groups-rings-fields

[delicate orchid]

I'm presuming you're using rho for a 2-cocycle. Which is like using "\omega" for a vector space

[lone niche]

Not yet, we have the Inverse Galois Problem for example. https://en.wikipedia.org/wiki/Inverse_Galois_problem

[hidden wind]

right right i've some reading queued up on this

[median pawn]

glad i made it here

so i'm wondering how it follows that ker(phi) is a normal subgroup

[grave sedge]

It's always a normal subgroup

The kernel of any homomorphism

[lilac sluice]

guys, can you help me?
If I have a group of order 4p and it is not simple, how can I show it is a solvable group? I mean, the approach. Because I have no ideia
I know that a group of order pq (p and q prime numbers) is solvable, but this does not help me

[grave sedge]

Take a normal subgroup

(proper)

It has to he solvable

And the quotient has to be solvable

[lilac sluice]

ok I will think about it
thank you very much!

[barren sierra]

I've got everything except f is irreducible

I've ruled out the possibility that a = p, b = p^2, p prime

But if that's not the case idk how to apply Eisenstein for example

[median pawn]

sorry my bad i meant to ask why is it non trivial

like how does the 11 and 8! argument work/help

[delicate orchid]

have you tried the rational root theorem? I don't see how the gcd nonsense would come into that though

[barren sierra]

G / ker is a subgroup of S8 is what that argument needs @median pawn

[barren sierra]

Yea I got nothing from that

[median pawn]

ahh cool! thanks

[barren sierra]

If you wanna be really concrete assume ker is trivial and go from there

[delicate orchid]

I'm dreadful at ENT so I couldn't tell you what that gcd condition is giving us

that if p^2|a then p cannot divide b and vice versa?

like ok?

[barren sierra]

I'm also dreadful lol

[delicate orchid]

these sorts of questions just require a way of thinking I do not possess

[rotund aurora]

isn't it irreducible by Eisenstein

[delicate orchid]

is it?

[barren sierra]

if p^2 divides a then we know p^2 doesn't divide b

[barren sierra]

I've been trying but how

[delicate orchid]

if it were that simple then why have they given a paragraph of nonsense

[median pawn]

what if you can't find p such that p^2 divides a?

[delicate orchid]

then a would have to be a factor of b

[median pawn]

no?

a = 3 x 2 and b = 3 x 5

gcd(a,b) = 3, can't find p such that p^2 divides a

that's probably the obstacle

[rotund aurora]

I guess the problem is that b could be divisible by gcd(a,b)^2. But in that case maybe you can contradict that the discriminant is a square?

[delicate orchid]

good point tbf

[rotund aurora]

you don't need p^2 to divide a

[barren sierra]

Been trying that for a while but couldn't find a suitable contradiction

[median pawn]

oopsie

[delicate orchid]

wait

gcd(a,b) is a product of primes

[barren sierra]

If this comes up on my qual I'm cooked

[median pawn]

but yeah this seems to be the move

[barren sierra]

Ya and in particular we have an odd prime

If that helps

[delicate orchid]

what's that stupid result about roots mod p

[barren sierra]

¿

[median pawn]

same

[rotund aurora]

yeah, it does help

I think you can just write things down, but I can try to write it to make sure I'm not yapping

[barren sierra]

So you mean b?

[rotund aurora]

||Let g=gcd(a,b). If you can't apply Eisenstein is because g^2 divides b. Say a=gx, b=g^2y then 4g^3x^3-27g^4y^2=g^3(4x^3-27gb^2) should be a square. This means that g should divide 4x^3-27gb^2, but then g divides 4x^3. g and x are coprime (because g is square-free), and g>2||

[lilac sluice]

here you mean the quotient is abelian, right?!

[delicate orchid]

abelian implies solvable

[lilac sluice]

oh no, it is indeed solvable
sorry

[delicate orchid]

yeah this seems correct. But what does this get us

[rotund aurora]

that f(x) is irreducible

[delicate orchid]

like is it a contradiction?

ok it is

mfs who can solve these questions always impress me way more than mfs who waffle on about abstract nonsense

[rotund aurora]

to show that the Galois group is Z/3Z I believe you just need that the discriminant is a square

@barren sierra hint: ||Maybe try proving in general that if F is a field of char!=2, P in F[x] is irreducible of degree n and with discriminant a square then the Galois group is a subgroup of A_n||

[barren sierra]

Ahhh interesting

[delicate orchid]

oooh

[dull ginkgo]

More like just a “twisted” semidirect product

[rain grove]

So I have been solving problems from the rings chapter and the geometric series came up. I know the sum of geometric series is $1+q+q^2 + \dots = (1-q)^{-1}$ But that only holds for $|q| < 1$. Why can we use this result in abstract algebra if q is some element of the ring and in ring I don't have ordering (as far as ik)

[cloud walrus]

OHHELLNAH

[quiet pelican]

Where are you using the result?

(I suspect you may be using it in a case where that sum is actually finite)

[chilly ocean]

Because in rings there are only finite sums, not infinite sums

You need a topological ring or something like that to talk about infinite sums. Or, you can also make sense of this particular thing in a ring of power series because the elements are formal infinite sums.

[rain grove]

So first problem was to show:

$x$ nilpotent element of the ring $\Rightarrow (1-x)$ is invertible
And here the result is nice cause in geometric formula x^n = 0.

But the next one is $1-xy$ invertible $\iff 1-yx$ invertible. I could not solve it and I found this (https://mathoverflow.net/questions/31595/how-would-you-solve-this-tantalizing-halmos-problem) and he used the infinite geometric series

[cloud walrus]

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

[quiet pelican]

For the nilpotent one, the sum is actually finite so it doesn’t matter

[rain grove]

Yes

[rain grove]

I don't understand sorry

[quiet pelican]

Yeah, for the latter it’s kinda “extend to a formal power series esque ring, it holds there, so it holds in the ring”

[barren sierra]

Yea modifying my proof for the n = 3 case was easy for the general n case

[chilly ocean]

What does something like 1+2+3+4+... mean?

[rotund aurora]

-1/12

[rain grove]

"formal sum"? but idk what that means just that ig. I know it doesn't have a value

[chilly ocean]

Exactly, it has no meaning for rings in general

Rings only tell you how to do finite sums

To define "infinite sums" you need more than just a ring

[rocky cloak]

The infinite sum is more of a tool to guess what the solution should be, rather than something actually need for the argument.

Alternatively, you could think of it like, imagine xy was nilpotent, what would the inverse of 1-yx be then? Does that also work if it's not nilpotent? Great, you're good to go

[rain grove]

Yeah okk, I buy that

[rocky cloak]

Like if z is the the inverse of 1-xy.
(1 + yzx)(1 - yx) = 1 - yx + yzx - yzxyx = 1 - yx + y(z - zxy)x = 1 - yx + yz(1 - xy)x = 1 - yx + yx = 1

No infinite sum needed, except for how would you guess that 1 + yzx is the solution

[next obsidian]

You get that equality by considering a power series and showing that (1-q) has a power series inverse the geometric series. The power series is a polynomial when q is nilpotent so it actually makes sense in the ring

You can make this precise by dealing with topologies for the ring and using the universal property of a power series ring and the fact that the (q)-adic topology is complete because q is nilpotent

[hidden wind]

can i have a hint for this one

do i have to go through C somehow ?

[quiet pelican]

I think it’s most natural to go through \bC

[next obsidian]

Yes

[hidden wind]

oki thanku

right i can use that the complex roots come in complex conjugate pairs

[next obsidian]

Yeh

[dull ginkgo]

Let M be a module over commutative ring R. Then the tensor algebra T_R(M) is a graded ring (R-algebra). If J is a homogenous ideal of T_R(M) disjoint from M, then does the quotient algebra T_R(M)/J have the property that each linear endomorphism of M allows endomorphisms of the n-th levels / modules of the quotient algebra?

[hidden wind]

i am quite fond of this book (allan clark) but it could be a bit more explicit about some things, like this :S

[next obsidian]

I don’t think this is true because you would need some sort of condition like having that the endomorphism preserves J\cap M right?

Like just thinking on the part of the tensor algebra in degree 1

[dull ginkgo]

I said it’s disjoint from M

Well

Trivial intersection ofc

[dull ginkgo]

Well I can impose the condition that it fixes the intersection to generalize it a bit, thanks

[next obsidian]

Oop

[rain grove]

${(x,y,z)\in \mathbb{R}^3 | x=y=z^2 }$ This is a set which is not a vector space over field $\mathbb{R}$.

In my book it says this set is a vector space over field $\mathbb{Z}_2$. But this set has elements $(x,y,z)\in \mathbb{R}^3$, so when say over field $\mathbb{Z}_2$ does the definition of the set change to $(x,y,z)\in \mathbb{Z}_2^3$?

[cloud walrus]

OHHELLNAH

[coral spindle]

Yes

[hidden wind]

the proof for this feels like cheating

assuming a is a complex root of the pol f with real coefficients, then we use that f(a) = 0 = conj 0 = conj (f(a)) = (conj f)(conj a) = f(conj a)

[next obsidian]

This is how it goes

This is something pretty unique to R and C and requires the analytic parts

[glad osprey]

Agree! And it begs the question (or even proves?) whether complex conjugation is the only function that distributes over addition and multiplication

That doesn't sound right, but those are the only properties we use, right?

