groups-rings-fields

[rustic crown]

but $\beta = \frac{1 + \sqrt{3}i}{2} = \frac{1+\alpha^3}{2}$

[cloud walrus]

det

[rustic crown]

alpha is \sqrt[6]{3}*e^{pi i}/6

so you can generate beta using alpha, which mean splitting field is just Q(alpha)

which definitely has degree 6 as t^6+3 is irreducible

yea weird things could happen 😛

[chilly ocean]

is there any algebraic double extension that isnt equivalent to a simple one?

[hot lake]

F2(x²,y²) inside F2(x,y)

[rustic crown]

Zef beat me to it 😶

[chilly ocean]

what about over Q

[hot lake]

xD

[rustic crown]

finite + separable = simple!

any algebraic extension over a characteristic 0 field is separable!

as no irreducible can have repeated roots

othewise gcd(f, f') would give a non-trivial factor of f

[chilly ocean]

oh i think u talked about this before

[rustic crown]

oh my bad, we need the field to have infinite size

and it does

nvm

XD

[hot lake]

in the case of characteristic 0 and finite fields iirc every finite extension can be generated by one element

if that's what you were asking

[chilly ocean]

perfect fields or smtg right

[hot lake]

yeah I guess

[rustic crown]

for a second i thoguth that we might need infinite elements in F2 so as to guarantee that there are infinitely many intermediate fields, but we actually needed F2(x^2, y^2) to be infinte

in this case we could verify by hand though, as the generator would look like a + bx + cy + dxy for a, b, c, d in F2(x^2, y^2)

but its square lies in the base field

[rustic crown]

[hot lake]

yeah it's pretty spooky having infinitely many intermediate fields between those two

wait are there really infinitely many intermediate fields

i know there are if you take your coefficients in the algebraic closure of F2

[rustic crown]

yea F(x+cy) where F = F2(x^2, y^2) are different for differnt c in F

[hot lake]

finite fields are the easiest fields

[rustic crown]

yea finite fields are very cute!

i only like the fields where i can describe the algebraic closure without zorn

[hot lake]

um

I would disagree haha

[chilly ocean]

F_p is easy, but i think i have had to work with F_{p^2} on a few occasions. if i have to work with F_{p^3} i would go in a corner and cry

[hidden haven]

Proving primitivity by checking intermediate extensions

Why tho

[rustic crown]

isn't that just easier?

[hot lake]

for your polynomial x⁶+3 since you already know that given one root, all the others are polynomial in it, that forces the factorisation mod p to have all their factors with the same degree

[hidden haven]

Ig if you don't assume theorems

But finite fields are perfect and finite extensions are then always primitive

By primitive elt thm

[rustic crown]

but we were talking about Fp(x^p, y^p) --> Fp(x, y)

[hidden haven]

Which also gives a pretty constructive way to find that element

oof hold on

[rustic crown]

we had to show non-primitivity

[hidden haven]

ohh I see

[rustic crown]

and its much eaiser to show that by exhibiting infinitely many distinct intermediate fields

like you don't need to worry how the elements look like and all that

[hidden haven]

You can also do it by contradiction. If there were some z of degree p² then it would be a rational function in x and y but then raising z to the pth power you get a rational function in x^p and y^p

[chilly ocean]

yo erminant

[hidden haven]

Since frobenius map is a homomorphism

[rustic crown]

[hidden haven]

lol

F

[sullen island]

heya, i understand statement (ii) (it is immediate from the orthogonality of the columns of a character table), but im not understanding statement (i)

if we interpret the character table as a matrix, $\sum\limits_{\chi} \overline{\chi (g)} \chi(g)$ can be seen as the "square of the length of the column vector corresponding to $g$" in the character table, but how do we get that it is equal to $\frac{|G|}{c(g)}$?

[cloud walrus]

xy

[hot lake]

you can try to fudge the entries of the matrix in order to make the rows orthonormal and not just orthogonal

and then the columns will be orthonormal too

and that falls out

[sullen island]

huh, cool, i was trying to think of it in terms of the norm of a vector w.r.t to the inner product defined on class functions

ill try that i guess

[languid meteor]

Hey all, ive solved (a) and (b) and im trying to make some kind of dimension/cardinality argument for (c) but im not sure what direction im going in

I dont have any theorems as far as I can see about whether a polynomial has a root in an extension field depending on dimensions/cardinality

I know the dim(E) over k is 5

I know z^2 + z + 1 is irreducible, so if I were to construct a field extension with this polynomial it would be degree two

what does this actually tell me about the possibility of a zero?

[chilly ocean]

If L is a finite extension of K, f irreducible over K and degree of f is prime with [L:K] then f doesn't have roots over L either.

[rustic crown]

so say there was such a z, then [F2(z):F2] = 2 and this will 5 = [E:F2] = [E:F2(z)][F2(z):F2] = even

[languid meteor]

is F2(z), F_2/(q(z))? I couldnt quite get the notation K(z) from how my lecturer presented it. So the idea here is that if q has zeroes in E, then F2(z) < E which implies 5 is even which is a contradiction

[rustic crown]

if you have a field extension F --> E and z is an element of E then F(z) is defined as the small subfield of E which contains both F and z

Notice then we'll have the map F[x] --> F(z) sending x -> z. If z is algebraic, then this map will be surjective. And the kernel would be generated by the minimal polynomial of z over F say q(x).
First isomorphism theorem will give you the isomorphism F[x]/(q) --> F(z)

[rustic crown]

by < does you mean less than?

[languid meteor]

I mean subfield

[rustic crown]

yea then okie, because 2 will have to divide 5 which is just false!

so your question, F2(z) and F2[x]/(q(x)) aren't same, but are ismorphic

[languid meteor]

am I choosing z such that it's a zero of z^2 + z + 1 or is it any algebraic element?

[rustic crown]

to get this isomorphism any algebraic z works with minimal polynomial q(x)

[languid meteor]

gotcha, sorry my notes are a bit messy I think I have it now, thank you!

[rustic crown]

[languid meteor]

I think im missing the surjectivity condition for algebraic elements from my notes

I have injectivity for transcendental

[rustic crown]

yea think about it!

[languid meteor]

I cant figure out how we can guarantee the evaluation map gives us an inverse for z

[rustic crown]

if degree of q(x) is n, then notice that 1, z, ..., z^(n-1) is a basis for F(z) over F.

[languid meteor]

yeah but dont we only get this once we know F(z) is iso to F[x]/(q)?

which we got from the fact that the map is surjective

[rustic crown]

okay what about this

we had the map F[x] --> F(z) sending x to z

the kernel of this map would be (q(x))

This will give F[x]/(q(x)) --> F(z) which is a map between fields

now the image of this map is a field which contains F and also z! so the image should contain F(z) implying this map is also surjective!

[languid meteor]

gotcha, that makes perfect sense. Thanks for all the help I really appreciate it!

[rustic crown]

[chilly ocean]

Hi guys. Do u know what could this be?

https://gyazo.com/85dc09057e6f17ee1859c0118c23f04f

Talking about infinite matrixes

for $\alpha = 1/2, 1/3, 1/4$

[cloud walrus]

nosequepasa

[quartz lake]

Hey all, would someone mind helping me with a group theory problem?

Is the set H = <3> x <5> a cyclic subgroup of Z6 x Z10? What about H = <4> x <5>? If it is cyclic, find a generator, if not, explain why

For H = <3> x <5>, I know the answer is no, its not a cyclic subgroup, because H isn't cyclic. I also know the answer for H = <4> x <5> is that it is a cyclic subgroup, and a generator is (2,5), but how do I get to these conclusions?

am i correct in assuming that <3> x <5> is the same as {0,3} x {0,5}?

[chilly ocean]

How can you know the answer to those questions and not know how to conclude them?

[quartz lake]

the answers were given to me by the professor

I'm just trying to figure out how to reach these answers

[old lava]

but uh, <3> x <5> is cyclic

so I think you may not know what you think you know

[chilly ocean]

Tell me a generator for <3> x <5> in Z6xZ10

[old lava]

<1, 1> ?

3 and 5 are relatively prime

oh in Z_6 x Z_10

I do be trolling then

[quartz lake]

@chilly ocean Why isn't <3> x <5> cyclic? My instinct is to agree with @old lava

[old lava]

because it's in Z_6 x Z_10

<3> = <5> = Z_2 (= is iso here)

Z_2 x Z_2 is well known to not be cyclic

[quartz lake]

Z6 is {0,1,...,5}, and Z10 is {0,1,...,9}, isn't 1 in both of these though?

