groups-rings-fields

[sharp sonnet]

maybe try to construct a counterexample

[formal ermine]

||what about S3/A3 oplus A3||

||this is iso to Z6 not S3||

[lethal dune]

are you familiar with cat theory

[formal ermine]

only remotely

it's the direct limit of something something adjunct functor there isn't it

or something like that

[lethal dune]

not exactly, it's much easier

it's a coproduct in a 'suitable' category

categories like Abelian (groups), R-mod and all that

[formal ermine]

what's a coproduct

[lethal dune]

co product of A,B is an object C with maps i: A → C and j: B → V s.t. for any other object Z with maps f:A →Z and g:B →Z, there is a unique map ψ: C →Z s.t. ψ ∘ i = f and ψ ∘ j=g

easier with pictures

you can verify that the definition of direct sum satisfies the universal property given

[chilly saddle]

In general you can't choose an element from aB, if you choose at random (like a) it most probably won't work

[marsh goblet]

Right, thanks guys

[chilly saddle]

For example with Z6, you have Z6/Z3=Z2, but if you try {0,1,2}×{0,1} -> Z6 by addition it won't even be a homomorphism, it's more subtle

[south patrol]

Well also like direct sum is a biproduct which means it is ~~ a product attracted to >=2 genders~~ simultaneously a coproduct and product

[chilly ocean]

I'm confused on calculating Ext^n_Z/4(Z/2,Z/2)

I know we have a free resolution given by ... Z/4 to Z/4 to Z/2, where each of the maps Z/4 to Z/4 is multiplication by 2

Ext^n_Z/4(Z/2,Z/2) should then be H^n(Hom_Z/4(F_bullet, Z/2)

I'm confused on this Hom chain complex

Hom(Z/2,Z/2) is isomorphic to Z/2

and I think Hom(Z/4,Z/2) is also isomorphic to Z/2? but im not certain

and then im not sure what the maps in this chain complex should be?

[south patrol]

You have an explicit description from applying the hom functor

[chilly ocean]

ah yes they are indueced by the maps from the resolution!

thanks

[south patrol]

Yuppp

Np

But yeah everything you've said thus far seems correct so dw

[chilly ocean]

thanks!

[rotund aurora]

How do I show this part?

wait

I think this is just Newton's theorem

since if you have p(r1,...,rn)=0 where ri are the roots, this is symmetric in the roots, which implies it can be written in terms of the coefficients. I think?

wait no it doesnt make sense

idk

[rustic crown]

i think that should work

say t1, .., tn are the roots, then the a_k are symmetric polynomials in t_i

so consider the automorphism of F(t1, ..., tn) which permutes these roots

definitely these a_k are fixed

so it's an element of Gal(F(t1, ..., t_n)/F(a1, ..., a_n))

then we just proved S_n is a subgroup of this galois group

so equal

[rotund aurora]

how? I don't see it

any permutation of the roots of any polynomial will leave the coefficients fixed

[rustic crown]

oh i might have assumed that these t_i are algebraically independent among themselves

[rotund aurora]

yeah, you cannot assume that

it's like the converse. Given that the coefficients are alg independent, show that the roots are also algebraically independent

[rustic crown]

right, my bad

[rotund aurora]

I think you can just solve for the coefficients in the algebraic expression

[rustic crown]

you could prove that directly using some trancendence basis stuff, but i don't think you want that

n >= tr.deg(F(t1,..., t_n)/F) >= tr.deg(F(a1, .., a_n)/F) = n

[rotund aurora]

oh yeah

that's how in the book they proved how s1,...,sn, the elementary symmetric polynomials in the variables t1,...,tn are algebraically independent

thanks

[rustic crown]

wait, i thought we were trying to prove the other way

like using galois group is S_n to get info about algebraic independence

or something

[rotund aurora]

mmh yeah in the exercise they seem to take that route

[chilly radish]

Doesn't this follow from the fundamental theorem of synmetric polynomials

[rustic crown]

wait so you're saying if i have a dependence relation among the t_i, then i also have a symmetric dependence relation?

from there the fundamental theorem would say that there's a polynomial in a_i which is 0, that would be a contradiction

[chilly ocean]

I seem to be making a mistake here

[chilly ocean]

I make this cochain complex

Hom(Z/4,Z/2) is isomorphic to Z/2 and consists of the "inclusion on cosets", and the zero map

I think that multiplication by 2 from Z/4 to Z/4 induces the zero map on Hom(Z/4,Z/2) to Hom(Z/4,Z/2)

opps this is okay

all the maps are trivial

but that doesn't mean the cohomology is trivial

[rustic crown]

Hom(Z/4, Z/2) = Z/2

so your cochain complex is Z/2 in every non-negative degree and all maps are 0 like you say

[formal ermine]

how do we show that Q(isqrt(2)) and Q(i) are not isomorphic as fields?

[rustic crown]

and you wanna know it's non-trivial in each degree

[rustic crown]

because x^2+1 is irreducible over one and not over other :3

[formal ermine]

how does that show that they are not iso

[rustic crown]

if they were iso then this wouldn't be the case right

[chilly saddle]

in other words, "there exists x such that x^2 = -2" is a field invariant

[formal ermine]

why

[rustic crown]

say Q(isqrt(2)) = Q(i) then the left side has a solution to x^2 = -1

[chilly saddle]

you could transport the x from one to the other

[formal ermine]

but why should it be a solution to it

[chilly saddle]

In Q(i sqrt(2)) take x = i sqrt(2)

It satisfies x^2=-2

Suppose the fields were isomorphic, h: Q(i sqrt2) -> Q(sqrt2)

then h(x)^2 = -2

[formal ermine]

why is h q invariant

[chilly saddle]

because h respects division

and natural numbers

[formal ermine]

can you prove that h is q invariant pls

[rustic crown]

h sends 1 to 1 by ring hom assumption

so fixes Z

[chilly saddle]

h(p/q)=h(p)/h(q) =p/q

[rustic crown]

another way to say this is by using that Q is the initial field in the category of char 0 fields

so any map between fields gives you a triangle
Q --> E --> F

which says Q is invariant

[chilly saddle]

h(p)=h(1+1+...+1 (p times)) = h(1)+...+h(1) (p times) = 1+...+1 (p tines) = p

I like concretness lol

[rustic crown]

me like abstractness

[formal ermine]

ah

ok thanks

[chilly saddle]

I love category theory still 💀

Computing limits in finite concrete categories is my job

[rotund aurora]

Now idea how to even start

I can't even see why $\text{Tr}(\dfrac{1}{p'(\alpha)})=\dfrac{1}{[E\colon F]a_0}$

[cloud walrus]

Croqueta

[rotund aurora]

if sigma alpha=r then alpha*sigma(p'(alpha))=-a0 for r!=alpha

where sigma is an automorphism

[formal ermine]

if f in R[x] has odd degree, then it has at least one real root

why cannot we prove this algebraically

like if we look at its root inside the algebraic closure

the complex ones appear pairwise

and if all were complex then there would be a complex number s.t. z = z bar

ah wait

I think I know where the mistake is

I'm assuming that they are all complex

but what if they lie outside of C

right

[chilly ocean]

[formal ermine]

tterra

this is without fta

I'm trying to prove fta purely algebraically

[formal ermine]

I just gotta prove this using algebra now

[chilly radish]

I'm pretty sure it's impossible

[rotund aurora]

some help is appreciated, please. So if p(x)=(x-a)q(x), then p'(a)=q(a), and we should have Tr(q(0)/q(a))=1, I think. But I have no clue why this should be true

[rotund aurora]

I could prove it for alpha an nth root of a rational number, but this case is nearly trivial

[south patrol]

Good luck lol

You'll need to use smth special about R compared to Q for example

So like the proof will use smth like completeness

[void cosmos]

yes

[sly rain]

yes to the first or second question?

