groups-rings-fields

[ancient night]

I'm interested in the former

Just to make sure

[oblique river]

yes

[ancient night]

How to differentiate between the two?

[oblique river]

they are literally the same polynomial

[ancient night]

Which object would preserve the computational details?

[oblique river]

you can use a polynomial to define a function on R

I dont know what you mean "computational details"

(x+1)^2 and x^2 + 2x + 1 are the same polynomial

they are exactly the same and there is no difference between them

[ancient night]

A polynomial decomposition may enable more efficient evaluation of a polynomial. For example,

$${\begin{aligned}&x^{8}+4x^{7}+10x^{6}+16x^{5}+19x^{4}+16x^{3}+10x^{2}+4x-1\={}&\left(x^{2}-2\right)\circ \left(x^{2}\right)\circ \left(x^{2}+x+1\right)\end{aligned}}}{\displaystyle {\begin{aligned}&x^{8}+4x^{7}+10x^{6}+16x^{5}+19x^{4}+16x^{3}+10x^{2}+4x-1\={}&\left(x^{2}-2\right)\circ \left(x^{2}\right)\circ \left(x^{2}+x+1\right)\end{aligned}$$
can be calculated with only 3 multiplications using the decomposition, while Horner's method would require 7.

[cloud walrus]

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

[oblique river]

it's fine, I can see the example on wikipedia

I'm not sure what your point is

[ancient night]

Should I introduce my hand-crafted algebraic structure for describing the details of polynomial evaluation?

Or can I refer to some available algebraic structure that can be rarely used but still present in literature

[oblique river]

I'm not sure what kind of algebraic structure you are looking for

do you mean like "a polynomial ring but where you can do composition"?

[ancient night]

And where the degree of a polynomial would reflect the power of factors instead of monomials

[oblique river]

https://en.wikipedia.org/wiki/Composition_ring

[ancient night]

Thank you!

[oblique river]

it's not really in common use anymore but if you're just talking about composition of polynomials

I would just say "composition of polynomials"

[ancient night]

I have to do some proofs about the new notion of degree

[oblique river]

"new" notion?

[ancient night]

Because polynomials with high exponents of factors can be relatively easy to compute in non-expanded form

[oblique river]

yes I certainly agree that factoring polynomials can be useful

[ancient night]

Thank you once again!

I'm really grateful 😄

[oblique river]

did I do anything?

[ancient night]

Yes, you clarified me the border between what is at least initially taught to be polynomial and what it actually is

[oblique river]

oh ok :)

[chilly ocean]

a polynomial is a sequence with finite support with terms in a ring

[ancient night]

Oh, that's what confused me the most. In Russian not a single grown person will ever say something like this. I perceive it as a substitution of the definition by its model.

We usually accompany such things with "can be thought of", "can be intuitively understood as", "can be modeled by" etc

In English I encounter such things very often and it makes me cry because often I end up not getting the definition that is just ``true". Let it be Bourbaki-style (like definition of a function), but objectively and in all cases true

Classes (proper classes + sets) = ❤️

cry from the heart

[oblique river]

that was fast

[ancient night]

@chilly ocean And I thank you for your intentions! It is admirable that you help people here

MIPT's lectures have spoiled me

@chilly ocean And thank you too

[chilly ocean]

intuitively a polynomial can be written in form an x^n + ... + a1 x + a0

So you can think this as an infinite sequence which is always 0 after some terms

(a0, a1, ..., an, 0, 0, 0, ...)

so here x = (0,1,0,0,...) and x² = (0,0,1,0,0,...) and a0 = (a0, 0, 0,...)

[unique juniper]

:o

[chilly ocean]

You are welcome. I mean, saying that "in Russian not a single grown person will ever say something like this" is wrong in all possible senses. Literally every algebra textbook on russian defines polynomials in that way (and it would be completely strange if they not)

[ancient night]

I mean that is a definition

Bourbaki-style with all its shortcomings

But still a definition

[unique juniper]

russian looks cool

[chilly ocean]

its not 'bourbaki-style'

Its a regular style for mathematical textbooks

[ancient night]

It is, meaning that it is understood through contrived models

https://math.stackexchange.com/questions/792063/what-is-bourbakis-style-in-mathematics

(without implied negative connotation)

[chilly ocean]

And which mathematical definitions is not "Bourbaki-style" on your flavor?

[ancient night]

"In mathematics, a group is a set equipped with a binary operation that combines any two elements to form a third element in such a way that conditions called group axioms are satisfied, namely associativity, identity and invertibility." It is natural to talk about groups as sets endowed with such structure.

P.S. I think we go off-topic so we should chat in DM.

[unique juniper]

what is bourbaki style lol?

[ancient night]

@unique juniper I've sent a link

[unique juniper]

ok ill read it

[ancient night]

I perceive it as a social construct with some polysemy (existence of several meanings caused by contiguity of semantics)

[unique juniper]

[chilly ocean]

It's strange that it is not a Bourbaki-style definition for you, regarding that this is literally definition which the Bourbaki gave.

[ancient night]

Yes, because it works for such concepts

[old lava]

it's not literally the definition bourbaki gave, since I doubt bourbaki wrote in english

[ancient night]

But it is bad idea to say that bee is a set of 2 wings, a body and some other things that would not differentiate a bee from a fly or some other insect or currently existing sorts of bees

[chilly ocean]

Ok, you think too broadly for me

[ancient night]

Haskell and FP does nasty things to imagination

[chilly ocean]

I see

[cursive temple]

In jacobson it is an exercise to show that if a^2 = 1 for all elements in a group, then the group is abelian. After this it is asked if the same holds for when a^3 = 1 for all a in the group. It turns out that this does not necessarily hold. What about when the condition a^n = 1 holds for all elements a in our group for some n > 3?

The counterexample in the a^3 = 1 case is this

Apparently its called a heisenberg group

[old lava]

pretty sure it only holds for a^2 = 1

[cursive temple]

For primes you can just expand on this idea with the heisenberg group i think

Yeah im inclined to agree

https://en.m.wikipedia.org/wiki/Heisenberg_group#Heisenberg_group_modulo_an_odd_prime_p

Wikipedia gives it in that case

[slender sable]

can anyone guide?

[carmine fossil]

Are you considering matrices over C or R?

[warm holly]

Is a dimension of a basis of $S_3={M \in M_3(K) : M^t=M}$ (subspace of squre matrices $3\times 3$ over a field $K$) 4?

[cloud walrus]

Dаniil

[carmine fossil]

You have {E_11,E_22,E_33,E_12+E_21,E_13+E_31,E_23+E_32} as a basis for that

Where E_ij is a matrix with everything except the (i,j)th Element being 0 and that element being 1

[warm holly]

so of dimension 6?

[carmine fossil]

Yes

[warm holly]

ok thx

I computed basis found only 4, thank you

[carmine fossil]

In general it's n^2+n/2

[slender sable]

in general...for the option a is it true or false?

[carmine fossil]

Where n is the size of matrix

[warm holly]

nice thank you

[carmine fossil]

a is true,mb

[warm holly]

oh i'm stupid I considered an element of a basis E_11+E_22+E_33 thats why i had 4 instead of 6

[carmine fossil]

Should be only c,I think

[warm holly]

of course on diagonal elements can differ

[slender sable]

yes coz its never the same?

