groups-rings-fields

[rocky cloak]

I mean, you don't need to know what self-adjoint is to know what positive definite is.

[amber burrow]

wikipedia only defines positive definite for hermetian matrices

For context, i was looking at the solution to this problem

the first two parts are quite logical

[elfin wraith]

I think proving the spectral theorem goes beyond a level where you could understand the notion of being positive definite. But like you could also just run a course without it, it’s a perfectly valid choice

There’s a lot of linear algebra out there and only so many weeks in a semester

Just wait until you do anything physics adjacent and they use slightly different words for every property of a matrix lol

[amber burrow]

ive seen a lot of memes/ppl making fun of math used in physics

[elfin wraith]

Meh, a lot of that’s just like first year UG memeing, there’s a lot of serious maths used in physics

Anyway, moral of the story, don’t be upset about not being aware of a definition, it happens, there’s lots of maths and not much time

[rocky cloak]

Weird. I would have just defined it as x^TMx > 0

[amber burrow]

this might be a bit of a silly remark, but i feel like theres just so much in linear algebra

if that makes sense

like theres the sideway <

whatever that is

always looks fancy

and then direct sum, but times

and idk just a bunch of stuff

and it all feels distinct

like analysis didn't really feel like all that much content imo

and group theory (although i dont know that much) doesn't seem that broad either?

[elfin wraith]

I mean theres a few things to say about that I think. Linear algebra is the study of vector spaces, its going to be broad, that is a broad goal. It perhaps feels more like disjoint topics then analysis because in some sense it is, the study of any vector spaces is argueably a more broad goal than understanding functions on a specific set (the real numbers).

More so though, I think algebra is just more unfamilar to you, stuff like the wedge sum (the sideways < you mention), the direct sum etc etc occur all over algebra. These ideas do repeat, and the do generalise. Vector spaces are modules over a field, so you can relax the field requirement to say commutative rings, which is where algebraic geometry lives. You can relax that further to non-commutative rings which is another broad fun area. But also all rings have an abelian group structure, and they all have a monoid structure etc etc

But even at the course level, I do imagine LA will be one of the most content dense courses you take in UG for the simple reason that LA appears everywhere

No matter what you do with maths you will need linear algebra at some point

[harsh citrus]

Can anyone explain the preposition part ie why Gf lies in Sp to me

[elfin wraith]

I think that is probably the key point in all honesty, it likely just is a bigger course than anything else youll take in your UG, there is a lot to say about any area of maths you can think about, but it just starts to get a little specific. Everyone benifits from more LA

[harsh citrus]

All of math is some form of generalisation of linear algebra in a way thats why it is vast
Even the question i asked right now (Galois theory) is an attempt to generalise linear algebra

[velvet hull]

well for galois extensions, any automorphism is uniquely determined by its action on the roots of f

and the automorphism must send a root of f to another root of f

[harsh citrus]

Oh got it

[velvet hull]

so combining these two facts we get that the galois group must embed into Sp

so it suffices to show that G generates the rest of Sp

[harsh citrus]

Thanks @velvet hull

[velvet hull]

it's a really nifty proof

albeit nonconstructive

[harsh citrus]

Can anyone explain proof 4.19

@velvet hull can you see this one

[velvet hull]

yeah what's the issue

[harsh citrus]

First paragraph
Why is it cyclic and what is E×

[velvet hull]

given any ring R, R^x denotes the group of units of R

so for fields, F^x is just F \ {0} because everything nonzero is a unit

[harsh citrus]

So why is E^× cyclic

[velvet hull]

ah, I can prove a stronger statement for you:

Let R be an integral domain.
If G is any finite subgroup of R^x, then G is cyclic

Since G is finite and abelian, the fundamental theorem of finitely generated ab groups tells us that G is isomorphic to a product of a bunch of cyclic groups Z_{i1} x Z_{i2} x ... x Z_{ij}.
Let k = lcm(i1, i2, ... , ij),
Then g^k = 1 for all g in G by Lagrange's theorem.
But then every element of G is a root of the polynomial x^k-1 = 0, so |G| = k.
But then this means that k = |G| = i1*i2*i3* ... * ij, or in other words all the i's are coprime.
Thus their product is isomorphic to a cyclic group.
(unfinished)

[next obsidian]

It’s actually equivalent teehee,

Let G be a finite subgroup of R^x, and F the field of fractions of R. Then G is a finite subgroup of F^x

[harsh citrus]

OK

[velvet hull]

truee

[south patrol]

They're equivalent as they're both true

[next obsidian]

Go fucking die in a death

[south patrol]

Let R be a ring in which x^|G| = 1 has no more than |G| roots

[velvet hull]

the proof I originally had in mind uses a lemma that's annoying to prove, but I think this works

[thorn jay]

I would argue the contrary, depending on how far you're willing to take the definition of linear algebra

[wraith cargo]

People who study division rings should be observed like little critters

[thorn jay]

But then again, depth or width of a branch of math really isn't comparable as every branch of math can be extended indefinitely, especially widely used subjects like group theory or LA

[velvet hull]

but for the sake of posterity, here it is:

Lemma. If G is a finite abelian group such that for all n, there are at most n elements of order n, then G is cyclic.

Let |G| = n.
Then every element of G satisfies the polynomial x^n-1=0 by Lagrange's Theorem.
Then for every 1 <= m <= n, there are at most m elements in G of order m because they must be exactly the solution set to x^m-1=0.
Thus G is cyclic.

the lemma is annoying to prove look it up

it's a pretty useful lemma regardless, as you see here

[rocky cloak]

I mean, it kinda follows from the classification of finite abelian groups

[velvet hull]

oh true, with the FT it becomes really easy

[thorn jay]

And you can just use the classification of fg modules over a PID to get that

[rocky cloak]

Without that I guess it's Möbius inversion time

[thorn jay]

So you're basically done

[thorn jay]

It seems easier to just prove that classification

[rocky cloak]

Agreed

[thorn jay]

By all accounts, for abelian groups it's not too hard

Not sure for PIDs though

[velvet hull]

the PID classification haunts me in my nightmares

[thorn jay]

It's very clean, I like it

[velvet hull]

the result is clean sure

[thorn jay]

But?

[rocky cloak]

Is it not exactly the same?

[thorn jay]

Idk I've never gone through the proof for a general PID

[harsh citrus]

All pure math is an attempt to generalise linear algebra

In some way or other

[thorn jay]

Topology?

[harsh citrus]

Generalisation of metric spaces

[thorn jay]

I wouldn't count that as LA

[rocky cloak]

Metric spaces is linear algebra now?

[harsh citrus]

It is LA . The most obvious metric space is Rⁿ
And much of its properties come from the fact that it is a NLS

[rocky cloak]

Does that mean set theory is linear algebra because R is a set

[thorn jay]

Lol

Closure operators are field theory because they generalize the fundamental theorem of Galois theory

[potent condor]

The setup of 2.20 is just that R is a pid and P = R[X]. How is it true that n = 2 if the chain is maximal and R has infinitely many primes? isnt (2, X) in Z[X] a maximal chain of length 1 despite Z having infinitely many primes?

[elfin wraith]

As far as linear algebra is concerned it’s mostly the inner product structure you care about no? At the very least the fact it’s a normed vector space rather than just the metric properties

[elfin wraith]

Explain the generalisation for analytic number theory and you’ll have me convinced

[potent condor]

either youre misunderstanding the question or im misunderstanding your answer

[velvet hull]

oh, wait I think I may have

hold on

[potent condor]

no

solve it now or i will email ur advisor and get you fired

[velvet hull]

lol I wish i had an advisor

[thorn jay]

Bro calling the manager

[velvet hull]

but aren't you missing (2) / (x)?

[potent condor]

oooh

i was thinking we were only considering subchains that are heads of longer chains

[velvet hull]

since we're working over PIDs the explanation is pretty intuitive.
any chain has at most 2, because the base ring can contribute at most 1 prime, and then the indeterminate itself can also only contribute at most 1 prime

[twin merlin]

do yall think someone can explain conjugation in the symmetric group

[velvet hull]

sure, what about it?

[twin merlin]

okay so like I see this yea

(1452), (134)(25), (2541) are all members of S_5 yea?

how does it have 4 indices then

if it maps 5 symbols

[velvet hull]

well you just leave the last one unchanged

so if you really want to write it out it's (1452)(3)

but we usually omit the 3

[twin merlin]

so (12345) * (1452)(3) = (14352) ?

[velvet hull]

I got (1534)

[twin merlin]

wait hol on

(12345) * (1452)(3) = (1534)(2)

is that how it works

as in I was under the impression that each symmetric group number served as like an index

yk

[velvet hull]

i dont know what you mean but as long as you got it now

[twin merlin]

like (12345)(21345)(31245) = (21345)(31245) = (32145)

each number in the element of the group indexes the previous set of numbers

and thats how they sorta work

[velvet hull]

I still don't understand, just move on

what next

[twin merlin]

okay wait can you explain how you got this

[cloud walrus]

HChan

[twin merlin]

ohhh

i got it

thanks so much

i was thinking of the stupid permutation group 😭

[woeful sage]

chat I have an issue with part b) of this exercise, I don't think it's entirely accurate

am I wrong? or was the condition (n, a) = 1 missing in D&F?

[rose prism]

how is that a contradiction

gxg^{-1} = x^2 has no solution right

the group is abelian

so its equivalent to x=0 which isnt possible

[woeful sage]

You're right

But in a general finite group G, if we had gxg^-1 = x^2 (where |x| = 12), then it would be a problem would it not?

by that i mean g couldn't be in the normalizer then

[rose prism]

why not

[woeful sage]

hm wait nevermind

we're looking at the normalizer of the cyclic subgroup generated by x in G

not the normalizer of G

I was just confusing the two

thanks

[rose prism]

np

[woeful sage]

but wait here we don't know if g commutes with x, since x, g \in G (which may not be abelian)

so gxg^{-1} = x^2 might be possible

in which case g <x> g^{-1} = <x^2> so it wouldn't be a normalizer

I should find a concrete counterexample

[vocal pebble]

You cant choose what a equals

[woeful sage]

that was supposed to be a counter-example

but it's wrong now

[vocal pebble]

The correct way to give a counter would be to give G,x and g

[woeful sage]

That is true

Now I was having a hard time finding that, because there is no counterexample

[woeful sage]

@rose prism @vocal pebble Okay I've proven it

Now I can rest easy, and proceed with the rest of the proof

it was kind of annoying to prove that g<x>g^{-1} = <x> without knowing a priori that (a, n) = 1

I mean it's perfectly valid

I just don't like it

[woeful sage]

If you want a much simpler proof, here is Bezier's proof:

if (a,n)=d, then x^(an/d) = 1. Hence 1 = (gxg^-1)^(n/d) = g^(n/d) x^(n/d) g^-1(n/d). Hence x^(n/d) = 1, meaning d=1

[woeful sage]

Chat

I know there's a relatively simple classification theorem for finite abelian groups

But what about finite non-abelian groups?

is research being done on that ?

