groups-rings-fields

[iron vessel]

Let K be a field with char(K) $\neq$ 2; define $K_1 = K(\sqrt{K})$. prove that $x \in \overline{K}$ is in $K_1/K$ if and only if $K \subset K(x)$ is a finite Galois extension for which $G = Gal(K(x)/K)$ is abelian of exponent 2.

[cloud walrus]

Évariste Galois

[iron vessel]

I was thinking for the for the first inclusion I could create a chain of extensions; i.e., $ K \subset K(\sqrt{\alpha_1})\subset K(\sqrt{\alpha_2}) \subset \dots \subset K(\sqrt{\alpha_n}) = K_1$ with $x = \sqrt{\alpha_i}$ for some i, which then means that $[K_1:K] = 2^n$. But to be honest im not sure if this is correct.

[cloud walrus]

Évariste Galois

[iron vessel]

As x is in $K_1/K$, then $x \in K_1$ but x is not in K; so $x = \sqrt{\alpha}$ for some $\alpha \in K$. Then I would use my argument above. Also, the finite Galois part is alright.

[cloud walrus]

Évariste Galois

[iron vessel]

Any hints are welcome 🙂

[iron vessel]

Actually, the chain is wrong, the following is what i meant:
$K \subset K(\sqrt{\alpha_1}) \subset \dots \subset K(\sqrt{\alpha_1})(\sqrt{\alpha_2})\cdots(\sqrt{\alpha_n}) \subset \dots \subset K_1$

[cloud walrus]

Évariste Galois

[dull root]

Is there an easy way to see that $\sqrt{2 - \sqrt[3]{2}} \notin Q(\sqrt[3]{2})$

[cloud walrus]

MasakaBakana

[dull root]

It is easy to see that it has degree at most 2, but I need to see why it can't have degree 1

[iron vessel]

Consider its minimum polynomial, which is precisely $x^6 -6x^4 + 12x^2 -10$. So $[\mathbb{Q}(\sqrt{2+\sqrt[3]{2}}: \mathbb{Q}] = 6$. Thus, if it WAS contained, then we would have $\mathbb{Q}(\sqrt{2+\sqrt[3]{2}}) \subset \mathbb{Q}(\sqrt[3]{2})$, which is impossible by their respective degrees.

[cloud walrus]

Évariste Galois

[dull root]

So the degree 6 of that element is actually what I want to find... However, I don't see how you were able to calculate the minimum poly so easily

[iron vessel]

Yeah i used a calculator lol, im sure there is a better way

[dull root]

Ultiamtely, I want to see that $[\mathbb{Q}(\sqrt{2 - \sqrt[3]{2}}: \mathbb{Q}] = 6$.

It is easy to see that $Q(\sqrt[3]{2}):Q$is deg 3, so I needed that $Q(\sqrt{2 - \sqrt[3]{2}}) : Q(\sqrt[3]{2}) > 1$

[cloud walrus]

MasakaBakana

[dull root]

if I can show it is > 1, I know it must be deg 2 and I have the deg 6 I want

[iron vessel]

Im not sure you can argue it that way

[next obsidian]

You can

[iron vessel]

Ok, thanks

[next obsidian]

It satisfies a degree 2 polynomial over Q(cbrt(2)) so it has degree at most 2

[cloud walrus]

MasakaBakana

[iron vessel]

Yeah but you still dont have that $\mathbb{Q}(\sqrt{2 - \sqrt[3]{2}})$ is an extension of $\mathbb{Q}(\sqrt[3]{2})$ right?

[next obsidian]

Yeah you do

[cloud walrus]

Évariste Galois

[dull root]

we can take the square

[next obsidian]

Take the square, subtract 2, then multiply by -1

And you have cbrt(2)

So it’s an extension

[iron vessel]

Fair enough

Oh nice, and then you just have the minimum polynomial of the nested root over cbrt(2) and so by tower law you get the 6. Right?

[next obsidian]

Idk I’d just try to write sqrt(2 - cbrt(2)) as like

a + bcbrt(2) + ccbrt(2)

The latter is what a generic member of Q(cbrt(2)) looks like

Try to find some contradiction

Maybe square both of them, then start equating stuff and you run into problems)

¯_(ツ)_/¯

[iron vessel]

Hey @next obsidian could you help me with this by any chance?

[next obsidian]

Wait

Hmm

Ah nvm the extension probably isn’t Galois

[next obsidian]

I have no idea

I am not good at field theory

[iron vessel]

Ahh thats too bad

[next obsidian]

Ultimately would the goal not be to show x^2 is in K?

If I understand what K(sqrt(K)) is

But that can’t be right

[iron vessel]

Yeah, the definition of K(sqrt(K)) is just weird.

[next obsidian]

Is it just taking square roots over and over?

If so, what you proposed works I think

[next obsidian]

Every degree 2 extension in not char 2 is generated by a square root

Using the structure of an abelian 2-group you can get subgroups of order 2^n, 2^n-1,…,1

[iron vessel]

But then with my argument I get that $|Gal(K(x)/K)| = 2 not 2^n

[next obsidian]

this gives you a chain of extensions of fields of degree 2

Yeah idk

[iron vessel]

Yeah, but the thing is that if, say, $x = \sqrt{ \alpha_{j-1}}$, then $|Gal(E_j/K)| = 2^j$ not $|Gal(K(x)/K)| = 2^k$. Because $x^2 \in K$, so $|Gal(K(x)/K)| = 2$

[cloud walrus]

Évariste Galois

[iron vessel]

Also, just to make sure I am not messing up, if $x \in K_1/K$, then $x \in K_1$ and x is not in K, right?

[cloud walrus]

Évariste Galois

[iron vessel]

What is the definition of a group being of "exponent 2"? As I understand it, it is just the smallest number a for which g^a = 1 for all g in G.

[untold basin]

Can someone give me a small hint for this ? :
Let A be an integrable commutative ring. Show that A contains a sub-ring which isomorphic to Z or Fp for a prime p

[proud bear]

there is a homomorphism Z->A where 1->1. consider the possible kernels

[chilly ocean]

i helped you with something similar a while ago in the group case

take the idea from there and do the same thing here

[untold basin]

Ok I understood

It's the exact same thing than in groups in reality

I've made note with "groups" and "rings" I should mix them up

[untold basin]

Ok I just have to figure out why the order of an element in a domain is always prime

[chilly ocean]

good one to think about

[wild solar]

I am gravely confused by the reasoning of why f(x) cannot be factored as a product of polynomials of deg 2. Why can one conclude it like this?( I mean the polynomial of deg 2 could be of the form x^2+ax+b, so why can one just plug it in and conclude like this?

[next obsidian]

It's because the proof is wrong

As far as I can tell it wants you to plug in t = x^2 to get t^2 - 10t + 1 and then notice that this has no rational roots, as t would need to be +-1 y the rational root theorem

but this doesn't suffice

Take for example
(x^2 + x + 1)(x^2 - x + 1)

If I computed this correctly, this turns into x^4 + x^2 + 1

plug in t = x^2 to get t^2 + t + 1 which has no rational roots

but clearly x^4 + x^2 + 1 factors as the product of two quadratics because I literally started off by writing it that way

[wild solar]

okay, thanks for the explanation....so I just assumed it could be factored and arrived at a contradiction. Does this look legit to you?

[next obsidian]

I didn't fully look over it, but this sort of thing seems right

[wild solar]

cool thanks 🙂

[next obsidian]

I'm not sure if you can actually WLOG b,d = 1, I think you might technically have to handle them both being -1

but this is definitely what you should be doing, this sort of thing is how I've done this many times in the past

[wild solar]

ohhhh okay, not a big problem

[raw flame]

Send the question here

[hidden haven]

You mean [ℚ(π³) : ℚ] = 3?

What is π here?

If it's the usual circumference/diameter π then this is not true

If π is transcendental, degree would be infinite

Because all powers of π are linearly independent over ℚ

That is also transcendental

[chilly ocean]

Those should be algebraic. At least of rational multiples of pi

https://math.stackexchange.com/a/1933276/476484

[south patrol]

Weird q but for this sort of thing, does exhibiting a factorisation of x^4 - 10x^2 + 1 over R work (using the fact R[x], Q[X] etc are UFDs)?

Like x^4 - 10x^2 + 1 = (x^2 - 5)^2 - 24 = (x^2 - 5 + sqrt(24))(x^2 - 5 - sqrt(24))

etc

[chilly ocean]

I think you can prove [Q(cos(p*pi/q):Q] <= q pretty easily, but stating a general theorem probably would require us to study Chebyshev polynomials like in the answer (or explicit calculations)

tldr I'm not sure

[strong yacht]

What was your thought process for this?