[warm wyvern]

Does this mean $N\cong M_1\oplus M_2$?

[cloud walrus]

DarQ

[warm wyvern]

[elder wave]

yes

[warm wyvern]

And if so, why the apostrophe in the second sequence?

[elder wave]

this is how they were introduced to me

i think it shouldn't matter

because you can swap the terms in a direct sum with isomorphic ones

but maybe someone else knows a concrete reason why your book phrased it like this

[warm wyvern]

I see

Thx for the sanity check

[elder wave]

yw

[warm wyvern]

There's an MSE post about this btw

Or was the MSE post proving the FTA assuming only odd polynomials have a root?

One sec

https://math.stackexchange.com/a/166012

Seems there are none

[elder wave]

yeah we told him that but he didn't want to believe us

[delicate orchid]

5 lemma innit

[rotund aurora]

the proof using Galois theory is neat, but you require to show that every odd degree polynomial has a real root, and you can't do that without analysis probably.

Is there any relationship between the trace of a matrix A and that of its inverse ?

[vagrant zinc]

Guys for exercise nine how could I check this, in one example I was given the matrix that this, but I don't know what to do exactly.

[chilly ocean]

one way to look at it is that you're asking if there's a nice relation between x + y + z + ... and 1/x + 1/y + 1/z + ..., but that might be too shallow (x, y, z, etc. eigenvalues)

[vagrant zinc]

:c

[delicate orchid]

ohhh yeah heisenberg group time

[delicate orchid]

determinant is a homomorphism

[sly rain]

Sorry, posting this here again (deleted my post here, because I used the math-help function)
Does "a zero divisor free, commutative, finite ring" imply that it cant be the zero ring?
Isnt the zero ring zero divisor free, commutative and finite?

[vagrant zinc]

Here it is in another Springer book but he takes it for granted D:

[vagrant zinc]

I don't know about that

[chilly ocean]

if you know what the determinant is then certainly you know that det(AB) = det A det B

[vagrant zinc]

yep

[rotund aurora]

mmh I guess since the inverse of a matrix is a multiplicative concept, the two don't fit quite nicely in general

[delicate orchid]

so you do know that the determinant is a homomorphism

[vagrant zinc]

I don't know what homomorphism is

[rotund aurora]

https://en.wikipedia.org/wiki/Homomorphism#:~:text=In algebra%2C a homomorphism is,form" or "shape".

Determinants are an example

[delicate orchid]

wack ass pedagogy

[rotund aurora]

[vagrant zinc]

I have already read something about it from my classmates, but I haven't seen the course and I can't understand that term either, but ok, I know about that property of determinant.

I am the guy who is studying the course in a self-taught way.

[rotund aurora]

did you just sully yourself?

[vagrant zinc]

no, I'm going to see the introduction to abstract algebra soon, and I want to take topics ahead of it.

[fallow bronze]

can anyone explain the underlined part of this proof

[next obsidian]

You want to find an element of order p

If y^p = e, then y is that element

Else, y^p has order q meaning y^pq = e, and that’s the smallest such number for which it’s true

But then y^q has order p cuz (y^q)^p = y^qp = e

[fallow bronze]

I cant really still see where the order q comes from

[next obsidian]

It’s just, the order

If y^p isn’t e, then it has some order not equal to 1

call it q

Oh this is a little different than what I said sorry

y^p is in <x> and x has order q, a prime number

So y^p = x^k for some k, 0<= k <= q-1

Because <x> = {e,x,x^2,…,x^q-1} since x has order q

Now if y^p = x^0 = e, then y is your element of order p

Else, y^p = x^k with 1<= k <= q-1

But all these elements have order q because x had order q, and q is prime

Then because p and q are coprime it will follow that y^q has order p

I guess the easiest way to do this is to show y has order pq

Clearly y^pq = e so the order of y has to divide pq

This leaves only that y has order 1,p,q,pq

It can’t be 1 by assumption, we already handled the order is p, so we just need to rule out q

But the order of y is divisible by p because y<x> has order p, indeed if n is the order of y then y^n<x> = e<x>, so the order of y<x> divides n.

Thus the order is pq

[flint crater]

How does one show irreducibility of the polynomial in the screenshot in Q[X, Y]? I thought about using

[void cosmos]

is Q the rationals

[next obsidian]

Eisenstein on (x-1)

[void cosmos]

and you can think of this

[next obsidian]

As a prime ideal of Q[x]

Considering this as polynomials in y over Q[x]

[void cosmos]

as a polynomial ring with only 1 variable over the ring of polynomials of the other variable

[void cosmos]

is this the same thing im saying?

right?

[next obsidian]

Yes

[void cosmos]

cool

[next obsidian]

It factors as Y^4 + (X-1)Y^3 + (X-1)Y^2 + (X-1)X

[void cosmos]

yeaa

[flint crater]

Yes sorry

[next obsidian]

It doesn’t have to be over the rationals

[vagrant zinc]

I have already tried det(AB) :c

[next obsidian]

This doesn’t factor so long as the base ring is a UFD

[flint crater]

Right I thought of using eisenstein considering polynomials in Y over Q[x] haha. Asked in one of those help channels first but thought I would confirm here

[next obsidian]

[flint crater]

Alright got it. Thank you :)

[vagrant zinc]

[next obsidian]

Isn’t SL defined as things of determinant one

And det is multiplicative

So

1•1 = 1

No?

[vagrant zinc]

what

no

det(AB)=1

[next obsidian]

Wut

[vagrant zinc]

Then it would be a homomorphism, as I was told above.

[next obsidian]

What are you saying

Aren’t you trying to show the product of two things in SL_2 have determinant one

I’m saying that det(AB) = det(A)det(B)

And by assumption if A and B are in SL_2

Then you’re just showing 1•1 = 1

[coral shale]

SL_2 is a group, qed

[vagrant zinc]

i saw it wrong i thought it was 1-1 =1 xde

my bad

[next obsidian]

Bruh

[vagrant zinc]

Let's see, I asked if the product of two matrices in SL_{2}(R) has determinants one.

So I was told to try
det(AB)=det(A)det(B)

I have already tried what I have been told so I could conclude that the product of two matrices that are in SL_{2}(R) have determinant 1.

[delicate bloom]

I don't understand this comment, are you still confused or are you summarizing what you've done

[vagrant zinc]

Sorry, I use the translator
is what I have done

[vagrant zinc]

this would be the most formal test

[warm wyvern]

Longer=more correct

[rotund aurora]

Can I tell wolframalpha to write the gcd of two polynomials as a linear combination in this two polynomials ?

