groups-rings-fields

[summer path]

Well you can tell if it is a unit easily by using the norm

So just find some non-trivial unit (ie u != pm 1)

[rocky cloak]

Feels like you need some argument about continued fractions of sqrt(2)

[delicate orchid]

are you sure boss

[minor wraith]

I am stuck and am pretty sure I'm going in the wrong direction again

[summer path]

I don't think you do

[minor wraith]

x is a unit if $N(x)=a^2-2b^2=±1$

correct?

[cloud walrus]

Kalgar

[summer path]

There are definitely ways to do this very elementary

[delicate orchid]

yes

[minor wraith]

then can I derive restrictions on a and b?

[coral spindle]

congratulations, you have the restrictions on a and b

[delicate orchid]

don't need a restriction, just an example

[minor wraith]

i.e. $a=\sqrt{1+2b^2}$

[cloud walrus]

Kalgar

[minor wraith]

this feels very wrong

[delicate orchid]

you literally just need any example such that a+bsqrt(2) isn't 1 or -1

[coral spindle]

Just try some numbers

[minor wraith]

can I take zeros

[topaz solar]

Well, you showed that x unit implies N(x) = \pm 1, but not so much that it implies it’s a unit but…, probably should have an idea how that could work

[minor wraith]

b = 0 a=1

[topaz solar]

[minor wraith]

b = 2, a = 3

[topaz solar]

Ok that’s not just 1

[rocky cloak]

I mean the solutions to a^2 - 2b^2 = ±1 is exactly the continued fractions of sqrt(2), so they are somehow involved in any argument that there are infinitely many

[delicate orchid]

we do not need continued fractions

[coral spindle]

You don't need that here.

[minor wraith]

and so our unit becomes $3+2\sqrt{2}$

[cloud walrus]

Kalgar

[minor wraith]

btw good sirs, are you all graduate students? If so, how do you all have so much time to help me on Discord

lol

[coral spindle]

Bank holiday

[delicate orchid]

yeah I'm taking the day off lol

[rocky cloak]

Procrastination 🌈

[coral spindle]

also that lol

[dim widget]

my fav jazz star from the 30's

[hybrid pike]

I have the field extensions $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$ and $\mathbb{Q}(\sqrt{6})$. The degree of the first two extensions seems to obviously be two. Am I correct in saying that $\mathbb{Q}(\sqrt{6}) = \mathbb{Q}(\sqrt{2}\sqrt{3}) = (\mathbb{Q}(\sqrt{2}))(\sqrt{3})$, which would imply from the tower rule that the degree of this one is 2 * 2 = 4?

[delicate orchid]

ok so do you know about the result that the product of two units is a unit

[cloud walrus]

NotAPenguin

[minor wraith]

yesh

[dim widget]

no

[hybrid pike]

hm

[coral spindle]

It is true that Q(sqrt(2)sqrt(3)) = Q(sqrt(2))(sqrt(3)), but you need to be smarter about the tower rule.

[minor wraith]

wait I swear I proved it somewhere

[delicate orchid]

I believe you dw

[minor wraith]

easy proof doe

just algebra right

[dim widget]

noooo

[coral spindle]

[dim widget]

parentheses are wrong

[delicate orchid]

this is all just algebra

[minor wraith]

ok

[coral spindle]

bleh

[delicate orchid]

anyway so we have our unit 3+2sqrt(2)

so what if we multiply it with itself?

[coral spindle]

Anyway my point is you need in particular to prove [ Q(sqrt(2))(sqrt(3)) : Q(sqrt(2)) ] = 2

[crystal turtle]

Everything is algebra

[minor wraith]

like with the N(xy)=N(x)N(y) and unit * unit = unit results, am I supposed to have derived them on my own ... by playing around with the symbols mindlessly until I notice something?

because, assume I know no abstract algebra at all

[dim widget]

So the difference is that sqrt(3) is in Q(sqrt(2), sqrt(3)), do you see why it's not in Q(sqrt(6))?

[minor wraith]

and onlyhave elementary number theory methods

• info given in Q

[delicate orchid]

it's anything but mindless, you can prove those things by just following the definitions

[minor wraith]

yes, but like

that's "after the fact" proof

easy to proof after I have the form

but how would I have come up with them in the first place?

[delicate orchid]

practice

by the end of undergrad you will see that f(xy) = f(x)f(y) comes up a quadrillion times

[coral spindle]

Oh wait I see lol the parentheses are wrong I misread lmao

thanks tteg

[delicate orchid]

no problem!

[coral spindle]

wew >:(

[delicate orchid]

anyway, lets get back on track

[hybrid pike]

its not some sort of multiple of sqrt(6) i suppose

[delicate orchid]

so we want to find infinite units, and we have one unit 3+2sqrt(2), and we know that we can multiply units together to make more units

[minor wraith]

so just multiply by some general c+d\sqrt{2}?

[dim widget]

so Q(sqrt(6)) is contained in Q(sqrt(2), sqrt(3))

what is the size of the second extension?

[hybrid pike]

im tempted to say 3 but that sounds wrong

[delicate orchid]

well no because a general c+d\sqrt{2} isn't always a unit

but we do know a certain c+d\sqrt{2} that is a unit

[delicate orchid]

tower law

[hybrid pike]

oh

that one is 4 following my earlier reasoning

right

[minor wraith]

because of the xy=1 condition aye

xy=1 => N(x)=±1

[delicate orchid]

I literally just want you to multiply 3+2sqrt(2) with itself

I have no idea what you're talking about

[minor wraith]

I meant like

[delicate orchid]

right yeah we can find one using that - in fact we have

[minor wraith]

in the original question, it defined a unit as some a+b\sqrt{2}

[delicate orchid]

yeah

[delicate orchid]

no, those are just elements of Z[\sqrt(2)]

2 is in that ring and it isn't a unit

[minor wraith]

wat

so "unit" specifically means x such that N(x) = ±1?

[delicate orchid]

N(2) = 4

no a unit is an element x such that there exists an element y with xy = 1

[dim widget]

so what are the possible degrees of Q(sqrt(6))?

[delicate orchid]

in this specific ring this does imply that N(x) = \pm 1

[minor wraith]

ok yeah

all clear now

thanks

[delicate orchid]

we haven't finished the proof lol

[minor wraith]

right

[hybrid pike]

1, 2 or 4 since its inside Q(sqrt(2), sqrt(3)), but its not the same so only 1 and 2 remain, definitely not 1, so 2 remains I think?

