groups-rings-fields

[next obsidian]

You then use the fact that these have expressions as those power series using Taylor’s theorem

[coral shale]

bro this is cursed. isnt this the algebra channel

whys there no way without calc

[next obsidian]

So on the domain where the power series converge you have the identity

[formal ermine]

the thing I'm worried about is that this is in Q_p[[X]], so I wanna know that if I assume that it's true in Q[[X]] then it's also true in Q_p[[X]], surely right?

[next obsidian]

That includes the part of Q where the power series converges

[crystal turtle]

I'm gonna be honest I'm not even sure how to formally justify all this I just take it for granted

[next obsidian]

It should hold as long as the expression you’re trying to show is equal to whatever is continuous in Q_p

[formal ermine]

wat

[topaz solar]

Don’t conflate the function and the power series, but they’re very closely related with analytic functions where they converge

[formal ermine]

I just wanna know

if the formal power series equation

is true

by my reasoning

arithmetic doesn't change for Q in Q_p

[topaz solar]

I’m pretty sure he said it’s true because density multiple times

[next obsidian]

The argument I’m saying does not work for formal power series

[topaz solar]

[next obsidian]

I mean as far as I know

[formal ermine]

but that's for the actual functions

no?

[next obsidian]

Perhaps there’s an identity theorem thing

[topaz solar]

Because you go back and forth between power series and analytic functions w/ Taylor’s

[delicate orchid]

actual functions are just evaluation maps from R[[x]]

[formal ermine]

let f(X), g(X), h(X) in Q[[X]]. assume that f(g(X)) = h(X) (in Q[[X]]). does it follow that if we embed f, g, and h into Q_p[[X]], we still have f(g(X)) = h(X)?

[next obsidian]

Oh yes

Lmao

I mean the coefficients are in Q

[formal ermine]

Yeah because arithmetic doesn't change for Q in Q_p right?

Yeah

that's what I've been wanting to know this entire time

[next obsidian]

The arithmetic of Q inside Q and Q inside Q_p are the same

[formal ermine]

Yeah

in germany we would say complicated birth

schwere geburt

[delicate orchid]

embeddings are homomorphisms no way

[topaz solar]

The only thing that would maybe change is where it converges

[next obsidian]

I feel like this doesn’t really change for reason

Which I don’t know

[crystal turtle]

nor do i care to actually know

[next obsidian]

Actually I do believe it changes

Possibly

Hmmm

No

[topaz solar]

I mean it’ll definitely change somewhat in the sense that it presumably works for a non Q p-adic

[next obsidian]

Hmmm

Idk

Oh sure but like the radius of convergence in Q is given by some < r

[topaz solar]

Ah yeah that idk

[next obsidian]

If you just take the same ball in Q_p for the p-adic metric it should converge in that ball for reason or something

And I don’t think it enlarges because then you get a too big Q-number it converges for or something

Idk

I’m not good enough at this shit

[topaz solar]

Can’t intuit it

[next obsidian]

I WANT MY RINGS OF CONVERGENT POWER SERIES FROM RIGID ANALYTIC GEOMETRY

TATE ALGEBRAS

[formal ermine]

fuck tate algebras

why does straßmann only work in those

[crystal turtle]

Please just let me intuit it I promise to only prove reasonable stuff

[topaz solar]

Cauchy proving convergent sequences of continuous function are continuous

[next obsidian]

This is true

🙈

[rocky cloak]

With the right topology everything is true

[crystal turtle]

jagr spitting

[white oxide]

sorry to post this problem again lol, just returned to it, why doesn't x -> g work?

where g is not the identity

[crystal turtle]

thats what I thought at first too

What's the kernal of that map?

[rocky cloak]

g is a unit

[topaz solar]

More directly, what does x^2 send to

[white oxide]

e?

[topaz solar]

What jagr said is big but I mean just directly looking at where it sends x^2 should give a good idea

[topaz solar]

Is that 0

[white oxide]

yeah?

[crystal turtle]

e = 1 (multiplicative identity of the ring)

[white oxide]

wait i'm lost

i thought G was a group

[topaz solar]

That’s what you’re failing to see here

G is a group yes

The map is to k[G] tho

[crystal turtle]

1e + 0g is the multiplicative identity, if it helps to see it this way

[white oxide]

o

[crystal turtle]

i think

[void cosmos]

if R is left artinian then the jacobson radical is a nilpotent ideal

let J > J^2 > J^3 > be a stricltly descending chain

there exists k such that J^n = J^k for all n>=k

[topaz solar]

As I said, think as polynomials

[void cosmos]

oh this is taken fuck

[topaz solar]

x^0 is not 0

[rocky cloak]

We can interleave

[white oxide]

nah go for it bro

[void cosmos]

yea you can cuz ur literally a genius lmfao

okay

[void cosmos]

now

[rocky cloak]

Embrace the chaos

[void cosmos]

if i can show that this J^k is 0 then im done

so umm

[rocky cloak]

Hint: ||nakayama||

[void cosmos]

okay wait

im new to this stuff

J is the intersection of all maixmal ideals

hence contained in every maximal idela

so our hypothesis is there reight?

right*

[rocky cloak]

Which hypothesis are you thinking of?

[void cosmos]

that the ideal is contained in every maximal ideal

[rocky cloak]

Then yes, the radical is the intersection of all maximal ideals

[void cosmos]

if A is an ideal contained in every maximal ideal , then 1-a is a unit for all a then whenever JA=J for f.g j then J= 0

lmfao wait

J^k * J^k = J^k tho right

cuz 2k > k

?

[next obsidian]

Dawg this is descending

[void cosmos]

yea mb

[rocky cloak]

Indeed

[void cosmos]

i did not attend this grade school class

okay but why would J^k be contained in every maximal ideal tho

J^k being the product of J(R)

like

wait

[rocky cloak]

Because J is the intersection of all maximal ideals

And the intersection of stuff is contained in stuff

[void cosmos]

yea obv

lmfao

mb

i need to show that J^k is finitely generated right?

is it tho

[rocky cloak]

Yeah, guess you need to show that

[proud spindle]

i've got this seemingly obvious problem/proof in a book i've been reading through and it's got me stumped

[void cosmos]

yea it is cuz J^k is just generated by the products of elements

of J up till k

ig

lmfao

[proud spindle]

let G and H both be abelian groups; prove G x H is also abelian

[void cosmos]

thats just how ideal multiplication works

[rocky cloak]

Yeah, so if you already know J is finitely generated, that works

[rocky cloak]

Just write the product of two elements in both orders and compare I guess

[void cosmos]

i cant tho

like

J may not be finitely generated

pick any like radical ring ig

[rocky cloak]

It's related to an assumption you made about R

A certain "finiteness" assumption

[void cosmos]

yea idk tbh

ik its some kind of

minimiality argument

[rocky cloak]

So when R is artinian it's also Noetherian, hence every ideal is finitely generated

But maybe you haven't proven that yet, so maybe you prefer going another route

[void cosmos]

??

wait

i didnt know that

thats so op?

[proud spindle]

that makes a lot of sense

[delicate orchid]

the proof is fairly easy at least

[rocky cloak]

Is it? I feel like there are some complicated steps, or maybe you have a different proof in mind than me

[void cosmos]

wait

this is over a commutative ring tho right?

