groups-rings-fields

[solar glacier]

That’s what I wrote in mine

[fleet pelican]

Same.

I started learning math at 19.

[rotund aurora]

based

[rapid junco]

Cayley Hamilton

[solar glacier]

I “started” at 18 but didn’t take it serious till I was 27

[tawny cloak]

I'm thinking to try doing all the exercises from a textbook (or certain parts of it) and then say I did it

[fleet pelican]

I went to one of the top universities in the world for math.

[rotund aurora]

that's what people was talking about, so thought it was a fun exercise

[solar glacier]

Where’d you go if u don’t mind my asking

[fleet pelican]

haha dm

[formal ermine]

I started taking/learning math seriously a couple of months ago

[rapid junco]

Are you in University? Or your own free time

[formal ermine]

I'm still in high school

[rapid junco]

Nice, what have you studied so far?

[formal ermine]

some complex analysis, some analytic number theory, group theory, some commutative algebra, field theory, galois theory, linear algebra, intro to real analysis, numerical analysis, point set topo, a bit of alg top, a bit of rep theory, currently doing alg geo

[rotund aurora]

When you go to uni classes, do you skip hs classes? Just curious about the education system in other countries

[formal ermine]

yeah but it's like

I can only take one class per semester at uni

and then I have to ask the teacher of the class I would miss

[rapid junco]

So what are you taking right now

[formal ermine]

if it's ok with them

it's usually not

[rotund aurora]

Do you go to regular math classes at hs?

[formal ermine]

yeah here in germany you don't get to pick what classes you take

everyone has to take math

each and every year

[formal ermine]

rn it's semester break but next sem I'm taking the geometry of newton okounkov bodies

[rotund aurora]

Are they interesting to you? (The normal math classes at hs)

[formal ermine]

I either sleep or read something

my teacher is chill

[rotund aurora]

Or like you would preffer not to go

[formal ermine]

attendance is mandatory

[rotund aurora]

Why wouldnt you take like general basic algebraic geometry course xD

The wikipedia page of that is like 8 lines

[formal ermine]

my uni doesn't offer an ag course next sem

[rotund aurora]

Ah k makes sense

[formal ermine]

these are all courses for next sem

the only other interesting thing would've been diff top

or maybe lie groups

[fleet pelican]

If you're early in your studies, it might be best to try and get some more general/intro topics.

[formal ermine]

like?

[rotund aurora]

I dont exactly agree. I think Illum is fine doing this kinda things

[fleet pelican]

Well, definitely focus on what you enjoy most. But if you have two courses you're considering, I'd more heavily weight the more general one the earlier in your career, and the more specialized the later in your career.

[formal ermine]

it's whatever

I still have like 3 years till I graduate high school

so plenty of time to do a wide variety of things

[rotund aurora]

arent the kurven courses like algebraic curves or something

[formal ermine]

it means curves and surfaces but it's diff geo

[rotund aurora]

you could do

Im doing algebraic curves things, and there is a lot of differential geometry and complex analysis going on

[formal ermine]

idk the semester starts in a week and I already picked that one course lol

[rotund aurora]

i dont really know diff geo nor complex analysis tho, but Im trying to keep up with the definitions and basic results

the course you are doing is fine yeah

[formal ermine]

I definitely wanna do analysis in the future too though

not just algebra

probability theory also looks interesting, but it needs measure theory

[thorn delta]

okay, i don't quite have a full answer. i found something wrong with the idea i had last night, but here's some ideas:

if phi : Q(sqrt(d1)) -> Q(sqrt(d2)) is the field isomorphism, then phi must fix Q, and therefore phi sends roots of a polynomial f to roots of f. In particular, phi(sqrt(d1)) is a root of x^2 - d1, and therefore phi(sqrt(d1)) = \pm sqrt(d1). Write phi(sqrt(d1)) = a1 + a2sqrt(d2) for some a1, a2 in Q. Then we can write sqrt(d1) = a1 + a2 sqrt(d2).

Now from here, I'm not exactly sure what to do. During a failed attempt, i noticed that if sqrt(d1 d2) is an integer and d1, d2 are square-free, then d1 = d2 (think about why). I feel like this might be the "trick" that you are supposed to see at some point.

For example (x - sqrt(d1))(x - sqrt(d2)) = x^2 - (sqrt(d1) +sqrt(d2))x + sqrt(d1 d2). So for this polynomial to be the irreducible polynomial of some element of an extension F/Q, then we essentially need sqrt(d1 d2) to be an integer. However, the best we can say at the moment is something like (x - sqrt(d1))(x - phi(sqrt(d1)) = (x - sqrt(d1))(x - a1 - a2sqrt(d2)) is in Q[x], and I'm not sure if this is strong enough to imply sqrt(d1 d2) must be an integer without imposing a1 = 0 maybe.

someone better at field theory surely knows how this works tho

[livid willow]

Hi, I had a test yesterday on Algebra and there was a question which I was not quite sure of the answer. They asked for Dim (C : R) [Field of complex numbers over the field of real numbers]. I was not entirely sure of the answer but since C can be written as a + bi where a,b belongs to R, the basis is (1,i), so I wrote the dimension as 2. Would the answer be correct with my reasoning?

[fleet pelican]

That is correct.

If we want to be a tiny bit more careful, we can write the basis as <1+0i, 0+ 1i>.

[livid willow]

Ah okay, thank you

[vagrant zinc]

Can anyone tell me the order of study of abstract algebra?

[delicate bloom]

I'd open up the table of contents of an undergrad algebra book like artin or something and look at that

[fleet pelican]

Artin is a great source. He has a bit of everything.

[vagrant zinc]

I am handling Thomas W. Judson's abstract algebra book and I have seen books that handle different order.

I have been reviewing and studying in this order so far.
Groups
subgroups
Cyclic groups and subgroups
Permutation groups
Lateral classes and Lagrange's theorem

and I don't know what comes next, I guess isomorphism.

[thorn delta]

aluffi's order in his new book if u want to be based

[fleet pelican]

That seems a pretty natural progression. I would hope you at least manage to get a good look at group orbit-stabilizer stuff. Later on comes things like rings, those are a almost a whole new subject.

[vagrant zinc]

Group action?

[fleet pelican]

The isomorphism theorems should teach you about normal subgroups and quotient groups. It seems like the book is hitting all the major topics.

[fleet pelican]

A group action is probably going to show up in a different chapter. It's a bit more fancy, you'll get there. It really helps you calculate a lot of group properties fairly quickly.

[vagrant zinc]

In these days I will be asking about permutation groups and side classes, I hope I don't bother asking so much, I don't know much either but I have studied, and I feel good to face the exercises.

[fleet pelican]

I think this is the right channel for that. I think if you put in the effort people would be happy to supply hints as you work through problems.

[vagrant zinc]

group orbit stabilizers, in which topic do you see that?

[fleet pelican]

Group actions.

[solar shore]

heyo, i was wonderin if my proofs for this entire theorem was correct; there are 6 parts and i was wondering if i could get some other eyes to check if they are indeed correct for them. (dont mind 2 and 3, we had proven them in class from our instructor; it was left up to us to do 1, 4, 5, and 6)

[kind temple]

in part 1 (a) g_1 = g_2 implies that bar(g_1) = bar(g_2) because phi is injective. If you show that phi^{-1} is a homomorphism first, then there is an alternative proof for injectivity by showing that its kernel is trivial.

in part 1 (b), to show surjectivity of phi^{-1}, start with an element g in G and produce an element x in bar(G) such that phi^{-1}(x) = g.

when you write, "for some phi(a) in phi(Z(G))" it is confusing, since it is meant to be fixed throughout that section of the proof. try writing, "let phi(a) in phi(Z(G)) be given" or, "fix phi(a) in phi(Z(G))" instead.

the same goes for when you "fix" an x in bar(G) and write x = phi(z).

otherwise the proofs are fine

@solar shore

[solar shore]

okay thank you so much!

i appreciate the feedback a ton

these are supposed to be notes that im going to look back on so i want them as clean as possible

tysm <3

[sonic coral]

could you post theorem 6.1 for me please?