[delicate orchid]

so 3+2sqrt(2) is a unit, and we know that a unit times a unit gives a unit - so use these two facts to make a new unit

[minor wraith]

$(3+2\sqrt{2})(3+2\sqrt{2})=9+4=13$

what am I doing

stop

[delicate orchid]

,w (3+2*sqrt(2))^2

[dim widget]

yes also a generator of the field satisfies a quadratic polynomial, so it could never be 4

[minor wraith]

great

very helpfuil

[delicate orchid]

whatever this is, we don't care

[minor wraith]

ok

[summer path]

x is a unit if there is why such that x y = 1

[crystal turtle]

why

[delicate orchid]

we just know that (3+2sqrt(2))^2 is also a unit

[hybrid pike]

alright yeah this is making more sense I think, thanks for the pointers :)

[summer path]

Auto correct moment

[minor wraith]

17+12√2

[delicate orchid]

now use a similar idea to generate an infinite amount of units

[minor wraith]

induction?

[delicate orchid]

sure you can use induction if you want

[minor wraith]

or just simply state that "it is possible" to keep squaring our unit

[delicate orchid]

that also works

[hybrid pike]

would you not have to prove then that that process doesnt repeat

[delicate orchid]

now there's one slightly annoying cavet with this proof

exactly

[minor wraith]

What would we be inducting on?

[crystal turtle]

How do you know it won't ever equal 1? You'd need to justify that x^n ≠ 1 for all n

(this is true and there's a simple justification)

[delicate orchid]

we'd be inducing on n in the statement (3+2sqrt(2))^(2^n) is a unit

[minor wraith]

I see

[delicate orchid]

and then as people are now pointing out we'd have to show that there isn't an n such that (3+2sqrt(2))^(2^n) = 1

which is pretty straight forward

[minor wraith]

contradiction?

[delicate orchid]

sure

[minor wraith]

technically it's (3+2sqrt(2))^(2^n)≠(±1)^2n isn't it

but the -1 dies from the squares

just checking

[delicate orchid]

if you hit -1 on the "next step" you'll hit 1 so it's ok to just show it's never 1

[crystal turtle]

Why would you need the exponent on 1

[minor wraith]

idk

wait why do we check ≠1 again?

so what if our unit = 1

why can't we have the trivial units

[delicate orchid]

because 3+2sqrt(2))^(2^n) = 1 then 3+2sqrt(2))^(2^{n+1}) = 1

so we don't get an infinite number of new units

[summer path]

Try taking an exponent of a trivial unit and see what happens

[delicate orchid]

we only get n

[summer path]

Or that

[minor wraith]

right

Intuitively, how does that make sense though..

3+2\sqrt2 is clearly positive

[delicate orchid]

it doesn't make sense... we're proving that it can never happen

[summer path]

You want to find a unit whose exponents diverge to infinity, that will give you infinite many of them

[minor wraith]

wait so why not for -1 too again?

[delicate orchid]

_ _

[minor wraith]

next step as in n+1?

[delicate orchid]

yur

[minor wraith]

"if you hit -1 on the "next step" you'll hit 1" I don't get this lol

[delicate orchid]

-1 squared is 1

[minor wraith]

yup I get that

[delicate orchid]

ok then you get it

if 3+2sqrt(2)^m = -1

[minor wraith]

I don't get your sentence lol

yup

[delicate orchid]

then 3+2sqrt(2)^{2m} = 1

could you just try proving it?

[crystal turtle]

(2m but yes)

[minor wraith]

yes

[delicate orchid]

we're going in circles

[minor wraith]

couldn't you just have said

any even integer power 2n of -1 is always 1

[crystal turtle]

That's basically the same thing

[minor wraith]

ok

Why can't I just do $\lim_{n\rightarrow ∞}(3+2\sqrt{2})^{2n}=∞$

[cloud walrus]

Kalgar

[delicate orchid]

yeah that works

since |3+2sqrt(2)| > 1 then |(3+2sqrt(2))^2| = |3+2sqrt(2)||3+2sqrt(2)| > |3+2sqrt(2)||1| > 1

[crystal turtle]

prove directly with epsilon-delta

[minor wraith]

I don't remember how to hahaha

will take Analysis next sem

Tell u wat I will attempt if you give me the defs I need though lol

[crystal turtle]

Don't bother lmao it's pretty obvious here

[minor wraith]

wait

then how have we answered the 1/x in Z/[√2] part

[delicate orchid]

these are all units

so by definition there is some y such that xy = 1

[minor wraith]

right

and 0 isn't a unit so we can do y = 1/x

[crystal turtle]

I mean that's kinda tautological

Since that makes sense in R as written, but in an arbitrary ring it's more of a formal expression

[delicate orchid]

the 1/x bro I want to cryyyy

[minor wraith]

for this argument btw, did you use contradiction?

becuase I can't find a contradiction

or did you just do ≠1 directly and used induction on n

[delicate orchid]

if x^n = 1 then |x| = 1

[crystal turtle]

By passing to the norm you can just use a property of the integers like wew said

[minor wraith]

wat is |x|?

the order right?

how is that a contradiction though?

[minor wraith]

you mean like $N(x^n)=\underbrace{N(x)\cdot N(x)\cdot N(x)\cdot...\cdot N(x)}_{\text{n times}}$

[summer path]

um

[cloud walrus]

Kalgar

[summer path]

yes since norm is multiplicative (you can check this)

[minor wraith]

So why is |x|=1 a contradiction?

What is it contradicting?

and to confirm, we're talking about order, not absolute value right?

[summer path]

|.| here probably means norm

although i have not scrolled up much to check context

[minor wraith]

we were using N(x) to depict norm above

[crystal turtle]

Wait what am I talking about

[minor wraith]

I have no idea lol

[summer path]

[minor wraith]

we're trying to show why our unit (3+2\sqrt{2})^2n ≠ 1

So suppose x^2n = 1 and we try to find a contradiction

wew said that $x^n = 1\implies |x|=1$

[cloud walrus]

Kalgar

[minor wraith]

and that is supposedly a contradiction

but I don't know to what

and what the notation '|x|' is supposed to be referring to either

I presume it's the order of x modulo m

but what is "m"?

[topaz solar]

Or

Hear me out

It’s the usual absolute value since subset of C

[crystal turtle]

I am not sure why one would guess this

But yeah do what Sharp said

[summer path]

of the things to reasonably guess, that is definitely not one of them

[crystal turtle]

Sorry I confused myself for a second

[south patrol]

Maths is fun

[rotund aurora]

can you do this without the structure theorem?

