groups-rings-fields

[chilly ocean]

8

[spark plank]

wut even is C_8

[chilly ocean]

8 elements that satisfy a^8 = e

[spark plank]

like

[chilly ocean]

In practical terms

[spark plank]

what's C_1?

[chilly ocean]

(Z/8Z, + mod 8)

[spark plank]

wait no

C_1 = {e}

C_2

what's that

[solar wyvern]

cyclic groups are those generated by a single element.

[spark plank]

so like C_2 = {e, a} where a² = e?

[chilly ocean]

C8 is the cyclic group with 8 elements

[solar wyvern]

yes

[spark plank]

C_3 = {e, a, a²} where a³ = e?

[chilly ocean]

a^6 too

[solar wyvern]

yes

[spark plank]

fam y u no say this

[solar wyvern]

i sorta did

[spark plank]

a has order 3

in that one

[chilly ocean]

Ye

[spark plank]

okay

[chilly ocean]

(Z/nZ, + mod n)

For ezy brain time

[solar wyvern]

1:32 PM] Katzen Aktion: $a$ is a generator for the cyclic group $C_8$, that is $\langle a|a^8=e\rangle \cong C_8$

[cloud walrus]

flimflam:

[spark plank]

u think I know wut that means?

[chilly ocean]

before we further confus

[spark plank]

so I got C_8 = {aⁿ : 0 ≤ n < 8} where a^8 = e

[chilly ocean]

Have you heard of a generator?

[spark plank]

apparently I have but I have no clue wut it is

I assume

that it generates all other elements

[solar wyvern]

@spark plank $n \in \mathbf Z$

[spark plank]

when raised to powers

[cloud walrus]

flimflam:
Compile Error! Click the reaction for details. (You may edit your message)

[spark plank]

[chilly ocean]

Just start with a

If a in generating set then a^n etc. until a^8 which is e

[spark plank]

so basically what I said no?

[chilly ocean]

Eyup

[spark plank]

kk

so we want to look at maps of this

and find which ones are endomorphisms?

[solar wyvern]

yes.

[spark plank]

or something more specific than that?

cuz I feel like there'll be a few of 'em

[solar wyvern]

nope ujst that 🤷

[spark plank]

kk

[solar wyvern]

try it out, there can only be so many

if you want, you can guess how many

[spark plank]

8

[solar wyvern]

okay now try cranking out them endos

if you haven't for whatever reason confirmed that $a \mapsto a, a \mapsto a^8=e$ are 2 of them, do so now too

[cloud walrus]

flimflam:

[spark plank]

oof

wait

1, 2, 4, 8

[solar wyvern]

explain urself

[spark plank]

$f:x\mapsto a^nx$

[cloud walrus]

Simple_Art:

[spark plank]

hm

[solar wyvern]

what's x

[spark plank]

if n=4 then it satisfies f(ab) = f(a)f(b)

x are elements of C_8

it also satisfies inverse

[solar wyvern]

what's a^n x

[spark plank]

but doesn't map to identity

sed

a^n left multiplied onto x

well obviously identity and inverses are endomorphisms

[solar wyvern]

oh ya, continue 🤦

[spark plank]

wait do I need to map the identity to itself?

[solar wyvern]

(yes)

u better

[spark plank]

k

[solar wyvern]

can u see how that rules out a bunch of things already?

why not indeed

🤔

[spark plank]

well if it need to map to identity

wait hm

fock

ur right

🤦

no not that

I just did it wrong

$f:x\mapsto x^n,n=1,2,4,8$?

[cloud walrus]

Simple_Art:

[spark plank]

this is probably what I was thonking

[solar wyvern]

i am okay with this, but I'd like a bit of explanation how/why you got there

[spark plank]

wait but inverses

so you can do negative n too

[solar wyvern]

yeah

wait does 6 work?

[spark plank]

no

[solar wyvern]

why not

[spark plank]

no not u

kek

$(a^ma^n)^k$\$=(a^{m+n})^k$\$=a^{(m+n)k}$\$=a^{km+kn}$\$=a^{km}a^{kn}$\$=(a^m)^k(a^n)^k$

[cloud walrus]

Simple_Art:

[spark plank]

for natural k, m, n

e^k = e for all natural k

so we just need inverse

but we know that a^8 = e

so...

$((a^m)^k)^{-1}$\$=a^{-mk}$\$=((a^{-1})^m)^k$

oof that was ez

[cloud walrus]

Simple_Art:

[solar wyvern]

why doesn't $a \mapsto a^6$ work tho

[cloud walrus]

flimflam:

[spark plank]

so it works for all natural k

[solar wyvern]

O:

[spark plank]

but for k > 8 we basically get the same thing

[solar wyvern]

same how

[spark plank]

that is, only k = 0, 1, 2, ..., 7 matter

$a^{k+8}=a^ka^8=a^ke=a^k$

[cloud walrus]

Simple_Art:

[spark plank]

for all a in C_8

cream

no

🤦

🤦

okay so

let a be the generator

$e^6=e$\$a^6=a^6$\$(a^2)^6=a^{12}=a^4$\$(a^3)^6=a^{18}=a^2$\$(a^4)^6=e$\$(a^5)^6=a^6$\$(a^6)^6=a^4$\$(a^7)^6=a^2$

[cloud walrus]

Simple_Art:

[spark plank]

u can verify that these multiply and stuff

and that it an endomorphism

and that it works for any power

wait, go do ur analysis

$(a^ma^n)^k$\$=(a^{m+n})^k$\$=a^{(m+n)k}$\$=a^{km+kn}$\$=a^{km}a^{kn}$\$=(a^m)^k(a^n)^k$

$((a^m)^k)^{-1}$\$=a^{-mk}$\$=((a^{-1})^m)^k$

[cloud walrus]

