groups-rings-fields

[rapid junco]

Here are my thoughts

X is a subspace of R^8, thus R^8 forms a direct sum of X and Z where Z has dimension 3. Z being a subspace of Y has 2 dimension difference which means by the pigeon hole principle we must have that at least 2 basis vectors are already spanned by the rest

[dusty verge]

R for reals and not R for ring yeah?

By the way the reason is cause the dimensions don't add up good

5+5>8

[dusty verge]

Why is Z a subspace of Y?

I mean if you don't already know you can add dimensions, I don't see why you can take that for granted either

[rapid junco]

yeah I know its because the dimensions dont add up but idk how to prove there is no non zero intersection

[dusty verge]

Oh here's how you do it

[rapid junco]

i was thinking because they will share basis vectors

[dusty verge]

There's a canonical map into the intersection yeah?

[rapid junco]

but again that can be a little trippy

[dusty verge]

Cause it's a subspace of both

[rapid junco]

Q: R^8 to X int Y?

[dusty verge]

Nah X to XnY

[rapid junco]

okay

[dusty verge]

Hmm does this work actually?

[rapid junco]

not sure

[dusty verge]

Basically my thought process is that you get two maps, kernel are disjoint

Oh, just go contradiction lmao

[rapid junco]

I was thinking about just using direct sum decompositions and quotienting everything

[dusty verge]

The span of their bases is dimension ten of they're disjoint

[rapid junco]

yes

oh

lmfao

bro fuck why is it that easy

[dusty verge]

Because it's only easy if you know why it works

All math is hard until it's easy

[rapid junco]

I wanna kms

[dusty verge]

Don't

That's just the nature of math my guy

Plus it's a bit harder than it sounds cause you gotta do some finessing with the span and shit

[pastel cliff]

idk how obvious this is but

why is a-bi also a root of the minimal polynomial of a+bi

also iss that backward direction okay

[lethal dune]

conjugates

yes ⟸ is fine

[pastel cliff]

awesome

[lethal dune]

[pastel cliff]

Prove that $x^3 - 2$ and $x^2 - 3$ are irreducible over $\Q(i)$. What is $[\Q(i, \sqrt[3]{2}) : \Q]$? can i get hints on this por favor - most direct seems to be to just find the roots but is there a better/smarter way

[cloud walrus]

not sebbb not stμ₂dying

[faint fractal]

So I've gotten to the point where I showed that a_ng(x) gives you a polynomial of lesser degree. Can someone explain why a_ng(x) = 0 then?

[wicked patio]

for the degree over Q one

[wicked patio]

Use Q(cbrt(2)) in the middle

so find the degree of that over Q and then the degree of Q(i) over that

[wicked patio]

If you have a non-zero polynomial g s.t. p*g=0, you can find another such polynomial with smaller degree

Imagine repeating this process until you get the 0 polynomial

[faint fractal]

sorry, can you explain why you can find another polynomial with smaller degree?

so a_ng(x) removes the first term

why does that mean it removes all terms?

[wicked patio]

it doesn't

it removes the first term

that makes the degree smaller

[faint fractal]

right

[wicked patio]

Sorry, this isn't right

[wicked patio]

Because g(x) is explicitly chosen to have minimal degree

So finding another polynomial with smaller degree would be a contradiction

[faint fractal]

wait sorry

I still don't understand why a_ng(x) = 0 then

oh

I understand

ah ok

cool thanks haha

wait

so if a_ng(x)p(x) = 0, is the idea just that if you look specifically at (a_ng(x))p(x) = 0, since a_ng(x) = polynomial of lesser degree, then since g(x) is a minimal polynomial, then the only solution would be for a_ng(x) = 0?

[wicked patio]

yeah

[faint fractal]

cool, thanks

[wicked patio]

np

[white oxide]

is this equivalent to saying that if p(x) is an irreducible polynomial, then any two splitting fields for p(x) are isomorphic?

[lethal dune]

Yes

[solar shore]

are there two groups? i cant think of the other one except D_n

[tribal moss]

Hmm, some n are easy enough, for example n = 2^k-1.

[solar shore]

oh

cant i just answer D_n for everything then?

bcs D_n would have exactly n elements of order 2 since its just all reflections

also since its odd it wouldnt contain R_180

[lethal dune]

you can maybe try Z/2xZ/2x...Z/2 x G

ah nvm it doesn't always work

[solar shore]

i would but the thing is this part of the book hadnt covered those kinds of groups

ive been doing some old exercises in preparation for my final

[tribal moss]

If n+1 = 2^m·k with k odd, how about the product of m copies of C_2 with D_k?

[dim widget]

I mean this is a bit dumb but I don't see why you cant just give D_n and D_n \times Z/5?

The question is not well-specified enough to make you have to find interesting answers

[solar shore]

yeah..

[tribal moss]

Oh, duh.

[solar shore]

i was thinking about skipping this exercise altogether

[lethal dune]

examples will somewhat involve D_n's

[south patrol]

or D_2ns

[tribal moss]

At least to the extent that two different order-2 elements of either generate a subgroup isomorphic to some D_n or infinitely many different order-2 elements of the form abab...ba.

[formal ermine]

@dim widget speedrunning moderator

[dim widget]

NEW white name to mod any% wr (glitchless)

[lethal dune]

yeah fastest I've seen

[chilly ocean]

how do I find a plane equation if I have 2 vectors? and no known points

[tribal moss]

Oh! I think you were expected to say "the dihedral group with 2n elements" and "the dihedral group with 2n-2 elements".

[solar shore]

right yeah

at least i think so

i ended up just putting the dihedral groups as my answer for that one

couldnt force myself to skip it

[tribal moss]

I think you should go with two different variants of TTEG's solution just to spite the teacher.

[wooden ember]

A well deserved promotion 👍

[solar shore]

your average math major spiting the book

"fuck you , you cant tell me what to do!"

[tribal moss]

And the n>3 assumption must be because they're not sure you'll consider V_4 a dihedral group. (Algebraically it ought to be, but geometrically it's not, say, the group of isometries of a set of 2 points 180° apart on a circle).

[solar shore]

what is V_n?

[tribal moss]

V_4 is the Klein 4-group, presented e.g. as < a,b | a²=e, b²=e, ab=ba >.

(There are no other numbers used as subscripts for the V!)

[solar shore]

oh i see

very cool

swear everytime i do more abstract algebra, i can understand the allure of the subject

[tribal moss]

These are easier to do (at least for me) if you consider them as "the only solutions to x^2 == 1 mod p^n are 1 and such-and-such". Then use induction on n to get all but the first of the n base-p digits of x from the induction hypothesis.

[solar shore]

do i also induct on p?

[low void]

Could someone please motivate the idea behind definition 2.7 in this book http://math0.bnu.edu.cn/~liuym/Book-algebras-and-representations.pdf
It relates to modules of polynomial algebras.

The K[X]-modules correspond 1:1 to tuples (V,a) where V is a K vector spce and a an endomorphism V-->V. You set X*v:=a(v). But what is the motivation behind it?

[rapid junco]

Prove if the inner product of Ax, and y equals zero for all x and y then A equals zero

Shouldnt it be just <Ax, y> = 0 for all x, y iff

<Ax, y = Ax> = 0

iff Ax = 0x

But this is for all x, thus A must be zero

Ye?

[tribal moss]

No.

Inducting on primes is rarely helpful for anything.