[minor wraith]

You mean subset of $\mathbb{C}$? Why is that significant?

[cloud walrus]

Kalgar

[crystal turtle]

complex absolute value

norm

[topaz solar]

[crystal turtle]

modulus

usual one

|x + iy| = sqrt(x^2 + y^2)

[minor wraith]

but x isn't a complex number

[hybrid pike]

[minor wraith]

what 😂

[crystal turtle]

it is

[minor wraith]

my conception of a complex number is a+bi

[hybrid pike]

Z[sqrt(2)] is a subset of C

[minor wraith]

am I using the wrong interpretation

[hybrid pike]

5 is also a complex number

[topaz solar]

bro

[minor wraith]

yes I know

[rotund aurora]

absolute value on R

then

[topaz solar]

yeah that also works but

Why isn’t it complex

[minor wraith]

I meant like

[topaz solar]

Yes

[minor wraith]

$(3+2\sqrt{2})^{2n}=1\implies |x|=1$

[cloud walrus]

Kalgar

[minor wraith]

why is that a direct implication

[topaz solar]

Yeah

[minor wraith]

am I missing a step

[crystal turtle]

if you have a complex number a+bi

[topaz solar]

|x| is complex norm

[crystal turtle]

then if |a+bi|^n = 1, then |a+bi| = 1

[topaz solar]

Please recall how that behaves under multiplication

[rotund aurora]

https://math.stackexchange.com/questions/438707/how-to-construct-a-homomorphism-from-an-abelian-group-to-mathbbq-mathbbz

this answers my question

[crystal turtle]

of course it sounds absurd when you write it specifically with 3 + 2sqrt(2). Becuase we know this is false

[minor wraith]

ok so why is |x|=1 a contradiction

[crystal turtle]

Because |3 + 2sqrt(2)| is not 1

So in total

[minor wraith]

right

[crystal turtle]

If there was an n such that (3+2sqrt(2))^n = 1, then we would have |3 + 2sqrt(2)| = 1. But we know this isn't true, so there is no such n

done

I think you are trying to rush something here. It would be helpful to review some earlier stuff from whatever book you're reading, as well as some properties about the complex numebrs

[rotund aurora]

ugliest gif in existence

[proud spindle]

someone please put this at the end of a completely wrong proof and put it on vixra

[crystal turtle]

they do that quite frequently

[white oxide]

here is k[G] the free vector space over G over the field k?

[crystal turtle]

remember: 2 > 0

[white oxide]

lol

nah but is $k[G] = {k_ig_1 + k_jg_2 \mid k_i, k_j\in k}$

[cloud walrus]

okeyokay

[crystal turtle]

I would imagine so. And in this case, since |G|=2 (i assume C was supposed to be a G), we know G is iso to Z/2

so, say, g_1 is identity, g_2 is the single genereator

[white oxide]

oh wait bro would you mind putting spoilers if ur gonna continue lol

[crystal turtle]

no that's it

lmao

[white oxide]

oh ok i'll take a look if i'm stuck

thanks

[crystal turtle]

that should just be describing k[G]

but here is a hint if you get stuck: ||first iso theorem sounds like it should help?||

[white oxide]

ok thank you

[proud spindle]

when i learned about ||first iso theorem|| it kinda blew my mind

[dense raven]

whats an example of an uncountable simple group

[sly crescent]

SO(3)

[topaz solar]

They say G there so is it the group algebra?

[white oxide]

oh okay, is that equivalent to group ring?

[topaz solar]

That is the group ring

But over a field

But yeah just first iso

[white oxide]

ye

[topaz solar]

You have that this is a ring again too

[white oxide]

time to try something stupid; gonna send k_ig_1 + k_jg_2 to its square and see what happens

[topaz solar]

Why

[white oxide]

idfk

im just playing

around

with the problem

[crystal turtle]

play around with it

[rocky cloak]

PSL(n, K) is simple for any field K

[crystal turtle]

to simplify your life, ||take g_1 to just be the identity, g = g_2, so you can write it as k_1 + gk_2||, and also another hint ||consider maps k[x] --> k[G] instead of the other direction, to apply first iso||

[sly crescent]

False

[white oxide]

will look in about 30ish minutes if still stuck, thank u

[minor wraith]

WAIT

[sly crescent]

PSL(2,2) and PSL(2,3) are not simple

[minor wraith]

how do you prove this again?

[crystal turtle]

https://math.stackexchange.com/questions/683296/is-mathrmpsl-2-mathbbq-a-simple-group#:~:text=Yes.,K|%3D2 or 3.

[minor wraith]

how do you prove that the product of two units is another unit

Before, I thought that it would be straightforward with algebra

[white oxide]

say a is a unit and b is a unit, then ax = 1 and by = 1 for some x y, then abyx = 1

[rocky cloak]

Any field with more than 4 elements then.

[crystal turtle]

you know they inverses. Consider what the inverse of xy is

[rocky cloak]

It works with uncountable fields is the point

[sly crescent]

I’m pretty sure PSL(1,K) is never simple

[minor wraith]

this is tehcnically two existential statements right?

[crystal turtle]

"As I said in my comment, PSL(n.K) is simple for all n≥2 and all fields K, except for n=2, |K|≤3"

[minor wraith]

i.e., for each unit a, there exists some x such that ax=1

[crystal turtle]

that's the definition

[crystal turtle]

point jagr was making I think is that under reasonable assumptions, you can take n, K to make it work

aside from a few boundary cases

[sly crescent]

Hmm

[white oxide]

oh wait no it has to fix k

[sly crescent]

Is PSp(2n,K) simple for all infinite K?

[white oxide]

OOPS

i was trying to define an epimorphism from k[G] to k[x] 💀

[topaz solar]

It is

Just directly do it

[white oxide]

well there goes 30 minutes of my life

[topaz solar]

Why would you try this

[white oxide]

idk

[topaz solar]

I mean just looking at the problem statement

[white oxide]

good question

[rocky cloak]

Hmm, is SO(3) isomorphic to PSL(2, R)? Seems plausible

[sly crescent]

No

[rocky cloak]

Alright

[crystal turtle]

alright

[white oxide]

let's see, so we probably want to make everything linear.... so maybe taking the nth derivative?

wait nvm

[topaz solar]

Maybe look at the problem statement

Or ya know

Map x to something

[white oxide]

homie has no chill

[topaz solar]

I mean it usually has good info

[white oxide]

nah i know im just jokin

[dim widget]

In fact for any field except for the obvious 3 exceptions

Which are the 2 which come from Gl2: PSp(2,2) and PSp(2, 3) and PSp(4, 2)

You can show this by letting Sp act on projective space

[white oxide]

wait i'm dumb is the kernel here elements that get sent to the identity of G or the additive identity of k

oh right the kernel is the same as the group homomorphism so it would be e??

the identity of G

?

