groups-rings-fields

[chilly ocean]

not sure

[tired gyro]

ok where would you have to send (1,1)?

[crystal vale]

No I don't want to use the coset stuff

[oblique iron]

Is there any reason you want to avoid coset stuff? It's a pretty fundamental tool

[crystal vale]

I am doing a napkin and he hasn't introduced coset stuff yet

And in addition, we have additional condition that group is abelian

[oblique iron]

Hmm I don't think there's a great way that isn't just using cosets in disguise (or essentially just proving Lagrange's theorem, which isn't too difficult), let me think more though

[chilly ocean]

well like
2x+3y to satisfy the homomorphism properties. though to make it satisfy the other requirements i think we would need for
3(2x) + 2(3y) == 0 mod 18
where like the x comes from the left side and y comes from the right side

so 6x + 6y cong 0 mod 18
so like x = 1, y= 2 ? would satisfy it right
so like we can send (1,1) to 2+6 i think

or like i think that would make our kernel atleast a subset of the desired kernel

wait i think i mixed up the numbers

[crystal vale]

Okay

[oblique iron]

oh here's an idea sidestepping Lagrange's Theorem:
Let n = ord(G). write out all the elements of G as G = {g_1, ..., g_n}. for your g for which you have g^n = e, consider the set {gg_1, ..., gg_n}. try to prove this set is equal to G, so you can say {g_1, ..., g_n} = {gg_1, ..., gg_n}. Do you see where to go from here? (this next step uses G is abelian, everything up to here is true if G is nonabelian)

if not I can give a further hint (provided you indeed prove these sets are equal, for which I can also give a hint)

[tired gyro]

let's think about the image of (1,0)

we need to make this 9

so that the first component will be zero for multiples of 2

[chilly ocean]

if we made that 9 then (2,0) would be in the kernel?

but (2,0) isn't in the requested kernel

[tired gyro]

oh yeah sorry

[chilly ocean]

it's okay dw about it, i have to go to sleep now

[tired gyro]

hang on i'm in the middle of a phone call

[chilly ocean]

like i gotta wake up early tomorrow i can't stay up any longer lol or I'll be groggy.
if you want to continue the discussion on it though you can dm me and I'll try to get back to you tmrw morning

[crystal vale]

Yes, {g_1,...,g_n } = {gg_1,...,gg_n}.

Got it

Thank you ❤️

gg_1gg_2...gg_n = 1 so g^ng_1...g_n = 1, and g_1...g_n = 1 so g^n = 1

Is it good exercise?

@oblique iron I am interested to know how it hits you ?

[grave sedge]

Yeah this is right, you should only do column operations

(because x,y generate the same thing as x+ny,y)

Doing only column operations you get that it's also generated by (1,0,1), (0,1,1), (0,0,2), which is isomorphic to Z^3 (and an isomorphism could be given by the row operations) but it's actually a proper submodule of Z^3

(i.e. it's the one where all coordinates have even sum)

[winter shore]

Any tips for this? I'm stucked in showing injectivity

also wondering if it holds in any abelian category

[quiet pelican]

I would do this by
“N -> M1 \oplus M2 is an isomorphism by the 5 lemma, and then the morphism you want is M_2 -> M_1 \oplus M_2 -> N”
Which also proves it holds in an abelian category, because the 5 lemma does

If you don’t have 5 lemma, then it’s just a slightly messy diagram chase that I’m not sure it’s particularly nice to describe in words

[winter shore]

ahhh right, thanks! completely forgot about that lol

[rocky cloak]

A simple way to see that it works in any abelian category is just to use the commutativity of the diagram.

If you have a map X -> N such that the composition X -> N -> M1(+)M2 is 0. Then also the composition
X -> N -> M1(+)M2 -> M2
which equals
X -> N -> M2
is 0.

So since M1 is the kernel of N -> M2, we must have that X -> N factors through M1.

Then 0 equals X -> N -> M1(+)M2 equals X -> M1 -> M1(+)M2. And since the inclusion of M1 is mono, X -> M1 is 0. So X -> N is 0.

This is in fact just a proof of (half of) the 5 lemma. And it's not necessary for the bottom row to be split.

[vague granite]

This is also a typo. It’s supposed to be an extension of M2 by M1

[glad osprey]

Is the idea here that the determinant, being a polynomial, has n roots, and C minus those n roots is still path connected? So we can let lambda be a path from 0 to 1 avoiding those roots?

[quiet pelican]

Yes

[icy totem]

I have the field extension F/K, where F is the algebraic closure of K.

Given α€F\K, I know there are (at least) n ways to include K(α) in F, where n is the number of distinct roots of the minimal polynomial of α.

Are there exactly n? Id say yes, following a reasoning that uses the FHT, am I right?

[chilly ocean]

What is FHT?

[icy totem]

First Homomorphism Theorem

[chilly ocean]

There can be less than n if K is not a perfect field

[icy totem]

What? Our lecturer proved that there are always at least n

[chilly ocean]

Are you working over fields of characteristic 0?

[icy totem]

No

Generic fields

n is the number of DISTINCT roots of μα, not its degree

[chilly ocean]

Oh my bad

[icy totem]

Np

Then do u know if there are exactly n?

[rocky cloak]

There are exactly n yes

[icy totem]

Thanks

[rocky cloak]

Just take the map K[x] -> F sending x to a specific element.

Then this factors as K[x]/(f(x)) -> F, where f is the minimal polynomial of x

So you have one map for each root of f

[dull ginkgo]

In general if A embeds into B, and we have an automorphism of B, we can compose the two to have another embedding

[rocky cloak]

Conversely a map from K(alpha) is determined by where alpha is mapped, and it can only map to other roots of its minimal polynomial

[icy totem]

Yes, I got this, I was having trouble understaning how an embedding K(α)-->F extends to an homomorphism K[x]-->F, but now I think I understood

[icy totem]

Oh yea this makes sense thx

[narrow zinc]

is normal subgroup a transitive property? i.e; if G is a group, A, B, subgroups and B normal subgroup in G, A normal subgroup in B, then A normal subgroup in G?

I'm sure I knew whether that was right at some point but I've forgotten

[chilly ocean]

Can you tell me of an example of a non-normal subgroup?

Actually, no need

[narrow zinc]

the Normal property is that you can define a quotiënt-group with it, even if G is not commutative

or N is a normal subgroup if, for any g in G, g * N = N * g

[chilly ocean]

The answer is no

[narrow zinc]

right...

do you have a counterexample?

I've been looking for one for a few hours

[chilly ocean]

I was trying to think of a simple example

But there are some on the internet

[knotty badger]

You can use dihedrals

The rotation subgroup is a normal subgroup of a dihedral group

And since its abelian, any subgroup of rotations is normal in the rotation subgroup

So you just need to find a subgroup of rotations which isn’t normal in the full dihedral group

Hmmmm wait maybe I’m misremembering this

[coral spindle]

No, that won't work, because the automorphism that the reflections act by preserves all subgroups.

[knotty badger]

Ah this one might’ve been permutations actually

[narrow zinc]

ah goddammit I just found an example using the alternating group (A_4) and I have been trying with that one for so long...

[knotty badger]

Ah it was dihedral

But not the rotation one

You take a square, and consider the subgroup of symmetries preserving a particular diagonal

And then you take a reflection subgroup of that, which has size 2

[knotty badger]

This is generated by a reflection and a 180 degree rotation

[narrow zinc]

I am so mad at myself

while checking A_4 with the normal subgroup H= Id, (12) (34), (13) (24), (14) (23)

it took A to be that subgroup of 4 elements, normal in A_4, and A_4 the normal subgroup of S_4

and look H makes a good normal subgroup for S_4

but the subgroup Id, (12) (34) is not a normal subgroup of A_4, I even noticed that...

but didn't realise that it is one for the subgroup of 4 elements above...

it must be, because its index in that subgroup is 2... you don't even need to check...

