G/K...
Z4 x Z2 has two subgroups Z4 x 1 and 2Z4 x Z2 which are nonismorphic, but which give isomorphic quotients (Z2)
elfin wraith
Z4 x Z2 has two subgroups Z4 x 1 and 2Z4 x Z2 which are nonismorphic, but which ...
Yah this is the example I put above
tardy hedge
i just dont think its useful to think on the level of sets and elements lol
But u need to for that dont you
thorn jay
K1/N = K2/N is thinking on the level of sets and elements
tardy hedge
Context was for showing the bijection in the correspondence theorem
tardy hedge
Yes for the same reason, we can just think in terms of sets and elements ? No algebra?
thorn jay
yes
elfin wraith
Context was for showing the bijection in the correspondence theorem
If its of interest at all, I had to painfully spell this out in my first ring theory class lol
tardy hedge
Haha
elfin wraith
We were specifically told to be extremly specific and clear so it does stick in my mind, which I guess is/was helpful
languid trellis
Hello good ppl
I am struggling with part c
Wait
hold the phone
Max ideals of Q[t]/(t^2(t-1)) correspond to max ideals of Q[t] that contain t^2(t-1). As Q[t] is a PID, its maximal ideals are genereated by a signle irreducible polynomial, so this ideal looks like (f). The condition (t^2(t-1)) \subset (f) means that there exists a g such that gf = t^2(t-1). So surely f = (t) or f = (t-1) ?
hmmmmmmmm
Would appreciate any hint
Please note this is a homework problem
elfin wraith
Looks good to me so far. Is this the comalg hw?
languid trellis
yeah
elfin wraith
Ive heard people are struggling haha
languid trellis
the first homework sheet hasbeen tough icl
elfin wraith
Q1 is striaght up just a classic example in AG
languid trellis
I've spent like 3 or 4 days on it
just about broken through
I think he's trying to get rid of ppl
elfin wraith
Yeah I dont think youre alone, it does seem tough though
languid trellis
It's been really satisfying when the right argument has come to my head though
elfin wraith
I think he's trying to get rid of ppl
I think he just assumes people are cracked, he wants me to work on some open K theory problem 
languid trellis
oh awesome
elfin wraith
Well, he suggested that to me, wants me to is maybe strong
languid trellis
lol
i'd prefer the lecturer assume im competent rather than assume im stupid
like in galois theory
actually kill me dawg
elfin wraith
Tbf, people gave him reason to assume were dumb
elfin wraith
The first lecture I went to bro was pissed
languid trellis
also q3 is disproportionately easy lmao
languid trellis
The first lecture I went to bro was pissed
oh yeah LOL
the first lecture he gave a high level overview of galois theory, got a few emails and then went back to basics
elfin wraith
I cant blame him, ive never heard people talk over a lecturer like that and it seemed that people emailed him all sorts of dumb ass questions
languid trellis
yeah
tardy hedge
Yeah I dont think youre alone, it does seem tough though
Bruh u guys are at the same school?
elfin wraith
Yeah
tardy hedge
Bruh
languid trellis
I cant blame him, ive never heard people talk over a lecturer like that and it s...
I did fall asleep in this one icl
like i was so bored
elfin wraith
Yeah Ive been to that, and I techinically attended last friday and thats been it
languid trellis
Yeah Ive been to that, and I techinically attended last friday and thats been it
todays lecture was good. shame you missed it
elfin wraith
Im only doing it because theres fuck all else sem1 algebra wise
languid trellis
it was gareth tracy doing some group theory. he's so enthusiastic and kind and nice
and we proved a nice reuslt about S_5
well two actually
elfin wraith
todays lecture was good. shame you missed it
Where on earth are we? Theres nothing to suggest where we should be from week to week
rapid cave
and we proved a nice reuslt about S_5
A5 being simple?
languid trellis
no
it was that S_n has a unique subgroup of order n!/2 and that (12345) and any transposition generate S_5
the argument is super neat
big fan
languid trellis
Where on earth are we? Theres nothing to suggest where we should be from week to...