[rocky cloak]

True for noncommutative as well, but sure

[void cosmos]

wow

okay so im done i guess

hopefully the grader doesnt like remove all points from the problem ( if comes on the exam )

[rocky cloak]

An alternate approach, without going through noetherianess could be to consider the annihilator of J^k

[topaz solar]

Well you might wanna prove Artinian -> noetherian though

[white oxide]

okay I'm just gonna clear it up because I'm actually stupid; the additive identity of k[G] if k is a field and G is a group is the additive identity 0 of k. the multiplicative identity 1 of k[G] is the multiplicative identity 1 of k.

[south patrol]

sure, i mean you can phrase this as saying k embeds in k[G] as a subring

[dim widget]

k*Id is the image of k in k[G]

[crystal turtle]

specifically, with coefficients of the non-identity coefficients all 0 (i.e., k |--> k*e + 0*g)

[lime warren]

F × F → F can anyone explain this notation? the arrow makes sense but I'm not sure what the cross means

[crystal turtle]

cartesian product

think of pairs (f_1, f_2), where f_1 and f_2 are elements of F

[white oxide]

yuh

OHHHH it's just polynomials in G with coefficients in k

@topaz solar

😹

[topaz solar]

[crystal turtle]

we been saying this

[topaz solar]

Why do I even try sometimes

[lime warren]

perfect explanation, thanks my guy

[white oxide]

cuz you're a g

[crystal turtle]

anyways

x |--> g doesn't work (||it's kernal looks something like x^2 - 1, since x^2 would be sent to 1||). How do you think you can remedy this? (||char 2, btw||)

[south patrol]

i like how they said characteristic 2 "> 0"

just to remind you that 2 > 0

tf lol

[crystal turtle]

i forgot about that for a hot second eariler, even tho I pointed that out as well

[south patrol]

lol perhaps my funny tip is just to consider k = F_2

there you have very few options for a surjective map k[x] -> k[G]

[white oxide]

ok i'll try

[white oxide]

ohhhh now this makes sense, x^2 gets mapped to e and we can consider e to just be 1 times e which is just symbollicaly 1?

sorry for being slow, this idea of considering a polynomial ring over some abstract indeterminate (or whatever the phrase is) is kinda confusing to me

so in this case x^0 in say Q[x] would be analagous to e

[crystal turtle]

you're fine lmao, I also had a few misconceptions when I first read it

but yeah, x^2 gets mapped to e (or 1). So the kernal would almost look like x^2 (from which we would be able to apply first iso), but not quite

Instead, it looks like x^2 - 1 = (x+1)(x-1). But we are in char 2! What does this mean for us then

[white oxide]

so the kernel would be x^2 - 1

ohhh okay wait hold up

[crystal turtle]

holding

[white oxide]

sorry you're gonna have to hold for a long time

x^2 - 1 = (x + 1)^2?

[crystal turtle]

So let's change the map accordingly. Try mapping x to 1 + 1g? (i think this works and will have kernal <x^2>: try to double check me!)

[white oxide]

wait sorry for the moment i'm trying to see how you got to mapping x to 1 + 1g

sorry i'm going to think out loud here

nvm i'm not

[crystal turtle]

good question let me think of a good justification for why other than "it sounds right?"

[white oxide]

ohh

[crystal turtle]

||I mean I guess it's kinda going back to that whole "treat it like polynomials in G" thing. We're off by a factor of + 1, so we can try to correct by factoring x^2 - 1 = (x+1)^2 (which is sent to 0). And then treating x and g as if they're the same, I'd like x^2 to map to (g + 1)^2 = 0, so there's a natural choice for this (namely, x |--> 1 + 1g)||

[white oxide]

okay sorry just for the moment i'm going to ignore the paragraph you just posted, just cuz i wanna figure it out on my own lol, but i'll definitely look at it if i'm stuck; so x maps to g implies that x^2 maps to 1, sure. but we want x^2 maps to 0, which is equivalent to assigning x to a nonzero value of k[G] such that x^2 = 0. i can use the fact that k is of characteristic 2 to produce a nonzero element of k[G] such that when squared gives 0.

this is my thinking process so far

but (g + 1)^2 is nonzero

[crystal turtle]

lemme just hide it then 🙃

[white oxide]

thank uuuu lol

[delicate orchid]

absolutely cracked that you're considering group algebras over positive char before being comfortable with them

[crystal turtle]

wew coming in to save the day

[white oxide]

OH WAIT

[delicate orchid]

"save"

[white oxide]

but 1 is -1 in char 2

😹

wait so g^2 + 2g + 1 = g^2 - 1 = 1e - 1 = 0

[crystal turtle]

uhh I hope you already had that realization when you sent this...

[white oxide]

o

[crystal turtle]

But yes

that's why this should work lol

[delicate orchid]

g^2 = 1 => g^2+1 = 0 => (g+1)^2 = 0 => k[C_2] \cong K[x]/((x+1)^2) \cong K[x]/(x^2)

swag

[proud spindle]

apparently conjugacy is an automorphism????

[crystal turtle]

yurr

[delicate orchid]

a very nice class of automorphisms

[crystal turtle]

special kinds

[white oxide]

sweet wew is a math lord

[crystal turtle]

called inner automorphisms

[white oxide]

next i have to show the existence of a unique maximal ideal!

whoo!

but thank you so much @crystal turtle

[delicate orchid]

huh yeah I guess that is true

only one simple rep after all

[white oxide]

classic wew

got it instantly

[delicate orchid]

I hope you're fully aware that I am a moron

[white oxide]

if you're a moron then I'm unconscious

[delicate orchid]

I have more machinery in my mind to work with it's not a fair comparison

[crystal turtle]

wew are you phd

[delicate orchid]

yur, just going into my 2nd year

[crystal turtle]

swag

[karmic moat]

i know a guy at my school with a british accent is it you wew

[delicate orchid]

yeah

I'm the essophysical embodiment of the phrase "innit bruv"

[proud spindle]

yeah i'm reading about the inner automorphisms, and it's being explained with D_6 (which does make sense considering it's talking about permutations and dihedral groups are pretty permutation-esque) but it pretty much gives a question related to D_6 and then says that conjugacy builds inner automorphisms for every group (as far as i can tell)

[topaz solar]

Same

[crystal turtle]

wait didn't you already work that out a few days ago?

I thought we'd been over this about conjugation before

[proud spindle]

oh yeahhhhh

[karmic moat]

now i need to buy dreaming discs vinyl

[proud spindle]

i have horrible memory

[delicate orchid]

dihedral groups are pretty permutation-esque
all groups are lol

[white oxide]

oh god wait showing that this is a homomorphism involves binomial expansion stuff doesn't it....

[karmic moat]

hahaha bahahahaha hahaha

[white oxide]

UGH

[delicate orchid]

can you not take that (x+y)^p = x^p+y^p in char p lol

it seems like you're passed the point where you'd need to show that

[white oxide]

o

[crystal turtle]

i mean this case it directly and easily calculable lol it's 2

[white oxide]

you're so right

[karmic moat]

shhhhh

shhhhhh

[crystal turtle]

also this for char p

[delicate orchid]

it's actually true over Z if you don't mind being sullied

[white oxide]

wait wdym

[coral spindle]

Just expand the brackets and see

[delicate orchid]

well I mean you directly computed it lol

[white oxide]

i'm trippin

[crystal turtle]

also something something binomial theorem generalizes something

[crystal turtle]

me too

[karmic moat]

binomial thm generalizes monomial thm

[summer path]

multinomial theorem generalizes binomial theorem

[karmic moat]

this is easily seen by taking b=0 for (a+b)^p

[crystal turtle]

:o

[proud spindle]

this is definitely a very obvious question but if conjugacy is an automorphism then does that mean an element of a group G that can be described as a conjugation is, in a sense, related to an automorphism of G

[crystal turtle]

what do you mean by "described as a conjugation"?

