groups-rings-fields

[ivory cosmos]

based

[next obsidian]

????

That looks like it's french but...

This is a Swiss uni right?

[sturdy marsh]

switzerland has a french part

[thorn delta]

is this channel open?

[next obsidian]

yeh

[ivory cosmos]

right, so EPFL is in the French part, and some of the courses are in French

less and less as you advance in the degree

for example, I no longer have any French courses at all in my 3rd year

[thorn delta]

ah, i think i figured out my question. Let $F$ be a free module of infinite rank. Then $F$ is not reflexive since a map $$f : \bigoplus_{i \in I}R_i \to \prod_{i \in I}\operatorname{Hom}(R_i, R)$$ is not even bijective when $I$ is infinite.

[cloud walrus]

kxrider

[thorn delta]

R_i being copies of R

[latent anvil]

nice!

[chilly ocean]

but

""""subset""""

in a SES 0 -> A -> B -> C -> 0 it might not be the case that A is literally (i.e. set theoretically) subset of B, but the second arrow is an injective map of A into B

so saying "subset" is a little rough but is understandable, especially if you're making some identifications

(im not sure if that's a common convention, so please excuse me if im wrong)

i don't know enough about ring theory to answer that, sorry

well id guess so

im just uncomfortable answering

[next obsidian]

okay wait

what's the origianl qeustion

this is where I come in

I see SES and Noetherian

O_O

yup

In an SES 0 -> A -> B -> C -> 0?

oh yeah, so this is true. So Noetherianness is stable under "sub object"

basically by definition

It should be equivalent to saying all submodules are finitely generated

then the injection 0 -> A -> B idetnifies A with a submodule of B

so A is Noetherian

And then actually like you said,

C is iso to B/A

by this SES

and then you can apply the correspondence theorem to show ascending chains stabilize

The interesting thing is that

if A and C are Noetherian

then so too is B

Umm, maybe

I'm trying to find the proof in Matsumura haha

I think this might be right

the proof in the book uses some result which is like

kinda messy to type

but basically comes down to some intersection thing, and I think the second iso

Actually, I guess I'll type it up,
M' < M, N_1,N_2 < M such that N_1 < N_2, and N_1\cap M' = N_2\cap M'. If N_1/M' = N_2/M' then N_1 = N_2

Where N_1/M' is the image under the quotient map

So it suffices to show N_2 < N_1 since we assumed the other direction, for that you take x in N_2, then we know that x + M' = y + M' for some y in N_1, then since N_1 < N_2, x - y is in N_1. Note that by that equality, we also know x - y in M', so that x -y in M'\cap N_1 = M'\cap N_2. Thus y has to be in N_2, else x - y couldn't be in N_2

(since x is in N_2)

from this, you can apply this result to an increasing / decreasing chain (for Artinian) of submodules of M

So if you have N_1 < N_2 <... in M, you also have N_1 \cap M' < N_2\cap M' <... < N_k\cap M' = N_{k + 1}\cap M' = ... by Noetherianness

Similarly, you have N_1/M' < N_2/M' <... < N_j/M' = N_{j + 1}/M' =...

so you can apply the previous result for all i >= max{k,j}

and then you get that the N_i are equal past that point

The same argument runs through for a decreasing chain and Artinian

[next obsidian]

You might be able to do it a different way!

I'd explore your idea too

if you feel like it

[latent anvil]

You can use the ascending chain condition to

If f : B -> A is the injection and M0 <= M1 <=… is your chain of submodules in B then f(M0) <= f(M1) <=… is an ascending chain in A, so stabilizes since A is noetherian. Injectivity of f implies the original chain stabilizes at the same point

[next obsidian]

So let's say A^ is the I-adic completion of I

if you have an element x in A^, is there some a in A such that x - a is in IA^?

so like x looks like (,...,a_2,a_1) where a_n is in A/I^nA

under the condition that a_i = pi_i(a_{i + 1})

when pi_i is the quotient map A/I^{i + 1}A -> A/I^iA

I believe an element of A is to be interpeted as

(,...,a,a)

where all the components are the same

wait...

for power series this seems mad easy

if you let a = a_1 you're done

...

gaaaaaaaaaaaaaaahhhhhhhh

[sturdy marsh]

can't you just do (..., a_1, a_1)

[next obsidian]

yes

but

[sturdy marsh]

x minus that

[next obsidian]

I was thinking this

but like...

I was struggling to see how this was in IA^

but this seems obvious for pwoer series

but for a general ring for some reason my brain doesn't see it

like...

a_2 - a_1

err

so the element

(...,a_2 - a_1,0)

should live in IA^

but why is that?

So for power series a_i is the (i - 1)th coefficient

so like an element

a_0 + a_1x + ...

turns into (,a_1,a_0)

in this case I = (x)

so if you subtract (...,a_0,a_0)

you can write this as

[latent anvil]

We have π1(a2 - a1) = 0, so a2- a1 is in I(A/I^2), right ?

[next obsidian]

umm

ohhhh

fuck

[latent anvil]

Feel like that does it?

lol

[next obsidian]

I was stuck on this

for hours

like 2 months ago

[latent anvil]

lol

[sturdy marsh]

[next obsidian]

and then stopped doing CA

becasue of this

meme

[latent anvil]

owned

[next obsidian]

Hmmm

[latent anvil]

Bro I can't want to do comm alg next quarter

[next obsidian]

today I will fucking k-word myself

[latent anvil]

Very hyped

[next obsidian]

can't want

[latent anvil]

Seriously!!

It's fun stuff

[next obsidian]

can't want

[latent anvil]

I've been neglecting algebra

[next obsidian]

want

[latent anvil]

> want

[next obsidian]

Can't wait for you to become better than me at CA

man :/

I don't like completions

[latent anvil]

horse

[next obsidian]

it feels unwieldy

[latent anvil]

I have done one problem about completions in my life

in g&w

[next obsidian]

I wish I could not dive into the actual construction so often

Like... if it followed by "is a direct limit" or something maybe it would be better

or is it inverse limit

I think the latter

but I find myself having to go into the construction as a subset of the product so often

and it makes me want to k-word

[latent anvil]

Isn't there a topological way of thinking about it?

[next obsidian]

yes but

for the stuff I've done I don't know how to get around it

fwiw I try to handle the case of k[[x]]

as the completion of k[x] at (x)

to try to get intuition

[latent anvil]

Like, A is I-adically complete if it's complete wrt the metric d(x, y) = least n such that x-y in I^n, right ?

[next obsidian]

Yeah

but I'm working with non I-adically complete stuff at the moment

[latent anvil]

Ah okay

[next obsidian]

Especially for the text

like to prove the theorems and little comments

[latent anvil]

I was wondering whether you could prove the thing you just did via cauchy sequences

[next obsidian]

I had to go into the construction so often

I hope!

If you could that would be better I think

I'm scared I'm thinking about it the wrong way

but Matsumura won't tell me that haha

[latent anvil]

So what was the exact statement?

