groups-rings-fields

[south patrol]

https://en.wikipedia.org/wiki/Euler's_sum_of_powers_conjecture is a good example

[thorn jay]

But that's neither a famous conjecture nor a nasty counterexample 😔

[south patrol]

there's polya's conjecture too

like it was proven there was a counterexample of size ~ 10^361

lmao

[tribal moss]

Later it was shown that 906,150,257 is also a counterexample, though.

[south patrol]

Though that is still pretty huge ig

Just not insanely so

[rotund aurora]

there's also Skewes' number, but it's kind of the same principle

[sonic coral]

the splitting field of x^{q} - x is F_q right? since all roots of this polynomial are just the elements in F_q?

[vast stump]

yes since |F_q^x| = p^n - 1 all elements of F^x satisfy x^(q - 1) - 1 = 0

[sonic coral]

I thought so. I am really trying to show that x^4 + 1 is irreducible modulo every prime p, but the reviewer giving me feedback on it keeps insisting that x^{p^2}- x does not split completely over F_{p^2}. The counter example they gave was p = 3 and x = 2, but they are working modulo 3^2 which I think is the mistake they are making. Like maybe the are thinking every extension of F_p has characteristic p^n and not characteristic p?

[glad osprey]

Splitting fields should really be called splitting extensions - you can't just talk about a splitting field, you need to specify the base field. For example, the splitting field of x^2 - x over GF(4) is GF(4), not GF(2)

Yeah, sounds like the reviewer is just wrong, x^(p^2) - x definitely splits over GF(p^2)

[rocky cloak]

What kind of reviewer are you getting feedback from?

But yeah your argument is correct

[sonic coral]

here was his first comment on the matter lol

I think he’s just working modulo p^2 instead but I just wanted a sanity check for this, so thank you all.

[rocky cloak]

Yeah, it's not equivalent to being zero modulo p^2 as they say.

But who is this person to you is what I'm curious about

Like are they grading your homework?

[barren sierra]

if that's from a journal / conference reviewer saying "that was surely known to Euclid..." is insanely unprofessional

unless you mean like reviewer of homework (i.e. grader)

but even then that's just rude phrasing

I guess this does read like a grader

[rocky cloak]

I'm also curious what was known "a thousand years before Euclid"

[sonic coral]

this was what he was responding to there.

I havnt met them yet. I am starting my PhD in august so I havnt been to the school yet. I am submitting practice problems for qualifying exams to the and they happened to pick this one up and give feedback on it

[rocky cloak]

So they have some sort of forum to help other students?

[sonic coral]

yes

[rocky cloak]

That's cool I guess. Though can lead to these issues

[sonic coral]

I don’t understand how it was known a thousand years earlier but he tells me it’s wrong?

[tribal moss]

That situation does entail a certain risk that they're just smack-talking for the sake of their own ego.

Eventually you'll have to decide for yourself whether you find the feedback helpful or not.

And I don't think we have any historical sources on the state of number theory 1000 years before Euclid, so that reflects rather badly on the feedback giver.

[sonic coral]

not this persons lol. as long as my argument works I guess it’s fine. I was just super frustrated with this bc we’ve went back and forth for two days now

[rocky cloak]

There is some evidence that the Babylonians new about Pythagorean triples around a 1000 years before Euclid.

But I don't think Diophantine equations and modular arithmetic can be claimed.

[glad osprey]

Btw, a related question: what's the splitting field of x^p - x over GF(q) if p is not a power of q? Like, x^5 - x over GF(3) for example?

[tribal moss]

Okay, fair.

[rocky cloak]

Anyway, I think it's just a classic case of Dunning Kruger, them thinking they're helping you

[tardy hedge]

Piazza?

My undergrad school used that all the time

Was great

Not sure why it seemed to only be used at UBC

[sonic coral]

this is my first experience with it

i’ve had good conversations with the other reviewers but this one was claimed by him. i would hate to go out of my way to someone else and it get back to him, or to like show some supervisor, since I havnt even been to the school yet or met anyone lol.

[rocky cloak]

Yeah probably best to just move on

[tribal moss]

Is it an option for you just to thank the guy for his feedback and move on, or do you need his formal approval?

[rocky cloak]

x^n - x would split when there is a primitive (n-1)th root of unity, which happens iff n-1 divides the order - 1.

So it should be the multiplicative order of q mod n-1. At least assuming q is relatively prime to n-1

[sonic coral]

I think I might, because it keeps coming back saying needs fixing. The other two have been fine and said completed. I don’t know if there is a punishment or consequence for just leaving it sitting there. If they say something I can just dive into the weeds of the issue with him I guess.

[glad osprey]

That makes sense, thanks

[rocky cloak]

So for x^5 - x it should be F9.
x^5 - x = x(x^2 + 1)(x^2 - 1)

[rocky cloak]

But these were just practice problems right? This is not like a formal submission of something is it?

[tardy hedge]

Why is he so rude

[tardy hedge]

“We need a full rigorous argument for this claim” … lol.

[tribal moss]

By the way, to say it explicitly: the reviewer is clearly in the weeds when they talk about "mod 9" for F_9; if you can at all, you should ignore what else they have to say.

[tardy hedge]

Damn that wasnt even all of it.. bro said “we need a full rigorous FORMAL argument for this claim”

Who is that guy?

[sonic coral]

He ask why x^{p^2}-x splits over F_{p^2}, I say that it is a basic Galois theory result and that it follows quickly from finiteness of the multiplicative group in this case. He says that it is not a Galois theory and that it was know 1000 years before Euclid, and, and then tells me it’s wrong? I wish I could send bigger screenshots with the complete conversation but I can’t crop out his name doing

[tardy hedge]

Just scribble the name out

[tribal moss]

We believe you. Kiand was just expressing amazement over how full of themself they seem to be.

[tardy hedge]

Yeah lol

[tribal moss]

It may be true that it would have improved your solution to give an explicit argument for x^p^2-x being the product of all the monomials, but the rest of the conversation looks like it's off the rails and best ignored.

[worthy trout]

Hungerford's Introduction to Abstract Algebra vs Fraleigh's A First Course in Abstract Algebra as an exposition self study for AA?

Most professor's at my school usually use one of these two books for the course. If I go with the latter, i can supplement with an instructor's manual I found.

[sonic coral]

okay, thank you all for the advice and reassurance

[next obsidian]

You should tell them to get good

[glad osprey]

Fraleigh is great for self-studying. I haven't read Hungerford, but I think I have heard it's a bit difficult for beginners. It's probably a decent choice too tho

[worthy trout]

Thx I'll probably study from the 7th ed for fraleigh then since I found a solution's manual for it

[hidden wind]

mactutor has a good article on the topic https://mathshistory.st-andrews.ac.uk/HistTopics/Babylonian_Pythagoras/

[vast verge]

Can someone explain how just analysing a subgroup is enough to show no isomorphisms?

[sonic coral]

if they were isomorphic they would have the same elements number of elements of order 2

[wooden osprey]

oooh that's interesting, is that in any way linked to lagrange's theorem?

[dull marsh]

Not really, view that as a consequence of order being preserved by isomorphisms

[wooden osprey]

ah i see, thanks, so having the same number of elements of a certain order is a sufficient condition for an isomorphism to be had?

[dull marsh]

I meant it as a necessary condition, I am pretty sure it’s not sufficient though, let me think of a counterexample

[wooden osprey]

ah yeah in the context of tropical greens' question it makes sense

[rocky cloak]

Heisenberg group mod 3 and (C3)^3

Both have all non-identity elements of order 3

[wooden osprey]

what is a heisenberg group?

[rocky cloak]

This group of matrices
https://en.m.wikipedia.org/wiki/Heisenberg_group

a, b, c being integers mod 3 in this case

[wooden osprey]

ill come back to this later, my shift just started

[velvet hull]

Very interesting, is there any meaningful thing that can be said about two groups if they have the same amount of elements of every order?

[rocky cloak]

I'd say that's already a meaningful statement

[velvet hull]

Other than they share a quotient by a free group

And I guess they’re different up to some set of relations that don’t collapse any elements?

[rocky cloak]

I'm not entirely sure what you mean.

Like you want to look at the map from the free group on all the elements and then compare relations?

[velvet hull]

I’m not even sure if my idea is correct

But if you take a group G, forget its structure, then G is going to be F(G) quotient out by the relations of G

[rocky cloak]

Sure

[velvet hull]

So if G, G’ share the same # of elements for every order, F(G) and F(G’) are going to be the same, and will still remain the same even when you just quotient out by the order of every element

So my idea was that you would just need to investigate the remaining relations you would need to quotient out by, to get back to G and G’

[tribal moss]

Why is the third line crossed out with a green line? It looks correct to me, and is not contradicted by the (also correct) line in green text.

[vast verge]

The solutions did not include the crossed out part, give me a minute to find it

[tribal moss]

Did you write both the black and green parts?
Did you write the black part and the green part is someone else purporting to correct you?
Did someone else write the bkack part and the green additions are your proposal for what it should rather say?

[vast verge]

Page #1 is the question
Page #2 is the solution from the textbook

Pages #3-#5 are my working, green is what I wrote after checking the textbook solution

Black and green are both me

[tribal moss]

So why do you think that line should be crossed out?

[vast verge]

It's not wrong but since the answer didn't include it, I got rid of it

I know that it's pretty cut-throat and simple minded but I mainly want to follow the more elegant solutions that the textbook provides

[tribal moss]

The printed solution is just cutting a corner. "We know the first group has 7 elements of order 2, but now we've already found more than seven order-2 elements in the second group, so there's no need to keep counting -- the final numbers cannot possibly be the same".

Since you've actually found the full count of 27, there's no need to go backwards from there and look at the subgroup instead.