I think we're far behind is where we are
rapid cave
the second result can be generalised to Sn
elfin wraith
Im at 6.6 just now, im guessing were around there?
elfin wraith
Oh
elfin wraith
Didnt he cover splitting fields in lecture 1 lmao, whats happened
languid trellis
yeah bro idk
im just as confused as youa re
also there were about 20 ppl at the lecture today
lmfao
rapid cave
weird lectures you got there
languid trellis
just this module
elfin wraith
I mean im not fussed, ill take an easy course because my next semester is cooked
elfin wraith
just this module
Lean is worse, not a clue whats happening there, weve still not done anything
languid trellis
Lean is worse, not a clue whats happening there, weve still not done anything
average lean moment
Okay i actually need to find the maxmial ideals of R/I now
then q3 is easy and im done yay
elfin wraith
Yeah sorry, and I should get back to doing some galois since ive not attended any of the lectures
languid trellis
q1 took me so long and then i realised everything is just the evaluation map
lmfao
languid trellis
Yeah sorry, and I should get back to doing some galois since ive not attended an...
lol no worries. you're way ahead dont worry about it
elfin wraith
q1 took me so long and then i realised everything is just the evaluation map
Yeah people on the warwick maths discord seemed to be struggling, I wouldnt worry lol, its also a pretty classic example in AG though so its good to understand
languid trellis
oh i should actually use that server
elfin wraith
Its not like super serious from what ive seen so far tbh, but its nice just for info about courses
languid trellis
yeah
thorn jay
q1 took me so long and then i realised everything is just the evaluation map
free objects my beloved
languid trellis
Yeah people on the warwick maths discord seemed to be struggling, I wouldnt worr...
I've honeslty really been enjoying the difficulty of commalg so far. Like the level of sophistication is a breath of fresh air and has let me see the objects in a different light yk
like when he defined a subring of a ring R to be a ring S such that there is an injection S to R
same with quotient
and also being forced to work with quotient rings is actually really nice
like I needed someone to shove C[X,Y,Z]/(some bullshit) down my throat
and i feel more comfortable with them now
also this is the first time i've actually used maps to study algebraic objects
at least properly anyway
my first instinct for any question is 'what map can i use to make this easy'
its been really good
big big fan
hopefully my request to take this module gets accepted, my tutor and the lecturer already agreed so just need dugs to agree
and im now free
assignment has been done
yippee
elfin wraith
hopefully my request to take this module gets accepted, my tutor and the lecture...
Oh are you just in year 2
But also yeah I know how you feel, my noncom course last year was the exact same and it taught me more maths than any course I can think of in my UG, like more than just the content it really helped my maturity and how I look at algebra
I unfortunately cannot say the same of my comalg class, that was a joke. I did get highest in the year and an almost perfect result though so I’ll take it
languid trellis
hm
elfin wraith
And just RIP comalg2
rapid cave
do number theory
elfin wraith
I mean it’s hard, all miles Reid courses are hard, but I would’ve actually got to do some ring theory (the thing I want to study)
I’m instead now just taking algcurves with miles meaning I’ll get cooked on a subject I’m worse at
elfin wraith
gl
Also taking Lie algebras, category theory and cohomolgy and Poincaré duality that semester, can only end well
languid trellis
i've heard lie algebras is okay
elfin wraith
Yeah I’ve also been told that, apparently it’s quite chill
thorn jay
do number theory
no
rapid cave
yes
elfin wraith
And I’m trying to pre-study cohomology, but I’ve heard that John sets a fair exam
If only the man could respond to an email as well
languid trellis
john slightly terrifies me
he strikes me as very old english
especially his voice
idk
maybe thats just me
languid trellis
If only the man could respond to an email as well
lmfao. this reminds me i need to start emailing round for urss
elfin wraith
Maybe I’ll get to find out one day if he ever responds to my email asking to chat
My other options for supervision I’m not currently in love with, but I’ve got another meeting next week so we’ll see how that goes, it’ll probably be an alggeo topic though so meh
languid trellis
not k theory with marco?
