groups-rings-fields

[delicate orchid]

yes, actually

[hybrid pike]

😕

[delicate orchid]

they'd all be set to 0 because 0x = 0

[formal ermine]

with no I meant yes but that's not all it does

[hybrid pike]

hm

[delicate orchid]

what else does it do lol............

oh right yeah the actual operations are now all mod 2 as well

fair point

[formal ermine]

penguin what other rings do you know where 2 = 0

[rocky cloak]

It's equal to (Z/2)[x]

[hybrid pike]

So what im left with is $\mathbb{Z}_2[X]$

[cloud walrus]

NotAPenguin

[rocky cloak]

Not sure what would be a satusfying description

[delicate orchid]

also if you want a ring that's isomorphic to it : Z[X]/(2) is isomorphic to itself

trolled once more

[hybrid pike]

lol

with the operations being mod 2 this starts making sense

since now all coefficients are 0 or 1

If I wanted to prove this, I could simply use the first isomorphism theorem with a morphism $\phi(P) = P\mod 2$ right?

ok the latex formatting is a bit messed up but oh well

[rocky cloak]

Use \mod

[hybrid pike]

aha

[cloud walrus]

NotAPenguin

[rocky cloak]

But yes, you could use first iso

[rocky cloak]

Another nice one is \pmod

I.e $1 \equiv 3 \pmod 2$

[hybrid pike]

$P \pmod 2$

[cloud walrus]

NotAPenguin

[hybrid pike]

oh thats nice

[cloud walrus]

jagr2808

[hybrid pike]

So then if I get this correctly, $\mathbb{Z}[X]/(2, X) \cong \mathbb{Z}_2$. To prove this a morphism $\phi(P) = P(1) \mod 2$ can be used. Then $ker(\phi) = (2, X)$ and $\phi(\mathbb{Z}[X]) = \mathbb{Z}_2$

[cloud walrus]

NotAPenguin

[delicate orchid]

yeah this is correct

[hybrid pike]

Thanks

[hybrid pike]

Okay I have one more

$\mathbb{Z}[X, Y, Z]/(X - Y, X^3 - Z)$

[cloud walrus]

NotAPenguin

[hybrid pike]

My thought process was as follows

This makes it so $X = Y$ and $X^3 = Z$, so if we apply a morphism $\phi(P) = P(2, 2, 8)$, then $\phi(P) = 0$ if $X = Y$ and $X^3 = Z$.

[cloud walrus]

NotAPenguin

[hybrid pike]

But now im stuck actually

[delicate orchid]

it should be clear that Z[X,Y]/(X-Y) \cong Z[X]

[hybrid pike]

Right

That makes sense

[delicate orchid]

so we can reduce this down to Z[X,Z]/(X^3-Z)

[hybrid pike]

That one still confuses me to solve though

im thinking Z = X^3

[delicate orchid]

it doesn't really help that you don't have a definitive answer in mind

[formal ermine]

what's Z[X, X^3] = ?

[hybrid pike]

Yeah

[formal ermine]

besides Z[X, X^3]

[hybrid pike]

isnt that just Z[X]

[formal ermine]

yeah

[delicate orchid]

yup

[hybrid pike]

ah, so then if we replace Z with X^3 then the result is Z[X, X^3], which is just Z[X]

[formal ermine]

now prove it rigorously

it always helps to think about these kind of things intuitively at first

[hybrid pike]

yeah knowing the answer helps a lot 😄

[delicate orchid]

if you want to think about it like this then instead of just \phi(P) = P(2,2,8) you can do ||\phi : Z[X,Y,Z] -> Z[X] by \phi(P(x,y,z)) = P(x,x,x^3)||

[hybrid pike]

ah thats smart

[delicate orchid]

up to you to show that this actually has the kernel you want though

[hybrid pike]

Right

[formal ermine]

perhaps a stupid question

but why is this true

[delicate orchid]

the cosets of H form a partition of G

so if you union all of the ones that aren't H you get G-H

[formal ermine]

ahh

I forgor that either aH = bH or empty intersection

[void cosmos]

what good can come out of looking at annihlator sequences

that is if A-->B-->C-->0 is some SES

How does Ann look like

[delicate orchid]

I don't think Ann interacts nicely with SESes

[solar vessel]

my brain reading these messages

[delicate orchid]

if it's split you get that Ann(B) = Ann(A) \cap Ann(C) iirc

but you don't really need a SES to view that fact it's just kinda obvious from being a direct sum lol

[void cosmos]

yeah ig

[solar vessel]

well in general

[delicate orchid]

this is due to Ann(-) not being a functor really

[solar vessel]

the dual of C can be thought of as annihilating A

specifically if you consider it as a submodule of B* then it's the kernel of the transpose B*->A*

[coral shale]

[rocky cloak]

For a short exact sequence 0 -> A -> B -> C -> 0

If r is in the annihilator of B, then it also annihilates A and C. If r annihilates C and s annihilates A, then sr annihilates B.

So the annihilator of B is somewhere between the intersection and the product of the annihilators at least.

[delicate orchid]

wonder if there's some connection there, probably not

we know that if the sequence splits then it's exactly the intersection

something something cohomology something something

[void cosmos]

An ideal P ofa ring R is left primitive ifand only if P is the left annihilator of a simple left R-module.

i can prove this but is this true if we assume R is commutative

so primitive <--> ann of some simple module?

[white oxide]

can somebody help verify if this solution for the other inclusion $A \cap B \subset AB$ is correct? it feels too easy/idk if I'm missing something. Let $w \in A \cap B$. Then $w \in A, w \in B$. Since $w \in R$ and $R = A + B$, we can write $w = x + y$ for some $x \in A, y \in B$. But then $w \in AB$ by the definition of $AB$.

[cloud walrus]

okeyokay

[white oxide]

rings are assumed to be unital btw

because here I didn't even use the fact that A or B are ideals, so I'm pretty sure I'm missing something right? or is this fine

[void cosmos]

can u

write out the statement

what ur trying to prove

?

[white oxide]

sorry it's at the bottom, I replaced A with R and ab with A and B respectively

[delicate orchid]

are these ideals?

[white oxide]

if A + B = R then A intersect B subset AB

yea

[delicate orchid]

ah

[white oxide]

my b

[delicate orchid]

no it's fine I'm just too lazy to read the image you snet

[void cosmos]

huh

[white oxide]

oh ok cool lol

[delicate orchid]

yeah this seems fine

[void cosmos]

do you mean

equal

i mena yea this is true but

[white oxide]

oh sorry like one of the inclusions was obvious

[void cosmos]

yea

[white oxide]

so i just did what lang told me to do

[delicate orchid]

oh you want AB in A \cap B

right got it

[white oxide]

oh wait nah i ignored the part where he said "then trivially", i was trying to show A \cap B in AB

[void cosmos]

where did u use that w is in the intersection

[white oxide]

i didn't at all lmfao

that's what i'm kinda unsure about

[rocky cloak]

What do you mean it's in AB "by definition"?