[tribal moss]

I think this is a somewhat wrong mindset. There are usually several valid ways to get through such a problem, and just because the book picked one of them to present doesn't mean that other valid solutions should be avoided.
When the book presents a different solution than the one you came up with, its definitely worth it to try to understand that solution too and see if you can learn some new tricks from it. But it seems harmful to think of it as "oh, it was wrong of me to do it another way", as your use of crossing-out suggests.

[vast verge]

Ohh okay now that makes sense, if a subgroup of the second group has more elements of order 2 than the first group then the two groups obviously won't have the same number of element of order 2 when compared with one another

[tribal moss]

Yeah, an element of a subgroup has the same order in the subgroup as it does in the outer group.

[vast verge]

I guess crossing it out isn't that bad because I've already done those type of methods many times that it's unlikly for me to correcct

[tribal moss]

Eh, your method looks completely correct to me.

[vast verge]

Well besides counting the number of elements in D12 x Z2 incorrectly

... But I digress

[tribal moss]

Oops, yes, 2+2+2=8 looks a bit fishy.

[vast verge]

Okay now that makes sense

That adds up to 48 elements in D12 x Z2

Thank you

[vast verge]

Just finished the chapter on external direct group products

Now I'm beginning normal subgroups and factor groups

[velvet hull]

I mean, there isn't a proof needed here though

it's just computation

this is true, but not relevant for the question

the question statement itself already assumes that this theorem is true

well S_G is isomorphic to S_|G|

that's a corollary of my statement, sure

what I'm trying to say is, if I permute a set of things then it shouldn't matter what I call the things

i.e. if I have a bijection from A to B then S_A is isomorphic to S_B

do you know what the left regular representation is

forget all about H, it's a confusing way to understand Cayley's theorem

when you have a collection of functions, that permute a collection of things

then that collection of functions embeds into the symmetric group of that collection of things

that's it

sure

when H is 1, you are just looking at the original group

it's just the group action of S3 x S3 to S3

anyways, putting that aside, can you tell me what the left regular representation is?

because now I'm starting to think that you don't know

yes

what is the homomorphism?

permutation representation of what?

what's the group action?

so what is it?

you're giving me different names but I don't think you've told me what it is

so it's a group action, okay, how is the group acting on itself?

but what is the arrow?

I agree that there is one

but there are more than one

I'm trying to ask you to describe how the function is defined

not what the function is

the left regular representation of a group, is the representation induced by left multiplication

left multiplication

so there you go, that's the function

[south patrol]

https://tenor.com/view/i-love-curry-vindaloo-korma-madras-tikka-masala-gif-4560598066228923646

[tardy hedge]

me too

[velvet hull]

curry curry curry

so now that you have defined the function, you just compute it for every element in S3

[south patrol]

I sjould say tha tthe joke is that one takes S3 x S3 -> S3 and "curries" it to get a map S3 -> Hom(S3, S3)

[velvet hull]

there's nothing more to it

[tribal moss]

I think you're overthinking this. It's not a "come up with a proof" exercise, it's just a "verify that you've understood the definition" exercise.

[velvet hull]

that's not the image

we're looking for an element of S_6

[thorn jay]

Holy Hom(A x B, C) ≈ Hom(A, Hom(B, C))

[tribal moss]

Careful with just saying "the image" because there are several different functions in play that could take (1 2) as an input. Ah, you're right, one of the S_3 has been relabeled to {1,2,3,4,5,6}.

[velvet hull]

so we have f(x) = g*x, and that gives a map from S_3 to Sym(S_3)

but Sym(S_3) is not S_6

look at the question again

it is

but isomorphic does not mean the same (in terms of set elements)

that's a good question

look at what the original question tells you

no, because there are more than one (and there is only one correct answer)

why did the question tell you to label the elements with the integers 1 through 6

right

so how do we jump from S_s3 to S_6

can you tell me, for example, what (12) does to (1, (12), (23), (13), (123), (132)) in order

no, I want you to list it out fully

this is going to be part of the answer anyways

and simplify all products into basic cycle notation

now is there a way to convert this into a permutation in S6?

there you go, that's what the left regular representation of (12) is

[tribal moss]

Uh, the representation of (12) shouldn't be a 6-cycle -- the order of (12) is 2, so the order of its representation should be 2 too.

[velvet hull]

oh, right, that's not cycle notation lmao

[barren sierra]

I think that's 1 line notation

[velvet hull]

"1 line notation"

[barren sierra]

although even then that's not a 2 cycle

[tribal moss]

Ah, that makes better sense.

[barren sierra]

wait yes it is I'm dumb

Tropo meant 6-cycle as in an order of element 6

so your understanding here is still correct

just notational differences in writing out permutations

[tribal moss]

I misunderstood that notation (215634) -- since that kind of notation would usually indicate cycle notation.

[velvet hull]

your answer is still correct, this is just not the correct notation.
So what we have here is that the representation of (12) is some element x of S_6, such that x ⤵ (123456) = (215634).

what is the cycle notation for x

[thorn jay]

This is why computing examples is important lol

[tribal moss]

Let's have mercy and just say it's (12)(35)(46), since that's not supposed to be the point of the exercise, and there are 5 more elements to compute anyway.

[velvet hull]

so that's how you compute the representation for (12).
this might seem like a lot of work, but I actually claim that you only need to do it twice, and everything else can be done inside S6.
can you see why?

[thorn jay]

Oh I'm guilty of not doing exercises when I self study

Always bites me in the ass because I just forget the stuff i learn after not doing it for a bit

Still best to do the exercises and get a good grasp I think

Just gotta pick out just the right amount of exercises

[velvet hull]

the left regular representation is a homomorphism from S3 to S6

S3 is generated by 2 elements

for the purposes of your question I'm calling it S6

[thorn jay]

I think it's a little poor wording that they used "left regular representation" when referring to an action on the set { 1, ..., 6 } lol

[tribal moss]

$(12)(35)(46)$ is cycle notation for the permutation in $S_6$ that would be notated $$\begin{pmatrix}1&2&3&4&5&6\2&1&5&6&3&4\end{pmatrix}$$ in two-line notation and that permutation is the representation of $(12)\in S_3$ you just computed.

[cloud walrus]

Troposphere

[tribal moss]

Yes -- probably just to allow you to notate your results more compactly. This is not supposed to be particularly deep.

When you wrote $(2 1 5 6 3 4)$ I thought you meant that to be cycle notation, that is, you had computed the representation of $(1 2)$ as
the 6-cycle $$\begin{pmatrix}1&2&3&4&5&6\5&1&4&2&6&3\end{pmatrix}$$

[cloud walrus]

Troposphere

[tribal moss]

Hmm, that can't be it. When you say 2 goes to 4, that would mean that (23)(12) = (13) -- since (12) and (13) are what the numbers 2 and 4 have been defined to represent, but actually (23)(12) = (132).

So your permutation in S6 should send 2 to 6.

[tribal moss]

The first line there has no formal meaning -- it just says "here's another variation of a vibe you've already seen an example of".

[thorn jay]

"itself" meaning the group

Have you never seen the notation for the powerset before?

Did you start with group theory without learning about sets at all?

[tribal moss]

For any set X, the "power set" P(X) means { A | A is a subset of X }.

[thorn jay]

Unrestricted comprehension..

Do you know families of sets? Union/intersection over a family of sets?

That's very required for group theory

[tribal moss]

Yes -- there's a dedicated axiom in ZFC promising that this particular instance of comprehension does produce a set.

[thorn jay]

Not if I have any say in it

[tribal moss]

You don't.

[thorn jay]

I will invent a new math

I'll call it.. new calculus...

It's a collection of sets, where every set is indexed using another set, called an index set

[tribal moss]

The practical usage of the word is not quite consistent about whether a "family" is supposed to be indexed or not. Sometimes we just say "family" as a synonym for set because we're going to treat its elements openly as sets (rather than numbers or pairs or whatever).

[cedar vault]

Can you provide more context? What role is C_a playing here

[spark veldt]

Is a class in module theory worth taking? I’ve taken algebraic structures, andy next year they’re offering Group Theory (like symmetry groups, Sylow subgroups, orbit-stabilizer typa stuff from the syllabus) and also module theory (torsion modules, module structures over Noetherian rings, Schur’s lemma from the syllabus). Honestly idk what I should take, can’t take both of them unfortunately.

[velvet hull]

it is worth it if you are looking towards doing algebraic geometry

but if you haven't done any group theory yet then it is most definitely the more important one

[spark veldt]

Ah I see okay.

I'm not really looking forward to algebraic geometry right now, so ig group theory is the better choice then

Thanks!

[minor gazelle]

A have a question is there some nuances between a subgroup H of group G with operations closed under multiplication and a subgroup H of group G with operators closed under multiplication plus closed under arbitrary power

And it is also said that the subgroup is defined if it satisfies containing inverse, identity elements and closed under multiplication is not necessarily a stable subgroup

In which case a subgroup is not stable subgroup?

[rocky cloak]

So I guess you're talking about groups with operators?

It's true that not every subgroup is stable yes. Being stable is an extra property a subgroup can have

[minor gazelle]

I am a bit confused though I read 2 lectures note from different school with different definitions of what subgroup is

One without the set Omega one with Omega

What even is an action 🤕

[rocky cloak]

So a "group with operators" is as the name implies a group with some extra structure.

And then a stable subgroup is a subgroup that also respects this extra structure

So the definition of "stable subgroup" is the definition of subgroup + an extra condition

[minor gazelle]

Now I got it the words are so abstract 🤕

[rocky cloak]

I would recommend trying to learn about groups and subgroups first

[minor gazelle]

Yes I will definitely do that

[rocky cloak]

Before moving onto something more abstract with extra structure

[minor gazelle]

Shouldn’t have searched those lecture notes, it really causes headaches when there’s a difference 🤕

I ordered a book so probably gonna be easier

Some terminologies are layered with different definitions 😵‍💫😵‍💫

[thorn jay]

Groups with operators were a first step to universal algebra

[rocky cloak]

Like historically?