[hidden haven]

Nope definitely not.
cos(π/3) is 1/2, so the degree is 1.
cos(π/6) is √3/2, so the degree for that is 2

(and this pattern surely doesn't continue)

[chilly ocean]

What's the number then

Does anyone know

[hidden haven]

I hope I'm not getting ligmad but what number

[chilly ocean]

f(p, q) = [Q(cos(p*pi/q)):Q]

[hidden haven]

That's a function though

[chilly ocean]

Yeah.

[hidden haven]

[chilly ocean]

how familiar are you with computing Galois groups of binomials?

is that a standalone exercise or is it part of something bigger?

[hidden haven]

@steel crane To compute the degree of an extension of the sort Q(b):Q, you have to find the dimension of Q(b) as a Q vector space. So there are a few things you get immediately from ring theory if b is algebraic over Q:

• There is a map f: Q[x] → Q(b), where x is a variable, defined by mapping x to b.
• ker f is a principal ideal.
• ker f is a prime ideal.
• Since Q[x] is a principal ideal domain, prime ideals are maximal, so Q[x]/ker f is a field.

• By the first isomorphism theorem, Q(b) then contains im f which is a field containing Q and b so contains Q(b), so Q(b) = im f so f is surjective.
• So you need to compute the dimension of Q[x]/ker f as a vector space over Q.
• ker f is generated by a unique monic polynomial, which is called the minimal polynomial of b over Q.
• 1, [x], [x^2], ... , [x^(deg(min poly of b) - 1)] is a basis of Q[x]/ker f.

This gives you the degree of the extension

Now try computing the degrees you were asking for. You just need to find the minimal polynomials of all the things mentioned. The minimal polynomial is, immediately from the definition, the least degree monic polynomial that the element b satisfies

[long obsidian]

Does anybody have a motivating example of why one would care about an algebra over a field?

[lethal dune]

you wouldn't have online payment systems without it

or discord

[raw flame]

How do you define Algebra over a field?

[chilly ocean]

polynomial rings

[long obsidian]

Maneta, I would basically say it's a vector space with a bilinear product.

Ryu, I'm not sure if you're replying to me but that's actually why I'm curious about (associative) algebras. I've seen them pop up in various type of "nonlinear algebra" topics. For example, I skimmed something about an algebraic version of signal processing. I've seen them other places too.

Is there like a nice structure theorem about them? I have a very small amount of representation theory that I know but I'm under the impression that the representation theory of algebras is studied and that's another thing I'm curious about

[chilly ocean]

all finitely generated algebras are quotients of polynomial rings, that's the first thing i can think of for "structure theorem". it's also not particularly deep, so maybe not what you're looking for

[raw flame]

Hmm, I see. I think Lie Algebras (And Lie Groups) have to do something with it and that's related to Physics if I'm not wrong

[chilly ocean]

lie algebras and algebras are different things

[raw flame]

Hmm, I remember that Lie Algebras are algebras too, I think the bracket is billinear too

[chilly ocean]

the bracket is antisymmetric, so it couldn't be the product for an algebra.

that said, there is an intersection between the two in what's called a poisson algebra

algebra equipped with an extra lie bracket structure satisfying a leibnitz rule

[raw flame]

This is where I remember what I said 🤔

That's Introduction to Lie Algebras of Karin Erdmann and Mark Wildon

Maybe is the definition of "Bilinear map"

[chilly ocean]

i see where i was wrong

i was thinking of algebras as being associative to begin with

you are right

[raw flame]

And Symmetric Billinear map, right?

The map is certainly not symmetric by definiton

[chilly ocean]

symmetric, not necessarily. that i was wrong about

maybe i spent too much time thinking of algebras as polynomial rings lmao

[hidden haven]

As an example, the representation theory of algebras is important because the representation theories of groups, quivers, and lie algebras can be stated as special cases of it

The category of representations of a group (resp. quiver, lie algebra) over a field is equivalent to the category of representations of the group algebra (resp. path algebra, universal enveloping algebra) over that field

Associative algebras are just rings which are also vector spaces over the base field at the same time. So this is just extra structure over being one of the 2 things and can be helpful. For example, in topology you can study homology and cohomology, and the construction of cohomology has an extra step, but it is easier to compute just because there is extra structure on the cohomology (cohomology is an algebra, while homology is just a sequence of vector spaces)

[raw flame]

What's a quiver? 👀 I have never heard of that algebraic structure

[hidden haven]

It is not an algebraic structure. It is just a directed graph but you are allowed to have multiple edges between pairs of vertices and you can also have loops at a single vertex

The way associative algebras arise is often as the set of endomorphisms of a vector space V. The prototype for a ring is the ring of endomorphisms of an abelian group A, which has + as pointwise addition, and * is composition (non commutative ring in general). Algebra is sort of a generalization where instead of abelian group, we talk about a vector space, and instead of group endomorphisms, we talk about linear maps

You can replace vector space in the above by module over a ring R, and then you'll get the notion of an algebra over R

So they are also natural things to study, since they arise in a similar way as rings do

@long obsidian

[woeful flint]

can anyone explain to me the intuition behind the differences between chain complexes and cochain complexes

like I know that they 'go' in the opposite direction but what does that mean precisely

a way of visualising it

[hidden haven]

That is it. The only difference is the numbering

[delicate orchid]

just reverse the arrows bro

[hidden haven]

wew did you watch robot

[delicate orchid]

I can't post my chitti gifs in here moldi don't taunt me like this

[raw flame]

Like a Category 👀

[next obsidian]

The difference is basically just convention, you could always write it the other way, but sometimes one is way more natural

Like you might start with an object A, then form a cochain complex by embedding A into an A^1

Then something like A^1/A embedded into A^2

[hidden haven]

ye it's the thing underlying a category, you take a category and forget that there a an arrow composition

[delicate orchid]

finite categories are just straight up quivers

[next obsidian]

So it becomes natural to have the indexing go A^0 -> A^1 -> …

[delicate orchid]

are infinite categories also just quivers moldi

[next obsidian]

Other times you start with A and have A^1 with a map A^1 -> A

[hidden haven]

yes

[next obsidian]

And then it’s natural to have a chain complex

[hidden haven]

In the same way that infinite abelian groups are just sets

[delicate orchid]

but the adjacency matrix isn't finite I can't do funny lin alg to it

[hidden haven]

[raw flame]

With finite category, do you mean n-category?

[delicate orchid]

I have absolutely no idea what that means

but!

no

I don't

I mean a category with a finite number of objects and isomorphisms

[raw flame]

Oh, ok

Lol

[hidden haven]

n-categories also have underlying quivers

[raw flame]

hahha

[delicate orchid]

spooky

[raw flame]

I thought that because when you refer to "Infinity category" you mean this https://ncatlab.org/nlab/show/infinity-category

[delicate orchid]

ah yeah, that's something different

and you're right tbh, it's bad nomenclature

[next obsidian]

Wait bruh when you said infinity category you meant a category with infinitely many objects?

Oh you said infinite category

[delicate orchid]

I said infinite

[lethal dune]

@delicate orchid why do u have chitti in yr dp

[delicate orchid]

we stan

[lethal dune]

[hidden haven]

[long obsidian]

Thank you so much! That's something along the lines of what I was looking for

[long obsidian]

Actually this might be very useful thank you

[long obsidian]

Thank you also! I have to retake my smooth manifolds qual so hopefully I can get some more intuition out of looking at lee algebras lol

[chilly ocean]

what is this theorem called?

it looks a little bit like snake lemma

but the conclusion is completely different

I guess this is like part of the zig-zag lemma

[raw flame]

Python Lemma

Just kidding

[simple mulch]

What does "multiplication in K[G] extends that in G" mean?

is it that $a_s s \in G$ ?

[cloud walrus]

mns

[chilly ocean]

$$\sum_{s\in G} a_ss\cdot \sum_{s\in G} b_ss = \sum_{s\in G} \left(\sum_{s_1s_2 = s} a_{s_1}b_{s_2}\right)s$$

[cloud walrus]

Blitz

[chilly ocean]

@simple mulch

It's what you get if you formally multiply everything

[simple mulch]

Ok I see the extension

its weird to explain with words

wtf

[chilly ocean]

G can be replaced by any monoid

If you take G = {1, x, x^2, ...} then you get polynomials over K

[simple mulch]

I would say that product in our algebra is another element in our algebra ?

since as_1bs_2 is also an element in K

[simple mulch]

nice one, thanks!

[chilly ocean]

Yes. In general we assume a_s and b_s are equal 0 for almost all s

But here G is finite so

We don't have to assume that

[simple mulch]

Roger that, thanks!