[dull ginkgo]

there are other ones

They are not well behaved though

Over a commutative ring R, is a set of elements linearly independent iff the annihilator of the collection of all finite wedge products of elements in the set is trivial

[glad osprey]

Ah, we also use the fact that the reals are unchanged under conjugation. And presumably this property along with distributivity uniquely characterizes complex conjugation

[grave sedge]

Thats basically like saying that the galois group of ℂ/ℝ has one nontrivial element (i.e. order two)

Which is indeed true

(and you're right in that you need to fix the reals)

For a proof of that just use the fact that f(i)²+1=0 if f is such a function

[glad osprey]

I haven't done galois theory yet, but that sounds interesting!

[hidden wind]

i still haven’t gotten to the meat in galois theory

exam in three days

[long obsidian]

I see. If 0 is absorbing than by your argument 0=1 so 1 is absorbing and then from there it follows the entire semifield is a single element. Thanks!

[dull ginkgo]

What’s the easiest way to show that there exists an $r : r \bigwedge_{n = 1}^{N}{m_n} = 0$ if and only if the $m_n$ are linearly dependent for general modules over a commutative ring

[cloud walrus]

The Library of Babble

[dull ginkgo]

I keep coming back to this article and it feels flat out wrong

because multiplication by R is NOT an R-Endomorphism

I don’t think by itself you can say anything about D besides it being a division ring

[grave sedge]

Multiplication by R on the right is a D-endomorphism

(where you see U as a left D-module)

[dull ginkgo]

Well yes I know that but how can you say anything else about D itself

As i said besides D being a division ring

[grave sedge]

That depends on R i guess

(i dont know if there's something specific you want since you seem to be mentioning previous discussion)

[dull ginkgo]

https://en.m.wikipedia.org/wiki/Jacobson_density_theorem

Several parts of this page feel wrong

[grave sedge]

Such as?

[grave sedge]

This should be DX i guess

[dull ginkgo]

No like conceptually

[grave sedge]

Everything else seems fine

[dull ginkgo]

This part is seems faulty

[grave sedge]

Why

[dull ginkgo]

I still don’t see how you can say anything about it being in the span of D

Without other assumptions on the ring R

[grave sedge]

Assuming the density theorem is true, do you see why this makes sense? (I agree that the theorem in itself is quite a surprising result)

[dull ginkgo]

The density theorem doesn’t make sense to me because we are using statements about the simplicity of the module to state more info about the endo ring

[dull ginkgo]

If x and y are linearly independent then their annihilators have trivial intersection

But that’s D-linear independence

Not R-linear

[grave sedge]

What the theorem says is basically saying that the action of R on X as D-endomorphisms is basically the same as the action of the whole End_D(X) ring

[grave sedge]

If R were just End_D(X) then this would be trivial linear algebra, wouldn't it?

[dull ginkgo]

I hate the right module thing ngl

Because thus far I tend to just assume U is an End_Z(U) module, which gives it both the structure of an End_R(U) and R module

They just commute, and End_D(U) is the double commutator of R

[dull ginkgo]

It’s just I’m unsure how to logically assert stuff about D as I said

Like for instance

[dull ginkgo]

Assume x and y in U are D-linearly independent

[grave sedge]

This is the reference they give btw

[grave sedge]

Well, U is a simple R-module

[dull ginkgo]

Well I know that lol

[grave sedge]

So you know R acts "transitively" on it

[dull ginkgo]

Yes, it’s cyclic (but not necessarily faithful unless R is primitive)

[grave sedge]

This means you might expect R to be quite a large subring of End_D(U) (and the theorem quantifies this "large")

[dull ginkgo]

For artinian rings you can conclude that it’s actually equal from this theorem

Which was my end goal

[grave sedge]

Well then i dont get what the issue is

[grave sedge]

This is a consequence of the density theorem (if R acts like the whole End_D(U) ring then given a bunch of vectors you can find an element of R which kills all but one of them, given the one it does not kill is linearly independent from the others)

And the density theorem is true because it is (again, quite surprising to me and the proof kinda reflects that - not very insightful, more of a "manipulation" type of proof if you know what i mean)

[dull ginkgo]

Oh so we’re kind of forcing commutativity

Between R and D

[grave sedge]

It's probably better if you've seen some examples

If R is a finite dimensional k-algebra for algebraically closed k then you get D=k

So it's basically a statement about rings of matrices

(if you've done some representation theory this should ring a bell)

[dull ginkgo]

Like here’s what I mean

Let’s say for a general module M instead viewed as a map from R to End(M) by left multiplication (where M is an abelian group)

then the centralizer of R is End_R(M) (additive maps where we can factor out R)

So by that logic, any central element of R is also an endomorphism by left multiplication

[grave sedge]

Yes, but there might be more (since the centralizer might be small)

[dull ginkgo]

Wdym

[dull ginkgo]

Centralizer of R_l is by definition End_R(M)

[grave sedge]

Centralizer of centralizer of R contains R but might be larger

[dull ginkgo]

Well yes ofc

But my point is

We are narrowing down our scope here

To considering D-linearly independent sets, namely their annihilators

For artinian rings, ofc gives us the fact that descending ideals (annihilators) terminate so we can extend our conclusion to all of R

[grave sedge]

Yes

What's bothering you then?

[dull ginkgo]

No longer bothering me

[grave sedge]

Nice

[dull ginkgo]

I am more comfortable with rings evidently

So just handling it within the scope of End(M) helps me

[dull ginkgo]

You can handle it absurdly if you tried to force A to commute with some r

[old hollow]

What is the unit group of $\mathbb{Z}[\sqrt{-5}]$

[cloud walrus]

a.b.s._.0.

[old hollow]

I feel like it’s only {-1, 1} but I am having trouble proving it

[void cosmos]

take norms

@old hollow

[old hollow]

Hm

That’s nice

Curious, is there a more algebraic way to see it

Like Z —> Z[x] —> Z[x]/(x^2 + 5) type of thing

[void cosmos]

idk

[surreal dagger]

You mean: An element a/b is invertible in the subring iff 1/a is in the subring iff a/1 is invertible. So if we consider all these elements in Z, they form a multiplicative system S such that xy in S implies x and y in S. The complement of them is closed under multiplication by ring elements and if two elements are not in S their product will not be in S?

So the complement of S seems to be related to prime ideals, but I dont know exactly how. I could look it up, but If I dont have a long way to go to the answer, I would like a hint.

[rain grove]

Show that a finite-dimensional complex algebra A without zero divisors is 1-dimensional.

How do I solve this?

Rn im thinking about vector space R^2 over C and yeah if I look at complex numbers as vectors then multiplying them can give you 1 dimension...

[chilly ocean]

How did you define complex algebra?

[rain grove]

When we say complex or real algebra we mean algebra over the field C or R

[chilly ocean]

Right, how do you define algebra?

[rain grove]

And i guess finite-dimensional would mean that the vector space (V,+,x) is finite dimensional

Algebra is a vector space with inner multiplication

[lean sail]

Do cyclic rotation groups always have an odd number of sides?

[rain grove]

or idk if that the correct term but that and the proprety x(uv) = (xu)v = u(xv)

[chilly ocean]

What do you mean by sides?

[rain grove]

Yeah but that is a requirement for a vector space

[lean sail]

sorry, not sides, vertices

something like this:

where we have only rotational symmetry, no reflective

oh, i just answered my own question

[chilly ocean]

Sorry that was a mistake. You don't require associativity or unitality?

[lean sail]

but what about this shape makes it only have rotational symmetry?

[chilly ocean]

There are shapes with only rotational and not reflective symmetry, yes

[lean sail]

i can tell by looking at it, but it's just intuition

[rain grove]

Ok so with these labels:
F field, A a set
x: scalar multiplication F x A -> A
*: multiplication A x A -> A
+: addition A x A -> A

A is algebra over F if

(A,+,x) is vector space
(A,+,*) is a ring
for all x in F and u,v in A: x(uv) = (xu)v = u(xv)

I guess associativity and unitality comes with the ring

[chilly ocean]

Do you require your rings to have identity?

Alright

So let A be a finite-dimensional complex algebra

What can you say about it when the dimension is 1?

[rain grove]

that its one dimensional?

for some $a\in A$ $a = x1$ for some $x\in F$

[cloud walrus]

OHHELLNAH

[chilly ocean]

for all a actually

and it will be a = a1

So you can say that as a vector space A = C

[rain grove]

Okk yeah

[chilly ocean]

Can you say what will the vector product be?

[chilly ocean]

I don't know about your shape, but maybe it's easier to prove it for a shape like this (hopefully it's clear what I mean by this drawing).

[rain grove]

It will be the same as scalar multiplication?

[chilly ocean]

Up to isomorphism, I believe so

Anyway, let A be a complex algebra of dimension >1 now

[lean sail]

im not really struggling with the idea that shapes can have only rotational symmetry, i am just wondering if there is a proof somewhere i could look at or if someone here knew the reason. i tried searching for something like this, but the results were somewhat obscure. if you google rotational symmetry there are a lot of different results.

[chilly ocean]

If you look at my drawing, any reflection that doesn't align with the two arms will send the arms somewhere where there is not arm. And if the reflection aligns with the arms, the hands will become upside down

[chilly ocean]

Any complex algebra of dimension >0 will have a copy of the complex numbers, which we will identify with C. Pick some x outside C. Then C and x generate a subalgebra B of A

[rain grove]

So if the dimension is n then x is a vector from C^{n-1}?