Like

[next obsidian]

This is an intermediary step in a proof

What I wanted to show was exactly that

[latent anvil]

If x in A^ then there's an a in A such that x-a is in IA^

[next obsidian]

If A^ is the I-adic completion, x in A^ then there's an a in A such that x - a in IA^

yeah

But A isn't I-adically complete so like

[latent anvil]

So x is like a limit of a cauchy sequence in A

[next obsidian]

right

[latent anvil]

x = lim an

[next obsidian]

I suppose so yeah

[latent anvil]

Choose N such the an-am in I for n, m >= N

[next obsidian]

also

limits might not be unique...

pretty sure

[latent anvil]

That's annoying

is d not actually a metric?

It's definitely symmetric and d(x, x) = 0

d(x, y) > 0 if x ≠ y by like Artin Reese but I guess not in general?

[next obsidian]

Something like that I think

yeah so

if you have I < J(R)

then they'll be unique IIRC

I worked this out

and that condition also let's you Artin-Rees

[latent anvil]

as long as cap I^n = 0 I expect

[next obsidian]

Yeah I belive so

[latent anvil]

hmm

[next obsidian]

that's the kernel of the map A -> A^

IIRC

[latent anvil]

So anyways

[next obsidian]

it's never spelled out explicitly

but I worked this out I think

[latent anvil]

This

[next obsidian]

sure

[latent anvil]

Let a = aN

Them x - a = lim an - aN

It's a limit of stuff in I

hmm

Are the sets I^k A^ closed? Or even just I A^?

[next obsidian]

soooo

for the problem I wanted all ideals are closed

by assumption

[latent anvil]

Ah nice

[next obsidian]

but in general ideals aren't closed

[latent anvil]

Then this is a proof

[next obsidian]

and this is a general thing I think

but it might be true that IA^ is alwasy closed

[latent anvil]

Idk I just like cauchy sequences

[next obsidian]

yeah i would try to work with them when A is I-adically complete

but when it isn't things get kinda spooky

[latent anvil]

Sure but you can always work with A^ that way right?

[next obsidian]

also I read the section like

ages ago

yeah

so I'm a little shaky on the topological stuff :/

[latent anvil]

Fair

[next obsidian]

also my brain is still pretty blasted

I was gonna msg you but didn't

[latent anvil]

Yeah haha me too

[next obsidian]

since you said you were exhausted too

[latent anvil]

Lmaooo

[next obsidian]

buuut...

haha

[latent anvil]

Seriously have a headache rn

[next obsidian]

oof

take a paink-worder

[latent anvil]

I think I just need to sleep

[next obsidian]

👍

Also

general tip

maybe you have realized this at this point

Finitely presented is useful in 1 really crucial way I figured out

If you want to show that F(M) = something for M finitely presented, usually it works like this

F(A^n) is clearly isomorphic to what you want (usually this is a natural isomorphism, so you want to show F(M) = G(M))

You have a pretty clear natural morphism F -> G

Then you just take the exact sequence A^n -> A^m -> M -> 0

apply F, and you apply 5 lemma

[latent anvil]

ahh cool

That makes sense

[next obsidian]

in my brain this is why finite presentation matters

like... a ton of things boil down to this

and you can maybe think of a few examples, the Hom thing with tensor

the thing about pulling out direct limits I think from Hom

all of them hold for f.p. and the proofs are just this

[latent anvil]

Nice

[next obsidian]

chmonkey tip

[outer estuary]

Anyone learning Cech cohomology and sheaf cohomology rn?

[next obsidian]

I learnt a good bit ~ 2 months ago

[old hollow]

this is probably dumb question

Normal subgroups are important because they (and only they) can be used to construct quotient groups of the given group. Furthermore, the normal subgroups of G are precisely the kernels of group homomorphisms with domain G, which means that they can be used to internally classify those homomorphisms.

it seems like normal subgroups have tons of ties to quotient groups but the "conjugation" requirement really throws me off

[carmine fossil]

Basically if N is a normal subgroup of G,G/N is a group

[mild laurel]

The conjugation requirement is kind of like saying that the equivalence relation of cosets is compatible with the group operation

[carmine fossil]

That's the motivation behind normal groups

[old hollow]

All the examples in my head of quotient groups are like Z/nZ and R[x]/(some polynomial)

but like, since those are abelian doesn't that trivially satisfy the requirement

[carmine fossil]

Consider the set G/N whose elements are left cosets where aN represents the set {an|n in N}.

The "natural" way to define group operation on G/N is aN.bN=(ab)N
now if aN=bN, aN.cN=bN.cN (for operation to be well defined)
Rewriting this gives you the conjugation condition

[old hollow]

ok I understand everything up to the last line, I am so sorry

I am like, rly tired rn

how do you rewrite it to get conjugation? like where would the g and g^-1 come from

[carmine fossil]

if aN=bN, a=bn for some n in N,i.e,(b^-1a) is a member of N

aN.cN=bN.cN implies (ac)N=(bc)N
Implies ac=(bc)n' for some n' in N
Now this implies (bc)^-1(ac) =n' is in N,i.e., c^-1 (b^-1a) c is in N

Now you can choose b^-1a to be any value in N(since aN=anN for all n in N)

Implying c^-1 n c is in N for all n in N is a necessary condition

Well,you can clearly choose a different c

So,This should be true for all c in G and all n in N

[old hollow]

Ah i see, thank you so much

[carmine fossil]

You still have to prove that the condition is sufficient

[cloud walrus]

Fiwam

Fiwam

[mild laurel]

Not exactly subsets, but ideals of Z in a certain sense

[plucky flicker]

Hi. Can someone give me an example of a ring, in which there are zero divisors, invertible elements and at least one element which is nor zero divisior nor invertible

some matrix or polynomial ring would fit this condition i guess, but i don't know how to build it up

[solemn rain]

i would guess u need a 'nonfinite' ring

for in a finite ring any element is either a zero divisor or non invertible

thats a start ig?

[plucky flicker]

yes

exactly, it should be infinite

finite rings have the nice property that only zero divisors and invertible elements are in there

[solemn rain]

so maaybe

M_2(Q)

should work

[chilly ocean]

how about Z (neither invertibles or zero divisors) -> Z[1/2] (has invertible elements) -> Z[1/2][x, y]/(xy) (has zero divisors)
well there is no reason to introduce y i guess, just mod by some reducible polynomial in x

[carmine fossil]

You have A(adj A)=(det A)I

So,If A is not invertible,A is a zero divisor

[solemn rain]

uh oh

yea

[plucky flicker]

what do you mean by Z[1/2]

oh i guess you mean the ring &{a + b/2 : a,b \in Z}& right?

[chilly ocean]

yeah

[golden pasture]

wait

isnt Z[1/2]

elements

that are like a/2^n

"localization away from 2"

[plucky flicker]

so Z[1/2] = {a + b/2 + c/4 : a,b,c \in Z}?

[golden pasture]

another very direct example is R[ε]/ε^2 "infinitesimal deformation of R"

1/128 is in Z[1/2]

[plucky flicker]

what is R here? an arbitrary ring?

[golden pasture]

yea some ring

[plucky flicker]

so R[ε]/ε^2 contains equivalence classes right<

?

