groups-rings-fields

[runic galleon]

wait

i have another question

they didn't say that the binary operation is multiplication

and this can't be valid on other operation but multiplication

[coral spindle]

It is standard to use a^p to represent any repeated operation.

Except those which are written additively.

[runic galleon]

so its like (aaaa)(aaaaaa) ?

i mean this

[coral spindle]

Yes.

[runic galleon]

so the second doesn't mean power ?

or does it ?

[coral spindle]

I explained here

[runic galleon]

yes sorry

thank you

[steel pulsar]

I was finally able to find my method ( that I wrote here #help-8 )

[grizzled crane]

just started rings, can I say the fundamental reason why minus times minus is positive is because integers form a ring under addition and multiplication?

more precisely also a consequence of distributivity of integers axiom

[coral spindle]

Idk if that's the fundamental reason per se, but it does mean that it is a consequence of requiring all the other nice properties of rings that we want.

It is not just a consequence of distributivity!

[grizzled crane]

I mean, it just looks impossible without distributivity

definitely we require ring axioms to prove it

[coral spindle]

Yes but it is equally a consequence of all the other ring axioms

[grizzled crane]

I was actually looking for an answer to this question from a very long time, say 9th grade

[slim kayak]

I mean without it you could just glue on an idempotent monoid onto your abelian group

[grizzled crane]

all i saw all these years was just unnecessary analogies

[slim kayak]

Distributivity + abelian group structure imply (-1)^2=1

[grizzled crane]

you require all ring axioms right?

multiplication associativity

[slim kayak]

Idk?
a=1a=(1+0)a=1a+0a=a+0a -> 0=0a

[grizzled crane]

you need to show (-a)(b)=(a)(-b)=-ab right?

[slim kayak]

Then 0=0(-1)=(1-1)(-1)=(-1)+(-1)(-1)

So the the first uses the cancellation property and uniqueness of inverses and the second uniqueness of inverses

[grizzled crane]

makes sense

What are some ring structures contains objects which aren’t numbers? Eg: Matrices

[daring nova]

functions

polynomials

[grizzled crane]

set of all real valued functions right?

[daring nova]

for example

[grizzled crane]

under pointwise addition and multiplication

is there somemore you can recall

i just wanted to see this notion of (-a)(-b)= ab in those contexts

We been dealing that with integers for years

[grizzled crane]

this is helpful

please add if you can recall more

[daring nova]

Z/nZ is very numbers

[grizzled crane]

they are sets right?

the elements are cosets or residue sets modulo n iirc

[daring nova]

yeah

as in any quotient group
this one happens to be a ring

[daring nova]

endomorphisms of an abelian group for addition and composition

[grizzled crane]

I’m unaware about the term sorry

[daring nova]

group morphisms

or vector space linear functions

[grizzled crane]

i’ve seen group homorphisms and isomorphism

[daring nova]

an endomorphism is a morphism from G to itself

[grizzled crane]

sort of like automorphism ?

[grizzled crane]

not surjective ?

[daring nova]

an automorphism is an isomorphic/bijective endomorphism

[grizzled crane]

this is any structure preserving function into itself?

[daring nova]

it's a morphism from G to G which is bijective

[slim kayak]

Morphism is a generalization of stuff like homo, homeo, diffeo, etc. -morphisms

Endo always means the domain and target are the same

[steel pulsar]

Is the signature of a 3-cycle 1 or -1?

I don't understand, there are sites that tell me 1 and others - 1

[slim kayak]

it would a nice exercise to check yourself

check the definition of your textbook and then try to see how many inversions a 3-cycle decomposes to

[steel pulsar]

There is no need to do this

[slim kayak]

why would that be

[steel pulsar]

the definition for a k-cyles is just (-1)^(k-1)

[slim kayak]

well apparently those sites use differing definitions, so the rule of hand of (-1)^inversion=(-1)^(k-1) would depend on those definitions

[steel pulsar]

but here they say that this is the signature of a 3-cycle

[south patrol]

I'm confused how the signature could differ

All definitions I've seen are equivalent

[south patrol]

Oh lol

[steel pulsar]

?

[rocky cloak]

Where do they say that?

[steel pulsar]

the justification is that: (because there are n − 3 orbits reduced to a singleton and an orbit of size 3, by definition of a 3-cycle)

[rocky cloak]

So the sign of a 3-cycle should be 1. So I guess were do they say it's -1 is a better question

[steel pulsar]

except that I don't understand the result

[steel pulsar]

.

what formula do they use here?

not (-1)^(k-1)

[delicate orchid]

an incorrect one

[steel pulsar]

jagr says it's correct, you incorrect, i'm lost

[rocky cloak]

I mean n - (n-2) is just 2= 3-1, but I'm not sure why they're writing it like that

Either way you get 1...

[delicate orchid]

but if you plug n = 2 in this says that 2-cycles are even, which is nonsense

[rocky cloak]

Like the formula could depend on more than just n, but without any context you can't really say what formula they're using

[steel pulsar]

I give you the complete correction

Lemma 1. The 3-cycles generate the alternating group An (when n ≥ 3).

Demonstration. The hypothesis n ≥ 3 ensures the existence of 3-cycles.
The signature of a three cycle is (−1)^(n−(n−2)) = 1 (because there are n − 3 orbits reduced to a singleton and
an orbit of size 3, by definition of a 3-cycle). So the set of 3-cycles is included in An.
Every element of An is the product of an even number of transpositions. (Indeed, the transpositions
generate Sn and as the signature of a transposition is worth −1 (because we have (−1)^(n−(n−1)) = −1, all
the elements of An are written as the product of a transposition number which must be even for
obtain the correct signature.

A product of two transpositions is either a 3-cycle or a product of 3-cycles:

(i, j)(k, l)=(i, l, k)(i, j, k)
(i, j)(i, k)=(i, k, j)
What concludes the proof

[delicate orchid]

(-1)^(n-(n-2)) is always 1, this seems like nonsense

what's the full formula for signature, not just 3 cycles

I don't know what n is, and thus I don't know what you're confused about

although considering the fact you're talking about An I'm presuming it's just the sign of the permutation

[rocky cloak]

So it appears they raise -1 to the power of n - #oforbits

So that's the formula they're using

Interesting formula, but I guess it works

[delicate orchid]

ok sure, I think I can see it

I still don't know where the problem lies. it says clear as day in the proof you've provided that it's 1

[steel pulsar]

way I will use the formula (-1)^(k-1) and for a 3-cycle I replace the k by 3

But if we had found that the signature is -1 we could not say that the set of 3-cycles is included in An?

[delicate orchid]

An is the kernel of the signature homomorphism, so a permutation is in An if and only if it's signature is 1

[steel pulsar]

ah ok

thank you

[frank cosmos]

if R is a ring, and A is a set, then what is the distinction between $R^{\otimes A}$ and $R[A}$

[cloud walrus]

Hello1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[frank cosmos]

if you let j:A->R^A be j(a)(x)=1 iff a=x then everything in R^A is just r1j(a1)+r2j(a2)+....

[rustic crown]

what do you mean by R^(⊗A)?

if A is finite, this should mean R⊗_Z ... ⊗_Z R

the description you're giving fits more to R^{⊕A} as a Z-module. (it's no longer a ring if A is infinite, since it won't have an identity)

[frank cosmos]

set functions from A to R that dont vanish on finitely many a \in A

[delicate orchid]

[comically loud blink]

[rustic crown]

yea that's usually called R^{⊕A}

⊗ is something totally different

[frank cosmos]

oh yeah i mixed up the symbols mb

[rustic crown]

[vivid birch]

Hey

[frank cosmos]

oh, is the only difference the constant terms?

[rustic crown]

nope, the multiplication is different

and R[A] as wayyyy more elements

[daring nova]

remember to ask your question

[rustic crown]

for example if A = {x} is a singleton, then R^{⊕A} = R*x
but R[A] = R[x] = R⊕Rx⊕Rx^2⊕...

[vivid birch]

oh yeah, I zoned out for a bit xD

I'm having some trouble with this;

Let $A$ be a commutative noetherian ring, $M$ a maximal ideal and $Q$ a $M$-primary ideal ($\sqrt{Q} = M$). Prove that there exists $n\ \in \mathbb N$ such that $M^n\subseteq Q \subseteq M$.

[cloud walrus]

Eduardo

[vivid birch]

does anyone have any idea how to solve this?