[old lava]

I misread it as Z_3 x Z_5

[chilly ocean]

1x1 isn't in <3>x<5>

[old lava]

it's not Z_3 x Z_5, it's <3> x <5>, I made the same mistake

[quartz lake]

<3> x <5> is the same as {0,3} x {0,5}, and there isn't an element in this that can generate Z6xZ10

Is that a correct understanding?

[chilly ocean]

no

there isn't an element in {0,3} x {0,5} that can generate {0,3} x {0,5}

that's why that subgroup, {0,3} x {0,5}, is not cyclic

[quartz lake]

why doesn't (3,5) generate {0,3} x {0,5}?

in z6 x z10

[chilly ocean]

what does it mean to generate?

[quartz lake]

in this case, applying addition modulo 6 and addition modulo 10 to 3 and 5 respectively. 3 +6 3 = 0, 0 +3 3 = 3, that's where I get {0,3} from, and then applying addition modulo 10 to 5 gets me {0,5}. by applying +6 and +10 to (3,5), I get {0,3}x{0,5}

[chilly ocean]

er (3,5) + (3,5) = (0,0)

[quartz lake]

but then (3,5) + (0,0) = (3,5), and then it loops back to (0,0)

[old lava]

yes

exactly

that is exactly what happens and why the subgroup not cyclic

and not generated by (3, 5)

[quartz lake]

I think I get i'm starting to get it now

I'm sorry I'm so slow at this, this branch of math is so foreign to me

[old lava]

algebra is pretty tough to first get

especially group theory, considering its abstractions are so far removed from what you're used to

[quartz lake]

so applying this to H = <4> x <5>, (2,5) is a generator becasue (2,5) + (2,5) = (4,0), (4,0) + (2,5) = (0,5), (0,5) + (2,5) = (2,0), (2,0) + (2,5) = (4,5),.... etc.

that makes sense now why <3> x <5> wasn't cyclic, because it would be impossible to get (3,0) and (0,5), so not every "possible combination" of {0,3} and {0,5} was achievable

Thanks so much @chilly ocean and @old lava

[chilly ocean]

thanks for thanking me

[cloud walrus]

Moth In Shambles

[maiden ocean]

i was trying to do this w/ the universal characterization of local rings: let S = {f^n, n > 0} and define h: A -> A_g by a |-> a/1

so i have to show that 1. h(f^n) is a unit 2. h(a) = 0 implies hf^n = 0 for some n 3. every element of A_g is of the form h(a)h(f^n)

1 isnt very hard you just do contradiction and then pull back the maximal ideal of f^n/1, done

im not sure how to do 2

h(a) = a/1 = 0 implies that hg^m = 0 for some m > 0

does V(f) = V(g) let us show that g = f^n for some n or something?

oh im silly

since V(f) = V(g), X_f = X_g

then r((f)) = r((g))

this g^m = f^n for some n

[unique juniper]

$x^p -2$ and p is prime

[cloud walrus]

Yes

[unique juniper]

trying to find the galois group

[rustic crown]

over Q?

[unique juniper]

|Aut(K/Q)|=[K : Q] = p(p-1), where K is the splitting field. but there are p roots, there would a total of (p-1)! possible automorphisms?

[unique juniper]

yes

slightly confused, are some of those possible automorphisms not actually automorphisms?

[chilly ocean]

if p=2 there'd be 2 automorphisms not 1

[oblique river]

you are right that there are p(p-1) automorphisms

not all permutations of the p roots (which would be p!, not (p-1)!) are actually field automorphisms

[unique juniper]

oh

it should be (p-1)! ? because the minimal poly for a pth root of unity has degree p-1 ?

[oblique river]

the splitting field of this polynomial is generated by the p roots of this polynomial, but you can also find a simple generating set with two elements. if you do that, the automorphisms will become clearer

the roots of x^p - 2 aren't pth roots of unity

[unique juniper]

i know, i mean

[oblique river]

the galois group of the splitting field of the pth roots of unity is cyclic of order p-1

[unique juniper]

what i mean is, pth root of 2 will only get mapped to itself right?

[oblique river]

so much smaller than (p-1)! (except of course if p = 2 or 3)

no

x^p - 2 has p roots

and is irreducible

so you know that the galois group will be transitive on those roots

it will permute them around

it will atke one pth root of 2 to another pth root of 2

but it's not going to fix all of the pth roots of 2

[unique juniper]

${\sqrt[p]{2}, x\sqrt[p]{2}, \dots, x^{p-1}\sqrt[p]2}$

[cloud walrus]

Yes

[oblique river]

those are the roots of x^p - 2, yes, assuming that by x you mean a pth root of unity

the galois group is going to permute that set

[unique juniper]

hmmmm

it does

i was thinking of something else....

mb

so its p! :D

[oblique river]

what is p! ?

[unique juniper]

if we were to permute those roots i mean

p! possible permutations

[oblique river]

there are p! permutations of those roots

but they are not all field automorphisms

[unique juniper]

yeah

[oblique river]

you have already identified that there are p(p-1) total automorphisms

[unique juniper]

yeah

now i figure out which are automorphisms

[oblique river]

that's probably not the best way to go about it

[unique juniper]

oh

you mentioned something about writing as a field generated by 2 elements

[oblique river]

that's right

[unique juniper]

Q(pth root of 2, x)

[oblique river]

that's right

now, let g be an element in the galois group. g is completely determined by what it does on this generating set. How many options are there for g(2^(1/p))? how many options are there for g(x)?

(by the way please call this z or something other than x, it's confusing because you've already used x to represent a variable in a polynomial)

[unique juniper]

sorry

[unique juniper]

p options

[oblique river]

yes, p options for hte first. and what about for the second?

[unique juniper]

yeah p-1

[oblique river]

great. so this gives us p(p-1) candidate automorphisms. we don't know that they are all in fact automorphisms, but we know that the true automorphisms must be among this set of p(p-1) which we've identified as candidates

but wait!

we know that there are exactly p(p-1) automorphisms!

so therefore the p(p-1) candidates we identified must all be true automorphisms

[unique juniper]

yup

[oblique river]

and the only true automorphisms

so there we go, we've identified them. we can call them g_(i, j) where g_(i, j)(pth root of 2) = z^i * pth root of 2 and g_(i, j)(z) = z^j

I suppose i should run between 0 and p-1 and j should run between 1 and p-1

(in reality, you want to think of i as an element of Z/pZ and j as an element of (Z/pZ)*, the units in Z/pZ)

now we've found all of the automorphisms. if you want to determine the group structure, you have to figure out what g_(i, j) * g_(k, l) is

[unique juniper]

yeahp

[oblique river]

and you can do that just directly

write down g_(i,j) ( g_(k, l) (2^(1/p)) ) and it should be equal to z^a 2^(1/p) for some a depending on i, j, k, and l

similarly plug in z and you should get z^b for some b depending on i, j, k, and l

and there's your group structure

[unique juniper]

thanks :D

[oblique river]

imo this is 100000% better than just starting with all p! possible automorphisms and trying to rule them out

[unique juniper]

yeah it really is lmao XD

[oblique river]

notice that we never had to prove that anything was an automorphism

we just said "we know that there are p(p-1) total, and we also know that the automorphisms must lie in this set of p(p-1) potential ones we found, so therefore all the potentials must be true automorphisms"

[unique juniper]

yes, much easier

[oblique river]

i gotta get back to work now, good luck

[unique juniper]

thank you :) good luck too!

[oblique river]

thanks

[chrome hinge]

Hey there, got a question about the definition of Boolean sigma-algebra

It is a boolean algebra in which every countable set has a supremum

and my question is: with respect to what order?

[chilly ocean]

Maybe $x\leq y$ iff $x \lor y = y$ ?

[cloud walrus]

Carla_

[chilly ocean]

@chrome hinge

[chrome hinge]

Ive found it defined like that on the internet, just want to make sure thats indeed the definition

[chilly ocean]

It would make sense, then it coincides with the usual partial order on the subsets of a set

[chrome hinge]

It does make sense

Thanks!

[chilly ocean]

Also makes sense with the usual definition of sigma-algebra: it is closed for countable union

[chrome hinge]

how does the fact of every countable subset having a supremum imply closure respect to countable union?

i mean why the supremum need to be in the algebra too?