[knotty badger]

Mhm index 2 being normal is pretty convenient

[narrow zinc]

"I'm missing the very obvious.." may as well become my catchphrase >_<

[knotty badger]

Happens all the time dw

[narrow zinc]

the worst part is is was checking S_4 and A_4 because I half-remembered trying that sometime in the past, implying that it was on the right track

I'm now fairly certain I've repeatedly made exactly that mistake (now at least twice, but I'd bet up to 4 or 5 times actually)

[knotty badger]

This is partially why I used flashcards a lot during my degree

It included things like this

[dull ginkgo]

Glad I’m not doing a degree in meth

[still dew]

After I've derived the Jordan decomposition of A into A_s and A_n how do I show that they are polynomials in A

I am essentially using the structure theorem over PIDs and have done all the usual stuff

But it's not clear to me how looking at the ring K[x]/(x-a)^n tells us that A_s is a polynomial in A

[shell pilot]

If $|F| = q$ is finite, prove that $|GL_n(F)| < q^{n^2}$

So far, I have that since we are working with nxn matrices, there are $q^{n^2}$ possible nxn matrices. So how do I prove that there are some matrices which do not fall into $GL_n(F)$?

I saw a solution that said there is at least one matrix without a nonzero determinant. But then if that is the case, it is not an element of $GL_n(F)$, right?

[cloud walrus]

Soap_Opera

[topaz solar]

Yes

[dull ginkgo]

zero matrix

[topaz solar]

Determinant 0 means not invertible

[shell pilot]

In dummit and foote, it says that GL_n(F) = all nxn matrices with nonzero determinant and entries in the field F

[dull ginkgo]

|M_n(F)| = q^n^2 and GL_n(F) is a proper subset of it (actually it's the group of units) lol

M_n(F) is free over F, and it's basis is the matricies that are just 1 somewhere, and 0 elsewhere

basis of size n^2, so it has |F|^n^2 elements

[shell pilot]

I think I'm confused since the definition of GL_n(F) says nonzero determinant and the solution is saying there is a matrix in GL_n(F) that has a zero determinant

[dull ginkgo]

there isn't

by definition GL_n(F) is the set of matricies with nonzero (unit) determinant

[shell pilot]

It's just saying to exclude that so that's why it's < q^n^2

[rocky cloak]

Not in GLn, but there are matrices with 0 determinant. Hence not all matrices are in GLn

[topaz solar]

Not nonzero means zero

[dull ginkgo]

there's q^n^2 many nxn matricies over F

at least one of them ISN'T invertible

[shell pilot]

Ah, ok, I understand it now. Thank you all!

[topaz solar]

(Namely, 0)

[dull ginkgo]

You can also calculate the size of GL(F,n) by understanding that there is a bijection between invertible matricies and basises of F^n, i.e linearly independent sets of size n

A vector in a linearly independent set of size n is not a linear combination of the other vectors in that set, of which there are n - 1 many. That kicks out F^(n-1) many choices

[void cosmos]

hungerford

idk tho can't we like literally just count them like

if F has p elements then GL_n(F) would have (p^n-1)(p^n-p)(p^n-p^2...)

assuming GL_n(F) means nxn matrices with nonzero det

[still dew]

Bump?

[void cosmos]

more context please

[still dew]

Oh A is a vector space endomorphism over a algebraically closed field

A_s , A_n are semisimple and nilpotent endomorphisms respectively

It's basically an excercise in Jordan decomposition of A in End_k(V)

[void cosmos]

okay so u have basically a linear operator ,A, on V

right

[still dew]

Yeah

[void cosmos]

what about it

what do u wanna do

[still dew]

And A is expressible as a sum of diagonal and nilpotent matrix

Wrt some basis

[still dew]

I got this part

Now this nilpotent and diagonal matrices are also polynomials in A

That is what I have to prove and I am stuck

Also for more context this is an excercise in modules over PID

And it's applications to vector spaces

[void cosmos]

can u show that they commute with A

[still dew]

Well I was trying to get them to be polynomials in A first so that would lead to this statement

[void cosmos]

ur field is algebrically closed and you have n distinct eigenvalues and hence ur minimal polynomial equals ur characteristic polynomial so the converse holds too

[still dew]

Wha

I didn't understand

Converse of what

[void cosmos]

converse of if one matrix is a polynomial of the other then they commute

now if u show that they commute then ur done

the proof iirc was like

the minimal poynomial is of degree n so there must be a vector v such that (v,Av,A^2v,..,A^n-1v) is a basis and then since B commutes with A then writing B in this basis is ur poly

[still dew]

Perhaps, but I am restricted to using modules here

V is a k[x] module where x acts via A

And using structure theorem and CRT I was able to find it as a direct sum of k[x]/(x-a)^i

[void cosmos]

yes

these summands are ur factors

of ur minimal polynomial

[still dew]

No

They have other factors too

Remember after the decomposition we have f_1|f_2|...|f_n

[void cosmos]

my bad

[still dew]

Where f_1 is ur minimal poly

[void cosmos]

f_is are ur

yeah

[still dew]

Now a basis of a particular k[x]/(x-a)^i is 1,(x-a),..,(x-a)^i-1 so using this basis to write a I get the Jordan stuff

But after that I am stuck

How do I get A_s and A_n to be polynomials in A

[void cosmos]

ur field is k

V is k[x]-module with f*v = T(v) where T is ur linear operator

decompose V into cyclic modules

[still dew]

.

[void cosmos]

wouldnt this imply that

V is cyclic

as a k[x] module

[void cosmos]

it's reallly the same as this but just writing it out ig

ur minimal polynomial when restricted to each summand should be a factor of the next summand

the last summand is ur minimal polynomial but the characteristic polynomial and the minimal polynomial are the same

u must have only 1 summand which is the last summand

once u have ur module is a cyclic module, and ur endomoprhism commute together you can write one as a polynomial of the other

okay let me rephrase

V is a k[x] module

V = k[x]/(f1) (+) k[x]/(f2) (+)....

each f_i annihilates V which means (by our action) we have the minimal polynomia divides the f_is

the final summand (call it f_m) would be equal to the minimal polynomial

but the char polynomial is the product f1f2...

so u must have V =k[x]/(fs)

[still dew]

What is V=k[x]/f's?

You mean direct sum?

[void cosmos]

no its just only 1 summand

by what i just said

by the fact that the min polynomial equals ur char polynomial

[still dew]

It doesn't

We are doing this for any matrix A

[void cosmos]

idk then honestly my bad i randomly thought A has n distinct roots

eigenvalues*

what i can maybe throw is

maybe same argument but on the semisimple operator only

since the semisimple operator has minimal polynomial = to the irred factors

[still dew]

I see

[void cosmos]

may i ask what textbook is this

[still dew]

It isn't a textbook

It's my assignment lol 😭

[void cosmos]

i see

good luck

i hope tho this argument of like (min poly = char poly --> V is cyclic --> commutes <-> is a polynomial) is useful in the future

[light canyon]

Is there an intuitive explanation of why characters are so useful for studying reps of finite groups? I understand that they provide an orthonormal basis for class functions but I don’t have intuition for why that is the case (maybe because I don’t have good intuition for the Great Orthogonality Theorem)

[delicate orchid]

instead of having to decompose modules you add and subtract little lists of numbers

their main benefit is uniquely characterising isomorphism classes of C[G]-modules

the orthogonality theorem is easy to gain intuition for: Let x, x' be two characters associated to simple modules U, U', then <x, x'> = rk_C(Hom_{C[G]}(U, U')). Since U, U' are simple, by Schur's lemma this homspace is either 0 if U is not isomorphic to U' (iff x is not equal to x') or if they are isomorphic then it is spanned by the identity matrix, hence rank 1

[light canyon]

Hmmm I’ll think about that thank you

I think the main thing that’s bothering me is the feeling that we are discarding information when we take a trace

[delicate orchid]

but that's the amazing thing! We're not

[light canyon]

But I guess the extra information is redundant for the irreps?

[delicate orchid]

not just irreps, all reps (via semisimplicity)

[light canyon]

Right

[brisk coral]

is this for harvard university

[delicate orchid]

no I'm at a mid rate UK uni

[tidal torrent]

hello

im kind of stumped on this problem

let me pull it up

#49

i know that the size of G is not finite

but how can I prove that

So group G is isomorphic to Zp correct?

[void cosmos]

it says that this is true for every prime p

[tidal torrent]

yeah thats given

the |Zp| would be p correct

if so

[void cosmos]

how can it be isomorphic to Zp thrn

[tidal torrent]

wait

[tidal torrent]

group G is infinite because the homomorphic image Zp cant contain more elements than G itself right

[tidal torrent]

i think this statement is still true |G| >= |Zp| ?

[tidal torrent]

meant to say group G is infinite

[void cosmos]

yeah

[void cosmos]

What

[tidal torrent]

my point is uh

how would i come up with an example

the size of G has more than Zp though

[void cosmos]

yeah

Img(G) must be Z_p for any prime p

for some homomorphism

What do u think

like for each prime p

[tidal torrent]

the homomorphic image of group G, Zp is given?