[rustic crown]

say M = (m_1, ..., m_r), then for each i we can pick n_i such that (m_i)^(n_i) lies in Q. what if we take n = n_1 + ... + n_r?

(also avoid asking questions is multiple channels >.<)

[vivid birch]

I know, I asked at #advanced-algebra but then @daring nova pointed out that this channel would suit this question better. my bad

[rustic crown]

sowwy then ><

[daring nova]

I don't know the topic, I just redirected to what's basically a 50/50 to me
Either way, it's more appropriate than a help channel

but yeah, should have mentioned #advanced-algebra and let him choose

[south patrol]

I also think it should be in #advanced-algebra tbf lol

I mean it's a standard comm alg problem and comm alg is meant for advanced alg generally

[rustic crown]

.<

[daring nova]

because it said commutative noetherian ring ?

[south patrol]

Primary ideals, noetherian rings, etc

This is a standard thing for studying primary decompositions

which is firmly in comm alg

[vivid birch]

My abstract algebra exam is next week and I've been using this server to get help on some exercises. I always have trouble figuring out if a question is more suited to this channel or to the adv-algebra one xD

[rustic crown]

ask in either one 🙈 (e.g. whichever one is inactive )

[daring nova]

I guess I'll just go with
Looks complicated and definitions I never seen/don't remember -> #advanced-algebra then
I guess I just progressed

[vivid birch]

also, why do you think my question is not appropriate for the help channels?

[daring nova]

Advanced helpers don't frequent the help channels as much

So you have low visibility for the relevant helpers

[vivid birch]

that's nice to know. Thank you

and @rustic crown @south patrol thank you for your help!

[frank cosmos]

is this ideal even in R? since it has infinite sums

[south patrol]

Who says it has infinite sums?

The ideal generated by a set of elements is, by definition, an ideal

This is all finite sums of the elements x_i and their multiples

[frank cosmos]

(x1,x2,...)=(x1)+(x2)+....=Rx1+Rx2+..... which contains eg x1+x2+x3+....

[south patrol]

No

Rx1 + Rx_2 + ... would be all finite sums of elements of each Rx_i

[frank cosmos]

[coral spindle]

An infinite sum is typically undefined for a given ring. A binary operation only defines finite combinations of elements!

[delicate orchid]

something something transfinite induction. Watch me go.

maybe someone should invent some kind of maths that deals with this problem, making infinite sums rigorous if they get close and stay close to some element of the ring

it's probably not possible...

[south patrol]

Wait that's kinda awesome

Should we work on this together wew

[tardy hedge]

has anybody here listened to sunbather by deafheaven

amazing album

[stark helm]

Can someone explain why it states [Q(2^(1/3), 3^(1/4)):Q(2^(1/3))] is at most 4? it seems that we have (1, 3^(1/4), 3^(1/2), 3^(3/4)) as the basis for Q(2^(1/3), 3^(1/4)) over Q(2^(1/3)). so I am wondering if it is possible that Q(2^(1/3), 3^(1/4)):Q(2^(1/3))] is less than 4?

[slim kayak]

this is offensive to my eyes

The degree is given by the minimal polynomial with coefficients in Q(third root of 2)[X], so since x^4-3 has it as a root the minimal polynomial must divide x^4-3

So it has degree at most 4

[slim kayak]

I would accept it if it was written in comic sans

[cloud solar]

A finite commutative ring with at least 3 elements. Let a= sum x^7 when x is in A and b=sum x^8 when x is in A. Prove that at least one element a or b is non invertible

If I assume that a and b are both invertible I can look at their product which is also invertible

In that product sum there is at least an element y to the power 15

Idk how to manage this problem

Any hint

[summer path]

I've always wondered, do textbook writers actually keep a list of numbered theorems and lemmas next to them when they type out the proofs

[rustic crown]

latex should handle the referencing

[rustic crown]

if u is a unit in the ring A, then show that (u^7 - 1) * a = 0 and (u^8 - 1) * b = 0. if a and b were invertible, show that this implies only unit in A is u = 1. ||With this, A is an F_2-algebra, and moreover, if x is in kernel of the Frobenius F : A --> A, then x^2 = 0, so 1+x is a unit. Show F is an iso, and thus b = F^3(sum of elements of A). since A is an F_2-vector space, it has size |A| = 2^n. and as F_2 is a proper subgroup, an individual coset looks like {x, x+1} which has sum = 1. Thus total sum = [A:F_2] which is even, and hence 0 since char A = 2. but then b = 0||

this was my first thought, so probably there are better ways

[celest furnace]

May I ask why? Is there a shorter way to do this?

[delicate orchid]

I mean just look at it

[celest furnace]

Like is the LaTeX bad?? Bad spacing?? Notation and work looks fine to me..

[south patrol]

Both

[celest furnace]

Ah

[south patrol]

Looks super ugly

[celest furnace]

$cos(x)$ moment

[cloud walrus]

The Ultimate Chad

[south patrol]

I think basically all the letters are too tall and similar heights

Also the Q should be mathbb

Or at least mathbf

Also not nicely written imo

[celest furnace]

How so?

[south patrol]

Idk just super dense

[glossy crag]

What's a Brauer pair 🤔? Is it a modular representation thing?

[delicate orchid]

yus

[glossy crag]

You do reprtheory right?

[delicate orchid]

maybe

[south patrol]

Fusion

[glossy crag]

What, like dance?

[south patrol]

Nah like cuisine

[slim kayak]

😋

[glossy crag]

I was recently reading a lot about the basics of modules over non-commutative rings (semisimplicity and such), but for some reason I find it hard to connect to the concrete problem of group representations, idk why. Not that I've done a lot of reading on that, just something that came up a bit.

[delicate orchid]

every representation is in corrispondence with a module over the group algebra

the group algebra is commutative if and only if the group is abelian

||assuming you're doing R[G] with R commutative. If R isn't commutative I don't want to talk to you anymore||

[south patrol]

R[G]

[dire siren]

check out the solutions for Alexandru Myller 2010 (https://www.mategl.com/download.htm)

[unkempt stream]

wew noncommutative algebra jumpscare

[slim kayak]

tf is a non-commutative ring

[delicate orchid]

don't worry about it sweaty

[slim kayak]

I would guess it is an empty set joke if it wasnt non-empty by definition

okay pookie

[cobalt heath]

Even when we are only dealing with commutative algebras, noncommutative rings somehow appear and demand to admit their existence

[barren sierra]

"Non-commutative polynomial rings aren't real, they can't hurt you"

Non-commutative polynomial rings: https://link.springer.com/article/10.1007/s00037-018-0171-9

polynomials over the free non-commutative polynomial ring

Learned about this in a special topics course I took last spring

[vagrant zinc]

Guys, what is the difference between the generator in line algebra and in a group?

[mighty kiln]

In a vector space, the generated subspace of a subset S is the collection of linear combinations of elements in S

In a group, the generated subgroup of a subset S is the collection of products of elements in S and their inverses

Both can be described similarly:

The subspace generated by S is the minimum subspace containing S, and also the intersection of all subspaces containing S

The subgroup generated by S is the minimum subgroup containing S, and also the intersection of all subgroups containing S

(This is usually what "generated" means in broader contexts)

[steel pulsar]

Hello

Suppose that G contains a product σ := (ab)(cd) of two transpositions with disjoint supports, and let e be the element of {1, . . . , 5} which
is not in {a, b, c, d}. Calculate (bde)σ(bde)^(−1) σ^(-1)

Solution:

I can't understand the calculation steps, can someone explain it to me?

[rocky cloak]

If you conjugate a permutation, the result is given by relabeling the elements.

I.e.

s (e d b) s^(-1) = (s(e) s(d) s(b))

[steel pulsar]

This is the last step that I don't understand

(bde)(eca)

on the other hand, what you wrote is not what was written in the correction

you wrote : s (e d b) s^(-1) = (s(e) s(d) s(b))

[rocky cloak]

Yeah, I used s instead of sigma

Otherwise it's the same as written

[steel pulsar]

.

[rocky cloak]

For the last step you just compute the composition

Where does b map, it maps to d, where does d map it maps to e, where does e map, it maps to c, where does c map it maps to a, where does a map it maps to b

So that's the cycle (bdeca)

[steel pulsar]

Why does e map to c? Forget the correction. Act like you didn't see the answer.

you have that

where do you see "c" here?