[golden pasture]

so basically it contains elements of a form r+sε with the multiplication rule (a+bε)(c+dε)=ac+(ad+bc)ε

think of ε as an "infinitesimal"

[chilly ocean]

ah sorry lmao

what is the notation for what i am thinking of? (1/2)Z?

[plucky flicker]

(1/2)Z = {1/2 * z : z \in Z} yes

[golden pasture]

it isn't a ring tho

[chilly ocean]

ah you're right, i'm being a dumbass

[golden pasture]

you may be thinking of like

Z[(1+sqrt(5))/2]

kinda thing

[chilly ocean]

ah right this ring does have invertibles huh

[golden pasture]

We usually call them units but yea

and units are precisely the solution to some pell equation

[chilly ocean]

i recall something like this 🙂

but now that i think about it some more, Z[1/2] (in the proper interpretation of this notation as elements of the form a/2^n) is also a proper ring with units

[golden pasture]

yesh

aka localization away from 2

cuz you are basically forming a ring that specifically excludes the prime (2)

[chilly ocean]

i should probably have studied more algebraic geometry, or math in general, i think i have some very basic familiarity with the terminology, but for example i never picked up why it is called a localization to take fractions like this (and now it is too late to get back to studying without significant effort)

[solemn rain]

why

[chilly ocean]

why, as in why it is a significant effort? one thing is that on weekdays there is work, then from a more practical perspective there are other things that i should be studying, then this means i have little time to spend learning random things, which is not good in the sense that you need a decent chunk of time for context switching/getting back up to speed on your previous level of understanding. and at this point it is not exactly worth it

I cant even get out of bed without sginificant effort

[golden pasture]

the idea is that localization allows you to study "local behavior"

which really means

what happens near a prime

so like in context of say

functions on C

localizing at some point essentially means the ring of all functions that doesnt explode at that point

[glossy wing]

if I want to find elements of order $2$ in $\mathbb{Z}_{30}\times \mathbb{Z}2$, if $(a,b)\in\mathbb{Z}{30}\times \mathbb{Z}_2$ then the order is $\text{lcm}(a,b)$ since we want the elements of order $2$, we have $\text{lcm}(a,b)=2$ that means that $a=1,2$ and $b=2$ $\text{lcm}(1,2)=\text{lcm}(2,2)=2$ both $(1,2)\in \mathbb{Z}2$ so either $1$ or $2$ works for the choice of $a$. Now I want to find the elements of order $2$ in $\mathbb{Z}{30}$ which is just $\varphi(2)=1$, hence the elements of order $2$ is just $1\times 2=2$?

[cloud walrus]

please request a new nickname

[rigid cave]

Hello! So I am trying to solve this problem but I am struggling with the very last part of this question (marked by the red underline). Any hints?

[rustic crown]

did you do the first two parts?

[rigid cave]

yes I did

[rustic crown]

so we know that degree of the extension is 4, and for beta = alpha^3 - 14 alpha, we know that degree of minimal polynomial of beta is [Q(beta):Q] which must divide 4. If you calculate beta^2, you'll see that this cannot be equal to a*beta+b for rational a and b (else alpha would satisfy a cubic which isn't possible). So minimal polynomial of beta has degree 4.

From there it shouldn't be too hard to find a degree 4 polynomial that has beta as a root.

[rigid cave]

Oh that's actually nice!

[rustic crown]

Indeed, on calculating beta^2 = -120 alpha^2 + 48. You know the minimal polynomial of alpha^2.... rest is just a simple scaling/translation

[rigid cave]

Oh okay, thank you so much!

[simple valley]

having a moment...

S_4 has 1 element conjugate with (), 6 elements conjugate with (12), 3 elements conjugate with (12)(34), 6 elements conjugate with (123), 6 elements conjugate with (1234)

that's 22, what did I miss?

oh

8 elements conjugate with (123)

[rich ravine]

@rigid cave In general you can find a polynomial annihilating a polynomial $Q(\alpha)$, knowing that $P(\alpha)=0$ for a certain polynomial P, by computing the resultant $\mathrm{Res}_X(P(X),T-Q(X))$

[cloud walrus]

Othenor

[rustic crown]

What is that last expression? Could you share a link about it?

[rich ravine]

I don't have any good resource on it off the top of my head

The wiki page is a bit of a mess but you could look at it

[weary terrace]

Hi, I have the questions in the pic.
Take question (a) for instance. Say the map from M to L is f. So to prove (a), I defined \bar{f} which takes a \otimes m to a \otimes f(m). The same for the other maps.
My question is, in (b), why wouldn't this method work? What makes the cases different?

[rustic crown]

consider this,
0 --> Z --> Z --> Z/2Z --> 0
first the map Z --> Z is multiplication by 2
what happens if you tensor with Q?

[rich ravine]

Let R be a commutative ring and P,Q two polynomials in R[X]. There exists an element $r\in R$, computable explicitly as the determinant of a certain matrix (the Sylvester matrix), that we call the resultant of P and Q, noted $r=\mathrm{Res}_X(P,Q)$. One of the properties it satisfies is a "Bézout identity" : there exists polynomials H(X), K(X) such that $P(X)H(X)+Q(X)K(X)=r$. If you take R=K[T], the polynomials in R[X] are actually elements of K[X,T], i.e. bivariate polynomials over K, and computing the resultant in X "kills" the variable X, so the resultant in X is a polynomial in T.

[cloud walrus]

Othenor

[rich ravine]

Suppose now we have a relation $P(\alpha)=0$ and we are trying to find a polynomial that annihilates $Q(\alpha)$. Denote $r(T)=\mathrm{Res}_X(P(X),T-Q(X)) $. By the Bézout property, there exists bivariate polynomial H, K such that $P(X)H(X,T)+(T-Q(X))K(X,T)=r(T)$. Now evaluating at $X=\alpha, T=Q(\alpha)$, we obtain $0 + 0 = r(Q(\alpha))$, as claimed.

[cloud walrus]

Othenor

[rustic crown]

I need to look up the Sylvester matrix...
Thanks for telling this cute thing

[rich ravine]

This is really useful if you have some symbolic computation tool at hand

For calculating by hand the Sylvester matrix often gets bigs really fast so it's not that practical

[rustic crown]

oh this matrix looks very cute

[rich ravine]

Another method you can use is to express the powers of Q(alpha) in the basis (1,alpha, alpha^2, alpha^3) and try to find a linear combination that sums to zero, i.e. find the kernel of the matrix you obtain

[mystic jungle]

For direct product of groups if I have a group G = (x, id) and x^2=id and im asked to list the element of G x G, am i just multiplying (x,id) * (x,id) ?

I get G x G = (xx, id*id, xid, idx)

but xx=id and id*id is just id, and xid = x and idx = x

so G x G = (x, id)?

[rich ravine]

Elements of GxG are tuples

What are all the possibilities for tuples ?

