groups-rings-fields

[rocky cloak]

I mean specifically product groups.

Like the order of (a, b) in AxB in terms of the order of a and b in A and B respectively

I mean, 1800 is divisible by 25. But maybe I'm misunderstanding what you're saying

[fiery dirge]

No I don't know that

[fiery dirge]

I don't understand the task well

How can there be many groups of order 1800? I can only think of Z_1800, D_900 and maybe a few. And then there's as you mentioned group products that can create further variations

But I don't recall when a group product is valid

I can't make a product out of anything like Z_2 x Z_900 might not be isomorphic to Z_1800

[delicate orchid]

got me curious so I just checked, there are 749 groups of order 1800 up to isomorphism

[elfin wraith]

Fewer than I expected tbh

[wraith cargo]

D_900 isn't abelian

[delicate orchid]

Z_1800 has an element of order 1800, Z2 x Z900 doesn't they can't be isomorphic

[wraith cargo]

also do you know the structure theorem for finite abelian groups?

[rocky cloak]

They are indeed non-isomorphic.

I'm not sure what you mean by them being "valid" though...

[fiery dirge]

💀

[fiery dirge]

Ahh I see

[elfin wraith]

Thankfully that doesn’t really matter because there’s a very strong theorem which tells you how to do this, the classification for finite abelian groups

[fiery dirge]

I don't know what's that

[delicate orchid]

it's ok, very few of those are abelian and you don't actually need to list a single group to do this

[fiery dirge]

So what's the theorem for abelian classification?

I might know it but I don't know

[rocky cloak]

Luckily there's only like 12 abelian ones

[thorn jay]

whats the asymptotic growth of the number of groups

[proud vigil]

https://en.wikipedia.org/wiki/Finitely_generated_abelian_group

[wraith cargo]

you can write any finite abelian group as the direct sum of groups of the form Z/p^jZ

[delicate orchid]

think about what Jagr is saying about the order of (a,b) and try and deduce when Z_a x Z_b is isomorphic to Z_ab using the fact that Z_ab has an element of order ab

[delicate orchid]

didn't want to say that exact number as this kinda spoils the problem a bit lol

[elfin wraith]

I don’t actually know how the number of groups grows with the size of the set tbh

[rocky cloak]

I think the bound is something like
n^(2/27 log(n)^2 + something)

[proud vigil]

thats so precise :0 i thought it would be a little prettier if an asymptotic existed

oh i guess a bound isnt a tight bound

[wraith cargo]

combinatorialists in the year 3025 when they find an explicit formula

[rocky cloak]

There are more than
p^(2/27 k^2(k - 6))
groups of order p^k appearantly

Which is less horrible at least

[proud vigil]

oo

i dont understand the 2/27 😭

[delicate orchid]

is that true for p = 2

oh more than

it's a lower bound duh

[rocky cloak]

You and me both

[proud vigil]

my first thought was something like

order n corresponds to k! ish for some k, and then maybe just count the subgroups of Sk? cycle structures or something ,,, maybe i dont want to think abt this anymore

[elfin wraith]

This looks perplexing and weird but that does sum up combinatorics so sure I buy it

What’s the whole 5/8ths thing about groups as well? I think wew told me about that at some point

[rocky cloak]

Yeah, a group is abelian if more than 5/8 of it commutes

[elfin wraith]

Yeah if you have a compact group the probability two elements compute is 5/8

Crazy

That’s just whacky

[rocky cloak]

I mean, sounds pretty reasonable to me

Not even a whacky number, 1/8 more than a half

[thorn jay]

yeah was about to ask what the FUCK is 2/27 doing in both the formulas

[elfin wraith]

I’m sure it’s reasonable enough if you know any of the theory, but I don’t and that just seems surprising

[thorn jay]

popping up once in a formula - sure coincidence ig

[rocky cloak]

I think it's very easy to prove

[elfin wraith]

I’m guessing it involves some measure theory so probably not for me lol

[rocky cloak]

I mean, just stick with finite groups if you don't want measures

[thorn jay]

finite groups is just with the counting measure smh

[elfin wraith]

That’s true I suppose, I guess I never really considered it for finite groups to get a sense of why it would be true

Still, a cool result nonetheless

[proud vigil]

wat

is there a counter example for when 34/56 of the group commutes 💔

[rocky cloak]

And then Q8 achieves the bound.

Two elements have exactly 5/8 probability of commuting.

[proud vigil]

oo I guess that makes sense

[rocky cloak]

If the probability is more than 5/8, then it's a 100%

[proud vigil]

right that makes sense

[delicate orchid]

By sheer chance you’ve chosen a denominator exactly 1/3rd of the order of PSL(2,7) which amuses me

[proud vigil]

my brain isn’t understanding how that’s possible but that’s okay

[delicate orchid]

Maybe 102 elements commute who knows 🫨🫨🫨

Jagr run the numbers what’s the prob for PSL(2,7) or as I like to call it GL(3,2)

[thorn jay]

crazy trivia

"what common group has order 3 times the denominator of 34/56"

[delicate orchid]

It’s the second smallest non-abelian simple group it’s not that crazy

[elfin wraith]

And what fool wouldn’t just know that off the top of their head

[delicate orchid]

Well you know the smallest I hope

[twilit wraith]

i mean 168 is many mathematicians favorite number

[delicate orchid]

And the third smallest is A6

[twilit wraith]

number of hours in a week, number of primes between 1 and 1000, order of second smallest nonabelian simple group

[delicate orchid]

So u might as well know the one in the middle

[elfin wraith]

I knew the smallest but I’m not sure I knew how the ordering went past that

Or at the very least I’ve never thought about it

[thorn jay]

yes thats basically the extend of my ordering knowledge

[delicate orchid]

Past A6 idk myself. I think you get more PSLs arising as finite groups of lie type and then it gets real messy

[delicate orchid]

PSL(2,7) is a good group to know

[thorn jay]

i think for me at this moment it would be more useful to remember the classification of minimal algebras

[twilit wraith]

looks like PSL(2,8) is the fourth smallest

[thorn jay]

why?

[twilit wraith]

168 comes up a lot in group theory

[elfin wraith]

So you can win the quiz

[delicate orchid]

It’s got D_8 as a Sylow 2 and we all love a good D_8

[proud vigil]

I don’t really get how mathematicians were able to prove that there’s no other simple groups outside of 3 huge families and 26 random looking ones

I guess I should read through the thousands of pages

[delicate orchid]

In fact it’s the smallest group with Sylow D8 where all involutions are conjugate

CRASZZYYYY 🫨🫨🫨🫨

[thorn jay]

D_8 also has D_8 as Sylow 2

[delicate orchid]

Mods can we fact check ts

[elfin wraith]

What is PSL(2,7)? Projective special linear group over F_7?

[thorn jay]

that is pretty interesting

[delicate orchid]

Yes, it’s iso to GL(3,2) if you prefer

[fiery dirge]

I don't care really, I get dopamine hit from understanding how to solve it, not knowing the solution

[thorn jay]

not sure why thats interesting to group theorists though

[fiery dirge]

When a and b are coprime right?

[elfin wraith]

I’m sure if I think about that I’ll agree

[thorn jay]

is it a nicely described isomorphism or an accidental one

[delicate orchid]

In that case. Number of abelian groups of order p_1^(n_1)p_2^(n_2)… with p_i prime is the product of P(n_i) where P denotes the partition counting function

[delicate orchid]

It’s a bullshit one 💔

[thorn jay]

ugh

not even anything that reveals cool structure or implications?

[delicate orchid]

Theres loads of bullshit ones https://en.wikipedia.org/wiki/Exceptional_isomorphism. the worst is PSL(2,4) = PSL(2,5)

[delicate orchid]

All exceptional isomorphisms arise from a cool structure that only appears cause things are small

[elfin wraith]

I don’t like that at all

[proud vigil]

omg stop the accidents

[thorn jay]

i guess

I suppose cool implications arise exactly because of the isomorphism

[rocky cloak]

s = 0
for g in G:
    for h in G:
        if g * h * (g^-1) * (h^-1) == matrix(GF(2),[[1,0,0],[0,1,0],[0,0,1]]):
            s+=1
s````

Sagemath says 1008/168^2 = 1/28

[south patrol]

Fusion systems.

