groups-rings-fields

[novel parrot]

Yeah

[maiden ocean]

Its going to be sent to (1)

like thats its extension in A_p

[novel parrot]

ah

[maiden ocean]

unless im unclear, i think youre referring to a prime ideal properly containing p right

[novel parrot]

yes

makes sense

[maiden ocean]

such an ideal is going to contain some s in A - p yeah

and then the image of s under the localization map is a unit

[novel parrot]

yea

yah

[maiden ocean]

[novel parrot]

So its the opposite of quotienting

[maiden ocean]

exactly

[novel parrot]

very cool

[cloud walrus]

Spamakin🎷

[barren sierra]

is that correct?

[mild laurel]

Yes

[barren sierra]

dope

[latent anvil]

let $\varphi : A \to B$ be an epimorphism of rings. what can we say about $\varphi$? can it be decomposed into a composition of localizations and quotients?

[cloud walrus]

goblin shamrock

[latent anvil]

inspired by @maiden ocean

thoughts: let $I = \ker \varphi$ and let $S = { x \in A : \varphi(x) \text{ is a unit } }$

[maiden ocean]

Poggers

[cloud walrus]

goblin shamrock

[maiden ocean]

oh

this is an MO question

https://mathoverflow.net/questions/109/what-do-epimorphisms-of-commutative-rings-look-like

i read this trying to fix the proof

[latent anvil]

oh nice

sorry i got disracted

ahh the geometric pov

i should have considered this

your question was about $B \otimes_A B \cong B$?

hmm

[cloud walrus]

goblin shamrock

[latent anvil]

let $Y = \mathsf{Spec} A$ and $X = \mathsf{Spec} B$, we're saying that the diagonal map $X \to X \times_Y X$ is an isomorphism

hmmm

[cloud walrus]

goblin shamrock

[latent anvil]

assuming that the map $X \to Y$ is a monomorphism?

[cloud walrus]

goblin shamrock

[latent anvil]

hmmmm

oh wait maybe this is category theory

it totally fucking is lmao

@maiden ocean let $\mathsf{C}$ be a category with pushouts and suppose $f : A \to B$ is an morphism in $\mathsf{C}$. Then the canonical map $B \amalg_A B \to B$ is an isomorphism iff $f$ is epi

[cloud walrus]

goblin shamrock

[latent anvil]

for the direction we care about, assume $f$ is epi

[cloud walrus]

goblin shamrock

[latent anvil]

[\begin{tikzcd}
A & B \
B & B \
&& C
\arrow[two heads, from=1-1, to=2-1]
\arrow[two heads, from=1-1, to=1-2]
\arrow["id"', from=1-2, to=2-2]
\arrow["id", from=2-1, to=2-2]
\arrow["\beta"', from=2-1, to=3-3]
\arrow["\alpha", from=1-2, to=3-3]
\end{tikzcd}]

[cloud walrus]

goblin shamrock

[latent anvil]

by contemplating this diagram we see that $\alpha = \beta$

[cloud walrus]

goblin shamrock

[latent anvil]

since they agree when precomposed with an epimorphism

[scarlet estuary]

kevin buzzard malding rn

[latent anvil]

so we can get a map $B \to C$ by using $\alpha = \beta$ and the diagram commutes

lol

[cloud walrus]

goblin shamrock

[latent anvil]

does he like labeled diagrams?

[scarlet estuary]

kevin buzzard doesn't like. he just dislikes.

[latent anvil]

seems right

[terse crystal]

It’s unlikely to be an isomorphism since the kernel is the ideal generated by elements of the form $b\otimes 1 - 1 \otimes b$

[cloud walrus]

Cogwheels of the mind

[maiden ocean]

Its not an arbitrary A-algebra

its f: A -> B an epimorphism

[maiden ocean]

im guessing we take C = B otimes_A B?

and let alpha and beta be the inclusions in the first and second coordinate

[latent anvil]

ah sure

[maiden ocean]

well

hm

[latent anvil]

but I was showing the universal property

for B

[maiden ocean]

for epis yeah

[latent anvil]

wym?

I was showing that B satisfies the universal property of a pushout of A -> B and A -> B

[maiden ocean]

Uh isnt that what it means for A -> B to be an epi pretty much

maybe i am confused

[latent anvil]

well yeah lol, that's the point

this shows that A -> B epi implies that the map B \coprod_A B \to B is an isomorphism

because B satisfies the universal property of B \coprod_A B

[maiden ocean]

Oh lmfao im dumb

Right

[latent anvil]

that's what you were asking about earlier right?

bc pushouts in CRing are tensor

[maiden ocean]

Right yeah i see

[latent anvil]

sorry i initially thought of this for pullbacks & schemes and so while dualizing back to ring I kind of forgot to explain what I was doing lol

but yeah this is just category theory

[maiden ocean]

Ok makes sense

[next obsidian]

I knew the epimorphism of ring thing hahaha

Owned!

Chmonkey flexes on a HSer again!

[vestal snow]

Can someone explain what it means by "where N_i denotes the trace acting on..."?

[terse crystal]

I haven’t learned that far but I know that H^i (X) is a linear space over F_p

[vestal snow]

Yeah I know that

Wait I think I get it

[terse crystal]

Therefore π induces a linear map from H^i (X) to itself

[vestal snow]

A map of curves induces a map on the cohomology modules

Is this it?

[terse crystal]

Yes. the linear map between cech cohomology groups induced from morphism between algebraic varieties

[vestal snow]

Got it

Thanks!

Also, if anyone would be interested in reading through Serre's Rational points on curves over finite fields, let me know.

[next obsidian]

Does anyone have an idea on what to do here?

Suppose you have an $\mathbb{N}$-graded Noetherian ring $R$ with $R_0 = k$ a field and let $m = R_1 + R_2 + \cdots$ the irrelevant ideal. If $R_m$ is an $n$-dimensional regular local ring then $R$ is a polynomial ring $k[y_1,\ldots,y_n]$ where the $y_i$ are homogeneous of positive degree.

So I can take $n$-generators for $m$ because $R_m$ is an $n$-dimensional regular local ring, and then it’s easy to show that these generators are algebraically independent over $k$ so we can write $R = k[y_1,…,y_n]$ but the issue is the generators of $m$ don’t need to be homogeneous. I don’t think any homogeneous ideal necessarily has a minimal generating set by homogeneous elements and so I’m kinda unsure what to do lol.