[solar shore]

the proofs for the specific parts?

or the entirety of theorem 6.1

heres what i wrote

[tender wharf]

these look fine

[quartz quiver]

what does it mean here by $G_{i+1} / G_{i+1} \cap (G_i N)$ is an image of $G_{i+1}/G_i$? i assume image of some homomorphism, but which homomorphism?

[cloud walrus]

Darylgolden

[quartz quiver]

<@&286206848099549185>

[untold cloud]

Hi, guys, given a group G, how to show that Inn(G) char Aut(G)?

[coral shale]

Invoke the defns?

huh. 2, 4 are natural, gcd(2, 4) isnt 1.

At this stage you should try figuring these out yourself before asking for hints.

Also.

yes it is...

I see you asking a lot and I get the impression that isnt happening

ic

In any case, it is hard for anyone to assist with a wall of text like this. Its hardly readable

Use @cloud walrus

And try not to ask for full blown proof verification here

Just my opinion

On getting yourself help better. In the end its only a waste of your own time if no one ends up helping

[formal ermine]

I thought it was like a short proof lol

[coral shale]

In any case - have some confidence in what you write

only ask the specific parts you are unsure about

rather than a review on the entire thing

thats the job of a marker

and yes, once youre done , you can compare to other proofs

[elder wave]

wonder where you got that one from

[grand cliff]

Hi, I just learned about Cayley graphs.

I'm curious if there's a name for some sort of "generalized" cayley graph for infinitely generated groups.

Take for instance (R, +) which can be generated by the interval [-1, 1].
We could then define a cayley "graph" G with vertex set R in the same way we do in the finite case.

Obviously the resulting object is not a graph, since there's an uncountably infinite number of edges and vertices, but I'm not sure exactly what to call it.

Some sort of topological object

[coral shale]

do graphs have to be restricted by cardinality...

surely theres a generalized defn that is not

[wooden ember]

I think the only restriction is that graphs be locally finite?

[coral shale]

this fails

[wooden ember]

Yeah for this example it doesn’t work

I mean nothing stops you from defining such a structure

[coral shale]

is it a cw complex? idk

[wooden ember]

Does seems like a very odd construct though

To me it feels like for this structure to make sense you just end up getting R lol

Under the right identifications

But okay technically speaking nothing is stopping you from making a disgusting CW complex with 0-skeleton given by R and 1-cells attached between any two points that represent points a distance <=2 away from each other. Topologically speaking though I don’t see how you will recover any information about R

Exercise: compute the fundamental group 🤮

[green walrus]

exercise left to the author

[cloud walrus]

LeftySam

[molten viper]

Checking my intuition: with a local ring, it has a unique maximal ideal. This would be the set of all non-units in the ring yes?

I'm at a very basic level but I'm seeing a lot of correspondences between local rings and modules, so I'm gonna say it's homomorphic to an R-module

in some way

the ideal itself?

so (R,m), m is an ideal and a module

[cloud walrus]

LeftySam

[elder wave]

? what they said is correct

oh it's a troll apparently

Galstaff what you said is true, it follows directly from the fact that every non unit in a ring is contained in a maximal ideal

[molten viper]

I see

Over time I'm trying to build up foundation for understanding some papers I've been recommended, so I'm trying to find holes in my knowledge as I start learning beyond my ug level

[formal ermine]

pending g+

[fleet pelican]

Local rings are a pretty central topic. Seems like you're going in the right direction.

[elder wave]

Local rings are very cool

what local properties there are, what additional assumptions you need for other properties to be local etc

all plays nicely together

[fleet pelican]

I was taught to think of it as a way to "zoom in" on a prime

[elder wave]

i'm mainly referring to flatness, injectivity and projectivity here

[molten viper]

I'm working on some stuff with modules for my ug thesis, so I'm starting with the basics of different properties of rings and basics of modules rn

which essentially means "hey here's this graduate course textbook, read it and have fun!"

I'm loving it

[fleet pelican]

I found Atiyah Macdonald to be a great treatment of modules.

[elder wave]

what are you reading

[molten viper]

AM is one of the ones I've been recommended lol

I've got a few texts my director is having me look through (or will have me look at)
AM,

[fleet pelican]

It really complemented my learning style. I find it a bit Rudin-esque in style.

[molten viper]

"Commutative Algebra with a View Toward Algebraic Geometry" is another from the springer graduate text series

She's also gonna have me look into a graph theory text

It's very exciting, I think I've finally gotten over some of the imposter syndrome

<@&268886789983436800>

thanks

anywyas

I'll come back when the troll gets remoived

ah excellent, thanks mods

There's something to be said for the 3rd or 4th year undergrad realization that you really know nothing

actually delving into something slightly more specialized has kind of helped me get over the sort of "imposter syndrome"

Alrighty well, I'll get back to it, I have under 11 hours to send a thesis proposal to the college

[unique topaz]

Yo im stuck if y=0 then whats the root of y=3x(squared)-11x-4

[rustic crown]

#prealg-and-algebra maybe?

[celest furnace]

moment

[molten viper]

Maybe!

[molten viper]

I think there's a typo in part v) here, I figure professor means u \nin Q

[chilly ocean]

Yes, otherwise Q(u)=Q

[wooden ember]

gonna soon give a talk about fourier transforms on finite groups but there's a piece of intuition im missing that im sure exists and id love to convey. I can see formulaically why the fourier transform changes convolutions to products, but why does it do so intuitively (it's a very convenient property to get the isomorphism C[G] = sum End(V_i) but I'm struggling to find an intuitive explanation for it)

[delicate bloom]

the way I see all convolutions is basically as just a shortcut trick, like my intuition comes from multiplying polynomials -> convolution gives you a way to get the coefficients directly. I dunno if that's the sort of intuition you want or that's still considered formulaic

but there are probably other, better intuitions than that for it

[wooden ember]

In the context of the talk convolutions are mainly used to compute the probability distribution of an iterated process and ideally I’d like to somehow link the property of the Fourier transform to that

But good point for polynomials

It’s a bit weird though cause I went from functions on a group to C[G] where we see that convolutions become products, and now I go from C[G] to sum EndV_i where I want a different interpretation for how a convolution is turned to a product

Your interpretation seems better for for the first case

[delicate bloom]

lmk if you find something I'd be curious to know. when counting solutions for varieties over finite fields I've used them, and they sorta ended up being like gauss sums, but I never really got any good intuition about that either

[wooden ember]

Will do

[rotund aurora]

I agree with this

[glossy crag]

From Morandi's Field Theory. He uses up almost an entire page just to show that g(x) is in F[x] (and I don't feel like bothering with those details), isn't it enough that min(a) factorises as \prod (x-\sigma_k(a)) in Fbar[x] (where \sigma_k are embeddings with distinct images of a)? This means that g and min(a) have exactly the same roots in Fbar (just with different multiplicities), then scaling things appropriately we see that g=min(a)^{n/m}=char(a) in Fbar[x], which by Viete implies the formulas for norm and trace. Is this not sufficient?

OK, nvm, I think some of the details are not adding up, so it's probably necessary to show g\in F[x].

[molten viper]

So, how would I go about proving the first part of ii? My first guess is writing it as some a + bu + cu^2 + ... + fu^5

but that seems like a mess of algebra

which feels wrong for a homework problem

Actually, I guess the fact it has degree 2 is telling

[rustic crown]

what's 1/z?

[molten viper]

z^6

[rustic crown]

in terms of cosines and sines

[elder wave]

literally all of field theory

[molten viper]

lemme think uhhh, cos(5pi/7) + isin(5pi/7) I think

mmmm no

[elder wave]

[molten viper]

it's not

12pi/7

[next obsidian]

use the complex exponential

[rustic crown]

hewwo

[rustic crown]

and that's -2pi/7 (modulo 2pi)

[molten viper]

yerp

[rustic crown]

so what's z+1/z

[molten viper]

so the previous problem has me look at... exactly that lol

Update: much simpler than the previous problem lol

Thnaks for the help

[chilly ocean]

How should one think about the difference between a solvable vs nilpotent Lie algebra? Like what’s the intuition for the difference of the derived series vs the lower central series

Ik that’s sort of a vague question, hopefully makes sense though

[fleet pelican]

Well. You can think of a nilpotent lie algebra as a wing which decays as you move along it. A solvable one is like a body with a wing attached. The body is the good bit which saves the wing from decaying. That’s my mental image anyhow.