[fiery dirge]

So we partition the group of order 1800 into a lot of smaller groups of prime orders. I didn't quite understand the last part of it though

[elfin wraith]

I started reading about fusion categories tonight, too many words, category theory is word soup

[south patrol]

aren't these p different beasts

lol

[elfin wraith]

Possibly lol

[south patrol]

fusion cats should be like the math phys thing

[elfin wraith]

Very

[delicate orchid]

They’re entirely unrelated

[south patrol]

indeed

[elfin wraith]

These are very different ideas yes

[south patrol]

people just can't give good names jk

[thorn jay]

theres a thing called "sesquimorphisms"

[delicate orchid]

Example. 160 = 2^5*5 so there are P(5)P(1) = 7*1 =7 abelian groups of order 160 as there are 7 ways to partition 5: 5, 4+1, 3+2, 3+1+1, and so on. You can derive this formula directly from your Z_a x Z_b = Z_ab <=> (a,b) = 1 observation

[delicate orchid]

I secretly mentioned a fusion system fact earlier see if u can spot it

[fiery dirge]

What is P()? Is it Euler's function or something else? It doesn't look like Euler's function

[delicate orchid]

I told you what it is and just explained what it is. P(n) is the number of partitions of n

[fiery dirge]

That's exactly the part I was confused about in your last message that's why I had to ask again

[rapid cave]

Emmy Noether

Emil Artin

[delicate orchid]

Think about abelian groups of order p^n to start. we can see that abelian groups of this form are of the shape
C_{p^(n_1)} x C_{p^(n_2)} x … x C_{p^(n_m)}
with the sum of the n_i = n. By your observation, each different choice of n_i gives a new group (up to iso, so just reordering the factors doesn’t count), hence the number of abelian groups of order p^n is P(n)

Im not reformatting it on mobile

add in the _s using the power of the human mind

[rocky cloak]

I mean, Emil is his dad right

[fading acorn]

yes

[delicate orchid]

To this day I still think emil should be a girls name 🪫

[rocky cloak]

People with the same last name related, pretty crazy when that happens

Never heard of the guy. Sounds cool though 😎

[rapid cave]

His phd advisor was zariski

[delicate orchid]

https://tenor.com/view/riddler-paul-dano-new-rockstars-does-he-know-gif-26833694

[fading acorn]

yea

[delicate orchid]

That’s CRAZY that they let a topology have students

[rapid cave]

And he is 91. Still alive!

[rocky cloak]

Im an (academic) descendent of Felix Klein

Love the math genealogy project

[delicate orchid]

I should actually look at my genealogy

Jagr what’s ur erdos number

[rocky cloak]

Will need to check

[thorn jay]

my birthname is somewhat close to emil and im AMAB

[elfin wraith]

I just found out the guy I might do my masters with is a descendent of cartan, that’s pretty sick

[rocky cloak]

5

[elfin wraith]

Assigned male at birth

[thorn jay]

got more testosterone than approximately 50% of humans

[delicate orchid]

I think mine is 3

[rocky cloak]

I think it's a pretty common boys name many places in Europe

[thorn jay]

yeah

Emiel

[elfin wraith]

No?

[thorn jay]

is very Belgian, but a boys name

[elfin wraith]

I mean they may be but not necessarily

[rapid cave]

Enpeace is also a common first name in europe.

[thorn jay]

well I am technically but not because of that

[rocky cloak]

enpeace lore dropping

[delicate orchid]

To offset it, the most notable person in my genealogy seems to be Phillip Hall who will be known to me and me alone

[thorn jay]

(this is a very recent development)

[rapid cave]

I know someone with erdos number 2

[thorn jay]

the same Hall as from the Hall subgroups?

[rapid cave]

Great proffesor

[rocky cloak]

My chain goes through McKay, so I'm happy with that (not John McKay, but I'll take it)

[delicate orchid]

Indeed

Although I have managed to get to ISAAC NEWTON directly so that’s fun

[elfin wraith]

Going to pick my PhD advisor based solely on their genealogy to maximise my aura

[thorn jay]

slowly, little by little, my whole life shall be revealed

[thorn jay]

erdos number farming

[quiet pelican]

One of my friends has basically 0 option for that lol (combi in Western Europe) iirc

[delicate orchid]

He’s my academic great great great great great grandfather’s second advisors great great grandfather I think lol

[elfin wraith]

That’s pretty close, you basically studied under him

[elfin wraith]

I won’t have many options either but that’s more for reasons of incompetence

[rocky cloak]

He never taught any students

[delicate orchid]

Click button on website a few times

[elfin wraith]

Didn’t he die at like 25?

[delicate orchid]

No excuses

[rocky cloak]

26 i think

[glad osprey]

Speaking of people with the same last name, I recently found out that the Conway who wrote the complex analysis book is not the same Conway from Conway's Game of Life. On the other hand I also found out that the Cox who cowrote Ideals, Varieties and Algorithms is not only related to the Cox who wrote the goated boon on Galois theory, it's the same guy!

[elfin wraith]

Real

[mint seal]

I wish the other Conway wrote an analysis book

[rocky cloak]

Conway is a great writer

[mint seal]

also feels strange referring to him as the other Conway

the analysis book Conway is the other Conway

[delicate orchid]

Throw a few more greats on the end and I’ve gotten to Galileo lmfao

Adam

Who’s gonna tell em

[elfin wraith]

Not to flex but in related to Gengas Khan

(Probably)

[delicate orchid]

Khan extensions.

[thorn jay]

Kan academy

[delicate orchid]

Of course named after Dr. Extensions

[thorn jay]

introduction to limits

[rapid cave]

I am related to myself

[rapid cave]

And thats what really matters

[glad osprey]

My great great great grandmother was killed by ghengis khan when she was a baby

[rocky cloak]

My lineage ends at friedrich leibniz which is Leibniz's father

[elfin wraith]

Look at mr history knower over here, didn’t realise you could double major in Switzerland

[thorn jay]

(i love sheafification)

[elfin wraith]

Is there no more data or did he just start handing out PhDs?

[tardy hedge]

Yayyy limits yayyy yayyy

[delicate orchid]

I guess this means equivariant group theory is astronomy and quiver reps are calculus

[rocky cloak]

It says advisor unknown

[thorn jay]

categorical limits >= analytic limits

[tardy hedge]

Dat moment when i still havent learned categorical limits

[elfin wraith]

Neither, I really need to patch up the holes in my category theory

I think I get the vibe of limits but I’ve never done it formally

[thorn jay]

its just the definition of inf but categorical

[tardy hedge]

Ya its just that like duh

[delicate orchid]

Limits is taking fixed points

Colimits are forcing your points to be fixed

[rocky cloak]

Know what products are, check
Know what kernels are, check
Yup know limits

[delicate orchid]

They’re called EQUALISERS

[rocky cloak]

I'll have you know they're name is difference-kernels

[delicate orchid]

Im gonna crash out

[elfin wraith]

I seem to remember the very nlab brain rot definition of CW complexes being some sort of colimit

[thorn jay]

direct limits have to do with finitary properties and inverse limits have to do with patching local information

[elfin wraith]

Idk if that’s a hallucination or if that’s something

[delicate orchid]

Even more vaguely. Colimits are gluing limits are extracting

[delicate orchid]

How else do you define CW complexes lol

[elfin wraith]

Well I do remember reading it and going yeah that’s what a CW complex is

[elfin wraith]

The way Hatcher does (badly)

[rocky cloak]

I mean, an attaching cell is a pushout

[delicate orchid]

I never got the point of CW complexes. Go straight to simplicial

[elfin wraith]

I don’t even know the definition of limits in a category so like this isn’t even a case of I just didn’t realise

It’s just never really came up for me

[thorn jay]

id argue limits to be gluing but of another kind, rather than the whole object being glued together its that every "element" of your object consists of glued together information

[delicate orchid]

I vibe with this

[elfin wraith]

I should learn it, and I’m taking an actual introductory category theory course next semester so I will, but I’ve currently been on a very need to know basis with category theory hence learning about abelian and monoidal categories but not really knowing yoneda or limits

I started to read about semisimplification tonight and gave up

[delicate orchid]

I bet they introduce it using the universal arrow bullshit I hate it when they do that

[elfin wraith]

Turns out Verlinde categories are an annoying thing to define

Or we’ll at the very least a very technical thing to define, so many words

[thorn jay]

(hence why a B-sheaf can be extended to a sheaf and how you get the correspondence sheaves <-> local homeomorphisms)

[delicate orchid]

This is what limits and colimits look like in any category that matters: https://stacks.math.columbia.edu/tag/002U @elfin wraith

[thorn jay]

lmao, it gives a nice link between adjoint functors and limits though

[delicate orchid]

Very explicit formulation for limits as a subwhatever of a product

[delicate orchid]

Ts. Pmo. The limit is an adjoint to the diagonal functor is something you can just prove directly. I don’t get why this would assist

[elfin wraith]

Oh yeah that’s fine, sure I’m happy enough with that

[rocky cloak]

It shows that limits commute with limits

[delicate orchid]

GRRRRRRRRRRRRRR

Wait it shows adjoint functors (on the appropriate side) commute with limits or limits commute with limits

[knotty badger]

limits commuting is a little hard to state precisely

[delicate orchid]

It would help if I could remember the precise details, but right adjoint commuting with limits essentially falls out of properties of the Hom functor right?

[knotty badger]

yeah that's how i think of it

if you understand the representability def of limits you're good

[delicate orchid]

So if you just directly show that the limit functor is an adjoint then you have limits commutes with limits?

