groups-rings-fields

[topaz solar]

I think that’s easier to do directly?

[crystal turtle]

is it?

If so that shouldn't be terrible to show directly

[formal ermine]

it's really easy

like

we use the iso to pZ_p

from exp_p (or ln_p)

then you can immediately see that that contains p^nZ_p

because pZ_p here is additive

[crystal turtle]

okay sure let's say that works I'm not gonna check it. so then we've got that

[void cosmos]

guys when ur done please ping me i just have a quick proof check

[formal ermine]

@void cosmos

[coral shale]

just post it

[void cosmos]

okay king

[formal ermine]

yeah and then we're done

[crystal turtle]

okay swag

[formal ermine]

Thanks guys

[coral shale]

what have i done to feather...

[formal ermine]

y'all are real life savors

btw ryx are you an undergrad or grad

[crystal turtle]

not sure exactly why we needed the topology part but gg wp ez

[crystal turtle]

entering last year of undergrad

[delicate orchid]

yet another undergrad showing me up

[crystal turtle]

sir you are a god and i revere you as such wdym

[void cosmos]

if R is a ring , I is left primitive then I is prime. proof: suppose ab is in I but a is not in I , then since I is primitive its the annihlator of some simple left R-module call it A , we have ab in Ann(S) --> ab(s) = 0 --> a(bs) = 0 ( this is for all s) --> a(bS) = 0 --> a(S) = 0 ( by simplicity )

contradiction

does this work

[delicate orchid]

I actually have no idea what I'm talking about and am just REALLY good at googling

[formal ermine]

Ah nice

[formal ermine]

me irl

[topaz solar]

Relatable but not really since 2 years (including this one) since masters concurrent

[steel light]

what

[delicate orchid]

wait you're not phd either

[steel light]

uh

[formal ermine]

do any of you guys play trackmania?

[steel light]

I don't think there is one for I

[formal ermine]

what does this medal mean?

[delicate orchid]

bro's first author medal hahahahah

[formal ermine]

I started playing this game 2 days ago

[delicate orchid]

and no I don't play trackmania

[steel light]

Since AI = IA = A which is only zero if A is zero no?

[delicate orchid]

what we doing? finding zero divisors

[crystal turtle]

bro's proof was just skipped over (i don't know what a primitive ideal is)

[delicate orchid]

same

oh wait yeah I do

[void cosmos]

a primitive ideal is an ieal such that R/I is a primitive ring

[delicate orchid]

it's like half a prime ideal

[void cosmos]

a primitive ring being a ring that has a faithful R-module

faithful R-module being has trivial annihlator

[steel light]

Yes

[delicate orchid]

xy in I implies y^n in I or something

[coral shale]

yeah lmao. Was a half troll from me

[void cosmos]

this is primiary ideal

[delicate orchid]

fuckkkkkkkk

[void cosmos]

diff thigns

things*

[crystal turtle]

that's primary yeah, I got them mixed up before

[steel light]

I should've realized that instantly LMFAO

wp shushu

[coral shale]

bruh find the pair yourself obvsly

[delicate orchid]

you just need two non-zero matrices that multiply to give the zero matrix

[coral shale]

[steel light]

Yes

Okay I will find a pair myself

[topaz solar]

Who

[delicate orchid]

you, nerdy bob junior

[topaz solar]

What

Literally what gave that impression

[delicate orchid]

do I need to link any conversation we've had in #advanced-algebra

[coral shale]

i have a genuine hint, but yh, try a bit first

[topaz solar]

I mean that’s like

The preliminary chapter of a textbook

Unironically

[delicate orchid]

not the most recent one - I could actually follow that one so yeah it's trivial

[topaz solar]

(Not an undergrad book in the SLIGHTEST) but still

[delicate orchid]

but the one before where you were banging on about fusion systems

[topaz solar]

I meant the finite Morley rank one yeah

That was chapter 1

[delicate orchid]

bullshit

[topaz solar]

I kid you not

[crystal turtle]

I think rather you want to note that, since a is not in I, aS =/= 0. So then since S is simple, aS = S, hence since abS = 0, it must be that bS = 0, hence b is in I
Don't take my word for this, I just learned these definitions 1 minute ago, and I do not normally care about noncommutative rings

but uhh this might work?

[topaz solar]

The book is essentially a long proof of the main result tho

[void cosmos]

wait did i write something wrong?

[topaz solar]

So

It’s def not an undergrad book

But it’s literally 3 sections of the first chapter

[delicate orchid]

so ur chatting about grad content and then wonder why I think ur grad

[topaz solar]

masters sure but

[crystal turtle]

idk and idk that what I wrote made sense either lol. But this is my understanding?

[delicate orchid]

bro I was learning character theory in my master's year

[topaz solar]

I am in no way phd material

[void cosmos]

tbh i got lost

[crystal turtle]

(might be worth posting in #advanced-algebra where it will not be drowned out)

[void cosmos]

cuz i thought whats diff from what im saying

yea okay sure

ty boys and girls

[crystal turtle]

I think it's fine in there

[delicate orchid]

I learnt what a free group is in my master's year

[void cosmos]

if some sad guy or girl pings me or pings this

[topaz solar]

Bro what

[void cosmos]

im going to get banned over him

or her

fuck them

[crystal turtle]

[topaz solar]

Idk characters tho

So

Beyond like a vague premise

[delicate orchid]

you know the trace map

[delicate orchid]

that's it

[topaz solar]

Based

[crystal turtle]

nah you did not think I was a PhD did you

💀

[topaz solar]

I need to ascend to PhD tier this year

[crystal turtle]

same

[solar vessel]

I need to shave

[void cosmos]

yoo i am seriously not this guy at all like i love spamming and shit

[crystal turtle]

even though I will be doing a masters the year after but still

[formal ermine]

I need to grow a beard

[void cosmos]

but can u guys like move to chill or discussion

😄

[topaz solar]

Same

[void cosmos]

no canc i swear

[delicate orchid]

is it really that unusual to assume people who know more than me are further ahead than me?

[topaz solar]

In actual on topic discussion, how fast should I blitz Lang & D&F?