[broken stirrup]

what does it mean by saying left action is affine?

i searched google but didn't get many hits for that

[lilac trench]

whats th difference between Aut(K) and Aut(K/F)

arent K and K/F the same field

[chilly ocean]

Aut(K/F) probably means the automorphisms of K which fix F

i couldn't imagine any other meaning for it

[woven delta]

By default automorphisms fix the prime subfield (Q in the case of characteristic 0 fields, Z/p in the case of char p) so you can think of Aut(K) as Aut(K/Q) in the characteristic 0 case.

[hidden haven]

K/F isn't a field, it's a field extension

[small bison]

K/F is a fraction

[chilly ocean]

By definition, a Ring (R,+,×) is a non-empty Set R equipped with two Binary operations '+' and '×' such that it forms an abelian group under '+' operation and a semi group under '×' operation, and '×' is distributive over '+'.

My question is do '+' and '×' specifically define addition and multiplication operations, or could they be any two different binary operations?

[prisma shuttle]

they are two arbitrary binary operations

like in the def of a ring there are certian properties that must hold, such as assocsiativy, ,distribuitivyt, etc

between the + and x operation

but there are + and x operations besides teh usual notions of + and x that satisfy these requirements

u can think of the + and the x as more o f just lables

for me personally i like to think of a ring as just an abelian group with an additional multiplication operation attached to it

and this multiplication operatiokn acts nicely with the addition operation nin the group

note however one warning that usually if a group is considered (without a ring), the bin operation of the group will be denoted by multiplication

but if the group is part of a ring structure then it will be denoted as addition

just note that so that u don't like get confused

but both the + and the x mean arbitrary binary operations

[chilly ocean]

Thank you for the detailed info !

[chilly ocean]

By definition, a cyclic group G is a group where an element a in G can be used to express other elements in G as the integral powers of a.

I don't really get the "integral powers of a" part.

Consider a Cyclic group of rotational symmetries of a triangle. The symmetries are at 0°, 60°, 120°,... which can be simply expressed as {0°, 60°, 120°} since 0° is same as rotating the triangle by 180°, rotating by 60° is same as rotating by 240° and so on.

Now how does the concept of "integral powers" come in this situation here?

[obsidian sleet]

here the group is written additively

so the meaning of integral powers is not to take powers but instead as multiplication by some integer

so 2*60 is 120 and so forth

in additive groups a cyclic group is one in which every element may be written as an integer multiple of a generator

[chilly ocean]

Ohh I see

[obsidian sleet]

Usually a group is written multiplicatively so the meaning of integral powers is indeed taking element to an integer power

[chilly ocean]

Tysm for the clarification 👍

[obsidian sleet]

Np

[hidden haven]

rotating by 0° is the same as rotating by 180°

[chilly ocean]

(Rotation in clockwise direction)

My freehand drawing sucks

[mint seal]

^ that's not right

they're 120, 240, and 360 degree rotations

[obsidian sleet]

oh Truee

yeah why is it written 60 120 180

[chilly ocean]

Oh yeah

Ty for the clarification and sorry about that 😅

[woeful flint]

what is a chain complex with zero differentials?

[hidden haven]

Do you know what a chain complex is?

And what the differentials are?

[woeful flint]

So i have a chain complex of free abelian groups with differentials d_n

I now have a chain complex (Z_n) where Z_n is the kernel of d_n

a question says let (Z_n) be the chain complex consisting of the groups Z_n with zero differentials

but I can't quite understand what zero differentials means

do the maps just stay the same

[hidden haven]

The differentials map everything to 0

[hidden haven]

Yes

And they are 0 on the entirety of Z_n

Because Z_n is the kernel of d_n

[woeful flint]

gotcha

ok for some reason I read exact

instead of complex

[hidden haven]

oof

[woeful flint]

so I was getting wrong kernals

kernels

[hidden haven]

ye

[woeful flint]

nice

cheers mate

[hidden haven]

[woeful flint]

quick follow ub

up

If i define B_n to be the image d_n+1(C_n+1) and consider the chain complex (B_n) then the differentials here are also zero no?

I mean possible differentials

[terse crystal]

Yeah, im(d_(n+1)) is contained in ker(d_n)) restriction on it is 0

[wooden ember]

So we proved the fundamental theore of finitely generated abelian groups in linear algebra using the smith normal form. Is there a smith normal form over PIDs so that this proof can be generalized to the fundamental theorem of finitely generated modules over PIDs?

[delicate orchid]

I believe it generalises directly to general PIDs

[hidden haven]

Yes

SNF works exactly the same way over any PID

[delicate orchid]

Thank you Moldi I was doubting myself

[hidden haven]

Anything for you bro

[chilly ocean]

Given an additive functor on R-modules, do we just introduce a chain complex structure when applied to a chain complex in the obvious way

[wooden ember]

Thanks I wasn’t sure since we found it using euclidian division

[hidden haven]

Ye but you'll notice that all you actually need is the fact that gcd(a,b) is a linear combination of a and b

Euclidean division is a way to explicitly compute the linear combination

But existence is enough for SNF

[hidden haven]

Didn't get the question. Are you asking if an additive functor A → B induces an additive functor Ch(A) → Ch(B)?

[delicate orchid]

The construction for a general PID uses a chain of ideals rather than direct computation of the gcd doesn’t it?

I really need to revise com alg

[hidden haven]

We might have seen different constructions then lol

In the end you get a chain of ideals ye

Of the ideals generated by the diagonal entries in the SNF

[delicate orchid]

Yeah, which for Z-modules naturally corresponds to the method using the gcd

Cool

[hidden haven]

I don't get it

That's just the end result

How can an end result be equivalent to a proof approach

[delicate orchid]

I can’t remember moldi it was 2 years ago

I’ll look into it

[hidden haven]

ok lol

[chilly ocean]

That was a stupid question tbh

I mean mine

[hidden haven]

F

[delicate orchid]

I did not remember the construction being 4 pages long

[cloud walrus]

Kraft Macaroni

[rigid cave]

is this true?

[woeful flint]

I think so

I've seen something similar with G in the other argument

[rigid cave]

iirc Hom is not exact

[rigid cave]

I think this is right

[woeful flint]

Wait so it's correct

[hidden haven]

This is correct

[woeful flint]

It seems to hold because the sequence is split from what I read

[hidden haven]

Yes

Split SESs are characterized by equations

If the first map is i and the second is j

You have
ji = 0
And there exists a left inverse i' of i, a right inverse j' of j, such that
ii' + j'j = identity on B

The maps (i', j') constitute a splitting of the SES

Check that if F is a functor which respects addition of homomorphisms (so that F(f+g) = Ff + Fg) then (Fi', Fj') constitutes a splitting of the sequence you get after application of F

Does that make sense?

[woeful flint]

yeah kinda

[rigid cave]

oh sorry I meant the first statement

[woeful flint]

I swear homological algebra is the justification for abstract on this thread

[hidden haven]

Also the if G is projective part is wrong

[rigid cave]

G injective right?

[hidden haven]

It is true for (not necessarily split) SES if G is injective

[woeful flint]

oh okok

[wooden ember]

Ah that’s fair

[chilly ocean]

I have a cyclic group of Integers mod 4.
H is a subgroup of G here.

How can I start making cosets of H without them overlapping

,rotate

[chilly ocean]

The group and subgroup is under addition btw

[hidden haven]

H isn't a subgroup

1+2 = 3 isn't in H even though 1 and 2 are

[chilly ocean]

Wait so

Wait shouldn't it be 0?

I'm confused

Ohh wait

It's mod 4

Yeah

Is this right

,rotate

[hidden haven]

ye

[chilly ocean]

Ty o7

[chilly ocean]

Previously subgroup H had both left and right cosets fully identical, it is a Normal Subgroup of G.

Now using this I tried creating the quotient group. Is my approach right here?

Q is the quotient group here

[next obsidian]

Yes it is

If it splits you have a map going the other direction, so applying functionality tells you that the map you need to know is surjective has a one-sided inverse

Alternatively you can identity the middle with the direct sum and then it also kinda becomes obvious

[rigid cave]

Oh shit I missed that the ses is supposed to be split

[young fiber]

I don't think so. Because the cosets of H in Z/4Z are ${\overline{0}, \overline{2}}$ and ${\overline{1}, \overline{3}}$.

