groups-rings-fields

[vital quail]

(this is routine to verify)

[south temple]

I mean, I can see that if f is an isomorphism and F is a field then R is a field by just proving it directly

[vital quail]

sure

[south temple]

but I'm trying to see why your reasoning holds

[vital quail]

do you understand my line of reasoning?

are you trying to get intuition?

[south temple]

so I is ideal of R, meaning that preimage of I is an ideal

this ideal can only be {0} or all of F

[vital quail]

yes

if it's {0} then I is just {0} in R

so then if it's F

[south temple]

because homomorphisms map 0 to 0 right?

[vital quail]

yes

then if the preimage is F

the thing is, our map is surjective

so the image of F is just R

i.e. I = R

[south temple]

ahhh ok

so we're proving any nontrivial ideal is all of R

this means that R is a field

[vital quail]

yes

[south temple]

ok I get your reasoning entirely now

[vital quail]

you can also show this by explicitly finding an inverse for any nonzero r in R

since f is surjective, we can write r = f(a) and then f(a^-1) is our desired inverse in R

[south temple]

thank you for the help and the explanation, I really appreciate it

[vital quail]

np

[cyan marten]

Can someone explain why we care about regular orbits of a group under the action of some other group?

I am studying of M. Isaacs' Finite Group Theory and he mentions many conditions that ensure that a subgroup L of Aut G has a regular orbit in G.

[vital quail]

what do you mean by regular orbit? @cyan marten (i haven't seen that term 'regular' before)

[cyan marten]

The action of G on X is said to be regular if it is transitive and fixed-point-free. This is equivalent to saying it's equivalent to the action of G on itself by left multiplication.

I think of it as follows: If G is a semidirect product of N and H (N being the normal subgroup) and H has a regular orbit in N, then there is a subset of N consisting of H-conjugates and no element of H centralizes any of them. This seems rather convoluted to me.

[chrome hinge]

Hii i was trying to show $(x^2+x+1)$ is a maximal ideal in $F_{5}[X]$

[cloud walrus]

Maikel

[chrome hinge]

Thought it could be simple using the Euclidean Algorithm

But if im not wrong i ended like proving every polynomial in F_5[X] was divided by x^2+x+1...

[next obsidian]

So

F_5[x] is a euclidean domain, so any prime ideal is maximal (this is true in any PID)

So it suffices to show that ideal is prime, aka that x^2 + x + 1 is irreducible

You can do this two ways, pretend it was able to be written as (ax + b)(cx + d) then expand that, and show this doesn’t work

Or, in any ring at all, a degree <= 3 polynomial is reducible implies it has a root

So there’s only 5 values you can plug in (since we are working in F_5) so you can plug in all of those values and show the polynomial doesn’t vanish on any of them

Either way you’ll show that x^2 + x + 1 is irreducible <==> prime (we are in a Euclidean domain) implying that (x^2 + x + 1) is a prime ideal <==> maximal ideal (once again, Euclidean domain)

[steep hull]

x^2+x+1 is reducible mod p iff p is 3 or 1 mod 3

More generally

[chrome hinge]

Hmm

In a PID maximal ideal iff prime ideal correct?

And iff the generator is prime iff it is irreducible

Sorry, im just getting started with this stuff, trying to keep the pace

I mean i get you can show the polynomial is irreducible and youre done (if my statement obove is correct, which im mostly sure it is)

But can you show the ideal is maximal by using the euclidean algorithm?

Like, taking an ideal I which contains (x^2+x+1) and is not the whole ring, and showing every element in I must be divided by x^2+x+1

[golden pasture]

uh kinda hard tbh

like try to think of it over say

Z[x]

would your idea work

[sturdy marsh]

It isnt very complicated to show that nonzero primes are maximal in a PID

[golden pasture]

ok have awoken now by euclidean alg do you mean finding rhe gcd of 2 polynomials

if so right there is a alg to actually do it which is along the lines of taking gcd with x^{p^d}-x or smt

[sturdy marsh]

if I = (f) contained (x^2 + x + 1), then f divides x^2 + x + 1

which implies that f = x^2 + x + 1 or f is a unit

so I = (x^2 + x + 1) or I = R

so the ideal is maximal

(note PIDs are UFDs)

I don't think the Euclidean algorithm is in play at all

[golden pasture]

i think the qn is how to show it is irreducible

[sturdy marsh]

oh

[tacit saffron]

i gots a proof "prove that the ideal <x^2 + 1> is prime, but not maximal in Z[x]"
ik Z[x] isn't a field and x^2+1 is irreducible in Z[x],
and for <x^2 + 1> to be prime, Z[x]/<x^2 + 1> has to be an integral domain,
and for <x^2 + 1> to be maximal, Z[x]/<x^2 + 1> has to be a field, so I have to prove Z[x]/<x^2 + 1> is an integral domain

i was told to assume it isn't prime, because it implies x^2+1 isnt irreducible?
and its a ring, and i just need to show it contains unity and has no zero divisors? if the ring has zero divisors means that there exists an a and b not equal to 0 where ab=0. im not really sure what to do with all this

[sturdy marsh]

just do it directly

assume you have polynomials f and g such that fg is in (x^2+1)

oh wait

do you know that Z[x] is a unique factorization domain?

[tacit saffron]

im not familiar with that term

[sturdy marsh]

Do you know that you can factor a polynomial over Z into irreducibles uniqueley

up to units

[tacit saffron]

from what google saying, i think so? yes

[sturdy marsh]

okay, then factor f and g

and see what happens

[dim escarp]

Just got to this term (perhaps used nonsensically), "algebraic" and "co-algebraic".

What field is it originated from and what do these mean?

Just wanna see if someone actually used math terms correctly or if it is just another wank

[sturdy marsh]

algebraic has a lot of meanings

depends on the context

i assume the origin is algebra lol

[dim escarp]

Oof.. yea

[tacit saffron]

wdym factor, like factor a f(x) = anx^n+a(n-1)x^(n-1)+a1x+a0 and g(x) = bmx^m+b(m-1)x^(m-1)+b1x+b0

[sturdy marsh]

no, f = (p_1)(p_2)...(p_n) and g = (q_1)(q_2)...(q_r)

p_i and q_i irreducible

then (h)(x^2 + 1) = fg = (p_1)(p_2)...(p_n)(q_1)(q_2)...(q_r)

and now use that the factorization is unique

[tacit saffron]

am i trynna show that this is isomorphic to Z which is an integral domain

[sturdy marsh]

the quotient?

it is not iso to Z

[tacit saffron]

frig

[sturdy marsh]

do you see why?