[cloud walrus]

Chmonkey

[terse crystal]

I am thinking that you can choose those homogeneous y_i having degree 1 since any linear basis y_1+m^2,…,y_n+m^2 of m/m^2 over A/m where A is a regular local ring with maximal ideal m ( in this case A=A_m and m=mA_m), y_1,…,y_n are generators of m

So maybe you can just cancel the monomials of degree greater than 1 in those original y_i since monomials of degree greater than 1 are contained in m^2

[next obsidian]

I don’t think you can get degree 1 in general

I’m almost certain of that

[next obsidian]

This isn’t true, m isn’t generated by stuff in degree 1

That’s an additional assumption you need to impose (commonly imposed by say Hartshorne for stuff involving Proj)

As an example you can quite literally just take a polynomial ring and weight the different variables differently. So like k[x,y] with x in degree 1 and y in degree 2. Then y is a monomial of degree 2 not in m^2

[terse crystal]

Yeah I am wrong

Wait if you have some generators of m, then replace those generators with their homogeneous parts or even with their monomials then they still generate m

Like replacing x_1+x_2 with x_1, x_2

[next obsidian]

But then it isn’t minimal anymore

The number is important

[terse crystal]

Eventually there will be n elements remained

[next obsidian]

I’m not sure how you expect the number to stay the same?

Or to be able to reduce the number of elements

When you take an inhomogeneous element and add in all its homogeneous parts you might increase the number of generators

How do you go back down?

[terse crystal]

Any set of generators y_1,…,y_k of m, y_1+m^2,…,y_k+m^2 span m/m^2 over A/m then choose the basis from them

[next obsidian]

-_-

Fuck

Lmfao

Right

Gaaahhhhhhh

That’s really simple

[terse crystal]

Using Nakayama

[next obsidian]

Yeah I used it earlier to lift generators in the first place

I just forgot that’s a vector space LOL

So it does it automatically for you

[terse crystal]

LOL

[next obsidian]

I was trying to like

Do the homogeneous crap up in R

But over A/m it’s just linear algebra

☠️☠️

Thanks, I was going crazy lol

[terse crystal]

😂

[next obsidian]

I was like

Writing down realllyyyy messy stuff to try and like reduce to homogeneous stuff manually

And it seemed impossible lol

Way too many coefficients

[terse crystal]

Yeah…

[novel parrot]

we dont really need the sequence to be exact?

any module homomorphism is enough no?

or is it to keep the assencidng chain in M

[chilly ocean]

Im 99% sure module hom is not enough but wait for moldi or chmoke

[novel parrot]

ig in any module homo

the chain may not be kept

[chilly ocean]

ye

that's my intuition kinda

[upper pivot]

You need exact for the <-

[hidden haven]

M' → M could be trivial, then a chain in M' gives rise to the trivial chain in M but that doesn't tell you anything about the original chain

Similarly M could be the trivial module inside M", in which case chains of M" pull back to the trivial chain in the trivial module which again tells you nothing

So you don't need the whole sequence yeah, but you need either the thing to sit inside M via injection into M or be a quotient of M by having some surjection from M

[upper pivot]

exactness is roughly telling you that M decomposes into M' and M'', which is kinda why you need it. you can think of weaker conditions ofc

[novel parrot]

i didnt understand

so if we had f : M -> N

any module homomorphism

M_1 < M_2

then f(M_1) < f(M_2) is true ?

surely ... ?

[hidden haven]

No, different things could map to the same

So M_1 could be {0}

M_2 = ker f

Both have 0 image

[upper pivot]

also if you have a chain in N

[novel parrot]

but since M2 contains M1, f(M2) has atleast the image of M1

[hidden haven]

Yeah I'm saying strict inequality need not remain strict

[upper pivot]

there is no reason it is induced from a chain in M right

[novel parrot]

oh

[hidden haven]

So even if the image of a chain stabilizes

[upper pivot]

like as moldi said, take a trivial map

[hidden haven]

How do you deduce that the original does too

[upper pivot]

then non-trivial chains are not induced by the map right

[novel parrot]

so the condition that image of alpha = kernel of beta is required ?

[hidden haven]

No

[hidden haven]

This

Injection into M or surjection from M is enough

M is noetherian → all submodules and all quotients of M are noetherian

And the theorem is also saying that the converse holds

For the converse you need both conditions together, so there you'll need exactness

If you are more comfortable, you can remove exact sequences and take it as M' ⊂ M, and M" = M/M'

That's really what a short exact sequence says by first isomorphism theorem

[gritty sparrow]

The theorem is a little stronger than the converse, bc it speaks of a particular submodule quotient pair

[hidden haven]

True yeah

[novel parrot]

ah im confused

[novel parrot]

do we have this

[hidden haven]

From the exactness at M' and at M"

The map M' → M is injective

M → M" is surjective

[novel parrot]

oh righty

[upper pivot]

are just a injection and surjection enough? couldnt you map the 0 module into M and then surject it onto the 0 module

[hidden haven]

Yeah, submodules of M and quotients of M are noetherian if M is

In your example 0 is both

And is noetherian regardless of whether M is or not

For the converse you need exactness

[upper pivot]

yeah right sorry misread what u said

[novel parrot]

why is surjection enough

like beta: M -> M'' is surjective

so an infinite chain in M''

how does the inverse give an infinite chain in M

if M1 < M2

why does it remain strict if surjective

how do we know that M2 - M1 will not just be sent to 0

[hidden haven]

Correspondence theorem for submodules in a quotient

M" can be identified with M/ker surjection

So taking inverses is an injective map from set of submodules of M" to submodules of M

[novel parrot]

you mean M/ker is isomorphic to M''

?

and thats injective

i can see it working if thats what you meant

[hidden haven]

Isn't that what I said

Identified means isomorphic but you treat things as equal by abuse of notation or whatever

[novel parrot]

just trying to be clear

for converse why do we need both conditions?

if an infinite chain in M, sending it back to M', it will stabilize there, and then it would also stablize in M then?

[gritty sparrow]

Here is an example, see where your logic goes wrong: Let M be some non noetherian module, let M’ be 0 and M’’=M

M’ is noetherian obviously, but M is obviously non noetherian

[novel parrot]

yeah

ok

im wrong

[gritty sparrow]

You can come up with similar sort of constructions to show that M” being noetherian does not imply M is noetherian. You need both

[novel parrot]

"it follows that Ln is stationary"

how then?

[gritty sparrow]

Let both the chains be stable after N, say. Now taking x in L_(N+1) we see that beta(x) is in beta(Ln). So we can say that x-l is in ker(beta) for some l in Ln. but now ker(beta)=im(alpha) so x-l=alpha(m). where m is in alpha^-1(L_n+1))= alpha^-1(Ln) . But that means alpha(m)=x-l is in Ln hence x is as well

[hidden haven]

Idea is that Ln's intersection with each coset of M' in M looks the same as the intersection with M' itself (or is empty). Thus knowing which cosets it intersects + how much it intersects M' is enough

[novel parrot]

ok

next question

alpha inverse may not always exist?