[coral shale]

found this https://www.wolframalpha.com/widgets/view.jsp?id=18ca12c754852a11ca8ffbbeeee77b76

,w PolynomialExtendedGCD(x^3+x^2, x^2+2x)

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Shuri2060#9525

[coral shale]

oh there we go

https://reference.wolfram.com/language/ref/PolynomialExtendedGCD.html

[rotund aurora]

thanks

Q(something algebraic) is a field, but calculating inverses is ???

so painful

[coral shale]

?

nvm i c

[coral shale]

this is the process to do it

[rotund aurora]

I mean something like calculate the inverse of 6x^5-24x^3-18x^2+24x-36 where x=sqrt 2+cbrt 2 expressed in the basis (1,sqrt 2, cbrt 3, cbrt 9, sqrt 2 cbrt 3, sqrt2 cbrt 9)

[rotund aurora]

yes but I meant that the calculations are annoying to do by hand

[rustic crown]

calculations are always annoying to do by hand >.<

[rotund aurora]

but gcd with integers is so easy

and bezout

[rustic crown]

i would often just reduce it to something a computer can easily do and call it a day

polynomial gcd is the same then

[sonic coral]

is this good or am i missing a step in the last line

[long geyser]

hint: do (x^-1y^-1)^-1

[coral shale]

this proof can be a one liner

[long geyser]

although you can take the eqn you got and multiply by xyyx or sth

[sonic coral]

this gives yx

[long geyser]

yeah and it also gives xy

which is the point

[sonic coral]

since ((xy)^-1)^-1) = xy?

[long geyser]

uhh

[sonic coral]

im not sure where we get the xy from

[long geyser]

yx holds in general xy is what you get by the assumption

use the assumption

I can't hand hold any more than this

[sonic coral]

lol okay let me think

is it not just since we assume (xy)^-1 = x^-1y^-1

then we take the inverse of each, the rhs gives (x^-1y^-1)^-1 = yx

[coral shale]

theres a few ways to do this

[sonic coral]

the left gives ((xy)^-1)^-1 = (y^-1x^-1)^-1 = xy

so were done

[coral shale]

this works also
(xy)' = x'y' = (yx)'
xy = yx

[long geyser]

you are making it very complicated

for some reason

I had in mind, (x'y')' = x''y'' by assumption

and also (x'y')(yx) = e so you get xy = yx

but as aforementioned there really are too many ways to do this

[sonic coral]

is the way i did it fine though

although probably not the most obvious

[coral shale]

probably because its plaintext, i find it hard to read

and i think u did some unnecessary things

[long geyser]

I barely understood what you did so

[coral shale]

theres no mistakes, probably

[long geyser]

oh yeah that is very clean too, (xy)' = x'y' means (xy)'' = (x'y')' = yx
I assume this is what you wanted to do more coherently

[coral shale]

also you need to prove both directions, so if your steps are clearly bidirectional, that would be easiest

[long geyser]

that is true, while you already did the other direction you could do both directions at once if you are smart enough about it

not necessary though

[sonic coral]

yes this is what i was getting at

sorry i’m not the best at explaining my thoughts

since inverse is denoted with ^-1, how should i notate the inverse twice

like i typed it previously or

[long geyser]

yes

,,(x^{-1})^{-1}

[cloud walrus]

lems

[sonic coral]

perfect, thank you both

[somber thorn]

any idea on how to solve this with group theory?

[tender wharf]

this looks like a number theory problem

are you sure it needs to be solved by group theory

[somber thorn]

its one of the exercises in a group theory chapter of a textbook

so yeah

[delicate bloom]

Binet's formula for Fibonacci numbers has $\sqrt{5}$ in it at worst, so the formula is valid in $\mathbb{F}p$ or $\mathbb{F}{p^2}$ and since it's a difference of powers raised to a multiple of $p^2-1$, it simplifies down to $1-1=0$. Special care has to be taken for $p=2$ and $p=5$ since they're in the denominator, probably isn't too bad

[cloud walrus]

Merosity

[somber thorn]

i feeel like it's an application of lagrange's theorem for the orders of group elements

[delicate bloom]

it is

p-1 and p^2-1 are the orders of the multiplicative group of the finite fields

[somber thorn]

so we're basically considering $(\mathbb{Z}/p\mathbb{Z})^\times$?

[cloud walrus]

sean

[delicate bloom]

only when 5 is a quadratic residue

otherwise I'm adjoining it to the field and getting a field with p^2 elements

0 is not in the multiplicative group, so there are p^2-1 elements

[somber thorn]

the thing is, they havent mentioned binet's formula and have only barely introduced groups and lagrange's theorem, so using facts from field theory is a bit of a stretch i think..

[delicate bloom]

bummer

who's "they" btw

[somber thorn]

evan chen

this is from evan chen's napkin book

[delicate bloom]

the spicy pepper makes me think it's supposed to be hard like using things like this for all I know

that book is for competition math isn't it?

You could be expected to know Binet's formula

https://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html it's not really that complicated if you've studied linear recurrence relations you can get it pretty simply too

[somber thorn]

no, but it is aimed at students trained in comp math looking to get into college math

[somber thorn]

hmm i see

ill try using this then

[delicate bloom]

something just occurred to me, since my proof was more general and yours was pretty specific

ah nevermind, I was thinking the 2 in there would make the sqrt(5)^2 = 5, but it's squaring the golden ratio nvm

[plush wasp]

Does Z[x]/(x + 1) denote the ring of all polynomials of the form p(x)/(x + 1)^k where p(x) is from Z[x] and k from Z \ {0}?

[maiden ocean]

No

[delicate orchid]

nope, it denotes the quotient ring {p(x)+(x+1): p(x) in Z[x]}, where (x+1) is the ideal generated by x+1

[maiden ocean]

You should read up on the definition of a quotient ring

or quotient group

[plush wasp]

oh also, what's Z(x)?

[delicate orchid]

Z[x] but with x^-1

[analog zephyr]

Hi

[plush wasp]

so if I were to consider a ring that contains elements of the form p(x)/(x + 1)^k where p(x) is from Z[x] and k from Z \ {0}, is there no notation to define it?

[maiden ocean]

There is

[delicate orchid]

there is, if you include k = 0, that's the localisation of Z[x] at x+1

[maiden ocean]

Well

Yeah you have to include k = 0

Also it is the localization at the element x + 1

[delicate orchid]

yes, sorry

[maiden ocean]

Not the prime ideal x + 1

very different though related

[plush wasp]

wait (@_@;) I meant Z[x]/<x+1>

[maiden ocean]

Again that notation denotes a quotient ring

you can find definitions of it online

[plush wasp]

alr! I'll check

[maiden ocean]

localization is an extremely important construction but you dont really need to learn about it if you don't have a handle on the content of an intro algebra course (including quotient groups and rings) yet

[untold cloud]

Hi, guys, we can show that any vector space has a basis by zorn's lemma, but it is not true that any module has a basis (e.g. Q is a Z-module). But why module is not the case?

what stop modules having basis

[chilly ocean]

units

[runic hemlock]

the point where the proof over fields fails is in the statement that if S is a linearly independent set of vectors and v is not in span(S), then v u S is also linearly independent. this is no longer true over non-fields.

[chilly radish]

This is why I like to saying that you localise at an element but away from a prime

I know I know there's a geometric meaning here also but idc

[untold cloud]

Thank you!

[rotund aurora]

How do you read this?

I don't know how to read sequences lmao

K is the field of fractions of O

J_K are the fractional ideals of K

[elder wave]

i'm not sure what the question is, are you asking what the maps are

[rotund aurora]

What are the 1 s?