[chilly ocean]

x is a vector from A

if the dimension is n, you can see A as having underlying vector space C^n not C^{n-1}

[rain grove]

Ok I see, so then this subalgebra B is one dimensional

[chilly ocean]

It's not

It would only be one dimensional if x was in C

Since x is outside C, 1 and x are linearly independent

So it is at least 2 dimensional

(it could be more, it could happen that 1, x and x^2 are all linearly independent for example)

[rain grove]

Oh sorry, I thought we are talking about the product Cx.

[chilly ocean]

We are talking about the subalgebra of A generated by C and {x}

(Actually we can just say generated by x since we are talking about unital algebras so C will automatically be there too)

[rain grove]

so x is just an element of A

[chilly ocean]

yeah but also outside of C

else we'd just get C

[rain grove]

Ok wait this is in my brain idk if its right:

x = [c_1,c_2,c_3 ... c_n] if A is n dimensional. and x would be in C if all components are the same

So x is just some element where at least 2 components are different

[chilly ocean]

components depend on a choice of basis

[chilly ocean]

for example, if your basis contains 1, that is not true

I don't think it's useful to think about this in terms of basis rn

[rain grove]

Could you show me how x outside C looks like in an example

I think it might help me cuz rn idk what is x

[chilly ocean]

Well, since we'll need this for later. Consider the following "cyclic" C-algebra

$\frac{\mathbb C[x]}{(x^2)}$

[cloud walrus]

Bequi

[chilly ocean]

Does this notation make sense to you?

[rain grove]

no

well idk any cyclic C algebra

[chilly ocean]

You don't know about quotients of rings?

[rain grove]

No

[chilly ocean]

It's very weird that you're doing stuff about algebras without knowing basic ring theory..

Well, hmm, you at least know what polynomial rings are, right?

[rain grove]

Well this is a chapter called "introduction to ring theory" and its just definitions and some examples and then fermat little theorem

next is homomorphisms and then quotients

[rain grove]

Yeah

[chilly ocean]

Which book is it?

[rain grove]

Well its not a published book yet but its a book my professor is writing but its not in english

[chilly ocean]

This might not be the intended way to prove this, but its an adaptation of what I was going to say without directly mentioning quotients

For an example of a complex algebra. Consider the polynomials in one variable with coefficients in C

Does it make sense that that is a complex algebra?

[rain grove]

Yep

[chilly ocean]

I won't be using it, but the idea of what I am doing could be explained using the algebra of complex polynomials and quotients of that

Anyway, again, let A be a finite dimentional complex algebra

And x an element of A outside of C

And consider the subalgebra C(x) of A generated by x

The elements of C(x) will be everything obtainable by evaluating polynomials in x with coefficients in the complex numbers

[chilly ocean]

Does this make sense?

[rain grove]

That is always just C?

[chilly ocean]

No

Maybe my phrasing was not the clearest

I'm not replacing the x by complex numbers

It doesn't have to lie in C

[rain grove]

Then what?

[chilly ocean]

It doesn't need to be an integral domain

[chilly ocean]

By itself

Like you have your element x

You can do stuff like i*x^2

You will get a new element on your algebra

All the ways you can use the operations on your algebra to obtain new elements from x

Is the subalgebra generated by x

And those ways are called polynomials

Depending on your x, it could satisfy something like x = x^2

That's what I mean by evaluating your polynomial

as polynomials, x and x^2 are different

But as elements of your subalgebra, it might be that x = x^2

[chilly ocean]

(It's not obvious a priori because your algebra could be noncommutative but it turns out that singly generated algebras are always commutative)

[coral spindle]

x is just a symbol that we are adding to a set to form a new structure

[chilly ocean]

Hey Boytije, do you know a way to prove this without using quotients of rings?

OHHELLNAH doesn't know about quotients yet apparently

[chilly ocean]

I was trying to rephrase it without directly using the word quotient but seems a bit messy and confusing for him so maybe my approach is not the best

[rain grove]

Thank you a lot for helping me, Ill take some time to re-read the things about ring polynomials and look what you said again, cause now i don't wanna waste anyone time

[chilly ocean]

Basically the idea (for the way I'd prove it) is

Let α be an element of your algebra outside of C (using a different letter so that maybe that's less confusing)

If the algebra C(α) generated by α has vector basis (1,α), it means α^2 = x+yα so α satisfies a polynomial equation

If not, then α^2 is linearly independent from (1,α)

So we ask again if (1,α,α^2) is a basis

We can't keep going forever because our algebra is finite dimensional (subspaces of finitely generated spaces are finitely generated. We are also using that C(α) is commutative, else there could be a non-polynomial way to obtain new elements. It turns out that algebras over a field generated by a single element are commutative and it's not hard to prove it but this is already long.)

Therefore we will have some basis (1, α, .., α^n)

So α^(n+1) = c_0 + c_1α + ... + c_n α^n

Therefore α satisfies some polynomial equation!

To be precise, it satisfies x^(n+1) - c_n x^n - ... - c_1x - c_0

But C is an algebraically closed field

[chilly ocean]

Therefore this already had n+1 complex solutions

So it now has at least n+2 complex solutions (x is a new one)

Therefore C(α) is not an integral domain

Therefore it has a zero divisor

Therefore since A contains C(α), it also has a zero divisor.

@rain grove Idk if this is helpful and maybe there's a more elementary way to prove it I don't know of that your professor intended. And don't worry, you're not wasting anyone's time, I'm here because I like talking about math :)

[rocky cloak]

I think you can sidestep it quite easily.

Like take x some element in your algebra and consider
1, x, x^2, ...
Since the algebra is finite dimensional, these most be linearly dependent. So there is a polynomial f with f(x) = 0. And take f to be of smallest degree.

If f has degree > 1, then f = gh for polynomials of smaller degree, and then 0 = f(x) = g(x)h(x), so g(x) is a zero-divisor.

I guess this is what you already said

[chilly ocean]

Yeah, you wrote it much more concisely and cleanly tho

[old spire]

Is there anything that can be said about uncomplexifying an algebra over the complex numbers?

For instance, $M(2, \Bbb C)$ uncomplexifies to $\mathbb H$ or $M(2, \Bbb R)$.

[cloud walrus]

wlad 🌻

[surreal sluice]

Let $R$ be a topological ring. Let $C$ be the space of all sequences in $R$ that converge. What more I need so that $\lim:C\rightarrow R:(x_{i})\mapsto \lim_{i\rightarrow \infty}x_{i}$ continuous?

[cloud walrus]

J

[chilly ocean]

That is not a function unless every sequence converges uniquely

And what topology are you putting on C?

[cloud walrus]

J

[surreal sluice]

Ok that is true, we do have convergence but not unique convergence. So let us assume that.

[cloud walrus]

J

[dull ginkgo]

For a simple module

Is there a criterion that you can use to determine when there is an endomorphism sending x to y

[delicate orchid]

maybe

[grave sedge]

It should always be possible, unless x=0

If one of x and y is zero then it's easy to determine, else ann(x)=ann(y)=ann(U), and so sending 1→x gives an isomorphism between R/ann(x) and U, do the same thing with y and them compose one of them with the inverse of the other to get an U→U isomorphism mapping x to y

[grave sedge]

Ah no this is just wrong

Ann(x)=Ann(y) is false in general (what is true is that both R/Ann(x) and R/Ann(y) are isomorphic to U as modules)

But you can save this

There should be an endomorphism sending x to y iff Ann(x)=Ann(y)

(in the case x,y≠0, at least)

Since automorphisms (i.e. non zero endomorphisms) preserve annihilators

[rocky cloak]

Mattedy is correct that Ann(x) = Ann(y) works (at least if y is nonzero).

An equivalent condition that also works for y=0 is that for an r such that y = rx, you have that Ann(x)r is contained in Ann(x). (I'm assuming left modules)

[rain grove]

I asked the professor and the way he intend it was with the endomoprhism L:A->A Lx = ax

[dull ginkgo]

Thanku! That makes sense

[lean sail]

does anyone know of a good python library for dealing with groups?

[rocky cloak]

I guess sagemath, though I guess it's a little more than a library

But you can run it like a normal library through conda I think

[distant summit]

um

this does not strike me as rigorous definition of a symmetry

hi micose

[quiet pelican]

A symmetry transformation is an automorphism of the set

(Where automorphism is just bijective structure preserving, as usual)

[distant summit]

that is significantly more meaningful ty

[delicate orchid]

it will be rigorous once u define "action on a set" properly lol

[chilly radish]

What book is this?

[delicate orchid]

this tbh, the wording is sus

[distant summit]

lecture notes

he is a theoretical physicist in my maths dept

presumably this is the explanation for waffley definitions

[distant summit]

see above

oh and he has not defined action

[chilly radish]

Yea, this was extremely physicist-coded

[delicate orchid]

wunderbar.....

[chilly radish]

That's why I asked

[delicate orchid]

I can quickly define it now if you want

[distant summit]

ngl i dont get why they ask physicists to teach pure maths courses

yes please wew

[delicate orchid]

I don't either it's really dumb

[chilly radish]

That sounds like a super bad idea

[distant summit]

like it makes perfect sense to get a maths dept theophys to teach maths student a qm module, but abstract algebra is used by pure and applied mathematicians so it shld default to pure

[delicate orchid]

actually wait I don't need to define group actions in general for this

Let Sym(X) be the set of all bijections from a set X to itself, an action on X is just an element of this set

the definition was so waffly I thought it was more general than it was

[distant summit]

how is a symmetry different to an action as you've defined it here?