[void cosmos]

yes

Isnt that whats said

[tidal torrent]

yeah

[void cosmos]

im on bed watching netflix so

Sorry if im stupid

[tidal torrent]

im the one whos asking for help

[void cosmos]

so Z_p is the homomorphic image of G

so for each p we have there exists a homomorphism such that im(G) is Z_p

what example comes to mind

infinite group as u saud

Said

[tidal torrent]

oh

[twin token]

Hiii
Is there a way I can know if a certain polynomial is solvable by radicals over Q without finding its Galois group? Specifically 3x^7-15x+5. I don’t know if it’s even possible to find its group

[rotund aurora]

It's S7 according to Sage but idk how to prove it

[rotund aurora]

it has 3 real roots, this implies it is not solvable since 7 is prime (Galois' theorem)

[dull ginkgo]

so here is a part of Jacobson

The intent is to find a base of a submodule generated by a given list of vectors in D^n, where D is a PID

And we can express the base using a relations matrix M. Using Elementary operations / Gauss-Jordan elimination, you can get to Row-Echelon form which we can call M'. So M' = KM for some invertible matrix K. But i don't get what you do from there to try to strip out any extra generators

[dull ginkgo]

hey chat, am I going insane too

Assume we have a ring R, where we have a unit b, and some element a such that (1 - ab) is a unit, then it's conjugate by b would be (1 - ba)?

b(1 - ab)b^-1 = b( b^-1 - a) = 1 - ba?

[crystal vale]

In a finite ring with unity, right inverse of an element implies that existence of left inverse, right?

Let a be an element in R such that ab=1. Now b cannot be left zero divisor so we have c such that bc = 1, so a = c.

Is it correct?

[dull ginkgo]

Yes conceptually, but here's an argument that proves the assumptions

Assume ax = 1 in finite ring R.

Lemma: a^n x^n = 1 for each n
Proof (Induction): a^(n + 1) x^(n+1) = a (a^n x^n) x = ax = 1.

Assume there exists an n such that: x^n = 0. then a^n x^n = 1 = 0. Thus for any r in R, r(1) = r = r(0) = 0, implying our ring is trivial.

Now, assume our ring isn't trivial. By pigeonhole principle, we have some n, m (n > m wlog) such that x^n = x^m.
If n - m = k, then x^m = x^(m + k), thus a^m x^m = 1 = a^m x^m x^k = x^k. Therefore 1 = x * x^(k-1), and thus x is a unit (with torsion).

[crystal vale]

Okay thank you

[dull ginkgo]

Also here’s a silly proof of something else i just thought of lol

[lean sail]

what kinds of things did you make flashcards on? (abstract algebra specifically)

[knotty badger]

Definitions, theorem statements, proofs, explanations, intuitions…

[dull ginkgo]

Let (1 - ab) = u, (1 - ba) = v
Then a - aba = av = bu

Assume xu = 1.

v = 1 - ba = 1 - bx(ua) = 1 - bxav
(1 + bxa)v = 1

So if u is a left unit then v is too,

[oblique iron]

Sorry, just saw this. There's a slight gap but the right idea. Because {g_1, ..., g_n} = {gg_1, ..., gg_n}, multiplying both elements of the sets together you get g_1...g_n = gg_1...gg_n, so using abelianness you get g_1...g_n = g^n g_1... g_n. cancelling g_1...g_n (multiplying by its inverse on the right), you get 1 = g^n. your solution doesn't quite work initially because you can't assume gg_1...gg_n = 1 right away without further justification (which amounts to proving the problem)

[dull ginkgo]

It’s basically verbatim the proof that a finite (or where two distinct powers of x equal for each x) left (right) cancellative monoid is a group lol

[crystal vale]

{gg_1,...,gg_n } = {g_1,..,g_n } = G, so gg_1...gg_n = 1, right?

[crystal vale]

Yes

[twin token]

Hi there 🙂
I wanna figure out why d divides phi(p^d-1) where phi is Euler’s function. Do I even need field theory for this?

lol sorry for interrupting

[oblique iron]

Yeah that works, just remember in your proofs you should include that bit of explanation

Hopefully the hint and such helped, best of luck on the algebra studies!

[crystal vale]

Thank you

[crystal vale]

You mean phi(p^(d) - 1) not phi(p^(d-1))

[twin token]

Yes

[crystal vale]

Interesting

[knotty badger]

You can use Lagrange’s theorem

[crystal vale]

How?

[twin token]

You mean find a subgroup of order d in U_(p^d-1)? I thought about it but how do I find one

[knotty badger]

Yeah

Wait what is that notation

[twin token]

The group of units in Z/(p^d-1)Z

[knotty badger]

Then yeah, try to find a subgroup of order d in that group of units

[twin token]

I don’t know how
It’s supposed to be related to field extensions somehow

[dull ginkgo]

Use Euler’s theorem

[rotund aurora]

One way of doing this is to consider the field F_p. How can you construct an extension of degree d?

[twin token]

The simplest is F_p^d

[knotty badger]

Wait huh

I had something a lot simpler in mind

[rotund aurora]

That's not even a field

[knotty badger]

Maybe my example doesn’t work…?

I can’t see why though

[dull ginkgo]

Phi being the order of the multiplicative group implies a^\phi(n) = 1 (mod n) if a and n are coprime (cuz then a mod n would be in the group of units)

[knotty badger]

I feel like you’re overthinking this

[dull ginkgo]

Soooo what happens when \phi(a^m - 1) has a nonzero remainder mod m :3

[knotty badger]

Maybe try to find some units in that group?

I really don’t think you need any fancy tech for this

[rotund aurora]

Well I said how it relates to field extensions, but I agree there is a much simpler manner

[lone niche]

are you mixing up Z/p^dZ with F_p^d?

[rotund aurora]

It doesn't even require Euler's theorem

[knotty badger]

Mhm

[dull ginkgo]

Euler’s theorem just allows you to go straight from the “order of group of units” def

[knotty badger]

I don’t even see how euler’s theorem helps you here

[dull ginkgo]

a is coprime to a^d - 1

[knotty badger]

You don’t need euler’s theorem for that

[rotund aurora]

Did you use that a is prime?

[dull ginkgo]

Oh shit yeah

[knotty badger]

This holds for any p

[dull ginkgo]

But not for any a

[knotty badger]

I’m not sure what you mean exactly

a is always coprime to a^d - 1

[dull ginkgo]

Oh brain fart

Also 2:30 am

Anyway my point being

[rotund aurora]

Not if d=0

[dull ginkgo]

oh I realized that instantly gives it away sorry

[twin token]

😭 thanks for the help
I think I lost it

[knotty badger]

Lost it…?

[twin token]

Lost you I meant

I’m not strong in number theory

[knotty badger]

This is still my suggestion

[knotty badger]

And you don’t really need any number theory for this beyond knowing modular arithmetic

[crystal vale]

I don't want a hint, I just want confirmation that v is unique so is it true in (1), if we want to prove converse 1 implies existence of inverse so can I use v is unique?

Please don't give hint

[quiet pelican]

You don’t need uniqueness in (1)

[pure forge]

can someone explain to me how we get bc-1 in the line where it says the remainder lies between 0 and bc-1.

[quiet pelican]

$0 \leq rc < bc$
and $rc$ is an integer
So $0 \leq rc \leq bc - 1$

[pure forge]

would it be then accurate to just say remainder lies between 0 and bc?

[cloud walrus]

Micose

[quiet pelican]

Kinda
But it can’t be $bc$

[cloud walrus]

Micose

[pure forge]

understood thank you!

[glad osprey]

Maybe I'm misunderstanding your reasoning, but g_1g_2...g_n = 1 is not true in general

[crystal vale]

But if {g_1...g_n} is a group which is abelian then it is true

Yes in non-abelian it is not true

[glad osprey]

hmm, what about (Z/3Z)^x? 1*2 = 2

I don't think it's true even for abelian groups

It's true if G is abelian and no element is its own inverse

[crystal vale]

Now I got it, my mistake thank you for pointing out my mistake ❤️

[crystal vale]

If I have 1, then we have uvu = u and vu^2v = 1.

So vuuvuv = uv so by using uvu = u we get, vuuv = uv => uv = 1.

And then vu = 1.

Is it correct?

And if I have 2, we have u(vu-1) = 0 and (uv-1)u = 0.