I think they preceed modules historically too

Like the Krull--Remack--Schmidt--Azumaya theorem was proven for groups with operators by one of the names (I'm thinking Krull, don't remember)

[thorn jay]

I yeah, I'm pretty sure

Ah, nevermind

Universal algebra, apparently already existed in 1898

Man, this shit's old

[vast verge]

For some reason the factor group of cosets I'm getting isn't cyclic

Can someone explain to me what I'm doing wrong?

[rocky cloak]

So I'm not familiar with this notation, but if U8(40) is supposed to be those elements congruent to 1 modulo 8, then
U(40)/U8(40) should be isomorphic to U(8) which is not cyclic.

So then either the problem have which would be cyclic and noncyclic swapped, or you have misunderstood what the notation means...

[vast verge]

Okay thank you

[rocky cloak]

In either case your computations are correct

[vast verge]

Currently it looks like $U_8(40) \sim \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$

[cloud walrus]

Tropical Greens

[vast verge]

So I was getting concerned when all the element of the coset group had orders 1 and 2

[rocky cloak]

You mean
U(40)/U8(40) = Z/2 x Z/2 ?

[vast verge]

Yeah you're right

My bad

[candid patrol]

U_8 ?

[minor gazelle]

🐈are the proofs done correctly though I found it very abstract in the sense that the theory uses language twist to make a proposition to contain more information.

[slate fulcrum]

well for 1.10 really you want to verify that the * in (H, *) is a well-defined operation from H^2 to H. The other axioms you'd need to verify yourself too. It is skipping steps but the actual work is not bad.
For the second paragraph they should really say for any x,y since y^{-1} is in H, by condition (ii) on x, y^{-1} we see that xy=x(y^{-1})^{-1}\in H.
But those are just minor nitpick it is otherwise fine.

[minor gazelle]

I shall be working on to fix it 🥰🥰🥰

[swift tundra]

I haven’t heard of groups with operators before. Where are they used?

[thorn jay]

I've looked online a bit and, it looks like it's just a unification of results? But if Noether extensively studied them I think they're more important than that

[minor gazelle]

I revised my proof yesterday is it better?

🫣🫣

Still not right

Is this better ill refine it till it’s perfect

So any suggestions are welcomed like as nitpick as possible

[rocky cloak]

So supposing H has an identity element eH seems a bit circular / unnecessary.

Would be better to just prove that e is an identity for H.

Similarly you seem to be doing a lot of calculations to establish something about inverses, but
x * x^-1 = x^-1 * x = e
is the definition of inverse, so you don't need to establish more than that.

[minor gazelle]

Thanks I am quite confused by all the abstraction which is kinda playing with my logic like it’s a gremlin🫣🫣😭

[limber tapir]

in the second proposition, you're proving the stronger statement that the axioms given in proposition 1.11 are equivalent to those given in definition 1.16. But the statement of the proposition is only that these "other" axioms imply the original axioms

Imo it would be better style to either change the statement of the proposition appropriately or remove the part of the proof where you prove the other implication

[minor gazelle]

I’ll be fixing it

[minor gazelle]

Should i actually just add some parts then would it be good? 😵‍💫😵‍💫

I’ll rephrase it like that 😵‍💫😵‍💫

[limber tapir]

The statement you proved is something like "H is a subgroup if and only if it satisfies (i) and (ii)"

[minor gazelle]

Omg this is actually hard 🫣🫣

If the revised version of P1.1 correct… I actually feel this like word game so I kinda don’t know what to prove

[limber tapir]

This is basically the first paragraph of your proof, just in other words. You proved "the proposition is equivalent to the definition of a subgroup"

But the original statement of the proposition is "the proposition is a sufficient condition for H being a subgroup", so what you proved is stronger

[minor gazelle]

😵‍💫😵‍💫😵‍💫now I realize it? Is it true? Or I did completely wrong🫣

[limber tapir]

It's true

[minor gazelle]

Then I’ll rephrase it 😭😭

And is this part still wrong? Becasue it’s hard for me to understand the logic of group language

It has too many definitions and I felt like I don’t have a clear target to prove 😭

[limber tapir]

Both proofs look correct

Also I think getting used to the language is just a matter of time, so don't worry too much about it

[minor gazelle]

Another proof, is this like rigorous already? I kinda got the idea of how this is done now in basic sense

[ivory talon]

Hi, I've got I have a normal sub group N of some group say G and another sub group of G say K then and say that G is of order g, N is of order n and K is of order k and I know that g = n*k I also know that the intersection between N and K is an empty set does this gurantee that the semi direct product of N cross K will give me G? If not what else should I consider?

[thorn jay]

Well by the second isomorphism theorem
|NK|/|N| = |K| / |N ∩ K|
|NK|/n = k/1
|NK| = n * k = g
So by size considerations, NK = G, meaning that indeed G is the semidirect product of N and K

[minor gazelle]

I made a slip I shouldn’t have said multiply on the left

Since even on the right I can justify it with associative law

[ivory talon]

So it's correct according to the second isomorphism theorem.
I don't know if it's stupid but just thinking about it it's hard for me to imagine how every element in G will specifically be represented by elements of K and N I mean what if some elements will be repeated?

[limber tapir]

Associativity doesn't allow you to swap elements

[minor gazelle]

You’re right

There still f(x^{-1})

[thorn jay]

That's what the intersection being trivial (not empty!!) guarantees

[limber tapir]

Also there's some implicit steps here where it's not clear if they are intentionally missing

[minor gazelle]

But it shows some interesting stuff those f(x)^{-1}= f(x)f(x^{-1})f(x)^{-1}

[thorn jay]

The intersection of N and K being trivial means that the group NK is a semidirect product of N and K

[minor gazelle]

Please tell me 🥰🥰

[thorn jay]

Then by size considerations it follows that G = NK, meaning that G is a semidirect product of N and K

[minor gazelle]

Isn’t this that matrix stuff?

[thorn jay]

I don't see how matrices appear here

[minor gazelle]

Like the matrix manipulation we have these

[thorn jay]

You're thinking of diagonalization / similar matrices , which is just an instance of conjugation in groups

[minor gazelle]

A^{-1}=B^{-1}A^{-1}B or something I usually find those commutative stuff very weird 🫣

[thorn jay]

AA^{-1}B is just B? It just cancels

That's only true of B commutes with A

[swift tundra]

Yeah totally. I might check out the “Groups with Operators” section of Lectures in Abstract Algebra I. It seems like they extend certain results in group theory to modules, rings, and such.

[minor gazelle]

Yes it only makes sense today for me

I usually found this weird concept

[limber tapir]

At the start, if you multiply f(ee) = f(e)f(e) by the inverse of f(e)^{-1} you get f(ee)f(e)^{-1} = f(e), only after noting that f(ee) = f(e) does it follow that the left side becomes e'

[minor gazelle]

I time it with f(e)^(-1) and since G’ is a group so its identity element

Since both f(e) f(e)^{-1} are in G’

[limber tapir]

Just so there's no misunderstanding: after multiplying the right side of the equation (f(e)f(e)) by f(e)^{-1} the result is f(e)

And after multiplying the left side the result is f(ee)f(e)^{-1} which is of course equal to e' after plugging in ee = e

[minor gazelle]

Yes

I did it correctly did I?

I am kinda scared of hidden paradox 🫣 I have those too often

[limber tapir]

Then my only point was not mentioned in your proof (which is fine) but by concluding that f(e) = e' it seems like you identified the right side with e' and the left side with f(e) instead of the other way around

But it doesn't matter much

[minor gazelle]

I will emphasize on that

Yes this is actually quite important, proof writing is very hard I generally felt

[limber tapir]

What maybe matters slightly more is to add e' = f(e) to the start of the second chain of equalities so that the left hand side becomes f(x)^{-1} after multiplying

[thorn jay]

I've looked and they don't mention applications

Just the basic theory

[minor gazelle]

How’s now the charity🫣

Multiplicative group

[limber tapir]

This is personal preference, but i always try to have the two things I want to equate on the opposing sides of such a chain, for example i would write something like e' = f(e)f(e)^{-1} = f(ee)f(e)^{-1} = f(e)f(e)f(e)^{-1} = f(e)

[minor gazelle]

I will change it, I felt it’s better that way too because it’s not visually clear still the algebra there of mine 😭

How’s now I feel much clearer

G’

[soft tiger]

The set of all invertible square matrices over a field form their own group when equipped with matrix multiplication…good catch.

[soft tiger]

I don’t think you have to reference (H,*) as a magma.

Also what you’re proving in the second picture is called the One-Step Subgroup Test sometimes.

[swift tundra]

Ah rip. Well I guess Ill keep it in mind as I study. Thanks for checking it out

[minor gazelle]

Yes I was originally confused by the logic I felt the abstract concepts density many thing so i was directed into completely wrong direction 🫣

[minor gazelle]

That handwriting !!!

[thorn jay]

Sadly. Would like to know how they're used in topology but nowhere I can find it lol

Eh they're generalised by algebraic structures where the congruences can be faithfully represented as the equivalence class of a fixed element anyways

Much cooler

[cinder onyx]

me when the one-step subgroup test has two steps

🤯

[tardy hedge]

WTF.