[white oxide]

like i didn't use the fact that w was in the intersection nor A and B are ideals

[white oxide]

cuz it's the set of all sums x1y1 + ... + xnyn with xi in A and yi in B

[void cosmos]

AB is generated by (ab)

[white oxide]

yea

so if w = x + y where x in A and y in B then w in AB

[rocky cloak]

Okay, but you haven't shown that w is a sum of products...

[rocky cloak]

What, why?

[void cosmos]

i think what okeyokay is doing

[white oxide]

wait but A + B = R so if w is in R then w = a + b right

[void cosmos]

is thinking of linear combations

haha

[void cosmos]

yea

[rocky cloak]

Yes, that's true

[void cosmos]

but this does not imply it being

iin the product of the two ideals

it would imply if you were to show that w = x_1_y1+x_2y_2+...

[white oxide]

wait i'm so confused though can't we just take x1 = x2 = 1 (since R is unital) and then x1a + x2b is in AB

[void cosmos]

x1 is a generator

if 1 is a generator

[rocky cloak]

No

[void cosmos]

then ur ideal is the whole ring

[white oxide]

oh right

fuck

yea i knew i was missing something lmfao

[rocky cloak]

One way to show it:

||Let a+b = 1||

[void cosmos]

you were but its okay

its not trivial

[rocky cloak]

Then ||w = w*1||

[void cosmos]

try to forget about the problem and be more comfy with products of ideals

[white oxide]

okay i'll look at this after like an hour if i'm stuck

[void cosmos]

forget about bezouts theorem and try to think of product of ideals over Z

or Z_n

how do elements look like

etc

you can try to think of the product

[rocky cloak]

then finally ||w = wa + wb||

[void cosmos]

as the smallest ideal containing all the possible products

also as a general hint whenever u have "access" to the whole ring ( as in with R= A+B) always try to include the identity element

like try to think of it as your key

lol

[white oxide]

yea i like to think about it as like being closed under scalar multiplication

[white oxide]

oki

or something something submodule

[void cosmos]

yea

you did injective modules right?

[white oxide]

yea but for like 2 days lmfao so i didn't even digest the information or learn it well

bc school started

[void cosmos]

okay nvm

good luck

[white oxide]

thx

[rocky cloak]

Don't know if it helps, but in R=Z, ideals are positive integers, A+B=R means they are relatively prime, intersection is lcm and product is product.

[white oxide]

ohhh right i keep on forgetting that an ideal is an additive subgroup and not necessarily subring

[rocky cloak]

So you can think about why the lcm of relatively prime integers is their product

[white oxide]

hmm okay i'll think about that

[void cosmos]

can someone stop me when im wrong?

okay here goes

let I be a regular left ideal ( R is a ring )

then I is contained in a maximal regular left ideal

now the intersection of all left regular maximal ideals is a left quasi-regular left ideal

and if we have a quasi regular left-ideal , call it P , then we have P must be in the intersection of Ann(M) where the M runs over all simple left R-modules

so we have the intersection of all left regular left maximal ideals ( which is quasi-regular ) is contained inside the intersection of all annihilators of simple left R-modules

is this correct?

i can give proofs

of any implication if someone asks for it

( actualy i prefer it haha )

[white oxide]

I'm pretty sure this is right: Let any $w \in A \cap B$. Since $w \in R$, $w = a + b$ for some $a \in A, b \in B$. Since $R$ is unital and $R = A + B$, write $1 = a_1 + b_1$ for $a_1 \in A$ and $b_1 \in B$. Now [w \cdot 1 = (a + b) \cdot (a_1 + b_1) = aa_1 + ab_1 + ba_1 + bb_1 = b_1(a + b) + a_1(a + b) = a_1w + b_1w] Since $R$ is commutative. But since $w \in A \cap B$, it follows that $w = a_1w + b_1w \in AB$.

[cloud walrus]

okeyokay

[void cosmos]

yea gj

[white oxide]

wait what

latex error

oh ok

that's good to know, lol w \cdot 1 is not equal to a_1 lmfao

[void cosmos]

yea

[white oxide]

i used ur hint that i could write 1 as some a + b lol

as you can see, good shit

[void cosmos]

yup

gj king

[white oxide]

thanks lol, didn't mean to interrupt ur question my fault

[void cosmos]

np

[white oxide]

Can I get a hint for this problem? Obviously $(a, p) \subseteq R$. So my strategy is to let any $r \in R$ and show that $(r - xa)a$ is in $(p)$ for some suitably chosen $x \in R$. this will show that (r - xa) is in $(p)$ (since $a \notin (p)$ and $(p)$ is prime), upon which $r - xa = yp \rightarrow r = xa + yp$ for $x, y \in R$ and the proof will be complete. idk if this is the right direction or not, but here's what I got so far - if I choose $x = p$, then $(r - pa)a = ra - paa$ and obviously $paa \in (p)$. the problem is showing that $ra \in (p)$, lol. also I haven't used the fact that $R$ is a PID and I have no clue where to use that

[cloud walrus]

okeyokay

[white oxide]

oh could I consider the ideal $R + (p)$ and since that's principal maybe I can do something with $a$, idk maybe that's dumb

[cloud walrus]

okeyokay

[delicate orchid]

how does the proof of PID => bezout domain go again chat

[white oxide]

tf is a bezout domain

[formal ermine]

wtf is a bezout domain

anyway

in a pid

prime iff maximal

[crystal turtle]

UFD + bezout identity for finding GCDs

[delicate orchid]

a bezout domain is a ring such that you can always find x,y for any two a,b such that xa+yb = gcd(a,b)

yur

[formal ermine]

and if a isn't in (p) but (p) is maximal, then (a, p) is...

[delicate orchid]

oh yeah lol that's way easier

[formal ermine]

this time I beat your wife

[delicate orchid]

forgot about this

[formal ermine]

real

[white oxide]

fogoar tabout htis

[delicate orchid]

MODS! CYBERBULLYING

[summer path]

Fog goat tablet itis

[formal ermine]

I remember having to prove this for a homework sheet once

took me like 1 hour to figure out lmao

[summer path]

I thought the proof was kinda just follow your nose

[formal ermine]

yeah

[rocky cloak]

So you know that (a, p) is a principal ideal. Say it's generated by b, then p is in (b)

What can you conclude?

[white oxide]

oh wait so R + (p) is an ideal and since R is a PID R + (p) = (s) for some s in R, but (p) is maximal so therefore R + (p) = (p) or R + (p) = R, if R + (p) = (p) then we can write r + ap = yp for some y in R and then r = -ap + yp so r is in (a, p). if R + (p) = R then r + yp = a for some y in R and hence r = a + -yp and r is in (a, p), idk if that works

[formal ermine]

What

[rocky cloak]

R + I = R, for any ring R and any ideal I

[formal ermine]

what is the definition of a maximal ideal again?