Nevermind thats a stupid question im sorry haha, its obvious

[chilly ocean]

yea if it was not in the algebra it wouldn't be a supremum

[chrome hinge]

exactly

so er

an actual sigma algebra would be a particular case of boolean sigma algebra correct?

where you take a collection of subsets of a given set

and intersection, union operations

I dont even know if this is the right channel lol, probably not

[chilly ocean]

i mean they're algebraic structures

a sigma algebra is a (S, A) where A subset of power set of S that is non-empty and closed for complement and countable unions

so yea

[chrome hinge]

yea youre right, they re

well

Thanks!

[delicate bloom]

for $a\in \mathbb{F}_q^\times$ is $x^q-x+a$ irreducible over $\mathbb{F}_q$?

[cloud walrus]

碳水化合物

[delicate bloom]

I'm able to prove it when q=p, and that if it factors over F_q it must factor into polynomials with degree divisible by p

I can explain those proofs if anyone's interested

but of course I'd like to narrow it down further and that it can't factor at all

[cursive temple]

https://math.stackexchange.com/questions/579659/irreducible-factors-for-xq-x-a-in-mathbbf-p?rq=1

this might be of interest

[chilly ocean]

ah i did it again

[delicate bloom]

ah perfect thanks, just what I needed

that proves it's irreducible iff q is prime, otherwise it factors into q/p different degree p polynomials

or when q=4 😬

[next obsidian]

You can go by first adjoining one

then you have to see if those irreducibles stay irreducible

@gleaming egret

if they stay irreducible, then you adjoin that root

and continue on and on until you have all the roots

I'm not certain what you mean

by "didn't work"

I think maybe you mean that t^2 + 2t + 3 still doesn't have a root over that extension?

yeah, so now you adjoin the root of the second one

and see if the third one is irreducible over that

and you know degrees of field extensions are multiplicative

so step 1 made you degree 2, and then step 2 was deg 2 over step 1

so in total you're now degree 4

yup

...

next question

It is

F_p^4

yes

no worries

no

I thought you meant how to get from your deg 4 thing

to F_p^4

and the answer is I don't know

cuz I forgot how to prove every finite field is F_p^n

lmfao

yeah but

you don't wanna convert to F_p^4 anyway

it'll be easier to see if your third polynomial is irreducible over like

F_p(sqrt{2}, blah blah)

Like uh, the point is just that elements of F_p^4 are hard to write down

but when you have it in this form where it's like F_p adjoin two roots

you can write everything kinda simply

yeah

anyway, the last step might be a bit tricky

I think it's true that this is still irreducible

but in the degree 4 case you don't know every element looks like

a + bsqrt{2} + c(-1 + sqrt{2}i)

because as a vector space you're degree 4, but this says you have a 3 element spanning set

The only reason I think you don't have that last root is because

let's say you did, then you'd have
sqrt{3}/2i and sqrt{2}i inside your field

but then you can like multiply them together to see that you have
sqrt{3}sqrt{2}/2 inside your field

and then you can like multiply by 2 again, so you should get sqrt{6} inside
F_p(sqrt{2}, -1 + sqrt{2}i)

and this seems super not possible

Oh, actually this is easy

divide by sqrt{2}, which you can do

and ultimately you get sqrt{3} in F_p(sqrt{2}, -1 + sqrt{2}i) or w/e

and this is just not true

maybe you do a little more arguing to make it airtight, but there's no way it exists there

yeah, you can show they're equal

just show both are contained in the other

this amounts to showing the generators are in each other

clearly both contain sqrt{2}

so just show that i lives in the other one, and -1 + sqrt{2}i lives in the other one

if that makes sense

yup

so now we want to show show the last root doesn't live in there

yup

so assume it did right

well we can multiply by 2i

err

first add 3/2

so now we know that sqrt{3}/2i is in there right?

F_5(sqrt{2},i)

now multiply by 2i

now we know sqrt{3} is in F_5(sqrt{2},i)

then show this just isn't possible

because 3/2 is in the field

I mean

it is right?

3 is there

2 is there

their quotient is there

3/2 isn't like 1.5 mod 5

It's just like

whatever multiplies into 2 to give you 3

which I think is 4 mod 5?

but 3/2 exists as like

an abstract element I guess

so it exists

That doesn't make much sense

My point is it exists, as like 3 * 2^-1

Yeah, but that doesn't matter

the fact it is like 4 mod 5

doesn't matter

Like for the element

-3/2 + sqrt{3}/2i

exactly

well we want to get to the point it would

be in our field

and then go "that's impossible"

yup

yup

It probably will

[delicate bloom]

I didn't read what came before but this doesn't look right

you can think of i being in F_5 since 2 and 3 both square to -1

similarly since 3=-2 mod 5 we have sqrt(2) = sqrt(-3) = i sqrt(3) so we already have one if we have the other

so it simplifies down to F_5(sqrt(2)) which is 2 dimensional

🤣 whoops

yeah true, that's how it goes, the more painful the more you remember it too I think haha

I guess as a general tip, don't forget the multiplicative group of F_q is really q-1 roots of unity since it's a cyclic group

of course representing them as e^{2pi i/(q-1)} is a sin

[delicate bloom]

remember I said this

so I don't know if what came before but like I can catch up to that if what I said doesn't really apply to your problem for some reason

your last factor seems off

ok I'll believe it

so let's see that middle one, we need it to split

one sec lemme work itout

ok it has roots -1 +- sqrt(3)

but remember we can represent sqrt(3)=sqrt(-2) = sqrt(-1)*sqrt(2) = i sqrt(2)

here, i can be 2 or 3

so your roots are

$-1 +2 \sqrt{2}$ and $-1 +3 \sqrt{2}$

[cloud walrus]

碳水化合物

[delicate bloom]

so you can check to be sure, $(x+1 +2 \sqrt{2})(x+1 +3 \sqrt{2}) = x^2+ 2x+3$

[cloud walrus]

碳水化合物

[delicate bloom]

haha

yeah like in terms of stuff I try to picture like all the q-1 roots of unity in F_q and then a bunch of stuff related

yeah hopefully they simplify in similar ways

like in my mind I just interchanges between writing 1,2,3,4 and writing 1, i, -1, -i whenever it's convenient

[delicate bloom]

also I probably should mention you so you can catch up on this too

doesn't matter yeah

well, like if you write the symbol i, it's a convention

does i mean 2 or 3

well whenever you square root you get two choices

+sqrt(-1) and -sqrt(-1)

so if sqrt(-1)=2 then the other will be 3

and vice versa yup

you're welcome 👍 lol

[next obsidian]

oh fucq

fuq*

I just trusted that those were irreducible

TFW you help someone but it was all wrong

[next obsidian]

:(

[rustic crown]

.<

[boreal pecan]

Can anyone suggest a good place to find some basic stuff about filters and filters of identity neighbourhoods? By basic stuff I mean mainly definitions and maybe some important properties or criteria for something 😄

[untold sapphire]

TIL what a filter is

The Wikipedia definition seems pretty good

[rustic crown]

its like saying 6 factors through 2

as 6 = 2*3

so when you have group map say f : G --> H
and N is a normal subgroup contained in the kernel, then the map f factors through the map p : G --> G/N

which mean that there is a map g: G/N --> H such that f = g * p

you factorized the map f into the product g * p

so this is just your old definition of factoring

[boreal pecan]

Did anyone study from this script ? :/

[unique juniper]

im trying to find the basis of Q{2^{1/8}} over Q(2^{1/2})

but i cant decide which powers 2^{1/8} are in Q(2^{1/2})

wait

ive got an answer, can you check too see if im correct or wrong before you tell me?

i think its $1, \sqrt[8]{2}, \sqrt[8]{2}^3, \sqrt[8]{2}^7$

[cloud walrus]

Yes

[unique juniper]

oh

[cloud walrus]

squirtlespoof

[unique juniper]

oh

i did that think that at first

im not sure about \sqrt[8]{2}^2

one sec, let me go over what i tried

yeah

:o

i dont like this :D

SAME LOL

[final pasture]

linearly independent in Q(1, sqrt[3]{2}, \sqrt[3]{2}^2), right ?

wut

[unique juniper]

i think this is helpful?

[final pasture]

what's a field extension of F_7 that contains sqrt[3]{2} @gleaming egret ?

huh ?