[delicate orchid]

there is a homomorphism sending any element g to it's conjugation map h |-> ghg^{-1}

[south patrol]

That is slightly vague but there is a canonical map G -> Aug(G)

[coral spindle]

This is not an obvious but rather a very vague question

[proud spindle]

it can be described as $aba^{-1}$ for $a, b \in G$

[south patrol]

That's every element nonpost

[delicate orchid]

well

[coral spindle]

Every element can be described in that form

dammit potato >:(

[crystal turtle]

this is either a dumb question or young blud's about to figure out Cayley on his own

[delicate orchid]

yeah, set a to be the identity lol

[cloud walrus]

nonpost

[karmic moat]

not me

[delicate orchid]

cat king cat king

[coral spindle]

And we typically call that "written in the form..." rather than "described".

[proud spindle]

i feel like i should've thought of that

[topaz solar]

Is it definable though

[coral spindle]

Now if you want to have something nontrivial, it's interesting to note that typically not all elements can be written as a commutator. A commutator is something of the form $[a, b] = aba^{-1}b^{-1}$

[cloud walrus]

Schur thing boss

[coral spindle]

[a,b] is notation defined here.

Exercise, if you wish: find all elements of dihedral groups expressible as commutators.

[delicate orchid]

slight_smile

[crystal turtle]

[coral spindle]

Not a hard exercise

[delicate orchid]

a tedious one

[crystal turtle]

true that's not bad for D_n lol

[coral spindle]

Not tedious either, surely you've done this yourself lol

[delicate orchid]

I've done it just now in my head

[coral spindle]

[delicate orchid]

it was tedious

[proud spindle]

ngl i can't do one tedious thing in math for >18 minutes without making a significant error and not noticing

[crystal turtle]

me

[coral spindle]

The subgroup generated by commutators is a very important one called the derived subgroup.

[wraith cargo]

every exercise is hard to someone
We try to be inclusive on this server please
Except to set theorists
I will never forgive them for the continuum hypothesis

[coral spindle]

Dude I set the exercise

Anyhow

[proud spindle]

every time i "prove" something incorrectly (i did not prove it) it is always wrong at the end bc by the time i get there i've lost all train of thought

[coral spindle]

If you'd like to know more about the derived subgroup and its importance, the relevant things to look at are the Abelianisation of a group and the concept of a solvable group

[wraith cargo]

that is why I reread my proofs 30 times a minute

[south patrol]

i just don't prove stuff

[wraith cargo]

I didn't know you did differential geometry

[proud spindle]

i like how i forgot the identity exists while doing group theory

[crystal turtle]

bruh

that's a pretty useful one

[delicate orchid]

boss ur going off on a tangent

[crystal turtle]

moreover

that's the only one you know exists generally

trivial group go brrrr

[delicate orchid]

now let me talk about the F-abelianisation

[coral spindle]

Bro I'm telling the dude about a thing

[south patrol]

i thought diff geo is fine w proofs lol

[delicate orchid]

fairs...

[crystal turtle]

(the dude is also 15 for reference)

[coral spindle]

Yeah and? I was about 15 when I started group theory too

Am I supposed to not tell people about relevant things?

[delicate orchid]

I was 18

[wraith cargo]

sir when you start group theory the first thing you talk abt isn't the intrinsic importance of the derived subgroup

[coral spindle]

And?

[crystal turtle]

fair, but for someone who is young and just learning groups, talking about abelianizations and derived subgroups will go over their head?

Like I get your point

But?

[coral spindle]

yeah that's why I'm saying it's important bruh

This is so silly

[delicate orchid]

it's really quite simple, it's the focal subgroup associated to the inner fusion system.

[wraith cargo]

but to a beginner it's not???

[coral spindle]

It is

[wraith cargo]

it's really not lol

[crystal turtle]

It really isn't

[coral spindle]

🙄

[delicate orchid]

how else will they apprechate the beauty of pontygain duality without it...

[proud spindle]

"can every element of a group be written in the form of a conjugation between two elements a, and b? sure would be nice if there was an element, idk 'e' that was its own inverse and had the property ea = ae = a: for all a in g, and was in every group"

[crystal turtle]

Like someone is first learning basic real anal, probably not a good idea to be like "oh this actually is a statement about connected locally compact metric spaces" like??

[wraith cargo]

Gelfand duality for life

[coral spindle]

Stop pretending what I'm saying is stupidly complicated.

[crystal turtle]

It's true and no doubt important. Just not relevant at their level

[south patrol]

ye it's not that complicated but there are probs more important things to learn first

[coral spindle]

Anyway, I just pointed out that although what was written was every element, there's a variation that is interesting 🙄

Thanks for saying that I should simply not mention that

[topaz solar]

Bro barely comprehends conjugation idt he knows normal subgroups

[crystal turtle]

I didn't mean for it to come off as rude btw

Sorry that it did

[coral spindle]

I didn't mention normal subgroups

[delicate orchid]

everybody who has a problem with this can go write a 40000 word essay on their myspace maths pedagogy blog

[topaz solar]

Ok now time to talk about ultragroups

[crystal turtle]

yes sometimes it helps to think of trivial cases

[proud spindle]

abelianization, i heard about once; derived subgroups, never heard of them; the real question is: why and how would a beginner like me need to know what something called a "derived subgroup" is

that name is inherently daunting and complex

[coral spindle]

Because you will soon see them, and I thought it would be nice to tell you what to look forward to.

I cannot believe people are acting like mentioning a name of a concept is like summoning a devil

[proud spindle]

i don't really care that much ngl

[coral spindle]

Noted.

[proud spindle]

i started learning group theory off wikipedia so tuning out jargon has become second nature to me

[wraith cargo]

boytije idk why you have a hard on for abelianization but that isn't really something you'll see that much until maybe like you start talking abt solvability of groups and stuff

[coral spindle]

I don't "have a hard on" for it

[delicate orchid]

can everybody chillax

[wraith cargo]

I need to win this online argument to show that I am the "alpha" of the group
My manhood depends on it

[delicate orchid]

I am about to rant about capitalism

[coral spindle]

It's one of the first examples you can use quotient groups for. In Fraleigh, the book I learned from when I was a beginner, it was given as a relatively early thing that he mentioned as a teaser. Oh no Fraleigh must also have a hard on for the Abelianisation 😱

[coral shale]

no, modular arithmetic.

is the standard one

[delicate orchid]

you don't motivate quotients

[coral shale]

true

[delicate orchid]

the winds of time will carry the grains of truth to you

[coral spindle]

Lord protect everyone from a windvane

[proud spindle]

yeah can this not be a thing, i do not care about hearing unexplained jargon as i am used to hearing much more of it than i have here so i don't see a good reason for anyone else to care that much

[delicate orchid]

if you got any more questions about conjugation or anything I'll happily answer them tho

[wraith cargo]

if you don't have a negative reflex towards a lot of unexplained jargon the first paper you write will look like IUTT

Hodge theater my ass

[proud spindle]

i know you said all groups are permutation-esque; what exactly makes groups in general like permutations

[coral spindle]

It's Cayley's theorem

Every group is isomorphic to a group of permutations – that's the theorem.

[proud spindle]

i used to know that then i forgot it

[coral spindle]

Well, here's an opportunity for you to try and rederive it yourself!