[void cosmos]

yes @rapid junco

choose any x=y

with x non zero

then u get A*Abar<x,x> = 0

or no

just A

[dire siren]

As an alternative without induction to @tribal moss's idea, for 84 you can solve that congruence by noticing that for any integer k, the numbers k-1 and k+1 can't be both divisible by p. Therefore p^n | (k-1)(k+1) is equivalent to p^n | k-1 or p^n | k+1.

[rapid junco]

?

[dire siren]

For p=2 things are a bit different: gcd(k-1,k+1) divides (k-1)+(k+1)=2, so it is either 1 or 2
If gcd(k-1,k+1)=1 we have the same conclusion as above.
If gcd(k-1,k+1)=2, the relation 2^n | (k-1)(k+1) is going to be equivalent to 2^(n-1) | k-1 or 2^(n-1) | k+1

[dim widget]

That's what a K[X] module is. An R module is just an abelian group M with a map of unital rings $R \to End(M)$. So a K[X] module is just a K-module $V$ (vector space over K) with an endomorphism (the image of $X$ in $\text{End}(V)$) of the endomorphism of the abelian group $V$ which commutes with the action of $K$: that's just a $K$-linear endomorphism.

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

This comes down to just showing that x is 0 iff <x, y> = 0 for all y.

[tribal moss]

Ooh, much nicer! And a slight extension of that deals with 85 too.

[low void]

Is it "what they are" or a way to characterize it? Is it like saying "the epsilon delta definition is continuity"? We can also characterize continuity by preimages of open sets.

[dim widget]

It just comes from unpacking the definition.

[low void]

Is it a way to make them more concrete or mathematically fruitful?

[dim widget]

More concrete than "a vector space over K with an endomorphism?"

[dim widget]

Maybe you could try to work out this: suppose the $K$-vectorspace $V$ is finite dimensional with an endomorphism $T$, can you work out what the $K[X]$ module structure is explicitly when $T$ has a full set of distinct eigenvalues? What about if the Jordan normal form is more complicated?

[cloud walrus]

Topos_Theory_E-Girl

[slim kayak]

How would I show that Z[x,x^-1] (Laurent polynomials over Z) isn't a PID without using krull dimension? Apparently (2) is a prime ideal and (2) isn't the same ideal as (2,1+x) but I don't know how to show that.

Specifically showing that 2 isn't the product of two Laurent polynomials

[dim widget]

Have you tried reducing mod 2?

[indigo ridge]

I’m trying to see for question (g) weather this relation is transitive or not however I’m having trouble with the absolute value inequalities

[rustic crown]

another idea is to find a non-zero prime ideal which isn't maximal

[slim kayak]

You mean showing that F_2[x,x^-1] has no zero divisors?

[dim widget]

Yep.

That would be enough to show that it is not a PID!

[rustic crown]

(wait did you mean not using krull dimension at all , ig then i was just cheating)

[slim kayak]

Yeah, I know. Same Problem

At least idk how to show that the sum a_ b_j with i+j=k equalling 0 for all k implying that a_i and b_j must be zero

[dim widget]

So let's ignore the x^{-1}

If F is a field

can you show that F[x] is an integral domain?

(e.g. if $F = F_2$...)

[cloud walrus]

Topos_Theory_E-Girl

[slim kayak]

Yeah, maybe I need a refresher but I know how to show that F[x] is euclidean, then F[x] is a PID and then (X) is a maximal ideal

[dim widget]

Okay, but how would you show that it is an integral domain?

That's all very nice but it's not the question that I asked.

[slim kayak]

Yeah alright, using degrees.

[solar shore]

i actually ended up doing this!

i appreciate it a ton

[dim widget]

Great! So why is F[x, x^{-1}] a domain, can you generalize this argument?

[slim kayak]

Nope

[dim widget]

Okay, well what does an element of F[x, x^{-1}] look like compared to a polynomial?

[slim kayak]

You can represent it as a the sum of polynomial in F[x] and F[x^-1] (or their isomorphic subrings) ig

[dim widget]

So maybe it's better to say that it is P(x)/x^m for some m and some polynomial P(x).

[slim kayak]

Oh?

Parsed as lowering the indices of x^i by m?

[dim widget]

Yes, and making them negative if necessary

[slim kayak]

Fair, alright.

[dim widget]

In general something like $\sum_i a_i x^i $ where $i \in \mathbb{Z}$

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

and $a_i = 0$ for all but finitely many $i$

[cloud walrus]

Topos_Theory_E-Girl

[slim kayak]

Thats the definition I was initially given, yeah

So I would think of my product of elements in the Laurent polynomial ring as f(x)g(x)(x^(-n-m)) ?

And so it follows that it is an integral domain since Z[x] is

[slim kayak]

Where f(x)/x^(-n) and g(x)^(-m) with f and g from Z[x] represent my two elements from Z[x,x^-1]

[dim widget]

Yes

Okay so now can you prove that it's an integral domain?

[slim kayak]

ye

[dim widget]

Okay great, so now do you see why Z[x, x^{-1}] is not a PID? Hint: can you show that the ideal (2, (x-1)) can't have a single generator?

[slim kayak]

For the original argument I'd still have to show that the ideal generated by 2 doesn't include 1+x.

Although expressing elements using Z[x] ought to make that easy

[dim widget]

Yep

[slim kayak]

Yeah, expressing elements as f(x)/x^m is night and day when it comes to tractability of calculation 😅

Thanks again

[dim widget]

No worries! Glad you worked it out.

[frigid lark]

Is it true that for distinct $a_1,..,a_d\in K$ a field, that $(a_i^n)_{n=1,...,d}$ for $i = 1,...,d$ form a basis for $K^n$?

[cloud walrus]

Parrot Tea

[delicate orchid]

this smells like a vandermode matrix

yeah if it ran from n = 0... d it would be a vandermode matrix

[frigid lark]

Oh, I never liked that matrix

[delicate orchid]

so the matrix with rows consisting of those vectors would have non-zero determinant, so yes they'd be linearly independant

[frigid lark]

Ok, LI is enough for me

[wraith cargo]

Here at the bottom of the page I assume they meant a^{-1}b \not \in (a)?

Or wait oooh

if b was in (a) then it would be of the form b = ca for c \in A
But since this isn't possible we have that a^{-1}b can't be any element of A

[untold cloud]

Hi, guys, why this discriminant is in K?

[lethal dune]

fixed by all automorphisms, though you need Galois extension to claim it

actually nvm you don't need Galois

[delicate orchid]

huh, it's the vandermode determinant again

two times in two hours

bizzare

[rustic crown]

did someone say det

.<

[lethal dune]

ye det for non commutative matrix ring

[untold cloud]

how the det relates to the question? do we put coefficient of f in matrix and by doing row operation, we have the det of this matrix is the discriminant?

[frigid lark]

It's a polynomial of elementary symmetric functions which are in K by vieta's formulas

Or are in K as they are the coeff's of f

[wraith cargo]

The discriminant is the determinant of a certain matrix

[dire siren]

Because the expression for the discriminant is symmetric in a_i's and we have by the fundamental theorem of symmetric polynomials that it can be written in terms of the elementary symmetric polynomials in those variables, which are all in K by Vieta.

[untold cloud]

Thank you guys!

I see it now

[cloud walrus]

MyMathYourMath

[rustic crown]

iso thms are your friends

k[a_1, ..., a_n] is just k, so you're good with what you say

[solar glacier]

ok define a map $k[x_1,...,x_n] \to k$ via $f(g(x_1.,,,.x_n)):=g(a_1,...,a_n)$

[cloud walrus]

MyMathYourMath

[solar glacier]

thats what i thought to begin with!!!