elfin wraith
Thats probably the current front runner. The other option I have sounds really cool but im not sure that I can get much done with it in 3 months, and I didnt have the most amazing first impressions of the supervisor
elfin wraith
Im metting with Gavin Brown next week though and although its AG, it seems like cool AG and he seems like a really nice guy
languid trellis
gavin brown is a top bloke
and it is cool ag
its like blowups and cuspodial points and stuff
I had linear algebra with him and i went to every lecture jsut because he was so sound
elfin wraith
Which like meh I dont love, but he does cool non-com deformation theory stuff
crystal vale
and it is cool ag
So this semester you have commutative algebra
languid trellis
yes
rapid cave
you are making me jealous 😵💫
crystal vale
Does your professor maintain notes ?
rapid cave
you are making me jealous 😵💫
cannot take more then one course this year
crystal vale
I want to do commutative algebra too
languid trellis
Does your professor maintain notes ?
yes but they're mega terse. like a whole semester of content in 46 pages
its fucked
rapid cave
thats very dense
crystal vale
🥲
rapid cave
my proffesor's CA notes were 300 pages
elfin wraith
Its likely similarly, or possibly less, dense compared to A&M
languid trellis
the lectures are good though
white oxide
Could I please have a hint for <=? I am assuming that x is not a zero divisor, so that x \notin (0:a) but x \in p. I want to produce some ideal q such that (0:a) \subseteq q \subseteq p, which contradicts the minimality of p. My instinct was to look at (x) or (ax) or something but that didn't work
tribal moss
What is the notation (0:a)?
white oxide
Just the annihilator of a
rocky cloak
What is the notation (0:a)?
x such that xa in (0)
tribal moss
I see.
white oxide
I feel like this would be a lot easier of the zero ideal was decomposable
rocky cloak
Could I please have a hint for <=? I am assuming that x is not a zero divisor, s...
I mean what would make sense would be to look at a maximal ideal between (0:a) and p that doesn't contain x and show that this is prime.
Haven't checked if that works, but that would make sense
white oxide
I mean what would make sense would be to look at a maximal ideal between (0:a) a...
Sorry wdym aren't maximal ideals prime
rocky cloak
Usually yeah, that's why that would be my guess
white oxide
hm
thorn jay
Usually yeah, that's why that would be my guess
usually?
rocky cloak
_usually_?
Yeah, if you pick the poset completely arbitrary (say just containing one ideal), then obviously being maximal says nothing about primality
thorn jay
oh lol like that then yeah
balmy python
does every group have a set of generators?
wait that's a stupid q
😭
ofc every group does
i guess you could also have an infinite group with infinite generators with finite order right?
rapid cave
i guess you could also have an infinite group with infinite generators with fini...
yes
take a direct sum of a countable amount of copies of Z/2Z
then all the elements of the form (0,......., 0, 1, 0, 0, ....) generate this group and they are all of order 2
proud vigil
not every group has a minimal set of generators though
balmy python
take a direct sum of a countable amount of copies of Z/2Z
direct sum?
rapid cave
direct sum?
tuples where a only a finite number of their elements is non-zero
this forms a group where the operation is done element-wise
south patrol
In other words: any vector space over Fp can be viewed as an abelian group where each element has order 1 or p (in fact this is the same data). If you take an infinite dimensional vector space over Fp then this gives you an abelian group as you desire
tribal moss
In other words: any vector space over Fp can be viewed as an abelian group where...
(in fact this is the same data)
Hmm, in which sense?
Consider F9 = F3[i] as a vector space over itself. The abelian group structure is just (C_3)^2, and that doesn't seem to preserve knowledge about how to scale a vector by i.
south patrol
Okay sorry I mean Fp
tribal moss
Okay, then. :-)
elfin wraith
In other words: any vector space over Fp can be viewed as an abelian group where...