[thorn jay]

We have the classification of all simple groups

Which is like, thousands? Of pages long all research combined

Group extensions in the abelian case are already messy enough

Group extensions in the nonabelian case are horrid

[south patrol]

I do wonder well just

What this classification will look like eventually lol

[thorn jay]

I don't think there'll be a nice classification of finite groups

Or at all

[delicate orchid]

It’s absolutely insane to even consider, think about classifying just 2-groups and how impossible that is.

It’s a miracle the simple groups are nice enough to completely write down

[thorn jay]

Classifying only two groups seems doable

Here, C6 and C12

[delicate orchid]

Why I oughtaaaaaaa

[thorn jay]

But yeah, amen

[south patrol]

Oh I mean simple ones lol

Just like afaik it hasn't been written up nicely and people have found errors and stuff which have then been fixed or whatever

Wasn't there meant to be smth but one of the authors died lol

[thorn jay]

That would be such a thing to happen

[woeful sage]

that's the thing with the monster group and E_8 lattice and stuff right?

monsterous moonshine conjecture?

or I suppose that's different

oh I guess the monster group is one of the simple groups

[woeful sage]

2-groups? groups of order 2? that's easy! every group of order 2 is isomorphic to Z_2

where's my nobel prize?

[crystal vale]

not noble, Abel prize

[woeful sage]

that's part of the joke

[vivid kestrel]

is a group where every element is of the same prime order always abelian?

[south patrol]

No, e.g. the Heisenberg p-group (of order p^3) has this property

[vivid kestrel]

I see. Thanks!

[delicate orchid]

exercise: show that if this particular prime is 2 then your group does indeed need to be abelian

not true

[vivid kestrel]

Just did that that's why I was wondering :D

[south patrol]

oh

[delicate orchid]

there's an extenstion of C_p^2 by C_p which is exponent p^2

[south patrol]

good catch thanks

is there a nice bigger family of examples for this

e.g. is there an example of order p^n for all n

[delicate orchid]

good question. I do know of a larger one as a sylow of some nonsense

I'm trying to remember how the classification of p^4 odd p goes

[south patrol]

i think i have a silly question lol

what is < e_1,...,e_p | e_i e_j = e_{j+k} where subscripts taken mod p lol>

inb4 trivial group

[delicate orchid]

looks blackburn-group esque but I'll have to check

[south patrol]

this feels like a natural construction though

[delicate orchid]

there is one for p^4

it's an extenstion of the heisenberg group by C_p surprise surprise

[delicate orchid]

it's abelian whatever it is. e_ie_je_i^(-1)e_j^(-1) = e_ie_j(e_je_i)^(-1) = e_{i+j}e_{j+i}^{-1} = 1

[south patrol]

this is true lol

actually yeah this is silly cause e_p e_p = e_p so e_p = 1

and then yeah just note that in a commutator the sum is 0

[delicate orchid]

and everything is order p lol

[south patrol]

lol

yeah

okay i didn't want it to be abelian rip

but thank

[delicate orchid]

np

[delicate orchid]

I think this is a more fruitful way of trying to prove this

just keep extending

actually wait are we silly. If you have He_p then just take He_p*C_p^n to get a non-abelian exponent p thing of whatever order you want

[south patrol]

ur welk

[south patrol]

oh lol

very true

[thorn jay]

eepy

[south patrol]

Eepy

[tough raven]

Assuming you mean e_{i+j}, isn't this just a multiplication table for the cyclic group of order p?

[south patrol]

Very true lol, the relations imply e_k = e_1^k

[glad osprey]

This is the solution to an exam problem asking for the Galois group of x^3 - 2. Am I wrong, or is this 10 times longer than necessary? Like, we know the Galois group is a subgroup of S_3, and |G| = 6, so G = S_3, right?

[rocky cloak]

Seems like they're avoiding to use that we know the Galois group is a subgroup of S3.

[glad osprey]

I hate stuff like that Just use the theorems we have available, why make things difficult!

surely the fact that the Galois group is a subgroup of S_n is pretty elementary, I don't see why you can't use that

(not ranting at you btw, just ranting in general)

[south patrol]

it is super elementary and i would just do this lol

in my experience model solutions etc are often suboptimal lol

[hard hearth]

just the existence of idempotent not equal to 0,1 is enough to disprove that it is a field, right? what is the all the yapping after that for?one can simply say:

e = \delta - 5 is not 0,1, so e and 1-e are not 0 but their product is 0, which dont happen in fields...

[glad osprey]

Yeah that's enough to conclude it's not a field, but I think the following yapping is just to show more specifically what kind of ring it is

[hard hearth]

ok

so e is like a "projection operator"

so if there is nonzero nonidentity idempotent eS x (1-e)S is not degenerate and we have isomorphism from S -> eSx(1-e)S

but is the fact that its isomorphic to F11 x F11 already not apparent from the fact that (1,\delta) is a basis for the ring S = F11[\delta]?

[woeful sage]

Let $G$ be a group and let $A, B \subseteq G.$ $A \subseteq H \iff B \subseteq H \ \forall H \leq G$ is enough to ensure that $\langle A \rangle = \bigcap_{A \subseteq H \leq G} H = \bigcap_{B \subseteq H \leq G} H = \langle B \rangle$ right?

[cloud walrus]

Neamesis

[chilly ocean]

what is the actual question

[woeful sage]

My reasoning is that

[woeful sage]

I just want to check if my reasoning is correct with this one

[woeful sage]

my reasoning is that if A is a subset of H iff B is a subset of H \leq G, then taking the intersection over all such subgroups should yield the same subgroup

i.e both the sets must generate the same subgroup

[chilly ocean]

looks fine as subgroup containing A has bijective correspondence with subgroup containg B

[woeful sage]

oh right yea you can state it in terms of a bijective correspondence as well

alright thanks

This is useful to restate a subgroup generated by a subset in terms of generating it by some other easier to work with subset

[tough raven]

Yes. You can also put H = B to see A ⊆ B and H = A to see B ⊆ A.

[chilly ocean]

H is subgroup when B may not be

[tough raven]

Oh subsets

I see

[thorn jay]

Then put H = <A> and <B>

😎

[chilly ocean]

it is el classico solution

[near gazelle]

Let $L/K$ be a finite field extension. Show that for any $a \in L$, the minimal polynomial of a over K coincides with the minimal polynomial of the K-vector space homomorphism $\phi_a : L \rightarrow L, x \mapsto ax$.

[cloud walrus]

Ante0417

[near gazelle]

I don't understand this question. In particular, what is the minimal polynomial of a K-vector space homomorphism? I have only seen it defined in terms of algebraic elements.

[velvet hull]

it's the minimal polynomial of the linear map phi.
in simplest terms, it's the radical of the characteristic polynomial det(A-xI), I'm trying to figure out a better way to explain it

[near gazelle]

Ye. I think i get it

Was just reading about it on internet

ty 🙏

[twilit creek]

Hey, I would like to show that f(x) = x^4 + 3 cannot is irreducible over F_5, why I could not use eisenstein criterion, what can I do ?

[velvet hull]

eisenstein's criterion relies on the fact that f(x) being irreducible in Z[x] implies irreducibility in Q[x] (Gauss's lemma)

and that relationship arises from the fact that Q is the field of fractions of Z

or in other words, factoring numbers in Z is "as good as" factoring numbers in Q

that is no longer true between F_p and Z

[twilit creek]

yep, thanks, It was a stupid idea

[velvet hull]

hint - ||fermat's little theorem||, ||x^4 is identically 1 in F_5||

[rocky cloak]

In general for any element x of a K-algebra, its minimal polynomial is the monic polynomial of smallest degree with x as root.

[delicate orchid]

I don’t buy this. Polynomials are not polynomial functions. And the function x -> x^4 is not the identity, it’s 1 on the units by fermat’s little theorem. Even if it was, what would this say about x^4+1? x+1 is certainly irreducible but x^4+1 isn’t.

Instead you could use Rabin’s algorithm or just make deductions about possible quadratic factors by comparing coefficients to arrive at a contradiction

[velvet hull]

first part is a typo, otherwise you're right

[amber burrow]

wanna make sure im understanding what they mean by conjugacy classes right

yeah i dont really get what they mean by conjugacy classes here

would you take a group of these homomorphisms, where your operation is multiplication?

i.e. $\tau \star \sigma(x) = \tau(x)\sigma(x) = \tau(x) \circ \sigma(x)$

[cloud walrus]

donut123

[mighty kiln]

It says conjugacy classes under Sn

[amber burrow]

yeah, so its what i was saying right?

[mighty kiln]

So conjugation here is probably φ → (y → x^-1 φ(y) x)

[amber burrow]

ok i think that makes sense

[chilly ocean]

as 3 and -3 are quadratic non residue , now try to write this as product of two qudratic with undetermined coefficiants

[woeful sage]

if 3 and -3 were quadratic residues it would mean that x^2 \equiv 3 (mod n) has a solution right (similarly for -3)?

[chilly ocean]

yes

[woeful sage]

I'm glad I still remember some basic NT

[chilly ocean]

you have to take n= p prime

[woeful sage]

ah okay

[vocal pebble]

it need not always form a group? since inverses needn't exist

a_n is just the number of orbits when S_n acts on the set of homomorphisms by conjugation im guessing

[amber burrow]

Wait so f and g would be in the same class if there exists some h (homo G to Sn) such that for all x in G, h(x)f(x)h^{-1}(x) = g(x)

Oh actually nah h would just be any permutation

Oh that makes sense

So Sn acts on the set of homomorphisms from G to Sn by conjugation

Oh that’s literally what you just said

[chilly ocean]

whats the meaning of studying all this when we have to go to war

[mighty kiln]

So that you might have seen something beautiful before being pushed into the killzone

[chilly ocean]

at the end that beutiful thing doesnt matter

that will be left with the body

[mighty kiln]

Is "at the end" everything that matters

The sun will eat the earth in some years anyway

[chilly ocean]

in 2026

math is beutiful but then why people says that it is language of universe

[mighty kiln]

People say that because math is the language of science and statistics

But science and statistics are very earthling things

[chilly ocean]

yes

[chilly ocean]

so it is language of science developed on earth

[mighty kiln]

By definition they are an earthly attempt to probe the cosmos

[chilly ocean]

so it is mere attempt , to live in delusion

[mighty kiln]

To call something delusional implies there is something else that is more real

[delicate orchid]

I love war

[thorn jay]

Bro, what the fuck

[chilly ocean]

sure, beyond

[thorn jay]

What's the meaning of doing anything besides making yourself angry at the people you're going to war with then???