[white oxide]

wait i didn't even use the fact that R is a PID lol

[coral shale]

its exactly what you think it is?

[white oxide]

ohhh yea let me think about this

[coral shale]

no proper ideals contain it except self

[formal ermine]

I was asking okeyokay...

[coral shale]

oh im an idiot

[formal ermine]

but great to know that you know the definition!

[coral shale]

bro this lacks a reply and 'again' sounds like u asking for uself

[white oxide]

are you guys related

[formal ermine]

I hope not

[white oxide]

ry(s or i)

[coral shale]

u can be ryo

[white oxide]

nah

[formal ermine]

Yes

this is an order

[rocky cloak]

Then you should prove that first though

[coral shale]

ryj

[white oxide]

wait what

wait i thought only maximal implies prime

[summer path]

This one is actually a first aa course statement

[white oxide]

wait what prime ideal doesn't imply maximal

[summer path]

For pids

Yes

[white oxide]

oh right for pids

actually what am i saying i never learned that in my first aa course

[summer path]

[rocky cloak]

Yeah, it's essentially the thing you're proving now

[white oxide]

we didn't even get to most of field theory lol

[formal ermine]

[white oxide]

so i should first prove prime iff maximal in a PID?

actually i'll just assume it for the sake of this argument and then prove that later lol

[delicate orchid]

That’s basically what we’re proving anyway

[white oxide]

oh ok

[summer path]

This problem follows immediately if you know prime iff maximal and vice versa

Bruh autocorrect

[delicate orchid]

The shadow government making my phone autocorrect in order to make me look quite the fool

[summer path]

[topaz solar]

Ayo wew what categorical properties do fusion systems have

Particularly for 2-subgroups

[delicate orchid]

you're FUCKED

uhhh everything is a monomorphism, they're connected uhhh

[topaz solar]

Solomon

[delicate orchid]

wait you actually were asking

THERE'S SOMEONE ELSE?!?!?!

joyous day! oh joyous day!

yeah the solomon fusion systems are sick

[topaz solar]

Nah it was mentioned in regards to some model theory thing

[delicate orchid]

...

[topaz solar]

I’m not diving into the deep end yet

[delicate orchid]

gonna need a moment chat

[topaz solar]

It came up in uhhh

[delicate orchid]

basically, take ur subgroup lattice

[topaz solar]

Showing all groups of finite Morley rank are algebraic groups

[delicate orchid]

pretend every line is an inclusion map going upwards

then add in Hom_G(P, Q) P, Q \leq S \in Syl_p(G) if it's not exotic

[topaz solar]

This doesn’t sound so bad

[delicate orchid]

otherwise just add in loads of isomorphisms you noob

no the definition is very easy

it's the erm saturation conditions which are slightly more involved

[topaz solar]

the book mentioned how some system found by Solomon shows not all are realized

Saturation related to model saturation?

[delicate orchid]

no

maybe?!

I don't know model theory

[delicate orchid]

yes, the solomon fusion systems are "exotic" and are the only known exotic fusion systems on 2-groups

they weren't the first found historically though - I believe that was on 7_+{1+2}

[topaz solar]

So they’re fusion systems but don’t have a finite group that realizes em?

Or a group at all?

[delicate orchid]

ngl I haven't thought about that question because all groups are finite

[topaz solar]

Fair

[delicate orchid]

not even sure how it would work in the infinite case - is the sylow subgroup of a profinite group pro-p?

what does that even mean

who know!

[topaz solar]

The thing mentioned fusion systems since uh we know it’s either algebraic or has a maximal elementary 2-subgroup which is of finite dimension

We have Sylow and Sylow^\circ stuff so

[delicate orchid]

there's probably an analogue I don't know

[topaz solar]

But they’re very similar

[delicate orchid]

if there's a notion of a sylow subgroup then you can define a fusion system on it for sure I just have no idea if it's saturated or anything nice

[topaz solar]

If G is finite RM, P a Sylow* 2-subgroup w/ maximal 2-torus T, then N_G(T) controls fusion in P

[delicate orchid]

oh right so it is similar to the finite case

[topaz solar]

Ye

That’s what finite RM does

[delicate orchid]

the case I'm thinking of is I think a theorem of burnside(?) that states if S is abelian and sylow in G then F_S(G) = F_S(N_G(S))

inch resting

[topaz solar]

We have uh

G finite rank and two involutions i, j, then either they’re conjugate or there’s an involution commuting with both

[delicate orchid]

lemme consult the grimoire

[topaz solar]

In uhh d(<i, j>) in the former, d(<(ij)>) in the latter

The smallest definable subgroup

Called “basic fusion lemma”

[delicate orchid]

oh

yeah this is directly linked to the solomon systems

[topaz solar]

Ye

[delicate orchid]

the exotic systems is exactly when you have the conjugation for uhh

Spin(q)? or something?

who knows

[topaz solar]

This holds for all finite RM groups

Including finite groups

[delicate orchid]

yeah but the solomon systems are specifically over some spin group which I really should know

[crystal turtle]

bobby fusions

[topaz solar]

I see

[delicate orchid]

it's always 7 bro what is it with 7

[topaz solar]

Real

[delicate orchid]

smallest known exotic systems? over a group of order 7^3 of course

[topaz solar]

I wonder how saturated relates here but uh

[delicate orchid]

saturated just means "nice" morally

[topaz solar]

Gimme a sec to find the Sylow definition here

[delicate orchid]

e.g. the fusion system where you only have inclusion maps and Aut_F(P) is trivial for all P < S is not saturated

[topaz solar]

Ok so it only really works nicely for 2

This is a horrifying statement

Considering how uh

2 ruins everything

[delicate orchid]

send an actual screenie of the thing

I wish to peep with my own peepers

[topaz solar]

Sylow is maximal p-subgroup, Sylow^\circ is connected component of a Sylow p, which isn’t assumed definable

[topaz solar]

It’s physical

[delicate orchid]

I'm about to get physical

ok so maximal finite p-subgroup

[white oxide]

well since p is in (b), (p) is in (b) right

[delicate orchid]

that makes sense

or does it

maybe if C_S(G) is infinite yeah

nah it makes sense anyway

cool

[topaz solar]

It doesn’t say finite

[delicate orchid]

https://media.discordapp.net/attachments/1025970402662563866/1123062919618367667/goofy3.gif

[rocky cloak]

That's true, but I was thinking more that it implied a certain way to write p

[topaz solar]

If it’s finite, it’s definable

[delicate orchid]

right

[topaz solar]

This isn’t necessarily

[delicate orchid]

ohhhh it's a choice thing

fuckin zorn's

[topaz solar]

Not so much choice as in like

[delicate orchid]

you can have an unbounded chain of p-groups

[white oxide]

ye, b = pr for some r in R, i went along with that but i'm a little stuck, will continue working on it

[topaz solar]

You can’t get a polynomial carving out an arbitrary subset of C^n

[delicate orchid]

I have no idea why that is related

[topaz solar]

To guarantee one exists yeah, but not to say it’s maximal if it exists

It probably uses choice though ofc

[delicate orchid]

hell I don't even know what RM means I've just been replacing it with "finite rank" in my head

[white oxide]

like i think what i'm struggling with here is where to introduce an arbitrary element of r

[topaz solar]

It basically is lol

[rocky cloak]

That would be for b in (p), so you mixed that up a bit

[white oxide]

oh write p = br my b

[delicate orchid]

what source are you reading sharp

[rocky cloak]

And p is prime...

