groups-rings-fields

[south patrol]

Anyway no I mean you can pick the prime subfield k, pick transcendence bases for any fields K,L of that characteristic, then one of those will have cardinality <= the other and you can just consider the algebraic closure of the bigger one

I don't see a way to do this avoiding something slightly more subtle like transcendence bases

[glad osprey]

If F and E have the same char, then they are both vector spaces over the base field. Does it work to just take the direct sum of these vector spaces?

[south patrol]

No, as that needn't be a field

[delicate orchid]

that might not be a field

[glad osprey]

Hmm, okay

[delicate orchid]

potato what's wrong with this one

[south patrol]

Well how do you do that

Without what I said

[delicate orchid]

brother I do not know what you said

just do K(L) the rest will figure itself out 😌

[south patrol]

For that to make sense you need to have embedded them both in a common field already tho

[delicate orchid]

do u? Just take polynomials with variables in every element of L and then formally adjoin inverses. A little K_0 if u will

if u use the woke definition where u intersect a bunch of shit then sure

[south patrol]

How do you map into that?

Oh ok

As in take K and then do a big ting for each el of L

But then how do you map from L into that

Anyway what I said is this: every field extension K/k can be cut up as k( {x}) for some big set of indeterminates followed by an algebraic extension K/k({x})

So in particular every extension embeds into the algebraic closure of k({x}) for some set of indeterminates

Now just take that set big enough for both the fields that you started with and both will embed in the alg closure

[glad osprey]

I think I get it, thanks

[delicate orchid]

ah good point

I will simply embed it.

[tough raven]

For an answer that sounds different but is probably the same, take the tensor product of the two fields (which is a commutative ring) and quotient it by a maximal ideal.

[thorn jay]

How do you prove that neither of the fields vanish when quotienting? Genuine question

[south patrol]

Ah nice, i did wonder if you could tensor stuff lol

this is very nice!

[tough raven]

Good question.

[south patrol]

I suppose the choicy part of it is then transferred from picking a transcendence basis to picking a maximal ideal aha, just the latter is much more familiar

[tough raven]

... actually, wait, I don't know that. 💀

Let me think.

[south patrol]

Any map from a field (into something nonzero) is injective anyway

So it'll still be an iso onto image

[tough raven]

Right yes.

[thorn jay]

Ohh right cuz there is no 0 map

Haha my brain is kinda funky atm

[tough raven]

My argument was that if k (⨯) 1 ↦ 0, then 1 = (k (⨯) 1)(k^{-1} (⨯) 1) ↦ 0, so the ideal of K (⨯) L would be improper.

[thorn jay]

Yeah thats the usual argument hehe,

[tough raven]

or as a slogan, "the pullback of a proper ideal is proper".

[thorn jay]

Really elegant, wow

[south patrol]

And yeah like

any nonzero element of K remains a unit in K (x) L, so it's not in a max ideal

[thorn jay]

Mhm

[south patrol]

i guess that is just identical to what you are syaing though

[thorn jay]

Hence doesnt get yeeted into the void

[south patrol]

but really these arguments are all the same lol

[thorn jay]

Yeah

[rocky cloak]

Every homomorphism from a field is injective

[thorn jay]

And all because of the fact that 1=0 <=> R = 0

How nice for you guys
cries in universal algebra

[thorn jay]

Yeah but i was thinking about a 0 map, not considering the fact 0 maps do not exist in ring theory

[rocky cloak]

I guess one would actually have to argue that the tensor product isn't 0, which is what happened when the characteristics are different

But then you just need to consider them as vector spaces

[proper tide]

Example of a subset of a ring that is a subgroup under addition but not a subring.[give 3 or 4 example]

[rocky cloak]

Zx < Z[x]

Ri < C

[proper tide]

Z[x] means?

Polynomial with Z as its coefficients?

[lusty marlin]

Polynomials in one variable with coefficients in ℤ

[proper tide]

Why Zx is not a subring of Z[x].

[delicate bloom]

you tell us

[proper tide]

Let ax and bx belongs to Zx for a,b belongs to Z
Then ax-bx=(a-b)x this is also belongs to Z since a-b belongs to Z
ii. ax.bx=ab.x² this is also belongs to Z

Am I wrong?

Please guide me..

[lusty marlin]

ax-bx ∈ ℤx
ax.bx = abx² ∉ ℤx

[proper tide]

Yeah got it...thank you so much..

[burnt stirrup]

hi! for determining the number of homomorphisms between Z -> Z_9 x Z_12, am I correct in thinking that, as Z is cyclic (generated by 1), effectively, a homomorphism is "dictated" by where f(1) goes? given |Z_9 x Z_12| = 108, I'd imagine this means there are 108 of them

the very next problem is how many homomorphisms g: Z_36 -> Z_9 x Z_12 exist, but I'd imagine this is exactly the same thing given Z_36 is cyclic, right?

[rocky cloak]

It's correct that f(1) determines all of f, but it's not clear that f(1) could be any element of the group apriori

[burnt stirrup]

oh, mmh

not sure how i'd go about justifying that

[rocky cloak]

Well, given f(1) are you able to define f for the other elements? Is this definition well defined?

[lusty marlin]

Try to find a relation between the order of an element in the domain and the order of its image under a homomorphism

[frigid epoch]

Is Z the limit of all C_n or is it some adic thing

I feel like I could construct some coherent series of elements thats, say, 1 mod every multiple of 11

So either that yields something not in Z or its forbidden by some other coherence condition

Nvm it literally just yields the number 1 lol

I feel like it's either [yes, by Chinese remainder theorem] or [no, by counterexample]

[feral kraken]

hi I'm new to abstract alg can I get good resource recommendations?

hopefully something available online

I wanna get into group theory

from a CS pov (if that matters)

[frigid epoch]

If you're at a uni/college and it's free to just attend lectures in your country, you can do that

At my uni in Germany, you can just walk into any lecture without being enrolled

[feral kraken]

hmmmmm

yk what that's a good idea thanks

[frigid epoch]

U enrolled in compsci?

[feral kraken]

yea

but I don't think I'll take this stuff in cs

so if I do that I'll prolly have to figure out at which college in my uni It's taught and attend there

things are pretty lax so I can prolly just attend in a college that isn't mind (depends on the prof)

one of the profs I was attending a lecture for made a whole deal and took the id of a student that wasn't in cs that was attending so idk tbh ☠️

[rocky cloak]

The limit of the finite cyclic groups is called the profinite integers. It's an uncountable group.

[frigid epoch]

Ah right I've heard of that

I was wondering if you could relate the natural logarithm in complex analysis to the n-th roots, cause their riemann surfaces have fiber Z resp. C_n over the complex plane (punctured)

But I guess what you'd get instead is a profinite gluing of all the roots

[feral kraken]

oh also any idea where I can find practice problems with solutions?

[frigid epoch]

And maybe the logarithm maps to that gluing if it's lucky

[frigid epoch]

Well the idea behind attending a lecture is

You'd get access to a bunch of exercises, proper educational pacing, peers to study with...

[feral kraken]

weeeell

idk about that tbh

my uni is kinda mid

I'd mostly attend sometihng like that for the lectures

[frigid epoch]

Idk if a unis midness is reflected in algebra exercises tho

Depends more on the midness of the professor

[feral kraken]

hm

I see

[frigid epoch]

@rocky cloak ?

[rocky cloak]

I'm not entirely sure what you're asking.

How are you gluing these Riemann surfaces in the first place?

[frigid epoch]

By taking a limit, of top spaces or G-bundles or something

And lifting all roots simultaneously

Over each point, the limit is the profinite group, so you just need to topologize it

[cedar vault]

how is HOM(E,F) a commutative ring? it isnt even a ring, no? given two elements, there is no multiplication structure unless F=E??

i think im just tripping really bad but i dont see where

[south patrol]

Here they seem to be using pointwise multiplication

Wait, hm, no, i guess i assumed E,F were fields when i saw that notation lol

[cedar vault]

there is no multiplication structure for vectors either

[south patrol]

Yeah okay i have no idea what they are doing lmao

[cedar vault]

oh okay this book is pretty thoroughly written so i thought this must be a mistake on my part

[vocal pebble]

also how is F^E a product group? (how commutative?? maybe they mean commutative under addition but then what does product group mean)

is this bourbaki

[digital seal]

most likely a typo ring -> group

[boreal inlet]

May anyone suggest me an easier way to prove that A_4/V_4 is Z_3?

Finding the cosets for everything is taxing

[coral spindle]

Something has gone wrong. S_4 has order 24, and V_4 has order 4, but Z_3 has order 3, not 24/4=6. So what you claim is impossible

[boreal inlet]

Omg

Sorry

I meant A_4

[coral spindle]

This is still impossible

A_4 has order 12

[vocal pebble]

that'd give you order 2 iso Z2

[boreal inlet]

Wait what

[coral spindle]

Oh, A_4 instead of V_4?

[boreal inlet]

12/4 = 3 right

Yeah

Also V_4 is the commutator I think

[coral spindle]

There is only one group of order 3.

So you’re done by numbers.

[boreal inlet]

I mean I actually want to prove it's the commutator so I want to use the isomorphism A_4/V_4 is Z_3 to show the commutator is a subgroup of V_4 first

[south patrol]

True, especially as this is probably translation lol

[south patrol]

Actually there is a proper class of groups of order 3

[boreal inlet]

....that is true.

But if someone asks me to provide an explicit isomorphism? Or atleast a surjective isomorphism from A_4 to Z_3 so that I can use the first iso thm

[vocal pebble]

you gotta quotient 😢

[south patrol]

There should be some geometric way to do this lol

Well, really it's probably easier the other way round

Defining a map out of Z/3 is super easy

[boreal inlet]

Oh shit so this is about some magic stuff permuting some points in some way

[south patrol]

Just send 1 to an element of order 3 in thjs case and you'll be done

[boreal inlet]

Oh wait

I think I know

[boreal inlet]

Yeah

That

But does that automatically force a product of transpositions to be mapped from 0

I think we need that as well

Because we have to show the kernel is V_4

...you know what i'll stick to "every group of order 3 is abelian"

This is fucking with me

Thanks guys

[serene dune]

sit for an weekend and classify up to order 20

[boreal inlet]

[fluid kelp]

given that the sign of 3 permutations are all even. how do you show no ordering of these 3 permutations can result in an odd one?