[cloud walrus]

MyMathYourMath

[rustic crown]

yep

[solar glacier]

cool thanks!

[formal ermine]

how do I understand the characteristic polynomial of an element in a field

[rustic crown]

whut do you wanna understand

[formal ermine]

what is it

I only know it for matrices

[rustic crown]

if F/k is a finite field extension and a is an element of F, you can look at F as a k-vector space and then multiplication by a is k-linear, so look at char poly of this linear map

[formal ermine]

thanks

[rustic crown]

[formal ermine]

when does it coincide with the min poly

[rustic crown]

when F = k(a)

[formal ermine]

what do we need for the primitive element thm again

[rustic crown]

else it is min poly ^ [F:k(a)]

[formal ermine]

just separable?

[rustic crown]

nope

not even separable

[formal ermine]

I keep forgetting the conditions

ah separable and finite

[rustic crown]

(to see this pick a k-basis of k(a), and a k(a)-basis of F, write the map as a block matrix and play a little)

[wraith cargo]

It doesn't need to be separable for this to work

[formal ermine]

ic

[wraith cargo]

Just finite

[rustic crown]

det of this map is called norm wrt F/k

[formal ermine]

yee

we did that in alg nt today

and I was confused what char poly meant

[rustic crown]

you'll need a bit of separability to play with the equivalent definition of norm which is multipliy all conjugates

[wraith cargo]

Separability just gives you the condition that your linear term in the char poly is the sum over conjugates

(same for constant term)

I.e. the trace and norm

[formal ermine]

lol we just assumed everything is finite and galois

[wraith cargo]

If it wasn't separable the you'd have to multiply/raise the elements of the sum/product by the degree of inseparability

[rustic crown]

is trace useful at all in char p?

because like inseparability would just make it 0 right

[formal ermine]

I hate linear algebra tbh

probably because I suck at it

I mean I suck at everything math

[rustic crown]

[formal ermine]

but especially at linear algebra lol

[rustic crown]

dw, you'll exhuast all possible ways to suck at it soon enough, then you'll be

[wraith cargo]

I mean for finite fields at least they are all separable extensions
So i don't think it's a problem here
Plus if the degree of inseparability isn't a multiple of the char there's no problem

[rustic crown]

insep degree is always a p-power where p is the char

[wraith cargo]

Oh really

Interesting

[rustic crown]

but yea, so ig it's only useful in separable situations

[dim widget]

If the degree of inseparability is not 0 then the trace is always 0

[wraith cargo]

I didn't know that

[dim widget]

That's what the trace measures really

It gives a perfect pairing on a field iff it is separable over the base field

[rustic crown]

whuts the pairing

[dim widget]

x, y \to Tr(xy)

[rustic crown]

[dim widget]

Ramification measures the extent to which this perfect pairing fails to extend integrally.

Trace pairing ties up a lot of pretty ideas in basic number theory/algebraic geometry.

[formal ermine]

oo we also defined this today

it's symmetric bilinear and non degenerate

right?

[dim widget]

Yes

It is by definition symmetric since fields are commutative, bilinear because Tr is linear and multiplication is bilinear

It is nondegenerate exactly when the field is separable

[lethal dune]

can you elaborate?

[dim widget]

The discriminant of an extension $A'/A$ of Dedekind domains (so $Frac(A') = L$ and $L/K = Frac(A)$ is finite separable, and A' is the integral closure of A' in L) with A' free, then if $a_1, \dots, a_n$ is a basis for A' as a module, $disc(A'/A) = det(Tr(a_ia_j)_{i, j})$

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

So the primes which ramify (those which divide the discriminant) are exactly the places where the trace pairing fails to be perfect.

[dusty verge]

What's a Dedekind domain?

I've seen trace before and it was gross, but I haven't run into that

Oh it's almost a pid, got it

[dim widget]

If $A'$ isn't free, then you can still make this argument locally at a prime and you get the same conclusion. But it's more visible when A' is free because you can make global definitions.

[cloud walrus]

Topos_Theory_E-Girl

[dusty verge]

What's the initial question here?

[lethal dune]

what does it mean for the pairing to fail being perfect?

[dim widget]

It means that the dual of the ring of integers under the trace pairing is not the ring of integers itself.

[lethal dune]

okay

[dim widget]

You know a priori that it gives some lattice inside of L, but the discriminant is telling you roughly which fractional ideal you get, or at least how far it deviates from O_L and at which primes.

[dusty verge]

What exactly is a perfect pairing?

Okay I googled it

[dim widget]

A bilinear form $B: M \otimes M \to R$ over a ring $R$ induces a map $M \to \text{Hom}_R(M, R)$ by $m \to B(m, -)$, and this is called perfect if this map is an isomorphism.

[cloud walrus]

Topos_Theory_E-Girl

[dusty verge]

After a quick Google, a bilinear form is just a bilinear function into the field?

Or oh, the underlying ring cause modules suck and I hate them

[south patrol]

Lol

But yes that is the specialisation to linear algebra

[dusty verge]

Linear algebra is nice because everything is nice and then modules were like what if you do linear algebra but everything is not nice

[south patrol]

True

But then there are some beautiful things even about linear algebra you can only see by allowing more general objects

Things like canonical forms for example

[dusty verge]

What is B(m, -) here?

Oh wait lol nvm

So it's invertible matrices?

Wait nah, it's different

This is reminding me of semi direct products, but idk if it's an actual similarity, or if the form is just similar

Hey so what the hell is a determinant?

Like sure I know I can use it for shir, and that it's like a homomorphism and stuff, but seriously what the hell is it?

In all my math I've never figured it out

[dim widget]

@rustic crown

[dusty verge]

Thanks

I guess I'll never know

[dim widget]

There isn't really a definitive answer. It measures the oriented area of a set of vectors over the integers. It's the universal algebraic homomorphism from Gl_n to an abelian group, there are lots of answers and it depends on what you want to use it for.

[dusty verge]

Hmm, I've actually convinced myself why the matrix thing works though

The determinant of a 2x2 matrix embeds a test for linear dependence right?

[rustic crown]

yee det has lots to do with invertibility

[near star]

I mean as far as I know in terms of linear algebra its an n-linear alternating map with D(I) = 1

[rustic crown]

and not just that, also the actual inverse... since you have A * adj(A) = adj(A) * A = det(A) * I

[dusty verge]

Alternating map?

[near star]

if u flip it go negative and if you have 2 of the same its 0

[dusty verge]

What's adj(A)? Looks like adjoint but I don't actually know what an adjoint is

[near star]

atleast thats what it said in H&K iirc

adj(A)_{ij} = (-1)^(i + j) det(i|j)

[rustic crown]

it's called adjugate (sometimes also adjoint, but idk why that's the case)

[rustic crown]

oh transpose of that

[dusty verge]

Like if f is a 2 linear alternating map, f(x, x) = 0 and f(x, y) = -f(y, x)?

[near star]

sry

[near star]

yes

ill ss a definition from a book

more reliable source cause i read it a long time ago might be missing a few details

[dusty verge]

Oh. Right I've heard adjoint wrt functors, so disregard that. What's it do though?