[elder wave]

trivial groups

[rotund aurora]

uh ok, I think I was mixing rings and groups

[glossy crag]

First 2 arrows are inclusions, 3rd arrow sends a to aO, 4th arrow is the canonical projection, last arrow is the constant morphism.

[chilly ocean]

show that every group of odd order is solvable

[formal ermine]

show that $|2^X| < |2^Y|$ implies $|X| < |Y|$

[delicate orchid]

show that if $g \sim h$ then $gz \sim hz$ for all $z \in \mathcal{Z}(G)$

[cloud walrus]

Wew Lads Tbh

[chilly radish]

show that if Ext^1(A,Z)=0 then A is free abelian

[coral spindle]

Try finding all groups that have no nontrivial normal subgroups. Shouldn't take too long

[formal ermine]

lmao

[delicate orchid]

finding the groups with no non-trivial subgroups 😌
finding the groups with no non-trivial normal subgroups

[chilly radish]

that might be a bit tricky for someone who hasn't done a lot of algebra. Try and just find the finite ones, that's more reasonable

[formal ermine]

try finding a galois extension over Q for every group

nice challenge for an evening

[rotund aurora]

As a warm up, show that every solvable group occurs as a Galois group over Q

[elder wave]

where did all that come from

[wooden ember]

This question came up on my group theory exam and I was stuck for the last 2/3 of the exam in it so I’d appreciate a hint. Let G be a group of even order acting on a group K of order 3^57^313^2. Show that for p=3,7,13 there is a Sylow p subgroup P such that G fixes P (fixes in the sense of the induced action of G on the set of sylow p subgroups)

[next obsidian]

Wtf is that number

[wooden ember]

All I managed to find was a stabilizer of order 2^km where |G| = 2^kn with n>=m but I didn’t manage to get anything with n=m (tried an argument where I said this stabilizer contains all sylow 2 subgroups of G and thought I might be able to increase the size of the stabilizer this way but it was a no go)

[wooden ember]

The number itself is probably irrelevant

But I’m not sure

[next obsidian]

Is that ((3^57)^313)^2

[wooden ember]

Oh sorry lol

[formal ermine]

I assume it's 3^5 * 7^3 * 13^2

[wooden ember]

Yep

[next obsidian]

Oh lmfao

[wooden ember]

Forgot * behaved weirdly on discord

[next obsidian]

Yeah I don’t get it

Do you have more assumptions on the action?

I don’t see why a group action would permute subgroups

Like G acting on itself by left mult doesn’t send a subgroup to a subgroup

[wooden ember]

It’s a group action on a group

So elements are sent to automorphisms

[next obsidian]

That’s a little ambiguous, but sure

[wooden ember]

Yeah in the context of the question it was clear should’ve made that explicit

[next obsidian]

Oog

[formal ermine]

what does the notation $\bL(\bK)$ for $\bL, \bK$ fields mean?

[next obsidian]

@wooden ember is the action faithful?

[wooden ember]

Not specified anywhere no

[next obsidian]

Then sham brought up a point

Which is

If G was any group acting on K, you can make an action of G x Z/2Z

By acting trivially with Z/2Z

Now you have an even order group acting on K, which I think still acts via automorphisms

So it’s trivial

But then G’s action is trivial

[wooden ember]

G is an even order group already

But why should its action be trivial

[next obsidian]

Err

Oh

You’re right

[chilly radish]

you're trivial

[next obsidian]

Oh sorry

What I mean is

There’s a Sylow p called P

Which is fixed by the G x Z/2Z actuon

So it’s fixed by the G-actuon

[wooden ember]

Why?

This is what I’m trying to show 🤔

[next obsidian]

Because G x Z/2Z is an even order group

Right I’m saying

As written

This theorem implies any group acting on K

Will get a fixed point

[wooden ember]

Oh yeah I agree

[next obsidian]

So this seems fake

[wooden ember]

I also think it’s false if that’s what you’re getting at

[next obsidian]

Actually I can prove

It’s not true

[wooden ember]

I just couldn’t get a counterexample because of the size of K but this is smart

[next obsidian]

Because if as written this is true

Then Aut(K) has a fixed point

But as long as K has non-unique Sylows

This is impossible by some Sylow theorem cuz of conjugation

[wooden ember]

Yeah I agree

[next obsidian]

So wait

was the problem a prove or disprove?

[wooden ember]

Wait no

Why would Aut(K) have a fixed point

[next obsidian]

BecUse if you have the theorem for even order groups

[wooden ember]

No it was just prove but I feel there’s a mistake in the question

[next obsidian]

Let Aut(K) x Z/2Z act on K

Letting the Z/2Z do nothing

[wooden ember]

Ah right yeah ofc

[next obsidian]

Then this action has a fixed point

[chilly radish]

why are you assuming that all the automorphisms are inner tho?

oh wait

[next obsidian]

Shin it won’t matter

[chilly radish]

the entire action has a fixed point

[next obsidian]

Yeah

[wooden ember]

Yeah that sounds good to me

[chilly radish]

yea you're right

[wooden ember]

well fuck that question then

[agile burrow]

I'm confused

Oh never mind, I think I understand

[wooden ember]

I think the teacher might’ve gotten mixed up cause another question temporarily assumed G to have order 2^k and in that case it’s easy

Tried asking a TA if the exam had a mistake and the dude legit said « I don’t know the exam can’t help you »

[chilly radish]

bruh

[elder wave]

TA
i don't know the exam

[next obsidian]

Yeah

Then you do class formuler

[next obsidian]

Good job Walter

[flint crater]

Perhaps a bit of a dumb question. Is a PID a domain by definition, meaning we take the ring to be a domain, or is it because when every ideal is a principal ideal it somehow implies it is a domain?

[runic hemlock]

it's part of the definition

for example Z/4 isn't regarded as a PID even tho every ideal is principal

[flint crater]

Ahh yes good example. Thank you!

[formal ermine]

when is an ideal in Z/nZ not principal

[delicate orchid]

quotient a PID
look inside
principal ideals

[formal ermine]

explain

is R/pid always a pid

[delicate orchid]

no clue if this is true so I'm going to go purely off of intuition and det can correct me

I is an ideal in a PID R, say I = (x), then mapping I into R/J through the canonical map should give (x+J)

this probably breaks in some case, which det will now tell you about

[elder wave]

never

being a principal ideal ring is preserved when taking quotients

the difference between PID and PIR is important here

[formal ermine]

hm

how do we show that Z is a pid?

by induction?

[elder wave]

no

Z is an euclidean domain

[formal ermine]

how do we show that it is euclidean

[rustic crown]

by giving a euclidean valuation

:p

[elder wave]

you define the absolute value as said valuation

[formal ermine]

is every non finite ring a domain

[elder wave]

wat

no

[formal ermine]

yeah but how do we show that it works

[elder wave]

take R x R for any non trivial R

and infinite R

[formal ermine]

multiplication defined component wise?