[topaz solar]

How do you define rank?

[topaz solar]

“Simple groups of finite Morley rank”

[white oxide]

what do you mean the element is prime? ig intuitively what i'm picturing now is that either b or r is equal to p (in my head i'm seeing a prime number)

[topaz solar]

Also, if it’s of finite rank

It satisfies descending chain condition of definable subgroups at least

Maybe not general subgroups but still

[rocky cloak]

By assumption (p) was a prime ideal, aka p is prime

And indeed it is true that one of b or r will be a unit

[white oxide]

is that p = ab implies that a or b is a unit

oh ok

[rocky cloak]

You should prove it

[topaz solar]

Chap 1 sections 2, 3, and 6 are of particular interest here ofc

[white oxide]

i didn't know this so i'll prove this too

[delicate orchid]

yeah I'll take a look

I haven't seen fusion systems much outside of the study of them if ykwim

other than block theory

[topaz solar]

Fair

Chap 7 section 2 is about “Fusion analysis” in relation to “Standard components of type SL_2”

But most of the fusion seems to appear in “controlling fusion” and “fusion of involutions”

[delicate orchid]

yeah so controlling fusion is basically just like

[topaz solar]

Fusion systems & Solomon are mentioned in the appendix notes as a possible future route

[delicate orchid]

finding a subgroup H < G such that F_S(H) is isomorphic to F_S(G)

maybe equivalent? I forget the details

[next obsidian]

me when I’m on r/legobuilderspro

[delicate orchid]

it's an isomorphism according to bobby which is bullshit

who gives a fuck about the size of the stablisers you fraud

[topaz solar]

The definition here is H, K < G, subgroups then K controls fusion if for any X, Y < H conjugate (in G) subsets of H we have k in K with X^k = Y

[delicate orchid]

yeah that's equivalent

[topaz solar]

Might be nicer here since not finite idk

They don’t really say stuff about fusion systems

[delicate orchid]

since when has not being finite made anything nicer

[topaz solar]

But they mention them

[topaz solar]

As in a nicer definition, since size of stabilizers is uhhh

[delicate orchid]

well you wouldn't give one about the stailizers directly you'd just show that the fusion systems are iso

ig you can't really do that for infinite groups? maybe you can

[topaz solar]

They mention definable cohomology and cite something about p-local groups, saying it can be rephrased model theoretically

[delicate orchid]

yeah p-local groups are fusion systems with an extra topological structure

[topaz solar]

And since sporadic simple groups are “sporadic” saturated fusion systems, they suggest fusions might be the way to go for the sorta nasty possibly existent simple finite rank groups

[crystal turtle]

what the fuck are these words

[delicate orchid]

yes! exactly

one of the main motivations behind studying fusion systems is a vast simplification of the classification of finite simple groups

[indigo ridge]

so quotient groups is a set of equivalence classes, is the relation R defined to be a R b if a is in the coset of H in G containing b?

[delicate orchid]

yur

[indigo ridge]

huh?

[topaz solar]

Any saturated fusion system over S can be realized in a group G which contains S, and every p-subgroup in G is conjugate to a subgroup of S

Replace S with p-unipotent

[delicate orchid]

is this for S sporadic

[topaz solar]

For any S

[delicate orchid]

well the first part is false, exotics absolutely exist

[topaz solar]

Ah, S in Syl_p(G) here

[delicate orchid]

right, never mind wrong

[indigo ridge]

im guessing that means yes then

[topaz solar]

But it mentions possibly infinite groups

[crystal turtle]

yurr

[delicate orchid]

7_+^{1+2} \in Syl_7(PSL_3(7)) but there are exotics over it

aka fusion systems that are not realised by an- oh

G can be infinite

[topaz solar]

Well, you can’t realize it in a finite one

But

[delicate orchid]

yeah I get you now

that's insane

[topaz solar]

Finite RM is finite-looking

[delicate orchid]

BARMY even

[topaz solar]

There’s a descending chain condition of definable subgroups

[topaz solar]

Theorem by Leary and Stancu

[delicate orchid]

hey I know one of those guys

[topaz solar]

If S is a p-unipotent subgroup of G (which is finite RM), “or possibly a 0-unipotent in the sense of Burdges,” and demand all the subgroups are definable and Hom(P, Q) are uniformly definable families of def. homs

We have a good theory of 2-Sylow theory

[delicate orchid]

yeah cause there's fuck all exotics over 2-groups

the solomons are almost certainly the only ones

[topaz solar]

G of finite RM and even type with a 2-Sylow subgroup gives a saturated fusion system akin to Alperin Fusion Theorem it claims

And realizability is gluing them

[delicate orchid]

making me remember the group theoretic definition of Alperin's theorem...

cruel and unusual punishment

[topaz solar]

Just take whatever formulation

[delicate orchid]

right so it's generated by automorphisms of "nice" subgroups

[topaz solar]

The claim they make here is replacing subgroup and homomorphisms with definable ones is essentially “finitizing” it

[delicate orchid]

cool

[topaz solar]

So it should look like a fusion system of finite groups

[delicate orchid]

yus

[topaz solar]

At the very least, there’s no infinite descending chains

[delicate orchid]

the ermm... lattice is ermm... bounded from erm... below

[topaz solar]

And when we intersect families of definable subgroups, we can just look at a finite sub family

In particular, there’s a minimal definable subgroup of finite index

That is, G^\circ, the connected component

[delicate orchid]

you're kinda losing me here

[topaz solar]

(This taking ^\circ is idempotent)

[delicate orchid]

dw about it

[topaz solar]

So there’s no infinite descending chain, so take your arbitrary family and just start intersecting then to make a chain

Think of it like a compactness thing

[delicate orchid]

I think the problem is that the idea of a "non-definiable" subgroup just doesn't make sense to me

I'm guessing it's a model theory term

[topaz solar]

Yeah, definable means you can write out a formula for it

[delicate orchid]

here's a formula nerds: H \leq G

[topaz solar]

Finite Morley rank is that we have finitely many times we can split it into countably many disjoint pieces by formulas

[white oxide]

here's what I got; if b is a unit, then (a, p) = (b) = R and we are done. if r is a unit, then since p = br, either b in (p) or r in (p). if r is in (p) then r is in (a, p), and consequently (a, p) = R. is this the right direction or

[topaz solar]

Doesn’t mean we have finitely many subgroups but they kinda are of bounded depth?