[steel light]

This is a pain in the ass Shuri

[solar vessel]

good work shuri

[void cosmos]

u cant blitz DF

[crystal turtle]

I do not know more than you

[void cosmos]

u can certainly blitz lang

lang is just def theorem proof

[crystal turtle]

also ya D&F is omega long for some reason

[topaz solar]

Mortal

[crystal turtle]

bro's having a whole ass conversation with you

[void cosmos]

also just pick one

why both

[topaz solar]

Yeah idk maybe I’ll just do Lang

[void cosmos]

why would you do lang tho

like whats the purpose

dont u already know the stuff

[topaz solar]

I have never heard of primitive ideals

[steel light]

I guess there’s like easy examples using almost-0 matrices

[topaz solar]

And I’m missing several ring properties

So it’s to fill in gaps

And learn Galois

[coral shale]

right tell us wtf u doing

[delicate orchid]

I will never learn galois

[topaz solar]

Or fill in holes, say, in my group theory

[coral shale]

yh...?

sounds legit

[delicate orchid]

so give us two almost 0-matrices

[coral shale]

(just sold rycs pfp in discussion)

[solar vessel]

are these easy examples in the room with us right now

[coral shale]

$$$

[delicate orchid]

ryc nfts

[steel light]

I just came up with a matrix and tried to replace numbers on a second matrix until it worked

[coral shale]

just checking. 2x2 was what u doing right

[delicate orchid]

what if that matrix was inverible

[steel light]

Yes

[delicate orchid]

invertible matrices are dense so it's probably was invertible

[topaz solar]

Relatable

But basic stuff is useful

[delicate orchid]

I've learnt the basic stuff via osmosis in this channel

[steel light]

If it was invertible you’d get I when you take the product, then you subtract I, it looks kinda similar to the eigenvalue problem but I didn’t go further with it because it didn’t make sense to me

[topaz solar]

Galois stuff shows totally transcendental fields are alg closed

[delicate orchid]

it's literally the study of a particular lattice isomorphism who cares

[topaz solar]

Because like

[steel light]

Anyways uh

[delicate orchid]

when will I ever care about this

[topaz solar]

Particular char q stuff and Kummer extensions

[delicate orchid]

ok can I put u out of ur misrey

[solar vessel]

no

[coral shale]

feather whats this one matrix uve set ur eyes on

[steel light]

Let me use (1, 0; 0, 0). (0, 0; 0, 1) is an easy example

[coral shale]

ur idea sounded good but run it by us

[topaz solar]

This is roughly the statement that non-closed fields you can have super infinitely ramifying variety-things in non alg closed ones

[coral shale]

yayyy

[steel light]

For right multiplication

[coral shale]

it looks good

[topaz solar]

As in, it can ramify into countably many pieces for each ordinal

[steel light]

I just used (1, 2; 3, 4) for my original test lmao

[topaz solar]

I think

[delicate orchid]

I literally don't know what any of that means

[topaz solar]

???

[steel light]

But gave up after messing up my guessing

[solar vessel]

my brother in christ

[delicate orchid]

determinant non-zero, aka invertible, aka not a zero divsor

[solar vessel]

that is invertible

[coral shale]

opencry

[delicate orchid]

I knew it I called it

I called it

[topaz solar]

AG over non-alg closed fields = no

[steel light]

Am I on to something?

[coral shale]

if its invertible

it like cant be the thing

[topaz solar]

Non commutative AG uses a wholly distinct route tho

[coral shale]

zero divisor

[steel light]

What

[coral shale]

prove it

[steel light]

Oh wow

Okay

[coral shale]

prove invertible => not zero divisor

for any element of any ring

[steel light]

ANY element of a ring?

[coral shale]

apart from the 0 ring ig

[delicate orchid]

can I please just post the matrices I have the latex in my clipboard

[steel light]

Huh

[solar vessel]

no

[coral shale]

Just prove for a general ring invertible => not zero divisor

[steel light]

Ok sure but like

[coral shale]

show us the goods sir

[topaz solar]

yes

[delicate orchid]

$\begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}$

[cloud walrus]

Wew Lads Tbh

[topaz solar]

Wew are you PhD student?

[delicate orchid]

yeah I'm 2nd year

[solar vessel]

wait wtf

[topaz solar]

Dang they let anyone in huh?

[delicate orchid]

they really do

[topaz solar]

Jkjk

[delicate orchid]

I'm not getting funded but yk

[coral shale]

Also, for a general ring. An invertible element is called a unit.

[topaz solar]

Oh

[solar vessel]

oof

how are you surviving

[topaz solar]

Just collab w/ me in adapting fusion systems to finite Morley rank groups

Gg ez fields medal

[steel light]

How is that related to what I was trying to do? I was trying to find a zero divisor FOR A = (1, 2; 3, 4), not show that it WAS a zero divisor

I’ll try to prove what you suggested

[topaz solar]

(This is a relatively longstanding conjecture solving thing it is horrible)

[coral spindle]

...how the fuck

🫂

gl

[steel light]

Just curious why A invertible <=> not a zero divisor is relevant to my problem

[open sluice]

wtf 1k messages in 3.5 hours

[steel light]

I was trying to find a zd for A

[delicate orchid]

I'm teaching for my funding while giving all the money I earn back to the university in fees

[coral spindle]

Christ wew that's unfair

[delicate orchid]

is it?

[topaz solar]

I’d need to learn this stuff first but dw about it

[coral spindle]

I think so!

[void cosmos]

lmfao i wish

they would lt me in

without me being a math majo

[delicate orchid]

I had to pay for undergrad why wouldn't i have to pay for post-grad?

[coral shale]

bruh if a is zero divisor and ab = 0.
Then so is b

[steel light]

Oh

[topaz solar]

I am significantly less worried about PhD apps

[coral spindle]

You do, but I think most of us get funding orgs for postgrad stuff. Anyway I won't interrupt any longer. Gl my dude

[topaz solar]

Thanks wew

[delicate orchid]

I applied to 12 places and only got one interview

[topaz solar]

But ur good at fusion systems smh

[steel light]

That was actually my guess but I doubted myself :(

[coral shale]

so yeah u had to start with a non invertible else u got no chance

[steel light]

Okay

[delicate orchid]

the most useful skill in the world and when I was applying I didn't know what a fusion system was

[topaz solar]

ok fair

[crystal turtle]

[coral shale]

this may seem a waste of time but u example hunt a lot in algebra

[topaz solar]

Fusion systems of infinite \omega stable groups

Well, uh

[steel light]

I don't see why it's any different than finding pathologies in topology or analysis

[topaz solar]

I don’t think that makes sense nvm

[steel light]

So I don't think it's a waste of time

[topaz solar]

But finite RM are in particular \omega stable and totally transcendental so

And they 100% can be made to work

[coral shale]

other examples of zero divs came from quotients

which u might wanna explore later

[topaz solar]

what do you even do with fusion tbh

[coral shale]

oh and ofc they wanted u to find a commutative ring example

before that

[topaz solar]

They’re the native landscape for sporadic groups but

[delicate orchid]

I love how everyone bleaks at my situation but I literally have nothing to compare to so I just get on with it

[topaz solar]

???