[tardy hedge]

probably not

[thorn jay]

Yeah, they don't come up that often, and if they do, they're not different from left group actions really

[south patrol]

it's only a notational difference

[minor gazelle]

I just realized my prior definition of submonoid was wrong (I didn’t specify e_T=e_S)

And I found such a crazy counter example

[vast verge]

No issue here, I really like these questions where you find isomorphisms of factor groups

I just think they're neat

[tardy hedge]

Any polynomial ring over k viewed as a k-module would be infinite dimensional vector space right?

[thorn jay]

Yes, because all of them contain the polynomial ring in one variable, which is infinite dimensional, as a linear combination of x^n is simply a polynomial, and therefore cannot be the zero polynomial without all the coefficients being zero

[tardy hedge]

I’m learning about monomial ideals (just general facts about them) but for some odd reason they havent really been sticking in my head

[thorn jay]

What're monomial ideals?

[tardy hedge]

Just ideals that are generated by monomials

This section in a book im reading thru just talks about their facts, their primary decompositions etc sort of thing

[thorn jay]

Ah nice

[tardy hedge]

Ngl its pretty boring but i need to understand them well for the combinatorics side

[thorn jay]

Often in math lol

[tardy hedge]

Lol yeah

[barren sierra]

what book are you using?

[tardy hedge]

Monomial ideals by herzog and hibi

[barren sierra]

ah yea I should look at that

although my impression of that book is that it's more useful as reference

keep it around to look up facts as needed

[tardy hedge]

Yea it does feel like a reference book

[barren sierra]

rather than actually work through a large chunk of it

this is a thing I'm realizing with more and more books I want to learn material from: they're more references than learning texts

[tardy hedge]

Bruns and herzog feels like that. Like damn something about that text is a pain to read

[white oxide]

Sorry to respond so late, but doesn't this also require that both tensor products are over the same ring?

[rocky cloak]

Well, no. But you have to be a little careful what exactly the property of multilinear maps is

Like the specific property should involve all the rings

[white oxide]

Sorry I'm a little confused by what you mean

Would you mind clarifying?

I just don't really know where to start

Actually nvm I'll save you the burden and probably just read a proof of this lol

[cinder onyx]

oh my god this absolute nerd 0-indexed his rules

[barren sierra]

everyone says this but it's still dumb because saying "n rules" refers to the total number not the largest index

[white oxide]

Silly question, ,but is the reason that this is well-defined is because it's bilinear?

Wait never mind the map isn't out of the tensor product

How can I see that this is well-defined

Oh wait I see

If (t_1, c_1) = (t_2, c_2), then c_1 = c_2 and t_1 - t_2 is in the submodule X of the free module on A x B generated by all bilinear elements. But this map kills X, hence it's well-defined I think

[tardy hedge]

The highlighted part?

Why do you need to check well definedness there?

[white oxide]

N sorry

[tardy hedge]

Oh mb

[white oxide]

I meant two sentences after that

[rocky cloak]

The general rule is that a bilinear R-balanced map
i.e. a map f: MxN -> A such that
f(mr, n) = f(m, rn)
f(m+m', n) = f(m,n) + f(m', n)
f(m, n+n') = f(m, n) + f(m, n')
corresponds to a linear map
M(x)N -> A
for any abelian group A.

This generalizes to arbitrary products over arbitrary rings in the obvious way

[shrewd onyx]

Can anyone help me

[rocky cloak]

This is indeed a picture of the Riemann zeta function for Re(s)>1

Np yw

[shrewd onyx]

Thanks

[tardy hedge]

Aut(H) means the set of automorphisms of H. An automorphism is an isomorphism H->H

[soft tiger]

Can you explain 3 again? I’m confused about that.

I’m trying to clear up a proof that this is true but one particular step is not making sense to me

[tidal schooner]

Are you asking for an explanation of why that's the answer? If so: Apply the first isomorphism theorem to the map $\det: GL(n,\bR) \to \bR^\times$ to show that $G$ is isomorphic to $\bR^\times$. So, it suffices to count torsion elements of $\bR^\times$, but the only ones are $\pm 1$

[cloud walrus]

harmacist

[untold torrent]

I see

Thanks

[tidal schooner]

Np!

[untold torrent]

Could you elaborate more on how it works?

@tidal schooner

I meant how can we sure that all the kernel will be special linear

[tidal schooner]

The identity element of $\bR^\times$ is $1$, so $\ker(\det)$ is the subgroup of $GL(n,\bR)$ whose elements have determinant $1$, i.e., $SL(n,\bR)$

[cloud walrus]

harmacist

[tidal schooner]

@untold torrent

[untold torrent]

So identity element would be 1

How did you say -1?

@tidal schooner

Actually I have not read it properly

[untold torrent]

What are the minimum elements of cyclic group?

[tough raven]

Cyclic groups are rarely ordered, so there is (usually) no such thing.

[tribal moss]

Let $x$ be some element, and let $A$ and $B$ be sets such that ${x}\triangle A=B$ and (therefore) ${x}\triangle B=A$. Without loss of generality, let's say that $x\notin A$, so $B={x}\cup A$.

Now let $M = {x}A$. Since ${x}{x}$ is the identity, we have ${x}M = {x}({x}*A) = A$, which requires that $A \subseteq {x}\cup M$. Since $x\notin A$, this means $A \subseteq M$. On the other hand, $M={x}*A$ is a subset of ${x}\cup A=B$, so in total
$$A \subseteq M \subseteq B$$
Since $A$ and $B$ differ only in a single element, there's no set that is strictly between them, so $M$ must be either $A$ or $B$. But it cannot be $A$ because then ${x}A = M = A = \varnothingA$, and we'd be able to cancel the $A$ and get ${x}=\varnothing$. So $M=B$.

Thus ${x}*A = M = B = {x}\triangle A$, and ${x}*B = {x}*M = A$ (which was already noted above).

(This should then be repeated from the right instead, since we don't yet know that the group is abelian).

[soft tiger]

I was actually able to figure that part out, but thanks a lot. This next part of this proof kind of has me stuck.

[cloud walrus]

Troposphere

[tribal moss]

Oh, wasn't that (3)?

[soft tiger]

Right that was (3), I was just saying that I understood your original argument now.

Assume that $*:\mathscr{P}(X)\rightarrow\mathscr{P}(X)$ and $\cdot:\mathscr{P}(X)\rightarrow\mathscr{P}(X)$ are group operations on $\mathscr{P}(X)$ such that

[
(\forall A,B \in \mathscr{P}(X)),, A*B\subseteq A \cup B
]
and
[
(\forall A,B \in \mathscr{P}(X)),, A \cap B \subseteq A\cdot B
]

[cloud walrus]

calebuic魏凯布

[soft tiger]

I'm trying to finish the proof for Lemma 3.9, but I'm trying to figure out how to draw a contradiction from assuming A \cdot A \neq X.

[tribal moss]

If you've already done the parts for *, I would just say:

Let Y = X and define f:P(X)->P(Y) by f(A) = X△A. This is a bijection. Now if we define
#: P(Y)×P(Y)->P(Y) by S#T = f( f^-1(S) · f^-1(T) )
then # is automatically a group operation on P(Y) and f is an isomorphism.
It also satisfies (as one can see by a bit of set algebra)
S#T subseteq S\cup T
so we can use what we already know to conclude that S#T = S△T.
Then A·B = f^-1(f(A·B)) = f^-1(f(A)#f(B)) = X△((X△A)△(X△B)) = X△X△X△A△B = X△(A△B), and you can prove lemmas 3.7, 3.8, 3.9 just by computing from that.

[soft tiger]

I did do the parts for *:

[tribal moss]

Sure you could also take the entire argument for * and demorganize it step for step, but it's just as instructive to transfer that already known results in this way.

[soft tiger]

This other operation \cdot corresponds to the "co-symmetric difference"

Which is just the complement of the symmetric difference in X.

[tribal moss]

(I don't think you need your lemma 3.6, by the way, if you end up showing A*B = A△B element for element, as in my steps 4,5,6).

[soft tiger]

I would like to try a more elementary proof

[tribal moss]

It's hard to think of any more elementary way to prove two sets are equal than to show they contain the same elements.

[soft tiger]

Right, and your proof makes sense. I thought that I should be able to just flat out prove 3.9 using only the hypothesis and Lemmas 3.7 and 3.8

As in supposing I knew nothing about the * operation

[tribal moss]

I think it's because you're missing an analogue of your lemma 3.5.

[untold torrent]

Every Sub group of quotient group order 8 is normal?

[rocky cloak]

Not necessarily no, unless you mean proper quotient.

For example the symmetry group of a square has order 8 is a quotient of itself and has subgroups that are not normal

[tribal moss]

... where "for example" means "this is the only counterexample, up to isomorphism".

[candid patrol]

What is your exercice ?

Let f € Aut(G), then f(H) is a subgroup of G of order |H|

=> f(H) = H

[cloud walrus]

UGOBEL

[candid patrol]

It is

np !

[amber burrow]

What are the identity and inverses in the free group on n letters?

I assume an empty word is identity

But what about inverses?

[tribal moss]

Reverse the word, and then swap generators for their inverses.

[amber burrow]

So each generator has to have some inverse?

Could I get a hint for 10.1.3

Also, I why is $<A_{m,n}>$ where $A_{m,n} = { g^m | g \in F_n}$ a subgroup.

[cloud walrus]

donut123

[amber burrow]

Identity and inverses make sense

But how is it closed?

[rocky cloak]

<X> is by definition the smallest subgroup containing X

[rotund dragon]

try working this out for some (small) examples. when you transpose a and b, which pairs do you gain/lose in the set? can you generalize to n?