I mean. In particular. Think of raising a solvable Lie algebra to a power. Like look at how solvable doesn’t imply nilpotent. There’s a certain lack of decay

[fleet pelican]

I guess to clarify a bit. I mean the difference between The function acting on the whole space. So like a solvable but not nilpotent will have [L_0, L_1] doesn't go to 0 but [L_1, L_1] does, where L_1=[L_0, L_0] Somehow the L_0 prevents you from losing dimensions

[chilly ocean]

I see, so some of the elements in L_0 that are not in L_1 give you non zero elements when bracketing

But when restricting to L_1 you can’t find any nonzero elements and thus it is zero

Or for instance n, n-1 instead of 1,0

Okay so the subalgebra kills itself but the whole space preserves some of the sub space

I’m seeing

[fleet pelican]

In particular, it is worth knowing that if L is solvable. [L,L] is nilpotent

[chilly ocean]

I will try to prove that..

[fleet pelican]

GL.

I will pray for your soul.

[feral agate]

For (b), I don't really understand how I can show that alpha +2 is a primitive element, i.e. that (alpha +2) is of order 48, without calculating all 48 powers of it.

Probably I have to use the calculations made with all the ks given, I could see for instance why k = 24 is useful, because it must give -1.

But for the other ks I don't really know how to use them

This is what I have so far:\\
\begin{align*}
(\alpha+2)^2 &= \alpha^2 + 4\alpha + 4 \mod (x^2+1) &\equiv 4\alpha+3\
(\alpha+2)^4 &= 2\alpha^2 + 3\alpha + 2 \mod (x^2+1) &\equiv 3\alpha\
(\alpha+2)^8 &= 2\alpha^2 \mod (x^2+1) &\equiv 5\
(\alpha+2)^3 &= 4\alpha^2 + 4\alpha + 6 \mod (x^2+1) &\equiv 4\alpha+2\
(\alpha+2)^6 &= 2\alpha^2 + 2\alpha + 4 \mod (x^2+1) &\equiv 2\alpha + 2\
(\alpha+2)^{12} &= 4\alpha^2 + \alpha + 4 \mod (x^2+1) &\equiv \alpha\
(\alpha+2)^{24} &= \alpha^2 \mod (x^2+1) &\equiv 6\
\end{align*}
Note that the calculations above assumes $mod 7$ in the coefficients.
To prove that $\alpha + 2$ generates $F^*$, we need to show that $(\alpha+2)$ is of order 48 as $\alpha + 2 \neq 0 + (x^2+1)$. Begin by observing that $$(\alpha+2)^{\frac{48}{2}} = (\alpha+2)^{24} = 6 \equiv -1 mod 7.$$

[cloud walrus]

FrankF

[fleet pelican]

You also know that if some lower power equals 1, that power must divide 48, these are all the factors of 48, so you're done

[feral agate]

I see very smart

[feral agate]

For 10(b), how are there zeros for y^2+1? We need to have that y^2 = x^2+x+1 for it to be a zero in the quotient ring, but the determinant of x^2+x+1 is negative so it does not have roots in the reals and it thus cannot be factored.

Am I missing somethign here?

[next obsidian]

3 = 0

You aren’t over the reals

[feral agate]

oops

[feral agate]

I am a bit confused about what I need to do, because in the end I want to find a $y \in Z \backslash 3Z$ such that $0 \equiv y^2 + 1 \mod y^2+y+2$. Thus I am not allowed to say that we need to find y such that $y^2+y+1 = 0$ right? So how else can I simplify the problem?

[cloud walrus]

FrankF

[next obsidian]

No

You want a polynomial f(x) in Z/3Z[x] such that plugging it into y^2 + 1 = 0, but you get to use the relation x^2 + x + 2 = 0

Let me illustrate an example with a different polynomial

So you can see the idea of what’s going on

Give me a moment

So I can cook up a polynomial

Okay, take y^2 - y + 2

Plug in (x+1)

You get
x^2 + 2x + 1 - x - 1 + 2

= x^2 + x + 2

= 0

[feral agate]

Ah I see so y needs to be in Z/3Z[x]

[next obsidian]

Right, but here actually

I didn’t even use that 3 = 0

But if instead of +2 I did say, -1

I would have ended up with -3

Which is actually 0

[feral agate]

Yeah

[next obsidian]

(I mean 2 = -1 in Z/3Z so of course)

But now you can maybe try to find a root

[feral agate]

Yess thanks!

[next obsidian]

[coral steeple]

Quick terminology question: If every group that satisfies xyz property is isomorphic to some other group (say (Z,+)), then can we say that "the group that satisfies xyz property is unique up to isomorphism"?

[next obsidian]

That makes sense. Maybe say “groups satisfying xyz property are unique up to isomorphism”

[coral steeple]

Ah yes that does sound better, thanks!

[feral agate]

Is it always the case that the kernel of a ring homomorphism is a principal ideal?

[barren sierra]

no

I think?

lemme think of an example

[next obsidian]

Every ideal is a kernel

Of the quotient to R/I

[feral agate]

yeah, but I mean whether it can be generated by one element

[next obsidian]

If it were true then every ideal is principal

[barren sierra]

is there a good notion of a ring generated by a set?

I want a ring generated by (1, x) or something lol

[next obsidian]

Holy hell this isn’t hard lmfao

Just take (x,y) in k[x,y]

Or any ring that isn’t a PID

[barren sierra]

yea

[next obsidian]

And yes, you consider the elements of the set as formal variables and take Z[S]

[barren sierra]

ah right

[feral agate]

For (b), I have shown that there exists an isomorphism between K and L. However, the exercise asks for a explicit one? So I interpret it that the definition of the mapping should also be found. Is there a general approach to finding an explicit isomorphism from the first isomorphism theorem? This is what I have so far:\\

\newcommand{\Z}{\ensuremath{\mathbb{Z}} }
Begin by noting that we need to find a $y \in (\Z \backslash 3\Z)[x]$ such that $0 \equiv y^2 + 1 \mod x^2+x+2$. In other words, it holds that $y^2$ is a multiple of $x^2+x+1$. Note that for $x = 1$ it holds that $x^2+x+1 = 0$ and for $x=0,1$ it holds that $x^2+x+1 \neq 0$. To verify, we see that $$(x-1)^2 = x^2+x+1.$$ Hence, we have found the root at $y = x-1$. Do note that this $1$ is a representative of the class $1 + 3\Z$, so all elements in that class is a root for $x^2+x+1$.\\
Consider the ring homomorphism $f: (\Z \backslash 3\Z)[x] \rightarrow L$ mentioned in the exercise. We know that by the first isomorphism theorem, there exists a map $\psi: (\Z \backslash 3\Z)[x]\backslash ker(f) \rightarrow im(f)$ that is an isomorphism. So we need to show that $ker(f) = (x^2 + 1)$ and $im(f) = L$. For polynomial $p(x)$, the map $f$ is defined by $f(p(x)) = p(x-1) \mod x^2+x+1$. We observe that for $p(x) = x^2+1$, it holds that $p(x-1) \mod x^2+x+1 = 0$. Since, in the first part of the exercise we have proven that $x = 0, 2$ are not roots for $x^2+x+1$, we have that the kernel is a principal ideal, hence $ker(f) = (x^2 + 1)$. Furthermore, let $ax + b + (x^2+x+1) \in L$ where $a,b\in \Z \backslash 3\Z$. Consider the representative $ax + b$. We know it is always possible to pick $p(x)$ such that the coefficient of $x$ in $f(p(x))$ is $a$. To get the constant in $f(p(x))$ to be equal to $b$, we just need to tweak the constant in $p(x)$. Therefore, $im(f) = L$.