The key idea is that if you look at how an element x affects other elements, you can figure out what element x was all along

[proud spindle]

i think the way i first heard it described was something along of the lines of: every group of order n is isomorphic to a subgroup of the symmetric group S_n but i forgot the exact definition i was given so don't quote me on if that statement is true or not

[coral spindle]

This is the same statement

[proud spindle]

ohhh

[delicate orchid]

bro is ego searching ⚠️

[proud spindle]

and so that actually explains how conjugacy makes automorphisms

[coral shale]

this again

[dusky burrow]

hi

im new

[coral shale]

will be memed for decades to come

[crystal turtle]

bro is still on this

[proud spindle]

i'm just saying that it circles back to where it started

[coral spindle]

Unfortunately Cayley's theorem has very little to do with conjugacy. The fact that conjugation is an automorphism one just has to see by inspection.

[crystal turtle]

not you

[delicate orchid]

exercise: prove it's an automorphism

[crystal turtle]

this

[proud spindle]

what i'm saying is that cayley's theorem makes "conjugacy is an automorphism" make a lot more sense to me

[coral spindle]

OK

[delicate orchid]

but it shouldn't? they're not really related

[proud spindle]

no big connection as far as i can tell

[delicate orchid]

or at least a homomorphism

[proud spindle]

it's hard to explain but it just clicks

[crystal turtle]

They're related in that they're about automorphisms maybe?

[proud spindle]

yeah that's it

[coral spindle]

Can you elaborate on how Cayley's theorem is about automorphisms

[delicate orchid]

^

[crystal turtle]

Iirc that's how one proves it: you take the symmetric group on G, and map g \in G to it's action by left multiplication (roughly speaking)

[delicate orchid]

left mulitplication is not an automorphism

you can't take the permutation arising from conjugation as your map for cayely's because it must be injective

[crystal turtle]

Also, symmetries of a set ~~ automorphism of the set

[coral spindle]

It doesn't even preserve the identity

OK.

[crystal turtle]

This is moreso what I had in mind I think

the multiplication structure being a set automorphism, sorry

[white oxide]

can somebody give me a hint to see how the kernel is contained in (x^2)? i wrote phi(f(x)) = 0 and expanded f(x) but i've gotten nowhere lol

and i don't think contradiction is the way to go either

[proud spindle]

i vaguely know what an automorphism group is and i'll bet a whopping $0 that if bijections on sets are considered automorphisms of sets then you could probably say the symmetric group is an automorphism group on a set

[crystal turtle]

Again, I don't disagree that this is a stretch for how they are related

[coral spindle]

The map is phi(x) = g+1 right?

[crystal turtle]

I was just trying to perhaps voice what nonpost was maybe thinking

[coral spindle]

Just checking

[white oxide]

ya

[proud spindle]

my brain will think two things are somewhat related if it looks like there's a decent relation between the two if you squint hard enough

[crystal turtle]

well there may be connections, but you gotta have a reason for the connections lol

like ofc once you learn rings, you'll notice that there are some similar (basic) statements. That's a legitimate connection (say, universal algebra)

[proud spindle]

i don't do this for everything i see, and it's not like i'd ever use "they look kinda related" in math

[coral spindle]

Well, every power above x^2 gets annihilated, and they indeed are in the ideal (x^2). So you only need to verify that linear polynomials aren't killed by this map

[topaz solar]

I use “hm this looks like XYZ” but I don’t try and blindly connect em, I try to see if there is indeed a connection there first

Yeah something “looks like” something else, doesn’t mean like anything about them is related

[delicate orchid]

alternatively could we use the corrispondence theorem some how?

[proud spindle]

me seeing two things and finding a probably indescribable relationship between the two is just a thing i do bc of my autism; it's just second nature and i do it because it's fun

[white oxide]

yea this was what i thought, i tried to assume for contradiction f(x) not in (x^2) which would imply that f(x) is not equal to g(x)x^2 for any g(x) in k[x], and i thought that would imply that f(x) is of degree 1 or less, which would lead to the contradiction. i thought that was wrong

[coral spindle]

What precise part of that did you think was wrong

[delicate orchid]

this would get us that (x) is the only non-trivial proper ideal of k[x] that contains (x^2) so the only proper non-trivial ideal of k[x]/(x^2) is (x)? am I smoking the zaza here

[coral spindle]

Oh I thought okey just wanted to verify that the kernel is as it claims

[delicate orchid]

we did that an hour ago

they did that an hour ago

[white oxide]

idk, my brain went "oh x^2 not being a factor of f(x) doesn't imply that f(x) is of degree 1 or less". like i thought of f(x) = 1 + x + x^2 + x3 or some shit, but i forgot that we're in a field of char 2

i guess i should try to also prove that formally lol

[coral spindle]

So here's an easier way of thinking about this

You have some polynomial, let's assume it isn't killed

we can ignore all the parts that are killed, since we may ignore them anyway because they will be sent to zero – so we are left with a polynomial of degree at most 1

[white oxide]

ohhhh that makes a lot more sense i think

[delicate orchid]

n >= 2, x^n = x^2x^{n-2} = 0x^{n-2} = 0

[coral spindle]

So this shows the kernel is contained within (x^2) after a little more thought

[white oxide]

wait i'm confused though, 1 + x + x^2 + x^3 is not sent to zero and it doesn't have x^2 as a factor, and is of degree > 1

[delicate orchid]

it's sent to 1+x

[white oxide]

my brain is not working properly i think

[delicate orchid]

at absolutely no point in human history have we ever divided by x^2

that's not what a quotient is

[coral spindle]

1 + x + x^2 + x^3 is sent to the same thing as 1+x is sent to. Cool w that?

So you can focus attention just on 1+x instead of a yuge polynomial

[white oxide]

yes

[coral spindle]

Now as it happens you can quite easily just look at all the degree <= 1 polynomials and see where they land to confirm that they don't end up getting sent to zero

So if it ain't in (x^2), it ain't in the kernel

There are only three such things of course

[white oxide]

huh ok. i think i've spent too much time on this problem and will return to it at some point, but for now i'll assert that and try to prove it later

my brain is like

literally non functional now

[coral spindle]

Schleep time

[white oxide]

but thank you boy

(tjie)

[coral spindle]

uwu

[proud spindle]

what i'm saying is that the pseudo-connection i made between conjugacy and Cayley's theorem is indescribable bc the "connection" is rather abstract due it being a quirk of my neurodivergent mind's slightly muddy, occasionally abstract idea of how two things can be considered "related"

[coral spindle]

You spelled my name right... this is a rare occasion

[white oxide]

do people normally say boytijie

[coral spindle]

I've heard "bowtie" and "boytoy" and all sorts

[proud spindle]

me proving something for >20 minutes

[proud spindle]

it's best to keep the discussion at this

[crystal turtle]

(it gets much worse than that btw)

[topaz solar]

I spent a few days on my lil dumb project just to prove

“Yep if this covering hits everything infinitely often it’s a transcendental”

[delicate orchid]

me spending a week to show one result

[crystal turtle]

me not noticing a mistake in a proof for a month

[delicate orchid]

only to discover that along the way I found a much better characterisation of the property I wanted to study

[crystal turtle]

[coral spindle]

prove a result about thing, write it down
work on different thing
get stuck
go back and look at first result
itisveryuseful.jpg
apply result and solve problem
derive false fact
look back on first result
spot mistake

^ 'bout two weeks

[delicate orchid]

thankfully none of my results have actual applications

[topaz solar]

*this didn’t even show it had everything transcendental, just that it had one

[crystal turtle]

mistake 8 pages into 35

[karmic moat]

yeah i’ll spend a good 10 minutes on a book exercise

if im feeling confident, 15

then to stack exchange i go

[delicate orchid]

see I absolutely cannot do book exercises

I lose motivation so quickly

[karmic moat]

me too

i see the next chapter

and just wanna get to the next chapter

[delicate orchid]

but if I think about a question myself (especially a combinatorial one) I will hyperfixate on it

[karmic moat]

so real

[delicate orchid]

which is a useful thing to have when doing research - undergrad not so much

[karmic moat]

spent all day the other day trying to remember why the cantor set was uncountable

[delicate orchid]

cause it's in bijection with R

[karmic moat]

only for the proof to be some base 2 decimal thingy that i wasnt expecting at all

[delicate orchid]

I would have thought it would be base 3

meh

[karmic moat]

oh yeah something like that

[delicate orchid]

ok thinking about this now

[karmic moat]

well it was like you get some decimal and u do enough subtraction so its only 0’s and 2’s then just change all the 2’s to 1’s

[delicate orchid]

you keep removing the middle third so what does that do to the base 3 expansion? it means you can never have a 1 right?