[white oxide]

but i feel like assuming b is in (p) is gonna be a pain in the ass

lets see

[topaz solar]

This isn’t too interesting for plain groups but like

Consider uhh groups inside another structure

Like multiplicative groups of fields

[rocky cloak]

So by assumption (a, p) = (b) was bigger than (p), so we cannot have b in (p)

[topaz solar]

If your field has finite rank, then it’s algebraically closed

[delicate orchid]

yeah now you've completely lost me

fields have rank?

since when

[topaz solar]

Well, the rank as in Morley rank

It should agree with group rank for groups

But in general it’s something like splitting a definable subset into countably many definable pieces of smaller rank

“Every totally transcendental field is algebraically closed” so it just needs to be bounded above by some ordinal

According to Hodges’ book

[topaz solar]

Dw I’m lost too that’s why I came asking about fusion

[delicate orchid]

that's ok!! I'm going to go watch slop on youtube now

[topaz solar]

I have multiple 600 page books open in front of me trying to decode this first bit

[crystal turtle]

[topaz solar]

Cherlin-Zilber conjecture moment

[white oxide]

right, got it, thank you!

is it bad that i have to get hints for all of these problems and they all take me around a couple of hours

[delicate orchid]

nope

[rocky cloak]

Nah, as long as you're learning

[topaz solar]

If you got it immediately you’d already know it, kinda thing

[rocky cloak]

Did you manage to prove that one of b and r is a unit when br is prime btw?

[white oxide]

okay, I'm just concerned that they're be a problem like this on the exam which is like 3 hours and then i wouldn't be able to solve one of them 💀

[white oxide]

nope, bouta work on that now

[crystal turtle]

I have had quite a few "stare at a problem until I feel stupid" moments. Everyone has

[white oxide]

that's good to know :)

[delicate orchid]

Joker (2019) scene: all I have are "stare at a problem until I feel stupid" moments

leave a like if you get it

[rocky cloak]

You'll have to use that you're in a domain btw

[last spoke]

If you have (E/F) as an extension where (\alpha, \beta \in E) are algebraic over (F), then is (\alpha) algebraic over (F(\beta))? I'm pretty sure it follows but I am not at all confident in this area

I just tested you, come on man

If you have (E/F) as an extension where (\alpha, \beta \in E) are algebraic over (F), then is (\alpha) algebraic over (F(\beta))?

[cloud walrus]

StarvinPig

[last spoke]

There we go

[south patrol]

Yes

This should hopefully be clear just immediately from the definition of what it means to be algebraic

[glossy crag]

Are there fission systems too?

[last spoke]

Yea but this is my first time doing any actual work with algebraic stuff and we covered this at the beginning of August so getting used to it

I get why, it was just said in a textbook example with root3 being algebraic over Q(root2) without any explanation

[south patrol]

Well okay what does it mean for sqrt 3 to be algebraic over Q(sqrt(2))

And why is sqrt 3 algebraic over Q

[last spoke]

because (x^2 - 3 \in Q[x]) but since (Q[x] \subseteq Q(\sqrt{2})), ( (x^2 - 3 \in Q(\sqrt{2})[x]) so (\sqrt{3}) is also algebraic over (Q(\sqrt{2}))

[cloud walrus]

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

[south patrol]

Yup, perfect

Minor typo though

Q[x] subseteq Q(sqrt(2))[x]

But yes perfect

[last spoke]

Same difference

I gotta figure out how to build galois groups but that's a future me problem

[void cosmos]

If R is simple and semisimple, then R is primitive.

just a pretty bad question sorry im stupid

why does every R-module must be simple?

wtf am i saying

nvm

let A be a simple R-module ( and now this exists cuz if it werent it would contradict that J(R) = 0 )

then just Ann(A) = 0 or R but it cant be R cuz it contradicts that the intersection of all annihlators of simple modules is 0

is that correct?

[white oxide]

sorry I'm confused as to how you would define the other ys, for example for y2 would you just keep a2 constant and cycle over the bis for i \neq 2

[white oxide]

can somebody help me understand how surjectivity follows from CRT?

[south patrol]

lol the corollary is what i'd call teh CRT

[void cosmos]

yea

thats how it is in dummit and foote

[south patrol]

Uh unwind definitions and you'll see it's exactly the CRT

like okay

What does it mean for that to be in the image of f

[cloud walrus]

Most Likely To Ponyo

[terse crystal]

I_i+I_j=A for any i,j gives you
I_i+(intersection of I_j: j doesn’t equal i)=A for any i, so a_i+b_i=1 for some a_i from I_i, b_i from the intersection. So the image of a_i is (0,…,0,1,0,…,0)

[white oxide]

I'm pretty sure I got it, correct me if I'm wrong; so we know that there exists $x \in A$ with $x \equiv a_i (\text{ mod } \alpha_i)$ for each $i$. that is, $x = b_i + a_i$ for $b_i \in \alpha_i$. therefore x maps to $(a_1 + \alpha_1, \dots a_n + \alpha_n)$ since the $b_i$s are absorbed

[cloud walrus]

okeyokay

[south patrol]

Okay your use of alpha is confusing me lol

[white oxide]

sorry

[south patrol]

But yeah i mean you can probably write that even simpler lol

[white oxide]

i don't know how the fuck to do the squarey a

[rotund dragon]

\mathfrak

[south patrol]

$x \equiv x_i \mod \mathfrak{a}_i$ is equivalent to saying $x + \mathfrak a_i = x_i + \mathfrak a_i$

[terse crystal]

Proved by induction, for that I_i+(intersection) part, but isn’t that contained in the proof of CRT in your reference?

[cloud walrus]

Most Likely To Ponyo

[south patrol]

so like it follows immediately

[white oxide]

oh right yea

[south patrol]

is a beautiful theorem

[last spoke]

For (\forall x \in \mathbb{Z}/p\mathbb{Z}), it's not true that (x^p = p) correct?

[cloud walrus]

StarvinPig

[last spoke]

It's iff x is a unit

[white oxide]

can somebody help me see this result? here I'm assuming that phi is the euler phi-function

[south patrol]

No

I mean " = p " is a bit moot, since p = 0

[dim widget]

u can do this!

[chilly ocean]

if G and H are finite groups, then what's the order of G x H?

[south patrol]

But for all x besides 0 yes

[last spoke]

Oh = x, typo lao

[south patrol]

oh lol

x^p = x holds for all x

[white oxide]

oh

oops lol

[south patrol]

fermat's lil theorem

[white oxide]

thanks whoosp

whoops

[last spoke]

What about like 2 in (\mathbb{Z}/4\mathbb{Z})

[cloud walrus]

StarvinPig

[south patrol]

well lol

you said p

which I assumed was a prime

Then the correct statement would replace p with φ(n) + 1

[last spoke]

I feel like I'm gonna email my lecturer about this and he's gonna go "You're right it isn't true all the time, but isn't it true most of the time?"