Forget about the multiplication for the moment

Also you might have a confusion coming from your notations. Let us write G={x,id}, meaning that as a set G has 2 elements x and id

Then by definition GxG is the set of tuples ${(a,b) | a\in G,b\in G}$

[cloud walrus]

Othenor

[rich ravine]

with a certain law for multiplication

given by multiplication coordinate-by-coordinate

That is, $(a,b)\star(a',b')=(a\star a',b\star b')$

[cloud walrus]

Othenor

[mystic jungle]

so a tuple for that group would be (id*id,xx) ?

[rich ravine]

There's no multiplication involved if you want to enumerate the tuples

GxG is a set

before being a group

and you're asked to enumerate the elements of the set GxG

knowing that G is a set with two elements

[mystic jungle]

so G x G should have 4 elements since G is order 2

[rich ravine]

yes

[mystic jungle]

im dumb sorry for the trivial question haha

[weary terrace]

I'll get a sequence of zeros.
So you mean that in the example you gave, using the above method, the induced multiplication map wouldn't be injective?

[rustic crown]

yea, you don't get a sequence of 0s

[weary terrace]

Sorry, is not zero s

Yeah.. I thot of Z/nZ

Thought

[rustic crown]

Z tensor_{Z} Q = Q and Z/2Z tensor_{Z} Q = 0

[weary terrace]

Ok nice.
So since Q is not free as a module over Z, we get that the induced multiplication map is not injective, resulting in the nonexactness of the sequence. Great example!

Well, I guess even without the injectivity thing, the sequence we get is obviously not exact..

Thanks! @rustic crown

[rustic crown]

[next obsidian]

So to be clear here, this doesn’t follow because Q isn’t free, you CAN tensor with a non-free / non-flat thing and preserve injectivity, it’s merely a requirement that Q be not free to even have a chance to not preserve it. In order to show the induced map isn’t injective you’ll have to do a calculation

For this specific example though... I’m pretty sure injectivity is preserved. The resulting map should be multiplication by 2 from Q -> Q which is still injective...

It also happens to be surjective which is why the Z/2Z turned into a 0

The sequence you get is
0 -> Q -> Q -> 0 -> 0 which is exact (the map Q -> Q is multiplication by 2)

For an example where injectivity isn’t preserved,
0 -> Z -> Z -> Z/2Z -> 0
Where Z -> Z is multiplication by 2.
Tensor this with Z/2Z, and then this will turn into the following sequence
0 -> Z/2Z -> Z/2Z -> Z/2Z -> 0
Where the first map is multiplication by 2, but this is the 0-map on Z/2Z, and the latter map is identity.

[tawny shell]

how are they getting this mapping? i understand that SO(n) has det=1 and Z_2 are the elements {0,1}, and since O(n) has det=/=0 we pick ad-bc=1, but still dont get how they've constructed this map, can anyone help?

[sturdy marsh]

is that K supposed to be an M

[tawny shell]

yes

[sturdy marsh]

okay, assuming K=M

[tawny shell]

I just found it to a solution stack exchange, and im just stuck

[sturdy marsh]

det M = \pm 1

[tawny shell]

yep

[sturdy marsh]

umm

is n supposed to be odd

[tawny shell]

yes

n is odd here 😄

[sturdy marsh]

okay great

[tawny shell]

sorry i forgot to mention it

[sturdy marsh]

then det(det(M)M) = 1

[tawny shell]

yeah im confused where they're getting this det(M)M from

[sturdy marsh]

what exactly are you confused about

the fact that det(M)M \in SO(n)?

[tawny shell]

yeah

since we have pm 1

[sturdy marsh]

right okay

so if you have a matrix M

and if you multiply any column by a constant c

the determinant of the new matrix is (c)(detM)

in this case, you are multiplying all the columns by c = detM

so the determinant of the new matrix is (c^n)(detM)

but as n is odd, c^n = c = det M

so the determinant is (c^n)(detM) = (det M)^2 = 1

which implies that det(M)M is in SO(n)

did that make sense?

[tawny shell]

yeah that makes sense

our n

runs from

n can only be 1?

[sturdy marsh]

no

[tawny shell]

so how is it c

c^n=c

[sturdy marsh]

c = \pm 1

clearly 1^n = 1

and if n is odd, -1^n = -1

[tawny shell]

right

[sturdy marsh]

so c^n = c

when n is odd

[tawny shell]

ahhh right ok

makes sense, and how does Z_2 just have

oh

right

think it just clicked

the (..., det(M)) part has to be like that since we exclude ad-bc=0

for Z_2

[sturdy marsh]

det(M) = \pm 1

[tawny shell]

yeah and -1mod2 is just 1

[sturdy marsh]

{1, -1} with multiplication as the operation is Z_2

[tawny shell]

and this is well defined because det(detM*M)=det(M)^n det(M)=(detM)^n+1 = 1

since n+1 is even, and det M = pm 1 for M in O(n)

ok ok ok this makes way more sense now, thank you

had a moment

[prime cloak]

Can someone help me showing that X^3+X+1 is prime in Q[X]

[mystic jungle]

How do I find the subgroup of a symmetric group of letters S_n given some n say 3 or 4

where the subgroup is defined by some f in S_3 and f(3)=3

what does it mean for f(3)=3 in the subgroup? Does that mean 3 maps to 3?

[mild laurel]

Yes

S_n can be considered as the set of all bijective functions from {1,2,....,n} to itself

[mystic jungle]

Ok so for example S_3 is my symmetric group, and i define f in S3 with f(2)=2

[mild laurel]

Well, that doesn't define an element of S_3

there are multiple elements of S_3 that satisfy that

[ivory cosmos]

An entire subgroup's worth

[mystic jungle]

would my subgroup just be the identity and alpha ''

where alpha '' = (1 2 3) ( 3 2 1) since f(2)=2

[mild laurel]

yeah

[ivory cosmos]

In general, the permutations that fix an element in S_n form the subgroup S_{n - 1}

[mystic jungle]

mmm not quite sure what you meant by that

[next obsidian]

if you consider S_n as the group of permutations on {1,...,n}

if you take all permutations which don't affect some fixed k in {1,...,n}

this forms a subgroup of S_n which is isomorphic to S_{n - 1}

you can think of it as simply being the group of permutations on {1,...,n}\{k}

[celest brook]

I'm having trouble thinking where to start. I know I need 3 things for a group? I'm just having trouble trying to think of the binary operaiton relating them? I would assume it is distance, so I'm just having a bit of trouble what I would define as a 'inverse distance'

[mild laurel]

The operation is composition

[celest brook]

huh..?

oh

[mild laurel]

If you have two isometries S and T, then you can compose them to get another isometry

Like S(T(x))

[celest brook]

ok yeah I get that

I guess I just got confused since it didn't explicitly outline that, thanks for clearing it up

[simple valley]

say p is the smallest prime divisor of |G|, how do I show that |G : H| = p implies H normal?