[karmic moat]

i think yeah u start w like .02333 or smth like that i forgot

yea

you’ll get 2’s and 3’s and 0’s i think

[delicate orchid]

yeah so you get a thing with a bunch of 2s and 0s

which is basically binary

so there's a bijection to [0,1]

[karmic moat]

man im reclined in my office chair with a beer in one hand do not make me get up and read the proof from my notes

yeah

[delicate orchid]

cool!

[karmic moat]

mhm yea

[proud spindle]

it gets worse?

[crystal turtle]

some proofs take hours

some take days or weeks, even months

big boy proofs take years even

(of course, broken into many parts for those, but still)

[topaz solar]

I’ve spent a few days trying to prove 1 group nontrivial

Or, find the condition where it is

[crystal turtle]

spent many hours trying to show that one function was just a rotation of another for complex analysis

[proud spindle]

honestly i like conjugacy (in the context of non-abelian groups)

speaking of conjugacy, the phrase "inner automorphism" implies the existence of an "outer automorphism"

[coral spindle]

An automorphism that isn't inner (i.e., isn't given by conjugation by some element of the group) is called outer.

[proud spindle]

makes sense

[coral spindle]

For an example, take any nontrivial automorphism of an Abelian group.

A famous fact is that S_6 is the only symmetric group with an outer automorphism.

[proud spindle]

just found out about soluble groups

[white oxide]

any tips on how to find a unique maximal ideal? right now i'm just blindly testing ideals and i feel like there's a better way to do this

like

[coral spindle]

As Wew said, the correspondence theorem. The ideals of k[x]/(x^2) are in bijection with ideals of k[x] containing (x^2). So what are those ideals.

Hint: there's only one nontrivial ideal, full stop.

[white oxide]

oh wait wtf i didn't even know this was a thing

[crystal turtle]

[white oxide]

is that something you learn in a second course in abstract algebra

[crystal turtle]

Same as with subgroups of quotient groups, if you know that one

[coral spindle]

This is something you learn in a first course in abstract algebra

[white oxide]

wtf

[coral spindle]

You should try to prove it.

[white oxide]

am i just misinterpreting what you're saying lol

[crystal turtle]

This is like, extremely important to know btw

for anything with quotients

[next obsidian]

Wat

This is like, covered at the same time you do iso theorems

[crystal turtle]

For groups it would be "for a normal subgroup N of G, there is a bijection between the subgroups of G/N and the subgroups of G containing N"

[next obsidian]

It’s like requires to make sense of third iso

[white oxide]

it's this right

not in fraleigh i'm pretty sure

[crystal turtle]

(+ some other stuff about preserving index, normality, ...)

[next obsidian]

Nah it 100% is

[crystal turtle]

how wtf

[next obsidian]

You must be smokin dope

[white oxide]

or maybe it was an implicit thing

[proud spindle]

that's interesting

[next obsidian]

I never read the book but

[crystal turtle]

(this correspondense also preserves primeness of ideals btw)

[next obsidian]

If it isn’t in it the book is FRAUDULENT

[next obsidian]

And normality of subgroups btw

[white oxide]

nvm i'm disabled

[crystal turtle]

[next obsidian]

[white oxide]

oh it was a lemma so i forgot that nobody theorem

[crystal turtle]

I already said that, copy🐱

[crystal turtle]

how is one of the isomorphism theorems just a lemma 💀

[next obsidian]

What the fuq

[white oxide]

ok but fr we didn't even get to the other isomorphism theorems in our aa, we didn't even get to the fundamental theorem of field theory i had to learn all the other shit

so i prolly just forgot this nobody lemma

i'm actually dumb, $(x^2, x) = {f(x)x^2 + g(x)x \mid f(x), g(x) \in \mathbb{F}[x]}$ right

[cloud walrus]

okeyokay

[crystal turtle]

One of these elements is in the ideal generated by the other

So you can simplify it to just (a) for one of them

[white oxide]

agh you're right

FUCKKK

wait

oh ya (x^2) is in (x)

silly me!

[crystal turtle]

So it's just (x)

[white oxide]

yea

i don't think this is maximal tho rip

[crystal turtle]

(reminding you of this!)

[white oxide]

o

right

[crystal turtle]

(you have a nontivial ideal, after passing through the correspondence)

[white oxide]

okay now i don't understand why there's only one nontrivial ideal but i'll take boytjie's word for it! haha!

[crystal turtle]

something something polynomial ring over field maybe? idk

[white oxide]

damn this problem is turning me inside out

this problem is gutting me alive

[white oxide]

huh ok

this problem's putting me through the fucking blender!

[next obsidian]

Suppose you are an ideal containing (x^2) [which just means you contain x^2]

Case 1: you are (x^2)

Case 2: you aren’t (x^2)

In this case you contain an element not in (x^2) which means a polynomial f(x) which isn’t divisible by x^2. You can thus be written as
f(x) = x^2•g(x) + ax + b

Where a or b must be nonzero else you’re in (x^2)

Subtracting away x^2•g(x) you remain in this ideal, so the ideal contains ax + b

If a is 0 then you contain a unit and so you’re (1). So assume a nonzero. Multiply by a^-1 and it contains x + b/a

If b is 0, then you contain x, and the ideal (x) is maximal so the ideal is (x)

Now multiply x + b/a by x, you contain x^2 + xb/a

Subtract away x^2, you contain xb/a

Multiply by a/b and you contain x

So you’re (x)

The only ideals containing (x)^2 are (x^2), (x), and (1), QED

[crystal turtle]

[white oxide]

holy shit bro you're a monster

aight lemme read through this

[white oxide]

sorry, why are you allowed to subtract away x^2? that is if we're assuming that a and b are nonzero

[topaz solar]

Im not sure my course last semester covered first iso even, or Sylow

[topaz solar]

You contain (x^2)

[white oxide]

i'm disabled

[next obsidian]

x^2 is in the ideal so if we subtract x^2•anything it remains in the ideal

[white oxide]

thanks forgot about that

[next obsidian]

That’s a lie

No shot

[white oxide]

ah the proof makes complete sense man

[topaz solar]

We 100% didn’t touch Sylow

[white oxide]

damn idk how you even come up with that stuff on the spot 💀

[topaz solar]

And I’m 50/50 we didn’t do first iso

[next obsidian]

It isn’t that hard

[topaz solar]

100% didn’t do second iso

[next obsidian]

Idk

I just assumed it isn’t the ideal and then that gives you an element

[white oxide]

damn ok

[next obsidian]

And from there you just hack away at it, each step is pretty obvious I think

I knew the goal is to get x in there

[white oxide]

ah yeah that's true

still dude good stuff damn

this server's cracked

[topaz solar]

I don’t think it’s the server

[next obsidian]

I mean also yeah I live and breath commutative algebra so

[crystal turtle]

What is there to do with goops without iso theorems

[next obsidian]

I’ve considered this ring hundreds of times

[topaz solar]

I think the only ring we considered was like, Q[sqrt 2]

[next obsidian]

Did you go to a university for toddlers(

[topaz solar]

It be like that sometimes

(Yeah it’s not a big math uni, which is related to me being as braindead as I am)

[crystal turtle]

Real

Mine is big uni but like 10 people in my program lmao

[next obsidian]

Program?

