groups-rings-fields

[heavy dagger]

Oh!

Thank you very much people

[tranquil parcel]

Seems rather silly to try to do so, right?

[heavy dagger]

yems

[tranquil parcel]

🙂 glad that helped

[spiral wolf]

For primes p, q, show that {1, sqrt(p), sqrt(q), sqrt(pq)} is linearly independent over Q.

I tried some brute force approaches but nothing worked out. Anyone have a concise proof?

[tribal moss]

You probably need to prove it in stages.
First prove that {1, sqrt(p)} is linearly independent.
Then use that to prove that {1, sqrt(p), sqrt(q)} is linearly independent -- you already know that a nontrivial rational relation must have a nonzero coefficient for sqrt(q), so you can divide that coefficient out and get that sqrt(q) = a+b·sqrt(p) for some rational a,b. Square both sides of this, giving q = (a²+b²p)·1 + 2ab·sqrt(p), but since the LHS is an integer, 2ab must be zero (recall that 1 and sqrt(p) are linearly independent), but a=0 or b=0 each lead to contradictions.
Finally {1, sqrt(p), sqrt(q), sqrt(pq)} should go similarly, just with more involved algebra.

[pastel cliff]

what is the significance of normal subgroups

[tribal moss]

They are the subgroups you can take quotients by.

[pastel cliff]

[tribal moss]

Equivalently, they are the subgroups that can be kernels of a homomorphism, which is often useful for figuring out whether homomorphisms with such-and-such properties are possible.

[pastel cliff]

moldi sticker

if i have a group and i know its subgroups are abelian can i conclude that the group itself must be abelian too?

i know an abelian group implies abelian subgroups but idk about the other way around

[tranquil parcel]

Yes, since it is a subgroup of itself.

[pastel cliff]

oh that's cheesy

then subgroups of lesser order

hmmm that actually probably doesn't work in general

[tranquil parcel]

Oh an example is the quaternion group $Q_8$

[cloud walrus]

Michael Harp

[tranquil parcel]

Every subgroup is abelian but it itself is very much not

proper subgroup*

[pastel cliff]

yup i realized, merci

[tranquil parcel]

de rien

[deep sky]

Wait yes

D_3?

[tranquil parcel]

D_6 but yeah that works too

[deep sky]

Ew

[tranquil parcel]

eh?

[deep sky]

using the 2n subscript instead of n

[next obsidian]

Any group of order p^3

Will only have abelian proper subgroups

p prime

[tranquil parcel]

oh yeah because any group of order p^2 is necessarily abelian

[next obsidian]

Yeah

[tranquil parcel]

and p is obviously cyclic

[next obsidian]

Yeh

[deep sky]

Wait, why is p^2 abelian?

[tranquil parcel]

It will have a nontrivial center by the class equation is the gist of it iirc

well that's true for any p group but in the case of p^2 you get a cyclyc quotient by the center

[next obsidian]

You don’t need to do it that way but

You can show it’s either C_p x C_p or C_p^2

If you have developed internal direct products

If there’s an element of order p^2 it’s the latter

Else once you know Lagrange, take two elements of order p which aren’t in the same cyclic subgroup

The product of the subgroups they generate is all of G cuz order p^2

Trivial intersection

Both are normal because index p

So it’s C_p x C_p

[celest mantle]

according to a teacher of mine, there aren't any (at least that he knows of)
though he made an interesting remark aside : G is abelian iff #(H^1(G,Q/Z)) = #G, thought it might be good to share it too

I stumbled onto something interesting though : even if H isn't normal in G, we still have an isomorphism H^n(H, A) ~= H^n(tHt^{-1}, A) for t in G

so that may be why at first sight it might be hard to find such a way to do so

[deep sky]

Oh direct prod of abelian groups is abelian

Why isn’t a group of order p^3 necessarily abelian then?

[next obsidian]

It doesn’t necessarily decompose like this

I think you can write it in terms of semi direct products maybe

But you can just take the quaternion group Q8

Or the dihedral group of order 8 as a counterexample

[tranquil parcel]

What does the matric representation of a frobenius automorphism look like?

I want to find the characteristic polynomial of the map $\theta : a \mapsto a^p$ over $F_{p^m}$ and im not sure if there is a clever trick to doing so.

[cloud walrus]

Michael Harp

[tranquil parcel]

I got $\sum_{0}^{p^{r-1}-1}x^i$ as the minimal polynomial

[cloud walrus]

Michael Harp

[delicate orchid]

my thoughts are that F_{p^m} is char p and so we might be able to use a fermat's little theorem meme

[rough drift]

hello,
so this is kind of bothering me. 
x^G is an orbit of G in respect to element x (by my understanding that means that x^G is invariant under G and G|S is transitive over S (which means that for every x,y in S there exists g such that x^g = y)).
Now G can only map x to the elements of x^G (by definition). That means we can partiton G ("main" set of G) in |X^G| partitons based on where they map x. Now i don't understand why does each coset contain |Gx| elements? Is that maybe because all g ∈ G that map x to some y form right cosets on Gx? How do i know that each of those cosets has the same cardinality as Gx?```

okay so i check my abstract algebra notes and it seems that i forgot that cosets share the cardinality of H (or in this case of Gx)

is my interpretation okay?

[rapid bramble]

I'm smooth braining a good place to start on proving $m\mathbb{Z}\subseteq n\mathbb{Z}\implies n|m$

[cloud walrus]

Mosh

[rapid bramble]

Cause I know there are integers p and q st x=pm and x=qn, but that doesn't seem to get me anywhere

[next obsidian]

You just need to show that m is in nZ

But that turns exactly into the divisibility thing

[deep sky]

If mZ is a subset/subgroup of nZ, then every multiple of m Is expressible as n times another integer, right?

[rapid bramble]

oh duh

m in mZ so m in nZ

so m=nk

n|m

[next obsidian]

Yah

[rapid bramble]

went too general ty

[fast stratus]

When we say a+b… How do we read cayley diagram… is it like first go to node a and apply b or go to b and apply a

[stone fulcrum]

Each node should have an element attached to it

And each edge will have a generating element

[fast stratus]

No, in non-abelian groups; how do we read a+b

Is it do first a and then b
Or do first b and then a

[stone fulcrum]

The action ab is generally thought of as "do b then a" but you lose nothing by reversing that

I feel like this is an xy question. Did you have a project in mind?

[fast stratus]

Yeah; thank you

[fast stratus]

?

[stone fulcrum]

Nvm, happy to help. Feel free to ask if you have anything else!

[chilly ocean]

Consider a group of a polygon symmetries. When it acts on that polygon (on the sets of vertices and edges respectively), then can we say the resultant y (g.x = y for an element g in Group G, and x in a Set X) belong to X?

[delicate orchid]

a group action by definition maps from GxX -> X, so yes

[chilly ocean]

Thank you

[languid walrus]

if B is an A-algebra, q a prime ideal of B and p the contraction of q, is B_q isomorphic to localizing the A-module B at p?

[gritty sparrow]

No, for example A=Z, B=Z[x], q=(x)

[trim grove]

What is the splitting field of cyclotomic polynomial $\phi_{p}(x)$?

[cloud walrus]

Solution

[celest mantle]

Q(zeta_p) ?