[rocky cloak]

sigma(d) = c

[steel pulsar]

Yes it's written in the correction

but what makes you say that when looking at this line

[runic galleon]

Lagrange theorem in group theory says that let (G, * ) be a group and ( H, * ) a subgroup of G , then o(H) divides o(G) , ... i'm condused here, i felt like its missing details

[rocky cloak]

Alright, I have no idea what you don't understand, so let's just start from scratch.

You want to determine what
(bde) s (edb) s^-1 is

This is a function mapping the letters abcde to themselves. So let's investigate where c is mapped.

To compose functions, what you do is apply one after another. So first apply s^-1, s^(-1)(c) = d.
Then apply (edb), that gives us b. Then apply s, that gives us a. Then apply (bde), that just leaves a. So the function maps c to a

If you do the same for the other elements, you can determine the entire function

[steel pulsar]

In fact I have the impression that something important is missing in what you say, it is that in the statement they say that:

σ := (ab)(cd)

and I don't see that you mention it unless I'm wrong

[rocky cloak]

That's the definition of sigma, given yes

Did you mean for sigma to be something else?

[steel pulsar]

And I think that to do the calculations you have to use this definition

[rocky cloak]

Yes, you have to use the definition that's given

[rocky cloak]

What kind of details did you have in mind

[rocky cloak]

So are you confused why sigma maps c to d and d to c for example?

[steel pulsar]

σ = (ab)(cd) this means that σ sends a -> b | b -> a | c -> d | d -> c

Let us calculate σ(bde):

bde sends b -> d | d -> e | e -> b

σ(bde) = σ(bd)(de)

no I don't understand, too bad,

let's forget, thank you

[rocky cloak]

We can forget it if you want, but it's like you said

(bde) sends b -> d

And sigma sends d -> c,

So the composition sigma(bde), send b -> c.
b -> d -> c, b ends up at c.

[steel pulsar]

σ(bde) sends b -> d -> c | d - > e -> b -> a -> b | e - > b -> a -> b -> a -> b .... ?

[rocky cloak]

d is sendt to e, and then sigma just sends e to itself, so
d -> e -> e

[steel pulsar]

ok actually I understand, there is no need to worry about sigma(bde)

it's simple and short

σ = (ab)(cd) this means that σ sends a -> b | b -> a | c -> d | d -> c

σ(edb) = σ(e)σ(d)σ(b)

For σ(e) : e-> e

For σ(d) : d-> c

For σ(b) : b -> a

Finished.

The only things need to know here are:

1. σ c σ^-1 = (σ(a1)....σ(ap)) if c = (a1....ap)

2. σ = (ab)(cd)

[grizzled crane]

Can you explain evaluation homomorphism? and why do we study those?

rather it’s importance

[south patrol]

Do you mean for polynomial rings? @grizzled crane

[grizzled crane]

I suppose this is with in groups

This is the definition I’m looking

[unkempt stream]

it has to do with the fact that your “cosets”, i.e sets of the form xH partition or split your supergroup into a number of cosets. All the cosets are the same size, so the subgroup order divide the order of the main group in order to split the supergroup up. the quotient of the |G|/|H| is the number of cosets, i.e the index

[unkempt stream]

i mean it kinda just evaluates the polynomials by setting them to their output. It’s a homomorphism

[vagrant zinc]

Very similar

[vagrant zinc]

name of book ?

[grizzled crane]

Fraleigh

[runic galleon]

can someone give me an example

[coral spindle]

Pick a whole number

Call it n

Then nZ is a subgroup of Z

[runic galleon]

for example n = 3

[formal ermine]

3Z is a subgroup of Z

[runic galleon]

3Z is a subgroup of Z

[delicate orchid]

3Z = {0, 3, -3, 6, -6, 9, -9,...}

we don't really know what you're looking for here

[runic galleon]

i'm just looking for this

thank you

[delicate orchid]

well there you go then

[rose prism]

what about 0Z

[delicate orchid]

0 is in N sweaty

[runic galleon]

he said a whole number i think 0 is not a whole number

but still

why cant 0 be a subrgoup

[delicate orchid]

well 0 is definitely in Z so

[coral spindle]

0 is a whole number.

[mighty spade]

it is

[coral spindle]

{0} is a subgroup

[delicate orchid]

it is a subgroup

[mighty spade]

by whole number they mean natural number

[rose prism]

0 = 0/2 so its a fraction

[mighty spade]

i presume

[runic galleon]

the set is a singleton but that doesn't mean it cant be a group

[delicate orchid]

1 = 1/1 so it's a fraction. What an asinine statement

[runic galleon]

what does Z/2Z represents

[rose prism]

let’s agree to disagree

[mighty kiln]

i think zero is not a whole number
Perhaps Z^+

[delicate orchid]

fine, but only because you're transparently baiting

[rotund aurora]

mmh what if 0 was an invention of the devil to make mathematicians waste time discussing whether 0 is natural or not

[runic galleon]

in french system 0 belongs to N

[mighty spade]

roughly speaking, arithmetic of integers if 2=0

[runic galleon]

but idk about other systems , i think they disagree with us

[formal ermine]

in french 0 is both positive and negative

dont listen to those weirdos

[formal ermine]

modular arithmetic

[coral spindle]

Have you seen quotient groups yet? If you haven't, then do not worry about this yet.

[unkempt stream]

dude is going to shit himself learning about groups

[formal ermine]

lmao

[unkempt stream]

Kernel more like colonel of satan

[coral spindle]

OK let me tell you now: 0Z = {0} is a subgroup of Z. I don't care about various other conventions for the natural numbers, but this is simply a fact.

[formal ermine]

whole numbers = ganze (Z)ahlen in german. thats where the Z comes from. as a fun fact

[runic galleon]

i found it in an exam exo, i have to know it but its not in my lecture

[mighty spade]

i always hated this explanation because i didn’t understand modular arithmetic until after i learned about subgroups and subrings of Z

[unkempt stream]

same

[unkempt stream]

multiplicative groups and rings

[formal ermine]

i did programming before math so modular arithmetic was something i was aware of

[mighty kiln]

When subring of Z

[unkempt stream]

modular arithmetic is more ring theoretic

principal roots of unity and ring theoretic fourier transforms

[mighty spade]

thanks mizalign, that was a very helpful reminder

[unkempt stream]

de nader

[unkempt stream]

they are kinda fun to throw images in shit into, then just compress them, then undo it

[runic galleon]

i will try to translate this, i need to understand the set i will be working on

• is the addition between two equivalent classes of Z/2Z ( how does Z/2Z look 😭)

[wraith cargo]

Z/2Z is the integers mod 2

[vagrant zinc]

I don't know, I know a little Portuguese, but it says a body ?

[runic galleon]

thank you so much

[steel pulsar]

Statement : Suppose that G contains a 5-cycle τ := (abcde). Calculate(cde)τ (cde)^(−1) τ^(−1)

Correction :

I want to know why (de)(ed) cancel each other out?

[mighty kiln]

• is the addition on the two equivalence classes of Z/2Z
  × is the multiplication on the two equivalence classes of Z/2Z
  1a) Why is (Z/2Z, +, ×) a division ring (field?)?
  b) Draw the addition table of Z/2Z.
  (*I think corps is division ring and corps abélien is field)

[steel pulsar]

here

[coral spindle]

Is it clearer if I were to point out that (de) = (ed)?

typo

[steel pulsar]

even if you say that the two are equal I still don't understand how they cancel each other out

[coral shale]

what do they do...

what is (de)

[coral spindle]

Is it clear why (1 2)(1 2) is the identity?

[steel pulsar]

yes

[coral spindle]

So...

[steel pulsar]

okt hank you

[runic galleon]

so Z/2Z = { -2, 0, 2 , 4, 6, ...} ?

[delicate orchid]

no

[mighty kiln]

That's 2Z

[delicate orchid]

Z/2Z = {{0,2,-2, 4, -4 ...}, {1, -1, 3, -3, ...}}

or just {2Z, 1+2Z}

[runic galleon]

if yes then its the same thing as 2Z

[mighty kiln]

Or you can imagine taking Z and doing
… = -2 = 0 = 2 = …
… = -1 = 1 = 3 = …, so that there are only 2 elements

And addition/multiplication works the usual way

This is something called a quotient ring (quotient of Z by 2Z)

[coral spindle]

I find it disquieting that you are expecting to be able to answer questions about things you don't know the definitions of!

[runic galleon]

we can say that Z/2Z consists of two sub sets , one which contains even numbers and the other odd ones rihght ?