[knotty badger]

well you have to say what you mean by "limits commute with limits"

[delicate orchid]

Fix a pair of indexing categories, then you can swap the limits and its naturally iso

[knotty badger]

yes that works

lots of examples of "commuting limits" in practice are... awkward to express in that form

[south patrol]

Nice

[south patrol]

Idk they are more flexible in a way that can be useful

In my mind they serve different purposes

[elfin wraith]

I went straight to simplical stuff when I started learning algtop

[south patrol]

Ig also depends what you mean by simplicial stuff lol

[elfin wraith]

Tbh now that I’m revising algtop (and patching some holes) is really the first time I’m properly learning to deal with CW complexes, I kinda just ignored them on the first pass

Like I was aware of them but never gave them much thought

[south patrol]

One clear advantage of cw complexes is that it lets you say stuff like this: CP^n has one cell in dimensions 0,2...,2n, so its cohomology is Z in each of those degrees

And cell by cell arguments etc

Wait this is #groups-rings-fields lol

[elfin wraith]

I was going to say this place is whatever we want it to be but I guess you have to be somewhat responsible since you’ve got the mall cop badge and all…

[thorn jay]

shhhh not in front of the mods

[south patrol]

You mean uhh

Shopping centre police

[unborn magnet]

Hey all! I'm new here. I see there's no Lie theory / representation theory channel. Is this channel the right place for such discussions?

[rapid cave]

#advanced-algebra is best for that

[unborn magnet]

sweet, thanks

[tardy hedge]

whats the context?

[vocal pebble]

What is R?

[tardy hedge]

im not really sure how to interpret Rm+mZ. Like Rm is ideal generated by m in R and mZ is ideal generated by m over Z?

[rocky cloak]

Presumably R isn't unital, so Rm + mZ is just the ideal generated by m.

Why what is needed?

[elfin wraith]

You really have to start providing more context when you ask questions lmao

[rocky cloak]

It is enough, and you can prove it by induction

It's just multiplying out and see that you get a multiple of m^l

Yeah.
Prove that (Rm+mZ)^l is a multiple of m^l by induction on l

[glad osprey]

oh wow, so Rm doesn't even necessarily contain m 💀

Moderators non-unital rings, ban him please

[tough onyx]

hi

[true bolt]

hi

yk what I agree, b&

[glad osprey]

wait, did that ping go trough?

sorry!

[true bolt]

Yeah, it still pings us even you put a backslash before hand

[coral spindle]

Uh oh!

[true bolt]

so uh don't do that

[coral spindle]

How funny

[chilly radish]

I'll be banning you for nonunital ring slander

Any last words?

[glad osprey]

oops, so sorry!

[true bolt]

Everyone gets a ban

[coral spindle]

If you say @Senior Moderators instead it doesn't ping btw

protip

[tough onyx]

how can you have last words in a ring if there's no order

[coral spindle]

it's also way funnier

Sorry!

[glad osprey]

damn, mods are swarming this channel now 💀 what have I done

[coral spindle]

Maybe, like kiand suggested, you could tell us the whole context for the question so that nobody has to collate various messages

and therefore dig the question back up again :)

[elfin wraith]

You did after prompting, but you have a habit of posting an absolutely minimal screen shot of a formula with no larger context or explanation of what the symbols mean (unless someone teases it out of you)

It’s not like an attack but you’ll just get better answers if people have enough information to actually help

[glad osprey]

Is the Noetherian assumption actually used here btw? Or do you just need m^l = 0?

[white oxide]

Sorry can you explain more why this implies n = n'

I understand that prime ideals of S^{-1}B are in one to one correspondence with prime ideals of B which don't meet S

Oh wait I think I get it

Yeah I think I get it?

lol

I forgot that q^c = p = q'^c

So then we can just extend to obtain m

and since the diagram commutes it must be that the extension into B_p then contraction is also equal to m

[thorn jay]

when you pass n and n' to A_p, they are maximal and thus equal. Then going back to B_p it follows that n and n' are maximal (by some integrality argument probably) so must be equal. Then using the correspondence of certain prime ideals of A and A_p youve got that q = q'

[rocky cloak]

The hypothesis is that there are no nonzero nilpotent ideals

[elfin wraith]

Not the Wedderburn-Artin theorem

[knotty badger]

oh nice you're norwegian now

[elfin wraith]

Assigned Norwegian by Shedow

[elfin wraith]

Other famous results include the Roch-Reimann theorem and the Borel-Heine theorem

[knotty badger]

the Zucker-Cox machine

[thorn jay]

Im much more of a fan of the elegant sum-Gauss

very simple yet beautiful

[tidal schooner]

Convergence domination theorem

[thorn jay]

the arithmetic theorem of fundamental

[elfin wraith]

Getting dangerously close to just inventing French

[glad osprey]

I kinda like Wedderburn-Artin, I think it flows better than Artin-Wedderburn, atleast when saying Artin the french way

[thorn jay]

it feels wrong

[tardy hedge]

cox zucker

[rapid cave]

why did you changed a2 to a2 * e

so e^2 = e, not e =1

but most likely its not 1

other then this yeah sure

[vapid vale]

?

you can certainly have idempotents which are not 1

[rapid cave]

ok

I am still not convinced about that part

[vapid vale]

what are you doing from the 3rd to the 4th line

oh

lol

[rapid cave]

you need to make that clear

so yeah your proof is valid

[tardy hedge]

Im not sure why i havent seen them or dealt with them much

I havent studied that theorem

[thorn jay]

then its just badly written lol, ideally you'd break it up into lemmas

[tardy hedge]

“A semisimple ring may be characterized in terms of homological algebra: namely, a ring R is semisimple if and only if any short exact sequence of left (or right) R-modules splits”

Oh, thats cool

[thorn jay]

more lemmas then!

it is a very deep result so of course it is going to be complicated

i mean if each of the lemmas say something interesting

[elfin wraith]

I proved Artin Wedderburn as a homework problem and I’m pretty sure we broke it up into 4 lemmas

The proof was pretty short as far as I remember as well

[tribal moss]

Does the classification of finite simple groups count as a theorem? It must take dozens of pages to even state what the claim is.

[crystal vale]

I don't understand, authors sometimes state some exercise as theorem

For example, Munkres state the FIP[ compactness] as theorem in his book, i don't understand why?

[glad osprey]

Is that Wedderburn Artin too? what's the deal with theorems in algebra having 100 equivalent formulations?

[glad osprey]

You mean they state some theorem as an exercise? I guess it makes sense if it's not important enough to put in the main text, but still interesting to note

[crystal vale]

No, some exercise as a theorem

Why is it a theorem, a closed subspace of compact set is compact?

[thorn jay]

thats way easier than the usual statement

lol

i dont think thats the artin wedderburn theorem its just a nice fact about semisimple algebras

[glad osprey]

Wdym some exercise as a theorem? You mean it's too simple to be a theorem, so it should be an exercise instead?

[crystal vale]

Yes

[glad osprey]

Presumably they need to use that theorem later, so they can't just demote it to an exercise

[tribal moss]

I don't think there's actually such a thing as "too simple to be a theorem".

[glad osprey]

Yeah, too useless to be a theorem maybe, but I don't think it can be too simple

[tribal moss]

Then again, I'm perhaps too influenced by formal logic, where "theorem" means anything whatsoever that has a proof. :-p

[glad osprey]

It's funny how you call it "some exercise" btw, as if it's common terminology for useless propositions

[thorn jay]

for me id call something a theorem if its a fundamental result about something

generally still based on vibes

[tiny jolt]

Any hints on how to do this?

[thorn jay]

im not sure if this is closed under taking inverses?

oh wait no it is nevermind

[tiny jolt]

Why not?

[thorn jay]

forgot that [0, 1] is a total order

[tiny jolt]

So far, I know that f is definitely not conjugate to any function with a fixed point in (0,1)

[tardy hedge]

Funky question

[tiny jolt]

It's so frustrating for how easy it is to state xD

[thorn jay]

i wonder what, in general, the conjugacy classes of this group are lol

in correspondence with subsets of (0, 1)?

[elfin wraith]

Kinda question that makes me freak on an exam

[tiny jolt]

there might be continuity issues?

[tardy hedge]

Hahaha real

[vapid vale]

consider the intervals given by [f^n(x), f^{n+1}(x)] (and similarly for g). maybe you can describe the conjugating function in these intervals

[glad osprey]

mq's theorem

[tribal moss]

Even Russell and Whitehead never reached such complex results.

[vapid vale]

theres this story where maxim kontsevich was giving a talk on whatever mathematical physics, and he had a bunch of 4's on the board. an elderly professor raises his hand asks, why all the 4s? and kontsevich replies "3+1" and the professor goes "ohh right"

[south patrol]

True

[tribal moss]

For physics that makes perfect sense: 3 space dimensions and 1 time.

[vapid vale]

ya lol

but out of context it is a funny anecdote

[tribal moss]

That's not even a question until you have a definition of "is a dimension".
Physical theories since the early 1900s generally make heavy use of vector spaces with dimension 4 where -- given suitable choice of basis -- three of the coordinates correspond to space than the fourth to time.

[knotty badger]

It is

[tribal moss]

In special relativity you can't get away with considering space coordinates to be separate from the time coordiante in that way. The fundamental postulate of SR is that laws of physics must be invariant under a family of transformations where the space coordinates change differently depending on what the time coordinate is, and vice versa. That's not really possible to work with without considering it all single 4-dimensional space.

(And basically all of modern physics builds on special relativity).