[delicate orchid]

an action isn't an actual thing in the physcists mind

you must understand that physcists dislike actually defining things

[chilly radish]

"Symmetry" is just a catch-all term for a group action by automorphisms, generally

[sly crescent]

What other kinds of group actions are there?

[delicate orchid]

group actions that aren't explicitly described via an automorphism ig

[sly crescent]

Such as?

[delicate orchid]

like I ain't writing out D_8 as pairs in C_2 and C_4

even though it's defined as a semidirect product

[chilly radish]

You can have a group action that doesn't preserve all of the existing structure

[rapid junco]

What is the canonical Lie group action on S^(n-1)?

[chilly radish]

It's just not very useful

[sly crescent]

I wouldn’t call that a group action

[delicate orchid]

it's literally a map C_2 -> Aut(C_4)

idk what a group action is if it isn't that

[chilly radish]

Like I can have a group action on a group that doesn't act by group automorphisms

[sly crescent]

Hmm

[delicate orchid]

shin's point is better anyway

left multiplication is only a set-wise automorphism but it's still an action

but I think it would be disingeuous to call it a symmetry of the group

it's like how you can have a group action on a pointed space that doesn't preserve the point but that is certainly not a symmetry of the space

[chilly radish]

It's an element of Sym(G)

[delicate orchid]

(or more generally, G-complexes vs complex with a G action)

[delicate orchid]

so true bestie...

[sly crescent]

So uh
Which of {Pin+(n),Pin-(n)} embeds into Spin(n+1)?

Or does it depend on n?

[dull ginkgo]

A group can act on itself by left/right multiplication or conjugation. Furthermore, it can act on its own subsets (which always preserves the sizes of these subsets) by multiplication

For instance, if you consider order / stabilizer of subsets of prime power sizes (where the prime power divides the order), we get the ideas from Sylow Theory

[rain grove]

Let $A$ be a finite-dimensional algebra. Prove the following:
(a) Every nonzero element $a \in A$ is either a zero divisor or invertible.
(b) If $a \in A$ has a left (or right) inverse, then $a$ is invertible.
(c) If $A$ is a division ring, then every subalgebra of $A$ is a division ring

[cloud walrus]

OHHELLNAH

[rain grove]

Need help with (c)

I got it nvm :))

[dull ginkgo]

Does this generalize a tad bit

Like to cyclic or finitely generated modules

Like assume we have cyclic left R-module U where Rx = U

Then for some element y, is it sufficient for y to be in the orbit of x under End_R(U) if Ann(x) is a subset of Ann(y)

[rocky cloak]

So Rx = R/Ann(x), so a map taking x in Rx to y in Ry is just a map from R/Ann(x) to R/Ann(y) sending 1 to 1. This is well defined iff Ann(x) < Ann(y).

[chilly ocean]

This comment is supposed to prove that if -1 is a quadratic residue modulo p, then p-1 is divisible by 4

I don't understand the second step. How can they just carelessly divide by 2?

Actually I can understand it if p is not odd and that's one of the assumptions.

But what about the last step

[grave sedge]

p-1 Is always equal to 2 times (p-1)/2

[chilly ocean]

It is senseless to raise to a fractional power

But we are assuming p is odd so that's actually ok

[grave sedge]

Well the theorem is false if you allow p=2

[chilly ocean]

Yes

[grave sedge]

So (p-1)/2 Is always an integer

[chilly ocean]

Yeah, forget about that, I realised it after writting

Why does 1 = (-1)^{(p-1)/2} imply (p-1)/2 is even tho?

[grave sedge]

What is (-1)^k?

[chilly ocean]

I see

Thank you.

[rapid junco]

If G acts on X, is there a natural action for G x G on X x X?

just defining (g, g) (x, x) = (gx, gx)?

[south patrol]

Yes

Well

Correct idea

But you can't assume it is of the form (g,g) or (x,x)

[rapid junco]

g, h.

sorry

[south patrol]

Yeah and (x,y)

But yes

I assume that is what you meant

[rapid junco]

and if so, does it follow that Stab_(x, y)(G x G) is equal to Stab_x(G) x Stab_y(G)?

[south patrol]

Yes

[rapid junco]

Nice

so then this means that product homogeneous spaces are naturally defined this way

[delicate orchid]

I really hate to ummm aktually you here but it isn't

the thing you described is a G x G -set

[south patrol]

Sorry

yes

Lol

This is the external product ig

[delicate orchid]

X x Y with the diagonal action is the product of G sets

[south patrol]

Ye

[rapid junco]

how does this transfer to the permutation representation

[delicate orchid]

not very well I don't think

[rapid junco]

oh okay

[delicate orchid]

disjoint union of G-sets corresponds to direct sums of G-reps

[rapid junco]

what are G reps?

[delicate orchid]

composition of (G,G)-bisets is the tensor product

maybe there is a nice way of viewing it though there has to be

oh duh it's just the product of course it is

I was way overthinking it

[rapid junco]

sorry could you clarify

[delicate orchid]

the corresponding permutation representation of the product of two G-sets is the direct product of the representations

i.e.

$R[X \times Y]\cong R[X] \times R[Y]$ as $R[G]$-modules

[cloud walrus]

Wew Duck Tbh

[rapid junco]

Okay

Then also, of course if we have a haar measure on G then the natural haar measure on G x G is just the product measure.

[delicate orchid]

R characteristic 0 btw

it should hold in positive characteristic but I'm scared

[rapid junco]

lol

so then I have this following integral transform.

[delicate orchid]

ok I'mma be real ur on your own now. I'm a character theorist

[rapid junco]

S^{n-1} is a homogeneous space, so we can write SO(n)/SO(n-1)

[delicate orchid]

I do not deal with infinite groups

[rapid junco]

Oh okay haha

[distant summit]

How much work do you actually need to do?

Set $A^n B^m \mapsto a^n b^m$, and then per that construction (obviously bijective) we have $\phi(A^n B^m) = a^n b^m = \phi(A^n) \phi (B^m)$.

[cloud walrus]

Douglas

[distant summit]

I was kinda expecting there to be more work involved but apparently not

[quiet pelican]

What I would do is probably a bit more formal and something you won't have necessarily been taught (this is kinda the problem with working with presentations before formally defining them)
I'd say that reasonably works at the level you want given that you prove that A^3 = B^2 = E and AB = BA^2

[distant summit]

What's your more formal method?

[quiet pelican]

Working with free groups and quotients
Basically a lot of universal property mess, if I want to use the presentation given

[distant summit]

I've seen quotient groups but not free groups

On the topic of quotients...

I'm not sure how to approach the final part of this, i.e. finding $H\setminus G$. I know that $H\setminus G={Hg | g\in G}$ but in the lecture notes the only examples are of finite groups, whereas this matrix group has got infinitely many elements so you can't just list it exhaustively in the way you might do with $S_3$ for example

[cloud walrus]

Douglas

[coral spindle]

Maybe you can list it exhaustively ;)

It’ll take some thought but you can describe representatives of the cosets in a nice way

[quiet pelican]

Essentially, if A, B are in the group, you want to consider when AB^-1 is in H
||So let A(l, m) be the matrix given at the top
Then A(l, m)^(-1) = A(1/l, -m/l^2)
So we want to know when A(l, m)A(k, n)^(-1) = A(b, 0)
Now we want A(l, m)A(1/k, -n/k^2) = A(b, 0)
So A(l/k, -nl/k^2 + m/k) = A(b, 0)
So nl = mk, or l/m = k/n
So we can pick a unique representative of the class where l = 1
Ie it is in bijection with the set A(1, m)||

(This looks messy, but it's the intuitive result we get by only caring about the matrix up to a constant multiple)

[delicate orchid]

that looks good to me

[distant summit]

Uh so I'm with you until penultimate line.

1. Why are we setting l=1?
2. Given that we have set l=1, are you saying that H\G is the set whose elements are those of the form H A(1,m)? H\G={H A(1,m) | m in R}

[quiet pelican]

2. Yes

1. It's an arbitrary but convienient representative

[distant summit]

Uh so setting this allows you to take any element in G and then place it in exactly one coset depending on the upper right entry?

[quiet pelican]

Yes

[distant summit]

Okie, I think that makes more sense. I will try doing some practice questions along these lines to consolidate

[toxic zephyr]

is it true that $R/(a,b)\cong (R/(a))/(b)$?

[cloud walrus]

inconspicuous old man & mime

[toxic zephyr]

trying to prove it but i wanna make sure im not on a wild goose chase proving something false

the map r+(a) -> r+(a,b) doesn't have any issues does it?

[knotty badger]

Yes

I’d do this with the universal property ig

[toxic zephyr]

im still rusty with rings. so like... what exactly does it mean for r+(a) to be in (b)? that r+(a)=b(r'+(a))?

[knotty badger]

Yeah by (b) here they really mean (b + (a))

Sniped lol

[rain grove]

What is C[a,b]?

[toxic zephyr]

okay yeah that makes a lot more sense

[knotty badger]

Polynomial ring

[rain grove]

Could it mean something else if C is not $\mathbb{C}$ but just regular C?

[cloud walrus]

OHHELLNAH

[tribal moss]

We'll need more context, I think.

[rain grove]

Show that the ring C[a,b] has divisors of zero.

But C has no divisors of zero So ring of polynomials also has no divisors of zero right?