[delicate orchid]

I'll let u know when I figure it out

[topaz solar]

Based

[delicate orchid]

spetses!!! woahhhHH wooohhh

what if we did weyl groups but just made them general coexter groups woahhhhhh

[steel light]

This class is too much Ill do it after too 😭

[topaz solar]

Well, I know some gigachads suggested finite RM simple 2-Sylow based stuff

And that might be native for uhh

“”Sporadic””/bad groups in this sense

Issue is, sporadic finite groups are still sporadic

So idk how natural it is for the “not algebraic” simple finite RM groups which may or may not exist

And the goal is either to find one via realizability via saturation

Or crush their existence

Either one is h u g e

[crystal turtle]

Nah I'm just scared of that happening to mme when I apply for grad schools. I got rejected everywhere in the states when I applied to UG, all 8 or 9 places 😭

[topaz solar]

I applied to 3

[delicate orchid]

gl boss

for ug I applied to 5 and got rejected from 3(?)

hard to remember it was 5 years ago

[topaz solar]

I got rejected by one

Common MIT L

Imagine how cracked I could be with a good math program

[void cosmos]

im the MIT admission guy

lemme accept u real quick

[topaz solar]

Is Hartshorne still the go to AG intro?

[crystal turtle]

I think so? That or Vakil is what I'll be reading

[delicate orchid]

like wise, believe it or not I did mainly applied for most of ug

because there were just 0 algebra courses

[topaz solar]

Insane

I did pure math but uh

“Prove 0a = 0” for a class I have this semester

[coral shale]

in non comm, do we distinguish about left or right zero divisors?

https://media.discordapp.net/attachments/844681108473118750/1146167596845117510/wolf.png

[solar vessel]

for ug I applied to 5 and got in 5
for phd I applied to 3 and got in 2 one of which was my ug where I am going

[coral shale]

cus this works the other way round, say

[delicate orchid]

huh interesting

[solar vessel]

so I'm gonna be in the same place for 7 years

[topaz solar]

Still not a unit

So

[delicate orchid]

I guess we do

or

hmm

[topaz solar]

Annihilators of (a)

[delicate orchid]

that's the word

[topaz solar]

That can depend on if it’s a left or right ideal

That is

[coral shale]

lemme see lemme see...

[topaz solar]

A left or right module

But

“Zero divisor” not so much

[solar vessel]

well in this case it's the same thing

[topaz solar]

They still kinda suck

Either direction

No unit moment

[icy bear]

is this a place to ask about simple group theory questions too and stuff?

[formal ermine]

yes

[icy bear]

oh ok

so

in a group, is the identity element in each of it's subgroups the same as the general group identity element

?

[hybrid pike]

Yes

[icy bear]

fr?

sure?

[hybrid pike]

You can prove this

[icy bear]

that's weird

[steel light]

The identity is unique

[hybrid pike]

It's not too difficult to prove

[steel light]

Think about it intuitively

[icy bear]

that's not enough tho

[steel light]

Why not?

If a subgroup contains an identity

Then the larger group contains that identity

But the identity of the larger group is unique

So they must be the same

[icy bear]

(intuitivelly) I think that like, the limitation of the subgroup can make it's elements have a coincidental identity that has nothing to do with the general identity

[steel light]

Otherwise your group would have two identities

[icy bear]

I cannot give any examples obviously because it's not true but my intuition says that

[hybrid pike]

If youre not convinced, try to construct a proof

[icy bear]

I will

that will be my exercise to the reader

ty

[hybrid pike]

In fact I think thats an exercise to the reader in my algebra textbook :P

[icy bear]

lmao

[hybrid pike]

I solved it last week when going through the capter lol

[icy bear]

oh actually it's obvious now lmao

[formal ermine]

it's the other way around

[hybrid pike]

Yeah

||e is an identity in G, so e is also an identity in H. Since H is a group, e is the only identity in H||

[icy bear]

actually not

I need to prove that the inverse is the same in groups and its subgroups

(I think it is, is it?)

[hybrid pike]

no

well, it is

[icy bear]

fuck

[hybrid pike]

but you dont have to

[icy bear]

what

ah

a.b = a (can be done in the group and the subgroup since both are closed)
a^-1.a.b = a^-1a (can be done in the group and subgroup since in both all elements have inverses, but here's the problem, idk if the inverse is the same)
b = e (the argument supposedly is the same in group and subgroup so b = e, just e, the same e)

my problem is being the second step

[hybrid pike]

Youre making it more complicated than it should be

[icy bear]

sorry if it's too detailed, I really struggle with groups and always need to call back axioms and simple theorems to be sure I'm right

oh wait actually, if the operation is the same, the inverse works the same way always

shit

[next obsidian]

Prove that if h has the property that g•h = g for all g (or even one g) then h = e

[delicate orchid]

oh I can do this, sir!! professor!! pick me sir!!

[icy bear]

always struggle with this type of simple stuff, whyy

[hybrid pike]

Takes some practice in my experience

[icy bear]

my head was burning

ty non-penguin creature

[hybrid pike]

Once you've made enough exercises using the group axioms they come a lot more naturally

[icy bear]

also that's the second time I struggle to remember how operations are defined

[icy bear]

hope so

[hybrid pike]

I failed my first attempt at my abstract algebra exam badly because all I did was spend 4 days studying the book

[icy bear]

abstraction is hard 😅

[hybrid pike]

Im currently taking a lot more time going through it and solving as many exercises as I reasonably can and it's going a lot better

[icy bear]

that's good

[hybrid pike]

I have uh, 7 days left until the exam

or 8

counting is hard

either way, practice is the way imo

[icy bear]

uhh just associate a subset of the natural numbers with the days one-to-onely

[hybrid pike]

what if I associate the days left with a cyclic group

That way the exam never comes

Because I always loop around

[icy bear]

also, does that thing mean the intersection of all subgroups of a group has at least e?