[coral spindle]

What do you mean by ordering here

You surely don’t mean product?

[delicate orchid]

all categories are skeletal

[thorn jay]

Blasphemy

[coral spindle]

Patience is a Virtue
Virtue is a Grace
Grace is a little girl
Who would not wash her face

[vocal pebble]

here the transitive property of being gives us some very interesting equivalences

[thorn jay]

Why won't patience just wash her face man

TwT

[coral spindle]

She is very young and doesn’t know any better

[empty perch]

Hey folks, just a quick reminder that the Abstract Algebra Lecture series is meeting regularly again ^_^
We've worked out our technical issues and will convene to discuss the Sylow theorems in about 20 minutes over in the ⁠⁠#1055201711679082516. More information can be found in our thread: ⁠⁠#1317307081535000606

[frigid epoch]

@feral kraken check this out

[feral kraken]

thanks

[crystal sundial]

I suspect that a, b !=0 in an Euclidean Domain implies that one of (a, b/(a,b)), (b,a/(a,b)) is a unit. Is there a way to prove this without using Euclidean Domain -> UFD?

[empty perch]

heya, for what it's worth, if you're interested in coming along with us i'd definitely recommend joining the Simple Group server— we're missing one of our admins so the thread doesn't get updated as much as it probably should

https://discord.gg/5VzAbcwt

[wooden fulcrum]

Hello

I am trying to prove that ring R sinple => R primitive, is it true that if I take I to be some maximal left ideal in R, then R/I is a faithful and simple left-R-module because if I were to find an element of r that annihilates all a' in R/I, then that would mean that ra in I, hence bra in I hence the 2 sided ideal generated by r is under I and so it has to be the zero ideal by R being simple?

It seems like so to me but I thought I had it like twice already so I'm not quite so sure anymore lol

[royal root]

hi

[rocky cloak]

Yes, any non-zero R-module is faithful for this reason

[glad osprey]

Any hints for 2.2? I think I should prove that L is algebraically closed by showing that every polynomial in L[x] splits, but I'm not sure how to do that

[south patrol]

Haven't they told you that that is the case?

[glad osprey]

Every polynomial in k[x] splits in L, but I don't see how to prove it holds for every polynomial in L[x]

[glad osprey]

Wait, I think this works: any a in k-bar is algebraic over k, and therefore the root of a polynomial p in k[x]. But p splits in L, so every root of p is in L, in particular a, so k-bar <= L

[fluid kelp]

i don't mean product

so here's the question

given that the sign of 3 permutations are all even. how do you show no ordering of these 3 permutations can result in an odd one?

i mean to say

that

say you have 3 even permutations a,b,c. i want to show that no matter what order you do a,b,c or to what extent, you can't have an odd permutation

[cedar vault]

here i would want to show that $A_i \cap A_j = {e}$ for all $i\neq j$, right? if that is the case, given $i$, we know that for all $j < i$, $A_i \cap (A_1 A_2 ... A_j ... A_{i-1}) = {e}$ so in particular, since $A_j \subseteq A_1 A_2 ... A_j ... A_{i-1}$, we have $A_i \cap A_j = {e}$. by symmetry, we get that $A_j \cap A_i = {e}$ for all $j > i$. so we are done?

[cloud walrus]

bathroom mug

[velvet hull]

you need a little more - in particular to finish the proof you need an isomorphism from $\varphi: \bigoplus_{i=1}^k A_i \to \prod_{i=1}^k A_i = G$

[cloud walrus]

HChan

[velvet hull]

but that isn't too hard

[cedar vault]

i suppose its the most natural one? (a,b,c...) -> abc..

so what exactly is the definition? that we have G = A1A2... and that we have the above isomorphism?

i feel like having that isomorphism should somehow be equivalent to trivial pairwise intersection

[velvet hull]

the distinction I'm trying to make here, is that the product of the normal subgroups is a subgroup of G

while the direct product is completely outside of G

and although we know that they are the same, a priori you need to prove that

[cedar vault]

ah, okay

i thought the definition of direct product is what you call the product of the groups

[velvet hull]

not sure which definition you're working with here, but that's not a hard proof

just prove it's a group homo plus 2 of the 3: injective, surjective, |G| = |\bigoplus A_i|

you also have freedom to chose which direction you want your isomorphism to go from

choose which one is easiest for your proof

[cedar vault]

this has trivial kernel, because all the groups intersect trivially so that (a,b,c,...) -> abc... = e =>a = b = .. = e . it is surjective, since G = A1A2...

[velvet hull]

yeah checks out

[feral kraken]

thanks

[left meteor]

Hey can you please help me with this problem:
There are 2 operations defined on a subgroup of a symmetry group given by cycles (1 3 2) and (3 4 5). Determine if cycle (1 3 4) is a member of this subgroup

Sorry if this is trivial

[kind temple]

what do you mean by operations

[left meteor]

Permutations*

[thorn jay]

Still what do you mean by "permutations defined on a subgroup"

[left meteor]

Sorry, I mean there are 2 types of permutations allowed (and their multiples)

I wanna know if using those permutations you could get the same effect as cycle (1 3 4)

[kind temple]

so you are asking if (1 3 4) is in the subgroup generated by (1 3 2) and (3 4 5)

[left meteor]

Yes

[kind temple]

what do you think?

[left meteor]

I would think probably yes

[kind temple]

why?

[left meteor]

it should be consistent with the orbits

[kind temple]

not sure exactly what that means since there isnt an explicit group action here…

however, you can compute all of the elements in the subgroup generated by (1 3 2) and (3 4 5)

do you see a way of doing this?

[left meteor]

Is there no better way of doing this?

I could try conjugating stuff and multiplying them out

[kind temple]

there are only so many ways to multiply the two elements since they have order 3

so it’s not terrible

[left meteor]

Damn I mean. Alright I got you. Thank you!

[left meteor]

Solved. Yes it is an element. Thank you again

[cedar vault]

(see question below)
so i want to find the order of G here and just divide by 2. that just means i have to count the number of solutions to ad - bc = 1 in F_p. There are a few cases i can think of here:

1)one of the terms ad, bc = 0
say a = 0. then d can take p values, and say b can take p-1 values and c is forced by b. so we have p(p-1) solutions. adding up for b = 0, c=0 and d=0, there are 4p(p-1) solutions

2)both terms are nonzero:
so here we can just isolate one element, say c = (ad-1)/b. here b can take p-1 values, and a,d can take p-1 values as well, so in total (p-1)^3 solutions.
thus we have in total:
(p-1)(p+1)^2 solutions.

since quotienting by C identifies a matrix and its negative together, and since X is a solution => -X is a solution, the order of G/C is (p-1)(p+1)^2/2

did i miss anything here?

i apparently have overcounted somewhere hmm

ohh d cant be the inverse of a i see

so the second case needs to be modified to (p-1)^2(p-2) so in total there are (p-1)(4p + (p-1)(p-2)) solutions and so G/C has order (p-1)(p^2 + p + 2)/2

[vocal pebble]

perfect

[vestal bobcat]

Hi I had a question……if k is a field will any maximal ideal of k[x,y] always contain a quadratic irreducible polynomial

[rocky cloak]

Don't think so.

What about something like (x^3 - 2, y^3 - 3) in Q[x, y]

[vestal bobcat]

Yes

[rocky cloak]

Btw don't post the same question in several channels

[vestal bobcat]

But can it be shown true for algebraically closed fields without referring to any form of Hilbert Nullstellansetz

[vestal bobcat]

It is true but I want a proof without using HN

[rocky cloak]

Hmm, I don't have any good idea for how to do it without using something like at least Zariskis lemma, but maybe it's possible...

[keen arch]

I cant find a proof of this theorem is it true?

[wide surge]

Can you clarify what the function ν is?

Oh thank you

[keen arch]

but this theorem just lets me check very quickly if a group is cyclic

It would really help me for my algebra exam If this theorem is true and Im allowed to use it

[wide surge]

I mean given a p-group you can decompose it as a direct sum of cyclic groups. If you know that every group of order n is cyclic and p^{k} is the greatest prime power of p that divides n then every p-group of that order has to be cyclic as well, which implies that k = 1 for every prime divisor of n

[velvet hull]

It sounds true, but also not an interesting result

A cyclic group of order n always exists for any n, so if you know that there’s only one group of order n then of course it’s cyclic

But actually finding how many groups of order n there are up to isomorphism is generally pretty hard

[wide surge]

The fact that all such groups are isomorphic is a direct consequence of the fact that two cyclic groups are isomorphic if and only if they have the same order.

[velvet hull]

This is not a conclusive test, but the way - just because v(n) != 1 doesn’t mean a group of order n can’t be cyclic

Cyclic groups have really, really degenerate structure - it shouldn’t be that hard to determine if a group is cyclic

(On a test, anyway)

[wide surge]

This is certainly not the case. A simple calculation is much easier

Also the theorem that you stated is not a test for cyclicity of a group, rather for a property of a positive integer. The order of a group might not be free of squares, and yet it can still be cyclic

[keen arch]

I should proof or disproof if Groups of Order 15 are always cyclic or not.

[wide surge]

You can use Sylow's theorem

[keen arch]

And what that theorem its a 1 liner

[keen arch]

yeah I think they want me to use those

[wide surge]

It is the standard way and much more useful to understand Sylow's theorem than the theorem you provided.

It is intersting as an excercise, but a direct corollary of more fundamental results.

[rocky cloak]

For m>1 note that CpxCp is a non-cyclic group.

So for n square free you can go by induction. By some Sylow counting argument you can show that G has a normal sylow subgroup. Then by induction the quotient group is cyclic, so G is a semidirect product. Then the fact that p is not 1 modulo any of the primes means those primes don't divide p-1, so the semidirect product is direct.

Conversely if p is 1 modulo q then you have a non-cyclic group of order pq given by semidirect product

[rocky cloak]

Actually you might need a little more work than just Sylow counting. But G will have a nontrivial normal subgroup at least.

[wide surge]

I mean we have outlined the proof of the stated theorem. If your goal is to prove that any group of order 15 is cyclic, there is nothing quicker than Sylow

As a second exercise perhaps try and write in more detail what @rocky cloak outlined for you!

[rocky cloak]

To me their just two ways to describe the same thing.