[crystal vale]

My college also provides post graduate, phd course in physics, so what I notice is that if you feel bad about you then you can teach maths to physics people then you will feel good about you

A few minutes ago, someone questioned me how commutator algebra can be different from commutative algebra?

Commutator algebra used in quantum mechanics

[quiet pelican]

You can, you’ll just be wrong 🙃

[tribal moss]

"Classical" is a bit ambiguous. From the point of quantum theories, special (and general) relativity are classical theories.

Pre-relativistic settings are simpler to work with and good enough for a very large range of situations. Just because that range is not "everything" doesn't mean we should just discard the entire theory.

[quiet pelican]

It’s a good approximation for basically any situation you’ll experience in your everyday life (especially everyday life pre-1900)
Unless you’re observing mercury’s orbit very closely

[tribal moss]

And learning relativistic physics without first having an understanding of how the math works in the Newtonian case would probably be insanely difficult.

[quiet pelican]

Learn QFT before any other physics 🙃

[tribal moss]

Anyway, this is not really #groups-rings-fields material ...

[quiet pelican]

It’s my usual “not serious” indicator

[crystal vale]

Btw Troposphere, are you a PhD student or you complete your phd already?

[tribal moss]

My PhD was long ago (and not in mathematics).

[crystal vale]

Then in which subject? What you do now?

[tardy hedge]

Bruh and BRUH

I think I speak for us all when I say that 😎

[waxen tinsel]

Is this legal lol
$gN = Ng \implies gNg^{-1} = N$

[cloud walrus]

bellamy

[waxen tinsel]

where N is a set, and g is an element of a group I feel like its a abuse of notation

[noble nexus]

no its not an abuse

[kind temple]

just prove the identities. as long as what you are writing down is well-defined and you can prove what you need from definitions, it's perfectly "legal"

[noble nexus]

multiplication of sets by elements is pretty nice overall it tends to satisfy most of the reasonable things you think it should satisfy

[chilly ocean]

I may be having a bit of a brain fart, but I don't see the N subset gNg^-1 direction

gNg^-1 subset N is pretty clear because, for every n in N, there is some m in N such that gn=mg, hence gng^-1=m, another element of N, but I'm missing how you get that any element of N are equal to gmg^-1 for some m in N

[thorn jay]

watch out! or the math police will come to your door and arrest you!

[kind temple]

gN = Ng iff g^-1N = Ng^-1, so n = gg^-1n = gmg^-1 for some m in N

[waxen tinsel]

I mean my justification is right multiplication by g^{-1}

[tough raven]

The hypothesis is gN = Ng, which also comes with two sides. So e.g. for all n in N, ng = gm for some m in N, so n = gmg^{-1} ∈ gNg^{-1}.

[tulip otter]

Yes

Ng={ng| n in N}. So
Ngg^{-1}={ngg^{-1}| n in N}={n| n in N}=N

[flint cave]

the obvious homomorphism F2[x] -> F2[X,t]/(t^2-X) is epi right?

[chilly ocean]

Ohh thanks

Idk how I missed that

[tardy hedge]

Im not sure what you mean by prove the identities

[kind temple]

yea, sure, ig its not an identity. i just mean that everything there is well-defined, so just prove the statement from definitions

that shows you the that the syntactic manipulations you are making are "valid"

[tardy hedge]

Yes if the statement you are assuming from that equation is true then … its a valid move, i guess

[rocky cloak]

If the obvious one means the one sending x to X, then no

The codomain is just F[t^2, t] = F[t] and you would be mapping x to t^2

[tribal moss]

But it might not be immediately clear mapping to x to t^2 is not epi, given that we'll be stuck in characteristic 2. (So the "obvious" strategy of negating t won't work).

[tribal moss]

Suppose f: F2[x] -> F2[x,t]/(t²-x) is defined by f(x)=x.
Then consider g and h: F2[x,t]/(t²-x) -> F2[u]/(u²) defined by g(t)=g(x)=0 and h(t)=u, h(x)=0.
Now gof = hof yet g!=h.

[fiery dirge]

how do I make elementary and normal form?

Take $\mathbb{Z}_{360}$ for example, I will first write it in elementary form: $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_5$

What is the normal form? Is it $(\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_5) \times (\mathbb{Z}_2 \times \mathbb{Z}_3) \times \mathbb{Z}2 = \mathbb{Z}{30} \times \mathbb{Z}_6 \times \mathbb{Z}_2$

So we try to take all primes until we run out of them and rank them in that order?

[rocky cloak]

I'm not sure what you mean by being stuck in characteristic 2, but the maps
F[t] -> F[t]/(t^2)
sending t to 0 or t respectively works

Which is the obvious strategy

[tribal moss]

I just meant that there's only a homomorphism F2[x,t]/(t²-x) -> R if R has characteristic 2.
And yes, that was the same maps I used in my next post.

[cloud walrus]

HMD

[rapid cave]

Z360 is not iso to this elementary form

[tribal moss]

The "obvious" strategy for me in a different characteristic would have been to compare id o f to [t -> -t] o f.

[fiery dirge]

hmm

[fiery dirge]

how did you come to that conclusion? Is it because Z_360 has elements of order 360, 180 etc? How do I figure out if those smaller group products come to that?

[tribal moss]

I'm not clear on the terminology, but surely one of the distinguished forms should be Z8×Z9×Z5?

[rapid cave]

Mb

Yes

[rapid cave]

In the case of Z360 you use the fact that
Za x Zb is iso to Zab if and only if a,b are co-prime

[fiery dirge]

I see, what about Z_90? It's Z_18 x Z_5 and 18 and 5 are coprime so we can proceed with elementary/normal form?

[tribal moss]

Well, every element of Z2×Z2×Z2×Z3×Z3×Z5 satisfies x^30=e, because that is true in each of the factors. So it cannot have order 360.

[rapid cave]

So what you have here is the elementary form (product of cyclic groupa of prime power order) and the normal form would be Z90

[fiery dirge]

I have a question though, how is that allowed?

splitting into Z_18 x Z_5 allows for that sure, but what about Z_30 x Z_3

they're not coprime

[rapid cave]

Yes so its not iso

[gentle crow]

it's not obscure; it's been used as a course text at my university.

[proven stone]

My group theory professor sent some abstract algebra exercises on groups one question is the same as this question. But how do I prove the direct inclusion?https://math.stackexchange.com/questions/336080/prove-that-langle-s-rangle-is-the-intersection-of-all-subgroups-containing

I have proved that $\langle S\rangle\leq G$ and also that the intersection of all the subgroups of $G$ that contain $S$ contains the group generated by $S$

[cloud walrus]

schrysafis

[proven stone]

if I prove for a subgroup $H\leq G$ with $S\subseteq H$ that $\langle S\rangle\subseteq S$ I'm done

[cloud walrus]

schrysafis

[rapid cave]

this will give you the same inclusion as before

call the intersection K

then you have showed K contains <S>

now you want to show <S> contains K

but that is easy because <S> is a subgroup containing S, and so it is part of the intersection defining K

[proven stone]

oh man I'm so stupid

okay thanks!

[tardy hedge]

Its definitely more obscure than other algebra texts

[flint cave]

Sorry, what if I insist on the codomain to be F2(sqrt(x))? Somehow thats what I intended but I messed up the notation

[tribal moss]

That is, the codomain of your morphism is now the field of rational functions F2(t), and you map x in F2[x] to t²?

[flint cave]

Yep

[rocky cloak]

Yeah there's only one map from F2[t] to F2(t) that maps t^2 to t^2

Because t would need to map to a square root of t^2

[flint cave]

Right, so the map F2[x] -> F2(root X) is epi right?

[rocky cloak]

Well, depends which category you're in

[flint cave]

In Ring, morphism take 1 to 1

[rocky cloak]

Then I don't think so no

[flint cave]

Do you have any example

[rocky cloak]

I mean the canonical example if it exists would be the pushout of the morphism along itself. So can try to compute that

[tribal moss]

(to HChan) Yeah, but here the embedding in question is not the canonical one.

[velvet hull]

lol, i dont feel like reading the entire convo so ill just shut up

[flint cave]

Pushout is colimit right?

[mint seal]

yup

[flint cave]

Colimit of the diagram F2[X] -> F2(rootX) is F2(rootX) itself tho?

[tribal moss]

For a pushout you need two separate copies of your F2(rootX) in the diagram, each with its own morphism for F2[X].

[flint cave]

Ohh fiber coproduct

[rocky cloak]

So I'll write it as
F2[t^2] -> F2(t).

Then consider R = F2(t)[x]/(x^2) then
F2(t) -> R by mapping t to either t or t+x should work.

[flint cave]

Very nice, thanks. How did you get the idea for this construction?

[tribal moss]

Are we sure t+x generates a subfield in R?

[rocky cloak]

Yes, is you can think about the image when you reduce modulo x

[rocky cloak]

Well, like I said before I thought about the pushout.

F2(t) is a little complicated, so I just did F2[t] first. Then the pushout is
F2[t, x]/(x^2 - t^2) = F2[t, x]/(x^2).

