groups-rings-fields

[sonic coral]

i’m kinda confused bc the polynomial is irreducible already

so it has an irreducible factor of degree n

oh i see k might not be n

[fading acorn]

<@&268886789983436800>

[little ocean]

ik it's pretty stupid, but how do i prove:
For every $k \in \mathbb{Z}: o(g^k) = \frac{n}{\text{gcd}(k,n)$ and $<g^k> = <g^{\text{gcd}(k,n)}$

[cloud walrus]

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

[little ocean]

Damn

For every $k \in \mathbb{Z}: o(g^k) = \frac{n}{\text{gcd}(k,n)$ and $<g^k> = <g^{\text{gcd}(k,n)}$

[cloud walrus]

Nico

For every $k \in \mathbb{Z}: o(g^k) = \frac{n}{\text{gcd}(k,n)$ and $<g^k> = <g^{\text{gcd}(k,n)}$
```Compilation error:```! File ended while scanning use of \frac .
<inserted text> 
                \par 
<*> 956940045711999046.tex
                          
I suspect you have forgotten a `}', causing me
to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.```

[little ocean]

For every $k \in \mathbb{Z}: o(g^k) = \frac{n}{\text{gcd}(k,n)}$ and $<g^k> = <g^{\text{gcd}(k,n)}$

[cloud walrus]

Nico

[little ocean]

For every $k \in \mathbb{Z}: o(g^k) = \frac{n}{\text{gcd}(k,n)}$ and $<g^k> = <g^{\text{gcd}(k,n)}>$

[cloud walrus]

Nico

[little ocean]

Oh yeah, and also we're talking about the group <g> of finite order n

[twilit citrus]

okay sanity check, the number of irreducible monic polynomials in polynomial ring $\mathbb{F}_5[X]$ is 10 right?

[proud vigil]

it might be easier to prove that o(g^k) = LCM(k, n) / k

that's more of an intuitive argument imo

then you can use the fact that GCD(k, n) LCM(k, n) = kn for any two integers k, n

[cloud walrus]

anonloki

[balmy vector]

i'm wondering - if G and H are groups for which there exists an injection and a surjection G->H, must G and H be isomorphic?

[rocky cloak]

A polynomial ring always has infinitely many irreducible monic polynomials.

There are 10 quadratic ones though

[rocky cloak]

Take G = Z/2 x (Z/4)^N and H = (Z/4)^N

They're not isomorphic, but you have injections and surjections between them in all directions

[kind temple]

consider the case when G and H are finite. from the surjection G —> H and the first iso theorem, G/ker = H. but since there is an injection G —> H, then |G| = |H|, so the kernel of the surjection must be trivial, therefore the groups are isomorphic

[crystal vale]

if K is a field extension of F and if a \in K, such that [F(a) : F ] is odd then F(a^2) = F(a). So i am trying to write a = p(a^2) for some polynomial p over F, but no i don't get anything useful, any hint?

[quiet pelican]

What is [F(a^2): F]?

[crystal vale]

that's what i have to find

[quiet pelican]

[F(a): F] = [F(a): F(a^2)][F(a^2): F]

[crystal vale]

yes i know

but [F(a) : F ] is odd so how it helps me?

i have to show [F(a) : F(a^2) ] = 1

[quiet pelican]

There are two possible things [F(a): F(a^2)] can be. What are they?

Hint: ||it’s a quadratic extension||

[crystal vale]

k is odd, not prime

k = [F(a) ; F]

[crystal vale]

i don't what is quadratic extension

[quiet pelican]

I know

[quiet pelican]

What is the degree of [Q(sqrt(2)): Q(2) = Q]?

[crystal vale]

2

[quiet pelican]

Why?

[crystal vale]

because minimal polynomial over Q which has root \sqrt 2 is x^2 - 2

[twilit citrus]

oops yea that's what I was asking, thanks

[quiet pelican]

Do the same for the example we want

[crystal vale]

ah, thanks i got it, i am dumb

[little ocean]

This may be a bit of a weird question, but I have my algebra exam tomorrow. Is there an 'easy' way to prove Sylow's theorems?

[knotty badger]

i'm also interested in an answer to this Q

[elfin wraith]

I’m pretty sure Jagr has a pretty slick, short proof he’s posted before but I’d be lying if I said I remembered it at all

[rocky cloak]

You mean this?

[little ocean]

I'm mainly struggling with remembering proofs for:
For every prime divisor p of |G|, G has a Sylow p-subgroup
n_p(G) = 1 (mod p)
Every two Sylow p-subgroups are added

Idk if added is the right word, but I mean that if P and Q are two Sylow p-subgroups, then P = g^-1Qg for some g in G

[knotty badger]

normally that's called "conjugate"

[crystal vale]

You can say conjugation

[little ocean]

Yeah, idk much English terminology

[crystal vale]

Have you seen Artin's proof on the Sylow's theorem?

[little ocean]

No

[crystal vale]

You can look at it, I don't know if it is easy or not but i found it interesting ( no induction needed )

[little ocean]

I can't find much about it, but I understand that it's using group actions of some sort

[rocky cloak]

That's the one that is on Wikipedia right?

[little ocean]

I don't use wikipedia

So idk

[crystal vale]

So I proved that if F1 and F2 are isomorphic fields then every vector space V over F_i have same cardinality so Q(√2) over Q(√2) has dimension 1 but Q(√3) over Q(√2) has not dimension 1 otherwise √3 will be in Q(√2).

Is it correct?

I think when I am saying V is vector space over F, do I have to mention their operation too, does if operation change implies dimension different?

[rocky cloak]

I mean, you're sort of assuming Q(sqrt(2)) is contained in Q(sqrt(3)) for this to make sense.

But it is true that any field isomorphism would take sqrt(3) to a square root of 3.

So if you can show Q(sqrt(2)) doesn't have a square root of 3 you'd be good

[crystal vale]

Yes i showed that

But does my argument make sense?

[rocky cloak]

Here under proof
https://en.wikipedia.org/wiki/Sylow_theorems

I don't really like this proof though, to much thinking about what powers of p divide stuff

[rocky cloak]

Like Q(sq3) is not a vector space over Q(sq2), and showing that it can't be made into a vector space just reduces to showing your original problem, that they're not isomorphic

[crystal vale]

no, but i am assuming Q(\sqrt 3) and Q(\sqrt 2) are isomorphic as field, so then i can made Q(\sqrt 3) as vector space over Q(\sqrt 2)

[rocky cloak]

Well, then there's no contradiction. If they're isomorphic then they would both be 1d vector spaces over the same field

[crystal vale]

and then Q(\sqrt 3) has dimension 1 as vector space over Q(\sqrt 2), can't we deduce some contradiction from here?

[rocky cloak]

I don't think so.

It's perfectly possible to put a Q(sq2) vector space structure on Q(sq3) (it's just Q^2 after all). You have to actually compare their field structures

[crystal vale]

okay

[lethal cipher]

Yea, it's basically saying the tail of the polynomial os "Eisenstein like". So what does that mean about the whole polybomial?

[crystal vale]

how do i show if A is algebraic closure of Q in C then [A, Q] is infinite?

okay A is countable

hint

[rapid cave]

by [A,Q] you mean the degree of the field extension?

[crystal vale]

yes

[rapid cave]

A contains every cyclotomic field

well you just need to argue there is an irreducible polynomial of infinitly many degrees

[crystal vale]

why it is enough?

[rapid cave]

If f \in Q[x] is irreducible and alpha is a root then A contains Q(alpha)

[crystal vale]

yes

[rapid cave]

and hence assuming by contradiction that A is finite over Q we get [A:Q] >= [Q(alpha):Q]=deg(f)

[crystal vale]

yes got it

and we can make x^2n + 1 irreducible polynomial over Q for any n

thank you

[crystal vale]

I have no idea, its getting complicated too

[little ocean]

Can anyone explain this to me:
Let R be a PID and M a nonzero finitely generated R-module. Then there exist unique positive integers s and t and unique ideals I_1 >= I_2 >= ... >= I_s, such that M = R/I_1 + ... + R/I_s + R^t

With I_1 >= I_2 mean inclusion and the sum R/I_1 + ... + R/I_s + R^t is the direct sum

[rocky cloak]

Like you want a proof or you don't understand the statement?

[little ocean]

A proof

Because the proof in my syllabus, I don't completely understand

If you want, I can try to write down the proof and tell you the steps I don't understand

[rocky cloak]

That would probably be better yeah

[little ocean]

Let (v_1, ..., v_n) be a finite set of generators for M. We know that M = F/N, where F is a free module of rank n and N a submodule of F. We know that N is free of rank m <= n.
consider the inclusionmap a: N -> F. We know that there exist bases C = (w_1, ... ,w_m) for N and (z_1, ..., z_n) for F, such that a(w_i) = d_iz_i for all i <= m, where d_1 | d_2 | ... | d_m, and all d_i's are unique, spare a unit.
Now we write down F and N (or rather a(N)) by means of their coordinates with respect to the basis D. We get
CO(F) = {(r_1, ..., r_m, r_m+1, ..., r_n)
CO(N) = {(d_1r_1, ..., d_mr_m, 0, ...)
where all r_i are elements of R.
We conclude that M = F/N = CO(F)/CO(N) = R(d_1) + ... +R(d_m) + R^t

This is the proof given in my syllabus

Is it a correct enough proof, first of all?

(Probs yeah, otherwise it wouldn't be in an algebra course)

The main thing that I don't get here is why CO(F)/CO(N) = R(d_1) + ... +R(d_m) + R^t

[crystal vale]

I want example such that [L1L2 : F ] < [L1 : F] [L2 : F ], i think 4 = [ Q( 2^1/2, 2^1/4 ) : Q ] < [ Q(2^1/2) : Q ] [ Q(2^1/4) : Q ] = 8, is it correct?

because [Q(2^1/2, 2^1/4) : Q(2^1/2) ] = 2

[quiet pelican]

Yes

[crystal vale]

can i get any hint?

[quiet pelican]

Given F-bases a_1, …, a_n of L_1 and b_1, …, b_m of L_2, find an F-spanning set of L_1L_2

[crystal vale]

i tried in that way, but somehow i got so notations and in next exercsie i have to show if their gcd is 1 and equality hold, so i am trying in that sense

[quiet pelican]