[south patrol]

The roots are precisely (by definition) the primitive roots of x^p - 1

[glacial moss]

I'm starting with ring and field theory, and I peeked into something which looks like integers decomposing into factors "smaller" than the usual primes: $(6)=(2,1+\sqrt{-5})(3,1-\sqrt{-5})(2,1+\sqrt{-5})(2,1-\sqrt{-5})$. I can vaguely guess what it means, but maybe someone can clarify some thoughts:
Is this really a decomposition of an integer into a unique factorization? I know $(6)$ isn't exactly the integer $6$, but can I treat it like that if I only consider multiplication? Maybe assume these non-principal ideals provide one layer of decomposing and then there is another prime decomposition within?
What is the most general name for this type decomposition ("integers into factors smaller than usual prime numbers")? Is it UFD or Dedekind domain or Ring of integers?
Can I generate these "factors" from the examples that are mentioned in https://en.wikipedia.org/wiki/Ring_of_integers

[cloud walrus]

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

[tribal moss]

If you need to "vaguely guess" what it means, then you shouldn't expect to reach any useful conclusions. These things have precise definitions.

[coral shale]

what tropo said.

You also need to know what an Ideal is

[tribal moss]

This most important fact to be aware of here is that "unique factorization" doesn't really mean anything in and of itself -- it depends heavily on which kind of factors we're considering for the factorization.

In the example above, I think the author must have decided to work in the ring Z[sqrt(-5)] instead of the ring of usual integers.

That makes more numbers available as factors and therefore changes both which factorizations we have at all, and also which of the factorizations we already know are unique in the new setting.

[glacial moss]

i know the definition of an ideal. but see that the above example specifically is not Z[sqrt(-5)] because that would be incorrect

[tribal moss]

In this ring 6 can both be made as 2 multiplied by 3, and as 1+sqrt(-5) multiplied by 1-sqrt(-5). It turns out that neither of these factorizations can be further refined as long as we're talking about honest multiplication in the ring. So Z[sqrt(-5)] doesn't have unique factorizations into products of irreducible elements.

Please elaborate on "that would be incorrect".

[glacial moss]

that was the whole point of the story. Z[sqrt(-5)] is not a unique factorization for 6. you have to consider Z[(1+sqrt(-5))/2] or so.

[coral shale]

no....... you are misunderstanding

Z[sqrt-5] is not a factorisation

it is a ring

it is the ring we decide to work in

In this specific ring, (6) has unique factorisation into 4 factors as stated above by you

[glacial moss]

no. that's the whole point that this ring does not allow a unique factorization. you have to do something else to get unique "factor". it's only the last part i don't understand well

[coral shale]

This specific ring happens to be a Unique Factorisation Domain

All those 4 factors written down are Prime Ideals

[tribal moss]

(No it doesn't).

[coral shale]

You havent seen the definitions and are making guesses in the wrong direction I'm afraid

[glacial moss]

i'm sure Z[sqrt(-5)] is wrong and not what is used here

[tribal moss]

I'm very sure Z[sqrt(-5)] is exactly what the example is about.

But factorization of ideals is not the same as factorization of elements of the ring.

[coral shale]

You need to know :

• Ring
  (- unit element, prime element, irreducible element, not necessary, but will enlighten WHY we're doing this)
• Ideal
• Prime Ideal
• Unique Factorisation Domain
• Ring formed by adjoining elements

And finally how to multiply ideals.

[tribal moss]

It's close enough that it can be useful for several of the same purposes in applications (and it is often better behaved), but it is not the same thing.

[glacial moss]

for multiplication $(6)=(2)*(3)$ kind of works?

[coral shale]

(a, b)
Do you know what this object is in what you weote?

[cloud walrus]

Gerenuk

[coral shale]

yes but (2) and (3) are reducible

[glacial moss]

yes

[coral shale]

and not prime

[glacial moss]

i know that object. but indeed, i do not understand this type of factorization

[coral shale]

(ive gotta revise some definitions myself smh)

[glacial moss]

i guess i'm mixing up "types of factorization"? is there a link where i can read up on this?

[coral shale]

cant remember if we're interested in irreducibility or primeness

or both

The point is, anyway, (2) and (3) can both be split up

[tribal moss]

is there a link where i can read up on this?
The book where you found the above example might be a good choice if you read it from the beginning instead of skipping ahead :-)

[coral shale]

We want to uniquely factorise ideals

[glacial moss]

i'm interested in "decomposing integers into parts smaller than the usual primes". not sure how to define that. i've understood ideals for now, but i'm not much further

[coral shale]

(a, b)

Do you understand what this is

[glacial moss]

i think i do

[coral shale]

this is vital because you need to then realise what
(a, b)(c, d) is

[glacial moss]

i think i can work this out

[coral shale]

using this defn

[glacial moss]

but it's not related to the usual factorization of elements?

[coral shale]

We arent talking about factorisation of elements no

this talk from your originally clipped passage is all about ideals

[glacial moss]

so the whole point is that my equation factorizes ideals?

[coral shale]

About how we can uniquely factorise ideals

not elements.

Elements do not uniquely factorise in that ring for sure.

[tribal moss]

It's related to factorization of elements, but not the same thing.

[glacial moss]

hmm, i need to understand the way it is related then...

[coral shale]

Aleph0's video on this was good iirc

[glacial moss]

that's where this question comes from

i'm just not there yet. i didn't consider factorizing ideals so far

ok. i'll go over it again. i was misled thinking it is related to factorization of elements

[coral shale]

This is important to know as well

Z[rt-5] is not a unique factorisation domain

We instead consider a ring of ideals in Z[sqrt-5]

[glacial moss]

thanks. i guess to need to look into factorization of ideal. i get the Z[sqrt(-5)] not being unique part

[coral shale]

in this ring of ideals, addition and multiplication are defined like this
edit: not a ring, mb

[glacial moss]

can the natural numbers be "part" of this ring of ideals?

[coral shale]

(n) is an ideal

Ah wait not a ring

additive inverse sounds like a problem

But regardless ^ this construction is useful to think about

[glacial moss]

hmm, ok. it's not really a ring

where are these screenshots from?

i need to look into the operations on ideals more closely. didn't consider that before

[coral shale]

If K is a number field, Ok is the ring of integers, which has a specific definition

my notes in algebraic number theory

im taking this course rn

[glacial moss]

oh. where i find notes or a course of that quality?

[coral shale]

🤷‍♂️

idk sry

i think aleph0

makes good suggestions for where

[glacial moss]

ok. i'll have a look

[coral shale]

This is algebraic number theory stuff, you will probably briefly brush on it when first starting Algebra

On the fact stuff like Z[sqrt-5] is not a UFD

And why

But you likely won't do stuff with Ideals as above I think

[glacial moss]

i do understand Z[sqrt(-5)] not being UFD. and watching the aleph0 video i didn't realize that it was not factorization of elements anymore. do you use a particular latex package to type your notes?

[coral shale]

Not my notes lol its my lecturers

If you watch the video again

I believe they briefly explain what an ideal is

And that they were multiplying these together

they had a visual thingy of a number line

[glacial moss]

i know what an ideal is. just didn't get they were being multiplied as well. but i rewatch either way

i thought there was a way to identify $(6)$ with $6$ in the end

[cloud walrus]

Gerenuk

[coral shale]

The text below the definitions is important in terms of computation

There is cus (6) is the multiples of 6 🤔

in that ring

6(a+b rt(-5))

[glacial moss]

you can identify when they start obeying similar algebra. (6)=(2)*(3) works in a way. but no additive inverse

[coral shale]

For integers a, b, we have
(a)+(b) = (a, b) = (gcd(a,b))

apriori gcd may not exist for non-integers

[glacial moss]

hmm. good to see that

[tribal moss]

So switching to ideals gives a nicer multiplicative structure (where eg factorizations are unique) but at the cost of no longer aligning with an additive structure.

