groups-rings-fields

[kind temple]

the argument for why we have such an X is given just above Corollary 1 in the paper

[golden turtle]

but M is not an element of R_n so I don't see how the corollary applies

[kind temple]

M is an n x n matrix, no?

[golden turtle]

yes

is not R_n the free module of rank R?

[kind temple]

R_n is used to denote the space of n x n matrices over R in this paper. it’s in the intro

[golden turtle]

oh, that's weird notation

[kind temple]

it’s a bit compact, but i think it works for the paper since they don’t ever make use of R^n. also, the subscripts and superscripts never interact, so it’s fine notation

but yeah, def needed to look it up in the paper

[golden turtle]

seems similar to this proof

then

[kind temple]

wait, isn’t this exactly what u want to prove

[golden turtle]

yes the one I posted is a proof

I just don't quite understand all of it

and I didn't want to copy

[kind temple]

ah, i see

i haven’t read the argument in the McCoy paper, i would guess it’s similar

[golden turtle]

It omits entirely the reason that detM being a zero divisor implies that some minor admits a common annihilator

[kind temple]

the McCoy paper? or the proof u posted

[golden turtle]

I think they are the same right

I mean the proof I just posted

I didn't think to go read McCoys paper

[kind temple]

i think it talks about this

[golden turtle]

Do you know where I can find it, it doesn't cite the paper in the proof

[kind temple]

this link

[golden turtle]

Oh wow

Sorry

I didn't even realize that my proof was citing your paper

[kind temple]

lol all good

[golden turtle]

Is this why we can say that there is some minor admitting a common annihilator

its my attempt at replicating the beginning of the proof

[kind temple]

i honestly have no idea

my algebra is pretty weak

i just found the paper

[golden turtle]

Lol oh

[kind temple]

but uh. if it’s any indication of how useful that paper is for this problem, i was able to piece together an argument for u without even knowing what an adjoint matrix is. i would scan it again for any details u might be missing in case

[golden turtle]

I would have to define nearly everything and then reprove theorem 1 to use the paper

the other proof atleast uses all vocabularly we've covered, albeit it is less explanatory

[kind temple]

try to translate between the proof u posted and the paper. it looks like they are saying similar things in different language

[late sequoia]

H is normal group of G, [G:H]=m, |H|=n, (m,n)=1, prove that H is the only one subgroup of G whose index is n?

I have no idea how to do it.

Oh I get it

[golden turtle]

I have this question and

say that F_s is generated by {x_1, ... x_s}

I want the isomorphism from Fs mod it’s commutator to Zs to be x_1^(r_1)...x_s^(r_s) maps to (r_1, .., r_s)

since the quotient by the commutator subgroup is abelian we can always write it this way

but I have to show its well-defined

what I tried is

supposed two elements of the quotient were equal, then did some manipulations

I am wondering if once I get to the step circle in green I am allowed to conclude what I did, by "freeness"

and then if that actually gives well-definedness

[stark moth]

I'm wondering how we get the "in particular" line from the line above

I understand why the "in particular" line is true, but what's the relation

[rocky cloak]

A representation of G on M is exactly a homomorphism G -> Aut(M).

So you're just replacing M by something specific.

[crystal vine]

I'm interested in the definition of algebraic integers and their minimal polynomial. If alpha is an algebraic integer, then there must be a monic polynomial with integer coefficients that has alpha as a root (as per the definition of integral elements).
Why does it follow from this that the minimal polynomial of alpha also has integer coefficients?

[void cosmos]

cuz u can multiply by the lcm of the denominators

this is basically gauss's lemma

[crystal vine]

but if I have an integral element, and I explicitly find a minimal polynomial for it, how do I know that (without multiplying by anything, assuming it is monic) its coefficients are integers?

ok I think I found it

[thorn jay]

The minimal polynomial of an element is always with respect to some base ring

[crystal vine]

base field you mean, no?

here I take the rationals

[thorn jay]

Commutative unital rings work too

[serene dune]

i adore that handwriting, btw

[crystal vine]

the base field of alpha is the rationals here

so the minimal polynomial has rational coefficients and is monic

[thorn jay]

Well i dont think you can assume coefficients are integers if it's monic

2x - 3 is a nonmonic minimal polynomial over Q

But you cant "make" it monic with the coefficients still in the base field

[crystal vine]

but it isnt the minimal polynomial of an algebraic integer

[thorn jay]

Ok but thats something else

Thats the root of a monic polynomial in Z[x]

Not Q[x]

[crystal vine]

maybe I'm confused

[thorn jay]

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

[crystal vine]

what I'm having trouble with is the computation of the ring of integers of Q[sqrt(d)]

[thorn jay]

Thats the monic polynomial
x^2 - d

[crystal vine]

the argument at some point is that for a = m + n*sqrt(d), the minimal polynomial is (x - a)(x - conj(a))

where a is an integer of Q[sqrt(d)]

and then the claim is that the coefficients of this polynomial must be integers

[thorn jay]

calculate a * conj(a)

wait im stupid

[crystal vine]

the argument uses the fact that the coefficients should be integers to determine what m and n can be

I just want to know why we can claim that they are in fact integers, the definition of integers of a number field requires there to exist a monic polynomial with integer coefficients, not specifically the minimal polynomial for instance

[thorn jay]

A monic polynomial is the product of two monic polynomials

In particular, over the integers, if p(x) is a monic polynomial in Z[x] then all the minimal polynomials which divide are also monic

[crystal vine]

what's the base ring for your minimal polynomials here?

[thorn jay]

Because if
p(x) = q(x) h(x)
Then p(x) is monic => q(x) is monic and h(x) is monic or both negated monic

[thorn jay]

Z

[crystal vine]

that's not what I'm interested in

[thorn jay]

Because thats what youre using for your algebraic integers

[crystal vine]

(x - a)(x-conj(a)) is not a minimal polynomial in Z, otherwise we would already be assuming that it has integer coefficients

[thorn jay]

But if there any monic polynomial p(x) with root a in Z[x] there exists a monic minimal polynomial q(x) with root a in Z[x]

[crystal vine]

sure, by multiplying the denominator

once again though, I don't care about any minimal polynomial, I want this specific one since I'm interested in its specific coefficients, without multiplying them by the denominators

[thorn jay]

No-

[crystal vine]

I think I misunderstood your statement

[thorn jay]

We're working in Z[x] for now

Because thats what your question is about

[crystal vine]

Let me change my question

I actually found what I was looking for in my course script

this is the claim that I wanted a proof for

[thorn jay]

Can multiplying an irreducible polynomial by a constant yield a reducible polynomial?

suppose p(x) is a monic minimal polynomial of a in Z[x]
And n * p(x) = h(x) q(x) for some h(x), q(x) in Z[x]
Then either h(a) = 0 or q(a) = 0
Sps wlog that h(a) = 0
=> deg gcd(p(x), h(x)) > 0 has Z[x] is a UFD
=> p(x) | h(x) as p(x) is irreducible in Z[x]
=> deg h(x) = deg n * p(x)
=> q(x) = m for some integer m

Suppose now p(x) is a monic minimal polynomial of a in Z[x] but can be factored through h(x) * q(x) in Q[x]
Then there exists integers n, m such that n * h(x), m * q(x) in Z[x]
=> nm * p(x) = n * h(x) * m * q(x) in Z[x]
But this wlog means that m * q(x) = k for some integer k
=> q(x) = k/m
=> p(x) = k/m * h(x)
Hence p(x) is also minimal for a in Q[x]

[thorn jay]

Sorry that took a bit im not the greatest at ring theory

But here you see p(x) minimal in Z[x] => p(x) minimal in Q[x]

In particular h(x) | p(x) => h(x) in < p(x) > the ideal generated by p(x) in Q[x]

[somber sleet]

Why is this not an integral domain? Can somebody tell me how one can generally prove that such rings are not integral domains?

[thorn jay]

Notice that
2^3 + 2 * 2 + 3 = 8 + 7 = 15 = 0 mod 5
Thus (x - 2) divides (x^3 + 2x + 3) in F_5[x] and therefore that quotient ring contains zero divisors

[somber sleet]

oh yes, thank you, I'm a bit rusty on algebra, so didn't really see the reason

[thorn jay]

Ofc!

In quotient algebras like this there's a zero divisor <=> the polynomial isn't irreducible

So try to see if you can either find a root or factor the polynomial

[somber sleet]

So does it mean that if I have R[X]/(f) where f is a polynomial, then the quotient is not an integral domain if f is not irreducible

[thorn jay]

Exactly

and when f is irreducible then R[x]/f is an integral domain

(Provided R is an integral domain)

[somber sleet]

It makes sense now, need to brush everything up a bit haha

[thorn jay]

Good idea

Im also just getting properly into polynomial stuff cuz i wanna learn galois theory

[somber sleet]

Spoiler: I did, was damn satisfying, but forgot EVERYTHING

[thorn jay]

Im sure itll be fiinee

[somber sleet]

I'm actually studying commutative algebra rn

[thorn jay]

Oh? How's that going

[mighty kiln]

Commutative algebra is cringe

[thorn jay]

Ikr universal algebra is so much better

But its still kinda fun

[rocky cloak]

Thought it was cring

[rocky cloak]

Think you would need it to be a UFD actually. Otherwise you can have irreducibles that are not prime

I guess if f is always assumed monic there's no issue

For example f = 2x^2 - 2x + 3 is irreducible over Z[sqrt(-5)], but 2f factors as (2x - 1 - sq(-5))(2x - 1 + sq(-5))

So (f) is not prime

[thorn jay]

ohh okay thanks :>

[somber sleet]

I have another question, If I want to show that a =>b, I know that R has only one prime ideal, therefore $P = \sqrt{(0)}={a \in R \mid \exists n \in \mathbb{N}: a^n = 0},$ so I know that P contains all nilpotent elements. But what does hold for R \ P? I only know that there are no nilpotent numbers there

[cloud walrus]

damn_guuurl

[inner steppe]

Can you say that this is true for all groups?

[rocky cloak]

I guess a hint: ||maximal ideals are prime||

[somber sleet]

I have no clue how that should help me

[rocky cloak]

Well, what do you know about units / non-units?

For example, do maximal ideals usually contain units?

[somber sleet]

invertible elements I guess

[rocky cloak]

Unit means invertible element yes.

[somber sleet]

I'd say no.....