Don’t think about the second part at all until you’ve done the first part (the way I can think of to do the second part won’t help with the first)

[crystal vale]

okay

[rocky cloak]

So F is just the direct sum of the
Rz_i and a(N) is the direct sum of Rd_iz_i, so F/a(N) is the direct sum of
Rz_i/Rd_iz_i = R/(d_i)

The tricky part is really in choosing this basis

[rocky cloak]

For an easy example, consider L1=L2

[crystal vale]

[little ocean]

Why is F/a(N) the direct sum of Rz_i/Rd_iz_i = R/(d_i)?

My friend says that F is isomorphic to R^n

In this particular construction

[rocky cloak]

In general if C is a submodule of A and D is a submodule of B, then
A(+)B / C(+)D = A/C (+) B/D

[little ocean]

Why

Sounds logical, but why

[rocky cloak]

Well, you can write down a map A(+)B -> A/C (+) B/D and use the first isomorphism theorem for example

[little ocean]

Like this, then?

Excuse my handwriting

[rocky cloak]

The kernel of exactly pairs with a in C and b in D, so C(+)D

[little ocean]

And so a(N) = d_1R + d_2R + ... + d_mR?

Thanks so much

So in particular, a(F) = R^n and so we can divide every R by (d1), so a(F)/a(N) = R/(d_1) + ... + R/(d_m) + R^{n-m}

?

[rocky cloak]

I'm guessing K is Q adjoin the mth roots of unity (?)

It's clear that the Galois group is a subgroup of (Z/m)^x, but should "todlers" really know why they're equal? / is there an easy proof?

[little ocean]

Thanks so much!

[thorn jay]

I mean you can explicitly write down the maps and show they're automorphisms no?

[rocky cloak]

Maybe you can, I'm not sure I can

Liker writing down the maps is fine, but how do you show they're well defined?

[thorn jay]

because you've got a Q-algebra homomorphism Q[x] -> K sending x to \zeta^k, and then you show that it factors through K \cong Q[x] / (x^m - 1) if k is coprime to m

[rocky cloak]

K is not Q[x]/(x^m - 1) though

[thorn jay]

oh yeah I forgot lol

ignore me then

[rocky cloak]

Like whatever you do it has to be very specific to Q, since it's not true over other fields

[thorn jay]

what fails in other fields?

[vocal pebble]

I don't fail in other fields(maybe I do), but I think he is talking about the galois group of K(zeta_n)/K not being (Z/nZ)^*

Example would be K=Q(zeta_n) itself

[supple ice]

you’re right here K = Q(ζ_m)

every Q-automorphism must send ζ_m to another root of X^m−1 so σ(ζ_m)=ζ_m^a for some a mod m
σ must preserve the order so ζ_m^a is primitive iff gcd(a,m)=1 giving an injective map Gal(Q(ζ_m)/Q) → (Z/mZ)× for surjectivity for any a with gcd(a,m)=1 ζ_m ↦ ζ_m^a extends (uniquely, since ζ_m generates K) to a Q-automorphism of Q(ζ_m) hence the map is onto and Gal(Q(ζ_m)/Q) ≅ (Z/mZ)×

[coral steeple]

I'm stuck. I've found that D^3/K has the form in the second screenshot, but I don't know how to find its size.

and the e_i bar are the cosets of the standard basis vectors

Well I guess I also know that it suffices to take the product of the sizes of the two direct summands, so the question is how to find those

[quiet pelican]

M does not look like that?
Did you mean it’s generated by those
Like, as written that second screenshot doesn’t make sense

[coral steeple]

hm what's wrong with that? It's bracketed kind of badly

better?

[quiet pelican]

Actually looking at that I agree that it’s plausible but I’m also not sure that’s the part of the working you want to extract

[coral steeple]

yeah I also have the Smith normal form

I have heard you can just take the determinant of the SNF but have no idea why that should work instead of finding this form for M

[rocky cloak]

This assumes zeta_m and zeta_m^a have the same minimal polynomial, which is pretty hard to prove

[But I agree this is the only part that isn't essentially oblivious]

[rocky cloak]

One possibility is that the field already has mth roots. Another could be the the cyclotomic polynomial isn't irreducible (finite fields for example)

[coral steeple]

Oh sorry I'm just realising that you were pointing out that I should indicate that these are cyclic modules generated by these factors

[supple ice]

the minimal polynomial of a primitive m-th root of unity is the cyclotomic polynomial Phi_m whose roots are exactly the primitive m-th roots so if gcd(a,m)=1 then zeta^a is also primitive hence it is another root of Phi_m so it has the same minimal polynomial as zeta

[rocky cloak]

Yes, if you can prove that the cyclotomic polynomial is irreducible you're done

(But that is hard as far as I'm aware)

[naive lance]

hello, i need to find a citable source for the following statement, if sb knows where i find it it would be cool. maybe it is in [1] Hilgert, J., Neeb, K.-H.: Structure and Geometry of Lie Groups. Springer, New York, (2013).

thank you in advance

Proposition. Let G be a connected non-compact Lie group that is a semisimple Lie group and
U:G∋g↦Ug∈B(H)
(B(H) being the set of bounded operators A:H→H) a continuous unitary representation over the finite-dimensional Hilbert space H

. The following facts hold.

(a) U

cannot be faithful.

(b) If G
is a simple group or, more generally, if G does not contain non-trivial proper normal closed subgroups, then U is the trivial representation U:G∋g↦I.

[supple ice]

not hard but not obvious either

[rocky cloak]

I think it's pretty hard, but maybe you know a trick

[crystal vale]

I think no, because if there exists such automorphism then since -i√3 also a root and it maps to -1 - i√3, but under such automorphism roots of f should maps to roots of f, but -1-√3 is root of f.

Correct?

[scenic pendant]

helloo ^^ was wondering if anyone could explain to me what a lie group is? my professor kinda brushed past it, he ended the lecture with “oh yh u can think of SO(3) as a sphere” and smth smth manifold, i have genuinely no idea what any of that means ;-;

[quiet pelican]

Do you know anything about topology?

[scenic pendant]

no not really, like surfaces and weird objects?

i’m taking that next year methinks

[karmic moat]

the bottom line is that a lie group is just a special kind of topological object with a group structure on it

[scenic pendant]

how can an object have a group structure

[karmic moat]

SO(3) is group of rotational symmetries of an object, so it is like a sphere

well first you define an operation on it

in the case of SO(3) the operation is multiplication

[knotty badger]

A group operation on a set X is a map X x X -> X satisfying certain properties

[karmic moat]

you can read the wikipedia article for lie groups. the intro section is nice

[knotty badger]

Instead you could have a manifold M, and a map M x M -> M satisfying certain properties

This is the strategy you take to generalise groups to other kinds of objects

Category theory gives a nice treatment of this with the notion of “group object”

[scenic pendant]

okay so i’m putting everything i’m doing with group elements and mapping it to an object

with SO(3) it happens to become a sphere

[scenic pendant]

i’ve never even heard of that wowie

[knotty badger]

It allows you to generalise the notion of “group”’ to any category

[scenic pendant]

i like the sound of that i’ll look into it

[knotty badger]

For example, a Lie group is a group object in the category of smooth manifolds

A topological group is a group object in the category of topological spaces

An ordinary group is a group object in the category of sets

[scenic pendant]

to me why i think im so confused is that i cant bridge the connection in my mind

i was tryna youtube an animation but they have just made me so confused

[karmic moat]

it would probably be useful to learn some topology first

[knotty badger]

Yeah category theory is in large part designed to help with bridging connections between different fields of math

[scenic pendant]

why is it specifically a sphere and not any third dimensional object

[scenic pendant]

i would love to but this module is weirdly constructed to bring up manifolds without us needing any topological knowledge, tryna not go on a big side quest before my exam

[karmic moat]

in my mind i think of SO(3) as being defined as the group of rotations of a sphere but maybe that isn't accurate

[karmic moat]

your class has a topic on lie groups without a topology prerequisite?

[scenic pendant]

they kinda drop it loosely to make u understand matrix groups better but it actually just made me more confused, i think my only option is to go down this rabbit hole sadly

[karmic moat]

that seems kind of ridiculous

[knotty badger]

Roughly speaking a manifold is a space you can do calculus on

[scenic pendant]

it’s not so much a topic, a learning objective is “understand that matrix groups/lie groups have an underlying geometric structure”

[knotty badger]

Well SO(2) is a Lie group

That might be a good place to start

[karmic moat]

i wonder how far one can get before having to use topology

[scenic pendant]

i just finished a vid that explained 2-dim manifolds which is cool stuff i liked it

[scenic pendant]

i shall

i didn’t realise this learning point on the spec was actually huge

i thought it was a fun fact 💔

[karmic moat]

https://en.wikipedia.org/wiki/3D_rotation_group#Topology

[knotty badger]

The theory of Lie groups is quite deep

[karmic moat]

maybe useful for intuition

[scenic pendant]

thank u sm i will read this

[quiet pelican]

you can make up a lot of theory if you’re lying about groups /silly

[rocky cloak]

SO(3) is not actually a sphere. It's homeomorphic to real projective space.

[knotty badger]

SU(2) is the sphere

[scenic pendant]

had to google what homeomorphic meant