[candid patrol]

?

yes

[cloud walrus]

mqinsweden (ping on reply)

[candid patrol]

with alpha = v_p(|G|)

[amber burrow]

Oh yeah…

[cloud walrus]

mqinsweden (ping on reply)

[amber burrow]

Idk what I was thinking

[candid patrol]

A p-Sylow is a p-group

[velvet hull]

no, not just a p subgroup

a p subgroup is called a p subgroup, that's it

a sylow p subgroup is a maximal p subgroup

[candid patrol]

What ?

Let |G| = mp^s with lcd(m,p) = 1

Then S is a p-Sylow subgroup of G iff S is a subgroup of G with |S| = p^s

[limber tapir]

No, Z/2 x Z/3 has a 3-sylow subgroup but its order is not a power of three

[cedar vault]

lcd?

You probably meant gcd

[candid patrol]

gcd of course mb 😅

[rocky cloak]

Largest common divisor ¯_(ツ)_/¯

[cedar vault]

Okay, never heard that abbreviation though

[tardy hedge]

In Z, the primary ideals are (0) and (p^n) for prime p. Are the primary ideals in any pid of this form?

[rustic crown]

yee

this easily follows from chinese remainder theorem

an ideal I in a ring A is primary if and only if A/I is a coprimary A-module

and coprimary means it is "sitting on a single prime" i.e. has a single associated prime

CRT+pid says A/I = ⊕ A/(p_i^e_i), so it is coprimary if and only if there is a single i

[tardy hedge]

I know I is primary if all zero divisors in A/I are nilpotent, is that related to that?

Not sure what a coprimary module is

oh, you said it there lol

[rustic crown]

yee in some sense coprimary is the more fundamental notion

and more conceptual

and primary-ness is the notion of the inclusion I ⊆ A than just I :p. more precisely it doesn't make sense to say when a module is primary, but it does make sense to say when a submodule N⊆M is primary. (it is when the quotient M/N is coprimary)

[rustic crown]

strictly speaking that's not the definition, but is equivalent to it in the nice noetherian+f.g. situation

[tardy hedge]

shoot this sounds good but it will take me some time to parse through it

[rustic crown]

(btw tbf it's easier to see it directly by this)

the equivalence of the two definitions is pretty routine but not very obvious

[cloud walrus]

mqinsweden (ping on reply)

[south patrol]

I think it's funniest to say 1 thougy and heckle the problem setter

[rocky cloak]

You kinda need to solve (a) the intended way to solve (b) anyway, so might as well

[dense crown]

I am trying to learn Galois Theory (only at a high level) and I am struggling to understand this (almost) final part. In the screenshot you can see the statement of Lemma 23.29 (which I understand) and the IMPORTANT theorem 23.30. My problem is in the very first step, where it says "By Lemma 23.29 we can assume the E is the splitting field [of] f(x)" (I am assuming it was supposed to say "of").

I don't understand how the Lemma guarantees that the splitting field of a solvable polynomial is itself a radical extension of the base field. By definition there is a radical extension which contains the splitting field, but how do we know that there is a radical extension which is the splitting field?

TLDR; How do we know that the splitting field of a solvable polynomial is a radical extension of the base field of coefficients (if this is even true)?

(I was told to ask this in here)

From here: https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Abstract_Algebra%3A_Theory_and_Applications_(Judson)/23%3A_Galois_Theory/23.03%3A_Applications

Another source saying the same thing: this page from Cambridge University says "If the polynomial p(x) is soluble by radicals, then the splitting field F of p(x) is a radical field extension of Q (can you see why?)."

My problem is that I cannot see why.

[south patrol]

seems to be a mistake here as those aren't cyclotomic extensions

[dense crown]

You can get to Q{√{1+√2}} by adjoining √2 and then √(1+√2) so it is not cyclotomic but it is radical (which is what it says)

[south patrol]

it says "which you can get to only using cyclotomic field extensions"

which seems a typo

[dense crown]

why

[south patrol]

but i think this is a good question you've asked

[south patrol]

well like nowhere else are cyclotomic extensions mentioned

[dense crown]

they define them in the previous part Il left out

[south patrol]

Anyway uh i think it is false that polynomials solvable by radicals have radical splitting fields

[south patrol]

Well this is wrong

and i am surprised that error has been made in a book or smth

[dense crown]

what is the correct definition

[south patrol]

I have only ever seen cyclotomic extension to mean an extension of the form Q(a)/Q where a is a root of unity

or more generally just adding roots of unity

[dense crown]

ok thanks,

can you shed any light on this part.

I’m just confused where it says “by the lemma we can assume that E [some arbitrary radical extension] is the splitting field”

that is, why can we assume that there is a radical extension which is the splitting field (not just which contains it)

[dense crown]

this part sorry

[south patrol]

I think like you can have polynomials solvable by radicals whose splitting fields aren't radical

However it shouldn't really matter here because uhhh

[dense crown]

That was what I thought but ai found 2 places saying it

[south patrol]

Sure but AI hallucinates references ig

i found proofs of counterexamples ig

I think you can basically embed your splitting field into a normal radical extension and then use the fundamental theorem of Galois theory to deduce that Gal(K/F) is solvable from Gal(L/F) being solvable

Like since K/F is Galois itself, Gal(K/F) is a quotient of Gal(L/F)

So if Gal(L/F) is solvable then so is Gal(K/F)

And then their proof works

But yes this is a good question!

[dense crown]

Oh okay. I think I can see why Gal(L/F) would be solvable, I just need to work out that last detail

thanks for your help!

[south patrol]

Nppp :)

Sorry for being negative about these resources aha, i think it's important to poont out mistakes tho

[rocky cloak]

Wait, what does solvable by radicals mean then?

[tidal schooner]

Splitting field is contained in a radical extension (not necessarily that it is a radical extension)

[rocky cloak]

Okay, I'm just not seeing how that's possible I guess

Like something where the roots are of the form
root(x) + root(y), but the splitting field doesn't contain root(x) or root(y), is that the point?

[tidal schooner]

Yeah, though not all choices of x and y will work of course. The usual example of this phenomenon is $\bQ(\zeta_7+\zeta_7^{-1})/\bQ$, where $\zeta_7=\exp(2\pi i/7)$

[cloud walrus]

harmacist

[barren sierra]

We have the word minor for determinant of a submatrix

is there an analogous term for the Pfaffian of a submatrix of a skew-symmetric matrix?

[candid patrol]

Something missing ?

Wath’s the full proposition ?

[rocky cloak]

I guess just apply the sylow theorems to the group PN

[somber sleet]

Hey guys, I was solving this exercise, but I do not understand the last point, why can we write f that way?

[coral spindle]

If there were any cross-terms, they would be zero in J since J contains every monomial with at least two distinct multiplicands.

Sorry I mean zero in R*

not in J

[somber sleet]

okay that's good I also understood it that way, what are thos fi's then in the polynomial down

[coral spindle]

Some polynomial

[somber sleet]

these should be just some powers, right?

[coral spindle]

No

They could be, for example f_1(x_1) = 1 + x_1 + x_1^5

and similar things for f_2 and f_3

[somber sleet]

okay yes exactly

[coral spindle]

They can be any polynomials

[somber sleet]

I mean they are always still in K[x_i], because otherwise they would be 0 right?

I mean they are not mixed

ever

[coral spindle]

That is the point, yes

[somber sleet]

okay that's is good, thank you for clarifying!!

[coral spindle]

No problemo

[glad osprey]

You mean xHx^-1? That's just the set of elements of the form xhx^-1 where h is in H

[coral spindle]

It is quite common notation to, given some function f : X -> Y, write the image of some subset A of X as f(A). This is the same principle.

[tribal moss]

It must be used with a bit of caution if there are multiple such subsets around in the same expression -- otherwise you may end up with things such as Z+Z != 2Z which can be confusing to the reader who isn't in on it.

Yes.

Yes, with the standard group self-action by left or right multiplication.

If the group acts on itself by conjugation, though, stabilizers of single elements can be more interesting.

[tardy hedge]

For R and S rings, if S is an R-module, does there necessarily exist a ring map R->S?

[rocky cloak]

This is true if S is an R-algebra, but not necessarily if it's just an R-module.

The map
R -> S
r |-> r*1
becomes a ring homomorphism assuming the multiplication of S is bilinear

[south patrol]

"is an R-module" is a bit oddly phrased cause (1) that structure isn't used in the ring map and (2) it only concerns the underlying abelian group of S. i guess you can phrase it as compatibility between some existence things but feels like a slightly funny question

idk what a counterexample would be though

[tribal moss]

Yeah, there doesn't seem to be a reason why it would be true, but it's tricky to find a counterexample either.

[rocky cloak]

S = Z^2, R = Z[sqrt(2)] I guess

[tardy hedge]

yea was a bit weird but I was thinking about it cause for S an R-algebra you have the map R->S that makes S an R-module, so I guess I was trying to think of if the "other way" was true too or something

[tribal moss]

You can map (a,b) in Z^2 to a in Z[sqrt(2)] and have a ring morphism, can't you?

[rocky cloak]

Goes the wrong way

[tribal moss]

Whoops.

[tribal moss]

Okay, I accept that.

[rocky cloak]

It says pk divides p!

[tardy hedge]

I'm not really sure on what you said here tbh

[rocky cloak]

That k divides (p-1)! is not just arithmetic

[tribal moss]

The point is that you need to explicitly assume some connection between the S×S->S multiplication and the R×S->S multiplication.

Such as r·(s1·s2) = (r·s1)·s2 and probably also r·(s1·s2) = s1·(r·s2) which would make S into an R-algebra.

[tardy hedge]

oh, and the assumption jagr said was that if the ring multiplication in S, SxS->S was R-bilinear, then the map R->S r -> r*1 is a ring hom?

[tribal moss]

Yeah.

[rocky cloak]

You just need the first condition to make the ring map, but might as well throw in the second one to get an algebra (or have S be commutative to get it for free)

[tribal moss]

Yeah, I didn't bother to worry about whether it was a minimal requirement.