[mighty kiln]

Ye

[delicate orchid]

have I been blocked or something

[mighty kiln]

Those are the elements of Z/2Z

[coral spindle]

You are confusing subsets with elements here

[runic galleon]

i'm not trying to prove anything, i'm trying to understand

[mighty kiln]

[runic galleon]

sorry, a lot of ppl are typing and i only take one message and try to understand it

[runic galleon]

yes then this is true so for example Z/3Z it contains 3 subsets 3k , 3k+1, 3k+2 ?

[coral spindle]

Not subsets — elements

But yes

Z/3Z is a partition of Z, in other words

[surreal sluice]

In a commutative ring, how does one prove that if $b\in (a)$, then a divides b? This is part of an equivalence.

[cloud walrus]

jaspreet.sidhu

[coral spindle]

You should recall the definition of (a), after which it should be clear.

[delicate orchid]

mfs type too fast

[coral spindle]

Get fucked you slow typers,

[mighty kiln]

[coral spindle]

jk love u guys

[delicate orchid]

you have a space bar tbf

[surreal sluice]

Wait, how? Isn't the definition just (a) is the smallest ideal containing a? It would be nice to have the other characterization using linear combinations.

[delicate orchid]

(a) = {ra : r in R}

[surreal sluice]

Note R is not unital.

[delicate orchid]

well now I leave

[coral spindle]

I wish you had stated this previously!

That's, you know, important context!

Yes I suggest you think about other possible characterisations.

It should be something like (a) = {n.a | n in Z} u {n.ar | n in Z, r in R} (typo)

[surreal sluice]

I stated that R is a commutative ring. Sorry, did not know if that needed to be stated separately 🙂

[coral spindle]

Stating that R is nonunital is important.

You will also need to state your definition of divisibility, because it's going to be weird too.

[mighty kiln]

You still have {ar : r ∈ R} ⊂ (a) tho?

[surreal sluice]

Another question, why do people assume a commutative ring is unital? Is there something in the terminology like ring vs rng?

[coral spindle]

Because people like unital rings

Because they're nice, and not weird.

[delicate orchid]

because almost all rings encountered are unital

[coral spindle]

Is there something in the terminology like ring vs rng?
my brother in christ you just stated the difference in terminology

[delicate orchid]

simmer down chat...

[mighty kiln]

Wait how does one define divisibility if neither ar = b nor b ∈ (a)

[coral spindle]

That's what I was sayin'

[delicate orchid]

the former would be strange for non-unital. you wouldn't always be able to say that a divides itself

[mighty kiln]

But for the latter the original question also wouldn't exist

Or perhaps (b) ⊂ (a)?

[coral spindle]

Huh quick thought, there should be a left adjoint to the forgetful functor CRing → CRng right? Like if our rng is R, we could have the set Z x R with the operation (a, r) + (b, s) = (a+b, r+s) and (a, r)(b, s) = (ab, a.s + b.r + rs). Then it has unit (1, 0). In fact I don't seem to need to require commutativity here. Interesting! I'd have to check if it works as an adjunction and indeed as a ring.

[delicate orchid]

could we not just append a 1 to our rng

[coral spindle]

Well yeah lol I'm working out how that would work

[delicate orchid]

I get you want the adjointness but hmm

[mighty kiln]

Free indeed sounds like ×Z

[delicate orchid]

yeah

[coral spindle]

I'm gonna quickly check this is a ring, a functor, adjoint etc

[delicate orchid]

https://ncatlab.org/nlab/show/unitalization

you were indeed correct, boytjie

[coral spindle]

[mighty kiln]

[south patrol]

Indeed it's pretty cute cause like

[steel pulsar]

What is the difference between a q-Syllow and a p-Syllow ?

[south patrol]

If we extend to like CRing -> CRng -> Ab (all the forgetful functors) then we know a left adjoint of the whole thing is given by sending an abelian group A to $\mathrm{Sym}(A) = \bigoplus_{i \ge 0} A^{\otimes i}$ and a left adjoint of the right hand side is sending to $\mathrm{Sym}{>0}(A) = \bigoplus{i > 0} A^{\otimes i}$

[cloud walrus]

potato

[south patrol]

So like in this very special case you are just adding in an extra copy of Z lol

But then that is true more generally etc

[south patrol]

Context

(*Sylow)

But if p is a prime then you say p-Sylow etc

I imagine q is also a prime in the other situation

[runic galleon]

Boytije could u tell me what i should be familar with to start solving exos ?

[coral spindle]

You should read the definitions of whatever the exercises talk about, usually

I can't possibly describe everything you'd need to know to solve all exercises. That's just not feasible. Since you seem to be working from some set of notes, you should probably just read the notes.

[rustic crown]

and try to feed boytjie's hungry kids

[coral spindle]

Who knows we may see them again...

[slim kayak]

Whats a nice characterization of polynomials g(x,y) with the property that g(x,g(y,z))=g(g(x,y),z)) ?

[mighty kiln]

Linear

x, y, x+y+c, and constants, in particular

[slim kayak]

how do you see that? Degree?

[mighty kiln]

Ye

[slim kayak]

i dont really have a feel for them in multiple variables

is there like a mult and comp formula there still?

[mighty kiln]

It works pretty much the same

Well if you consider degree of one of the variables, that is

[slim kayak]

actually does the degree behave as if I have a map R[x1,...xn] -> R[x] and just send anything everything to x?

all the x_i

[mighty kiln]

Ye ig

[slim kayak]

okay yeah

[mighty kiln]

But I don't think the total degree helps much here (?)

Oop forgot constants

[slim kayak]

hm did you arrive at it being linear that fast then?

[mighty kiln]

^

[slim kayak]

huh, so R[x1,...,x_n] and the degree of x_k is the degree of f as an element of R[x1,...., skip x_k, x_n][x_k] yada yada

or like, treat the other variables as "constants"

[south patrol]

x + y + cxy should work

Source: formal group laws lol

I can actually find the proof online of a characterisation like this

[mighty kiln]

Oh I forgot xy has degree 1 in each var rip

[slim kayak]

ah shit

nvm its fine

[mighty kiln]

cxy too

[slim kayak]

A_Z remains a group scheme 🥳

[runic galleon]

thank you

i love u

if u ever need a wife4let me know

[slim kayak]

as long as you are not mean to him

[coral spindle]

Uh I'll take that as a compliment, thanks

[south patrol]

But ye I can find u the proof

If u want

Well it is for formal group laws which have more constraints than what you gave tbf

But the idea is that you can classify the fgls coming from polys

[slim kayak]

the fgls?

[south patrol]

Formal group laws

[slim kayak]

ah

Yeah in my case I just needed something that played nice with the other choices for group object structure, from the name I take it they are related

Speaking of them, to show that Hom(X,G) has group structure for a group object G and any X I would need to know that X has a corresponding X x X with the appropiate diagonal map, no? The statement I have in front of me doesnt mention them

[frank cosmos]

if in a UFD, x=u*i where u is a unit and i is an irreducible, how can i factor x?

[slim kayak]

you cant meaningfully, you basically described an irreducible element

[frank cosmos]

if u is a unit, and u^(-1)a is a unit, then is "a" a unit?

[agile burrow]

You shouldn't need a diagonal map X -> X x X. Given two morphisms f: X -> G and g: X -> G, you get a unique morphism X -> G x G making relevant stuff commute, which you can compose with multiplication to get a morphism X -> G

[frank cosmos]

of course it is

so this means if x is a unit times an irreducible, then x is irreduble

cool

[slim kayak]

Ah yeah. The induced map should behave just like (f,g) \circ diag_G so the argument ought to be the same

[agile burrow]

That's right, and the axioms for the multiplication morphism in the group object ensure that this multiplication in Hom(X, G) satisfies the group axioms

[slim kayak]

yeah I already checked them, just panicked a bit when I noticed i did an illegal move

[celest cairn]

May someone give me an outline or provide a different example on how to do this?

[coral spindle]

Have you tried anything yet?

Just start by trying to prove it

and if it gets hard, try to find counterexamples.

[celest cairn]

Ok thanks.

[coral spindle]

In general, I strongly suggest you give exercises a good try before asking for help with them. You don't learn if you don't try things independently, and it's a bit disrespectful to ask for someone else to just do it for you if you're not willing to put in the work.