[cloud walrus]

FrankF

[barren sierra]

For the connecting morphism from the snake lemma

suppose we're in the category of R modules

[cloud walrus]

Spamakin🎷

[barren sierra]

(delta is the connecting morphism)
EDIT: nvm got it it's just an annoying chase

[prime sundial]

is there any straightforward way to find a generator of the multiplicative group of F[x]/(p(x)), p irreducible? if it helps, F = Z/2Z

my p is cubic and i really don’t want to show the order of my proposed generator is greater than 13

would the field necessarily be isomorphic to Z/26Z? can i just find an isomorphism and let the generator be w/e maps to 1?

i have the result that x is of order > 5, so maybe i can just go up to 13 without much issue

ok that was it

an easier question, if E = F[x]/(p(x)), then the roots of p(x) in E are all of F right?

or am i oversimplifying

sorry i meant all of E

[next obsidian]

It’s F_8 no?

If you’re over F_2 and quotient by a cubic

You’re degree 3 so order 8

[prime sundial]

F_3

sorry if i said F_2

[next obsidian]

You typoed >_<

[prime sundial]

yeah my bad

[next obsidian]

Anyway, all I’m gonna say is 🫥

Sorry bruh this seems cursed

Luckily I think like uhh

[prime sundial]

nah it wasn’t that bad actually

[next obsidian]

,w phi(26)

[cloud walrus]

Results provided by WolframAlpha

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

Requested by Chmonkey#9767

[next obsidian]

Almost half of them generate so

[prime sundial]

ord(x) > 5 and the last couple parts had to do with the representative of x^5 in E

[next obsidian]

Maybe just pick a random one and say you should get 12/26% credit because you’re that likely to be right

Ah

[prime sundial]

so it simplified quite a bit

well

it was the representatives of some powers of x^2 + 2

but they worked out to be the same

so you’re really just doing (x^2 + 2)^2 = x^10, which was also a precious part

and then you just show multiplication by (x^2 + 2)^3 is not one

[prime sundial]

this is more theoretical though and i feel like i over simplified

p(x) in E is = 0, so p(alpha) for any alpha in E must be 0 as well right? thus any element of E is a root

[prime sundial]

i’m given that R is an integral domain, F a sub ring that is also a field, and we define the VS over F with R as elements and scalar multiplication to be left multiplication by elements of F. i need to show if dim_F R is finite, then R is a field

i assume i want to pick a basis and abuse it but i need a hint

maybe i put the multiplicative identity in the basis and name drop graham schmidt?

maybe i don’t need or can’t even use graham schmidt

[cloud walrus]

LeftySam

[coral shale]

unravel defn, looks like geometric progression on factors maybe

[cloud walrus]

LeftySam

[coral shale]

.

idk

u can also try converting the RHS into that

looks easier

[cloud walrus]

LeftySam

[coral shale]

huh. expand it using this definition?

[coral shale]

rhs of this

[cloud walrus]

LeftySam

[coral shale]

very confused.

\phi_{dn}(t)

expand this...

and see if it gets u anywhere

sry if it doesnt

[cloud walrus]

LeftySam

[coral shale]

thats what i thought

[coral shale]

then try turn it into this

try for simple m n

maybe theres a trick

idk whats good...

nvm 2 and 3?

???

what

try for small m and n

and see if theres a rearrangement trick

i mean small when i say simple

for example 2 and 3

...

[hot lake]

doesn't he just show that they have the same roots ?

[hot lake]

well he isn't detailing his reasoning much tbh

yes they are monic polynomials with no repeated roots

yes because then they are both the product of the (x- ri) over the same set of roots {r1,r2,...}

yeah or over the field algebraic numbers

uh

that if P is a polynomial over a field F and r is a root of P then P is divisible by (x-r) ?

yeah by induction and using that r is a root of P <=> P is divisible by (x-r)

more generally, K[x] is a unique factorisation domain (even if K isn't algebraically closed)

I'm not completely sure how you would define multiplicity without knowing that K[x] has unique factorisation

yeah because 1 * 1 = 1

the product of monic polynomials is monic

[cloud walrus]

LeftySam

[hot lake]

no

they are the primitive nth roots of unity

the polynomial whose roots are the nth roots of unity is x^n-1

[cloud walrus]

LeftySam

[hot lake]

if t^m is an nth root of unity then t^mn = 1

[cloud walrus]

LeftySam

[hot lake]

yeah so among the mn-th roots of unity, you have to find out among the possible orders they can have, which ones are a root of the LHS

and yes you do that with chinese remainder theorem

that is, which ones have an mth power that are primitive nth root of unity

because the group of mnth roots of unity with multiplication is isomorphic to (Z/mnZ,+)

and you can translate questions about order of particular roots into order of some element in (Z/mnZ,+)

[cloud walrus]

LeftySam

[somber sleet]

What is the difference between inner and outer semiproduct?

[rustic crown]

inner is when you have a big group and you wanna decompose it as a semidirect product of smaller subgroups

[fleet pelican]

Whether they are subgroups of group you're constructing.

[rustic crown]

outer is when you have two weird groups, and you put them together in weird way to get a big group

[fleet pelican]

They basically work the same, it's more about how you arrive at the construction. Inner is a decomposition and outer is a construction.

often ppl will just call it a semidirect product, and not bother specifying

[somber sleet]

Ohh okay, thx guys

[shell brook]

I have the following set up $G, H \supset F$ are all fields. What does it mean to say that $G$ and $H$ are algebraically disjoint as fields over $F$?

[cloud walrus]

jesse

[fleet pelican]

it means that the polynomials which generate the extensions have gcd 1

[shell brook]

Oh okay

this is nice

thanks

[south patrol]

G a field

😭

[shell brook]

not my notation sry

[formal ermine]

H is even worse

[shell brook]

yeah idk why they use this notation

its in that hellish 90s mathscript too

[formal ermine]

bro those aren't fields you're doing sheaf theory

[elder wave]

only cowards delete messages jesse

[shell brook]

i cant be mean in the public chats

[coral shale]

[formal ermine]

what did he say

[south patrol]

mathscr

[elder wave]

some crazy stuff

jesse you can't be doing illum like that

[formal ermine]

I don't quite understand why sigma(K) is a strict subset of K

[prisma ibex]

I mean strictly speaking this isn't true

[formal ermine]

but then how does this work

[rotund aurora]

I think you just have equality?

cuz the sigmas that act on L fixing Q also act on K fixing Q, but K is Galois

[formal ermine]

this is perhaps a stupid question but what does K/Q being galois change

[rotund aurora]

nothing can escape from K

cuz if a in K, all the galois conjugates of a are in K (galois conjugates of a = roots of the minimal polynomial of a)

and sigma sends galois conjugates to galois conjugates

[formal ermine]

what is a galois conjugate

[rotund aurora]

(galois conjugates of a = roots of the minimal polynomial of a)

[formal ermine]

rightt

galois requires being normal

[south patrol]

ye

[formal ermine]

ok yeah that's what I forgot

thanks

[south patrol]

anyway yeah you should get equality just cause like

[formal ermine]

it's an automorphism in K

fixing Q

[south patrol]

σ is an injective Q-linear map

[formal ermine]

yeah

[south patrol]

Actually tbf easier to just think in terms of roots lol

no vector space theory involved

[formal ermine]

doesn't the inverse limit consist of sequences/tuples/whatever you want to call it?

[fleet pelican]

Usually a limit is the unique object satisfying a property. It'll be initial or final in some construction.