Then the analogous thing worked for F2(t)

[flint cave]

Somehow it doesnt add up to me since in my mind the image of F2(t) has to be a field and the image of t^2 has exactly one square root- the image of t... whats wrong with this intuition

[rocky cloak]

It has one square root in the image

Both t and t+x square to t^2, but a subring containing both t and t+x is not a field

[flint cave]

Oh right!

Thanks jagr!

[tribal moss]

Hmm, not quite getting it.

[tardy hedge]

im not sure about this

[tribal moss]

Ah, perhaps I do get it.

[rocky cloak]

So an element of R is a unit iff it's constant term is a unit.

When multiplying the constant term is not at all affected by the x-term

[tribal moss]

Hmm, yeah, a nonzero polynomial in t+x has the form a+bx where a,b are polynomials in t and a is unit in R. Then (a + bx)(1/a - bx/a²) = 1

[rocky cloak]

I guess an easier way one might come to the same conclusion:

The problem is that t^2 only has one square root, so just add another one.
F2(t)[x]/(x^2 - t^2) = F2(t)[x]/(x-t)^2 = F2(t)[y]/(y^2)

[crystal vale]

what is nil ideals?

[karmic moat]

nilpotent implies nil. dont need noetherian

[rocky cloak]

See if you can prove it.

Here you will need Noetherian

[south patrol]

Ah I'd have called that locally nilpotent

Maybe this is an AG thing

[thorn jay]

probably

how does that translate to local properties though

[rocky cloak]

I mean, the proof looks good, but I thought you wanted to prove that nil implies nilpotent?

[white oxide]

Sanity check: If A is a ring and q an ideal of A, then A[x]/q[x] is isomorphic to (A/q)[x]

[rocky cloak]

I mean, you never proved T was nilpotent

Okay, then the proof works I guess. But you're kinda hiding the most important step

[tardy hedge]

Not sure why i never thought before how q[x] would be an ideal of A[x]

[white oxide]

Lol yeah me too but it comes up as several exercises in AM

[tardy hedge]

q[x] would be finitely generated if q is i guess

[white oxide]

yeah, and if q is prime, q[x] is prime, if q is p-primary, q[x] is p[x]-primary, etc etc

[tardy hedge]

Oh (2)[x] in Z[x] is just (2)

Not sure why this is kinda tripping me up lol

[rocky cloak]

q[x] just means the ideal generated by q in A[x], right?

So then (2)[x] is generated by 2

[tardy hedge]

I guess generally we cant write q[x] as something finitely generated

Idk u saying generated by q (where q is the ideal) made me feel weird

[white oxide]

Yeah, or I guess equivalently all polynomials with coefficients in q

[tardy hedge]

Ideal generated by an ideal

Oh I guess like subset of A[x] generating an ideal

[thorn jay]

interesting, the class of rings where (0) is primary is also closed under taking polynomial rings?

[white oxide]

I have no clue tbh

I'm just in the midst of doing this ex

[tardy hedge]

Why u mention (0) @thorn jay

[thorn jay]

#proofs-and-logic

[thorn jay]

coherent conditions

:P

[tardy hedge]

Broski

[thorn jay]

what! they pop up everywhere in ring theory I'm telling you

[tardy hedge]

I mean like ik (0) is important cuz if da shi prime it be integral domain lol

But what else r we talking about

[rocky cloak]

You're describing proving that
A and B and C
implies a contradiction. So then one of A, B or C must be false. Not necessarily A

[thorn jay]

suppose I < q < R. Then q is primary iff q/I is primary

in particular q is primary iff R/q contains no non-nilpotent zero divisors

[tardy hedge]

Brb

[rocky cloak]

Many properties of ideals turn into properties of rings, by reformulating (0) having the property

[thorn jay]

i.e. many classifications of ideals are coherent conditions

:>

[white oxide]

I'm too lazy to think about polynomial multiplication but p[x] is contained in the radical of q[x] because if f is a polynomial with coefficients a_i in p, then a_i^{n_i} in p since q is p-primary, so we f^{max{n_i}} in q[x]?

[thorn jay]

and this is nice because it means that the preimage of an ideal by a projection map satisfying that property again satisfies that property, and quotients too

[rocky cloak]

If you've proven B and C are true then it's not really an assumption anymore

[thorn jay]

why is this discussion here lol

this proof is correct, given that A, B, C prove E and A, B, C, E prove F yes

[rocky cloak]

No, it could be that it is B or C that is wrong

[thorn jay]

but if you prove B and C then you've proven not A

[rocky cloak]

Like, are you trying to prove (B and C) implies (not A)?

[thorn jay]

I'm loving the context for this question lol

like what do you want to know this for

and you decide to ask it in groups-rings-fields lol

[proven stone]

Let $G$ be a group, and $H, K \leq G$. Let $HK := {hk: h\in H, k\in K}$, and $KH := {kh: h\in H, k\in K}$. Suppose $HK = KH$. 1. Show that $HK \leq G$. (This is fine)
2. Show that $\langle H, K\rangle = HK$. (I struggle with this one)

[cloud walrus]

schrysafis

[proven stone]

I'm sorry for posting again just preparing for group theory lesson

[tough raven]

Recall how <H, K> is defined.

[proven stone]

Oh yeah

$\langle H, K\rangle={g | g\in H\cap K}$

[cloud walrus]

schrysafis

[thorn jay]

what

[proven stone]

wait I'm drunk

let me fix this

[vapid vale]

thats me

[thorn jay]

uponthewitnessing

[proven stone]

$\langle H, K\rangle={g_{1}g_{2}...g_{n}: n\in\mathbb{N}, g_{i}\in H\cap K \forall i\in{1,2,...,n}}$

I=1,2...N

[thorn jay]

those are the same

👍

[proven stone]

i=1,2,...n

[thorn jay]

H \cap K is already a subgroup

[cloud walrus]

schrysafis

[proven stone]

so?

[thorn jay]

so <H, K> = H \cap K

[proven stone]

oh yea I get it

I noticed it too

then?

How can I use this hint prove it?
Let me first introduce a standard notation in computer science. Given a subset $S$ of $G$, let $S^0 = {1}$ and $S^{n+1} = S^nS$ for all $n \geqslant 0$. Finally, let $S^* = \cup_{n \geqslant 0} S^n$.

Hint. First prove that $(HK)^$ is equal to $\langle H, K\rangle$. Next, using the relations $HK = KH$, $HH = H$ and $KK = K$, show that $(HK)^ = HK$.

[cloud walrus]

schrysafis

[thorn jay]

I think <H, K> means the subgroup generated by H \cup K

not H \cap K

[proven stone]

yeah

thats the correct thing

[thorn jay]

Well write out what (HK)^n is and use those relations underneath to simplify that to HK

use induction

[proven stone]

I did that

then the union isn't just HK?

[thorn jay]

then the union of all the (HK)^n is HK yes

[proven stone]

idk if that helps

[thorn jay]

and then you just need to prove that (HK)* is <H, K>

[proven stone]

this is my struggle

[thorn jay]

well this (replacing H \cap K with H \cup K) basically solves it

[proven stone]

what allows us to do that? I'm losing something probably

[thorn jay]

can you write out as a set what (HK)^n means?

[proven stone]

$(HK)^{n}={h_{1}k_{1}h_{2}k_{2}...h_{n}k_{n} | h_{i}\in H, k_{i}\in K \forall i=1,2,...,n}$

[cloud walrus]

schrysafis

[thorn jay]

notice how every element of this set is in some (HK)^n

[proven stone]

hm

I still don't get where this is going

we want to prove $\bigcup_{n=1}^{\infty} (HK)^{n}=\langle H, K\rangle$

[crystal vale]

So HK = KH is given, right?

[cloud walrus]

schrysafis

[proven stone]

yep

[crystal vale]

So HK \subset < H, K >, it is clear, right?

[proven stone]

why that?

[crystal vale]

How do the elements HK look?

[proven stone]

$hk, h\in H, k\in K$

[cloud walrus]

schrysafis

[crystal vale]

And < H, K > is subgroup generated by H \cup K

[proven stone]

alright

[crystal vale]

So both H and K in < H, K >, so h and k in < H, K > implies hk in < H,K>

You know how the subgroup generated by arbitrary subset S looks ?

[proven stone]

yes, $\langle S\rangle={s_{1}s_{2}...s_{m} | m\in \mathbb{N}, s_{i}\in S \forall i=1,2,...,m}$

oops

[crystal vale]

< H, K > is an intersection of all such subgroups which contains H \cup K, so < H, K > \subset HK, because HK is subgroup you proved in part i), and it contains both H, K

[cloud walrus]

schrysafis

[crystal vale]

In general subgroup generated by S is the intersection of all subgroup which contains S.

[crystal vale]

No, it is not correct

What about inverses?

[proven stone]

our professor defined it this way

and defined the set of their inverses too

wait

no

[crystal vale]

Say S = {1} in group Z then what will be < S > ?

[proven stone]

you get the union of S and S^{-1}

you want their inverses too

[proven stone]

${1,-1,2,-2,...,n,-n,...}$

[cloud walrus]

schrysafis

[crystal vale]

$\langle S \rangle = { s_{1}^e_1 s_{2}^e_2\dots s_{m}^ e_m | m\in \mathbb{N}, e_i = 1, -1 }$.