[void cosmos]

yo just a quick question

if I is maximal then its left primitive

do i go consider R/I and let A be any R/I-module ( with action (r+I)a = ra) then A is simple as R/I has no proper ideals , so we have if (x+I) is in Ann(A) then xa = 0 for all a --> x = 0 ( its an integral domain )?

the thing is

i used a very similar argument

when showing that if R is a ring such that every left R-module is free then R is a division ring

by showing this I maximal ideal to be 0

and idk why i cant do that here ig cuz i could just say I is a subset of that Ann

but more like

I is a subset of the Ann over the ring R

not over the ring R/I

elements of R/I are equivalence classes not just ideal I

so it would be wrong to say I subset Ann_R/I(A) = 0

right?

this would just say 0 is in the Ann

[sly crescent]

And the B_n and D_n groups (which don’t seem to actually have names)?

[dim widget]

If only someone had given names to the orthogonal groups

The poor things

[delicate orchid]

holy nerd emoji

they do have names, they're called C_2 \wr S_n and the other one

[dim widget]

Those are weyl groups

[delicate orchid]

no way there are two different B_ns

what's next? two different A_ns?

[void cosmos]

any1

?

[void cosmos]

yea king but i already asked

[void cosmos]

.

[crystal turtle]

[dim widget]

Yeah that’s fine

Fields are simple over themselves

So I is left primitive

Ez

[void cosmos]

yea

now

can u give a counterexample

for the reverse

like a left primitive ideal that is not maximal

should be easy ig with matrix rings

but

i cant think of one

[white oxide]

wait wouldn't it just be g_ 1 + g_2k_2 since if g_1 is the identity then kg_1 = g_1 for all k in K?

[crystal turtle]

What

You aren't directly multiplying the k's and g's

[white oxide]

oh you aren't

huhh

wdym by that

[rotund aurora]

can someone tell me a hard cool field/Galois theory problem

but like solvable in a reasonable amount of time

[crystal turtle]

Just notationally, there's no need to put the identity, so if you scrap that, it might make the required map more obvious

Like how we don't put x^0 along with the constant term

[void cosmos]

wdym

whats reasonable for you

like 2-days

@topaz solar

[void cosmos]

XD

[rotund aurora]

uh

like the solution shouldn't be longer than a page

lmao

I said that just in case someone suggested unsolved problems

[white oxide]

wait there's no action of k on G? like there's no map from k x G --> G

i'm lost

[topaz solar]

If it’s solvable in a couple days I would feel sad but understand

[white oxide]

like i thought it was just a free k-module on the group G

[topaz solar]

Why are you asking for this ngl

[white oxide]

i'm just confused as to how to write the elements of k[G]

[topaz solar]

Bro just define a map k[x] -> k[G]

[rotund aurora]

lol Ill just go to Langs book

[topaz solar]

And G has 2 elements

[white oxide]

and how we define k x e for the additive identity in G and any k in K

it should be e right

?

[crystal turtle]

provided I'm understanding this right, the elements should be representable with (a + b*g), with a,b \in k and g \in G. Explicitly, we could write an e with the a, but there does not seem to be a reason to

[topaz solar]

My guy

[dim widget]

Maybe like classify degree 4 extensions

[topaz solar]

It’s polynomials in G

[white oxide]

ohhh i didnt know that

[topaz solar]

Why would you expect different from that notation

[crystal turtle]

So the multiplication is (a + bg)(c + dg) = ac + bd + (ad + bc)g

[cloud walrus]

Dragonslayer Sharp

[dim widget]

Of q

[crystal turtle]

kinda like writing complex numbers with i, but instead of i^2=-1, we have g^2 = 1

[rotund aurora]

cool

[cloud walrus]

Dragonslayer Sharp

[topaz solar]

That’s it

[rotund aurora]

I also thought about trying explicit cases of the Kronecker Weber theorem

[topaz solar]

That’s how you multiply em

[rotund aurora]

liket he quadratic case, which is easy

but I did no more

[crystal turtle]

along those same lines then, k[x] is just k[Z], since Z is the free group on a single variable (written here as x)

[dim widget]

I guess if you haven’t seen the ANT proof of quadratic reciprocity you could do that, but it’s not that hard

[topaz solar]

Well, that’s Laurent polynomials, the usual one is k[N]

[crystal turtle]

fuck

that's what I mean

damn it

[topaz solar]

But you get the idea

[delicate orchid]

we talking about group algebras and I wasn't invited

[topaz solar]

It’s just polynomials bro

[crystal turtle]

polynomials, but if x^2 = 1

[topaz solar]

Wait a minute, what was the problem statement

See the connection

[crystal turtle]

this

[topaz solar]

I was being dramatic

[crystal turtle]

fuck you then

[delicate orchid]

idk but I haven't seen anyone mention the fact that it's just k^{|G|}

[formal ermine]

hmmm I wonder if there's a nicer proof of showing that sin_p and cos_p aren't periodic

[delicate orchid]

with multiplication given by stuff in G

[formal ermine]

besides using straßmann

[delicate orchid]

this absolutely isn't algebra. leave

[formal ermine]

Bitch

[topaz solar]

I said it was polynomials and the multiplication rule on the monomials

[sly crescent]

How about this?

[delicate orchid]

I mean sure you can think about ANY ring like that via universal ploperty (ok not any ring, cause of the monomials)

I just find it much easier to think of them as formal sums of group elements

[topaz solar]

Well I also suggested x^g for g in G

which is the same

[delicate orchid]

hmm

[proud spindle]

i have a question and i'm not sure if it's abstract algebra, elementary algebra, or nonsensical and meaningless

[crystal turtle]

share

[topaz solar]

Superscript time

[delicate orchid]

please only post it if it is nonsensical and meaningless to match everything else in this channel

see above ^

[cloud walrus]

Mecejide

[delicate orchid]

I may be skill issuing but I'm not buying that they're isomorphic

[crystal turtle]

that's a skill issue my guy

[topaz solar]

Polynomials have finitely many nonzero coefficients

It’s literally the same

[crystal turtle]

it's just notational?