[cloud walrus]

giannis_money

[hidden haven]

That looks correct to me

[hidden haven]

That's what Pencil wrote too

[young fiber]

Yes but it doesn't match what they wrote for Q

[hidden haven]

It does

They're not using bars for elements of ℤ/4

[young fiber]

this is what they wrote

rather than Q = ${{\overline{0}, \overline{2}}, {\overline{1}, \overline{3}}}$

[cloud walrus]

giannis_money

[north sand]

well they use $\overline{0}={0, 2}$ to denote the coset of 0. that's pretty standard. and they don't use the bar to denote general elements of Z/4Z

[delicate orchid]

{0, 2} looks like the equivalence class of 0 to me, like wise for {1, 3}

I don't see the issue ngl

[cloud walrus]

Denascite

[young fiber]

but the elements of ℤ/4ℤ are themselves cosets. because here we're finding the quotient group of a quotient group

[north sand]

so?

it's just notation

pretty standard to not use the bar for Z/nZ

or generally whenever from context it's obvious what is meant

[young fiber]

hmm.. fair enough i suppose. i guess i'm not used to this sort of notation

[north sand]

I feel like using $\overline{a}$ or $[a]$ is just stuff you use when learning these concepts. later on you just drop that

[cloud walrus]

Denascite

[north sand]

maybe a different example from analysis, do you know L^p spaces?

[young fiber]

no

[north sand]

ok then forget it

[delicate orchid]

tbh I still don't know why we don't just use x*N for cosets of G/N from the start

I haven't used overlines or [] since first year as Denascite says

[young fiber]

ngl i'm fairly new to group theory so this explains it

[north sand]

in a few weeks or so you'll notice that you don't need the overlines or [] anymore to know what you are talking about. only early on it's useful to not get confused

[lilac trench]

how are subrings containing the base field a subfield of the extension

[tribal moss]

They aren't in general, unless they happen to be fields.

(which will automatically be the case if the field extension you're talking about is finite)

[chilly radish]

It suffices that the extension is algebraic

[tribal moss]

Right.

[chilly radish]

But yea in particular this is true for finite extensions

[lilac trench]

i know that if its finite then it is algebraic

[chilly radish]

What is the context of your question

[lilac trench]

For some algebraic field extension of F show that every subring of the extension containing F is a subfield

my approach is to find the inverse of some algebraic element

and showing that this element is an element of the subring

I figured I could use the minimal polynomial in a clever way

[little root]

well, the smallest subring containing some element alpha is F[alpha]. But F[alpha] = F(alpha)

and i guess since your extension is finite you can induct kinda

oh its an algebraic extension nvm

actually this works to find inverses of any algebraic element

I have a question. If 1, s, ..., s^{n - 1} are distinct and s^n = 1, how might I go about showing that

(1 - s)(1 - s^2) ... (1 - s^{n - 1})

is a nontrivial linear combination in 1, s, ..., s^{n - 1}?

assuming no relations among the s^i except for s^n = 1

my current solution is if it were a trivial linear combination then setting zeta = root of unity would be a contradiction. but i was wondering if there was a more straightforward way to show this

[tribal moss]

That's an excellent strategy. Note that if a is your algebraic element then the existence of a minimal polynomial tells you that F[a] has finite dimension as a vector space over n.

[tribal moss]

It this happening in a field? Or integral domain?

And linear combination over what?

[little root]

well, the s's are elements of a galois group, so they are a multiplicative group but as far as i know they dont have any nice properties in regard to addition other than the fact that 1, s, ..., s^{n - 1} are linearly independent

and i guess i mean linear combinations in the integers

actually wait no, the characteristic of the field could be nonzero, so linear combinations over the integers or some Z/pZ I think?

[tribal moss]

So we're in a Galois field extension F/K and your polynomial is computed in the ring of K-linear transformations of F?

[little root]

oh yes

[tribal moss]

Probably K-linear combinations too, then.

[little root]

actually i think I realized this is false in nonzero characteristic. If char F = 2, then
(1 - s)(1 - s^2)(1 - s^3) = (1 - s - s^2 + s^3)(1 - s^3) = 2 - 2s^2 = 0

but i think its at least true in characteristic 0

[wise beacon]

I have a question.
Let G be a group. Show that Inn(G) cannot be nontrivial cyclic.

[chilly ocean]

what is Inn(G) isomorphic to?

[wise beacon]

G

[chilly ocean]

no, that's not the case, take any non-trivial abelian group for a counterexample

[sonic parcel]

how do you show the if part of this holds

G = S_5 and H = A_5

[strong yacht]

The roots of X^n-1 are 1, s, s^2, ..., s^{n-1} so have the factorisation X^n-1 = (X-1)(X-s)(X-s^2)...(X-s^{n-1}). Divide by X-1 and have 1+X+X^2+...+X^{n-1} = (X-s)(X-s^2)...(X-s^{n-1}), sub X=1 and then you're done. Except the claim obviously isn't true for some characteristic and some n as you already said. I am kind of guessing what your question is a bit since it lacked some details

[tribal moss]

I don't think that works in the context Frank gave afterwards. The ring s lives in is not nice enough to be sure we have the only roots of X^n-1.

[strong yacht]

you saying I can't just solve the problem I want instead of what was asked :p

[delicate orchid]

Consider that Inn(G) is isomorphic to G/Z(G) where Z(G) is the centre of G, see if you can come to a contradiction by assuming this quotient is non-trivial and cyclic

[chilly ocean]

Ty

[chilly ocean]

I see thank you for giving me an insight on this.

[lilac trench]

is it ok if we have lines crossing in a subgroup lattice

[next obsidian]

Yeah for sure

[chilly ocean]

After reading this definition (source: Wikipedia), is it necessary that the group has to be additive [(R,+) here], resulting (I,+) to be additive too?

I previously asked a question here and thanks to JustKeepRunning's answer, a Ring can have two arbitrary operations other than addition and multiplication

In that sense, can we define an Ideal with an operation other than the addition operation?

[chilly ocean]

Oops here is the screenshot

[hidden haven]

How do you define addition and multiplication?

In your own words

[chilly ocean]

As two different binary operations:
+: S×S -> S
. : S×S -> S

Acting on a Set S

Basically maps an ordered pair to an unique element in S

[hidden haven]

Right. Then I don't get your question. You can define a ring in which the "addition" operation is denoted £ and the "multiplication" operation is denoted ✓, as the symbols used to denote these operations are irrelevant

[chilly ocean]

Ohh
Basically, I asked does it has to be the usual addition and multiplication operations in math

[hidden haven]

Nope. You can define rings such as
R = {True, False}

• = Exclusive or
  . = And

Boolean operations

And you can check that this forms a ring

[chilly ocean]

Ohh

[hidden haven]

But we don't have any numbers around

[chilly ocean]

I see, thank you for the clarification 👍

[hidden haven]

[lilac trench]

how do you say abelian group in the group presentation

like I have <a,b,c|a^2 = b^2 = c^2 = 1>

am i just supposed to include all the ab = ba

and what not

[chilly ocean]

That would get tiresome fast but that's certainly a way to do it

[lilac trench]

is there a better way

[chilly ocean]

You could always write [a, b] = 1 instead

[lilac trench]

hmm im not familiar with the notation

[chilly ocean]

And not include "=1" part since it's not needed

[chilly ocean]

It's the commutator

[lilac trench]

ahh ok

[chilly ocean]

<a, b, c|a^2, b^2, c^2, [a, b], [a, c], [b, c]>

[charred flume]

my task is to find the distinct cosets for

$K=\mathbb{Q}[x] /\left\langle x^{2}-1\right\rangle$

i use the division algorithm applied for F[x]. and get

$K=\left{a+b x+<x^{2}-1>: a, b \in \mathbb{Q}\right}$

is this what it mean of distinct cosets?

[cloud walrus]

Ji

[lilac trench]

all the polynomials in this quotient ring are of degree 1

the way i wrote this down in my homework assignment is that x^2 -1 = 0

in the ring

and so x^2 = 1

and thats exactly what you have

so yeah

i would agree with your answer

[charred flume]

oh nice, thankss

[chilly ocean]

@charred flume yes, each element of Q[x]/(x^2-1) can be written uniquely as a+bx+(x^2-1) using Euclidean division

Those are all the cosets, with no repetitions

[charred flume]

how do you find the basis of $\mathbb{Q}(\sqrt{3}+\sqrt{5})$ over $\mathbb{Q}(\sqrt{15})$

[cloud walrus]

Ji

[charred flume]

i can solve it if it's over Q

but i dont understand how to do it in like $\mathbb{Q}(\sqrt{15})$

[cloud walrus]

Ji

[charred flume]

is it the same?

[chilly ocean]

[Q(sqrt(3)+sqrt(5)):Q(sqrt(15))] = 2

[charred flume]

meaning the first 2 basis of Q are the only basis?

[chilly ocean]

So basis is just 1, sqrt(3)+sqrt(5)

[charred flume]

over Q(sqr15)

ohhhh

i thought it is something more complex

[cloud walrus]

please request a new nickname

[lilac trench]

How do we find the minimal polynomial for this?

