groups-rings-fields

[south patrol]

Lol

Smh

[next obsidian]

You didn’t say that

[south patrol]

Left out half the method

Oh joe

A nice alternative apparently for composites of separable tings is like

You can sps u get smth non separable and play w the min poly since we know we must have char p and smth of form g(x^p)

[warm shoal]

the invertible elements in AxB are those a,b s.t. (a,b)*(a',b')=(1, 1), yes?

Does that emoji mean yes?

=[]

[south patrol]

Uh lemme reas the original source again

[warm shoal]

could just as easily mean no

[south patrol]

But it's on stack exchange

[warm shoal]

@rustic crown am i right

[rustic crown]

yee lol

[warm shoal]

yay!

eahahahheuehuehehheheeeeeeeeeeeeeeee

true

is there anything else i should state about that invertible elements of the direct product of rings?

[rustic crown]

yea just (A x B)* = A* x B*

group of units functor is right adjoint to group ring functor so preserves products

[warm shoal]

ahhhhh

[rustic crown]

.<

i gave a stupid fact which is more painful to verify >.<

than directly proving it lol

[south patrol]

Lol

[warm shoal]

is (0, y) a nonzero element?

or just (0,0)

[rustic crown]

yes if y is non-zero

[warm shoal]

in a direct product of rings

[south patrol]

When is (0,y) = (0,0)

[rustic crown]

is 0 a 0 divisor

cause i don't think i've seen people say "non-zero zero-divisors"

[void cosmos]

isnt in the definition its assumed its nonzero

[rustic crown]

ah probably

[south patrol]

I have heard that

And I've seen people state that the set of zero divisors is a union of prime ideals too

[rustic crown]

cause if i take it too literally, then every element divides 0

[void cosmos]

also you can define it with the map x-->ax is injective

but it wont be injective iff the ker isnt trivial

so ig it wouldw ork

[south patrol]

Why?

Well I assumed ur only interested in finite but the comment seems to agree it is correct

Hm ill look more lol

Well it says like any irreducible factor is of that form but yeah hm

[warm shoal]

why isn't <5> an ideal of Z12?

or wait it is

but how is <1> = <5>

[south patrol]

5*5 = 1 for example

[warm shoal]

ah

thank u potato

[south patrol]

Np

But yeah more generally like

Note that if m is coprime to n then m is a unit in Z/nZ

[sonic coral]

is the length of a permutation written in disjoint cycles the number of elements in the cycle, or that value minus 1?

[south patrol]

Hm is length of a permutation even a standard term

[sonic coral]

of the cycles

[south patrol]

Okay sure yeah it's the number of elements in the cycle

[sonic coral]

that’s what i thought. i was seeing it what i thought to be both ways in different places, but i think it’s because the cycles weren’t disjoint

[tribal moss]

It can sometimes lead to confusion that a cycle of even length is an odd permutation and vice versa.

[wicked zephyr]

An R-Module is projective iff it is a direct summand of a free module. Direct sum of flat modules are flat. Why do this two facts imply that projective modules are flat

?

[sonic coral]

this is only true if the cycles are disjoint right?

i guess if they weren’t, we could still find the parity of the composition, two evens is an even, two odds is even, and even odd is odd.

unless i’m thinking about that wrong

[tribal moss]

I'm saying a single cycle of even length is an odd permutation. There's nothing for it to be disjoint from in that statement.

[thorn delta]

Suppose F = K \oplus P where F is free and P is projective. Then a map f : A -> B is injective if and only if the map g : K \oplus A -> K \oplus B defined by g(k,a) = (k, f(a)) is injective. Tensor g with P, and then use natural isomorphisms.

oh, and you'll need to use the fact that free modules are flat

[wicked zephyr]

zenc yu @thorn delta

[thorn delta]

npnp

[thorn delta]

i think i said this slightly wrong. Tensor f with P to get a map A \otimes P -> B \otimes P, and then define g : K \oplus (A \otimes P) \to K \oplus (B \otimes P), and then use natural isomorphisms and flatness of F to see that g is injective, and therefore the map A \otimes P -> B \otimes P is injective