Simple_Art:

Simple_Art:

[spark plank]

and obviously e^k = e

wut

$a^{30}=a^{24}a^6=(a^8)^3a^6=a^6$

[cloud walrus]

Simple_Art:

[spark plank]

$a^8=e$

[cloud walrus]

Simple_Art:

[spark plank]

but it's basically exponential rules

exponential rules satisfy group properties

hence they form endomorphisms

$a^ma^n=a^{m+n}$\$(a^m)^k=a^{mk}$

[cloud walrus]

Simple_Art:

[spark plank]

The second one is to make everything into one power

and the first one is to show that f(ab) = f(a)f(b)

[solar wyvern]

if u can show me which $n,m$ define the same maps $a \mapsto a^m,a\mapsto a^n$ i be happy--also you'd exhibit a partition of $\mathbf Z$

[spark plank]

well it simplifies stuff at least

Okay so

[cloud walrus]

flimflam:

[spark plank]

since a is our generator

every x, y in C_8 is of the form a^m and a^n for m, n in Z

right?

[solar wyvern]

ya

brb after i finish shitparacompactness

[spark plank]

ur set is literally {e, a, a², a³, ...}

it's all powers of a

okay so let f(x) = x^6

$f(a^ma^n)$\$=(a^ma^n)^6$\$=(a^{m+n})^6$\$=a^{6(m+n)}$\$=a^{6m+6n}$\$=a^{6m}a^{6n}$\$=(a^m)^6(a^n)^6$\$=f(a^m)f(a^n)$

[cloud walrus]

Simple_Art:

[spark plank]

$f(xy)=f(x)f(y)$

[cloud walrus]

Simple_Art:

[high bobcat]

cool

$bruh$

[cloud walrus]

𝔎𝔄𝔇𝔦𝔰𝔠𝔬𝔯𝔡:

[chilly ocean]

@high bobcat Kys

[solar wyvern]

pls no disturb, this warning#1

[spark plank]

also since $(a^m)^n=a^{mn}=(a^n)^m$ we get that $f(x^{-1})=f(x)^{-1}$

[cloud walrus]

Simple_Art:

[spark plank]

also disgusting username

plz change

or request nickname

[high bobcat]

@chilly ocean kys fgt

im not harming anybody

[spark plank]

joins math server
goes to channel where people are clearly having a math discussion
proceeds to be an ass

words

idk fam

actually

ya they do

kek

well anyways

we get that $x\mapsto x^k$ is an endomorphism for all natural $k$.

[cloud walrus]

Simple_Art:

[spark plank]

did u get the k=6 case?

[solar wyvern]

@chilly ocean try to cook up a counterexample if you think this wrong

[spark plank]

[solar wyvern]

also stirring up shit when mod literally warns u

[spark plank]

dam this sl||im||e ED 2 is really growing on me

so obviously

if you raise $a^m$ to a power

[cloud walrus]

Simple_Art:

[spark plank]

you get $a^n$ for some other $n=0,1,2,3,\dots,7$

[cloud walrus]

Simple_Art:

[spark plank]

that is, it's in $C_8$

[cloud walrus]

Simple_Art:

[spark plank]

okay

so do you get how $(xy)^6=x^6y^6$ for all $x,y\in C_8$?

[cloud walrus]

Simple_Art:

[spark plank]

tbh they're just powers of a, so you actually get abelian stuff I think

well we can easily show that this is the case

did you get what I wrote earlier tho?

$f(a^ma^n)$\$=(a^ma^n)^6$\$=(a^{m+n})^6$\$=a^{6(m+n)}$\$=a^{6m+6n}$\$=a^{6m}a^{6n}$\$=(a^m)^6(a^n)^6$\$=f(a^m)f(a^n)$

[cloud walrus]

Simple_Art:

[spark plank]

this bit

that $f(xy)=f(x)f(y)$ for all $x,y\in C_8$

[cloud walrus]

Simple_Art:

[spark plank]

where $f:x\mapsto x^6$

[cloud walrus]

Simple_Art:

[spark plank]

then it's an endomorphism no?

yeah, the 6 in the above proof has no significance

but

there are actually only 8 endomorphisms that come about from this

$(a^m)^{k+8}$\$=(a^m)^k(a^m)^8$\$=(a^m)^ka^{8m}$\$=(a^m)^k(a^8)^m$\$=(a^m)^ke^m$\$=(a^m)^ke$\$=(a^m)^k$

[cloud walrus]

Simple_Art:

[spark plank]

$x^{13}=x^{5+8}=x^5x^8=x^5$ for all $x\in C_8$

[cloud walrus]

Simple_Art:

[spark plank]

since $x^8=e$ for all $x\in C_8$

[cloud walrus]

Simple_Art:

[spark plank]

Yeah, so we basically only have mod 8

so now we probably need to show there are no other solutions

ya u can always bring it down to [0, 7]

oh hm so I think I can show these are the only solutions

let $f(a)=a^k$

[cloud walrus]

Simple_Art:

[spark plank]

Then $f(a^2)=f(a)f(a)=a^ka^k=(a^2)^k$

[cloud walrus]

Simple_Art:

[spark plank]

in general $f(a^m)=f(a)^m=a^{km}$

[cloud walrus]

Simple_Art:

[spark plank]

and hence we must have $f(x)=x^k$ for all $x\in C_8$ for $f$ to be an endomorphism

[cloud walrus]

Simple_Art:

[spark plank]

any natural k

it could be 0 too

[chilly ocean]

Splintegers fren

[spark plank]

The key point is that $f(a^m)=f(a)^m$ if $f$ is an endomorphism

[cloud walrus]

Simple_Art:

[spark plank]

this comes from repeatedly applying $f(xy)=f(x)f(y)$

[cloud walrus]

Simple_Art:

[spark plank]

If it's an endomorphism, we know that $f(a)\in C_8$ so $f(a)=a^k$ for some $k$.