[mighty kiln]

We all have existential crises sometimes 🫂

[thorn jay]

Hell, you're gonna die in war anyways, so why not just kill yourself immediately right now

[chilly ocean]

develop science for more intense war , so that we have to build things again

[mighty kiln]

What is realest is what's in front of you

[chilly ocean]

a coward words are those, i would die as a warrior

[thorn jay]

Absolutely indoctrinated

[chilly ocean]

that i am looking for

[mighty kiln]

If it weren't for H*tler we wouldn't have spectral sequences

[thorn jay]

I would frankly like the moderators to step in

[amber burrow]

wait im still confused

[glad osprey]

one? 🙋‍♂️

[thorn jay]

Two!!!

Wait

Holy shit we live in the zero ring

[chilly ocean]

I have solved upto part H already, but couldn't start with part J

[amber burrow]

How can Sn act by conjugation on homomorphisms from G to Sn

Do you like fix some element of G, and conjugate on phi(x)

[chilly ocean]

The group structure, I have proved is D4

[chilly ocean]

thats because galois group has order less than 8 and cant be abelian(if it is then K will be galois which is not galois) so D_4 is write ans for part (H)

[amber burrow]

Let $S_n$ be the symmetric group on $n$ elements (we agree that $S_0=1$). Compute $h(z) := \sum_{n=0}^\infty a_nz^n$, where $a_n$ is the number of conjugacy classes (under $S_n$) of homomorphisms of $\phi\colon G\to S_n$,where $G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$

[cloud walrus]

donut123

[amber burrow]

So I'm still struggling to understand what conjugation means here

but focusing on the homomorphism part, you can always consider the associated group action ($g\cdot x = \varphi(g)$)

[cloud walrus]

donut123

[amber burrow]

If you split up $[n] = {1,\dots,n}$ into orbits, each orbit is quotient of $G$ by stabilizer.

[cloud walrus]

donut123

[amber burrow]

Your stabilizer is a subgroup of $V_4$ (same thing as $G$), so its either the group itself, one of the three order two subgroups, or trivial

[cloud walrus]

donut123

[amber burrow]

Now the logical leap I don't get in the solution is that they just say

The stabilizers are $1, G$, and the three versions of $\mathbb{Z}/2$. So the number of conjugacy classes is the number of solutions to the equation $n_1 + 2(n_2+n_3+n_4) + 4n_5 = n$

[cloud walrus]

donut123

[chilly ocean]

because we have nothing to loose

[south patrol]

Uhh

[stone elbow]

Hi all, I need help with an excercise from Pinter: "Let $c \in \mathbb{C}$ be a root of a cubic $f \in \mathbb{Q}[x]$. Then $\mathbb{Q}(c)$ is the splitting field of $f$."

[cloud walrus]

a-square

[south patrol]

This isn't true

[stone elbow]

I need a hint. I tried looking at the various fields involved and only managed to prove that the splitting field is of degree at most 6. I then tried calculating the roots directly and got a quadratic discriminant that doesn't look like it's a perfect square

[south patrol]

Or is the question to see if it is true lol

[stone elbow]

C4

[south patrol]

Hm pretty sure this is just false. For example the "root field" [usually people say splitting field] of x^3 - 2 is not of this form

For any choice of root

Indeed Q(c)/Q is a degree 3 extension for any root, but the splitting field is a degree 6 extension of Q

The other exercises seem fine for what it's worth

[stone elbow]

I noticed errors in some other exercises in the latter chapters of Pinter, but this is the first time when the result is flat out wrong :\

[south patrol]

Ah OK rip

[stone elbow]

Should I switch to another textbook for Galois theory maybe? I chose Pinter because it has lots of exercises, but I understand the basics of category theory (but not homological algebra) so a more advanced treatment is okay as long as it has exercises for both computations and proofs

[south patrol]

For Galois theory textbooks you won't need any category theory or homological algebra anyway so you should be more than fine. It sort of depends on what you want to do with Galois theory I guess, but I quite liked Rotman's book (quite popular iirc) personally. The stacks project stuff is very clean too but a lil unorthodox

[stone elbow]

I'm doing it as a hobby. My goals are:

1. Plug gaps in my undergraduate education, which was in applied maths,
2. Prepare for taking on classic algebraic geometry (I've already started working through Harris's First Course in parallel).

[south patrol]

Ah very nice

Oh another standard textbook is Stewart's iirc but yeah I would recommend either that or Rotman heh

[stone elbow]

Thanks!

[stone elbow]

Is it this one? https://www.amazon.co.uk/Galois-Theory-Universitext-Joseph-Rotman/dp/0387985417

[south patrol]

Yes exactly

[amber burrow]

Help with this pls

[next obsidian]

Tteppa

[rocky cloak]

So I'm assuming conjugacy classes here just means that two homomorphisms are the same if they differ by an inner automorphism of Sn.

So we're essentially just picking two commuting elements of order 2 (or 1), so they will be a product of disjoint transpositions, and you want have any halfway overlap of transportations between the two elements.

Other than that conjugation means you can relabel the transpositions however, so it's just about the amount of transportations and how they overlap.

So it would be something like
sum[i=0 to m] (i+1)(m-i+1)
where m = floor(n/2)

So m is the max possible disjoint transpositions, i is the number in the first element, then the overlap is between 0 and i, so i+1 possibilities and then between 0 and m-i for the rest.

I guess you can use the formula for triangular and pyramid numbers to make a nice formula

Feels like there should be some generating function methode to do this, but idk what it would be

[amber burrow]

solution

i don't really get anything after when they say partiion [1,n] into orbits, then get the stabilizers

like they seem to suddenly jump to the equation

also in your definition of conjugation, what you do you by they "differ" by an inner automorphism

[rocky cloak]

The orbits is equal to G/H where H is the stabilizer

[amber burrow]

i got that

[rocky cloak]

So you can write [1, n] as a disjoint union of G/H for various choices of H

[amber burrow]

yup

[rocky cloak]

So then you get it?

[amber burrow]

well i dont get the jump from this to conjugation?

because this stuff all applies to a single homomorphism, right?

[rocky cloak]

Not really no

[amber burrow]

we're saying that this homomorphism gives some group action, defined by $g\cdot x = \varphi(g)$

[cloud walrus]

donut123

[rocky cloak]

Yes

And then conjugation is the same as giving an isomorphism of G-sets, so you're just counting G-sets up to isomorphism

[amber burrow]

sry, whats a g-set 🥲

[rocky cloak]

A set on which G acts

[amber burrow]

oh so like in this case, [1,n] is a G-set for G

[rocky cloak]

Yeah, or [1, n] together with a specific action

[amber burrow]

so when we conjugate in this case, we are conjugating every element of this G-set?

so we would be conjugating $\varphi(x_1),\varphi(x_2),\dots,\vaprhi(x_k)$ for all $x_k \in G$ individually

[cloud walrus]

donut123
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rocky cloak]

Say
psi: G -> Sn
and
phi: G -> Sn
are two actions on [1, n].

Then what it would mean for these to be isomorphic G-sets is that there is a bijection
f: [1, n] -> [1, n]
that commutes with the action, i.e.

f(psi(g) x) = phi(g) f(x)

Which means
f o psi(g) o f^-1 = phi(g)

I.e. phi and psi are conjugate

[amber burrow]

ok that makes sense

[amber burrow]

Ok I want to clarify my understanding. Say I have homomorphisms $\psi$ and $\phi$, and I have the $G$-sets $([1,n],\sigma)$ and $([1,n],\tau)$ respectively. If we already suppose $\psi$ and $\phi$ are conjugate, there exists some permutation $\rho$ such that for all $x \in G$, $\rho \circ \psi(x)\circ \rho^{-1} = \rho \circ (a_1a_2\dots a_k)\circ \rho^{-1} = (\rho(a_1)\rho(a_2)\dots\rho(a_k) = \phi(x)$. So if $H_\psi$ is the stabilizer of $x$ for the action $\sigma$, and $H_\phi$ is the stabilizer of $x$ for the action $\tau$, then left multiplication in $G/{H_\psi}$ and $G/{H_\phi}$ are the same, so $H_\psi = H_\phi$. On the other hand, if we suppose $H_\psi = H_\phi$, then the actions act the same, up to the elements of $[1,n]$ we choose to associate with this particular stabilizer. We can map these choices onto each other by conjugation, since conjugation in $S_n$ is kinda just relabelling the numbers in the permutation. Thus, $\psi$ and $\phi$ are conjugate if and only if the associated stabilizers are the same. Any choice of these stabilizers specifies a good action, so it suffices to count choices of these stabilizers. We can pick any subgroup, so that gives the given equation ($n_1$ is for $G$, $n_2,n_3,n_4$ are for $\mathbb{Z}/2$, $n_5$ is for ${e}$)

[cloud walrus]

donut123

[amber burrow]

My last question is why does any choice of stabilizers create a valid action?

Like I get that the action comes from left multiplication on the cosets of $G/H$ for some subgroup $H$

[cloud walrus]

donut123

[amber burrow]

yeah idk i just dont get how you start with some choice of subgroups and get an action

if the action is transitive it makes sense

also is it just me or is the solution lacks enough depth?

[rocky cloak]

The action is just left multiplication on the cosets.

For multiple groups you just take the union of G/H1 and G/H2

Like let's call the elements of G 00, 10, 01, and 11.

Let H be the subgroup generated by 10 and K the subgroup generated by 01.

Then G/H u G/K is a set with 4 elements. 10 swaps the two cosets of K while leaving the cosets of H as they are.