[topaz solar]

I think there are 2 professors who do algebra here

[topaz solar]

Wdym

[next obsidian]

They said “10 people in my problem”

[crystal turtle]

Oops

[topaz solar]

0 who do any models or logic stuff, certainly no commalg/AG

[next obsidian]

Rip

[topaz solar]

They do have applied math and PDEs tho

[next obsidian]

Bro goes to Le Courdon Bleu

[topaz solar]

So I’ll give em that

Nah it’s funny American still but

Definitely not top uni in math

[next obsidian]

Le Courdon Blue is (was) an American university

[topaz solar]

oh no

[next obsidian]

Oh wait

Hmmmm…..

I swear there was a US campus

Or maybe those fucking ads on TV was actually for people to go to France

In which case ????

[topaz solar]

(Btw the issue with the definability I mentioned is because defining the transcendence basis is the issue, not extending once you have it)

[next obsidian]

In the United States, 16 schools used to operate under the "Le Cordon Bleu North America" name through a licensing agreement with Career Education Corporation (CEC), a for-profit education company based in Chicago, Illinois.[6] In 2009, the license was estimated to be worth $135 million.[7] In 2014, Le Cordon Bleu North America generated $178.6 million in revenue and $70.6 million of operating losses.[8] However, in light of the gainful employment rules implemented by the US Department of Education in 2015, CEC made the decision to sell the 16 campuses. When CEC failed to find a buyer[6][9][10] it announced on 16 December 2015 that all 16 campuses in the United States would close by September 2017, giving enrolled students time to finish their programs.[9][11][6] The last new students were accepted in January 2016.[9][6] In June 2016, the Securities and Exchange Commission requested documents and information regarding Career Education's fourth quarter 2014 classification of its Le Cordon Bleu campuses.[12]

Le Cordon Bleu has continued to maintain a presence in the United States through its New York office, Le Cordon Bleu Inc., which places students in the locations abroad.[13]

Aha

[topaz solar]

Interesting

[topaz solar]

This is kinda like how totally transcendental groups have descending chain condition on definable subgroups

If it ain’t definable it ain’t seen and totally transcendental fields are algebraically closed

[topaz solar]

70.6 million losses

Whuh

I mean that’s still +100 mil but

[last spoke]

yea we got 20K people but about this many math grads

[crystal turtle]

Think mine is like 40k

My real anal 2 course was pretty much the entire honors module group, that was 11 people

[topaz solar]

Macintyre is a weird theorem btw

[crystal turtle]

But we have like 30 math grads maybe? Still tho a lot of our classes are pretty baby

[topaz solar]

As baby as “show 0a = a0 = 0” in rings, or 2 days on basics of limsup of sets?

[crystal turtle]

Maybe not that baby I guess

Compared to what I see others posting about though, yeah

[topaz solar]

That showing 0a was literal, assigned homework

[last spoke]

my galois theory class has 6 people

[topaz solar]

To be turned in

[crystal turtle]

I've had some questionabke hw

[last spoke]

I've seen it on both the first and second year algebra exams

[proud spindle]

i'm pretty sure my proving skills are just a bit less baby than "show a0 = 0a = 0"

[last spoke]

I might have seen it on two separate exams for the second year case

[crystal turtle]

10% of my complex anal 2 course (senior year/split graduate course, btw)

[topaz solar]

People having trouble with how a basis generates a topology in the grad topology course

[next obsidian]

I got docked points on my grad manifolds hw because I used the third isomorphism theorem for vector spaces

[topaz solar]

And considering that they have their pic on the student list, PhD students

[last spoke]

I mean we have no undergrad topology classes here

[topaz solar]

What

[crystal turtle]

what

[next obsidian]

And it wasn’t covered in the linear algebra part of ism

[topaz solar]

Third is the riji one?

[next obsidian]

Yeah

[topaz solar]

I can never remember names oop

[next obsidian]

I had to do dimension count and just used that for ease

[topaz solar]

docked

[next obsidian]

I dropped that class cuz the grader was a cunt

[topaz solar]

What mentality would lead someone to dock on that

[topaz solar]

Also, we have one but that’s because they just toss em in w/ the grads

[proud spindle]

i don't know if proving is hard or if i'm just really bad at proving

[crystal turtle]

same, it's 4th year here

[topaz solar]

They can both be true, but you get better with practice

[crystal turtle]

[topaz solar]

Though, then again, im still clowned by chmonkey so

[last spoke]

it's a finicky one that can make you go tf when you first get asked so like, ehh

[proud spindle]

i'm pretty sure if someone asked me to prove the first iso theorem i would blank for like 30 minutes (hyperbole)

[last spoke]

theorems with names aren't meant to be proven, just accept them at face value

[crystal turtle]

me when my group theory class spends like an entire day proving first iso theorem (we had previously seen it in an abs alg course as well)

[last spoke]

I think that's how long we spent on it the first time we saw it

[topaz solar]

I wouldn’t want to prove Macintyre offhand because I am small baby at Galois stuff

But

[crystal turtle]

like bro just give the map and check bing bong oop it's an isomorphism like damn

[last spoke]

I don't even know what Macintyre is

[proud spindle]

i can't even imagine what a proof for it would look like

[crystal turtle]

(it's not that bad)

It's just giving an explicit map that's an iso. Perhaps not the most enlightening or obvious thing when you first see it, but eventually you'll be like "oh yeah, that's the obvious choice for the map"

[last spoke]

I'm busy staring at galois theory homework that I left yesterday thinking it'd be easy but now I am facing the consequences of my actions

[topaz solar]

It kinda boils down to

Totally transcendental
-> additive group has connected component
-> connected component an ideal
-> it’s the whole thing because field
-> 1 generic type over it, and therefore over multiplicative group
-> multiplicative is connected
-> it has all roots of unity

-> either it’s algebraically closed, or an algebraic extension is Kummer (can’t exist here) or order q or smth for char q, adding a root of x^q - x = a

So totally transcendental fields are algebraically closed it seems

[last spoke]

We didn't cover algebraically closed

[crystal turtle]

[topaz solar]

I’ve seen this referenced for finite rank -> alg closed, but Hodges has the theorem statement for any ordinal rank, so idk

Finite rank is particularly interesting since \omega stable or wtv

[last spoke]

we finished galois theory at f(x) has solutions by radicals -> Gal(F(f(x))/F) is solvable

[topaz solar]

And any finitary shenanigans on it are of finite rank, including varieties ofc

[proud spindle]

the only part i could see being easy to prove is that the image of a homomorphism is a subgroup of its codomain

[crystal turtle]

that's not the iso theorem, but yes that's relatively easy

[proud spindle]

from what ik of the iso theorem it's part of the explanation of it

[crystal turtle]

(another thing you'll see is common in algebra, for many algebraic objects btw. But not always true!)