Could you expand on that? I’m not sure I’m seeing what you’re saying
tribal moss
Potato was generalizing the answer to Pentium's question "you could also have an infinite group with infinite generators with finite order right?"
balmy python
take a direct sum of a countable amount of copies of Z/2Z
Z/2Z is just the group consisting of residue classes of 0 and 1 right?
tribal moss
AKA the (unique up to isomorphism) group with 2 elements.
balmy python
i hate this weird notation 😭
it's so similar to quotient vector spaces
oh wait
quotient groups
are quotient vector spaces quotient groups 😭
south patrol
Like point was you can view this as a question about vector spaces where the answer is clear
south patrol
are quotient vector spaces quotient groups 😭
Yes
balmy python
ohhh
south patrol
The underlying group of a quotient vector space is the quotient of the underlying groups
rapid cave
i hate this weird notation 😭
this comes from ring theory, where we quotient the ring Z by the ideal 2Z
tribal moss
And quotient rings are also quotient groups.
rapid cave
yes xD
tribal moss
In each case, with additional structure that is preserved under certain conditions on the subgroup you're quotienting by.
rapid cave
just forgetting the ring structure and keeping only the abelian group structure
balmy python
take a direct sum of a countable amount of copies of Z/2Z
okay so these form an infinite dimensional vector space right
thorn jay
yes
balmy python
Can someone tell me if my proof is right and if there are any other nice ways of targeting this problem?
Prove that every infinite group has infinitely many subgroups.
So I split this into the case where the group either has a finitely many generators or infinitely many generators.
In the case where the group has finitely many generators, at least one of these generators must generate an infinite group which we can call <g>. However, we have infinitely many subgroups of <g> being <g^n> for each n \in N.
In the case where the group has infinitely many generators, we can take the cyclic group generated by each individual generator and then we have infinitely many subgroups.
south patrol
Sounds good, just I would be careful with the infinitely many generators bit as you need to justify why you get infinitely many distinct subgroups
mint seal
right, I was about to ask why that's so
also for the finite generators part, how do you know you end up with infinitely many distinct ones from the <g^n>
south patrol
I would write it like this: suppose every element has finite order. Consider the finite subgroups H. Since they're finite, each omits some element g, but then <H,g> is some bigger finite subgroup, so there must be infinitely many of them
And I would have said like if you have an element of infinite order then you reduce immediately to Z where it is obvious. But this is just a rephrasing lol
I just think it is cleanest to divide like this (by orders of elements) rather than the "number of generators"
thorn jay
Let X be the set of <g> for g ∈ G. Suppose X is finite. Considering any element g, then by finiteness there must be some n, m for which <g^n> = <g^m>
=> g^n = (g^m)^k = g^mk for some k
=> g^{mk-n} = 1
=> g has finite order
=> all elements of G have finite order .
Thus X is a finite set of finite groups which cover G (every element of g is indeed contained in a cyclic subgroup), so G must be finite.
no cases needed
south patrol
I would write it like this: suppose every element has finite order. Consider the...
Or even just like, consider the finitely generated subgroups. If there are only finitely many of them, then chuck all the generators together to get a maximal finitely generated subgroup, which is nonsense
tribal moss
How about a middle way:
Consider all the subgroups of the form <g>.
If there are infinitely many of them, then we're done.
However, if there are only finitely many, then at least one of them must be infinite, because together they cover G. But if there is an infinite cyclic subgroup, then there are infinitely many of them, a contradiction.
thorn jay
basically what i did lol
tribal moss
I just wanted to get rid of the nitty-gritty about exponents.
elfin wraith
I would write it like this: suppose every element has finite order. Consider the...
I like this, this is pretty slick
I think I would’ve come up with enpeaces proof but this one is nice and simple
tribal moss
I would write it like this: suppose every element has finite order. Consider the...
Hmm, that goes a bit fast for my taste. It's not assumed that the group is abelian, so a finite number of elements each of finite order can generate an infinite subgroup. So how do we know <H,g> is finite?
thorn jay
no, take the extreme example of C2 \ast C2
elfin wraith
Oh yeah good point
south patrol
Hm isn't it a finitely generated group where every element has finite order, hence finite
tribal moss
I mean, it probably is finite since we have assumed all elements have finite order -- but that assumption doesn't seem to be used explicitly on the way to <H,g> being finite.
south patrol
I am using that implicitly
thorn jay
whats an infinite nonabelian group where every element is torsion and the set of orders of the elements is finite?