I figure I gotta act on subgroups of G by conjugation but like how does p come into this?

oh hmm, the stabilizer of H includes at least H

so it's either G or H. If it's G then it's normal

if it's H then the orbit has p elements

not sure what next

[latent anvil]

Let G act on the cosets of H by left multiplication. We get a map f : G -> S_p via this action. Let K be the kernel. Observe that [G : K] = |im f| by the 1st iso theorem. In particular |im f| divides |G|, so any prime dividing |im f| is >= p. But also |im f| divides |S_p| = p!, and all the primes dividing p! are <= p (and there's only one copy of p). This forces |im f| = p or |im f| = 1. Observe that if x in K then xH = H, so x in H. Thus K <= H, so the index considerations force K = H. Thus H is a kernel

Does that make sense to you?

This trick of turning actions into homomorphisms and looking at the kernel/doing number theory is very common

Conjugation is a natural action to try but it turns out it just doesn't give enough information

[celest brook]

@mild laurel so the elements are functions and the binary operation is function composition essentially?

[latent anvil]

wait wtf where is

[mild laurel]

yes

[celest brook]

alrighty

ahhh

[simple valley]

hmmm

interesting

[celest brook]

so I'd need to prove all of those things, yes?

\begin{center}
Let $S$ represent the set of isometries in $\mathbb{R}^3$ We wish to prove 3 things.
\begin{itemize}
\item There exists a binary operation across $S$ which is closed across S. The binary operation is also associative (binary operation is function composition)
\item There exists an identity element in $S$, call it $e$ s.t. $\forall g \in S$ $e(g(x))=g(e(x))=x, x \in \mathbb{R}$
\item $\forall g \in S$ $\exists h \in S$ s.t. $h(g(x)=g(h(x))=e(x)=x$
\end{itemize}
\end{center}

[cloud walrus]

Moosey

[celest brook]

those are the things I would need to prove, all in all,yes?

[simple valley]

you used e(x)=x before defining it as such

[celest brook]

oh

[simple valley]

otherwise yes

[celest brook]

I thought I did

w/ identity element

[simple valley]

"identity element in S" doesn't automatically mean x mapsto x

[celest brook]

yes

ahhh

i'm confused

oh

ohwait I should also say x in R^#

R^3

i'm dumb

What should I define as an identity element then

\begin{itemize}
\item There exists a binary operation across $S$ which is closed across S. The binary operation is also associative (binary operation is function composition)
\item There exists an identity element in $S$, call it $e$ s.t. $\forall g \in S$ $e(g(x))=g(e(x)), x \in \mathbb{R}^{3}$
\item $\forall g \in S$ $\exists h \in S$ s.t. $h(g(x)=g(h(x))=e(x)$
\end{itemize}

[cloud walrus]

Moosey

[celest brook]

like that?

[simple valley]

are semidirect products fibrations

Im getting fiber bundle vibes

in the short exact sequence $A \to A \rtimes B \xrightarrow{\pi} B$ the kernel ($A$) acts as the fiber of of $\pi$

[cloud walrus]

mniip

[simple valley]

hmm, that might just be quotients in general

quotients are fibrations, semidirect products are fiber bundles?

[latent anvil]

yes

err

I think you are getting at something good

[next obsidian]

Cease this conversation immediately

[latent anvil]

semidirect products are quotients which admit a global section (which is a homomorphism)

[next obsidian]

Do not the sanctity of group theory

[latent anvil]

ie short exact sequences which split

[next obsidian]

With your filth

[latent anvil]

check aluffi

topologically G -> G/H is an H-fiber bundle for H a closed subgroup of a lie group G

(issues can happen in the purely topological category)

so you're definitely getting some things right

H need not even be normal

but I'm not sure about the fibration = quotient, fiber bundle = semidirect product thing, that feels wrong

[simple valley]

right, I'm not sure how to reconcile the algebraic structure with the notion of a fiber in general

but kernel is a kind of fiber

other fibers aren't subobjects, but they're equivalent to the kernel in a way

[latent anvil]

this feels like what you want to think about tbh

[simple valley]

hmm

[latent anvil]

modulo topological pathologies what you are saying is true

[simple valley]

yea na I'm just getting dependent pair vibes from semidirect products

and dependent pairs are obviously fibrations

tfw foundatioomer

[latent anvil]

I think I sort of see what you mean

the fibers of the map decompose G into a partition of sets bijective with H

so G is noncanonically in bijection with $\Sigma_{x \in G/H} H$

[cloud walrus]

Shamrock (not a furry)

[latent anvil]

or I guess you're just saying $G = \Sigma_{x \in G/H} x$

[cloud walrus]

Shamrock (not a furry)

[latent anvil]

(which is true)

[simple valley]

set theoretically yes

but can we lift anything useful to the category of groups?

the SES and split SES I guess

[latent anvil]

yeah, not really

I guess I think of this as a topological phenomenon almost

[simple valley]

you can probably construct a semidirect product of two groups as single-point 1-types in HoTT

transport is the homomorphism from H to Aut(N)

[prisma ibex]

If R is Artinian than any finitely generated R module is Artinian. If R is not Artinian there are counterexamples

(There are Artinian, non Noetherian modules)

[next obsidian]

This is mad weird

Such a module is necessarily non-finitely generated since in the fg case the Artinian => noetherian for rings gets it for you

[prisma ibex]

the usual example is Z[1/p]/Z

which is isomorphic to Z(p^\infty)

🙂

[vestal snow]

k(x) has no gap numbers right?

[delicate hawk]

why isnt the ideal (x, y) the entier set Q[x, y]

is there a counterexample i cant think of

xy and all the constants in Q can be written as ax + by (a, b \in Q[x, y])

[chilly ocean]

are you sure the constants can be written that way

[carmine fossil]

How can a nonzero constant be written as ax+by

[delicate hawk]

oh yeah i messed up

if we pick c \in Q

why is it that (c) is either the entier Q[x, y] or (c) is just the 0 ideal

since (c) = ac (a \in Q[x, y]), cant we miss some elements?

[wind parrot]

no, if c is nonzero and in Q, and p is in Q[x,y], then p = pc^(-1)c, which is in (c) since pc^(-1) is in Q[x,y]

[delicate hawk]

oh, i see

[ivory cosmos]

In general, the principal ideal of a unit (u) is the entire ring

[chilly ocean]

[cloud walrus]

mirzathecutiepie

mirzathecutiepie

mirzathecutiepie

[shell brook]

What is the transcendence degree of a field? And why does transcendence degree + characteristic describe algebraically closed fields upto isomorphism

this is not homework or anything I'm just working through a note set

[next obsidian]

So transcendence degree is a propery of an extension

So let L / K be an extension

[shell brook]

mhm

[next obsidian]

Call a subset S < L algebraically independent if any finite subset of S does not satisfy a polynomial over K

so here's an example

{sqrt{pi}, 2pi} are both trascedental over Q

[shell brook]

right

[next obsidian]

but if you take the polynomial 2x^2 - y

if you plug in (sqrt{pi},2pi) this equals 0

so that set is not algebraically independent over Q

So it's stronger than just saying a single element is transcedental over the field

[shell brook]

ohh okay I see

[next obsidian]

So the transcendence degree of an extension is the size of a maximal algebraically independent subset

This is well-defined and blah blah blah

[shell brook]

okay I think I get that well enough

[next obsidian]

And you have something called a transcendence basis

the size = transcendence degree

and the point is that the basis elements basically act like variables

so say you have tr. deg 2

and transcendence basis alpha, beta

then basically the subfield of L generated by alpha,beta acts like K[x,y]

the alpha and beta have no relations over K

so they act like as if you adjoined variables

[next obsidian]

guess for subfield it should be K(x,y) not []

[shell brook]

right

that makes less sense but I'm following

[next obsidian]

And then the point is

If S is a transcendence basis

Then L is algebraic over K(S)

[chilly ocean]

jesse is an algebraist?!?!?!!?!?!