[topaz solar]

ayo Chmonkey what tf is a Kummer extension

[last spoke]

So mild galois theory question

Nvm my question was a dumb, still stuck on this question though

[proud spindle]

a map is like a morphism, right?

[crystal turtle]

map = function = morphism ( = homomorphism = continuous map = ...)

Depending on the contect

Here, I meant homomorphism

[proud spindle]

makes sense

[last spoke]

Any of you have any suggestions on c btw. My brain ain't working today

Got 19/20 on my last assignment though, so we take those

[crystal turtle]

[last spoke]

This is the course with my best grade this tri which should not be the case

[last spoke]

I don't think c is true because (\varphi) is obviously not identity

[cloud walrus]

StarvinPig

[crystal turtle]

(Lagrange)

[last spoke]

I have not seen lagrange pop up at all

[crystal turtle]

\phi is not identity if E is not F_p

but Lagrange should give you \phi^n is for some n, since E is a finite extension

(just need to show that all F-autos are of the form of the form \phi^n then, idk how to do that lol, idk any Galois theory)

[last spoke]

yea idk what Lagrange theorem you're citing here

[crystal turtle]

for groups

multiplicative groups

of the field

ya know

[last spoke]

still ain't ringing a bell

I have a feeling we haven't named it that here

[crystal turtle]

the order of a subgroup H of G divides the order of G

[proud spindle]

the book about groups i was reading actually only has one chapter dedicated to groups

[last spoke]

yea I just don't think we've named that

[crystal turtle]

In particular, a consequence for the multiplicative groups of finite fields F_{p^n} is a^{p^n} = a (classically for n=1, this is Fermat's ittle theorem)

[crystal turtle]

odd, since this is a standard theorem/name

[last spoke]

this I don't think we've seen at all

Well fermats theorem ofc but bigger powers no

[proud spindle]

yeah Lagrange's theorem is one of the few group theoretic theorems i know

[crystal turtle]

well it generalizes (just a simple cardinality argument)

[cloud walrus]

Ryx(Just a dragon by this point)

[last spoke]

That also looks new

[crystal turtle]

bro 💀

it's a field

with p^n elements. Of these, all but 0 are multiplicatively invertible

hence p^n-1 invertible elements

[proud spindle]

ohhhhhh

[last spoke]

Fair, we haven't used (F_{p^n}) that way, we just use (\mathbb{Z}/{p^n}\mathbb{Z})

[cloud walrus]

StarvinPig

[crystal turtle]

Please tell me your class is not using Z/p^nZ for field of order p^n

Please tell me that

[last spoke]

We've never done anything higher than n = 1

[proud spindle]

isn't Z/nZ a ring?

[crystal turtle]

okay good because that would be false if that was true

[crystal turtle]

yes, and is a field iff n is prime (exercise for whenever you learn rings/fields!)

[last spoke]

oh wait that (p^n) where it's some irreducible of degree n?

[cloud walrus]

StarvinPig

[crystal turtle]

uhh you quotient F_p[x] by some irreducible of degree n (or take an appropriate splitting field)

but yeh that's how you get them

[last spoke]

Oh yea found the theorem in my notes that basically finishes this question

Maybe?

So I need to show (E = (\mathbb{Z}/p\mathbb{Z})/(p_n(x))) for some appropriate (p_n)?

[cloud walrus]

StarvinPig

[crystal turtle]

(Z/pZ[x])/(poly)

but I don't think you'll need to show that for this here

[last spoke]

Well if I have E in that form then I get (Gal(E/F) = <\varphi>)

[cloud walrus]

StarvinPig

[last spoke]

Which is what I want

[crystal turtle]

:o

[crystal turtle]

I think this is true. Surprised you would not have shown this at some point? (in particular: that all finite fields of the same order are iso?)

This showed up in my abs alg course

[last spoke]

It doesn't look familiar

Well proving that's true at least

I mean the finite field thing sure, makes sense and I may have seen it

[crystal turtle]

Well then there's probably a proof without using that then

[last spoke]

I mean the theorem gets me everything I want if I can show that

[crystal turtle]

yeah... gl with that one lmao

Not exactly short iirc

[last spoke]

I mean how else do I do this question, I at least have something that's a direct result of a theorem we were shown in class

[crystal turtle]

(idk galois theory enough to answer that part, sorry)

[last spoke]

I think I have something that works if you don't read it

[rocky cloak]

I'm not sure how it helps you that E is of that form, but you can prove that the Galois group is cyclic with just a few facts

• the order of the Galois group is the order of the extension
• x |-> x^p is an automorphism
• a polynomial of degree n has at most n roots.

[last spoke]

If E is in that form then (E = GF(F_{p^n})) so (Gal(E/F) = <F_{r_p}>) which is (\varphi)

[cloud walrus]

StarvinPig

[rocky cloak]

I'm not sure I follow, you've already proven what the Galois group of a finite field is?

[last spoke]

We did that in class I believe

[rocky cloak]

Alright, then I guess the exercise is done

[last spoke]

Also ignore it being rotated, lovelyt

[hybrid pike]

Is the field extension $F(X)$ the same as the field of polynomials $F[X]$?

[cloud walrus]

NotAPenguin

[hybrid pike]

(or isomorphic to)

Wait, F[X] isnt even a field 🤔

I guess that answers my question

[hybrid pike]

I need to compute $[\mathbb{Q}(\sqrt{3}, \sqrt[3]{5}) : \mathbb{Q}]$. I figured the minimal polynomial of $\sqrt{3}$ in $\mathbb{Q}$ was $X^2 - 3$ and the minimal polynomial of $\sqrt[3]{5}$ in $\mathbb{Q}(\sqrt{3})$ is $X^3 - 5$. That led me to concluding $[\mathbb{Q}(\sqrt{3}, \sqrt[3]{5}) : \mathbb{Q}] = 2 . 3 = 6$. But in a second question I need to compute a basis for this field as a $\mathbb{Q}$-vectorspace. I came up with ${1, \sqrt{3}, \sqrt[3]{5}, (\sqrt[3]{5})^2}$, but this only has 4 elements, so now I'm confused.

[cloud walrus]

NotAPenguin

[hybrid pike]

I feel like I could have gone wrong with the second minimal polynomial but I don't immediately see how extending Q with sqrt(3) could help me find a lower degree polynomial with root sqrt_3(5)

Wait, I might simply be missing $\sqrt{3}\sqrt[3]{5}$ and $\sqrt{3}(\sqrt[3]{5})^2$ as basis vectors

[cloud walrus]

NotAPenguin

[dim widget]

Yep

Good job again!

[hybrid pike]

[dim widget]

Happy you are having success with discord qua rubber duck

[hybrid pike]

It helps to organize my thoughts a bit :P

[south patrol]

The number of times I've typed smth out only to have to delete it lol

[coral shale]

im a rubber duck im a rubber duck, rant your math to me

[teal vessel]

honestly, discord as a rubber duck is a huge reason I even visit this channel, lol. I type stuff up, solve it halfway through, and finish it out anyway to share the thought process.

[coral shale]

penguin -> duck

[last spoke]

for finding the galois groups, does it not being separable complicate things

[teal vessel]

knowing what little I do about separability and the techniques it allows for, probably.

[last spoke]

I gotta find the Galois group of (\mathbb{Q}((x^3 - 1)^2)/\mathbb{Q})

[cloud walrus]

StarvinPig

[dim widget]

All extensions over Q are separable, but what you wrote doesn’t really make sense, do you mean adjoint a root of that polynomial equation?