[tardy hedge]

the second one so that it maps f(R) into the center of S?

[somber sleet]

Guys if I have an irreducible affine variety of dimension 0, why does it have to be only a point?

[coral spindle]

What definition of dimension are you working with?

Well actually in any case I'll point out an argument

[coral spindle]

OK yeah that's perfect

Well note firstly that a point is a closed set (i.e. a variety!) and it has no nonempty subvarieties, so it's hopefully clear that points are zero-dimensional irreducible varieties.

Just thinking about the converse, one moment

For showing other way round, observe that every variety has an irreducible subvariety in each of its points. Indeed it's a union of these (possibly infinitely many!) irreducible dim-0 subvarieties

If there were an irreducible dim-0 variety other than the point one, it would have a dim-0 subvariety given by any one of its points

So that's contradictory -- it would be dim 1

(at least)

Does that make sense?

[rustic crown]

det doesn't understand this sentence

[coral spindle]

Also hi det, it's been a while

[rustic crown]

haiii

[coral spindle]

If V is a variety and v in V, then {v} is an irreducible, dimension-0 subvariety of V

[rustic crown]

ah okie

not a subvariety right

unless you take closures

(idk i'm assuming varieties here have non-closed poitns)

[rocky cloak]

The closure of a point is an irreducible closed set, so since you can't have a chain of two irreducible closed sets, the closure of any point must be everything.

That means your space has the indiscrete topology.

Then something something, affine varieties can't have indistinguishable points

[coral spindle]

I'm not sure what you mean, it is a subvariety of V

(@ det)

[coral spindle]

Nice proof very slick

[rustic crown]

subvarieties need to be a locally closed subsets right

[coral spindle]

Point sets are indeed closed in the Zariski topology

And therefore also locally closed I think?

[rustic crown]

okie so we doing classical

<

[coral spindle]

Oh yes I am assuming classical lol

[coral spindle]

Also, are these Gathmann's notes? Awesome if so, they're great

[rustic crown]

that what me was confused here ><

[coral spindle]

Ah ok I see

I think we were talking past each other

my bad det

[south patrol]

Lol perennial problem with ag convos here

[somber sleet]

meaing that it would strictly contain the point, right?

[somber sleet]

what do you mean with: "since you can't have a chain of two irreducible closed sets"? (sorry this question might be silly)

[somber sleet]

yes, they are, I really like them, but I just began studying algebraic geometry and am a little bit struggling ngl

[coral spindle]

Exactly yes

And then by definition of dimension above, it wouldn't (after all) be dim 0

[coral spindle]

It's definitely hard!

[karmic moat]

if you had a chain of two irreducible closed sets then your dimension wouldnt be zero

[coral spindle]

(it's essentially the same argument but in slicker language)

[rocky cloak]

That's the definition of 0-dimensional

[somber sleet]

pff hahah, yes of course, thank you!

thanks guys, it's much clearer now!

I asked because I was tryin to show that if I have a non-empty open subset U in X for X irreducible affine variety, then dimU = dim X and in the exercises given by my professor it's done by induction. Honestly, I don't really like the proof, do you think there is some other nice way of proving this statement?

[coral spindle]

I think that's frankly the way I would do it

[south patrol]

What is the dimension of the empty set

[somber sleet]

is it even defined? haha

[tribal moss]

Does that mean that R^n with the good old metric topology has dimension 0?

[south patrol]

I would take it to be -oo

[south patrol]

Yes

[somber sleet]

that proof is really ugly though

[tribal moss]

Statements dreamed up by the utterly deranged.

[south patrol]

But it does give you e.g. that k^n with the Zariski topology is n-dimensional for an alg closed field k, which is nice

[rustic crown]

people have come up with crazy definitions of dimension that work good in many contexts

there's this BMU-dimension that works well with both varieties and manifolds

[coral spindle]

I think inductive proofs are quite nice and pretty usually but fair enough, it's up to your taste really

[rustic crown]

oh more is true

oh okie no i lied

what i was gonna say was the elementary statement that there is a bijection between irreducible closed subsets of X intersecting U and irreducible closeds of U

the maps are simply Z --> Z∩U for the forward direction, and Z --> cl(Z) in the backwards.

but yea we need some version of "going down" for the thing you want to prove

and that kinda depends on stuff being finite type over the field k

an example to say why shit can go wrong is this:

look at A^1(k) and localize at the point with coordinate 0.

this will have coordinate ring given by k[x]_(x) and the Spec has two points {(0), (x)}. so it is one-dimensional and the (only non-trivial) open subset {(0)} sadly is 0-dimensional.

if your argument didn't use something specific about being finite type over k, then it would have worked for this nice ring k[x]_(x) where shit is not true.

[rustic crown]

in any case, a nice statement one should prove is that for (irreducible) affine varieties X over a field k, we can give simpler definition of dimension. You look at the coordinate ring Γ(X) which is an integral domain. take the fraction field Frac(Γ(X)) and look at the transcendence degree of this over k. this agrees with dim X.

(if you do a careful induction, you can avoid using going down.)

since this function field Frac(Γ(X)) knows the dimension, we can show the easier statement that Frac(Γ(X)) = Frac(Γ(U)).

[tribal moss]

Shouldn't all this be in #algebraic-geometry, by the way?

[rustic crown]

true 🙈

[next obsidian]

Boooo

[tardy hedge]

Saying S is an R-algebra doesn't necessarily mean we are considering S as an R-module by rs = f(r)s , but its just that we can do such a thing?

[rustic crown]

it's saying that there are two things happening at the same time

S has a ring structure, and we can also do scalar-multiplication where scalars come from R

[karmic moat]

to say "S is an R-algebra" means you have a ring S with a ring hom f : R \to S

[rustic crown]

so yea, it's not asking you to forget the ring structure on S.

[tardy hedge]

is the tensor algebra of M an R-algebra from the ring map R->T(M) just being inclusion?

[south patrol]

ye

if you want lol R = T(0)

[tardy hedge]

ok thx. im trying to learn about the exterior algebra

[south patrol]

noice

[tardy hedge]

How do I think of how elements of T(M) look like? Its an infinite direct sum so can I think of an element as tuple (r, m, m1 (x) m2, m1 (x) m2 (x) m3 , ...) etc but finitely many entries nonzero?

what does the ring multiplication in T(M) look like?

[tardy hedge]

Or should I think of it as "sums" instead

[coral spindle]

Your picture is incomplete because the entries need not be simple tensors

[tardy hedge]

yeah

[coral spindle]

But yes, T(M) as an R-module is indeed just R (+) M (+) M(x)M (+) ...

[coral spindle]

Either way works

After all, it is literally a sum of embeddings of things from each of the components.

People usually just write them as sums of the things, so e.g. 1 + v_1 (x) v_2

[south patrol]

I just think of an element as a linear combination of stuff in each degree

Indeed this is how i would think of most grsdrd rings ig

[south patrol]

It is the "tautological" thing

Like to multiply simple tensors, put a (x) symbol between them

[tardy hedge]

Yeah I was thinking something wrong by thinking some ring multiplication is like componentwise along a tuple (..)

[south patrol]

In other words the multiplication V^m (x) V^n -> V^(n+m) is the usual isomorphism

Where i am using powers for tensor products

[south patrol]

Yeah like key thing is this is a graded ring

So like there are pieces R(m) and the multuplicayion comes from maps R(m) x R(n) -> R(m+n)

[coral spindle]

Oh I immediately forgot to tell you about the multiplication AGH

[south patrol]

Also like there is a way to make this precise: if you have an R-algebra S and a map M -> S of R-modules then you get a unique extension along M -> TM which gives a ring map TM -> S

[coral spindle]

I saw that 👁️👄👁️

[south patrol]

Chungus

[tardy hedge]

precise in the sense that that is the correct ring multiplication on TM required to make that universal property true?

[tribal moss]

Polynomial rings seem to be the one exception to this -- which makes it a bit weird that they're also the prototypical example of graded rings.

[coral spindle]

So obviously potato has explained it but you can think of it as being 'polynomials in M'. We introduce a completely formal multiplication on elements of M and let the multiplication be as free as possible. The universal property formalises this

[coral spindle]

Are they necessarily? I mean they are literally T(R)

[south patrol]

Idk cause you usually write polynimuals as sums a_n x^n

[coral spindle]

x^n just keeps track of the degree

it's like a degree marker

[south patrol]

Which is explicitly writing it as a sum of things in each degree

[tribal moss]

But still it feels a lot more natural to think of a polynomial as a single thing, than it does for, say, a mixed element of a tensor algebra.

[south patrol]

Ig yee

[cloud walrus]

mqinsweden (ping on reply)

[coral spindle]

Yes yes, single variable :P

I get u

[south patrol]

I am mostly being silly there dw

It is Lagrange applied to G/K as a subgroup of S_p

[coral spindle]

Lagrange states that |K| x |G/K| = |G|, so by definition |G/K| divides |G|.

Oh wait Potato is right

[tribal moss]

Of course that gets blurred a bit when things like Clifford algebras are defined as grade-mixing quotients of tensor algebras, because that requires viewing the mixed elements of the tensor algebra as single things after all.

[south patrol]

p!

[coral spindle]

p! is not p

[south patrol]

indeed

What is it you are proving lol

[coral spindle]

This is Sylow isn't it

I would say in general that people do not think of the proofs of the Sylow theorems very often lol

[south patrol]

Boytjie this thing about poly rings being symmetric algebras has been quite interesting to me recently, like it has been subtle

Very nice fact

[south patrol]

Basically thing is that theres a construction which takes in a ring and you are interested in its valhes on poly rings. But if you write the poly rings as Sym(V) and use "canonical" equivalences then you only get smth natural in maps coming from vector space maps

Probably just from computing examples

This is one of the main ways to easily know a subgroup is normal

There is also an easier proof for p = 2

[tribal moss]

Or alternatively they're condensed down from proofs of more immediately meaningful properties that are themselves the result of wild desperate trying-everything.