[hidden haven]

For any function f: A → B, between sets or any kind of structure, if X is a subset of B, f inverse(X) is defined to be the set of things in A that map to X

[novel parrot]

oh

okay

i think i got now

ty

[hidden haven]

:catKing:

[novel parrot]

[novel parrot]

what exact sequence are they considering?

in 6.6

[gritty sparrow]

a->A->A/a

[novel parrot]

okay

was slightly confused becuz a was automatically noetherian

[novel parrot]

dont get that last part in bold

i understand everything else but not how it implies its larger than or equal to k

[novel parrot]

i dont know

do y mean l(Mk)

that is 0?

oh nvm

atiyah mcdonald making me feel real dumb lol

[topaz leaf]

that's how you know you're reading it right 🙃

[novel parrot]

[novel parrot]

is there not an overlap of 1 module here

$M'=M_0 > \cdots > M_n = 0$ and $M'' = N_0 > \cdots > N_m = 0$

[cloud walrus]

ActiveChapter

[novel parrot]

wouldnt N_m overlap with M_0

wouldnt beta inverse on N_m = alpha on M_0

nvm figured it out

[wooden ember]

bruh proving there were no simple groups of order 420 was actually kinda hard

[chilly ocean]

dude weed lmao

[urban acorn]

imagine caring about pure finite group theory

420 = 42 * 10 = (21 * 2) * (2 * 5) = 2^2 * 3 * 5 * 7

then Sylow stuff

[south patrol]

looks like there are enough prime factors you should be able to force there to be enough sylow p groups to contradict group size

[mild laurel]

I feel like you need a lot more than just Sylow stuff to do this

[lunar spruce]

if H is subgroup of (Z/nZ)*, but H is not equal to (Z/nZ) *, I know with lagrange that |(Z/nZ) *| = sumNumber * |H|, with sumNumber > 1...does that mean every strict subgroup is max half as big always?

[chilly ocean]

ye

order of the subgroup always divides the order of a group

[lunar spruce]

damn thats sick. If I can only proof one element of (Z/nZ) * not being in specific H, and I know atleast half of the elements are not in H xd

thats important for ma rabin miller prime test

[south patrol]

just looked up the problem and people did it using sylow

[wooden ember]

it's not an easy sylow though

you show there's no element of order 14

then consider the normalizer of a sylow 7 subgroup

and show it must have an element of order 14

[latent anvil]

what's the problem?

[wooden ember]

show there are no simple groups of order 420

[latent anvil]

ah

[wooden ember]

it's done but it was interesting to see it wasnt basic sylow

[latent anvil]

Okay I want to try this

No spoilers pls

[wooden ember]

at least not basic to me

aight

[latent anvil]

420 = 2^2 * 3 * 5 * 7

so n7 must be 15

If I did the mental math right

[wooden ember]

yah

[latent anvil]

15*(7-1) is sadly only 90

not there yet

n2 could be many things, all we really know is n2 >= 3

We know that |N_G(P)| = 420/15 = 28

where P is a sylow 7

n3 could be 4, 10, or 28

so n3 >= 4

n5 could be 6 or 21

n5 >= 6

just counting up the total number of elements in the 3,5,7 sylows is way too small

[wooden ember]

i did say not basic

[latent anvil]

I didn't disbelieve you!

Just want to work through all my options from simplest to least simple

[wooden ember]

of course, id do the same

[latent anvil]

We could act on the 7 sylows by conjugation

[wooden ember]

plus no information is useless

[latent anvil]

This gives map into S_15

But 7 < 15

So sadge

[small bison]

god i hate sylow

[wooden ember]

that was me once

[small bison]

he created his theorems so that grad students could struggle on algebra quals

[wooden ember]

i now love sylow

[latent anvil]

We could act on the cosets of the normalizer of the 7 sylows

Er no

We just did that

also sylow theorems are great

I haven't done a problem like this in years

feel rusty

[wooden ember]

this one may not be the best one to start on then haha

[latent anvil]

hahaha

Okay I seem to remember some trick with like

Normalizes of intersections

But the 2 sylows are small...

[wooden ember]

bruh my ps4 is ventilating like hell

i hope it doesnt explode

[latent anvil]

like, if P, Q are 2 sylows then they're abelian so P \cap Q is normalized by both P and Q

So N_G(P cap Q) contains P and it contains Q

But that's still only like, order 8 at worst

hmm

I know a trick in the case that n2 = 3

But I don't see why that would hold

[wooden ember]

what's the trick if i may know, im not familiar

[small bison]

there's also a trick with like extensions

where you can count morphisms n stuff

[latent anvil]

If |G| = p^k m, p and m coprime, and G has exactly p+1 sylow p subgroups, then their union has cardinality p^(k+1)

[wooden ember]

huh interestin

[latent anvil]

You can use this to show that, if k ≠ 1, G is not simple

[wooden ember]

is the proof of that complex?

[latent anvil]

It's tricky

Doesn't use machinery but it's not super easy

Well like machinery beyond standard sylow stuff

[wooden ember]

ill think about it at some point then

[wooden ember]

wait but the m is useless here:isnt this just saying the union of sylow p subgroups is p^(k+1) when the sylow p subgroups have order p^k? That doesnt sound right

[latent anvil]

So I guess I can assume n2 >= 5

[latent anvil]

Yeah it's just saying "let p^k be the highest power of p dividing |G|"

[wooden ember]

that really doesnt sound right, i mean the size of the intersection of sylow p-subgroups can vary

weird

[latent anvil]

it is indeed weird

The fact that there's exactly p+1 subgroups is very restrictive

[wooden ember]

oh right

that's a condition i forgot lol

[latent anvil]

Yeah haha it's extremely not true in general

[wooden ember]

yeah lmao

[novel parrot]

slightly confused about how it forms a topology

because union of closed sets are closed only if, the sets were ideals?

[latent anvil]

Oh you're wondering how (iv) is sufficient when it's defined as V(E) for any subset E?