[delicate orchid]

hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

[topaz solar]

I’m using x^g instead of g as the basis

[delicate orchid]

how on earth is it just notational

[crystal turtle]

hmm all you want

[topaz solar]

That’s it

[delicate orchid]

oh yeah R[Inn(G)] is isomorphic to R[G]

what?

[dim widget]

Get wrecked wew

[topaz solar]

that’s literally it

[delicate orchid]

no I'm not backing down, this isn't true

[topaz solar]

$x^g x^h = x^{gh}$ is the same as $g\cdot h = gh$

[cloud walrus]

Dragonslayer Sharp

[delicate orchid]

ok not R[inn(G)] but the R[G]-module given by the conjugation representation

[crystal turtle]

where the fuck did this come from

[dim widget]

@delicate orchid they lied to you about groups having centers

[topaz solar]

It’s not exponentiation as conjugation you fusion system crack addict

[dim widget]

They all lied to you

[delicate orchid]

WHAT?!?!?! I knew it

[topaz solar]

It’s a mf superscript

That’s it

[delicate orchid]

who the hell denotes basis vectors with a SUPERscript

x_g?? sure!

[formal ermine]

geometrists

burn them

[delicate orchid]

yeah lemme just $e^1 \otimes e^2$?!?!?

[cloud walrus]

Wew Lads Tbh

[coral shale]

[topaz solar]

Because it’s polynomials and R[N]???

I am not crazy here

[delicate orchid]

R[N]?? no maybe later though

[topaz solar]

Classic wew

[sly crescent]

@dim widget is this guaranteed simple for n>4 and K an infinite field?

[delicate orchid]

your notation does not align with the general literature I win you lost bye bye good night don't let the bed bugs bite

[formal ermine]

https://tenor.com/view/chad-gif-25546777

[topaz solar]

I can’t tell who is the brick wall but yes

[delicate orchid]

anyway what were we talking about

[crystal turtle]

we all are

[delicate orchid]

right multiplication in R[G]

[crystal turtle]

don't remember

nonpost are you typing up a novel my child?

[topaz solar]

okay boys was trying to show k[C_2] = k[x]/x^2

[delicate orchid]

$x = \sum_{g \in G} x_gg, y = \sum_{g \in G} y_gg$ then $xy = \sum_{g' \in G}\sum_{g \in G} x_gy_g'gg'$ ya feel me

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

oh right, this is actually cool

did you actually explain it before I uhhh did what I do best

[crystal turtle]

yeah

[proud spindle]

i had a shower thought: you know how when doing plain, elementary algebra, you manipulate equations by doing something to both sides (subtract 3 from both sides, half both sides, etc). i've heard that conjugacy is an equivalence relation; does that mean you could do a sort of algebraic manipulation with conjugacy instead of equality?

[delicate orchid]

ah cool

[crystal turtle]

sure

[topaz solar]

I tried

[delicate orchid]

yeah sure, the point of equivalence relations is that they're "basically" like equality

and you can turn them into literal equalities by "quotienting"

[topaz solar]

That’s the whole idea, though you want it to be compatible

[crystal turtle]

that's kinda what quotienting is

consider two things up to some equivalence via a (normal) subgroup

[topaz solar]

You want additional hypotheses like a ~ b and x ~ y means a+x~b+y or similar

[crystal turtle]

or could also do up to conjugation or something idfk

[delicate orchid]

"u have g^2 = Id in C_2 so g^2 = 0 in k[C_2] so u adjoin order 2 nilpotent... hip hip hurray!!!!"

[delicate orchid]

yeah like saying two dudes are equal "up to isomorphism"

basically means they're equal when you quotient out by the corrisponding equivalence relation

[topaz solar]

Bro mentioned nth order derivatives so idk how well my explanation got across

[delicate orchid]

... how

[crystal turtle]

"actually g^2 = 1 in k[Z/2]" 🤓

[formal ermine]

wait guys I'm not going insane right?

[crystal turtle]

you are

[topaz solar]

.

[formal ermine]

v(1/x) = -v(x) for a valuation v

[delicate orchid]

oh yeah good point... why is it k[x]/(x^2) again

[crystal turtle]

that's right

corresponds with |1/x| = 1/|x|

[topaz solar]

.

[crystal turtle]

first iso baby

[delicate orchid]

consider the v(xx^-1) = v(1)

[crystal turtle]

|| x |--> g ||

[delicate orchid]

I mean I know deep in my bones that it’s K[x]/(x^2)

[crystal turtle]

wait huh

[topaz solar]

x^2-1

[crystal turtle]

am i being stupid

[topaz solar]

Hmm

[proud spindle]

wait does this mean that the phrase "equal up to something" mean that something is an equivalence relation

[crystal turtle]

yeah

[delicate orchid]

It splits into two dim 1 spaces

[delicate orchid]

Like K[x]/x^2-1 by CRT

[proud spindle]

ohhhhhhhh

i had a feeling isomorphism was an equivalence relation but i wasn't sure

[crystal turtle]

hmm i am feeling stupid

[crystal turtle]

prove it

[topaz solar]

Char 2

[crystal turtle]

for some reason i thought it was just first iso applied to this map

hmm

silly goofy me

[topaz solar]

The kernel is indeed not x^2

Try zero divisors

Also, char 2

[delicate orchid]

Any thoughts on this? It splits as the two irreducible reps of C_2 are blah blah

[delicate orchid]

SINCE WHEN

[crystal turtle]

oh yeha i forgot abuot that

[topaz solar]

Bro didn’t read the problem statement

[delicate orchid]

yeah it’s k[x]/(x^2) then you nerds

[delicate orchid]

Of course I didnt??

[crystal turtle]

|| x |--> 1+g then||

[crystal turtle]

I did it's just been a hot 2 hours

of accomplishing nothing, but still

[topaz solar]

Genius

[void cosmos]

let K/k be a finite separable field extension, and let L/k be any field
extension. Show that K ⊗_k L is a product of fields.