[tacit saffron]

sadly no

[sturdy marsh]

assume X mapped to some integer

under the iso

what would happen?

[tacit saffron]

i meant Z[x] is that still wrong

[sturdy marsh]

yup

again, do you see why?

[tacit saffron]

no

idk what would happen

[sturdy marsh]

isomorphisms need to be injective

ring map takes 1 to 1

so n to n

where can X go?

[tacit saffron]

injective other way to say one to one right

[sturdy marsh]

yup

[tacit saffron]

uh itself?

[sturdy marsh]

aren't we talking about a map from X to Z

Z[X] to Z sorry

[tacit saffron]

sorry i didn't recognize just X, but i still be confused

im sorry if my monkey brain make stroke

[sturdy marsh]

nah this stuff can be very confusing when you're just starting out

strokes/brainfarts are normal and expected

[tacit saffron]

i mean more make u stroke

u v patient

i don't know where X Z[X] goes

[sturdy marsh]

it has to go to some integer right

as those are the elements of Z

[tacit saffron]

cuz Z?

ok

i was just over thinking fric

[sturdy marsh]

so not injective

[tacit saffron]

why not

[sturdy marsh]

because n in Z[X] also goes to the same thing

[tacit saffron]

are some elements of the codomain the image of more than one element of the domain

oh ok

[sturdy marsh]

I guess it's pretty useful to get this early on

there's a unique map from Z to any ring

[tacit saffron]

i have hard time internalizing some of it which is quite a handicap when i need to review basics everytime

[sturdy marsh]

as a map from Z is determined by where 1 goes, and 1 must go to 1 (in the other ring)

and mapping out of the polynomial ring is sorta similar to mapping out of a vector space

the image of X determines the map

(in the same way the image of basis elts determine a vector space map)

aluffi's chapter on rings is pretty good at explaining a lot of this

[tacit saffron]

i'll take a look into that we use artin in my class

[sturdy marsh]

havent read artin

dunno anything about it

[tacit saffron]

ye i got no reference point

[tacit saffron]

idk how i'd use this

[sturdy marsh]

try doing it just for integers first

prove that if 3 divides nm, then 3 divides n or 3 divides m

and repeat the same proof for polynomials

[tacit saffron]

why couldn't 3 divide n and m

[sturdy marsh]

it could

[tacit saffron]

ok yeah

[sturdy marsh]

or can mean both

[tacit saffron]

right

idkhow id prove that cuz it makes logical sense

if it divides nm part of nm's prime factorization must have 3, which would then be a factor of n or m

wait does h divide fg?

from what u defined?

[sturdy marsh]

yup

[tacit saffron]

and h divides f or h divides g

but h is the factors of f and g

it has more factors

[sturdy marsh]

we dont care about h

the fact that (x^2 +1) divides fg implies that there is some h such that (x^2+1)h = fg

[tacit saffron]

ohHHH

ok so (x^2+1) divides f or g

[sturdy marsh]

you need to prove that

[tacit saffron]

does that prove that its prime

[sturdy marsh]

yes

[tacit saffron]

what about "not maximal in Z[x]"

[sturdy marsh]

try to find a prime ideal that contains it

[tacit saffron]

im still kinda stumped on the (x^2+1) divides f or g proof

[steady axle]

x^2+1 is irreducible in Q[x]

[tacit saffron]

but using that is basically using what the proof is asking, because that should mean its prime? "prove that the ideal <x^2 + 1> is prime, but not maximal in Z[x]"

[south temple]

@tacit saffron have you learned about the Rational Root Theorem?

[tacit saffron]

like the rational root test for irreducibility? where if a/b in Q is a root of f, then a divides the the x^0 coefficient, and b divides the x^n coefficient?

[south temple]

yes

use that to show that x^2 + 1 is irreducible in Q[x]

[chilly ocean]

Can't you argue by saying that in C[x] the only roots are +/- i, which aren't in Q?

[south temple]

you could do that, you should first show that the only factors of $x^2 + 1$ have degree 1 and also show that $\pm i$ are indeed the only roots of $x^2 + 1$

[cloud walrus]

vov&sons

[south temple]

then it follows that $x^2 + 1$ is divisible by only $x + i$ and $x - i$ which are not elements of $\mathbb{Q}[x]$ or $\mathbb{Z}[x]$

[cloud walrus]

vov&sons

[south temple]

but Rational Root Thrm is easier imo

[tacit saffron]

confused how this leads me to prove that ideal <x^2+1> is prime

[south temple]

it shows that $x^2 + 1$ is irreducible

[cloud walrus]

vov&sons

[tacit saffron]

so then wouldn't my proof just be "x^2+1 is irreducible so its prime"

[south temple]

have you shown in class that if $f\in\mathbb{Q}[x]$ is irreducible then it is prime?

[cloud walrus]

vov&sons

[south temple]

if yes, apply this theorem

if not, you should try proving it --- it's very similar to the same theorem for integers ($p\in\mathbb{Z}$ is prime iff it is irreducible)

[cloud walrus]

vov&sons

[tacit saffron]

we have shown that in class

[south temple]

perfect, so use this fact to finish your proof

[tacit saffron]

but where does it the proof start

this part of the proof is literally just applying that one theorem

(proof being "prove that the ideal <x^2 + 1> is prime, but not maximal in Z[x]")

[south temple]

you've been explained how to start several times

start by showing that your polynomial is irreducible

[tacit saffron]

but that just means its prime

and bam i've down that first half

my confusion comes from the fact that i would have thought theres a bit more to it idk

[south temple]

suppose that $fg \in \langle x^2 + 1 \rangle$. then $x^2 + 1$ divides $fg$

[cloud walrus]

vov&sons

[south temple]

what can you say about either $f$ or $g$

[cloud walrus]

vov&sons

[tacit saffron]

either is divided by x^2 +1

[south temple]

Ok so what does that tell you about your ideal

I guess let's first suppose (without loss of generality) that x^2 + 1 divides f

What can you conclude about f?