I'm confused, probably 'cause idk much field theory either lol

but I don't get what does F_7(sqrt[3]{2}) mean, the definition of K(a) I know only makes sense if we already know some field extension of K that contains a 🤔

yes

atleast that's what I understand when I read F_7

you meant smth else ?

[unique juniper]

@gleaming egret i just checked, it seems youre correct

thanks

[unique juniper]

XD

[languid meteor]

Does anyone have any tips for finding the degree of this over Q?

I was trying to use the tower theorem but its not working out how I planned

[carmine fossil]

The degree is definitely less than or equal to 9

[languid meteor]

yeah I thought that much, im pretty sure it actually is 9

I know that it's divisible by 3 as well by the tower theorem

if I can find [Q(cuberoot(3) + cuberoot(4)) : Q(cuberoot(3)]

then I have it, but its proving slightly difficult

[unique juniper]

can you show that it equals Q(cubroot 3, cuubroot 4)

will make it easier

@languid meteor

[languid meteor]

I can show containment in one direction but showing Q(cuberoot(3)+cuberoot(4)) contains cuberoot(3) and cuberoot(4) individually is quite tricky

I can keep trying though

[unique juniper]

yes

[languid meteor]

cool, ill work on that so. Thank you!

[unique juniper]

(i dont know if this is the best way to do it, but its what i would have done)

[rustic crown]

one way is to just bash by matrices, not the best one and often pretty bad.
multiplication by that element in Q(cbrt(4), cbrt(3)) could be represent by a matrix. Now you can compute this matrix and its characteristic polynomial. the minimal polynomial is an irreducible factor of this so just computing some stuff would do the job.

i don't know any nice ways of doing this, but usually simple computations work.

[cloud walrus]

det

det

det

det

det

[rustic crown]

squaring that again,

[cloud walrus]

det

[rustic crown]

This shows,

[cloud walrus]

det

[languid meteor]

wow thats perfect, I just learned a lot about computing these things thank you! Surely then this implies that [Q(alpha):Q(cuberoot(3)] = 3? I can just adjoin the extra term with x^3 - 4

and then I have my degree 9 extension overall

[rustic crown]

there is still some work left though

if we can show that x^3-2 or x^3-4 remains irreducible over Q(cbrt(3)) then we're done

[languid meteor]

small degree irreducibility test for x^3-4 shows its immediately irreducible since we have no roots in Q(cbrt(3))

[rustic crown]

yea you need to show that cbrt(4) doesn't lie in that... other two roots are complex so they definitely don't.

[languid meteor]

how would I go about doing that? Im already thinking to come up with some contradiction but I dont know where i'd find it. I dont know if there are any elements excluded from Q(cbrt(3)) except for complex numbers so far

unless I show [Q(alpha) : Q] > 3 then [Q(alpha):Q(cbrt(3)] cant be 1

[rustic crown]

lemme think, its the first time i'm seeing something where both are cuberoots, usually the coprime degree comes pretty handy and we don't need to work that hard

[languid meteor]

I think if cbrt(4) is in Q(cbrt(3)) then I should be able to write it as cbrt(4) = a + b*cbrt(3) right?

[rustic crown]

there could be a +c*cbrt(9) as well

[languid meteor]

hmmm

I think this still gives us our contradiction

cube both sides and you find that the sum of a bunch of irrational numbers is rational

but maybe the a^2, b^2 and c^2 terms give an out to the contradiction

[rustic crown]

yea i think we can use that 1, cbrt(3), cbrt(9) is a basis of Q(cbrt(3)) over Q and so that might work... but i wanted to see if i can use some theory to answer that... i was trying to look at the splitting fields of (x^3-2) and (x^3-3)

cubing something with 3 terms is a nightmare

[languid meteor]

yeah its a pretty brutal problem, I thought it should be easy but it really isnt haha

there must be something to do with the fact that 3 and 9 are coprime to 4

that we cant express cbrt(4) as the sum of cbrts that are coprime to it

[rustic crown]

lets wait for someone who tells us a nice solution 😛

@hidden haven can you take a look at this?

[languid meteor]

haha fair, I was thinking of how to show Q(cbrt(3)) is not isomorphic to Q(cbrt(4)) but I legit dont think ive been taught enough theory to work with it

[final pasture]

inb4 🍎

[rustic crown]

like a told, the stupidest idea would be to use matrices to get a degree 9 polynomial and plug that into a calculator to factorize it

and hope that its irreducible

but doesn't tell you much about the theory which is sad

[verbal aurora]

does anyone know how to implement this function:
f(1,2,1) = 3
f(1,2,2) = 5
f(1,2,3) = 8
f(1,2,4) = 13

f(2,6,1) = 8
f(2,6,2) = 14
f(2,6,3) = 22
f(2,6,4) = 36

[hidden haven]

oh hello @rustic crown

[rustic crown]

hello!

[hidden haven]

Is this the problem?

[scarlet estuary]

what do you mean by "implement"?

[rustic crown]

yea

[verbal aurora]

Namington: implement = define the function

[chilly ocean]

you already defined it

[scarlet estuary]

(on an 8-element domain)

[hidden haven]

I'd suspect that it will be Q(cbrt(3), cbrt(4)), so the claim is that that alpha is a primitive element of this extension. Primitive elt thm says (or at least its proof does) that for Q(a,b)/Q, a+cb is a primitive element when c is not of the form (b-b')/(a-a') where a' and b' are any conjugates of a and b, and 1 doesn't have that form

[rustic crown]

yea you can check above that, using a few computations we were able to settle that the field is Q(cbrt(3), cbrt(4))

but still checking cbrt(4) not in Q(cbrt(3)) doesn't look obvious

[hidden haven]

So just need to prove that Q(a) ≠ Q(b), ig this was the hard part?

Hmm let me think

[rustic crown]

btw how do you see that 1 doesn't have the form?

it looks pretty hard to say some weird relations between cbrt(3), cbrt(4) and w don't hold

[hidden haven]

You can see graphically, magnitude can never be 1

cbrt of 4 - a conjugate of cbrt of 4 would be bigger in magnitude than same difference for cbrt 3

[languid meteor]

can I ask about this implication actually, is this immediately true as a result of the basis?

[hidden haven]

Because they are chords of different radii circles but subtend the same angle

[languid meteor]

like I agree cbrt(18) is in Q(alpha)

[rustic crown]

subtract 3*alpha from that as well

[hidden haven]

I think I should also just read the chat once lol I was trying to avoid that but idk what all you've tried

[languid meteor]

ah ofc, thank you

[rustic crown]

i was wondering if you knew some cool ways to handle small degrees

by that i meant non-computational

[languid meteor]

im thinking that there must be some simple way to show cbrt(4) is not in Q(cbrt(3)) because we havent done anything about primitive elements, galois fields etc

[rustic crown]

yea that's why i was trying to avoid those stuff

[verbal aurora]

this is the progress i've made so far

f(2,2,1) = 4 * 2 - 2
f(2,2,2) = 6 * 2 - 2
f(2,2,3) = 10 * 2 - 4
f(2,2,4) = 16 * 2 - 6
f(2,2,5) = 26 * 2 - 10
f(2,2,6) = 42 * 2 - 16

f(3,3,1) = 6 * 2 - 3
f(3,3,2) = 9 * 2 - 3
f(3,3,3) = 15 * 2 - 6
f(3,3,4) = 24 * 2 - 9
f(3,3,5) = 39 * 2 - 15
f(3,3,6) = 63 * 2 - 24

[hidden haven]

Yeah but most are very situational, I have something I could try but I'll only know if it works after trying to work it out once

[rustic crown]

the most useful tool for me was usually that degrees are coprime so the degree of the big extension is atleast the product

but here both are 3 and thigns are sad

[hidden haven]

Yeah

[rustic crown]

then i thoguth maybe lets find the galois group of (x^3-4)(x^3-3) to help me understand the intermediate fields, but i don't know very well how galois groups of degree 6 polynomials work nicely so it was going pretty computational as well

[old lava]

Suppose cbrt(4) is in Q(cbrt(3)), then Q(cbrt(4) \subseteq Q(cbrt(3)). Then, 3 = [Q(cbrt(3)) : Q] = [Q(cbrt(3)) : Q(cbrt(4))] [Q(cbrt(4)) : Q] = 3 [Q(cbrt(3)) : Q(cbrt(4))]. So, Q(cbrt(3)) = Q(cbrt(4)), but that's obviously not true, since the minimal polynomials of cbrt(3) and cbrt(4) are different

doesn't this work?