You can show that V(E) = V(I) where I is the ideal generated by E

(since "ideal generated by E" is the same as "intersection of all ideals containing E")

[novel parrot]

ah okay thank you

[latent anvil]

Oh right, that's what (i) says

Okay back to sylow

I'm thinking

Let P be a 7 sylow

[south patrol]

also since people were talking about finite groups - i often see problems solving problems by letting a group G act on a set S (often the Syl_p(G)) and considering the induced homomorphism G -> Aut(S) and doing stuff with that that often seems like black magic - is there anywhere in particular you can learn that stuff from or do you just need to have seen loads of tricks from experience lol

[latent anvil]

little bit of column A, little bit of column B

[wooden ember]

tricks from experience i think

[south patrol]

yeah sure

[latent anvil]

Dummit and foote has a good section on sylow tricks

[wooden ember]

yeah it's the chapter im on

[latent anvil]

but I mostly learned by doing guided problems

[south patrol]

i mean i've seen a bit from Artin and yeah just little things like er

[wooden ember]

it's a really good section

[south patrol]

lol someone did showing S_6 only non-cyclic group order 6 using it as well even

[south patrol]

oh thanks i'll have a look at that

[latent anvil]

What's the section number narwhal?

[south patrol]

why do i forget the sylow proofs every time i do them lol

[latent anvil]

So I said |N_G(P)| = 420/15 = 28 I think

[south patrol]

too many similar but slightly different group actions :p

[wooden ember]

yes

[latent anvil]

This means N_G(P) contains a 2 sylow and a 3 sylow

Say a 3 sylow Q maybe

[wooden ember]

you can do a lot by just left multiplication and conjugation

[south patrol]

ye

[wooden ember]

ive never really used other actions

[latent anvil]

Then PQ is a subgroup, since PQ = QP

|PQ| = 21 since P, Q have coprime orders

[south patrol]

also 4.5 it seems

[latent anvil]

Wait

[south patrol]

has proofs and then loads of tricks

[latent anvil]

3 doesn't divide 28

lmfao

I'm so fucking bad at math holy shit

Okay so N_G(P) contains a 2 sylow Q

oh wait!

hmm

[wooden ember]

this is on the right path

[latent anvil]

How many 2 sylows does it contain

Either 1 or 7

I don't know the normalizer is simple, is the problem

in fact it can't be

By burnsides lemma

[wooden ember]

it's not simple

[latent anvil]

yeah

[wooden ember]

since P is normal in N_G(P)

no normalizer can be simple lol

[latent anvil]

Oh lmfao that too

but also by burnsides lemma

I think my proof is pretty good too

The real thing I want to know is if n2 can be 1

Oh in that case it's a direct product

wait...

It's a semidirect product of a 2 sylow acting on a 7 sylow anyways

PQ is

[wooden ember]

^^

this is good

[latent anvil]

ohh and PQ = N_G(P) by order reasons

Right?

[wooden ember]

yeah

[latent anvil]

Cool

So how does C4 or C2×C2 act on C7

Aut(C7) = (Z/7Z)^* = Z/6Z

So a potential C4 action factors through C2 anyways

the only nontrivial map C2 -> C6 sends 1 to 3

3 being uhh

6 in Z/7Z^×

So the only way for C2 to act nontrivially on C7 is by x |-> x^-1

Ofc

[wooden ember]

sorry was busy getting my ass kicked by an old man

[latent anvil]

This feels like I'm getting lost in the weeds

that's okay

[wooden ember]

this is good but wont get you anywhere without another piece of information

which requires a different approach to get

[latent anvil]

Yeah

Thinking

I mean the most important bit rn is whether the 2 sylows are cyclic

Maybe not actually important

But that's the vibes

Okay, so here's another tack

I wanted to know if n2 = 1 within N_G(P)

this is true iff N_G(P) is commutative iff Q acts trivially on P

So does the 7 sylow commute with the 2 sylow?

[novel parrot]

in a topology can we have a set be open and closed?

[wooden ember]

yes

[novel parrot]

i see

[wooden ember]

a clopen set

@latent anvil hint:||show there are no element of order 14 in G||

[novel parrot]

even in prime spectrum?

this feels counter intuitive

[wooden ember]

idk what that means

i only know raw basic topology dont ask me

[latent anvil]

sorry im taking a break to catch up on the drama in #「ivory-tower-emeritus」

I'll return to this problem in a minute

[novel parrot]

this topology ^

[wooden ember]

ooo drama

[latent anvil]

the zariski topology is very very bad

[novel parrot]

why

[latent anvil]

it will almost never satisfy the axiom "{x} is closed for every point x"

[wooden ember]

damn

[novel parrot]

by points u mean the primes in X?

[latent anvil]

yes, those are the points of the space Spec A

[novel parrot]

so why are they not closed?

[latent anvil]

try figuring out which points are closed in $\mathrm{Spec} \mathbb{Z}$

[cloud walrus]

goblin shamrock

[latent anvil]

as an exercise

[novel parrot]

okay

the points will be contained in themselves no?

(2) in spec Z

V(2) = (2) no?

super confused rn

[latent anvil]

yes, V((2)) = {(2)}

and V((3)) = {(3)}

[novel parrot]

but um 0?

[latent anvil]

yup

what's the closure of {(0)}?

[novel parrot]

closure meaning the closed sets that contain 0?

(idk topology)

[south patrol]

closure of A is intersection of all closed sets containing A (hence 'smallest' closed set containing A)

[novel parrot]

oh

ok

so closure doesnt exist?

[latent anvil]

it does exist

[novel parrot]

all of spec Z i mean

?

[latent anvil]

yes

say C is a closed set containing {(0)}

C = V(I) for some ideal I

the fact that (0) is in V(I) means I is contained in (0)

so I = (0)

so V(I) = evertything, since every prime contains 0

[urban acorn]

This seem simple enough, but I can't seem to figure it out right now, I might've taken too much downers

so suppose A is some abelian group

and the direct sum of A with B is isomorphic to A

then inductively, we can say that about the direct sum of A with any finite number of copies of B is also isomorphic to A

can we conclude this about A with countably many copies as well?

[latent anvil]

yep

oh sorry

was yeping the penultimate post

[urban acorn]

yeah lol

[latent anvil]

vibes are a yes

thinking

hmmm

okay so

say we have $\varphi : B \oplus A \to A$ an iso

[cloud walrus]

goblin shamrock

[latent anvil]

then $\varphi \circ (\mathrm{id}_B \oplus \varphi)$ is an iso $B \oplus B \oplus A \to A$

and so on

[cloud walrus]

goblin shamrock

[latent anvil]

we have these subgroups iso to B in A

are they like, nested?

no, I don't think so

no no no

they're part of different direct summands

we can define a map from the infinite direct sum into A

right?

like, call our sequence of isos $\varphi_k$

[cloud walrus]

goblin shamrock

[latent anvil]

defined inductively by $\varphi_{k+1} = \varphi \circ (\mathrm{id}_B \oplus \varphi)$, I guess

[cloud walrus]

goblin shamrock

[urban acorn]

It's gonna be a chain of subgroups, each a proper (assuming B is nontrivial) subgroup of the previous, and each isomorphic to A, with a complement isomorphic to B

[latent anvil]

right

well, with a complement isomorphic to B^n

[urban acorn]

wait, right

yes

[latent anvil]

so we can define a map from the infinite direct sum into A

[urban acorn]

B^n

[latent anvil]

right?