[tacit saffron]

eeee its reducible?

[south temple]

Sure, is f in our ideal though?

[tacit saffron]

ye

[south temple]

f is not necessarily reducible though, my bad

It's possible that f is x^2 + 1

[tacit saffron]

oh yeah

[south temple]

Anyways, finish your argument and conclude that x^2 + 1 generates a prime ideal

[tacit saffron]

idk if its me overthinking things or the fact that i haven't slept yet but i don't understand how i take x^2+1 being irreducible then being prime and that if x^2+1 divides fg, then either f or g is divided by x^2+1, and therefore in the ideal and connect them to that end goal

probably both

[golden pasture]

you know that x^2+1 must split into lower degrees if it is reducible

lower degree = linear factors

but there are no integer roots

[tacit saffron]

but its irreducible?

[delicate bloom]

if it has a root then x^2=-1 means x has no power of 2 in its numerator or denominator, so it's odd

but x^2+1 != 0 mod 4 when x is odd

so it's irreducible

[tacit saffron]

i still confused how this brings me to it being a prime ideal

[golden pasture]

so whats a prime ideal right

Z[x]/I should be a domain

so essentially it is saying you cant find 2 polynomials multiply to become a multiple of x^2+1

but because Z[x] is unique factoring domain

it implies that if we find fg a multiple of x^2+1, then we can divide away appropriately to find f'g' = x^2+1

it is like

Show 3 is prime

If we have ab in (3), we can divide ab by ab/3 appropriately to get a'b'=3

so for instance

86 is a multiple of 3
we have
86/16 = (8/8)(6/2) = 3

or say like 1530 is a multiple of 9 then we essentially get 33

rough idea

[tacit saffron]

wait then how is it possible to have an f(x) and g(x) where fg=n(x^2+1), theres no way f or g could be of form (ax+b) cuz its irreducible, can't be a multiple

[golden pasture]

f=x^2+1, g=2

[tacit saffron]

is that where the if x^2+1 divides fg, then it divides f or g comes from?

like cuz x^2+1 is irreducible in Z thats the only possible way for f and g to be

[golden pasture]

yea another way to state (x^2+1) is prime is if fg=n(x^2+1), then x^2+1 divides f or x^2+1 divides g

[south temple]

You should really review your definitions. I think all the trouble you're having is due to not being familiar with the terminology

[tacit saffron]

yes ik this is something i've struggled with majority of my academic career

[golden pasture]

the main idea here is that Z[x] is unique factoring domain, so we only need to show x^2+1 is irreducible

maybe a more direct proof is like

suppose fg=n(x^2+1)

f = prod p_i^e_i
g = prod q_i^e_i

and show that if x^2+1 is irreducible iff it occurs in at least one of the p/q

[tacit saffron]

what is prod

[golden pasture]

product

$\prod p_i^{e_i}$

[cloud walrus]

可爱的猫（＾∇＾）

[tacit saffron]

eeee never seen that symbol

[golden pasture]

wait

srs?

[tacit saffron]

mhm

[golden pasture]

ever seen like

$n!=\prod_{i=1}^ni$

[cloud walrus]

可爱的猫（＾∇＾）

[tacit saffron]

i have enough intuition to read what that means but i've never seen that no

[golden pasture]

ah

i thought most ppl take calc before alg🤔

[carmine fossil]

Are you familiar with loops?

[golden pasture]

loops?

[tacit saffron]

i've taken multi never seen that

[golden pasture]

oh

strange

[tacit saffron]

did a quick scroll through my textbook, its first instance is in galois theory in a couple chapters

[golden pasture]

ye it is used to define the norm

[tacit saffron]

f = prod p_i^e_i
g = prod q_i^e_i what do these mean

given i unfamiliar with notation

[golden pasture]

$f=p_1^{e_1}p_2^{e_2}\dots p_n^{e_n}$

[cloud walrus]

可爱的猫（＾∇＾）

[tacit saffron]

are p's polynomials?

and e different degrees?

[golden pasture]

for instance

$1728=2^63^3$

[cloud walrus]

可爱的猫（＾∇＾）

[tacit saffron]

ahhhhh

how you know that one off top of head

[golden pasture]

it is super commonly used

[tacit saffron]

ah

ok i also gotta prove <x^2+1> is not maximal in Z[x], so ik i gotta make an ideal that contains <x^2+1> that isn't the entire ring and isn't equal to <x^2+1> but im not really sure what that entails

[golden pasture]

show that prime ideals are maximal for rings with unique factorization

essentially it has dimension 1

[tacit saffron]

but im showing they aren't maximal

wait what

[golden pasture]

har

oh wait

im dumb

lol

ignore me

ok anyways

show that Z[x]/(x^2+1) is not a field

so like

say 1/2

is 2 invertible?

probably not

[tacit saffron]

ye integers don't have multiplicative inverse so the ring isn't a field

[golden pasture]

alternatively you can show that like (n,x^2+1) is an ideal containing (x^2+1) for any n not -1,0,1

[golden pasture]

probably way easier lol

[tacit saffron]

what does it mean when the ideal has two entries

[golden pasture]

uh

so what does (x^2+1) mean

[tacit saffron]

its likea

subset

thats like

3 ideal for set of multiples of 3

[golden pasture]

so what is it exactly

[tacit saffron]

i don't really know

i think of them like what normal groups are to groups

[golden pasture]

essentially it is the ideal generated by x^2+1

so it is (x^2+1)Z[x]

(f,g) as an ideal in R is just fR+gR

which in other words

all elements of the form fr+gr' where r and r' are arbitrary elements in R

[tacit saffron]

so for (n, x^2+1) is that same as (n)r + (x^2+1)r' for arbitrary elements r r' in Z[x]?

or it not work like that

[golden pasture]

yes

[tacit saffron]

how is (x^2+1) contained

if n isn't -1,0,1

r would be 0?

[golden pasture]

yes

essentially

[tacit saffron]

so is that enough to say (x^2+1) isn't maximal because its contained by (n,x^2+1) doesn't seem rigorous

and i think i'd need to show that the (n,x^2+1) is contained by the Z[x]?