[cloud walrus]

Simple_Art:

[spark plank]

It thus follows that\$f(a^m)$\$=f(a)^m$\$=(a^k)^m$\$=(a^m)^k$

[cloud walrus]

Simple_Art:

[spark plank]

That is, $f(x)=x^k$ for all $x\in C_8$

[cloud walrus]

Simple_Art:

[spark plank]

and this has nothing to do with 8 btw

aside from the mod 8

Hence C_n has n unique endomorphisms

|Endo(C_n)| = n

yeah it works for any natural k

We could have $f(a)=a^0=e$

[cloud walrus]

Simple_Art:

[spark plank]

and hence $f(a^m)=f(a)^m=e^m=e$

[cloud walrus]

Simple_Art:

[spark plank]

which forms a pretty trivial endomorphism

The key point was to write elements of C_8 as a^m and use exponential rules here

lesson noted for dealing with cyclic groups

idk

think this it

got learnt tho

[stone fulcrum]

Well, what are you missing? Anything you wanna know?

[chilly ocean]

Just gotta grind

[spark plank]

[chilly ocean]

I mean - we've effectively taught you from the ground up

So it's not that bad

[stone fulcrum]

I imagine quotients are pretty intuitive by now. The cosets of a normal subgroup form a group under coset multiplication.

The group they form end up being an image of a homomorphism, where the subgroup is the kernel

Iso

[spark plank]

=pup monomorphism

[mellow vapor]

Do more with Wolfram|Alpha pro

Query made by

@spark plank

[spark plank]

o no

cats already

I'm nit ready yet

[stone fulcrum]

Mono if isomorphism from group to itself

[spark plank]

o right

[stone fulcrum]

Nono isomorphism

Mono if f:G → G

[spark plank]

[stone fulcrum]

Am I drinking?

[solar wyvern]

@spark plank making progress

[spark plank]

[stone fulcrum]

I am drinking

[spark plank]

how 'bout u tho

[solar wyvern]

can you show the subgroup of isomorphisms now

[chilly ocean]

I'm trying not to drink

[spark plank]

what's an iso

[chilly ocean]

bi homo

[solar wyvern]

invertible maps

[chilly ocean]

invertible structure preserving maps

[solar wyvern]

in endo(C_n)

hence structure preserving 🤷

hint

a to a is one, a to e usually isn't

so right now $Endo(C_n)$ isn't really a group, consider $f \in Endo(C_n), f: a \mapsto e$ \

does $f$ have an inverse?

[cloud walrus]

flimflam:

flimflam:

[solar wyvern]

it's actually a monoid O:

it doesn't, that's the point

[chilly ocean]

monoid mean no inverse

[solar wyvern]

I'm asking for the largest subgroup contained in Endo(C_n)

[spark plank]

So...

x^n

for n coprime

To 8

[solar wyvern]

that is, Aut(C_n)

[spark plank]

what's an automorphism

[solar wyvern]

invertible endomorphism

[spark plank]

[solar wyvern]

@spark plank can u prove this

ummm

@chilly ocean where u at

[covert vector]

hi

[solar wyvern]

hi woog

math building locked today

[spark plank]

ya

[covert vector]

uh oh !

[spark plank]

not too hard tho

[covert vector]

u should look for a secret entrance

[spark plank]

[solar wyvern]

my roommate has keys but I'm not friendly enough to ask

😦

[covert vector]

thats what I did when I was locked out of the math building

[spark plank]

3, 5, 7 are the only powers we lookin' at

[covert vector]

I just took that time to search the whole campus for secret entrances

until I found one

[solar wyvern]

prove for all n

[covert vector]

then used that to enter instead

[solar wyvern]

sneaky maus

oh

[spark plank]

x is coprime to y iff x+y is coprime

u can't be givin' me these numbre theory questions

[solar wyvern]

incidentally, the formal proof is on pp44-45

Automorphisms of Cyclic Groups

[covert vector]

what do u mean x+y is coprime

[spark plank]

[covert vector]

u need to compare

[spark plank]

coprime to y

Otherwise you could pull out a factor n of y

But then u can pull out a factor of n from x

[solar wyvern]

actually we should probably do CRT first

can u do dat @covert vector

finding the structure of the automorphisms may be a bit 2 hard tbh at this point

even for C_p, the literal ezest case

[oblique river]

what are you trying to do?

[solar wyvern]

oh yeah, we were just finding normal subgroups as kernels of endomorphisms

so do that guis

[bleak abyss]

You know for a half second I interpreted that as \mathbb{C}_p

And was like wut

[solar wyvern]

algebraically complete p-adics tho

[oblique river]

algebrically closed and topologically complete

two for the price of one

[solar wyvern]

[spark plank]

PARACOMPACTNESS

[solar wyvern]

buncho

for $C_n$ are the distinct endos of the form $a \mapsto a^n, n \in n\mathbf Z, 1+n\mathbf Z,...,(n-1) +n\mathbf Z$

[oblique river]

do you mean n \in Z?

[cloud walrus]

flimflam:

[oblique river]

i'm not sure exactly what you mean by that last part, but I'd just write "a \mapsto a^k for k = 0, ..., n-1"

oh I see

yeah I mean the slick way to write it is "a \mapsto a^k for k in Z/nZ"

[solar wyvern]

okay, but just so everyone's on the same page

[oblique river]

yes

there are n endomorphisms of C_n

corresponding to taking a generator

and sending it to wherever you want

[solar wyvern]

was hoping to instantiate how each normal subgroup can be expressed as the kernel of a group morphism, then realized @chilly ocean and @spark plank may not really get kernels, group homomorphisms, normal groups etc so went back to cyclic groups

then got distracted with automorphism groups

[spark plank]

veri confuzzled

[oblique river]

showing that kernels are the same thing as normal subgroups isn't too hard

which way are you having trouble with?