[crystal turtle]

idk man

god i really should read the questions before i spout nonsense huh

[proud spindle]

i think the best i could do is write the things that make it an equivalence relation and be unable to prove any of them

[crystal turtle]

try it at least

good practice

yeah @white oxide the fact that it's characteristic 2 also helps lmao, i forgot about that

[topaz solar]

It’s still first iso

[formal ermine]

Why am I so fucking stupid

[dim widget]

This is not an algebra question

Please see #discussion

[crystal turtle]

I ask myself this every day

[proud spindle]

two groups, $A$ and $B$, are said to be isomorphic if there exists an isomorphism $\phi\colon A \to B$. the existence of an isomorphism $\phi$ implies the existence of an isomorphism $\phi^{-1}\colon B \to A$, thereby making isomorphism symmetric. for every group A, there exists an isomorphism $f\colon A \to A$, making every group isomorphic to itself, meaning isomorphism is reflexive. transitivity is a property of equivalence relations that can be proved via symmetry (a ~ b ~ c is the same as a ~ c ~ b, therefore a ~ c)

[cloud walrus]

nonpost

[crystal turtle]

:o

[proud spindle]

my tildes were stolen

[crystal turtle]

Kinda but missing a lot of details

Yeah Tex doesn't like ~

[next obsidian]

\sim

Also I thought this was chatgpt at first

😭

[crystal turtle]

For symmetry, it works but would be better stated along the lines of "since phi is an iso, with inverse phi^{-1}, then so is phi^{-1}, with inverse phi" or something like that idk.
For reflexive, what isomorphism always works? For any group A?
For transitivity, you just need to show that if f and g are isos, then so is gf. That is, f and g bijective implies gf is, and f and g homomorphism implies gf is (this suffices in this case)

[delicate orchid]

the proof of transitivity is icky

[crystal turtle]

Like you have the ideas but details are

[proud spindle]

forgot to write that i was referring to the isomorphism that maps every element to itself

[crystal turtle]

Good

This also makes sense in any category btw

[next obsidian]

Wait, yeah, I don’t buy the proof of transitivity?

It seems like to make that work you already need transitivity

[delicate orchid]

there's a reason we list it seperately from symmetry

[delicate orchid]

le skeleton has arrived

[south patrol]

or like lol associativity of -> or smth

which doesn't hold

[proud spindle]

i would've sworn that transitivity would be defined in terms of symmetry (it looks like it atleast )

[crystal turtle]

Nope

[south patrol]

if transitivity were defined in terms of symmetry etc there'd be no reeason to demand equivalence relations to be transitive

[delicate orchid]

you've shown that symmetry and already knowing that a ~ b ~ c imples a ~ c ~ b

I think

[proud spindle]

yeah my thought process was "ok if symmetry means that the placing of the two elements in the relation doesn't matter, then it seems to me like you can just switch b and c"

[topaz solar]

Symmetry should look like demanding inverses

Reflexivity is like demanding identity

And transitivity is demanding composition

And this is for every equivalence relation not just isomorphism

[crystal turtle]

Calculus of equiv relations

[topaz solar]

Transitivity is separate from symmetry

Since having inverses doesn’t really mean composition makes sense

[proud spindle]

ohhhh that makes sense

[crystal turtle]

Don't educate the children. We cannot let them become too powerful

[proud spindle]

i was thinking of transitivity in terms of symmetry and not like composition

[crystal turtle]

unless we are making HSCTs

[proud spindle]

i wish i was an HSCT

[delicate orchid]

a~b~c => a~c looks like composition so it must be!

wait

[crystal turtle]

Usually (tho not always) we don't add redundant statements to definitions

[delicate orchid]

can I take the nerve of this

[crystal turtle]

I.e. ones that can be derived from the other requirements

[topaz solar]

I mean it kinda literally is

[delicate orchid]

yeah I suppose

you can shove it into a category if you really want

unique morphism between a, b if and only if a ~ b

[topaz solar]

I was thinking type-y

[delicate orchid]

and then take the nerve of that - cool

[crystal turtle]

That's equiv to a discrete category then

Since skeleton is equiv classes

[delicate orchid]

oh true

yeah

[crystal turtle]

Hurr durr

[delicate orchid]

stfu don't bully me nerd

[next obsidian]

Dumbassssssssss

[topaz solar]

I was thinking a -> b is proofs of P(a, b)

[next obsidian]

Just waiting for Eric to tell me off for this too

[proud spindle]

honestly i'm just happy that i managed to get it mostly correct and not make it completely unreadable

[delicate orchid]

yeah 2/3 is pretty good

[crystal turtle]

Don't kid yourself tho transitivity was unreadable

But decent otherwise

I rate 5.75/10

[next obsidian]

I rate 5/7

[topaz solar]

29% is enough to be a good model, according to music theorists

[crystal turtle]

Geoffrey

[wraith cargo]

what do musicians have to do with model theory?

[topaz solar]

Agustín

[topaz solar]

Nothing good

[proud spindle]

i suppose that's a given when you don't fully understand how transitivity works lmao

[next obsidian]

They have a lot to do with @dim widget though: https://link.springer.com/book/10.1007/978-3-0348-8141-8#:~:text=Topos of Music is an,different levels of musical description.

[wraith cargo]

we should bring one of these people to an empty stadium and watch tteg choke them out

[karmic moat]

can i just consider submodules of the form (0, …, 0, h_i, 0, …) where h_i in H_i and is in the i-th spot

[next obsidian]

I am in favor

[karmic moat]

then each such submodule is isomorphic to its corresponding H_i

ik i can do this in the finite case but what abt arbitrary?

[delicate orchid]

yur sounds right

[karmic moat]

sick

[next obsidian]

Yeah, big swag

[karmic moat]

sunglasses emoji

[next obsidian]

This also IMO is the dumbest shit ever and this notion of internal direct sum should be slaughtered

[topaz solar]

This is what im clowning with the 29% btw

[delicate orchid]

use the definition of internal direct sum using the uhhh funny trivial intersection and whatever

[crystal turtle]

This seems like a fucking dumb question

[karmic moat]

:(

[proud spindle]

i think the notion of external semidirect product should be erased from history

[crystal turtle]

Not you

[karmic moat]

:(

[next obsidian]

Your mom is a funny trivial intersection

[karmic moat]

🙁☹️

[topaz solar]

I’m not convinced the grammar is right here it’s hard for my logibrain to read

[crystal turtle]

Like it's basically just that you can identify H_i with 0 x 0 x ... x H_i x ... 0

[delicate orchid]

intersection of u and the set of bitches is trivial

[next obsidian]

Cuz I’m not a bitch

Yeah

[topaz solar]

It’s absolutely a cringe question though

[crystal turtle]

Like why is this notion of internal direct sum needed. No use for it

[karmic moat]

im gonna screenshot this convo and send it to my prof

[next obsidian]

No

[delicate orchid]

same goes for internal anythings lol

[karmic moat]

im gonna tell him that mathcord thinks his question is cringe

[karmic moat]

he will retire

[delicate orchid]

quick what's the definition of an internel semidirect product

[karmic moat]

and probably just delete his cv

[crystal turtle]

Direct sum should be categorical

[next obsidian]

G = NH with N normal H a subgroup and N\cap H = {1}

[topaz solar]

Categorical logic in shambles

[karmic moat]

mmm yes mmmm direct sum of modules is a coproduct mmmm yes mmmmmmd

[delicate orchid]

NERDDDDDDDDDDDDDDDDDDDDDDDDD

[karmic moat]

delicious mmmmm

[next obsidian]

CV mention!!!!!!!!!!111?1!111!!1111

[karmic moat]

OMG COLLATZ REFERENCE????? 2n+1 referenxe????????