[elder wave]

yes

[rustic crown]

because euclid's division thingy. say a is an integer and b is a non-zero integer, look at the smallest natural in the set a + bZ

[formal ermine]

ah

ah

how do we show that Z is a domain

[rustic crown]

absolute value

or square

ab = 0 then a^2 b^2 = 0

[formal ermine]

wdym

[rustic crown]

but these natural

if both non-zero then they positive

so prod is positive

[formal ermine]

but how do we know this

[next obsidian]

Bruh it’s Z you learned it was a domain in like 3rd grade

[rustic crown]

[formal ermine]

yeah but how do we show it formally

[rustic crown]

that's kinda the definition of order in Z

[next obsidian]

What’s your formal definition of Z?

[rustic crown]

Z is an ordered ring, such that positive elements are well ordered

[formal ermine]

N u -N

[next obsidian]

What’s your definition of N

What’s your definition of multiplication

My point is that these are things you need to be able to answer if you want to show it formally, but you also don’t gain anything doing this

It’s just an exercise in tedium

[elder wave]

[next obsidian]

It works exactly how you learned arithmetic works in like, 1st grade even

It’s in a sense made exactly so that it works that way

But if you believe it’s N u -N, then it’s a domain because the Cartesian product of nonempty sets is nonempty

Because you’re probably going to write numbers as certain sets defined recursively with 0 being the empty set, and multiplication being the Cartesian product basically

You could also use Peano arithmetic

[formal ermine]

ah von neumann

[elder wave]

oh god

[next obsidian]

If you accept that the sum of positive things is positive

Then you can define 1•m = m

(n+1)•m = n•m + m

To recursively define multiplication

[knotty frigate]

oh this is smth i wanted to ask

[next obsidian]

This shows that positive numbers multiply to nonzero things

[formal ermine]

thanks chmonkey

[next obsidian]

Then treat - as like, a symbol

[knotty frigate]

how can you set up a first order theory of the integers

[rustic crown]

you could use a lot of stuff for constructing Z, but these will involve a lot of unnecessary choices. best thing to do is use your favorite construction but later give a list of properties which uniquely characterize Z, that way you won't have to think again about how an integer is defined

[next obsidian]

It’s just tedious to do this and all you get after trying is

[knotty frigate]

in a peano esque way

[rustic crown]

this is a cute one line description of Z

[next obsidian]

The same exact thing you had when you were in 1st grade learning about + and •

So I don’t think it’s that useful lmao

[rustic crown]

it can be useful if you're learning rigor, but beyond that... nothing :p

[formal ermine]

I was just curious

[next obsidian]

You have to pickup a like

Elementary set theory book

Or look into Peano arithmetic or something

The problem to me is that you probably don’t know a formal model for Z

[coral spindle]

Fwiw you can show Z is a domain by showing multiplication in N is cancellative, which is a fairly tricky thing to do from the Peano axioms and definitions

[formal ermine]

I knew once but I forgor

[next obsidian]

So it becomes impossible to actually dissect why it’s a domain

Yeah so you have to pore into the construction of a model and what x is

To answer it

[coral spindle]

N.b. this is also why we simply assume that R is a field lmao

[next obsidian]

Yeah

I think this is less fundamental in a sense though

[coral spindle]

It definitely is

[next obsidian]

Because showing you complete a field to a field

Is useful in other parts of math

And might need to be reproduced again for eg p-adics

I think an actual formal construction of N is…

Not that important

[rustic crown]

yea, and another thing is eventually you're going to accept a few axioms anyway. like construction of N would boil down to the axiom of infinity + a lot of work. might as well change axiom of infinity to "N exists"

[next obsidian]

I guess my last point is

I think Peano arithmetic is reasonable to have someone new to proofs go through

But if you’re doing abstract algebra and field theory and stuff at that point I think you’re past the point you get much development from it, so it’s only something I’d say you should go through if you just feel like it

[next obsidian]

Tbh in my head it is

I know it just asserts there’s some infinite set and then you do crap to get N but

¯_(ツ)_/¯

[rotund aurora]

what is a set? 🤓

[next obsidian]

Please ask #foundations such questions will not be heralded in #groups-rings-fields

[rustic crown]

people in#foundations probably have a lot of patience

i always want to learn some model theory, but never have enough motivation to do so >.<

[chilly radish]

Hello wildberger

[chilly radish]

A collection of elements

I will not be taking further questions

[warm wyvern]

A thing that obeys the ZFC axioms

[simple mulch]

I understand but there's something that is not connecting in the following question: IS the set of the three reflections about the vertices of an equilateral triangle a transformation group?

A transformation is a bijective map from a set onto itself

A group if considering all transformations in a family of transformtations has a group structure

Now, suppose we have something like /_\ with the upper vertice being A

[formal ermine]

you probably mean symmetric group rather than transformation group

[simple mulch]

Well yeah but we do not have all the symmetries since we do not have rotations

[formal ermine]

Sn can be defined as all bijections from a set of n elements onto itself

[simple mulch]

This specific case would be a subset of Sn

[coral shale]

where n is some cardinal i hope.

[formal ermine]

ah

[simple mulch]

We should only consider reflections

[rotund aurora]

I think mns wants to see if those three reflections from a group

[simple mulch]

yeah

But I think it does not

[delicate orchid]

they do, it's S_3

[rotund aurora]

no

he/she asked about the set of three reflections

not the set generated by those reflections

I think

[delicate orchid]

ah I see

[rotund aurora]

two reflections is a rotation

[simple mulch]

ye!!

[delicate orchid]

same thing to me

[simple mulch]

reflect on vertice A

[rotund aurora]

reflections are not closed under composition

[simple mulch]

We can't go back to this initial using only reflections?

[delicate orchid]

reflections are not closed under composition

[simple mulch]

ohh

yeah

[formal ermine]

reflections are not closed under composition

[simple mulch]

good point

xD

[rotund aurora]

ok

[south patrol]

reflections are not closed under composition

[rotund aurora]

my message was not responding to the picture, obviously

[simple mulch]

thanks

[chilly radish]

C

[rotund aurora]

reflect again...

[warm wyvern]

Cope

[chilly radish]

You don't need choice to define sets

[elder wave]

☝️

[warm wyvern]

People define sets?

[formal ermine]

you don't?

[warm wyvern]

Informally ig

[delicate orchid]

yeah a set X is a set containing all elements of X

[sonic coral]

If i have a group G of order 2n, how can i show that there is atleast one element of order 2

so far i have that there is an even amount of elements in G

and so we have an odd amount of non identity elements, 2n-1

but i feel like we would have to have an even amount of elements in G with an odd order

but im not really sure if thats true or even the correct direction to go

[coral shale]

try to construct that element

im guessing.

[sonic coral]

what element?

[coral shale]

.

[south patrol]

That is not how the proof usually goes lol

The proof is typically nonconstructive

[simple mulch]

FYI

[coral shale]

oh nvm use this fact

[formal ermine]

did an analyst write that

[sonic coral]

i think i should pair the elements in some way

[coral shale]

yh

[delicate orchid]

guys can someone help me with my lin alg homework

[south patrol]

Yeah, sure

[delicate orchid]

can u prove this for me??

[formal ermine]

!help

[flat tree]

Please read #❓how-to-get-help

[formal ermine]

wtf

[simple mulch]

Vladimir Arnold

[formal ermine]

,,\varpi

[coral shale]

bet its induction

[delicate orchid]

yur varpi_n is a nth root of unity in C and omega_n is a nth root of unity in Z/p^2Z

[formal ermine]

the conjecture is false

a counter example is trivially found

[sonic coral]

cant i like pair each element in G with its inverse, since theyre all unique.