[solar wyvern]

where did i go wrong @oblique river

[oblique river]

one direction: show directly that the kernel of a homomorphism is normal: Let H = ker f where f: G --> K. Then for any a in G and h in H, aha^(-1) is also in H, because f(aha^(-1)) = f(a)f(h)f(a^(-1)) = f(a) e f(a^(-1)) = f(a*a^(-1)) = e

for the other direction: If N is a normal subgroup then N is the kernel of the canonical map G --> G/N

basically by definition

okay which part of that are you confused on

do you believe that the map G --> G/N is a homomorphism?

[solar wyvern]

@spark plank do u believe this

[oblique river]

so just as a general piece of math advice: you should believe that it's a homomorphism before you even check it. Because shit would be fucked up if it weren't a homomorphism

like, G/N is constructed from G. and when you defined G/N as a group (i.e. when you defined the group operation on G/N), you used the group operation on G as an input

[solar wyvern]

if u don't believe it then why haven't you checked it

[oblique river]

so you better get a homomorphism haha

[spark plank]

sees many things SA doesn't know what means

[solar wyvern]

@spark plank what don't u get

[oblique river]

but yes, just check it: the map f: G --> G/N is defined by f(g) = gN, i.e. send an element g to the left coset gN

what

[spark plank]

[oblique river]

do you know the definition of homomorphism?

[solar wyvern]

For $H < G, G/H := {aH: a \in G}$

[oblique river]

okay

do you know the definition of G/N?

[spark plank]

kernels, G/H

[cloud walrus]

flimflam:

[spark plank]

kek no

kek

[oblique river]

I think we are getting distracted @chilly ocean

[cloud walrus]

creamy shits:

[oblique river]

so

a homomorphism is just a thing that preserves the group operation

that's it

[spark plank]

So basically set of cosets

[oblique river]

like, you have groups

and you have functions between them

[spark plank]

And I assume other direction triangle are right cosets

[oblique river]

but like those functions had better respect the group structure

[solar wyvern]

Then for $H \leq G: (\forall a \in G) aH = Ha$, we (could) write $H \triangleleft G$, and define multiplication as follows

[
(aH)(bH) = (ab)H
]

Then u should check this makes G/H a group for H a normal subgroup \
(ie G/H has identity, inverses, multiplication)

[oblique river]

or else, why are you looking at them lol

a homomorphism is literally just a map that satisfies f(gh) = f(g)f(h)

what it says is that "it doesn't matter if you take the prodcut in the first group or in the second group"

that's all a homomorphism means

okay so if you know what a homomorphism is

then prove that the map G --> G/N is a homomorphism

what?

there is an obvious function

you even said it yourself earlier

just think about it. if you start with an element of G, how could you get an element of G/N?

[cloud walrus]

flimflam:

[oblique river]

there is an obvious map

if you take an element of G

let's call it g

you want to produce an element of G/N

i.e. you want to produce a coset of N

[spark plank]

(aH)(bH)

[oblique river]

what would be a reasonable way to take g and turn it into a coset of N?

[spark plank]

tfw ur cosets be multiplied

[oblique river]

what does that mean

what?

okay let's try an example

let G = Z and N = 3Z

YES

that's literally all it is

take an element g and send it to the coset gN

[spark plank]

tfw u feel like u need actual symbols to differentiate between different multiplication

[oblique river]

I was gonna ask you where you thought the element 4 should go under the map Z --> Z/3Z

uhhhh

[spark plank]

e.g. $(a\cdot H)\times(b\cdot H)=(ab)\cdot H$

[cloud walrus]

Simple_Art:

[oblique river]

yes

you send each element in G

to its own coset

shouldn't you prove that?

uhhh

that doesnt prove it's a homomorphism

to prove it's a homomorphism you need to show that f(ab) = f(a)f(b)

those are literally the same thing

yes

great

see what I meant earlier?

this map is literally a homomorpism basically by definition

like there's nothing to prove

the group operation on G/N was defined in order to make the map G --> G/H a homomorphism

it sounds like you've already proven that the map G --> G/N is surjective

although i'm not erally sure how you did given that you seemed to not even know what the map was until just now

it's surjective again by definition

every element of G/N is some coset of N

let's call that coset gN

then gN = f(g)

so the map is surjective

and finally, the kernel of the map is N by definition

the identity element of G/N is the coset N (or you could write eN if you want)

so the kernel of G --> G/N is the set of all g in G such that gN = eN

but gN = N if and only if g is in N

and now you're done

[solar wyvern]

if u appreciate the help put in the work to understand

[oblique river]

ok gl with that

[spark plank]

[chilly ocean]

Welcom

Want to do something different?

k

[chilly ocean]

ye\

You're getting the point though

ker(f) = [e]

Cosets must be of same size

And |ker(f)| divides |G|

So the partition is equal and you can use the representatives to define coset multiplication.

Show it now

,$ gN = {gn : n\in N}

[cloud walrus]

Jichael Mackson:

[chilly ocean]

bearing in mind the N is a subgroup

Does it make sense that some g could send two elements of n to the same element of G - making |gN| < |N|

ye

aH = {ah: h in H}

ye

Couldn't have an ah_1 = ah_2?

Is it something in the axioms?

Axioms of a group

I have h_1 and h_2 in my subgroup

Why can't ah_1 = ah_2 for some a outside the subgroup and h_1 =/= h_2?