[carmine fossil]

I think it's always the same

[slender sable]

no..its never the same..after you apply row transformation to get echloen form..its never the same

[carmine fossil]

mb

A is true

Take A^-1=-A

B is definitely true

Av=0 implies A^2v=0

D is weird since 0 is not seen as an eigenvector and that always satisfies

[slender sable]

cool

[carmine fossil]

I guess it's D then

[tepid thunder]

http://prntscr.com/132mkk7

hey i am confused on part c

any ideas on how i would go about this

[hot lake]

what's confusing about it ?

[oak perch]

yxxy = xxxyxy = xxxxxxyy --> ?

[tepid thunder]

I just don't really know where to start to be honest

@hot lake

[hot lake]

do you know what a group homomorphism is

[tepid thunder]

yes sir

[hot lake]

do you know what Z4 is

[tepid thunder]

all integers mod 4

{0,1,2,3}

[hot lake]

yeah

so if you know that phi(x)=1 and phi(y)=0, do you think that determines phi completely ?

is it possible to know what phi(x²) is from that information ?

[tepid thunder]

oh yea it is

so i just gotta compute like that?

[hot lake]

well you can try pulling as much information as you can from what you are given

[tepid thunder]

so phi(x^2) would be 1

[hot lake]

why would phi(x²) be 1 ?

what's the group operation of Z4 ?

[tepid thunder]

I am confused again

[chilly ocean]

its okay

[delicate bloom]

a group is more than just a set like 'integers mod 4' you also need to specify the binary operation, which is addition

[tepid thunder]

the question does not state the binary operation

oh, okay my bad i didnt read the last bit

[delicate bloom]

I'm assuming that's why you're saying this, so phi(x*x) = phi(x)+phi(x)

[tepid thunder]

yea that is

I just got confused cause like a few questions before hand I was multiplying as it had the binary operation of multiplication

thank you

[delicate bloom]

yeah, as a hint you can derive a contradiction by playing with the generating relations a little

or maybe not, I might have made a mistake lol

[languid meteor]

hey all, im looking at the quotient group dZ/nZ and I know that it's generated by d + nZ. What I can't immediately accept is that the order of d + nZ is n/d. How can I prove that there isnt a smaller integer that sends d + nZ to zero?

zero here being nZ

[carmine fossil]

n should divide kd,where k is the smallest number that sends d+nZ to 0

kd=na
k=(n/d)a

[languid meteor]

ah, perfect thank you!

[outer monolith]

Hi, i'm a little bit confused for the notation used in finite fields. For example, I see the symbol:
$(\mathbb{F}_2^4)^4$, what do the 2, inner 4 and outer 4 refer to?

$(\mathbb{F}_2^4)^4$

[cloud walrus]

Daddy

[outer monolith]

I assume the 2 refers to elements taking the values 0 or 1? (i.e, binary of length 4 perhaps?)

[golden pasture]

huh ive never seen this before

could give context

do you mean like

$$\mathbb F_{2^4}^4$$

[cloud walrus]

ari 十年生死两茫茫，不思量，自难忘。

[outer monolith]

No:

[golden pasture]

hm could send where you seen that

[outer monolith]

M is an MDS matrix

[golden pasture]

huh

[outer monolith]

minimum distance separable matrix

[golden pasture]

i suspect it is actually this, the 4d vector space over F_16

[outer monolith]

maybe this is too computer scienc-y

basically it is a matrix that is applied to a 16-bit number

a b c d are 4 bit numbers

[languid meteor]

If we have if |G| = 155, x,y in G with |<x>| = 5 and |<y>| = 31 and <x> intersection <y> = e do we know that |<x,y>| = 5*31 = 155?

I cant find anything about the order of a generating set

[oblique river]

yes but for much simpler reasons. the order of <x,y> must divide 155 because it's a subgroup. but the order is clearly larger than 31 since x isn't in <y> and therefore it must be 155

[languid meteor]

oh god, of course. Thank you

[kindred mist]

Is there a way to show using exact sequences that for any ring $R$, module $M$ and submodule $N$ that $M/N\bigoplus N\cong M$? Is it even true? I was trying to show that the exact sequence: $0\to N\to M\to M/N\to 0$ splits, but I can't seem to find a way. The canonical projection $\pi$ is surjective and thus has a right inverse map but idk if it has a right inverse homomorphism (I believe that is equivalent to the statement that $M/N\bigoplus N\cong M$).

[cloud walrus]

𝓒𝓸𝓾𝓷𝓽𝓪𝓫𝓵𝓮

[next obsidian]

This isn't true

[final pasture]

M = Z/4Z and N = Z/2Z as Z-modules is a counter-example, no ? @kindred mist

[hidden haven]

Also M=Z, N=2Z

[next obsidian]

As you've noted, the projection map splits set theoretically

but it will not always split as a module map

[hidden haven]

M/N is Z/2Z but direct sum N would give Z/2Z + Z

[kindred mist]

N isnt a submodule of M?

[hidden haven]

You can embed it

As 2Z/4Z

[kindred mist]

yeah ok that makes more sense

and this is all going on in ZMod right?

[hidden haven]

Ye

[next obsidian]

A related thing

One could attribute this to the fact that the object you can put in the middle of this exact sequence

0 -> N -> ? -> M/N -> 0 up to isomorphism isn't unique

this leads to the notion of Ext which classifies, up to isomorphism, the various objects which can fit inside there

[kindred mist]

This would be Ext^1(N,M/N) right?

(the set of equivalence classes?)

[sturdy marsh]

the other way around

but yes

[vestal snow]

Is there a quick way of doing this using the colimit definition of stalks?

It's pretty straightforward using the explicit construction

But I'm trying to get better at category theory arguments

[chilly ocean]

can someone give me an example of reducible polynomial with no root ?

[hidden haven]

(x^2+1)^2 over R

[chilly ocean]

thanks

what about x^3+x ?

[hidden haven]

That has 0 as a root

[chilly ocean]

that's right

thanks

[obsidian sleet]

🤨

[next obsidian]

I think this is going to be hard

The category you’re taking the colimit over has to be in Ab, because you’re taking a colimit of modules over different rings

[vestal snow]

That's what I was thinking too

[next obsidian]

But you can in general describe how a colimit of modules over a system (A_i) is a module over the colimit of the A_i, and you could maybe do this by means of the fact that a module structure is a map of the form

I hope I have this right, but it’s like

A (x)_Z M -> M

In Ab

This describes left multiplication by scalars

So I think you can take the module structure for the various M_i in terms of

A_i (x)_Z M_i -> M_i

And try to take colimits and get a map A (x)_Z M -> M

[vestal snow]

That seems too much work lol

[next obsidian]

Yeah, it’s not really worth it IMO

[vestal snow]

I think I'll just go with the concrete way

[next obsidian]

I’ve thought about this sort of thing for far too long

I think it’s not worth it at all

[vestal snow]

When does AG get difficult in terms of actual problems instead of just difficult because of abstractness?

[next obsidian]

Around II.3 of Hartshorne

[vestal snow]

Which is...?

[next obsidian]

When they introduce fibered products

And then makes you do hard exercises

[vestal snow]

Got it

Vakil gang rise up

[next obsidian]

💤

[vestal snow]

You should read Vakil like a novel Chmonkey

Once you're done with Hartshorne

so you can help me

[next obsidian]

Vakil is even less like a novel than hartshorne

Idk what novel has you stop twice a page to fill in the details

It’s like an evil madlib

[vestal snow]

So far, its been pretty good at communicating the intuition behind things

[next obsidian]

¯_(ツ)_/¯

[vestal snow]

For example, stalks were pretty weird when I encountered them in Liu, but Vakil's motivation of it through differentiable functions drove the point home for me

[golden pasture]

honestly learning about schemes in context of complex surfaces helps a lot for intuition

[unique juniper]

the last paragraphy

g(X) is the polynomial obtained by removing multiple factors

i dont really see how the splitting the field for g and the product of p(x)'s would be the same

and why g(x) is seperable

what if there was a root of multiplicity >2?