[delicate orchid]

indeed you can

[sonic coral]

meaning that since one element has order 1, we need another to also be its own inverse, therefore being order 2 since there can only be one identity element

idk if im explaining it right but thats how im thinking about it

[next obsidian]

Is this like

Sum of all elements is the unique involution

[delicate orchid]

group of order 2n has unique involution

[next obsidian]

That’s not true

The number of involutions is odd is all you can get

[sonic coral]

oh god

[delicate orchid]

am I forgetting what an involution means

[next obsidian]

Klein 4 has 3

Involution is self-inverse

[delicate orchid]

right yeah but that's order 4, not 2n

n is coprime to 2 - I presume

[next obsidian]

Bruh

[south patrol]

what

[next obsidian]

Br.uh

Okay well again not true

S_3 has many

[delicate orchid]

ah nevermind I misread the question

[next obsidian]

And by many I mean it has like, 3

[sonic coral]

well if its odd its fine, since 1 is odd

and i needed to show at least 1

[next obsidian]

So what the statement?

Ah yeah

Then you do what you did that’s exactly right

[next obsidian]

In general because of that pairing up thing, you can show that |G| and # of involutions have opposite parity

[sonic coral]

that makes sense to me

[next obsidian]

Nice

I knew I should’ve ignored wew

All he does is sow chaos

[delicate orchid]

I would never

[sonic coral]

im glad you didnt i learned what an involution is

[next obsidian]

🗿

I just like to call them involutions because my TA who made our algebra psets

Just included the word randomly over and over

So we just thought it was really funny

[sonic coral]

its fun to say

[next obsidian]

Yeah

[sonic coral]

i was anxious about learning abstract algebra bc every math major ahead of me who has taken it made their worst grade in it

but i think im getting it slowly

[next obsidian]

Swag

True story:

When I started learning algebra I thought it didn’t make sense

And said “I think my brain just is built for analysis”

[south patrol]

When I started algebra I thought it made sense and now I think it doesn't make sense

[delicate orchid]

another fun involution question: show that if a group has exactly one involution, that involution has to be in the centre

I nearly failed first year group theory

[sonic coral]

group theory is its own class? bruh

[delicate orchid]

group theory is its own sprawling area of mathematics

[next obsidian]

Wew lives in England-land

Logic doesn’t apply

[sonic coral]

we will cover ring theory im sure

[delicate orchid]

why wouldn't it be it's own class?

[next obsidian]

I like rings

[sonic coral]

i saw all of this stuff when i took the gre and just used my guess letter for them lol

[next obsidian]

Tbf my first exposure to algebra

Was a qtr of group theory

And then 1 qtr of…

Rings, modules, field theory, and Galois theory

Wait wtf

[chilly radish]

I don't believe you

[delicate orchid]

it's true

[chilly radish]

oh yea?

[delicate orchid]

yeah, nerd

[chilly radish]

prove it

[next obsidian]

ShiN here’s a hint: ||get good||

[delicate orchid]

well that would spoil the fun wouldn't it

[south patrol]

here's a hint: ||praise the sun||

[chilly radish]

shhh chmonkey I am gaslighting wew into thinking he got it wrong

[next obsidian]

Oh okay ShiN, proceed

[delicate orchid]

I never think I'm wrong (I am delusional)

[south patrol]

I think I'm always wrong (I am correct)

[chilly radish]

btw I am saving these 2 questions and will give them to my students

[warm wyvern]

Is all notation in grad math such a messy abomination?

Wtf is sideways Beta??

[next obsidian]

Lol

That’s a varpi

[delicate orchid]

you don't want to see the notation before I cleaned it

[next obsidian]

Also you haven’t even seen nG-math

[coral spindle]

Nice this is good exercise

[warm wyvern]

This is cleaned???

[south patrol]

One of my friends' papers for example

[warm wyvern]

This is a parody, right?

[coral spindle]

Oooooh sometimes things are complicated ooooh so scary

[chilly radish]

noncommutative algebra has neat notation

[warm wyvern]

This does look neat tbh

Tho, the composition is a bit tilted

5/10

[dawn latch]

for an operation to be a binary operation, does it need to be closed?

[chilly radish]

that is usually implicit in the definition of 'binary operation', yes

[coral spindle]

Now we hope they weren't talking about topology

[chilly ocean]

discrete topology

[vivid tiger]

in my opinion, differential geometry has a very clean notation system...and another very clean notation...and a million others, all used in the field. there's a reason why it's sometimes said to be "the study of objects that are invariant under changes in notation"

[vagrant zinc]

Guys for this exercise I have to show that it is a group.

My idea is the following
a_i+b_i=(a_1+b_1,...,a_n+b_n)

now let c_i=(c_1+...+c_n) which belongs to Z_2 for all i=1,2,...,n
then we prove that it is commutative
therefore

(a_i+b_i)+c_i=(a_1+b_1,...,a_n+b_n)+c_i
=(a_1+b_1+c_1,+...+a_n+b_n+c_n)
=(a_1+(b_1+c_1)+....+a_n(b_n+c_n))
=(a_1+...+a_n)+(b_i+c_i)
=a_i+(b_i+c_i)

the identity would be e which belongs to Z would be 0

[tender wharf]

you haven't showed there's an inverse

[rigid cave]

lmfaoooooo

[vagrant zinc]

It would be as follows if a_i belongs to Z_2^n then there exists an inverse such that -a_i belongs to Z_2^n, therefore
a_i+(-a_i)=(a_1+(-a_1)+...+a_n+(-a_n)
=(e_1+...+e_n)
where e is the identity belonging to Z_2^n then
a_i+(-a_i)=(0,...,0)

[south patrol]

Weird question: say we have A in M_n(Z). We know that A satisfies a monic polynomial with integer coefficients, but is there any obvious way this works without using Cayley Hamilton? Over a field one can just use linear dependence of some big set of powers, say

But idk if there's some like super easy way to see this

[next obsidian]

M_n(Z) is a finite free Z-module

In fact it’s an algebra over Z

[south patrol]

Ye

[next obsidian]

It follows that any submodule is,

And so there hs to be some dependence between the powers of A

Otherwise you’d have a free submodule of unbounded rank for example

[south patrol]

Yeah sure, it's from there on that it seems harder I mean

[next obsidian]

This is like some general ring theory

In fact no

Like

[south patrol]

Nah I mean like I'm fine w what you said, just how do you know you get a monic thing

[next obsidian]

Yeah so

Fuck the free thing

You know you’re finitely generated

So the Z-span of {A^n} has a finite spanning set

You can see from some finite spanning set you can extract that {1,A,A^2,…,A^k} is a spanning set for some k

[south patrol]

Yeah sure so then you have polynomials in A^n generating it and just pick a massivenpoeer of A

And write as a Z lin comb of smaller ones

[next obsidian]

Well no, like

There’s finitely many

[south patrol]

Sorry that was vague language I meant

[next obsidian]

So the degree is bounded

So pick the powers of A at that degree and lower

Then take an even bigger A^k

[south patrol]

You have polys in A generating it as a Z-module

[next obsidian]

That is written as a Z-linear combo of the smaller stuff

[south patrol]

Which is equiv to what u said

[next obsidian]

Yeah

[south patrol]

Yeah okay nice

Thats good and feels silly now cause I did that argument earlier anyway for smth slightly different

Thanks

[next obsidian]

ChmonkaS

[south patrol]

lol

I just didn't realise oops I forgo

[dawn latch]

ok thanks

i thought that a binary operation's closure is optional

[dreamy fiber]

If I have a map f: R^m \to R^n over some commutative ring R with unity, I can in particular represent f as a matrix, what do people mean when they say the "rank of f"?