[glacial moss]

interesting. i haven't reached that part yet. that's why my confusion

[coral shale]

I think factorisations are only unique under certain cases even within ideals or am I remembering this wrong

Class number comes to mind

[glacial moss]

you don't remember the technical term for when that is unique?

i read something about dedekind domains?

is that related somehow?

[coral shale]

probably

[coral shale]

i remembered this wrong

[coral shale]

.

At least with Ok we have unique factorisation, but in general nope

[glacial moss]

i haven't learned Ok yet. Is that related to things like Z[x]/...?

[coral shale]

see above i pasted all definitions that i referred to

[glacial moss]

ok thx

[deep sky]

When determining orbits and stabilizers for symmetry of points on an n-gon, if we’re just asked for the “stabilizer of a vertex” are there specific conventions about where, say, the reflection elements’ line is on the polygon?

I’m finding the stabilizer of a vertex on a square, but it seems like in the first place there are two different ones depending on if the vertex starts on the line of reflection or not

And that the reflection line could be set in 3 different places

[coral shale]

Transitive group action, primitive group action, blocks

what course would this stuff be studied? Rep theory? Group theory???

Is there a course dedicated to group actions somehow?

[stoic rose]

I guess they're usually studied in a group theory course

[chilly ocean]

rep theory course

i learned about transitive group actions cus they came up in alg top

and then diff top

[deep sky]

What does it mean decompose an object of a group action into orbits?

[chilly ocean]

group actions are multiplication defined on a set by left or right multiplication by a group element

so g*p is well defined

if you do Gp which is the set of gp for all g in G then we get an orbit of p

if Gp=Gp’ they are in the same orbit

[deep sky]

For a fixed p, right

[chilly ocean]

yup

There is an equivalence relation for two points being in same orbit precisely p~p’ if Gp=Gp’

[deep sky]

So if a question is asking about decomposing a vector V in Rn bring acted on by GLn(R), would that literally just be that each entry in V has an orbit of all real numbers

[chilly ocean]

im assuming so

besides 0 maybe

because GLn(R) is invertible

so you dont want the transformation to make V into 0 vector

[deep sky]

That’s where I was a little unsure too, but Can’t an individual entry become 0 as long as overall the number of nonzero entries stays the same?

[chilly ocean]

yeah

but they cant all be zero

[deep sky]

Like, the rules of invertibility are about the nature of the overall product, and we can’t really say anything about an individual entry

Yeah of course, but any given element could be 0 as long as at lease one other element is not zero

[chilly ocean]

yes

[deep sky]

Awesome, thanks

[chilly ocean]

what class is this for

[deep sky]

Abstract

We just started symmetry

[chilly ocean]

cool

So the orbit space might be Rn\{0}

May I ask a question

[deep sky]

0? Wouldn’t it just be R?

[chilly ocean]

If you have the group action on a vector space

You said GLnR acting on V in Rn

[deep sky]

Oh the total orbit space yeah

No I getchu

But decomposition means they’re asking about any one given entry ?

[chilly ocean]

no clue on that side

[deep sky]

I guess it’s a petty point as to whether to avoid talking about the total orbit space while I’m at it

Makes sense at any rate

[chilly ocean]

mhm

[deep sky]

Just go ahead

(Thanks for waiting)

[chilly ocean]

So uh I'm first time dealing with group actions, so I just thought doing this example would give me some sort of intuition on this topic. Is my notion of understanding correct? [G is a group of rotational symmetries of triangle, X is a set of labeled vertices of the triangle]

[warm holly]

hello, I need to show the following: Let $K$ be a field and $\lambda_i,i=1,..,r\in K$ eigenvalues of matrix $A \in K^{n\times n}$ and $m_i,j=1,..,r$ it's algebraic multiplicities. Let $m_1+..+m_r=n$. Show that $\det (A)=\prod_{i=1}^r \lambda_i^{m_i}$. As $lambda_i$ are eigenvalues of $A$ I can simply write down it's characteristic polynomial as $p(x)=(\lambda_1-x)^{m_1}\cdot..\cdot (\lambda_r-x)^{m_r}$ (as eigenvalues are only squareroot of characteristic polynomial and the previous polynomial is of degree $n$) and then setting $x=0$ I get the determinant of my matrix. Is it correct to do this way?

[cloud walrus]

𝔻аniil

[south patrol]

Yeah that's correct

Similarly you can find a formula for the trace of the matrix in terms of the eigenvalues if you're interested

[west violet]

how would one approach this?

[coral shale]

f : G -> H

Let's say you want to find homomorphisms in general

If you have a set S which generates G

can you see how this will be handy?

G = <S>

@west violet

[west violet]

oops forgot to also put this there

[coral shale]

yh sure

[west violet]

so i understand this

i can create 3 conditions based on those conditions for $D_8$

$r^8=1,s^2=1 \text{ , and } (rs)^2=1$

[cloud walrus]

suck2015

[coral shale]

Sure sure, but in general - have you been shown how thinking about generators helps for constructing homomorphisms?

[west violet]

kinda

i guess you can define f(g) = k in H

[coral shale]

🤔

no no - there is an important result

that tells you what f(1) has to be.

f(1) = 1 (see if you can prove this if you havent seen it)

[west violet]

yeah because homomophism send the identity of G to identity of H

[coral shale]

yh ok.

If we have S generates G

what does that tell you about every element of G?

[west violet]

S is a set? or an element?

[coral shale]

set

a generating set

have you come across this?

[west violet]

i guess Dihedral group is generated by {r,s}? would this be a generating set

[coral shale]

the dihedral group is generated by a rotation and reflection (but you can't just pick any old rotation)

a 90 degree rotation and any reflection does generate D8

But in general --- we say S generates G

If any element in G can be written as a (finite) product of elements in S (including powers of and inverses)

[west violet]

ah yeah ok

[coral shale]

If S = {a, b} for example

every g in G
g = a^n b^m

n, m integers

Now - can you see how this helps in terms of the homomorphism?

[west violet]

uh our homomorphism is f:G ->H?

[coral shale]

yes

$g=\prod^n_{i=1} s_i^{\alpha_i}$

[coral shale]

And every g in G

can be expressed as some 'product'

of things in S

(not necessarily uniquely, but that does not matter)

[west violet]

every s_i is a product of a and b right

[coral shale]

No, I'm talking about a general example

$S = {s_i : i\in I}$

[coral shale]

I some indexing set.

[west violet]

oh yeah of course

i understand that part

[coral shale]

And note this product is not necessarily commutative

aba may be different from aab, and so on

If we have a homomorphism f : G -> H, though

we can still say something based on the fact S generates G

Take this --- any ideas what you could do?

[west violet]

for my question?