[tribal moss]

The generic hand-wavy meaning ought to be something like "the smallest ring that contains C (whatever that is) and a (whatever that is) and b (whatever that is)".

[tribal moss]

Not enough context yet. Scan upwards to find somewhere that explains what C and a and b are.

[chilly ocean]

It could also be the ring of continuous functions from [a,b] into the reals or into the complex numbers

[tribal moss]

Oooh.

[rain grove]

Ohh yess

[grizzled spindle]

I thought this referred to continuously différentiable functions on a, b

Ig I got sniped oop

[chilly ocean]

That would probably be denoted C^1[a,b] instead

[tribal moss]

I'd have expected C([a,b]) for a function ring, but that makes sense.

[dull ginkgo]

Let $R$ be a ring, $U$ a left R-module. Assume we have two R-linear maps $\phi, \psi$ of $R^N$ to $U$, where $\mathrm{Ker}(\phi) \subseteq \mathrm{Ker}(\psi)$ and $\psi$ is endomorphic.

Does $\psi = \rho \circ \phi$ for some endomorphism$\rho$ of $U$?

[cloud walrus]

The Library of Babble

[toxic zephyr]

so the example i'm working with is Z[x]/(bx-a,p(x)) which is apparently isomorphic to (Z[x]/I)/(p(x)+I) where I=(bx-a). if im not mistaken Z[x]/I is isomorphic to Z[a/b]. so what happens to the p(x)+I? (Z[x]/I)/(p(x)+I) is isomorphic to Z[a/b]/(?) what ideal are we quotienting?

[chilly ocean]

What does endomorphic mean?

[dull ginkgo]

Rather incredibly, if C(R) is the ring of continuous functions of R to R, then a function is a zero divisor iff it's preimage of 0 is not nowhere dense

[toxic zephyr]

is it p(a/b)?

[dull ginkgo]

silly whacky proof

[rain grove]

ty for this but this picture is enough for me

[dull ginkgo]

The idea is that f(x)g(x) = 0 iff whenever f(x) =/= 0, then g(x) = 0 and vice versa

[rain grove]

yepp

It does not contain idempotent elements that are not 1 or 0?

f(x)(f(x)-1) = 0, but I can't find a non zero continuous function that satisfies this

[grizzled spindle]

I know that finitely generated groups are countable, but is the converse true?

If not what’s an easy example

I’d expect it to be false but I can’t think of something easy

[agile burrow]

The free group on countably many generators is countable since its elements are all finite words

[grizzled spindle]

That’s very simple yeah

Alright thx

[chilly ocean]

For an abelian example: Q

[toxic zephyr]

if we have a constant n in the ideal J=(p(x),q(x)) of Z[x], then can we say Z[x]/J is isomorphic to Zn[x]/J? where we're abusing notation and J is like appropriately mapped into Zn[x]

[barren sierra]

Yea and I think you should be able to prove this with some work + isomorphism theorems

[toxic zephyr]

I was messing around and I think I stumbled on a proof but I wanted to sanity check. thanks ❤️

[old hollow]

Let $R$ be a subring of $\mathbb C$, and let $U \subseteq R$ be the set of elements with norm 1. Is $U$ a group under multiplication?

[cloud walrus]

a.b.s._.0.

[old hollow]

I think this is true, but I’m not sure how to prove it

[mighty kiln]

You just have to prove each of the group axioms

[old hollow]

Inverses is the tricky part

The inverse is just the conjugate

But I can’t explain why the conjugate would need to be in the ring

If U is finite, then we’re done, since it’s multiplicatively closed and has identity

@mighty kiln

[mighty kiln]

I'm wondering whether this is actually true

I don't think so actually

[old hollow]

Yeah…

It’s a bit hard to invent a counterexample

You need U to be infinite

[mighty kiln]

You can pick any transcendental element with norm 1 and generate a ring

[old hollow]

Oh hm

That will be infinite

And … it won’t have inverses?

[old hollow]

Generate a ring by doing what?

Did you mean group

[mighty kiln]

Under addition, subtraction, multiplication

[old hollow]

Like smallest ring containing it?

[mighty kiln]

Yea

[old hollow]

What would that look like

[mighty kiln]

The collection of values P(x) where P is an integer polynomial

And x is your transcendental element

[old hollow]

Oh shi

So that set would certainly be larger than the unit circle

[mighty kiln]

?

[old hollow]

Well if it has norm 1, add it to itself over and over

You get bigger norms

So it’s an unbounded subset of C

Why would its restriction to the unit circle not be a group

[mighty kiln]

Well x has no multiplicative inverse

[old hollow]

Ok I’m a bit rusty lol

I’m trying to think why

Wait but does this ring have 1? @mighty kiln

There’s no way right

Cuz then x wouldn’t be trans

[mighty kiln]

Yea cuz constant polynomials

[old hollow]

Oh…

I’m stupid lol

why doesn’t it have an inverse

[old hollow]

Oh 😭 that makes sense

implying x is algebraic

ahh ok hm

With this conversation I realize any subring of C must contain Z, and therefore be an extension of it

[tardy hedge]

Lol someone put trans flag

Dats funi

[old hollow]

I wonder what kind of condition would be necessary and sufficient for U to be a group

[void cosmos]

yo guys

consider the free module Z^n over Z

and consider K to be the submodule generated by elements f_i, where f_i = sum(a_i,j*e_i)

e_i is the standard basis

such that det(a_i,j) !=0

why is |Z^n/K| = det(a_i,j)

|det(a_i,j)|*

[eager willow]

do you mean f_j = sum_i(a_i,j*e_i)?

[void cosmos]

sum(a_i,j*e_j) as the sum taken over j

sorry yes

[eager willow]

and what is meant by |Z^n / K|? It's possible the order of the group is finite or infinite

not sure what |det(a_i,j)|* means either

[void cosmos]

yeah fuck that problem whatever

ty

[eager willow]

Any finitely generated abelian group is of this form, and the subgroup K is also finitely generated and free of a rank at most n. For example Z/2 has order 2, and is a quotient of n=1. The matrix defining is simply the number 2 but it is also the number -2. In absolute value these are the same.

oh wait not all finitely generated abelian groups have this such that the deterimant is nonzero

I guess I mean when adding enough elements to make it a square matrix

And it's certainly true if it's nonsingular that the quotient is a finite group, because tensoring with the field Q will always make it 0, and one applies the structure theorem for finitely generated abelian groups

And it is also easy to show by taking the quotient directly, for any upper triangular, that the order of that group is going to be the absolute value of the determinant, because that's the absolute value of the product along the diagonal values

of course also for lower triangular

Going from this to an arbitrary matrix feels tricky, maybe there's another approach

[void cosmos]

is this easy?

define R-->Rx with r-->rx then Rx is R/Ann(M)? Ann(M) being prime

idk where i used the fact that R is a pid tho, i really just used that its a domain

this is not true.

Ann(M) being prime

not so easily

or is it idk

i just remember proving b4 that M is simpple --> Ann(M) is prime so yeah

Rx is tho

[quiet pelican]

Well PID => noetherian
So now assume that Ann(M) isn't prime
Then there is a, b such that a, b \notin Ann(M) but ab \in Ann(M)
So now consider the module M_1 = R(ax), which has <b, Ann(M)> \subset Ann(M_1)
Continuing in this way, we get a chain of ideals of R which must stabilise
Now what can we say about the "final" module?

[void cosmos]

im sorry

what "way"

can u like show the chain

[quiet pelican]

Let x_n generate M_n
Suppose Ann(M_n) isn't prime, take a_n, b_n\notin Ann(M_n) but a_n b_n \in Ann(M_n)
Consider the module M_{n+1} = R(a_n x_n), which has <b, Ann(M_n)> \subset Ann(M_{n+1})
Then the chain of ideals is I_n = Ann(M_n), and x_{n+1} = a_n x_n

[distant summit]

my groups lecturer says "one to one" and "onto"

i dont like this

"one to one" sounds like it means bijective but it actually means injective

and sticking to -jective means there's consistency

Isn't saying "$\psi$ is onto its image $\psi(G)$" somewhat tautological? Like uh, he's basically saying "were the range of $\psi$ to be $\psi(G)$, then it would be surjective", but that is literally just the definition of surjective

[cloud walrus]

Douglas

[quiet pelican]

It's trivial, but it is something that should be remarked upon

[distant summit]

Hmmmm. And how does $\psi(G)\subset S_n$ work? If $\psi$ is an arbitrary element of $\mathrm{Perm}(G)$ and $\mathrm{Perm}(G)=S_n$, shouldn't $\psi(G)=S_n$?

[cloud walrus]

Douglas

[quiet pelican]

\psi isn't arbitrary

it's the permutation induced on the underlying set of G by some element of G

[distant summit]

Right so something like $\chi_g (h)=g h g\inv$ would be in $S_n$ but is clearly not in general an element of $\psi(G)$?

[cloud walrus]

Douglas

[quiet pelican]

Yes

(Under the identification of S_n with Perm(G))

Well, under a chosen identification of the two

[distant summit]

Wdym? If G is a group of order n, isn't S_n always equal to Perm(G)?

Becuase the permutations are bijections from G to G, and that is equivalent to reordering its elements

[quiet pelican]

There is a (set of) non-canonical bijection(s) between the two

Because you need to choose a numbering of the elements of G to get an identification of S_n with Perm(G)

[lean sail]

does a group have to be nonempty?