[amber burrow]

ok yeah i see how that works out

ok ty for the help, i think i got it all now

[amber burrow]

also jagr what do you recommend for a group theory book

or just algebra in general

ive been using dummit and foote because i feel like artin goes through things fast

but once im done with the group theory chapters (1-6) i might switch back to artin cuz it covers some extra stuff

but i was thinking about aluffi

but i dont wanna be this guy

[tough raven]

If A is an integrally closed domain and K is its field of fractions and L is a finite-dimensional K-algebra, is it true that b ∈ L is integral over A iff its characteristic polynomial has coefficients in A?

[tough raven]

I haven't really read Aluffi, but I don't think it's a mistake or even "category-pilled" to see a lot of commutative diagrams and exact sequences while doing algebra. That much is just getting used to a good (but not the only) language/notation.

At least for abelian groups/modules. For groups and rings, I think a very generous use of isomorphism theorems would probably suffice.

[woeful sage]

hell yeah D&F gang

Also nice blender donut

I just hope they release a 4th edition of D&F where they cover groups with operators

They are both in their 70s, so I'm sure it's possible 🙏 maybe

[dim widget]

yes if b is integral then its minimal polynomial has all of its roots in the integral closure of A inside L, thus the roots of the characteristic polynomial are in the integral closure of A inside of L. Then the coefficients lie in the same ring, but the coefficients of the char poly are in K so we win. The other direction is obvious

[tough raven]

To be clear, all roots of the characteristic polynomial are roots of the minimal polynomial because a and (multiplication by a) have the same minimal polynomial?

[dim widget]

yeah

[tough raven]

Actually, L might not be normal, so I'm just going to add the assumption that it's a field extension IG.

[tough raven]

Cool. Thanks!

[dim widget]

Oh I didn’t read that L was not a field

[tough raven]

Eh I mean this is just out of my curiosity so it doesn't really matter

I think in general we can just work in the copy of L in M_n(K) via "multiplication by" embedding

and use that char poly divides some power of min poly for matrices (should follow from rational canonical form)

[dim widget]

anyway I suppose as you say as long as L is finite it doesnt matter

the only thing we used about L was that the integral closure char poly etc were well defined

[chilly ocean]

algebraic curves?

[woeful sage]

Is this fine?

[minor fulcrum]

it's valid, maybe a little wordy

[woeful sage]

How can I make it less wordy?

and by wordy do you mean I'm using more words than necessary?

Does this help

[minor fulcrum]

hmm, so I think the key point in the argument here is that any ascending chain of subgroups in a finite group must terminate at G

[woeful sage]

ye ye

it seems that even (Q, +) doesnt have maximal subgroups, when it comes to infinite groups

[minor fulcrum]

so rather than separating the proof into two steps, you can just show that (like the last part of your proof), and then this means that H is contained in a maximal subgroup pretty trivially

like I said, this is a style thing not correctness

[chilly ocean]

yes right

[woeful sage]

But there aren't two parts, everything after "If not, then there is a proper subgroup K_1..." is one cohesive unit

everything before the 2nd paragraph is me constructing the ascending chain of subgroups

[woeful sage]

[chilly ocean]

so we have to use part (i) , this means we have to find all intermediaqte subfields, still i dont see how it will help

[woeful sage]

Holy crap! Apparently every non-trival finitely generated group has a maximal subgroup!

I just saw this in an exercise in D&F

that's really cool so even if the group is infinite, if it's finitely generated it must have a maximal subgroup

[chilly ocean]

did you proove?

using chains

[woeful sage]

yeah that's the exercise

I'm doing it rn

[chilly ocean]

i think it is a proof by contadiction , if no maximal subgroup, group isnt finitely generated

[woeful sage]

ye

[chilly ocean]

nice steps are given here

[woeful sage]

Now I have

How I felt writing "By Zorn's Lemma S has a maximal element, hence G has a maximal subgroup":

https://tenor.com/view/fire-writing-gif-24533171

[chilly ocean]

then imagine how zorn felt in his time

[chilly ocean]

https://tenor.com/view/writing-fire-bro-how-bro-felt-gif-8015564281503585223

[cloud walrus]

Aguacate

[vocal pebble]

They mean the same thing

[echo vine]

:o

how?

wait i wrote the question wrong

[woeful sage]

Lol

[cloud walrus]

Aguacate

[echo vine]

this is what i meant haha

[vocal pebble]

Polynomials of degree 2 dont form a ring tho

[rocky cloak]

Coefficients in Z/2

[vocal pebble]

So it must be the latter

[echo vine]

ah right beause you have the product

[vocal pebble]

Even if i didnt, i would have addition

[rocky cloak]

I guess you might see something like R_2 for the degree 2 homogeneous component of a graded ring, but if so it would be (Z[x])_2

[woeful sage]

In general $R[x]$ denotes the ring of polynomials of any degree with coefficients in $R$

R may be a ring or a field

well I guess every field is a commutative ring

commutative division ring

[cloud walrus]

Neamesis

[echo vine]

ah, I see. I got confused because with the reals, if I'm not wrong, $\mathbb{R}_n[x]$ denotes the set of polynomia of degree less or equal than $n$ with real coefficients right? So its the same notation for very different things

[cloud walrus]

Aguacate

[woeful sage]

I don't know if that's standard notation, but that would form a vector space of dimension n + 1

[echo vine]

Its the notation i used last year in linear algebra maybe its a thing used more in my country?

[woeful sage]

Or maybe the book your course was following

I typically use $P_n(\bR)$ to denote the set (vector space) of polynomials with degree less than or equal to $n$

[cloud walrus]

Neamesis

[woeful sage]

which I think is in Axler's LADR

[woeful sage]

But yea $\bZ_2[x]$ denotes the ring of polynomials with coefficients in $\bZ_2$ and similarly $\bZ_p[x]$ denotes the ring of polynomials with coefficients in $\bZ_p$

[cloud walrus]

Neamesis

[echo vine]

thank you for your help!

If im not wrong, in an integral domain, if $ab=0$ then $a=0$ or $b=0$.
But I also found a theorem that says that the characteristic of an integral domain is either 0 or a prime number.
How can it be different from 0? If it has characteristic $p>0$, because its an ID then $pa=0 \Rightarrow p=0$ or $a=0$!

[cloud walrus]

Aguacate

[vocal pebble]

Z/(p) is an ID with characteristic p, for example

[echo vine]

I see. Then what did I say that was wrong about IDs?

[vocal pebble]

Nothing is wrong

p=0 is true in Z/(p), for example

[echo vine]

oh..

[tardy hedge]

ive always been a little confused when they say "R-module complement" here

the last sentence i guess

C gives a complement to psi(A) in B. that doesnt mean the set-theory complement to psi(A) in B is C right

i guess im just thinking complement in what sense?

[rocky cloak]

It's in the same way the word is used in linear algebra.

W is a complement to U in V if W(+)U = V is an internal direct sum

[jade mason]

The notation na for elements a ∈ A, n ∈ ℤ means repeated multiplication (for positive n, its just a+a+...+a with n summands), so its not really multiplication in the ring (but scalar multiplication in the ℤ-algebra). To view it as multiplication in the ring, you can identify an integer n with the element n * 1 (where 1 is the multiplicative identity of A), then na = (n * 1)a is multiplication in the ring. So in the case where A has char p, even though neither of the factors in the scalar product p * a are zero, in the corresponding ring product (p * 1)a the factor p * 1 is zero (and this is why Asteroid said that p = 0 in Z/(p))

[echo vine]

That makes a lot of sense, thank you for explaining

[amber burrow]

yeah this is why i was thinking about aluffi

[glad osprey]

Be careful not to confuse ring multiplication with repeated addition. The definition of characteristic is the smallest n such that $n\cdot 1 = 0$, where $n\cdot 1 = 1 + 1 + \dots + 1$, while the definition of an integral domain talks about ring multiplication. In Z/3Z for example, let's call the elements ${0', 1', 2'}$, we have $3 \cdot 1' = 1' + 1' + 1' = 0$, but $3$ is not an element of Z/3Z, it's just an integer, so "normal" ring multiplication of $3$ and $1'$ doesn't make sense. I see Jussari already answered you, but I didn't want to delete what I had already written

[cloud walrus]

sheddow

[glad osprey]

Also, you can define the characteristic of R as the kernel of the unique ring homomorphism from Z to R, which is sometimes nicer to work with

[next obsidian]

Characteristic -37

[somber thorn]

why are free modules important

[next obsidian]

They nice

[somber thorn]

convince me

[rocky cloak]

Every module is a quotient of a free module, so taking a free presentation allows you to understand finitely generated modules through manipulating matrices. See for example the classification of fg modules over a PID for an example of this.

Free modules have the nice property that Hom(R^n, M) = M^n, so you can express equations using maps of free modules.

Free/projective resolutions give rise to derived functors, which is a very strong invariant for understand representation theory.

They're also useful for showing existence of things defined by universal property. See construction of tensor products for example

[somber thorn]

by R^n and M^n you mean RxRx.... n times right

[rocky cloak]

Yes

[somber thorn]

right i see i see

the first statement is kinda just like groups ig

group presentations and then the classification of finitely generated abelian groups

this is getting too categorical for me as an ex analyst

it's too hard

universal property more like universally pissing me off

[glad osprey]

If F is a finite field and p(x) in F[x] is irreducible, does p(x) split in F/<p(x)>? i.e. does adjoining one root give you all? I feel like it should be true, since finite field are nice, with cyclic galois groups and everything, but can't see how to prove it

[rocky cloak]

It is true yeah.

One way to see it is if you look at the splitting field of p, then by the Galois correspondence gives that F[x]/(p(x)) corresponds to some subgroup. Since every subgroup of a cyclic group is normal it is a normal extension

Or I guess with much less machinery every finite field is the splitting field of x^q - x, so are Galois extensions

[glad osprey]

Nice, thanks

[lyric turtle]

could someone help with this proof? I know you have to use the correspondence theorm somewhere but I'm a bit unsure how exactly i'd go about applying it. Suppose that H ⊂ G is a normal subgroup. Show that if G/H is abelian then every subgroup
of G containg H is normal.

[rocky cloak]

Do you know what subgroups of G containing H corresponds to as per the correspondence theorem?

[lyric turtle]

i don't think i do

[rocky cloak]

Okay, what do you know about the correspondence theorem then?