[cloud walrus]

Notknow🙇
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[proven stone]

I think the one index goes to e

[crystal vale]

Yes, latex typo

[proven stone]

yep

[crystal vale]

So that's how your group looks

[proven stone]

that

that's a better way to define it

[crystal vale]

So you can show it is equivalent to say it is the intersection of all subgroups which contain S, you understand this?

[proven stone]

yes I proved it at home

because our teacher gave it as an exercise

[crystal vale]

Then you can just use this and show that < H, K > \subset HK

Use part i)

[proven stone]

oh yea

[crystal vale]

Now you are done

[proven stone]

I'm stuck on the inverse inclusion

[crystal vale]

Which one ?

HK \subset < H, K > ?

[proven stone]

but opposite

yea

[crystal vale]

.

[proven stone]

wait you showed it

sometimes I forget where I started

[crystal vale]

No problem

[proven stone]

just for fancy could you show me the proof with this hint?

I want to get experience with proofs

[velvet hull]

start from this definition

and try to work from there

[peak sedge]

could anyone give me some intuition on how the quotient group is defined like it is

and why it’s called a quotient ?

[proven stone]

Let us write each set: $(HK)^{*}=\bigcup_{n=1}^{\infty} (HK)^{n}, \langle H, K\rangle={h_{1}g_{1}h_{2}g_{2}...h_{m}g_{m}: m\in\mathbb{N}, g_{i},h_{i}\in H\cap K \forall 1\leq i\leq m}$.

[cloud walrus]

schrysafis

[proven stone]

For space save I don't write my proof that $\langle H, K \rangle\subseteq (HK)^{*}$

[thorn jay]

its a quotient because you divide the group into parts and take those parts to be a new group (and normal subgroups are precisely when you can turn the cosets into a new group)

[cloud walrus]

schrysafis

[proven stone]

any ideas on how to use $HK=KH$ for the inverse inclusion?

[cloud walrus]

schrysafis

[karmic moat]

think of quotienting as analogous to the "mod" operator in the integers

[peak sedge]

i see

[velvet hull]

wait, which inclusion have you shown?

[karmic moat]

hence why people usually say "G/H" as "G mod H"

[proven stone]

^

[proven stone]

$\langle H, K \rangle\subseteq (HK)^{*}$

[cloud walrus]

schrysafis

[velvet hull]

the other direction should be straightfoward

by reading off the definitions for both

[proven stone]

alright I got it

cheers

[thorn jay]

this idea of dividing into parts is best seen in the first isomorphism theorem, where the "parts" are the sets f^-1(h) = { g ∈ G | f(g) = h }, and then you divide by these parts to get something naturally isomorphic (the same) as the image of your homomorphism

[tardy hedge]

When someone who doesnt study math sees this: omg math is just about squares and dotes and lines 🥹

[thorn jay]

simplicial objects

[tardy hedge]

UPONTHEWITNESSING

[glad osprey]

non-mathematician seeing this: if you have a square of area 9, then you can divide it into 9 squares of area 1, and 9 squares = 9 dots

[tardy hedge]

so true

[tardy hedge]

Thats actually a dub diagram

[thorn jay]

did you know that if you have an algebra A and an algebraic set X of A, then the tame congruence theoretic localisation A|_X is nothing more than the algebraic theory formed by the full subcategory of algebraic sets of A of the objects 1, X, X^2, X^3, ...

[tardy hedge]

I didnt know that brochacho

Its nothing more than that hey? 🧐

Did you know: 1+1 is nothing more than 2

[thorn jay]

using duality this means that the localisation by an algebraic set is a coalgebraic theory in a prevariety, meaning it can be extended to arbitrary coherent conditions

:0

[thorn jay]

and because these are polynomial clones, they carry a natural model of themselves so this means that tame congruence theory really can be studied using algebraic geometry

[tardy hedge]

I just still dont see why i would care to study the universal algebra framework

Personally

Like i wont lie thats probably in part because i think its too above me in terms of cognitive effort required of me to understand wtf you are talking about

So im not saying it’s uninteresting necessarily

[proud vigil]

That’s so real 😭

[tardy hedge]

😂

Im glad i resonated with someone

[proud vigil]

personally I kind of find things as annoying until I spend time to understand it and then it becomes more interesting

maybe that applies to most things but idk I also notice that things become boring after I understand it so maybe everything is just boring

[tardy hedge]

Yeah youre def right with that

[thorn jay]

honestly that's pretty real

[tardy hedge]

But if universal algebra really doesnt have many applications i cant see myself wanting to put the effort into learning it

[thorn jay]

I've gotta stop thinking about something just to appreciate its beauty again

[tardy hedge]

Wdym exactly

[thorn jay]

like I had to take a break from my research to appreciate how cool everything about it was again

[tardy hedge]

Oh yes thats what i suspected u meant

I feel the same way too

Sometimes i like going as silly as ooh cool words and diagrams 😁

[thorn jay]

idk at this point for me its super obvious to be interested in universal algebra

like it's literally the study of functions on sets

alternatively, the study of representations of generalised monoids

[tardy hedge]

In what sense? Id probably like it if u used more dumbed down language at first lol

[thorn jay]

well, an algebra at its core is a set A (called its universe) with a collection of operations F, where an operation is defined as a function f : A^n -> A, n being called the arity of f

you can look at indexed algebras, which are algebra over some signature T, where T kind of fixes the symbols and arities of the operations (like for rings you'd have the signature { 0, 1, -, +, * }) and the look at classes of indexed algebras which have certain properties etc

so any time you have a situation of an object where you can look at it as something with operations on it, or as a collection of functions parametrised by a fixed other object (like modules), then you naturally are in the domain of universal algebra

[tardy hedge]

Not sure abt the signature for rings

[thorn jay]

no? how so

[tardy hedge]

What is 0,1 there for

[thorn jay]

nullary operations, they are constants

basically distinguished (special) elements of your algebra the must be preserved by homomorphisms and be present in subalgebras

[tardy hedge]

Ok. And u wrote - + * for what reason

[thorn jay]

they're respectively unary, binary, binary operations

(one input, two inputs, two inputs)

[tardy hedge]

Where in the signature does it say that

Unary binary binary

[thorn jay]

I've said it a little informally, but the actual definition is basically just said once and then never thought of again: a signature is a function ar : I -> N, where N is the natural numbers 0 inclusive, and I is an index set

anyhow duality theory says that algebra is anything dual to geometry so that's also a nice way of looking at it

(half joking)

[chilly ocean]

Is there a universal algebra definition of a groupoid? Or any useful perspective of it from UA or anything? 🤔

I guess not because the operations have to be functions from A^n

[thorn jay]

username checks out

lmao

oidification is something purely categorical though

you can see it as a partial group but I think that's not super fruitful, as the reason why UA is so powerful is precisely because of the fact every operation is total

[chilly ocean]

Ic

[thorn jay]

total algebras (so every operation is total) are always models of algebraic theories. This, though, fails for partial algebras, because if you have partial operations t1, ..., tn defined on the domain U and an n-ary operation f defined on the domain V, then f(t1, ..., tn) is suddenly an operation defined on a domain that might be different from U. This means that you can't categorify it neatly

[white oxide]

Wait sorry to bother you but I just wanted to sanity check this logic for $\mathfrak{n}^c = \mathfrak{n}'^c = \mathfrak{m}$; The following diagram commutes:
[
\begin{tikzcd}
A && B \
\
{A_{\mathfrak{p}}} && {B_{\mathfrak{p}}}
\arrow[from=1-1, to=1-3]
\arrow[from=1-1, to=3-1]
\arrow[from=1-3, to=3-3]
\arrow[from=3-1, to=3-3]
\end{tikzcd}
]
By assumption, $\mathfrak{q}^c = \mathfrak{q}'^c = \mathfrak{p}$. Thus $\mathfrak{q}^{c^e} = \mathfrak{q}'^{c^e} = \mathfrak{p}^{e} = \mathfrak{m}$. Since the diagram commutes, we have
[\mathfrak{n}^c = \mathfrak{q}^{e^c} = \mathfrak{q}^{c^e} = \mathfrak{m} = \mathfrak{q}'^{c^e} = \mathfrak{q}'^{e^c} = \mathfrak{n}'^{c}]

[cloud walrus]

okeyokay

[white oxide]

:Sob:

about q^{e^c} = q^{c^e} I am a little bit ucnertain about but I'm sure it's true since the extensions are generated by the images

[thorn jay]

the extensions are into the localisations right

man I hate that notation it pmo

[white oxide]

Yeah

[thorn jay]

yeah okay

[white oxide]

I think I get the rest of the proof

that's where we use the correspondence theorem

between prime ideals of localization and prime ideals of the original ring

[thorn jay]

(prime ideals disjoint from A \ p)

[thorn jay]

yeah looks good

you kinda alr implicitly use the correspondence theorem for localisations in the second and first to last equalities but that's besides the point, its still correct

[white oxide]

ohh ok got it

[thorn jay]

(usually you can only this diagram chasing when you go against every arrow in the diagram, here that would be going from B_p to A, but because going against the natural map f : A -> S^-1A is injective on prime ideals, what you did is allowed - and of course necessary for the proof - here)

[elfin wraith]

Completely unrelated to the maths, but ya know you can just let tikzcd worry about the placenta and lengths of the arrows right? It’s like the whole reason to use tikzcd

[white oxide]

wdym placenta

[white oxide]

ah I see

[thorn jay]

✨ placenta ✨

[elfin wraith]

Like you can just do this \begin{tikzcd}
A \arrow[r] \arrow[d] & B \arrow[d] \
C \arrow[r] & D
\end{tikzcd}

[thorn jay]

he meant placement I think

[cloud walrus]

Nope

[elfin wraith]

ya, typo

[white oxide]

Oh I just use https://q.uiver.app/ lol

I don't actually type up the code myself

[elfin wraith]

Fair lol, there also this site (which quiver kinda copied) which I prefer, does the same job though https://tikzcd.yichuanshen.de

[alpine island]

Is there a way to show that for coprime $a,b$, $(\mr a)\times(\mr b)\cong \mr{(ab)}$ without using chinese remainder theorem? I'm actually trying to go go backwards and imply crt with this, but this part isn't coming easily

[cloud walrus]

nHail

[white oxide]

people actually type up their diagrams?????