[clever coral]

The axiom of neutral element says that there exists an element in the group such that ... so it must be non-empty

[lean sail]

yeah, that's why i was asking - thank you

that was also how i interpreted it

[surreal dagger]

Is this the universal property of product and coproduct for Modules just stated in Homset language? I have only seen it in words like "the product MxN is universal in the sense, that given the two projections for every Module A and homs from A to M and N, there is a unique hom from MxN to P that makes the diagram commute".
I think the picture is saying the same thing, is it just a different perspective or is there more to it?

[knotty badger]

Yes it is the universal property

This is the “representability” perspective

You’ve likely seen the “initial/final” perspective before

[surreal dagger]

Right, iirc you like the representability perspective better, why is that?

[knotty badger]

I think this is not quite true

They’re both useful perspectives to have, and indeed some universal properties are much easier expressed in terms of initial/final

That said, I have often found that it’s psychologically easier for people to understand the representability perspective when meeting universal properties for the first time

I also think it’s a lot more “freeing” in terms of letting you discover universal properties

Since some of them are rather awkward to express in the initial/final perspective

[lean sail]

quick question, in the group $GL(2, \bR)$, i noticed that the determinant of the members must not be $0$... is this because the members need to each have an inverse?

[cloud walrus]

proofman

[knotty badger]

Yes essentially

[knotty badger]

Initial algebras of endofunctors come to mind for this

[knotty badger]

And the tangent bundle comes to mind for this

[surreal dagger]

Thanks for the clarification. I dont know these examples yet, but I can just compare the perspectives when I come across more universal properties.

[knotty badger]

Mhm!

Cat theory taught me that there’s not One True Perspective on anything

It’s often useful to instead have multiple perspectives, so long as you have the ability to translate between them

Indeed, there’s a systematic way to convert a “representability” formulation into an “initial/final” formulation, using something called the category of elements

You can go the other way too but it’s kind of dumb

[rain grove]

Let K be any ring, then prove that for odd n there exists such polynomial f(X) from K[X] that f(X)(1+X) = 1+X^n

I proved this by "guessing" with geometric series but whats wrong with even n?

[grave sedge]

Choose your favourite ring K and show that it's not true that you can find such an f

(or evaluate both sides at X=-1 and find out for which rings you can have your f)

[rain grove]

I see, it has to be characteristic 2

[chilly ocean]

(or characteristic 1 aka the zero ring)

[dull ginkgo]

So here’s some batshit insanity:

Assume we have a ring R, and a left R-module U that is finitely generated (say by N elements)

Let (x_n) be a generating set of size N

Then we have an epimorphic map R^N to U sending e_n to x_n

Now let’s say that we have another map from R^N to U, psi, who’s kernel contains the latter map’s kernel

Then you’d think that psi factors through an endomorphism of U and the first map right, it’s well defined.

But does every endomorphism of U arise this way, as if two morphisms are equal on the generators, then they are equal as maps

[chilly ocean]

I don't understand the question. If two morphisms are equal on generators then surely they are equal as maps right? Because you can obtain any element from the generators using the module operations and that commutes with morphisms

[dull ginkgo]

It was a criterion to determine wether there’s a map that sends the generators to a specific tuple

But it turns it that the whole kernel inclusion thing hints that it’s due to the isomorphism theorems lol

[surreal dagger]

Im not sure I understand, what exactly is kind of dumb?

[knotty badger]

It’s just “initial/final iff it represents the singleton functor”

Which idk I think that’s kinda silly

[dull ginkgo]

Let’s say we have two maps, f and g, of A to B, where f is surjective/epi, and Ker(g) contains Ker(f).

By FIT, u: A/Ker(f) ~ B provides an isomorphism, of which contains Ker(g)/Ker(f). Thus g factors through this isomorphism, and you can conjugate by it to get the endomorphism

[surreal dagger]

I dont know what the singleton functor is, but it doesnt matter right now. Ill look into it at some point and see why its silly. The keywords are: category of elements and singleton functor?

[knotty badger]

It’s just the constant functor that sends every object to a singleton set, and every morphism to an identity

That’s the singleton functor

But yeah those are the keywords

[dull ginkgo]

Let $A, B$ be left R-modules. Let $S \subseteq A$ and $H_S = {f \in \mathrm{Hom}(A,B) : S \subseteq \mathrm{ker}(f) }$, then I’m pretty sure $H_S$ is also a left-R module, and that if $\phi : A \twoheadrightarrow B$, then $\mathrm{End}(B) \cong H_{\mathrm{Ker}(\phi)}$

[cloud walrus]

The Library of Babble

[dull ginkgo]

This is true actually

[chilly ocean]

Do you mean \phi is a morphism from A to B?

[dull ginkgo]

If $\phi$ is an epimorphism of $A$ to $B$, then there is the isomorphism $\lambda: B \rightarrow A/\mathrm{ker}\phi$. Now if $f$ also maps $A$ to $B$, and $\mathrm{ker}\phi \subseteq \mathrm{ker}f$, then $f$ factors through $A/\mathrm{ker}\phi$ as $f = f’ \circ \pi_\phi$. Then $f’ \circ \lambda$ is an endomorphism of $B$ determined uniquely by $f$. Furthermore, this provides a bijection as if $\rho$ is an endomorphism of $B$, then $(\rho \circ \phi)’ \circ \lambda = \rho$ \\
To show uniqueness, $\lambda$ and $\pi_\phi$ are both epi and thus are right-cancellative under composition. So if $\rho_f = \rho_g$, then $f’ \circ \lambda = g’ \circ \lambda \Leftrightarrow f’ = g’ \Leftrightarrow f’ \circ \pi_\phi = g’ \circ \pi_\phi \Leftrightarrow f = g$. All of these are two-way. \\
One can show this is actually $R$-linear using the fact that all of the above maps are $R$-linear and using the definition of $\mathrm{Hom}(A,B)$ (excuse the bad Tex, I am on my phone lol)

[cloud walrus]

The Library of Babble

[grave sedge]

H_S is not a left R-module btw

[dull ginkgo]

If both f(x) and g(x) vanish on S, then af(x) + bg(x) vanish on S

[grave sedge]

af is not well-defined

[dull ginkgo]

what is not well defined?

[grave sedge]

Ok not precisely this

[dull ginkgo]

H_S is the set of R-homomorphisms between A and B that vanish on S, which is definitely a module since it is closed under sums and scalings (consider the valuation morphisms too if you want)

[grave sedge]

It's not closed under scalings

Is what i'm trying to say

I mean it is but the scaling action is not a module action

(it's not associative)

[dull ginkgo]

If f(x) vanishes on S, then so does rf(x)

[grave sedge]

(rf)(sx)=rsf(x), if rf were a module map then (rf)(sx)=s(rf)(x)=sr(f(x))

[dull ginkgo]

Oh wait yeah it’d be a right module

[grave sedge]

Nope

Unless you have a right module action on the codomain (i.e. it's a bimodule)

[hidden cairn]

i need some hints for 2 implies 3

[dull ginkgo]

Oh it’s an abelian group

[grave sedge]

Or on the domain as well

[grave sedge]

Yeah

[dull ginkgo]

I forgot, it’s like the tensor product

[grave sedge]

Yeah

[dull ginkgo]

adjoint go brrr

So it’s an abelian group isomorphism

Which is a module with R commutative

Thank you

My brain flat lined on that fact

[dull ginkgo]

This proof makes proving Jacobson density theorem almost trivial

Since every set of elements in a simple module is a generator, so we have an epimorphism from R^|S|

And we can use End_R(U) linear independence

[delicate orchid]

the only simple cyclic groups are of prime order

[hidden cairn]

so i need to extract more subgroups in between each H_i+1 and H_i until the consecutive groups give a simple quotient group?

[distant summit]

Probably a stupid question...

Suppose f:G->H is a surjection and that ker(f)=K. Then by the homomorphism theorem, G/K=H.

Does it follow that G is isomorphic to the product group K*H?

[hidden cairn]

so something like H_0 < G_0 < G_1 < ... < G_k < H_1 < G_(k+1) < ... < G_t < H_2 < ... < H_s = G

[chilly ocean]

Consider f: Z -> Z/2Z given by f(x) = x (mod 2).

[distant summit]

Yeah thought so

[grave sedge]

U is not necessarily simple as a End_R(U) module though

[delicate orchid]

basically yeah. Whenever you just have a cyclic quotients in a subnormal chain , when you extend it out to a full composition series the quotients will remain cyclic but must be simple (as it's a composition series) and so you'll end up with prime order factors

[hidden cairn]

That makes sense

Thank you

[dull ginkgo]

And?

My point is that being generators of U allows you to determine, in a sense, what kind of endomorphisms there are

[dull ginkgo]

There is something called the “splitting lemma” that addresses when this is true

[distant summit]

oh nice

[dull ginkgo]

Oh wait for abelian groups

Yes

[chilly ocean]

I believe it only addresses when G is a semidirect product of K and H (if you don't asume abelian)

[knotty badger]

Ooh interesting

[dull ginkgo]

I misread it as semidirect lmao

I usually assume semidirect with regular groups

It’s a quotient by a normal

[chilly ocean]

Quotients are always by a normal

[surreal dagger]

For the canonically part: R/m is a field, but a R/m Module has to kill every element of m, hence we have to take the quotient M/mM to make sure?

[dull ginkgo]

That’s my point

[rain grove]

In the ring of formal sums R[[X]], when is f'(x) = f(x)?