[rocky cloak]

And can you back up that claim?

[somber sleet]

I mean I see that all the units have to be contained in R \ P, but I do not understand why it has to be an equality

[thorn jay]

suppose a unit u is in a maximal ideal I
what is the defining property of an ideal?

[rocky cloak]

Well, let's think about what else we know.

What do you know about maximal ideals, are there rings without any?

[somber sleet]

nope

[rocky cloak]

Alright, so then for example R/P would need to have a maximal ideal

What could that ideal be?

Maybe you know something about what the ideals in a quotient ring is....

[thorn jay]

correspondence theorem 🔥

[I, R] \cong Con R/I

[serene dune]

are we winning ?

[inner steppe]

Can someone help me with this, what groups wouldn't this be true in

[somber sleet]

I think we are having a misunderstanding because I'm wondering about R without P

[rocky cloak]

I'm not misunderstanding you at all

[thorn jay]

any even commutative group

[somber sleet]

Why did you write thenR/P?(This would be the quotient right)

[inner steppe]

not following

[somber sleet]

Just trying to make sure we are talking about the same thing

It is clear to me that R\P also needs to have a maximal ideal

OH WAIT

[thorn jay]

oh right I'm kinda dumb lol nevermind

[rocky cloak]

R\P is not a ring. If you want to understand the structure of the elements you should start with R/P

[thorn jay]

it is true what am I on

[inner steppe]

now I am real confused lol, what is true?

[rocky cloak]

Do you see what is being used at each step here, and do you know the axioms of a group?

[thorn jay]

ba^2b^-1 = 1 <=> a^2 = 1

that's true in general

[somber sleet]

I mean it has a maximal ideal, which is itself, because otherwise R would then have more then one prime ideal, since very maximal ideal is also prime(?)

Since also P is prime, R/P is an integral domain

[rocky cloak]

Yes, so a maximal ideal of R/P would correspond to a maximal ideal in R containing P. Since maximal ideals are prime, and P is the only one, P must be maximal!

[inner steppe]

Let a, b, c ∈ G , a group.
if bc = a^−1b, is c = a^−1?

Yes I know what is being done at each step, I did it given this question but I am not sure whether this holds for all groups

The axioms are associativity, closed, invers and identity

[somber sleet]

I get it now, sorry for taking so long and thank you very much ✨ ❤️‍🔥

[rocky cloak]

Take the time you need to figure things out

[thorn jay]

no, because groups do not commute in general

[inner steppe]

wrong question sorry

[rocky cloak]

So bc = a^-1 b does not imply c = a^-1

But that is not used anywhere in your image

[inner steppe]

This was the question

if (bab−1)^2 = 1, then a^2 = 1?
A: Yes, B: No, not in all groups

[coral spindle]

No, in general this will not hold. Simplifying your question, you are asking:

if ab = bc, does a = c?
And this is not true.

[thorn jay]

this is true

[coral spindle]

Yes, this holds in all groups.

[inner steppe]

that is what I meant by wrong question

[inner steppe]

that is what I thought because there is no reason for it, but it's lecture note so I have no way to be sure when I asked chatGPT it said no

[dense root]

whats nilpotent i forgot

[inner steppe]

and gave a weird examples with symmetric groups

[coral spindle]

Do not ask ChatGPT for help with math.

I can imagine it was weird considering it was false.

[inner steppe]

But it actually gives good help on theory

[rocky cloak]

nil = zero, potent = exponent
it's x such that x^n = 0

[coral spindle]

it literally just told you something that was wrong

[dense root]

oh okay

thanks

[inner steppe]

Yeah I don't believe it for questions or just chatGPT answers as true. But if you know what is going on and have a good basic knowledge I really think it can help. I sometimes just let it explain theorem from the book worded in weird way and it really does help.

OR I'm just learning crap

[dense root]

from my experience i only used it to learn not to apply

but it can give you tips

[inner steppe]

True, I think it is only bad if you just post the slides and be like explain and just take it for facts. I read the book but usually it is really worded in werid way so when I ask chatGPT it gives me simple examples of the theorem and it clicks

[elfin wraith]

Ok but it literally just makes things up and if you already don’t understand the material you won’t pick up on that and you’ll just end up more confused

[inner steppe]

I agree but you can't really access live humans 2:00 am at night

[elfin wraith]

People are on here 24/7

[dense root]

yeah and thats even better

[inner steppe]

I thought you meant TAs but yeah you are right, discord is really awesome in that way.

[thorn jay]

ikr I barely sleep

[inner steppe]

Hey my fellow humans, I have another question.

Is the symmetry group G_triangle ​of the triangle isomorphic to (Z 7 ​ ∖{0},×)

How should I think here?

Isomorphism is a bijection β(g ∗ g ′) = β(g ) ◦ β(g ′).

but I really find doing computation with triangle group confusing

[thorn jay]

you mean the dihedral group D_3?

that's noncommutative

but (Z7 \ {0}, x) is commutative and isomorphic to Z6

so an isomorphism can never be made

[inner steppe]

not really following

[elfin wraith]

If two algebraic objects are isomorphic then they share all the same algebraic properties, they’re essentially indistinguishable

So if you can find a property that one group has, and the other doesn’t, this suffices to show there can exist no isomorphism between them

In Z6 it doesn’t matter which way round we add our elements, but this isn’t true in D3, it matters if you do a reflection before or after a rotation for example

[thorn jay]

in fact, the center of D3 is trivial so it's as noncommutative as it gets lol

[kind temple]

i don't think this is the way to go about it. here is what i think you should do:

if s = t, then Fs and Ft are isomorphic. this much is clear.

suppose for contradiction that Fs and Ft are isomorphic but s != t. since Fs and Ft are isomorphic, then Zs and Zt are isomorphic. but Zs is simply the direct sum of s copies of Z and Zt is the direct sum of t copies of Z, so the two cannot be isomorphic as groups

[elfin wraith]

True but probably doesn’t mean much to him haha, quite easy to picture why it matters the order in which you apply your rotations and reflections to a triangle though

[thorn jay]

yeah I mean the center being trivial is a direct consequence of that

as the subgroup of rotations is prime

[elfin wraith]

Yeah I mean I know this, but I can’t imagine that any of these terms mean anything to someone just learning what an isomorphism is lol

[thorn jay]

yeah

ah well, something too look forward to understanding

[bitter rover]

What sort of computations are you being asked to do with D_3?

If you're not to the point where you're comfortable thinking of "group-wide" properties like commutativity, you can look at element-wise properties, too.

For example, (Z 7 ​ ∖{0},×) has elements of order 6, namely 3 and 5, while the "triangle group" has no elements of order 6.

Also worth looking at the "Groupprops" website: https://groupprops.subwiki.org/wiki/Groups_of_order_6

[bitter rover]

You believe it clicks, but ChatGPT is optimized to make you nod your head in agreement.

Make sure you put that feeling to the test by attempting some relevant exercises. If you really do understand then the exercises should be straightforward. If they're not then that "clicking" feeling must have been a false positive.

[inner steppe]

that was the whole question

[bitter rover]

I see — my mistake!

When you said "but I really find doing computation with triangle group confusing" I thought that was a more general remark about manipulating the elements of the triangle group.

[inner steppe]

True but I meant it in a sense, I don't ask chatGPT hey explain what "Group" is. I just post the entire theorem and tell it to give me an example when I feel unsure and it gives me an example which feels in the void, not always tho. But if it is math, I always do exercises regardless who I learn it from.

[bitter rover]

As long as you are able to check for yourself whether the example is correct!

(And you do check)

[thorn jay]

Yeah i learned a lot just by trying to prove theorems

[golden turtle]

Right.. but I have to show that Fs mod it’s commutator is isomorphic to Zs to use that, and that’s what my question was about

[kind temple]

use the universal property instead:

show that for any map out of {x1,…,xs} into an abelian group A, there is a unique map out of the quotient of the free group which factors through the standard inclusion

[golden turtle]

That gives me a homomorphism from Fs to Zs but not an isomorphism

[rocky cloak]

Tbh, I think it would be easier to just solve this exercise more directly without going through the abelianization, but I guess that's part of the exercise...

[golden turtle]

Yeah

[rocky cloak]

Fs is not supposed to be isomorphic to Zs though

[kind temple]

it also shows u that the quotient of the free group is uniquely isomorphic to Zs

[golden turtle]

I mean mod it’s commutator

[golden turtle]

How does it show this?

[rocky cloak]

If something satisfies a universal property it's unique up to unique isomorphism.

[golden turtle]

I thought the universal property for free group’s guaranteed a unique homomorphism

[rocky cloak]

Talking about the universal property of abelianizations

But you can also think about the universal property of Zs.

Like A = Fs/[Fs, Fs] is generated by (the image of) s right. So just take the map from Zs mapping onto s.

This is an inverse to the map A -> Zs, so it's an isomorphism

[golden turtle]

By the image of s? Where s is the generators of Fs and the map is the quotient map?

[rocky cloak]

Yup

[golden turtle]

I’d also have to show that that is well defined which comes back to my original question

I can write any element in Fs/F’s as
x1^(p1)…xs^(ps)F’s
I map this to Zs as
(p1, … , ps)
And this is an isomorphism
It only remains to show that it’s well defined and all representatives of the same coset get mapped to the same thing in Zs

[kind temple]

you don't have to show anything.

there is a unique map out of F(S) into Z(S) whose kernel is exactly the commutator subgroup [F(S),F(S)] by using the universal property of the free group over S.

now you can apply the universal property of the quotient group to F(S)/[F(S),F(S)] to get a unique isomorphism from F(S)/[F(S),F(S)] to Z(S)

[toxic zephyr]

so conjugation of any subgroup by any element gives another subgroup (maybe not distinct) of the same size. and so being normal means that it's always fixed by conjugation. what is the significance of this? like conjugation keeps coming up over and over, so it seems like conjugation is more than just something you can do with elements.

[golden turtle]

How do you know about it’s kernel being the commutator? I’m imagining the triangle for the universal property
S

F(S) Z(s)

[bitter rover]

Normal subgroups are the kernels of group homomorphisms

[golden turtle]

We get a unique homomorphism from F(S) to Z(s) from this

But idk how to see it’s kernel is the commutator

[toxic zephyr]

sure but I'm asking more generally about conjugation

[golden turtle]

Conjugation is a group action

And lots comes from that

[bitter rover]

Write out a permutation using cycle notation and conjugate it by another element. What happens to the original permutation?