thank u for the clarification guys 🫶

also any book recs on the matter? preferably accessible as i’m just an undergrad, i wanna go down this rabbit hole but maybe not for my exams

[quiet pelican]

Isn’t it S^3 not S^2
(If you say the sphere, I’d assume it’s 2d)

[karmic moat]

https://arxiv.org/pdf/2201.09397

[knotty badger]

It’s S^3

[karmic moat]

the first few chapters of these notes should be okay for now

i think later on in the notes etingof starts skimping out on details which is kind of annoying

but the first few chapters are fine for your needs imo

[scenic pendant]

thank you sm!! 🫶🫶

[thorn jay]

Just use sheaves on sites smh

[karmic moat]

https://tenor.com/view/true-feel-that-thats-true-gif-7410050725709839510

[rocky cloak]

If it's your goal to avoid it you can probably go your whole life

[quiet pelican]

Just don’t do maths

[karmic moat]

though doesn't sound like a very fulfilling life

[scenic pendant]

circle = SO(2) = U(1) because circles/rotations = multiplications of e^iθ 🤯 guys this is so cool why did they not teach me groups like this to start with

never knew it was so geometric

[knotty badger]

groups are very geometric!

they originated as symmetries after all

[scenic pendant]

it’s so coooool

maybe i don’t hate this module after all

[knotty badger]

:D

[round anvil]

@molten atlas

we can use this channel?

[molten atlas]

Ok

Thx for the help

[molten atlas]

yes

[round anvil]

Permutation group has to do with the arrangement of elements

Since they are generic elements, I will call them a_1, a_2, ...

[molten atlas]

ok

[round anvil]

in this order

and we can shuffle the elements however we want

In the image, there is one possible shape, but there are many more ways to shuffle the elements.

[molten atlas]

what do the arrows mean?

[round anvil]

To which 'place' in the order will the elements go after being shuffled

[molten atlas]

ok

[round anvil]

There is a proof that guarantees us there are n! ways to shuffle n elements

[molten atlas]

ok

[round anvil]

We can take all these ways of shuffling and call them: shuffle 1, ...

And then, store everything in a set

[molten atlas]

how do you store the ways of shuffling? like as lists

[round anvil]

like functions

note that each shuffle can behave like a function. See in the example, a2 was moved to a1 ...

like f(a2) = a1

[molten atlas]

so is each change its own function?

[round anvil]

yesa

yes

[molten atlas]

like if i swap a4 and a1 then swap a2 and a3 would i have a set of f(a1) and f(a3)

[round anvil]

but

[round anvil]

With this, you have only one permutation, and that permutation is just a function

[round anvil]

is just one element of that set

we can work with just 3 elements to simplify

[molten atlas]

its ok i get what you mean now

[round anvil]

here 2 permutations of 3 elementos

each of that permutation is a function

with 3 elements, we have 6 permutations

so, if we want a set with all permutations of 3 elements, we have a set with 6 elements

ok?

[molten atlas]

ok

i get it

[round anvil]

Since each permutation is a function, we can operate on these permutations using function composition. And thus, we have the permutation group.

[molten atlas]

how would that work function composition on the permutations?

[round anvil]

look at that permutation 1 and permutation 2. Note that permutation 1 loads the second element to the third position and permutation 2 moves the second element to the first position.

so, this is the result of the composition of the two permutations

[molten atlas]

ok thank you

[round anvil]

A good question you can ask yourself is whether these permutations are commutative.

You already understand the group of permutations, so we can ask ourselves about its subgroups (the alternating subgroup)

Look at the red arrows. When they cross, we will count one inversion. Okay?

If a permutation has an even number of inversions, we will call it an altered permutation.

The alternating group, then, is the group formed by permutations with even inversions.

[molten atlas]

ok i get it now

[round anvil]

🙂

[molten atlas]

so in our example the composition would be in the alternating subgroup?

[round anvil]

yes

Here you got a good grasp of the group's intuition, and you can dive deeper into books in cycle notation, where you will understand all of this better.

[molten atlas]

ok thx

[crystal vale]

b, I done when char K ≠2, how do I show if K(√a) = K( √b ) then √(ab) in K when char K = 2?

[lapis latch]

is that even true for char 2. Seems like K=F_2(x) provides a counterexample with K(sqrt{x+1}) and K(\sqrt{x})

[wicked patio]

Are those extensions equal? I don't see it

[lapis latch]

\sqrt{x}+1 squares to x+1

[wicked patio]

Oh, strange

But can you actually write one of those as a K-linear combo of 1 and the other?

Suppose the fields are just isomorphic, and there's no larger field L that can contain both

[lapis latch]

Yeah just semantics of field theory. You can think of field extensions which are isomorphic as equal. If we took any L to include both square roots and the extensions K(sqrt{x}) K(\sqrt{x+1}) to live inside L then they are equal. Basically we’re just saying a square root of x+1 exists within K(\sqrt{x}) and vice versa.

[wicked patio]

Agreed if there exists such an L

[lapis latch]

L=K[t]/(t^2-x)

Square root symbol no good

[wicked patio]

ah yeah ok I see

oof

so the linear combo is just √(x+1) = √x + 1

[lapis latch]

yep

[sweet ether]

@rough kettle

[crystal vale]

sorry i don't get it

[rocky cloak]

If K has characteristic 2, then K(sqrt(x+1)) = K(sqrt(x)+1) = K(sqrt(x))
even though sqrt((x+1)x) is not in K

[crystal vale]

i see

[crystal vale]

How | L | < |S | helps me here?

I think it helps me make a injective function L -> S such that restricted to M is identity

[quiet pelican]

Yeah

[thorn jay]

I really dont get this

Is \mathcal{A} the set of all subsets of S containing F with a distinguished field structure such that F is a subfield of it??

[quiet pelican]

Yes

This is slightly messy because you’re trying to avoid issues where A is not a set

[delicate orchid]

I agree it's not worded very well

[thorn jay]

Yeah but i see now

The minimum size thing is just making sure you get a representing element of every algebraic extension

The cardinalities of those are bounded by |(F x N)^n| so you take it to be strictly bigger than either |N| (in the case F is finite) or |F|

[sour rain]

I'm a little confused by this exercise. My solution is: consider the embeddings of K over F as characters from K into the algebraic closure of F. Then, since Tr(alpha) = sum of embeddings, which are necessarily distinct, we cannot have this sum equal to 0 because of the linear independence of characters (all coefficients of the linear combination are 1). Thus there exists some input alpha for which the trace is nonzero.
but where did i use the fact that K is galois over F?

as far as i can tell, this argument works for any finite extension K over F

which apparently is not true because if K is inseparable then the trace is 0...

[quiet pelican]

which are necessarily distinct

Try looking at this for the “standard” example of an inseparable extension adjoining a pth root of t to F_p(t)

[sour rain]

hold on im really confused, this is how dummit and foote introduces the norm (and the trace is introduced as just the sum instead of product)
dummit and foote says "set" of coset representatives so why would we ever be taking two of the same embedding in the first place?

[quiet pelican]

You’ve only defined what the norm (and presumably, the trace) actually mean for extensions contained in a galois extension
Hence, you haven’t defined it for an inseperable extension

[sour rain]

oh

ok

[crystal vale]

Yes, so I need A to be a set? For applying Zorn's lemma I need A to be a set or can I apply on collections ?

[quiet pelican]

You need it to be a set

[crystal vale]

so then how do i show \math{A} a set?

[quiet pelican]

It’s a subset of the power set of S

(Actually technically not but like
Usually if something is smaller than a set which you can give explicitly, it’s a set)

[crystal vale]

i see

thank you

[crystal sundial]

Hi, I am reading the following proof of the Cauchy Scwarz inequality. I understand the technical details of the proof, but don't understand the motivaiton of it. Namely, I don't see why it is natural to consider the function <x-tz, x-tz>. I have seen a similar technique being used within this text (Folland) to prove an orthogonal decomposition result for Hilbert Spaces. Any insight would be greatly appreciated.

[south patrol]

(Not groups/ring/fields - maybe #linear-algebra or #real-complex-analysis idk)

[cerulean swan]

I'm working on this problem. By a theorem I have that $\mathcal O_K\cdot\mathcal O_L\subset \mathcal O_{KL}\subset \frac{1}{gcd(D_K,D_L)}\mathcal O_K\cdot\mathcal O_L$. I have shown that we have $\mathbb Z[\sqrt 3,\sqrt{23}]\subset \mathcal O_{KL}\subset \frac{1}{4}\mathbb Z[\sqrt 3,\sqrt{23}]$. By following the hint I see that we cannot have $\frac{1}{4}\mathbb Z[\sqrt{3},\sqrt{23}]$. However, I am struggling to further determine it from here...

[thorn jay]

looks like multivar analysis

[cloud walrus]

Sleepybear

[cerulean swan]

I think I got it actually

[white oxide]

Can I please have a hint for the first part of 12? I’m not sure whether I should try to show the coefficients of the polynomial they gave me are G-invariants

[quiet pelican]