[tidal schooner]

It's well-known that subgroups of index 2 are normal. So this is a natural way of generalizing that fact

[south patrol]

But yeah this is very interestung for smtu

[coral spindle]

You shouldn't expect to be able to come up with every fact about groups. It took us all a long time to work it out lol

[coral spindle]

Hm that's interesting

Is it hard to figure out why they coincide or are you saying that you haven't yet demonstrated it?

[south patrol]

Well the point here is that like you have a construction on commutative rings (actually simplicial commutative rings) which is left Kan extended from polynomial rings, and on polynomial rings you can compute its values quite easily given the identification of Z[x_1,...,x_n] with the symmetric algebra on Z^n

[tardy hedge]

extended to sums by distributive law in the sense of , the multiplication given there is how to multiply simple tensors in T^k(M) by ones in T^j(M), so to define multiplication on the whole T(M) we can just extend that formula by a distributive law?

[south patrol]

But then one can't use this to control the values in a functorial way because i used the fact that it came from sym on an abelian group

[south patrol]

Yeah like you just define it so that (a+b)(c+d) = ac + etc

like it's enough to define it on simple tensors and then everything works out by writing stuff as sums of simple tensors and "expanding everything out"

[tardy hedge]

yea so theres like two distributive laws here right, over simple tensors and then over sums in T(M)?

[stable ravine]

Is anyone in here familiar with wreath products? I'm writing up some notes about twisty puzzles for fun, and I'm a little concerned that I might not be formalizing something quite right. I would love to have someone to bounce some stuff off of

Brand new to this server. Hope this is the right place to ask

[coral spindle]

I did indeed see your question and I am indeed familiar with wreath products, but I confess I was completely unclear on what your problem actually was

You seem to be saying that your construction is a wreath product, and... ok?

[stable ravine]

Oh yeah, I'm deleting it from the other page. I think that is the wrong one for it

Well, I'm unsure if I'm actually understanding wreath products correctly. I only found them after coming up with my construction

Trying to redo my construction in the way wreath products seem to be normally notated does not feel natural

But I did not completely explain what it was in the first comment I made

[coral spindle]

OK would you like to talk through what you think wreath products look like?

[south patrol]

idk what your sentence means

[coral spindle]

As a quick preliminary question, do you know what a semidirect product is?

[south patrol]

The proof given there seems clear enough so please lmk what confuses

[stable ravine]

Yeah. I'm a little rusty as I haven't taken an algebra class in several years

[coral spindle]

OK

Let's start then. Maybe let's make a thread

it says that since pk divides p!
k must divide (p-1)!
Do you see why this is true, firstly?

OK

Great ok

So let's say q is some prime dividing |G| and p < q

Here's a fact

to show that q does not divide k, it suffices to show q does not divide (p-1)! by transitivity of divisibility

So why would q not divide (p-1)!?

(p-1)! is a product of 1, 2, ..., p-1

And by the definition of primality, if q divided (p-1)! it must divide one of 1, 2, ..., p-1

Now can you tell me why this means that q <= p-1? You can probably see this now

[south patrol]

Found theorem I wanted a proof of without any proofs

Rage.

[coral spindle]

A little imprecise as proofs go, but your reasoning is correct

Indeed you're done

So q does not divide (p-1)!, so q cannot divide k.

then k must cannot contain a prime smaller than p
That is what you needed to prove

[cloud walrus]

mqinsweden (ping on reply)

[tardy hedge]

if A and B are R-algebras then the tensor product A(x)B is also an R-algebra. A(x)B would already be an R-module by r(a(x)b) = ra(x)b = a(x)rb, but this action is different from how R acts on A(x)B when considered as an R-algebra?

in the first case I guess it would be r(a(x)b) = f(r)a(x)b but the second action would be r(a(x)b) = (f(r)(x)g(r)) * (a(x)b) = (f(r)a(x) g(r)b))?

[south patrol]

No, the two should agree. If the maps are f: R -> A, g: R-> B then h: R -> A (x) B sends r to f(r) (x) 1 = 1 (x) g(r), so you can check it works out

[coral spindle]

Remember that the tensor product is over R, and that elements of R commute with all other elements in an R-algebra

So in particular, ra (x) b = ar (x) b = a (x) rb = a (x) br

[tardy hedge]

in a/m they have the map h be r -> f(r) (x) g(r)

[south patrol]

maybe i'm cooked...

[coral spindle]

Mind posting the definition?

[coral spindle]

This is surely wrong, right Potato?

I mean we would want there to be maps B -> B (X) C and likewise for C

and this should preserve the algebra structure

so it must be a |-> f(a) (x) 1 = 1 (x) g(a) via that

[south patrol]

Hm thing is if you do like the tensor product R (x)_R R[x] then this gives you r -> r (x) r and that doesn't seem to be the correct thing

[coral spindle]

Indeed

What's the book again?

[south patrol]

atiyah macdonald

commutative algebra

[coral spindle]

Oh ofc

[coral spindle]

WAit yeah that commutative diagram below literally says this

I'm betting on typo.

[south patrol]

https://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald

[coral spindle]

Yupp

[south patrol]

Phew I'm not insane!

[tardy hedge]

thanks

[karmic moat]

world's best PS ever

[coral spindle]

That would be at least two of us

[south patrol]

and haven't been misunderstanding tensor products of algebras for the last few years

lol

[coral spindle]

Dear Tim,

[south patrol]

np

but this means u asked a good question

/ thought well about it

[south patrol]

yeah note that the map they give isn't even additive in general

lol

[coral spindle]

what CLOWNS

how DARE they

I could give you the definition right now, it's very easy to write down

I don't know the page but it's very simple

Would you like that?

@sacred wharf

forgot to ping

I have never opened Dummit and Foote, so I don't know that either

Are you familiar with the Cartesian product of sets? So if I had sets A and B and I wrote A x B, would you know what I mean?

In TeX: $A \times B$

[cloud walrus]

Boytjie

[coral spindle]

Yes, it's the set of ordered pairs.

[cloud walrus]

Boytjie

Boytjie

[coral spindle]

That is the group structure of the direct product of groups.

That's all there is to it.

It's likely to be an exercise in D&F, but you should work out the identity and inverses in this group

[white oxide]

Does this proof for c) show that $B$ is free abelian? Let ${\mathbf{x_1}, \dots, \mathbf{x_n}}$ be the set of nonzero generators for $B_n = B$. Then ${\mathbf{x_1}, \dots, \mathbf{x_n}}$ is a basis for $B$. Moreover, the subgroup generated by each $\mathbf{x_i]$ is infinite cyclic, for otherwise the subgroup $\pi_i (\mathbf{x_i})$ of $\mathbb{Z}$ is not infinite cyclic.

[cloud walrus]

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

[coral spindle]

What do you mean?

OK so... yes you get a new set

so...?

What's the question?

I wish you had started with the thing you actually wanted to ask

Anyway this follows immediately by orbit-stabiliser

N_G(S) is the stabiliser of the element S of the G-set P(S) under the conjugation action

The bijection between the cosets of N_G(S) and the conjugates of S will be given by gN_G(S) ↦ gSg^-1

OK, this is too small a snippet for me to understand

Can you try explaining what you're trying to ask

OK it's a bit simpler

Simply apply the action of h^-1

Then $g \cdot a = h \cdot a$ implies $h^{-1} \cdot (g\cdot a) = h^{-1} \cdot (h \cdot a)$. Can you see how to continue?

[cloud walrus]

Boytjie

[coral spindle]

Same thing. Apply the action of h instead of h^-1

$h^{-1}g \cdot a = a$ implies $h \cdot (h^{-1}g \cdot a) = h \cdot a$. Can you see how to continue?

[cloud walrus]

Boytjie

[coral spindle]

Is it? That surprises me, my old mac worked fine with it

but each to their own I guess

British English. I suppose the issue is language then

Anyway please go on

[cloud walrus]

mqinsweden (ping on reply)

[coral spindle]

Sure

That works

[karmic moat]

(i use a macbook air and i dont notice a diff between it and windows)

[coral steeple]

How can it be possible that the f_j's make a basis for Rm? They are elements of Rn

[velvet hull]

Presumably it means to say that they generate a ring / module isomorphic to R^m

[coral steeple]

R^m is the free left R-module of rank m

This book uses row-vector convention, which I find very confusing

Is the point that we need a R-module homomorphism \eta: R^n -> R^m whose matrix is B so that the \eta(f_j) form a basis for R^m?

[velvet hull]

A here is the base change matrix from R^m back to R^n

Which corresponds to a R module homomorphism

So if it has a two sided inverse then it defines an isomorphism

m here may or may not be distinct from n (it should be the same as n, but I don’t think the proof here requires that fact)

[coral steeple]

Ok, but if I use the definitions, then I have to find linearly independent elements of Rm spanning Rm. Where do I get these?

[velvet hull]

Sp you’re shown that G has trivial center , but that’s not enough to conclude that G has a unique class equation

[velvet hull]

Well you get it from the f_m’s

[coral steeple]

These are linear combinations of e_i's, which are in Rn

[velvet hull]

I would be careful about using the word linearly independent when working with modules

I’m not saying that the f_m’s themselves are the basis

But you do something with them

[coral steeple]

Is there something different about linear independence in a module rather than a vector space? $\sum_i c_ix_i=0\implies c_i=0$ does not use the assumption that any $c_i$ is a unit

[cloud walrus]

person2709505

[velvet hull]

Actually you’re right I don’t think there is explicitly anything wrong with it it’s just a personal preference of mine

[coral steeple]

I see

Actually I might have figured it out. I just don't like thinking about what matrix multiplication does to rows

[velvet hull]

Can you figure out what the conjugacy classes are?