[rocky cloak]

Which map is it you're questioning being well defined

[golden turtle]

This one

[winged void]

Theere is a tip that i can give you think of isomorphism as the same elements but having diffrent names:
Three ways to check if that is the case:

1. Cardinality of both is the same so not then clearly its not an isomprphism:
2. if the set is abelian then the other set must be as well abelian
   3)Both have the same number of elements of the same order

that is a tip

@rocky cloak correct me if im wrong but those are tips that i use most of the time

[golden turtle]

They’re correct

[rocky cloak]

But you already know that you have a map Fs -> Zs and that this maps Fs' to 0. So then you get a map Fs/Fs' -> Zs

[thorn jay]

https://youtu.be/0ob1m4XnVwQ?si=VPgjFoSYc2k_yS1E

[golden turtle]

How do we know it maps Fs’ to 0? I was wondering this earlier

I know we’re guaranteed a unique homomorphism by the universal property of free groups but I don’t know why it’s kernel must be Fs’

[rocky cloak]

Well, Fs' is by definition generated by commutators. The image of a commutator is a commutator, and commutators in Zs are 0

[golden turtle]

Ohhh

So it’s very similar to this

N is normal and in the kernel of alpha

So we get an isomorphism

This is like obviously different groups and sets and stuff

But same idea?

[rocky cloak]

Sure

[golden turtle]

Oh okay

[thorn jay]

Where is this from?

[golden turtle]

I’ll try working it out

[thorn jay]

Did you write itb

?*

[golden turtle]

My homework

Yes

[thorn jay]

Nicely formatted

[golden turtle]

Thank you I’ll try working this out now

And thank you @kind temple

[kind temple]

the way that you get exact equality is a bit lengthy, which is what i was trying to write out. but one thing should be clear: the kernel certainly contains the commutator subgroup, and maybe that is enough. it seems like you figured this out tho.

[toxic zephyr]

this is incredibly helpful. so it seems like it really is just sort of change of basis.

[golden turtle]

Yes I think in the universal property for quotients we only need N subset ker(alpha)

[thorn jay]

Change of basis is conjugation :3

[toxic zephyr]

yeah exactly

my brain works on linear algebra so that's how i think of everything

[thorn jay]

Mm, fair

Linear algebra is used almost everywhere anyhow

Representation theory for example

We went too abstract and now we're backtracking and trying to link everything to linear algebra loll

[bitter rover]

Strongly recommend doing what I suggested: conjugate a permutation written in cycle notation. That conjugation acts like a change of basis will be immediately apparent.

[toxic zephyr]

yeah i did that, it's pretty cool

[bitter rover]

Other reasons it's important...

[bitter rover]

In general, if you have a structure, it's worth looking at automorphisms of that structure. So for a group G, the set of all isomorphisms from G to itself forms a group denoted Aut(G).

Conjugation corresponds to an "inner automorphism": https://en.wikipedia.org/wiki/Inner_automorphism

[thorn jay]

The only properly computable automorphisms 😔

G/Z(G)

[cloud lynx]

is there a generel form for zero more than 1 variable?

like for R[X] f=(X-a)g

[thorn jay]

What

[cloud lynx]

I have to show that a set is algebraic or not

but its in C^3

and during my thought I wonder how the general form look like for (x,y,z) zero of a polynomial f

just interested

there is a set which is not algebraic and I thought it would help if I know the generel form

do u have any tricks to show that a set is not algebraic?

are there some helpful tricks or tools

[south patrol]

There isn't really a general form since there are so many different sets

Often you just want to use intuition about polynomials (e.g. they are continuous) and restrict to lines so they become polynomials in one variable

[cloud lynx]

ah ok ty

and do u know some tricks to show that a set is not algebraic?

[inner steppe]

is {id, (124), (142)} ⊆ S4

I answer no because (124) * (142) = (1 4) which doesn't belong to the subgroup {id, (124), (142)}

[coral spindle]

Well it's certainly a subset, and you used the subset symbol

I think you miscalculated the product in any case.

I can tell you right now that that's definitely a subgroup.

[rocky cloak]

I guess the easiest would be to check if it fails any of the properties that all algebraic sets satisfy.

Like if it isn't closed, or if when you intersect it with a line you get something infinite (that isn't the entire line)

[golden turtle]

I think that I may have been wrong about the commutator being a subgroup being enough

It may need to be the entire kernel

[inner steppe]

1->4->1
2->1->2
4->2->4

Damn it you are right

[coral spindle]

:)

[inner steppe]

thx

[cloud lynx]

ok ty^^

[inner steppe]

{(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} ⊆ S4

Is there an easier way to check or do I have to check all combinations?

no wait

there is no idenitity

nm

[coral spindle]

Again you're using the subset symbol

But you mean subgroup

[inner steppe]

Copied it from slides

[inner steppe]

but if we say there was an indentity element in the subgroup

is there an easier way to find out? Or do I have to check each combination to check if it belongs to the subgorup?

[bitter rover]

For groups of small order, you can figure out which groups it has to be isomorphic to.

For example, there's exactly one group of order 3 up to isomorphism.

[inner steppe]

not following, can you explain a little more

[bitter rover]

Have you learned about group isomorphisms?

[inner steppe]

Yes

bijection f : G1 -> G2 where for all a,b in G1, f(a+b) = f(a) * f(b)

aha wait

Are you telling me to try and find a group that the potential "subgroup" is isomorphic too?

If yes, isn't that harder than just computing the x*y for all x,y in subgroup?

[bitter rover]

That's a very specific way to write it and makes me unsure whether you understand, but that's correct if you think of + and * as two different group operations without any other association to the operations we usually denote with + and *.

[inner steppe]

yes I was gonna use the circle notation

but didn't know how to use that on my keyboard

so I used + instead

[bitter rover]

Any two groups of order 3 are isomorphic to each other. So in particular any group of order 3 is isomorphic to (ℤ/3ℤ, +), i.e., the integers-modulo-3 with addition.

So that tells you exactly how the products need to behave in order for things to work out.

[rocky cloak]

Guess that depends wether you think it's actually a subgroup or not.

If it isn't a subgroup it might be a relatively fast way to show it isn't. But if it is a subgroup it probably won't save you much time.

[bitter rover]

In particular, if (1 2 4)(1 2 4) = (1 4 2) then you're good.

[inner steppe]

Shoudn't it be the other way around, that if it is a subgroup it is easy to show by just finding an isomorphic, but if it isn't.

I mean how can we make a conlclusion that there are no isomorphic groups to a group just because we have not found one

[bitter rover]

Because it's not isomorphic to any group of the same order.

[inner steppe]

yeah

[bitter rover]

And each of those groups has a particular element structure that you can check.

[rocky cloak]

Because for small numbers we know what all the groups are

[inner steppe]

but aren't there like endless of groups with the same order

[thorn jay]

there are finite cayley tables of a given finite order

so no

[rocky cloak]

There's usually quite few, but for orders like 2^n there are quite many

[bitter rover]

For example, add id to that second example:

{id, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}

Well, the other elements all consist of disjoint 2-cycles. That means if you square any individual element you get the identity element.

So, you are looking for a group of order 4 where each element is its own inverse.

[thorn jay]

ah, my favorite group of all time

[inner steppe]

We didn't go thru it in class but I watched a YT video on it where a guy explained that in cayley's table there is like this pattern to see if two groups are isomorphic. I am gonna rewatch it but Ig what you guys are saying is that if I find an actual group of that order and if I can prove that that my "subgroup" isn't isomorphic to that group then I can conclude that it is not a subgroup.

Otherwise I am lost

[bitter rover]

Well, there are only two groups of order 4 up to isomorphism:

ℤ/4ℤ and ℤ/2ℤ×ℤ/2ℤ

[thorn jay]

this is actually a subgroup of S3 embedded in S4

as in, take the subgroup of S3
{ id, (1 2 3), (1 3 2) }
(you can check this is a group, one interesting thing to note that if we notice that S3 is isomorphic to D3, then this subgroup is precisely the subgroup of rotations of a triangle)
this looks a whole lot like the subset which you defined above and, indeed, the map
id -> id
(1 2 3) -> (1 2 4)
(1 3 2) -> (1 4 2)
is an isomorphism

[inner steppe]

btw guys you know that I am talking about this right

{(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} ⊆ S4

if it had an identity element

[thorn jay]

then it would be my favorite group of all time

[bitter rover]

Can this be isomorphic to ℤ/4ℤ? Why or why not?

[kind temple]

yea, this is what i was hoping u would be able to avoid. but maybe there is a slick way to see it

[thorn jay]

and yet four reduced Latin squares of order 4, how come? Z4 greedily takes 3 of them leaving Z2xZ2 to be left with 1
at least it's normal, unlike Z4

[inner steppe]

let me work on it and get back to you

[bitter rover]

All you need to know is that each element in your set would have to be its own inverse (why?) and then ask whether that's also true of ℤ/4ℤ: #groups-rings-fields message

[bitter rover]

If two (finite) groups are given by their multiplication tables then determining whether they are isomorphic is equivalent to the graph isomorphism problem, which is not trivial: https://en.wikipedia.org/wiki/Graph_isomorphism_problem

It's certainly not something you can do by just "looking".

[inner steppe]

I honestly have no clue how to do that

[inner steppe]

so can I answer Idk?

[thorn jay]

😔 better start permuting those rows and columns

[bitter rover]

So, while I believe you know the definition of a group isomorphism, I suspect you don't really understand the consequences of it.

If f: G → H is a group isomorphism and an element g in G has order n, what is the order of f(g)?

[inner steppe]

I guess n?

[bitter rover]

Sure. Can you prove it?

[inner steppe]

no

[bitter rover]

If the order of g is n, what is f(g)^n?

[tardy hedge]

the only cyclic groups that have every element as a generator is Zp right?

[inner steppe]

Idk, I would guess the identity element e

[thorn jay]

every element except the identity

and yeah that's saying gcd(n, m) = 1 for all 1 < m < n
which is only true for n is prime

[thorn jay]

I mean I suppose the trivial group

[glad osprey]

That's Z_0, yet another reason why 0 is prime

[thorn jay]

i hate that 0 being prime makes sense with the definition of prime ideals

[bitter rover]

Say the order of g is 2, so that g·g = e.