Now I will prove that u is not zero divisor.

Let uw = 0. Then u(v+w)u = u, but v is unique therefore w = 0.

Hence, uv = 1 and vu= 1.

Is it correct?

[quiet pelican]

How do you get vuuv = uv => uv = 1?

[crystal vale]

By second

[quiet pelican]

I would do it like this
1: || uvu = u so uvu^2 v = u^2 v
So u = u^2 v so then as v u^2 v = 1, vu = 1
Then again as vuuv = 1, uv = 1||
Your argument for 2 works

Ah right yeah

[crystal vale]

vu^2v = 1 so uv = 1

[crystal vale]

Thank you

[dull ginkgo]

One has to wonder

Who the hell comes up with some of these extremely specific problems

[rain grove]

Suppose $|G| = $55. Show:
(a) Every real subgroup of $G$ is cyclic.
(b) $G$ is either Abelian or has a trivial center.

[cloud walrus]

OHHELLNAH

[rain grove]

a) by Lagrange theorem every nontrivial real subgroup can be either of order 5 or 11. Both of them have prime order so they are cyclic (and Abelian). How to do b)?

[delicate orchid]

consider the possible sizes of G/Z(G)

[rain grove]

But how do I know what is the Z(G)?

[delicate orchid]

G/Z(G) is cyclic iff G is abelian

if you haven't seen this before, try and prove it

[rain grove]

Yeah I proved that

[delicate orchid]

see if you can use it together with a to prove b

[narrow zinc]

Give the "ideals" := 'normal subrings' of F_q [X]/ (X^q-X) where q is a power of a prime?

I do not know how well I translated that

We're looking for the ring version of a normal subgroup for groups (all examples)

[delicate orchid]

you translated it well

[narrow zinc]

Now X^q-X is completely divisible

All roots are in F_q

That's all I got 😆

[delicate orchid]

I'd use the corrispondence beteween ideals J of R that contain I and ideals of R/I

[narrow zinc]

Oh and that means the rign clearly has zeor divisors

[narrow zinc]

Ah yes

So X must be in it?

As it is a group for the first operation?

Ah wat that doesn't imply the statement

[rain grove]

So I can do quotient group G/N only if N is a normal group, so it has to be a subgroup. So I got 4 options for Z(G).

G/{1} ≅ G (trivial center)
G/Z_5 ≅? group of order 11 -> cyclic -> G abelian
G/Z_11 ≅? group of order 5 -> cyclic -> G abelian
G/G ≅ {1} (center is whole group -> G abelian)

Im not sure about case 2 and 3 tho, to me it looks like that would be a nice but Idk the argument

Ohh is it just lagrange theorem? |G| = [G:N]|N|?

yeahh it is omg thats so cool

[amber burrow]

Can someone explain the second step

I get that you can write it as a product of cyclic groups in that form

But why only powers of 2

[delicate orchid]

the product is a power of 2

[amber burrow]

Well the product of the orders of the elements is 2

[delicate orchid]

so all of its divisors are also powers of 2

hence all of the elements have order equal to a power of 2

[narrow zinc]

That works, because F_q is a field, F_q [X ] is a HID

[amber burrow]

But isn’t the order of some element of the product the max of the order of some element on the list

So as long as the max order is a power of 2 it’s fine

[delicate orchid]

dunno what this means

[narrow zinc]

Thus any ideal can be generated by a single element

[delicate orchid]

if some element has an order that isn't a power of 2, there is some odd prime that divides its order, and thus some odd prime that divides the product of all of the orders

immediate contradiction

[narrow zinc]

As X^q-X must be in it, I'm thinking the generators must be the factors

[amber burrow]

When you say element, do you mean if G or an element of one of the cyclic groups in the product

[delicate orchid]

does it matter? it's a statement about prime factors of integers

it holds for any group

there is zero group theory in that statement

[amber burrow]

I was mainly thinking that $|(g_1,g_2,\cdots,g_n)| = max(|g_1|, \cdots,|g_n|)$, so not every individual $|g_i|$ must be a power of 2, so long as the max is

[cloud walrus]

donut123

[delicate orchid]

yes that's true once you know that G is of the form they stated

using it to prove G has that form is circular

[amber burrow]

OOOH IM SO STUPID. I like completely misinterpreted the structure theorem in the first place

I got it now thank you

I just ignored the fact that the orders of each cyclic correspond to some factor of G

And by Lagrange theorem, G must be a power of 2, otherwise an element can have an order that is not a power of 2

[knotty badger]

Is it correct to say that orbit-stabiliser gives the following statement:

Every action of a group G on a set X is isomorphic to the coproduct/disjoint union of a family of “coset actions”, meaning an action where X is the set of (left) cosets of some subgroup of G?

[south patrol]

Yes

In detail 1) every G-set splits as a sum of its orbits, which are transitive G-sets tautologically 2) every transitive G-set is of the form G/H by orbit stabiliser

But 2) is a slightly stronger form than what is sometimes stated as orbit stabiliser ig, since often it is expressed as a bijection rather than an isomorphism of G-sets, but the map involved is clearly equivariant

[dull ginkgo]

Real quick are G-sets subsets of G or sets G acts on

[knotty badger]

Cool cool

I spent some time thinking about whether this was anything like decomposing a representation into irreducibles

But “simple objects” in [G, Set] don’t correspond to actions on sets of cosets

So it’s not really semisimplicity or anything, but it is a decomposition

[south patrol]

Sets equipped with an action of G

[dull ginkgo]

Thanku

[south patrol]

How are you defining simple objects? As in not having any nontrivial proper subobjects?

[knotty badger]

Oh I mean

[nocturne sail]

Can someone tell me the answer

[south patrol]

This doesn't go here or really anywhere

[knotty badger]

https://ncatlab.org/nlab/show/simple+object

That definition

[south patrol]

Maybe #prealg-and-algebra

[south patrol]

Sure

Though one important thing is like yeah this does give a decomposition of permutation reps but they can split further

[dull ginkgo]

I mean if we define a G-set to be simple if any subset is stabilized (viewed as G acting on P(G)) then it is empty or the main G-set, then we have that the action is transitive

[south patrol]

Obvious example being ZG lol

[south patrol]

Yeah if you use subobjects then you get these

[dull ginkgo]

Because the orbit is like a cyclic submodule

[knotty badger]

Oh I wasn’t thinking of the representations here, just the actual group actions

[dull ginkgo]

Just different idea with group G to Sym(S) versus ring R to End(M)

[delicate orchid]

it's exactly the same

there's even character analogues called "marks"

[south patrol]

These were important but then became far less valuable in 1920s Germany

Specifically around 1923

[delicate orchid]

it's woke gone mad

they're just representations over F_1 :smoked:

[rain grove]

Let $N \triangleleft G$. Is $G$ finitely generated if $N$ and $G/N$ are finitely generated?

[cloud walrus]

OHHELLNAH

[rain grove]

Im thinking it is possible that G is not finitely generated but i can't find an example. Was going in the direction of some polynomial structure but they are all rings... but maybe im just wrong

[quiet pelican]

G is necessarily finitely generated

Do you want me to give you hints towards that?

[rain grove]

Oh wow... Ig if N and G/N are finite G is finite, can that help me somehow?

[rain grove]

Yes

[quiet pelican]

Let g_1N, ..., g_kN be a generating set of G/N, and n_1, …, n_i a generating set of N
Hint 1: ||I claim g_1, …, g_k, n_1, …, n_i generates G||

[void cosmos]

if G/N has generators and N has generators then

oh okay micose got u

[quiet pelican]

Hint 2: ||Given g \in G, I claim there is some g’ \in <g_1, …, g_k> such that g’^-1 g \in N||

That should be enough to start

[rain grove]

got it ty

[south patrol]

The canonical map G -> G/N has finitely many fibres (preimages of points) and each is finite

[dark wave]

How do you calculate character tables for groups?

[quiet pelican]

There are a couple different ways

What information do you have about the groups?

[dark wave]

Like for dihedral and other finite groups for example

[knotty badger]

Finding the abelianization helps

Since that helps with the 1D characters

[quiet pelican]

This is a hard question to answer in general
Because there’s an algorithm that works for any group, but it’s pretty messy to calculate with
For most of the groups you’ll look at in courses, generally the best strategy is to spot a handful of characters, then essentially sudoku out the rest

[dark wave]

What do you mean sudoku out the rest?