[thorn jay]

isn't this literally the chinese remainer theorem?

[elfin wraith]

Ill type up triangles sometimes yeah, its often quicker than loading up quiver and stuff, but yeah for anything mildly complicated im pulling up quiver

[alpine island]

I mean yeah

it is

[thorn jay]

so proving this is proving the chinese remainder theorem

[elfin wraith]

or well, the tikzcd online edittor

[white oxide]

is crt related to classification of f.g. abelian groups

[alpine island]

well okay, I don't want to prove this by just applying the number theoretic formulation of crt for modular arithmetic

that would be the sane option

[elfin wraith]

Its true over any rings so you can just look up that

[thorn jay]

yes

[white oxide]

isn't it just a special case of the classification

or wait no i g

if you consider that every fg abelian group is Z

or something I forgot

[thorn jay]

yes every fg abelian group is indeed Z the cyclic group 👍 👍 👍 👍 👍 👍

[alpine island]

every fg abelian group is Z? this is a much stronger classification than I remember

[elfin wraith]

Group theory is the study of Z

[thorn jay]

every R-module is isomorphic to R

[south patrol]

No that's number theory

[thorn jay]

type shit

[quiet pelican]

The problem is that it’s usually too nice

[south patrol]

Lol

[thorn jay]

its boring

[quiet pelican]

It’s like using a field as an example of a ring

[south patrol]

Well also like group theory like mostly focuses on noncomm stuff

[thorn jay]

that's like saying "let's go listen to electronic music!" and putting on slap house

[thorn jay]

yeah because the commutative stuff is called module theory

[elfin wraith]

Hey this ring theory stuff is really easy, everything’s injective and the ideals are pretty simple

[thorn jay]

I can't believe it! every R-module is free!

[tribal moss]

FREEEDOM!

[thorn jay]

I'm finally free!! ohhh martha I'm coming home sweetie

please tell me anyone catches this reference

[alpine island]

Figured it out. Look at the natural morphism into Z/aZ x Z/bZ, get its kernel, and then use first iso.

[white oxide]

Isn't that just the proof of the CRT?

[thorn jay]

yeah

just in the language of group theory

[alpine island]

yes but very cleverly disguised.

[quiet pelican]

Ok what are your subfields?

[thorn jay]

look at the galois groups wrt the prime field, because we can assume every field extension is Galois

[elfin wraith]

I thought for sure there’d be a Galois gif for me to respond with but apparently not, boo hiss I say

[thorn jay]

https://tenor.com/view/evariste-galois-bereute-sehr-attentat-von-zug-und-bundes-brief-gif-18957572

this is the ONLY thing that pops up when you type in "evariste galois"

[elfin wraith]

I know, not helpful

[thorn jay]

I'm sorry qwq

wtf does it even say

[elfin wraith]

No I’m sorry that came off so wrong lmao

I meant I know, the gif is not helpful

[thorn jay]

NOT HELPFUL, YOURE FIRED

[alpine island]

Okay I will admit this whole exercise was just an excuse for me to put a commutative diagram in my intro algebra homework

I kinda wish I wasn't taking this class, because I should already have credit for it under the old system

but that's a long story

[thorn jay]

no way product diagram

[alpine island]

exactly lol

[tardy hedge]

there was something that was bothering me earlier today. it is very basic. i never understood it properly.

[elfin wraith]

Uhhhh my German is pretty bad but the first part is like from a letter from Galois, something about heaven

But it gets gradually less understandable for me lol

[elfin wraith]

What was it?

[thorn jay]

google translate:
"From a letter by Evariste Galois. Heaven is my witness. To have spoken such a disastrous truth to men of such limited ability."

German is so fancy

[tardy hedge]

ive never been comfortable with modular arithmetic. In particular I was trying to understand why if a is coprime to n then a has a mult inverse in Z/nZ

[rocky cloak]

$\mathfrak{GERMAN}$

[cloud walrus]

jagr2808

[elfin wraith]

I need to start practicing more again, I want to get better but it’s so hard to find the time (and effort)

[tardy hedge]

so real

[tardy hedge]

i like it

[rocky cloak]

Ich hat Deutch in der Schule gelehrent

[tardy hedge]

i no understand

[white oxide]

Here, is $x^{q_i^{e_i - 1}}$ supposed to be $x^{q_i^{e_i} - 1}$? My first thought for the highlighted sentence was to show that $g_i^{q_i^{e_i} - 1} \neq 1$. But under these conditions I'm not sure that's true or it might require some thought

[cloud walrus]

okeyokay

[rocky cloak]

Aber sehr slecht

[thorn jay]

I dropped any language besides Latin and English as fast as I could

[rocky cloak]

Same actually

Wait, latin?!

[thorn jay]

Went to a gymnasium, so you either have to take Latin or Greek

but in the end Latin was pretty fun

[white oxide]

Wait no then because then q_i^{e_i} - 1 doesn't divide p - 1 which we need for congruence 2 to have more solutions than congruence 1 hmmm

[thorn jay]

congruence

[rocky cloak]

Think it should be q_i^{ei - 1} as it appears

[white oxide]

hrmmmm

[elfin wraith]

That just follows from the definition of modular arithmatic and Bezouts identity

[rocky cloak]

Like x has order dividing q^e, but not dividing q^(e-1)

So order q^e

[elfin wraith]

I wish my school offered german, I only got to do french and I hated it

Do not speak a word now

[tardy hedge]

oh yea thanks. i guess i should review why bezout is true

[elfin wraith]

This is just the division algorithm as well, its all pretty elementary but defiently easy to forget. I think I need to remind myself how the division algo works about once a year

[white oxide]

Wait I'm lost 😂

[tardy hedge]

yeah i definitely looked at this many times but for some reason i remember it not being intuitive to me

[rocky cloak]

Where'd you fall off

[white oxide]

how do we know that g_i has order dividing q^e? g_i is an element of the units of U(Z/pZ) which has order p - 1, and the order of g_i must divide p - 1, but that's all I can see

I know that they have provided the prime factorization of p - 1 too

[glad osprey]

I usually think of this in terms of Bezout's identity: if a is coprime to n, then ax + ny = 1 for some x, y in Z, then just reduce mod n. More intuitively, I think of it like, when gcd(a, n) = g, and you look at the set { ka mod n | k in Z }, then every element of that set is gonna be a multiple of g, so the only way for a to have a multiplicative inverse is if g = 1. Like, if you add a to itself multiple times, while reducing mod n, you're never gonna get smaller than g, just the same way how if you start at 12 PM and jump around the clock, let's say, 9 hours at a time, you're never gonna reach 1 PM. Dunno if that helps

[rocky cloak]

In a group, if g^n = 1 then the order of g divides n

[white oxide]

ah ok, so if that's true then the order of g_i divides the order of q_i^{e_i} as you stated. so the only other case where the order of g_i is not q_i^{e_i} is when the order of g_i is q_i^{j} for 0 \leq j < e_i... and I'm struggling to see why this can't be the case...