What is f(g·g)?

[tardy hedge]

Is there a difference between R/I as a R-module and R/I as a R/I-module?

[thorn jay]

not much

if we're talking universal algebra then in fact they're polynomially equivalent

(which means basically basically the same)

[inner steppe]

f(g)*f(g)

[prime arch]

i mean its the same set, but Z/pZ as a Z-module is just an abelian group, but Z/pZ as a Z/pZ-module is now a vector space . i mean its kind of trivially different still

it was already a vector space, since it was a field

[thorn jay]

if we think about the modules as representations of rings (namely the slice category (Ring V End_Z(R/I)) where R/I is taken to be the underlying abelian group) then there exists an epimorphism (R, m) -> (R/I, m')

that causes the modules to be basically basically the same :>

[bitter rover]

What is f(e)?

[inner steppe]

Can you explain to me where we are going at cuz I'm a bit lost atm.

Didn't you say f is an isomorphism

[bitter rover]

Yes. What do isomorphisms do with group identities?

[inner steppe]

I would assume it maps it to the identity elem

in the other group

?

[bitter rover]

Don't assume; prove it. Or look at a proof.

It's almost certainly one of the first things you prove about group homomorphisms in general.

Anyhow, if you don't know many basic facts about homomorphisms or isomorphisms, then no, you won't be able to use those facts in other situations.

It's fine if you don't! Your class will cover them, I'm sure.

[inner steppe]

I don't know what to say. I am not at that level of understanding. I cannot prove it

[bitter rover]

Can you prove the identity element of a group is unique? That is, if e and h are two elements with the property that eg = ge = g and hg = gh = g for all g in a group, then h = g?

[inner steppe]

we have never even talked about homomorphism to begin with, it is fast paced course with multiple "math" parts unrelated to abstract algebra and we are done with group theory/ abstract algebra module so I am pretty sure we won't

[thorn jay]

Im impressed you had a group theory part without mentioning homomorphisms

Which is one of the most central parts of group theory

[inner steppe]

I did watch a YT video about it tho but don't really remember so gotta rewatch it

[bitter rover]

Interesting. Why were they introduced?

Is the course like a sampling of different math topics?

And now you're on to the next thing, never to talk about groups again?

[inner steppe]

I am taking a course in discrete math, we have graph theory, abstract algebra, error-correcting codes and modular arithmetic

[bitter rover]

Anyhow, many "basic" proofs in abstract algebra look like this one, which proves the identity is unique. Let e and h be two identity elements, then:

e = he   (because hg=g for all g)
  = h    (because ge=g for all g)

[prime arch]

interesting version of discrete math , i haven't seen one like this before

[bitter rover]

If f is a group isomorphism then

f(e) = f(e·e)
     = f(e)·f(e)

and you get f(e) is the identity in H by multiplying both sides by f(e)^-1. etc.

[inner steppe]

but earlier when you were saying

f(g)

what were you expecting me to do?

[bitter rover]

Sure, well, if the order of g is 2 then that means g·g = e

[inner steppe]

f(ge) = f(g)f(e)

[bitter rover]

So...

e_H = f(e)
    = f(g·g)
    = f(g)·f(g)

where e_H is the identity of H

So if g has order n then f(g)^n = e_H

If f is an isomorphism then, in fact, the order of f(g) is n.

A homomorphism is a function that preserves the structure but isn't necessarily a bijection. An isomorphism is a bijective homormorphism.

So if I can see from the cycle structure that every non-identity element of my set of permutations has order 2, then every non-identity element in a group it's isomorphic to will also have order 2.

[golden turtle]

So somehow I need to prove that if you’re in the kernel of alpha you’re a commutator of Fs

[bitter rover]

And that narrows down the possibilities of what groups it can be isomorphic to.

[golden turtle]

Seems not at all clear how is do that since I don’t know what alpha is explicitly

[bitter rover]

Anyhow, never mind this stuff unless you're interested. Your class wouldn't skip over it if it is critical later.

I'm curious if groups come up again, though!

[thorn jay]

Ah you see they take small but discrete chunks of math and pile them together

[prime arch]

hahaha

mine was like that too, but they felt somewhat more related i guess. i mean it also covered way fewer topics

(it was at a community college)

[thorn jay]

Hmm

What topics did it cover?

[prime arch]

extremely basic propositional logic and truth tables, extremely basic elementary number theory things (even/odd proofs, divisibility proofs), induction, the well-ordering principle, and basic set theory and basic abstract definition of functions stuff

my gf's covered some of those topics but she also did some graph theory, i think? i later went to UT Austin, i don't know what their discrete math class was like, but it probably covered graph theory

[inner steppe]

I really apprecitate the help, I need a little break but will check it out letter.

[inner steppe]

No we are done with groups, that is that was the last part of the course

[thorn jay]

I like graph theory

Especially the algebraic side of it

:p

[inner steppe]

I like graph theory too

[prime arch]

literally never taken a course on graph theory , i think its too late for me now. maybe one day i'll end up reading up on it

[inner steppe]

kinda my favorite part of the course,

[thorn jay]

Same

[inner steppe]

no offense to yall but abstract algebra is my least lol

[golden turtle]

graph theory is top tier

[inner steppe]

We are also working on algorithms and graphs on another course it really helps with the understanding on this module

[prime arch]

it's kind of ironic, because i have a publication from an REU i did (with others!) basically about graph theory and combinatorics (and group theory). but graph theory and combinatorics are like the areas i know the least about, other than probably stats and PDEs

[thorn jay]

But i fucked around with it when a yt'er asked me a question about graph theory lol

Mattbatwings the goat

[prime arch]

rip

[golden turtle]

So, Fs' is a subgroup of the kernel of this map but how do we know it is the entire kernel to actually get an isomorphism?

[thorn jay]

cries and sobs

[thorn jay]

Always better than stats

[inner steppe]

I honestly don't know what to say about that. I took stats on the most stressful times which made me hate it but I really found certain aspects of very interesting. But again no offense to yall but I really have not found any joy in abstract algebra so far haha.

[thorn jay]

I dislike stats and probability simply because unlike algebra i am bad at it

Lmao

[rocky cloak]

Well because of the inverse map
Zs -> Fs/Fs'

[inner steppe]

stats is interesting tho specially if you connect it with machine learning

probability part is boring

[thorn jay]

I dont find machine learning to be terribly interesting either haha

[golden turtle]

I'm sorry but I don't know what the inverse map is

[inner steppe]

Honestly yes but that was the only part that I liked from stats, we did a little lab on data training which was interesting but tbh now that I think of it I would prolly pick abstract algebra over it

[thorn jay]

Yesss ill count that as a win

[rocky cloak]

The one where you map s in Zs to the image of s in Fs/Fs'

[inner steppe]

I had to give yall something lol

[rocky cloak]

Zs is free abelian, so you can map s wherever you like in any abelian group

[golden turtle]

"the image of s in Fs/Fs' " meaning the coset sFs'?

[rocky cloak]

Yup

[golden turtle]

So I understand completely to here

And then you’re saying we get an inverse homomorphism so it must be an isomorphism

Is that right?

Is it specifically the inverse of alpha-bar

[rocky cloak]

Yes that's right.

If we can it beta, then for x in s
alpha-bar(beta(x)) = alpha-bar(xFs') = x
So that composition is the identity. And beta is surjective, so then the other composition will be aswell

[thorn jay]

Loosely related to this channel? Notation i used when studying Latin squares as sets of permutations

How did i not go insane? Maybe i did and this was the result..

[golden turtle]

just to clarify, can we also just show easily that beta(alphabar(xF's))=Beta(x)=xF's

so beta really is the inverse of alpha-bar and it thus must be an invertible homomorphism hence an isomorphism

that is so clean

[golden turtle]

I am trying to follow the direction of "Det A is a zero divisor implies A is singular"

I am specifically confused about the k=n-1 case, (k defined as in the above image)

Why does such a column v exist?

is it because, if it did not, then all columns v of adjA would be annihilated by alpha?

I am concerned that in this case all of the columns v of adjA could just be 0

[toxic zephyr]

whats the definition of singular here? not surjective or not injective? i don't remember if they're equivalent for a matrix over a generic commutative ring.

[golden turtle]

not injective

it says at the beginning "we have to agree what singular means first.." and says for this it means not injectiv

[toxic zephyr]

so the idea of the adj when the rank is n-1 for A over a field is that we can suppose one column of A is a linear combination of the others vi=c1v1+...+cnvn (i not included), but the rest are independent. the cofactors of the ith column can't all be nonzero, because at least one of them is the determinant of an (n-1)x(n-1) invertible matrix (otherwise, they would be dependent). for the cofactors of the other columns say col j, we can do column operations to isolate the linear combination cjvj in the dependent column. so you get +-cj times the cofactor column of the ith column. so that's why it's rank 1. it becomes alpha wu^T where w generates ker(A) and u annihilates col(A) (i.e. u generates ker(A^T)).

so maybe you can generalize that to a commutative ring. i don't remember just how directly we can apply it.

[golden turtle]

why is this problem so hard 😭

[toxic zephyr]

i just don't know how finnickey determinants are in a commutative ring.

like could you take the det of independent columns and get a zero divisor? oh... wait that's what you want to prove

[golden turtle]

I can even pretend to understand the step that I was stuck at

but the next part

is just worse

I've found, maybe 2/3 proofs of this? and they all look the same as this one

well 4 if you include a paper that has 26 pages and 25 pages of wedge product stuff and then a clean proof on pg 26

[toxic zephyr]

im curious to what extent the determinant is zero or just a zero divisor. like is it a matter of how dependent they are?

[golden turtle]

🤷‍♂️

[toxic zephyr]

i would conjecture that a zero det means linearly dependent with like non-zero divisor coefficients (like vector space independent). but perhaps if you could only get a nontrivial zero linear combination using zero divisors, perhaps the determinant reflects that.

so i'm wondering if you can use that idea to guarantee that the entries of the adj in the n-1 case are not all strictly zero, but simply zero divisors (with at least one nonzero).

but idk im not even really helping anymore sorry

[golden turtle]

its ok

my HW was hard but this problem is just really annoying me

its like way harder than the rest

[tough cairn]

what's a nontrivial zero???!??

[rocky cloak]

If all are zero then they definitely have a common annihilator

[inner steppe]

Is (S3, ◦) a (non-trivial) direct product of groups?