[next obsidian]

so like if you have transcendence degree n

[shell brook]

i'm reading about model theory over fields tterra

[next obsidian]

you are algebraic over K(x_1,...,x_n)

[shell brook]

so sort of i guess

[next obsidian]

is the idea

it's easier when it's finite

[shell brook]

ohh okay

[next obsidian]

since then it looks more like something you might be used to

[shell brook]

this makes sense I think

[next obsidian]

So like

geometrically there's sort of neat stuff going on

Here you're using () instead of []

so I'm a little less clear on the picture

but I think it's trying to say something like inside of L you have affine n-space over K

and then over that you're algebraic

so what does this mean really? idk

[shell brook]

okay

[next obsidian]

Tfw do AG for 1 year

know no geometry

Also fwiw

actually nvm you said your reason

I was gonna say this stuff isn't usually all that important unless you do number theory type stuff usually

[shell brook]

the other important part was this for me: "why does transcendence degree + characteristic describe algebraically closed fields upto isomorphism"

[next obsidian]

Ah okay

I think this is mainly the "Is algebraic over K(S) thing"

I'm gonna assume char 0 just to try to reason for this one

Actually also

is this transcendence degree over Q?

[shell brook]

just over an arbitrary field I think

[next obsidian]

mkay

[shell brook]

last par

[next obsidian]

hmm

Alright I'm gonna do char 0 and over Q

just for now

[shell brook]

okay

[next obsidian]

so I can think about why it's true

so assume K, L are fields, algebraically closed, and transcendence degree n

Ah okay

so both are algebraic closures of Q(x_1,...,x_n)

so they're iso

they're both algebraic over this field, and algebraically closed

[shell brook]

x_1,... x_n is the transcendence basis?

[next obsidian]

well...

not exactly so

Inside of K you have n elements, say alpha_1,..., alpha_n (this is the transcendence basis)

but the subfield of K generated by those is isomorphic to Q(x_1,...,x_n)

so you have an embedding of Q(x_1,...,x_n) inside of K (by means of the transcendence basis), which K is algebraic over

[shell brook]

why is the subfield iso to Q(x_1,...x_n)?

[next obsidian]

You can map x_i to alpha_i

the fact that alpha_i are algebraically independent says this is injective

or something like that

Let me think a bit harder lol

I think you might be looking at the Q-span of them not the entire subfield generated by it

[shell brook]

and K is algebraic over the embedding b/c of this?

[next obsidian]

Right

[shell brook]

I can approximately see why this works

[next obsidian]

http://www-personal.umich.edu/~alexmw/TranscDeg.pdf

Here's a short pdf

which should have a rundown on it

it's like 6 pages, and it isn't too dense

[shell brook]

awesome

I think I got enough to let me continue

thank you so much!!

[next obsidian]

also for the characteristic thing, I think this is basically just the following fact

If F is a field

You get a map from Z -> F

and the kernel is (p) for p the characteristic of F

this gives rise to a map Z/(p) -> F

if char p > 0 then this means a map from F_p -> F

and I think the transcendence degree you're looking at is over F_p

since that's initial among char p fields

if char p = 0 you have an injective map Z -> F

which gives rise to a map Q -> F by localization

[shell brook]

This part is much more confusing but I think I lack the algebra knowledge to understand it

[next obsidian]

Z is initial in ring

you have a map given by 1 -> 1 and then extending linearly

the characteristic of a field F is th eorder of 1 in F

this is the same thing as the kernel of this map Z -> F

if the order is p, then p maps to p = 0

so the kernel of Z -> F is (p)

By taking a quotient by the kernel you get a map Z/(p) = F_p -> F

but if p = 0 (aka F is char 0)

then the map Z -> F is injective

then you get a map from Q -> F given by sending m/n to m/n inside of F

[charred pewter]

Does anyone here know what the group denoted by B_n is called? I have an assignment that has to do with this group but google keeps giving me unrelated results when I look it up

And it has to do with matrices

[latent anvil]

braid group?

oh matrices hmm probably not

[chilly ocean]

i googled "Bn matrix group" and i found something that referred to the group of invertible upper triangular nxn matrices as B_n

[oblique river]

that's the borel subgroup, yes

it could be B for borel

although I think "borel subgroup" is something used much more generally and I'm not sure if B_n canonically refers to the borel subgroup of GL_n

[latent anvil]

strong lie subgroup theorem: any subgroup which is a borel set is a lie subgroup

[chilly ocean]

[charred pewter]

My assignment had to do with B_2 specifically, and it looks something like the Borel subgroup

[celest brook]

bijective homomorphism

[chilly ocean]

homomorphic bijection

[wild sapphire]

Hey guys, If I have a set A and set B a subset of A. With S a set of permutations p() of A such that for b in B, p(b) in B

Is this S a group?

So basically if we select a subset B of A, and we have a set of all permutations of each element in B. I feel like this does form a group because every permutation can be expressed as a product of disjoint cycles, and so we could say we need to have all related cycles present in the set, so every permutation that is in the group must have it's inverse also present.

Is this the right line of thinking?

[next obsidian]

I don't think it's closed?

If you extend this to say S is the set of all permutations of A which fix elements in B then yes, you're right

err sorry not fix elements

[viscid pewter]

does it necessarily have an identity?

[next obsidian]

but like

p(b) in B when b in B

[viscid pewter]

i don't think it necessarily has to have an identity

[next obsidian]

If you take all of them it will, since the identity satisfies p(B) = B

[viscid pewter]

yeah ofc

[next obsidian]

but if it's just some amount of them, you have issues with identity and with being closed even

[ivory cosmos]

I believe if you have a set of size n + m

and you take the group of permutations moving those m elements only among themselves, you get a group isomorphic to S_n x S_m

[next obsidian]

you are indeed correct

because of bijectiveness

so the m elements move amongst themselves

but so too does the other n, by bijectiveness

and then you just look at what it does to each subset simultaneously

useful for Galois theory

by looking at what it does on the roots

[ivory cosmos]

I think there may even be a functor here

from the skeleton of finset^OP to grp

[next obsidian]

D:

I mean there's no reason to look at finite sets anyway

[ivory cosmos]

countable symmetric groups

[next obsidian]

really you can state this as saying it in terms of looking at permutations of a set S

and any subset T

[ivory cosmos]

[next obsidian]

you get something iso to S_T x S_T^c

[ivory cosmos]

yeah, I think you want Set^OP?

well, or a contravariant functor, depending on your terminolgy

the idea is that coproducts become products

Well, actually I think the category you want is S -> T whenever S contains T

wait nvmd

the permutations of A union B aren't the permutations of A times the permutations of B

[next obsidian]

yeah haha

by what we just observed

the latter has no cross pollination

[ivory cosmos]

What you can do is the banality that if n <= m then S_n <= S_m as groups

but uh, yeah obviously

I wonder if this defines a functor though

I guess I can send cycles (ab) to (f(a) f(b)) right

i.e., that's the homomorphism you get out of a set map

I don't know if this is immediately a well defined homomorphism though

you probably need some condition on f

[wild sapphire]

🤔

[deft oasis]

if it's injective then it'll be a homomorphism

proof: trivial

[latent anvil]

what's the proposed functor?

bijections on a set ?