[slim kayak]

how does a pullback of idk what relate to the the group structure on Hom(X,G) being functorial in X?

like showing the rhs is really straightforward I am just wondering what they meant

[agile burrow]

I mean showing that pullback preserves the group structure is equivalent to functoriality, so I'm not sure what the confusion is. Like to spell it out, pullback commutes with multiplication as we've defined it on Hom(X, G). So if you have a morphism f: Y -> X, then the morphism f*: Hom(X, G) -> Hom(Y, G) taking g to g \circ f is a group homomorphism.

[teal vessel]

are there divisible finite groups? I'm showing that there are no divisible finite abelian groups, but the track I'm looking at seems not to depend on the abelian property.

[slim kayak]

Under pullback I just picture the pullback of a fiber product

[agile burrow]

Oh, I think in this case pullback is just meant as precomposition

[slim kayak]

Oh

[delicate orchid]

How would you define nG for G non-abelian. They’re not Z-modules

[teal vessel]

I have no idea what those are

[agile burrow]

How are you defining a divisible group?

[slim kayak]

why are cat theorists so extra sometimes

[delicate orchid]

Agree with this, and I was with you on the “what the fuck does this mean” train

[teal vessel]

a group is divisible, if, and only if, for all members g of the group, and for all non-zero integers k, there exists a member x of the group such that x^k=g

[delicate orchid]

Take k = |G| then

By Lagrange x^|G| = 1 for all x in G

[teal vessel]

sure, but that doesn't seem to depend on being abelian

[slim kayak]

appreciated

[teal vessel]

I have this abelian component here that seems to matter else why mention it? but I don't seem to require a dependence on it.

[agile burrow]

I've typically only seen divisible groups in the context of abelian groups because it's relevant for homological algebra, but sure this definition works. But yeah, wew has shown that there are no non-trivial finite divisible groups as you've defined divisibility

[teal vessel]

yeah, it's a nicer version of the thing I was trying, but it just weirded me out that I was particularly proving abelian group properties when this seems irrelevant to being abelian by the definition given.

[agile burrow]

Sure, yeah. If you haven't seen modules yet then homological algebra may be a bit out of reach at the moment. But divisible abelian groups are relevant there and I haven't seen a particular use for non-abelian divisible groups so that's probably why the abelian condition is being imposed

[teal vessel]

well then I guess "no finite groups exhibit this property, therefore abelian finite groups definitely don't exhibit this property" is a good enough proof. Having the extra context for why we only care about abelian groups seems nice, but if they're not going to talk about it, it feels like the extra legos in the bottom of the bag.

[delicate orchid]

yeah I don't know why you're talking about divisible groups at all then

[daring nova]

I'm trying to understand this section of the lecture notes about quotient groups of invertible matrices and how the determinant induces a certain morphism:

Notations, though they're mostly standard
So we have GLn the invertibles, SLn the ones of determinant 1
k*, the field's multiplicative group, is also used to represent the set of homotheties/dilation since the isomorphism is clear
µn is the nth roots of unity
PSLn = SLn / µn, PGLn = GLn / k* are groups divided by (a subgroup of) the homotheties

That's fine, so is the injection of PSLn into PGLn, but I don't understand how the determinant "induces a surjective morphism from GLn to k* / k^{x, n}, of kernel k* SLn" giving the exact sequence at the bottom
the determinant outputs a scalar, so why is k* quotiented by whatever k^{x, n} is ? I assume it's either the vectors, or the diagonal matrices, but neither makes sense

[icy totem]

Is it true that Z[i]/(x) is a field iff x is an irreducible element of Z[i]? Z[i] is the domain of gaussian integer, while (x) is the ideal generated by x. The professor taught us that it is true if i replace Z[i] with K[x], with K field, but intuitively it has to be true with Z[i] too, right?

[delicate orchid]

yes, because the Gaussian integers are a principle ideal domain

[slim kayak]

the determinant gives you map into the units of k, so I guess you can try to work out in reverse what k^x,n is meant to be

[daring nova]

the units of k ?

but it's a surjection

[delicate orchid]

more generally, R/I is a field if and only if I is a maximal ideal

[icy totem]

Thank you

[daring nova]

oh the invertibles

[slim kayak]

Yeah

[icy totem]

Ahh I see, thats interesting

[daring nova]

I'm sure it's meant as (k*)^n but idk how to interpret it here

[icy totem]

Even tho we didnt talk about maximal ideals (I know what they are tho)

[daring nova]

actually I'm not sure

[delicate orchid]

fun exercise to try and see why

[icy totem]

Ill try it then

[slim kayak]

probably the set of n-th powers of invertible elements in k. You'll have to see why quotienting those out gives you a surjective map

[daring nova]

but the determinant is naturally surjective ?
Why would we quotient anything ?

[slim kayak]

its naturally surjective on GL

[delicate orchid]

the determinant is different here, but they use the same thing

it's the factoring through the quotient by k^{\times}, I imagine

[daring nova]

so he typoed and meant PGL ?

[delicate orchid]

the main reason I can see behind quotienting out by that is the fact that if you take a multiple of the identity matrix, say 2I, then in GL_2 this has determinant 4
but in PGL_2 we want this to agree with the determinant of just I

so we have to quotient out by 4

so on for all x, we have to set x^2 to 1

[slim kayak]

Typo where

[daring nova]

2nd to last line, when he says GLn(k) -> k^x / k^{x, n}

[slim kayak]

probably not

but it sounds like using the property of quotient groups then

[delicate orchid]

I'll draw the diagram

[slim kayak]

As in, the kernel of GL->PGL has imagine contained in (k^x)^n

[daring nova]

cause it's just the homotheties
So isn't it actually an equality ?

[delicate orchid]

what's "it"

[daring nova]

the kernel Kerr just mentioned

[delicate orchid]

my main issue is that (k^x)^n isn't a subgroup so what are we actually quotienting by

[daring nova]

if it's the nth powers it is a subgroup

[delicate orchid]

it being in french is really messing with me as well

[daring nova]

actually, PGL doesn't have a nice set of representants does it ?

each coset is just matrices such that one is a (scalar) multiple of the other, but you can't favor one easily

[slim kayak]

oh

[delicate orchid]

I might be wrong on that, I just don't buy that it's a subgroup

something very very wrong with that claim to me

[daring nova]

the nth powers ?

[delicate orchid]

holds for k finite by fundemental theorem of f.g. abelian groups

[daring nova]

yeah, k is abelian

btw

[delicate orchid]

k is a field, of course it is

I just can't convince myself this is true for Q

[slim kayak]

wait so like why isnt it a subgroup?

[daring nova]

it's a nth power iff the numerator and denominator are, once reduced ?

[daring nova]

x^n y^n = (xy)^n
1 = 1^n
(x^n)^-1= (x^-1)^n
is very subgroupy to me

[delicate orchid]

I'm thinking of like, amego groups here

[slim kayak]

yeah

[delicate orchid]

continue on without me anyway

[daring nova]

ok so let's go back here

[slim kayak]

Yeah

Do you have the map GL->PGL on hand

[daring nova]

he didn't make anything explicit,

[slim kayak]

do you even know what PGL means 😭

[daring nova]

I'm guessing A -> A k*

[daring nova]

it's defined on top, as the quotient by homotheties

[slim kayak]

thats pretty dry. Homtheties?

[daring nova]

which amounts to this

[daring nova]

diag(a, ..., a) for some a

[slim kayak]

Ah, which have determinant a^n

[delicate orchid]

I'm being doxxed.. modddsss

[daring nova]

then det(A k^x) = det(A) {det(diag(a, ..., a)} = det(A) k^{x, n}
So I see why it works with PGL as domain, not GL

[slim kayak]

yeah so with det = a^n for the elements in the kernel this becomes pretty standard quotient stuff

[daring nova]

but ker(pi) = k^x is not a subset of ker(det) ?

[delicate orchid]

this is essentially what I wrote up here lol

I believe it's a subgroup now, I forgot that the multiplication map A^n -> A restricted to the diagonal subgroup was a group homomorphism for abelian groups lol

[slim kayak]

huh?

[daring nova]

the homotheties (dilations if you prefer) don't have determinant equal to 1

[slim kayak]

oh i shouldnt have written det, pretend its det star or det bar. Det tilde if you are 💅

[daring nova]

which is ?

[slim kayak]

anyways the kernel of pi is just the the matrices diag(a,...,a). Those have determinant a^n. The kernel of the det in my pic is any matrix whose determinant has the form a^n. So the former is deffo a subgroup of the latter

the actual map is det and then the quotient map k^x to k^´x/(k^x)^n

[daring nova]

which is A -> det(A k^x) ?