But A x B doesn't need to be isomorphic to S3 tho?

[grave sedge]

"S3 is a direct product of two groups" is equivalent to "there are two groups the product of which is isomorphic to S3"

[inner steppe]

For groups G , H, {(g , 1_H ) | g ∈ G } is a subgroup of G × H?

How do I verify that, I mean how do I know if it contains the identity element

[rocky cloak]

Like columns being independent is equivalent to the matrix being injective. Which Austin is showing is equivalent to determinant not being a zero divisor.

So if you take your ring and quotient out the annihilator of the determinant, the determinant now becomes a non-zerodivisor. So if you had some linear combination that resulted in 0. Since the vectors now become independent, the coefficients must have been in the annihilator.

So I'd wager if you take the ideal of coefficients that results in 0 linear combination, you'd exactly get the annihilator of the determinant. At least one direction is clear

[rocky cloak]

Well, if you look at the definition, it tells you which elements it contains.

Namely those of the form (g, 1). And (1, 1) is indeed on that form

[thorn jay]

There is no nontrivial direct product of two groups

There is, however, a nontrivial semidirect producr of two groups

[inner steppe]

but

how do we know 1 is the identity element

[thorn jay]

Because it acts like the identity element and identity elements are unique

[inner steppe]

but 1 is not always the identity element tho, for example 0 is the id when operation is addition or etc.. I am confused

[rocky cloak]

That's what 1_H means "the identity element of H". That's just what that notation means

[inner steppe]

aha

okay so now it makes a little more sense but they wrote just one without 1_H

then we cannot make the same conclusion either

[thorn jay]

1 in abstract groups are always implied to be the identity element

Because we are not working with rings

Unless you're explicitly working in Z/nZ you can assume that

[inner steppe]

but what aobut the case I mentioned about addition where 0 is the identity

[thorn jay]

Then 1 usually doesnt have any meaning

[inner steppe]

Wouldn't that contradict what you said earlier tho

[inner steppe]

this

[thorn jay]

Yeah, i guess

When you denote the operation using • then the identity is 1, or e
When you denote the operation using +, then the identity is 0

[inner steppe]

So I guess unless we are specifically talking about numbers and + where it means "adding", then 1=e which is a symbol for identity

If yes, I can live with that.

[rocky cloak]

An abstract group here mean one you don't really know anything about.

Like if we say H is a group, then we need some symbol to refer to it's identity.

If we know that H is the integers under addition, then 0 is the identity, and we can just call it 0.

[inner steppe]

Yes but they didn't specify whether G or H was groups under addition or anything so how can we make the conlusion that 1 is idenitity was my confusion

[thorn jay]

I also use + and 0 for general abelian groups

[rocky cloak]

It's always a thug of war between having a new notation for every concievable setting and having some notations overlap / contradict eachother.

It's common to use * and 1 and x^-1 for the operation, identity and inverse in an arbitrary group.

It doesn't mean that it has anything to do with multiplication or the number 1.

You could introduce new symbols for abstract groups, but we only have finitely many symbols, and it's much harder to get people to agree on completely unfamiliar symbols

[finite turtle]

how does one know that group isomorphism preserves properties like abelian, normal etc without explicit proof?

[tropic spade]

Try and prove some and see I guess?

It's not uncommon to prove these kinds of things in an intro course but a lot of the times people take them as obvious too.

[hollow topaz]

YO does anyone have full solutions for abstract algebra by pinter? theres sols for chaps 2-4 and 15-28 online but the one for chaps 2-4 uses a different edition(diff exercises at least)

[tropic spade]

Might as well just do the problems and ask for hints when you get stuck.

[hollow topaz]

ohhh thats what you meant

alright

[hollow topaz]

here? or in a help forum?

ill just try asking lol, thank you

[ashen heron]

in #groups-rings-fields

oh

[hollow topaz]

context, x * y = |x + y| and were asked if * has an identity in R
attached proof(?) is from https://yurrriq.codes/abstract-algebra-pinter/exercises.pdf.
isnt this wrong though? |-5 + 0| = 5 != -5

[ashen heron]

i thought this was discussion

[sharp ice]

yeah its wrong. not every element has an identity

[hollow topaz]

thank you!

but just to clarify, its wrong for the reason i stated or smth else?

thanks a lot

im gonna keep asking questions btw

[sharp ice]

for the reason u stated. -5 theres no element that fixes it

[glad osprey]

Why do textbooks only define irreducible and prime elements in integral domains? Do they not make sense in more general rings?

[rotund aurora]

This is not true. Usually one considers prime ideals, which are defined for any commutative ring with unit (you can probably define them for other more general rings too, but prime ideals for commutative rings with unit are used a lot)

[glad osprey]

I guess commutativity is important, because otherwise you would have to differentiate between left and right ideals

but D&F specifies that R must be an integral domain before defining what irreducible and prime elements are:

[finite turtle]

is there a symbol for <= but with nothing in between

[rocky cloak]

Once you have zero-divisors it becomes a little weird.

Like say xy = 0, then x = x(1+y). So either zero-divisors are basically never irreducible or you would need to modify the definition. In either case the concept becomes less useful.

Prime I don't really see a problem with, but prime elements are less important than prime ideals anyhow.

[coral spindle]

Wikipedia suggests < with a dot in it for the covering relation. I don't know if this is standard, I haven't seen it before. I don't know what the appropriate tex would be.

[rocky cloak]

It's $\lessdot$

[cloud walrus]

jagr2808

[glad osprey]

I see, thanks

[tardy hedge]

Ive always kind of had a hard time interpreting what the difference is when you consider R as a ring and R as an R-module

Because i know R-mod homs and ring homs are not always the same here

I sort of have this feeling of ok in the module case the R multiplication is more some ambient thing versus more tightly connected to the structure in the ring case… or something like that

But yea as u can see i dont really have a good feel for what the difference really is

[tardy hedge]

Question 10) is like an application of lattice theorem for submodules right

[rocky cloak]

Yes, you will have to use that

[rocky cloak]

I guess it makes sense to just look at what homomorphisms will look like.

Like a ring homomorphism should satisfy f(xy) = f(x)f(y), whereas an R-module homomorphisms should satisfy f(xy) = xf(y).

And that's the difference really.

The ring structure of R is about how elements multiply, the module structure is about how we can multiply by elements of R from the left.

Maybe also worth thinking about why kernels of ring homomorphisms are two-sided ideals, while module homomorphisms only left ideals:

If f(x) = 0 then also f(x)f(y) = 0, but you don't necessarily expect xf(y) to be 0.

[glad osprey]

The ring structure is slightly stronger than the module structure in a sense, isn't it? Like if you have a ring hom from R to S, then you can form S as an R-module, and you get an R-module hom from R to S, but I don't think you can recover a ring hom if you have an R-module homomorphism from R to S

or more generally, a ring R gives rise to an R-module, but I'm not sure you can create a ring from an R-module?

[rocky cloak]

Well, the definition of an R-module already contains a ring (namely R) so I'm not sure it really makes sense to relate them in that way.

But yes a ring homomorphisms R -> S gives S a canonical R-module structure. But if for example S=R, then that R-module structure is probably different from the usual one

So I guess what I'm saying is that they are fundamentally different things, and you don't really need to say one is stronger than the other

[thorn jay]

I guess it kinda of seems similar as both deal with addition and multiplication which distribute

The thing is that modules have no inherent multiplicative structure

[tardy hedge]

Yeah

[thorn jay]

It is instead the ring structure of it's endomorphisms which are studied

[glad osprey]

True, it doesn't make sense to compare them directly like that. But I think you can say that you lose (or atleast change) some structure when you turn R into an R-module over itself; for example, there is only one ring hom from Z to itself, but there are infinitely many Z-module homs from Z to itself. And between modules there is always a trivial homomorphism, which isn't the case for rings

[tardy hedge]

I know there are connections between rings and modules over the ring, but i havent studied or seen too many of them yet. Ive heard of an R-module M equivalent to a ring homomorphism from R to End(M) i think it was?

[thorn jay]

Well you're also fundamentally changing your context aren't you?

[thorn jay]

Yes, thats called a ring representation

I played around with them a bit, really fun

[tardy hedge]

Yeah its cool

[glad osprey]

yeah, which is why it doesn't make 100% sense to compare them like that. But you can compare their categories, and move between them somewhat

[thorn jay]

What i mean, it's different from the relation between modules and their underlying abelian groups

[thorn jay]

True..

I dont know if that would lead to anything useful though

But eh who cares really

[tardy hedge]

@thorn jay you only recently joined this server right ?

[thorn jay]

Yeah, how so?

[tardy hedge]

Just asking. You have become a mainstay haha

I am here a lot too

[thorn jay]

What can i say, im passionate about math

[tardy hedge]

🥳

[rocky cloak]

Maybe you can say that, but I still think it's a bit of a weird relation. Like the structure of an R-module contains the ring R, so it would be kind weird to say that it has less structure.

On the flip side, if you're just given a ring S there isn't really a way to turn it into an R-module. S would need to have some relationship with R first.

[thorn jay]

Slice category then

(R V Ring) could have an induced functor to R-Mod?

That's interesting because (Ring V End_Z(A)) is the category of ring representations in an abelian group A

Aka, modules with A as underlying abelian group

[glad osprey]

Yeah, I'm mostly just relating what I read in Aluffi, I know you can't compare different algebraic structures like that. But in the context of the original question, I think it's fair to say that there are more morphisms out of R as an R-module vs out of R as a ring

[thorn jay]

Oh that's cool im reading Aluffi rn

[rocky cloak]

Maybe, but for example if R = Z[x1, x2, x3, ...] is the polynomial ring in infinitely many variables. Then the are only countable many module homomorphisms R -> R, but there are uncountably many ring homomorphisms.

[thorn jay]

Permuting the variables?

[rocky cloak]

For example

[thorn jay]

Right

[glad osprey]

interesting, I need to think about this example

[glad osprey]

Nice, chapter 0? I'm still on Notes from the Underground, to avoid overdosing on category theory 😅

[thorn jay]

Yeah, im a halfway through the group theory chapter

If im motivated ill finish it this month

But school stuffs 😔

[thorn jay]

Countably many module endomorphisms, but uncountably many module homomorphisms

[rocky cloak]

But yeah, in general modules are nicer and less rigid structures. But things are not so black and white

[thorn jay]

Rings feel like such weird algebraic structures to me

I dont know why

Maybe its just coming from group theory

[ripe sage]

Can someone help me with this problem?
Let W = all nxn matrices with trace=1
Is W a subspace of V?