[rustic crown]

idts

Q(cbrt(2)) = Q(cbrt(4)) but their minimal polynomials are differnt x^3-2 and x^3-4

[old lava]

oh true

is there another way to show that Q(cbrt(3)) isn't equal to Q(cbrt(4))

[rustic crown]

writing things out and then cubing and then showing that doesn't have rational solutions looks pretty hard as well

sad thing is we don't get the help of eisenstein

it would be nice if we could find a subring of Q(cbrt(2)) which is a ufd

[chilly ocean]

is there generalization of eisenstein for non Q?

[languid meteor]

I could go the long computation way with the doesnt have rational solutions method, but I have to try like 9 things to have a complete argument

[rustic crown]

yea you can do eisenstein over UFDs

[chilly ocean]

woa nice

[rustic crown]

if you have a polynomial of degree n and a prime ideal such that modulo that prime ideal, the polynomial looks like x^n, and the constant isn't in the square of prime ideal then we can say the polynomial is irreducible

[verbal aurora]

f(2,2,1) = 4 * 2 - 2
f(2,2,2) = 6 * 2 - 2
f(2,2,3) = 10 * 2 - 4
f(2,2,4) = 16 * 2 - (4 * 2 - 2)
f(2,2,5) = 26 * 2 - (6 * 2 - 2)
f(2,2,6) = 42 * 2 - (10 * 2 - 4)
f(2,2,7) = 68 * 2 - (16 * 2 - (4 * 2 - 2))
f(2,2,8) = 110 * 2 - (26 * 2 - (6 * 2 - 2))
f(2,2,9) = 168 * 2 - (42 * 2 - (10 * 2 - 4))
f(2,2,10) = 278 * 2 - (68 * 2 - (16 * 2 - (4 * 2 - 2)))
f(2,2,11) = 446 * 2 - (110 * 2 - (26 * 2 - (6 * 2 - 2)))

[rustic crown]

numbers are scary

[old lava]

maybe can use this https://math.stackexchange.com/questions/1246404/techniques-to-prove-that-two-field-extensions-are-distinct ?

[hidden haven]

Your question wasnt well defined

[chilly ocean]

you are living a lie LiTHiuM

[nova plank]

How did you go and upgrade from #prealg-and-algebra to #groups-rings-fields ?

[old lava]

x^3 - 3 mod 3 factors into 3 irreducibles, while x^3-4 mod 3 doesn't so we might be able to use the same reasoning as that mse post to conclude that Q(cbrt(3)) is not equal to Q(cbrt(4)) (and aren't even isomorphic)

[rustic crown]

my alg nt knowledge is pretty low

yea fields are bad you don't get nice modular contradictiony arguments

[languid meteor]

it seems like it should be very obvious that cbrt(4) != a + b cbrt(3) + c*cbrt(9) for any a,b,c in Q(cbrt(3)) just because of the fact that 3 and 9 are coprime to 4

[old lava]

it does seem obvious

[languid meteor]

but thats probably a deeper result than can be shown easily

[old lava]

but proof by obviousness isn't a real technique

[hidden haven]

what ive been trying to do is use automorphisms of Q(cbrt(3)). For example if you want to show that sqrt(3) is not in Q(fourth root of 2), you can write sqrt(3) as a cubic in fourth root of 2, then apply the automorphism which negates the fourth root of 2 to the equation to get 2 of the coefficients negated on one side and you get +-sqrt(3) on the other. Take 2 cases and adding and subtracting the equations easily gives a contradiction

[old lava]

sadge

true, constructing homomorphisms might work, if Q(cbrt(3)) = Q(cbrt(4)), then there must be an isomorphism between them

[hidden haven]

but here its cbrt so negation isnt an automorphism multiplying by omega now

[old lava]

and you can so there's no isomorphism between them?

idk

[rustic crown]

yea i tried adding omega and some stuff

but then i wasn't sure how i would apply that automorpihsm to cbrt(4) without knowing if i can write it as something in Q(cbrt(3))

[hidden haven]

yeah but i dont see how thats easier, we are essentially trying to show that the identity is not an isomorphism

you will get omega cbrt 4 or omega^2 cbrt 4

2 cases still

[rustic crown]

can we use some trace-norm magic?

[hidden haven]

because root of x^3-4 has to go to another root

oof i dont remember that stuff properly

[rustic crown]

there are 3 ways to embed Q(cbrt(3)) into the closure

[hidden haven]

ye

[rustic crown]

trace will be sum of the 3 images and norm would be product

this is all i remember

[languid meteor]

I can show this actually using the cubed thing, a,b and c have to be powers of cbrt(3) >=1 because otherwise you immediately get a rational = irrational contradiction

[hidden haven]

so trace = 0 and norm = 3?

[languid meteor]

then you can work from left to right and show that a has to be cbrt(3) and b has to be cbrt(9) but then once you get to the 3rd term you get an irrational

giving you the contradiction

[rustic crown]

of cbrt(3) ig

[languid meteor]

not as nice as a theoretic way of doing it but it works I suppose

[hidden haven]

any way that doesnt do too much computation is nice

[rustic crown]

i think you'll just get 3 weird equations between 3 rationals and you'd be trying to see why that doesn't have any rational solutions

i was pretty interested in looking at the galois group of (x^3-2)(x^3-4) because if we show that has size more than 6, then we're done!

it would be subgroup of S3 x S3 but i can't see any nicer things from there

both the projections should be S3

and its size is atmost 18

and divisible by 6

[hidden haven]

call $\sqrt[3]{3}$ $s$ and $\sqrt[3]{4}$ $t$. Suppose $t\in \bQ (s)$ then $\exist a,b,c$ such that
$$a+bs+cs^2 = t$$
In the splitting field of $\bQ (s)$, apply the automorphism $s\mapsto \omega s$ to get
$$a+b\omega s + c\omega^2 s^2 = \omega^i t$$
where $i = 1$ or 2. But now the magnitudes dont match because magnitude of left side is reduced but magnitude of right side is the same, unless $b=c=0$, but then $t$ would be rational, which is a contradiction

[rustic crown]

\sqrt >.<

[hidden haven]

lol what is the command for that then

[rustic crown]

your didn't put a \

[hidden haven]

oh shit

argh

[rustic crown]

never use ()

always go for {}

[hidden haven]

lol

I give up

[rustic crown]

call $\sqrt[3]{3} = s$ and $\sqrt[3]{4} = t$. Suppose $t\in \bQ (s)$ then $\exist a,b,c$ such that
$$a+bs+cs^2 = t$$
In the splitting field of $\bQ (s)$, apply the automorphism $s\mapsto \omega s$ to get
$$a+b\omega s + c\omega^2 s^2 = \omega^i t$$
where $i = 1$ or 2. But now the magnitudes dont match because magnitude of left side is reduced but magnitude of right side is the same, unless $b=c=0$, but then $t$ would be rational, which is a contradiction

[cloud walrus]