[formal ermine]

wikipedia says this

[fleet pelican]

that is correct

[agile burrow]

Yeah, by specifying a map Gal(L/K) -> Gal(M/K) for each M, you get a unique map from Gal(L/K) to the limit. Presumably the isomorphism is induced by the unique map to the limit

[fleet pelican]

note that A is an object which satisfies a universal property i.e. is initial or final in the relevant construction

[formal ermine]

ahh

so it's like

[agile burrow]

If you want to think of it through the explicit construction of the limit, you're just sending an automorphism of L/K to the sequence of automorphisms where you restrict to M for each M/K

[formal ermine]

we send it to subset Gal(A/K) x Gal(B/K) x Gal(C/K) by sigma -> (sigma|A, sigma|B, sigma|C) ?

yeah ok

thanks

not exactly the cartesian product, a subset of it or whatever

but yeah I get what you mean

oke thanks

[agile burrow]

Are you doing infinite Galois theory

[formal ermine]

no

I'm reading this

https://people.math.harvard.edu/~smarks/mod-forms-tutorial/mf-notes/galois-reps.pdf

[agile burrow]

Fun

[formal ermine]

yee

[fleet pelican]

oh word i know sam

[agile burrow]

Word

[fleet pelican]

he's a great guy, one of the best lecturers ive known

i think hes about finishing up his phd now

[formal ermine]

now we know 👀

[fleet pelican]

i was never in school with him lol

i think he did princeton for ug?

[void cosmos]

yo if u dont mind asking what text is this

[formal ermine]

.

lol

[void cosmos]

oh mb

is this for a number theory class?

[formal ermine]

no worries

[formal ermine]

no idea

just reading it for the funnies

found it while searching for something else

[void cosmos]

oh okay

[formal ermine]

and it seemed interesting

[void cosmos]

cool

good luck and enjoy

[hollow dune]

Does anyone know a trick to find the inverse of a 3x3 fast?

or at least to find the matrix of minors fast since thats what takes the most time when finding the inverse.

Thanks!

[formal ermine]

#linear-algebra ?

[hollow dune]

oh mb

[gleaming quarry]

This probably won't fit here, but seems outside the scope of Early University
What is the difference between gradient matrices and gradient strength maps?

[next obsidian]

This has nothing to do with abstract algebra

[formal ermine]

what even is a gradient strength map

[next obsidian]

If anything this sounds like more of an analysis thing

[gleaming quarry]

good question, I don't know

196 Google results for it

[celest furnace]

so few?

[gleaming quarry]

https://www.google.com/search?q="gradient+strength+map"+&client=firefox-b-d&ei=Yj8jZPs9lpqSBceBl8AJ&ved=0ahUKEwj79pSvrf_9AhUWjaQKHcfABZgQ4dUDCA8&uact=5&oq="gradient+strength+map"+&gs_lcp=Cgxnd3Mtd2l6LXNlcnAQAzIICAAQogQQsAMyCAgAEKIEELADMggIABCiBBCwAzIICAAQogQQsANKBAhBGAFQhwZYhwZggghoAnAAeACAAQCIAQCSAQCYAQCgAQHIAQTAAQE&sclient=gws-wiz-serp
jep

got it from

where Hh and Hv are the horizontal h and vertical v gradient matrices, respectively. Gh
and Gv respectively represent the horizontal and vertical gradient strength maps.
-Blind Additive Gaussian White Noise Level Estimation from a Single Image by Employing Chi-Square Distribution

[solar shore]

could someone enlighten me with the last statement?

$|a_iH| = |H|$ for each $i$

[cloud walrus]

blanket

[fleet pelican]

the size of G is the sum of the sizes of the aH

and the size of aH is the same as the size of H

[solar shore]

oh wait...

i get it now woops

thank you!

[fleet pelican]

so that the size of G is the size of H time the number of cosets

no doubt

lagrange is one of my favorite theorems

i once presented it during a speed class, where the goal was to prove a theorem within two minutes

[solar shore]

oh damn thats kinda cool

[pastel cliff]

is there any intuition for minimal and characteristic polynomials? do they only exist for JCF and RCF or is there something more intuitive for them existing on their own

[next obsidian]

Eigenvalues

Square matrix also is an endomorphism

So it tells you that any endomorphism satisfies a relation

You can eg prove Nakayama’s lemma using this

[pastel cliff]

this is with regards to characteristic polynomial right

[south patrol]

No

Minimal polynomial and char poly share the same roots

[pastel cliff]

ok my definitions are as follow

the minimal polynomial of T \in End(V) (where V is a k-vector space) id the monic polynomial of minimal degree such that f(V) = 0

the characteristic polynomial of T is c_T(t) = det(t*id - T)

[south patrol]

Sure

[pastel cliff]

how is it apparent they have the same roots then?

[fleet pelican]

the matrix itself is a root of both

[south patrol]

We mean over the base field

Which is much simpler

[fleet pelican]

not for showing that minimal is factor of char

[south patrol]

I'll write it up lol

[south patrol]

That isn't what we're doing tho right

Anyway so

Suppose $m_{T}(\lambda)=0$ for some $\lambda \in k$. Then $x-\lambda \mid m_T(x)$, so we can write $m_T(x) = (x-\lambda)p(x)$ for some $p(x) \in k[x]$. $p(T) \ne 0$ (think why) so there's $v$ with $w:= p(T)v \ne 0$. But $0 = m_T(T)v = (T-\lambda) p(T)v = (T-\lambda I) w$, so $w$ is an eigenvector and $\chi_T(\lambda) = 0$

[cloud walrus]

potato

[south patrol]

The converse is easier

if $\chi_T(\lambda)=0$ and $v$ a corresponding eigenvector, then $p(\lambda)v = p(T)v$ for any polynomial $p \in k[x]$. In particular, $0 = m_T(T)v = m_T(\lambda)v$.

[cloud walrus]

potato

[south patrol]

:)

[pastel cliff]

with regards to think why, if it was then this wouldnt be a linear transformation right

im lacking the words but it would have a row of 0's

[next obsidian]

No

[pastel cliff]

oh well i guess by defn m_T already has to be of minimal degree

[next obsidian]

Yes

[pastel cliff]

ok i get what potato said before

and this connects to companion matrices since the characteristic polynomial of those guys is a thing

ohhhh okay okay

so using structure theorem we can break a k[t] module into the direct sum of a bunch of things modded by ideals

each of those k[t]/(a(t)) has a matrix with a(t) as its characteristic polynomial

(not sure exactly what that companion matrix represents for a given k[t]/(a(t)) in the decomposition though) - (is it to do with the basis of the component of the decomposition?)

but then using the companion matrix for each module in the decomposition, and putting those all in a block diagonal way in a bigger matrix we get the RCF of the k[t] module V_T (where V is a k vector space)

and this is useful becauseeeeeee

ok im stuck at that because

something something encodes information

[south patrol]

Hm im unsure exactly what ur asking

Like

Imo the way to view RCF and its proof is that
-instead of just looking at decomposing V as a k-vector space, we incorporate T into the structure and get a k[t]-module as you describe
-By RCF, the k[t]-module V is isomorphic to a direct sum of modules k[t]/(a(t))
-Thjnking of k[t]/(a) as just a vector space, we see that a basis is given by 1,t ,... ,t^(n-1) with n = deg a, and that "multiplication by t" has a very nice form. In fact 1 is a cyclic vector.
-This corresponds to a T-stable subspace of V with a basis given by v, Tv ... T^(n-1) v with the same nice matrix
-So overall we have a decomp into cyclic T-stable subspaces, and cyclic spaces r super easy to study

[charred crescent]

can someone help me understand the following factor group?:
G is a group and N, H are both normal in G, with N subset of H.
Group in question: (G/N) / (H/N)

[fleet pelican]

H/N is a subgroup of G/N

can you verify that it is normal in G/N?

if it is, then you can mod by it..

Is that isomorphic to G/H?

[charred crescent]

well we do want to show that it is isomorphic to G/H, but i don't understand the set (G/N) / (H/N)

[fleet pelican]

do you understand G/N?

do you understand H/N?

can you verify that H/N is a subgroup of G/N?

[charred crescent]

yes

and the third one is a "possibly"

[fleet pelican]

can you verify that H/N is normal in G/N?

[fleet pelican]

ok, so go through the requirements of subgroup

closed, inverses, identity

[cloud walrus]

LeftySam

[fleet pelican]

isn't that #elementary-number-theory ?