[golden pasture]

yes

that is why i included n isnt 1 or -1

[tacit saffron]

and this enough to show it isn't maximal

[golden pasture]

yes

[tacit saffron]

there be a follow up proof after this

"prove more generally that no principal ideals in Z[x] are maximal"

couldn't i use the same logic for this

[golden pasture]

yes you can

[tacit saffron]

so if i have an f, and its degree 0, and not -1,0,1, (n,x) would be a bigger ideal thats not the whole ring (consists of only constant terms divisible by n), and if f has degree ≥1, then given a prime p that doesn't divide the coefficients of f, (p,f) would be a bigger ideal thats not the whole ring?

[golden pasture]

i dont get the (n,x) part but looks like idea is there. 2 cases to look at

ideals (p), p is in Z
ideals (f), f is nonconstant polynomial

[tacit saffron]

i have the n part just if i like f=2

[vital quail]

no the x in (n, x) will give you all the constant polys

since ideals are closed by multiplication from all elements of the ring

or rather the degree 1 ones lol

so like you will have polys of the form 2a + xP(x)

[tacit saffron]

wdym two cases

[golden pasture]

(6)

or (x+1)

[tacit saffron]

so p just being a constant?

[golden pasture]

yes

[tacit saffron]

isn't that kinda what i did

[vital quail]

by the way, you can show that R/I a field <=> I maximal, so what you are trying to prove is saying "Z[alpha] is not a field" where Z[alpha] is Z[x]/<g(x)>

[tacit saffron]

ok cool i gots good grasp of all this thank you all

cuz there like 7 of you that helped

i sleep now

[distant rain]

Hi everyone. Wanna ask something. Is it true any two cyclotomic fields of the same degree are isomorphic as fields? For an example the cyclotomic field obtained by adjoining primitive 3-rd root of unity and the one obtained from adjoining primitive 6-th root of unity have same degree. But are they necessarily isomorphic as fields?

[delicate bloom]

I think it might be easier to find a counter example with the 4th root of unity and 3rd root of unity

like try to construct where the isomorphism sends an element of the field with the 3rd root to where it sends i

square it, and see if you can pull out a contradiction

[vital quail]

you can

letting f(w) (w = e^i2pi/3) be a+bi you get

f(w^2) = (a^2 - b^2) + (2ab)i = f(-w-1) = -a - bi - 1

thus a = -1/2 from comparing Im parts but then you get b = sqrt(3/4) from subbing that in for Re parts

[distant rain]

Oh okay. I see

Thanks

[delicate bloom]

hey you weren't supposed to do it for him lol

[vital quail]

oh oops lol

[golden pasture]

more generally right

the degree is φ(n)

you can find many many numbers with φ(a)=φ(b)

simple way to see is

φ(n)<n

φ(n) is never perfect prime power

[delicate bloom]

depends on what a is, so yeah

k[a] is contained within k(a)

k(a) lets you divide by polynomials too

it might not be the case that f in k[a] implies 1/f is in k[a]

for instance 1-a in k[a] if you try to put 1/(1-a) into k[a] you might have to do 1+a+a^2+... which isn't a polynomial

yeah, well that's part of the distinction I'm describing, it depends on what kind of extension you're doing

do you know what the norm is

so just take the norm of an element you want to invert, then that's really a product of stuff that equals something in k

so you can divide out the piece of it you want to invert

which shows you have your inverses in k[a] which means it's the same as k(a)

like here, for instance

(a+b sqrt(d)) (a-b sqrt(d)) = a^2 -b^2 d

the field k adjoin the element a

oh k[a] means polynomials with a

I see your comment earlier was a misconception that it's separate I see

ok so like you can do R[i] let's say, and write out any polynomial with coefficients in R

it just happens we can always simplify down using i^2 = -1

so degree 2 polynomials and higher collapse down

because we have this relation

a+bi+ci^2+di^3 for instance

can just be rewritten as A+Bi

since i^2 is just -1 and i^3 = -i

yeah

yeah

yeah

when it's a transcendental extension

like, Q adjoin pi

there's no algebraic equation it can simplify down with

I believe so yeah

well now you have 1/(1-pi)

this is not an element of Q[pi]

so that shows they're not the same

I guess I'm sort of cheating, I haven't super rigorously shown that

[vital quail]

pi is transcendental so

it's the same as adjoining x

so yes

well

when you adjoin anything

you're really just taking a quotient of Q[x] / (minimal poly of x)

now when you adjoin something like pi this is just Q[x]

right, it is not algebraic

[delicate bloom]

yeah

like technically we should prove pi is transcendental maybe

but that's another thing lol

[vital quail]

idk if i've even seen such a proof

i think it involves like

integrals

🤮

[viscid pewter]

yeah it was very contorted

absolute madness

dunno how anyone coulda come up with it

[delicate bloom]

speaking of absolute madness

https://mattbaker.blog/2015/03/20/a-p-adic-proof-that-pi-is-transcendental/

I'm considering learning this if anyone wants to join me lol

[viscid pewter]

this is beyond my paygrade

[chilly ocean]

@delicate bloom tell me how to start learning p-adics

[golden pasture]

usually if k[a] is a field we write k(a)

[golden pasture]

join me in neukirch!

[chilly ocean]

whats neukrich

[delicate bloom]

I was gonna say pick up a book like Gouvea but that could work too

[chilly ocean]

alg nt ok

[golden pasture]

jacobson gal theory chapter has it

[vital quail]

ah okay

[chilly ocean]

which chapter has p adics?