[last spoke]

This is what I got

[south patrol]

That isn't (x^3 - 1)^2

[last spoke]

I'm dumb but I'm pretty sure I'm not that dumb

[south patrol]

Constant coefficient

[last spoke]

Yep

[south patrol]

Anyway it not dumb lol just a lol mistake

[last spoke]

Pain

It is 1am

But man that hurts

[last spoke]

bing has the same answer so now it's funny

[south patrol]

Dw I have done this many times lol

But realising it is quadratic in x^3 is a useful enough observation anyway

[last spoke]

Oh same, but also like I don't wanna find the roots

Though I'm pretty sure I need to?

[hybrid pike]

So for the field $\mathbb{Z}_3[X]/(X^2 + X + 2)$ I can see that it's isomorphic to the field with 9 elements, which according to my notes is ${0, 1, 2, \alpha, 1 + \alpha, 2 + \alpha, 2\alpha, 1 + 2\alpha, 2 + 2\alpha}$. Is there a nice way to figure this out? And how would I know what for example $\alpha^2$ is?

A similar thing was done for the field with 4 elements, which was done by elimination

But that seems really tedious for larger fields :P

[crystal turtle]

Z_3[x]

Not just Z_3 (and this is quite overload notation anyways :p)

[hybrid pike]

Right

[cloud walrus]

NotAPenguin

[hybrid pike]

Fixed :)

[karmic moat]

i've already defined an R-mod hom j : C \to A in the obvious way, but i'm not sure how to define the composition g o j

since j has codomain A but g has domain B

[next obsidian]

Lmfao

It’s supposed to be j:C->B

Bruh moment

[karmic moat]

oh okay

yeah i'll reach out to my prof lol

just wanted to make sure i wasn't missing something

[next obsidian]

You’re missing a correctly stated problem

☠️

[karmic moat]

[last spoke]

felt

[last spoke]

Am I right that the gal group is (D_6)?

[cloud walrus]

StarvinPig

[delicate orchid]

a^2 is pretty clearly -a-2 = 1+2a by just looking at the quotient

[hybrid pike]

Ooooooooooh

I think I see why

[delicate orchid]

because we're setting x^2+x+2 = 0

[hybrid pike]

Yeah, so x^2 = -x - 2 so a^2 = -a-2

How does that give 1+2a though

[delicate orchid]

you're in Z_3...

[hybrid pike]

Right

Okay

Yeah this is making sense now, ty

[void cosmos]

IfR is a ring, then the quotient ring R/J(R) is semisimple

any hint

[delicate orchid]

interesting question

I can get that it's semiprimitive cause J(R/J(R)) = J(R)/J(R) = 0

[void cosmos]

wait tahts it

[delicate orchid]

so if we have that it's artinian then we have it's semisimple right

[void cosmos]

no

my semisimple is just zero radical

[delicate orchid]

oh

[void cosmos]

but

[delicate orchid]

then yeah

[void cosmos]

how did u do that

[delicate orchid]

if I <= J(R) then J(R/I) = J(R)/I

[crystal turtle]

🤯

[void cosmos]

is that ez to prove

lilke i can try it myself

[delicate orchid]

probably - I don't remember the proof myself

probably follows from the corrispondence theorem? maybe?

[void cosmos]

yea it looks like that

[rocky cloak]

Maximal ideals of R/I is exactly maximal ideals of R that contain I. If I is in J(R), then it's contained in all maximal ideals, so R and R/I have the same maximal ideals, thus the intersection is the same.

[void cosmos]

yea i figured it out 😄

thank you so much

once jagr pings u u know ur problem has been solved

lmfao

also tysm wew

[rocky cloak]

In a perfect world, the Chinese remainder theorem would also hold for an infinite set of ideals, but alas. Better just assume all rings are Artinian

[void cosmos]

what

there is another section

which is the final one for my exam

called semisimple rings

which iirc defines them as the direct sum of simple ones

so i thought 0 radical <-> this

but i think what ur saying is

0 radical + artinina <--> this

[delicate orchid]

yes, that's correct

[void cosmos]

yea

[delicate orchid]

hence why I was struggling

[rocky cloak]

I forgot you're working with nonunital rings

What a horrible life that must be

[crystal turtle]

yeah I was like how is J(R) = R

[delicate orchid]

J = 1 😹

[void cosmos]

actually

i am liking the problems where like

everything is going to shit

like no commutativity no units

no nothing

idk why

more fun to me

its like working with assembly language type of vibe u know

[delicate orchid]

world's first magma theorist

[rocky cloak]

I can agree with this assessment.

[lethal dune]

sucks to have no unit then

[rocky cloak]

That feeling when you've found the 1 ❤️

[void cosmos]

but what about non-two-sided ideals?

like J(R) is the intersection of left regular maximal ideals

do they correspond as well?

regular ideal I meaning there exists e in R such that r-re is in I for all r

[lethal dune]

wym? it's also the intersection of right maximal ideals

[void cosmos]

yea hbut i meant

regular

[lethal dune]

so many made-up terms

[rocky cloak]

Yeah, left and right ideals also fit in the correspondence. And for regular I think you proved the other day that any ideal containing a regular ideal is regular right?

[void cosmos]

any left regular ideal is contained in a regular left maximal idea true

by considering the quotient

yeah

[steel light]

I’m having trouble seeing the difference in definition between prime ideals and “regular” (informal) ideals of a commutative ring

Intuitively I think I see why they are different

[chilly ocean]

use the integers to see the difference

[steel light]

If I think about it as a prime factorization, we’re basically saying that any product has at least one factor in P i.e a “prime” factor, so we get a prime factorization of sorts

But in an ideal J is the same not true? If we have a ring R and ideal J and consider the product rj (for all r in R and j in J), which is by definition in J, then we can clearly see one of the factors belongs to J, so what is the difference?

[steel light]

Sure

I suppose the goal is to construct a prime ideal and a nonprime ideal and compare the two to see how they’re different?

All of the rings I’m considering will be commutative and unital in case that matters

[crystal turtle]

yes

and in the integers, (in which all the ideals are generated by a single element, so they all look like nZ for some integer n) there are maybe some good choices for n to get a prime ideal

[steel light]

Gee I wonder

:p

[void cosmos]

aaaaaaaaaaaaa

[rocky cloak]

If j is in J then rj is also in J, but the point is that when J is not prime it's possible to find a and b such that ab is in J, but neither a nor b is.

[void cosmos]

for number 7

is there any nice way that the jacobson radical interacts with localizations?

( just a yes or no i want to try it out my self boys and girls )

i just think 7 is just (2Z+1)^-1[Z] or something

or id

idk

or maybe i would use quasi-regularity or something lmfao

[rocky cloak]

I don't think so, maybe depends what qualifies as nice

[steel light]

Oh

10Z

20 = 5*4

Yes?

[crystal turtle]

yurr

[white oxide]

how would you define multiplication in k[[x]]? I'm assuming for addition it's just term by term addition of the series

[void cosmos]

okay

[steel light]

Nice

[crystal turtle]

You can think of it as an idealization (idk if that's a word lol) of the property that prime numbers have, where if p divides ab, then p divides a or p divides b

[white oxide]

my b didn't mean to interrupt

[void cosmos]

i just attempted the worst proof ever, who wants to check it out?

[steel light]

I think I can vaguely see why Nullstellensatz is true now

[rocky cloak]

Maybe specifically for localizations of the form (1 + I)^-1 R, where I is an ideal of R you can say something nice