[cloud walrus]

LeftySam

[fleet pelican]

maybe #advanced-number-theory

because it's about field extension solutions to polynomials

also, when you ask teh question, it helps to explain the terms you're using, it isn't obvious what the notation refers to.

no. i dont know things.

galios theory is number theory though

if by ring you mean feild

and if by field you mean Q

then sure

[charred crescent]

okay i verified H/N is a normal subgroup of G/N

[fleet pelican]

amazing

[charred crescent]

🥲

[fleet pelican]

then you can mod out by the normal subgroup

[charred crescent]

i guess my confusion arrises when i try to think about what is means to mod out by an already modded group if that makes sense

[fleet pelican]

well rthe good thing is....

it's just G/H

so if you can just verify that, you never nee dto thing about it again

[charred crescent]

now are they exactly equal or just equal up to isomorphism

if the latter is the case then im having trouble defining what an element from (G/N) / (H/N) looks like

well in both cases actually😔

[fleet pelican]

there's a difference?

[fleet pelican]

starty with an element of G/H then

[charred crescent]

well by exactly equal i mean set equality

mapped to what exatly

[fleet pelican]

ok, but we have group equivalence, which is enough in group theory

do yoiu know what the element looks like

[charred crescent]

maybe im being dumb or am just not seeing something obvious, but i am under the impression that i need to define a mapping between elements, but the problem is i don't know the form of the elements of (G/N) / (H/N)

[fleet pelican]

it can be tricky

an element of (G/N)/(H/N) looks like the coset gHN

which you may be able to verify is equal to the coset gH

[charred crescent]

but how do you know just by looking at it

[fleet pelican]

you don''t

but remember that 1 is in N

and N is a subset of H

[fleet pelican]

you kind of use the definition tbh

[charred crescent]

cuz G/N = {gN | g in G} and H/N = {hN | h in H}, but when i think about (G/N)/(H/N), i think {ghN | g in G, h in H}, but that isn't quite right

[fleet pelican]

more like g{hN}N

which, you'll notice, is the same as gH

i think we should be precise and explcit

[charred crescent]

okay i think it makes more sense, im going to try to think through it some more

thank you for your help!

[fleet pelican]

so... it's "really" (G/N)/(H/N)={(gN){hN|h in H}| hN in H/N, g in G}

BUT, recall gN={gn|n in N}

I think you can work it out from tehre

[chilly ocean]

Why (1,1), (0,3)??

[fleet pelican]

whats LT

[cloud walrus]

LeftySam

[chilly ocean]

Leading term

[cloud walrus]

LeftySam

[fleet pelican]

I think this usually comes down to choice of definition tbh

[cloud walrus]

LeftySam

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

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

LeftySam

LeftySam

[coral shale]

no. do not ping users

[tender wharf]

no one is under obligation to help you

[solar shore]

how does one generate subgroups of $S_n$?

[cloud walrus]

blanket

[solar shore]

im trying to look for examples to practice calculating orbits and stabilizers, and there's only one given example subgroup of $S_8$ in my textbook

[cloud walrus]

blanket

[tender wharf]

that's gonna be fun

you can probably generate the cyclic ones already

[solar shore]

it sorta is

but im trying to generate the ones generated by a set S

i dont think they're subgroups

correct me if im wrong, but consider $S_5$ and the subgroup $\langle (123), (12)(345) \rangle$

[cloud walrus]

blanket

[solar shore]

or

so..

@tender wharf

ive done it this way so far

i have 18 cycles

do i have to do the calculations with the ones where the powers of (12)(345) go first?

[tender wharf]

uh

wow that looks really painful

why are you doing this

[lusty marlin]

What are you trying to gain by undertaking these huge computational exercises? If you want to understand stuff better, then I suggest you look at a standard book and try out examples.

[tender wharf]

you mean a cat?

shuri has been rude to you? how so?

[tender wharf]

well, would you want to be randomly pinged for help?

pinging people directly for help suggests you think they are your personal helpers

people here who help you are volunteers. if you are struggling you might want to go to office hours with your professor

[rustic crown]

okie whut about this... there's definitely a rule saying not to ping helpers... definitely not for for latex help.

[tender wharf]

also helper pings are more for the help-n channels rather than this lol

rip abstract algebra it has turned into chill again

[rustic crown]

the hint here is that, if (m, n) = 1, then any divisor of mn can be split into a divisor of m and one of n. namely given d | mn, take d1 = gcd(d, m) and d2 = gcd(d, n). then d = d1 * d2

this is some elementary-number theory... like how you prove that the divisor counting function is multiplicative

[coral shale]

I don't see how my advice yesterday on attempting the proof for m=3 n=2 was unhelpful

Far more helpful than spoonfeeding yourself the answer from SE before having completed the question in my view.

[tribal moss]

The server rules indeed don't explicitly say not to ping people out of the blue. That ought to be common politeness and is not something that is specific to this server. No wonder people are dismissive towards you if you treat them like that.
As moderator I will hereby explicitly tell you not to ping people who you have no reason to ping.

[coral shale]

Also I didn't help you specifically because I was pinged but more like it wasn't happening regardless being the first thing I see in the morning in my notifs.

[untold cloud]

Hi, guys, suppose F_q is a field of order q, where q is a prime power, then why if q is odd, 1 is not equal to -1?

[coral shale]

the only case when 1 = -1 is when ur in char 2?

[untold cloud]

Oh i see, thank u!

So if F has odd order, it cannot have char 2?

[coral shale]

hmmm

I can't remember how to justify but like

[untold cloud]

Ohoh i see

Because if F has order p

[coral shale]

your order has to be a prime power or infinite right?

[untold cloud]

Then char F is p

[untold cloud]

Prime power

[coral shale]

either, i mean.

and specifically, a power of the char when it is

[untold cloud]

Yeah i see it now

Thanks!

[fleet pelican]

That’s going to be a subgroup of S_5. Which is reasonable to work with. As opposed to S_8, which you shouldn’t work with explicitly at all. It’s just too damn big at like 40,000 elements.

Note that the order of a subgroup of S_5 will divide 120

[coral shale]

Then ask and I can clarify

I can't remember, but you were proving something for arbitrary m and n. I was suggesting you proved the statement for m = 3, n = 2

If looking up the solution on SE isnt spoonfeeding idk what is

[fleet pelican]

As others have said. It’ll be better to think of this more abstractly. I’d suggest first trying exercises in S_4. With only 24 elements is much more reasonable.

[coral shale]

Because I thought the polynomials would turn into cubics and quadratics which are easier to work with. Using this perhaps u could see a general method

[cloud walrus]

LeftySam

[formal ermine]

if F has char 2 then it is a Z/2Z vector space thus resulting in an order of 2^n for some n > 0

(assuming F is finite)

[fleet pelican]

odd order means finite

[formal ermine]

some people here like to nitpick

[fleet pelican]

ew

[rustic crown]

yea it follows immediately from FTA

yea something like that

or you can prove it by using stuff like bezout

cause writing out prime factorization is a tiny bit messy... can be made simpler by introducing things like multisets but still not that clean to write

if d1 = gcd(d, m) and d2 = gcd(d, n) then you can see (d1, d2) = 1 but since d1 and d2 both divide d, we get d1d2 divides d.
conversely we know d divides mn, so d/d1d2 divides (m/d1)(n/d2). now since d1, d2 were gcds, (d/d1, m/d1) = (d/d2, n/d2) = 1 which means d/d1d2 divides 1 so d = d1d2

oh so this fact a|bc and(a, b) = 1 then a|c

d/d1d2 divides that stuff, but both are coprime to it

[rustic crown]

proof of this is simple bezout ax+by = 1 means a(cx) + (bc)y = c

and clearly a divides the left side now

d = gcd(a, b) means gcd(a/d, b/d) = 1 else you'll get a bigger common divisor than d

okie mb, we were proving d = d1 * d2... so i was trying to show both sides divide each other.... first i prove d1d2 divides d... and "conversely" d divides d1d2

now that i think about it, "conversely" wasn't the best word :p

yee

yee

(d/d1d2) divides (m/d1) * (n/d2)
and (d/d1d2) is coprime to (m/d1), which gives you (d/d1d2) divides (n/d2) * 1, applying it again it tells you (d/d1d2) divides 1

[coral shale]

You are currently taking group theory? Did you take number theory or no

Are you writing all these arguments down on paper and trying them out

(these are all number theory arguments, really)

[fleet pelican]

Yeah. I told him to take it to that channel...