[vital quail]

wait what

[golden pasture]

it is really like

a lot of sums

[vital quail]

why does he

have that

[golden pasture]

just sum sum sum

[chilly ocean]

ok 2nd

[vital quail]

why does he include that lol

[golden pasture]

cuz it is somewhat elementsry

[steep hull]

How’s Neukirch going now

[golden pasture]

like lindermann weiestrass

[vital quail]

yeah but a lot of things are elementary

doesn't mean they should be included

🤔

jacobson is already very concise

that's kind of surprising to me

[golden pasture]

at chapter 2.10ish rn havent rlly worked through like every exercise but poking at interesting ones and implementing things in sage lol

[golden pasture]

but linderman weiestrass is super powerful as well

[vital quail]

speaking of sage, i have started using it

[golden pasture]

nice

become a number theorist

[vital quail]

solving systems of equations via J.ideal() go brrrrrr

[golden pasture]

HAHAHA ikr

[vital quail]

lol

[golden pasture]

what is that called ah

[steep hull]

Cool

[golden pasture]

gröbner basis

op af

[vital quail]

is it

🤔

[golden pasture]

ye

[vital quail]

i just saw an article

write your system as an ideal ( , , ,) of dim 0 and then sage tells you

oh right you jsut go J.variety()

lol

[chilly ocean]

@golden pasture do you like that book and what chapter you on?

[golden pasture]

sage computes the gröbner basis

at chapter 2 rn

chapter 2 is much easier than chapter 1 for me lol

[chilly ocean]

@delicate bloom why gouvea or should I neukirch

[golden pasture]

cuz it is just algebra

[chilly ocean]

wtf chatper 1 is 100 pages

btw I was reading this different alg nt book

[delicate bloom]

well depends on what you wanna do, gouvea is just p-adics, neukirch is more than that

[chilly ocean]

lemme check whats the author

[golden pasture]

wait was chapter 1 100 pages

maybe that is why i took so long

[chilly ocean]

kenneth s williams introductory algebraic nt

[golden pasture]

at some point

you just get lost

neukirch's proofs are also super short lol

oh wow neukirch is long

[chilly ocean]

how fast did u get through the 1st chapter?

[golden pasture]

like

1.5 months?

[chilly ocean]

[golden pasture]

i mean i had a lot of background beforehand soz

[chilly ocean]

exactly

[golden pasture]

chapter 2 is like

breeeeeeze

[delicate bloom]

I read the beginning of chapter 1, I really liked what I read so far

[chilly ocean]

and it still took 1.5months

[golden pasture]

background is essentially sufficient haha

[delicate bloom]

like you said, short proofs are nice

[golden pasture]

i dont reach neukirch full time

[chilly ocean]

wait so what are prereqs for the book?

[golden pasture]

essentially nothing

neukirch intros it for you

[chilly ocean]

ok page 35 already looks scary and I havent seen much of this stuff

[delicate bloom]

yeah, read neukirch if you're fearless and don't have a problem with reading slowly

[chilly ocean]

k nvm its too hard

[golden pasture]

wdym did you read page 1

[chilly ocean]

first few pages are ez but it gets harder exponentially

@delicate bloom whats the connection between flowers and p-adics?

[delicate bloom]

lol they're both things I like

that's about it lol

you could consider p-adics to be like the leaves of an infinite tree, and trees are plants, and flowers are a kind of plant

haha

[chilly ocean]

mero discussing p-adics again

[golden pasture]

i consider p adics as completion of Z at p

[chilly ocean]

is neukirch good

[sturdy marsh]

yup

some of the problems are pretty hard

[gleaming quarry]

https://cdn.discordapp.com/attachments/486844829209198613/783410389592702976/unknown.png

I have to find the zero crossings of the polynomial in finite field F5[X], what'd be the right way to do it?

Inserting each element results in y = 0.

e.g. f(0) = 255 = 0

[delicate bloom]

not sure I know what you're asking here, sounds like you're asking how to factor this polynomial

[viscid pewter]

they want the roots

[delicate bloom]

first off factor out an X, end up with X* (X^4+4) so now you can try to think how to factor that, well 4=-1 so

hopefully you can factor X^4-1

[viscid pewter]

they put every element in and got 0

idk anything about fields but to me that suggests all the elements are a root

[delicate bloom]

well, I interpreted them as asking if there's a better way than brute forcing

so I'm showing how to factor

[gleaming quarry]

I know both ways

I'm confused by the amount of roots

[delicate bloom]

what do you mean

there are 5 roots

how are you counting them

[gleaming quarry]

shouldn't there be 9 roots?

[viscid pewter]

why

[gleaming quarry]

sec

well

[delicate bloom]

5th degree polynomial can't have more than 5 roots

there aren't more than 5 elements in F_5

[gleaming quarry]

because the area of the function extends into the negative aswell?

[viscid pewter]

doesn't matter

[gleaming quarry]

?

[delicate bloom]

-1=4 these are the same element

are you aware of this

2=-3

[gleaming quarry]

well yes...

[delicate bloom]

-100 = 0

[gleaming quarry]

It's kinda hard to accept

Yeh, makes sense. Thanks

[golden pasture]

yes

stop doing olyms

and do neukirch

[golden pasture]

ikr

[vital quail]

@golden pasture that's the same definition smh

p-adics as a completion and p-adics as a tree

[golden pasture]

thonkkkkk

no

[vital quail]

just v i s u a l i z e

[golden pasture]

thonkkk

wait

i visualize p adics as like

the point at infinity of a line

[vital quail]

each level corresponds to a residue class

[golden pasture]

lol

[vital quail]

of p^k

each p-adic is just a walk down the tree

with p options at each level

[chilly ocean]

@ merosity

oh he was right there

😌

[golden pasture]

p adics is a limit really

the true visualisation

[chilly ocean]

this is what algebraic geometers say their abstract nonsense's "geometry" is

[golden pasture]

look

it is literally true

you rarely care about the individual elements

you really only care about like

even in number theory ideals

a small point that is thickened really gives the DVR feel

[chilly ocean]

thicc points

[golden pasture]

mmmm

[vital quail]

@golden pasture there seems to be a mistake in your drawing

Spec Z is a 3-manifold after all

[golden pasture]

spec Z is a line

[vital quail]

https://ncatlab.org/nlab/show/Spec(Z)

[golden pasture]

stop this fake news

[vital quail]

[sturdy marsh]

ah yes, primes are knots in spec Z something something

[vital quail]

primes are knots

[golden pasture]

idk lol

[vital quail]

no need to say the rest

[golden pasture]

im not agpilled enuf

[vital quail]

R is just the terminal coalgebra of a certain endofunctor on Pos

[golden pasture]

thonk?

england

[sturdy marsh]

tmf is the connective cover of the global sections of the sheaf of E∞ ring spectra on the compactified moduli stack of elliptic curves

[maiden ocean]

stop posting "spec Z is a 3 manifold"

thats not even what that article says

[vital quail]

[maiden ocean]

it literally isnt

read it

[vital quail]

i know, but the joke is to say that it is

[maiden ocean]

the joke is bad

ur bad

cease

[vital quail]

look, this did not originate with me

[maiden ocean]

[vital quail]

i am simply copying [redacted]

[viscid pewter]

just to check

[golden pasture]

arch cringe

[viscid pewter]

algebraic geometry is where you give points coordinates right

:3

[golden pasture]

uh

very no

[maiden ocean]

hurb

[golden pasture]

ok you know like

[viscid pewter]

i meme

[golden pasture]

uh

[viscid pewter]

expend no effort

[golden pasture]

ok

specgluspec

[viscid pewter]

fianto eset?