[white oxide]

can somebody give me a hint for a)? my first strategy was to show that the cancellation laws hold in R which is equivalent to saying that R has no divisors of 0, but i couldn't really get anywhere for that. i then tried multiplying both sides by b and i got that ab is idempotent, but again nowhere with that. maybe contrapositive or contradiction is the way to go?

and i'm not so sure how i can do a direct proof, since if a is nonzero then i must show that ab = 0 implies b = 0, but here a is nonzero on the right side of the equation aba = a

[opal osprey]

Do you guys have any examples of a finite dimensional associative unital k-algebra that do not admit a finite dimensional faithful representation.

[white oxide]

i suppose i could do aba - a = 0 so a(ba - 1) = 0 and if i could get ba = 1 then that would show that the cancellation laws hold

but that's assuming R has divisors of 0, but it seems like they want me to go in that direction?

nvm i got it

[white oxide]

what? wouldn't we have to show that ab = ab and ba = ab since ab is the unity in a ring R and a unity is unique?

well i suppose they have ab = 1

and ba = 1

[rustic crown]

whut about the left regular rep? A acting on itself by left multiplication, since A is unital it should be faithful.

[rustic crown]

not sure i understand what you asked >.<

[formal ermine]

I'm trying to show that the absolute galois group of Fp is the profinite integers

so we have invlim Z/nZ

I was thinking about using crt "to get something with primes"

but not sure how I would continue from there

[rustic crown]

show that the galois group is inverse limit of galois group of finite galois extensions of F_p

[formal ermine]

yes I've shown that already

I got it down to invlim Z/nZ

[rustic crown]

oh okie

[formal ermine]

my definition is prod_p Z_p

how do I show that those are equivalent

is my question

[rustic crown]

universal property bash it :p

but yea CRT is the only ingredient needed

[formal ermine]

am I missing something obvious

we have invlim Z/p_1^i_1Z x ... x Z/p_n^i_nZ

what's the next step

[rustic crown]

and this breaks apart

but instead of doing it like that, better to construct maps both ways

[formal ermine]

right I'll try that

[formal ermine]

but for this, do we have invlim A_n x B_n = invlim A_n x invlim B_n or something like that in general?

[rustic crown]

that's true but here you don't have the same indexing n for both right

[formal ermine]

yeah

invlim A_i x invlim A_j

interesting

yeah then it's obvious

[rustic crown]

(ignore what i just wrote)

[solar shore]

why is it isomorphic

[formal ermine]

what did you write

[rustic crown]

something wrong 😂

[silent oxide]

try writing out the obvious map

[formal ermine]

I didn't even see it lol

thanks for the help tho

[solar shore]

how do you mean?

wait

im mf stupid

.

bru

disregard this questoin, i forgot its 10...

[kind jacinth]

is this D, B, D?

or do i get negative points

[rustic crown]

when was this

[frigid lark]

Helping some friends with algebra

[rustic crown]

is L2 context free?

doesn't feel like it... but i don't remember much of this stuff

there was also a pumping lemma thing for CFGs

[kind jacinth]

im not entirely sure but for context free it means i can think of a grammar for it right?

[rustic crown]

yee

but here i don't think you can

[kind jacinth]

how would i prove this by contradiction

so we take z = b^x c^3x b^x (cb)^x?

ok i think i can see why its not context-free. in that case I would go with C?

[rustic crown]

oh wait

i dind't see j was bounded >.<

ofc then it's context free

[kind jacinth]

only j is bounded

i is from 62 upwards

[rustic crown]

yea, so it's a finite union of context-free languages

which will be context free

[rustic crown]

that's a minor thing, doesn't affect much

iirc context-free intersect regular is context-free

although intersection of two context-free may not be context free

simplest example is {a^nb^nc^m} intersected with {a^nb^mc^m} which is {a^nb^nc^n}

[kind jacinth]

alright thanks for the help!