Alright then but the idea of a coset as a transformation of an element on subgroup is an idea I want you to hold in your brain

So we agree cosets must be all of the same size

Do you believe that they must also be disjoint

Alright well there's something else we can try

Suppose we have cosets aH and bH

Show that if $aH\cap bH \neq \emptyset$ then $aH=bH$

[cloud walrus]

Jichael Mackson:

[chilly ocean]

Yep

Why not?

Well

What if I had some element

h that was in both aH and bH

That is

if x in aH and y in bH

then ax = h = by

Oof

Hol up

x,y in H

ax = h = by

That the cosets aren't disjoint

(((try multiplying by an inverse)))

Inverses are unique

ax = h = by

axh^-1 = e = byh^-1

So ax = by

So if there is an intersection - all members of the two cosets are identical

So either they're disjoint or the same coset

Kewl

Cosets are the same size, cosets are disjoint.

So do cosets cover the group?

Ie. for each element of G is there a coset that it belongs to?

Why?

gH = {gh : h in H}

H is a subgroup.

How can I guarantee that every g has a home in a coset of H?

Awesome

So now with these three pieces of info

1. Cosets are the same size

2. Cosets are disjoint

3. Cosets cover a group

Prove Lagrange's theorem

Well number of elements of all cosets = |G| right

Because cosets cover the group

The cosets must all be the same size |H|

The cosets are disjoint so you can count how any elements are in 3 cosets because its 3|H|

So |G| = n|H|

Nooo problems?

With that proof of Lagrange theorem?

No there isn't

I'm wondering if there's something about it that you don't understand or like

I find my informal one gets me by

But ye das Lagrange

We notice the same thing for right cosets

etc.

Idk what you were doing in your wall of text up there

Trying to show that Ker(f) is always a normal subgroup?

Proved G/Ker(f) = Im(f)?

Probably should hit the hay

Let it sink into your brain pan

That's me in 10

All the info magic happens in your sleep

I'm convinced

Nah dude

That's where you crystallize that memory

[spark plank]

[errant drum]

Wdym "wym"

[sonic current]

@chilly ocean weren't you done with abstract algebra?

[chilly ocean]

Yaaay 😄

[worthy kindle]

pls come back to analysis

D:

[chilly ocean]

2 + x = 5

smh

[proud warren]

Hey

I studied linear algebra and oh god I forgot soooo much😂

[worthy kindle]

it's hard to remember everything when you stop practicing for a while

[proud warren]

So what I did is that I designed a projection application in the E = R³ space, projecting on F = Vect [(-4,0,1),(2,2,2)] opposed to G = Vect [(1,0,0)]

[worthy kindle]

intéressant

[proud warren]

Français ?

[worthy kindle]

Yup 🥖

[proud warren]

Incroyable

[jagged gate]

🥖

[proud warren]

Je suis francophone oui, mais pas de France

[worthy kindle]

(c'est Vect qui t'a trahi)

[proud warren]

Ah on dit comment en anglais ?

[worthy kindle]

span

[proud warren]

Je ne savais pas

[worthy kindle]

moi non plus avant de venir ici xd

[proud warren]

XD

En gros, je me suis amusé à me faire des exercices pratiques

Pour repratiquer

Et retrouver la théorie

J'ai aussi refait des exos de diagonalisation

C'était ma partie préférée des mathématiques de prépa, l'algèbre linéaire

[raven widget]

ewww, l'alg lin... Quelle horreur

[solar wyvern]

oh, u wanna do varieties?

[stone fulcrum]

Wtf is a variety?

I hear that all the time and I don't get it

You haven't done ring theory yet. You'll like it soon

[raven widget]

@stone fulcrum Variety has many meanings. At least, algebraic varieties are just zeroes of polynomial systems

[stone fulcrum]

It is what I meant

And I would argue it's easier, since it uses a lot of ideas group theory already put in your head

There's a quotient of rings, it's easier than the quotient on groups

[raven widget]

field stuff 🤢

[stone fulcrum]

@raven widget
Thx for the info. I'm not at algebraic geometry yet

[raven widget]

K^^

[stone fulcrum]

And then there's fields, which is like two groups squished together

Two groups, one has +, one has ×

× distributes over +

For fields

Oh, and everything is commutative

[covert vector]

field is a set where you have PEMDAS

or BEDMAS

or whatever order of operations acronym you learned

yes

[stone fulcrum]

Indeed, you've been working with fields for your mathematical career. You'll find them natural

[covert vector]

inverse is what division is

everything commutes

and u can use brackets and distribute stuff

it works as you'd expect

that's what fields are

yes

yes

then you can think real hard and try to generate perhaps less obvious examples

integers modulo a prime number form a field

[stone fulcrum]

Rings are more interesting.

• is an abelian group
  × is closed and associative
  × distributes over +

× don't have inverses! So you can't divide in a ring. The integers are a ring, for example.

[covert vector]

no

quotient groups have only 1 operation

if you look at multiplication of Nonzero numbers in the field

everything is invertible and associative

so that forms a group

and everything including zero wrt addition is invertible and associative, so that forms a group

it's more natural to go from a field to looking at those two operations as groups

than to go from two groups to a field

(in my opinion)

what I'm saying is that you probably shouldn't think of fields as two groups working together in a nice way

makes things less intuitive

youve been doing multiplication and addition on rationals and reals your whole life, much more experienced

[solar wyvern]

fields are 0D rings.

[oblique river]

(that is the "reason" why module theory is so much easier over fields than over general rings!)

[long beacon]

equivalently it's because the theory of fiber bundles over discrete topological spaces is easier than over nondiscrete spaces

almost tautologically

[stone fulcrum]

Fields are 0D rings.
Wat that mean?

[long beacon]

literally not important

[stone fulcrum]

I'm studying commutative algebra so if there's anything helpful for me there, let me know!