[coral shale]

Well yeah, it's worth brushing up on number theory if u want to get to the bottom of these kinds of arguments

Some of these Q's you're asking are facts from there you generally learn fairly early on

[rustic crown]

yee

ig you can phrase the argument differently

d | mn implies d/d1 divides (m/d1) * n

so d/d1 divides n

which gives d | d1 *n

then doing it again, d/d2 divides (n/d2) * d1

so d/d2 divides d1

so d | d1d2

a | bc and gcd(a, b) = 1 implying a | c

this statement is equiv to FTA

idk what it's called

i just call that FTA as well

if you think about it, this is purely a statement about the multiplicative structure of N or Z

but to prove it, you need the additive structure

namely bezout

you can only do it within like 5 minutes of posting the message

[tough raven]

Let Γ(n) be the set of 2×2 integer matrices with determinant 1 and congruent to the identity matrix modulo n, i.e. the kernel of the “reduction modulo n” homomorphism from SL_2(Z) to SL_2(Z/nZ).
Is it true that the composite of the subgroups Γ(m) and Γ(n) is Γ(gcd(m,n)) for all positive integers m, n?

[rustic crown]

yea that's true

[coral shale]

with regards to the roots of unity and cyclotomic polynomials, are you taught them as visualisations in terms of the unit circle?

[tough raven]

Also hoping for a proof 👉 👈

[rustic crown]

hehe cute

i don't have the proof at the top of my head, but i can give you leads

so you have to prove some weird facts

the weirdest was this reduction map is actually surjective

[tough raven]

That I know (as in, have seen the proof of)

[rustic crown]

once you have that, with some work you can calculate the index Gamma(n)

looks similar to the formula for phi(n)

[tough raven]

Wait the proof is by subgroup indexes?

[rustic crown]

after that first show intersection of Gamma(m) and Gamma(n) would be Gamma(lcm(m,n))

[rustic crown]

yea you show one inclusion and show they have equal indices

at least that's how i remember doing it

[tough raven]

That seems anticlimactic somehow
Why didn't I try that

[rustic crown]

hehe :p

[coral shale]

https://i.imgur.com/a6vtSQy.png

x^6 - 1 those are the roots

[rustic crown]

do you know the same proof i know?

i thought there would be a cute proof

but i could only come up with some weird argument

[coral shale]

So one thing to visualize about cyclotomics are that the 1, 2, 3, 6 cyclotomic roots are some partition of these 6 roots

[rustic crown]

manually tweaking some 2x2 matrix which had det 1 mod n

[tough raven]

Yep, the index as a function of m is multiplicative so index_{gcd} index_{lcm} = index_m index_n

[coral shale]

If you look at the definition of the nth cyclotomic polynomial, hopefully that makes sense?

[tough raven]

That should do it

Thanks!

[rustic crown]

[tough raven]

That seems natural enough to me
Although yes, not cute

[rustic crown]

oh like i meant, the way i did it wouldn't easily generalize to higher dimensions

so kinda weird

idek if the reduction thingy is surjective for > 2

[coral shale]

I like to think of the process of 'adding more roots' so x^6 -> x^12 for example and consider what happens.

You get 6 more roots, and these correspond to the new factors, so 4th and 12th cyclotomics

I mean - it's not necessarily helpful to the original question, but a useful visualisation to understand what's going on at times

[coral shale]

When I suggested 'try m = 2, n = 3' as a hint I was considering this diagram

As I said before though, that might be a dead end, but it was something to try

Algebraically at least, you can verify the equality for that example and maybe gain something from it

The hint given in the SE answer relates to that diagram.

[pastel cliff]

what did you mean by T-stable

[fleet pelican]

that means the subspace contasins its T image

[cloud walrus]

memeke

[fleet pelican]

It is likely in fact that T(W)=W, i.e. it need not be a proper subset

[south patrol]

Ye and T-stable subspace of V is equivalently a k[t]-submodule of V with the standard thing

[fleet pelican]

i love linalg, it's so elegant

[south patrol]

It's p cool

The algebraic side of it gets really cute lol

[rustic crown]

Fun Things are Fun!

https://tenor.com/view/kon-fun-things-are-fun-anime-cute-yui-gif-16787421

[south patrol]

What anime is that

[rustic crown]

K-On!

[south patrol]

Oh K-On!

I only watched a few episodes and wanna watch the rest

😭

[rustic crown]

me is rewatching

[south patrol]

I have dvds for most of them lol

[rustic crown]

so cool

[lethal dune]

you have dvd for animes

[rustic crown]

i wanna be that cool

oh

but then i'll need a dvd drive to watch the actual animu

first dvd drive, then ethernet port, then headphone jack...... i don't like where this is going

.<

[coral shale]

#groups-rings-fields-anime

[rustic crown]

[lethal dune]

I'm good with Netflix

[rustic crown]

cwunchy woww twake mew hwome >.<

[lethal dune]

what does that even mean

[formal ermine]

crunchy role take me home

[kind jacinth]

why is the second one yes?

is it because it can be contained in either L1 or L4?

and all words are in L1?

[rustic crown]

pretty much by definition union of regular languages is regular

[kind jacinth]

wait so L4 is regular??

why is L4 regular?

[rustic crown]

yea

only L3 isn't regular

answer to first happens to be Y because it only contains like the empty word

[rustic crown]

you could explicitly write a regex, (a+b)^3((a+b)^6)*

since number of letters is 3 mod 6

[formal ermine]

regex uwu

[formal ermine]

does $\hat{\bZ} = \prod_p \bZ_p$ have a specific name?

[rustic crown]

profinite integers, completion of Z, etc

[formal ermine]

ah ic

[rustic crown]

it's also iso to End_{Ab}(Q/Z)

[primal tusk]

There is one groups of order 6 in |Zmod2 x Zmod12| due to the fundamental theorem of cyclic groups right

[south patrol]

what are the bars for

[primal tusk]

order

[south patrol]

But like

[primal tusk]

wait i typed that wrong lmao

[south patrol]

Oh lol

[primal tusk]

Zmod2 x Zmod12

[south patrol]

There is more than one subgroup of order 6

[primal tusk]

wait but the order is 24 and 6 divides 24 so shouldnt there only be 1?

[south patrol]

Well Z/2Z x Z/12Z isn't a cyclic group

[primal tusk]

ah true

thank you

[south patrol]

Np

hm tryna determine when $\mathbb Q\Big(\sqrt{\sqrt{5} + a} \Big)/ \mathbb Q$ is normal for $a \in \mathbb Q$. Are there any more general techniques one can apply or is this just a bit of a pain?

[cloud walrus]

potato

[south patrol]

Hm okay so I think i've shown that this is always a degree 4 extension and then I guess it suffices to see whether that splits

which corresponds to when a - sqrt(5) is a square, or (in turn) when a^2 - 5 is a square

I imagine that is kinda the end of the line for what one can do in general though

[rustic crown]

you can classify such a if you're asking that

we want to classify rational points on the hyperbola x^2-y^2 = 5. since you know (3, 2) is a rational point, any other rational point is then an intersection of a rational line through (3, 2) and the hyperbola

if you call the slope of this line t, then doing some calculations shows that the other intersection point is x = 3 + 2(2t-3)/(t^2+1)

only point we probably could have missed is when the line is vertical, but that point also has x = 3.