My classmate was telling me of a way using a system of equations, but it doesnt really make use of the theory.

The original approach I had was to use the automorphisms of $Q(2^1/3)$ but this extension isnt galois

imma just spit

Q(2^1/3,p) is Galois though for p cubed root of unity

and 1 + 2^1/3 +4^1/3 is in this extension

[strong yacht]

does the identity (1+x+x^2)(x-1)=x^3-1 help?

[hidden haven]

You can find all the conjugates of this in ℂ, that gives you the polynomial in factored form

[hidden haven]

Yeah kind of what I did

The polynomial comes out to be ||x³ - 3x² - 3x - 1|| idk why I expanded it out by hand but I checked with Wolfram this is correct

[strong yacht]

It's not so bad is it? $1 + \sqrt[3]{2} + \sqrt[3]{4} = 1/(\sqrt[3]{2}-1)$. The candidate for the min poly is immediate by multiplying through by $\sqrt[3]{2}-1$, re-arranging and cubing. The result needs to be checked that it is irreducible (e.g. Eisenstein)

[cloud walrus]

Greenman

[strong yacht]

In fact since it's a cubic, don't even need Eisenstein. Rational root theorem will do

[hidden haven]

That is nicer than what I did lol

[hidden haven]

Don't need either, since we know that this element has degree a factor of 3, and it is not rational

[strong yacht]

It's not immediately obvious to me that 1+cuberoot(2)+cuberoot(4) is not a rational number, but is my brain lagging?

[hidden haven]

If it were, then degree of cbrt2 over Q would be 2 or less

because 1+x+x^2 = p/q would then be an equation it satisfies

[strong yacht]

Ah yes, you're right. Alternatively what you can do is just check x=+/-1 as well for irreducibility if your brain lags like mine

[chilly ocean]

I'm stuck how to go ahead and frame the quotient group.
Should I use equivalence classes and check if one of them is a normal Subgroup of Z and create the quotient group?

[hidden haven]

There is a subgroup mentioned in the problem statement itself

You have to spot it

[chilly ocean]

You quotient by the equivalence class of 0

That's how you obtain equivalence between congruences and normal subgroups

[tribal moss]

In yet other words, the definition of your relation R already has the general shape of "how to make an equivalence relation from a (normal) subgroup".

[chilly ocean]

Ohh I see

Thank you all for the help 👍

[chilly ocean]

Consider an Additive Group of Integers (Z,+) with (H,+) being the subgroup of Z, where H = {...-2,0,2,...}

H is a normal Subgroup since its left and right cosets are fully identical

After finding the Quotient Group Z/H, I found that for z in Z, when z = 1 and z = -1, I get the same set. Will the Quotient Group for this actually be {-1,0,1} or {0,1} or {-1,0} (with bars on top of each element)

[next obsidian]

1 = -1 (mod H = 2Z)

So it’s either of the latter two

Well I mean

Okay it is also the first one, but this is using the convention that you just remove duplicates in a set lol

But usually you want to write down distinct elements, so you should stick with one of the latter two

[chilly ocean]

I see, thank you 👍

[chilly ocean]

well actually it'll be {0+H, 1+H} = {0+H, (-1)+H}

[chilly ocean]

Ohh okay 👍

[chilly ocean]

I have a question about derived functors. What's L_0 and R^0 supposed to be. I'm given a definition which plugging n = 0 should mean that I have some map d_0, but I don't have any map like that

The map d_1 is supposed to map P_1 into P_0

[next obsidian]

It’s just F

[chilly ocean]

and I only have map epsilon which maps P_0 into A after that

[next obsidian]

However you compute it

It should become the cokernel of the map FP_1 -> FP_0

But the fact that you have P_1 -> P_0 -> A -> 0 exact tells you that the FP_1 -> FP_0 -> FA -> 0 exact

This exactness tells you FA is the cokernel of the map

You’re looking at the homology of the chain complex
FP_1 -> FP_0 -> 0-> 0 -> …

And that ends up being the cokernel of that last non-trivial map

[chilly ocean]

yeah that makes sense, thanks

[chilly ocean]

I mean, they don't assume that the functor F is additive or even anything about its exactness, but I guess it works the same way
H_0(TP_A) where P_A is the deleted complex, so I guess d_0 is just the map into 0

[next obsidian]

I don’t get what you mean

If you take homology at FP_1 -> FP_0 -> 0

You just get FP_0/Im FP_1

And the fact that this is FA follows solely from right exactness of F

Oh wait are you trying to do this for F not right exact?

If so then you’ll get garbage

You can try to left-derive an arbitrary functor but it won’t be even close to anything useful

[chilly ocean]

Well, the book didn't tell me anything about exactness yet

[next obsidian]

Then well it’ll still be the cokernel of FP_1 -> FP_0 by construction

But that isn’t going to be anything tractable unless you assume F is right exact

[chilly ocean]

as for additive, I was wondering because if you have deleted complex for example, then isn't it interpreted like P_1 -> P_0 -> 0 -> 0 -> ...
So don't we need F to be additive so that F0 = 0?

[next obsidian]

I mean

I think no?

0 is uniquely determined by maps in and out of it

It being final and initial and all

Well… idk maybe

But just assume it is additive or take the definition to be the homology of
-> FP_1 -> FP_0 -> 0 -> …

Or something

Or really you should probably just assume F is additive since you’re doing this in an abelian category or maybe an additive one

[chilly ocean]

hmm... well this is just to introduce Tor and Ext functors so I guess I can assume everything is additive anyway

[sage wagon]

What's the difference between Aut(X) and Sym(X)?

I've seen group actions being defined both as a homomorphism G -> Aut(X) and G -> Sym(X)

[next obsidian]

Kinda depends on what they’re defined as but if I had to take a guess, Aut(X) = structure preserving automorphisms. So if say X was a group, then group automorphisms. Then Sym(X) = all bijections

But this is a guess, it really just depends on what that author has defined those to be

[sage wagon]

Ah alright, ty!

[cloud walrus]

please request a new nickname
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lilac trench]

this is my lattice, its missing generator symbols but each element is order 2 and the group is abelian

apparently im missing one more subgroup in the second row, but I cant find it

second row from the top

[hot lake]

Look at those elements in the third row that have less lines going up than the others

[lilac trench]

ahhh

wait, whats the connection between the number of lines and what they're connected to

like, did you immediately look for that

[hot lake]

Because on this group every non identity element should be indistinguishable from the others

The group has lots of symmetries

[lilac trench]

ahh

whelp time to retype my entire lattice

[lilac trench]

id say this works

[woven delta]

Yeah that's good

[chilly ocean]

Couldn't we do that with any even equivalence class of R?
Because, [0] = [2] = [4] = .....

[woven delta]

R there means an equivalence relation, not the reals

[frank fiber]

Let $G$ a groups and $H,K$ subgroups, if i know $d(H)$ and $d(K)$, what can i say about $d(H \cap K)$?

[cloud walrus]

Or x1

[chilly ocean]

what does the d mean?

[frank fiber]

minimal number of generators

[long obsidian]

I haven't seen that notion for the number of generators before. Maybe d is for dimension.

But idk if you can say anything about the intersection of the two groups. Its definitely the case that the number of generators is bounded above by d(H) and d(K) lol

[latent anvil]

why is that true?

It seems unintuitive to me

[woven delta]

It's not true

[latent anvil]

i've been trying to find a counterexample hahaha

i was pretty sure it was wrong

ah, found one [incorrect]

[woven delta]

The free groups on 2 generators and 3 generators

I think

[latent anvil]

what do you mean?

[woven delta]

Hmm wait I'm being silly

[latent anvil]

like F(a,b) in F(a,b,c) is not an example

[woven delta]

Yeah I realized lol

Don't you want the opposite?

[latent anvil]

wait

yeah

whoops lol

i was also trying to do something with free groups

bc of the embeddings of large rank free groups in smallrank ones

[woven delta]

Yes

[latent anvil]

i wonder if you can get like, C2 x C2 x C2 as the intersections of two subgroups each generated by 2 elements

wait

isn't this like obvious

no no no

i was wrong

mb

[woven delta]

Lol

[latent anvil]

HMMM

[woven delta]

Oh wait no

My example worked

Lmao

[latent anvil]

oh wait what was it?