[golden pasture]

wat

[viscid pewter]

u first

[golden pasture]

spec glue spec

pretty self explanatory

[viscid pewter]

does this

does this actually mean anything

or am i getting eulered

[golden pasture]

learn more algebra to find out!

[viscid pewter]

my one was gobbledygook btw

i thought we were posting randomness

[golden pasture]

take my free course today for $8.88 every 8 minutes!

[sturdy marsh]

you dont need to know much algebra

[golden pasture]

oh

[sturdy marsh]

try reading some classical AG kaisheng

[viscid pewter]

look

[sturdy marsh]

or riemann surfaces

really fun stuff

[viscid pewter]

rn i'm cutting my teeth on normal subgroups

[golden pasture]

i consider that some algebra

[viscid pewter]

and lagrange

[golden pasture]

like yea

[viscid pewter]

and isomorphism theorems

[golden pasture]

cuz compared to us

[viscid pewter]

i think that's gotta be later

[golden pasture]

it is really a lot more algebra

xd

[sturdy marsh]

there's a riemann surfaces proof of abel-ruffini

[solemn rain]

whats algebraic topologyh

[sturdy marsh]

tmf is the connective cover of the global sections of the sheaf of E∞ ring spectra on the compactified moduli stack of elliptic curves

[viscid pewter]

it's like when you have a piece of paper with letters on them

and then you tape it into a mobius strip

[golden pasture]

literally wrong channel

#point-set-topology

[golden pasture]

wai what

[chilly ocean]

@solemn rain yo homeboy where did you learn algebraic topology stuff from?

[sturdy marsh]

yeah arnold wrote a book on it

[golden pasture]

:o

[sturdy marsh]

for high schoolers

[viscid pewter]

no

[golden pasture]

woa

[viscid pewter]

cartography

[maiden ocean]

lmao

[viscid pewter]

i think

[golden pasture]

whats the book called

[sturdy marsh]

abels theorem through problems or something like that

(not to be confused with the other abels theorem)

[chilly ocean]

ariana make me a list of books one should read (ordered) to learn algebra

[golden pasture]

ah

do you like

literally use deck transforms

or is it more involved

[golden pasture]

everyone is different

[sturdy marsh]

there is more

[chilly ocean]

im you

[golden pasture]

oo

[viscid pewter]

look

do as i say

not as i do

[sturdy marsh]

it isnt just a flipped version of the usual galois theory thing

[golden pasture]

icicic

[viscid pewter]

actually don't do what i say do what i meant to say

[sturdy marsh]

(I've been told, I havent read the book)

[golden pasture]

oo found it

lemme just scan through

the section before it is literally monodromy group

[sturdy marsh]

yup

[golden pasture]

ya it looks like a transformed version of galois' argument

instead of looking at field extensions

you look at monodromy groups

[sturdy marsh]

instead of field extensions you look at riemann surfaces ye

[golden pasture]

kinda cute tho tbh

[sturdy marsh]

arnold writes really good books

and the dude solved one of hilberts problems at 19 🤯

[golden pasture]

🤯

omg

[sturdy marsh]

drinfeld proved GL2 Langlands at 20 lmao

[chilly ocean]

hey guys can I ask a question, do credit card numbers, transactions and or isbn numbers involve group theory in anyway?

[sturdy marsh]

maybe, at least that's what I tell my dates

[chilly ocean]

lol

[sturdy marsh]

(that statement may be vacuously true)

[golden pasture]

geez wtf

feeling super insecure now time to quit math

/j

some ppl are just born chads

i dont even get langlands

[sturdy marsh]

Anyway, @chilly ocean groups do turn up in ecryption algorithms

[sturdy marsh]

same

[chilly ocean]

what do you mean by encryption?

[golden pasture]

how tf does some 20 year old solve it ???? scary

essentially diffie hellman protocol

you do use like F_{2^n} in some areas as well

[sturdy marsh]

you should probably just look it up, there are quite a few easy to read articles and videos online

RSA might be a good starting point

[sturdy marsh]

ye im 20 and cant do shit lmao

[vital quail]

i recommend seeing a bowel doctor

[sturdy marsh]

I was trying to come up with a joke about lax monoidal functors but failed

bet Drinfeld could

ffs

[golden pasture]

sameeee

im essentially 20 now

sad

[vital quail]

@golden pasture

no ur not

[golden pasture]

yes i am 3 years wht can i do

[vital quail]

bru

3 years is like 3/2 the time i've spent doing math

[golden pasture]

by the time i finish the books i want to read im 20+ already

[vital quail]

okay

it's till 40 for fields tho

[sturdy marsh]

you can do a shit ton is 3 years lol

[golden pasture]

too much pro ppl tbh and too much stress

should prob delete discord

productivity increase by 300%

[sturdy marsh]

same (maybe a 3/1.7 idk)

[golden pasture]

urgh can i delete high school

[vital quail]

@sturdy marsh how old are you

[sturdy marsh]

20

[vital quail]

so how did you come about math

[sturdy marsh]

read random shit online

[golden pasture]

same

[vital quail]

no but llike

how did you first get into math

1.7 years ago

[golden pasture]

wait 3/2*3=9/2=4.5

thonk?