[coral shale]

well yes, but in general, really

this is an arbitrary group G, arbitrary generating set S of G

and arbitrary homomorphism f : G -> H

[coral shale]

Try applying f to both sides of this

[west violet]

well $f(g) = f(\prod s_i^{\alpha_i}) = \prod f(s_i)^{\alpha_i }$

[coral shale]

dont know why you brought alpha down

And sorry - just realised this notation is a bit crappy

$g=\prod s_i^{\alpha_i}$

[coral shale]

This is better (and specify the product is finite)

But I think you have right idea

[west violet]

oh no , i mistook alpha as a power of s_i lol

but yeah i understand

[coral shale]

$f(g)=\prod f(s_i)^{\alpha_i}$

[coral shale]

We have this because of the homomorphism property

f(a)^n = f(a^n) and f(a)^-1 = f(a^-1)

so we can conclude this

This is important because it tells us if we know where the generating set S maps to

this determines where everything else maps to

Does that make sense?

[coral shale]

Ah one last step --- taking the power of alpha outside (property of homomorphism)

And my mistake for including indexes (its best not to generalise)

We want to describe a general product including abbaba^-1abba^-1ba^-1babba^-1ba because our group is not necessarily commutative

[cloud walrus]

suck2015

[west violet]

ah ok

[west violet]

yeah i think i understand this

[coral shale]

If you have done linear algebra

it is a bit like a basis

(or just a spanning set, really)

If you decide where this maps to

it determines your linear map

[west violet]

ah like a basis with linear combinations

[coral shale]

The same idea applies here. If we decide where the generating set maps to, it determines our homomorphism

This scales down the problem a lot

For any dihedral group

We can have a generating set of 2 elements

So rather than deciding where the entire group maps to for a homomorphism

You just need to decide where the generators go

and check whether the resulting thing is a homomorphism

So back to this, that cuts down a lot of possibilities

But that's not all for generators --- the generators of the codomain are also worth looking at

Can you see why, perhaps?

f : G -> H

Let A generate G, B generate H

(This is more applicable to constructing isomorphisms, tbh, less so for homomorphisms, but still relevant to a certain extent)

[west violet]

why looking at the generators of the co-domain is important?

[coral shale]

hmmm I will adjust this statement slightly for a homomorphism

f(G) are you familiar with this notation for the image of f?

this will be a subgroup of H (result)

[west violet]

yeah

[coral shale]

Right - I claim, if A generates G

f(A) generates f(G)

This is something worth thinking about/trying to prove

So just by knowing where the generators map to, you can determine what the image of the map is.

And hence also deduce the size of the kernel, I believe.

(By lagrange?, not sure)

|ker f| |im f| = |H|

If I haven't remembered/thought this wrongly 🤔

[west violet]

i think thats from the first iso theorem

[coral shale]

could be too 😄

yh, sounds right

[west violet]

hmm, so I should try and think where the generators maps to?

[coral shale]

So basically yes - the general approach is to think about generators only

There is still some casework to be done, but it reduces work a lot

Thinking about Lagrange, Isomorphism thm will give you information about divisibility of sizes

[west violet]

my initial thought was f: D_8 --> Z_12 is our homomorphism, and we know D_8 is generated by the set {r,s}, then 2f(s) =f(1)= 0 (mod 12)

[coral shale]

which further cuts down on stuff

Like for example im f cannot be of order 6

We have order 6 subgroup of Z_12

But the order of the image needs to divide the order of the domain

I'm just spouting facts, can't think explicitly why atm --- i'm sure its 1st iso thm 😅

[coral shale]

and yh thats true

Order of f(a) needs to divide order of a for any a in G in general (lagrange)

[west violet]

ah ok

so then if i do the same for r, and rs, I get only 4 possible homomorphism

[coral shale]

Once you narrow down which are possible, it is important to double check they actually are homomorphisms

[west violet]

well actually let me rephrase that

I get possible values for f(s) and f(r)

so i check if these are homorphisms?

well with the possible values, I know every element in D_8 can be written as (r^a)(s^b)

[coral shale]

yh i think you got it down

Just make sure you argument is rigorous for why those possible values you have down are necessary for a homomorphism

And then show which ones are sufficient for a homomorphism via checking

[west violet]

how do you check? Don't i just have to show f(ab) = f(a)+f(b) mod 12?

[iron vessel]

How can i do part c? I thought about the fact that finite fields with the same number of elements are isomorphic but in all fairness im not confident whether they are finite

[leaden scaffold]

well yeah but forall AB. I think the key is to use the first isomorphism theorem which tells you the kernel has to be a normal subgroup and G/ker(f) has an isomorphism to H for a homomorphism f: G to H

[west violet]

@coral shale thanks for earlier(i apologize if this is an unnecessary ping)

[coral shale]

no problem, was afk

[west violet]

ah yeah, i kinda figured it out, it was an interesting problem. I mean the kernels i got for all of them were subgroups of G

[indigo ravine]

I have somewhere in my textbook that for a given field, (X^n - 1) is separable if characteristic is 0 or prime with n. I'm not sure I understand why

oops sorry did not want to answer your post not related

[next obsidian]

@maiden ocean

You can like

Ask questions here if you have some

I know the idea behind what this stuff is

And what you can do

[maiden ocean]

How do i read the front matter

Oh nvm

im silly

[next obsidian]

Yeah haha

[maiden ocean]

i can download it in full

[next obsidian]

Yup

[deep sky]

I get matrices as motivating/familiar examples, but they are just the least enjoyable problems to do

[pastel cliff]

i feel that pain greatly and have yelled about it in this channel substantially

yeahhhh uhhhh let's learn the fucking uhhh symmetric group in terms of uhhhh fuckin matrices hurrr ddurr

[kind temple]

the largest exponent in f(x) with non-zero coefficient

so if f(x) = -4x^5 + 3x^2 - 2
then deg(f(x)) = 5

[prisma shuttle]

can someone explain why the kernel of a ring homomorphism from an infinite ring to a finite ring must be infinite

I know that $R/\text{ker}(\phi)\equiv \text{im}\phi,$ of which the right side is finite and the $R$ on the left side is infinite, so it makes sense that $\text{ker}(\phi)$ shoudl be infinite as well but is there rigorous way to prove

[cloud walrus]

JustKeepRunning

[woven delta]

The union of the cosets (of which there are finitely many and they are all bijective) is R. If they were all finite then their union would be finite

[barren sierra]

I'm stuck on a seemingly simple problem but having trouble working with definitions

so I need to show that x^2 + y^2 - 1 is irreducible in Q[x, y]

I should be able to apply Eisenstein

with prime y - 1

but not sure how Eisenstein works with multivarible polynomial rings

if my prime is y - 1

do I treat everything in terms of x, so then y^2 - 1 is my "constant" term and so we have that
y - 1 does not divide 1
y - 1 divides y^2 - 1
(y - 1)^2 does not divide y - 1
and so we're done, Eisenstein applies????

so I'm thinking of Q[x, y] as Q[y][x] I guess

[woven delta]

What is the assumption for Eisenstein?

[woven delta]

Oh it's just integral domain

[barren sierra]

yea

[woven delta]

That's good

[barren sierra]

Q is integral domain so Q[x]/Q[y] are integral domains that's all good

so then that works?

[woven delta]

Yeah that seems good

[barren sierra]

cool

[heavy dagger]

How do I show that if n ≥ 3, then the center of Sn is of order 1? I've tried induction, and have shown it for S3, How can I show Center(Sk) = (1,2....k) => Center(Sk+1) = (1,2....k+1)?