Show that the entire polynomial is a g-invariant
This entails that each coefficient is

[rocky cloak]

I guess a hint is that you can extend the action of G to act on A[x]

[dull ginkgo]

One of those fuckin things I can’t extend as much as I try

[waxen plover]

Can anyone help me with problem 12 please please pleaseeee?

[kind temple]

,rotate

[waxen plover]

anyone up to help now?

[velvet hull]

hint: ||is there anything that can be said about the set of all elements that is sent to their inverse?||

[waxen plover]

Tried to think if it can form a subgroup

But I can’t prove it

[velvet hull]

what do you have?

[waxen plover]

T^2 =I for the restriction

If it is a subgroup, then it is Abelian

I tried taking all product combinations with powers 1, using the set of elements. The count will exceed the order of the group

Some product must be in the subset

That much I can say

Nothing more

[velvet hull]

can you recall what the definition of a subgroup is?

what do we want to show?

[waxen plover]

It is a subset which is a group

I’m unable to prove that subset is closed under the operation of the group

[velvet hull]

so what do we want to show

can you spell it out for me

[waxen plover]

that the set of all elements which get sent to their inverses is the whole group

That’s the end goal

T(G) = G

[velvet hull]

we want to show that if T(x) = x^-1, and T(y) = y^-1, then T(xy) = (xy)^-1

[waxen plover]

Yes

It is (yx)^-1

Stuck there to be precise

[kind temple]

i don't think so

[kind temple]

,rotate

[waxen plover]

Sorry I don’t know how to rotate before sending 😅

[waxen plover]

What next?

I want to investigate the maximum size of the subset of elements which get sent to their inverses.

To arrive at 3/4

Why only three quarters?

But I gotta solve the problem first

[kind temple]

you are showing that the set K = {x in G : Tx = x^-1} is a subgroup
so if x and y are in K, we want to show that xy is in K, that is, T(xy) = (xy)^-1

[waxen plover]

Yes

I want to show that

[waxen plover]

And I’m here

I’m sorry if I’m stuck at something silly

Happens to me sometimes

And I’m really clumsy today

Thank you for your patience

[kind temple]

no, ur good. i don't remember the trick. im thinking too

[waxen plover]

I forgot to put detergent in washing machine, after I put I forgot to close the lid for a good while. That’s how clumsy I am today 😭

That bound is where the answer lies, the three quarters

Because if you see the next problem, you’ll realise it

My prof told me this problem is due to miller

A bit tricky with a laugh emoji

And left the chat

😭

[waxen plover]

Bruh, the subset can’t be a subgroup by lagrange’s theorem

Yes

At least

I has to be the entire group

That’s what we wanna prove

[next obsidian]

Lmfao. Hot water bath

[waxen plover]

Cold

Please tell me if there is anything wrong

Never mind

This seems right

Thanks

@patent harbor , you were typing?

[patent harbor]

Yea, I didn't follow your proof...I see the implication that the intersection of H with aH is nonempty...I agree with this but for a reason I don't see written down. Since H and aH have the same number of elements (greater than 3/4 of G) their intersection must have more than 1/2 of G (just a short counting argument which is easy to visualize--- picture the places of aH that do not intersect H).
So in particular the intersection is nonempty.
This argument works for any a in G (and not just elements in G without H).

I didn't understand the next implication? If b is in the intersection why does that force a to be in H?

[waxen plover]

b is in H and aH

ba^-1 is in H

Since b is in H

a is also in H

Hope that makes it clear

Or gives rise to some counter example

To improve my proof

You are only talking about cardinalities

But you also need to see the closure

Under multiplication

[patent harbor]

I'm struggling to see why ba^-1 is in H...When I apply T to ba^-1, I want to show that this is ab^-1.
Sorry, it doesn't follow readily for me!
Since b is in H, I know T(b) = b^-1
Since b is in aH, I know there is some x in H such that b=ax and hence T(b) = T(a) x^-1.
But I cannot reconcile the two to what you said.

[waxen plover]

b = ax, right? You said it

b and x are in H

Then where does a lie?

[patent harbor]

But H is a set, not necessarily a subgroup at this stage.
Even if I write a as b x^-1, I get stuck!

[waxen plover]

Oh yes, thank you

😁

Saved me

[patent harbor]

Perhaps, from this discussion above, we know b^-1 = T(a) x^-1 and since b =ax, we have the following result:
For each a in G, there exists an x in H such that T(a) = x^-1 (a^-1) x.
If we had commutativity, then we would be done (interchanging the orders to cancel x^-1 and x).
So the challenge is now showing G abelian.
It is a tough problem, but I think this might be a step in a good direction.

[waxen plover]

Yes yes

[terse crystal]

What book is this? Looks like all exercises are of high quality

[waxen plover]

Herstein

The man

The myth

The legend

One of the grandmasters of algebra

[wicked patio]

Showing G abelian given x |-> x^-1 is an automorphism looks like the easy part

[waxen plover]

Yeah yeah

Showing x maps to its inverse is the actual problem

[wicked patio]

Yeah

This reminds me of uhhh

Let G be finite, S,T be subsets of G such that |S|+|T| > |G|

[waxen plover]

Go on, I think I’m interested

[wicked patio]

Show ST := {st | s in S, t in T} = G

[waxen plover]

Really?

That true?

Wow

[wicked patio]

Yeah the proof is elementary but tricky to find

This is from the Apostol analytic number theory book

[waxen plover]

Imma note it down as an exercise.

Thankssssssss

Now the problem still remains

[wicked patio]

I'd be curious under what circumstances a group automorphism could send exactly 3/4 of its elements to their inverses

[waxen plover]

The quaternion group of order 8

Now help me with the problem 😭

[wicked patio]

I'm trying to build my intuition for why this might be true by looking at the limiting cases

[waxen plover]

[wicked patio]

Any time we have a set a,b,ab that all gets sent to their inverses, a and b commute

We can certainly start by finding such a pair: choose a and b randomly from G-{e}, and use a probabilistic argument to show the expected value of the number of elements from {a,b,ab} which c get sent to their inverse is >2. Thus, there is some example where all 3 get sent to their inverses

Now we know <a,b> is Abelian, and every element in it gets sent to its inverse

[waxen plover]

I’m blank

Sorry

[wicked patio]

Hmm

Maybe we just start with an element a which gets sent to its inverse

[waxen plover]

And regarding this, I have a doubt

[wicked patio]

Then, every element in <a> gets sent to its inverse. Also, trivially, <a> is Abelian

[waxen plover]

Isn’t this a counter example?

[wicked patio]

If K is G\H, then |K| + |H| is not greater than |G|

[waxen plover]

Shit my bad

Sorry

[wicked patio]

no worries

[waxen plover]

And yes we are here

[wicked patio]

Ok, I want to find a new element b not in <a> such that b and ab get sent to their inverses

actually

I want to find an element a' from <a> and an element b not in <a> such that b and a'b get sent to their inverses

eh

my goal is kinda to build G by adding elements like this

[waxen plover]

T^2 = I for T restricted to H

I want to write K as a union

[wicked patio]

Oh wait

[waxen plover]

Yes

What about K?

[wicked patio]

Ah but we don't know T^2 is a homomorphism

Ok

Or no

[waxen plover]

T^2 is an identity only restricted to the set H

[wicked patio]

T(T(x)) is an automorphism

Right

[waxen plover]

Yes

It’s a group

[wicked patio]

How big is its kernel?

[waxen plover]

Obviously

Just e

[wicked patio]

Kernel is the set of elements that gets sent to the identity

[waxen plover]

Yes

Just e

[wicked patio]

oh lol

right

idk what I was thinking

but

can you show the set of elements fixed by T is a subgroup?

[waxen plover]

It can’t be

Our goal is to prove it is the whole group

Yes

[wicked patio]

The whole group is a subgroup of itself

[wicked patio]

You can show it?

[waxen plover]

No

That’s the question

😂

[wicked patio]

If T fixes a and b then T fixes ab

Shouldn't be hard to show

[waxen plover]

Please help

[wicked patio]

T(a) = a, T(b) = b

What's T(ab)?

Use properties of homomorphisms

[patent harbor]

T doesn't fix elements in H; it sends those elements to their inverses.

[wicked patio]

oops

you're right

T^2 is what I meant

[vapid vale]

for each element h of your group, you have the homomorphism sending g -> gh. suppose g is one of those elements that is sent to its inverse by T. then there are two possibilities, gh is another one of those elements, or it isn't

show that over half of your groups elements are in that first category

[waxen plover]

Group homo?

How?

[vapid vale]

yes, an automorphism

multilpication by a fixed element is always an automorphism

[waxen plover]

How?

Isn’t that just a permutation?

[vapid vale]

how do you check something is an automorphism

[wicked patio]

What

[waxen plover]

It’s an isomorphism from a group to itself

[wicked patio]

Multiplication by a fixed element is never an automorphism, unless that fixed element is e

[patent harbor]

it is a bijection

but that is it :(

[waxen plover]

A mere permutation of group elements is not an automorphism

[patent harbor]

i think you mean inner automorphisms?

[waxen plover]

Automorphisms are isomorphisms

[vapid vale]

an inner automorphism is an automorphism