south patrol
Ah i am still wrong. Interesting
Burnside's problem
Thank you for pointing this out
thorn jay
Burnside's problem
oh lol ofc this turns out to be an open problem
south patrol
Nah it was answered but in the 60s
I just forgot about it lol
But yeah there are weird examples of fg groups where every element has finite order but the group is still infinite
I guess it sounds plausible if you have crazy relations lol
thorn jay
nontame groups huh
south patrol
I should not do nonabelian groups
elfin wraith
Ah i am still wrong. Interesting
Well damn, I thought your solution was nice but I guess it was too nice haha, for the best though because I wasn’t aware of burnsides problem, that’s a really cool result
And from a brief scan of the Wikipedia the counter examples are weird
Or perhaps not weird, but like, big
thorn jay
lol it was solved by a theorem in noncomm homalg
what's it with stuff of 3 elements and breaking everything
south patrol
Yeah this is something I am like happy to be wrong about as I have learnt smth lol
thorn jay
i wonder if burnside problem is true for groups generated by 2 elements
balmy python
Sounds good, just I would be careful with the infinitely many generators bit as ...
two different generators generate different groups no?
balmy python
also for the finite generators part, how do you know you end up with infinitely ...
isn't that kinds trivial
thorn jay
two different generators generate different groups no?
no
g and g^2 generate the same group if o(g) is odd, f.e.
balmy python
wait
how do you get o(g) if you have infinite order
(im half asleep my bad)
its a finite order set right
thorn jay
if your group has finitely many subgroups, its easy to see that every element must be of finite order (although you still have to prove that ofc), the hard part starts there
balmy python
g and g^2 generate the same group if o(g) is odd, f.e.
say o(g)= 5
so we have
waittt
they do generate eht esame group
tribal moss
i wonder if burnside problem is true for groups generated by 2 elements
The Wikipedia article seems to say it's an open problem whether "2 generators and every element has order 1 or 5" forces a group to be finite.
Ah, but it also says that there are infinite groups with two generators and all elements orders <= 8000.
vapid vale
xd
elfin wraith
Ah, but it also says that there _are_ infinite groups with two generators and al...
I really wonder how you actually work these things out, like presumably there’s at least some amount of pruning done, certain properties the groups either must or couldn’t have, but beyond that, do you just make GAP chug away?
I should look some stuff up about this tomorrow, that’s a cool problem
thorn jay
The Wikipedia article seems to say it's an open problem whether "2 generators an...
is the rest solved? weird
thorn jay
Ah, but it also says that there _are_ infinite groups with two generators and al...
thats fucky
why cant groups just be niceeee
thorn jay
I should look some stuff up about this tomorrow, that’s a cool problem
do share if you find smt interesting abt it
fading acorn
why cant groups just be niceeee
theorem: all groups are free, and they admit a natural lie group structure
thorn jay
preach
in fact, every algebraic structure is in fact free in every variety it belongs to
i will not take any questions
karmic moat
Theorem: all rings are unital and commutative and noetherian 🤤
thorn jay
approved!
karmic moat
Lfg
thorn jay
theorem: if φ is a Noetherian coherent condition such that Fφ is closed under taking direct limits, then F√φ is too
🙏 🙏 🙏 🙏 i beg you
karmic moat
@annals can we look into this
thorn jay
i am missing lore i think 
karmic moat
The math journals
Like that theorem was so awesome we should get the big math journals involved
thorn jay
oh lmao
well i do suppose the theorem says a lot
a corollary is that, if K is closed under embeddings, equationally Noetherian, and closed under direct limits, then ISP(K) = Q(K)
determining when ISP(K) = Q(K) is a pretty important problem Id say
thorn jay
Tbh idk what those mean
ISP(K) = all algebras which can be embedded into an arbitrary product of members of K
Q(K) = smallest quasivariety containing K. A quasivariety is a class of algebras axiomised by formulas of the form
∀x1 ... ∀xn ( p1(x̅)=q1(x̅) ∧ ... ∧ pm(x̅)=qm(x̅) → t(x̅)=s(x̅) )
a great example is reduced rings
tardy hedge
Bro is speaking but no one’s hearing 👂
thorn jay
crying tbh
thorn jay
axiomised as the class of commutative rings satiafying the set of sentences ∀x (x^n = 0 → x = 0) for all n = 1, 2, 3, ...
thorn jay
ISP(K) = all algebras which can be embedded into an arbitrary product of members...