[lyric turtle]

so what i was going to do for the proof using the correspondence theorem set up a function phi

and like show it's injective

and then invoke the correspondence theorem

oh wait ig i could just say A/H is a subgroup of G/H since every subgroup of an abelian group is normal A/H is normal then by the correspondence theorem since A/H is normal in G/H A is normal in G

would this work?

[rocky cloak]

Yup

That's it

[lyric turtle]

niceee ty

[rocky cloak]

So then the answer to this question is that it corresponds to subgroups of G/H

(which are all normal)

[lyric turtle]

Do you think this write up is proper? Since H in G is normal and G/H is abelian for some subgropu A of G containing H, by the correspondence theorem A/H is a subgroup of G/H. G/H is abelian so its sugroups must be normal thus X/H is normal so all subgroups of G containing H are normal

[tardy hedge]

this has never made sense to me ever

specifically F|psi(L) I have no idea how to interpret

the original sequence is L -> M -> N

[woeful sage]

Categories are cool though

[woeful sage]

D&F spotted

[tardy hedge]

For Hom functor, how does one interpret the exactness in the middle entry?
0->L->M->N->0 to
0->Hom(D,L)->Hom(D,M)->Hom(D,N)

I am trying to review this stuff to get better intuition for it and im wondering if its worth going through the proof of the exactness in the middle spot again

[rocky cloak]

It's not making any sense to me either if that helps.

There would be nothing natural about such a bijection F <-> (g, f). If the sequence splits you could at least construct an additive one, but even then it requires a choice.

So this seems mostly like nonsense

[rocky cloak]

Exactness in the middle is just kernel equals the image.

So if D -> M -> N is 0, then D->M should factor through L

[rocky cloak]

What is this from?

[tardy hedge]

dummit and foote page 388

[rocky cloak]

Errata

[south patrol]

Lol that screenshot seems slightly oddly phrased

[tardy hedge]

Where in the proof is the fact that P is a direct summand of F(S) important?

I know its kind of the main point but

[south patrol]

Idk what you mean like you have used the projection explicitly

A lot

[rocky cloak]

The definition of the map P->M is to go though the splitting of pi

[tardy hedge]

Yeah sorry im trying to think about what my question is better

[south patrol]

Also unsure what the statement they are proving is

[tardy hedge]

If P is a direct summand of a free module then it has that lifting property

So P projective

[south patrol]

Sure

Yeah I mean ig idk what ur question should be but the other direction should be insightful

[tardy hedge]

Probably a silly question but what if P just injects into a free module but its not a splitting?

[rocky cloak]

So the structure of the proof is that you have a map
P -> N, then composing this with the projection
F(S) -> P you get a map F(S) -> N for which you can apply the lifting property.

If P is just a submodule of F(S) how would you get a map F(S) -> N

[tardy hedge]

I meant to ask if you have F(S) -> P surjection and P injects into F(S)

But no splitting

[rocky cloak]

Well, try to step through the proof yourself and see where it fails.

For simplicity, assume we're working over the ring R=Z/4 and P=Z/2, F(S)=R and M -> N is the surjection
Z/4 -> Z/2

But morally why it fails ||if the surjection and injection are unrelated, then you essentially just have a random map F(S) -> N, and the lifting property won't help you||

[tardy hedge]

Ty

[hard hearth]

instead of the bit enclosed in red brackets, can i write the following?:

suppose f(a1,a2,...,an) = 0 (eqn: 1). now divide f(x1,x2,...,xn) by x1 - a1, treating the rest of the variables as constants. so after division we get f(x1,x2,...,xn) = (x1-a1)q1(x1,x2,...,xn) + r(x2,x3,...,xn). by induction we can write f(x1,x2,...,xn) = (x1-a1)q1(x1,..,xn) + (x2-a2)q2(x2,..,xn) + ... + (xn-an)qn(xn) + c where c is a constant in C. we know c = 0 from eqn:1 , so f \in ideal generated by (xi - ai)s. The other inclusion is easy. Hence M_a is generated by the (xi - ai)s

Is this argument fine?

it seems fine to me but wanna make sure. I feel like the books proof is a little handwavey

[hard hearth]

ok thanks for verifying

[stone elbow]

I have a question about Appendix A in Eisenbud

[rocky cloak]

Are you asking how one would prove this for extensions with infinite transcendence degree?

[stone elbow]

I think the proof in the book is incomplete even for finite degree

Since we can rescale by elements of k, the ground set shouldn’t even be finite if k is infinite, right?

[rocky cloak]

What is "the ground set"?

[stone elbow]

But without L being finite, we need to assume that each B is finite for the proof of the Lemma to work

[stone elbow]

Ground set of the matroid

[rocky cloak]

Well, you just need one of them to be finite

You just compare your basis to one of smallest cardinality. And if that cardinality is finite, then the proof goes through

[stone elbow]

Good point

So I just take L = K then?

Confusing AF 😅

[soft tiger]

Is the requirement Q being a subset of the ring R mentioned in the last paragraph necessary here? Or can R just be any a division ring?

[dire turret]

im unsure if this is the right channel to ask, but i will ask here because idk where else

is there a name for lattices with internal "implies" terms? like a /\ b ==> c iff a ==> (b -> c)

nevermind! i have found that what i was looking for was probably heyting algebras

[chilly ocean]

Can anyone verify this, here D_4 is galois group of polynomial x^4-44x^2+464=0

[rocky cloak]

You need to be able to divide by n!, so that means you contain Q.

[thorn jay]

Yes, they are to intuitionist logic what Boolean algebras are to propositional logic

[chilly ocean]

https://tenor.com/view/got-dam-gif-26287696

[warm badge]

if O(G) = p^n, p is prime, then G has a subgroup of order p^a for all 0 ≤ a ≤ n.
i proceed by induction. supposing the result holds for all groups H of order p^m < p^n, i want to show it holds for G. G has a non-trivial center, let a be in Z(G). let |a| = p^k where 0 < k < n-1, for k = n-1, n the result holds trivially. since p divides |a|, there is a cyclic subgroup of G of order p, say generated by b, G has an element of order p. ive looked at a hint that says look at G/<b>. <b> is normal in G, so G/<b> make sense and it is of order p^n-1 so it has subgroups of all order but how do i relate this to G?

[chilly ocean]

there are cases, like center is full group

[chilly ocean]

if G is abelian that case must be treated separately, for non abelian you have to take inverse of projection hom of subgroup H of G/Z(G) this inverse image has will have desired prop.

[rocky cloak]

There is a relationship between subgroups of
G/<b> and certain subgroups of G. Usually called the correspondence theorem. Are you aware of this connection?

[warm badge]

no i am not. what is this theorem

[rocky cloak]

So the subgroups of G/N are exactly of the form H/N where H is a subgroup of G containing N

It's a very important and foundational theorem, so I'm a little surprised if you haven't heard of it

[warm badge]

i will have to look it up then

[chilly ocean]

maybe he doesnt know that it has a name but used it many times

[warm badge]

no i don't think i do really lol

[rocky cloak]

Anyway, once you have this you're done. Since then you get subgroups of size
p * p^k for k<= n-1

[warm badge]

yes indeed thank you

[somber thorn]

this is probably trivial but if M and N are R modules and f: M -> N is a surjective module homomorphism, it is true that we can find a restriction f': M' -> N where M' is a submodule of M and f is an isomorphism of modules right?

[uneven sage]

Show that if $G_1,G_2$ are two finite groups with $\gcd(|G_1|,|G_2|)=1,$ then show that $Aut(G_1\times G_2)\cong Aut(G_1)\times Aut(G_2).$

[cloud walrus]

Franklin244

[uneven sage]

Any idea how to prove this?

[chilly ocean]

how the elements aut(G_1*G_2) looks like

[rocky cloak]

This is would mean N is a direct summand, so it's not usually the case.

Consider for example Z/4 -> Z/2

[rocky cloak]

Consider an automorphism of G1xG2, what do you get if you restrict it to G1?

[uneven sage]

No idea...

[chilly ocean]

They are just homomorphism on product of group , so it is tempting to define a map which takes $\sigma\in Aut( G_1\times G_2)$ to $(\sigma|{G_1\times {0}},\sigma{G_2\times {0}})$

[cloud walrus]

Akhi Mishra(Riemman Slayer)

[somber thorn]

oh wow i see ur right

[tough raven]

Does anyone have a reference with a precise definitions and statements of decomposition of torison modules over a Dedekind domain into primary torsion parts?

Something like: for R a Dedekind domain, T a torsion R-module (not necessarily finitely generated), T = (+)_p T_p, where T_p is the "p-torsion" of T.

[crystal vale]

Can I show that if H has order n! /2, H is a subgroup of S_n, then H is A_n.
Without using A_n is simple for n≥ 5?

[tough raven]

H does not contain all transpositions lest it be S_n. But it is normal (index 2). So H contains no transpositions. But S_n/H ≡ C_2, with all transpositions mapped to the non-identity element, so any product of an odd resp. even number transpositions is non-trivial resp. trivial in S_n/H, i.e., is not in H resp. in H. But that is exactly the criterion to not be resp. be in A_n.

[crystal vale]

What is resp. ?

[tough raven]

respectively

ie, odd product is non-trivial hence not in H, even product is trivial hence in H

[crystal vale]

Yes

Now I got it, thank you ❤️

[hard hearth]

am i heavily mistaken, because this seems heavily false. take the point 1/2 + 0i \in C with a ball of radius 1/2 around it. it contains no points of Z[1/2i], right?

[tough raven]

Z[a] ≠ Z (+) Za

in this case

[hard hearth]

oh, wait a minute you are right we are talking about Z[x]/(2x^2 + 1)

[tough raven]

Well, really it's image in ℂ. But yes, it includes (1/2 i)^n for all non-negative integers n, e.g. it contains -1/4.

[hard hearth]

yeahh

thanks for the help

[tough raven]

Rambling about annihilators and primary decompositions

[tough raven]

Let A be a Dedekind domain and S a saturated multiplicative set. Let T be the set of prime factors of elements of S. Then S = {a ∈ A : all the prime factors of a are in T} and S^{-1}A ⊆ ∩_{p ∉ T} A_p. Is the reverse inclusion true?

[next obsidian]

This equality is going to be true iff the primes not hitting S is exactly the complement of T

I’m like 99% sure

For a normal ring, you have A = \cap A_p where p ranges over height 1 primes of A, and I am like 99% sure in full generality that the inclusion is proper if you don’t include even a single A_p

[tough raven]

What's your definition of hit?