[long beacon]

rings can be given a geometric incarnation like in algebraic geometry

[stone fulcrum]

Oh no not algebraic geometry

[long beacon]

so you can say "Spec R" for any ring, which will be a topological space whose space of functions is given by R

yeah, ikr

anyway, for fields k, Spec k contain a single point

so they're 0D

[stone fulcrum]

I'll have to look into that

[long beacon]

it's also sorta wrong though

i mean, people who say that fields are 0D rings

here's a 5 minute primer to algebraic geometry:

[stone fulcrum]

Finally

[long beacon]

there's a philosophical duality in mathematics between "spaces of functions" and "geometry"

when you think geometry, what do you think of?

topological spaces + structure that might make you measure things

or some other shit

[stone fulcrum]

Metric spaces and natural shapes that go into them

/coffee cups

[long beacon]

if you were like, a 10 year old

someone drew a circle on the board and you were like, fuck yea, circles and shapes and shit

[stone fulcrum]

Sounds like me

[long beacon]

and then 5-7 years later, in high school someone wrote down x^2 + y^2 = 1

and you were like, fuck yea, circles and shapes and shit

but those two things are so philosophically different

one was a "function" of sorts, and another a geometric object

some smart piece of shit a few hundred years ago named Descartes made that up-- the cartesian plane was a way to unify these fields of "algebra" and "geometry" together

okay, fast forward a few hundred years, and now we get to philosophize some more

[stone fulcrum]

He sounds like a pretty lit guy

[long beacon]

see, suppose someone gave you a few smooth manifolds

you might want to figure out what smooth functions exist on said manifolds

right?

[stone fulcrum]

Not skilled in differential geometry yet, when might a smooth function not exist on a smooth manifold?

[long beacon]

if you played with this long enough, you realize that you could distinguish these manifolds pretty much exactly by what functions existed on it

(there always will be)

[stone fulcrum]

But okay, knowing which functions fit is enough to determine manifold, I'll take your word for it

[long beacon]

and it turns out that in many situations, people started realizing that the geometry of the manifold could be determined completely by what kind of functions exist on it

(note note: the space of functions on a <insert adjective> manifold is a RING)

this is a sort of philosophical thing that turns out to be really useful-- you can study a geometric object not by staring at the object, but by studying it's FUNCTIONS

just like how in representation theory, you study a group by how it acts on stuff

here you study spaces by which functions live on it

so it became a thing to study these ALGEBRAIC RINGS OF FUNCTIONS as actual geometric objects in their own right

as a sort of mirror to geometry

this is effectively what algebraic geometry aims to do

[stone fulcrum]

Okay cool, so knowing functions is knowing manifolds

[long beacon]

right

so things you can do with manifolds and geometry, you can try to do in algebra

for example, you could try to take the cohomology of a ring and play with that

then you find that when people say "fields are 0D", it's because they probably haven't heard of etale cohomology yet

[stone fulcrum]

Ultimately, things that can happen on rings, correspond to things that happen on manifolds

[long beacon]

pretty much

[stone fulcrum]

Cool. I can't wait until I have anywhere near the mathematical maturity to even read this stuff

[solar wyvern]

what's a >0D field ooc

dimension of a ring just means length of a sequence of prime ideals ordered by inclusion @stone fulcrum, and since fields (typically) only have one nontrivial proper ideal {0}, it's said to be of krull dimension 0

idk etale coho

[long beacon]

well, sorta

dimension is an overloaded word in geometry

[solar wyvern]

idk any other dim used for rings tho

[long beacon]

there are a bunch of random ones in arithmetic geometry

krull dim is pretty general, but people in arithmetic geometry might be more interested in, say, etale cohomological dimension or motivic homotopical dimension, etc.

but yes, you're right that I probably should have just said krull dim

[solar wyvern]

also welcome to the server i guess

hope you enjoy your stay

[long beacon]

danke schon

[solar wyvern]

are you gonna explain etale coho & motives to us at some point

[bleak abyss]

Ayy welcome @long beacon! Now I have yet another person I can piss off with my random algebra questions

[primal mango]

if there's a set of 3 objects, then why does that imply an infinite amount of sets?

[chilly ocean]

Try formulating that problem again

[errant drum]

3 + 3 = 2 3's =  Ɛ3 = 8 = ∞

[long beacon]

@solar wyvern if you want :3

[daring wolf]

Is there an irreducible polynomial in Z_2[x] with exactly four nonzero terms?

trying to understand what does it mean by non-zero terms...

isn't there only 1 in this case since Z_2[x]={0,1}?

why would there be 4?

I am so confused

[stone fulcrum]

Remember that Z2[x] are all polynomials with coefficients in Z2

[daring wolf]

ah

[stone fulcrum]

x² + x + 1 is an element in Z2[x] for example

I'm not sure if that's reducible

[daring wolf]

yea i can figure out if things are reducible or not

4 non zero terms

does that just mean x^4+x^3+x^2+x

some sort in that form

with coefficients?

[stone fulcrum]

I think so.

Not completely sure but it makes sense

[daring wolf]

i mean powers can be other numbers too

hmmm

[stone fulcrum]

Note your only non-zero coefficient is 1

[daring wolf]

yea

lel english words are my worst enemies in math...

and definitions...

[solar wyvern]

@stone fulcrum u can brute force to check for low degree stuff like that

but there are various rules for reducibility

[stone fulcrum]

Yus. There is no root in Z2 so it's irreducible

[solar wyvern]

[stone fulcrum]

Note that in Z2
x² = x

[daring wolf]

yea

sigh... so confused about what non-zero terms mean lol

[solar wyvern]

it means

nonzero coefficients

[daring wolf]

wait

is that it?