[toxic zephyr]

just want to make sure i got the correct answer that the only solution of this is x=[1]

[kind temple]

take n >= 3. if f is not the identity, then f(i) = j for some i != j. take some k with k !=i, k!=j (why does one exist?) and consider the transposition (k i) or (k j)

[chilly ocean]

The equation reduces to (x - [1])² = 0

[kind temple]

you could also just check all [n] for 0 <= n <= 10

[carmine fossil]

Which implies (x-[1])=0 since 11 is prime

[carmine fossil]

Use the fact that transpositions generate S_n and consider centralizers of each transposition

[heavy dagger]

Thank you, how did you come up with this construction?

Like I see that it works and why, but how did you get this idea

[heavy dagger]

I haven't really learnt generators yet : (

[carmine fossil]

Ok,That idea is a refined version of mine I think

[kind temple]

i honestly don’t know. it just worked. maybe somebody else can explain the reasoning behind why it works better than i can.

[carmine fossil]

Actually ignore the generators part, that's not needed

So,take the transposition (i,j)

Now let x be a centralizer of (i,j). x is either one of 2 things. It fixes i and j

Or x(i)=j and x(j)=i

Now suppose A is an element of centre that's not identity,then A(i) is not i for some i,then if you take a pair (i,j) A(i) has to be j and A(j) has to be i

Take another pair (i,k) and you get a contradiction

Ok same thing I think

[carmine fossil]

Well n>=3 would imply you need to consider things that change atmost 3 elements

Transpositions are a natural thing to consider

[chilly ocean]

Basically, an element in the center is an element such that zg = gz for any g, or g⁻¹zg = z. Clearly if z is an n-cycle with n >= 3 and g is a transposition then this is never true
Edit: ignore this

[next obsidian]

There’s a simpler way to sort of induct

Or well…

Nvm

My idea didn’t work :(

You can by induction only have to prove it when g is an element which permutes every element from 1 to n

Because if not, it lives inside a copy of S_{n-1}

If g fixes i, you can look at all elements which fix i, and this is isomorphic to S_{n-1}

And so there’s another element in that guy it won’t commute with

[chilly ocean]

Right. I realised that my cycle idea doesn't work

[next obsidian]

Yeah so cycles are bad because

[chilly ocean]

So uhm... ignore me *slides away*

[next obsidian]

z could fail to commute with g

But that won’t mean it doesn’t commute with gh

Like if h = g^-1 for example

So I think you basically have to just do what c squared did and like just construct an element it won’t commute with

If you’ve developed a lot more theory of symmetric groups and group actions there’s a really simple proof tho

[chilly ocean]

Wait, does the class equation actually work?

I am shook

[next obsidian]

Namely, write things in cycle notation, then ghg^-1 is the thing where you take h and apply g (considering it an automorpjism) to all the elements of h

Like if h = (123) then ghg^-1 = (g(1)g(2)g(3))

And gh = hg iff ghg^-1 = h

So you can take any h and just cook up a g which won’t have ghg^-1 = h

But this requires a lot more theory lol

I think this might be a way to see what csquared did more naturally

Maybe

[kind temple]

yea idk. i was kinda forced to take this approach. i wanted to make it as simple as possible, so naturally that means a cycle of some sort.
taking g = (i j) doesn’t always work, since f might be a transposition. but if you just mess up another element, the thing works

[next obsidian]

Yeah

This is exactly like

When you do the computation of what (ij) does on the left vs right

You’re basically figuring out the thing I said

[heavy dagger]

Why is group theory associated heavily with the study of symmetry? I understand that every group is isomorphic to a group of transformations (on itself) and how transformations that preserve some property on an object (like those regular polygons) sometimes form a group but it still doesn't feel natural to associate group theory heavily with symmetry yet.

[sharp sonnet]

its how groups originally appeared

as symmetries of objects, geometrical but also otherwise (like polynomials)

even today groups are often studied via their actions on other objects

and they act as symmetries on those objects for some definition of symmetry

[heavy dagger]

Does this refer to groups that are transformations, since something like the group of additive integers doesn't really act on objects or does it?

[upper pivot]

It does

So one of the things it acts on is the real number line by translations

And it turns out this is actually a pretty important action

[heavy dagger]

But aren't the elements of the group, integers and not the translations themselves? Or are you referring to them as the same because they are isomorphic?

[upper pivot]

The quotient space of this action for instance is just a circle, and I won’t go into a lot of topological stuff here but this gives you crucial info about the circle

I mean the point of a group is regardless of how I write it down

To make it act on something

To think of it as maps on something

And his is one example of Z being maps which is useful

[sharp sonnet]

a group action lets you identify a group as a subgroup of Sym(M) for some object M

[heavy dagger]

Wait can you rephrase that

[sharp sonnet]

the set of all bijections from an object to itself is a group, right?

[heavy dagger]

Yeah

[sharp sonnet]

this is the symmetric group

and a group action is a group homomorphism G -> Sym(M) for some set M

if this is injective it lets you identify G as a subgroup of Sym(M)

[heavy dagger]

Oh yeah

[sharp sonnet]

in fact you can always do this

for any group G and some M

which maybe explains this as well

[heavy dagger]

Because of Cayley's theorem?

[sharp sonnet]

yeah

my point was that originally groups were studied only as this

like, the original notion of group was "subgroup of Sym(M) or GL_n"

then it was abstracted

but the original idea is still there

[heavy dagger]

Right!

I see

[sharp sonnet]

and group actions play an important part in the theory

(and linear representations)

[heavy dagger]

Gotcha

[coral shale]

Let L/M/K galois tower of extensions (L/M, L/K, M/K all galois)
Then
Gal(L/K)/Gal(L/M) = Gal(M/K)

By fundamental thm of Galois Theory

This looks like 3rd isomorphism in some ways... Is there some link with Yoneda in cat?

[hidden haven]

I doubt

[lethal dune]

.

[coral shale]

no link at all?

to something in cat

[simple mulch]

Why do I feel like this isn't a transformation group?

Notice the functions in G are not bijective as for f(x) = 3x + 2 and g(x) = 2x+1 we have f(1) = 5 = g(2) but 1 != 2

[lethal dune]

?????????

are you checking bijectivity across multiple functions ?

[simple mulch]

oh fk

LOL

[thorn delta]

f(x) = 3x + 2 would be an element of your group for example, and this is a bijection

[simple mulch]

I see I see

lol

Its because I am having an hard time finding the inverse of each function that I was trying to find an excuse

like we want g such that fog(x) = id(x) = x = g(ax+b) = gof(x)

[lethal dune]

y=3x+2 => x=(y-2)/3

[simple mulch]

https://tenor.com/view/the-simpsons-homer-simpson-bush-hide-bye-gif-16150806

[pastel cliff]

first iso today in class

are stickers gone wtf

[pastel cliff]

i cant use stickers on any server apparently so that's cool

[next obsidian]

Yeah I can only do it on my phone

Lmfao

[dull root]

If I have a subfield of C, with an element a, is it true that the complex conjugate of a is also in the field?

[next obsidian]

Yes

Or uhhh

[dull root]

Why is that so, I ceratinly see it for normal extensions of Q

but not in general

[next obsidian]

Yhhh

Well if it contains R

It’s true

Because C is degree 2

So the only intermediary fields are R or C

But what if it doesn’t?