ISP(K) is called a prevariety
these are also axiomisable but by implications where the LHS can be an AND that ranges over an arbitrarily large set of equations over an arbitrary amount of variables
(you may also run into size issues with proper classes and such? but i think i saw a theorem somewhere that says its all okay)
thorn jay
these are also axiomisable but by implications where the LHS can be an AND that ...
but this is horrible to work with
lets be real
karmic moat
So in the case of reduced rings, what does ISP(K) = Q(K) intuitively mean
I.e. why is it important to know
thorn jay
So in the case of reduced rings, what does ISP(K) = Q(K) intuitively mean
it means that reduced rings are decided by a property that can be checked ""locally"" i.e. on finite sets (or finitely generated subrings)
karmic moat
Oh nice
thorn jay
it turns out that, when K is nice (so-called equationally Noetherian), then Q(K) is actually the local closure of ISP(K), that is, the class of algebras A where every finitely generated subalgebra of A is in ISP(K)
I am pretty sure™ that this local closure property is equivalent to the φ-radical being an algebraic closure operator
but thats beside the point
tardy hedge
axiomised as the class of commutative rings satiafying the set of sentences ∀x (...
Could there be different sets of sentences that axiomize the same class
thorn jay
oh yeah def
tardy hedge
Yeah id imagine so
tardy hedge
it means that reduced rings are decided by a property that can be checked ""loca...
Wow thats pretty cool
thorn jay
thats why, in theory, you usually want to work with the set of all such sentences that are satisfied by the class.
tardy hedge
You dont want to leave one out?
thorn jay
thats why, in theory, you usually want to work with the set of all such sentence...
This yields a "Galois connection" between classes of algebras and sets of quasiidentities (sentences of the form described above). I put it in quotations because the collection of classes of algebras is of course not a set
tardy hedge
axiomised as the class of commutative rings satiafying the set of sentences ∀x (...
And is the formalism here like in propositional logic or smth?
karmic moat
I am pretty sure™ that this local closure property is equivalent to the φ-radica...
here an "algebraic closure operator" is analogous to a closure operation on an ideal?
thorn jay
it then turns out that the "closed classes" are exactly the quasivarieties, and the corresponding closure operator is given by ISPP_U, which is very nice if you know model theory lol
thorn jay
here an "algebraic closure operator" is analogous to a closure operation on an i...
algebraic just basically is a lattice-theoretic characterisation of the "being able to be checked locally"
tardy hedge
This seems like some hardcore shi ngl
karmic moat
algebraic just basically is a lattice-theoretic characterisation of the "being a...
ah okay
thorn jay
And is the formalism here like in propositional logic or smth?
formalism here is model theory
tardy hedge
Oh ok, idk about that
thorn jay
but you dont need to know anything abt it haha
karmic moat
sharp knows model theory
and ultraproduct
like the user ultraproduct not the thing
thorn jay
all the class operator stuff (the I, S, P, P_U, Q things) is mainly universal algebra to me
karmic moat
oh i guess you're in #foundations so you already know that oops
thorn jay
all the class operator stuff (the I, S, P, P_U, Q things) is mainly universal al...
the connections with logic are nice but the moment im studying them algebraically
or... geometrically in a sense?
karmic moat
👀
karmic moat
thorn jay
(technically)
tardy hedge
Bruh
karmic moat
(universal^{TM}) algebraic geometry
thorn jay
its called algebraic geometry because youve got a contravariant functor to Top okay
tardy hedge
At a certain point do you know still what youre studying
I guess u do if you have intuition for it
thorn jay
its hard to characterise
karmic moat
its called algebraic geometry because youve got a contravariant functor to Top o...
is it a sheaf 👀
thorn jay
not yet sadly
karmic moat
neuron deactivation 💔
thorn jay
havent found a good substitute for "local quotient of polynomials"
frustratingly, general algebraic structures do not come equipped with a multiplication which is commutative and distributes over all the other operations
thorn jay
neuron deactivation 💔
im sorry 😔
karmic moat
frustratingly, general algebraic structures do not come equipped with a multipli...