[next obsidian]

Guys if you use base 60 collatz is easy

The mayans did this

[karmic moat]

shoutout to the mayans

[topaz solar]

Bro clowning wrong it was Babylonians

[next obsidian]

😭😭😭😭

Mayans on that base 20 tho

[crystal turtle]

I wonder if he's still in this server

[next obsidian]

☝️🤓

[karmic moat]

i thought torsion was gonna be some super cool sick definition

because its got a sick name

:/

[delicate orchid]

yeah it's just "zero divisor but module"

[karmic moat]

booooo

[wraith cargo]

torsion is cool wtf are u on about

[crystal turtle]

No

[karmic moat]

i thought the definition would be way cooler

[next obsidian]

When your testicles are an abelian group with elements of finite order: 😱😱😱😱😱😱😱😱

[wraith cargo]

Have you ever heard of a little book called torsion theories by Golan 🫦

[delicate orchid]

the coolest things have the simplest definitions

[wraith cargo]

venture bros reference

[next obsidian]

????

[wraith cargo]

What lol

[delicate orchid]

what

[next obsidian]

I was just free styling a testicular torsion joke

[wraith cargo]

https://www.youtube.com/watch?v=slobhI2HXhA

[karmic moat]

venture bros invented testicular torsion

[crystal turtle]

🫦

[karmic moat]

they are the cause of my suffering

[topaz solar]

Torsion free simple group of finite Morley rank

[delicate orchid]

u keep making up these words bro

[topaz solar]

Btw if anyone knows one dm

(This is a counterexample to a conjecture)

[delicate orchid]

the uhhh the erm...

[topaz solar]

Cer-something-Zil’ber conjecture

Relating stable groups and algebraic groups

In particular, finite Morley rank ones

https://en.m.wikipedia.org/wiki/Stable_group

it’s the omega stable simple groups so uhh

[white oxide]

Using only the axioms of a field, do we have to prove that 1 + 1 = 2? We don’t right, since that’s the definition of 2?

[hidden haven]

Yes that's by definition

[topaz solar]

Basically if you have a countably infinite subset, then there’s only countably many complete types (as opposed to larger, like the upper bound of 2^N)

So there’s countably many kinda characterizations of things you can do using those elements

(Complete types meaning you have either p or -p in it for each formula p, so it’s like deductively complete)

Think restricting the cardinality of Spec B for the Boolean algebra associated with the formulas

[proud spindle]

what a cool defn, let's never define this externally

(i hate the external semidirect product)

[topaz solar]

Complete types are like characterizations of what something should be (if it exists) I think?

Very polynomial root-ish

[warm ember]

let G be a finite group and H_1 and H_2 be isomorphic subgroups

then is G/H_1 isomorphic to G/H_2

in other words is there an automorphism of G mapping H_1 to H_2

[coral shale]

try prove it.

[warm ember]

o

[delicate orchid]

fusion moment

this isn't true btw

[formal ermine]

fusion deez nuts

what do I do if my prof is ghosting my emails

[warm ember]

o

[coral shale]

wait this isnt, wew?

hmmm

[warm ember]

hmm

how do you do 8 then

[delicate orchid]

oh lawdddd what's a circulation again

[warm ember]

uh

flow is negative in the reverse direction

and kirchoff

[delicate orchid]

I have absolutely no idea what that means

[warm ember]

o oops

[delicate orchid]

it's a me problem not a you problem

[coral shale]

im confused whats a counter

[formal ermine]

physicist spotted

blocked

[delicate orchid]

oh EWWWWWWWWWWW

[warm ember]

bruh what

this is graph theroy

[formal ermine]

kirchhoff
bro it's physics

[delicate orchid]

ok gradually reding through this and it almost smells like graph homology

[karmic moat]

u smell like graph homology

[delicate orchid]

yeah probably

[formal ermine]

uhhhhhh

I have an algebra question

but it's high school algebra

[warm ember]

idk homology bruh

its this

[formal ermine]

how do I show $$\sum_{n = 1}^\infty (-1)^{n+1} \frac{\left(\sum_{k=0}^\infty \frac{x^k}{k!}\right)^n}{n} = x$$

[delicate orchid]

oh wait yeah this IS just homology, the B_H is exactly the (generating set for the) 1-boundaries

[formal ermine]

I'm tryna show ln(exp(x)) = x

but x is a p adic number of norm less than wtv

[crystal turtle]

Holy shit lil bro stop trying to prove them directly

Infinite sums of infinite sums is never fun

[formal ermine]

hmm

wait

I want to use this

see first sentence for h(X) = f(g(X))

can I argue that it's true in R

and so the formal power series match up?

[warm ember]

can someone provide a counterexample pls

[formal ermine]

G = Z, H_1 = 2Z, H_2 = 3Z

[warm ember]

ye but thats infinite

[formal ermine]

oh

[coral shale]

well thats helpful anyways, interesting

[formal ermine]

Hmmm

[delicate orchid]

take G = Z_2 \oplus Z_4 and ur two subgroups to be (1, 0) and (0, 2)

[formal ermine]

fucking beat my meat to it

[delicate orchid]

quotients are Z_4 and Z_2 x Z_2

[warm ember]

oh

ty

[coral shale]

wth whys maf so hard

sigh

but at least like u said, if the subs are isomorphic thru automorphism, then ur good

mb

[solar vessel]

it's true for finite dimensional vector spaces tho

[warm ember]

oh

ye

thats all i need ty

[solar vessel]

wat

[delicate orchid]

"did you guys know that if n-a = n-b then a = b"

[coral shale]

no.