[ivory cosmos]

A -> S_|A|

f : A -> B goes to phi((ab)) = (f(a)f(b))

I don't know if phi is a homomorphism just like that though

[prisma ibex]

where did the finite field question go 😢

[chilly ocean]

the void

[languid meteor]

sorry I thought I solved it then I realized I didnt

I thought I was interrupting another question

back due to popular demand, the field is F_5[x]/(x^2-2)

Im not sure if direct calculation is supposed to mean find the generator and find all the powers of it to show it generates the group, or im just meant to do some kind of a proof using Lagrange's theorem and certain elements of the group

[stark sigil]

I've seen a few proofs that the nonzero elements of any field (especially finite fields) is cyclic

They don't seem that easy, but the most direct proof I've seen is the one which uses cyclotomic polynomials

[sturdy marsh]

do you mean finite subgroups of the group of units of a field

C^* is not cyclic

by cardinality reasons

[stark sigil]

yes

[sturdy marsh]

the argument I like is the one that uses the structure theorem for fin gen ab groups

or ig you don't need the full result in this case as you're only dealing with finite abelian groups

[stark sigil]

That is pretty high powered

[sturdy marsh]

ye you dont need it

you only need the sylow thms

[stark sigil]

The one I'm thinking of is: If the subgroup has order $n$, then $x^n-1$ factors into linear factors over the field, therefore so does $\Phi_n(x)$, and any root of $\Phi_n(x)$ has multiplicative order exactly $n$, QED

[cloud walrus]

Icy001

[sturdy marsh]

that's pretty neat

[stark sigil]

Now that I think about it, Saintscratch's problem for a specific field was meant to be solved by direct calculation, which probably means that you're meant to explicitly choose some element of $\bF_5[x]/(x^2-2)$ and calculate its powers

[cloud walrus]

Icy001

[prisma ibex]

right so x^d-1 can have at most d roots so there can be at most d elements whose order divides d

only cyclic groups have this property

[languid meteor]

its really annoying to find which element is the generator :/

gonna just have to try a bunch of them I guess

[prisma ibex]

I mean you won't be able to find a particular one, you can just say that there are \phi(n) order n elements in here and all of them have to satisfy x^n=1; so just pick a nontrivial root of x^n-1

[sturdy marsh]

why can't you find a particular one?

it's probably super tedious

but shouldnt be impossible?

[stark sigil]

I suspect nGroupoid is thinking the problem is to prove it for general finite fields

[sturdy marsh]

oh ok

[languid meteor]

is that phi(n) euler phi function?

sorry im taking a really weird set of modules where im doing rings 2 and groups 1 at the same time

and rings 2 requires more group theory than I have currently

[sturdy marsh]

yes

[wild sapphire]

Hey guys, If I'm talking about proof of sets and their related permutation groups and subgroups. Can I just talk about permuting the sets {1,2,3,...} or {1,2,3,...,n} ? Like surely the actual elements don't matter right?

[viscid pewter]

think so

[scarlet estuary]

this is formalized by group actions

specifically, permutations of an arbitrary set are simply actions of the symmetric group over {1, 2, 3, ... n} (of the appropriate size) on that set

and this gives you the symmetric group of the set

[wild sapphire]

group actions like if you said "the binary operation is composition" or something

I think I understand what you're saying

I'm trying to prove that if we have a set X and a subset Y, consider the set S of permutations p of X such that for all y in Y, p(y) in Y

is S a subgroup of S_X?

So I think its like taking an arbitrary subset Y, and saying the permutation group contains all cycles including elements in Y

I feel like this is a subgroup

[chilly ocean]

i think there could be some weird finiteness issue

is Y finite?

[wild sapphire]

This leads me to the question, are there permutations that aren't in cycles?

Like aren't cycles a key part of permutations, like if you just keep applying permutations doesn't it have to "loop back around"

or am I all wrong

[chilly ocean]

they do have to loop back around, if the set of things you are permuting is finite

er

i think that should be acacurate

[wild sapphire]

it feels right at least

🤔

[chilly ocean]

eg what if we are permuting N, including 0, and we shift all even numbers up by 2, we shift 1 down by 1, and we shift all other odd numbers down by 2

[languid meteor]

hey so uh ive found an element that generates 23 elements of the group I mentioned earlier, but im almost certain this is impossible

and I double checked all my calculations so I have no idea whats going on

[mild laurel]

the group should have 24 elements?

[languid meteor]

yeah

like I found the order of an element to be 23

[mild laurel]

okay yea thats not possible

[languid meteor]

yeah I have absolutely no idea whats happened because I just went through the calculations again, I guess ill check them again

[mild laurel]

if you can post i can check with a calculator

[languid meteor]

So its F_5[x]/(x^2 - 2)

So x^2 = 2 and everything is mod 5

Im missing the element 3

[proud bear]

i think you calculated (1+2x)^6 wrong

[median tree]

In my algebra course the professor was discussing the fact that if you have an orbit of X the action of G on the orbit can only produce elements of the orbit, so the orbit inherits an action of G.

He pointedly said that because if we have an element of the orbit and we act by an element of the group we must say inside the orbit, so we have a well defined action of the group. But that in order to say we have a group action on the orbit we have to check that the identity acts trivially and that it's associative.

I'm not entirely clear on what the distinction he's making here is between well defined action of the group and there being a group action on the orbit

How can we have a well defined action of the group... on the orbit... without there being a group action on the orbit?

[viscid pewter]

the only thing i can think of is like maybe there's a thing called an action of which group actions are a subtype?

like, an action just shuffles things, a group action shuffles things but retains like identity and group structure?

[median tree]

Hmm, well it was my impression that "action" generally isn't a well defined term, but "group action" is, which is my confusion

[viscid pewter]

i've never heard of an 'action' sans 'group'

i don't know much AA tho

[median tree]

Alright, well thanks for weighing in. Although this now reinforces my uncertainty as to what distinction he's actually making here, lol

[mild laurel]

By well defined, I feel like your Prof is just saying that the action of G is closed in some sense

So the action actually takes an element of the set X to itself

[median tree]

That would be the same as saying it's transitive, yes?