[rustic crown]

it's hard to say what "det" is... cause you can give many equivalent definitions. one of the nicer to work definitions is from exterior algebra

[near star]

and u prob know what n linear means, if not u think of determinant as a function that takes in the columns of the matrix like D(a,b,c) for each column a, b and c, and its linear in each of those

[rustic crown]

if you distill all the vocabulary from it, you'll end up with what @near star is telling

[dusty verge]

Oh holy shit yeah, if you perform gaussian elimination on a 2x2 matrix, the determinant is just what ends up in the bottom right corner

And I'm assuming you can induct on that

[near star]

also there's only one D having those properties (we call determinant, incase u thought there could be more then 1)

u can quite easily derive the permutation equation if u give it some thought

[dusty verge]

Permutation equation?

[near star]

let me get it rq

its called leibniz formula

[rustic crown]

(whut's rq? )

[near star]

real qwick

S_n can be thought of all the bijections of {1,2,...,n}

so like rearrangements basically

[dusty verge]

Okay, so that's the (transpose of the) adjugate? What's that I|j notation

[near star]

(aka permutations)

[dusty verge]

Oof I quoted the wrong one sorry

[rustic crown]

(i'm seeing that for the first time as well 🙈)

[near star]

det(i | j) means taking the determinant of A when u remove the ith column and jth row... maybe, I always mix rows and columns up with matrices, but its either ith column and jth row or ith row and jth column

[rustic crown]

but it shoudl mean take the matrix and delete the ith row and the jth column

[dusty verge]

Okay yeah that was my guess too

[near star]

ah so it was ith row and jth column xd

[rustic crown]

yea i had to memorize that in A_{ij} the first entry is for row

[dusty verge]

Nah you're just dealing with nega-matrix multiplication

Which does the rows onto the columns

[near star]

I had the same type of thing with my left and rights when I was younger

so hard for me to remember that kinda thing

[dusty verge]

So it's the matrix that contains the terms you add up to get the determinant in the recursive algorithm

And encodes where they are

[near star]

yeah basically, but the recursive algorithm only requires one row

you can use any row for it iirc

[rustic crown]

yea it's weird because if you're doing math, you're likely thinking of matrices as a bunch of columns (because of the right to left function notation thingy), so A_i might mean ith column, but the cs guys tell us that a matrix is a collection of rows A[i] :p

[near star]

lul

[dusty verge]

Yeah but cs matrices are kinda

[near star]

A + i + j

[dusty verge]

Learning abstract linear algebra is a journey btw

[near star]

a i

jiaia

[rustic crown]

[near star]

im reading dummits abstract algebra

it made me like

write a computer program

as an exercise

for computing gcd's

[dusty verge]

First your like vectors are collections of numbers, then you're like no they're not, and then finally you're like oh wait yes they are

[near star]

🗿

thought that would be number theory but oh well

[dusty verge]

For all x in range( 1, max (a, b)): if x | a and x| b, list.append x; return max(list)

[rustic crown]

that's a lot of math lol, lie algebras are like matrices with commutator, well let's generalize and see what are other objects that satisfy similar properties, oh we got subgruops of matrices again :p

[near star]

wtf

xd

mine was so much bigger

I made an array for all the remainders and quotients and did euclidean algorithm

[rustic crown]

linear complexity

[near star]

but tbf I had to also find the coefficients to a z-linear combination of the two numbers that resulted in the gcd

so like

gcd(a,b) = xa + yb

find x and y (which r integers)

[dusty verge]

That's cause mine does it the dumb way of just checking if every number is the gcd

[near star]

Pain.

[sonic coral]

love the euclidean algorithm

[near star]

I made a mistake on (f)

and had to start over again

after almost finishing

[rustic crown]

maybe they wanted you to conjecture and prove the number of steps taken is like (at most) 5 times the number of digits

[static temple]

you gotta do what you gotta do

there are 10000 ways of expressing this same algorithm for finding a and b

there is a matrix way

a way using continuous fractions

the standard way

the matrix way was the most flex

[dusty verge]

K = 0, while true: if (b|a +kgcd(a, b)): return(k, (a + kgcd(a, b))/b); else if (b | a - kg): return (-k, (a -kgcd(a, b))/b); else k++

[static temple]

but the continuous fraction was also cool

[rustic crown]

another cool one is this

it's a diff algo tho

[dusty verge]

Just do it the monkey way

[sonic coral]

this is neat

[dusty verge]

And check every linear combination

[static temple]

yeah but then you gotta find those x and y *

wait

yeah

[rustic crown]

also it's really uwu because computers do shit in binary

[sonic coral]

do you have a proof of the algorithm

[static temple]

its always the euclidean algorithm

there is no other ways X)

[dusty verge]

K = 0, while true: if (b|a +kgcd(a, b)): return(k, (a + kgcd(a, b))/b); else if (b | a - kg): return (-k, (a -kgcd(a, b))/b); else k++

[rustic crown]

yea you can do that in this algo as well

[static temple]

(big question btw: are there other ways)

[dusty verge]

Yes I just gave one

You check every number

[static temple]

wait

no its the same idea, you divide till you can't divide anymore

it's euclidean

[dusty verge]

Nope, it's not

[rustic crown]

the only idea in all algos are gcd(a, b) = gcd(a, b-a)

you just spice it up here and there

[static temple]

yeah

[dusty verge]

I don't think mine uses that

My algorithm for the gcd was just that gcd is the biggest divisor of both

[rustic crown]

(yea because that's not the fastest algorithm :p)

[dusty verge]

Well yeah lmao,my algorithm sucks but

But it's easier to write

[near star]

fastest algorithm for gcd

[rustic crown]

(i can't be the only one who thought that was a duck)

[static temple]

its actually the euclidean algorithm but where instead of finding the k directly, you do a k++

but you still do a a-kb to find a rest

[dusty verge]

Nah, that's my algorithm for finding the coefficients in bezouts identity

[lethal dune]

cringe

[rustic crown]

ryu

[dusty verge]

You already know the gcd there

[near star]

alright no more videos

[lethal dune]

det ◖⁠⚆⁠ᴥ⁠⚆⁠◗

[rustic crown]

[lethal dune]

gcd( a, b%a)

[dusty verge]

Like I guess it looks kinda similar, but it's not finding the gcd. It's just checking if ax +by = gcd(a, b) by checking if gcd(a, b) - by is divisible by a

If it is, then ax + by works

[near star]

the way we did it was by reversing euclidean algorithm

you just went back up the remainders and quotients and packed them all together

[rustic crown]

a bit better is
gcd(a, b) = gcd(b%a, a)
that way you don't infinitely loop if a > b :p

[static temple]

yeah that's the "standard" one (which is known to be the fastest)

[dusty verge]

Save the last bit, shift a and b over until a bit has swapped, and take the difference in the number of times you've shifted them, right?

Plus some other details

[lethal dune]

lol haven't written in a long time

[near star]

Let R - {0} be the group of reals over multiplication, what is the value of |2|

1. Infinity
2. 2

[dusty verge]

Just use your Turing oracle to make sure that doesn't happen

[dusty verge]

Define ||

[near star]

Order or abs 😈

you have to guess

[static temple]

it's |2|

i'm so good

[dusty verge]

Which reals?

[pastel cliff]

it's me

[near star]

basically how imaginary numbers were invited

[pastel cliff]

im so real they call me reals

[static temple]

what is real

[near star]

sqrt(-1) = sqrt(-1) but lets call it i 🧠

[dusty verge]

If you mean the group of the reals 0, 1, and 2, over multiplication, it's 2

[near star]

the reals without 0

the set of all possible decimal expansions, finite or infinite

(signed)

without 0

[dim widget]

hi, I'm the problem it's me.

[dusty verge]

Z_3^* don't have zero

[pastel cliff]

real!