not distributing is crazy
thorn jay
try not having any binary operation at all
karmic moat
i think people just call those sets
karmic moat
oh joy
thorn jay
yup
tardy hedge
Uhh
thorn jay
like the best i can do is the sheaf of local polynomial / term operations, but for varieties this is the constant sheaf 💔
tardy hedge
Oh Hell Nah!!!
karmic moat
this reads like no one other than Marshall Hall has ever used it
karmic moat
no i can feel it in my bones. something big is coming soon in the field of planar ternary rings
karmic moat
W
thorn jay
well it does seem really interesting
just, people dont care about ts anymore
its all AG now
those darn kids
karmic moat
as it should be
karmic moat
when in doubt just sheafify
thorn jay
well the problem is that the affine spectrum sheaf is so specialised to prime ideals
like it relies heavily on the properties of prime ideals
properties which would need to have an analogue for any class closed under embeddings both contained in and containing the class of integral domains
this shit is just too general 💔
why am I ranting about my shitty researxh here I should really go to sleep
karmic moat
cya
thorn jay
trick question ive been trying that for 3 hours alr
karmic moat
last night i watched a bunch of youtube videos on like super mario 3 tas and how some guy got it in 5 miliseconds
after going down that rabbit hole for a while i went to sleep satisfied
could be the move
karmic moat
that's how it often goes
thorn jay
theorem: if φ is a Noetherian coherent condition such that Fφ is closed under ta...
imma be fr ts is keeping me up
it feels so close
karmic moat
maybe you'll dream up a proof or counterexample
thorn jay
last night i watched a bunch of youtube videos on like super mario 3 tas and how...
5ms??
karmic moat
maybe you'll dream up a proof or counterexample
i had a prof who said that they dream about donaldson-thomas invariants
thorn jay
maybe you'll dream up a proof or counterexample
a counterexample would be detrimental
karmic moat
5ms??
ye
lemme find the video
it was super interesting
this was the first one i watched by this guy, he explains a 0.2 tas which was done some number of years ago
https://www.youtube.com/watch?v=fQYX_AVxGq0
then that guy made a 5ms tas and explains it in this one
https://www.youtube.com/watch?v=pK7hU-ovUso
karmic moat
damn i didn't even know that song had a name
thorn jay
toby fox music has been getting popular too
i am very normal about toby "radiation" "tricky tony" "mr annoying dog across time and space" "specimen" "SPAMTON G SPAMTON certified" "gave toriel a homestuck typing quirk" "tricky tony v2" fox
tardy hedge
well it does seem really interesting
it did?
thorn jay
okay i think ive ran out of funny things to put there
karmic moat
damn... fake ass fan...
tardy hedge
true
tardy hedge
Lol
karmic moat
i don't think i've played undertale nor deltarune
though i've watched playthroughs of undertale
thorn jay
you might ask "what hype" and id agree
tardy hedge
i havent played a videogame in like a year
thorn jay
i don't think i've played undertale nor deltarune
undertale and deltarune were life changing for me
tbh
tardy hedge
undertale is really that good?
karmic moat
i play old school runescape because it's mostly tedious like math
thorn jay
depends on your personality
karmic moat
i think my life changing games were probably super mario galaxy 1 and 2
tardy hedge
i think my life changing games were probably super mario galaxy 1 and 2
i wouldnt say life changing for me but i played a ton of those as a kid
thorn jay
personally it resonated with me because toby tox humor is like exactly my kind of humor, and im a sucker for metanarratives
and overall the whole story and music just worked for me
tardy hedge
is it turn based rpg?