[next obsidian]

Intersection is nonempty

Basically I’m saying that Spec S^-1A = Spec A \ T

Is I think equivalent to the claim

[tough raven]

This is true by definition of T.

[next obsidian]

Yeah

[tough raven]

Ah, I see.

[tough raven]

So we caan

[next obsidian]

Why you’re asking is simpler than what I was saying

[tough raven]

can just apply this to S^{-1}A

[next obsidian]

Yours can turn down to the even simpler statement that

A = \cap_p A_p = \cap_m A_m

When p ranges over all primes, m over all maximal ideals

[tough raven]

height 1 should be maximal in general, right?

[next obsidian]

For this case yeah

[weary frost]

Does this prove the third isomorphism theorem?:
Take the composition of morphisms (canonical projections):
G -> G/N -> (G/N)/(H/N)
The kernel of this is the preimage of (H/N) under the projection G -> G/N, but this is just H since N is a subset of H.
Since the map is surjective we have that (G/N)/(H/N) is isomorphic to G/H by the 1st isomorphism theorem.

[next obsidian]

Yeah

But I guess maybe you need to prove that H/N is normal or is a subgroup or whatever

[weary frost]

ah yea

[stone elbow]

Can you check my proof, please?

I’m confused because I had to use ideal quotients, which Lang hasn’t introduced at this point

[next obsidian]

Jesus Christ

I mean you are using localization

You can manually show by just using the definition of prime that if p is prime and missed S then S^-1p is prime

And that it’s an iff

[stone elbow]

Damn

[next obsidian]

Then you can turn your proof into just noting that S^-1p is maximal in S^-1A so it’s also prime in S^-1A

But I’m pretty sure you can also just argue by contradiction

Like if xy in p and x and y aren’t in p, then you can produce an even larger ideal, like clearly x and y can’t be in S or else xy is and p hits S

[stone elbow]

Is my proof at least correct? 😅

[next obsidian]

Then you can conclude that one of x or y isn’t in S, and one of x or y isn’t in p, and then because p\cap S is empty that WLOG x is not in p and not in S. Now look at (p,x) and you can argue this doesn’t intersect S

I am not gonna read all that sorry vro

I saw the word monoid and peaced out

[stone elbow]

Come on, pretty please 🙂

This exercise follows the chapter on localisation, so I assumed that it’s the right tool

[rocky cloak]

The map phi_s isn't surjective so you shouldn't expect the image of an ideal to be an ideal.

[next obsidian]

Sorry this is fake and untrue, you need an assumption of radicalness I think

[rocky cloak]

2nd iso only tells you it's an ideal in the image

[stone elbow]

Okay, I’ll see if it still works if I replace the image with an extension

[drifting mauve]

hi everyone, I'm super lost with this question. I was hoping someone could help give me some direction: Find all left, right and two-sided ideals of the ring of n × n complex matrices.

I mean, obvioulsy we have the trivial ones, but how would I even go about classifying all of them

[tough raven]

Multiplying by elementary matrices helps partially.

(Here elementary matrix means matrix with only one non-zero entry.)

[drifting mauve]

ah, how so? as in like row operation?

[tardy hedge]

Bro said pretty please

Pretty please with a cherry on top? 😜😘

[chilly ocean]

the only proper two sided ideal is 0 ideal

but it has non trivial left and right ideals

[drifting mauve]

I kinda managed to cobble this part together.

its the left and right ideals which I'm having trouble with

[chilly ocean]

set of matrices with all the entries 0,with possible exception at j-th column is simple left ideal

i gues any left ideal will contain these type of ideals

[drifting mauve]

how would the proof of this go though?

[chilly ocean]

it is left ideal is trivial, it is simple can be show using multiplying matrices as suggested above

these are special because these are simple left ideals

[drifting mauve]

right, but how can we know that this is the set of all leftideals?

[chilly ocean]

i did not say that

[drifting mauve]

sorry, im lost

[rapid gyro]

Hy any hint for this problem :¿

[crystal vale]

I have to prove every subgroup of D_n of odd order is cyclic.

So I am thinking about if H is a subgroup of odd order then is it possible that any reflection lie in H ?

Oh no

Because if H contains any reflection then order of H must be even

So only subgroup of odd order H must be a subgroup of < r >, so therefore H is cyclic

[vocal pebble]

Even if you dont have reflections in H the order of H can be even

[crystal vale]

Yes it can

Take {e, r^2} in D4

[chilly ocean]

it is true by some triviality

[vocal pebble]

I dont see how this is directly implied from your previous statement

[sonic coral]

cauchy

[vocal pebble]

What if the subgroup contains elements of the form sr^k?

[vocal pebble]

Reflections arent the only elements of order 2 in D_n(elements like sr, sr^2,..) . You need to take care of the other cases as well. It is trivial, but you need to show it in a proof

[crystal vale]

If the subgroup contains an element of the form sr^k, that means the subgroup has an order even

[vocal pebble]

Yes thats what i am saying you need to mention

[crystal vale]

.

[crystal vale]

Yes I proved this, you can verify my argument

[vocal pebble]

That does not account for subgroups like {e,sr} for example

[crystal vale]

Why?

sr is a reflection

[vocal pebble]

Well i usually call only s a reflection

Then sure that proof works yeah

[crystal vale]

Why?

[vocal pebble]

sr i would usually say is rotation reflection

[chilly ocean]

Rotating the mirror you mean

[warm badge]

im asked to determine all the subgroups of S4 generated by the elements of order 2. now the elements of order 2 in S4 are exactly the transpositions and product of transpositions which has no letters in common. if a product of transposition has a letter in common then that is a 3 cycle which has order 3. so in total we have 9 elements of order 2. now all transpositions generate S4, so that's one. each one of these 9 elements generate their own cyclic subgroups, so in total now we have 10 subgroups. if we try taking product of transpositions and there is a letter in common, we have a 3 cycle which has order 3. do we allow it? because then it is a subgroup of S4 generated by an element of order 3 not of order 2 or am i confusing? like even though it is a 3 cycle it's still coming from elements of order 2??

[crystal vale]

Can you post the question?

Because in general you can generate S_n by all the transpositions

[warm badge]

that is the question, Determine all the subgroups of S4 generated by elements of order 2

[crystal vale]

I think you have to consider cases,

1. Case when number of generators is 1
2. Case when number of generators are > 1

[warm badge]

yes that's what i am trying to do. the first case is done. now about the second case there is my confusion

[crystal vale]

What's the confusion?

Yes it can be generate element of order 3

(12)(13) = (132)

[crystal vale]

Questions says about subgroup generated by order 2, not that H = { x | |x| = 2 }

[warm badge]

ohkk thanks then all the cyclic subgroups of 3 cycles in S4 are also in the list

what else do i need to consider

[crystal vale]

What do you mean by cyclic subgroup of 3 cycles

[warm badge]

generated by 3 cycles

[crystal vale]

You cannot take {1, (123), (132) }

Because it is not generated by elements of order 2

[warm badge]

ok my phrasing is wrong

[crystal vale]

Also can you generate S4?

A4?

So A4 is not possible

[warm badge]

yeh, transpositions generate S4. also <(1,2,3,4)> = <(1,4)(1,3)(1,2)>, so we can include the subgroups generated by 4 cycles as well

[delicate orchid]

Well no you can’t. (1234) isnt order 2

It’s contained in <(14),(13),(12)> sure but isnt generated by an element of order 2

[warm badge]

no no the question asked by elements of order 2, there can be multiple elements

[delicate orchid]

Yes…. Which may contain cyclic groups of order 4 as subgroups…. but those cyclic groups themselves don’t count

As I just said

The problem is (14)(13)(12) is not order 2

[delicate orchid]

Exactly the same as up here

This group is contained in <(12), (13)> = S_3 but is not equal to it

[warm badge]

ohk i see back to square one then. i have only counted the subgroups generated by each of the elements of order 2 in S4

[delicate orchid]

If you could do that then you’d just get every subgroup, because S_4 is generated by those three transpositions

But remember you also have the elements (12)(34), (13)(24), and (14)(23) to use

Perhaps as a little helping hand: || the isomorphism classes of subgroups of S_4 are 1, C_2, C_3, C_4, V_4, S_3, D_8, A_4, and S_4||

[warm badge]

this would seriously help!! thankss for the hint

[tough raven]

Let F/k be a finitely generated field extension with trivial algebraic part (i.e., no f ∈ F \ k is algebraic over k). Must there exist a separating transcendence basis for F over k?

Hmm, is separability of F/k equivalent to dim_F (Ω_{F/k}) = trdeg(F/k)?

[rocky cloak]

Let k = Fp and K = k(x^(1/p^n) : n in N).

[tough raven]

Ah, that's good to keep in mind.

For me finitely generated means field of fractions of a finitely generated algebra though, i.e., a finite extension of rational functions in finitely many variables.

[rocky cloak]

Ah, I missed the finitely generated part

[tough raven]

Hmm, I'm not sure about some of the following (particularly for transcendental extensions), but: F/k separable ⇔ F/k, k(a^1/p)/k linearly disjoint for all a ∈k ⇔x^p-a irr over F ⇔ a^1/p does not exist in F ⇔ algebraic part of F/k is separable

[stiff gull]

determine the R-module structure on the abelian group M: R = Z/10Z and M = Z/3Z. I know there isn't one, but can anyone explain this

[hard hearth]

Let S = / = 0 be any subring that is discrete. we know that S must contain 1 and 0. if S contains any noninteger real, we may divide by 1 and obtain a number in (0,1) that is in S. then all its powers are in S, so in particular we get a sequence converging to 0, so it will not be discrete. Thus the subring must contain only integers. Since it contains 1 it contains all integers, so S=Z. does this make sense?

[rocky cloak]

For any m in M, what should 1*m be?
What about (1+1)*m, and (1+1+1)*m and so on?

What happens if you add 1 to itself 10 times?

[stiff gull]

1 * m = m, (1 + 1 ) * m = m + m and so on. If we add 1 to itself 10 times, we get m added to itself 10 times?

[rocky cloak]

And what is that equal to in this case?

And what is 1 added to itself 10 times in Z/10

[stiff gull]

0

[rocky cloak]

Yeah, and the first question?

What is 10m modulo 3

[stiff gull]

10m mod 3

[rocky cloak]

Yeah, what is that?