Uhm, no nevermind

different idea

But alright, so I suppose he's saying that just because the action is closed doesn't necessarily mean it meets the criteria of being a "group action"

[viscid pewter]

yeah

[median tree]

Tyty for helping my clarify that

[ripe basalt]

are elements of a finite-dim C*-algebra self adjoint?

nvm, I mean I can't see any reason they would be

[latent anvil]

can't you take any matrix which isn't self adjoint as a counterexample?

eg
0 1
0 0
in M(2, C)

[ripe basalt]

lmao yeah

where is edd at

we were proving that some map was completely positive last night, and I think we implicitly assumed that the stuff in M_n(C) was Hermitian

because we use rayleigh coefficients, which looks like you can only do if you know your thing is hermitian

[tall jay]

I was wondering if someone can help me with this. I wrote all of this out, so I can show my thought process, but I’m not quite sure how to proceed without listing all the integers in the equivalence classes and then play a matching game

[carmine fossil]

(I will use (p,q,r) to represent the integer which is p mod 5,q mod 7 and r mod 11)
Find integers a,b,c such that a=(1,0,0),b=(0,1,0) and c=(0,0,1)

Your required integer will be 4a+5b+7c

To find a, consider the number 7*11,This leaves a nonzero remainder with 5 and a zero remainder with other 2 numbers. You can then find a number p such that p*7*11 leaves a remainder of 1 with 5 since 7*11 is coprime to 5

Any doubts?

[tall jay]

If we replaced that $x\equiv 7 (\bmod 11)$ with $x\equiv [something] (\bmod 35)$, then we should still have the same solution set, right?

[cloud walrus]

beeswax

[carmine fossil]

x=7 mod 11 and x=19 mod 35 would be the same x as in the given system of equations

[tall jay]

Wait, how did you get the 19?

[carmine fossil]

x=4 mod 5 and x= 5 mod 7 implies x=19 mod 35

Well,I just checked manually

If you want a more methodical approach ,15=(0,1) and 21=(1,0) (first is modulo 5,and second is modulo 7)
So,your required x will be 155+214=75+84=89 mod 35=19 mod 35

[tall jay]

Is this correct?

[rustic crown]

Haven't seen the calculations, but notice that the congruences tell you (2x-3) is divisible by 5, 7 and 11. Hence 2x-3 is divisible by 385. hence 2x = 3 = 388 (mod 385) => x = 194 (385)

so you get 194 and 194+385 as the two incongruent solutions mod 770

[tall jay]

Thank you @rustic crown and @carmine fossil for the help.

[rustic crown]

[tall jay]

It's okay now I think

[rustic crown]

yep... more precisely, action of G on the set A, is same as a group homomorphism from G to Sym(A).

yep. but i like to think the other way.... "all the elements you can get from a"

[carmine fossil]

Perfect

[scarlet estuary]

this seems a bit backwards

[rustic crown]

for example, when you say the orbit of earth you mean all the points the earth can go (in this case the transformation is rotation around the sun)...

x = bh for some h is H is same as saying b = x * (h^-1) for some h in H

[scarlet estuary]

this can be written more succinctly as [x] = {h*x | h in H}

we often use the notation H*x instead of [x] to refer to orbits specifically

(typically with a \cdot, not a *, but discord)

b would be any element of A there

so its the set of all b in A such that there is some h in H with b = xh

oops

A

not X

sorry you were using x so i misremembered the symbol

but yes, the set

in other words, its the set of all action-products xh

sure.

huh?

im saying that the equivalence class [x] is the set of all products of the form x*h

for some choice of h

what do you mean by that

h can be any element of H

in either definition

like okay

can you see why the following sets are equivalent:

{x | x=5} and {5}

if so, same reasoning here

{x * h | for all h in H} doesnt even make sense

i mean the other definition doesnt make sense either

youre quantifying but you dont have a statement associated with your quantifier

okay heres another analogy:

ugh

internet dying

here are two ways we can write the even integers:

{b | b = 2n for some n in Z}

{2n | n in Z}

its the set of all products x*h, where we can choose any element of h.

so if H = {1, 2, 3}

it'll be

{x*1, x*2, x*3}

again youre not really quantifying over a statement

so that declaration doesnt quite make sense

its the set of all xh SUCH THAT h is in H

?

{A | B} = the set of all A such that B

i feel like youre missing a fundamental concept with set builder notation here

you can phrase this using quantifiers but you dont have to

thats my point

{A | B} = the set of all A such that B holds

so:
{b | b = 2n for some n in Z} is "the set of all b such that (there exists n in Z such that (b = 2n))"

{2n | n in Z} is "the set of all 2n such that (n is in Z)"

yes, those a would be the elements of your equivalence class

(assuming the a are meant to be distinct as well)

er wait

hold up

it should be the a's and b's

also it should be 1*b, not b*1

since you dont have abelianness necessarily

have you proved ~ is an equivalence relation yet?

it might help with intuition here

okay, so a ~ b iff there is some h such that a = hb

meaning that if H = {1, 2, 3}

a ~ b iff at least one of the following holds:
a = 1b
a = 2b
a = 3b

a and b are fixed here

but if a ~ b

then they belong to the same equivalence class [a]

so the equivalence class [a] is the set of all b such that a ~ b, i.e. such that either:
a = 1b
a = 2b
or a = 3b

idk what it means for the b to be "distinct" in this context

i get that

i still dont see how it fits in

why

what if A is the set of one element

then 1b = 2b = 3b

for sure

okay sure, the size of this equivalence class is the number of images of a under (group action) multiplication by elements from H

hence, at most the number of elements from H

you might

again, if your set has 1 element

or 2 elements

well you can come up with examples for any cardinality

depending on how your action is defined

its just "forced" for sufficiently small sets

[chilly ocean]

yo does anyone know about some book/paper that explains alternating groups well with describing its properties?

[ivory cosmos]

If R[x] is a euclidean domain, is R necessarily a Field?

[scarlet estuary]

yes

in fact, if R[x] is a PID, then R is a field

this is a decent exercise; it can be done in like 3 sentences

[ivory cosmos]

Let's see:

1. Irreducible elements in a PID are maximal
2. x is irreducible
3. R is iso to R[x]/(x)
4. R is a field

And I guess Euclidean => PID because you can do the standard proof where you pick the minimal degree polynomial in an ideal, and use the division algorithm

[scarlet estuary]

this is the slickest proof

rather, the end of it

and yeah euclidean -> PID

also this of course requires R a comm ring

otherwise R being a field doesnt make sense

[ivory cosmos]

do you ever talk about euclidean domains in a non-commutative setting?

[scarlet estuary]

yes i just mean

my claim it suffices to prove this for R[x] a PID

does require R to be a ring

but since youre discussing euclidean domains, thats given anyway

[ivory cosmos]

I can grok complicated conditions on Rings

but non-commutative variants of these require too much juggling for my head

[scarlet estuary]

yeahhhh working in noncommutative settings is kinda hellish

its not much HARDER in terms of like, actual reasoning skills required

its just a massive pain

and easy to get criss-crossed

[ivory cosmos]

you have to be careful about a lot of stuff I guess

idk, I feel like working with modules is the more natural setting?

this is a bit of a hunch, but I feel like the internal structure of a non-commutative ring is less well behaved then the external structure

that is to say like, the category R-mod