thorn jay
mixed with bullet hell
karmic moat
super mario galaxy 1 and 2 made me fall in love with video game soundtracks
tardy hedge
zelda for me^
thorn jay
as far as gameplay goes deltarune is far superior imo
thorn jay
nothing beats zelda music imo
deltarune and undertale solos
tardy hedge
ill look them up
thorn jay
not biased at all btw
karmic moat
in super mario galaxy 1 the sound as the game progresses
tardy hedge
but imma say no prolly
karmic moat
as u fix ur starship and it gets better and better
thorn jay
haha yeah thats fair
tardy hedge
as u fix ur starship and it gets better and better
tru
karmic moat
and everything starts lighting up and bro omg the way the music changes
tardy hedge
wow nintendo sucks ass now hey
thorn jay
the music works best in the context of the game when you recognize all the different leitmotifs
karmic moat
instruments get added every time you improve your ship
tardy hedge
i used to be big fan and still am but not if theyre bad
karmic moat
literally chills every time i listen to it now
i think i actually cry lol like it's so god damn awesome
thorn jay
wow nintendo sucks ass now hey
as do all big companies
tardy hedge
true
karmic moat
zelda botw was also good in that aspect
thorn jay
i think i actually cry lol like it's so god damn awesome
thats me but UT/DR music
and minecraft music
tardy hedge
botw soundtrack is good but not close to the other games
karmic moat
i can't remember the name of the town but it's the one you build yourself and u invite ppl from each of the four regions over
tardy hedge
terry town or smth
thorn jay
and minecraft music
volume beta >>>
karmic moat
and when someone new comes to the town the instruments associated to their region are incorporated into the theme
yea terry town
what y'all know about wii tanks tho
thorn jay
i love that vid
karmic moat
rivals beethoven and the like
i love scruffy's channel
super good high quality stuff
tough raven
i think my life changing games were probably super mario galaxy 1 and 2
Oh that's interesting; how so?
tough raven
super mario galaxy 1 and 2 made me fall in love with video game soundtracks
Bruh too real
karmic moat
they were a huge part of my childhood
thorny cobalt
you must listen to the celeste soundtrack 🙏
karmic moat
it was basically my first experience with mario games outside of the DS
and i used to play that one, among other mario games for the wii, with my dad a lot
idk where the wii went so we dont play it anymore
thorn jay
you must listen to the celeste soundtrack 🙏
ive heard its great
you two just popped out of nowhere
karmic moat
@delicate orchid knows a lot about celeste soundtrack
thorn jay
the fucking name xhange again 😭
karmic moat
it's even better
thorn jay
i should change my name
thorny cobalt
ive heard its great
yeah its peak
karmic moat
it's got me thinking like damn maybe i need to drink booze and gamble too
thorny cobalt
never got bored of it after like 200 hours on the game
thorn jay
yeah its peak
peak.. you say? i know peak...
thorny cobalt
~~althgouht tbf 100 hours of those were mods with diff soundtracks ~~
karmic moat
i played celeste obsessively for like a week or so
thorny cobalt
i played celeste obsessively for like a week or so
did you finish it
karmic moat
i tried getting the corner jump (?) down
thorn jay
a friend of mine is bugging me to play in stars and time apparently its music is great too
karmic moat
no i ended up raging so hard i stopped playing
thorny cobalt
i tried getting the corner jump (?) down
corner jump in the first week 💀 💀 💀
karmic moat
mega skill issue
thorny cobalt
yo
karmic moat
well i started playing celeste because i found out about it through this server
namely ryc and wew, both who liked speedrunning it
thorny cobalt
celeste speedrunning is so cursed 😭
karmic moat
it looks so awesome
im a sucker for games where u can min max everything
hence why i love old school runescape
thorny cobalt
its all like chained ultras, cboosts, spinner stuns, crouch dash, and other funny gm-level speed tech
karmic moat
yea tbh i dont think i even played the game that much i just spent a lot of time trying to do crouch dashes and corner jumps
thorny cobalt
noooo 💔
crouch doush u can keybind
corner jumps are js really annoying
i treid learning them but you need like good input timings and i couldn't get it realtively consistenl
karmic moat
for me keybinding takes some of the fun out of it
the fun is in the pursuit of precision
holy poetic
but yeah i like the process of learning those things more than actually doing them
thorn jay
the fun is in the pursuit of precision
what a quote
thorny cobalt
doing all of this as soon as you get the game is crazy work 😭
thorn jay
"there are cathedrals everywhere for eyes to see" ahh