[delicate orchid]

ODE moment

[rocky cloak]

I'm not sure what answer you're looking for, but if f(x) is given by
a0 + a1x + a2x^2 + ...
then the derivative is
a1 + 2a2x + 3a3x^2 + ...

So this happens exactly when a[i-1] = i ai

I.e. when f is a multiple of the power series for e^x

[rain grove]

Yeah exactly but thats never unless f(x)= 0 no?

[delicate orchid]

the e^x in question:

[rocky cloak]

I'm not sure what you mean

[hidden wind]

i’m having a silly moment

[rain grove]

what is an example where f is not a 0 function and f'(x) = f(x).

[rocky cloak]

e^x

Like I said

[hidden wind]

z mapsto y x^-1 z should work as a bijection between cosets xH and yH right

[chilly ocean]

If you start with an arbitrary a0, you can solve for a1, and then for a2, etc

[hidden wind]

my brain is currently refusing to verify this

[rain grove]

But e^x is not element of ring of formal sums

[hidden wind]

thank jagr

[chilly ocean]

If it is a polynomial (by which I mean it has finitely many nonzero terms) then it must equal 0. But otherwise it needn't be 0

[chilly ocean]

But its power series is

[rocky cloak]

It is, it is equal to
1 + x + x^2/2 + x^3/3! + ...

[knotty badger]

Any constant times that also works

[hidden wind]

my exam is in ~ 22 hours and i still haven’t to look at any galois theory luckily they really don’t expect much, i should just be able to do the bare minimum with the very simplest examples

which should be doable

[rain grove]

ohh ok

[rocky cloak]

Is it like miscellaneous abstract algebra?

[hidden wind]

yeah first course in abstract algebra, titled «grupper, ringer, kropper»

[rocky cloak]

I Oslo?

[hidden wind]

mhm

[rocky cloak]

Yeah, I've taken that course

[hidden wind]

though from what i’ve gathered (which is not a lot as i didn’t attend lectures, but i’ve glanced through the lecture notes), they haven’t talked much about rings at all

[hidden wind]

lovely

[rocky cloak]

I think just saying what homomorphism, ideals and the isomorphism theorems are is the extent one would expect from such a course

Then one can learn more about rings in another course

[hidden wind]

looking forward to it, if i can survive non-study things until that points

gonna maybe try giving tutoring another try next semester…

though the pay is pretty shit and much of the work that goes into it is not even counted as work hours

ahem but that’s off-topic sry

[hidden wind]

hmm i'll just gamble that they won't ask about fields of a non-prime number of elements

my shaky elementary number theory is really haunting me here ehehe

though simple examples in group theory has also increased my appetite for ent and provides some lovely new perspectives

[rain grove]

I showed that a formal power series from the ring K[[X]] is invertible <=> leading term is invertible element of K

does that mean K division ring <=> K[[X]] division ring?

[delicate orchid]

no? why would it

sorry that came off as a bit mean

[rain grove]

npnp, but I feel one side is ok. If K is division ring then all elements of K[[X]] are invertible because all have the leading term from K

[delicate orchid]

what do you mean by leading term here

the constant term?

[rain grove]

yeah a_0+a_1X+a_2X^2 + ...

[delicate orchid]

what if that term is 0

[rain grove]

ohh yeah then its not invertible

[south patrol]

Note x generates a non-trivial proper ideal

[rain grove]

Hmm ok Idk what ideal is yet.. A problem wants me to show that ring of formal Laurent series K((X)) is division ring <=> K division ring and thought I can just cheat it out with that lol

[dull ginkgo]

Is K((x)) the field of Puiseux series?

[hidden wind]

whew this is pretty fun stuff, sad to have put myself in a situation where i have to rush through it

[wary sorrel]

The way that I'd think about it is as follows. Let M be an abelian group, and let End(M) denote the set of all group homomorphisms M -> M. This is naturally a ring, where (f + g)(x) = f(x) + g(x) and (fg)(x) = f(g(x)). Then an "R-module structure on M" is the same as a ring homomorphism R -> End(M). In this case, if M is an R-module, then so is M/mM. Moreover, m is contained in the kernel of the map R -> End(M/mM), and so the map above factors through R/m, giving you a canonical R/m module structure on M/mM

[knotty badger]

Ooh neat

So just as a group action of G on a set X is equivalently a group hom G -> Sym(X)

An R-module structure on an abelian group M is equivalently a ring hom R -> End(M)

[dull ginkgo]

The centralizer in End(M) are the R-endomorphisms of M :3

[knotty badger]

Uh

I can’t parse what that means

[dull ginkgo]

Well the left module structure, so the right module structure is a map from the opposite ring of R

[knotty badger]

Sure sure

(Am not an algebraist)

[dull ginkgo]

R maps into End(M) if M is an R-module by left multiplication, right

[knotty badger]

Uh, sure

[dull ginkgo]

The set of additive maps (endomorphisms) of M that commute with the left-multiplications are R-linear endomorphisms of M

I.e if a_l(x) = ax

Then f(a_l(x)) = f(ax) = a_l(f(x)) = af(x) :3

[urban geyser]

Quick clarification: $\ G_{1} \times G_{2}$ forms a product group. call it $G$, and $p_{1}: (x,x’) \mapsto x$, where $x \in G_{1}$ and $p_{2} : (x,x’) \mapsto x’$ where $x’ \in G_{2}$. $p_{1}$ and $p_{2}$ are surjective, since they’re just projections. (Likewise, if we wanted to, we could exhibit homomorphisms which I imagine we could denote $p^{-1}{1} : G{1} \to G$ and $p^{-1}{2} : G{2} \to G$ and show they’re injective and can be used to identify $G, ; G’$ with subgroups $(G_{1} \times e). ; (e \times G_{2}), ; (G_{1} \times G_{2})$. The first two are normal subgroups of $G_{1} \times G_{2}$ since they’re the kernels $\mathrm{Ker ;} p_{1} = (e \times G_{2}), ; \mathrm{Ker ;} p_{2} = (G \times e)$). However, is $\Phi$ necessarily an isomorphism?, since $\mathrm{Ker ;} \Phi = \mathrm{Ker ;} \varphi_{1} \cap \varphi_{2}$ and it’s not a given that it would equal ${e }$, right? Simple example I was given would be $C_{6} \cong (C_{2} \times C_{3})$

[\begin{tikzcd}
& {\tilde{G}} \
{} & {G_{1} \times G_{2}=G} \
{G_{1}} && {G_{2}}
\arrow["\Phi"{description}, from=1-2, to=2-2]
\arrow["{\varphi_{1}}"', curve={height=12pt}, from=1-2, to=3-1]
\arrow["{\varphi_{2}}", curve={height=-12pt}, from=1-2, to=3-3]
\arrow["{p_{1}}", from=2-2, to=3-1]
\arrow["{p_{2}}"', from=2-2, to=3-3]
\arrow["{p_{2} \circ p_{1}^{-1}}"', curve={height=12pt}, from=3-1, to=3-3]
\end{tikzcd}]

[cloud walrus]

thimg

[urban geyser]

Oops sorry am I interrupting

(Lemme know and I'll delete)

[knotty badger]

I see

[delicate orchid]

\Phi definitely doesn't need to be an isomorphism, I'm not sure where or how you're concluding that

[knotty badger]

Ish

[urban geyser]

Just wanted to make sure

[delicate orchid]

centraliser of what exactly, the image of R in End(M)?

[dull ginkgo]

Yes

R_l

[delicate orchid]

let G_1 G_2 be trivial and G tilde be non-trivial

[rotund aurora]

Laurent

Puiseux is K{{x}}

[urban geyser]

I think what made me doubt myself for a bit was knowing that homomorphisms \Phi from G tilde to G1 x G2 are in bijective correspondence with pairs of homomorphisms phi-1 from G tilde to G1 and phi-2 G tilde to G2 (phi-1, phi-2)

But yeah nvm

[delicate orchid]

yeah so there's an isomorphism of hom-sets, not the underlying groups

[surreal dagger]

Very nice, this perspective is very slick. Just like you can show every abelian group is a Z-module in one way in a similar manner. These are cooler to state lol, but I guess it cant hurt having done it by hand aswell. Thanks

[urban geyser]

affirmative 🙂‍↕️

[dull ginkgo]

Let’s say we have a left R-module M. The Annihilator of M is the (two sided) ideal of R where xm = 0 for every m in M. It should make sense that then (r + Ann(M))m = rm for each m, r

So obviously we can interpret M as a R/Ann(x) module, or an R/J module if J is an ideal in the annihilator (because essentially Ann(x) is redundant). But the problem is that the maximal ideal here might be greater than our annihilator. Let’s call our maximal ideal K here because ASCII lol

So what we can do is consider the submodule KM, like the “orbit” of M under K. This is a submodule and we can quotient out M by it, forcing any elements that are a “K-multiple” to go to 0. Suddenly we have a module who’s annihilator IS K, so we can quotient out the redundancy in R to have a vector space if R is commutative :3

[surreal dagger]

Nice this makes sense

[dull ginkgo]

This is basically identical to the “left R module is when R maps into the endomorphisms of abelian group” approach because the kernel of this map is the “annihilator” as defined above. In essence we can view it as a “R/J” module because of that map factoring through R/J if J is in the annihilator by the isomorphism theorems

Though the quotient group M/mM being a module and that map can be a bit fucky but you can use FIT again

Some things might be easier to prove one way, some might be more intuitive another

Pick your poison lol, use both if you like

[surreal dagger]

I think its good to have many perspectives anyway

[noble lynx]

for this lemma can you not just set a_2, \ldots, a_n = 1 and then a polynomial in k[x_1, \ldots, x_n] becomes a polynomial in k[x_1] which we can use the root bound on that since k is infinite.