det

call $\sqrt[3]{3} = s$ and $\sqrt[3]{4} = t$. Suppose $t\in \bQ (s)$ then $\exist a,b,c$ such that
$$a+bs+cs^2 = t$$
In the splitting field of $\bQ (s)$, apply the automorphism $s\mapsto \omega s$ to get
$$a+b\omega s + c\omega^2 s^2 = \omega^i t$$
where $i = 1$ or 2. But now the magnitudes dont match because magnitude of left side is reduced but magnitude of right side is the same, unless $b=c=0$, but then $t$ would be rational, which is a contradiction
```Compilation error:```! Undefined control sequence.
<recently read> \exist 
                       
l.55 ... = t$. Suppose $t\in \bQ (s)$ then $\exist
                                                   a,b,c$ such that
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.```

[hidden haven]

check the argument tho

oh it is \exists

[rustic crown]

exists

ye lol

[hidden haven]

why cant discord have latex autocomplete

[cloud walrus]

det

[hidden haven]

yay

oh the possibilities at the end are that any 2 of those coefficients are 0

not that b=c=0

and that will still give a contradiction because cbrt(4/3) is not a rational

does the proof seem legit?

[rustic crown]

not sure how you claim the magnitudes don't match

a, b,c could be negative as well right

[hidden haven]

oh shit im kinda dumb ngl

[rustic crown]

[cloud walrus]

det

[rustic crown]

v sad

[hidden haven]

lol

[rustic crown]

why are their weird subgroups of G x G

can't just group behave like ideals and subgroups be like H1 x H2

[hidden haven]

ok lets say $$a+b\omega s + c\omega^2 s^2 = \omega t$$
then subtracting the 2 equations, $a$ cancels and we get
$$b(1-\omega) s + c(1-\omega^2) s^2 = (1-\omega) t$$
by linear independence of the 2 vectors $(1-\omega^i)$, you get $c=0$
if the RHS were $\omega^2 t$ instead, you would get $b=0$

[cloud walrus]

Moldilocks

[hidden haven]

so at least one of b and c has to be 0

[rustic crown]

oh yea this works

[hidden haven]

oh then you get a contradiction because with $c=0$, you would have $a+b\omega s = \omega t$ which gives $a=0$

[rustic crown]

in either case you get something like s/t or s^2/t is rational

[cloud walrus]

Moldilocks

[hidden haven]

ye

lets go

[rustic crown]

nice but i'm too sad at this point lol

[hidden haven]

[rustic crown]

is there not a nice way to do these instead of computing stuff

[hidden haven]

but hey @languid meteor we have something that works

idk i felt this was pretty nice lol

[languid meteor]

sorry for bringing the sadness of this problem into your lives hahaha

but thanks for all of the help, I feel like I just learned so much. I really appreciate it

[rustic crown]

yea that's true. but we still have to do the work to first show that the field is actually Q(s, t)

i don't wanna work 😶

[hidden haven]

primitive elt thm gives easy proof tho as long as you remember enough of the proof to remember the condition but not enough to think about it as part of the argument

[rustic crown]

yea pretty nice

[chilly ocean]

Im confused about associates. im asked to find an associate of g(x) = x^2 + x + 1 in C[x].

is h(x) = -x^2 - x - 1 in C[x] an associate?

[rustic crown]

yep

any non-zero constant in C[x] is a unit!

so associates of g(x) are precisely c*g(x)

[chilly ocean]

since -1*h(x) = g(x) and -1 is in C?

[rustic crown]

yep

[chilly ocean]

ok thank you!

[rustic crown]

(if you haven't done so, try to show that only units are these non-zero constant polynomials)

[median pawn]

what is the notation (b) and (a)

[rustic crown]

(a) is principal ideal generated by a

[carmine fossil]

ideal generated by a is (a)

[rustic crown]

so all the elements {a*r | r in R}

[median pawn]

oohh okay thanks!

[sharp jacinth]

Why order of all elements in (Z,+) is infinite
According to definition , order of element is smallest number of times it must operated with itself to get identity , so according to this if I add 1 any number of times , I won't get zero , so why order to said to be infinite ?

[bleak abyss]

So

If g\in G has order finite n

Then what can we say about the subgroup <g> of G?

[unique juniper]

are all normal extensions splitting fields?

[bleak abyss]

iirc yes

[unique juniper]

but not the other way around ?

[lavish gale]

what is iirc Dami?

[bleak abyss]

iirc means "If I recall correctly"

[lavish gale]

I see
ty

[bleak abyss]

Carla I was trying to socratically get him to that statement

[chilly ocean]

oops

[sharp jacinth]

No of elements of <g> would be n

[bleak abyss]

Yup muskan

So if an element doesn't have finite order then it generates an infinite subgroup

[unique juniper]

not all splitting fields are normal extensions right?

[bleak abyss]

Should also be

[unique juniper]

wiat

not all normal extensions are splitting fields i think

[bleak abyss]

You might need some technical condition on the field? But my memory is that splitting field = normal

[unique juniper]

i think a splitting field is normal but not the other way around

[rustic crown]

i think it depends on your definition of splitting fields. if you take splitting field of a single polynomial, then you definitely get a finite extension. so any infinite normal extension isn't a splitting field.

but if you define splitting field for a set of polynomials, then they are equivalent.

[bleak abyss]

Ah I guess I was thinking of just finite things

[unique juniper]

ohh i see

[rustic crown]

if the set of polynomials is finite, then you migth as well look at only one polynomial, that is the product of those finitely many...

so this is different only in the case when the set is infinite

[bleak abyss]

@sharp jacinth to wrap up our discussion: I prefer defining the order of g to be the order of <g>, where "order of a group" means its size

[sharp jacinth]

Can we have an element of order zero then ?

[unique juniper]

nop

[sharp jacinth]

Yup I know that one as well , I just got confused with the other one

[rustic crown]

one direction is very easy actually, if k --> F is a normal extension then look at the set of polynomials {f_alpha | alpha in F} where f_alpha is the minimal polynomial of alpha over k.

[bleak abyss]

Yeah they're both fine in the case of finite order but it's weird to process the "Which power is the identity" for things which don't have finite order

[sharp jacinth]

Oh I see , alright

[lavish gale]

also recall order is defined as the smallest positive number such that..

[sharp jacinth]

Can an element of a group has order zero ?

[unique juniper]

no

[sharp jacinth]

Okay , thanks

[chilly ocean]

||define g^0 = e and every element has order 0, qed||

[rustic crown]

but identity has order 1

[chilly ocean]

you have order 1

[rustic crown]

I'm trivial

[chilly ocean]

0 is the largest integer all other are smaller

i need helperino

I have this formula

[viscid pewter]

sorry, all we have here is helperinae

[chilly ocean]

Which is supposed to be the formula to calculate a momentum matrix with weights

but the exercice sais

calculate the momentum matrix of Legendre polynomios with f(x) = 1 between -1 and 1

So i do know what is what on my formula

[chilly ocean]

_i moved to #computing-software _

[uncut girder]

How many cubic polynomials are there over F_q with a double root but not triple root

[final pasture]

q² - q ?

[chilly ocean]

👌🏻😔

[final pasture]

a polynomial with a double root is of the form P = (X-a)²Q

now since P is cubic, deg(Q)= 1

so Q = (X-b) for some b

now you don't want triple root so you want (a, b) with a != b

oh and that only gives you the monic polynomials

so (q-1)(q² - q) should be the answer ? ig

idk I'm too tired to think

[chilly ocean]

Do the roots have to be in Fq? (I guess it is really only really the double root that this is applicable for, I think)

[old lava]

that would make it a bit harder

maybe

[final pasture]

If the roots doesn't need to be in Fq, the answer is easy, an infinity

wait

actually no

nvm

I'm just dumb

[chilly ocean]

I think all roots must exist in Fq^2

[old lava]

I'm 99% sure I've proved that if c(x) = (x-a)^2 (x-b) in \bQ[x], then a, b \in \bQ

does that hold true for F_q

[final pasture]

dunno why I thought alg closure is infinite => there's an infinity of polynomials of the kind we're looking at

also is my proof wrong ?

[chilly ocean]

Ah right I am a dumbass

I think I agree that roots must be in Fq

[old lava]

ok I found the proof, and I do a bunch of division by powers of 3, but nothing else that's specific to rationals, so I'm also 99% sure it holds true in F_q as long as F_q isn't characteristic 3

[final pasture]

[chilly ocean]

isn't that something about perfect fields actually

[old lava]

could be

oh true, I think it is

[oblique river]

it is true for all perfect fields

nothing about powers of 3 here

[old lava]

the powers of 3 was just a specific thing I did

before I had knowledge of perfect fields