[unique juniper]

[vestal snow]

Do an example

What is the splitting field of (x-sqrt(2))^2 and the splitting field of x-sqrt(2) over Q?

[unique juniper]

yes its the same Qroot 2

i mean tho

like

what happens if p1 has distinct linear factors?

then g wont contain any of those

what about then?

so for example

$(x - x_1)^2(x - x_2)$

[cloud walrus]

Yes ツ

[unique juniper]

g would be x-x1

then the splitting field wouldnt be the same

?

assumiing x2 isnt in F(x1)

[rustic crown]

no, in this case g would be (x-x1)(x-x2)

[unique juniper]

but g is obtained by removing multiple factors?

[rustic crown]

only remove the redundant part

[unique juniper]

wdym?

[rustic crown]

you want to find a polynomial which has same splitting field as p_1 . p_2 ... p_n

and which has no repeated roots.

[unique juniper]

yes

so we taking all factors once?

for g

[rustic crown]

yep

[unique juniper]

i see

[rustic crown]

lol

[unique juniper]

thank you XD

[rustic crown]

also this is kinda bad to say since sqrt(2) isn't in Q.

[unique juniper]

i understood what they meant

[daring ibex]

Okay so i'm rereading this proof and i realized i didn't understand one step

I don't understand how they got to $\sum_{c \mid d} \psi(c) = d$

[cloud walrus]

urdurdus

[daring ibex]

As far as i understand, what this result is saying that if $d \mid p -1$ implies $\psi(d) = d$

[cloud walrus]

urdurdus

[daring ibex]

so doesn't this mean that this reduces to $\sum_{c \mid d} c = d$

[cloud walrus]

urdurdus

[daring ibex]

but you can pull a d out of the left hand side which leads to this being completely preposterous

[daring ibex]

or well at least what the first few lines show

[hot lake]

well if you have a set A and a function f : A -> B, then A is the disjoint union of the f-1({b}) for b in B

here A = U(Z/pZ), f(x) is the order of x

and B is the set of divisors of d

and f-1({c}) has size psi(c) be definition of psi

so the size of U(Z/pZ) is the sum of the psi(c)

[daring ibex]

only if surjective tho, right?

[hot lake]

oh wait they have various d

[daring ibex]

[hot lake]

uuh let me reread

okay they pick A = the elements of order dividing d

[rustic crown]

no its saying psi(d) = phi(d)

[daring ibex]

i am referring to the first two lines

[hot lake]

in total it has sum for c dividing d of psi(c) elements

and proposition 4.1.2 says that there are d elements of order dividing d

so sum for c dividing d of psi(c) = d

[daring ibex]

I don't understand

[hot lake]

pick A = the elements satisfying x^d = 1

[daring ibex]

yeah

[hot lake]

aka the elements whose orders divide d

[daring ibex]

oh wait

yeah i think i see my mistake

[hot lake]

classify the elements of A according to their order

for each c dividing d, look at the set of elements of A of order exactly c

that set has size psi(c)

[daring ibex]

I assumed x^d = 1 implied the order was d

of x

which is very wrong

[hot lake]

very

[delicate bloom]

you could say psi=phi in that step if you take an element of order d there and look at its powers x^1, x^2, ..., x^d which are necessarily distinct

if one of those has power c that divides d, then it has order d/c

[unique juniper]

[chilly ocean]

is there a way to divide f by g to find gcd(f,g) ?

[golden pasture]

use euclidean algorithm

[delicate bloom]

there are a few tricks to make your life easier by hand probably just noticing some patterns

[unique juniper]

oof

[delicate bloom]

I directly factored it cause I think they designed it to be done that way

$f = (x^3-1)(x^2+2x+2)$ and $g=4(x^4+4)$

[cloud walrus]

Merosity

[delicate bloom]

at this point I guess $(x^4+4)=(x^2+2x+2)(x^2-2x+2)$ since we need to make 0 in the middle and 4 on the last term

[cloud walrus]

Merosity

[delicate bloom]

nothing more to check cause they're both irreducible by eisenstein and the other poly factors as (x-1)(x^2+x+1)

[chilly ocean]

what's the benefit for long division ?

[sinful mirage]

does this seem right?

I tried to write out all multiplications explicitly and I am not sure whether what I wrote is right or not

[chilly ocean]

the benefit is we can write uf+vg in a shorter form. nothing more. their gcd(f,g)=1

[delicate bloom]

but I just explained why the gcd is x^2+2x+2

[chilly ocean]

😄 I have actually crossed those common term to make long division ...

[winter vigil]

a quick question, is $\mathbb{R}$ an algebra ?

[cloud walrus]

Otoro

[hidden haven]

Algebra over what ring?

[chilly ocean]

{0}

[winter vigil]

so basically, Im looking at the derivation of an algebra, trying to link this back to the derivation of a point, cause I couldnt get it yet

so I was trying to look at some simpler examples to get the idea

[chilly ocean]

||space of vector fields / tangent vectors on a smooth manifold||

[unique juniper]

i dont get this part :3

specifically why there are that many elements equal when restricted

[hidden haven]

What's H?

You have those equalities because it doesn't matter if you restrict a function in steps or in one go

[unique juniper]

yes i understand those equalities

now

but

not the part under it

H is a subgroup of the codomain

[hidden haven]

It seems that H has to be a previously defined subgroup

Oh I'm guessing it's the set of tuples (sigma, tau) that agree on the intersection of K1 and K2?

[unique juniper]

yes

[hidden haven]

Right

So rephrasing that sentence, it's saying that given an automorphism sigma on K1 intersection K2, there are |Gal(K2/K1 \cap K2)| many extensions of it to automorphisms of K2

[unique juniper]

yes

[hidden haven]

Have you seen any relevant theorems on extending field homomorphisms?

[unique juniper]

probably

[hidden haven]

This just follows from K2/(K1 cap K2) being Galois really

You can induct on the number generators

So assume the extension is generated by one element alpha

[unique juniper]

hmmmmmmm

[hidden haven]

Then alpha can map to any root of sigma (minimal polynomial of alpha)

And each gives a unique extension, and the number of roots will be equal to the degree of alpha since it's a Galois extension

[unique juniper]

how do we know this?

K2/K1 cap K2/F

[hidden haven]

Yeah, K2 is galois over F

[unique juniper]

k2 is galois over F but not K1 cap k2 ?

[hidden haven]

So must be galois over intermediate fields

[unique juniper]

oh

yeah my bad XDD

but tbh i dont understand what you wrote afterwards xD

and how that helps

[hidden haven]

Which part?

[unique juniper]

everything after you saying k2/k1capk2 is galois

[hidden haven]

are you using some textbook for this?

I'm not sure what theorems you are assuming

[unique juniper]

any thereom

what theorems are you using?

or want to use?