[lusty marlin]

What is V?

[ripe sage]

Set of all existing matrices

I think I got it. It is not a subspace

[glad osprey]

I think I see my mistake now: if you have a ring homomorphism phi : R -> S, then you can realize S as an R-module M, and you get an R-module hom from R to M. If you have a different ring hom from R to S, then you don't necessarily have two module homomorphisms from R to M, it just means you can realize S as an R-module in two different ways.

So in your example there are uncountably many ways to realize R as an R-module, but not necessarily uncountably many module homs on the canonical realization of R as an R-module, is that correct? (And permuting the variables is ring hom, but not a module hom, since for example if phi is swapping x and y, then phi(x^2) = y^2, but x\phi(x) = xy)

[rocky cloak]

Yeah, everything you said here is correct.

Not that you really said anything wrong before

[glad osprey]

I might not have said anything wrong, but I think I was thinking something wrong

[elfin wraith]

Im trying to show that the ore extension given by a derivation ( R[x;\delta] = R<x>/(xr-rx-\delta(r)) ) is freely generated by {1,x,x^2...}

My approach for this was to show that we have an isomorphism \bigoplus Rx^i to R[x;\delta], and im pretty sure ive shown that theres a surjection here, just by the fact that the words xr_1xr_2... generate the free ring, so I think we get a surjection just through the canonical map R<x> to R[x;\delta], but I have no idea how to go about showing this map is injective, I mean its presumably by showing the kernel is trivial but I really dont know how to go about showing this, would appriciate any pointers to somewhere to begin

[thorn jay]

Yis.

[golden turtle]

In this, what are a, b, c and lambda elements of?

I’m trying to understand tensor products

[thorn jay]

left of the tensor operation is from V, right of the tensor operation is from W

Lambda is a scalar from the base field

[golden turtle]

Is it necessarily a field or just a ring

[thorn jay]

(V and W implicitly share a base field)

[golden turtle]

Or a commutative ring

[thorn jay]

V and W usually denote vector spaces

[golden turtle]

In class we were doing tensor products with rings, I think in this video hes doing vector spaces but I’d like to understand the more general version

[toxic zephyr]

@golden turtle did you figure out the det zero divisors injectivity problem?

I had an idea for a proof while I was driving

[golden turtle]

I pieced together a solution, would you like to see

[toxic zephyr]

sure

[golden turtle]

Your proof is probably nicer

The hard direction begins like a paragraph or 2 in

[toxic zephyr]

my idea was to do the contrapositive

• suppose A is injective and detA is a zero divisor and mu detA=0 for mu nonzero
• then im(mu adjA) is contained in kerA which is trivial so mu adjA=0. thus mu annihilates the columns of adjA
• claim: adjA has a nonzero column.
• the det of a sub matrix of A being zero implies that the columns are dependent (so maybe this is a proof by induction)
• suppose for contradiction that all the cofactors of a column are 0 (wlog the last column).
• if the first cofactor of a column is zero, then we can take a linear combination of the other columns to eliminate the first n-1 entries. the last entry cannot be a zero divisor or the kernel is nontrivial. thus a scalar multiple of en is in the image. we can repeat this for the whole column to get scalar multiples of all the standard basis vectors
• take all those columns into a matrix B, and then we can multiply B by a diagonal matrix to get it in the form k detA I=A(kadj(A))
  uhhhh that's as far as I got. Idk if that works but I have to do something. I'll keep thinking about it

[golden turtle]

seems very similar

it is by "induction"

[sage lodge]

Is there an easy way to see this? I'm not exactly sure how I can do it given an arbitrary set of generators and relators with only the knowledge that they define a finite group.

By "solvable word problem" it means that there exists an algorithm to tell whether a word in the generators represents the identity element in the group

[wraith cargo]

There exists finitely many groups of a given order up to isomorphism
So you can just brute force all the possibilities till you find the right one ig

[bitter rover]

Yeah, embed it in S_n and then solve it there.

[sage lodge]

hmm, how do we check that a group we have enumerated is isomorphic to the presentation we are given?

from my understanding all we are given to start is a finite presentation, so I am picturing trying to solve it just from something like <x,y | x^2 = 1, y^3 = 1, yxyx = 1>

[bitter rover]

Embed it in S_n, which you could do by enumerating every possible homomorphism (if worse comes to worse). Then once you have an embedding, where each element is represented by disjoint cycles, check whether the image of your word simplifies to the identity.

[sage lodge]

thank you, that makes complete sense. is it cayley's theorem (that every group can be embedded as a subgroup of a symmetric group) which guarantees that we will eventually find an injective homomorphism?

[bitter rover]

Yes

[sage lodge]

appreciate it

[thorn jay]

Cayley's theorem <3

[sage lodge]

I'm just running this through in my mind, how can I tell if a homomorphism is an embedding?

once I've chosen targets in a symmetric group for my generators such that the defining relations are verified, I thought I need to show that there are no words other than the defining relations that are in the kernel

[bitter rover]

You can just check if it's injective, yeah?

Every permutation has a canonical decomposition into disjoint cycles blah blah

This is like the dumbest possible algorithm.

[sage lodge]

This is where I am going:
Let $G = \langle a_1,\ldots,a_m \mid r_1(a_1,\ldots,a_m),\ldots,r_n(a_1,\ldots,a_m) \rangle$ be a finite group. We find an embedding of $G$ into a subgroup of a symmetric group $S_k$. To define such a homomorphism on $G$, it suffices to choose targets in the symmetric group for the generators of $G$ and verify that all the relations $r_1,\ldots,r_n$ are verified in the symmetric group for the chosen targets. For each $k = 1,2,3,\ldots$, enumerate the finitely many $m$-tuples $(\sigma_1,\ldots,\sigma_m) \in (S_k)^m$ and find an $m$-tuple that satisfies $r_j(\sigma_1,\ldots,\sigma_m) = 1$ for all $1 \leq j \leq n$. Since every finite group can be embedded as a subgroup of a finite symmetric group, this process always terminates for some finite value of $k$ and produces an $m$-tuple $(\sigma_1,\ldots,\sigma_m)$ such that the mapping $a_i \mapsto \sigma_i$ is a homomorphism from $G$ to $S_k$.

maybe I am missing an easier way

[cloud walrus]

Tushar

[winged void]

How can i prove something is a principle ideal

[thorn jay]

Find an element generating it

[winged void]

that is already the case

so idk if that is the good way

[wraith cargo]

you have two elements generating it

hint: find the gcd of 13 and i-5

[sage lodge]

or maybe it simply suffices to say that it is solvable in finite symmetric groups (due to the existence of cycle decompositions), hence it is solvable in finite groups, like you said?

and all of my work is just overcomplicating

[sage lodge]

think I got it, thanks

[tardy hedge]

lol this guy

this guy is a W

[golden turtle]

He’s a beast

[tardy hedge]

Fr

[tardy hedge]

Wait, not really even i think

R -> M by r-> rm has a kernel call it I so R/I = M, so then u can just compare submodules of M and R/I directly, if M is irreducible then R/I has no nontrivial submodules so I had to be a maximal ideal ?

Well I guess u first argue that if R/I has no nontrivial submodules then there is no submodule N with I < N < R, and then u can translate that to rings to say I is a maximal ideal

[rocky cloak]

I mean, that is exactly the lattice theorem yes? That submodules of R/I corresponds to ideals of R containing I

[thorn jay]

That is indeed just the lattice theorem rephrased in a very specific way

Lol

[tardy hedge]

if u have a R mod hom from M to AxB, do u have information about homs M to A and M to B?

AxB just elements of form (a,b) not some internal sum or something

Oh, thanks

[thorn jay]

And this isnt even the nontrivial way - going from two maps f : M -> A, g : M -> B to a map fxg : M -> AxB lol

[tardy hedge]

im trying to go from the fxg map to the f and g tho

[thorn jay]

Ye

Use the natural projections AxB comes equipped with

Like,
pi_1 : AxB -> A
pi_2 : AxB -> B

[terse crystal]

a,b in a ring R, 1-ab is invertible i want to find the inverse of 1-ba. I have a very not rigorous thought: (1-ba)^-1=Σ(ba)^k=1+b(Σ(ab)^k)a=1+b((1-ab)^-1)a. So this makes no sense but the result 1+b((1-ab)^-1)a indeed can be shown to be the inverse of (1-ba). I wonder if there is actually a way to formalize this idea so I don’t have to say i guessed the result.

[terse crystal]

Nvm it seems that lim R/(ab)^n
works, like p-adic integer

[terse crystal]

Still not done, if the intersection of all (ab)^n is 0 indeed this can be rigorous, but I don’t know whether this is always the case…

[cobalt heath]

Wow, this is a great way to intuit about it!

[rocky cloak]

Maybe you can think about the noncommutative polynomial ring S = Z<x, y>. Now the inverse of 1-xy lives in the noncommutative power series ring, so the localization S' = S[(1-xy)^-1] is a subring of Z<<x, y>>.

You have a map S -> R sending x to y, and by universal property it extends to the localization by mapping the inverse to the inverse.

Now inside the power series ring you can reason as before, and notice that the inverse of 1-yx is in S', so maps to an inverse of 1-ba

[terse crystal]

Thank you. I didn’t know non-commutative ring could have localization. But the inverse, 1+y(1-xy)^-1x, in S’, I tried to write it as 1/1+(y/1)(1/(1-xy))(x/1)=(1-xy)/(1-xy)+yx/(1-xy)=(1-xy+yx)/(1-xy), I multiplied it by (1-yx)/1 on the right but it didn’t become 1, what went wrong?

[rocky cloak]

You seem to have assumed that the inverse of 1-xy commutes with x and y, if I'm following what you've written

[terse crystal]

I thought (a/1)(1/s)=(1/s)(a/1)=a/s, I mean inverse of 1-xy in Z<x,y>_(1-xy) is 1/(1-xy) right

[rocky cloak]

Commuting with 1/s is equivalent to commuting with s.

So a * s^-1 = s^-1 * a is way too strong

[terse crystal]

Oh, when talking about localization I took the commutative case as granted….