but ya, I now see that it is true for all perfect fields

[chilly ocean]

poor F[x]-module

[next obsidian]

@scarlet estuary have you ever seen this before? Apparently the proof is via banned K-theory witchcraft.

If M is a fg projective A-module, then some finite tensor power of M being free is equivalent to some finite direct product of M being free.

[old lava]

secret K theory tech?

[delicate bloom]

might help to see if a is a root, then we can look at a conjugate $a^7=3a$ and so by induction $a^{7^n} = 3^n a$, since 3 is a generator of $F_7$ these are all the distinct 6 conjugates

[cloud walrus]

Merosity

[delicate bloom]

well ok I suppose it's easier to just realize all elements b of F_7^x satisfy b^6=1 so if a is a root so is ab, lol

[shut halo]

Does anyone know how to show <x+1, -x^3 +zy^2 -1> is not a radical ideal?

I've been trying to find a polynomial p^n in the ideal s.t. p not in it, but can't seem to find a way.

Using Maple, I worked out that z*y should work but can't see why

[old lava]

seems like a question where using grobner bases might make your life easier probably

[shut halo]

Haven't learned that in the course I'm taking so I wouldn't how to do it

[chilly ocean]

That sounds kinda right. My algebra isn't great, but probably something like, first, that ideal equals <x+1, zy^2>, if zy exists in the ideal, then p(x+1)+q(zy^2) = zy, where p and q are polynomials. So zy divides p, and you get x+1 and y are gcd=1, or something. Hm, now I'm not sure how to proceed though

[shut halo]

ideal equals <x+1, zy^2>

Could you explain why is this the case?

[chilly ocean]

-x^3-1 = -(x+1)(x^2-x+1)

[shut halo]

Ah I see yes

[verbal aurora]

f(2,2,4) = 16 * 2 - (4 * 2 - 2)
f(2,2,5) = 26 * 2 - (6 * 2 - 2)
f(2,2,6) = 42 * 2 - (10 * 2 - 4)
f(2,2,7) = 68 * 2 - (16 * 2 - (4 * 2 - 2))
f(2,2,8) = 110 * 2 - (26 * 2 - (6 * 2 - 2))
f(2,2,9) = 168 * 2 - (42 * 2 - (10 * 2 - 4))
f(2,2,10) = 278 * 2 - (68 * 2 - (16 * 2 - (4 * 2 - 2)))
f(2,2,11) = 446 * 2 - (110 * 2 - (26 * 2 - (6 * 2 - 2)))

derive function f(x,y,k)

[chilly ocean]

Ah, I think I recall an argument like suppose <x+1, y> = <1>, then p(x+1)+q(y) = 1. Plug in x=-1 and y=0, contradiction

[delicate bloom]

the splitting field is F(alpha) but that doesn't mean [K:F]=2, it's 6

well, why would it be 2 in the first place

not sure I follow what you mean by that last comment

if you try to adjoin a 3rd root of unity to F_7 you won't get a new field since it already contains a 3rd root of unity

if it's 3 then you're saying you can factor t^6-3 into two cubics

since [K:Q]=deg of irreducible polynomial

you shouldn't write e^{2pi i} for roots of unity in a finite field, since that really only works if your field is in C

technically you shouldn't really do it even for Q since you could put it in a p-adic field too

well ok I think we should just focus on proving it's irreducible

what I had in mind is once you know all the roots are of the form ka for k in F_7^* and a a 6th root of 3, then you know if you were to factor it, you'd end up with some polynomial that has constant term that's a power of a times an element of F_7^*

since F_7^* contains no sqrt or cbrt of 3, then that term of the polynomial we factored is really not in F_7 after all

so that means it's irreducible

I gotta go, I'll be back in an hour or two

[old lava]

I mean technically, e and pi don't even make sense if you're working over Q, so you should just not use it even if you consider Q as a subfield of C, no?

but notation abuse do be notation abuse

[delicate bloom]

well if you are visualizing some roots of unity adjoined to Q as a picture in the complex plane or looking at a lattice, then it makes sense that way since that's the same topology, so I don't think it's a big deal

[old lava]

true, I was just thinking in like a purely algebraic sense it makes no sense

but like it doesn't really matter at the end of the day, it's useful notation to have, and everyone understands what you mean by it when you're treating Q as a subfield of C

[delicate bloom]

I guess while we're here, you can technically have an element you call e in extension fields of the p-adics

although exp as a power series converges for values < p^{-1/(p-1)}

which means you can define exp(4) for p=2 and exp(p) for p>2 and then adjoin the roots of x^p - exp(p) lol

[restive star]

Hey can someone double check my intuition here for a sec?

so I have to find an ideal properly containing (2x-1) over z[x]

would just simply (3,2x-1) work?

[rustic crown]

okie

yea

[restive star]

thanks!

seemed a tad too easy so I figured it was best to double check

[rustic crown]

i mean (1) would also work

[rustic crown]

this looks like a maximal ideal as the quotient is F3

[restive star]

oh true lmao

[quaint tree]

A problem that's been killing me for a while: Let $G$ be a finite group, and $K$ a $p$-Sylow subgroup of $G$. Let $X={xKx^{-1}|x\in G}$ be the set of all conjugates of $K$. Let $\sim$ be an equivalence relation on $X$, where $C_{1}\sim C_{2}$ iff $C_{1}=aC_{2}a^{-1}$ for some $a\in K$, and let $[C]$ be the equivalence class of $C$. Show that the number of elements in $[C]$ is a divisor of the number of elements in $[K]$ for all $C\in X$.

[cloud walrus]

Bannanachair Monarch

[next obsidian]

Eh?

Isn't [K] just a singleton?

Like, I think [C] should be the orbit of C under the conjugation action of K on C, but shouldn't K be fixed by this?

[quaint tree]

Building up to proving that

This is part of a long, multi-part exercise, I get to proving that later, but I'm not allowed to use that [K] has only one element yet.

[next obsidian]

I mean... it follows like immediately

just unravelling the definition of ~

Like, I don't see how abstractly you can show |[C]| divides |[K]| if the latter is a singleton

unless you just show that |[C]| = 1 for all C

I at least don't see some group action-y way to show that |[C]| divides |[K]|

[quaint tree]

I'm meant to conclude that $|[C]|$ is either 1 or a power of $p$.

[cloud walrus]

Bannanachair Monarch

[next obsidian]

That follows because it's the orbit of something acting on a set of size p^n

err

no

Well that would show that |[C]| = 1

since actually X has size not divisible by p

since it has size [G:Normalizer{K}] which can't be divisible by p

I'd just mess around with group actions

and orbit-stabilizer a ton

Or wait am I getting orbit stabilizer wrong lmao, maybe it follows by orbit stabilzer that it hs size dividing p^n

[quaint tree]

I'm just looking at where in the book orbit-stabilizer is; there has to be a proof that's not orbit-stabilizer-y.

[next obsidian]

Oof

Yeah, idk

[quaint tree]

Okay, yeah, it has to be a typo in the book

One of the next questions is "show that $[K]$ is the only class with a single element"

[cloud walrus]

Bannanachair Monarch

[next obsidian]

Maybe it meant the other way?

That [K]’s size divides the others?

[quaint tree]

Oh wait I see now, I'm meant to show that $|[C]|$ is a divisor of $|K|$, not of $|[K]|$. So many brackets, this is obnoxious.

[cloud walrus]

Bannanachair Monarch

[next obsidian]

Oh

Okay, yes

This would follow from orbit-stabilizer, so it seems more believable to show this

[quaint tree]

I was stuck for literally 3 days on this tiny lemma

[next obsidian]

Rip

[quaint tree]

Because I misread the question

[mortal hill]

Can I get a hint as to how to solve this? And what is the motivation behind such a definition? It seems a bit hard to work with under multiplication to me.

[fickle brook]

?

is there an image that didnt upload yet

[carmine fossil]

You can show f and g one of which is non primitive => fg not primitive easily(See if p|a and p|b then p|ax+by for all x,y)

[mortal hill]

Oh right idk what happend sorry.

Just a sec.

[rustic crown]

Buncho answered one direction
for the reverse, say f and g are primitive but fg is not primitive. Then the ideal generated by coefficients of fg is a proper ideal, hence contained in a maximal ideal. Now go modulo this maximal ideal m and look at what happens to the product of f and g in the ring (A/m)[x]

[rustic crown]

i don't think looking via elements is good. we may not be in a pid.

just notice that if coefficient of f generate the ideal I then coefficients of fg also lie in the ideal I.

[carmine fossil]

If let's say coefficients of g are all divisible by p,you can write the coefficients as pa_i for some a_i

And when you do the product all terms will be of the form p b_i (you can get b_i if you do the computation)

[chilly ocean]

I think he is saying, converse of "whole ring" ideal is not "generated by 1 element" ideal

[carmine fossil]

mb

[rustic crown]

[rustic crown]

[unique juniper]

[chilly ocean]

Is C(x) isomorphic to C?

as fields

[final pasture]

dumb q 'cause I never get the notations right lol:
if A is a ring, A[x] denote the smallest ring containing A and x and A(x) the smallest field answering to the same condition ?