[urban acorn]

that's what I'm thinking

[latent anvil]

by saying like, send the nth copy of B to the nth of these complements

I think it works if you get really nitty gritty about the isos

so does this split?

[urban acorn]

so I also thing this is true for a reason like this, but I can't make it formal rn

but the reason I'm doubting it

is that I think I used it to show something that someone on mathoverflow said is false

[latent anvil]

ah hm

interesting

[urban acorn]

but their counterexample is too involved for me to work through

[latent anvil]

can you link it?

[urban acorn]

https://mathoverflow.net/questions/218113/a-is-isomorphic-to-a-oplus-mathbbz2-but-not-to-a-oplus-mathbbz

here's the link

[latent anvil]

Sorry, how does this cause problems with your question?

[urban acorn]

so the idea is, suppose A is isomorphic to A (+) Z^2, then it should be isomorphic to A (+) Z^2 (+) Z^2 (+) ..., which is in turn obviously isomorphic to A (+) Z (+) Z (+) Z (+) ..., and adding another copy of Z to the last obviously doesn't change it

which would imply that the situation OP is asking about there never happens

but people in the comments supposedly provide examples

which are too complicated for me to follow

[latent anvil]

no, we never get that A is iso to A (+) Z

we get that $B \cong B \oplus \Z$ where $B = A \oplus \left(\bigoplus_{i=0}^\infty \Z^2\right)$

[cloud walrus]

goblin shamrock

[urban acorn]

this B is isomorphic to A

[latent anvil]

oh, right

good point lol

hmm

Yeah, this makes sense

[urban acorn]

because A + Z^2 = A, and according to the statement we were thinking is true, this implies A = A + Z^2 + Z^2 + ...

[latent anvil]

yup

so I guess your claim is wrong

[urban acorn]

yeah, so I'm trying to figure out how that is

because it really looks right

[latent anvil]

I can try to understand the counterexample if you want?

[urban acorn]

sure, go ahead

[novel parrot]

does that mean that any other open set will be contained in the union of X_f ?

[wooden ember]

not sure if you should be studying the zariski topology if you havent done regular topology 🤔

[latent anvil]

this is a very complex counterexample

[urban acorn]

more so, any other open set will be equal to the union of a collection of X_f's

[wooden ember]

i mean i dont know anything about the zariski topology but it seems rather involved

[latent anvil]

not really

it's easy to define and purely algebraic

[wooden ember]

then nvm me

[novel parrot]

yeah i just need to understand the topological definitions

[latent anvil]

most things you learn in a topology course totally fail for the zariski top anyways

like

if you spend 90% of the time on hausdorff spaces or manifolds or cw complexes or whatever

that won't help you herer

[novel parrot]

a collection or all of the X_f's ?

[latent anvil]

@urban acorn I wonder if there's a simpler example for your problem then the MSE thing

hmm

[urban acorn]

the Zariski topology doesn't quite describe the topology of a geometric object the way the topology of something like a sphere does

but there is probably an analogy to be made

[latent anvil]

hmm i don't know if I agree

I would say it describes the topology of a geometric object but not the geometry

but maybe this is just preexisting ag brainworms

[urban acorn]

it's kinda like the topology you get when forcing polynomials to be continuous

[latent anvil]

sort of, although this is the one for affine schemes and not varieties

[novel parrot]

so was it that any other open set will be equal to the union of all the X_f's or just some of the X_f's?

[urban acorn]

in practice, when your space comes in a natural topology, even more functions from your space are continuous, and so the Zariski topology is weaker than your natural topology

[latent anvil]

just some of them

specifically, all of the X_f contained in that open set

[novel parrot]

okay

[latent anvil]

the union of all the X_f's is the whole space

since X_1 = X

[novel parrot]

yea

in dummit and foote zariski topoplogy is kinda different

[latent anvil]

how so?

[novel parrot]

oh im blue now lol

[latent anvil]

there are two things which are called the zariski topology

hahaha

one for affine schemes and one for varieties

[novel parrot]

but yeah they defined it with algebraic sets

[latent anvil]

right, that's the one for varieties

[wooden ember]

eyy you're active now

[latent anvil]

oh hey

i'm no longer emeritus

i was this morning

[wooden ember]

why give them the same name

[latent anvil]

they're kind of the same thing

[urban acorn]

they're the same idea really

[wooden ember]

yeah getting back to very active from emeritus is easy

thought so much

[latent anvil]

@novel parrot you can think of this as like the only in d&f, but we add an extra point for each irreducible algebraic set

but they're weird points

like the (0) in Spec Z

they're "generic points"

and, also, it works for all rings instead of just finite type reduced rings over an algebraically closed field

but if you think about say, $A = \mathbb{C}[x,y,z,w]/(xy-zw)$

[cloud walrus]

goblin shamrock

[latent anvil]

this is some irreducible 3d surface in C^4

er, the coordinate ring of such a surface

call that surface $X = {(x,y,z,w) : xy = zw}$

[cloud walrus]

goblin shamrock

[latent anvil]

we can give X the zariski topology, the algebraic set thing you were talking about

but there's also a map $X \to \mathrm{Spec}\ A$

[cloud walrus]

goblin shamrock

[latent anvil]

do you know what the nullstensatz is?

[novel parrot]

yes

[latent anvil]

right, so what are the maximal ideals of A?

what do they look like?

[novel parrot]

points

[latent anvil]

sure, they're of the form $(x-x_0, y-y_0,z-z_0,w-w_0)$ where $x_0 y_0 = z_0 w_0$

[cloud walrus]

goblin shamrock

[latent anvil]

so we have this map $X \to \mathrm{Spec}\ A$ sending a point to the corresponding maximal ideal

[cloud walrus]

goblin shamrock

[latent anvil]

and this is a bijection onto the set of maximal ideals

in fact, it's a topological embedding

so you can think of $\mathrm{Spec}\ A$ here as just the surface $X$ with the topology we're used to from dummit and foote, but we've added in a bunch of weird extra points

[cloud walrus]

goblin shamrock

[novel parrot]

i seee

[latent anvil]

like, there's a surface ${(x,y,z,w) : y = zw, x = 1}$ contained in our space

right?