A representation refers to the G -> End(M) definition and a module refers to the GxM -> M definition.

The answer is saying that the word module specifically refers to linear things, and that representation is more general. And maybe some people use it that way, but to me representations are always linear anyway.

But in that case a G-set would have a corresponding representation G -> Sym(X), which is not a module, since there is no linear structure.

Another example could be quivers, where a representation is usually thought of as a homomorphism of quivers to the underlying quiver of some category. Here I'm not familiar with any "module definition" for representations. But representations correspond to modules over the path algebra. This works fine for finite quivers, but for infinite quivers the correspondence breaks down. So that would perhaps be a significant difference.

(you can fix the correspondence, but then you have to talk about nonunital rings with enough idempotents and locally unital modules and horribleness)

[hard hearth]

aH -> aHa^{-1} is a well defined bijective mapping? so that [G: aHa^{-1}] = [G : aH] < infty? is this valid?

[south patrol]

I don't really get what you're doing

[rocky cloak]

How is [G:aH] defined?

And it's true that the index of aHa^-1 is finite, but I don't see how it helps the problem

[hard hearth]

sorry. number of elements in {aHa^{-1}| a in G}=[G:H]

[south patrol]

That equality certainly doesn't hold in general - suppose everything is abelian, for example

Indeed the assignment aH |-> aHa^-1 needn't be injective

[hard hearth]

perhaps i define the map the other way round?

[south patrol]

Well this means there cannot exist such a bijection

[hard hearth]

hmm, but then it isnt well defined necessarily

any hints?

[rocky cloak]

I mean it's well defined, just not a bijection.

[rocky cloak]

Think about when
aHa^-1 = bHb^-1

Can you frame that in terms of a relationship between a and b?

I think you're pretty close to something with approach you had actually

[wide surge]

Would it be productive to consider the action of the normalizer of H on the syzygy classes of H (by syzygy i mean the internal automorphism h -> ghg^{-1})?

(my explanation of syzygy is perhaps redundant but i am not that familiar with English terminology on group theory and my memory is disappointing me)

[hard hearth]

with my original approach i wanted atleast an injection into G/H. anyways this gives me that ba^-1b^-1a in H. im still not seeing anything

i could switch the roles of a and b and get another such similar relation. but what after that?

[rocky cloak]

So if you instead of trying to construct an injection into G/H you tried to construct a surjection from G/H you could have something.

Anyway, your relation is not quite right. It should be that b^-1 a is in the normalizer of H.

[hard hearth]

sorry, what is a normalizer?

[hard hearth]

is it elements that commute with elements of H? because if so, this does give me that

ah wait, yes i do have that relation here but not quite following from what i said

[rocky cloak]

x is in the normalizer iff xHx^-1 = H

[hard hearth]

so for any h in H, aha^-1 = bh'b^-1 for some h' in H. from this i have, b^-1aha^-1b = h' => b^-1aHa^-1b subset H => b^-1aH = Hb^-1a

[junior badge]

Here p is a prime, K a number field an O_K its ring of integers. Why does p | q?

[hard hearth]

right, so there is a surjection, the one i said initially, which is aH -> aHa^-1. so the cardinality is <=. my bad. anyways i have what i want now

thanks for the help, jagr2808, Prismatic Potato

[rocky cloak]

The norm of p1 divides the norm of p. And the norm of p is just p^n

So q is a nonunit divisor of p^n

[junior badge]

Ahh makes sense. Thanks!

[south patrol]

One thing to be careful though between linear reps and modules over k[G] is that many constructions are not what one might naively think

for example the tensor product of G-reps is not just the tensor product of the corresponding k[G]-modules

[rocky cloak]

Is there an actual terminology clash there? Like do people usually mean tensor product over kG if they talk about "the tensor product of two G-modules"

[south patrol]

I wouldn't say it's a clash, just that you should be a bit careful w stuff like that

[rocky cloak]

I agree you should be careful, but I'm not sure it's coming from a distinction between reps and modules.

It's just that there are two things "tensor product of G-reps" can mean

[glad osprey]

Whether a field is a splitting field depends on the base field, right? Like if k <= F then F may or may not be a splitting field over k, but F is always a splitting field as a trivial extension F <= F?

[tardy hedge]

When talking about a splitting field dont u need to specify the polynomial too

And specifying the polynomial means a polynomial over some field right

[south patrol]

Sure

[cedar vault]

am i tripping or is this not true? take R over R, and the set {...,-2,-1,0,1,2,...} is a stable subgroup but is certainly not a submodule over R

[south patrol]

Well it is still meaningful to say smth is a splitting field, but you should say it relative to a base. Like for example if F is characteristic 0 then an extension K is a splitting field over F iff K/F is finite Galois

[wooden fulcrum]

Hello, does all ideals finitely generated <=> no infinitely ascending chain of ideals depend on choice?

[rocky cloak]

A polynomial could be a polynomial over several fields at once I guess, e.g.
splitting field of x^2 + 1 over Q vs splitting field of x^2 + 1 over R.

[cedar vault]

oh wait bruh they mean something else by subgroup. the operators will be R here

[wooden fulcrum]

Thank you, so this holds for any nonzero R-module whenever R is simple?

[rocky cloak]

Well => should be clear at least.

But the converse seems to me you might need some weak form of choice. But maybe you can do it without...

[south patrol]

Wait I got stuck in my head uh

Okay yeah you can definitely do it with just dependent choice by building a countable infinitely ascending chain if it's not fg

Probably does need dependent choice

[glad osprey]

If k <= F then we call F a splitting field if it's a splitting field for some polynomial in k. So it's a property of a field extension rather than a field I guess

[south patrol]

Indeed for this reason some people say "splitting extension"

as there isn't a meaningful absolute variant

[wooden fulcrum]

Thank you guys

[glad osprey]

Yeah, that's a better term imo

[wooden fulcrum]

Hmm I'm not sure I see this completely still, if the module M is not simple then it might be not faithful right? If R isn't guaranteed to be commutative

[rocky cloak]

Any homomorphism R -> End(M) can only have kernel (0) because R is simple

The only exception being if M=0, when the kernel is all of R

[wooden fulcrum]

Ohhh right

So the proof I sent was not needed at all

Thank you

Riiiight annihilator is not just one but two-sided ideal

[surreal swan]

Suppose G is a finite group, A and B are subgroups, A intersect B is trivial, #A * #B = #G
Therefore each element of G can be uniquely represented as a*b, a in A, b in B
Consider b*a, apply previous statement to it
So we have a function f from BxA onto AxB such that:

f(b, a) = (a', b') iff b*a == a'*b'

Through f we can ask the question: given A and B, try to find G (how can subgroups be combined to get a bigger group?)
f has to satisfy a bunch of properties, for example we expect (b*b')*a == b*(b'*a) , which produces equations:
(denoting fA for the first component of f, fB for second, so f = (fA, fB) kinda)

fA(bb', a) == fA(b, fA(b', a))
fB(bb', a) == fB(b, fA(b', a)) fB(b', a)

Similar for b*a*a'
Also some simple properties due to 1B and 1A getting glued together kinda:

fA(1B, a) = a
fA(b, 1A) = 1A

Suppose we knew A was normal in G
Then

b * a == b * a * b^-1 * b = phi_b(a) * b; phi_b : A -> A isomorphism

We get semidirect products

So my question: is the more general construction through f fruitful? Is it explored somewhere?
Atm the functional equations feel pretty terse / impenetrable, though some more properties can be shown, like fA(fixed, varied) : A -> A is bijective, for every fixed b in B

[rocky cloak]

It's called Zappa-Szep product
https://en.m.wikipedia.org/wiki/Zappa–Szép_product

[chilly ocean]

I’ve been thinking about for this about 30 minutes, but I still don’t see why them being disjoint except at the identity implies the representations of elements are unique. I’ve been trying to find a contradiction, but I really need to learn more AA..

Anyone got any hints?

[surreal swan]

Uniqueness is 2 parts: each element in G has an ab representation, and the representation is unique
Did you show the first part?

Try figuring out how many elements the set AB = {ab : a in A, b in B} has

[chilly ocean]

I’m using the assumption that 1. Is true(which includes HK=G and H$\cap$K is trivial, and trying to derive a contradiction from also assuming there is some g in G with hk=g=h’k’ for h≠h’, so I’m already assuming every element have a representation

[cloud walrus]

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

[chilly ocean]

The same amount as G, since it’s equal to G

[surreal swan]

This is good approach, try to manipulate algebraically

[rocky cloak]

Write
a b = a' b'
Manipulate the equation to get
Something in A = something in B

Hence what you get must be in the intersection

[minor fulcrum]

You know how SU(2) is the universal cover of SO(3)? Is there some sort of generalization to higher dimensions?

[sly crescent]

Yes

Sp(2) is the universal cover of SO(5) and SU(4) is the universal cover of SO(6)

[minor fulcrum]

does that generalize to arbitrary dimension or is it just a few exceptional cases

[sly crescent]

It’s just a few exceptional cases

[minor fulcrum]

ah, alas

[rocky cloak]

In general you have the Spin(n) group is a double cover of SO(n), but I guess that's more or less by definition.

[minor fulcrum]

is there some general relationship between SU(n) and Spin(m)?

[rocky cloak]

Don't think so

[wooden fulcrum]

Hello, is it correct that every simple module is isomorphic to R/I as an R module for some maximal proper left I < R?

[rocky cloak]

Do you know the correspondence theorem?

[next obsidian]

The module has to be cyclic else you would get a proper submodule that is cyclic. From there you see it is R/I for some I, and then if I wasn’t maximal it has a submodule

[wooden fulcrum]

No

I arrived at it going R^{+J} = F -> M surjective and through F I can drag some nonzero R

[rocky cloak]

Alright, well without that it's quite hard to see.

But it says that every submodule of R/I corresponds to a submodule of R containing I.