[wicked harbor]

How do I show that $\sqrt[3]{2} \in \mathbb{Q}(\sqrt[3]{2} + i\sqrt{5})$?

[cloud walrus]

sunnyside1

[last spoke]

I'm stumbling my way through this and it ain't clickin'

[void cosmos]

which one

starvin

[last spoke]

I've done a to an adequate degree for now, mainly looking at b

Though b is super useful for a obviously if I've interepreted F-automorphism of E correctly

[void cosmos]

try out examples

take p = 5

or 3

[last spoke]

I mean see above, p = 4, x = 2

[chilly ocean]

p = 57

[void cosmos]

wait what

ohh

p is prime here starvin

ik its not written ot

out

but its prime

wait how did u a

do a*

[last spoke]

I love my lecturer

[void cosmos]

oh u took the hint for granted

yeah i see

or no but like

u forgot that a field only has prime characteristic or 0

so u just chose p =4 haha

its ok

[last spoke]

Oh characteristic means something I need to think about

[void cosmos]

do you know what the characteristic of a field is?

[last spoke]

If p is prime then b looks way less annoying

I probably learned it a year ago and forgot, it hasn't been mentioned in this course

[void cosmos]

it's okay

you know what a field is correct?

[last spoke]

Yea

[void cosmos]

koay

the characteristic of a field is just the smallest number of times u would have to add 1 to get 0

if it never reaches 0 , like in R for example then you say this has characteristic 0

what is the characteristic of Z/pZ?

(p is prime :d )

[last spoke]

Isn't it just 1, because p - 1 + 1 = 0

[void cosmos]

the characteristic of a field is just the smallest number of times u would have to add 1 to get 0

so like

[delicate orchid]

no, it's p, because p-1+1 = 0

[void cosmos]

its the smallest n ssuch that n*1 = 0

[last spoke]

Oh add 1 to itself

[void cosmos]

yea

[last spoke]

Yea p

[void cosmos]

so in a field of char 2

u would have 1+1 = 0

now

let F be a field

why does the characteristic of F must be prime or 0

try to go for a contradiction

like suppose its some composite number that is not 0

and see what u get

remember that fields are integral domains

brb

ik this isnt your problem but like

i just figured u have to know about this stuff u know

idk

:d

[delicate orchid]

one message. put it in one message.

[void cosmos]

i cant im sorry

[last spoke]

I don't think we covered Characteristic at all

[void cosmos]

it's okay its not that deep

[last spoke]

Which is absolutely fantastic for my lecturer to mention it

[void cosmos]

just try to think of this

[last spoke]

If p = uv for u, v < p then uv(1) = 0 where u and v are in Zp so uv = 0 therefore they're zero divisors

[void cosmos]

wdym p = uv

whats p

[last spoke]

Composite

I'm on the bus, lazy

[void cosmos]

oh np

so p is ur characteristic

and ur assuming is composite

[last spoke]

Ye

[void cosmos]

yea good job

now

1 last thing

do you know the binomial theorem

[last spoke]

I remember what it does, I'd need a sec to write it out

[void cosmos]

okay

basically

in a field of characteristic p

show that (a+b)^p = a^p+b^p where a and b are field elements

i dont know if this is do-able on the bus

but just use the binomial theorem

and use that ur in characteristic p

now do b) haha

[chilly ocean]

HEY WHTWHA

sorry my keybord glitched

[crystal turtle]

damn

bro got autobanned

[void cosmos]

yea np

wait he got?

why

wtf

[chilly ocean]

n word

[void cosmos]

where

[chilly ocean]

auto deleted

[void cosmos]

ohh

damn

u there king starvinpig?

u thinking about b?

cuz i wanna go sleep ( open netflix on bed )

[last spoke]

Uhh sure. Idk why I'd need characteristic for it

[void cosmos]

u dont

but the thing is

u choosing p=4 would just be a "red flag" for your lecturer you know

so like

u just had to know these little things so u watch out from these typos or mistakes ( that they did not say p is prime )

[last spoke]

I mean I know what he's like with counterexamples, even if they're pretty big ones

Like if we're proving something for all polynomials and it doesn't hold for ax, the answer is just keep going

[void cosmos]

huh

[void cosmos]

how are u proving something for all polynomiasl if it doesnt work for a polynomial

[void cosmos]

oh..

yeah

yeah ig

happens bro

[last spoke]

Like I get trivial counterexamples, here its "obvious" p should be prime but that one don't feel trivial

[void cosmos]

no imo it should be obvious that p is prime

cuz Z/nZ is not even a field

for n not prime

[last spoke]

Well it wasn't to me lmao. It's been a bit since I've stared at math questions

[void cosmos]

and thats okay

[last spoke]

It is mid-trimester break but be got a test first lecture back and this is on it

[void cosmos]

it is for you now as you just proved it

[white oxide]

can somebody help me understand why this part follows? so lambda is the inclusion map, and lambda(a) = 1 + p^rZ. then i got lost

[void cosmos]

bro doesnt lambda send elements to Z/p^rZ *

[white oxide]

dude idek at this point my brain is fucking fried

[void cosmos]

take a break then its okay

[white oxide]

yea imma take a break and come back to this ugh

[void cosmos]

yea fuck this shit

go play some tf2 or somethingf

anyways

starvin

[last spoke]

It's 2 for 1 meals at the pub so that's my break

[void cosmos]

i would like to give you advice

[last spoke]

But that's for later

[void cosmos]

for dealing with these lecturers

as i have ENORMOUS experience with them

1. do not get into an argument

2. do not get into an argument

[delicate orchid]

If lambda(a) = 1 in Z/p^rZ then isn’t it just immediate?

[void cosmos]

it is

[delicate orchid]

Ah no you’re backtracking

[void cosmos]

but he is tired

[void cosmos]

😠

[last spoke]

Oh this lecturers fine

But he'll go "I claim this isn't a big deal" if he's not bothered doing a finicky proof

[void cosmos]

i dont think so my friendo

this polynomial problem might be on a test

and if u showed that it doesnt work for ax

then u are right

hah

haha

[last spoke]

Oh it was an assignment question to show it holds for all polynomials

[void cosmos]

yeah but it doesnt soo

[last spoke]

It holds for all other polynomials so I went and did that

[void cosmos]

what

[last spoke]

Assuming f =/= ax, it holds

Well that are irreducible because that's what it asked for

Hopefully he'll mark it in time for the test

[void cosmos]

yeaah hopefully everything goes fine for you

anyways do you have any problems with b?