[static temple]

no, it's p-adic

[lethal dune]

trying to be anti heros

[dusty verge]

Two mathematicians are beefing in old timey times and one is like "wanna duel" and the other one starts crying

[pastel cliff]

this is the story of galois

[static temple]

no one cried X)

[dusty verge]

The future if Galois was a better shot

[static temple]

dunno but there were theorems that generalized some Jordan like theorems for matrices that were quite cool

but the first proofs are always some kind of weird recurrences on the dimensions

then it goes to the notion of eigenvalues and graphs, and everything goes brrr

[dusty verge]

I feel like the proofs people tell you about are really intuitive ideas that get refined down into insanity

Usually it's just like the dimensions don't add up

[static temple]

I should redo some lie algebra, have some nice memories about them

[dusty verge]

You lie

[static temple]

had some nightmares too not gonna lie X)

[rustic crown]

haskell code is nice sometimes :p

[dusty verge]

That was clearly python

[rustic crown]

it's like pretty much math

[static temple]

but lie algebra was the most down to earth thing I learned in abstract algebra X) (with finite groups)

[rustic crown]

oh yea that was ofc python lol

[dusty verge]

All code can be python if you download the right shit

[lethal dune]

fortran

[rustic crown]

like the one for gcd would be

gcd :: Integer -> Integer -> Integer
gcd 0 n = n
gcd a b = gcd (mod b a) a

[solar shore]

assembly

[dusty verge]

I knew a PhD physics that insisted on using fourtran

I think he might have been amish

[lethal dune]

mov eda, 10
mov edb, 5
add eda, edb
..

[dusty verge]

What the fuck?

[lethal dune]

shit I don't recall assembly

[static temple]

I learned coding without a computer by using the books my brother left me, so I learned fortran... Then I saw python when they bought me a computer and was like "oh, fortran wat is dat? Neva heard of that"

[lethal dune]

[solar shore]

i hope you dont

[dusty verge]

Oh got it

[solar shore]

assembly’s scary ppl who like it r scary

[rustic crown]

maybe i should change the name tho, gcd would be already defined :p

[solar shore]

[lethal dune]

fastest tho lmao

[solar shore]

true but ur scary if you fr know it

[dusty verge]

It's machine code for cowards

[static temple]

and assembly DEPENDS on the langage you are using X)

[barren sierra]

Faster != better

[lethal dune]

I used to

[static temple]

nasm ou x-asm are all differents

[solar shore]

YA

[dusty verge]

It is stronger and harder though

[solar shore]

[static temple]

texas instruments calculators had basic but you could upload some programs written in a weird asm

z asm?

[rustic crown]

i'm sure people who do stuff with arduino know it >.<

[lethal dune]

idk you can do audrino stuff in C

[dusty verge]

I programmed all my stuff into my calculators programming thing for the AP calc exam

[lethal dune]

even python has a library now

[dusty verge]

I got away with it, but also I didn't end up using it so shrug

[rustic crown]

what if you make a breadboard calculator

[solar shore]

i dont code as much anymore but i am starting to get back into it to make my minecraft mods

[lethal dune]

ngl assembly cool

[solar shore]

[static temple]

never programmed something for my math courses in highschool, except the ax+by=truc

[lethal dune]

makes exploiting easy

[static temple]

because it was really useful

[dusty verge]

Programming is math for cowards

[pastel cliff]

im an expert in desmos

[dim widget]

:hackermans:

[lethal dune]

ppl actually put desmos as a skill in their linkdin profile

[dusty verge]

A code is just a proof that the thing does the thing

But less sexy

[static temple]

learned magma because of a book on finite groups

[rustic crown]

ryu hecker

[static temple]

but it was the worth idea

[lethal dune]

[solar shore]

im an expert on command blocks

[static temple]

magma is so closed, like how do you even get access to it without going on its website

[solar shore]

[rustic crown]

det is an expert on

[dim widget]

ryu spending their time reading machine code and coming up with assembly level exploits

[solar shore]

eevee so cute though

[lethal dune]

knowledge of assembly is needed for reverse engineering actually

[static temple]

(ryu?)

[solar shore]

i jus realized i joined this server almost 5 years ago

[lethal dune]

I mean it's essential, not that you write exploits in assembly

[static temple]

well it's mostly in C i guess nowaday

[lethal dune]

yeah who tf is ryu, even I'm wondering

[near star]

decompiler try to be 80% accurate challenge (impossible)

[dusty verge]

Hadoken

[static temple]

oh this ryu

[solar shore]

https://tenor.com/view/shou-ryu-ken-gif-6057847

[near star]

🐒

[dusty verge]

https://youtu.be/49jKoMbJyEo

[static temple]

How do you guys remember stuff like "oh a module on a principal ring is x and y"

[rustic crown]

me didn't know existence of discord before 2020

[static temple]

?

[dusty verge]

Uh you don't remember it

Just think about it so hard that you see why it can't be anything else

[dim widget]

That is only true if it's a module over F_2, most modules have more than two elements!

[solar shore]

i dont remmeber joining it in 2018 tbh

[dusty verge]

I joined a week ago

[rustic crown]

wait 2018 was 5 years ago?

[solar shore]

i just remember going on it for a month when i first joined and u werent allowed to help people who were already being helped

[static temple]

no by x and y i meant some properties x and y and stuff

[solar shore]

[rustic crown]

wasn't it like 1 year ago

[solar shore]

[static temple]

and x could be y if it was the case X)

[dusty verge]

Most modules with more than two elements have more than two properties!

[lethal dune]

fuck it's been that many years

[dusty verge]

But leff, the trick is when you understand it you don't have to remember it

[static temple]

but yeah, I had problem memorizing those properties, of course it's nothing compared to learning the tissues in biology but yeah

[solar shore]

real

me learning neuroanatomy right now

[dusty verge]

The trick is to come up with analogies for the things. All modules are vector spaces , until they're not

And then get good at knowing when to use which analogy

[rustic crown]

i'm still living in when covid started

[static temple]

no I mean yes but there are so MANY fucking hypothesis sometime for a X or Y theorem

and you slightly change it and its no more true

[lethal dune]

yeah definitely feels like that ngl

[static temple]

when you have to remember 1000 small theorems I ignore the method to learn them

[dusty verge]

It's cause if you really make sure you get it, you eventually can see why they're there

[static temple]

dunno, I'm maybe stupid

[lethal dune]

I spent covid time doing shit

same now

[dusty verge]

Of course dommukos theorem only works on orange monkeys, it'd be absurd if they were brown

[static temple]

because I do get it, can prove it if you give me the premisses, but it takes time everytime

[dusty verge]

Being able to prove it means nothing

[rustic crown]

same >.<

[static temple]

played minecraft mostly

[dusty verge]

It's not a dick measuring contest

[near star]

he spent covid doing a shit

😱

[static temple]

and stopped math also during covid

[dusty verge]

Don't prove it, just try and understand what's happening with the thing

[dim widget]

#chillgebra moment

[static temple]

I think it's the bourbaki way of showing math that kills me everytime

[rustic crown]

what's the theorem... i'm interested

[dusty verge]

Bourbaki?