So in particular R/I is simple iff I is maximal

[wooden fulcrum]

But I wasn't sure if the conclusion was correct

[wooden fulcrum]

Ah that's this I didn't know it had this name

Okay thank you guys very much

[thorn jay]

Gotta learn UA better then smh

[next obsidian]

Dawg, 1 is in only a single ideal

The entire ring

😭

Also r^n = r^m doesn’t mean r^(m-n) = 1, you need to be in an integral domain to conclude this

Consider the element (1,0) in say

Any product R x S

This satisifies x^2 = x

For this, if 1 is in I, then for any x in R, x•1 = x is in I, so I = R

No

They are specifically not subrings

Unless they’re everything

They’re a submodule of R

Considering R as a module over itself

Ehhhh

Okay so

If your convention for ring doesn’t involve having a 1

Then it’s a subring

But then if you have a ring with 1 and wanna deal with those

A subring should not only have a 1

It should be the SAME 1

Again consider R x S

And the set R x {0}

This is a ring

In fact, even under the induced operation from R x S

It has a 1

But that 1 is the element (1,0)

Which is not the 1 of R x S

Very confusing no?

They r cool

I gave u an example

But it’s because you can have

xy = xz without y = z

You don’t have cancellation when you have zero dicisors

Yup

That was the example I gave u even

[rocky cloak]

Just to dot the ts here. You can have radical ideals.

Just take k a finite field and R = k[x]/(x^2) and I = (x) for example

[chilly ocean]

Is there any generalization of a field with exponents? E.g. like how a ring is a group with multiplication too, or a field is a commutative ring with division too?

[mint seal]

it sounds like it could maybe be covered by the concept of Lawvere theory or algebraic theory

but I haven't thought through the details

if you've seen how there's a certain finite graph category that encodes the axioms for a group

and that any particular group can be realized as the image of that category through a functor to Set

algebraic theories are like that but more general

I don't know if there's a type of algebraic theory that corresponds to your idea and has a name

[chilly ocean]

Alright, thanks!

[mighty totem]

Theorem: let G be a group. If the derivative G' is 0, then G is a constant

[kind temple]

what do you mean?

[mighty totem]

Like if you have a function whose derivative is 0 then it must be constant. Same is true for groups

[kind temple]

what do you mean by derivative of a group

[mighty totem]

The derived subgroup

[chilly ocean]

Commutator subgroup? Or what

Isn’t this true for any perfect group tho

Wait nvm

[spice whale]

broke: i only work with abelian groups they're nicer
woke: all abelian groups are literally trivial

[mighty totem]

woah is that the game with omega

[sly crescent]

Based

[thorn jay]

Division in a field is a partial operation, which is the problem.
What you can do is take the axiomatic definition of a ring and simply adjoint an exponential operator ^ and adding the rules
(a ^ b) ^ c = a ^ (b * c)
(a ^ b) * (a ^ c) = a ^ (b + c)
(a ^ b) * (c ^ b) = (a * c) ^ c
a^1 = a
Although the last one could be a little too strong perhaps

Using induction you can prove that indeed a^n aligns with the monoid definition of power

Ah wait, this would probably not even be defined propeely

Oh well

As long as you do not force both a^0 = 1 and a^1 = a

Because then you'd have a^-1 * a^1 = a^{1-1} = a^0 = 1 thus giving 0 and inverse, collapsing the ring to the 0 ring

[hard hearth]

is there a good reason to believe A4 can be a product of two proper subgroups? and is there a good reason to believe that there is no subgroup of order 6 of A4?

[hard hearth]

i know that if |H| = 6 and H < A4 then the only possibility can be S3 here

[charred vessel]

Given the ring A=Z/3[X] and the polynomial f=X^4+k, how do I find the characteristic of the quotient ring A/(f)? It'd have to be at most 3 I'd say because 3*1=0, but I'm not sure how to conclude that it is exactly 3.

[rocky cloak]

A subgroup of order 6 has indeed 2, so must be normal and hence the kernel of a surjective map A4 -> C2. But A4 is generated by elements of order 3, so any such map is trivial.

And I'd say there isn't a good reason to believe it can be written as a product mostly because it can't.

[rocky cloak]

So since it divides 3 it's either 3 or 1.

So you just have to prove 1 =/= 0.

I guess you can argue by degree, any element in (f) has degree at least 4

[charred vessel]

Oh right thanks!

[hard hearth]

hmm what about C3 and V4 subgroups of A4?

[hard hearth]

what about order 3 subgroups here?

[rocky cloak]

Ah, you don't mean direct product

[hard hearth]

oh yeah actually i was kinda confused about this too in my quiz

[rocky cloak]

Then sure, the fact that just two primes divide it would be a good reason

[hard hearth]

really?

i see

[rocky cloak]

Yeah, any such group will be a (not necessarily direct) product of its sylow subgroups

[rocky cloak]

It has order 3 subgroups. But the same argument will give that they can't be normal

[hard hearth]

also one more question: is the map h(H intersect K) -> hK a bijection H/(H intersect K) -> HK/K? its well defined, because if h(H intersect K) = h'(H intersect K) then hh'^-1 in K . its surjective clearly one preimage of hK is h. injective because if hK =h'K then hh'^-1 = k in H, so that h(H intersect K) = h'(H intersect K)

[rocky cloak]

But I guess this is also just true for any group that isn't cyclic (maybe depending on how you're defining product)

[rocky cloak]

Yes, this is one of the isomorphism theorems

[hard hearth]

doesnt that say something about H,K being normal subgroups?

[vocal pebble]

like one of H or K must be normal afaik.

[vocal pebble]

so you cant directly quote the theorem, but this proof works anyways

[hard hearth]

ah. i guess i didnt tell you where H and K come from. they are just subgroups of some G

[rocky cloak]

Yeah, I guess there is some subtlety whether HK = {hk} or if it's the subgroup generated by H and K.

[mint seal]

"Hua's identity": if $a,b$ are nonzero elements of a division ring, with $a \neq b^{-1}$, then
$$a - aba = (a^{-1}+ (b^{-1}- a)^{-1})^{-1}$$

how the heck did someone in the 20th century get their name on this thing??

[rocky cloak]

According to Wikipedia, this identity is a crucial step in proving Hua's theorem, which is a bit more natural statement to consider.

So I guess that's how they stumbled upon it / how it got named

[serene dune]

what am i looking at ?

[cloud walrus]

ManifoldCuriosity

[serene dune]

put \textbf or underline at the heading if you want

[mint seal]

fair enough, though I'd hope it has more than just one application to get named like that

[rocky cloak]

Maybe you're more hopeful than me

[thorn jay]

Me when i give a fancy name to a lemma I'll never use again

[warped oar]

Guys what's $2x2

[crystal vale]

localization at prime ideal p is a process of construct ring of fraction of ring R\p, correct?

high chances to be wrong

[tardy hedge]

Its constructing ring of fractions where your denominators are R-p

R take away p is the multiplicatively closed set

[crystal vale]

yes, isn't same as constructing ring of fraction of ring R\p?

[glad osprey]

I don't know much about localization, but R\p isn't a ring, so I don't think that definition works

[crystal vale]

yes i know they write R\p as set theory notation of difference but i am asking different question i have a doubt what if i take ring R\p and the construct its ring of fraction

i got it

[crystal vale]

annihilator of submodule generated by m in M is not necessary prime ideal, right/

let m in M, i know Ann(m) \subset Ann N, where N is a submodule generated by m, is Ann N \subset Ann(m)? I think no

[wooden fulcrum]

gotta learn everything better smh...

[crystal vale]

M is a R - module

[next obsidian]

They’re the same

[crystal vale]

oh i was right

but how can i show that

got it

[hard hearth]

what is a good motivation for the definition of a semidirect prodcut? i dont understand why it is defined the way it is AT ALL

[coral spindle]

OK

[hard hearth]

from my understanding, if G,H are groups, phi a compatible association of H with elements of Aut(G), then G semidirect_phi H is defined (g1,h1) (g2,h2) = (g1phi(h1)(g2),h1h2)???

[coral spindle]

Suppose $G$ is a group with a normal subgroup $N$. Now we have a quotient $G/N$. Note that typically, there is no subgroup of $G$ which is isomorphic to $G/N$.

Let's suppose that we are in a `nice' circumstance where there is. So there is a subgroup $H \leq G$ such that the map $H \to G/N$ defined by $h \mapsto hN$ is an isomorphism. Then $G$ is isomorphic (in a natural way) to the semidirect product $N \rtimes H$.

[cloud walrus]

$\mathbf{Boytjie}$

[coral spindle]

So there you go. Semidirect products are this 'nice' circumstance where the overall group looks like the normal subgroup and the quotient combined in a nice way.

[hard hearth]

oh huh i didnt know this i saw some book defining it this way and it confused the shit out of me like i felt like i was understanding what was going on but not at the same time

[coral spindle]

In general, if:
\begin{enumerate}
\item $N$ is a normal subgroup of $G$,
\item $H$ is a subgroup of $G$,
\item $NH = G$, and
\item $N \cap H = 1$,
\end{enumerate}
then $G \cong N \rtimes H$ in a natural way.

[cloud walrus]

$\mathbf{Boytjie}$

[coral spindle]

So this is good motivation I think.

[hard hearth]

oh wait is the external semidirect product creating a group where these nice circumstances occur with these specific groups being the nice gourps???

[coral spindle]

Yes

[hard hearth]

holy

thank you very much

[coral spindle]

C.f. the external and internal direct product

no problem

[hard hearth]

i think i can work out verifying that these nice circumstances occur (in the external case) on my own

[coral spindle]

That sounds worthwhile

[glad osprey]

Visual Group Theory? I've read a bit of it, but it's never really helped me understand much, except maybe parts of the first chapters with simple Cayley graphs. I wonder if it's even possible trying to get any actual visual intuition in a subject like algebra, it feels like an impossible endeavour imo

[vocal pebble]

nvm ignore that

[toxic zephyr]

so talking about rings of fractions... we have the denominators as a multiplicative system (closed under mult and has unity).
and for the def we basically have a/b=c/d iff ad-bc is a zero divisor.
is there an intuitive reading why zero divisor is natural? like it's not obvious to me atm. zero, sure, and it sounds reasonable it would more generally by zero divisor but like I'm not seeing the necessity

I guess like just an example of a ring of fractions where you actually need the zero divisors would be super helpful and I can try to come up with an example myself?

[rocky cloak]

So the condition isn't exactly it being a zero-divisor, but specifically that it becomes 0 when multiplied by something in your multiplicatively closed set.