[stiff gull]

10m = 3k?

for some k in Z

[tough raven]

If there was a ℤ/10ℤ module structure on ℤ/3ℤ, then 10m = (10 ∈ ℤ/10ℤ) ⋅ (m ∈ ℤ/3ℤ) would have to be 0 in ℤ/3ℤ for all m ∈ ℤ/3ℤ.

[rocky cloak]

So what if m=1 for example, what happens then?

[stiff gull]

well, there isn't such k exists

[south patrol]

Yes, very nice

Have you seen the fun result that any subgroup of R is either discrete or dense? This is fun

[amber burrow]

when do you want to think about the action by left multiplication?

like, i think its pretty useful when dealing with the index of subgroups, right?

[tough raven]

Let L = K(a) = K[x]/(f) be a finite extension of K with inseparable degree p (the characteristic). Is there a classification of field extensions E/K over which L becomes non-reduced?

ie such that f is not square-free in E[x]

[crystal vale]

Yes

[crystal vale]

In a prove of Cayley theorem

[hard hearth]

can we do better than just discrete here?

i feel like you should be able to say if a subgp is discrete then it must be cyclic

yes so if G\leq R is discrete define inf_{g\in G}(|g|) := s > 0. s must generate G. if some element x is not generated by s, then we may divide x by s to get 0< x - ns < s for some n and x-ns \in G which contradicts the minimality of s

all of these have roughly the same argument

so even the subgroups are isomorphic to Z

basically discrete subgps of R is the same as scaled copy of Z and discrete subrings of R is only Z? makes sense because subring => subgroup and 1 \in subgroup => scaling factor = 1

[stone elbow]

Can you check my proof, please?

[south patrol]

Sorry yes

[tough raven]

Never realised this before but there is a description of the base-change of infinite separable (algebraic) extensions:

For a pro-set $A = \varprojlim_i A_i$ (a formal directed inverse limit of sets) and an object $X$ of a category with products and directed limits, define $X^A \defeq \varinjlim_i X^{A^i}$. (In the category of sets, $X^A = \operatorname{Hom}(A, \iota(X))$, where $\iota$ is the natural embedding of sets in pro-sets.)

[cloud walrus]

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

[tough raven]

For $L/K$ an algebraic extension and $E/K$ a field extension, $\operatorname{Hom}_K(L, E)$ is naturall equipped with the structure of a pro-(finite) set (the inverse limit over all finite subextensions).

[cloud walrus]

Raghuram

[tough raven]

If $L/K$ is a separable algebraic extension and $E/K$ any extension splitting every finite subextension of $L$ (e.g., the separable closure of $K$; there is a smallest possible $E$, namely the splitting field of the minimal polynomials of all elements of $L$, which is also the normal closure of $L$), then the formula $L \otimes_K E → E^{\operatorname{Hom}K(L, E)} \colon l \otimes e ↦ (\sigma(l)e){\sigma \in \operatorname{Hom}_K(L, E)}$ defines an isomorphism of $E$-algebras.

[cloud walrus]

Raghuram

[tough raven]

Since both L (⨯) E and E^{Hom(L, E)} are the directed limits of l (⨯) E of E^{Hom(l, E)} over all finite subextensions l of L, the proof is trivial given the usual result for finite separable extensions. But this does identify a definition to understand for the codomain.

[hard hearth]

if a set R has binary operations +,* satisfying all ring axioms except addition commutativity, then by the assoc., distributive law for any a,b \in R, we have: 0 = (b-a)(-1 + 1)=-b + a + b -a => b + a = a + b

whats the point of including addition commutativity in the axioms, then?

is it so we can directly say: R,+,* is a ring iff R,+ is ab. group and R,* is monoid?

[tough raven]

You don't have to if you don't want to.

[tough raven]

(and * distributes over +) yes, pretty much.

[hard hearth]

distributive is most important 😅 (imo)

[tardy hedge]

Is a finitely generated projective module itself free because if it wasnt then it couldnt be a direct summand of a free module?

[elfin wraith]

Proving that the group has to be abelian was the first ever question in my intro ring theory class and it’s a problem which has weirdly stuck with me, random straightforward but fun fact

[thorn jay]

that is what makess a lot of things work yes

[tough raven]

This is just false.

Canonical example being k^n as an M_n(k)-module (more generally, any simple module of a semisimple ring, e.g., irreps of finite groups in characteristic 0).

Another example: any fractional ideal of a Dedekind domain R is a finitely generated projective module (in fact, for any fractional ideal I with inverse J, I (+) J is isomorphic to R (+) R, so is a direct summand of R^2), but is not free unless it is principal (more generally, two fractional ideals are isomorphic iff they are equal in the ideal class group: Cl(R) parametrises fg projective R-modules of rank 1 up to isomorphism).

[stiff gull]

what are some common methods to show a module is free?

[tardy hedge]

Oh my bad it was fg module projective then its a direct summand of a fg free module. Why is that the case then?

@tough raven

[tough raven]

What have you tried?

[tardy hedge]

I willl get back to u

[rocky cloak]

Just explicitly finding a basis works in many cases.

But it depends a lot on what information you're given about the module I think. Often it makes more sense to look at properties of the ring, like are all projective modules free, are all torsion free modules projective.

[stiff gull]

Let p1(x), p2(x) ∈
Z[x] be two nonconstant polynomials in the Unique Factorization Domain Z[x].
Suppose that 1 is a greatest common divisor of p1(x) and p2(x). Show that the
quotient Z[x]/(p1(x), p2(x)) is finite

can someone guide me through this one please.

[velvet hull]

do you have any ideas?

[stiff gull]

first using Bezout?

but then lost there

[tough raven]

It's not a perfectly usable condition, but a necessary and sufficient condition is that for some irreducible (or equivalently any non-trivial) factor g of f over E_sep, the coefficients of g are perfect p^th powers in E.

(This works for any inseparable degree of f.)

[sharp ice]

if u view p1, p2 as polynomials in Q[x], then there is a, b in Q[x] such that a p1 + b p2 = 1. there is some c in Z such that a' = ca and b' = cb in Z[x], and so a' p1 + b' p2 = c. so c in (p1, p2). u can carry on from there

[tough raven]

Let E_sep be the splitting field of the separable part of f (f_sep := g such that f(t^p^k) = g(t) and g is separable); then the fields E ⊇ E_sep making L non-reduced are those containing one of a^(1/p) for a a zero of f_sep. There isn't one such smallest field, but a set of them forming precisely the quotients of A_f := E_sep[x]/g(x^p). So at least over E_sep, the fields making f non-reduced/non-square-free are those admitting (local) homomorphisms from A_f (assuming f is inseparable to begin with).

[round zodiac]

So... what can you say about the quotient Z[x]/(c)?

[stiff gull]

my guess is that in the quotient ring c acts as 0, hence we have Z/cZ[x] / (p1,p2).

[tough raven]

Let F/k be a finitely generated field

[merry summit]

hey, i don't quite understand the ideal of a ring and reading the definition made me confused abt normal subgroups as well which i grasped well earlier

[rotund aurora]

You can define ideals as kernels of ring homomorphisms

likewise, normal subgroups are kernels of group homomorphisms

[mental silo]

If I want to show there is an isomorphism from Gl(2, R) / Sl(2, R) -> R does it suffice to show f: Gl(2, R) -> R by f(A) = det(A) is a surjective homomorphism with kernel Sl(2, R)

[lusty marlin]

Yes
SL(n, ℝ) is the kernel of the determinant map GL(n, ℝ) → ℝ^×

Which is indeed a surjective homomorphism

This also generalizes to any commutative ring with identity

[proven shadow]

Guys I have this problem in my homework:

There are 1001 hats numbered 1 to 1001 and 1000. People stand in a line, and each person is given a hat (so that one hat remains unworn). The people stand so that each person sees only the hats in front of them, not their own or those behind them, and no one sees the unworn hat. The person in the back (who can see everyone) is asked what number he thinks appears on his hat, he answers out loud, the next person in line is asked, and so on - only numbers between 1 and 1001 are allowed, and the same number may not be repeated twice. What strategy can the people coordinate in advance so that at most one person gets their answer wrong?

I'm literally dumbfounded as to how to solve it due to the constraint that the same number cannot be repeated, because I thought of using T as the sum of all hats, S_1 as the sum of the 999 hats person 1 sees, and then guessing S_1 mod 1001, but if it comes out to be neither the hat of person 1 or the hat without a person, it screws over the person that is supposed to guess this number... Anyways, will appreciate any help 🙂

[stone elbow]

Please 🙂

[merry summit]

ted ed has a similar video on this puzzle

[chilly ocean]

why dont you check commutative algebra by balwant singh

its there

[stone elbow]

I have this book, I just want to finish Lang’s chapters on Galois theory before going into CA

[chilly ocean]

it is there

[stone elbow]

What do you mean, what is there?

[chilly ocean]

your proof is there

[stone elbow]

Oh, okay 🙂 Which chapter?

[chilly ocean]

it is localization section

[stone elbow]

Thanks

[elfin wraith]

Is the difference between the internal direct product and external direct product the same idea as the Cauchy vs Dedekind completion of Q? I ask because its never been clear to me why some texts bother defining both just to instantly say theyre the same and just speak of the direct product.

I can see that theyre technically different constructions so at best give you isomorphic objects , so I guess this is the idea

This was prompted by an exercise in my group theory course to prove theyre both the same, and it reminded me of the, in my opinion, strange decision in Rotman's Homological Algebra to introduce new notation for the external product, only to like 10 pages later show theyre both the same, which leads me to believe there must be some reason to distinguish them

[thorn jay]

I guess the internal direct product naturally has the factors as subgroups instead of quotients?

That's about the only difference I can think of lmao

[tribal moss]

Basically it's a question of whether you already have a group that contains both factors (so you know how multiplication works, but it may or may not be a direct product you get), or they are unrelated groups and you need to construct a set with a fresh operation that can count as their product.

[rocky cloak]

I mean, the internal direct product is more if a property than a construction.

For two subgroups you have they're product, and then you call it the interval direct product if it satisfies certain properties.

Whereas the external direct product is an actual construction

[tribal moss]

Once you have constructed an external direct product, then internal direct product of (the images of) your factors in the group you just constructed that will turn out to exist and be the entire groip.

[rocky cloak]

It's certainly very different from the distinction between Dedekind and Cauchy completion.

They are quite different constructions the happen to be the same. One relaying on the order of Q and the other relying on the metric.