a red car is a car

[waxen plover]

Yes an inner automorphism is an automorphism

Automorphisms form a subgroup of bijections from a set into itself

[vapid vale]

this is a very very easy thing to prove, which you should do

[waxen plover]

Automorphism is not just any bijection

[vapid vale]

(not the statement you made, the one im making)

[wicked patio]

it's fucking false blud

[vapid vale]

correct

[waxen plover]

It must be an isomorphism

[wicked patio]

It's literally almost never true

[waxen plover]

Only if that permutation is an identity

Lmao

[vapid vale]

wait lol

[wicked patio]

Are you thinking of a bijection?

[waxen plover]

Yeah

[vapid vale]

lol ignore me its late

in any case the argument i was making for the problem doesn't rely on the fact that map isan automorphism

epic fail

[patent harbor]

@waxen plover I think i have a proof. It is not beautiful. Nor does it show any intuition. It is simply what was needed to push the proof tentatively built, as we had discussed previously, through.

We have already proven from an earlier post that it is enough to show G is abelian.

Let a be in G.
Consider the subgroup Ca consistsing of all those g in G which commute with a (the centralizer of a).
It suffices to show that Ca has more than 1/2 the elements of G.

To do this, the natural thing to do for me at least is construct an injection between a set with more than 1/2 the elements of G with Ca.

Now one such set we have established already with more than 1/2 the elements of G is the intersection H with aH.

Can you construct an injection from H cap aH to Ca?
As a hint, fix an element from the nonemptiness of H cap aH, call it b.
Using this element define an injection.

[wicked patio]

1. Show that the set of elements fixed by T^2 is a subgroup of G
2. Show that this subgroup has cardinality > |G|/2
3. Profit

[vapid vale]

the intuition is that you have some fixed 3/4ths of your group, and any bijection (in particular right multilpication) will keep at least 1/2 still in that 3/4, since there is only 1/4 of the group that can escape

[waxen plover]

T^2 fixes all the elements

Thanks @wicked patio

I think we are done

[wicked patio]

Np

[waxen plover]

Lmao it was so short

In the end

😂

[wicked patio]

Took way too long

[waxen plover]

Yeah

[wicked patio]

The set of elements on which f and g agree (in this case, T^2 and id) is called... either the equalizer or coequalizer

I forget which one

but it's always a subgroup

[patent harbor]

i think the equalizer. (limit of the double arrow diagram)

[waxen plover]

I’m stuck again. So clumsy I am today

[patent harbor]

I am stuck to. I agree wiht 1 and 2 with you Dreyuk. I cannot profit!

So T^2 is identity. But how can i even use that to prove T(x) = x^-1 for all x?

[waxen plover]

Yeah

[vapid vale]

you'd have the result if G was abelian, right?

[waxen plover]

Yes

[vapid vale]

the proof outline i was hinting at doesn't use the fact that every element gets sent to its inverse

you just need to show that for over half of g in G, g and gh get sent to their inverse for every fixed h

and show that if g and gh get sent to their inverse, then g commutes with h

[waxen plover]

The wonderful takeaway from the discussion here for me is that we are investigating the automorphisms of a group, of order 2.

That’s enough for me to start digging

[wicked patio]

Yeah I can't profit either

I must have forgot the difference between T = T^-1 and T(x) = x^-1

I can tell you more than 5/16 of the element pairs commute, because
> (3/4)² = 9/16 of pairs (a,b) will have T(a) = a^-1 and T(b) = b^-1
> 3/4 of pairs will have T(ab) = (ab)^-1
and when both of those are satisfied, ab=ba

If we can prove more than 5/8 of the pairs commute, we are done (a is in Z(G) at most 1/4 of the time, and when it's not, b is in C(a) at most half the time)

Interestingly, in the provided example Q_8, exactly 5/8 of the element pairs commute. This would seem to suggest there really is a strong relationship here.

This also shows that if T takes more than 7/8 of the elements to their inverses, then G is abelian. (A little less than 7/8 actually, around 86.9%)

[waxen plover]

Ahh I’m done with it

Thanks for the help

[wicked patio]

Np lol

I'll let you know if I come back to it

[patent harbor]

I couldn't sleep. Here's the proof sketch I have for it with holes left for you to fill in.

The summary of the sketch is:

1. Prove for each a in G that the cardinality of H intersect aH is greater than 1/2 the elements of G.
2. Prove for each a in G and b in H intersect aH, that T(a) = b^-1 a^-1 b.
3. Constrcut, for each a in G, an injection from H intersect aH to the centralizer of a in G. As a hint, you can construct an injection for each choice of element in H intersect aH. Since the set is nonempty you can pick an element from the get-go. You should use (2) here.
4. Use this to argue G is abelian. And use abelian to clean up the result in (2).

[wicked patio]

real

[round anvil]

Does an integral domain need to have a multiplicative identity?

[rocky cloak]

Yes, people also typically require 0 and 1 to be distinct

[round anvil]

ok

[waxen plover]

I mean, I got it

[waxen plover]

nice

Thank you so much, @patent harbor

😁

[tulip otter]

it might be obvious but why should any element in A not in p be a unit in A

[thorn jay]

An element is contained in some maximal ideal iff it is not a unit

And A is local

[dull ginkgo]

An element is a left/right unit if it lies outside of all (left/right) proper ideals

Maximal ideal is easier to interpret :3

[patent harbor]

because A is a local ring.
To expand on the answers above:

Let x be an element in A not in p.
Then look at (x), the ideal generated by x.
Supposing it is not a unit, then (x) must be contained in a maximal ideal (by Zorn's lemma).
But there is only one maximal ideal. So x is in p, a contradiction. Hence x is a unit.

[dull ginkgo]

This same sort of jazz is used to define the equivalence of Jacobson radical as intersection of all maximal ideals or x such that rx + 1 is a unit

[tulip otter]

Ohh I didn't know this

Or maybe I knew and forgot about it

[dull ginkgo]

(Since adding a unit to an element kicks it out of an ideal so if it’s in every maximal ideal then it is kicked out of all the ideals so it becomes a unit)

[tulip otter]

Ah yea right

Tysm everyone

have a great day/night

[crystal vale]

why in category of fields, products doesn't exists? any hint?

[knotty badger]

They sometimes do but not in general

Try some examples

[thorn jay]

think of the characteristic of fields

[rocky cloak]

A product should satisfy Hom(A, BxC) = Hom(A, B)xHom(A, C).

Try to plug in a few examples of A B and C into the right side and see what happens

[crystal vale]

what i am thinking, if there is product of Z/3Z and Z/5Z, then that field will be 1 = 0 , {0}, but we excluded this ring in our field's definition

[tall igloo]

I think the product of fields is the largest field that is a subfield of both (is this true?)

[thorn jay]

that's in posets I'm not sure this holds in general though

[crystal vale]

is it correct?

[thorn jay]

products are intersections anyways

if F1 and F2 are subfields of F then the fiber product of F1 and F2 over F is F1 \cap F2

[velvet hull]

Why does the field have to be 0?

[tall igloo]

hmm I see, yeah I guess field is not a full subcat of poset

[thorn jay]

that's not what I mean

in a poset regarded as a category, the product of two elements would be the meet if it exists

[tall igloo]

ahh wait I see what you mean lol

too many morphisms

[thorn jay]

yes exactly

I couldn't give you an exact example though of it failing over a common characteristic

[crystal vale]

because it has field homorphism to Z/3Z and Z/5Z which shows that additive order of 1 should be 1

[knotty badger]

https://youtu.be/4uJLvby7K9Y?si=j8wEcIePcjOMyEIO&t=1603

[rocky cloak]

Not really no.

Consider for example K = Q(sqrt(2)), then KxK is not equal to K.

[knotty badger]

this video has some nice examples for products in the category of fields

[crystal vale]

jagr, my counter example makes sense?

[knotty badger]

explicitly, ||the product of two copies of C does not exist in the category of fields||

[rocky cloak]

I'm not entirely sure I follow your reasoning about 1 being 0, but it is true that Z/3 x Z/5 can't be a field

[thorn jay]

interesting, ||would the algebraic closure of Q work too?||

[knotty badger]

idk maybe? it seems like the issue is ||C having nontrivial automorphisms||

[rocky cloak]

Any field with a nontrivial automorphism should probably work

[thorn jay]

I see

[thorn jay]

yeah that's kinda what I was thinking but struggling to properly formalize lol

[crystal vale]

if product exists, it will have field homomorphism to Z/3Z and Z/5Z, homomorphism are injective here because of field, so it shows additive order of 1 should divide 3 and 5 which gives 1, so 1 = 0

[rocky cloak]

Alright, then I buy the reasoning

[knotty badger]

[rocky cloak]

I'm gonna further conjecture that the prime fields are the only examples where K^2 exists

[knotty badger]

which ones are those?

oh, fields with order a prime?

[rocky cloak]

Z/p and Q

[knotty badger]

ah nvm

[rocky cloak]

Field with no subfield

[knotty badger]

i see

[thorn jay]

I conjecture \infty to be a prime

[knotty badger]

sounds like there'd be a galois argument you could use

[thorn jay]

Hmm
is it provable that every nonprime field has a nontrivial automorphism?