[oblique river]

yes; i would interpret "rank of f" to mean "rank of the matrix" in the usual sense, like maximal number of linearly independent columns or something like that

[dreamy fiber]

thanks!

[charred crescent]

quick question, if i show that a potential subgroup (call it H) is closed under the operation induced by the group in question (call it G), and show the existence of inverses, must i also show that the identity of G is in H, or do it get that for free?

it would make sense if you could get it for free since an element operate on its inverse is the identity and H is closed so that identity is defacto in H

[tender wharf]

yes you get that for free

[next obsidian]

No

[tender wharf]

wait

[next obsidian]

You need to have it be nonempty

[tender wharf]

{} is not even a subgroup

[next obsidian]

Anyway this observation leads to the following:

A nonempty subset H is a subgroup if and only if for any x,y in H, xy^-1 is in H

[charred crescent]

so would this work?

[tender wharf]

are you convinced by your own proof?

[charred crescent]

yes but that doesn't imply it is sufficient

[tender wharf]

that's a first sanity check

takes a while to read the proof gimme a sec

why do you spam so many brackets btw

[charred crescent]

as in ()?

[tender wharf]

yeah

like (p_1)(p_2)

feels unnecessary and just clutters the proof up

[charred crescent]

fair point

[tender wharf]

e_2 = phi(e_1) immediately lets you conclude e_2 is in phi(H) by definition

[tender wharf]

this is a faster way to do it btw

[charred crescent]

haha i did not see that until now

[tender wharf]

what's y phi(h^1)

how on earth did we get h^1? typo?

[charred crescent]

typo probably

[tender wharf]

this is fine

[charred crescent]

yup

[tender wharf]

a bit long winded

[charred crescent]

okay, fair enough. thank you for your time 🙂

[tender wharf]

do clean up the brackets though

it really makes it difficult to read

[woeful sand]

also inclusion morphism

[tender wharf]

they may have not reached that point yet

[woeful sand]

i feel it's ok to write long proofs

[woeful sand]

as long as

and you don't lose yourself with many assumptions leading to wrong conclusions

[woeful sand]

it's a fairly common habit of mine tbh... so i know how bad it's

[barren sierra]

long proofs are fine

but I think it's good to practice trimming down proofs

just like you're supposed to edit essays before submitting them, you should edit proofs

tweak wording, trim fat, etc

(this is much easier if you typeset your work)

also this closer reading and editing forces you to pay more attention to what you wrote, forcing you to proofread your work (never a bad thing)

[woeful sand]

agreed

I might have a somewhat personal question to you guys.
Are y'all taught abstract algebra, categories and universal algebra in a intuitive way?

[haughty hollow]

Could someone help me with this? If I let e + fi, and g + hi, then (e + fi)(g + hi) = eg + ehi + fgi + fhi^2 = eg - fh + (eh + fg)i. How do I show that c + d is nonzero? From the first two elements, e + f is nonzero and g + h is nonzero. If you did (e + f)(g + h) = eg + eh + fg + fh is nonzero. But, you can't determine if eg - fh + eh + fg is nonzero or not since fh is positive and in the other expression, fh is negative. So, is this a binary operation or not? I'm confused.

[woeful sand]

Isn't it defined above? c+d != 0

[haughty hollow]

Yes, c + d != 0, but I'm trying to show that (e + fi)(g + hi) belongs in this set, so I'm trying to show that c + d is nonzero for this possible element.

[oblique river]

Saying cd=0 but c+d nonzero is just a fancy way of saying “exactly one of c and d is zero”

So start with two elements of that type

And multiply them

[haughty hollow]

@oblique river That's what I did above.

I took (e + fi)(g + hi) where e + fi and g + hi are in G.

[oblique river]

You never used that ef = gh = 0

Im saying that you should do it by cases.

One of e and f is zero and the other is not.

One of g and h is zero and the other is not

That gives you four cases. Do them all separarely.

[haughty hollow]

I see. Okay, let me try this out.

@oblique river So, whenever you see cd = 0, then you have to split it in cases?

Like cd = 0 implies c = 0 or d = 0.

Then, see if (e + fi)(g + hi) is an element of G under the assumption that e = 0.

Then, f = 0.

[oblique river]

You dont have to

But in this case it’s very useful to do so

[haughty hollow]

Alright, makes sense.

Thanks

[oblique river]

For complex numbers, “cd = 0” is the same as “c = 0 or d = 0”

And that happens to be useful here

[haughty hollow]

Yep, ok that makes sense. 🙂

Ty

[oblique river]

Np and gl

[haughty hollow]

@oblique river Quick question, for e = 0 case, (eg - fh)(eh + fg) = e^2gh + eg^2f - feh^2 - f^2gh. You get - f^2gh, but that's not equal to 0. So, I'm confused.

Wait, I think I'm confusing how to show the element is in this set.

Nvm, I think I'm right. I'm trying to show how (eg - fh)(eh + fg) = 0.

[eternal plover]

Hello, I'd like someone to tell me if my proof would be rigorous enough for a test.
Given Z and Z_{10}, prove that f(x) = [x]_{10} is a ring homomorphism, and describe Ker f. Is Ker f a subring of Z?
Proof: we can assume that Z_{q} is an additive abelian group and that it's closed under multiplication, so it inherits the moltiplicative identity from Z.
Given some x, y in Z , x \equiv z_{x} (10), y \equiv z_{y} (10) for z in {0, ... , 9}. Then, by modular arithmetic x + y \equiv z_{x} + z_{y} (10) -> [x] + [y] \equiv z_{x} + z_{y}. We know therefore that there is a corresponding element in Z_{10}. We can follow a similar approach for multiplication.
Ker f is an ideal but not a subring of Z because it does not contain 1.

[lament dawn]

Can you just put it in latex

Youve pretty much already written it out that way

If not maybe just write it readably

[warm wyvern]

Why is this set closed?

[cloud walrus]

DarQ

[elder wave]

what's phi_A

[warm wyvern]

The characteristic polynomial of A

[elder wave]

i mean there you have it then

you're looking at the zero locus of a polynomial

btw what is that horrendous notation, fixing some A, looking at it's characteristic polynomial and then using reusing A for matrices that you plug in

i assume this is for some fixed A because otherwise this set is the whole space

[warm wyvern]

I don't think it is fixed

And we're trying to prove it's the whole space

We're trying to prove the Cayley Hamilton with the zarisky topology

[elder wave]

oh that's the context

[warm wyvern]

Yea

[elder wave]

ignore my messages above then i did not assume one would reprove basic linear algebra with the zariski topology

[warm wyvern]

Is it basic tho?

We're dealing with an arbitrary field

[rustic crown]

the characteristic polynomial of A would be something in K[a_ij][x]
and when you ask phi_A(A) = 0, its like asking n^2 many polynomial equations in K[a_ij] to hold true

[eternal plover]

what is the syntax? mathjax?