[rocky cloak]

Well g^(q^(e-1)) is not 1

[white oxide]

Oh lol

Ohhhh

Okay now that makes sense

ty

[elfin wraith]

It basically comes from the fact that you can pull out common factors, like gcd(453,102)=3 and 453=3x151 and 102=34x3. So then 453-102= 3(151-34), so 3 is still the GCD of 351 and 102, and then you can just keep doing this until the numbers are small enough that you can just read off your remainder. There is a really good explanation of the extended euclidan algorithm somewhere, Ill see if I can find the one im thinking of

:I saw that

[south patrol]

No I agree w ur ting

[rocky cloak]

    while a != 0:
        a, b = b % a, a
    return b```

[elfin wraith]

These number theory notes are just pretty good, I would say quickly skim through the first few sections if youre looking to brush up on your basic modular arithmatic @tardy hedge

[tardy hedge]

thank u everyone

[elfin wraith]

If it makes you feel any better it took me 5 minutes to prove sqrt 7 is irrational earlier

[tardy hedge]

it really be like that 😢

[thorn jay]

an easy application of the FTA

🔥

[rocky cloak]

Step one: prove the FTA

[elfin wraith]

I remembered you squared stuff and then you get a contradiction but it took me a few minutes to work out/remember wtf that contradiction was

[thorn jay]

trivially follows from the fact that Z is a PID

[vapid vale]

step zero: prove euclids lemma

[elfin wraith]

Step 1: trivial fact known by any mathematican

[rocky cloak]

Euclid's lemmarick:

When a prime a product divides
In one factor it solely resides
Won't split in two
Nope, no can do
That is what Euclid decides

[elfin wraith]

Thats great ive never seen that before, big fan

[vapid vale]

:D

[elfin wraith]

Unless you just came up with it, in which case, props to you

[rocky cloak]

I used to run a lemmarick column in the student news paper

[vapid vale]

proof : it rhymes!

[rocky cloak]

Just made this one right now

[vapid vale]

yoneda lemmarick

[elfin wraith]

God damn a man of many tallents jagr, very impressive

[rocky cloak]

I think the first one I ever wrote was

Take a partially ordered set P
where all of its chains can be
Bounded below
Then we will know
There's a minimal element g

[tardy hedge]

Lol

[thorn jay]

in appreciation of algebraists' favorite lemma

[elfin wraith]

Jagr may be the only person in the world qualified to teach both homological algebra and poetry for mathematicians

[karmic moat]

Jagr can you do Whitehead’s Lemmarick

[rocky cloak]

Take a natural transformation
From a functor with representation
All is done
When you map 1
What a great revelation

[proud vigil]

anapestic meter is fun

[karmic moat]

https://tenor.com/view/writing-too-fire-fire-writing-writing-fire-friendcord-gif-7801118907023267493

“What a great revelation”

[proud vigil]

the meter does feel a little flexible here :P but it seems hard to fit math terms around meter, not a lot of synonyms i imagine

[elfin wraith]

So when’s the quivers book written entirely in limericks dropping?

[thorn jay]

I would buy that

100%

[alpine island]

this is how all theorem statements in Mathematics Made Difficult should have been.

[rocky cloak]

Don't know if it turned out amazing, but...:

Representation of finite dimension
From Dynkin classification
Co-homology
In first degree
Is gone, went on vacation.

[karmic moat]

That’s awesome but it begs the question

If cohomology were to go on vacation, where would it go

[rocky cloak]

Probably just moved up a few degrees

[thorn jay]

bishop goes on vacation, never comes back

[karmic moat]

LMAO

That was really good

The corresponding wikipedia page usually lists different proofs

[elfin wraith]

Not to my knowledge, but often if you look on like math overflow there will be a few methods given, there’s also regular Wikipedia and the proof wiki

Besides that I guess just cross reference different textbooks I guess

[proud vigil]

deleting messages making everyone look crazy 😭

[swift tundra]

I am working through Atiyah Macdonald and I’m not too sure how they are getting the equality in 21 iii). Does anyone have a hint?

[alpine plank]

Which Inclusion have you shown?

[swift tundra]

Closure is a subset of V(b^c)

[alpine plank]

I guess this is not really helpful but try showing that any closed set containing phi^* V(b) contains V(b^c). I can't think of anything more that would be a hint and not a solution

[rocky cloak]

The explicit description of the closure as the V of the intersection might be helpful

[swift tundra]

Yeah I have been trying to do that for a while now haha

[swift tundra]

Well if I show that any closed set containing phi^*V(b) contains V(b^c), then V(b^c) is contained in the intersection right (which is the closure)?

[rocky cloak]

I mean that's true. But my hint is to think more explicitly about what the closure of phi*V(b) is.

And remember that intersections play nice with preimages.

You may want to first prove it for the case b is a radical ideal

[swift tundra]

Hmmm ok I'll think about this

[rocky cloak]

I wasnt talking about the intersection of closed sets containing phi*V(b) if that's what you're saying here

I'm saying closure A = V(intersection over A)

[swift tundra]

Where intersection over A is intersection of closed sets containing A?

[rocky cloak]

No, that wouldn't parse. You take V of ideals.

I mean intersection of the elements of A

[swift tundra]

Oh interesting

But does V(-) distribute over arbitary intersections? It certainly does for finite intersections

[rocky cloak]

Distribute in the sense that
V(intersect x) = closure (union V(x))?
Yes

[swift tundra]

Ah ok

I fear I still don't entirely know what you are saying. Maybe I have just been looking at the problem for too long, haha.

[rocky cloak]

I'm just saying, which ideals are in phi*V(b), and what is their intersection?

[swift tundra]

Wait wait I think I understand more now let me work it out a bit.

[swift tundra]

Oh wow I think I got it now. So $$\overline{\phi^(V(b))}=V(\bigcap_{x\in \phi^(V(b))}x)=V(\bigcap_{y\in V(b)}\phi^*(y))=V(\phi^{-1}\bigcap_{y\in V(b)}y)=V(\phi^{-1}(\sqrt{b}))=V(\sqrt{\phi^{-1}(b)})=V(\phi^{-1}b)$$

[cloud walrus]

original goober gamer (OGG)

[swift tundra]

Whoops it went off the screen a bit, but the rest was just that radical commutes with contraction basically

Thanks for the hint Jagr, I hadn't seen that closure identity for the spectrum before, Ill make sure to note it down. I do still wonder if there is a way to do it in a more topological fashion...

[rocky cloak]

Well you would need a way to interpret V(b^c) topological then I guess.

Like the relationship between b and b^c seems mostly algebraic to me

Like the closure of phi*V(b) will be V(a) for some ideal a, so it's just about determining which ideal that is

[swift tundra]

Yeah and I suppose showing that V(b^c) is contained in any arbitary V(I) which itself contains phi*(V(b)) isn't super straight forward

[swift tundra]

I'm actually coming around to this method more, it is very clean and intuitive (when I step back now and look at it haha)

Maybe a key idea in this proof is that there is a nice correspondance between closed sets and radical ideals?

But really, thank you so much for the help! I had spent like 3 hours on this part of this problem...

[elfin wraith]

Yeah, and it’s an integral domain

I don’t think every ideal being principal implies ID but it may do

Well a PID is a ring with no zero divisors and every ideal being principal, I presume the latter doesn’t imply the former simply because it’s always stated that way but I don’t think I’ve tried to come up with a counter example before

But yeah your hierarchy is correct, there’s a big one on Wikipedia which has loads of classes of rings ordered by inclusion

There’s also one for like Cohen Maculey, Gornstien, complete intersection etc, Wikipedia is quite good for that stuff

[elfin wraith]

Oh Z/4Z is an example duh

Yeah

Generally a domain is that, just without the commutativity assumption

Yur

[thorn jay]

any quotient of a ring with principal ideals has principal ideals

lol

[fading acorn]

All rings are PID

[elfin wraith]

Yeah if you scroll down a little I woke up a little bit lol

[thorn jay]

YOURE WOKE NOW???

[elfin wraith]

Always have been

[thorn jay]

what has this world come to..

whats next, youre going to identify as british???

[elfin wraith]

That would be significantly less woke, would never happen

https://tenor.com/view/braveheart-mel-gibson-scotland-gibson-scottish-gif-21954284

[thorn jay]

amen

[delicate orchid]

do the scottish do anything other than larp

[elfin wraith]

Heroin

[delicate orchid]

ain't he from minecraft

[thorn jay]

"youre a hero, brian!"

[elfin wraith]

https://tenor.com/view/herobrine-my-honest-reaction-my-honest-reaction-meme-reaction-honest-gif-235046893189116626

[thorn jay]

this is actually my 13th reason

[delicate orchid]

so glad I campaigned for this channel to exist

[thorn jay]

why the fuck is that the top gif for "herobrine honest reaction"

[thorn jay]

for this exact reason huh

[delicate orchid]

ts shit ain't exact

[thorn jay]

its exact because i say it is

if irreducible it cannot be written as a product of a unit and a nonunit

the product of units is always a unit

element called prime if nonunit and a|xy => a|x or a|y but this coincides with irreducibles in PIDs

[rocky cloak]

Such a ring is called a principal ideal ring, and they are exactly finite products of quotients of pids

[thorn jay]

oo a class of the form PH(K)

missed a condition :>

[elfin wraith]

Ah interesting, haven’t come across them (explicitly) before, but yeah when I woke up and realised Z/4Z is an example

It’s kinda clear how that structure theorem works for Z/nZ but how does it go in general? That seems harder

[thorn jay]

i guess, what kinda properties does a quotient of a PID have

lol the proof uses primary decomposition

this is looks like it employs a lot of heavy machinery lol

thats kinda funny

[tardy hedge]

this is by correspondence theorem right

[rocky cloak]

I'm not sure what you mean by how it works for Z/nZ exactly. But it's easy to see that products of quotients of pids are principal ideal rings.

The difficult part is showing the converse. A proof exists here
https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-25/issue-3/On-the-structure-of-principal-ideal-rings/pjm/1102986148.full