[static temple]

yé

[dusty verge]

I was asking for an elaboration

[static temple]

oh, it's simple

[dim widget]

Nicholas Bourbaki was the most prolific mathematician of all time

[static temple]

there is no deep explanation, it's just definition, properties, theorems

[dim widget]

Just like all rings

[static temple]

corrollary

[rustic crown]

https://tenor.com/view/kanna-kamui-pat-head-pat-gif-12018819

[dim widget]

corrollarryy

[lethal dune]

lol

[dusty verge]

An orange monkey is contiguous if and only if it's tail can be set parameters

[static temple]

dunno i'm french

corollaire we write

[dusty verge]

Man what a dumbass

[solar shore]

just write corr

[dim widget]

Imagine being french unironically

smh my damn head

[dusty verge]

Fuck that guy, he doesn't know shit

[static temple]

nah it aint latex

[solar shore]

[solar shore]

seeing corr makes me think correlation between two random variables

[static temple]

random what? never heard of that

[solar shore]

[dusty verge]

Anyone who thinks that's what math is is a moron

And a coward

[dim widget]

ITT Halliday challenges Grothendieck to "fight me you coward"

[dusty verge]

It's noticing a thing is hard to understand and instead of trying harder, giving up

Isn't he dead?

[static temple]

nah

[dusty verge]

I think I'd win

O

[dim widget]

That's presumably why you think you can get away with it

[static temple]

grothendieck lives on our heart

[dusty verge]

Oh yeah I'd totally win

[static temple]

(and mostly on cat maniacs)

[dusty verge]

Imagine being unironically dead

[dim widget]

couldn't be me

[static temple]

I love how there are arrows everywhere

[rustic crown]

cat theory

[dusty verge]

But yeah giving up on understanding things is for wimps

[static temple]

oh my god there is a cat maniac amongus

[dusty verge]

Wanna see my cat?

[rustic crown]

[static temple]

sure :>

[dusty verge]

Sorry someone said not this and made me angry

[lethal dune]

https://tenor.com/view/big-bang-theory-gif-3437399

[static temple]

oh MY GOD THERE ARE CATS

[delicate orchid]

wtf r u pepole talking about

[lethal dune]

cats

[static temple]

the guy seems a little crazy, that's a cat disease

[static temple]

CAT

[lethal dune]

nice

[dusty verge]

Me?

[static temple]

no, in the gif X)

[dusty verge]

Damn I have to try harder

But yeah ignore that grothendiecian pussy shit

He don't know nothing

[static temple]

yeah this antiscience man

yeah, he seemed like a good guy

[dusty verge]

Look, you can generally understand all the abstract stuff, sometimes you just have to try hard

Don't worry about proving things

A proof should be a byproduct of understanding, not the end goal

[static temple]

well, there is something called the Grothendieck effect

abstraction doesn't give results except if it's Grothendieck

it somehow work

[dusty verge]

Proofs are only the goal if your goal is to prove how smart you are by showing you can do the thing that people say is hard to do

Otherwise, they're just a way of making sure the math actually works, and that's more like crossing your ts yeah?

[dim widget]

The math actually working is the whole point of math.

[dusty verge]

No, the point of math is the things it does

Proofs just are making sure you're not being a dumbass

[static temple]

there is no point of doing anything, life is meaningless in itself

[solar shore]

the point of math is to see numbers

🫡

or letters

[dusty verge]

Like the point of groups is that hey, groups are cool

Look you can do this neat thing with them

Oh look it happens to work well with numbers eventually if you do like Galois shit, or go into rings instead or whatever

[static temple]

I see it like a natural structure that appears everywhere

[dim widget]

Hard disagree, the entire virtue of modern mathematics as a science is that it is logically coherent and all obtained knowledge is analytic. That's why we're so much more successful at discovering truth in our subject than other branches of science.

[dusty verge]

Nah it's just cause maths like way easier

[dim widget]

You don't go to math conferences where people ask "so you say you've proved that this is true, but in what sense is it really true?"

[dusty verge]

We're allowed to have a higher standard of truth cause the things we deal with are so much simpler

(I'm also ignoring the fact that the "axioms" of science are based off of observation and not assumption, but bear with me)

[dusty verge]

That just means all the people who go to conferences are wrong

[dusty verge]

Simpler is a better word than easier. Compare a number to a cell, or an atom, and there's a world of difference in the level of complexity / detail. Since math deals with simpler stuff, it's you can get farther with the same amount of work

[static temple]

it's the definition of "simple" that makes me wonder what is trully simple. Like take just GL2(Z) or Sn

in its definition, there seem pretty simple

but when you look it from a representationnal point of view

[delicate orchid]

they're both still simple :pack:

[dusty verge]

I mean, I'm using a loose definition here. But notice that a number doesn't really have an internal structure. There aren't really exceptions to the rules it follows, right?

[static temple]

their decomposition are quite complex, studying them require very complex tools

[dusty verge]

So you can make complex stuff out of simple things

[static temple]

well for Sn if you look the non-complex representations of them, you see that it's really not an easy task to classify them

[dusty verge]

Math is just the study of simple things, but the subjects difficulty comes from the fact that when you study simple things, you get to say a lot

[lethal dune]

rep theory

[dusty verge]

Like to describe a thing in biology in the same level of detail that we can describe things in math would take significantly more work

[rustic crown]

understanding math is also really important right... like even if math was decidable, that doesn't tell you how you should think about it, only that some weird statements are logically consistent
sort of like art, even tho AI can do art, doesn't mean it's useless and people won't care about it

[static temple]

a simple definition doesn't give rise to simple objects

[dusty verge]

Hmm, maybe

[static temple]

and knowing how to manipulate them doesn't give every information contained in them

well thats my point of view of course

[dusty verge]

But a more simple definition will give rise more simple objects, usually

Like a cell is much more complex than a mathematically ideal trianglen

[static temple]

actually that's something I find very counter intuitive in math

[dusty verge]

One has organelles and shit, and the other has three sides

[delicate orchid]

oh, yeah I mean that's true for pretty any group lol

*much

put a much in there somewhere

[dusty verge]

but like, we CAN classify (finite simple) groups

[delicate orchid]

barely

[dusty verge]

The fact that it's even possible says a lot

[lethal dune]

[dusty verge]

Name a thing outside of math that's completely understood, anywhere

[rustic crown]

eevee is cute

[static temple]

yeah X) and Sn is by far one of the easiest

[dusty verge]

But like, everything's super hard

[static temple]

but it still require some work

[dusty verge]

Math's just the easiest one

Imagine trying to understand something like psychology in the same way we understand math

It doesn't even make sense to suggest because of how absurd the idea is

[static temple]

well the precision and the imagination required to understand math is far more complex than biology for eg. You do not need to see the fractal patterns in biology to do an thesis about for eg only the actine filaments in the cytoskelette of a cell

I would say it's different type of reasoning that is required on all fields

[dusty verge]

But that's not something inherent to the subject of math. It's just that we understand math well enough that the subject is far larger than the subject of biology

We already got all the low, medium, high, super high, and even higher hanging fruit

[dim widget]

It's hard to imagine psychology even being studied at the level of rigor applied to, say, economics in the current state of the field...

[static temple]

or applied mat... nothing

[dusty verge]

Yeah exactly

It's cause psychology is really fucking hard to answer questions correctly in

Cause people are kinda complicated

[static temple]

yeah I more or less agree to this statement

[static temple]

there are so many theories for the simplest stuff in psychology

[dim widget]

I think it's just because people in that field constantly practice bad science

[static temple]

its quite complex too

[dusty verge]

I disagree. I think it's because good science is hard to practice in psychology

[static temple]

in psychology you gotta travel fast from a theory to another

and say which one explains better this behavior or stuff

[dim widget]

I don't think it's harder than all of the fields it underperforms in replicability, good statistical practices, etc...