[delicate orchid]

the infinite cyclic group in quesiton:

[hybrid pike]

Yes

[icy bear]

that'd be amazing tbh

[hybrid pike]

This is the trivial subgroup

[icy bear]

:0

[delicate orchid]

in fact, the intersection itself will be a group

wait no yeah it's always trivial lol

[crystal turtle]

hmm if the universe always loop after n days, perhaps we ought to model everything in Z/n?

[delicate orchid]

the intersection of two subgroups is a group though that's what I meant

[crystal turtle]

unions crying rn

[hybrid pike]

hmm perhaps

[crystal turtle]

(as they should)

[delicate orchid]

good

[icy bear]

that's exactly why I was asking those previous questions

[hybrid pike]

Looks like a good exercise

[icy bear]

it is, it's simple and tbh easy but made me check a lot of what I thought

[coral shale]

find an example where A u B not a subgroup

[formal ermine]

literally first thing that comes to mind

[rotund aurora]

isnt the converse more interesting

[coral shale]

i literally do not know

[rotund aurora]

like tell when A union B is actually a subgroup

[coral shale]

what is the first thing that comes to mind

[formal ermine]

2Z 3Z

[rocky cloak]

Is it even possible to find an example of AuB being a subgroup without one being contained in the other

[delicate orchid]

there is an awful theorem about when the union of three subgroups is a subgroup

[coral shale]

infinite groups are too big

[delicate orchid]

no I don't think so

[rotund aurora]

that's what I was thinking

[delicate orchid]

https://math.stackexchange.com/questions/959886/can-a-group-be-a-union-of-three-subgroups

[formal ermine]

beggers can't be choosers

[ashen heron]

span((0,1)) cup span((1,0)) are both subgroups of R^2 but union not closed hnder addition ?

[delicate orchid]

this is so disgusting

I hate this so much

[coral shale]

idk about u guys but istg my first examples of groups were Cn or some crap

1997?

wtf

[delicate orchid]

not that recent

[rocky cloak]

Seems like a law of small numbers type of thing

[ashen heron]

the thumbs up mean my example works?

[delicate orchid]

yeah

[hybrid pike]

lol wtf is this

[delicate orchid]

or just 7 being awful

people rag on 2 being an annoying prime but 7 is far more subtle

[rocky cloak]

But it fails for numbers bigger than 7 too though right? Or is it just 7?

[delicate orchid]

no clue, I'll check the paper

[rocky cloak]

If it's just 7, then that is indeed totally bizarre and disgusting

[delicate orchid]

no it isn't known for higher n

which is what I expected

[rotund aurora]

ig you would ask if it is possible for groups of arbitrary size

[hybrid pike]

who wants to spend their time proving that anyway 🙃

[delicate orchid]

free paper

[rocky cloak]

"free"

[rotund aurora]

it is interesting honestly lol

[next obsidian]

I tried for a while to prove the theorem about quotient V_4 and couldn’t get it lol

[delicate orchid]

it seems quite annoying

just cause unions don't really appear much

[hollow mica]

we had this discussion before I think

[topaz solar]

Somebody here is doing research on 1/7

[ashen heron]

there's one in linear algebra where vector space over an infinite field can't be a finite union of proper subspaces

[topaz solar]

As in like, the fraction

At my school

[delicate orchid]

what

[south patrol]

Good theorem

[topaz solar]

Im being facetious in my description but yes

[rotund aurora]

good

I liked proving that

[south patrol]

Can use it for the primitive element theorem

Sorta

[delicate orchid]

my research is on the map h -> ghg^{-1}

[rotund aurora]

ye

[south patrol]

We used it to develop galois theory lol

[crystal turtle]

please elaborate

I am curious on this supposed "1/7" number

[topaz solar]

@cursive spindle

Ok so

1/7

2/7

3/7

…

They’re permutations of each other in base 10

[cursive spindle]

why abs alg

.-.

[topaz solar]

My guy said

What if I found 1/7 looking things in other bases

Saw him working on lemma 30 in the lounge once

[topaz solar]

Number theory vaguely

[cursive spindle]

oh ic

I got scared

[topaz solar]

he hadn’t yet shown 1/7 numbers exist in other bases

[cursive spindle]

I was like "why would somebody ping me in abs alg?"

[topaz solar]

I don’t wanna clown him too much tho

[crystal turtle]

[topaz solar]

Anyway use ur calculator and see the n/7 being permutations of 1/7

[delicate orchid]

yeah but like

[topaz solar]

In particular, all 6 n

[delicate orchid]

why should I care

[rotund aurora]

by that you mean the decimal expansion?

[topaz solar]

ye

[crystal turtle]

how are there even 30 things to say about this

[topaz solar]

bro idk

[rotund aurora]

no one is saying you should care I think

[delicate orchid]

oh in that case I care a lot

[topaz solar]

He was stuck on showing infinite repeating expansions exist so

[crystal turtle]

I get it tho

[topaz solar]

(I walked over and said coprime it and do whatever/b^k-1)

[crystal turtle]

I like caring about things that have no use

[rotund aurora]

are there any cool theorems

[topaz solar]

no

[rotund aurora]

ok

[topaz solar]

Im of this group

[delicate orchid]

yeah he directly proved the hodge conjecture

[topaz solar]

Though if there’s other ones he can find then that’s cool

[delicate orchid]

yeah it sounds like a nice project

although

nvm

[rotund aurora]

typical of unemployed people

[crystal turtle]

I am well employed thank you

not actually I just got fired from tutoring for not returning after summer break

[delicate orchid]

I was thinking if you were in base m with m coprime to p then because 1/p has p-1 repeating things it'll have to be transitive but that's not true at all

[delicate orchid]

I cant keep tutoring cause I'm transfering unis and can't tutor at my new uni immediately cause they need to "train" me...

[crystal turtle]

oh it was online, I could have

I just didn't want to because eww trying to tutor grade schoolers online sucks

[hollow mica]

eeee

[crystal turtle]

e

[steel light]

Is the simplest example of a commutative ring with nonzero zero divisors Z/(4Z)?

[rotund aurora]

probably

[steel light]

2 is a nonzero zero divisor for 2 right

@coral shale Does that work

[delicate orchid]

yeah although "for 2" is a little redundant

and I'm not sure if that's the simplest one, defnitely the smallest

[ashen heron]

maybe the R^X thing where X has at least two elements

[steel light]

Why? 2 * 3 = 5 equiv 1 which isn’t 0 no?