[south patrol]

Hm okay this probably seems overkill given the context of this lol

Hm, weird

[void cosmos]

yo this trick always work right

like starting from one solution and noticing the others are the intersection of tangents or something

[rustic crown]

yea rationals points are far easier to understand than integer points :p

[south patrol]

Pain

Yeah I can't tell with Galois/field theory stuffs like

whether something will be a painful calculation or have a neat trick

[void cosmos]

math shoould have a new patch

where problems involve like certain tags

wether this should be easy or hard

[fleet pelican]

I'm a big advocate of math 2.0 hashtag-release-the-upgrade

[void cosmos]

yea srsly

mee too

[solar shore]

im fine with doin it abstractly but i was trying to see if i could calculate what elements were in the subgroup at all

i think im doing something wrong though because every calculation ive done ends up like

not being correct if that makes sense

[fleet pelican]

it does

that's been my experience in the past

its tough to do, easy to make mistakes

[solar shore]

but maybe ill go down one and try s_4

i think s_5 just has too many to keep track of

[fleet pelican]

deffo recoomend

[solar shore]

for me personally

[fleet pelican]

yeah, same

[solar shore]

and the s_8 one was just given so i didnt have to do anything

[fleet pelican]

i usually prefer more bstract treatments, it helps manage the difficulty setting

[solar shore]

right yeah

i wanted to get my hands dirty w some calculations though

[fleet pelican]

id say, i usually think of these groups in terms of their conjugacy classs now

[solar shore]

also i was using it as a practice for a test soon where we're calculating permutations by hand :p

[fleet pelican]

it feels much more natural

[solar shore]

i c i c

[fleet pelican]

that is a wonderful instinct

[solar shore]

also for the specific example i said before, i think i figured out where i was going wrong

for the cyclic subgroup $\langle (123), (12)(345) \rangle$, is $(12)(345)(123)(12)(345)$ in there?

[fleet pelican]

it looks lke you were only multiplying the base elemnts

[cloud walrus]

blanket

[solar shore]

like is bab in there as well?

[fleet pelican]

exactly!!

[solar shore]

oh..

thats why it was terrible to multiply out

ok i see what i've been forgetting then LOL

[fleet pelican]

in a subgroup, you can also multiply as much as youd like

[solar shore]

right okay

[fleet pelican]

gl!

[solar shore]

thats why i was like, wait why am i not having random elements pop up when i tried ot randomly pick the ones id already calculated

forgot abt the closure of it all

anyways, thank you! appreciate it @fleet pelican

[kind jacinth]

For L3 can i not have regex ((ab)^2)^*?

[rustic crown]

oh . means concatenation

and + is union

so (a+b)^2 means words with letters a, b of length 2

what you wrote means words which are made of abab repeated some number of times

[kind jacinth]

Ah but id want sth lik aa bb abab which cant be written in regex

Right?

[rustic crown]

hm?

that has length 8, and you want squares right

[kind jacinth]

Sorry aa, bb, abab

Separate

Things

[rustic crown]

oh oops

2 isn't a square either

[kind jacinth]

Woops

Is there a way i can figure out if its regular or not?

Ive hesrd of pumping lemma

[rustic crown]

yea

[kind jacinth]

Applied it for L3 but i cant think of how id apply it for L4

[rustic crown]

what's your definition of regular btw?

something that has a regex?

[kind jacinth]

I see it as a set thats closed

So i can form any words ajd it willl be in the set

[rustic crown]

another equiv definition is stuff you get from a DFA/NFA

[rustic crown]

closed under unions, concatenation and star?

[kind jacinth]

Yes

[rustic crown]

okie, so i think just show that the set of lengths of words in regular language is a finite union of arithmetic progressions

or even better

the set of lengths, if infinite will contain an arithmetic progression

which i think is pretty much obvious from definition

if the set is infinite, then you used * in the regex

so then it contains some arithmetic progression

and clearly set of squares doesn't contain any

ig i pretty much gave a weaker version of pumping lemma lol

[kind jacinth]

This makes sense for disproving L3. But for L4 how did u instsntly know its regular?

[rustic crown]

weird question >.<

the language looked "nice" enough to be regular would be the answer :p

more precisely i could imagine an automaton producing that language

because you only need finite memory to store the residue mod 2, 3

but giving a regex is equally easy

intersection of regular languages is also regular, but that may not be obvious from the definition you're using

[kind jacinth]

But this onr doesnt include bbb which could work right?

[rustic crown]

it includes

a+b means a or b

[kind jacinth]

Ohhhhh

Ok sry u said u use + for union

[rustic crown]

yea sorry i don't know the standard notation

.<

[digital raptor]

Is there a name for a module who's ring is countable?

[south patrol]

Tbh just say that lol

A module over a countable ring

[digital raptor]

Yeah okay, thanks

[rustic crown]

countable rings.... haven't heard that much

[kind jacinth]

So then bbb is in there. Then using this i can take S=bbb where y=b x=b and z=b then xy^2z gives bbbb but this isnt in the set so it shouldnt be regular based on this definition right?

[rustic crown]

ig because algebra is more sensitive with finite/infinite than with countable/uncountable

[rustic crown]

it only says there exists some pumping length... so what you just showed is pumping length is more than 3 :p

in our case, P = 9 works

do you know this statement? F/k algebraic with F = k(a1, ..., an) is separable if and only if each a_i/k is sep?

oh i had separable degree stuff in mind >.<

[south patrol]

One nice way is that L/K is separable iff each embedding of K into its algebraic closure can be extended to L in [L:K] ways iirc

[rustic crown]

yea given an extension F/k you define [F:k]_s as the number of such embeddings

[south patrol]

I think you can just replace the alg closure w like a galois closure

[next obsidian]

No you define it as the degree of the separable closure of k inside of F >_<

[rustic crown]

.<

[south patrol]

OwO

Yeah we had a cursed way to show that towers of separable extensions r separable

Hm

[next obsidian]

This is kinda just a cursed fact to show

[rustic crown]

why tho >.<

[rustic crown]

if you define sep-deg like this, then it's quite uwu

show this is multiplicative

then a finite extension F/k is sep iff [F:k] = [F:k]_s

and this gives an easy proof of both tower thingy and f.g. with sep generators is sep

[south patrol]

Fair

[rustic crown]

embeddings into alg-closure of k

[south patrol]

Oh the way I gave is the nice one joe

[rustic crown]

whut is the cursed one 🙈

[south patrol]

We gave one that does it in the galois case and then deduces from that

But I didn't understand the wording of the problem sheet question we had to do it for and got it wrong apparently (but marked correctly), weird

Lol

So idk full details

[rustic crown]

yea you probably can do that... but separable degree is such a useful tool when dealing with separability that it's just better to define alg closure before that lol

[next obsidian]

But you have to show that this agrees with actually being separable

_<

[rustic crown]

yee >.<

[south patrol]

Pain.

[rustic crown]

lol potato's pain arc

[south patrol]

Well im just doing some galois problems and they seem to require more tricks

[warm shoal]

what would the invertible elements of a direct product of rings look like>

[south patrol]

The number of embeddings like respects the tower law

Well

[warm shoal]

what do u mean?

[next obsidian]

Joe 1 have you ever met Joe 2

[warm shoal]

I reckon the identity element is (1, 1)

[next obsidian]

Pairs where both sides are invertible

So the units is the product of the units of R and S

[south patrol]

If L/K/F a tower and M an alg closure and you can extend F -> M to a map K -> M in [K:F] ways, then extend each of those in [L:K] ways to L, you've got [L:K [K:F] = [L:F] extensions of F -> M to L

[next obsidian]

[south patrol]

Because Joe Mama

Ye

But composition of normal extensions ain't normal

And I do not know any particularly nice examples lol

[rustic crown]

oh just use that deg 2 are normal

F/k deg 4 need not be normal

[south patrol]

Well that's what I was thinking lol but ye

What is a nice example of a non normal degree 4

[rustic crown]

Q(2^1/4)/Q

[next obsidian]

Yah dumbdumb

[south patrol]

The example I know from a question I did earlier is like Q(sqrt( sqrt(5) + 2))

[south patrol]

Oh okay fair lol

[next obsidian]

[rustic crown]

[next obsidian]

[south patrol]

Hm why is a real thing enuff

[next obsidian]

Well you also need an imaginary root

[south patrol]

Yeah