[lone niche]

h(x_1,1,...,1) could be the zero polynomial

[chilly ocean]

What is known about the asymptotic complexity of calculating a primitive root of the group of units of a finite field?

[hidden wind]

ok having looked at a couple of previous exams, they are definitely going to ask about finite fields

oh well time to look at finite fields

[dull ginkgo]

Is there a connection between Jacobson Density Theorem, Jacobson Bourbaki Theorem, and double centralizer theorem for artinian rings

[lean sail]

"tikzcd": what tex lib does that come from?

[urban geyser]

I just use quiver latex editor

[barren sierra]

what is $j_N$? Am I right in thinking it is just the map $n \mapsto 1 \otimes n$?

[cloud walrus]

Spamakin🎷

[wintry sluice]

does neutral element mean the same thing as identity element?

[barren sierra]

yes

[lean sail]

does anyone know why the elements in group $SL(2, F)$ must have a determinant of $1$? you might know this group by a different name, but it is the group of $2 \times 2$ matrices with entries from $\bQ$, $\bR$, $\bC$, or $\bZ_p$ (where $p$ is prime)

[cloud walrus]

proofman

[lean sail]

i am thinking that this has something to do with the elements of the inverse being fractional (if the determinant were not 1), but i am not 100% certain

[knotty badger]

Is this not by definition

The special linear group consists of those matrices with determinant 1

[lean sail]

i think there is a reason

[knotty badger]

You might be thinking of the general linear group

[knotty badger]

Which consists of invertible matrices

[lean sail]

the specialized linear group also consists of invertible matrices

[knotty badger]

I think the requirement to be in the special linear group is to have determinant 1

[lean sail]

i acknowledge that it is in the definition, i am not trying to disagree there, i am just wondering why that is the case

[knotty badger]

Oh yeah idk the etymology

[lean sail]

my hunch is that it has something to do with not being able to have fractional inverse elements due to allowing elements from $\bZ_p$ and $Q$

[cloud walrus]

proofman

[lean sail]

which obviously may be wrong!

[tribal moss]

Are you asking why the subgroup of GL(n) with determinant 1 is interesting enough to have a name of its own?
Or asking why that name happens to be "special"?

[lean sail]

sort of - why do texts go out of the way to say "the determinant is 1", as well as giving it its own specialized name?

[tribal moss]

Huh, that doesn't really tell me which of the two questions you're asking.

[lean sail]

"Are you asking why the subgroup of GL(n) with determinant 1 is interesting enough to have a name of its own?" - this is closest to what i am asking, an answer to this question may be helpful @tribal moss

thinking more about your question, this is more or less what i am asking

there's obviously some reason, why texts would go out of there way to separate out this group

[tribal moss]

I don't think there's a single answer to that -- the special linear groups turn out to be useful to consider for several separate reasons.
Having a common name for them is then useful for quickly connecting to their properties when you come across a group that happens to consist of determinant-1 matrices over such and such field (or ring).

[knotty badger]

I guess they’re the kernel of the determinant homomorphism

So a useful normal subgroup

For simple Lie groups you usually need determinant 1, for this reason

[lean sail]

i guess this answer makes sense, because there is $GL(2, F)$ does not require a determinant of 1, so it's not like groups that use elements from the sets i named necessarily require a determinant of 1

[knotty badger]

I think you mean nonzero determinant

Zero determinant would be non-invertible

Also, GL(2, F) does require nonzero determinant

[lean sail]

are you saying that these groups with nonzero determinants and determinants of 1 come up often later?

[knotty badger]

Both nonzero determinant, and determinant 1 (which is a special case)

[tribal moss]

Yes.

[knotty badger]

I remember doing the rep theory of GL_2(F_q) at the end of my course

Would’ve much preferred S_n I think…

[lean sail]

i goofed up a couple things

hopefully fixed now

can $SL$ be a subgroup of $GL$?

[cloud walrus]

proofman

[tribal moss]

Not only can, is a (normal) subgroup of GL(n).

[knotty badger]

Mhm, cause it’s the kernel of the determinant homomorphism

[lean sail]

going back to what you said originally, i think this was correct

it really is just taking that specific subgroup and calling it its own thing, for purposes which will become clear later

thank you all!

[distant summit]

What is a good way of finding the conjugacy classes?

That [e]={e} and [a]={a, a^2, a^3} is obvious, but then finding the other(s) seems like you just have to go case-by-case

[quiet pelican]

What conjugates a to a^2?

[distant summit]

Christ I hate algebra

It doesn't

I don't think

[quiet pelican]

Basically think about what this does on a square, and which ones “look the same from different angles”

[rocky cloak]

You have some pretty strong geometric intuition here though. Like D4 is the symmetries of a square, and two symmetries are conjugate if they're the same "type" of symmetry.

(Like reflection along a diagonal, reflection along a side, 90 degree rotation)

[quiet pelican]

Like 90 cw and 90 acw look the same up to flipping the square

[distant summit]

I don't really know a good way of doing these sorts of questions, because the way the conjugacy classes have been defined are as $[x]={ gxg\inv | g\in G}$, so for every $x$ you have to try conjugating by every element, and that doesn't seem like a very time-effective method

[cloud walrus]

Douglas

[rocky cloak]

Once you know that they should be it's not so hard to show it case by case

[distant summit]

@quiet pelican Is it correct that elements in the same conjugacy class must have the same order?

Or rather, elements are in the same conjugacy class if and only if they have the same order?

[dull ginkgo]

Yes

[south patrol]

Conjugation by an element is an isomorphism and so preserves elements and stuff

[south patrol]

This is very false. E.g. consider abelian groups

[distant summit]

Huh? If the group is abelian then isn't each element's conjugacy class only itself?

[south patrol]

Yes

[distant summit]

So what's the problem?

[a]={a}, and a has the same order as a

[south patrol]

Elements can still have the same order

[dull ginkgo]

Conjugation is an automorphism for each x you’re conjugating by. Assume the order of conjugate of g or g is greater than the other. Observe what happens.

[distant summit]

Oh I see what you're saying

[south patrol]

You said iff

Ye

[distant summit]

Yeah

[south patrol]

Like all nonzero elements of Z/pZ have the same order

[distant summit]

Hmmmm. Going back to D_4, I still don't know a good way of determining the conjugacy classes.

My best attempt would be finding the orders of each element (there are only 8 so this is practical if not ideal), and then that narrows down the possibilities

So you know that only a and a^3 have order 4, so either [a]={a, a^3} or they are on their own

And I think (ab)a(ab)^-1=a^3, so you shown that a^3 is conjugate to a

Hence [a]={a, a^3}

Then the order 2 elements are a^2 and a^n b

But yeah then it's just case by case

[distant summit]

And I feel like there must be an easier method than conjugating each of those 5 elements by 8 different elements

The "worked" solution just says "by computation [conjugacy classes]" without actually showing any of the computations

Which is massively unhelpful

[rocky cloak]

Well, you don't really need to conjugate by 8 different elements.

You know all elements are of the form a^n b^t, so you could just conjugate by that and see what you get.

For example you see that
a^n b^t a^2 b^-t a^-n = a^2
because b^t a^2 = a^2 b^t

Just using a^n b^t = b^t a^[(-1)^t n]

[knotty badger]

the main speed up I’ve found for these questions is focusing on what conjugating by generators does

but yeah these Qs always sucked for me

[chilly ocean]

Also you can't have conjugate elements with different properties (for the right notion of properties) so that can be a quick way to see two elements are not conjugate

[distant summit]

yh ngl i know D_n is the set of the symmetries in the regular n-gon, but ive never actually been taught this and havent got a great intuition ofr it

[knotty badger]

i think in my undergrad i remember being confused by this a little

for some reason it seemed like the “fundamental” description was preferred to be these generators with relations

and that the geometric picture wasn’t the “true” D_n

just found the bias towards thinking with symbols and disavowing geometric intuition weird

[chilly ocean]

I used to be confused too for a silly reason. They told me a symmetry is a transformation that keeps the shape invariant. So I thought it could be an arbitrary transformation (rather than one that preserves the metric) and that therefore there'd be infinitely many symmetries of any polygon

[knotty badger]

right yeah that makes sense

i hope aluffi can actually teach me algebra

if i ever get around to it

[hidden wind]

exam in 47 minutes

[dull ginkgo]

Best of luck!

[hidden wind]

can i have a hint for how to construct the field of four elements

as in, how could i have arrived at it myself

[knotty badger]

gl gl

[dull ginkgo]

Do you know about maximal ideals

[knotty badger]

uh hmm is that Z_2[X]/(X^2 + X + 1)

[chilly ocean]

irreducible polynomial of degree 2

[dull ginkgo]

Take any irreducible polynomial of degree 2

Shit

[chilly ocean]

That's not a hint, that's straight up the solution lol

[distant summit]

aluffi?

[knotty badger]

well yeah idk how you come up with this

[knotty badger]

it’s a textbook on algebra

[chilly ocean]

There aren't a lot of degree 2 polynomials over F_2

[knotty badger]

and probably my last chance to ever understand algebra