Let me actually be clearer about my struggles

[hidden haven]

If you have F[alpha]/F (any extension) and a homomorphism phi from F to some E, and phi(minimal polynomial of alpha over F) has a root beta in E, then theres a unique extension of phi to a homomorphism from F[alpha] to E sending alpha to beta

[unique juniper]

well yes

I understood that there are |Gal(K_2/ K_1 cap K_2)| many T in Gal(K_2 / F) that will fix K1capK2 but not why they are equal to sigma with the restriction

[hidden haven]

oh

so instead of fixing K1 cap K2, think of them as extending the identity function

instead you can extend the restriction of sigma

another way to view it is that any T whose retriction is the same as sigma's, is sigma composed with some element whose restriction is identity

[hidden haven]

try to see this

[unique juniper]

sso uhh

there are |Gal(K_2/ K_1 cap K_2)| many taos that are the identity when restrictyed to k1capk2

yes?

[hidden haven]

yep

[unique juniper]

but er

[unique juniper]

wdym by the last half??

[hidden haven]

lol

[unique juniper]

my brain cannot understadns this at alll

[hidden haven]

so call this subgroup N

[unique juniper]

okie

[hidden haven]

Then any T that has the same restriction as sigma, will be in sigma N

Because T = sigma (sigma inv) T

[unique juniper]

hmmmmmmmmmm

i kinda understabnd

[hidden haven]

right so the set of T that have the same restriction as sigma

is just the coset sigma N

and that has the same cardinality as N

[unique juniper]

alright

thanks

alot

[hidden haven]

[unique juniper]

you are the king!

[hidden haven]

[vestal snow]

@next obsidian about the problem we were discussing last night, is it possible to define a category where the objects are of the form (R, M) where R is a ring and M is an R-module and morphisms are (f,g) where f:R -> R' is a ring morphism and g:M -> M' is an abelian group morphism with the property that g(rm)=f(r)g(m)?

And then think of F_p as an O_{X,p} module by taking the colimit in this category

The colimit's existence would need to be verified

But that's easy

I guess my question is that will $(F_p, O_{X,p})$ be the colimit in this category of $(F(U),O_X(U))$ as $U$ ranges over all open sets containing $p$?

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

I'm asking this because the remark in this exercise seems to suggest viewing $(F_p, O_{X,p})$ as the colimit

[cloud walrus]

Have a Banana, Bitch

[next obsidian]

So you could try to do it that way

But idk if the colimits agree tbh

[latent anvil]

So morphisms like you want are morphisms M -> f^* M' where f*M' is the restriction of scalars

This makes me think of like

Hmm

[next obsidian]

really makes you think

[tight otter]

im trying to show that $x^4+3x+3$ is irreducible in $\mathbb{Q}(\sqrt{21})$ but i keep running into barriers

[cloud walrus]

panoramatopia

[tight otter]

i know the polynomial doesnt have any real roots so if it were reducible it would have to be factored into quadratics

[next obsidian]

Maybe use the fact that complex roots would come in conjugate pairs

you could try to compute the complex roots and then see something about how it would have to factor as a product of quadratics or something

In general, I've had success with these sorts of things by using analysis, and looking at how it operates in C

Also, if Z[sqrt{21}] is a UFD you could try to appeal to Gauss's lemma or something?

[tight otter]

we did some poking around and found that Z[sqrt(21)] wasn't a UFD, so we tried that

expanding it out using conjugate roots is not very friendly either

[rustic crown]

Z[sqrt(21)] isn't a UFD, but the ring of integers of Q(sqrt(21)) is

it has the integral basis 1, (1+sqrt(21))/2

but it could get sad if 3 gets ramified

i think i have a much simpler way tho

lemme think about it

[golden pasture]

can't you eisenstein on the prime p=(3-sqrt(21))/2

that proves irreducibility

(note that p^2=(3))

essentially you complete the field WRT the prime then the polynomial is irreducible in the completion by drawing newton polygon (a fancier way of saying "effectively by eisenstein")

[rustic crown]

no right.. 3 just ramified you want the constant to stay outside (p)^2

[golden pasture]

the p-adic valuation of the last 2 coefficients in this case is 2

it works cuz if your poly is irred in completion, it is irred in your base field

and good thing about completions is checking irreducibility is basically trivial

this is the entire proof :D

[rustic crown]

sounds pretty fun... i should read about p-adics

[delicate bloom]

wait I don't think this works cause 3 ramifies, the newton polygon I have in mind looks different

like it goes from 0 to 4 not 3 on the horizontal axis

and right at where 2 is, above that is a possible valuation, so it could be two polynomials with the same slope

like for instance x^4 - 21 would be a reducible polynomial with the same newton polygon to give an example

[tight otter]

i think i figured out a solution

if it factors into quadratics then u could have $x^2+3x+3=(x^2+ax+b)(x^2-ax+c)$ and by vieta's formulas u get $-a^2+b+c=0$, $a(c-b)=0$, and $bc=3$.

[cloud walrus]

panoramatopia

[tight otter]

then either $a=0$ or $b=c$. if you have the former then $b=-c$ and $-c^2=3$ which isn't possible given that $a,b,c$ are all real. If it's the other way then you get $b,c=\sqrt{3}\not\in\mathbb{Q}(\sqrt{21})$

[rustic crown]

a(c - b) = 3 😶

[cloud walrus]

panoramatopia

[tight otter]

fugg you're right

this is so sad!

o well

to office hours i go!

this small problem part is legit like the hardest problem i've had this whole term

i have no idea why

spent like two hours this evening spitballing solutions with classmates and we barely got anywhere

[hidden haven]

I miss being able to solve problems together with classmates in college discussion areas

Not that we did that very often but

[tight otter]

yeh i spent the first two terms of this year starting all my homework at the last minute and never interacting with my classmates and it made me very depressed and stressed all the time so this term im trying to engage more

[snow flint]

this is how ur supposed to do hw tho

by urself the night before

[daring ibex]

panoram nice pfp

[unique juniper]

so uhhhh

whats the difference between seperable extension

and galois closure?

seperable extensions are galois

[hot lake]

no ?

[unique juniper]

seperable extensions arnt galois ?

k/f is galois iff k/f is seperable

[hot lake]

I don't know where you are seeing that

[unique juniper]

well

[hot lake]

a galois extension is an extension that is normal and separable

but not every separable extension is normal

and hence we do need different words

[latent anvil]

Q(cube root of 2)/Q is separable but not galois

[unique juniper]

this tho

[latent anvil]

I'm not sure what this is supposed to show in reference to your claim

It seems to show that a galois extension is normal and separable

[unique juniper]

yes

[latent anvil]

And is the splitting field of a separable polynomial

But you're claiming that we can go the other direction

[latent anvil]

A separable extension might not be a splitting field

See Q(cuberoot 2)/Q

The minimal polynomial x^3 - 2 does not split

Q(cuberoot 2) only contains real numbers, but the other two roots of x^3 - 2 are not real, so they don't lie in Q(cuberoot 2), so the poly doesn't split

It is

[unique juniper]

nvm

[latent anvil]

Why do you say it isn't?

any extension in char 0 is separable

[unique juniper]

yes mb

[latent anvil]

also why do up keep deleting things lol

[unique juniper]

[latent anvil]

there's nothing to be ashamed of or anything

everyone on this server has misunderstand something and said incorrect things about math before

I've done it today

anyways, do you get the distinction now?

A separable polynomial might not split after eg adjoining only one of its roots

[unique juniper]

yes i understand

sorry

and thank you

[latent anvil]

don't apologize!

being mistaken is good

it's how you learn math

[unique juniper]

[daring ibex]

le me forgetting what an isotropy group is has arrived

and also me thinking that everything in a quotient ring is an ideal

it happens

[vestal snow]

There is an easy way to extend the notion of presheaves to categories which do not necessarily have an underlying set structure, but is it possible to do it for sheaves?