[woven delta]

Because it says and not or

[latent anvil]

oh lol yeah

[woven delta]

If you take the free group of rank 3 inside the free group of rank 2

It works

[latent anvil]

I was thinking of the stronger version with max(d(H), d(K))

yeah

[woven delta]

I guess that was what I was thinking originally

Lmao

[latent anvil]

i would like to say on the record I didn't say you were wrong before you called yourself silly :P

[woven delta]

Oh lmao

[latent anvil]

I thought at first you found like exotic copies in F(a,b) or something

[woven delta]

No lol

[latent anvil]

wait so

[woven delta]

The or question is still more interesting I guess

[latent anvil]

we can find an example where d(H cap K) < min(d(H), d(K)) and an example where d(H cap K) > d(H)

can we find an example with d(H cap K) > max(d(H), d(K))?

yeah

[woven delta]

I think taking intersections in free groups are pretty easy

Actually I'm lying

[latent anvil]

hahaha

also a cool thing this made me find out: the intersection of two subgroups of the fundamental group corresponds to the pullback of the associated pointed covering spaces

[latent anvil]

woah!

oh wait I actually did have a problem related to this I think

in my intro group theory course

that the kernel of the map G * H -> G x H is free of rank (|G|-1)(|H|-1)?

I think?

[woven delta]

Oh actually this tells us you can't have an intersection subgroup of rank 3 from intersecting two rank 2's

[latent anvil]

oh yeah!

cool!

[woven delta]

That's wild

Yeah shamrock that actually makes sense, pullbacks are intersections in a category of open sets

[latent anvil]

yeah

that's why

[woven delta]

Right

[latent anvil]

wait, open sets or subgroups?

i mean it is in both lol

[woven delta]

Subgroups lol

In a lattice

[latent anvil]

yee

I was thinking about something else involving pullbacks in lattices today actually but i forget what it was

(yes I know they're meets)

(I was thinking about some specific lattice)

[woven delta]

Right I guess it's just because the thing you're mapping too doesn't matter because of the join, so any pair of arrows is going to factor through the join

So you may as well do the pullback of the join

[latent anvil]

right

[woven delta]

Which is just the same thing as product in this case

Lol

[latent anvil]

i had another thought like this recently, this = "reframing something simple (intersections) in terms of something complex (category theory)"

you know how a scheme map X -> S can be thought as like a family of schemes parameterized by the points of S, namely the fibers?

like if X = Spec k[x,y,t]/(xy - t) and we project out the t-coordinate onto Y = Spec k[t] then the fibers are the different sets xy-c for c in k, and this is like a family of hyperbolas degenerating into two intersecting lines?

[woven delta]

Not really

[latent anvil]

oh

well then instead of what I was going to say, this is just a cool thing to think about

imo

[woven delta]

Okay cool

Oh okay I see

Yeah this makes sense

[latent anvil]

like X = Spec k[x,y,t]/(ty-x^2) over Y = Spec k[t] is a family of parabolas, but as t goes to 0 the two halves get closer and closer together

and then at t = 0 they fold up into a single line

[woven delta]

So the scheme that lies above a point c is Spec k[x,y]/(xy-c)?

[latent anvil]

For the first example, yes

and on the algebra side this is because $k[x,y,t]/(xy-t^2) \otimes_{k[t]} k[t]/(t-c) = k[x,y]/(xy-c)$

[cloud walrus]

Shamrock?

[latent anvil]

this is actually because of how fiber products work. the inclusion of c into the line A^1 = Spec k[t] corresponds to the ring projection k[t] -> k[t]/(t-c), and (using intuition from set theory/topology) the fiber of f over a point p should be the pullback with the inclusion of p

and pullbacks of schemes correspond to pushouts (tensor products) of rings

but yeah just thinking in terms of like setting t = c is also good

like that's the geometric picture

oh sorry I mean xy-t not -t^2

typo sorry :(

so the ty-x^2 thing is also really cool

the parabolas whose two sides degenerate into a line

and this is because at 0 you get x^2 = 0, which isn't just a line but a line with multiplicity two!

the ring k[x,y]/(x^2) sort of "remembers" that it's coming from the degeneration of of two halves of a parabola into one

[woven delta]

How do you define multiplicity?

That is really neat though

[latent anvil]

Yeah so there's not really a single definition of multiplicity

like "double line"

because we can have different ways of doubling it

like $\mathrm{Spec} k[x,y]/(x^2, y)$ versus $\mathrm{Spec} k[x,y]/(x^2, y^2)$ versus $\mathrm{Spec} k[x,y]/(x^2, xy, y^2)$

[cloud walrus]

Shamrock?

[latent anvil]

(note that (x^2, xy, y^2) = (x, y)^2)

wait

that last one is wrong umm

i think i have fucked up

okay first off for the yt-x^2 example we actually Spec k[x,y]/(x^2)

Spec k[x,y]/(x^2, y) would be a double point 🤦‍♂️

well, what I said is still true but it's about doubled points not doubled lines haha

[latent anvil]

the first one is like the origin in A^2 but it remembers a tangent vector in the x direction, the second one is like the origin which remembers a tangent vector in both the x and y direction, and the third is like the origin but it remembers the entire tangent space?

I think that would be right

okay i jumped ahead to something there

i think i'm too stoned to give this explanation justice emma sorry :(

[woven delta]

I'm also stoned

So it's all good

[latent anvil]

it is really cool and useful to think about scheme maps as parameterized families of schemes tho

also true in differential geometry and kind of for fibrations too

in differential geometry you have like smooth families of maps being important for transversality

[woven delta]

I actually am learning about sheaf toposes btw

So this is probably useful to me

[latent anvil]

very possible!

[woven delta]

I'm reading about intuitionistic logic in sheaf toposes

It's kind of neat

[latent anvil]

i really want to learn that stuff

there's this really cool paper I want to read

[woven delta]

I'm just writing out everything and it's kind of working

idk

I should do more of that actually

What paper do you want to read

[latent anvil]

I'm triyng to find it sory

something about proving basic things in ag about sheaves using internal logic

[woven delta]

I'm reading a paper by Steve Simpson and his student Basu about the muchnik topos

Yeah I've seen those sort of things vaguely

Although I don't understand it

The muchnik topos is basically the sheaf topos on the muchnik degrees viewed as upwards closed sets of turing degrees

So basically all of the upwards closed sets of turing degrees form a topology on the Turing degrees

And you just look at the sheaf topos on that

Basu got his PhD basically for coming up with the idea it seems

[latent anvil]

i think it was this https://arxiv.org/abs/2111.03685

cool!

lmfao

[woven delta]

Yeah I feel weird about that

I really need to know that theorem about turning classical theorems to intuitionistic

[latent anvil]

like the thing that says if you can prove P classically you can prove ~~P intuitionistically?

[woven delta]

Idk let's see

I'm looking at the theorem that says that a geometric formula holds at a point implies it holds on a neighborhood

That's really neat

Because that says that the set of points it holds on is open I guess

[latent anvil]

neat!

yeah

[woven delta]

And here's an interpretation

I think

It defines a map to the subobject classifier

And the preimage of false in that should be closed

[latent anvil]

oh

[woven delta]

So the complement should be open

I mean I'm kind of bullshitting rn

[latent anvil]

wait i think I'm confused by this

yeah

[woven delta]

But this seems plausible

[latent anvil]

like is this defining an interpretation?

it seems like the subobject classifier is like the object of truth values

if im using my meme brain

[woven delta]

Okay so the subobject classifier is the open sets I think

Is that true? I saw that in that paper I was reading

Like we are looking at sheaves on X

[latent anvil]

idk I'm also just reading slides rn lmao

oh that seems suspicious

umm what about the constant sheaf with stalks {0,1}?

that would make sense

so a map phi : F -> {0,1}_

should tell us a subsheaf

we could let G(U) = { f in F(U) : phi(f) is the constant 0 function }

is this a subsheaf?

yes!

[woven delta]

One sec

Yeah so we're basically saying the same thing I think

You view the constant sheaf as the set of open sets

The paper I was reading was defining sheaves to be their set of sections

Yeah this is good

[woven delta]

anyway I do think this is pretty reasonable now, the geometric formula could define a local evaluation map

and then the set of 0's are going to be the things which don't satisfy the formula

and then you would want the set of 0's to be a closed set

so you probably put the sierpinski space topology on {0,1}

I think this is actually why the theorem is true unironically

lol

although I should read the proof

[chilly ocean]

It can't be arbitrary, because then R/~ won't be a group or a set of left/right cosets

[wooden ember]

so i dont think we've actually shown in class that finite degree extension => finitely generated by algebraic elements so could someone check my proof of this? Let $K\subset L$ be an algebraic extension of finite degree $[L:K]=n<\infty$. We proceed by induction on $n$. There's nothing to do for $n=1$ so we consider $n>1$. In this case, $\exists \alpha\in L\setminus K$, and we have $[K(\alpha):K]=m | n$, $m> 1$. Hence $[L:K(\alpha)]=\frac{n}{m}<n$ and by induction $\exists \alpha_1,....,\alpha_r$ such that $L=K(\alpha)(\alpha_1,...,\alpha_r)=K(\alpha,\alpha_1,...,\alpha_r)$.