The reason for this is that you want
s/s = 1/1

so then if sa=0, then
a/1 = sa/s = 0/s = 0/1

[rocky cloak]

Consider C[x, y]/(xy) and {1, x, x^2, ...} as your set

[toxic zephyr]

ohhhhh. so like... an "S zero divisor"? or torsion element of S?

[toxic zephyr]

I'll try this ty!

[rocky cloak]

Yes that's right.

So then you want
a/b = c/d
iff
a/b - c/d = (ad - bc)/bd = 0

[lofty mural]

um

it would be a black jack

with 1 total deck left and 6 in game

[toxic zephyr]

just want to check: can we say
"I is a left ideal of R iff RI subset I"
basically just wanna know if "RI subset I" means that rx in I for all r in R and x in I

[south patrol]

Yes

though RI should really be closed under addition which isn't immediate if I isn't already

[toxic zephyr]

so basically we need that

• I+I subset I (closed under addition)
• RI subset I (closed under left mult by R)
  these are just more compact ways to say that?

[coral spindle]

Yeah

N.b you need nonemptiness

Otherwise the empty set is an ideal (it’s not)

You can express this by just saying 0 is in I

[rocky cloak]

It isn't immediate even if I is

[south patrol]

True yes

I guess I am finding it hard not to think of I as an ideal lol

I am used to writing RI when I is an ideal of a subring or smth

[rocky cloak]

Yeah, though I guess for the purposes of showing I is an ideal it doesn't matter that much how you define RI.

But the normal definition is sums of things of the form ri

[wooden fulcrum]

Hello, I came upon an exercise that asks for the initial object of R-Alg, in the general case where R is not commutative this is not some "known" object or is it? Would it be something like R/C where C is some kind of commutator ideal?

[south patrol]

Agreed

[south patrol]

How are you defining R algebra? For most notions can think of there is a clear choice

[wooden fulcrum]

Ring A with a f: R -> C(A) I guess

[wraith cargo]

what is C(A)?

[wooden fulcrum]

Center of A

All elements that commute with everything wrt multiplication

[wraith cargo]

why do you need to map it into something that commutes?

are your algebras commutative?

[thorn jay]

Because of the identity
r • (x • y) = (r • x) • y = x • (r • y)

I presume

(r in R, x, y in A)

[wraith cargo]

ah okay that makes sense

[thorn jay]

Maybe try R/I where I = ( xy - yx | x, y in R )

[wraith cargo]

isn't the initial object just R?

[wooden fulcrum]

Oh yeah I forgot about minus thank you

[thorn jay]

C(R) is a subring of R

[thorn jay]

:P

[wooden fulcrum]

That's what I meant by some kind of commutator ideal

[thorn jay]

That, like the commutator subgroup, is a specific case of smt from universal algebra

[wooden fulcrum]

I don't think so but I'm not good with noncommutative rings so I don't think I can come up with a counterexample xd but you have to specify the homomorphism too

[wraith cargo]

yeah lol I might be thinking too much in commutative algebra land

[wooden fulcrum]

Noncommutative rings at it again making things complicated

[thorn jay]

For an identity p≈q consider the congruence generated by the pairs
< p(a_1, ..., a_n), q(a_1, ..., a_n) >
For all a_i in A

Every homomorphism h : A -> B where B satisfies p≈q must factor through that quotient algebra, hence every homomorphism h : R -> C(A) must factor through R/I, as C(A) is commutative

[thorn jay]

At least you mfs have a notion of commutativity/noncommutativity smh

Be grateful

[wooden fulcrum]

I see thank you

[thorn jay]

:3

[wooden fulcrum]

But I don't think there is like a good way to describe the ring in ring terms at least as far as I've gone in the book xD

@aluffi

[thorn jay]

I believe this is a good description at least already

It's the same with the abelianisation

It has a universal property and a construction, but you cant get much better than that

[wooden fulcrum]

True

I think the author might have meant to say R was commutative but idk

[rocky cloak]

The initial R-algebra is simply R

[thorn jay]

consider noncommutative R

[rocky cloak]

There isn't anything called R-algebra for R noncommutative as far as I'm aware

[thorn jay]

R -> Z(S)
where Z(S) is the center of S

[rocky cloak]

Well R/[R, R] it would be then

Kinda silly though, since you would just replace R by something commutative

[thorn jay]

ye that's the conclusion we came to too :3

[thorn jay]

agreed, every homomorphism factors through R/[R, R] anyways

but still

[wooden fulcrum]

Is [R, R] the commutator ideal

[rocky cloak]

Yeah

[wooden fulcrum]

Oh truu lol

Oh I see my bad

[thorn jay]

I've also seen like.. "balanced" R-algebras or smt?

with
(x * r) * y = x * (r * y)

over a left- and right-module

[wooden fulcrum]

Definition 5.7. Let R be a commutative ring. An R-algebra [...]

Sorry lol

My eyes went over this when Id gone back to check before

[thorn jay]

nah it's cool

[rustic rapids]

Hello, Im wondering a little bit about the importance of rings. When studying group theory, I was able to understand it at least somewhat as "studying symmetries" and so some importance was immediately relevant, but im having a hard time seeing the main idea of rings. Most theorems just seem like "technical" details so im not really sure what we are building up to and why. I know its somewhat used in algebraic geometry but I probably wont be taking that soon (if ever) and so it's not very immediately satisfying. So, I was wondering why rings are important and perhaps some applications?

[coral spindle]

Sure ok

So you're saying you find groups easily motivated because they act on sets via symmetries.

Rings are quite similar. They act on Abelian groups, and by 'linear' (kinda) maps instead of symmetries.

Maybe you know about Cayley's theorem, saying that every group is embeddable in a group of permutations.

Rings have a similar thing, saying that every ring can be embedded in the ring of endomorphisms of an Abelian group

Do you know much linear algebra? Fields and vector spaces are a (very nice) first example of rings and modules. Rings can act in much more exotic ways than the nice simple structures of vector spaces, and this leads to a truly enormous amount of very cool mathematics.

If you end up doing some ring theory, it's actually very likely you will do at least a little algebraic geometry.

[rustic rapids]

ah I see, that actually does seem to be fairly interesting

[thorn jay]

the most popular application outside of algebraic geometry is in number theory, and particularly cryptography

[mint seal]

my 2 cents: rings are algebraic structures inspired by the integers and polynomials over the integers, and can be used to study things like divisibility and primeness in new places. So there are lots of applications to number theory. One classic theorem is that a prime number p can be expressed as a sum of two square integers exactly when p is congruent to 1 modulo 4. The most revealing proofs make use of the Gaussian integers: numbers a + bi where a,b are integers and i^2 = -1. This is a ring.

[coral spindle]

Lovely example

[south patrol]

Tbf rings are just pointless

Maths is a pyramid scheme

[mint seal]

The number 5 is prime as an ordinary integer, but in the Gaussian integers it can be written (2+i)(2-i), showing it to not be prime there

and multiplying that out gives 2^2 + 1^2

[coral spindle]

I think they're kind of incomparable in some ways, but I think it's not that wild to say that rings are much more general than groups. So the theory is less neat (in some ways) but a lot more varied.

[thorn jay]

well, at least if you look at the universal algebra side of things, then rings admit a much larger variety of "purely ring-theoretic" classifications, simply by the characteristic alone

[rocky cloak]

You should hopefully be familiar with a lot of rings already, even if you haven't explicitly thought of them in terms of ring theory:

Z, the integers forms a ring. As does Z/n, the integers modulo n. A lot of number theory can be expressed through these rings and related rings.

The set of all polynomials (with for example real coefficients) forms a ring. And ring theory can be useful for studying polynomials and their roots. (Studying roots of polynomials is in fact the origin of group theory, so rings and groups go hand in hand here)

The set of nxn matrices forms a ring. This gives an example of what Boytjie mentioned how a ring can act as symmetries on spaces with more structure. (Matrices are like the symmetries of a vector space in a sense).

The set of smooth or continuous functions (from some space to R for example) is a ring by point-wise multiplication. And this ring can contain a lot of information about the geometry of your space.

[thorn jay]

not to mention all the definitions of a more model-theoretic character, such as primality, reducedness, etc

[coral spindle]

Honestly I am speaking purely on vibes lmao, I think every group is a precious gem, whereas rings feel a lot more arbitrarily structured

But if there is a good technical justification, I'll take it

[thorn jay]

I do believe there is merit to what you said

[coral spindle]

Fields, perhaps, are more precious ;)

They do feel like gems

[thorn jay]

yeah, it's more often you see someone studying one group in particular than someone studying a ring in particular, maybe because of the fact a group is a monoid that it is a "set of functions" by cayleys theorem, and therefore is more easily related to other objects, lending itself to a more interesting/in-depth study

[rustic rapids]

i see. Thank you all for the explanations!

[rocky cloak]

Idk, groups can go kinda crazy too.

Fields I can get behind

[thorn jay]

groups are a more structured kind of crazy

[rocky cloak]

I'd say they are less structured, since they literally have less structure

[thorn jay]

for me they feel more structured

[coral spindle]

Well of course, Galois has told us that some fields are not too unlike groups

I think of groups as being more rigid. It's harder to quantify that, I realise

But when I see a finite group, I feel like it's a minor miracle it exists

Modulo the boring examples of course :P

[thorn jay]

terrible pun, I love it

[rotund aurora]

The empty set has no structure at all, but it is very structured, no?

[rocky cloak]

Is it? Does it even make sense to measure

Maybe, but I'm not so sure

[rotund aurora]

it has a very clear structure, because it has none

[rocky cloak]

Do the parts relate well to each other though?

[coral spindle]

All its parts relate exactly

[chilly ocean]

This makes sense in a non-Abelian way imo

[coral spindle]

I'm not sure if there is a nice way to make this precise

[chilly ocean]

Groups can have non-Abelian operations, the underlying additive group of a ring can’t

[rotund aurora]

there are no "proper" parts, so I'd say yes. The empty set is related to itself quite well

[rocky cloak]

And is the structure even that clear. I mean maybe the empty set has a million n-ary functions defined on it. Should they not contribute to the structure?