[golden pasture]

oh nvm

[sturdy marsh]

oh wait wtf

okay miscalculated

[golden pasture]

idk fractions anymore

[sturdy marsh]

I have been doing math for 2.5 yrs

[sturdy marsh]

wanted to do physics, read math, math cool

[golden pasture]

oh wait i actually havent been doing math for that long wtf

i only got jacobson at like last year probably september time

thonk?

[maiden ocean]

ari didnt u do analysis

for a while

[golden pasture]

well yea

true

[maiden ocean]

smh

[golden pasture]

i had a lot of analysis background lol

[maiden ocean]

misrepresentation!

[sturdy marsh]

If you have 3 yrs to 20, you'll know a lot of stuff at 20 if you keep going at the same pace

[golden pasture]

didn’t really follow books tho

[maiden ocean]

i started at the end of freshman year so 1.5 years

[golden pasture]

time feels slow and fast

[maiden ocean]

i started from i guess

not knowing calc

[golden pasture]

how am i already almost done with am insane

[golden pasture]

damn

[vital quail]

I came on this server but I didn't start my first math book until like

hmm

december 2019 maybe

in terms of pure math that is

like, random other stuff before that inconsistently

but not really pure math before that

my background in december 2019 was like, a tiny bit of computational linear algebra, up to calc 2

that's it

[golden pasture]

my anal was super like just reading through se and random lecture notes lol

idek if i actually know analysis tbh

[vital quail]

i am hoping that once i get help (im already seeing like a counselor so i suppose i am talking about medication) i'll finally be able to do math at the sort of pace i want to

[sturdy marsh]

good luck

[golden pasture]

i should too tbh

urgh difficult choices

[sturdy marsh]

I hope I can stop procrastinating lol

[golden pasture]

kek

delete discord twitter insta

and tiktok

[sturdy marsh]

maybe post corona

[golden pasture]

kinda scared if i get help i cant do math for a long time ngl cuz it happened who one of my friends who for forced long story should sleep instead

[chilly ocean]

[bronze trench]

Hey cool people! I'm struggling with something and I could use some help 🙂

I don't ask for a solution or anything, just for someone to help me understand the relevant things

(the prof uses normal series for a series in which each group is a normal subgroup of the previous one but not necessarily of the big one)

Ok for the first one
Fut from the fact that H is subnormal I can't quite prove the there is a composition series going up to H (so like G > ... >H) and one going from H to 1, and my intuition says that I'm supposed to find those and then "glue" them

I kinda want a result that goes like "If G is a group which admits a composition series then for every normal subgroup H of G, both H and G/H admit a composition series"

if that's true then I think the first one is done! But I can't quite prove that statement

is it even true? xD

[sturdy marsh]

it is true

[bronze trench]

ok that's a good start

thanks 🙂

[sturdy marsh]

it even goes the other way

if H and G/H have a compoisition series then G does

[bronze trench]

ok so it's like one of those G satisfies P iff H and G/H satisfy P?

[sturdy marsh]

yup

[bronze trench]

ok great those are cool results I guess

[sturdy marsh]

wait hold on

[bronze trench]

hm, ok, I think I have an idea but I'm not sure how to do it

basically given a normal series G > G_1 > ... > G_n = H > ... > 1 I want to say I can refine it to a composition series

[sturdy marsh]

i might have fucked up

[bronze trench]

ok that'd solve the question right away so this is too strong probably

[sturdy marsh]

composition series is where the quotients are simple right?

[bronze trench]

yes

[sturdy marsh]

and solvable is where you have a composition series where the quotients are abelian?

because the thing that I said is true for solvable groups

[bronze trench]

yep

and the thing with H and G/H in indeed true for solvable 😛

[sturdy marsh]

eh maybe the same idea works

[bronze trench]

we proved that in class

so yeah good idea

I'm going to look at the proof instead of crying about how I'm dumb and not even try... 😂

[sturdy marsh]

dont look

you have the right idea

paste the two composition series

[bronze trench]

no I meant at the proof for the solvable case and see if I can see if it's frue for composition too

[sturdy marsh]

that is the idea for the solvable case

[bronze trench]

and indeed it seems to be the exact same, I just gotta check the quotients are still simple but seems easy

[bronze trench]

ok so the gluing step is true I guess, now I need to try and do the hard part 😂

[sturdy marsh]

length?

nah that's easy too

what's the hard part?

part 2?

[bronze trench]

wait I'm very confused wait a sec

oh yeah I solved the first one already I gues...?

I'm not thingking straight sorry lol

yeah ok I'm done, I take the composition series for G/H (which exists by a proof like the one similar to the solvable case) and it's of the form G/H>G_1/G>...H/H and pull with the corrspondence theorem a thing going from G>G_1>....>H with simple quotients

then take the series for H and append it to the end

that's it right?

I'm basically just thinking out lout sorry bout that lol

[sturdy marsh]

yup

[bronze trench]

thanks for the help! Now I'll have to think about the second one for a bit and see if I manage to do it 🙂

[thorn delta]

where does the composition series for H come from?

[sturdy marsh]

pick a composition series for G

intersect with H

use 2nd iso thm

[bronze trench]

yeah

and corresp theorem for the quotient ig

[sturdy marsh]

wait hold on, H isnt normal in G

so no quotient

[thorn delta]

H is normal

oh, no it isn't lol

[sturdy marsh]

yeah you need to use the other normal series you have with H in it

[bronze trench]

I don't need H to be

I iteratively do it to each term in the series

G>G_1>....>H
I do one to get one in between G and G_1, then one between G_1 and G_2 and so on

But yeah I need to be careful

I wrote it wrong before

[cloud walrus]

slimvesus

[sturdy marsh]

find the roots

of the polynomial

you need to check if it is lol

there's no reason a priori

you just explained what needs to be done

now do it

they're asking you for an explicit map so that isnt going to cut it

(y^3 + y^2 + y + 1)(y-1) = y^4 - 1

might ease up computations

idk

you can always just bruteforce

every element of F' is a polynomial in y

of degree less than 3

so x maps to some polynomial

check relation to find coeffs

bruteforce is good

[white nymph]

can someone help me understand what I should be noticing once i show that f(X)/X^n is irred over Q[Y] via Eisenstein?

like how that helps me show f(X) irred over Q[X]