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

Simple enough, just wanna be sure im not making any mistakes

[woven delta]

Looks good

[lilac trench]

Yo

so.... rubiks cube.

its just a group of symmetries right

each turn is an element of order 4

and this somehow connects to all the possible combinations of the rubiks cube

which according to google is 43 252 003 274 489 856 000 combinations

rn im thinking that there are six sides so the group is generated by six elements

so 6^4 elements so far

[wooden ember]

https://en.wikipedia.org/wiki/Rubik's_Cube_group

[lilac trench]

maybe i can get more elements by just rotating the whole cube

[next obsidian]

Way simpler proof, let alpha1 through alphan be a generating set for L as a K-vector space. It is then obvious that L = K(alpha1,…,alphan), and each alpha is algebraic as L is algebraic over K

[wooden ember]

right of course

lol

[next obsidian]

If you want to remove the “finite algebraic” part and only do finite, then just notice that a transcendental element forces the degree to be infinite as if t is transcendental then 1,t,t^2,… are linearly independent

As a linear dependence gives a polynomial t satisfies

[wooden ember]

i mean finite => algebraic

[next obsidian]

Yeah

I wasn’t sure if you had that part yet

[wooden ember]

yeah i had that i just added algebraic cause im used to just saying finite algebraic

[next obsidian]

Yeh

[chilly ocean]

Yes
But my question was if [0] = [2] = [4] = ...

Why we only use [0] equivalence class here.
Ordered-pairs with (2,2n) also gives a number that is even when both are subtracted. Same for (4,2n) and (6,2n) and so on. Is it defined to use only the equivalence class [0] in my previously asked question, or are we even allowed to use [2] or [4] and so on

[hot lake]

yes, [0] = [2] = [4] = ...

[woven delta]

oh I misread the question

🤷‍♀️

[hot lake]

anywhere you write [0] you can write [18] without changing its meaning

if that pleases you

[chilly ocean]

What do you mean by only? You just wrote the same thing in different ways

We use the equivalence class of 0 specifically because 0 is the neutral element

[chilly ocean]

👍
Ty all

[modern hawk]

quick question about latticies, how is a lattice a set? if that is the definition

v is a basis of linearly independent vectors

[uncut girder]

Its a set of vectors

[modern hawk]

right, but doesn't the sum make it a single vector?

or am I miss understanding something

[uncut girder]

You are misunderstanding

[modern hawk]

right

figured as much

[uncut girder]

Its the set of all such sums

[modern hawk]

so a would look something like a = {0,1}

and then v = {{0,1},{1,0}}

[uncut girder]

Just draw a picture

[modern hawk]

yeah I know how it looks on a picture but I'm a little confused about the math behind it

is a infinite essentially?

[uncut girder]

The black points form a lattice

[modern hawk]

how does it become a set of vectors, as in my mind I'm seeing that above example, then it becomes a_0 * v_0 + a_1 * v_1

[uncut girder]

${nb_1 + mb_2 \mid n,m \in \mathbb{Z}}$

[cloud walrus]

nyamin

[uncut girder]

Thats the above pictured example

[modern hawk]

ohh right

so just to double check

when you say in math n in Z it means the set of all n in Z

basically just "iterating" over everything

[uncut girder]

This is called set builder notation

[modern hawk]

right

[uncut girder]

https://en.m.wikipedia.org/wiki/Set-builder_notation#:~:text=In set theory and its,that its members must satisfy.

Your question is not really about lattices, its about how to read set builder notation

[modern hawk]

yea perhaps

but what's a_i indexing into then?

it would make more sense if it was just a

[uncut girder]

No

[modern hawk]

oh

[uncut girder]

Do you understand this example?

[modern hawk]

I think I do

it would create the set of any integer * b1 + any integer * b2

[uncut girder]

Yes

And those integers don't need to be the same

Thats why there's 2 separate letters n and m denoting them

[modern hawk]

yeah

but what difference would it make to the example here if it was a instead of a_i

a is a vector I suppose

[uncut girder]

You could write it like ${\sum_{i=0}^n a_i v_i \mid a \in \mathbb{Z}^n}$ if you wanted

[cloud walrus]

nyamin

[modern hawk]

ohhh I think I get it now

and then it's the set of all integer vectors in Z * v

[uncut girder]

No

$\mathbb{Z}\cdot v_1 + ... + \mathbb{Z}\cdot v_n$

[cloud walrus]

nyamin

[uncut girder]

${v_1, ..., v_n}$ is a set of linearly independent vectors in $\mathbb{R}^n$

[cloud walrus]

nyamin

[modern hawk]

oh I see, thanks <3

I'm pretty sure I get it now

when drawing it was fairly easy to understand, but the set part wasn't very clear at first

[uncut girder]

Learn set builder notation

[modern hawk]

Yeah, I apparently need to do that

[fiery berry]

Let G be a finite group with a representation rho on V over C where dim V >= 2. The corresponding character fulfills chi(g) is real and >= 0. I have to conclude that rho is reducible. My approach so far was to say that reducibility would mean there is an invertible matrix P such that all Matrices rho(g) can be put into upper triangular form by P. However i don't know how to conclude the existence of such a matrix just by knowing that the sum of the eigenvalues of all the matrices is real and greater 0. I feel like i won't come very far with this approach but i also don't really have anotehr idea. Any hints would be appreciated

[stoic rose]

Hint : ||Assume that V is irreducible and derive a contradiction from the irreducibility of characters||

[lethal shoal]

Hello, guys, i have some difficulties with this exercice :
Let E be the set of continuous functions from [-1;1] to R which are affine on [-1;0], [0,1]. I have to proove that E is a 3 dimension R-vector space
but i don't really know how to proceed here and all my theorem looks unnecessary here

(i hope this is the right channel) i don't need answer, just a help pls

[woven delta]

What does it mean to be affine on [-1,0] and on [0,1]?

[chilly ocean]

well, an affine function from R to R is basically a so called "linear" one

[lethal shoal]

expression of the fonction in [-1;0] and [0,1] can be write like that : ax + b and cx + d

[chilly ocean]

You should be able to check if all the properties of a R-vector space hold, so your problem is only with the dimension, yes?

[lethal shoal]

yes with the dimension

it's a new concept that i saw a while ago^^

[chilly ocean]

Let f be an element of E. Doesn't T(f) = (f(-1), f(0), f(1)) give an isomorphism?

those are functions consisting of these three points, connected with lines

no additional information is given in the structure of functions in E

[lethal shoal]

yes

[chilly ocean]

I hope it's not "an answer"

I didn't know how to say it in another way

[lethal shoal]

I'll keep looking, with the supplement you brought, thank you

[lethal shoal]

ok just found^^

but now

I want to exhibit a base of E, how to do that?

[chilly ocean]

If T is an isomorphism, and (1, 0, 0), ... is a basis for R^3, then a basis for E will consist of functions T^-1(1, 0, 0), ...

So piece-wise linear functions f with f(-1) = 1, f(0) = f(1) = 0, etc

Hopefully this drawing makes it clear

[lethal shoal]

Oh i didnt saw your message, thx

Yes this drawing is clear

[hollow fjord]

Is there any theorem that ensures that given an element g in a group G there is a homomorphism f:G -> G such that f(g) = g and f is not the identity ?

[void cosmos]

homomorphism induced by conjugation group action

@hollow fjord

[latent anvil]

This is not true in general, but it's almost always true. If G = Z/nZ is cyclic of order 2 then the only homomorphisms G -> G are the zero map and the identity, and the 0 map doesn't fix the generator. If there is some non-central element h such that hgh^-1 = g then we can take f(x) = hxh^-1 (but if h is central then this will be the identity).

So if g is such that there is no f≠id with f(g) = g then C_G(g) = Z(g)

In particular g is central

[void cosmos]

when does conjugation fail?

[latent anvil]

Take G to be abelian

[void cosmos]

poh

ohh

ye

thats the identity

yup

[latent anvil]

Also, there is a family (for n >= 2) of groups Gn fitting into a short exact sequence 1 -> C2 -> Gn -> Sn -> 1 ("the double cover of Sn") which I am told by the internet has trivial outer automorphism group but a unique central element

Wait

I got flipped up again ugggh

This is a very confusing condition to me

I keep getting it negated in my head

Oh, if G has such an element g then G must be abelian

Because the condition on g forces C_G(g) = Z(G), so g is central and then C_G(g) = G so G is abelian

Now consider the family of maps f_n(x) = x^n. If G has an element h whose order k does not divide m = |g| then f_{m+1}(g) = g but f_{m+1}(h) ≠ h, as if h^(m+1) = h we would get h^m = 1 so k|m. Hence f_{m+1}≠id but f_{m+1} fixes g. This is valid when k or m = infinity with the convention that infinity is divisible by everything and divides only itself

So if g has infinite order this doesn't say a lot, but if we restrict to finite G this says exp(G) = g, ie x^|g| = 1 for all x in G

Oh well we know G is abelian, so if we restrict to the finite case we can just look at the classification

[fallow plume]

this be kinda cursed 💀

[tough raven]

Why is C_G(g) = Z(G)? The right-to-left inclusion is obvious.