[delicate orchid]

what

[steel light]

what

[rotund aurora]

🐒

[ashen heron]

2*2 cong 0 mod 4

[delicate orchid]

a zero divisor is a x such that there is a y with xy = 0

[steel light]

Oh existence

Okay

Lmfao

[ashen heron]

this isn't the simplest is it? what's counts as a simple example anyway?

[delicate orchid]

R^2 is simpler imo

[topaz solar]

Yeah I’ll just pass that to him next I see him lmfao

[steel light]

Oh I never considered this lol

[ashen heron]

also zero divisors are required to be nonzero are they not?

[steel light]

Because (-1, 0) I think

[hollow mica]

you're about to end their whole career with this one

[ashen heron]

2 is a nonzero zero divisor

[delicate orchid]

(1,0)*(0,1) = (0,0)

[steel light]

defn dependent

[delicate orchid]

how

the direct product of rings comes equipped with addition and multiplication automatically

if it didn't it wouldn't be a ring

[steel light]

I meantthis

[steel light]

Oops yes

[rotund aurora]

are there rings with zero divisors of any finite size

[ashen heron]

like 0 is just always a zero divisor

[rotund aurora]

no, like 2

but for which are there

[ivory trail]

https://en.wikipedia.org/wiki/Zero_divisor#Zero_as_a_zero_divisor
here wikipedia suggests 0 is a zero divisor iff the ring is nonzero

In a commutative noetherian ring R, the set of zero divisors is the union of the associated prime ideals of R.

[delicate orchid]

there aren't any for size 2,3

and I'd say size 1 as well

any higher than 4 makes me cry to think about

[ashen heron]

why vomiting? I say something wrong?

[coral spindle]

Interesting question

[rotund aurora]

if n is composite, just take Z/nZ

so we just have to worry about primes

[delicate orchid]

we won't be able to do products to get them obviously

[coral shale]

trust nlab. 0 is a 0 divisor

[ivory trail]

it's what i said

[delicate orchid]

the problem is the number of rings of a certain order is absolutely gigantic

[coral spindle]

So the underlying group is Z/pZ necessarily, and it must therefore be of characteristic p, so it is indeed the ring Z/pZ. So no, the primes oughtn't have such things.

[delicate orchid]

oh yeah Z is inital

swag

[ivory trail]

you also have that multiplication by an element is injective iff it is not a zero divisor

[topaz solar]

Well, unless you don’t have a multiplicative identity

In which case

[delicate orchid]

SHHHHHHHHHHHH

[topaz solar]

Ya know

[coral spindle]

SILENCE.

No more of this non-unital tomfoolery

[topaz solar]

For any infinite cardinality

R[x]/x^2

[coral spindle]

Trivial via compactness >:(

[topaz solar]

based statement

[coral spindle]

Thank u

[coral shale]

feld

[delicate orchid]

:breakfast:

[coral spindle]

Oh I didn't mean compactness although it helps. I meant Loewenheim-Skolem. Semi-based............

Compactness tells me there is an infinite ring with zero divisors

[rotund aurora]

[delicate orchid]

lord knows why

[coral spindle]

Here's a twist: what's the smallest ring with exactly n zero divisors?

[delicate orchid]

croqueta you got any idea what these geezers are talking about

[coral spindle]

Model theory

[delicate orchid]

Z[x_1, x_2,..., x_n]/(x_1^2, x_2^2, ..., x_n^2)

there is no smaller

[coral spindle]

SO TRUE...

[delicate orchid]

uhhh lemme think of a serious answer

[rotund aurora]

yes

[coral spindle]

1 zero divisor: Z/4Z

[delicate orchid]

in Z/nZ if ur not a unit ur a zero divisor, there are phi(n) units so there are n-phi(n) zero divisors

[coral spindle]

Not a bad first step

[delicate orchid]

so pick the minimal n such that n-phi(n) is what you want

I'm happy with this answer

[rotund aurora]

you have an infinite set of formulas. If for each finite subset of formulas you have got a model satisfying those formulas, then you have got a model satisfying all the formulas

or something like that

[coral spindle]

Yeah that's compactness

[delicate orchid]

why does everybody know model theory

[rotund aurora]

so like

a classic example is hyperreals

[hollow mica]

anyone know how many (unital, not necessarily commutative) rings of size 3 there are? it's at most 27, since the ring {0, 1, x} is determined by what x + 1, x + x, and x * x are... but for instance we can't have x + x = x, so that already rules out 9 of the 27

[coral spindle]

So the set of formulas, in english, would be {there is a zero divisor, there is one element, there are two distinct elements, there are three distinct elements, ...} u {axioms of rings}. Clearly there are finite rings of arbitrary size with zero divisors so compactness tells us there is an infinite ring with zero divisors

[rotund aurora]

consider the formulas that say standard things about ordered fields, now for each natural number n consider the formula there exists x suh that x>n. Then Q or R satisfies any finite subset of these formulas

[delicate orchid]

additive group has to be C_3

0 has to be 0

kinda forces it

multiplicative group has to be C_2

oh no not quite

could have three idempotents

[coral spindle]

Nah nah same argument as before

[delicate orchid]

oh I'm not talking to you guys

[coral spindle]

We have a unit, and the ring is of characteristic 3

so it has to be Z/3Z up to iso. It's all determined

[topaz solar]

Have however infinitely many symbols and say x =/= symbol or smth

[delicate orchid]

oh right yeah we have a unit

BORINGGGGGG

[hollow mica]

why not just use the single formula "for all natural n there are n distinct formulas"

[coral spindle]

That is not a formula in first-order logic

[coral spindle]

So yeah it's Z/3Z up to iso. Same thing for any prime; see the argument above

[topaz solar]

*these are all first order formulas

[hollow mica]

oh

that'd be pretty hard

Wait I guess

even if you could firstorderify it

[topaz solar]

Or uh however you do upward Lowenheim Skolem, I think something similar gets you downward?

[hollow mica]

you couldn't apply the compactness theorem in the same way

[coral spindle]

As it happens, you cannot express such a thing in first-order logic

Compactness proves so

[rotund aurora]

yeah otherwise you would be able to hack the universe

[topaz solar]

You can do some silly nonsense like

First order theories on V_\omega(S)

But uh

That’s still quite bounded, even if you can work with, say, continuity, open sets, etc

[hollow mica]

oh rings of prime order do that I didn't know

[topaz solar]

This is useful for some nonstandard shenanigans, even outside the reals

Since, ya know, functions live in V_3(S), where V_n+1(S) = PV_n(S) u V_n(S), and V_0(S) = S

Or maybe it was 2? 3 should get you sets of functions? nah I think it’s 3 since f is a set of objects in the form {{x}, {x, y}}?