[quiet pelican]

Using row and column orthogonality, as well as knowing that the sum of squares of dimensions is |G| and similar other facts

[fierce veldt]

GAYTTT

[rotund aurora]

mmh so a 1D representation factors through the abelianization, is there something analogous for GL_2? I mean I suppose you can consider the intersection of all the kernels of homomorphisms -->GL_2 but asking if there is a simpler description

[knotty badger]

Hm, not one that I know of…

[rotund aurora]

the structure of finite subgroups of GL_2(C) is quite rigid so maybe you can cook something artificial lol

idk much about GL

[chilly ocean]

There's a funny thing I've just noticed

In a group, for coprime m,n and element a, the equation x^m=a cannot have more than one solution of order n

To prove it, suppose x, y are such solutions, use Bezout to write 1=km+ln, then x = x^(km+ln) = a^k = y^(km+ln) = y

Is this correct? Would you prove it differently?

[hidden cairn]

it looks correct to me. maybe something like this also works:
assume x, y are two such solutions. as (m,n) = 1, we have $<x> = <x^m> = <y^m> = <y>$ so $y = x^k$ for some $k$ coprime to $n$. Together with the fact that $x^m = y^m$, we have $n | (k-1)m \implies y = x^k = x$

[cloud walrus]

pink_panther

[hidden cairn]

Does anyone mind checking my proof for this question? Let $Q$ be a Sylow $q$-subgroup of $S_q$. Show that $C_{S_q}(Q) = Q$.
We must have $|Q| = q$. So $Q$ is a cyclic subgroup of $S_q$. Let $Q = < \sigma >$ where $\sigma = (a_1 ... a_q) $ is just some q-cycle. Then $C_{S_q}(<\sigma>) = C_{S_q}(\sigma)$. Since q is a prime number and conjugation preserves the cycle type, we must have, for $\tau \in C_{S_q}(\sigma), (\tau (a_1) ... \tau (a_q)) = (a_1 ... a_q)$. This tells us that $\tau$ should shift the order of the numbers appearing in $\sigma$ by any amount, therefore $\tau$ must be an element of $Q$. Similarly, conjugation by any element of $Q$ shifts the order of the numbers by some amount and keeps the permutation $\sigma$ unchanged. So $C_{S_q}(Q) = Q$.

[cloud walrus]

pink_panther

[south patrol]

Yes that seems good.

Another quick and essentially equivalent way I believe would be to consider the conjugacy class of a generator of Q. ||It is conjugate to all q-cycles, and there are (q-1)! of those||

[dull ginkgo]

hey potato man

[delicate orchid]

Incredibly late but here’s some tricks, when I say character I mean irreducible character:

• lifting from the abelianisation
• the sum of the squares of the degrees is the order of the group, and the degree must divide the order of the group
• row and column orthogonality
• they come in complex conjugate pairs, and are only not real valued on conjugacy classes that are not closed under taking inverses
• If x is a linear character and y an irreducible, then xy is irreducible
• the square of the degree of a character is bounded by the order of the inner automorphism group

More generally, you can also look at tensor/alternating/symmetric powers of characters, induced characters, and lifts from quotients and factor them using the inner product

[south patrol]

Yo

[dull ginkgo]

dude ngl I feel like a moron because this section of Jacobson is not clicking for me

[south patrol]

Here I should say it is a beautiful and actually surprisingly elementary theorem that every rep arises as a subrep of a tensor power (p sure this sharpens to exterior)

So like if you are stuck sometimes just tensoring up is a nice way to find something new lol

[delicate orchid]

Did not know about the sharpening. That is nice if true

[south patrol]

It is likely false then

[delicate orchid]

Cause iirc the exterior power operator is nilpotent

[south patrol]

Wait what am I even saying

Tensor power of what

[delicate orchid]

The character/rep

[south patrol]

Yeah but as stated it is tautological

Lol

[delicate orchid]

U literally just square the character

Oh true

Uhhhhh what’s the actual statement again

One mo

[south patrol]

Oh isn't it like

You pick one faithful one

[delicate orchid]

I remeber

Ur thays it

[south patrol]

And then everything is a subrep of a tensor product of that

[delicate orchid]

It’s peak…

Does THAT sharpen to exterior power

[south patrol]

Do faithful representations come cheaply enough though lol

[delicate orchid]

Uh yeah

[south patrol]

Well obviously you can take CG

But like lol

[delicate orchid]

In any group that matters (finite simple)

[delicate orchid]

Does it have to be irreducible?

[south patrol]

Maybe this is false w exterior powers tbh

[delicate orchid]

Nope

[dull ginkgo]

Trying not to interrupt the convo so I'll wait :3

[delicate orchid]

I really don’t want it to be

[south patrol]

It is true for symmetric powers apparently

[delicate orchid]

Ughhhh not as nice

[south patrol]

If |G| = n then Everything is contained in kth symmetric power for some k < n

[south patrol]

But still quite strong

[delicate orchid]

Nvm if it’s bounded it’s just as nice as the thing I had in my head

Even nicer cause there’s no (-1)^k bullshit in the character formula

[south patrol]

Pog

[delicate orchid]

Do u know how it works btw potato (if that is ur name)

[south patrol]

Lol

Uhh no but I found the proof online, 1 sec

the tensor power one is easier but you've seen that

[delicate orchid]

The tensor power one is trivial

[south patrol]

Lol is it trivial

[delicate orchid]

Yeah kind of

What’s the trace of the kronecker product of two matrices

[south patrol]

Wait i think we are talking about different things

lol

I meaant the proof everything is a subrep

[delicate orchid]

Im talking about \chi_V^\otimes

Oh no that one isn’t

[south patrol]

okay that is trivial yes

lol

diagonalise

:chad:

[south patrol]

Tbh I have never worked through this and you have just made me interested

I think this actually came up in my diss last year in some form maybe

hmm

[delicate orchid]

Ok “library of babble” ur turn

[dull ginkgo]

yeah i am quite stupid and do not fathom what the fuck to do here :3

Like namely

We can use "elementary" row operations and some primitive form of gauss-jordan

but Idk how to actually take advantage of the Bezout property

because you can only remove scaled copies of other rows

[delicate orchid]

This really is a smith normal form moment

Full disclosure I saw a tuple of 3 integers and nothing else

[south patrol]

Wait lol

There isn't even a closed form right

Lol

[dull ginkgo]

dude idfk

[delicate orchid]

Ok upon re-reading I would smith normal form a) at least

[south patrol]

Do you want the exponential generating function

i'm tlking to wew sorr

[dull ginkgo]

dude idfk

[delicate orchid]

Iterative solution is fine cause that’s the only one I know of rn

[south patrol]

as in these

[dull ginkgo]

i've reread this like 5 times

[south patrol]

Oh lol

[dull ginkgo]

still not clicking

[delicate orchid]

I’m scared

[delicate orchid]

I’m even more scared but just Google “smith normal algorithm”

[south patrol]

Oh like a recurrence relaation type ting?

Hm

[dull ginkgo]

i tried reading it but it was even MORE confusing

[delicate orchid]

Wait is this true? I guess it is cause u can’t just stick it in the elementary sym/alt polynomial cause u have the chi(g^k)s AND the chi(g)^ks

[dull ginkgo]

just gives us this "Relations relative to another basis"

but doesn't elaborate at all wtf to do with it

or how to find a basis out of it

[delicate orchid]

Don’t read it numbnuts try performing it

Follow the steps without a single thought in ur little head at first

[dull ginkgo]

just the wording of it fries my brain

and I don't know what the hell I am doing

or how to show it gives a basis

[delicate orchid]

Yeah you can write them in terms of the character of the previous power iirc

[dull ginkgo]

I can compute the normal form, found it already but IDFK what to conclude

or how the fuck to find a basis from it

[south patrol]

Yeah apparentlyt this is basically nonsense with the Newton polynomials

Which makes sense lol

[dull ginkgo]

and frankly i feel like a moron enough because it's pathetic to be stuck this long on a section with basic fucking linear alg

[delicate orchid]

Have you see the proof of the structure theorem for modules over PIDs

The non-zero diagonal elements of the smith normal form describe the torsion part and the number of 0s tell u the rank. That’s what conclude from jt

[delicate orchid]

WHOton polynomials ykwim

[dull ginkgo]

that's in the next two sections

I really don't like skipping ahead

in jacobson mainly

[south patrol]

This is stuff I only hearad of through adams ops tbh lol

[delicate orchid]

The proof is basically “u compute the smith normal form and u can compute it”

[delicate orchid]

This still applies

[delicate orchid]

What the newton polynomials?