[latent anvil]

Suppose h was an element of C_G(g) not in Z(G)

Then the automorphism γ(x) = hxh^-1 fixes G but does not fix every element, ie is not the identity

This contradicts the assumption that any endomorphism of G fixing g is the identity

haha np

I'm writing out a classification rn

[wild solar]

I wanted to ask for some suggestion or hints for this hw problem but on the other hand I'm honestly clueless and just wanna give up >___< ...

never heard of this criterion before but the prof just casually put it in homework

[tough raven]

If g generates G, it's impossible. Otherwise you have a short exact sequence ‹g› -> G -> G/‹g›
Which might not split unfortunately
A solution must restric to the identity on ‹g›; in particular, preserve ‹g› so descend to G/‹g› -> G/‹g›
Is this process reversible?

[tough raven]

What have you tried?

Note that usually irreducible mod (something) => irreducible, because a factorisation that holds absolutely will also hold mod whatever.

[latent anvil]

lol my classification is 203 words over the limit

The only possible element $g$ of a finite group $G$ such that ``any endomorphism of $G$ fixing $g$ is the identity'' is (up to isomorphism) when $G = \prod_{k=1}^r\prod_{j=1}^{m_k} \frac{\Z}{p_k^{a_{k, j}}\Z}$ for distinct primes $p_1,\ldots,p_r$ and $0 \leq a_{k, 1} < a_{k, 2} < \ldots < a_{k, m_k}$ for each $k$ where $g = (g_{1,1},\ldots,g_{1,m_1},\ldots,g_{r,1},\ldots,g_{r,m_r})$ for generators $g_{k,j} \in \frac{\Z}{p_k^{a_{k, j}}\Z}$. Using the formula $\mathrm{End}(G) \cong \prod_{k=1}^r\mathrm{End}\left(\prod_{j=1}^{m_k} \frac{\Z}{p_k^{a_{k, j}} \Z}\right)$ and thinking about the element $(1,\ldots,1)\in \prod_{j=1}^{m_k} \frac{\Z}{p_k^{a_{k, j}} \Z}$ we can see such $g\in G$ are counterexamples. Now consider a general finite group $G$ and $g\in G$ such that any morphism $G\to G$ fixing $g$ is the identity. As I said earlier $G$ is abelian, so either $G = 1$ (which is in the family of counterexamples) or $G = \prod_{k=1}^r\prod_{j=1}^{m_k} \frac{\Z}{p_k^{a_{k, j}}\Z}$ for distinct primes $p_1,\ldots,p_r$ and $0 < a_{k, 1} \leq a_{k, 2} \leq \ldots \leq a_{k, m_k}$ for each $k$ where $g = (g_{1,1},\ldots,g_{1,m_1},\ldots,g_{r,1},\ldots,g_{r,m_r})$ for elements $g_{k,j} \in \frac{\Z}{p_k^{a_{k, j}}\Z}$. If for some $k, j$ we had a nonidentity endomorphism $\alpha : \Z/p_k^{a_{k, j}}\Z \to \Z/p_k^{a_{k, j}}\Z$ fixing $g_{k, j}$ then we could define an endomorphism $\varphi : G \to G$ which applies $\alpha$to the ${k, j}$ coordinate, and this fixes $g$ but is not the identity. So each $g_{k, j}$ has the property in $\Z/p_k^{a_{k, j}}\Z$. This forces $g_{k, j}$ to always be coprime to $p_k^{a_{k, j}}$, since if we could write $g_{k, j} = da, p_k^{a_{k, j}} = db$ for $d\neq 1$ then $(b+1)g_{k, j} = bg_{k, j} + g_{k, j} = g_{k, j}$, and the endomorphism $f(x) = (b+1)x$ is not the identity because $f(1) = b + 1 \neq 1$ as $1\leq b < p_{k, j}^{a_{k, j}}$.

So $g$ is a tuple of generators of the various $\Z/p_k^{a_{k, j}}\Z$ if we had $a_{k, j} = a_{k, j+1}$ for some $k, j$ then we could swap those two factors of $G$ and multiply by appropriate elements to find an automorphism fixing $g$ but not everything else in $G$. So this shows $(G, g)$ is of the form I described.

[cloud walrus]

Shamrock?

Shamrock?

[latent anvil]

So the kicker is that if G is finite and not of the form I described above here, which is a family of very special abelian groups, then every element g has such an automorphism

I'm not sure about the infinite case

I feel like there should be a more intrinsic characterizations of these elements btw

[tough raven]

Maybe that all of the projections are surjective on ‹g›?

[wooden ember]

For the infinite case if g isn’t in the center you could just take conjugation by g

[wild solar]

can anyone give me a hint on how to show eisenstein criterion follows from the schoenemann's criterion?

[chilly ocean]

@wild solar Eisenstein criterion as in, x^n+a_1x^(n-1)+...+a_n is such that p divides a_1, ..., a_n and p^2 does not divide a_n, then this polynomial is irreducible over Z[x] ?

[wild solar]

yes

and schoneman's criterion as this

[chilly ocean]

what if you take a = p and denote the polynomial in Eisenstein criterion by f(x)
Then f(x)-(x-p)^n has all coefficients divisible by p, is of degree less than n, and at x = p it's equal to f(p) where f(p)/p is not = 0 mod p

so f(x) is not irreducible modulo p^2

[wild solar]

I started off by assuming a polynomial satisfying the eisenstein criterion, but then I dunno what to do

[wild solar]

let me think >____<

[chilly ocean]

If f(x) were reducible in Z[x], then we could write f(x) = q(x)k(x) where q(x), k(x) are monic polynomials of non-zero degree. Then q(x) and k(x) have the same degrees when taken mod p^2 as well, so f(x) is reducible mod p^2

contradiction, so f(x) is irreducible in Z[x]

I think like I shouldn't have said all this, but I was solving as I was writing it

[wild solar]

sounds good,I appreciate it

[wild solar]

Hi, I tried proving it till the point I proved f(x) is irreducible mod p^2. But how do I use this to prove f(x) is irreducible over Z and hence over Q by Gauss's lemma?

[chilly ocean]

@wild solar well you can't prove that f(x) is irreducible over Z if you don't assume further something about the coefficients of f(x)

[wild solar]

why ：（

okay then I guess this approach doesn't work

🥺

[chilly ocean]

you can still prove it's irreducible over Q though

[wild solar]

oh wait

[chilly ocean]

because say, something like f(x) = 2x isn't irreducible over Z

the ring coefficients are what's blocking it from being irreducible

[wild solar]

it only asks for proof of eisenstein criterion follows from schonemann's criterion, but I just did that, right?

[chilly ocean]

do you want to prove it's irreducible over Q then?

[wild solar]

by proving that a polynomial satisfying eisenstein's criterion also satisfies the schonomann's criterion

okay I'm kind of confused by my own logic

[chilly ocean]

it only implies irreducibility mod p^2

[wild solar]

oh, right...

[wild solar]

then yes

but I thought irreducibility over Q or Z are the same thing

sorry, I'm kinda lost

[chilly ocean]

no, because in Q every non-zero element is invertible, in Z we don't have that

to study irreducibility in Z[x] you need the notion of so called primitive polynomials

[hot lake]

Gauss lemma here says that irreducibility is the same over q and over z

[wild solar]

I remember my prof said that Gauss lemma shows that irreducibility over Q and Z are somewhat similar, but I don't remember very clearly what he said

[chilly ocean]

it doesn't, something like 2x is not irreducible over Z

[hot lake]

This one has leading coefficient 1

[chilly ocean]

you need it to be primitive, so that the coefficients are coprime

2x can be reduced as 2 times x

[chilly ocean]

well originally I thought amelia wants to prove it for when the leading coefficient is 1, but it's not necessarily 1 here

[hot lake]

It is

F has degree less than n

[wild solar]

I still don't know how to proceed from irreducible over Z mod p^2

[hot lake]

The leading term is just x^n

[chilly ocean]

I meant the starting polynomial

[wild solar]

[hot lake]

If p is reducible in z then it's reducible mod anything

[chilly ocean]

what do you mean by p

[hot lake]

Ah sorry the polynomial