[last spoke]

I think I can show the forward direction

Not quite figuring out the other one

Like I can go x not in F -> x not in U(F) -> x^p-1 =/= 1 -> x^p =/= p, idk if that works though

[void cosmos]

it does

u shown the contrapositive

[last spoke]

Yea and the negation of that is the easier one

[void cosmos]

what

wdym

[last spoke]

Just negate every statement, keep the implications the same

x in F -> x in U(F) -> x^p-1 = 1 -> x^p = x

[delicate orchid]

That’s the converse not the contrapositive

(A => B) <=> (not B => not A)

You did the latter

[last spoke]

Yea I know, but I need to show the if and only if

So I need both

[delicate orchid]

Ah

[last spoke]

I was being lazy with my terms because it ain't the negation technically

I dont need formal logic, it's algebra

[dense raven]

heres a fun problem

Prove that there exists a group monomorphism f : R -> Sym(N).

(you need AC)

[rocky cloak]

Yeah, exactly. Just let a2 take the role of a1, and cycle over i not equal to 2 gives you y2.

[hybrid pike]

I am looking at a proof for the following property:

If R is a ring with $char(R) = n \neq 0$, then the order of each element in the group $R, +$ is a divisor of $n$. The proof uses a property that only applies to finite groups. Am I correct in assuming this proof is incorrect then? For example the ring $\mathbb{Z}_n[X]$ has a finite characteristic but is clearly infinite.

[cloud walrus]

NotAPenguin

[hybrid pike]

Wait, I misread something 🤦

It does apply to infinite groups

[hybrid pike]

I have made a proof for the fact that a morphism of fields is always injective, but I feel like I'm doing something suspicious. Can someone take a look?

Suppose $\phi: F \rightarrow E$ is a morphism of fields. Suppose $x \in ker(\phi)$. Then if we can prove that $x = 0$, then $\phi$ is injective. Suppose $x \neq 0$. Then $x$ has an inverse $x^{-1}$. Then $\phi(x) = 0 \implies \phi(x) * \phi(x^{-1}) = 0 \implies \phi(x * x^{-1}) = 0 \implies \phi(1) = 0$, but for a morphism of fields, $\phi(1) = 1$. So $x = 0$ and $\phi$ is injective.

[cloud walrus]

NotAPenguin

[dim widget]

nope, thats right!

[hybrid pike]

Thanks :)

[dim widget]

one other way to say this is that the kernel of a map of rings is always an ideal, and fields only have one nontrivial ideal which is 0.

[hybrid pike]

Oh you're right, that's clever

[hybrid pike]

So yesterday I asked a bunch of questions for finding quotient rings in the form of $\mathbb{Z}[X, Y, Z]/(X - Y, X^3 - Z)$, which turned out to be isomorphic to $\mathbb{Z}[X]$. Now the topic came up again with a friend and we were wondering how to solve cases where multiplication of variables is involved, for example $\mathbb{Z}[X, Y]/(XY)$. My first instinct said this would be simply $\mathbb{Z}$ but then I realized this was probably wrong. We tried finding a morphish with (XY) as the kernel but failed to come up with anything. Is there a nice way of solving these?

[cloud walrus]

NotAPenguin

[delicate orchid]

I think this one is just "it is what it is"

[hybrid pike]

Hmm I see, I recall the previous exam having a question like this. I'll see if I can find it

[delicate orchid]

you're adjoining two zero divisors to Z, maybe you can force it to be something else but I can't imagine it being a nicer form than this

[hybrid pike]

ah, I found it

$\mathbb{Z}[X, Y]/(5, XY - Y + 6, Y - X^2)$

• Prove that this is a field
• What 'known' field is isomorphic to this field?
• How many roots does T^3 + T + 1 have in this field

These were the questions on the exam

[cloud walrus]

NotAPenguin

[hybrid pike]

I guess it is a bit different here though

[delicate orchid]

ok so cause of the 5 we have that this is iso to $\bZ_5[X,Y]/(XY-Y+1, Y-X^2)$

[cloud walrus]

Wew Lads Tbh

[coral spindle]

Remember when you quotient by an ideal containing A - B, this should be thought of as setting A = B.

[delicate orchid]

Y = X^2 by the second term

[coral spindle]

So including Y - X^2 is just the same as setting Y = X^2, hence completely getting rid of Y.

[delicate orchid]

so this is Z_5[X]/(X(X^2)-X^2+1) = Z_5[X]/(X^3-X^2+1)

[hybrid pike]

ooh this is starting to make sense yeah

Thats a lot easier than it looks

[coral spindle]

Yes.

This isn't to say there aren't still quite hard ones...

[hybrid pike]

Naturally :P

[coral spindle]

The structure of Z[x,y]/(x^2 + y^2 - 1) is not totally easy to describe for example...

[delicate orchid]

circle

[coral spindle]

circle the part that confuses you sweaty

[hybrid pike]

If (X^3-X^2+1) is irreducible in Z_5[X], that would prove this is a field right?

[delicate orchid]

or this

[delicate orchid]

yus

[hybrid pike]

alright, that would solve the first question already then

[coral spindle]

This raises my cortisol levels significantly by simply looking at it

[delicate orchid]

and since it's a degree 3 irreducible polynomial you know what size the resulting field will be

[hybrid pike]

I might not have studied that part yet because that doesnt ring a bell

[delicate orchid]

F_125

[hybrid pike]

hmm

because the degree is 3 and we are working in Z_5, so the field is F_5^3 = F_125?

[delicate orchid]

F_{5^3} not (F_5)^3

if you meant the first one then yeah

[hybrid pike]

Yeah I meant that, my bad

[delicate orchid]

np

although if you haven't seen the proof of that fact maybe you should try and prove it urself

[hybrid pike]

I think it comes a little further in my course notes

So I'll probably stumble on it soon

If not I'll give the proof a try

[wooden ember]

i feel like i know the answer to this but having a bit of a bruh moment here. Does anyone have an example of an extension of UFDs for which coprime elements in the base ring are not coprime in the extended ring? If not what's the proof that this can't be done again

In the setting of Dedekind extensions this cant be done but im not sure about other settings

[dim widget]

what do you mean by extension?

a morphism R \to S with what properties?

[wooden ember]

injective

[coral spindle]

So we need a non-Bézout domain right, otherwise Bézout kills our chances

[wooden ember]

what's a bezout domain?

[coral spindle]

One in which Bézout's lemma holds

[wooden ember]

oh yeah i agree

same argument as for euclidian domains right

[coral spindle]

Yeah if I'm not mistaken every Euclidean domain is Bézout too

so an example of a non-Bézout domain is Q[x,y]

I was gonna suggest Q[x,y][z]/(x^2 + y^2 - z^2) so that Bézout gets us something but this doesn't actually introduce a common divisor

[dim widget]

just injective is too weak

you can, for instance, take Q[x, y, z]/(xz - y)

[wooden ember]

Q[x,y] injects here?