[delicate orchid]

this question is incredibly weird. The only group of order 15 is cyclic.

[velvet hull]

It only says at most one

[delicate orchid]

you've assumed G is non-abelian...

[velvet hull]

Sure, and if one can show that there’s still at most 1 under this assumption then it’s fine, no?

Doesn’t mean that such an nonabelian G must exist

[delicate orchid]

why are we calculating the centre of a group that doesn't exist

all of mqinsweden's proofs are fine but it's for an object which isn't real

just start with the class equation straight away and show that there's only one possible

[velvet hull]

Well if the centre is trivial then ||at most one class equation can exist as 14 must be a nontrivial combination of 3 and 5||

See that’s the thing though, I’m not sure if that’s possible without splitting into the abelian and nonabelian case first

Without sylow theorems, which is the assumption I’ve been working under

[delicate orchid]

I guess if you don't have sylow theorems

that too

[velvet hull]

Oh, at most 1

[delicate orchid]

now we use the hint about <g> being a subgroup of its centraliser to show that this class equation cannot arise from an actual group

what a STRANGE question I'm fascinated by it

[velvet hull]

I mean, as long as you can check that there’s no other way to combine them to form 14 then we’re done

[coral steeple]

Does this work for the backwards implication?

[sonic coral]

what book is this from? every group of order 15 is cyclic lol

[delicate orchid]

atp just prove it's abelian: Pick an element of order 5 x and consider the conjugation action of G on <x> (which must be normal as it is index 3, the smallest prime dividing |G|), then because 3 doesn't divide |Aut(<x>)| = 4 this conjugation action must be trivial, hence <x> is central in G, but by your proof earlier this implies that G is abelian.

If we absolutely must find a contradiction from the class equation. Then again we have that the subgroup <x> is normal in G and G cannot have multiple subgroups of order 5 by third Sylow theorem (idc atp 💔). Hence G has 4 elements of order 5 which must fall into conjugacy classes of shape 1+1+1+1 2+2, 4. None of which appear in your class equation.

[velvet hull]

Honestly I don’t fully understand the point of this question, a homomorphism f is an isomorphism iff there exists a homomorphism g that acts as both a left and right inverse for f.
So your question is merely observing the fact that R-module homomorphism between R^n and R^m can be naturally represented as a matrix transformation

[coral steeple]

Yes, it's the first question (so probably just meant to test basic understanding of definitions) following the chapter on the correspondence between R-module homomorphisms and matrices

[tardy hedge]

where is algechill

i dont see it anymore

[tardy hedge]

I wasnt sure why f(r) (x) 1 = 1 (x) g(r), but its because the r scalar can move across like r(1(x)1) = (r.1(x)1) = (1(x)r.1) right

[rocky cloak]

Go to advanced algebra -> threads, and also maybe change your settings for when threads get automatically hidden

[south patrol]

Ye

[full jungle]

Given the Hasse diagram of a lattice of subgroups for a group G, I need to show that it is indeed the correct lattice for G.

I'm not well versed in lattice theory though. After proving that the underlying set of the lattice contains exactly the subgroups of G, is it sufficient to show that the line segments in the diagram are right? (i.e. that those inclusions do hold between respective subgroups)

Dummit-Foote suggests checking that "their pairwise joins and intersections are correctly drawn" but that sounds like a lot of work and I'm not sure it's necessary.

[thorn jay]

Prove that a nonabelian group of order 15 decomposes into groups of order 16

[coral spindle]

Yes

[thorn jay]

Yes, if two lattices have the same underlying poset then they are the same lattice

[real cypress]

Hey, sorry quick question, would anybody know what the capital K with a double line (like with real or complex number symbol) generally means?

This

[thorn jay]

Depends on the context

[barren sierra]

I see that often used for a field but like

There's nothing stopping someone from using that for whatever

[rocky cloak]

I feel like I've mostly seen it to mean a field that is either R or C.

[thorn jay]

I'm gonna use it to denote a nonabelian group of order 6

And ridicule everyone using it otherwise

[tribal moss]

Good plan. Why not order 15 instead, though?

[real cypress]

Ah sorry didn't know these are called fields in English.
Ok, so it's just some default field, not something specific.
I see, thanks!

[thorn jay]

Are you kidding? That's incredibly redundant

[rocky cloak]

Well, either R or C is pretty specific.

But it could be it's being used more broadly. Hard to say

[real cypress]

Ah ok

Thanks!

[tardy hedge]

How can I show that 2(x)1 is nonzero in 2Z (x) Z/2Z? can I show it by making a bilinear map and showing (2,1) does not get mapped to 0?

[velvet hull]

what are you tensoring over

[tardy hedge]

Z

[south patrol]

Probably easiest psychologically just to note 2Z is iso to Z as a Z-module lol

so this becomes 1 (x) 1 in Z (x)_Z Z/2Z

[tardy hedge]

Ok, and then at this point would you just use that this is iso to Z/2Z and 1(x)1 goes to 1?

[south patrol]

Yup, or you can even just be lazy and just say there's a non-zero bilinear map Z x Z/2Z -> Z/2Z

[tardy hedge]

Sending (a,b) to ab

[tardy hedge]

To show 1(x)1 not zero we need there is a nonzero bilinear map that also has (1,1) not mapping to 0 right

[velvet hull]

if its nonzero then (1,1) can't go to 0

[tardy hedge]

Because Z and Z/2Z are generated by 1?

[south patrol]

ye

[velvet hull]

not just generated, it is straight up 1

[south patrol]

though here also just like you can again write down the example kinda simply

[velvet hull]

so you can multiply with any element you want and biliearity does the rest

[south patrol]

I mean like if you just write down the kinda obvious Z x Z/2 -> Z/2 multpilication that does the job (sends 1 (x) 1 |-> 1)

and indeed this is the map inducing the isomorphism Z (x) Z/2 -> Z/2 anyway

this is kind of a good question lol like a bit of a troll

[tardy hedge]

f(a,b) = abf(1,1) right

[minor gazelle]

Is this proof actually correct it felt pretty simple? Or I have to use homomorphism?

[coral spindle]

It is a simple thing to prove.

[minor gazelle]

Then it’s fine I thought it has to be delt with homomorphism

[coral spindle]

You are actually using a homomorphism, you're just not saying it

m ↦ x^m is a homomorphism Z → G

(in fact every homomorphism Z → G is of that form)

[minor gazelle]

Yes, you’re right.. I didn’t know 🫣🫣🫣

[coral spindle]

Something new every day

[minor gazelle]

I thought I could’ve crafted (and indeed crafted a contradiction easily) yes something new everyday🥰🥰🥰

[coral spindle]

Well I mean, you use proof by contradiction redundantly here

"Let me prove x. For contradiction, assume not x. But in fact unrelatedly we can prove x. Which is a contradiction."

Your proof directly shows that every element of x is torsion.

A proof that would actually use the contradictory hypothesis would (for example) argue that x, x^2, x^3 are all distinct elements, so that G contains infinitely many distinct elements.

I should have spotted this before, that was my mistake.

[minor gazelle]

I don’t have to assume like the order of it is infinite and then argue existence of m and n such that x^m = x^n it’s still fine?

[coral spindle]

Why would you have to assume that?

Where did you use the hypothesis that x was of infinite order?

If you didn't use an assumption, it's not necessary

[minor gazelle]

Because if x is not infinite I can’t apply pigeonhole principle

[coral spindle]

No of course you can

What's the pigeonhole principle to you? What's the precise statement?

[minor gazelle]

Like if you have more pigeons than boxes there is one box at least it has more than one pigeon

[coral spindle]

OK that's not a precise statement

Let me tell you the mathematical pigeonhole principle

[minor gazelle]

🥰

[coral spindle]

If $f : X \to Y$ is a function and $|X| > |Y|$ then $f$ is not injective.

[cloud walrus]

Boytjie

[coral spindle]

Note that you constructed a function f : N → G sending m ↦ x^m

This function is not injective, since N is infinite and G is finite.

This gives you precisely what you need

(N.b. I said function here, not homomorphism)

[minor gazelle]

Yes indeed, I just realize this part 🫣🫣 that’s such a brilliant brain twist

So essential Becasue G is finite should be good enough right?

[coral spindle]

Yes, you do not need to assume that x is of infinite order

[minor gazelle]

🥰🥰🥰🥰 that is such a nuance 🍀🍀

[coral spindle]

Ain't math great

[minor gazelle]

Yes🥰🥰🥰

[tardy hedge]

I like the enthusiasm

[minor gazelle]

I used to feel quite worried about delayed graduation though enthusiasm led me to think less over that (which is in fact trivial). So I like my own enthusiasm too

[charred iris]

your point? G is not assumed finite

[tardy hedge]

G being finite is different from it being finitely generated

[white oxide]

Silly question, but does meet usually mean nonempty intersection or that they're equal?

[rocky cloak]

Here it means having nonempty intersection at least

[white oxide]

Thank God that's weaker

[rocky cloak]

I would say that's what it "usually" means aswell, but idk, context changes words yk

[white oxide]

Also thanks for the hint on the other channel, I'm going to try to find out the iso on my own then read the rest of your message

[white oxide]

Yeah ikwym

[fading acorn]

atiyah ig

[white oxide]

Atiyah Macdonald

[karmic moat]

$10 on atiyah macdonald

let's fucking go

best book ever

[fading acorn]

lol