[rustic crown]

what's your logic for third being D?

that language feels more complex than halting, so feels like it's not RE

but idr this stuff

[civic valve]

is a the channel I can use for questions about linear algebra?

[silent oxide]

graduate linear algebra, yes

[warm shoal]

"list all the homomorphisms from Z2 to Z4"

aren't there so many like

im confused

[silent oxide]

no

0 must be mapped to 0

so you just control where 1 is mapped to

Can 1 be mapped to 1?

[warm shoal]

i think not but idk why

[silent oxide]

what happens when you do phi(1+1), where phi is the homomorphism?

[warm shoal]

u get phi(2)

but that

is out of domain

right?

[silent oxide]

well 2=0 in Z2 so it's the same as phi(0)

[warm shoal]

oooo

[silent oxide]

so 2*phi(1)=phi(1+1)=phi(0)=0, so phi(1)=0 or 2 in Z4

[analog zephyr]

Mm general, for Zm to Zn the prime numbers have something to do with the amount of homomorfismo between Zn and Zm?

[delicate orchid]

It’s the gcd of m and n

Any map into a cyclic group of prime order must be surjective or trivial as Zp contains no proper subgroups - hence there are only non-trivial maps when Zm contains a copy of Zp - maybe that’s what you’re thinking of

Likewise with maps going out of Zp actually, they have to be injective or trivial

Le schur’s lemma has arrived

[warm shoal]

so @silent oxide

[analog zephyr]

For two cíclic group there are only one isomorphism, no ?

[warm shoal]

we're talking ring homomorphisms btw

[analog zephyr]

Ohhh

I though you were talking about groups

[warm shoal]

sorry no

[silent oxide]

yeah me as well

lol

[warm shoal]

ring homomorphisms from

Z2 to Z4

sorry!

this is confusing the boy!

[delicate orchid]

Oh lol

[silent oxide]

well phi(1) can only be 0 or 2 as well, but if phi is a ring homomorphism, we also have
phi(1)=phi(1*1)=phi(1)*phi(1), so of phi(1)=2 we have a contradiction

[south patrol]

lol

[coral shale]

schur

[untold cloud]

Hi, guys, we know that if N, H are subgroups of G, and if N is normal( H< N_G(N)), then HN is a subgroup of G, but does the same condition imply that NH is also a subgroup?

[formal ermine]

the product of two subgroups is a subgroup iff they commute with each other

this is the case if either one of them is normal

if both are normal then their product is normal

[lethal dune]

under your condition HN=NH

[untold cloud]

Oh, I see, because we have hN = Nh for all h\in H

Thanks!

[kind jacinth]

what does this language even mean??

like w=aaabba so numa(w) = 4 and numb(w) = 2?

so a^4#b^2#aaabba is in my set?

[glossy crag]

So I know the natural irrationalities theorem for finite extensions (K/F finite Galois => KL/L Galois and Gal(KL/L) isomorphic to Gal(K/K\cap L) via restriction to K), but does it hold for infinite ones (i.e. only assumption K/F Galois)? It's easy to show that KL/L is Galois and that restriction to K defines an injective homomorphism Gal(KL/L)->Gal(K/K\cap L), but in showing it's surjective the proofs I know use the Galois correspondence for K/F and idk how to show the image is a closed subgroup for K/F general (is it?).

[rustic crown]

can't you say like for finite galois extensions k/KnL, you have the isomorphism, then you take inverse limits over all such k. the extensions kL/L are also finite galois and moreover are a cofinal subset as any finite galois extension l/L with l <= KL is contained in some kL/L for k/KnL finite galois.

[rustic crown]

yee

[rustic crown]

or in other words, to show surjectivity i'll pick any automorphism of K/KnL and look at the induced automorphism of k/KnL, that will give me an automorphism of kL/L, these automorphisms define an automorphism of KL/L because each element of KL is contained in some kL, and the family of these automorphisms all agree on intersections as the reverse map was just restriction.

[formal ermine]

in the case of topological spaces, does the inverse limit "loop over" the open sets?