[dull root]

yes, for example something like Q(a) where a is just some (nonreal )complex number algebraic over Q

[tribal moss]

Not necessarily. If you take an irreducible cubic with one real root and adjoin one of the complex roots to Q, you don't get any of the other two roots.

[dull root]

I see that if you ajoint the real root, you can not get any of the complex roots, but why is it if you ajoint a complex root, you can not get another root?

[lavish nexus]

seems 0 would be in D

But zero divisors are nonzero

[tribal moss]

The splitting field has an automorphism that takes the real root to one of the complex roots. So that takes Q(real-root) to an isomorphic field, but Q(real-root) obviously only contains one root, and so its image also contains only one root.

[dull root]

Wait, I don't think what you said is true. If we look at a 3rd primitive root of unity, then x^3 - 1 is its minimal polynomial, and ajoining a complex root gives you all the roots. That polynomial has 1 real root and 2 complex roots

[tribal moss]

x³-1 is not irreducible: x³-1 = (x-1)(x²+x+1).

[dull root]

yes i am dumb :

So is the summary of what is happening as follows? Since we are working over C, our fields has characteristic 0, so any splitting field is a Galois extension.

1. We know the Galois groups acts transitively on the roots of this cubic polynomial. This is why there exists an automorphism in the galois group that sends the real root to acomplex root.
2. This automorphism maps Q(real root) to Q(complex root) since the automorph maps real root to complex root and fixes Q.
3. Q(complex root) can not contain another other roots, since its inverse image is another root, but Q(real root) only has 1 root

[hidden haven]

root

[next obsidian]

🥔

Potatoes aren’t root vegetable

🥕

Maybe?

[hidden haven]

[dull root]

Automorphs sounds like some transformers shit

But does this idea hold in general regardless of the degree of the polynomial? It seems to be there is nothing special about a cubic polynomial in this case. We just need the irreducible poly to have a real root and complex roots

[prisma ibex]

maximal ideals, call that optimus prime

[clear fiber]

I had this thought while working on a different proof. Is this statement true?

Let $R$ be a ring and let $S$ be a subring of $R$. Then $0_S = 0_R.$

[cloud walrus]

Rmoney

[clear fiber]

I tried googling it and couldn't find an answer

[dull root]

Yes

[next obsidian]

Yeah it’s true

This is actually just about groups

Because this is about the underlying group structure

[pastel cliff]

what is the dihedral group of order 4

i figure order 6 is symmetries of a triangle

[dull root]

you mean D_4?

[tribal moss]

It's just V4, which can be realized in several different ways.

[pastel cliff]

im being asked to prove that it's isomorphic to that lol

[tribal moss]

Oh. Hopefully you have a definition of "dihedral group" to work with.

Otherwise, perhaps the group of geometric symmetries of a line segment.

[pastel cliff]

i'll go back and check, i just default to D_8 usually

also are there any elegant ways to prove isomorphism

i usually default to just a table if the groups are small enough but that feel scuffed

[lavish nexus]

There are only two groups of order 4

It clearly can’t be Z4

so it’s V4

[dull root]

Yes, its either C_4 or C_2xC_2 the klein 4 group

[lavish nexus]

It is useful to remember that table of groups of small order

[dull root]

I actually have a related question. If we know our group is order 4 how do we decide between K_4 or C_4? They are both abelian.

[lavish nexus]

is there an order 4 element

[dull root]

So since we have only four elements, the best way to do this is literally check element by element right?

[lavish nexus]

Only 3 to check

if you have one order 4 it’s Z4

[pastel cliff]

👉 👈

[dull root]

In general, try to show it is hom first and show surjective

[lavish nexus]

D4 is {1, r, s, rs}
V4 is {1, a, b, ab}
done

[pastel cliff]

is it enough to say that if two groups have the same order for each element theyre isomorphic

[lavish nexus]

Probably not

[tribal moss]

No -- Z and Z^2 both satisfy that every non-identity element has infinite order, but they are not isomorphic.

[dull root]

Double checking: K_4 has 3 proper subgroups right, <a>, <b>, and <ab>

[lavish nexus]

Yes

[dull root]

So I am having the following problem.

We have four complex elements: $\pm \sqrt{\frac{3}{2} \pm i \frac{\sqrt{7}}{2}}$ and I know that Q ajoint by those four elements is a Galois extension with Galois group $K_4$, the klein four group.

We only have three proper subgroups, so three proper subfields, but the issue I am having is finding the corresponding fixed fields under the Galois correspondence. Since none of the automorphisms are fixing any of the generators, I don't see at all what these fixed fields are.

What is the general technique to finding fixed fields under the Galois correspondence, if I know like in this case all the subgroups of the Galois group

[cloud walrus]

MasakaBakana

[dull root]

In this case what is nice about working with K_4 is that all the subgroups have index 2, so the field extensions over Q are also of degree 2, but it still doesn't show what these fixed fields are

[pastel cliff]

is it at all important to remember that the conjugation map is an automorphism

or it just like

yeah

oop you need that to show that conjugate subgroups are isomorphic

[lavish nexus]

What’s the kernel of conjugation

[pastel cliff]

identity right?

[lavish nexus]

yeah

also not hard to show it’s onto

[next obsidian]

It’s its own inverse

You only have to show it’s a homomorphism

[pastel cliff]

idk if this is silly to ask but

class has been a lot about normal subgroups

is the notion of a group being normal at asll useful

[next obsidian]

Normality is a property of subgroups

[pastel cliff]

ah nvm then

[next obsidian]

It doesn’t make sense outside of that context

[pastel cliff]

yeah i was tryna think of it but idk

[next obsidian]

Because you have nothing to conjugate by

[lavish nexus]

normal means you can take quotient

[next obsidian]

Yeah I mean this is a relative notion

It’s about how H lives inside G

[dull root]

Ignore normal subgroups, deal only with finite simple groups

[next obsidian]

Hurbed

Dead subject

[pastel cliff]

is there more to elaborate about this

cuz like i know that it's about that

but idk how the actual act of conjugation translates to that if that makes sense

[lavish nexus]

You want the set of cosets to have a group structure

[next obsidian]

It ends up translating exactly to the statement that H is a kernel for G

Normal <==> is a kernel of some map

[lavish nexus]

In order to have a group structure multiplication must be well defined using any representatives

then a little manipulation will show for {gH} to be a group the subgroup H has to be normal

And given a hom f, because G/kerf is iso to a group, kerf has to be normal

[bright marsh]

quick question, if i wanted to prove an isomorphism

and i already have a homomrophism, what adiditonal property do I have to show?

[next obsidian]

Just that it’s bijective

Assuming you’re dealing with algebraic objects

(Groups, rings, modules,, vector spaces, etc)

[bright marsh]

bijective as in

same number of elements?

[lavish nexus]

It is onto and ker = 1

[bright marsh]

o shit aight word

does these two properties imply isomorphism? i remember showing an isomorphism from like a dihedral group to like a symmetric group, D8 to S4 or something, and there was one special property we had to prove in particular

like in D8, ab^3 = ba or something

[lavish nexus]

D8 is not iso to S4

[bright marsh]

oh i mtrippin

leem look

[lavish nexus]

it is a subgroup of S4

[bright marsh]

oh 8 element subgorup of S4

yea

does showing bijective and homomorphism automatically satisfy these so called "hidden" properties of some groups?

for some reason, i'm not entirely intuitively convinced

[lavish nexus]

Yes

[bright marsh]

ohhhh

dang that's crazy

wait how does it do that though?