[cloud walrus]

goblin shamrock

[novel parrot]

yes

[latent anvil]

(in fact this is isomorphic to A^2)

well, this corresponds to the prime ideal $(x-1)$ of our ring

[cloud walrus]

goblin shamrock

[latent anvil]

this is prime but not maximal!

[novel parrot]

oh

[latent anvil]

and so in $\mathrm{Spec}\ A$, we get this point which is not closed

[cloud walrus]

goblin shamrock

[latent anvil]

sorry, a point of the spectrum is closed iff it's maximal

this is an exercise ig

but still

the closure of that point $(x-1)$ in $\mathrm{Spec}\ A$ is this point along with every point which lies on that surface

[cloud walrus]

goblin shamrock

[latent anvil]

so ${(x-1)} \cup {(x-1,y-y_0,z-z_0,w-w_0) : y_0, z_0, w_0 \in \mathbb{C}}$

[cloud walrus]

goblin shamrock

[novel parrot]

i dont think im ready for this ive lost track

[latent anvil]

sorry haha

I got too excited

the important thing is what I said at the start about the nullstensatz

[novel parrot]

i understood that part how we get the extra points

[latent anvil]

when you're focused on quotients of polynomial rings over an algebraically closed field, Spec A is just the algebraic set defined by A but with extra points

one extra point for each irreducible algebraic subset

[novel parrot]

yep

[latent anvil]

okay back to sylow

or wait all groups are abelian had a problem

but that problem seemed too hard

[urban acorn]

I have an idea for a direction on it

so for an abelian group A and a subgroup U, say that an A,B-appearance inside U is a subgroup of U isomorphic to A with complement in U isomorphic to B

if no subgroup U is specified, we assume U = A

so then A = A + B precisely when A has some A,B appearance (inside A)

and A = A + B + B precisely when A has some A,B appearance inside A, and then another A,B appearance inside the previous one

so the idea is to construct a group that has arbitrarily deep chains of appearances, but no infinitely deep chain of appearances

Here's a different problem I thought of which I feel like the solution to could share ideas with what we need, and it's probably easier.

If an abelian group contains a copy of Z^n for all n, must it contain a copy of Z^\infty?

[mild laurel]

I want to say that Q/Z is a counterexample to that statement

Or maybe, this depends on whether you mean direct sum or product by Z^\infty

[latent anvil]

presumably direct sum

esp. since the direct sum would be a counterexample otherwise

[urban acorn]

yeah I meant direct sum, so you're right, Q/Z is a counterexample

I decided to retire from math for the day, I'll just watch some youtube and try to sleep, then I'll think about this problem tomorrow

thanks for the help though

[cyan marten]

Is there a simple description of the units of Z[w], where w = exp(2pi i/p), p prime?

Let K = Q(w). I was thinking maybe I should calculate the field norm and see when it equals 1, but the ring of integers of K is usually larger than Z[w], so maybe that's just a necessary condition...

[urban acorn]

So basically Z[x]/(x^p-1) ?

[cyan marten]

You need to remove the (x - 1)

[urban acorn]

I'm not sure what you mean by this

but I suspect the units in that ring are just the obvious roots of unity

[cyan marten]

I mean Z[w] is smaller

X^p - 1 = (X - 1) f(X)

It's Z[X] / f(X)

[cyan marten]

The coefficients give more freedom here. For example you can have things like -w, where w is a third root.

[urban acorn]

ohh, I see

[urban acorn]

I wasn't paying attention last time, this actually isn't a counterexample. Z^n in my question referred to the n-fold direct sum of Z, not the cyclic group of order n.

which Q/Z can't contain because it's a torsion group

[mild laurel]

oop im the tired one ig

[cyan marten]

This seems fallacious: Take the subgroups G_n ~ Z^n. The inclusion maps into G are "compatible" and so it extends to a map from the direct limit to G, and this is an embedding.

At any rate, every element of the direct sum is in G_n for some n, so this is reasonable

[simple valley]

is the inclusion Z -> Q epi in the category of rings that might not have a unity?

the main catch seems to be that given f:Q->R, f(1) might not be a unity, but looks like it is one for the Im(f) subring?

just being wary of manipulating invertibles in the usual way in absence of a unity

[simple valley]

ah apparently there's a cute zig-zag proof

$$f(a/b) = f(1/b) f(a) = f(1/b) g(a) = f(1/b) g(ab) g(1/b) =$$
$$= f(1/b) f(ab) g(1/b) = f(a) g(1/b) = g(a) g(1/b) = g(a/b)$$

[cloud walrus]

mniip

[crimson falcon]

Ey guys, help with a type of problem
If 'im being given a group of transformation preserving some object, like an hexagon or other form (like, "pentagonal base prism whataver) and i find the order of it thorugh the orbit-stabilizer formula, how do i show isomorphism to stuff like the direct product of a symmetry group and integers modulo n?
A particular case would be this. G is a group of order 20 that preserves a right prism (prisma?) of pentagonal base. How do I show isomorphism to the tiedral group of 10 elements and integers modulo 2?

[wooden ember]

Wdym by preserves

What’s the action considered

[mild laurel]

I assume it's just the automorphism group of that object

[wooden ember]

Oh wait a prism

I was thinking a pyramid

Then yeah any symmetry of the prism is either a symmetry of its base, a flip between the two pentagonal faces, or a product of these

The former are D5 while the latter is Z2

To see that it’s a direct product (as this only shows every element can be viewed as a product of both) you need to show the two are normal in G and have trivial intersection

But that part comes out pretty obviously

[mild laurel]

In general, the idea of semi direct products and the Schur Zassenhaus theorem can be helpful for these sorts of problems

[next obsidian]

Schur Zassenhaus

Pls

[crimson falcon]

"Schur Zassenhaus "

[crimson falcon]

Ahh alright, this is the part i was missing

[mild laurel]

Wait what

Did I misspell I don't get it

[novel parrot]

in the first part

$X_I = \cup X_f$ for every f in I

[cloud walrus]

ActiveChapter

[novel parrot]

how do i show that

[fossil shuttle]

you could try demorgan's law

[novel parrot]

ive tried

[dusty river]

Try putting the definitions of X_I and X_f into words

[fossil shuttle]

which inclusion are you having trouble with

[novel parrot]

going from right to left

so if P is in the union

P doesnt contain some f

thats correct?

[dusty river]

ye

[novel parrot]

isnt it some f

[dusty river]

it is

[novel parrot]

hmm

so its in X_I like that

[dusty river]

So X_I is set of all primes that don't contain I. X_f is set of all primes that don't contain f. A prime doesn't contain I iff there is some f in I that the prime doesn't contain

This gives both directions

[novel parrot]

ok

[dusty river]

and is what I meant by try to put the definitions into words

[novel parrot]

is the prime condition necessary?