[south patrol]

Yeah lol

Well the terminology anyway

[delicate orchid]

I couldn’t tell you anything about them nor do I really want to

[dull ginkgo]

BUT H O W D O I

KNOW WHAT TO DO

[delicate orchid]

Idk put it in a computer

[dull ginkgo]

WITH THE NORMAL FORM

[delicate orchid]

I just told u boy

[dull ginkgo]

, SBCRHyga,bkjw e.ynk.hqEAKFNUD;M ealk <o{
w

[delicate orchid]

Ur cooked

The diagonal elements are literally the generators

[dull ginkgo]

Apparently according to Jacobson that's relative to another basis

so you would have to return to the original basis no?

[delicate orchid]

Doing it over Z and u got (3.4.5.0) diagonal elements woaahhh

[dull ginkgo]

so can you just "slice out" the null column/row of the diagonal

[delicate orchid]

We’re in Z there’s only one basis of a free Z module up to sign

Let M and M’ be in M_2(Z) in different bases iff exists a X in GL_n(Z) such that XM = M’ taking determinants we have det(M) = \pm det(M’)… ok maybe this isn’t a complete proof cause but like cmon now just visualise it 😭

Ur acting on a cube u can only reflect things by changing signs cause it’s like S_n

Point is I don’t care about this basis shit. Does it give the required submodule? Yes.

[dull ginkgo]

that's WHAT THE PROBLEM IS ASKING F

[delicate orchid]

Are the terms written in the standard basis e_i = (0,…,1,…0)? Yes

[dull ginkgo]

Also don't i have to invert the very last matrix

in order to find the actual basis elements

[delicate orchid]

Dunno what that means but good luck inverting something that will usually have both non-units and zeros on the diagonal

[dull ginkgo]

I mean

[south patrol]

The deal is like, fix $n \ge 1$ and put $p_k(x_1,\dots,x_n) = \sum_{i=1}^n x_i^k$ and $e_i$ the ith symmetric poly [in $n$ variables]. The newton thingy identities say $ke_k = \sum_{i=1}^k (-1)^{i-1} e_{k-i} p_i$. Now note that if $T: V \to V$, $\dim V = n$, has eigenvalues $\lambda_1,\dots,\lambda_n$ [i'll assume diagonalisability for ease...] then $\chi_{\Lambda^k V} g = e_k (\lambda_1,\dots,\lambda_n)$. So ig ur question then becomes like why the Newton identities and iirc the easiest waya is some identity with $1/1-t$ lol

[dull ginkgo]

LRX^-1 = D

[cloud walrus]

Prismatic Potato

[south patrol]

Uhhhh

[dull ginkgo]

so LR = DX is a basis?

i am super fucking confused

trying to go between what jacobson is stating and what you are

[delicate orchid]

I remember doing this stuff in third year ug

Would’ve been fun if I didn’t have to compute SN form by hand

[dull ginkgo]

i mean I did it for the matricies given but got stuck on the second one

Just using elementary invertible column and row operations

[delicate orchid]

We’re in U(n) it had better be diagonalisable

[south patrol]

i think like yeah $\prod_{i=1}^n (1-tx_i) = \sum_{k =0}^n (-1)^k e_k t^k$ and then uhhh how do you get the identities

[cloud walrus]

Prismatic Potato

[dull ginkgo]

but the problem is

it doesn't generalize apparently

[south patrol]

I think ur meant to take 1/ everything lmao

[dull ginkgo]

and you have to use Bezout property and IDFK how

In general I don't think you can write out every invertible matrix as a product of elementary matricies

[delicate orchid]

SN form uses bezout’s everywhere

[dull ginkgo]

it seems to be a Euclidean ring quirk where that is true

[south patrol]

Yeah

Doesn't even hold over PIDs

[dull ginkgo]

which smith normal form calculation DOES use it

so idfk

if I just have to

allow a fourth type of elementary matrix

[south patrol]

Appaaarently differentiate aand mess about lmao

[dull ginkgo]

in the standard gauss-jordan methods

[south patrol]

Which makes sense given the ke_k

[dull ginkgo]

Trying to think of a way to actually like

describe this fourth type of elementary matrix

[delicate orchid]

What the sigma is going on

The matrices in SN don’t even have to be invertible dawg

The matrix in the last question will have a kernel of rank 1

[delicate orchid]

Sorry, 2

[delicate orchid]

Ok? Who gives a shit

The matrix you start with doesn’t need to be invertible

[dull ginkgo]

yeah no shit

But the elementary ops

have to be

[delicate orchid]

Ur opps are elementary atm

[dull ginkgo]

and I am trying to describe a fourth type of elementary op that incorperates the Bezout Lemma

[delicate orchid]

U mean multiplying a row by a number

[dull ginkgo]

well not exactly

multiplying a row and removing another in an invertible manner

[delicate orchid]

If this is really bothering you just solve them as a set of simultaneous equations like ur in highschool lol

Should work

[dull ginkgo]

Messy as fuck and I'll only be more screwed over in the next section

[delicate orchid]

It’s the same algorithm you’re just putting the coefficients into a matrix I don’t get it

[dull ginkgo]

you can't just scale a row/column by a nonunit willy nilly

you can add a scaled row/column

[delicate orchid]

Uhhhhh yeah u can actually

[dull ginkgo]

nonzero determinant?????

[delicate orchid]

I’m kinda crazy like that I’m kinda insaneeeee

I might even make the matrix bigger for no reason who’s gonna stop me

[dull ginkgo]

very unhelpful :3

and once again

[delicate orchid]

Like ok it’s nice that the matrices turn out to be invertible but u should just try and diagonalise it all costs

Who cares just get it in the diagonal form

[dull ginkgo]

Let M be the relator matrix, then we have
R M C = D, where R is the composition of the row ops and C is the composition of the column ops, and D is the diagonal

jacobson says the matrix R M (C^-1)^-1 generates the submodules as the relation matrix relative to the basis vectors of C^-1?

so wouldn't RM be the basis then apparent fucking ly

[delicate orchid]

R M C^-1^-1 or a I like to call it R M C

[dull ginkgo]

C^-1 = P is how Im stating it's instead relative to the basis of the INVERSE

for my own sanity because this is already confusing as it is

[delicate orchid]

You’ve confused me about your confusion now

Is 2am so I’m just gonna once again reiterate. There is a algorithm. Use it.

There is a finite procedure with zero thought required that will generate you infinite explicit examples of this theorem. Use it

Do way more examples over Z

[dull ginkgo]

that defeats the fucking point of me trying to understand what I am doing

instead of just doing an algorithm willy nilly without knowing why

like the algorithm gives me a goddamn diagonal sure

but idfk how to find a matrix off of that given what jacobson states

[dull ginkgo]

USE IT FOR WHAT DUDE

I have a diagonal matrix cool yeah wtf am I supposed to do with it

i have a huge tension headache now so i am going to sleep

[long swan]

[tulip glacier]

A monoid represents the endomorphisms of an object in an arbitrary category, under composition.

A group represents the automorphisms of an object in an arbitrary category, under composition.

Do abelian groups represent anything in this sense too ?

[chilly ocean]

That is a good question. I've wondered about this a lot and haven't found a satisfying answer yet. One thing you can see is that it turns out that abelian groups are the same thing as group objects in the monoidal category of groups with cartesian product. But I don't think that's very satisfying or explains their importance very well

[rocky cloak]

I doubt there could be something like this.

Like being an automorphism is a specific property of a morphism. But a group being abelian is more of a global property. There's nothing that could set something that's an element of an abelian group apart, because every element of a group is also an element of an abelian subgroup.

Of course you can cook up some abelian groups associated to an object if you want though. Like the center of the automorphism group for example.

[south patrol]

Also minor quibble but I wouldn't rly say represents here - I assume you essentially mean the "endos naturally form a monoid"

Unless I misunderstand

[rocky cloak]

Another perspective is that when you're concerned with endomorphisms, you're really thinking about how monoids / groups acts on objects.

But the action of abelian groups is not really significantly different from general group actions. Abelian groups are more prevelant as the thing being acted on (i.e. modules)

[wide brook]

Hello everyone, I have a quick question. Why isn’t the ideal $\mathbb{Z} \oplus 0 $ of the ring \mathbb {Z} \oplus \mathbb {Z} a free \mathbb {Z} \oplus \mathbb {Z} module? Isn’t (1,0) a basis of it?