[lavish nexus]

It’s the definition

[bright marsh]

i know it's gonna be tough to show me

[lavish nexus]

Iso is a bijective hom

[bright marsh]

oh is it the homomorphism applies to ALL elements defn?

ah ok

[next obsidian]

The map is injective and surjective

You can just find a set theoretic inverse

[void cosmos]

in modules

is phi(0) =0 ?

iu am trying to show that

if phi:A-->B is a hom

then phi(A_tor) is a subet of B_tor

?

[next obsidian]

Yes

Every map of modules is a group homomorphism

[void cosmos]

so

this proof is right:

r1phi(a) = phi(r1*a)=phi(0)=0

rightt?

where r1 is the "torsion" of a

or like the element that fucks up a

@next obsidian

[next obsidian]

No

You use additicity

phi(0 + 0) = phi(0)

But also phi(0) + phi(0)

So you just subtract a phi(0) from both sides

[void cosmos]

i meant for proving torsion

[void cosmos]

.

subset*

[next obsidian]

Oh

I guess yeah

[void cosmos]

cool

bro

[bright marsh]

i have another quick question, for groups with |G| < 6, how can we conclude that every non identity element has order 2?

[next obsidian]

We can’t

[void cosmos]

idk

[bright marsh]

for |G| < 6 and |G| \neq 4

https://math.stackexchange.com/questions/1216827/show-that-every-group-g-with-g6-is-abelian

[next obsidian]

It’s false

[bright marsh]

im just working off of this

oh

shiet

wrong soln rip

[next obsidian]

But the main thing they’re using is probably lagrange

[void cosmos]

100% u use some fumdanetal theorem stuff

yea

and lagrane

[next obsidian]

Or something close to it

If g is in G

Then |g| divides |G|

This handles every group of prime order

It says groups of prime order are cyclic

So you only have to handle groups of order 4

And you can show those are abelian in a number of ways

[bright marsh]

wait wait could u stop at |g| divides |G|

how do you conclude that?

[next obsidian]

Well..

If you know what lagrange’s theorem is

You can use that looking at <g>

You can prove it adhoc like this though

Define a relation ~ saying that h ~ k if k = g^nh for some n

I’ll prove this is an equivalence relation

h ~ h is trivial taking n = 0

It’s reflexive because if h ~ k then k = g^nh, but then g^-nk = h so k ~ h

It’s transitive because it k = g^nh and r = g^mk

Then r = g^{n+m}h

So we can look at the equivalence classes of ~, this just means subsets of G defined like [h] = {k in G, h ~ k}

These partition G, so G is the union of these, and distinct ones don’t intersect

Note that for any h in G, that h is in [h]

So the union is G

Now if [h] and [k] intersect then take some r in [k] such that r is also in [h].

So that h ~ r.

But for any s in [k], k ~ s, and k ~ r, so we know that r ~ s. Now by transitivity as h ~ r we know h ~ s

So now s is in [h], so [k] < [h]

[bright marsh]

👀

[next obsidian]

Now run this same argument but like the other direction and you see [h] < [k] so they’re equal

Now the cool part

Note that the size of [h] = |g| for any h

Indeed, [h] is seen to be the set {g^nh} for n in Z

But you can see that g^nh = g^{n + |g|k}h for any integer k

So there’s at most |g| distinct elements, but h, gh,.., g^{|g| -1}h are all distinct

So there’s exactly |g| elements

So now, we know that G is the union of all the [h], but these are all disjoint

[bright marsh]

lemme screenshot this, gonna digest this for like an hr

[next obsidian]

So |G| = Sum |[h]| = Sum |g|

So |G| is just a multiple of |g|!

[void cosmos]

how do u show

that f(M) = M_tor

is a left exact functor

for category of R-modules

[next obsidian]

Just like

Do it by hand

[void cosmos]

so

suppose M-->N-->0 is exact

[next obsidian]

That’s not enough

You need one more module

[void cosmos]

wdym

okay

[next obsidian]

To show left exact

[void cosmos]

yea yea mb

let D be a module

M-->N-->D-->0 is exact

then 0-->M_tor-->N_tor-->D_tor

is exact

right?

im sorry if im wrong

[next obsidian]

Btw

I don’t think this is even true

Lol

[void cosmos]

it is

left exact

[next obsidian]

Like in particular

[void cosmos]

but not right

[next obsidian]

Oh left exact!

Sure

[void cosmos]

yes

[next obsidian]

So no

Left exact says

0 -> M -> N -> L

Then 0 -> M_tor -> N_tor -> L_tor is exact

[void cosmos]

okay

[next obsidian]

It fails exactly for surjectivity which is what I came up with a counterexample to

[void cosmos]

okay so

ssuppose this is exact

M-->N (f) and N--> L (g)

this means what

ker(f) = im(g) right?

nope the other way around haha

[next obsidian]

It means that M -> N is injective

And the image of M -> N is the same as the kernel of N -> L

[void cosmos]

yea

so

i already have

1 sec

so

i need to show what

M_tor-->N_tor image is the same as N_tor --> L_tor kernel?

[next obsidian]

Yes

Assuming that you know the kernel = image thing without the _tor

[void cosmos]

yup

okay so suppose M_tor-->N_tor is f dash and N_tor --> L_tor is g dash for example umm

they do have to be the same maps right?

not f dash

im sorry this is fuzzy for me

[next obsidian]

I mean you’re the one that has to show this is a functor

So you’re the one defining what it does to maps

But anyway yes it’s just the same map

But restricting the domain

[void cosmos]

yo

if 1-->G_1-->G_2.-->....---G_n-->1 is exact

then why is product(i=1 to n) |G_i|^(-1)^i is 1

do i do inequalities?

[next obsidian]

Umm pretty sure it should become 0

Also it shouldn’t be a product

Or oh Hurb

[void cosmos]

i did an example

in y head

it came 1 😦

fuck

[next obsidian]

No

Nvm

This is different from what I was thinking

[void cosmos]

so its right?

[next obsidian]

I thought you were doing the sum

Idk maybe

[void cosmos]

i thought like

[next obsidian]

I’d prove it for like

[void cosmos]

u can get some inequalities

[next obsidian]

I’d do it by induction or something

[void cosmos]

from the maps 1-->G_1 and G_n-->1

[next obsidian]

Yeah idk

It probably isn’t hard

[void cosmos]

can u think with me on it

?

[next obsidian]

I really don’t want to lol

I’d try to induct

[void cosmos]

np

ty

anyone else?

[void cosmos]

.

[lavish nexus]

Is there some generalization of the action of taking triple closure

Like V(I(V(I)))= V(I)
I^ece = I^e

and similarly for Galois group

[latent anvil]

@lavish nexus do you know what adjoint functors are?

I think you could think of this as the triangle identities for the unit and counit of an adjunction

[latent anvil]

So at a less abstract level, a "galois connection" is a pair of posets P, Q and monotone maps f : P -> Q, g : Q -> P such that f(x) <= y iff x <= g(y) for any x in P, y in Q. The extension/restriction is an example of this when P, Q are posets of ideals, f(I) = I^e, and g(I) = I^c, because I^e <= J iff (image of I) <= J iff I <= J^c. The galois group example and the algebraic geometry one are also galois connections, but you need to take the opposite of one of the posets because you have anti-monotone maps, so eg you have J <= I(C) iff C <= V(J).