[latent anvil]

Yeah, this is the idea of a "grothendieck topology"/"site"

You give a category enough data to know what collections of maps are "coverings" of an object

Ends up being important in AG

[vestal snow]

I think you misunderstood

Like say we have an arbitrary category C and a topological space X

Then is there a way to define a sheaf which sends open sets in X to objects in C

Such that things glue together uniquely (axioms of a sheaf)

From what I understand, you said that instead of a topological space X, we start with some other weird grothendieck topology/site

[steady axle]

How is the set of transitive actions of a group on n points 'same as' subgroups of index n?

[golden pasture]

oops i cant couint HAHAHAH

but newton polygon stuff still works for ramified p adic fields

[pine patio]

for part i) are the orbits R\0, (R\0)i, {0}, {R\0}(a+bi)

[golden pasture]

what do you mean by {R\0}(a+bi)

[pine patio]

so like for each a,b we have {R\0}(a+bi) as an orbit

but then 2a,2b and a,b should have the same orbit

im trying to say that for numbers of the form a+bi where a,b are not 0, the orbit is {r(a+bi): r is in R\0}

[latent anvil]

Ah yeah I totally did, sorry

You replace the unique gluing stuff with what's called the equalizer condition

If you have a presheaf $\mathcal{F}$ on $X$ it says that for any open set $U$ and any cover ${U_i}{i\in I}$ of $U$ the diagram
$$
\begin{tikzcd}
\mathcal{F}(U) \arrow[r] & \prod{i\in I} \mathcal{F}(U_i)\arrow[shift left=1]{r} \arrow[shift right=1]{r} & \prod_{(i, j)\in I^2} \mathcal{F}(U_i\cap U_j)
\end{tikzcd}
$$
is an equalizer diagram

[cloud walrus]

shamrock

[latent anvil]

Meaning that the first object and it's map into the second object is an equalizer of the next two maps

Implicitly we're assuming that these products exist, but you can just as easily include that as part of the sheaf condition

The first map is the one induced by the universal property of the product and all the restriction maps U -> Ui

The second one is a little weird

For the top map, the map from Π_k F(U) into the (i, j)th factor first projects out into Ui and then restricts. The bottom map projects onto Uj instead and then restricts

If you think about what an equalizer is in set/Ab/ring, you'll see that this condition encodes that any family of sections on the cover (an element of the product) which agree on intersections is the restriction of a unique section of U

[vestal snow]

I get how uniqueness of gluing follows from the diagram, but how does existence follow?

[latent anvil]

Do you know what the concrete construction of an equalizer is?

We have an isomorphism from F(U) to that which commutes with the restriction maps in an appropriate way (draw out a triangle)

This isomorphism is a bijection between elements of F(U) and gluing data (ie tuples of sections on each Ui which agree on overlaps)

[pine patio]

how do i find orbit(V4)

[hidden haven]

What's V4?

[viscid pewter]

Z2^2

[pine patio]

{e, (12)(34), (13)(24), (23)(14)}

[latent anvil]

the group of symmetries of a nonsquare rectangle

[viscid pewter]

so you know what conjugation in S4 does, right?

[pine patio]

yes

[viscid pewter]

ok, so the conjugation has to map those double-transposition permutations to double-transposition permutations

[pine patio]

yes

[viscid pewter]

so what are all the double-transposition perms in S4

[pine patio]

basically the ones that appear in V4

so like (12) and (34), etc

[viscid pewter]

so

list them all

[pine patio]

(12), (13), (14), (23), (24) (34)

oh lol

u didnt mean transpositions

those are the elements in V4

[viscid pewter]

what are all the permutations, in S4, that are the composition of two disjoint transpositions

[pine patio]

(12)(34), (13)(24), (14)(23)

[viscid pewter]

right

and that's just the elements in V4

[pine patio]

yes

[viscid pewter]

so any conjugation has to map V4 to V4

[pine patio]

yes

[viscid pewter]

ez

[pine patio]

thats what i got, but im still confused

we got stab(V4) = S4 right?

[viscid pewter]

yes

[pine patio]

so |StabV4|x |OrbV4| = |size of set of subgroups of S4|

[viscid pewter]

wait

isn't it just the size of S4

[pine patio]

i googled and apparently it's 30

[viscid pewter]

isn't it just: |StabV4|x |OrbV4| = |size of S4|

[pine patio]

but S4 is acting on the set of subgroups of S4

[viscid pewter]

yes

so?

[viscid pewter]

exactly

size of G, not size of S

[pine patio]

im blind

yes size of G, sorry

can u verify that stab(sym(1,2,3)) = sym(1,2,3) for the next part?

[pine patio]

here

[viscid pewter]

that sounds right

[pine patio]

ok thanks

[viscid pewter]

i mean the orbit will have order 4

so surely

[pine patio]

how do u know that directly

[viscid pewter]

so just intuitively the different versions of S3 will be on {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}

[pine patio]

yes

[viscid pewter]

then the stabiliser will have order 24/4

and only group of order 6 in S4 is S3

[pine patio]

so is this the orbit then

[viscid pewter]

not exactly

the orbit will be the set of symmetric groups on those sets

if you see what i mean? if i'm right?

[pine patio]

yes so sym(of those)

[viscid pewter]

{Sym{1, 2, 3}, Sym{1, 2, 4}, Sym{1, 3, 4}, Sym{2, 3, 4}}

[pine patio]

yes

[green stone]

where are you looking at group actions from? is that d&f?

[viscid pewter]

doesn't look like it

[pine patio]

whats d&f

i just read my lecture notes

[green stone]

dummit and foote

[viscid pewter]

d&f is a popular abstract algebra textbook

[pine patio]

do u recommend it?

[viscid pewter]

it's quite good

[pine patio]

how does armstrong compare

[viscid pewter]

no clue, lmao, d&f is the only one i know

[pine patio]

i have it but i havent read armstrong

[green stone]

i know a really bad one by Beachy & Blair

[pine patio]

ok then i'll try to get my hands on d&f

[viscid pewter]

if you want

[pine patio]

my lecture notes are written badly so it'll be good to have a proper textbook

[green stone]

you can't expect too much from lecture notes yeah.

[vestal snow]

So if $(x_i){i\in I}$ agree on intersections, then we can consider the inclusion map from the sub-(group/ring/module/etc) generated by $(x_i){i\in I}$ to $\prod_{i\in I} F(U_i)$ and this must factor through the equalizer, which is F(U).

[cloud walrus]

Have a Banana, Bitch

[latent anvil]

I think I agree with this, but it's not what I was thinking of

I think was thinking that the equalizer is $E = { (x_i) \in \prod_{i\in I} \mathcal{F}(U_i) : x_i|{U_i\cap U_j} = x_j|{U_i\cap U_j}}$

[cloud walrus]

shamrock

[latent anvil]

The equalizer of two maps is the subset/submodule/etc on which they agree

So we have an isomorphism from F(U) to this making a certain triangle commute, and commutativity of the triangle tells us that the isomorphism is given by restricting onto each component

[pine patio]

i got S5 for first 2, but how do i do the third one?

<@&286206848099549185>

[hidden haven]

Leave smallest for now, are there any n's that you know will work?

[pine patio]

8

[hidden haven]

Right, but you can tighten that further

Think of C_2 as a group that acts on a 2 elt set

[pine patio]

i dont think s4 works

[hidden haven]

so that product C2 x C2 x C2 will be a triple of 3 permutations which each act on 2 elt sets

[pine patio]

what do u mean by elt

[hidden haven]

can you think of this as an action on a single set with <8 elements?