[rocky cloak]

R does not (but RxR doesn't exist)

[knotty badger]

wait R doesn't have nontrivial automorphisms

[thorn jay]

guh

[knotty badger]

?

[rocky cloak]

It does not.

It does have endomorphisms though, which I guess leads to the same problem

[knotty badger]

huhhh i thought some choice bullshit would get you automorphisms

[rocky cloak]

It's a fun exercise

The key is that whether a number of positive or not can be determined just from the field structure

[thorn jay]

||let f: R -> R be an endomorphism. Then f(x^2) = f(x)^2 >= 0. Hence f(x) >= 0 for all positive x. Thus:
a > b => a - b > 0 => f(a - b) > 0 => f(a) - f(b) > 0 => f(a) > f(b)||

[knotty badger]

ah interesting

and then it's determined on the rationals

[thorn jay]

yes

[knotty badger]

wait

then how do you get endomorphisms

[thorn jay]

¯_(ツ)_/¯

no clue

[knotty badger]

your argument seems to suggest any endomorphism of R is the identity

https://math.stackexchange.com/questions/3382939/the-only-endomorphism-of-mathbbr-is-identity

yeah i'm not sure what jagr was referring to

[rocky cloak]

Right, I guess it doesn't have any endomorphisms either, but it has subfield that can embed in different ways (hence RxR can't exist)

[rocky cloak]

So there are exceptions to the conjecture, like Q(cuberoot(2)). But

Say K is a field that is not a prime field, and let F be its prime field.

If K^2 exists it must map onto K hence K^2 = K.

If K/F is not algebraic, then K contains F(x). Then F(x) and F(x+1) are different embeddings contraindication!

If K/F is algebraic, and F is finite K/F is galois, so has automorphisms.

So the only possible examples are algebraic extensions of Q

[cosmic lion]

are there groups of the same order, each of their elements are of the same order, yet they aren't isomorphic? i'm trying to prove something and my proof works if that when a group is of order 8, and it's elements are order 1,2,4,4,4,4,4,4, then it's isomorphic to Q_8.

[velvet hull]

in general that is a difficult question, but in your specific case it is not hard to show that it is true

hint to start with - ||since there is a unique order 2 element, any order 4 element must square to it||

[velvet hull]

but in general yes, it is possible for that to happen

[cosmic lion]

i'm gonna check that hint later, try it completely myself first

[rocky cloak]

The dihedral group has more elements of order 2 though. So might depend how the question is supposed to be interpreted.

The mod 3 Heisenberg group and (Z/3)^3 works though

[velvet hull]

yeah you're right my example doesn't work

[rocky cloak]

Well it works if you just think of the set of orders of elements (and n is even)

[cosmic lion]

yes, I think I could specify by saying I'm interested in the same sequence of orders of elements not just the set

[rocky cloak]

I guess the appropriate term would be multiset

[coral steeple]

An update. Is it clear how I conclude this?

Well actually I want the RHS to read $\frac{(\mathbb{Z}/(k))[t]}{(t^2+1)}$

[cloud walrus]

person2709505

[coral steeple]

So that this quotient consists of cosets of linear polynomials (in t) with coefficients in Z/k, meaning there are k^2 choices for the coefficients

[quiet pelican]

Yeah, I would write this instead

[coral steeple]

My reasoning isn't flawed?

Coincidentally, I'm also not sure if this is the best way to write that, since the parentheses mean both "ideal generated by" and "take the quotient before adjoining the indeterminate t"

[violet spade]

it's clear that we're talking about ideals. you can use i think 3nd isom. thm to see that $R/I \cong (R/J)/(I/J)$ where we take $J=(k)$

[cloud walrus]

joel

[river heart]

What's the smallest example?

[rocky cloak]

Mod 3 Heisenberg group and (Z/3)^3 have order 27.

I doubt you can go smaller than that

[limber tapir]

Imo this sort of obscures the point. The point is that Z[i]/(k) = (Z/k)[i] and after your argument you would still have to argue mostly the same thing with i replaced by t.

[limber tapir]

Like the reason this equality holds is just that there is a canonical isomorphism that is essentially just the identity

Although i should probably not say words like "equality" so let's pretend I said iso everywhere

[near tapir]

Let f(x) in Z[x] be a monic polynomial, and let g(x) and h(x) in Q[x] both be monic polynomials such that f(x) = g(x)h(x). Then g(x) and h(x) both have integer coefficients.

In this statement, why is g(x) and h(x) assumed to be monic. Does it not work without?

[rocky cloak]

Consider for example f(x) = x^2
g(x) = 2x, h(x) = x/2

[coral steeple]

@limber tapir would you say that the reasoning I give in this comment is clear? The point of these isomorphisms is I'm trying to justify why the ring has this structure: linear polynomials with coefficients in Z/(k) having no relations between them, i.e. two choices of coefficient

I couldn't justify this counting to myself just looking at the LHS

[coral steeple]

Isn't that what I'm doing in the second isomorphism? Do you mean something different?

[violet spade]

We see (with $ I = (k,t^2+1) $ and $J=(k)$) that $I/J \cong (t^2+1)$ and $R/J \cong \mathbb{Z}/(k)$

[cloud walrus]

joel

[violet spade]

So we get the quotient as desired

(manipulating the RHS)

[limber tapir]

Yes that's fine

[white oxide]

Sorry I can't really see why this implies that each coefficient is G invariant

[quiet pelican]

What does it mean for two polynomials to be equal?

[white oxide]

Ah right I understand now

Worked it out with a small example

Thank you

[thorn jay]

Very readable

[tardy hedge]

Mico loves :catno:

Wow i cant even use it

Thats so sad

[quiet pelican]

mishu like

[tardy hedge]

Truuuuuuue

I like cat no too

[thorn jay]

you forgot to capitalize the N

[tardy hedge]

:catNo:

Breh

[thorn jay]

damn

[tardy hedge]

😢

Discord b like :kianno:

Im going in date with irish german girl rn

She on exchange in halifax lol

I think she walked by to washroom i was like :0

European fashion sense man

[thorn jay]

nah def not, where I used to live before uni everyone has the plainest fashion sense ever

[tardy hedge]

Hmm ok

I really like good fashion

[thorn jay]

except like a couple people

[tardy hedge]

Makes big difference for me

[thorn jay]

I'm spending more and more time on it

it's fun

[tardy hedge]

Agreed

[thorn jay]

I'm def more on the alt side

shouldn't be too surprising lol

[tardy hedge]

Me too bruddah

I feel like a lot of ppl on mathcord are lol

[iron arrow]

Two p-sylows which aren't of order p, can have non neutral intersection right?

[quiet pelican]

Yes

[foggy axle]

My understanding of this is very basic but from what I understand $\CC = \RR[x]/(x^2 + 1)$ (by convention) is a field extension of $\RR$, which we make by coming up with an irreducible polynomial in said field and then including its solution (along with the necessary multiplied and added values for closure)

My question is: is there a such thing as a polynomial irreducible in $\CC$? and if not, does it mean that it cannot be extended (in this manner?)?

[cloud walrus]

Coolempire93

[fading acorn]

C is algebraically closed, so there is no nontrivial algebraic extension over C (i.e., you cannot have bigger fields by adjoining elements of a root of a polynomial with coefficients in C)
however there can be transcendental extensions of it, for example consider the field C(x) of rational functions over C

[foggy axle]

This makes sense

Until the rational functions part 🤔

[fading acorn]

transcendental here means: it cannot be a zero of a polynomial function with coefficients in C: can the element "x" be a zero for a polynomial f(y)=a_ny^n+...+a_1y+a_0 where a_i\in C? (while x\notin C, what happens if x\in C?)

[foggy axle]

Right, transcendental numbers by definition are the non-algebraic ones

But I couldn't see the connection to the rational functions over C 🤔

[fading acorn]

well if you just adjoin elements x,x^2,.... then you have the ring C[x] which is not a field
what happens if you consider the fraction field of C[x]?

oh lol if you mean extended in sense of algebraic extension then of course C(x)/C is not algebraic (which you have asked) but here im talking about transcendental extensions other than what i've answered above

[foggy axle]

Ah right the polynomial ring is a ring 😂 it helps when I actually think about the names

I guess what trips me up thinking there is that aren't fractions of complex numbers still just complex numbers

Is there actually any new element there

Unlike integers to rationals

Is there a better analogy for me to consider

[mint seal]

a rational function over C isn't a fraction of complex numbers

it's a fraction of polynomials (in variable x, say) that have C coefficients

if x is replaced with some complex number throughout the expression of a rational function, then you'll have a fraction of complex numbers

[foggy axle]

ahh

I was thinking a function with rational coefficients

🤦‍♂️

[mint seal]

ah, no worries

[brittle wharf]

suggestions for an abstract algebra book? the ones i know of are fraleigh, artin, aluffi's notes from the underground, and d&f

i have experience with linalg (notably from a comp linalg class and ladr)

[fading acorn]

just pick one of these and start studying lol

[twilit wraith]

I used gallian at first

[fading acorn]

anything here seems okay

[twilit wraith]

Its not bad but its weak with group actions

Its a good start if the others feel overwhelming