If you have a galois connection then f(x) <= f(x), so x <= g(f(x)) and similarly f(g(y)) <= y. Taking y = f(x) we see f(g(f(x))) <= f(x), but also because f is monotone x <= g(f(x)) implies f(x) <= f(g(f(x))), so we have f(x) = f(g(f(x))). Similarly g(f(g(x))) = g(x)

The relationship with adjoint functors is that posets are the same thing as categories where

1. There's at most one map between any two elements
2. Every isomorphism is an identity.
   Then functors between these categories are the same as monotone maps and an adjunction is a galois connection

[next fulcrum]

Let's replace these groups (right? I guess) by vector spaces, and consider the case n=3:
0→G_1→G_2→G_3→0
Then the (short-)exactness says G_2/im(G_1→G_2) is isomorphic to im(G_2→G_3), and since im(G_1→G_2)≃G_1 and im(G_2→G_3)=G_3 we get dim G_2-dim G_1 = dim G_3, or that ∑_i(-1)^idim G_i = 0.
Ofc if you consider the product of exp(dim G_i)^(-1)^i you'll get a similar formula, which tells essentially the same thing!
Is that what you've expected, Moamen?

[barren sierra]

dumb question

how is I_k an ideal of S

wait nvm I'm dumb

I was going to say "can you not multiply any polynomial by x^k and not be able to write it in terms of the desired form"

but x^k is in R and importantly is not in S

wait no but x^k y is in S and 1 is in I_k but x^k y* 1 = x^k y is NOT in I_k right?

and more importantly x^k y * y = x^k y^2 is also not in the ideal right?

[lavish nexus]

yes but I have not heard of the next part
I'll look them up tyvm

[dire summit]

i've got a homework question on getting the splitting field of a polynomial over various fields, i'm still not quite clear on how to obtain a splitting field in general tho

here's the problem in question:

Find the splitting field of $x^4+x^2+1$ over $\mathbb{Z}_2$, over $\mathbb{Z}_3$, and over $\mathbb{Q}$.

[cloud walrus]

Snodlop

[dire summit]

i know the roots of the polynomial are +- (-1)^(1/3) and +- (-1)^(2/3), i'm just not sure what to do with this information from here

my notes are not very helpful rn lol

[barren sierra]

yea can someone help me with part b here

I have some vague ideas but nothing too concrete

cause I can't say every element of I_k takes the form f(x, y) * (a_0 * y + a_1 * x * y + ... a_k-1 * x^k-1 * y) for constants a_i in F and some f(x, y) in S right?

also is this not just 2nd isomorphism theorem? (diamond)

nvm I'm dumb it isn't

is it just first isomorphism theorem?

[thorn delta]

It might follow from second isomorphism theorem, but it shouldn’t be any harder to prove it from the first isomorphism theorem really

Just find surjections M —> N2 whose kernel is N1 and M —> N1 with kernel N2

[barren sierra]

ye

I love the name of that sticker lmfao

[full panther]

how do you find a basis for a derived lie algebra given some basis for it? (i.e. basis for [L,L])

can you just do bracket of the basis elements with any element in the lie algebra or something like that

[void cosmos]

I already proved in the lsdt exercise that for vectors this is true

Idk how to do the product

[next fulcrum]

Oh alright

Let me LaTeX my solution, wait a moment

Well it'd be faster to write down

I don't think you can obtain something like Π_i(-1)^i dim G_i = 0

@void cosmos

Depending on $r$ being odd or even, the LHS becomes either $e^{-1}\prod_ie^{\dim G_i}$ or $\prod_ie^{\dim G_i}$

[cloud walrus]

shota___inoue

[void cosmos]

How do i get to the order of G

?

[void cosmos]

e^(dim(Gi)) what does this mean

anyone?

[next fulcrum]

I didn't understand why the order appears in your formula ..

And are the G_i's in your formula just groups? Or is there some additional structures

[void cosmos]

yes

[next fulcrum]

This is the exponential power (e is the Napier number)

[void cosmos]

yea yea ik

[next fulcrum]

Hmm

[void cosmos]

i meant how does it relate to order

[next fulcrum]

I didn't give the answer, but I've just made a formula for vector spaces, not for groups

so I should probably change the way to tackle this

For example, consider a group homomorphism f: G→H and the induced short exact sequence
0→ker(f)→G→im(f)→0

I want to yield the formula |G|=|ker(f)||im(f)| to get Π_i|G_i|^((-1)^i), where G_1=ker(f), G_2=G, and G_3=im(f)

When G is finite this is a consequence of the Lagrange theorem and the isomorphism theorem

[void cosmos]

yes

i got to that too

[next fulcrum]

And I suppose that your formula only makes sense for finite groups, so that'll be the answer

[void cosmos]

yes

how do i equate this to 1 XD

[next fulcrum]

You can use induction on n to produce the same result for any positive integer n!
But when you get stack feel free to ask here

[void cosmos]

okay so i got confused now

lets wrap this up

from first isoo and exactness

we get |G_i|= ker(fi)im(fi) right?

which is ker(fi)ker(fi+1) right?

[next fulcrum]

If you put the |...| on the RHS yes

[void cosmos]

okay

[next fulcrum]

Oh

Wait, what's g?

[void cosmos]

f_i+1

[next fulcrum]

Are you writing the map G_i → G_i+1 as f_i+1? Then |G_i|=|ker(f_i)||im(f_i+1)|

[void cosmos]

yes

then what

[next fulcrum]

Hmm

I guess the formula we want to get is
Π_i|G_i| = Π_j|G_j|
where i runs all the odd numbers and j runs all the even numbers appearing in the indicies of G_i's

and if you substitute the kernel identify

And divide both sides by a certain nonzero number

[void cosmos]

i am not following you tbh

how can i get to this

i have no idea haha

[next fulcrum]

What you get is the identify |ker(0→G_1)|=|ker(G_2n→G_2n+1)| (if the sequence is ...→G_2n+1→0) Ummmm

Sorry I'm not sure what I'm even doing haha

I'm on the train coming back home, so when I have a pen and paper I'll write down the solution I'm having in my head!!

Not a type of calc that can be down in my head lmao

[void cosmos]

okay

np

consider the functor F in the category of R-modules
F(M) = M_tor
why does this fail to be right exact for the sequence Z --> Z-->Z/nZ --> 0

is it because img(f) from Z_tor to Z_tor is Z_tor but kernel of the natural map to the quotient tor is 0?

[void cosmos]

elements of V3 are of the form rV2+V1 right?

so

they are of the form (a,b,0,c)(x,y)+(x,0)

right?

(a,b,0,c) is the matrix

[void cosmos]

a,b,c in C and x in C

im stuck with Hom(V2,V1)

is Hom(V2,V2) =V2

?*

[delicate orchid]

I’d still write V1 here instead of (x, 0)
And I believe it just acts like scalar multiplication as the top variable is quotiented out

so (a b 0 c)(x y)+V1 = (ax+by cy)+V1 = (0, cy)+V1 as ax+by is just a complex number

I hope that’s what you were after lol

I dunno where Hom comes into it

[void cosmos]

Question B