[slim kayak]

It sends A to det A (k^x)^n

[daring nova]

so it's det o pi

I guess

[slim kayak]

I mean, it also sends other matrices to the identity

[daring nova]

we didn't say it's injective

its kernel is k^x Sln

which makes sense to me as they have det x^n for some x

[slim kayak]

hm, right. You will have to check that the kernel of the induced map is PSL_n(k)

But I think that ought to be no trouble

[daring nova]

the kernel is the A such that det(A) is a nth power

so for PSLn, det(A µn) = det(A) = 1 so it's entirely in the kernel

if A k^x in ker(det), then det(A) = a^n for some a, so A/a is in SLn, so A k^x is in PSLn's image

right

[slim kayak]

your notation is a bit weird but that should be the idea, yeah

keeping to SL and GL the kernel of the induced det are those matrices which decompose as diag(a, ..., a) times a matrix from SL. But PSL are exactly those matrices under the equivalence relation of being multiple of diag(a, ..., a)

[lusty jungle]

Was told to repost this from help channel:

I would greatly appreciate if anyone could provide some insight on this (and pls ping me when responding)

[glossy crag]

What they mean here is K:=ker(...) is a subrepresentation. The elements of K are mapped back to K by all pi(g), so (pi,K) is a representation in its own right.

[lusty jungle]

Hmm yeah

Oh wait so here

Saying that some vector space V is a representation is really just an abbreviation for saying that (pi, V) is a representation right

This prob sounds slightly pedantic but I just want to make sure I have all the terminology right lol

[glossy crag]

Yes, it's standard practice in math to omit words/symbols whenever possible.

[lusty jungle]

Got it thanks!

[glossy crag]

Quick sanity check: if A,B are groups and f:AxB->AxB is a map, then to show it's an automorphism it's enough to show its restrictions to Ax1 and 1xB are its components f_1,f_2 are, right (where f(a,b)=(f_1a,f_2b))? Never mind, this doesn't make sense.

[celest furnace]

Yep you figured it out. You could do something where you take B = A and swap the factors and that’s an automorphism I think

Weird stuff can happen with the direct product

But you get this really nice theorem: if gcd(|A|, |B|) = 1, then every subgroup of AxB is of the form CxD for C <= A and D <= B (NOT true in general!!)

[vivid birch]

I need some help with the following:

Let $A\subseteq B$ be an extension of commutative rings and assume that $B\setminus A$ is multiplicatively closed. Prove that $A$ is integrally closed in $B$

[cloud walrus]

Eduardo

[vivid birch]

if c(A) is the integral closure of A, then if A is not integrally closed ver B, we can find x in c(A)\A and in that case, x is in B\A

this means that $x^n + a_1 x^{n-1} + ... + a_{n-1}x + a_n = 0$ for some $a_i\in A$

[cloud walrus]

Eduardo

[vivid birch]

and each power of x is in B\A because it's multiplicativelly closed

but this is as far as I got. Any tips on how to proceed from here?

this tells us that $a_1 x^{n-1} + ... + a_{n-1}x + a_n \in B$. maybe this is the way?

[cloud walrus]

Eduardo

[rustic crown]

say f is the minimal degree monic polynomial which x satisfies. then f(x) = x * g(x) - a for g monic with smaller degree.
argue that g(x) is also in A, hence g = 1.

[vivid birch]

hhmmm I'll try that. thanks

[rustic crown]

[vivid birch]

i mean @rustic crown

tapped the "tab" key way to fast xD

[copper smelt]

A ping?

[vivid birch]

my bad

[copper smelt]

Oh

Nvm

[rustic crown]

[vivid birch]

I don't get 2 things @rustic crown . 1) why is g(x) in A? 2)If it is in A, why it's the constant polynomial 1?

[vivid birch]

in this case, $g(x)= x^{n-1} + a_1x^{n-2} + ... + a_{n-1}$. why is it in A?

[cloud walrus]

Eduardo

[rustic crown]

x * g(x) = a is in A

[vivid birch]

oh

got it

[rustic crown]

and g is a smaller monic poly

g(x) = a'

then x satisfies g - a'

g cannot have positive degree

else f wan't be the minimal degree thing

rephrase it whatever way you like the most >.<

[vivid birch]

I think I get it

[rustic crown]

[rustic crown]

typo*

[vivid birch]

but isn't this a contradiction already?

because deg(g - a) < deg(f)

oh oh oh nvm

[rustic crown]

yea so g - a' is already the zero polynomial

[vivid birch]

i see it now

[rustic crown]

g = 1 = a'

[vivid birch]

thanks :)

[rustic crown]

(since g was monic)

[vivid birch]

let me just do one more question about this. How did you figure this out so fast? did you already do something like this? what was your thought process? @rustic crown

[rustic crown]

nah, saw this for the first time.

tried a few things and this worked

[vivid birch]

haha thanks xD

[rustic crown]

you threw away the leading term, i threw away the constant term

and it magically worked

[vivid birch]

algebra in a nutshell (at least for me)

[rustic crown]

hehe :p

i think first thing i tried was to come up with an example of this situation

that wasn't so easy

so tried looking at it directly

[rustic crown]

(tbh i only looked at Z --> Q, and realized 2/3 * 3/2 = 1, so gave up looking for an example)

[vivid birch]

Either way, this was very helpful. thank you again @rustic crown :)

[rustic crown]

[rotund aurora]

Maybe I'm clowning but what if you take G to be countable, take the discrete topology on G (so that G is second countable and locally compact) and some other topology giving another topological group G' (as groups these are the same), such that G' is locally compact and second countable. Then the identity f: G-->G' is an isomorphism of groups and is trivially continuous, but it's not open, otherwise the topology on G' would also be discrete, which is not.

💀

apparently this is not possible? lol

[rustic crown]

how do you know two distinct topologies on G exist (which are both locally compact and second countable)?

[rotund aurora]

Pick a suitable G lol

[rustic crown]

like?

[rose prism]

what

[rotund aurora]

you edited bruh chill

[rose prism]

you proposed a counterexample

[rustic crown]

.<

[rotund aurora]

Yeah, i realized this is probably not possible

I'm trying to prove

[rose prism]

what is probably not possible

[mighty kiln]

Second countable regular Hausdorff → metrizable

[rotund aurora]

I'm not writing it again, you can read up

[rose prism]

lol ok

i read your messages already

they dont make sense

[rotund aurora]

they do, because det apparently understood them

[rose prism]

ok

anyway countable topological groups are discrete

so idk why you say youre trying to construct a counterexample

[rustic crown]

that's not true, (Q,+)

[rose prism]

what

[rustic crown]

(locally compact )

[rose prism]

yes

[rotund aurora]

sure lol

[rose prism]

with the adjectives

[mighty kiln]

If there are no isolated points, then BCT gives contradiction

If there is an isolated point, then every point should be isolated

By transitivity

So you end up with discrete

[mighty kiln]

Actually even just locally compact Hausdorff is enough

[runic galleon]

is morphism the same thing as homomorphism

[dull marsh]

Group homomorphism is a special case of morphism

[rocky cloak]

In most cases yes.

The word morphism is used in category theory to describe the arrows in your category. So for example the morphisms in the category of groups are the homomorphisms of groups.

You could also have a category where the morphisms are not homomorphisms of course, but the word just comes from abbreviating homomorphism

[prisma ibex]

Morally speaking, yes

[runic galleon]

thank you everyone 🐬

[prisma ibex]

The only time I can think of this causing some confusion would be if it’s not clear what category you should be working in, like if you said “morphism G->H” to mean a function of underlying sets of groups, but then this is bad writing and you should say “function on underlying sets” or “morphism |G|->|H|” to disambiguate this for example

I guess morphism would also be bad if you ever talk about heteromorphisms but this is such a rarely used notion (I’ve literally never ever seen this even in pretty category theory heavy papers) that this kinda doesn’t matter

[cobalt heath]

Some ppl like to talk about isomorphism A -> B between sets

[mighty kiln]

When bijection is a word

[cobalt heath]

"We like fancy general word, so we gotta use isomorphism instead of bijection"

[runic galleon]

guys do the binary operators in a morphisme has to be not the same ? or they can be the same ?

[mighty kiln]

What are binary operators in a morphism?