[final pasture]

element

[pine patio]

6?

[hidden haven]

yeah, how?

[pine patio]

cause i kinda know s4 doest work

s5 doesnt either, but im not too sure

[hidden haven]

ok good guess

but try to justify it

[hidden haven]

like just that it embeds into S6, thinking from this you should be able to reason through it

[pine patio]

hmm i can see how it embeds into but i think my reasoning is still cloudy

[hidden haven]

So let's say the element of C2xC2xC2 is (a, b, c)

a, b and c all act on 2 element sets (say {1,2}, {3,4}, {5,6} respectively)

Then you can have the triple act on the 6 element set {1,2,3,4,5,6} with a acting on the first 2 things, b on the next 2 and so on

More concretely, it is the subgroup generated by (12), (34), (56) of S_6

[hidden haven]

This was all intuition, but you can also check that the subgroup I've given is isomorphic to that product

[final pasture]

(why can't it be embedded in S_5 ? 🤔)

[hidden haven]

It will need sylows theorems probably

[final pasture]

oh wait

I think I got an elementary argument

(C_2)^3 is generated by 3 elements and every element in (C_2)^3 is of order 2

I think we can list fairly easily order 2 permutations of S_5 (like maybe just say that they must be product of disjoint transposition ? Not entirely sure if this is true), and prove that we can't have 3 of them that don't "overlap" in a such way that every element in the subgroup generated is still of order 2

Ok that sounded better in my head

[gritty sparrow]

There is a unique sylow 2 subgroup of S5 up to isomorphism and (note that 8 is the highest power of 2 in 120) and (1234) and (13) will henerate a group of order 8 that is not C2xC2xC2

[final pasture]

yeah that works too, I was thinking that we could do it without invoking Sylows theorems, but nvm

[pine patio]

but then how do i do it without using sylows

we havent done sylows yet

[low surge]

think that works

[hidden haven]

I was trying something similar, I think you can first prove that it will contain something of the type (12)(34) because any transpositions that have non trivial intersections will compose to give order 3, so most you can fit is 2 transpositions. Then if you have (12)(34), anything else in the embedding of order 2 fixes 5, because otherwise the product of that with (12)(34) won't have order 2.

Also very wordy lol

[final pasture]

I feel like that should work, but I think we should be able to reduce the work that needs to be done by hand a bit 🤔

[final pasture]

yeah something like that ^

[hidden haven]

Once you know that some element is fixed, it should be easy (you reduce to S4)

[low surge]

you need 3 order 2 generators so must be transpositions. S5 has 5 things you can permute, so by pigeonhole principle, two of the generating transpositions must overlap at some element (choosing 3 trnaspostions = 6 things). WLOG say (12)(23)=(231), but all elements order 2 so contradiction?

[hidden haven]

Not necessarily transpositions

[final pasture]

the generators need not to be transpositions

[hidden haven]

(12)(34) has order 2

[low surge]

but still that will be an overlap

the transpositions is the best possible case right

[hidden haven]

(12)(34) overlaps (13)(24) but their product is order 2

[low surge]

hmmm

[hidden haven]

So you'll need slightly more justification

But yeah I think if you compose (12)(34) with any transposition involving 5 you run into trouble, but I'm not sure what happens if its a transposition involving 5 along with another disjoint transposition

[final pasture]

I feel like there's must be a slick way of handling all these cases properly, because the way we're putting it rn leaves a lot of hand checks if one wants to write down the argument properly

[hidden haven]

Sylows theorems

But yeah lol

[final pasture]

that's a bit too slick

maybe involving signature of the permutations somewhere ? To quote mirza, "I'm currently throwing everything I know at the problem rn and hope that solves it"

[hidden haven]

Perhaps

[low surge]

i might be wrong again but that might work

doesnt A5 have size 60 which isnt a multiple of 8

[final pasture]

order 2 permutation doesn't imply even signature

(12)(34) isn't in A_5 but has order 2

[hidden haven]

Is in A_5*

(12) isn't in A5

[final pasture]

yes this lol

[low surge]

oh yeah you have the single ones as well

lol

[hidden haven]

(12)(34)(15)(23) is a 5 cycle I think?

Time to draw

wooo it is

[final pasture]

Hey you spoiled me, I was still drawing

[hidden haven]

Embedding into S5 is possible iff embedding into S5 is

[final pasture]

possible

[hidden haven]

Because you can have at most 2 transpositions in the embedding, so you must have something of the form (12)(34). Now suppose you have something else of order 2 that moves 5. There are 3 cases (wlog)

1. (15) - gives an order 6 product with (12)(34)
2. (15)(34) - order 3
3. (15)(23) - order 5

So 5 is fixed by everything in the embedding

[delicate bloom]

yeah but it doesn't work in this case because of the other stuff I said

[hidden haven]

@pine patio

[golden pasture]

wait why tho

ohh icic

[delicate bloom]

cool cool 😎

also technically doesn't matter but I think you wrote your newton polygon backwards left to right

but maybe that's also a convention people just do cause f(x) and x^n f(1/x) both are reducible/irreducible

[golden pasture]

ahh icic why it fails

yea makes sende

[delicate bloom]

yeah that was also what I tried when I saw the problem and was sad it failed too 😢 no p-adic to the rescue lol

[golden pasture]

sad

[rustic crown]

I converted the problem to showing a degree 8 polynomial was irreducible over Z, but didn't know how to do that.

[delicate bloom]

spooky lol

[green stone]

maybe eisenstein's criterion?

[rustic crown]

Nah the usual stuff didn't work

I'll send that poly

[delicate bloom]

nah, newton polygon is a stronger version of eisenstein basically

[green stone]

but you can use a shift and sometimes that works

[delicate bloom]

I tried shifting, didn't find anything

[rustic crown]

sage: M = matrix(QQ, [ 
....: [0, 0, 0, -3, 21, 0, 0, 0], 
....: [1, 0, 0, -3, 0, 21, 0, 0], 
....: [0, 1, 0, 0, 0, 0, 21, 0], 
....: [0, 0, 1, 0, 0, 0, 0, 21], 
....: [1, 0, 0, 0, 0, 0, 0, -3], 
....: [0, 1, 0, 0, 1, 0, 0, -3], 
....: [0, 0, 1, 0, 0, 1, 0, 0], 
....: [0, 0, 0, 1, 0, 0, 1, 0] 
....: ])                                                                        
sage: M.characteristic_polynomial()                                             
x^8 - 84*x^6 + 6*x^5 + 2652*x^4 + 252*x^3 - 36279*x^2 - 7920*x + 196947

If this polynomial is irreducible, then [Q(sqrt(21) + alpha):Q] will be 8

which would mean that sqrt(21) is not in [Q(alpha):Q]

so [Q(sqrt(21), alpha):Q(alpha)] = 2 which would then say [Q(sqrt(21), alpha):Q(sqrt(21))] = 4.

modulo 2 that polynomial is (x^4 + x + 1)^2 and modulo 3 its x^8

sadly the constant term is divisible by 9

[rustic crown]

(sage does say its irreducible, but idk why without complete bash)

[golden pasture]

eisenstein on 3?

oh wait

right you can divide by 9 on that

lol

[rustic crown]

mod 2 tells me that if it reduces both factors have degree 4, but i don't know how to extract more information

[golden pasture]

i think you can newton polygon here right

lemme actually recap newton polygon make sure i didnt mess up

[golden pasture]

ahh i see the issue

newton polygon tells us the roots of this polynomial has valuation 1/4

which only lets us conclude that we can factor it like (deg 4)(deg 4)

[rustic crown]

ah sad

why is irreducibility so hard

[golden pasture]

it is hard for him and easy for computers

you can numerical irreducibility

[next obsidian]

Newton polygon :O

I only know of those in how you can write Puiseux series solutions to stuff

Which is pretty sick

[delicate bloom]

oh are you talking about factoring elements of the algebraic closure of F((X)) or something?