ANYWAY
This lets you do nonstandard nonsense topologically, taking “polysaturated models” and saying like X compact iff every y in X* is in \mu(x) for some x in X

[rotund aurora]

Factually true

[topaz solar]

(This should look like Bolzano Weierstrass, for sufficiently strengthened notion of sequences)

I don’t think you need polysaturated but you want at least as big as your neighborhood bases for this one

[coral spindle]

Nerd.... nerd......

If you know what the morely rank is, braces instantly appear on your teeth and your eyesight gets worse by 50%

[coral spindle]

(I should learn what the morely rank is to get free braces)

[topaz solar]

My eyesight got much worse in exchange for no braces

Morley rank is very algebra related though

I believe

Im not an expert though so

[hollow mica]

TIL learning model theory improves your smile

[topaz solar]

Totally transcendental field -> alg closed

And Morley rank is like the dimension of your varieties or smth

Morley degree is how many irreducible components

And appropriately generalized to generic model shenanigans

See here for Macintyre

And showing it has roots of unity again applies that Kummer extensions don’t exist in this context, in the proof I saw

Finite Morley rank also apparently implies omega stability (I don’t know how this one goes since I haven’t gotten there in my reading yet)

Presumably something about how you can’t infinitely ramify things

[coral spindle]

OK fr now we should probably stop talking about model theory in an AA channel

[topaz solar]

oh right

[coral spindle]

I'm curious about morely stuff; I might look at it another time

[topaz solar]

On a more algebraic note, this all connects back to algebraic groups

Via the Cherlin-Zil’ber conjecture

[crystal turtle]

Why would we ever talk about AA in #groups-rings-fields , that's not what we do here

[topaz solar]

Which is what I was talking with Wew about before

Which states that simple omega stable groups are algebraic groups

In particular, I was talking about the still horrible problem of simple groups of finite Morley rank

Which admit a nice theory of Sylow 2-subgroups

And other p are dead to us

We get uhh Sylow and Sylow^\circ subgroups

Since Sylow needn’t be definable, but Sylow^\circ are certainly and are the smallest definable subgroup of finite index (of the Sylow group in this case)

“Connected component”

And this somehow ties in with Fusion systems?

[coral spindle]

Yeah like in the theory of algebraic groups

[topaz solar]

I do not know anything about those

[coral spindle]

Given an algebraic group G there's a smallest closed subgroup of finite index, G^\circ

[topaz solar]

Ye

Here it’s smallest definable subgroup

[coral spindle]

This is just the irreducible component of G containing the identity

Yeah nice

[topaz solar]

By virtue of totally transcendental

As there is no infinite descending chain of definable subgroups

so any intersection is actually really just an intersection of finitely many or smth

Sylow subgroups can be undefinable but connected component isn’t

The idea is these things look like algebraic groups unsurprisingly

And as it turns out

They are

Almost

(The conjecture is they are, but we only have algebraic group over char 2 alg closed field or has finite 2-rank)

Where here 2 rank is dimension of maximal abelian 2-subgroup or smth like that

They actually seem to think all simple \omega stable groups are algebraic interestingly

Idk how that looks though so I can’t comment

[topaz solar]

(The proof of this is basically you make some infinite binary tree thing iirc and therefore unbounded rank)

[summer path]

why is it so common in commutative (at least what i've seen so far) to want to use the statement of things not belonging to an ideal \mathfrak{p} to show that it's prime

[south patrol]

well given an ideal p of (commutative) A, it's prime iff A/p is an integral domain and so in some sense not being prime is the more "positive" notion

like, to show smth's prime you just need to find non-zero a,b in A/p such that ab = 0

whereas to show it actually is prime would need you to show that's never the case which is possibly trickier

[summer path]

hmm i see

thanks potato

[south patrol]

np

may need to ask smth here myself soon lol

[void cosmos]

is Z --> Z/p^nZ ( the natural projection )

a counterexample to show that the homomoprhic image of a semisimple ring need not be semisimple?

[coral spindle]

Are you sure Z is semisimple?

[void cosmos]

yea i forgot tell to you

semiseimple for me means only having zero jac radical

need not to be artinian

[dim widget]

This is not a good notion of semisimple ring

The right notion should be that every ses of modules splits

[wraith cargo]

tell that to so many books that define it that way

[void cosmos]

Its not mine tbh

[void cosmos]

This is a theorem in the text but also assumes the ring is left artinian

is semisimple + 1 = artinian semisimple?

does problem iiic work with the text's definition or the more usual definition? ( that is R is semisimple iff R is left artinian with zero jac )

[white oxide]

what's the best way to find nontrivial orthogonal idempotents in C[G]? i wrote (z_1 + z_2g)^2 = z_1 + z_2g and compared coefficients, and got 2z_1z_2 + z_2^2 = z_2, z_1^2 = z_1. so I chose z_1 = 1, which implies that 2z_2 + z_2^2 = z_2 or that z_2^2 = -z_2, so I choose z_2 = -1. hence z_1 + z_2g = 1 - g, but (1 - g)^2 \neq (1 - g) lmfao

[night onyx]

I think 1/2 (1 + g) and 1/2 (1 - g) work

[white oxide]

wait i'm trying to see why 1 - g doesn't work, i must have done some arithmetic wrong

[white oxide]

oh ok

so g^2 = 1 i believe

[night onyx]

yeah, in a 2 element group g^1 = 1

[white oxide]

yea that seems right, for 1 - g (1-g)^2 = 2(1 - g) which is not equal to (1 - g) :(

[night onyx]

lol g^2

not g^1

[white oxide]

ye

[drowsy lantern]

Does anyone know what the actual answer is

[crystal turtle]

!help

[flat tree]

Please read #❓how-to-get-help

[crystal turtle]

this is the wrong channel for this

[drowsy lantern]

Idk how to get right channel

Can u explain

Please

[white oxide]

lmfao i wrote (z_2g)^2 = z_2^2g

im trippin

[white oxide]

i'm special, i thought (1/2 + 1/2g) and (1/2 - 1/2g) were the maximal ideals of C[G] but turns out they're both equal to C[G] lmfao rip.

now for the part that really matters

can i have a hint?