[chilly ocean]

(1,1)*(1,0) = (1,0)*(1,0) so there is no uniqueness

[charred iris]

Moreover {(1,0)} isn't linearly independent

Since over Z oplus Z you have non trivial (0,1).(1,0) = (0,0)

[wide brook]

Oh yeah you’re right I forgot about uniqueness

[wide brook]

This means that the anhilator of (1,0) is non zero right?

[wide brook]

But what do you mean it’s not linearly independent. If it’s one vector, how can it be linearly dependent?

[charred iris]

Well it's not a vector, which is how

I gave an example of $r.(1,0) = 0_R$ where $r$ was not $0_R$

[cloud walrus]

Edward II

[charred iris]

Where in this case $0_R$ is $(0,0)$

[cloud walrus]

Edward II

[wide brook]

Hmmm okay I think I got it, thanks!

[charred iris]

And I mean Z oplus Z by R

[wide brook]

Might be stupid but can we see (1,0) in another way like (1) oplus (0) right?

And you showed that (1) and (0) are linearly dependent? That’s what’s happening here?

[charred iris]

idk what you mean

[wide brook]

Yeah tbh I don’t know how to express it. Don’t worry I think I got it

So whenever a certain finite set like (1,0) here contains the 0 element of the module it can never be a basis right?

[chilly ocean]

{(1,0)} doesn't contain the 0 element

It has a nonzero linear combination that gives 0

That is what it means not to be linearly independent

[quiet pelican]

Is there a natural way to define A_\infty?
Like we can define S_\infty as the permutation group of a countably infinitely set

[coral spindle]

If you take S_infty to be the union of all the symmetric groups - which is to say, permutations of N with finite support - then A_infty is the obvious thing inside that.

This is more often what people refer to as S_infty

[quiet pelican]

Except I don’t want to do that (I’ve already thought about it and that’s an easy case)

MO is saying there is none, but just stated as a fact in the question

https://eudml.org/doc/217873
Reference, apparently

[chilly ocean]

You could still define the subset of permutations that are a finite even amount of transpositions away from the identity. Idk if that's natural or useful though

[quiet pelican]

That’s just the direct limit of S_n

[chilly ocean]

Is there any particular reason why you want to define something like this?

[naive lance]

i have a question regarding krulls principal ideal theorem and dimension theorem

in the dimension theorem: can we start the induction with the empty generated ideal?
in the principal ideal theorem if (x) isnt a unit or a zerodivisor, and p is a minimal prime ideal over (x) then hight(p) = 1
ht(p) leq 1 is clear by the dimension theorem but i dont get the hight of p is bigger then 0

[quiet pelican]

Because I want to

[cloud lynx]

what is the connection between units and invertible Elements?

isnt it just unit only if when its also invertible?

per definition

[dull ginkgo]

A unit is an invertible element basically. But there are also left and right exclusive units

Ones which have a right inverse or a left inverse (like retractions and sections for the function analogues)

[cloud lynx]

ahh ok tyyy

[cloud walrus]

clubsoda14

[nimble folio]

Am I doing this right? Seems kind of fishy to me

I don't even know that by showing this I answered the question

[chilly ocean]

What is Hom_F(F, F)? F-Vector space homomorphisms?

[nimble folio]

Yes

[chilly ocean]

You want to show that given an endomorphism T, T(v)=alpha*v for some alpha

Not show that T(v)=alpha*v is a endomorphism

[nimble folio]

I see

How do I do that?

Like a hint or something

[chilly ocean]

Can you think of any nice basis for F as a F-vector space?

[nimble folio]

Uhh

{1} ?

[chilly ocean]

Yes, that is a very good basis

[nimble folio]

Well

I'm just having trouble seeing why, in this problem, its necessary that I find a nice basis

I don't have that intuition

[chilly ocean]

You could with any basis

This one is just quicker

You will see

[nimble folio]

Why do I need any basis!

How does that help?

[chilly ocean]

Do you know about any relation between linear maps and bases?

[nimble folio]

Uh

No

Do bases map to bases or smth

[chilly ocean]

No

But given a basis of a vector space V

And another vector space W

Linear maps from V to W correspond bijectively to functions from the basis of V to W

This is a very important proposition of linear algebra

[nimble folio]

Okay I should have probably mentioned

I've taken two semesters of abstract algebra

But its probably safer for you to assume I've had 0 linear algebra

I took it a very long time ago and I don't remember the basics

[chilly ocean]

That's unfortunate

[nimble folio]

Yea...

[kind temple]

viewing F as a dimension 1 vector space over itself, you can write every element f in F as 1 * f

more generally, for any non-zero element b of F, {b} will be a basis for F as a dimension 1 vector space over itself

this is really all you need

as a suggestive hint, break it into two cases:
either T(1) = 0 or T(1) != 0

[nimble folio]

im not looking

plz delete

[dull ginkgo]

Sorry, misread

[nimble folio]

all good LOL

[knotty badger]

Not sure why cases are necessary here

[kind temple]

they arent

[kind temple]

its just to set up for use of this.
its not the slickest way to go about it

[knotty badger]

I’m not sure bases are necessarily how I’d think about this either

[dull ginkgo]

The idea of a basis is not necessary here yeah lol

[knotty badger]

There are kind of two separate Qs here

[dull ginkgo]

Unless you are seeking to generalize to like

Dual of direct sum is direct product of duals

sorta business

[knotty badger]

1. If T(v) = alpha*v for some alpha, how can we find alpha using T?
2. How can you show that any T : F -> F takes the form T(v) = alpha*v for some alpha?

[nimble folio]

did I not show 2) in my proof

[knotty badger]

What was your proof

[nimble folio]

this

[knotty badger]

That’s not a proof of “2”

You’ve shown that, for any alpha in F, the map sending v to alpha*v is linear

But this isn’t what I’m asking with 2

[nimble folio]

i see

[knotty badger]

“2” is essentially the following statement

$\forall T : F \to F \text{ linear, } \exists \alpha \in F \text{ such that } T(v) = \alpha \times v \space \forall v \in F$

[cloud walrus]

Pseudonium

[dull ginkgo]

@nimble folio if T is linear, then what is T(xy)?

[nimble folio]

Uhhhh

I'm just gonna say

xT(y)

Even though I think thats wrong

[dull ginkgo]

Yes!

[nimble folio]

Okay

Wait

[knotty badger]

Yeah that’s right

[nimble folio]

Wait

[dull ginkgo]

x and y are BOTH vectors and scalars

[nimble folio]

Why is it not xyT(1) then

[knotty badger]

It is lol

[nimble folio]

Wtfffffff

[dull ginkgo]

it is xyT(1) too

Because we “factor out” scalars

[knotty badger]

Yep

[dull ginkgo]

But the vectors are the scalars

[nimble folio]

Thats so weird

Wait before you guys continue to hold my hand through this problem can I post what I've written

Im working on a whiteboard

[dull ginkgo]

Epic

Whiteboards are the shit

[knotty badger]

(This trick is called “applying the yoneda lemma”)

[dull ginkgo]

Blackboards are too if the chalk isn’t like writing with a chunk of granite

[dull ginkgo]

Morita

[nimble folio]

Let $\alpha \in F$, $\alpha \neq 0$. Then ${ \alpha}$ forms a basis for $F$ (as an $F$ v.s.).

Let $v \in F$. Then $v = \alpha v_0$ for some $v_0 \in F$.

Hence $T(v) = T(\alpha v_0) = \alpha T(v_0) $

[cloud walrus]

clubsoda14

[dull ginkgo]

Well no

They are not linearly independent

Actually the exact opposite lol

[nimble folio]

Where did I assume that they are linearly independent

v = av0?

[dull ginkgo]

Oh

I thought {alpha} was over all of them

I am a dummy

[nimble folio]

Wait

I'd like to understand why you thought that

[dull ginkgo]

I am a moron

That is why

[nimble folio]

No but I'd still like to know for pedagogical reasons

[dull ginkgo]

Literally it’s me just magically interpreting set builder notation that wasn’t there

[dull ginkgo]

The biggest problem with making something foolproof is underestimating the ingenuity of complete fools

~Douglas Adams

[nimble folio]

Ok....