In my opinion though, d&f is just really good since it has so much content

[brittle wharf]

a lot of them vary wildly based on the table of contents

[fading acorn]

well my point still does stands

whats wrong with varying toc with various books

you can just pick one that fits you and learn other stuffs (that is not included) later if you need them

[brittle wharf]

true ig

[iron arrow]

May I ask, if a polynomial p(x) is irreducible in a ring K[X] where K is a field, then p(x+alpha) where alpha is from K is also irreducible? Mainly I ask this because I've been kind of stuck in a problem and this would solve it (I'm trying to see that Q(cube_root(p),fourth_root(p))=Q(cube_root(p)+fourth_root(p)) and therefore because I've already seen x^3 - p is irreducible in Q(fourth_root(p)) then (x-fourth_root(p))^3 - p would be irreducible there too, and it has root cube_root(p)+fourth_root(p), and because it is contained in the other field extension we have that they're equal)

[quiet pelican]

If p(x + a) = g(x)f(x) then p(x) = g(x - a)f(x-a) and the degree of f(x - a) is equal to the degree of f (and same for g)

So yes

[iron arrow]

This is so goated

I was stuck on this problem for a long time

Which is basically to show that Q(sqrt(p) + cube_root(p)+fourth_root(p)) = Q(cube_root(p),fourth_root(p)

[tough crow]

Do I need to learn abs alg on it's own to progress farther into mathematics? If so, what book should I use for that purpose?

[quiet pelican]

Where are you in maths, and what are your goals?

[tough crow]

I've done linear algebra from LADR, into analysis from baby rudin and just recently finished ch 2 of folland.

I don't have any specific goals yet, I just self study based on whatever seems interesting

Though I eventually want to go into pure math research (if that's relevant)

[quiet pelican]

Algebra (especially at the level of a first course) is useful pretty much regardless
Dummit and Foote is a standard recommendation for example

[tough crow]

The entirety of d&f?

[quiet pelican]

Like parts 1-4 excluding chapter 6

Common for that to be split into 2 courses
But like, they’re a series

[tough crow]

Alright, I'll take a look. Thank you!!

[tardy hedge]

Nah Brotha d/f for rings and modules and Galois theory is good

For group theory it does look a bit hard to me

[foggy axle]

Here to hate on fraleigh

Reading that book felt like I learned nothing

Well not nothing just leaving out a not insignificant portion

[karmic moat]

#book-recommendations message

[mint seal]

I'm a big fan of Aluffi's Algebra Chapter 0, and recently took a look at Notes from the Underground. No idea why he named it after a Dostoevsky novel, but it seems great

I also like D&F a lot. Never checked out Fraleigh or Artin

[brittle wharf]

g-g-g-g--g-grroup??????????????

[brittle wharf]

yeah chap 0 seems like soemthing i'd read if i wanteed a second look at abstrac talg in the future xd

im planning on reading artin and consulting notes from the underground as a second text

[vestal trench]

notes from the underground is also quite good i think

[foggy axle]

Is unitary the same as unary?

[thorn jay]

I don't think so, else the exercise would be kind of pointless lol

"show that the inverse of an automorphism is a unary function"

[jaunty magnet]

unitary operators are those satisfying T* = T^-1

[foggy axle]

ahh, this makes sense

Are these two results not trivial

[noble nexus]

they're pretty trivial yeah

[brittle wharf]

they are

[brittle wharf]

by definition, they're invertible isometries. their matrices wrt some orthonormal eigenbasis is diag{e^{i\theta_1}, \dots, e^{i\theta_n}}. their adjoint is their own inverse. etc etc

[tough crow]

Any alternates you'd like to suggest?

will look into it, ty!!

[swift prawn]

notes from the underground is great (imo) i use it lots as a reference for more basic material

it has a weird order (rings -> modules -> abelian groups -> groups -> fields) but otherwise it covers all the usual things

the only issue I think it has is that he does not do the structure theorem for modules over a pid, instead does it for modules over a euclidean domain which makes the proof easier

[copper kestrel]

hi all! i'm doing problems 12 and 18, i had 2 questions:

1. Question 12: When it refers to "usual multiplication" does that mean component wise? or are we foiling? i would think its component wise since we havent defined it to be a ring and therefore don't know if the distributive property holds.

2. Question 18: Would all the units be of the form 1 x Q x 1 and -1 x Q x -1?

[rapid cave]

The operations inherited from C

[tardy hedge]

Nah brotha im right u wrong

[copper kestrel]

sick

[rapid cave]

For 2, use (AxB)*=A*xB*

[tardy hedge]

Contemporary abstract algebra is hella friendly

Idk the author

[foggy axle]

Yes I loved contemporary for a handbook

When I needed to reference while researching it was easy to open and find stuff

Them promoting gaining an intuition for the theorems was also nice from a pedagogy standpoint

No idea who the author was either

[tough crow]

this?

[tardy hedge]

Yea

[tough crow]

alr guess I'll use this as well (Found this being resold for a very cheap price last week so I bought it)

[foggy axle]

Ah yeah gallian

[rocky cloak]

1. Usual multiplication just meaning the usual way to multiply real numbers

2. There are a few more units.

[copper kestrel]

ty!

[uneven bobcat]

Terrible cover though, most editions I've seen have a nice pretty picture on the front.

[tough crow]

aw 😔

well it costed like 2 usd

[uneven bobcat]

lol nice.

[tardy hedge]

True Dat

It usually has some swish swooshing woah design

Math textbooks love the swish swoosh

[thorn jay]

the best representation of how math looks

[tardy hedge]

Yas

[thorn jay]

aside from just straight up diagrams

[copper kestrel]

we actually have not proved this unfortunately so idk if i'm allowed to use it

[rapid cave]

prove it yourself

[copper kestrel]

:')

i think i figured it out actually

[rapid cave]

great

[copper kestrel]

nvm ur right i have to prove it myself LOL

[tough raven]

Just checking, if R is a ring and I is an ideal, there is at most one complement to I in R that is also an ideal of R, right?

Namely {j in R : Ij = 0} (and symmetrically, {j in R : jI = 0}).

[supple ice]

if you mean an ideal J with I+J = R and I∩J = 0 then there’s at most one such J
if it exists, it must be the annihilator Ann(I) = { r : I r = 0 }
also note Ann(I) doesn’t always give a complement it only does when I actually splits

[crystal vale]

Q(\sqrt3 ) and Q(\sqrt3 w ), where w is a non-real complex root, i think both are the isomorphic as field, isn't f(\sqrt[3] (2) ) -> f(\sqrt[3] (2) w) bijective morphism ?

wait maybe there can be a problem in homomorphism

[quiet pelican]

It is an isomorphism

[crystal vale]

okay

[quiet pelican]

The easiest way to see that is that both are isomorphic to Q[x]/(x^3 - 2)

[crystal vale]

Yeah

[mint seal]

does this proof - particularly the last part where I bring in a Dedekind cut - look acceptable?

I mildly feel like I'm being handwavy at the end

[quiet pelican]

I think it works, and I don’t think there’s another nice way to do it directly
But I think a contradiction proof might be nicer here

[mint seal]

I see what you mean, contradiction does seem to make it more convincing than how I said it

thankee

[violet spade]

there should be an alternative method of showing this via real numbers = equivalence classes of Cauchy sequences

[twilit wraith]

what is this question even asking

like i feel like i know but still im somewhat confused by it

it seems the answer is no for the intuitive reason that M would be some sort of 1-dimensional submodule of the Q module

but idrk

[topaz solar]

What do you even mean 1-dim

Anyway, suppose you had a Q module structure on M, what does that tell you

What is a Q module

What are properties of finite groups

[kind temple]

just to clarify the task:

an R-module is an abelian group together with an R-action R x M -> M that satisfies some distributive and unital constraints.

this is asking if, given a finite Z-module M, can we extend the multiplication by integer scalars (the Z-action Z x M -> M) to multiplication by rational scalars (a Q-action Q x M -> M) in a way that preserves the original Z-action

[twilit wraith]

Just analogizing submodules with subspaces in my head

[twilit wraith]

I see

[kind temple]

my strategy with these types of problems is to look at really small examples

[twilit wraith]

Ok so if I have like Z4 then

And I let Z act on it in the expected way

Oh then theres just nothing 1/2 can map to

I see it now

[kind temple]

yea, here is a minimal counter-example:
take M = Z/2Z. this has a natural Z-action Z x M -> M, (n,k + 2Z) |-> nk + 2Z.
if we could extend to a Q-action, put a = 1/2 (1 + 2Z). then a + a = 1 + 2Z, but no choice of a in Z/2Z satisfies this

[twilit wraith]

Awesome

Makes sense

[kind temple]

there is a cool adjunction between restriction and extension of scalars tho

[topaz solar]

Yep

Generally, if x has order n, then nx = 0, but (1/n)(nx) = (n/n)x = x

uh oh!

[limber tapir]

Also the only finite Q-VS is 0 right?

[topaz solar]

Of course it suffices to have a counterexample, but this is to say that the counterexamples are everywhere

[topaz solar]

That’s exactly what this tells you

[limber tapir]

Just you can only see this because a Q-VS with dim > 0 would need to contain the infinite set Q

[topaz solar]

(Kind of exactly the proof of it)