[runic galleon]

ok wait maybe im a bad explanator

[mighty kiln]

Binary operators in an algebraic structure?

[runic galleon]

let ( G, * ) and ( H, ¤ ) be two groups

f : G = > H is a morphism iff f ( x * y) = f(x)¤ f (y) while x,y (- G

my question is does * have to be diff than ¤ or the can be the same binary operator

[mighty kiln]

Ye they can be totally different groups with different operators, it's a homomorphism as long as the condition is satisfied

For instance, the exponential function f(x) = e^x, from the group (R, +) of real numbers under addition to the group (R+, ×) of positive real numbers under multiplication

[runic galleon]

no i know this

my question is does they have to be diff operators or they can be the same

[mighty kiln]

I'm not sure what you mean by "different operators". The operator is something that belongs to a group, and different groups will have different operators

[void cosmos]

if f: G-->H

then f(x) is an element of H

if the operation over H is whatever

then it wouldnt make sense to operate on elements of H other than with whatever operation H is equipped with

[runic galleon]

but like ( C, + ) and ( R, + ) are two diff groups with the same operator ?

[mighty kiln]

They're not the "same" operator

They are indeed related: the operator on R can be obtained d hy restricting the operation on C to the subset R

But they're not the "same"

[mighty kiln]

They can be something like some sort of "addition" to some sort of "multiplication", as here

[runic galleon]

i didn't understand

why they are not the same

• is called a binary operator right ?

• too

minus i mean

these are the operations

[mighty kiln]

• doesn't represent just one operator

• can represent many different things

[dull marsh]

Yeah it's overloaded notation

[mighty kiln]

Addition on complex numbers, addition on real numbers, addition of matrices,

[dull marsh]

Mfw gotta define + more than four times

[runic galleon]

what the hell

fr?

[mighty kiln]

Most of the "additions" one learns in high school are very similar in definition,

So we usually ignore the difference

But abstract algebra is the study of different types of algebraic structures

So naturally many operators

+, ×, etc are just notation

Just like x, y, z etc are used as notation for variables

Their meaning depends on context

[mighty kiln]

Symbols like 0, 1 also depend on context in abstract algebra

They don't exist on their own; they exist as elements of groups / rings / fields etc

So their meaning and their properties may change depending on the group / ring etc

[ivory trail]

there's also no reason to believe that two things with different names have to necessarily be different

if that's the point of confusion

[topaz solar]

<@&268886789983436800>

[teal vessel]

really dumb question to jog my memory. For any groups (A,+),(B,&), the group A x B is the set of ordered pairs with domain A and codomain B along with binary operation (+,&), yes? that is, the group operation of the direct product is element-wise using the original operations from the component groups

[mighty kiln]

Yes

[teal vessel]

cool. that makes life a lot easier

[runic galleon]

what does a sub group generated by an element mean ?

what do they mean by generated in this context ?

[south patrol]

If G is a group and x an element, you can think (slightly informally) as being what you get from x after multiplying it with itself, taking inverses etc and doing all possible operations

[teal vessel]

a "generator" is kind of like a seed. remember that all groups are closed under their operation, so if you say that element x is in a group, then by closure so is x^2, x^3, etc, as well as inverses

[south patrol]

Formally you can define the subgroup generated by any subset S of a group G to be the intersection of all subgroups of G which contains S, or basically the "smallest" subgroup contianing S (since G is a subgroup of G containing S, this intersection is well-defined)

[runic galleon]

isn't the intersection of all subgroups of G containing S is S ?

since G is also a subgroup of G , and we take the smallest subgroup, then it must be S ?

[south patrol]

No

For example every subgroup of Z containing 1 is Z

Because you get 1 + ... + 1 = n etc

[runic galleon]

i didn't understand

[runic galleon]

this kinda ryhmes

[runic galleon]

this doesn't

[teal vessel]

if you're only intersecting subgroups, shouldn't what you get at the end be a group? can it be a group if you only have S and not the inverse of S, as well as the identity?

or, another way to think about it: every subgroup containing S is a group so it should have the inverse of S as well as a pairing, and all groups need identity, so all subgroups will have that identity

[cobalt heath]

Yea like

Any subgroup of Z should contain 0, so a subgroup generated by S = {1} cannot be S

[slim kayak]

its an algebraic description justification of why any subgroup containing 1 is Z. If you have some subgroup containing 1, then it contains -1 and 0. It also contains 1+1=2 and -1-1=-2, and 1+1+1 and -1-1-1... so it ends up containing everything

As for the original question, "generating" works exactly the same as it does with vectors. If I gave you a handful of vectors I can ask you what subspace you can make from those. But an equally valid description is to look at any subspace that contains those vectors and then take the intersection of all of them, for groups this ends up the standard definition of "subgroup generated by S"

[frank cosmos]

how can i prove that if R is a UFD then if (f)=(c)(g) where f,g \in R[x], c\in R, g primitive implies (c)=(cont f)

also is this proof that (f) \subset (g) -> (cont f) \subset (cont g) right? we have f=p(x)g -> (cont f)=(cont pg)=(cont p)(cont g) -> (cont f) \subset cont(g)

[slim kayak]

What's your definition of cont f?

[frank cosmos]

gcd of coefficients

[slim kayak]

And primitive?

[frank cosmos]

not contained in PR[x] where P is a prime ideal of R

equivalent to saying (cont f)=(1)

[slim kayak]

Well, f is always divisible by it's content right

So consider if you have a prime factor of cont f, which is it dividing? c or g?

[frank cosmos]

i dont get it

what it boils down to is (c1)(g1)=(c2)(g2) where c1,c2 in R and g1,g2 primitive implies (c1)=(c2)

[slim kayak]

Yeah, that's one way to phrase it

[frank cosmos]

idk how to prove this lol

[slim kayak]

If p is prime and divides ab, then it either divides a or b. Yes?

You have a UFD, any element is it's unique factorization into primes

[frank cosmos]

oh wait irreducibles are prime in ufds right

[frank cosmos]

yes, since ab \in (p) -> a \in (p) or b \in (p)

[slim kayak]

Convince yourself of that fact

[frank cosmos]

ya i proved it earlier

[unkempt stream]

hello chat

[slim kayak]

Yeah so try to see where (if anywhere) the argument in Z[x] fails to be applicable to an UFD

[frank cosmos]

what is this in regards to

[delicate orchid]

Proving this

[slim kayak]

Yeah, or the formulation before

Hi wew

[delicate orchid]

Think about the fact that (c1)(g1) = (c2)(g2) is like two different factorisations into irreducibles

[delicate orchid]

Hello chat

[frank cosmos]

i havent done irreducibles in polynomial rings yet

nvm

[delicate orchid]

Irreducible => prime

[frank cosmos]

yeye

ok i think i got it: let (c1)(g1)=(c2)(g2), suppose p | c1. then (c1) \subset (p) hence (c2g2)=(c2)(g2) \subset (p)(g1) thus (c2g2) \subset (p) -> p | (c2g2) -> p|c2 since p|g2 -> g2 not primitive

[unkempt stream]

by primitive do you mean the ideal generated is primitive ideal

[slim kayak]

gcd ís a unit

[slim kayak]

sure, that works. A bit hard to read tho

[delicate orchid]

I thought primitive was just “Durr minimal polynomial”

Or does that coincide with the content being 1

[unkempt stream]

i’m tempted to go back on my tyrade of trying to analyze the ideal structure of R[X] based off R

[slim kayak]

what were minimal polynomials outside of fields again

[delicate orchid]

Wait nvm duh. They’re monic

[unkempt stream]

i.e Noetherian-ness and Jacobson-ness being inherited from the subring

[delicate orchid]

Oh yeah. Fair point

[unkempt stream]

hey wew

[slim kayak]

ill interpret that as a no then

[unkempt stream]

do you know of a simple-ish proof of Artin’s Lemma that doesn’t require vector space magic

[slim kayak]

any set of homomorphism into the units is linearly independent?

[unkempt stream]

or that if F^G is the fixed subfield under finite automorphism subgroup G, that [F : F^G] = |G|

[unkempt stream]

That proof can be done using Chinese Remainder Theorem lol

[slim kayak]

nvm that was dedekinds lemma, they should really work on their naming schemes

[slim kayak]

what?

[unkempt stream]

Yeah.