Im just gonna say

I don't fully understand why I was able to write v = a v_0

Im guessing its since {a} forms a basis and so every element of F can be written as a linear combination of the basis

[dull ginkgo]

Yes

[nimble folio]

Ok

[dull ginkgo]

y = (yx^-1)x

[dull ginkgo]

But that doesn’t help us determine that T(x) = ax, just that T(x)= (x \alpha)^-1 T(\alpha).

but \alpha^-1 is a scalar, so what can you do with it?

[nimble folio]

You can pull it out

[dull ginkgo]

Or in this case, put it back into T’s argument

Giving us what?

[nimble folio]

T(x) = xT(1)

[dull ginkgo]

:3

[knotty badger]

And there you go

[nimble folio]

And T(1) is just an arbitrary scalar Im guessing

[dull ginkgo]

T(1) UNIQUELY determines T

[knotty badger]

It’s an element of F

[nimble folio]

Woah

[dull ginkgo]

Think about if F(1) = T(1)

[nimble folio]

Yes

[knotty badger]

The idea is that F has a multiplication

[nimble folio]

I just got flashbacks from group theory

[knotty badger]

So given an element a of F

You can ask what it “does” to other elements of F

So you can look at the function f(x) = a*x

You’ve converted a, an element of F, into a function F -> F

You can also go in reverse - if you know what an element “does” you can deduce what it “is”

And the way you do this is by “following the identity” element for multiplication

You follow where 1 gets sent, and that gives you a, since f(1) = a * 1 = a

Of course, if g : F -> F is any function, you can always find whatever g(1) is

But there’s no reason to expect that g(1) determines all of g

[dull ginkgo]

This leads to an approach to show that the F-linear endomorphisms of F^n (maps to itself) are matrices, a vector space over the matrices who are 1 in one spot and 0 elsewhere

[knotty badger]

It turns out that if g is linear, then g(1) does determine all of g

[dull ginkgo]

Like constructively

Simple consequences of “functions equal over generators are equal”

[knotty badger]

So essentially, one thing that identity elements are useful for is allowing you to convert from “does” to “is”

And if you have a general function, there’s usually some analog of linearity you have to impose so that “following the identity” really gets you the whole function

Which in a sense is the “is” to “does” direction

[nimble folio]

Im so sorry

[nimble folio]

I don't understnad whats happening here

T(v) = T(a v_0)

[dull ginkgo]

yes but what is a

[nimble folio]

How do I get to T(v) = (v a)^{-1} T(a) ?

[dull ginkgo]

I just used different variables

If v = a v_0

What is a?

[nimble folio]

a = v v_0 ^{-1}

Oh

Lol

Okay

One second

[dull ginkgo]

Trust me it happens

[nimble folio]

I promise I'll read your guy's explanations after the fact

[dull ginkgo]

I was close to slamming my head on my desk yesterday due to my intro alg textbook

[knotty badger]

oh?

[dull ginkgo]

The end of it

[nimble folio]

okay

I think i figured it out

its really confusing since these are F-vector spaces over F

am i just able to decide what I want to behave as a "vector" and what I want to be able to behave as a "scalar"

$T(v) = T(\alpha v_0) = \alpha T(v_0) = v v_0^{-1} T(v_0) = v T(v_0^{-1} v_0) = vT(1)$

[cloud walrus]

clubsoda14

[dull ginkgo]

Shit gets conceptually fucky at first when your base scalar ring/field can be embedded right into your module / vector space

[nimble folio]

Yea thanks guys I appreciate it

I'm gonna reread the cool explanations you gave after the fact I just had to figure out what was going on

[dull ginkgo]

It takes a bit sometimes also to get used to the difference between inclusions and embeddings

Like an injective map from A to B let’s you say that “A is a subset of B” via A’s image but

You can have many other embeddings (especially with automorphisms of B) of A into B

[nimble folio]

Thats cool

Not quite sure what the difference between inclusions and embeddings are but I'll look it up

[dull ginkgo]

Dude it took me eons to process

[nimble folio]

I don't even know what an embedding is LOL

[dull ginkgo]

An injective map

[nimble folio]

How is that different from an inclusion!

What!

[dull ginkgo]

I say map because we want it to respect some structure

[nimble folio]

Oh

[dull ginkgo]

Exactly, an inclusion is being a “subset” of another

[nimble folio]

Ohhhhh

[dull ginkgo]

Like R is a subset of C (inclusion) But R can be embedded into R^2 via x to (x,0) OR (0,x)

[nimble folio]

Okay

I think I understand

Its like

If $A$ is just a set contained in a $B$, but $B$ has added structure (like if it were a group), then $A \subseteq B$ is just an inclusion

[cloud walrus]

clubsoda14

[dull ginkgo]

It is still an inclusion,

Like an injection is an embedding of sets

But if A is a subset of B that doesn’t mean it necessarily is the ONLY way it embeds

Like for instance, the evens are included in the naturals right

[nimble folio]

Yes

[dull ginkgo]

But the evens are bijective to the naturals, so the naturals embed into themselves

And sending the evens to the odds within the naturals by adding one is also an embedding

Namely the IMAGE is included

sometimes it is a canonical embedding, or a “special” one

Like a ring R into it’s polynomial ring R[X]

But that’s because that inclusion is critical to how R[X] behaves

[nimble folio]

I'm gonna have to really think about this more

Maybe we'll talk about it in my analysis class

[dull ginkgo]

It’s funny because you’re basically ridding yourself of what you’ve expected to be concrete

Slowly you just accept what you see and see how it’s useful and why, and work with it

[nimble folio]

Lol

Hopefully

This stuff is pretty complicated I have a long way to go

[dull ginkgo]

It’s complicated but you get used to it

It’s like learning a different way to think or a different framework

Oddly similar to object oriented programming

Which made me understand category theory a LOT better and vice versa

Unsurprising because programming is very similar

[dull ginkgo]

The problem is though sometimes you get used to these frameworks and start craving generality and understanding exactly why things happen which is frankly a luxury in algebra lol. Sometimes there is insight to dropping the specifics

I was super enthralled by an alternative way of handling Galois theory (Jacobson-Bourbaki) and frankly it makes more sense but takes a bit more effort

Also helping me understand the structure of modules and endomorphisms

[knotty badger]

Yeah so for example, A being a subgroup of B is the same as saying the inclusion map A -> B is a group homomorphism

[icy totem]

Find integers solution of x^2-2y^2=+-1

To solve this problem I should use ring theory apparently

Anyone manages to do it?

[wraith nexus]

consider Z[sqrt(2)]

[naive lance]

:/?

[rotund aurora]

You can identify the integer solutions of this equation with the units of some ring

[rocky cloak]

I think you can prove that every element of a minimal prime is a zero divisor.

If your ring is Noetherian then localizing gives an artinian local ring, so then it's straight forward. But it should be true in general as well.

Hmm, yeah I guess the argument generalizes quite nicely. If you have a dimension 0 local ring, and a non-unit that is not nilpotent, then you can invert that element and get a non-zero ring. But any ring has a maximal ideal, so then there would be a prime properly contained in your height 0 prime

[glad osprey]

Our lecture notes says that if R is a division ring then 0 is the only nilpotent element. Isn't this true even if R is an integral domain?

[chilly ocean]

Division rings are not integral domains. But it is true for domains, yes

Domains are like integral domains but without the commutativity requirement

[glad osprey]

Oh, right, forgot about commutativity. Thanks 👍

[chilly ocean]

If x was nilpotent then x^n=0 for n>1 so x(x^(n-1))=0. If x^(n-1) is not 0, then x is a zerodivisor. Else use induction until base case xx=0

Also should be noted there are non-domains without nilpotent elements like Z/6Z

[glad osprey]

Yep, that was the proof I found too (assuming you meant x^(n-1))

[chilly ocean]

Oops, of course

[south patrol]

Not being a zero divisor means multiplication by x is injective, and the composition of injective maps is injective

Lol

[chilly ocean]

True, that is a nice perspective

[rocky cloak]

In general such rings are called reduced

[glad osprey]

I see. Is there a concept of elements in a ring for which a^n = 1 for some number n? Since we have both the characteristic and nilpotency which talk about how elements return the additive identity after some number of operations

[dull ginkgo]

Torsion usually

But that implies that a is a unit

So yeah it’s torsion within the group of units

[glad osprey]

Aha, so torsion is technically a group-theory term, but we can just look at the group of units

Can you use terms torsion ring or torsion-free ring then?

[dull ginkgo]

However

If we have a ring such that, for each x, we have an n such that x^n = x, then our ring is commutative