[solar wyvern]

yeah

[daring wolf]

that'd make this question not too bad

[stone fulcrum]

I think you're right
x⁴ + x³ + x + 1
Is 4 non-zero coefficients

[solar wyvern]

the zero polynomial is \sum \alpha_k X^k but all the alphas are 0

[stone fulcrum]

Checking for irreducibility for polynomials that large is more difficult though

[daring wolf]

just brute force it 030

i mean f(0)=f(1)=1 then it is irreducible

[solar wyvern]

there are methods especially for specific forms

[daring wolf]

in this case

[stone fulcrum]

Actually is
x³ + x² + x + 1
Reducible?

[daring wolf]

wait

it is reducible

[stone fulcrum]

Yes, because x = 1 is a root

[chilly ocean]

Provide factor

[daring wolf]

4 nonzero terms is impossible

since your only options are 0 or 1

and 4 terms guarantees one of the brute force attempt to be 4 "1"s

so then 4 = 0 in Z_2

hence impossible

with 4 nonzero terms

[solar wyvern]

so what can u say about red/irred polynomials in F_2[x]

[stone fulcrum]

I suppose that's true. There's no irreducible members of Z2[x] with four terms

[daring wolf]

@solar wyvern

no idea...

[solar wyvern]

[daring wolf]

wait

F_2[x] is a field right?

[solar wyvern]

is it?

[daring wolf]

lemme check ..

[stone fulcrum]

F2 is a field F2[x] is a ring

[chilly ocean]

Wat

[solar wyvern]

f_2 is a ring too

also no spoilers

[stone fulcrum]

Wat

What's F2 am I derping?

[daring wolf]

I know it's a ring

[chilly ocean]

x^4+x^3+x^2+x

[daring wolf]

but I am not sure if it's a field

[solar wyvern]

u should check for urself then, and not listen to these 2

u know various things about which rings are fields I hope

like what's reducible, what's unit, what's irreducible

[daring wolf]

Q is

XD

[solar wyvern]

I'm assuming you know units are invertible elements

[stone fulcrum]

Is F2 not the same as Z2?

[daring wolf]

Z^x _p

i don't think F_2 is same as Z_2

[solar wyvern]

@daring wolf what are you defining F_p as?

[stone fulcrum]

Nobody was using the term F2

[daring wolf]

i am confused now XD

[solar wyvern]

I did

[daring wolf]

F_p is Z_p?

[solar wyvern]

so you can actually show there's a field with q=p^n elements unique up to iso

[daring wolf]

yea... that

we were given the definition F_q <-> Z_q

where q=p^n

[solar wyvern]

that's wrong.

[daring wolf]

n belongs to Natural numbers

[solar wyvern]

on multiple levels

[daring wolf]

o.o

[solar wyvern]

so you're saying Z_9 is a field?

that is Z/9Z?

this should be a pretty quick verification

[daring wolf]

trying to get the image over to discord

discord not liking it

[solar wyvern]

use words

[daring wolf]

o.o

[solar wyvern]

what's the characteristic of Z/9Z

[daring wolf]

sorry don't understand what it means by characteristic ><

man I have too many knowledge gaps lel

[solar wyvern]

number of times you add 1 before u get 0

[daring wolf]

OH

same as order?

[solar wyvern]

it's only for 1 in your unital ring, but that's the idea

[daring wolf]

oh ok nvm...

well that'd be 9 times wouldn't it?

[solar wyvern]

can a field have nonprime (or nonzero) characteristic?

[daring wolf]

no...?

[solar wyvern]

why not

[daring wolf]

no idea...? 😛

well reading it up right now

Oh no zero divisors

[stone fulcrum]

Yus, a field will have no zero divisisors because every element has an inverse.

Element is invertible ⇔ Element is not a zero divisor

[daring wolf]

Yea 0.0 oof

How did I forget about that lol

wait so Z_9 isn't a field

[stone fulcrum]

Nop
Z_p for a prime p IS a field though

You can also construct a field of 9 elements, but it's not Z9

[daring wolf]

yea i am looking at that right now

and it's confusing

so a field Z_11

sorry

[stone fulcrum]

Also useful, finite fields are unique. There's only one field of n elements

[daring wolf]

a field F_121 defined by Z_11

what does that mean

[thorny slate]

F121 is a field extension of Z11

[stone fulcrum]

Do you know how quotienting by an ideal works?

[daring wolf]

nope -.-

quotienting by an ideal?

[stone fulcrum]

You'll want to get there first, then making those fields will be easy to understand

[sonic current]

huh - quick question:
G group, H and K subgroups of G, in particular H normal subgroup of K. For any given subgroup L
Must show H ∩ L ⊳ K ∩ L

[daring wolf]

what's K o.o

[sonic current]

fixed!

I've done some work but I can only show, through the normality definition xH ∩ Lx^-1, where x is an element of K ∩ L

that from xH ∩ Lx^-1 I can conclude an element of H is in H

but I must have it in L as well

it falls apart when I use the H ⊳K hypothesis

because for g in H and k in K, the substitution kg = g' k doesn't guarantee the element g' should be in L anymore

[oblique river]

you only have to conjugate by an element of H intersect L

everything in sight lives in L

[sonic current]

sure

I can see that

[oblique river]

maybe I'm confused by the question: If h in H cap L and k in K cap L, you want to show that hkh^(-1) in K cap L

[sonic current]

hmm

just a sec

[oblique river]

it's in K because K is normal in H

it's in L because everything is in L

h and k are both in L, so hkh^(-1) is, too

[sonic current]

ok I'll try and write down what I did

we want to prove xHx^-1 ∈ H ∩ L, where x ∈ K ∩ L

[oblique river]

no

that's not what you want to prove

[sonic current]

H ∩ L ⊳ K ∩ L

is it not?