[topaz solar]

Do you know anything about having idempotents and splitting up ur ring?

[white oxide]

no i don't i'm pretty sure

this was like the second problem i've worked on that has involved idempotents

[topaz solar]

Why

[white oxide]

Why

[topaz solar]

Well I wonder why, of all things, it’s group rings you’re dealing with

But without like

Local rings, idempotent properties, etc

[white oxide]

yea idk we skipped that section in lang and now we're on modules lol

[topaz solar]

[white oxide]

it's a bit weird that we started with ring theory as well but alas

[topaz solar]

Well, here’s a small hint, what polynomial do idempotents satisfy

[white oxide]

as in they're a zero of it?

[topaz solar]

Yes

All idempotents

[white oxide]

x^2 - x and x - x^2

so i suppose i'll consider the principal ideals generated by (1 - g) and (g - 1) then

[topaz solar]

Well, what’s eR look like for an idempotent e

[white oxide]

ah this must be connected to the first third of this question then in someway

uh

wait i'm so dumb i didn't even consider the first two thirds of the question which i proved lol

[topaz solar]

Well, if you know it’s R1 x R2, that should give an idea of an ideal

[white oxide]

ya ok thanks for the hint, i'll keep that in mind

[topaz solar]

And to check it’s maximal, there’s that the quotient is a field

[white oxide]

yea yuh

[topaz solar]

What might your uhh R1 x R2 isomorphism look like

[steel light]

Would these be good to prove on my own or are their proofs really complex

[topaz solar]

Ez

[south patrol]

They should be fairly quick

In any case important

[steel light]

Yeah

[topaz solar]

As a hint, consider (1,0), (0,1), (1,1) = (1,0)+(0,1)

(Which makes it obviously necessary to have orthogonal idempotents)

[crystal turtle]

yes they are good exercises

[steel light]

Is the forwards direction of i) really stupid easy

Or am I missing something lol

[drowsy lantern]

Can someone help me

With my homework

[steel light]

If it’s an integral domain then we have no zero divisors besides 0, hence xy = 0 necessitates either x is 0 or y is 0

(Should I prove that {0} is an ideal or is it fine to skip that since it’s easy/obvious enough?)

I don’t think I’m allowed to call things easy or obvious though so perhaps I should

[crystal turtle]

It's really stupid easy yeah

that's definitionally true

[steel light]

What is

[crystal turtle]

the forward direction

of i)

[crystal turtle]

it's exactly this statement as you noted, is what I mean specifically

[topaz solar]

I can call em ez and it is

[crystal turtle]

[steel light]

Let x, y in R and suppose WLOG x neq 0. If {0} is a prime ideal then for all xy = 0 we have y = 0 by defn of {0} being a prime ideal. Done?

[crystal turtle]

x = 0 or y = 0

oh

[steel light]

Yes but I said x neq 0

[crystal turtle]

i'm blind

hehe

sowwy

[steel light]

Nah it’s all good I’d rather a mistake than be wrong

[crystal turtle]

yeah good then

[steel light]

Wow nice

[crystal turtle]

i) is easy, 2) is a bit more interesting (not hard, but this one is actually pretty useful)

[topaz solar]

1 is basically the definition

[steel light]

And just to make sure I understand right, the closet R/P is {r + p} for all r in R and p in P right

Closet?

Thanks autocorrect

Coset

Isn’t a coset defined for each fixed element in the set you quotient by

[south patrol]

No

It's the set of r + P where r is in R

[steel light]

Isn’t that this?

[crystal turtle]

{r + p | p \in P} is the coset of r

for an r in R

R/P is the quotient ring (not called a coset itself)

[steel light]

Ahhh that makes so much more sense

[crystal turtle]

which is the set of all r + P

[steel light]

Yes okay

[steel light]

It just inherits the ring operations from R?

[crystal turtle]

(identifying together ofc the sets r + P and s + P if they are equal as sets, i.e., if r and s are in the same coset)

[steel light]

Yeah that makes sense

I’ll do ii) later

[steel light]

No that doesn’t make sense

The elements are sets

[crystal turtle]

right

But they are all of the form r + P

[steel light]

Yes

[crystal turtle]

So the operation is inherited on the "r + " part

[white oxide]

well for the first part I showed R is isomorphic to Re1 x Re2, under r --> (re1, re2). so if that's the case then C[G] should be isomorphic to C[G]e1 x C[G]e2

[steel light]

So (r + s) + P?

[white oxide]

and then iguess i could consider the ideal embedded in the product

[topaz solar]

Well if you have that isomorphic, then yeah what kinda ideals does the product have

Since, ya know, iso

[crystal turtle]

Yeah that

And (r + P)(s + P) is then rs + P

[topaz solar]

Ah, in particular, does that splitting look like, say, ae1 + be2?

[crystal turtle]

Now, of course, there are a few things to check

one is that these are ring operations

[steel light]

Yes

[crystal turtle]

(meh okay kinda boring whatever it should be inherited from R)

Main thing to check is that these are well-defined

[steel light]

As in different representations of the same coset should sum to the same thing

[crystal turtle]

yup

[white oxide]

Wdym by splitting

[steel light]

Just regular equivalence class stuff

Okay

I don’t understand something about cosets

Nvm yes I do

[crystal turtle]

[steel light]

What is a

at the bottom

another ring element?

Oh yes duh

Ax

And principal ideals are ideals because if we take a product b*ax then we can write this ba*x and since rings are closed under multiplication ba is just another ring element and hence it’s a multiple of x and hence it’s in (x)

Well I’d still need to show additive subgroup

But is the logic for the absorption property right?

[crystal turtle]

absorption?

[long geyser]

yes

[crystal turtle]

oh yea

sorry I misread

hurr durr

[steel light]

Okay that’s easy enough

[crystal turtle]

having one of those days huh

[steel light]

So nZ are the principle ideals of Z

[long geyser]

no

they are the principal ideals

[steel light]

Oop

LOL

caught me lacking

[long geyser]

[topaz solar]

As in splits it up?