[dusty river]

No it should work for any set I guess

[novel parrot]

and now, what about part 5?

[dusty river]

where are you stuck with that?

[novel parrot]

the question, i need to show that if X is contained in a union of open sets, then its also contained in a finite collection of those open sets?

[dusty river]

yes

did you see the hint at the end

[novel parrot]

yes

im going to try it

[dusty river]

ok

[novel parrot]

ok dont really understand the hint

[dusty river]

which part

[novel parrot]

X < U X_f for some f's in the ring, are we saying that these f's form an ideal

and its the unit ideal

[dusty river]

no

To prove that every open cover has a finite subcover, it suffices to show that every cover by basic open sets has a finite subcover. Those X_f's are just some arbitrary basic open sets

You have to prove that the f's generate the unit ideal from the fact that X_f's cover X

[novel parrot]

ok

but afterwards how do we know that the generating set was finite

[dusty river]

You know how to explicitly write the ideal generated by some set in set builder form?

[novel parrot]

no

[dusty river]

hmm ok so suppose I gave you a singleton {a}

what is the ideal generated by this set

in a ring R

[novel parrot]

Ra

[dusty river]

yep, what if I gave you 2 elements?

a and b say

[novel parrot]

Ra + Rb

you mean that?

[dusty river]

Yep

What would it be for an arbitrary subset S of X

[novel parrot]

just doing the same with elements in S

[dusty river]

What if S is infinite?

You can't do summation Rs for s in S

[novel parrot]

no

[dusty river]

Right, so try to work from the definition of "ideal generated by S"

ie smallest ideal containing S

What elements must it contain?

[novel parrot]

hmmm 1

[dusty river]

Yeah

But you can say a lot more

well not 1 lol

0

I am not talking about the set in this particular case

[novel parrot]

ok

[dusty river]

So the ideal generated by S must contain all finite R-linear combinations of elements of S, right?

[novel parrot]

yes

[dusty river]

Need it contain anything more?

[novel parrot]

yes

[dusty river]

what?

[novel parrot]

nvm

lol

[novel parrot]

yes i agree

[dusty river]

nice

Right but it will be exactly that

ideal generated by S = set of all finite R-linear combinations of elements of S

easy to check that this is an ideal

[novel parrot]

yup

[dusty river]

So now you know that the {f_i} defined above are such that 1 is in the ideal generated by them

so you know that 1 must be of that form

[novel parrot]

havnt proved that yet

hm

[dusty river]

ah ok yeah so you have to deduce that from union X_fi = X

[novel parrot]

oh it equals?

hm

is it some X_fi will equal X ?

[dusty river]

well X_fi's are all subsets of X right

[novel parrot]

yh

[dusty river]

so their union is a subset of X by default

being a cover gives the other inclusion

[novel parrot]

alrighty

so theres no prime that contain every element

so it has unit

and 1

correct?

[dusty river]

yeah no prime contains all the f_i

[novel parrot]

still dont know how we know that the generating set is finite

[dusty river]

1 is a finite linear combination of the f_i

the f_i that occur in that generate 1

[novel parrot]

yes

[novel parrot]

wait no

i dont know why its finite

[dusty river]

this

infinite linear combinations don't make sense in an arbitrary ring anyway

you can't talk about convergence

[novel parrot]

oh okay

[novel parrot]

in part 2

what does that bar mean

i cant find a definition for it in the book

[hidden haven]

Closure

[novel parrot]

closure how?

[waxen hedge]

topological closure

[hidden haven]

S with a bar over it means S closure, ie smallest closed set containing S

[waxen hedge]

for the Zariski topology

[hidden haven]

And the definition makes sense because intersection of closed sets is closed

[novel parrot]

ok

[novel parrot]

what do neighborhoods mean again

a closed set?

[south patrol]

a neighbourhood of a point x is an open set containing x (or a set containing an open set which contains x, depending on who you ask :p)

[novel parrot]

okay

[hidden haven]

Second definition is superior so if you want the first one make sure to say open neighborhood or I'll shoot

[next obsidian]

You’ll have to kill me then

Although actually almost any time I use it, it doesn’t matter whether I say open nbd or nbd

[hidden haven]

💦 🔫

[next obsidian]

Since I usually just want a property to hold on some open around each point or sometbinf

[south patrol]

ngl i got worried answering that question because of differing terminology lol

[novel parrot]

do we even need the boolean ring condition for the first part?

i dont think so but just confirming

[hidden haven]

X_2 in Z should be a counterexample

Or any X_p for non zero prime really

They're not closed

[novel parrot]

but i think ive done it without using the fact that prime ideals are maximal

and not using x^2 = x

Xf is equal to V(A-f) ?

[next obsidian]

No

It’s the complement of V(f)

V(A-f) is probably gonna be empty almost always

[hidden haven]

This says "a prime ideal doesn't contain f iff it contains everything other than f"

[wooden ember]

bruh dummit and foote just gave me two consecutive exercises that were each near identical copies to previous ones

not very professional

[dusty river]

It's a test to see if you are still awake

[wooden ember]

and now it asks me to prove a false statement

waaaiiiit nvm

im just stupid

[oak grove]

i know this is a dumb question but i had to skip a day of class

aaaand i got it

[south patrol]

That is a theorem in artin so I think it's true xd

[wooden ember]

the proof of this is so neat

[wooden ember]

Bit stuck on showing that if $G$ is simple of order $p^2qr$ then $|G|=60$ (so $G\cong A_5$). All the solutions ive found online use theorems i havent proved yet like burnside's transfer theorem

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

there should be a proof using only sylow theorem machinery but im struggling to find it

[woven obsidian]

Without giving me the answer of the exercise, is the V^n supposed to be V tensored with itself n times?

Or is it a direct product?

[mild laurel]

It's direct product

[woven obsidian]

Thanks, although then I guess the universal property for tensors won't help much

Hmm or perhaps it will

[small bison]

It’ll help you define the map

[woven obsidian]

Yeah I think I got it now

Bashed my head for a few hours trying to find a map to the tensored product

[red fox]

I have a question about cosets in group theory. Is this the right channel for it?

[urban acorn]

yes

[red fox]

speak of the devil, my question is actually about representing non abelian groups with abelian groups lol

[urban acorn]

lol

[red fox]

you might be the perfect man

[urban acorn]

well, ask ahead

[red fox]

I'm specifically interested in the symmetries of the square

As you know, some subgroups are not normal, so produce different left and right cosets

There are two sets of conjugated subnormal cosets, but let's focus on these two subgroups

(does the notation make sense to you so far?)