[green stone]

is there a stable algorithm for determining irreducibility?

irreducibility over polynomials with rational coefficients that is

[golden pasture]

stable in what sense

[gritty sparrow]

There are algorithms for some fields, I don’t know what you mean by stable though

[golden pasture]

If you can embed to C

the solution is simple for computers

conceptually what you do is

You compute the roots numerically

the put the roots α^i in a lattice

and run some fast lattice reduction like LLL

[next obsidian]

I meant like, taking a polynomial f(x,y) and solving for y locally in terms of x

It’s like uh, implicit function theorem or w/e

[golden pasture]

the algorithm is polynomial in the degree and number of bits for poly

[green stone]

ah i see

[delicate bloom]

oh interesting, how does the newton polygon help with that

[next obsidian]

Uh, you can basically determine you get the thing because you look at the slope of something. You know you’ve solved it when it gets flat iirc

It’s in the first section of “lectures on resolution of singularities” by Kollár

And I think you can find the original paper by Newton

[golden pasture]

btw mero i figured why newton poly didnt work

cuz newton poly only tells us the valuation of roots

so valuation in the prev case is 1/2 and deg is 4, so we can have deg 2*deg 2

sad

[next obsidian]

This is a super shit explanationbex

[delicate bloom]

yeah, exactly

[next obsidian]

Because I read it like 7 months ago

[golden pasture]

newton poly works for any hensellian fields anyways

[delicate bloom]

sounds reasonable what you're saying I think I kind of have an idea to make it work

[next obsidian]

Okay cool lmao

[delicate bloom]

well plausible, I don't know what you're actually saying haha

[next obsidian]

Like I remember you worked in stages and you basically assumed WLOG that you had some point on the y-axis

Okay, this is not gonna go anywhere lmfao

[chilly ocean]

is there a fast way to find galois group of x⁴-p over Q?

[hidden haven]

The roots are ±r and ±ir where r is the 4th root of p. Assuming p is not a square (equivalently irreducibility by Eisenstein), you adjoin r which is a degree 4 extension, and that doesn't contain any imaginaries so you adjoin i next to get Q(r, i)

This is a degree 8 extension and I think the Galois group is D_4

Yes it's D_4

[simple valley]

do I understand correctly that the only semidirect product $\bZ_7 \rtimes \bZ_5$ is the direct product?

[cloud walrus]

mniip

[simple valley]

the only Z_5 -> Aut(Z_7) = Z_6 is 0 because of order reasons

[rustic crown]

yep

and in general there is a theorem than any group of order n such that gcd(n, phi(n)) = 1 is cyclic which is pretty cute

(i hope Z_5 means Z/5Z... idk much about p-adics)

[lunar coyote]

what kind of proof is this?

[tough raven]

A correct one?

[lunar coyote]

i meant like is it contraposition

[rustic crown]

not a complete one since part (b) is missing

[tough raven]

The last sentence is for part (b).

IG it's not very detailed

Oh wait

Nvm 🤦

[rustic crown]

nah its the other direction

[lunar coyote]

is this proof by contraposition?

[rustic crown]

so you know that A => B is equivalent to proving not B => not A

[lunar coyote]

yeah

[rustic crown]

this is called contrapostive

and we used it twice

[young coyote]

if and only if, bidirectional proof

[rustic crown]

A <=> B is same as proving not B <=> not A

[lunar coyote]

how comes it's only proved in one direction?

don't you have to prove it in both directions?

[rustic crown]

they did prove both directions!

first 2 lines are not B => not A and last line is not A => not B

[young coyote]

Since they didn't summarize these conclusions, it may have been unclear that they had done it on both directions. (One of the downsides of being a bit too terse.)

[lunar coyote]

yeah i think i get it now

[unique juniper]

i dont understand exactly

whats going on

"for any tao in H the elements (tao)(sigma) run over H as sigma runs over H, It follows that (tao)(alpha) = alpha"

Is it just that

tao x whole group = the same group?

thats what they mean?

so its fixed by anything in H

[cloud walrus]

det

[unique juniper]

yeah

Thank you!!!!!

[rustic crown]

[chilly ocean]

is there an usual notation for minimal polynomial of some algebraic element w over a field K? i just been writting p_K w

[solemn rain]

J_a

[rustic crown]

i usually use min_K(w) or m_K(w) or p_K(w) like you said. But yea ig its just nicer to write let p_K(w) be the minimal polynomial of w over K... that's what I usually did in my exams.

[languid moss]

Would anyone be interested in self-studying Algebra with me? Since I never took it as an undergrad. I'm looking for two or three so I can keep up and motivate me in learning the material. Not sure what book but I spoke with with the professor at my graduate school and he said is certain he will be using dummit and foote. Please DM or just ping if you are interested 🙂

[hidden haven]

Same as det, but (min_K(w))(x) when you want to specify that it is a polynomial in x makes the notation more fun

[rustic crown]

lol i just write min_K(w) = f in K[x].

[hidden haven]

Not as fun tho

[rustic crown]

that's evil bro 😳

[hidden haven]

Do you want hanny to enjoy grading?

[rustic crown]

lol

hanny doesn't call minimal polynomial minimal polynomial, but instead calls minimal polynomial irreducible polynomial.

[hidden haven]

hanny is just avoiding defining more things because he thinks your batch is gonna fail the course anyway

[rustic crown]

[frank fiber]

hi

what means that a equivalence relation is compatible with group structure?

[hidden haven]

Probably that if g1 ~ g2 and h is any other element then hg1 ~ hg2 (and same for right mult)

[carmine fossil]

Shouldn't it be more like
h1~h2 and g1~g2 then h1g1~h2g2?

[hidden haven]

Yeah that'd make sense

[carmine fossil]

Ok,It probably follows from that

[hidden haven]

Idk does it?

Prove it

[carmine fossil]

Let's say hg1~hg2 and g1h~g2h for all g1~g2

Then h1g1~h2g1 and h2g1~h2g2 and you are done

[tough raven]

g1h1 ~ g1h2 ~ g2h2

[hidden haven]

ooh smort

Raghuram failed to snipe

[nova plank]

No, Buncho didn't do it right

Well, he did after editing it to make my sully look silly.

[hidden haven]

[nova plank]

So the definitions are equivalent. Yall so smart. Both got the definition right.

[hidden haven]

I got it right first

[nova plank]

Tru

Mold > Dragons

[tough raven]

Yes
Was wondering if I should delete

[tough raven]

Mold =

In fact I think of as :moldilocks:

[hidden haven]

[tough raven]

Anyway
I feel like this patter occurs fairly often
In a sort of "changing multiple things can be done one component at a time"
And the one place it doesn't work is topology

[hidden haven]

Yeah

It does in metric topology when you have triangle inequality

[twilit pawn]

if I have a finite field $F_q$, and I have $f \in F_q[T]$ of degree $d$, do I know the size of $F_q[T] / (f)$?

[cloud walrus]

doubledual

[twilit pawn]

been way to long since I thought about these things lol

is it just q^{d-1} with a division algorithm argument?