[chilly ocean]

yea?

if in second case A is a field at least

[unique juniper]

i think?

[chilly ocean]

actually when youwrite x usually yiu mean polynomial ring or fiels of rational fractions

My guess would be that they are, C has so many transcendental numbers (over Q), how bad could it be to add another one?

i guess maybe could take a infinite chain of transcensental and shift them to make space for x

but if that works probably would work for R(x) too

independent transcendentals*

[final pasture]

ok thanks

[rustic crown]

my guess would be no?

[wind parrot]

They aren't, C is alg closed and C(x) isn't

[rustic crown]

yea

[wind parrot]

But an alg closure of C(x) would be iso to C I think

[chilly ocean]

Ah damn

Can we explicitly find which subfield of C is C(x) isomorphic to?

[rustic crown]

(we would probably need zorn? so that doesn't make it explicit?)

[chilly ocean]

yea that makes sense

[old lava]

Rip I love zorn and i hate zorn

Gives fun results but also half the time you cannot even construct a remotely decent example to see what's happening when you use zorn

[hot lake]

by "half the time" do you mean "always" ?

[rustic crown]

I feel kinda happy that construction of the algebraic closure of R, Q, and Fp don't require zorn.

[old lava]

wait, but wouldn't this imply that C is isomorphic to a proper subset of itself

which is like not true

[languid meteor]

why does (v,v) = 0 for a bilinear form in V mean that (u,v) = -(v,u)

I saw it somewhere and im trying to show it using the fact that its linear in each component but im not getting it

nvm I proved it

[old lava]

0 = (u - v, v - u) = (u, v - u) - (v, v - u) = (u, v) - (u, u) - (v, v) + (v, u) = (u, v) + (v, u)

[languid meteor]

nice, I proved it with (u+v,u+v) = (u,u) + (v,v) + (u,v) + (v,u) = 0 => (u,v) = -(v,u)

it feels interesting that the condition (v,v) = 0 leads to alternation

[old lava]

it's just that linear dependence means that its 0, and the "self-linear dependence" (for lack of better words) is removed when you expand it and have (u, u), (v, v) = 0

so the remaining terms must sum to zero

which gives that alternation

[wind parrot]

I'm pretty sure it is? I know for sure at least that any alg closed field of char 0 and of the same cardinality as C is iso to C

And I'm fairly sure that an alg closure of C(x) would have the same cardinality as C

[hot lake]

yep

[old lava]

but like C is a proper subset of C(x), which is a subset of its algebraic closure, and if the algebraic closure is C, then wouldn't we have like

$\bC \subset \bC(x) \subseteq \bC$

[cloud walrus]

F[x]-module

[rustic crown]

Consider the Frobenius map $\mathbb{F}_p(t) \to \mathbb{F}_p(t)$ and notice that image is $\mathbb{F}_p(s)$ where $s = t^p$. You can stuff a field into itself in a non-trivial way.

[old lava]

ya I know you can, but I thought you can't with C

[cloud walrus]

det

[old lava]

I know it is possible to have isomorphic copies of fields properly contained in themselves

just thought that was untrue for complex numbers

could be wrong

[rustic crown]

Find a transcendence basis B of C over Q. Then as B is infinite, there will be an injective map B --> B \ {a}.

This will give a map Q(B) --> Q(B\ {a})

Now Q(B) contained in C is algebraic, and say L is the algebraic closure of Q(B\{a})

Then we can lift this to a map C --> L. Now this would give a map C(x) --> L sending x to a.

So you were able to stuff C(x) inside C.

[old lava]

I see

very cool

[oblique river]

There's a theorem in model theory which says soemthing like

if kappa is a cardinality larger than countable and p = 0 or a prime number, then there is a unique (up to iso.) algebraically closed field of cardinality kappa and characteristic p

what this means is that C is the unique algebraically-closed field of characteristic 0 and with the cardinality of the continuum.

One can convince onesself that the cardinality of C(x) is the same as the cardinality of C, and furthermore, the cardinality of hte algebraic closure of C(x) is no larger

therefore the algebraic closure of C(x) must be isomorphiic to C

[rustic crown]

this is so cute

[oblique river]

and therefore C(x) is isomorphic to a subfield of C

[old lava]

I see, that's pretty interesting

my kneejerk reaction was that it feels very wrong

[rustic crown]

i wanna learn Model theory now

[oblique river]

it does feel wrong but it's so so right

[chilly ocean]

same

[rustic crown]

[oblique river]

Actually maybe you dont even need model theory https://math.stackexchange.com/questions/2756388/there-is-exactly-one-algebraically-closed-field-with-prescribed-characteristic-a

It just seemed like the kind of theorem which would show up in model theory

[chilly ocean]

What are the big bonuses of presentations of groups

[old lava]

easy representations of elements

easy way to write the group itself

[rustic crown]

easy way to find map from the presentation to a group

[old lava]

also nice way to see how the group is in fact a quotient of a free group

[hidden haven]

I saw that theorem in my model theory course but we used algebra to prove it that same characteristic and transcendence degree implies isomorphic for ACFs

We used this to prove completeness of ACF_p because this says that it is k-categoric for all k

[oblique river]

Idk what that means haha but it sounds nice :)

[rustic crown]

same

[hidden haven]

ACF_p is the first order theory of alg closed fields of char p. k-categorical for a cardinal k means that it has a unique model of cardinality k. k-categorical for some k large enough (larger than cardinality of language) implies completeness of the theory, ie every first order statement in the language of rings is either true in all models of the theory or false in all models of the theory

[rustic crown]

how do you write the axiom for it being closed? i can see how to do it for a particular degree, so does ACF_p have infinitely many axioms?

[hidden haven]

Yep infinite theory

You also need infinitely many axioms to say char 0

[rustic crown]

oh yes

[hidden haven]

(p is allowed to be 0)

[chilly ocean]

So there isnt a unique countable algebraic closed field in general?

[oblique river]

Not for any p

Q, Q(x), Q(x, y), etc

All have nonisomorphic algebraic closures

[old lava]

sadge

[hidden haven]

Yeah I think k needs to be uncountable

[oblique river]

Similarly if you replace Q with F_p

[hidden haven]

Yeah but we fix p

[old lava]

infinite fields of prime characteristic are weirdchamp

and I don't like them

[oblique river]

Lol

[chilly ocean]

0 is a prime number change my mind

[oblique river]

Youre wrong

Did i change your mind

[chilly ocean]

yes

[old lava]

but (0) is a prime ideal

hence 0 must be prime

[oblique river]

No

[old lava]

you're wrong

[cloud walrus]

det

[rustic crown]

(in an integral domain)

[hidden haven]

bro this is a no facts zone

pls leave

[rustic crown]

[chilly ocean]

don't be mean to the eevee

[vocal mural]

how to find y

if we know what a and b is?

[hidden haven]

Intuitively you want y to send 2 to 1, 3 to 3, 4 to 5, 1 to 6, so that when you apply a, the effect of a is to permute 2, 3, 4 and 1 instead of 1,3,5,6 and then applying y inverse you end up with those numbers back in their places with a 4 cycle done.

So it seems that any y that does what I said above would work, so for example (216)(45) should work

@vocal mural

y won't be unique because you could also send 2 to 3, 3 to 5, 4 to 6, 1 to 1 and the same reasoning would apply. So you can't hope to find y by just manipulation of the equation

If it's not clear how I get where to send what, I'm sending numbers in the beta cycle to the corresponding numbers in the alpha cycle

[wind steeple]

sigma(a1 ... an)sigma^-1 = (sigma(a1) ... sigma(an)), enjoy

[untold sapphire]

For real just draw the wiring diagram

It's like 10 times as easy to see things like this

[old lava]

don't reveal the secret tech

[next obsidian]

So secret every textbook contains it

[golden pasture]

[pine patio]

if G is a group, then is the intersection of the centralisers of all g in G equal to {e}?

it's for part iii) in this question

[pine patio]

true

so what do i write instead?

[old lava]

it's the center

she literally said it

[pine patio]

we havent done that

the definition of centralisers was mentioned in a previous question

[old lava]

it's the centralizer of the whole group

[pine patio]

so i just expressed the kernel as an intersection of all the centralisers of all g in G

[old lava]

you can easily prove inclusion both ways

to prove that centralizer of G = intersection of all centralizers

the center is just a name for centralizer of whole group

[cloud walrus]

google

[old lava]

using |

[pine patio]

oh ok

[old lava]

bad latex skills smh

use \mid

you're bad at latex

and it's my duty