[chilly ocean]

But the multiplicative group can so

[rotund aurora]

there are no functions is just one function related to the empty set tho, unless I forgot the definition of function. Like it should be some element subset of emptyset x X x Y x... where X, Y, ... are sets. But this product is again the empty set

[coral spindle]

Yes, there is a unique map {} -> Y for any set Y

[rocky cloak]

So the empty set is both a set and a function

Now that's structure

[thorn jay]

The structure of the empty set is induced by the controversy of whether or not it should be considered an algebra

[rotund aurora]

there's none

[mint seal]

Cayley's theorem feels like a heavily constraining kind of result on groups. Maybe that's a sense that they're more rigid or "structured". I don't know about an analogous result for rings

[thorn jay]

Making it the only superposed algebraic structure in existence

[coral spindle]

If you think of a function X -> Y as a subset of X x Y with some univerally quantified properties (which is indeed the definition), then trivially the empty set is a function {} -> Y for any Y.

This also works in type theory btw, this is how we define exfalso

[rotund aurora]

oh wait yeah it was a subset, not an element. mb you are right

[rocky cloak]

Any ring R is a subring of End_Ab(R).

The proof is the same as Cayleys

[coral spindle]

No problemo, it's confusing

Note this is why some people might argue that 0^0 = 1 in combinatorics

[thorn jay]

The thing is that for groups the group it gets embedded in is the same for groups which have the same size

[rocky cloak]

Well for rings it's the same for rings that have the same underlying abelian group

[mint seal]

oh nice. Is there a good intuition for what End_Ab represents in general? Along the lines of Sym being permutations?

[thorn jay]

Thats a way stronger condition though

[rocky cloak]

Not really, the classification of finite abelian groups is pretty close to just the classification of finite integers

[coral spindle]

I think it's best to just picture matrices it's not quite easy to imagine automorphisms of Abelian groups in the same way, but you get most of the right picture if you just imagine you're working over a field instead

So instead of End_Ab picture End_k where k is a field

Linear transformations are nice, right? That's basically the picture

[mint seal]

for sure they are. I'll have to think about this more

[rocky cloak]

Yeah if you replace ring with algebra, then the classification of vector spaces is literally by cardinality (dimension)

[thorn jay]

End_Ab(R) is still nowhere near as nice as S_n though

[coral spindle]

Hmmm I think it's more similar than you might think

[thorn jay]

I suppose

But guess which one can be made into a Latin square and which can't? Checkmate atheists

[coral spindle]

I think of M_n(k) as a bit like a linearisation of S_n. And of course the algebraic group SL_n has the same dynkin diagram and whatnot

[thorn jay]

Hmm

Okay

[coral spindle]

I wonder if I can make this precise.

Well ok I'm internally just identifying kS_n with its image in End_k(k^n), so it's kind of cheating.

Or is it? Not sure.

Identifying is the wrong word

Just shoving it through the map

[thorn jay]

Which should be injective right? Due to the existence of 1

[coral spindle]

Oh I don't think so

I did make a typo originally, I meant k^n not kS_n

the algebra kS_n has dimension n! and End_k(k^n) has dimension n^2

[thorn jay]

Right, okay

I forgot S_n has order n! And not n

[coral spindle]

I made the same mistake

[thorn jay]

Reason for that is that i always see S_n in the context of Latin squares, so order n and S_n are always paired together

[white oxide]

If F is a field and {p_i} is a finite set of irreducible polynomials is the ideal generated by the p_i maximal

[coral spindle]

Polynomials in F[x]?

Well if you have two distinct irreducible polynomials, they are coprime, so...

And Bézout does work in F[x]

[white oxide]

sorry bro ngl i forgot all my algebra 😂

so yes then right

[coral spindle]

Hm well is F[x] a maximal ideal of F[x]?

If you have two coprime polynomials in an ideal, then you have 1 in the ideal.

(Because of Bézout)

[south patrol]

What if the set is empty

Jk

[tough raven]

For ✨ reasons ✨, the determinant is supposed to flip sign when two columns are interchanged. This essentially forces the use of the sign factor because any permutation is a product of transpositions and a product of one (-1) for each transposition gives the sign of the permutation.

[crystal vale]

I done this exercise but I don't get the meaning of remarks here, the sets V(E) satisfy the axioms for closed sets in a topological space.

Oh so complement of V(E) are open set

Okay I done i) but what's the point of this i) exercise?

[kind temple]

for any two basis elements B1, B2 and for any x in B1 n B2, there should be a B3 contained in B1 n B2 with x in B3

[crystal vale]

Here in v), quasi - compact, isn't the same as compactness?

[kind temple]

i) shows that the intersection of two basis elements is again a basis element

[crystal vale]

I got it thank you ❤️

[crystal vale]

Yes

[kind temple]

somewhere in the text, do they define compactness differently?

[chilly ocean]

The exact definition of compactness varies some, especially in older texts iirc

[kind temple]

yea

[chilly ocean]

What they call quasi-compactness is mostly known nowadays as compactness.

[crystal vale]

No

[kind temple]

hmm. not sure then. what text is it?

[crystal vale]

Atiyah

And to show $X_f$ are basic open sets we have to show that $X \subseteq \bigcup_{f\in A} X_f$, so we can take $ f = 1 $

[cloud walrus]

Notknow🙇

[crystal vale]

Is it correct?

[kind temple]

i believe some texts, e.g., heartshorne, use compactness to mean compact + hausdorff, while quasi-compact is just compactness

[crystal vale]

Yes

[kind temple]

this makes sense in the given context, since the zariski topology is not hausdorff

[crystal vale]

Oh

[crystal vale]

This one?

[kind temple]

yea, by iii)

[crystal vale]

Yes

Any hint how I can show v)?

How can I show f_i generates the unit ideal?

[kind temple]

assume for contradiction that the f_i’s don’t generate 1

then there is a prime ideal containing the ideal generated by the f_i’s…

[rocky cloak]

Sn = Mn(F1) right

[slate fulcrum]

this is in hoffman kunze Linear algebra page 179, I want to check if the sgn(sigma) formula they wrote is correct, I am pretty sure it should be (-1)^{sigma(1)+...+sigma(r)+(r(r+1))/2}

because it take sigma(i)-i adjacent swaps to move sigma(i) to the ith position

[rocky cloak]

Yes, I guess the easy check is that your formula give the right answer when sigma is the identity.

[slate fulcrum]

yeah pretty sure hoffman's gives -1 when r=3 😂

[rocky cloak]

But you know, up to sign it's correct

[slate fulcrum]

ok cool thanks for confirming, ig this was never listed in the list of errors page for hoffman and kunze on stackexchange

[south patrol]

@rocky cloak lol have you seen the fun way to define sgn using graphs

[ruby ingot]

Hey there, how can I show, that the only units in Z[sqrt(-5)] are plusminus 1? I heard from an approach using something like a norm function, but we didnt introduce that in the lecture, so I am looking for another way to proof this 😅
I tried it like that but the b != 0 case does not work out in this case

[south patrol]

It's a bit easier if you note that it is easy to write down the inverse within Q(sqrt(-5))

You then just need to check if that is in Z[sqrt(-5)]

This helps avoid tedious calculations

[ruby ingot]

like so?

[rocky cloak]

Don't think so

[ruby ingot]

Is this argument valid?

bc a^2 and b^2 are positive and b^2 >= b in Z

[south patrol]

Ye though I wouldn't say it obviously implies 1/a is an integer

It's more that the numbers above have the wrong magnitude unless a = +-1 and b = 0

[ruby ingot]

Okey, thanks 😄

[elfin wraith]

Honestly feel like you deserve a thank you for helping me get these @rocky cloak, massive amount of help to me last semester

[crystal vale]

got it

[wooden rain]

Congrats!!

[twilit wraith]

Were these grad level courses? They look interesting

[tardy hedge]

There are many goats in this server for sure

[elfin wraith]

Topics in ring and representation theory was, math10 is for final year UG and math11 is for masters level stuff

[twilit wraith]

Nice

[tardy hedge]

Random question: When talking about the disjoint union of sets X and A, is that always equal to X x {0} U A x {1} by definition?

subset of (X U A) x {0,1}

[frigid epoch]

yesno

that definition can be used in all cases

it gives precisely a disjoint union

[tardy hedge]

My context is this:

(A)

ik wrong channel sorry

[frigid epoch]

yeah you can use the definition with {0} and {1}

[tardy hedge]

Thx

[frigid epoch]

nothing wrong with it other than that it's a bit long to notate sometimes

[hidden wind]

share plz

[coral spindle]

I’d want to see this too

[arctic trail]

does anyone have an example, which is not S_3, of a solvable group with only one minimal normal subgroup?

[coral spindle]

Haha well you won't like it, but I have infinitely many examples...

Z/p^2Z

[arctic trail]

Apologies I meant specifically not p-groups

[coral spindle]

As for noncommutative, I think you should be able to find examples amongst the dihedral groups perhaps?

Am I barking up the wrong tree here

[arctic trail]

and I want like two different primes for the order of the minimal normal subgroups

[elfin wraith]

Is there a group theory version of DaRT? I.e. somewhere you can search for known groups with a combination of properties?

[coral spindle]

Like doesn't Dih(2p) work in general?

S_3 is just Dih(6) after all

[arctic trail]

Dih meaning Dihedral group?

[coral spindle]

Indeed

Forgive me if I'm wrong, I haven't checked details

[south patrol]

Exceptional isomorphism

[coral spindle]

Potato explainnnnnnn

Nevermind I get it

[soft tiger]

Are rules (3) and (4) pictured here supposed to be immediately obvious

[coral spindle]

You can calculate them to be the same 3-cycle on both sides very easily

I'm not sure if they're immediately obvious, but they're extremely easy to verify.

[soft tiger]

Right

[rocky cloak]

For any prime, the semidirect product of Z/p and Aut(Z/p) should work I think

Also dihedral group like Boytjie said

[coral spindle]

I guess the dihedral group is just a proper subgroup of what you mentioned, so it's a lesser example

[candid patrol]

Hom

[rocky cloak]

A4 should be another

[arctic trail]

damn

[tough raven]

Let $L = K(a) = K[X]/(p(X))$ be a simple field extension. The map $f \colon E \mapsto \operatorname{min}_{E}(a)$ from intermediate extensions $L/E/K$ to polynomials over $L$ dividing $p(X)$ and divisible by $X - a$ is injective and satisfies $E_1 ⊆ E_2 \implies f(E_2) \mid f(E_1)$ (in fact, $E_1 ⊆ E_2 \iff f(E_1) \in E_2[X]$, see https://math.stackexchange.com/a/3960876).
\begin{enumerate}
\item Is the converse true? That is, does $f(E_2) \mid f(E_1)$ imply $E_1 ⊆ E_2$? (Equivalently, does $f(E_2) \mid f(E_1)$ imply that $f(E_1)$ also has coefficients in $E_2$?)
\item What is the range of $f$? In particular, if $p$ is separable, is $f$ surjective? In general, is the range of $f$ all factors of $p$ such that all roots have the same multiplicity, equal to some power of the characteristic exponent?
\end{enumerate}

[cloud walrus]

Raghuram

[tough raven]

https://math.stackexchange.com/a/3960876 for easy clicking.