[lament dawn]

Latex

Idk the compiler probably pdf

[elder wave]

doesn't matter

[chilly mantle]

mathjax is just latex put on websites basically lel so yes same syntax

[eternal plover]

Let's try:
Hello, I'd like someone to tell me if my proof would be rigorous enough for a test.

Given $\mathbb{Z}$ and $\mathbb{Z}{10}$, prove that $f(x) = [x]{10}$ is a ring homomorphism, and describe $Ker f$. Is $Ker f$ a subring of $\mathbb{Z}$?

Proof: we can assume that $\mathbb{Z}{q}$ is an additive abelian group and that it's closed under multiplication, so it inherits the moltiplicative identity from $\mathbb{Z}$.
Given some $x, y \in \mathbb{Z}$ , $x \equiv z{x} (10)$, $y \equiv z_{y} (10)$ for $z \in {0, ... , 9}$. Then, by modular arithmetic $x + y \equiv z_{x} + z_{y} (10) \implies [x] + [y] \equiv z_{x} + z_{y}$. We know therefore that there is a corresponding element in $\mathbb{Z}_{10}$. We can follow a similar approach for multiplication.

$Ker f$ is an ideal but not a subring of $Z$ because it does not contain 1.

[rotund aurora]

#latex-testing

[elder wave]

now i'm curious, what are you reading and why is it proving cayley hamilton like that @warm wyvern

[rustic crown]

i feel bad for merli now >.<

[warm wyvern]

It's John's stupid blog lol

[rotund aurora]

I'm just pointing out a place where merli can practice latex, in case needed

[elder wave]

oh i was actually on there just now

[cloud walrus]

Merli

[rustic crown]

(oh okie >.<)

[elder wave]

the first sentence says that it is a well known theorem and typically encountered in a first course in lin alg

[eternal plover]

Ok I think I'm done now, sorry x:

[elder wave]

[eternal plover]

I was wondering whether Discord uses MathJax or KaTeX or something else

but whatever it is it works I guess

[rustic crown]

what's KaTeX

[warm wyvern]

I've only seen it for K=C case

[elder wave]

it's an interesting read tho

[eternal plover]

https://katex.org/

afaik, it's a newer alternative to mathjax developed by Khan Academy folks

[rustic crown]

bash it with structure theorem of f.g. mods over pid

[warm wyvern]

I dunno structure theorem

[rustic crown]

do you see closed-ness from this?

[warm wyvern]

Not really

[rustic crown]

(if you know it for Z, then you know it for each ring :3)

[rustic crown]

okie, so you want each of these n^2 polynomial identities to hold. each of these will define a closed subset in the affine n^2-space, and so their intersection is also closed

zariski topology by definition had closed subset = zero locus of a bunch of polynomials

[warm wyvern]

I'm not really sure which nxn polynomial equations you're referencing

[rustic crown]

okie so lets take an example

A = [a b]
    [c d]

the char poly is

phi_A(x) = (ad-bc) - (a+d)x + x^2

so if you plug in x = A
you're gonna get a polynomial in each of the 2 x 2 entries with variables in a, b, c, d and coefficients in Z
and you want this matrix over Z[a, b, c, d] to be all 0s, which amounts to looking at the closed subset described as the zero locus of these 4 equations

(and since these coefficients will be integers, proving these identities hold over Z would automatically imply it holes over any ring)

(also since Z embeds into C, proving it for C implies it holds for Z)

[warm wyvern]

Oooooh

So like

In phi_A(A)

Everything depends on the entries of A

[rustic crown]

yep

[warm wyvern]

I see

That took a moment

Thank youu

[rustic crown]

it will get faster :3

[warm wyvern]

Hopefully

[rustic crown]

[wooden ember]

This is what I keep telling myself

[rustic crown]

(you're not alone )

[oblique river]

I think you are still really overthinking it. Elements of G look like either c or ci for c a real number

The product of any two elements of that form is also of that form

If you want to call your two elements e + fi and g + hi, then there are 4 cases:

1. e = g = 0. Then the product is (fi)(hi) = -fh = -fh + 0i which is of the correct form.

2. e = h = 0. Then we have (fi)(g) = fgi = 0 + fgi which is of the correct form

3. f = g = 0

4. f = h = 0

In your argument, you forgot that both elements must be in G

So one of g and h is 0, which means -f^2gh is 0

[upper pivot]

yeah the method can definitely be seen as overkill here ig, but its very useful to know, as in the future you can prove "continous" properties on just diagonalizables and not have to worry. (I have used this method a lot)

[elder wave]

Yeah I didn't see the bigger picture when i said that, after reading the entire thing i realized that it's more about the method than the theorem, which is an accessible example for said method

Sorry if that came across as me trying to bash the blogpost that was not my intention

[upper pivot]

oh dw i didnt feel that, i was just clarifying

this is an old blog post i wish i emphasized the method over thm a bit more

[south patrol]

lol a friend sent me that blog post recently and I'm like

😏

[upper pivot]

was i the friend lol

[south patrol]

No

[upper pivot]

damn getting popular fr

i should edit it to be better if ppl are actually seeing it lol

[eternal plover]

bump q_q

[south patrol]

I don't really understand your notation

[eternal plover]

What is unclear?

Although the fact that you don't understand it might be an issue in and of itself

[rotund aurora]

Let $p$ be the minimal polynomial of $\zeta$. Why is it the case that
[
\textup{Tr}_{\bQ(\zeta)/\bQ}\left(\frac{\zeta^{n-1}}{p'(\zeta)}\right)=1
]

where n is the degree of zeta

[cloud walrus]

Croqueta

[rotund aurora]

I kinda believe this is true now, the leading coefficient of p' is n and both numerator and denominator have the same degree, and the numerator is monic. So you should be able to write that fraction as 1/n+c_1zeta+...+c{n-1}zeta^{n-1}

and I think Tr(zeta^k)=0 for k not divisible by n (no lmao)

[simple walrus]

hi. considering (a,b)=greatest common divisor, and [a,b]=least common multiple, A= a commutative ring and domain (no divisors of 0). I am trying to prove this: if [a,b] exists, then (ca, cb) exists, $\forall$c$\in$A

[cloud walrus]

Maths Enjoyer

[simple walrus]

we note [a, b]=m, and have the property: a$|$m , b$|$m and $\forall$m'$\in$A , with a$|$m', b$|$m' $\Rightarrow$ m$|$m'and since we know that if both lcm and gcd exist, their product = product of the numbers, I believe I have to try to prove (a,b)=$\frac{ab}{m}$

in particular, I guess I also know $\forall$ a'$|$a $\Rightarrow$ a'$|$m. similar for b.

given that A is a ring, I also cannot use $\frac{ab}{m}$ = $a\frac{b}{m}$ ; $\frac{b}{m}$ doesn't even have to exist (in A)

thinking about it, maybe I can justify unique factorisation of a and b if [a,b] exists / that all divisors of a and b are prime atoms. then, it would be simple to prove (a, b) exists, and probably same for (ca, cb) with c $\in$ A.

hope it's ok for me to ask here.

[cloud walrus]

Maths Enjoyer

[rotund aurora]

I wonder if the following is true:
[
\textup{Tr}_{\bQ(\zeta)/\bQ}\left(\frac{\zeta^k}{q(\zeta)}\right)=0
]
for deg $q>k$, q any polynomial