[dusty verge]

The thing you're studying is a. the most complex thing we know to exist, b. knows your studying it, and c. requires a lot of resources to study in any kind of numbers

It's a lot easier to look at a billion atoms than a billion people

[static temple]

like just for the visual perception, you have like the gestalt theory, the theory of the whole, some kind of old theories that is even before the gestalt theory, the evolutionnary point of view

it's quite diverse, but are those theories very rigorous? Can't be because of the limitations of the empirical data

[dusty verge]

My point is that it's harder to make replicable experiments, because there are more variables to control for. It's hard to have good statistical practices because people are expensive

It's almost impossible to do psychology as good as you do physics, because with physics you can build a thing that smashes two atoms together at the speed of light to see what they're made of, and if you did that with people you'd get arrested

[static temple]

it's still good science, but it's often really limited, because of overspecialization, because of funds, because you can't find 10000 persons that have this specific cortical lesion, because of x and y

but psycho evo for eg. has so many criticism because it's very non rigorous

it's grounds are very hard to prove

[dusty verge]

What you're saying is exactly what I mean about it being hard to practice good science

I'm not drawing judgment on psychologist there

[static temple]

sorry i was rambling like an old man X)

But yeah, structures studied in math are far more complex than the structures that are considered in other fields. They do massive simplifications to be then able to do math on them

[dusty verge]

Yeah fosho

[static temple]

for eg just looking at the roots of the trees in biology, they consider so less parameters

they reduce it to highschool math

it's crazy

[dusty verge]

Statistics are for when you can't write detailed enoug hequations and have to look at averages instead

[dim widget]

I don't think this comparison even makes sense because the goal of math and the goal of an empirical science are so radically different. There certainly isn't an interesting comparison to be made.

[static temple]

(I mean by roots: how much soil is accessible by roots in a forest for eg)

[dim widget]

All I wanted to say is that psychology is a bad example because it's sabotaged by objectively bad scientific practices, whether or not it's a complicated field of study.

[dusty verge]

I will point out that you're the one that initially made it : "Hard disagree, the entire virtue of modern mathematics as a science is that it is logically coherent and all obtained knowledge is analytic. That's why we're so much more successful at discovering truth in our subject than other branches of science."

[dim widget]

Yes but that isn't trying to make a comparison between the two on some axis of "simpler vs harder" which is really facile

Just that because one can be studied analytically, the progress is faster, so I think we should stick to our strengths.

[dusty verge]

Yes, but I was providing a reason for why it's coherent

It's not like mathematicians are better than scientists or something

[dim widget]

You can then argue after the fact about why this is true or whatever.

[dim widget]

By no definition of scientist are 99% of mathematicians really scientists...

[static temple]

I think researchers in science in general should do more math, no, really

[dusty verge]

Well sure but that's just cause math is useful

[dim widget]

Probably also true! To the extent to which it is possible.

[static temple]

but except of that, I think they should do more collab

[static temple]

I think there is an overspecialization in science in general

[dusty verge]

My definition of scientist: something with a 99% chance of being a mathematician

[tribal moss]

Perhaps this general debate should move to, like, #math-discussion ? It doesn't look very specific to abstract algebra.

[static temple]

Sorry I was thinking the same thing X)

[dusty verge]

Uh sorry I was answering leff's question of how to remember what premises are needed for a theorem

[static temple]

(I gotta go, but it was nice discussing @dusty verge @dim widget )

[white oxide]

are splitting fields and simple extensions the same thing?

[dim widget]

No

Simple extension: Q(cuberoot(2)), splitting field: Q(cuberoot(2), \zeta_3)

All finite separable extensions are simple, not all are splitting fields.

[white oxide]

what's a separable extension?

[dim widget]

Simple extensions are generated by a single element, splitting fields contain all of the roots of a certain polynomial. All splitting fields of separable polynomials are simple, but the reverse is not true.

[dim widget]

An extension in which all the elements satisfy polynomials with no multiple roots.

[white oxide]

oh so it's different from an algebraic extension?

[white oxide]

ah ok

[dim widget]

Yep! an algebraic extension may or may not be separable, but all finite extensions in characteristic 0 or over finite fields are separable.

[white oxide]

ohh ok got it thank u!

[dim widget]

No worries!

[dusty verge]

what's zeta here?

Oh got it, I was confused as to why it wasn't \pm zeta, but then I realized I'm dumb

[dusty verge]

Think of field extensions as adding elements to the field. A simple extension adds one element, while the splitting field of a polynomial adds the roots of the polynomial (which means in that field, you can factor OR SPLIT the polynomial) . An algebraic extension is such that the added elements are the roots of a polynomial (a better way to see why this matters is to look at a non algebraic extension. Since pi is not the root of any polynomial with rational coefficients, the extension Q(pi) is not algebraic)

[wooden ember]

Ah neat equivalent definition is that an element is separable iff its associated multiplication map is diagonalisable over an algebraic closure

[dusty verge]

What's the associated multiplication map? Oh the matrix associated with the element if you treat the new field as a vector space?

[wooden ember]

K|Q a finite extension, z in K, the map *z from K to K as a Q linear map

[dusty verge]

What's it mean for an element to be separable then?

[wooden ember]

I should’ve specified for finite extensions

[wooden ember]

Minimal polynomial is separable

[dusty verge]

oh right

[dusty verge]

Okay so the matrix you get representing the linear map of multiplying by the thing if you treat the extension as a vector space?

[wooden ember]

No need to talk about matrices but yeah

[dusty verge]

"diagonalisable"

[wooden ember]

That’s defined for any endomorphism

No need to pick a basis

[dusty verge]

Well sure

[wooden ember]

Since the point of being diagonalisable is it tells you a nice basis exists

[dusty verge]

but matrices have diagonals

Wait is diagonalisable different from invertible? huh

[wooden ember]

1 0;0 0

Diagonal non invertible matrix

[dusty verge]

Are all invertible matrices diagonalizable?

[wooden ember]

No

[dusty verge]

I did a classic

[static temple]

nop

[wooden ember]

Also no

Very different things

[dusty verge]

Why would it mean that the element is seperable?

[dim widget]

What is its minimal polynomial?

[dusty verge]

Oh

[static temple]

love working on perfect fields

[dusty verge]

Okay, because rings are not fields and I hate rings

[white oxide]

oh ok

[white oxide]

ah that really helps

thx

[static temple]

but it's often used for doing studying galois groups, you need separability (which you do not need to study if the "base" field is perfect as any extension of a perfect is perfect) and normal

(i've never worked on non-finite galois theory but whatever im a noob)

[wooden ember]

Compute Gal(bar(Q)/Q)

Fun exercise

[static temple]

oof

the big problem

[slim kayak]

Can someone tell me how the irreducible polynomials in the rational canonical form are related to the characteristic polynomials? It just got introduced as something you obtain when applying the structure theorem on the derived K[X] of your K vector space and endomorphism of choice

[dusty verge]

So just contextually, I'm assuming the reason you asked me this is because the minimal polynomial ends up being the characteristic polynomial of this map (which is the determinant of yada yada), so it's just x - each element of the diagonal or so on. Why is the minimal polynomial that, though?

[wooden ember]

They are the irreducible factors of it if I’m remembering rational canonical form correctly

Is that where your blocks are companion matrices?

[slim kayak]

Yeah

[wooden ember]

Yeah so it should be that

[slim kayak]

That... sounds like an important detail to mention.