[toxic zephyr]

okay i feel like i'm going a bit crazy. consider $R=\bZ[x]/(x^2+5)$ and $\mu=\alpha+\beta x\in R$. let $\bar N\defeq N(\mu)=\alpha^2+5\beta^2$. if $\beta$ is coprime to $N(\mu)$, then $\beta^{-1}$ exists mod $\bar N$.

we can define the homomorphism $\phi$ from $R\to\bZ/\bar N\bZ$ via
$$\phi(1)=1,\quad \phi(x)=-\alpha\beta^{-1}$$
then the kernel of $\phi$ is precisely the ideal generated by $\mu$. thus, $R/(\mu)\cong \bZ/\bar N\bZ$ by the first iso theorem.

would this prove that if $\beta$ is coprime to $\alpha^2+5\beta^2$ then $\alpha+\beta\sqrt{-5}\in\bZ[\sqrt{-5}]$ is prime if and only if the norm $\alpha^2+5\beta^2$ is prime in $\bZ$?

[cloud walrus]

eigentaylor (STfFGMOaPID)

[toxic zephyr]

in my class, we had a question on our midterm where we showed that 6+sqrt(-5) is prime in Z[sqrt(-5)] step by step, and it very much came down to the primality of 41=N(6+sqrt(-5)) in Z and the equivalence of divisibility of a-6b by p and a+bsqrt(-5) by p. but like... this seems like a much more concise way to argue it. in that case we just have
phi: Z[sqrt(-5)]->Z/41Z via phi(a+bsqrt(-5))=a-6b mod 41.
which seems to tie it all together much more nicely? like... is this right?

[rocky cloak]

So it's certainly not surjective. Like if the degree of L is prime, then there are no intermediate extensions.

[rocky cloak]

It's right, but I guess there is some work in showing well definedness and that the kernel actually is (6+sqrt(-5))

[toxic zephyr]

good to know it's right. yeah i feel it's more obvious with the
phi(a+bx)=a-b(alpha beta^-1) mod (alpha^2+5beta^2) construction

[white oxide]

What's a way to tell if a degree 2 polynomial in F[x, y] where F is a field is irreducible

[rocky cloak]

Eisenstein + Gauss lemma is always something to try. But in general I think it's difficult

[white oxide]

Got it, thanks

[rocky cloak]

For degree 2 you can actually just brute force

But for bigger degrees it gets more anoying quick

[drifting sigil]

how is a two fold rotation axis different than a mirror plane?

[tough raven]

Hmm... that's not actually a problem unless there are intermediate factors. But we can ensure that if L/K is Galois cyclic. IIRC this means we either have a Kummer extension p(X) = X^p - a or Artin-Schreier extension p(X) = X^p - X - a.

I can't think of an obvious further condition to impose on the factors that will fix these.

[tough raven]

The former fixes a line (the axis) and inverts a plane (normal to the axis), while the latter fixes a plane (the mirror) and inverts a line (normal to the mirror).

[tardy hedge]

in (c), X^A is the collection of set functions A->X? So A^X is then the collection of set functions X->A?

[thorn jay]

Yes

It's fairly annoying notation for Hom_Set(X, A)

In general for any locally small category C, Hom_C(-, Y) is a contravariant functor from C to Set

Similarly, Hom_C(X, -) is a covariant functor

That's just exercise (b)

This gives a so-called bifunctor
Hom_C(-, -) : C^op x C -> Set

[tardy hedge]

A bifunctor is a functor on some "product" category?

[thorn jay]

Ye

(Here with C^op and C cuz the first variable is contravariant)

Everytime you've got something with the first variable of the Hom-functor there will be some kind of contravariance

[tardy hedge]

im not sure why u do C^op x C

[coral spindle]

A morphism f : A -> B becomes f^* : Hom(B, X) -> Hom(A, X). Note the change of direction.

That's why you need the C^op

[thorn jay]

It's kinda weird to think about at first

But eventually it starts to make sense

[kind temple]

you can leave it as C, its just convenient to have C^op there so that you dont have the bifunctor being covariant in one component and contravariant in the other

[thorn jay]

I like my functors covariant until proven contrarily

[proper jolt]

im trying to do this question from hungerford

anyone know what $E_1E_2$ means? is it like the field formed from taking the product of element from $E_1$ and $E_2$?

[cloud walrus]

thebirdsandbees

[prisma ibex]

compositum

usually E_1 and E_2 are taken to be subfields of a field E, then E_1E_2 is the smallest subfield of E containing both E_1 and E_2

you can write this using fields of fractions like E_1E_2=E_1(E_2)=E_2(E_1)

[proper jolt]

oh ok ok nice thanks

[tough raven]

Incidentally, the set of sums of products of elements of E1 and E2 is the smallest intermediate ring including both E1 and E2. It is equal to E1 E2 if and only if it is a field (in general E1 E2 is its field of fractions), which happens if (but not only if) one of E1, E2 is algebraic.

[true drum]

Hope I'm not intruding on a current discussion here. As an exercise I've been trying to classify all of the commutative rings R with the property that every subring of R is a (two-sided) ideal of R. when I got stuck I looked on stack exchange and found this post which was just baffling to me. https://math.stackexchange.com/questions/2262859/ring-such-that-every-subring-is-an-ideal#comment10801495_2262859
How can it be that Z and Z_n are the only rings with this property? Even if the class is reduced to unital rings and subrings which contain the identity 1_R, it seems like other examples exist like the finite fields F_p. What the heck am I missing? Thanks in advance

[tough raven]

The only ideal containing 1 is the whole ring.

So a ring such that every subring is an ideal must have the property that any subring (containing 1) is the whole ring.

In particular, the additive subgroup generated by 1 is a subring (the image of the unique (unital) ring homomorphism from ℤ), so it has to be an ideal, so it has to be the whole ring. That is, the ring is generated as an additive group by 1.

This implies that the ring must be ℤ/nℤ for some non-negative integer n.

[true drum]

Oh my gosh I feel like a goob. In the post, Z_n is denoting the quotient Z/nZ, not the direct product of Z with itself n times, huh

[tough raven]

Let a_n = sigma^{-d_n}(leading coefficient of f_n), i.e., a_n is the leading coefficient when f_n is written in the form ∑_{0 ≤ i ≤ d_n} X^i a_i. Then you can use
f_{m+1} - ∑_{i = 0}^m sigma^{d_{m+1}}(r_i) X^{...} f_i.

Do you know if the result is supposed to be true when sigma is not invertible?

IIRC it is not unless you pick between Xr = sigma(r) X + delta(r) and rX = X sigma(r) + delta(r) carefully.

[tawdry venture]

I can't think of any counterexamples though...

[tough raven]

The idea is along the right lines, although it's not a proof in and of itself.

[tough raven]

Try subrings of ℚ.

[tawdry venture]

I fail to see how any subrings of Q would fail the closure under multiplication. I was thinking of some wonky matrices

since when you add the elements from A and B, they will eventually share the same denominator

[sly crescent]

What subrings are you thinking of?

[next obsidian]

What is 1/2•1/3

[tawdry venture]

diagonal and/or upper triangle matrices

within the ring of 2x2 matrices

usually those mess up at the multiplication part

[next obsidian]

Wait this doesn’t work elegiggle

ChmonkaS

Turns out 1/2 - 1/3 = 1/6

[chilly ocean]

This is a little late but

is this correct?

[quartz wind]

theres an example with Z i believe

[tawdry venture]

If you add any 2 elements of Z, you get another element in Z. If you multiply any 2 elements in Z, you still get another element in Z.

thus it is still closed under addition and multiplication

oh wait I wonder if I can do quotient rings

It just bothers me knowing the statement is false but I cant come up with any counterexamples

[quartz wind]

wait does it have to proper

[next obsidian]

I came up with a counterexample with subrings of Q

[nimble folio]

I have a question

If $A,B \in M_{m,n}(\mathbb{C})$ and $v \in \mathbb{C}^n$, does $(A+B)v = Av + Bv$?

[cloud walrus]

clubsoda14

[nimble folio]

Nvm I think I convinced myself that this is true

[thorn jay]

Yes, matrix multiplication is just dot producting with the rows, and the dot product has this distribution property

[nimble folio]

That is true

I remembered that the matrices are just linear transformations from C^n to C^m

Pointwise addition

[lusty marlin]

This is true of matrices with entries in any field

Not just ℂ

[nimble folio]

👍

Thank you thank you

[chilly ocean]

Would this be the right channel for discussing crossed modules and crossed complexes?

[mint seal]

perhaps #advanced-algebra but eh I wouldn’t sweat it, no clear line separating these two channels

but I dont even know what those things are

sounds like homological algebra or something

[rocky cloak]

This is correct yes, but you're maybe missing the conclusion

[rocky cloak]

The easiest examples should be those with few relations. So maybe think of subrings of Z[x, y]

[frigid epoch]

Nonabelian groups are fucking weird

[ruby ingot]

Hey, there! Could maybe someone verify my tables? The assignment was to show that Z3[x]/(x^3+x^2+2) is a field, depict its + and * tables and find the inverse of alpha + 1

[prisma ibex]

any ring

[tawdry venture]

That’s a good idea! Thank you!

[elfin wraith]

Yes but that makes them more interesting!

[frigid epoch]

tbh i just can't handle normal subgroups all that well, i lack intuition for them

maybe if there were a group conjugacy index, with normals having index 1, it'd be a bit more chill

and you wrote that down very very frequently (for ALL subgroups, not just normals) to hammer it in your head

apparently already S4 has a normal klein subgroup, but also three nonnormal klein subgroups which are conjugate to each other 😁

[lusty marlin]

Ah right

[crystal vale]

let A be a Boolean ring and let X = spec(A), how can i show for each f in A, the set X_f = X\V(f) is closed ?