[dim widget]

yes

[wooden ember]

oh lmao yeah ofc

okay thanks

[coral spindle]

nice example

[wooden ember]

what other conditions would you impose on your morphism?

[dim widget]

I think basically if you have flatness this cannot happen

[wooden ember]

remind me, flatness is when R is flat as an S module?

[dim widget]

yes, whichever way around makes sense

[minor wraith]

Any tips for b)?

I am confused

consider a=b=0

then $N(x)=x\cdot \bar{x}=0\cdot0=0≠±1$

[cloud walrus]

Kalgar

[minor wraith]

a contradiction?

[summer path]

You don't have 0 is a unit

[minor wraith]

why not?

it's for ANY $a,b\in\mathbb{Z}$ right?

[cloud walrus]

Kalgar

[minor wraith]

or am I misunderstanding the def?

[summer path]

0 is not a unit in Z[sqrt(2)]

[delicate orchid]

wait didn't we have this question like 4 days ago

[summer path]

I don't remember

[minor wraith]

I made no progress / forgot / didn't understand lol

[delicate orchid]

fairs

[summer path]

There does not exist y such that 0y=1

Since 0y=0

[dim widget]

I think you're thinking of something else

[summer path]

Maybe

[dim widget]

0 is definitely a unit in Z[sqrt(2)]

[summer path]

I'm sleepy

[dim widget]

It's not divisible by any prime

[wooden ember]

what

[minor wraith]

oh but they have the added restriction of xy=1 right

[delicate orchid]

stop trollolololoing

[delicate orchid]

yes

[wooden ember]

lmao got so confused

[dim widget]

lmao i do little a gaslight

as a treat

[topaz solar]

What is N(ab) for arbitrary a, b

[wooden ember]

N(a)N(b)

[delicate orchid]

not asking you boss

[summer path]

I've been gaslit more than once in this channel

[wooden ember]

oh oop i thought it was a joke

[summer path]

[wooden ember]

i need to go sleep soz

[summer path]

Ok well I'm going back to sleep anyway

[minor wraith]

For context

this is from a Number Theory class

[topaz solar]

Yeah doesn’t matter

[minor wraith]

but I did some reading ahead on abstract algebra for fun

[topaz solar]

What’s N(ab) for arbitrary a, b

[minor wraith]

N(a)N(b)

[topaz solar]

What’s N(1)

[minor wraith]

uh

1*1

[topaz solar]

So xy=1

Combine

[minor wraith]

wat did i do wrong to earn the cat

[delicate orchid]

have some confidence

[wooden ember]

technically nothing

[minor wraith]

1 lol

[summer path]

I have none

[topaz solar]

I mean it’s a lil clown-y to write 1*1 but it’s right

[minor wraith]

oh

I thought you wanted me to see the N(a)N(b) thing

[topaz solar]

Yeah I do

Combine the two Holmes

[delicate orchid]

can I be watson

[minor wraith]

y = 1/x?

[delicate orchid]

N(xy) = N(1)

[minor wraith]

xy = 1

x = y = 1

[topaz solar]

Why

[delicate orchid]

that isn't even true in Z

[summer path]

-1 * -1 = 1

[minor wraith]

wat

I don't get why we're considering N(1) then

is this just for my own understanding

or is it supposed to be related to the Q above

[summer path]

You have xy=1

And you're taking N(.) Of both sides

[delicate orchid]

go from here

[minor wraith]

oh right we can treat it as a function

dat cool

[delicate orchid]

wdym

it just is a function

what else would it be

[summer path]

It's a norm function lol

[minor wraith]

a algebraic formula

that happened to be given in the Q

[delicate orchid]

so, a function

cool

anyway use the properties you know about N to manipualte N(xy) = N(1) into something a bit nicer

[minor wraith]

N(xy)=N(1)=N(1)N(1) = 1*1 = 1 ?

[topaz solar]

homie

[minor wraith]

[delicate orchid]

N(xy) = N(1)
N(x)N(y) = 1

[topaz solar]

Try not looking for 1*1

[delicate orchid]

N(x) and N(y) are integers

two integers that multiply to give you 1

[minor wraith]

x and 1/x

[summer path]

I thought you were gonna let him conclude this

[delicate orchid]

so the only possible values for N(x), N(y) are....?

[minor wraith]

x = y =1? ...

[topaz solar]

That it spits out an integer is kinda given?

[delicate orchid]

or

[minor wraith]

two integers multiply to get 1

[topaz solar]

my guy

Please consider if there are any other possibilities

[summer path]

It is but not explicitly in the problem

[minor wraith]

before we go on

[topaz solar]

Not everything is 1*1

[coral spindle]

Are you clear on what the word 'integer' means?

[minor wraith]

am I supposed to use only elementary knowledge]

[topaz solar]

Yes

[crystal turtle]

If people are telling you it's not just x=y=1 please try other things 💀

[delicate orchid]

the fact that (-1)^2 = 1 is pretty elementary and the fact you're being asked to show it can be \pm 1 in the question should tip you off to that

[minor wraith]

So you're telling me

oh my

oh

I expected many more intermediary steps to the result

that was like a 2 liner

[delicate orchid]

lord knows why we're only doing this in Z[\sqrt(2)] when this immediately generalises to Z[\sqrt(n)] but meh

[summer path]

We should do it again

But for n=4

[minor wraith]

Wait so do you have any tips for the next Q/

I'm thinking of fixing b and doing induction on a

[summer path]

What question

[delicate orchid]

what's the next question

[minor wraith]

oops

the 2nd bullet in B)

[delicate orchid]

oh yeah I remember this

[summer path]

Try to find a non trivial unit

[delicate orchid]

I really hate the notation 1/x here I must admit

use x^-1

there's no fractions involved

not directed at you kalgar just a general complaint

[delicate orchid]

anyway yeah, do this

and by doing this we mean find an element x of Z[sqrt(2)] such that there's another element y with xy = 1

and x isn't the obvious choices of -1 and 1

[hybrid pike]

Can I use in an exercise/proof that $\mathbb{Q}(a\sqrt{b}) = \mathbb{Q}(\sqrt{b})$ without further comment? This feels obvious but not very rigorous

[cloud walrus]

NotAPenguin

[hybrid pike]

I guess the answer to that is kinda subjective

[coral spindle]

It depends on what a is

[hybrid pike]

oh, its a nonzero rational number

[coral spindle]

Yup, it's true. Prove it.

[hybrid pike]

Fair enough

[coral spindle]

It is very easy to prove via the definition of Q(a)

[hybrid pike]

I feel like it follows directly from the definition of a field extension

Yeah

[summer path]

Q gobble gobbles up the a

[hybrid pike]

nomnom

[delicate orchid]

if a in Q then how a ouside Q in bracket?? how can both... cnanot...

[minor wraith]

so... any x and its multiplicative inverse?

x≠0