Chinese Remainder Theorem kind of lets you find central orthogonal elements in a quotient and I think it uses that from what i remember

[slim kayak]

our course definitely didnt use the remainder theorem, just some dense confusing sum vector space magic

[unkempt stream]

artin’s lemma does too

but it’s like, the fucking backbone of Galois Theory

[delicate orchid]

Wtf is Fartin’s lemma

[unkempt stream]

this

[delicate orchid]

Ah. I suddenly don’t care at all

[unkempt stream]

Stub your toe. Now

[delicate orchid]

Something something sums of orbits something something invariant irreducible elements something

Then the dimension is nice

Guessing that’s how it goes

[unkempt stream]

artin’s proof uses vector space magic

[unkempt stream]

i know the orbit-polynonial method works to show fixed point subfields are integral but

not really that they’re finite

[delicate orchid]

Oh wait that actually works?! Haha what a guess

[unkempt stream]

Yeah the finitivity has to do with it being a field

but I am unsure how to actually prove it without the vector space magic

[chilly ocean]

Why J二)R

[unkempt stream]

Are you asking why the Jacobson Radical is in the Nilpotent Radical?

[chilly ocean]

Yes

[unkempt stream]

The Nilpotent Radical is the intersection of all prime ideals, while the Jacobson Radical is the intersection of all maximal ideals (which are prime)

[prime sundial]

did you read the sentence right after that

[unkempt stream]

So the nilpotent radical is a more "specific" intersection,

[delicate orchid]

In their defense the sentence after implies that the Jacobson is an intersection over a larger set

[unkempt stream]

I had no fucking idea what the hell "Why J二)R" meant

[prime sundial]

i wasnt trying to shame them haha, i was literally asking if they read the explanation given

like whether they were confused by it

or if they just didnt have any idea whatsoever

[delicate orchid]

The explanation given is just wrong, that’s not what the Jacobian is

Or rather, it is, but that’s a stupid way of stating it. Why not just say “all maximal ideals “

It’s confusing ME reading it

[prime sundial]

i think what they said makes it make sense

you are intersecting over more sets

you should expect a subset

[delicate orchid]

Yes. You’re not reading what I’m saying. I have no problem with their definition of the nilradical

[prime sundial]

what

im talking about the jacobson radical

[delicate orchid]

There’s just an unneeded “and prime”

[slim kayak]

be careful with your "and" usage

[prime sundial]

what did i say that pointed to the nilpotent radical

[slim kayak]

the intersection over (prime and maximal) [for the love of god just say maximal] vs . the intersection over prime and maximal

[delicate orchid]

_ _

[prime sundial]

ah i see the confusion

[chilly ocean]

Why not R contains J..

[slim kayak]

there are less maximals than primes usually

J is the intersection over a subset

[unkempt stream]

I wonder if they just lazily sidestepped that maximals are prime lmao

[wraith cargo]

J is the intersection of all maximal ideals
While R is the intersection of all prime ideals
All the maximal ideals are contained in the set of prime ideals
So R is smaller than J

[prime sundial]

i just realized i was confused about this from the start

so i concur that this is indeed a confusing explanation

[slim kayak]

Maybe, but I cant imagine who leaves that as an exercise to the reader™️ and also adds "prime" in there

then again I cant imagine anyone doing latter

[unkempt stream]

The fact that if x lies in an ideal, then 1 + x cannot is obvious but at the same time not immediately obvious

[dim widget]

what about the ideal (1)??

[unkempt stream]

(1) is always the ring

[dim widget]

but it's still an ideal

[unkempt stream]

I meant strict

strict ideals

[slim kayak]

me casually forgetting the fact that maximal ideals either contain x or some element of the form 1+xy flashback

[unkempt stream]

yeah

[dim widget]

but yeah it's a standard use of the closure relation to show that if M is closed under an invertible operation $+$ and its inversion $-$, then $x \in M, y \notin M$ then $x + y \notin M$

[cloud walrus]

nG+

[unkempt stream]

I realized that the set of units is just the ring removing all the maximal ideals way later on than I should've

I used to be very anal about the proof of Hilbert Basis Theorem because I didn't like how the Leading Cofficient map used for the proof wasn't an actual ring morphism

[slim kayak]

I dont think I ever thought about the former, the closest thing is the variant for a local ring

[frank cosmos]

how do you show that primitive polynomials over R[x] where R is ufd, is irreducible?

this feels very obvious

[south patrol]

That's not true

[frank cosmos]

oops

[south patrol]

e.g. x^2 - 1 is reducible over Z[x]

What you more need is uh

R a UFD with field of fractions k

then primitive polynomials in R are irreducible over R iff they are irreducible over k

The proof is essentially the same as the case R = Z (and k = Q) where this is called Gauss' lemma

[frank cosmos]

yeah, im using my question as a way to prove this lol

irreducible over R -> over K first step is to suppose f is primitive since otherwise you can factor out the content

but then (f)=(content)(primitive) but im trying to see why this makes f reducible

i think showing (primitive) != (1) suffices? since obviously (content)!=(1)

but then i guess 1 \notin (primitive)?

ughh theres so many lemmas/theorems to keep track of in basic commalg lol

cool fact, should have been taught in high school alg (or maybe it was)

if a polynomial with integer coeffients cant be factored w/ a polynomial with integer coeffieints then it cant be factored with a factor having rational coeffieints

wait i think this is weaker than rational root theorem?

since this requires gcd(coeffieints)=1 while RRT works with gcd(an,a0)=1

?

[slim kayak]

thats essentiakky what you are trying to show

[rotund aurora]

why is this true?

[mighty kiln]

Convergence on product topology is pointwise convergence

Consider sequences of the form
(a, 1, 1, …)
(1, b, 1, …)
(1, 1, c, …)
…

[rotund aurora]

why should these converge?

[mighty kiln]

*index-wise convergence

[rotund aurora]

yeah, sure

but I don't see how this remark resolves the question

ah wait

[south patrol]

preimage of 1 right

well

[rotund aurora]

@mighty kiln I am not sure where you are getting at. The images of (a,0,0,...),(0,b,0,....) needn't have a limit point either since GL_n(C) (or just C^x) is not compact, if that's what you meant

[mighty kiln]

Pick a neighbourhood U of 1 ∈ GL_n(C) s.t. the only subgroup H in U is 1

[mighty kiln]

Assume the statement is false, and pick such a sequence whose image stays outside U infinitely often

Oh wait ig I assumed countable product

Something similar should work tho

[rotund aurora]

I don't understand what you are doing

You didn't prove anything

[mighty kiln]

This contradicts continuity since image does not converge

[rotund aurora]

? why should it converge

[mighty kiln]

^

[rotund aurora]

ah wait sorry, right, it converges to (1,1,1,....)

[rotund aurora]

sry but I don't see why this is true

[mighty kiln]

The image of (1, 1, …, 1, G, 1, …) is a subgroup

[rotund aurora]

yes? but how are you obtaining a convergent sequence from those

[rocky cloak]

So the point is that the preimage of U will look like some finite number of proper open sets and then the entire group for all other indicies

[rotund aurora]

why? You could have some elements of (G,1,....) inside U and some outside U

[rocky cloak]

That's what open sets look like in the product topology

[rotund aurora]

ah, yes

Thanks

so just to make sure, the compactness of the groups is unessential here?

[rocky cloak]

Seems so yeah

[south patrol]

Lel

Should just replace by an open about 1 ig

[rose prism]

im skeptical of this

suppose all the groups are compact abelian

[cloud walrus]

memorylessfunctor

[rose prism]

actually maybe this is just the same thing

i think compactness should matter though

[rocky cloak]

Shouldn't it be the other way, like dual of sum is product of the duals

Or I guess both work yeah

[rose prism]

why?

i am pretty sure that is not right

the product of duals need not be a discrete group

hence it cannot be the dual of a compact group

to write it more categorically

we can regard the operation of taking pontryagin duals as a contravariant functor F : CAb -> Ab, where CAb = {compact abelian groups}. this follows from the fact that a locally compact abelian group G is compact iff its pontryagin dual is discrete.

what i wrote is then saying

coproduct of F(G_alpha) = F(product of G_alpha)

[rocky cloak]

I don't see any contradiction.

If G_i is compact, then ^G_i is discrete, and Prod G_i is compact and Sum ^G_i is discrete.

At the same time Sum G_i is not compact and Prod ^G_i is not discrete.

The fact that
Hom( Sum G_i, U(1)) = Prod Hom( G_i, U(1))
is just the definition of the coproduct.

And the one you wrote follows from the preceding conversation.

[rose prism]

oops sorry