[hidden haven]

q^d

[twilit pawn]

oh yeah d ok

constant term lol

haha thanks

[hidden haven]

Yeah

[sturdy mirage]

That is only true if f is irreducible, right?

or minimal?

[tough raven]

It won't be a field if f is reducible, but it should still be a vector space with the same dimension and cardinality.

(and ring)

[final pasture]

Hi, I need some help with a linear algebra exercise, I've been stuck for a few weeks now

So let $(b_1, \cdots, b_n)$ and $(c_1, \cdots, c_n)$ be basis for a vector space $V$, and $m \in [![1, n-1]!]$.\
I want to show that there exists a $\sigma \in \mathfrak{S}n$ such that:\
$$(b_1, \cdots, b_m, c{\sigma(m+1)}, \cdots, c_n)$$
and
$$(c_{\sigma(1)}, \cdots, c_{\sigma(m)}, b_{m+1}, \cdots, b_n)$$
are both basis of $V$

[cloud walrus]

Shika-Blyat

[final pasture]

showing that I can pick vectors of (c_1, .., c_n) to complete (b_1, .., b_m) isn't hard, the hard part is to show that I can always do it while making (b[m+1], ..., bn, remanining ci's) a basis

[twilit pawn]

notation question: is sigma a permutation here?

[final pasture]

yes

[twilit pawn]

ok gotcha

ok so I believe I see how to do this

I'm pretty sure that if you choose your c_{sigma(i)} so that the top set is a basis, the bottom set is also forced to be a basis

[final pasture]

no

let B = (0,1), (1,1) and C = (1,0), (0,1)

[chilly ocean]

top set and bottom set 😳

[final pasture]

you have B' = (1,0), (1,1) which is a basis

but C' = (0,1), (0,1) is definitely not a basis

[twilit pawn]

hmmm ok

[final pasture]

[twilit pawn]

I was doing an argument in my head lemme just write it out and see what went wrong

[final pasture]

hi det

[twilit pawn]

you could write $c_{\sigma(i)}$ ($1 \leq i \leq m$) in terms of the top basis

[cloud walrus]

doubledual

[twilit pawn]

oh i see where I messed up lol

I was trying to use linear independence of the c's to do something

[rustic crown]

Anko is so cutee

[final pasture]

Also I tried to write it with matrixs:
Let $A = (a_1, \cdots, a_n)$ and $B = (b_1, \cdots, b_j)$.\
The matrix of $A$ in $A$ is obvsly:\
$$\begin{pmatrix}
1 & \cdots &0 \
\vdots &\ddots &\vdots\
0 &\cdots &1\
\end{pmatrix} = I_n$$
and say $B$'s matrix in $A$ is:\
$$\begin{pmatrix}
\lambda_{1, 1} & \cdots & \lambda_{1,n} \
\vdots &\ddots & \vdots \
\lambda_{n,1} &\cdots & \lambda_{n, n}\
\end{pmatrix}$$
What I want to know is how to pick $m$ columns of the first matrix and swap them with $m$ columns of the second matrix without changing the fact that both matrix are inversible

[cloud walrus]

Shika-Blyat

[twilit pawn]

ok so how about this

for each c_i, there are 3 possibilities

$c_i$ is in the span of $b_1, ..., b_m$

[cloud walrus]

doubledual

[twilit pawn]

$c_i$ is in the span of $b_{m+1}, \dots, b_n$

[cloud walrus]

doubledual

[twilit pawn]

or $c_i$ is in neither span

[cloud walrus]

doubledual

[final pasture]

so the first 2 cases are easy

like there's no choice to do

[twilit pawn]

I think it doesn't matter which side you put it on in the last case

you'll get two linear independent lists

[final pasture]

I'm not convinced by that🤔

First it's not clear that after adding the ci's that are in the span of b1, ., bm, you don't get new cj's that are in the new span

so you'd probably want to apply this process until that doesn't happen anymore

🤔

and even after that, if you add a new ci that is in none of the 2 new spans, you could get remaining cis that are also in the span

I'm not really clear sry lol

[twilit pawn]

nah it's ok

when you add a ci, you shouldn't be able to reach a new cj that you weren't already reaching before, because the cs are a basis

so let's try and break down the argument a little here

where know where to put the cs in the first two cases

so we just do that, and move onto cs falling into case 3

[final pasture]

it's not even clear that after doing that we still have a basis though

[twilit pawn]

if it's a linearly independent list that's enough

[final pasture]

like in the first 2 cases, we don't have any choices about where to put the cis

[twilit pawn]

im only going to force the lists to be LI

[final pasture]

yeah ok

[twilit pawn]

so we assign the cs which are forced. the other cs are still not in the span of anything

so we take one and put it in the first list if there's room

maybe that forces some cs into the other list

that should be ok

at no step are the spans going to overlap

and you have finitely many vectors so this process should terminate

this is a sketchy sketch, but im pretty sure it works

you could write it out as an algorithm and I think it would be pretty clear actually

[final pasture]

I feel like this shouldn't work, but I'll try to find a counter-example before asserting it doesn't

[final pasture]

So first I think this is wrong

let A = (0,0,1), (0,1,0), (1,0,0) and B = (1,0,0), (1,1,1), (0,2,1)
let's say we take A1 = (0,0,1), (0,1,0) and A2 = (1,0,0)
since (1,0,0) € A2, we add (1,0,0) in A1 and clearly, all the remaining elements of B are in the span of A1

[rustic crown]

spans of what are not going to overlap?

[twilit pawn]

yeah ok you are correct

[rustic crown]

as soon as you add a single ci, the spans will overlap i think

[final pasture]

that isn't a problem here, but the claim just like that is wrong

[twilit pawn]

ok so let's see if we can still save this type of approach

we add things that are forced

and maybe more things are forced

we don't want any c vector to be in the span of both, so we have to see if that's the case

[final pasture]

oh that should definitely work, just putting forced stuff every time it is forced, then do some arbitrary choice, check if anything is forced, do the forced stuff, and repeat

like if the exercise is right

[twilit pawn]

yeah exactly

[final pasture]

then this must work

[twilit pawn]

so i think the only thing to check is that this process never causes a c vector to be in both spans

[final pasture]

The hard part is to show that at any point of the process, we can't have an arbitrary choice that leads to a situation where this process can't end

[twilit pawn]

there are finitely many vectors, it will terminate

[final pasture]

ofc

I mean can't end as we can't complete the families with the remaining vectors without losing linear independance

[twilit pawn]

right

[final pasture]

(so actually this doesn't need to work, maybe only some arbitrary choices work)

[twilit pawn]

yeah im with you on that

[final pasture]

(oh also is this the right channel for linalg questions or was #linear-algebra more appropriate ?)

[twilit pawn]

it's fine there isn't other discussion here rn

it's a subtle enough question in my humble opinion

[nova plank]

No, it's not fine! Boo shika

[final pasture]

le french hater has arrived

[final pasture]

ok thanks

[twilit pawn]

ok now that i actually got paper and started writing things, my confidence this works has gone up

[final pasture]

I'm not convinced it doesn't work, but if I had to bet, I'd probably bet the naive algorithm I described earlier doesn't work, yeah

wait

does "has gone up" mean it augmented ?

[twilit pawn]

ok i'm going to write the full argument carefully

this is an inductive construction, and we can assume choices have been made so that the following happens

if $B_1 = (b_1, \dots, b_m, c_{\sigma 1}, \dots, c_{\sigma k})$, and $B_2 = (b_{m+1}, \dots, b_n, c_{\sigma m+1}, \dots, c_{\sigma j}$

[cloud walrus]

doubledual