[delicate orchid]

well now you do

[warm ember]

wait actually

[warm ember]

how would this work

cuz you need a field

obviously H^n/H^k=H^n-k

but is H^n/H'=H^n-k where H' is isomorphic to H^k

[solar vessel]

even better it's true for finite dimensional subspaces of arbitrary vector spaces

[delicate orchid]

holy guacamole!

[coral shale]

look i got a good one

its true for quotient sets

[solar vessel]

what is

[coral shale]

finite ones

um how do i write this...

this sht has no structure

[warm ember]

does Diestel even know group theory

[delicate orchid]

quotients of sets don't corrispond to a substructure

[coral shale]

plz, S/H is clearly when u have multiple disjoint copies of H

that partition S

[delicate orchid]

you can probably still do it using some stupid universal property

[topaz solar]

Bro just reason about power series for these

[formal ermine]

how?? #groups-rings-fields message

this? #groups-rings-fields message

[delicate orchid]

tbh I don't even know why ln and exp are inverses in R and I don't really care to find out

[topaz solar]

You can say things about power series in fairly general ways

[formal ermine]

how

[coral shale]

he just did.

[warm ember]

is this even true

[formal ermine]

where

[coral shale]

here...

[formal ermine]

but how

[coral shale]

nvm, im just pulling your leg

[crystal turtle]

yes by this

[topaz solar]

What’s the power series for ln and exp

[formal ermine]

that's the thing I'm trying to show

[topaz solar]

Notice: where are the coefficients

[formal ermine]

here #groups-rings-fields message

[crystal turtle]

it's a power series with coefficients in Q my friend. this equality should then make sense because it holds in R

[topaz solar]

I was trying to lead him to looking what the coefficients are

[crystal turtle]

The issue to check in convergence

That's your job now

[formal ermine]

so basically what I said here? #groups-rings-fields message

[topaz solar]

And how they aren’t essential Q_p

[crystal turtle]

if stuck check the book's proof

[formal ermine]

I DID

BUT THEY JUST SAY "yeah the formal equality is true"

[crystal turtle]

yeah sure but specifically you need that the coefs are in Q

[formal ermine]

Yeah

[topaz solar]

Power series are inverses

So just check convergence actually makes sense anywhere

[coral shale]

breh does this identity hold outside

like just formally

forgetting R, Q

[topaz solar]

Yeah because that’s what it should be

[formal ermine]

shuri gets what I'm asking

[coral shale]

like yeah if u expanded it formally things die

should die, i assume

[crystal turtle]

I mean sure you maybe could show it

but uhhh

infinite sum of power of infinite sum?

I'd rather not

[formal ermine]

but yeah the argument that it's true in Q works right?

[coral shale]

please its a little algebra.

[crystal turtle]

(idk why it works. Let's just trust that it does)

[crystal turtle]

okay then enlighten me

prove it with that

[coral shale]

just symbolab it...

[crystal turtle]

symbolab

proof

one or the other bestie

[coral shale]

if the first 1000 cancel id trust the rest would

[crystal turtle]

yes

[formal ermine]

ok good

[summer path]

neither, just give up

[topaz solar]

It works over Q gg, therefore just check convergence makes sense

[formal ermine]

Yeah convergence is ez

[crystal turtle]

i think thi sums it up well enough

[formal ermine]

strongest wins 💪

[coral shale]

but yeah, the proper way is to reason about a particular power and why it dies

right???

[next obsidian]

Is this not some shit about Q being dense in Q_p or some shit

• continuity or something idk

[crystal turtle]

idfk

[next obsidian]

Something something “holds in R cuz of calculus, these power series are the same so holds in Q so holds in Q_p by continuity and density”

[crystal turtle]

probably

[formal ermine]

Why do we need continuitiy and density

[next obsidian]

To extend the result from Q to all of Q_p

[formal ermine]

yeah but like why

what fails

[delicate orchid]

actually true

[next obsidian]

You’re asserting that fg=id on a dense subset and if fg is continuous this means it’s true everywhere

[topaz solar]

Z dense in itself moment

[delicate orchid]

I don't believe this

[formal ermine]

but fg = id is false for all of Q

ln(1+x)=power series doesn't converge everywhere

[delicate orchid]

he's got a point!!!!

[next obsidian]

Whatever, it holds on the part it converges in and that’s dense in the part of Q_p it converges in

[topaz solar]

I say it because ilum said it meant Z was complete and Ultra btfo’d him for saying he intuited analysis

[next obsidian]

Z is actually complete tho, he do have a point

NO CAUCHY SEQUENCES

[formal ermine]

uhh wat

ln(1+x) converges in the open disk around 0

in Q

[next obsidian]

Let Q_cool be the part of Q where it converges and the composition is equal to id

Then Q_cool is dense in (Q_p)_cool and then blah blah or something like that

[formal ermine]

Q_p_cool?

[next obsidian]

[delicate orchid]

pass through the completion

[formal ermine]

Ah

[delicate orchid]

I supose it would aktually be better notated as Q_cool_p 🤓

[topaz solar]

Why did that get through to him

[next obsidian]

I have no fucking idea

[delicate orchid]

bro is categorically minded

[formal ermine]

idk it's 11 pm

I slept 5 hours last night

[delicate orchid]

time zone FAIL

[formal ermine]

I am playing trackmania right now

and doing bicep curls

I need sleep

[topaz solar]

I learned a fair bit via category or type nonsense too so

HSCT really does change mfs

[delicate orchid]

I am more ok with continous functors commuting with limits then I am with continous functions commuting with limits

[formal ermine]

ok just to clarify this is another way of proving that f(g(x)) = h(x) right? using the theorem and arguing that f(g(X)) = h(X) because it's true in Q also works yes?

[next obsidian]

I don’t… I don’t know what you mean

My point is just that shit is continuous and so you can prove the identity on a dense subset

You can do that on the part that lives in Q cuz it’ll be dense

To prove it there you appeal to calculus

[formal ermine]

how do I show that ln_p(exp_p(X)) = X as formal power series? can I just argue that it's clearly true in Q?

[next obsidian]

You prove it in R

Using calculus or something

[formal ermine]

a priori I'm saying that we know that it's true in Q or R or whatever

[next obsidian]

You’re trying to show this identity right?

[formal ermine]

Yeah

as formal power series

[next obsidian]

You do it by saying that exp(ln(x)) = x as analytic functions or whatever the fuck

Using calculus