[terse crystal]

I think I got this now, thanks a lot. I have another question, is there also a way to justify the case when A , B are matrices of size m times n and n times m respectively? I forgot the set of matrices over a field don’t form a ring…

[rocky cloak]

Anyway, I guess this argument is slightly unhelpful. As you would have to prove that the localization is that subring of the power series ring. In which case you may have to prove that the localization satisfies such relations you're trying to prove in the first place...

[rocky cloak]

In that case it's not true right.

Like if A = [1/2, 0], B = [2; 2], then 1 - BA is invertible, but 1 - AB is 0

[terse crystal]

But 1-BA=(0,0;-1,1) also not invertible

[rocky cloak]

Ah, did I mess up the calculation

[terse crystal]

We can exam that this works for matrices, just I don’t know how to justify this reasoning in matrix case…

[terse crystal]

Oh non-commutative localization elements are no longer necessarily of the form r/s but sum of elements of the form r1 s1^-1 r2 s2^-1… right?

[rocky cloak]

Yes, they're horrible

[rocky cloak]

Right, so verifying the solution still works fine

[terse crystal]

Then S’ is contained in Z<<x,y>> should be trivial? Since we map f1 (1-xy)^-k1 f2 (1-xy)^-k2… to f1 (1+xy+xyxy+…)^k1 f2 (1+xy+xyxy+…)^k2…

[rocky cloak]

There's a canonical map S' -> Z<<x, y>>, but I guess you have to show it's injective

[terse crystal]

How is elements being equal defined in localization of non-commutative ring?

[rocky cloak]

So you just define the localization by universal property

But for Z<x, y> it's not hard to see it will be Z<x, y, z>/(z*f = f*z = 1)

(for f the thing you're inverting)

[terse crystal]

Oh I see. And can we have a direct description in general? Like commutative case r/s=r’/s’ when s”(rs’-r’s)=0 for some s” in S. Can we have a similar element wise description for non-commutative case too?

[rocky cloak]

One usually consideres multiplicative sets that satisfies the Ore condition. In that case there is a description similar to the commutative case

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

[terse crystal]

Thank you. I’d never thought about non-commutative case

[elfin wraith]

I think im missing something here but im not quite sure what, so id appriciate anyone pointing out where im going wrong, Im trying to show that assuming $k$ is a field of characteristic 0, then $A_n(k) \cong \Delta(k[x_1,\ldots,x_n]).\$

We get a surjection via the universal property by the map $\phi : k\langle x_1,\ldots x_n,y_1,\ldots y_n\rangle \to \Delta(k[x_1,\ldots, x_n])$ sending $x_i \mapsto \hat{x_i}$ and $y_i \mapsto \frac{\partial}{\partial x_i}$ so we just need an injection.$\$

My first thought was to use the first isomorphism theorem though, since the differentials and $\hat{x_i}$ commute, these are in the kernel of phi, and then a quick compution shows that $x_iy_j - y_jx_i$ and $y_ix_i - x_iy_i - 1$ is too, so we certainly have that the ideal in $A_n(k)$ is a subset of $\ker(\phi)$ but I dont think that weve used the fact that $\text{Char}(k) = 0$ anywhere, so Im guessing that it must show up in the other direction showing that $\ker(\phi) \subseteq I$ but im not quite sure how to do this, I think im just being dumb and missing some really obvious approach but im missing it none the less

[cloud walrus]

Nope

[tardy hedge]

Whats the difference between jacobson radical and nilradical again? Nilradical is a subset of jacobson right

I guess im not sure what the jacobson radical is supposed to represent or be

Yea i guess i just havent seen what its used for

[south patrol]

Yes, you're intersecting more things for the nilradical

[tardy hedge]

Thx

[void cosmos]

non commutative rings are cool

[rocky cloak]

Jacobson is x such that 1-yx invertible for all y. Nilradical is x such that yx is nilpotent for all y

[void cosmos]

jacobson is the intersection of all left primitive ideals in R

[rocky cloak]

The Jacobson is also the common annihilator of all simple modules

(meaning the intersection of the annihilators)

[void cosmos]

the jacboson is also the only ideal such that every element is quasiregular

left or right

@rocky cloak will never forget ur proof when u helped me with showing End_R(S) is finite dimensional

where S is a simple R module and R is left artinian

[void cosmos]

u used this to consider wlog instead of R, R/J(R)

[tardy hedge]

@void cosmos is that u in ur pfp

[void cosmos]

and then since this is artinain with 0 jacobnson this is semisimple

then u used some hom properties

it was so beautiful

[tardy hedge]

Im excited for that

[void cosmos]

yeah

[tardy hedge]

Algebra is my favourite

[tardy hedge]

Lmao nice

[void cosmos]

algebra is cool yeah

[tardy hedge]

I feel more comfortable learning algebra versus other math for some reason

Im in topology rn too and i do like it but it feels more daunting to me

[void cosmos]

i think im the opposite

but i can do more problems in algebra than analysis

[tardy hedge]

Interesting

[void cosmos]

yet with analysis i feel like i get quicker intuitiojn

but i can't solve shit

[tardy hedge]

I see

Yea i find it interesting how the flavours / feel of math can be so different

[void cosmos]

yeah im doing hatcher now and it diff can get very different

[tardy hedge]

Analysis has always scared me tbh

Oh sweet im super excited to get there

[void cosmos]

yeah its cool

[celest cairn]

Hi

[void cosmos]

very diff flavor for me ig

[tardy hedge]

Interesting

[void cosmos]

u can feel it once u just like

skim the textbook

and see how even the textbook reads

and pics and stuff

[elfin wraith]

Hatcher has the flavour of pain suffering and waffle

[void cosmos]

its totally diff

[tardy hedge]

Yea i bought Hatcher its just been sitting pretty on my bookshelf lmao

[void cosmos]

yeah hatcher is the like

[tardy hedge]

Yea cool, ill do that

[void cosmos]

hardest textbook ever

like its so funny how i struggled with covering spaces, read rotman and instantly figured out everything

[elfin wraith]

The man manages to say so much without actually saying anything it’s genuinely frustrating, and the decision to just have 80% of the book be walls of text is wild

[tardy hedge]

Lmao bruh

[elfin wraith]

But hey it’s the most approachable AT book I guess?

[void cosmos]

no way bro

like no way not for me

at al

i think for me its rotman

but its def the most "standard"

[void cosmos]

yes literally

[elfin wraith]

Most approachable in terms of lightest prereqs, rotmann takes a categories first approach no?

[void cosmos]

there is a certain like finnese people learn like

when skipping certain details in a proof or an argument

it's like a balance wehre u dont wanna skip a whole ass detail that very few readers can get

nor u wanna ike just write out everything

it's like playing pool

hatcher grabs the stick and shoves it

i think hatcher accompanied with pierrea lbins youtube lectures is great

• exercises

thje only good parrt

anyways this is #groups-rings-fields so i will just post a group theory problem for fun

prove or disprove G/Z(G) is abelian then G is abelian

[void cosmos]

no i dont remember that actually

[elfin wraith]

Here’s a random fun ring theory problem…

[void cosmos]

but it is definitely self ocntained in the sense that i learnt cat theory for the first time from this

its not like may or something

where van kampen is the colimit

??

xd

[elfin wraith]

If anyone has any suggestions as to why it has a K basis of monomials I’d also not be upset to hear some ideas

I don’t think it’s that because we’re not covering the diamond lemma for another couple of weeks I don’t think, could certainly be an approach but yeah…

I think I probably want to show that my maps injective rather than using first iso tbh, so I think I may need to show the monomial basis thing first

Because if I’m using the fact that k is characteristic 0 I’m going to want some sort of sum/product to claim is non zero

[tardy hedge]

My prof assigned question 4

But not q2

Q2 is prob important isnt it

For q4 and in general probably

[hidden wind]

oh god i’ve still only done one problem in AM and we’re over halfway through the semester now

[tardy hedge]

Thanks, yea ill work thru it so that i understand polynomials better

Q2 is or 4?

[glad osprey]

I'm pretty sure it's not true, but I'm struggling to come up with a counter-example Maybe some matrix group works?

[void cosmos]

yeah it isnt true

[bitter rover]

Look at p-groups

https://en.wikipedia.org/wiki/P-group#Non-trivial_center

[glad osprey]

I see, we can take D4, which is non-abelian, and unless I'm mistaken its center is {e, r^2}, so D4/Z(D4) is Z4 or Z2xZ2 (can't be bothered working out which, but they're both abelian)

nvm, D4/Z(D4) must be Z2xZ2, since otherwise D4/Z(D4) would be cyclic and therefore D4 would be abelian

[void cosmos]

yeah

[fossil shore]

Could anyone give a small hint on showing the last part?

[delicate bloom]

what does it mean if an element of K is fixed by an element of the automorphism group?

[fossil shore]

Yeah, I've tried that. The difficulty for me is showing that this would be fixed for all g in G. I was thinking about whether multiplying some g in G onto each of the sigmas would give a new partition of H in G as cosets, but I don't think so. Am I wrong in that?

Actually... maybe it does give a partition...

[wraith cargo]

why wouldn't it give a partition?

[fossil shore]

Indeed it does seem to do so. Since g sigma_i must be represented by some sigma_{k_i} and g sigma_j must be represented by sigma_{k_j} and clearly g sigma_i H has empty intersection with g sigma_j H. So multiplying the sigmas by g essentially just gives a permutation

[fossil shore]

Yeah, it does. Nevermind

[frail summit]

Technical question: when D&F originally defiend F(a), they defined it as F[x]/(p(x)) with p(x) being the minimal polynomial of a over F. But in this exercise, they say K1 != K2, which, by this definition they would be equal. I presume they here work a different definition? Should I understand F(a) as the smallest field containing F and a in the algebraic closure of F, or?

[delicate bloom]

maybe it helps to see a simpler example, you know that x^3-2 is irreducible over Q right? You can adjoin a real root or imaginary root from it and get distinct fields

[frail summit]

Yes, I understand how it intuitively works. But formally how does one define F(a) in this case?