[white nymph]

i dont follow at all sadly. @chilly ocean

i was able to show the hint - that f(x)/x^n is irred over Q[Y]

i also dont follow T^2 = 1 because arent you saying Tf(x) = f(x)/x^n so would TTf(x) = T f(x)/x^n = f(x)/x^2n?

so if f(x)/x^n irred over q[y], then we can write f(x)/x^n = g(y) for some irred poly g. then we have f(x) = x^n g(y) = x^n g(1/x) but i feel like im not making any progress/not understanding

and skip the hint?

kk

so we get h(x)g(x) = x^n g(1/x)

for f(x) = h g

f(x)/x^n = a_0Y^n + a_1Y^(n-1) + ... + a_{n-1}Y + a_n = g(y) = g(1/x)

yeah

i see now

right right

i was dumb and called 2 functions g

you mean the g(1/x)

k

right

constants?

okay so i follow. it feels like theres something to prove with "inverting f is the same as (invert g)(invert h)", yes?

ohhh interesting. never seen this before as far as i can recall

so it hinges on what i have in quotes above

because we have that invert f is irred

so if f was reducible, f = gh... then invert f = invert g * invert h but since g,h not units, their inverts are not units as well

i have to find contradiction

ok

yeah yeah gotcha

i get paranoid

right

that helps a lot - thanks for sticking it out with me

[hybrid cove]

Hi, I don't really understand why the following statement is true:

If $(S,m)$ is a local noetherian ring and $M$ is an $S$-module then $Tor_i^S(M,S/m)$ is a finite length $S$-module

[cloud walrus]

rhylx

[hybrid cove]

If someone is familiar with homological algebra, I would appreciate an explanation of this statement 🙂

[golden pasture]

im not sure if this is true? cuz we can take the infinite direct sum of S/m

(this is equal to saying the m-torsion submodule has finite length so it would be true if M is finitely henerated)

[hybrid cove]

I don't really know

[golden pasture]

You would have to assume M is noetherian

[marsh fractal]

if i have a field

how do i show that ab = 0 = ba iff a or b is 0? 0 being the zero element

at the moment i have a method but i use 0 = (1-1) which isnt necessarily true i think

[delicate bloom]

what's your definition of a field

[carmine fossil]

Are there different definitions of a field

[delicate bloom]

well there's nothing to prove since I consider this to be part of the definition in the first place

[marsh fractal]

A set with 2 operations where they both are commutative, associative, have identity elements, inverse elements for every element, and distributivity holds

[delicate bloom]

"inverse elements for every element"

what does that tell you

[marsh fractal]

hmmm

[delicate bloom]

if a, b are nonzero, they have inverses

[marsh fractal]

yes

[delicate bloom]

so multiply by the inverse of (ab)

ab=0
1=0

[marsh fractal]

yeah but then you assume 0 multiply inverse of ab is 0 right?

or more like using the claim to prove it

[carmine fossil]

There is no multiplicative inverse of ab if ab=0

[marsh fractal]

ah right i see its contradiction

[stone fulcrum]

It's common to outright say "there's no zero divisors in a field" as part of the def. I'm suprised your book doesn't go that way, haha

But yeah that's the same thing

[delicate bloom]

yeah to me the multiplicative structure is a group without 0, so multiplying two things to get 0 is breaking the closure of that group lol

[marsh fractal]

yeah that actually makes more sense

field => ring => group

group is for + or x

so they each in their own set and the ring is just like the union of the two

and fields just add commutativity

does that make sense?

[delicate bloom]

not really no

cause there are a lot of commutative rings that aren't fields lol

[marsh fractal]

yeah but just for considering the field

[golden pasture]

essentially any interesting ring isnt a field

[delicate bloom]

incoming joke about how all rings are commutative

aww I misjudged

[golden pasture]

sad

[stone fulcrum]

There's non-commuative rings?!

[golden pasture]

😯

[delicate bloom]

all matrices are diagonal

[golden pasture]

true

[carmine fossil]

[hybrid cove]

Yep, you're right.

[golden pasture]

yay my one brain cell is still intact

[vital quail]

5/8 braincells 😳

[golden pasture]

😳

[stone fulcrum]

Just enough to make the group abelian

[chilly ocean]

Can someone help me with a project?
When would you ever use group theory in real life?

[viscid pewter]

heavy applications in particle physics i've heard

also chemistry with molecule symmetry and things

[carmine fossil]

Somewhere in quantum mechanics

[oblique river]

I'm a mathematician and I use group theory in my life

[carmine fossil]

How do you use it?

[oblique river]

to do math

[viscid pewter]

but is the life strictly real or over the complex numbers

[oblique river]

specifically number theory

[golden pasture]

too little structure

*much

[chilly ocean]

Suppose that an Abelian $p$-group $G$ can be written as a direct sum of countably many finite cyclic groups. Let $a_1,...,a_k$ be generators of $k$ distinct summands of this direct sum, and let these elements have orders $p^{n_1},...,p^{n_k}$ respectively. Then for what values of $m$ are $a_1,...,a_k$ linearly independent over $\mathbb{Z}_{p^m}$?

[cloud walrus]

lugita15

[chilly ocean]

I guess m>=max(n_k)?

[prime cloak]

Yo guys can you help me on this

[cloud walrus]

bdobba

[prime cloak]

Question 10 we need to show that the application is bijective

Knowing that it is a morphisme from G to G and G is a commutative group.
We also know that there exist a k such that x^p^k= e the neutral element

Tell me if you don't understand

And also you need to know that m doesn't divide p

[vital quail]

what does 'p-primaire' mean

[untold sapphire]

I can't read french sorry

[prime cloak]

It means that : for all element in G there exist an integer k>0 such that x^p^k =e

[vital quail]

for x |-> x^m to be an isomorphism, this is equivalent to m being coprime to |G|

[prime cloak]

The application x^m is going from G to G

What is comprime ?

[cloud walrus]

bdobba

[vital quail]

it is of course a homomorphism (can you verify this?) and if m is coprime to |G| then there's a y such that my = 1 (mod |G|)

so can you show that there is an inverse homomorphism

[prime cloak]

What is |G| please

[vital quail]

the order of G

[prime cloak]

I am a beginner sorry

[vital quail]

the size