[rocky cloak]

I mean there is.

Like you can count the number of conjugates a subgroup has, or the index of its core.

In both cases a group is normal iff the number is 1

[frigid epoch]

yeah it should just be communicated more and mapped out for more examples

[rocky cloak]

Groupprops usually lists it as "size of conjugacy classes" in their tables for subgroups

[frigid epoch]

in studies, i mean

not on some site you aren't told about haha

[rocky cloak]

Well I can absolutely recommend groupprops for numerical facts about small groups

[frigid epoch]

mhm

[cloud walrus]

.enpeace_music

[elfin wraith]

Is there a groups version DaRT btw? Is groupprops the closest thing there is?

The ability to search for combinations of properties on DaRT is really cool but I don’t think you can do that on groupprop

But I could also just be dumb/blind

[crystal vale]

got it

[thorn jay]

I just realised this is isomorphic to the group of automorphisms \psi where \psi(x) = f(x) • x for some homomorphism f : G -> Z(G)

[tough raven]

Well, you don't know it's false yet.

[chilly ocean]

Since H and K are subgroups, $a’^{-1}a\in{H}$ and $b’b^{-1}\in{K}$

Oh yeah also forgot to write a, a’ and b, b’ belong to H and K respectively lol

[cloud walrus]

NAT Enthusiast

[toxic zephyr]

Let $R=\mathbb{Z}[j]$ for $j^2+5=0$, and suppose $p=\alpha+\beta j$ is such that $\bar N=N(p)=\alpha^2+5\beta^2$ is prime in $\mathbb{Z}$. Let $r\equiv-\alpha\beta^{-1}\bmod{\bar N}$. Since $\bar N$ is prime, then $\beta$ has an inverse mod $\bar N$ and $p$ is irreducible. We know that if $p$ is prime in $\mathbb{Z}[j]$, then $N(p)$ is prime in $\mathbb{Z}$. We aim to prove the converse.

We can define a ring homomorphism $\phi:R\to S$ where $S=\mathbb{Z}/\bar N\mathbb{Z}$ via
$$\phi(1)=1,\quad \phi(j)=r\implies \phi(a+bj)=a+br$$
By construction of $r$, we have $p\in\ker(\phi)\implies(p)\subset\ker(\phi)$.
$$\phi(p)=\alpha+\beta(-\alpha\beta^{-1})=0\bmod{\bar N}$$
In fact, $\ker(\phi)\subset(p)$ as well. $a+bj\in\ker(\phi)$ if and only if
$$a=b(\alpha\beta^{-1})+\bar Nk\iff a+bj=p(b\beta^{-1}+(\alpha-\beta j)k)$$
$$\implies p\mid a+bj\implies a+bj\in(p)$$
Further, this is clearly a ring epimorphism, so by the first isomorphism theorem, $R/(p)\cong \mathbb{Z}_{\bar N}$ which is a field (integral domain) so $(p)$ is prime so $p$ is prime.

Therefore, $p\in\mathbb{Z}[j]$ is prime if and only if $N(p)$ is prime in $\mathbb{Z}$.

[cloud walrus]

eigentaylor (STfFGMOaPID)

[toxic zephyr]

is this proof correct?

and, if so, for what other j does this proof work for? like the key is that N is a multaplicative norm function. so i assume j^2<0.

but like what else?

[wraith cargo]

there are three possibilities essentially

and they depend on the legendre symbol

either your prime is inert (which happens in your case), it splits or it ramifies

oh wait

agh okay I have different notation in mind

I thought p was a prime in Z

[wraith cargo]

anyways if N(p) is prime then p is prime
But the other direction doesn't hold generally
It can happen that N(p) is the square of a prime for some prime p

[rocky cloak]

The direction you've proven is correct, but p can be prime without N(p) being.

Like p=7 for example.

[toxic zephyr]

ohhhh uhhhh what if i specify that alpha and beta are both nonzero

but good point. i guess N(p) prime implies p prime is good enough

[wraith cargo]

maybe 2+j is a counterexample?
I think that's prime and N(2+j)=9

[toxic zephyr]

wait no

uhhh
(2+j)(2-j)=9=3(3)
2+j divides 3^2 but 2+j doesn't divide 3

[wraith cargo]

Yeah your ring isn't a UFD!!!

[rocky cloak]

So the intersection of a prime ideal with Z is again a prime, so for a+bj to be prime there must exist c + dj with (a+bj)(c+dj) prime in Z.

Taking norm yields that c+dj must have norm 1, hence is a unit. So either a+bj has norm a prime or a+bj is a prime in Z times a unit

And for this ring in particular ±1 are the only units.

[chilly ocean]

Hello! I wanted to ask for some guidance with the following question: Suppose that a group $G$ has at least three elements of order $4$. Can $G$ be cyclic? What if $G$ has exactly two elements of order $4$?

[cloud walrus]

s

[chilly ocean]

Not quite sure how to proceed with this, but I know that for $G$ to be cyclic, there must be an element of order $20$, i.e. every other element can be written as a power of this element.

[cloud walrus]

s

[chilly ocean]

I think the key is that $G$ has three or more elements of order $4$. But I'm just not sure how this relates to $G$ being/not being cyclic.

[cloud walrus]

s

[wraith cargo]

there's a condition on orders for G to be cyclic

[chilly ocean]

I don't think we've covered the Euler totient function as of yet

[rocky cloak]

Say G is cyclic of order n generated by g.

Can you see what the order of g^d is in terms of d and n?

Or more specifically what must happen for g^d to have order 4?

[chilly ocean]

the order would be ceil(n, d)? or lcm(n, d)?

no wait

that's stupid sry

[rocky cloak]

It's not stupid, but it's not quite right

[chilly ocean]

i'm tempted to say ceil(n/d)

since n/d is not guaranteed to be an integer

[rocky cloak]

So let's go specific.

How could g^d have order 4, or what does that even mean?

[chilly ocean]

that means g^{4d} = id

[rocky cloak]

So n divides 4d

[chilly ocean]

i have a feeling it's gonna be something like $n/lcm(n, d)$

[cloud walrus]

s

[chilly ocean]

the order of g^d

or gcd

[rocky cloak]

And we can say a little more.

Like for example g^0 does not have order 4 even though g^4*0 = id

[chilly ocean]

yeahj

[rocky cloak]

Well maybe test with a few cyclic groups and see if any of your guesses seem correct.

Then you can try to prove it

[chilly ocean]

i have a feeling that this will have to do with subgroups since this is the chapter on subgroups

specifically cyclic subgroups

what if we have three cyclic subgroups, one generated by a, b, and c all of order 4

[rocky cloak]

Yeah, so cyclic groups do have a very nice property when it comes to subgroups.

Namely ||a cyclic group of order n has exactly one subgroup of order d for each d that divides n||

[chilly ocean]

i did not know that

should i try to prove it?

[rocky cloak]

Sure

[chilly ocean]

so it doesn't feel like i'm using it for granted

or maybe i can use lagrange's theorem

[chilly ocean]

like if we have three groups: {a, a^2, a^3, a^4 = id}, {c, c^2, c^3, c^4 = id}, {d, d^2, d^3, d^4 = id}

[rocky cloak]

Well Lagranges theorem tells you the order of subgroups divides the order of the group.

This is like a converse to that for cyclic groups. Whenever you have something dividing the order you also get a subgroup

[rocky cloak]

But yeah for your original question, Lagrange tells you that n must be a multiple of 4 if you have an element of order 4.

So then g^(n/4) would be one element. Then you have to figure out if there can be any more

[chilly ocean]

i think i got it from here, thank you @rocky cloak 💟

[white oxide]

We argue by contradiction. Let $I \subseteq R$ be an ideal that is not finitely generated. Let $a_1 \in I$. Since $I$ is not finitely generated, there is $a_2 \in I \setminus (a_1)$. Similarly, there is $a_3 \in I \setminus (a_1, a_2)$. Continuing on in this fashion, we obtain an infinite ascending chain of ideals $(a_1) \subsetneq (a_1, a_2) \subsetneq (a_1, a_2, a_3) \subsetneq \dots$, which contradicts the assumption that every ascending chain of ideals in $R$ is finite.

Is this correct?

[cloud walrus]

okeyokay

[south patrol]

yeah that's great

[white oxide]

Thanks

[keen badge]

Let $X$ be a commutative $F$-algebra. To prove that $X$ is a field, what conditions should I prove? Only the inverse of multiplicity, or is there smth else?

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[thorn jay]

A commutative F-algebra is a commutative ring with an injection h : F -> X

so yes, only inverse of multiplication for non-zero elements

[keen badge]

Nice, thanks

[tawdry venture]

Does adding two subrings of a ring R always yield a subring within ring R?

for context - subrings require to be closed under addition, multiplication, negation, contains 0 and 1 element

[south patrol]

What does "adding" mean

[south patrol]

Sure okay just that

[tawdry venture]

I did subring of Z[x] and Z[y] but that also doesn't work

[somber bluff]

A homomorphism of a ring does not necessarily map identitys to each other but an isomorphism does. Correct?

[south patrol]

For most people, ring homomorphisms send identity to identity by definition

[somber bluff]

how do you prove all automorphisms of complex numbers send i to i or -i

[lusty marlin]

A good place to start is seeing where real numbers get sent under any automorphism of ℂ

[somber bluff]

f sends 1 to 1 and thus -1 to -1

implying f(i) = i or -i

right?

[kind temple]

this is as a field homomorphism, correct?

[somber bluff]

yes

[lusty marlin]

Yes

[tawdry venture]

If I know something is an ideal, can I conclude that it is closed under negation or no?

[kind temple]

negation, as in, the additive inverse?

[lusty marlin]

Yes

[tawdry venture]

yes additive inverse

so I wasn't sure if I could conclude it automatically has addictive inverse

[woeful sage]

quite addictive

[tawdry venture]

it's closed under addition, multiplication and has 0 element

[lusty marlin]

You should try and prove it then

[tawdry venture]

nothing about addictive inverse there

I noticed some people believe rings should not have multiplicative identity and some do

does it matter?

[woeful sage]

I don't see how that's relevant to the existence of additive identity in an ideal

hm wait it might