[delicate bloom]

that's just to say, adjoining a root from an irreducible polynomial does not get you all the roots

F[x]/(p(x)) is modding out by the ideal, it's not the same as adjoining a single root, so it's not equal

[delicate bloom]

you'd have to go back to the definition in the book

the ( ) notation means F(a) is taking all rational polynomials in a with coefficients in F

I'm about to head out for a walk, I'm sure someone can jump in if you're still workin at it, good luck

[frail summit]

Ah. I figured it out. I mistakenly said F(a) = F[x]/(p(x)) by definition. But it's simply the smallest subfield (of some extension K containing a) containing F and a

It's just that the fields F(a) and F(b) will be isomorphic if a and b are roots of the same irreducible polynomial which was what confused me cause I therefore thought one was supposed to think of the two fields as the same field F[x]/p(x), but I see that's not so

Thank you 🙂

[delicate bloom]

yeah youre welcome!

[dull ginkgo]

Bro got enlightened

I.e posted the question, realized it, felt like a dumbass

I get to bully Nope to pretend I don’t do that all the time!

[elfin wraith]

Either im being stupid or this problems quite easy and im inclined to think its the former so if anyone could point out where im missing something id appreciate it. Im trying to show that a ring (non-com, unital) is left Noetherian iff all finitely generated left R-modules are Noetherian. $\$

My solution goes: The reverse implication is trivial, every ring is finitely generated as a module over it's self by ${1}$, so if all finitely generated left $R$-modules are Noetherian then $R$ is Noetherian as a left $R$ module.\

Let $R$ be Noetherian. If $M$ is Noetherian it is certainly finitely generated, so assume that $M$ is finitely generated. We know that a left $R$-module $K$ is Noetherian iff both $N$ and $K/N$ with $N\leq K$ are, but viewing $R$ as a left Noetherian module we see that $M$ must be Noetherian.\

Now im confident that im missing something because im not actually using the finitely generated assumption, but the exact place where im being dumb isnt jumping out at me

[cloud walrus]

Nope

[elfin wraith]

bro forgot he had to hold shift when pressing enter to take a new line

I wish i had found enlightenment

[dull ginkgo]

Left ideals are “left submodules” of the ring

[elfin wraith]

Oh yeah thats where im being a dumbass

[dull ginkgo]

Spontaneous enlightenment

[elfin wraith]

M does not need to be a submodule of R viewed as a module, why did I think that

This homework has genuinley liquified my brain

[dull ginkgo]

TOO MANY WORDS OOOGA BOOGA ADD AND NO INFINITE CHAINS

Tbh I feel ya

[elfin wraith]

I didnt even think to use the ACC tbh, I know that sounds silly for a question about showing something is Noetherian but the previous question was to show the N\leq M, M noetherian iff N and M/N are so I kinda just lazer foucsed on trying to use that

Theres probably a nice argument using chain conditions though, since M is finitely generated im guessing you like write each each of the submodules as a sum of the generating elements for M and that seems helpful

oh is it just that thats a submodule of R^n and so then we get a chain in R^n which stabilises because R is noetherian?

Actually im not even sure if its true that the direct sum of Noetherian modules is noetherian, but it feels true

[dim widget]

it is

[south patrol]

Finite direct sum yes

[elfin wraith]

Ok well in that case i guess ive just about solved this? go prove the direct sum thing, work out the details in the R^n chain and hopefully be one question closer to never touching this homework again

I only have the actually difficult questions left after this

Ok the direct sum thing is like one line, thats easy enough

[dull ginkgo]

I think the direct sum/product of finitely many noetherian rings is noetherian if you break down the ideals/submodules and can probably use that I guess

And then just quotient the ever loving fuck out of those products to get all finite-generated submodules

[elfin wraith]

I just let M,N be Noetherian, took M+0 and quotiented that out of M+N which is just N then you use the lemma i mentioned earlier

M+0 \leq M+N etc and that should be enough

[dull ginkgo]

Pretty sure the finite chain conditions are preserved by direct sum and products?

@south patrol is this true?

[elfin wraith]

Ok I think I stopped being dumb

this should work?

[dull ginkgo]

I think so depending on how specific your prof is

[elfin wraith]

Yeah the end is a little hand wavey but honestly for how long and difficult this homework has been I think shell need to be fine with it

[dull ginkgo]

I would’ve gone through if M is finitely generated then there is a surjective morphism R^N to M, and the chain conditions pass through R^N

As a sketch

[chilly radish]

Rm_1 + ... + Rm_n in general is very much not going to be a submodule of R^n

what if M has torsion that doesn't come from R-torsion?

I would prove that noetheriantiy is preserved under quotients and use the fact there's a surjection R^n->M

[elfin wraith]

OK cool, ill go have another crack at it

[chilly radish]

e.g. Z/2 is a finitely generated Z-Module but it most certainly does not embed in Z

[elfin wraith]

I think the fact that i'd still say this is the 3rd easiest question on this homework really drives home quite how much fun ive been having doing this

not that I think this is a particularly tough problem, but im not exactly making an easy job of it

[dull ginkgo]

Yo homie the left/right ascending/descending chain conditions are preserved under finite direct product of rings right

[chilly radish]

The ideals of RxS are of the form IxJ for I<R,J<S

So yea

[dull ginkgo]

Alright sick yeah

All of that uses images and preimages of ideals under surjective morphisms I think

[elfin wraith]

Ok so yeah its just surjection sending r_1,...,r_n \mapsto r_1m_1 + ... + r_nm_n, this is a surjection, quotient out by the kernel of this map which is submodule of R^n and then universal property?

[chilly radish]

yes (first isomorphism theorem)

[elfin wraith]

I realise ive explained that horribly but hopefully still inteligibly

[dull ginkgo]

That’s what I meant by “Quotienting the shit out of the ring products” lmao

[chilly radish]

Eh you don't need that. If I is an ideal of RxS then obviously I\subseteq I_R x I_S = (1,0)I x (0,1)I. Now if r\in I_R,s\in I_S then by defn (r,s')\in I for some s'. Now (1,s)(r,s') = (r,ss')\in I. Similarly, there is r' such that (r',s)\in I, so (r',ss')\in I, so

(r,s) = (r',s) + ((r,ss')-(r',ss')) = (r',s) + (r-r',0)

[elfin wraith]

I genuinely think my brains been ruined today, that should not have been so difficult lmao thank you though

[chilly radish]

You could phrase this in terms of projections probably but it's unecessary

[dull ginkgo]

I tend to like to think of things from a surface intuitive level when I’m not actually solving problems i.e now lmao

I.e as much as I can feasibly get across without having to symbolically write out anything

[elfin wraith]

Ok time to go do the actualy hard questions, we got 2 options for one of the questions based on how much group theory we know and I think its easier to go learn about polycyclic groups real quick than do the other question

[dull ginkgo]

horror

The fuck is a polycyclic group

Not even close to what I was expecting from prefixes alone k

[elfin wraith]

Seems far nicer than,

[dull ginkgo]

I want to look into Galois theory eventually but the noncommutative side of it is way too interesting

[dull ginkgo]

Classifying ideals of monoid rings torture

[dull ginkgo]

This is more annoying to write out than anything

Conceptually understanding is easy but the carpal tunnel beckons

[elfin wraith]

I mean youve got to pick your poison and group rings seem nicer than the unholy fucking hell that is dealing with multi indecies

[dull ginkgo]

The Weyl algebra is defined by its relations anyway lo

[dull ginkgo]

Yeah this is probably way easier since it doesn’t need much group theory anyhoo

[chilly radish]

it's just a group built via extensions of cyclic groups

[dull ginkgo]

Solvability go brrr

I haven’t done much algebra actually since like June

[chilly radish]

there's a nice characterisation: Polycyclic groups are the noetherian solvable groups

[elfin wraith]

Yeah I mean I dont see it using anything that I dont know, when it refers to group theory its the 2nd course in group theory that im currently taking

[dull ginkgo]

I pick up most of my alg knowledge by osmosis and frankensteining facts together

Yo shin hear me out here

Division ring B containing integral domain A, let’s say B as a module is finite rank over A

Then End_A(B) is finitely generated as an A^op module right

Can’t you consider it as an even smaller-rank module over B^op

since the generating set is the maps sending b_i (basis module element) to delta_{i,j}b_k (gives an iso to the matrix ring)

[chilly radish]

is B free over A? Otherwise how do you define these maps

[dull ginkgo]

What do you mean by free

I forget that terminology

Oh wait shit

Yeah it’s free since I am explicitly defining the rank

[elfin wraith]

I really need to learn enough cat theorey to properlly understand what free objects are, my understanding is quite vauge and probably holding me back because everything I seem to touch these days is free

[chilly radish]

How is b*f given the structure of an A-homomorphism?

is A central?

[dull ginkgo]

Oh wait shit

Yeah if A was an integral domain in its center then it’d work out

[dull ginkgo]

Nah wait it is right multiplication by b

[chilly radish]

you need to define stuff more clearly

oh yea sorry you're looking for a right B-module structure

In that case then \delta_{ij} would generate it as a B-Module but how would you get a smaller generating set?

[dull ginkgo]

Inverses of the basis elements

[chilly radish]

wdym?

[dull ginkgo]

let b_i be an indexing of the basis

We have the maps sending b_i to \delta_{i,j}b_k as the basis as a right A module

[chilly radish]

Yes

[dull ginkgo]

well

b_k = (b_k b_i^-1) b_i

Actually got it switched up

b_k = b_i (b_i^-1 b_k)

So the “diagonal” ones generate it over B^op

And you can show linear independence

[chilly radish]

Sure

Sounds like that works

[dull ginkgo]

Alrighty cool

I don’t think it has any applications

I still refer to right R modules as “R^op modules” I need to kick that habit asap

@chilly radish btw is there any big applications of Ore localization or is it just a rarity that “sometimes” happens but in general has scarce application/?

It seems like sided multiplicative sets are quite specific

Unlike the Jacobson radical which is like the beating heart of noncommutative module theory afaik

[chilly radish]

It's pretty much the only good theory of Noncommutative localisation

You can define things more generally but then you aren't able to work with things explicitly almost at all

There's some statements about localisation that generalise like flatness

This kind of thing actually comes up pretty naturally a lot of the time

The construction of the derived category is via ore localisation

[dull ginkgo]

I know some flatness arguments but they seem “rare”

[hollow fjord]

I'm trying to prove that for an automorphism F of (R,+), we have that F(x) = F(1)x

Can it be done using only that F(x+y) = F(x) + F(y)?

[delicate bloom]

what's R, real numbers?

[hollow fjord]

Yes