[oblique river]

uh you have that triangle backward I think

[sonic current]

ah it's possible, sorry

[oblique river]

what you wrote means "the right is normal in the left"

but in any case, that's still not what you want to prove

[sonic current]

it's precisely the opposite

[oblique river]

(but btw if you look at my solution, I'm doing what you wrote with the arrow in the other direction)

[sonic current]

I want to prove H ∩ L is normal in K ∩ L

[oblique river]

write down what that means

[sonic current]

I suppose that, for any x in K ∩ L, then
xgx^-1 ∈ H ∩ L, where x ∈ K ∩ L, for g ∈ H ∩ L

[daring wolf]

@stone fulcrum to find total number of generators in F_121 defined by a quadratic that belongs to Z_11 we have to calculate phi(121-1)?

is that right?

[oblique river]

no, you have something backward

to say A is normal in B means that bab^(-1) is in A for all a in A and b in B

wait

nvm sorry I had it backward that time :)

yes that is correct :)

[sonic current]

alright

so I start with xgx^-1

and all of those are in L, right? I can see that

[oblique river]

yes

[sonic current]

what I did next was use the H normal in K hypothesis, to say xg = g'x

and so g'xx^-1 = g' which is an element of H

but not necessarily of L

[oblique river]

what?

you just said

that xgx^(-1) was in L

and that's what you care about

[sonic current]

but why would g' be?

[oblique river]

xgx^(-1) = g'

[sonic current]

I don't follow that

it seems like it could very well be an element outside of L

[oblique river]

what

xg = g'x

multiply by x^(-1) on the right

[sonic current]

yes, I can see that...but it feels so wrong

[oblique river]

w-why?

that's literally the proof

[sonic current]

😦

feels like it came out of the blue

[oblique river]

what?

that's literally what normality is

[sonic current]

I don't think I could argue g' is in L too

[oblique river]

normality means "you can conjugate by something and it stays in the subgroup"

[sonic current]

sure

[oblique river]

you don't even need to give that "something" a name

what do you mean? g' is literally equal to xgx^(-1)

[sonic current]

but couldn't this g' element be one in K but outside of L?

[oblique river]

which you already told me (correctly) was in L

[sonic current]

argh

I want to feel this is true but I feel there's something missing in the proof

[oblique river]

no

[sonic current]

even though I totally see your point

[oblique river]

ther'es nothing missing

[sonic current]

g' = xgx^-1 makes it very clear

but I feel like it is a leap of faith or something.

[oblique river]

what?

you literally just multiplied by x^(-1)

which is totally valid in a group

maybe don't give g' a name

here is how I would write it:

Let x in K cap L and g in H cap L. We want to show that xgx^(-1) is in H cap L. This means we need to show that xgx^(-1) is in H and in L.

Step 1: showing it is in H. Since x in K and g in H, and since H is normal in K, we know that xgx^(-1) is in H. [[This is the definition of normal subgroups]]

Step 2: showing it is in L: Both x and g are in L, so x^(-1) is also in L, and thus xgx^(-1) is in L.

[sonic current]

ok I thought about that too

[oblique river]

if you want to call that element g', you can

or you can call it 🍌

[sonic current]

but I still found it a little problematic

[oblique river]

but the glyph you use to represent it doesn't change the fact that it's equal to xgx^(-1) which is in both K and L

[sonic current]

sure

I guess I can live with that lol

but I thought that doing that in two steps would not imply necessarily they would be the same element

but oh well. I think I get your point

thanks so much, this was bugging me a lot

[oblique river]

what?

xgx^(-1) is equal to xgx^(-1)

it doesn't matter if you wrote one of them 3 minutes after the other one

they are the same element

[sonic current]

yes, but in the way I was doing it they would not be

[oblique river]

no

they are the same

it's not about "how you do it"

it doesn't matter if at first you write "xgx^(-1) = g' " and then a minute later you write "xgx^(-1) = 🍌 "

then g' = 🍌

they are the same

[sonic current]

lol

thanks so much, buddy

[oblique river]

so whatever you know about one

you knwo about the other

sorry I'm going on abotu this

I'm just concerned that you don't get how groups work

since you were worried about multiplying by x^(-1) lol

if you have more questions, feel free to ask

[sonic current]

nah, i'm good in that part

that multiplication was never a problem

idk. bit tired

[daring wolf]

I do have questions 030

But about another topic

[sonic current]

go ahead, @daring wolf

thanks so much @oblique river

[daring wolf]

Is a field defined by a quadratic equation that belongs to Z_p

Same as a quadratic field 0.0?

[oblique river]

np, glad it's all sorted out

[daring wolf]

Or am I confused

[oblique river]

uhh depending on the context, "quadratic field" could mean different things

but yes what you said sounds right, as long as that polynomial is irreducible

[daring wolf]

Yea I am needing to find the number of generators in it

And it’s abnormally large

It’s F_121

Do I just go about it with the phi function

[oblique river]

you want the number of generators of F_121 as a field?

No, the phi function is about cyclic groups Z/nZ

F_121 is not the same as Z/121Z

[daring wolf]

That’s defined by a quadratic function that belongs to Z_11

[oblique river]

here is the easiest way to count it:

pick an element of F_121. It generates some subfield of F_121. The only subfields of F_121 are itself and F_11

so if you pick 11 elements, you'll get F_11

and the other 121-11 must generate all of F_121

(well I guess technically, 10 of those elements generate F_11 and the 11th element is 0 which doesn't generate anything other than itself)

[daring wolf]

Oooooo

So confused sorry ><

basically this is what the question is asking

Consider the field F_121 defined by the quadratic x^2+x+7 belongs to Z_11 [x]