[tribal moss]

Conversely, when an internal direct product exists, it will be isomorphic to the external direct product (which always exists), but just looking at isomorphism classes loses information about where in the original ambient group the internal d.p. sits.

[elfin wraith]

Hmm yeah I see that from the construction vs property distinction

I suppose part of the reason I always feel unsure about the emphasising the difference is precisely because they do give isomorphic objects. Thank you both for the good answers though!

[weak prawn]

Hey, do you have any tips on how to show the following:
"Show that there only exists one group homomorphism $g:\mathbb Q \to \mathbb Z$, that is the trivial" (where Q and Z are additive groups)

[cloud walrus]

Michael

[south patrol]

Suppose Q sends 1 to a. What does p/q get sent to?

[weak prawn]

Not quite sure...

Only thing I can think of is that g(p/q)=g(1/q+1/q+...+1/q)=p*g(1/q), but I don't really see the use in that

[charred iris]

you can find g(1/q) by doing basically the same thing just not with p

[weak prawn]

Could you elaborate?

[charred iris]

1 is a multiple of 1/q

[weak prawn]

Ohh, so g(1/q)=g(1)/q. Then g(p/q)=pg(1/q)=p/q*g(1)=pa/q?

[tribal moss]

Perhaps first argue that f(nq)=nf(q) when n is a natural number.

[weak prawn]

And since a is an integer, and we need this to be true for all p/q, we need a=0?

[mental silo]

Is this the first isomorphism theorem

[tribal moss]

Hmm, I would just say f(1)=f(n/n)=nf(1/n) for all n>0, so f(1) is a multiple of everything.

[lusty marlin]

What do you mean by this?

Once we have checked that the determinant map is a homomorphism, it immediately follows that the subgroup of matrices of determinant 1 is the kernel of the map.

[slim kayak]

Yeah, a ends up having to be divisible by every non-zero number -> is 0.

[lusty marlin]

And that is the definition of SL(n, ℝ)

[weak prawn]

Yeah, that's how I got to g(1/q)=g(1)/q. But yeah, the answer is already there

[elfin wraith]

Potentially very incorrect, but does this "important fact" follow from the sequence $\begin{tikzcd} 1 \arrow[r] & K \arrow[r, hook] & \mathbb{Z}^s \arrow[r, "\theta"] & A \arrow[r] & 1\end{tikzcd}$ and the fact that $\mathbb{Z}^s$ is Noetherian? If I am correct, which im not convinced I am, is there an easier way to realise this? It feels a little overkill if this does work

[cloud walrus]

Nope

[rocky cloak]

Yes, an abelian group is Noetherian if and only if every subgroup is finitely generated, and that's exactly what is going on here

[elfin wraith]

I guess if they hid the proof it could be because it needs exact squences, Noetherianity and the 5 lemma lol but that just felt very overkill

thanks though!

Wait lol the kernel is a subgroup so it falls out even more directly

[rocky cloak]

I mean it's really just about being Noetherian, you can prove it with some simple induction.

Like if e_i is the standard basis for Z^s, then K\capZe_1 is generated by just a single element, then you can look at the image of K in Z^s/Ze_1 = Z^s-1. By induction this is finitely generated, so K can be generated by just s elements.

[elfin wraith]

Yeah I realised that if M is Noetherian and 0->M’->M->M”->0 then M’ and M” are Noetherian is insanely overkill, that definitely works

And also just the fact that the kernel is a subgroup and then remembering what Noetherian means lol

[mental silo]

like the fact that a homomorphism from g -> h is an isomorphism from g/k -> h if k is the kernel

[thorn jay]

I was about to say tf are you using exact sequences for

[elfin wraith]

Look man idk it was the first thing that came to mind

[thorn jay]

Brainrot

Happens sometimes, I get it lol

[elfin wraith]

I did at least acknowledge it seemed like insane overkill

[thorn jay]

Self awareness is the first step to recovery

[glad osprey]

Is it true that if a finite group is solvable, then it has a composition series where each factor is cyclic of prime order? I think you can take the derived series, then refine it to a composition series where each factor is abelian and simple, therefore cyclic of prime order, does that work?

[rocky cloak]

Yup

[lusty marlin]

I know what the first isomorphism theorem is, I'm just saying that one need not invoke it to conclude that the kernel of the determinant map is SL(n, ℝ).
The first isomorphism theorem in this case shows that the group ℝ^× is isomorphic to the quotient GL(n, ℝ)/SL(n, ℝ).
Also your statement as currently written is incomplete. The theorem states that G/K ≅ ϕ(G) (image of G under the homomorphism ϕ), so if the homomorphism is surjective, we get the special case where G/K ≅ H.

[mental silo]

Ok thanks

I think it's js a misunderstanding since I replied to u saying sl is the kernel but I meant like "once you have that information do you conclude there's an isomorphism based on the first isomorphism theorem"

Sl(2, R) being the kernel under the determinant map is like trivial

also yea my statement was wrong there I forgot surjective

Anyways ty

Also how do u type the mathbb R without latex do u have a macro or do u js copy paste it every time

[mighty kiln]

Have a pro orz keyboard

[lusty marlin]

I have a LaTeX extension for Gboard on my phone

[keen badge]

What group action can I define $D_n \curvearrowright \mathbb{C}^2$?

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[rocky cloak]

If we're talking linear actions, then all representations of Dn are classified
https://groupprops.subwiki.org/wiki/Linear_representation_theory_of_dihedral_groups

In short, either the trivial action, or the reflection acts as some kind of reflection and the rotation acts as some kind of rotation with order dividing n.

And I guess for n even you have extra one, where the reflection doesn't do anything but rotation rotates 180

[keen badge]

That is what I tried to do, but I was told that this action is wrong:

Because x,y in C and not in R

I dont know why it matters tho

[rocky cloak]

I don't know who told you that. I'm also not sure what it means for an action to be "wrong". Like are they saying it doesn't define an action, or that you haven't found all of them or... I don't understand what that means

[keen badge]

So there is nothing wrong(in your opininion) with this action(onto C^2)?

[rocky cloak]

No, it's one of the irreducible representations

[keen badge]

Great! Thank you!

[rocky cloak]

I guess if you want it to look more "complex" you can diagonalize the r-part instead of the s-part

[woeful sage]

wow nice

[keen badge]

Can we NOT diagonalize both?

[rocky cloak]

diagonal matrices commute with each other

So no

[keen badge]

yeah, cool.
Thanks

[woeful sage]

It seems groups of order 2^n are special?

Where can I read more about these?

[mighty kiln]

Not really, most groups are order 2^n

[glad osprey]

every group is special ❤️

a group of order 2^n is a p-group btw: https://en.wikipedia.org/wiki/P-group

[woeful sage]

what

[woeful sage]

Oh that's true

[glad osprey]

https://math.stackexchange.com/questions/241369/more-than-99-of-groups-of-order-less-than-2000-are-of-order-1024

[woeful sage]

woww

super cool

[rocky cloak]

It feels to me like a lot of the heavy lifting just comes from C2 being so small, so you can pack n composition factors into a group with just 2^n elements.

I wonder if one instead compared groups of length n, would p-groups still dominate?

I guess it's a little tricky to formalize since there are infinitely many groups of length n...

[tribal moss]

I feel it's mostly a matter of 2 being the smallest prime, so pretty much any reasonable enumeration of the finite groups will tend to reach groups with power-of-2 orders earlier than other groups with the same power of another prime.

[woeful sage]

hm yeah it does seem like that's the case

What do you mean by "length n" though?

[rocky cloak]

A group with a composition series of length n

[rocky cloak]

Yeah looking at the list of number of groups of order n, there are
5 of order 8
5 of order 12
5 of order 18
5 of order 20
5 of order 27
4 of order 30
2 of order 45
5 of order 50
3 of order 75
So it seems the number of prime factors seems more important than whether they're all the same prime

[glad osprey]

I want to prove that sqrt(2) + cbrt(3) is irrational. Assume that x = sqrt(2) + cbrt(3) is in Q, then sqrt(2) = x - cbrt(3) so sqrt(2) is in Q(cbrt(3)) and conversely cbrt(3) is in Q(sqrt(2)), which implies that Q(sqrt(2)) = Q(cbrt(3)), but this is impossible since they have different degrees. Does this work?

[tardy hedge]

In R/I, why can’t you have an ideal of R that is not in I so its not 0 in R/I but doesnt contain I?

[amber burrow]

Centralizer of a subgroup is always a subgropu of that subgruop, right?

wait atually no

nvm

[rocky cloak]

R can have plenty of ideals that are neither contained in nor contain I

If you have an ideal J in R and you consider it's image in R/I you get (J+I)/I, works for any ideal J

[tardy hedge]

Ok, I was getting confused with the correspondence theorem. i think i understand what i had wrong now

[hidden wind]

why should one care about group rings? like i’ve seen them mentioned but just a possible construction, but i haven’t seen any reason to comsider them or anywhere they show up naturally

[coral spindle]

Modules of group rings are reps of groups

In general it is a common pattern to want to study some kind of thing, therefore construct some algebra whose modules are that thing.

[hidden wind]

ohh so the reason i haven’t seen any reason to consider them is that i haven’t been given any motivation for rep theory

that checks out

[coral spindle]

The motivation for rep theory is that it's nice.

[hidden wind]

nice

[coral spindle]

nice

[hidden wind]

i have gathered as much as there being geometric things involved and i like geometric things

[coral spindle]

I wouldn't call it very geometric

Or geometric at all, I can't lie

[woeful sage]

Yes

the proof is trivial

[coral spindle]

No this is false

[woeful sage]

oh I missed the wording

"subgropu of that subgruop"

[coral spindle]

The centraliser in G = S_3 x S_3 of H = {1} x S_3 has trivial intersection with H, even

[woeful sage]

I thought it said subgroup of that group

[hidden wind]

geometric in the same way as for example many finite groups can be considered from a geometric perspective, they act on geometric things

[elfin wraith]

Group rings are also just weird and cool

[woeful sage]

ah nice

[tardy hedge]

Mathematicians love saying stuff like this

[woeful sage]

are they wrong though

[tardy hedge]

Nice things are nice 👍

[hidden wind]

it’s true

[elfin wraith]

I mean C[S_3] is weird and confusing, M_2(C) (+) C^2 is nice, so it’s kinda true

Though admittedly my knowledge of rep theory pretty much starts and ends with like Maschkes theorem for group rings lol

[rocky cloak]

I mean representation theory of groups is basically what motivates group theory