At least if you make a graded exercise on it

[dusty verge]

I mean it's not the professors problem if you get it wrong

[wooden ember]

Anyhow you should show it

Cause I’m known to say wrong things on this channel

[dusty verge]

What's the galois group? I don't actually recall it, and it feels important

[wooden ember]

Lol that was a troll

[dusty verge]

Isn't it like the permutation group of the roots of the minimal polynomial or some shit?

Yeah I figured

[slim kayak]

It's the group of all automorphism fixing the base field.

[wooden ember]

It’s the absolute galois group of Q which is a big topic of research

[static temple]

it's the automorphism group of the galois extension

[wooden ember]

Oh you were asking what a galois group is mb

[dusty verge]

Oh and for finite extensions its the permutation group of the roots of the minimal polynomial yeah?

[slim kayak]

Not quite sure about that one, often time it is.

[dusty verge]

Like gal(C/R) is Z/2Z

[static temple]

yé, which can also be seen as k-morphisms

[formal ermine]

if you ACTUALLY want a fun exercise

find gal(Fp bar/Fp)

[static temple]

yep, err no Z2

[dusty verge]

shh I never changed it

[dim widget]

The rcf is giving the decomposition V $\cong \oplus_i K[X]/f_i$, where usually the $f_i$ is the minimal polynomial on the corresponding subspace and $f_i|f_{i+1}$

[static temple]

for finite galois theory, the card is equal to the dim

[cloud walrus]

Topos_Theory_E-Girl

[dusty verge]

it's also true for infinite galois if you don't know what a card or dim is

[dim widget]

So the char poly is just the product.

[static temple]

I never studied infinite galois theory so I do not know

[slim kayak]

Hm... we only had the existence of such f_i infered by the structure theorem, no constructive info to go off on but maybe I haven't thought about it hard enough

[dusty verge]

half the things I say are completely meaningless, don't worry about it

[dim widget]

I mean yeah, it's a hard thing to compute. It's finding the decomposition of the vector space into cyclic subspaces for the action of the endomorphism, which is typically hard unless you know that it is diagonalizable or something.

But I'm answering your question.

[slim kayak]

Yeah true, thanks

[dim widget]

Pain

[pastel cliff]

today i am not become algebra and i am not become topology, but instead a secret and more mysterious third thing

[south patrol]

Statistics

[dusty verge]

algebraic topology?

topologic algebra?

[pastel cliff]

computer science

[slim kayak]

Combinatorics

c'mon i meant graphs and simplical sets and the other cooler stuff with combinatorics : (

[pastel cliff]

ive never put cool stuff and combinatorics in the same sentence

but ive not studied it at length yet so idk

[delicate bloom]

u just did 😎

[pastel cliff]

[formal ermine]

hey sebb

[slim kayak]

bro(sette)[non-broary] hasn't seen using the fundamental group to show the existence of the free group over n elements

[formal ermine]

b

[dusty verge]

To be fair, he had never when he wrote ive never

[formal ermine]

non-broary

I'm dying inside

[pastel cliff]

illum

[south patrol]

Why would you do it this way wtf

What lmao

[dusty verge]

what's the fundamental group?

[slim kayak]

loops and more loops

[formal ermine]

read the first 5 pages of hatcher chap 1

[dusty verge]

books are expensive

and I'm on here instead of working

[formal ermine]

there are pdfs online

he made one publicly available on his uni profile site

[slim kayak]

You take all topological paths with a certain starting point and consider the equivalence relationship of fixed-endpoint homotopy, it forms a monoid. You then consider the paths with starting pont = endpoint and see that its equivalence classes and obtain the fundamental group.

or if you are intellectual

[slim kayak]

functor

[pastel cliff]

functor???? i HARDLY KNOW HER

currently waiting on my prize for funniest person to ever exist

[dusty verge]

okay so its how holy the shape is

[slim kayak]

(Hatcher 16:4)

[dusty verge]

What's the group operation here?

[slim kayak]

Uh, dont have the lingo rn

[pastel cliff]

sometimes i look at math and it just looks like some shit the ancient egyptians could've put in some dude's tomb

[dusty verge]

That's just your handwriting

[pastel cliff]

copying notes

but still

[dusty verge]

That's just their handwriting

[slim kayak]

The paths are cts functions from [0,1] to your topological space. Since all loops have the same starting point you can just combine two loops y1 and y2 as the piecwise function y1(2t) from 0 to 1/2 and then y2(2t-1) from 1/2 to 1

[dusty verge]

Wait why do they start at the same point?

That seems non immediate

[summer path]

Because that's how they are defined

[slim kayak]

They are loops, same end and starting point

[dusty verge]

you loopdeloop

[slim kayak]

The fundamental group of a pointed topological space (X,x) is a group pi_{1}(X,x)

x just being the start and endpoint of your loops, usually your choice of x doesnt change the group so its often written without it

[dusty verge]

No what seems non-immediate is that it looks like the fundamental group is attached to the space, and not a point on the space. Which seems to imply that either I'm misunderstanding that, or you somehow end up with the same group no matter what

[dim widget]

#Algechill moment

[summer path]

Also why are we literally just going through basic definitions of alg top in here

[elder wave]

I would recommend unpacking definitions by yourself or moving to #point-set-topology

[dusty verge]

Nah I got it

[dim widget]

Halliday asks questions

[slim kayak]

finally now we can enjoy our algebraic objects free of applications

[slate tide]

the good of applications of algebra are just more algebra

[delicate orchid]

An entire intro alg top course in MY #groups-rings-fields…. Cringe

[summer path]

i feel like abstract algebra is by far the most chill-like advanced channel lol

[dim widget]

Not anymore! From now on we'll create 1000 threads and it will be pure like the first snow of winter.

Only galois theory homework in here from now on.

Thank god.

[summer path]

galois theory

[dusty verge]

What's a lie group?

[dim widget]

Hi @dusty verge, I think this might be better suited to #1100505349427036191, but that's just a guess!

[formal ermine]

@dusty verge how do you not know some definitions but are then able to prove some hard problems that use those definitions

[dusty verge]

Uh wikipedia

[formal ermine]

huh

[dusty verge]

Idk I look up the thing and think about it really hard until it makes sense

[formal ermine]

bro's the reincarnation of groethendieck

[dusty verge]

"reincarnation" yeah sure I died

[formal ermine]

I die inside every time I realize the question I just asked is stupid

especially during a lecture

[dusty verge]

Why?

If you don't ask the stupid question the stupid stays within

[slate tide]

That's poetry

[slim kayak]

galois theory was invented by mathematicians to create more homework problems

[formal ermine]

because I realize what the answer is right after I ask is

[dusty verge]

But that's a good thing, it means you get the answer

[formal ermine]

like this one time at the end I asked "what explains this one phenaminenan" and it was literally what we discussed during the lecture, I just wrote something wrong down

[slim kayak]

phenaminenan

[dusty verge]

That's great, means you wrote the right thing after

[slim kayak]

Are you allowed to mix that with ibuprofen?

[formal ermine]

I can't spell

😭

phenomenon

[dusty verge]

Being adverse to looking wrong doesn't make sense to me, because everyone is wrong sometimes

[dim widget]

🇩🇪 moment

[dusty verge]

Depends on what country you're from

[formal ermine]

good night chillgebra

I hate having to get up early

[valid night]

Let $G$ be a finite cyclic group of order $k$. Prove that there exist exactly $k$ distinct group homomorphisms from $G$ to $\mathbb{C}^{\times}$

[cloud walrus]

Kroros

[valid night]

I have no idea where to even start with this