[urban acorn]

Yeah, I think I can make sense of the notation

[urban acorn]

I don't know what you mean by "subnormal cosets"

[urban acorn]

but I do understand which two subgroups you are referring to here, and they are indeed conjugate

[red fox]

the subgroups that are not normal in the dihedral group order 8 are 2-subnormal, but it is not an important detail

[urban acorn]

I don't know what "n-subnormal" means, I don't believe that's standard notation

is it the size of the orbit of the subgroup under conjugation?

(in which case being normal would be precisely "1-subnormal")

[red fox]

I think that is right, but I'm not sure. It isn't necessary for my question about cosets, is it okay if we circle back and discuss that later?

[urban acorn]

yeah

so, go on with your question

[red fox]

So these would be the left and right cosets of subgroup 𝐻={𝓮, 𝓯}

so my first question is, can I reduce each of those cosets to the C₂×C₂ Klein four group? I know I'm losing some of the stucture of the group, but I have a plan to get it back

[urban acorn]

idk what you mean by "reducing those cosets to a group"

[red fox]

I mean forgetting the dihedral relations of the coset, and treating each as just part of the klein four group

specifically, you lose the rotation

If you start with an element of the left coset, and apply 𝓻𝓻, you will not end up at the same element, but in a klien four group, if you apply any action twice, you will get back to where you started:
(0,0) + (0,1) + (0,1) = (0,0)

does that make sense?

[urban acorn]

okay, so I need to ask a couple questions about your notation

[urban acorn]

over here, the arrows show that the rotation is counterclockwise, but then in the square "r" is seen rotated 90 degrees clockwise from the identity

[urban acorn]

it looks like what you're trying to do here is obtain klein four as a quotient, but this wouldn't work

and I'm not sure in what sense you would get klein four from here

how do you define multiplication of these cosets? because multiplication of representatives doesn't work

[red fox]

Imagine the square is like a combination lock with an arrow pointing to the selected value

[urban acorn]

okay, but this feels a bit backwards (pun absolutely intended)

r is what is needed to take the part labelled r to the identity, instead of what is needed to take the identity to the part labelled r

[red fox]

I see what you mean. I could redo the pictures another way if that would help

[urban acorn]

nah, it's okay

[red fox]

I tried a few ways of visualizing the group and this is the way I liked the best

but there are a few options

[urban acorn]

so I'm not sure how you're trying to obtain a group structure on the cosets of <f>

you labelled them with elements of C2 x C2, but didn't explain how you get that labeling

if multiplying one coset with another just means picking an element out of each and looking in what cosets the product of your elements lands, then it's not well defined in this case

[urban acorn]

we care about normal subgroups because they're the case when this works

i.e. if you tried this with <r^2>, then I'm pretty sure you would get C2 x C2 as a quotient

[red fox]

right, this is not a quotient

[urban acorn]

okay

[red fox]

but I am trying to achieve something similar

because a quotient is the same thing as a left or right coset of a normal subgroup

but these are not normal

[urban acorn]

personally, I don't think there's an interesting group structure to be found in the way non-normal subgroups sit in your group

if you think you can "see" something there

try to formally define how you multiplied these cosets

[red fox]

oh, there is definitely something there

one sec, let me draw out the other half of this

[red fox]

I can redraw these left and right cosets as columns and rows, where the intersection is in the normalizer {𝓮, 𝓯, 𝓻𝓻, 𝓯𝓻𝓻}

and then I can add the left and right cosets of that second subgroup {𝓮, 𝓯𝓻𝓻} into that table, which is a conjugate of {𝓮, 𝓯}

now I've made something that is like a quotient, because it is closed under left and right actions

but I have never heard of people talking about something like this

The columns are left cosets and the rows are right cosets

[urban acorn]

okay, so you noticed that the left cosets (which aren't also right cosets) of H2 are the same as the right cosets (which aren't also left cosets) of H1, and vice versa

[red fox]

no, they just overlap

[urban acorn]

how do you get this quotient-like-but-not-a-quotient group out of this?

[red fox]

there are 4 unique cosets

but they overlap into a closed system overall

[urban acorn]

yeah, okay

[urban acorn]

I still don't see this?

[red fox]

because it accounts for both left and right cosets, like a quotient does.

my claim is, if you reduce each of those 4 cosets into klien four groups, you can faithfully represent the dihedral group

with only C₂ groups!

you know how in computation, all of math can be reduced to 1s and 0s? I'm trying to do something similar for the dihedral order 8 group by reducing it to cyclic 2 groups

[urban acorn]

it's not clear what you mean by "represent" and "reduce" here, you have to make this formal

[urban acorn]

and by the way, some math can't be put on a computer

[red fox]

:0 really?

[urban acorn]

like, you can't represent arbitrary real numbers on a computer

[red fox]

you can't program in those properties?

[urban acorn]

most types of real numbers that one actually works with in practice are computable

like you can model rational numbers on a computer, you can even model all algebraic numbers, and like do arithemtic with them

you can also model arbitrary computable numbers, which are numbers whose digits have an algorithm for computing them, and you can do arithmetic with those, but equality is not in general decidable

[red fox]

I can appreciate the subtleties of pure math. At the very least, computers are an incredibly powerful tool for mathematics, even if they have to approximate some things

[urban acorn]

sure

[red fox]

I know quantum computers might be able to do different kinds of math

[urban acorn]

well, there's a thing called a decomposition series

[red fox]

tell me more

[urban acorn]

all groups can be in some sense "reduced" to simple groups, which in the abelian case are the prime cyclic groups

most non-abelian groups that you know of are reduced to abelian groups, and those are called solvable groups

the first non-solvable group is of order 60

[red fox]

so the dihedral group order 8 is definitely solvable

[urban acorn]

and it's done just with quotients, no weird quotient-like-but-not-quite-quotient objects

[urban acorn]

yes

[red fox]

not yet, they aren't

[urban acorn]

lol

well, you have an idea, clearly, but I don't understand it

[red fox]

the solution for the dihedral group order 8 can be represented in an elementary group order 16

[urban acorn]

so what you need to try to do is make it formal

[urban acorn]

you mean "elementary group order 16" as in C2 x C2 x C2 x C2 ?

[red fox]

yes

with 1 degree of linear dependence

the 1 degree of linear dependence is because some of the left cosets are not the same as the right cosets

again, I've never heard of such a thing

but I'm very sure it works for at least the dihedral group order 8, and I expect more groups

what would D8 (in terms of order) normally factor into? C2 x C2 x C2?

and then you could reconstruct D8 from the 3 C2s with the semidirect product?