[twilit wraith]

We can generalize this to all finite abelian groups as well right

As all of them have cyclic subgroups

[topaz solar]

Yep that what my comment said

[twilit wraith]

Oh oopsies

[topaz solar]

So obstruction 1 is torsion

[kind temple]

if $f : S \to R$ is a ring homomorphism, there is a functor $f^{\ast} : R\text{-Mod} \to S\text{-Mod}$ (i think of this as the pullback) specified on objects by taking an $R$-module $M$ to the $S$-module $f^{\ast}(M)$ with scalar multiplication defined by $s \cdot m := f(s) \cdot m$. this is called restriction of scalars functor. $\newline$

the extension of scalars functor is $f_{\ast} : S-\text{Mod} \to R-\text{Mod}$ given by $f_{\ast}(M) := R \otimes_S M$, since we can think of $R$ as an $S$-module. $\newline$

The extension of scalars functor is left adjoint to restriction of scalars functor, so there is a nice relationship between extending and restricting scalars.

[cloud walrus]

c squared

[kind temple]

maybe you will find this interesting

[twilit wraith]

Oh nice

[kind temple]

i guess in the infinite case, Z can't be made into a Q-module for the same reason, if a = (1/2)1, then 2a = 1, contradiction. but obviously Q as a Z-module can be made into a Q-module

[topaz solar]

In this case, being a divisible group is a key word

[kind temple]

oh cool

[topaz solar]

Divisible = for every x in G, n>0 in N, there is y with y^n = x

Which you can see as basically “there is something in there that is like (1/n)x”

[kind temple]

yea, writing the group operation additively makes it more clear why its called a divisible group

[topaz solar]

Now here’s the fun part, assume G is divisible, abelian, and torsion free

Then (1/n)x is unique

Since if a, b satisfies that, then a^n = b^n = x, so a^n b^-n = 0

So ab^-1 torsion

[kind temple]

right

[topaz solar]

so it’s a = b

This means Q-vector space = divisible abelian torsion free group

[kind temple]

oooh that is fun

i like it

[topaz solar]

It’s a fun example in model theory of adding conditions that don’t seem enough on their own to get a complete theory

(Infinite) Abelian isn’t complete, abelian torsion free isn’t, but divisibility finishes it off

[topaz solar]

Anyway it’s not a hard example but it’s the completion of that exercise right?

Since it’s purely group theoretic descriptions of when you can extend

[kind temple]

yea, i would agree with that

very cool characterization of Q-vector spaces

[topaz solar]

These kinds of things can appear elsewhere too in unexpected ways, like things “really just being these simpler objects”

[crystal vale]

how do i show SL_2(C) / { I, -I} is simple group, any hint?

[delicate orchid]

that's rather involved for an exercise I cannot lie

can you use Iwasawa's theorem? that's the only hint I can give

[crystal vale]

i don't know what is that

actually i have to show mobius transformation group is simple group

[delicate orchid]

a theorem telling you that when you have a faithful and primitive action satisfying some extra conditions then your group is simple

[crystal vale]

oh

[delicate orchid]

you prove it by assuming there's a normal non-trivial subgroup N, then using some Frattini argument nonsense show that G = N

[crystal vale]

i see

[delicate orchid]

maybe there's a simplier proof for C, cause the one I have in my head doesn't care about what field you're over as long as it's not F_2 or F_3

[delicate orchid]

sorry I should also add that you need to know that your group is perfect for this to work - which is obvious as perfection is preserved under quotients

[twilit wraith]

Just checking if this proof looks valid

I never invoked that F had to be a field in particular but i think that may not be necessary anyways

[south patrol]

Ye field shouldn't be needed at all

[south patrol]

But for the converse, you should show that it is never a submodule (i.e. for any n > 1) rather than just n = 2

Though there is a trick that immediately moves you from n=2 to any higher n lol

There's also a bit of a more conceptual way to look at all this, if you want, but maybe I can describe that after you're done aha, feel free to ping me!

[twilit wraith]

a counterexample doesnt prove that M is not a submodule of R as a right R-module?

[south patrol]

It's a counterexample for n = 2

[twilit wraith]

ah i see what u mean now

[twilit wraith]

heres the proof of the last part

[south patrol]

Okay nicee

[south patrol]

The thing I had in mind here was like

If you write a matrix B = [v1 ... vn] as a bunch of column vectors then you can write AB = [Av1, ...., Av_n] lol so here it's clear

[twilit wraith]

i see

[south patrol]

Literally the same lol

But now when you swap it, it gets messed up --- just transpose this and it now fails

[twilit wraith]

i cant think of a cleaner way to do the reverse

i mean considering that we are working over a field we could consider R as the affine space F^(n^2) i think

maybe theres something to that

[twilit wraith]

and then here we have a vector space so linalg can be done

[south patrol]

like now it acts on the rows on the right, but you can do something arbitrary to those rows e.g. swap them

ig yeah this is just like multiplication on the left is a bunch of row operations or whatever and on the right is a bunch of column operations?

[twilit wraith]

yeah

[karmic moat]

By B/J \subset A/I, does he mean pi^sharp(B/J) \subset A/I? Because B -> A is just an integral ring map and not an integral extension

[south patrol]

ye must be

[karmic moat]

Yea thought so thanks

[south patrol]

you can also always factor ur integral map as a surjection followed by an injective integral map, and the first bit clearly corresponds to a closed immersion

[karmic moat]

yea

[copper kestrel]

how would one show that two rings and two fields arent isomorphic if they both have the same cardinality :')

i'm trying to show that 2Z not iso to 3Z and R not iso to C

like i think for 2Z i can say that the only map that turns 2Z into 3Z is a transformarion but transformations are not ring homomorphisms

[delicate orchid]

Use the fact that C has many more roots of unity than R to arrive at a contradiction. 2Z and 3Z aren't rings under standard addition and multiplication and are isomorphic under the obvious modification so I don't know what's going on there

[south patrol]

Ig can view them as nonunital rings and then they are non isomorphic ||e.g. 2 has to be sent to +-3 but then consider where 4 goes||

[delicate orchid]

sure

thatt makes sense

[feral timber]

Suppose G is a compact connected lie group. I need to prove that any normal abelian subgroup N must be contained in the center of the group. Now I know that the center of the group is precisely the intersection of all the maximal tori in the group. So it suffices to show that for any arbitrary normal abelian subgroup N and any arbitrary maximal torus T, N is contained in T.

Now I took n from N. Then by conjugacy theorem there must be some g in G and t in T such that n=gtg^-1. By normality of n, I get t to be in N. But after this I'm stuck. I tried to show that I can get each element n as some t in T but I fail to do the same.

Firstly, am I going in the right direction? If I am, then any hints?

[rocky cloak]

I guess you want to first look at the connected component of N around the identity

[feral timber]

Okay, so I look athe the connected component N' of N around identity. Now I take its closure (let's call it N''). Then N" is closed in compact G so it is compact. Thus N" is a compact connected lie group in itself, hence will admit maximal torus. Conjugacy theorem proves that any two maximal tori would be conjugate to each other in N", but N" is abelian. This means that there will always be a unique maximal torus for N", let's call it T". Now this might not be maximal in G, so there will be a maximal torus of G containing it. Thats all I can think of with this right now though.

[rocky cloak]

So N'' is a torus, so contained in a maximal torus.

And one thing that is true is that all tori in G are conjugate

[copper kestrel]

how would one go about finding the number of homomorphisms from Z x Z x Z into Z?

[rocky cloak]

You can try to describe them by picking a set of generators for Z^3 and seeing where a homomorphism could send them

[feral timber]

Okay so this shows that N" is contained in T. How do I go from here to show that whole of N is in T?

[copper kestrel]

hm

[feral timber]

Taking quotient?

[rocky cloak]

Yeah, so now you have N'' in all the tori, so in the center. Next you should think about N/N'' inside G/N''

[solar kernel]

For finite fields, are all automorphism classified? I think i saw somewhere that for F_p, all automorphism are all in the form x^(p^k), is that correct? I dont have much experience with Rings and Fields yet, using field results often as im doing a summer research thing in Finite Projective Geometry, which uses Finite Fields very often.

[south patrol]

Yes

This is a standard thing in Galois theory if you want. Like if you look at the Galois group of finite fields

The map x -> x^p is called the Frobenius and you get all automorphisms of finite fields by iterating this

[solar kernel]

Amazing, thanks! Im taking a Galois Theory course next semester, so im excited to learn about it in a formal setting haha.

[knotty badger]

Wow that’s really unexpected

[south patrol]

It's like F_p^n/F_p is a degree n extension, so the Galois group has order <= n, and the Frobenius has order n, so gg