[steel light]

what

how can a) be true

[crystal turtle]

what is a)

[long geyser]

wow you just got filtered by a TFAE statement

[crystal turtle]

[long geyser]

the situation is rough

[steel light]

Oh wait

It’s because we’re going with all rings are unital for this

[open sluice]

there are no rings

[crystal turtle]

all rings are unital and commutative, so it does have a 1

[open sluice]

multiplication is always commutative and all elements have inverses

[crystal turtle]

you might not have any rings. I have many rings, like the great dragon I am

[white oxide]

I don’t follow, sorry

[steel light]

So there’s a multiplicative inverse for every nonzero element

[topaz solar]

I honestly don’t know what you were confused by here but

[steel light]

That’s really the only property separating rings and fields? Lol

Multiplicative inverses and no nonzero zero divisors?

[crystal turtle]

once you assume unit and commutative, sure!

[steel light]

Ah

[crystal turtle]

if something has an inverse, it cannot be a zero divisor

(prove it)

[topaz solar]

Does it literally put every element into re1 + se2 for some r, s?

There is not precise meaning to that split

[steel light]

Sure

In a bit

[topaz solar]

I’m saying it literally is breaking it up

[steel light]

If x has an inverse y then x cannot be a zero divisor because then we would have xy = 0, but that contradicts the fact that y is the multiplicative inverse

[topaz solar]

[crystal turtle]

why would we have xy = 0?

[topaz solar]

Bad

[steel light]

because x is a zero divisor

[crystal turtle]

nonono

There's a element b such that bx = 0

Not that all elements b satisfy bx = 0

It's existential

[steel light]

You could say I’m having an existential crisis right now

Let me review my definitions

[crystal turtle]

[steel light]

Let x be a zero divisor and suppose it has a multiplicative inverse. Then there exists a b such that bx = 0. Then, bxx^(-1) = b = 0. But we can’t have b = 0 by definition of x being a zero divisor

[topaz solar]

Yes

More like by definition of being an inverse

[steel light]

Huh?

[crystal turtle]

*Should add "there exists a nonzero b such that bx = 0"

Then that should be good

[steel light]

That’s implicit in the definition of being a zero divisor isn’t it :/

[topaz solar]

Yes

But say it

[steel light]

When can I expect for this to get hard? Exercises? I think someone mentioned after 2 or 3 chapters?

[steel light]

Alright

[topaz solar]

If you’re a Chad, never

But yeah

[steel light]

I am not a chad

[long geyser]

which book is this

[steel light]

AM

[long geyser]

hungerford?

oh

wtf

you really are reading a CA book

[steel light]

Yes I really am

[long geyser]

absolutely crazy

[steel light]

Maybe

[crystal turtle]

like i said, chapters 1 and 2 are baby, chapter 3 introduces a very important concept (rings/modules of fractions), and then after that it does get hard

[steel light]

Alright

Fractions?

[topaz solar]

Yes

[crystal turtle]

indeed

[long geyser]

is this related to the field of fractions? idk ca either

[crystal turtle]

generalization thereof

For the field of fractions of an integral domain, you essentially "formally invert" all nonzero elements. This needs it to be an integral domain so that this set is multiplicatively closed. We can generalize this to inverting any multiplicatively closed subset of a ring R (even with the possibility of it including 0, though this is much less interesting lol)

And many properties of interest are invariant under this process

[steel light]

Wait does AM claim that “ring” means “unital ring”? I don’t think so right? How do we know that x is a unit here?

[crystal turtle]

A particularly important case is inverting the complement of a prime ideal (exercise for feather: show than an ideal P is prime, if and only if A - P is multiplicatively closed: that is, if x and y are in A - P, then so is xy). There's a correspondence between the ideals of this ring of fractions (let's call it A_p), and the ideals of A that are disjoint from this mult. closed subset. Accordingly, since we have taken the complement of an ideal, there is a unique maximal ideal after passing to A_p, namely, the image of p itself. This is called "localization", which you've maybe heard before and might motivate why this is important

oop

that's a rant and a half already so I'll cut it there

[long geyser]

oo

[long geyser]

so are they "locally fields" or something?

[crystal turtle]

A ring with a unique maximal ideal is called a "local ring". A lot of things are simpler in local rings (no examples off the top of my head oop)

But an important idea is the concept of a "local" property, which is one that holds in A iff it holds in all localizations A_p. So it's useful because it sometimes allows you to consider only "nice" cases (those for local rings), and then use localization to generalize the result to all rings (satisfying whatever appropriate conditions)

[long geyser]

damn

[crystal turtle]

You are assuming 1), which is that A is a field

[long geyser]

does topology enter into the picture at all here?

[steel light]

Ohhh right they’re proving i implies ii

[crystal turtle]

uhh I am actually just getting to the chapter which starts to talk a bit about topology; chapter 10. Something about -adic completions? like p-adics

[agile burrow]

The first chapter also begins by saying they only consider commutative rings with identity

[crystal turtle]

feather 💀

[steel light]

Is this inspired by the term localization :p

[steel light]

What

[long geyser]

cool

[crystal turtle]

you gotta assume the assumptions my friend

[steel light]

I didn’t pay attention to the implication lolol

[open sluice]

it's an equivalence proof
1 is true iff 2 is true iff 3 is true
1 is false iff 2 is false iff 3 is false

[steel light]

Yep

[steel light]

Oh

I think

[topaz solar]

I-adics

[crystal turtle]

hmm I ought to rant more like Sharp does it

[steel light]

It’s related to the spectrum of a ring

[crystal turtle]

smaller messages

[steel light]

Which is related to the Zariski topology iirc

[crystal turtle]

than beeg paragraph

[steel light]

Yeah I remembered this

[crystal turtle]

Zariski topology is given on the spectrum of the ring yeah

[steel light]

I meant the concept of localization

[crystal turtle]

AM actually has a lot of exercises on Zariski topology btw

[steel light]

I didn’t read what you wrote fully since I’m kinda doing three things at once rn lmao

So maybe I’m totally off

Probably par for course for feather

[crystal turtle]

me too

[steel light]

What’s a good algebra reference text?

Quick and dirty “skip to the section I need to review” type book, efficient but clear

[crystal turtle]

I don't actually know. probably not D&F

[topaz solar]

Lang?

[steel light]

Doesn’t everyone hate Lang 💀

[topaz solar]

It’s a lot of “here’s a result”-y

[crystal turtle]

i think we hate lang as a person maybe

[steel light]

LMAO

[crystal turtle]

but I'll probably try reading lang next semester in my rings/modules course

[steel light]

Who is Serre or Sierre or however it’s spelt? Borcherds was sucking his dick (figuratively)