[agile burrow]

What do you mean

[formal ermine]

let T_i be a sequence of topological spaces, then proj lim T_i = subset of prod T_i

is this limit the topology

so prod T_i refers to the topologies as a product topology or whatever?

[agile burrow]

The limit will be a subspace of the product of the topological spaces

Also, maybe this isn't suited for this channel

[formal ermine]

sorry

that was a dumb question after thinking about it again

[empty rose]

is there a short english name/expression for this monoid? $\langle a,b | a^2 = a, b^2 = b \rangle$

wait no hang on that's the wrong monoid

[cloud walrus]

bee

[empty rose]

there

[glossy crag]

but an auto of K/KnL does not necessarily restrict to an auto of k/KnL, k would have to be normal (aren't we doing this for k finite arbitrary?).

[rustic crown]

oh we're doing it for k/KnL finite galois

so yea you have normality

[pastel cliff]

is "bilinear map of Rmodules that yields an Rmodule" appropriate intuition for what a tensor product is

[rustic crown]

it also yields a linear map

M x N --> M⊗N is the universal bilinear map, so any other bilinear map M x N --> P must factor uniquely through M⊗N giving a linear map M⊗N --> P

[pastel cliff]

in other words any bilinear map out of M x N is a tensor product composed with something else

[rustic crown]

yee

[pastel cliff]

Show that the element $2 \otimes 1$ is 0 in $\Z \otimes_\Z \Z/2$ but that it is not 0 in $2\Z \otimes_\Z \Z/2$.

[cloud walrus]

sebbb

[pastel cliff]

ok i have this as a practice problem

the definition of a tensor product that i have is like

in terms of the universal property thingy

but what's it like for individual elements

[south patrol]

Well

A key thing is that the tensor product map is a bilinear map A x B -> A \tensor B

in other words lol like

(a+b) (x) c = a (x) c + b (x) c and all that jazz

[rustic crown]

only two unicodes i remember are u+2295 and u+2297 for ⊕ and ⊗ :3

[pastel cliff]

\otimes

[charred crescent]

can someone help me with the following problem?
Let N be normal in G, such that [G:N] = m. Show that a^m is in N for all a in G.

[rustic crown]

whut have you tried

[charred crescent]

I first considered the natural projection p: G -> G/N by p(a) = aN

then p(a)^m = (aN)^m

but im not sure what i can and can't say about (aN)^m given [G:N] = m. For instance, is the order of aN = m? does it divide m? I'm not sure what can be correctly said regarding this bit.

i guess i don't know the rules well enough to proceed

[rustic crown]

when since N is normal G/N is a finite group of order m

so what does lagrange tell you

[charred crescent]

does order of aN divide m?

i don't know if im being honest

i know that any subgroup has order that divides the original group

[rustic crown]

yep, and you have an arbitrary group H of order n, and h is an element then you can look at the subgroup generated by h!

what can you say about the order of the element h vs the size of the subgroup <h>?

[charred crescent]

if h generates it then the order of h equals the size of <h>

[rustic crown]

yep, so order of h has to divide |H| :3

in particular h^|H| = 1

[charred crescent]

so the order of aN divides |G/N|

[rustic crown]

yep :3

so that gives you p(a^m) = identity in G/N

[charred crescent]

okay so (aN)^m = (a^m)N = N, thus a^m is in N

[rustic crown]

yep

[charred crescent]

alright so is it even necessary to talk about the natural projection here? or can we simply consider aN in G/N for any a in G?

[rustic crown]

yea not required

[charred crescent]

oh okay, that makes sense

thank you for your help!

[rustic crown]

[pastel cliff]

det can we talk about RCF again

wondering if this was all accurate

link to convo

ok more concretely im a bit confused at this lemma

given a decomposition of V_T (a k[t] module) into k[t]/a(t) for a bunch of a(t)'s

each component of the direct sum has the "standard basis" (or maybe more like the expected basis) of a polynomial ring modded an ideal generated by a polynomial

what's the k-linear operator

[rustic crown]

okie so when you have a k-vector space V and a linear operator, you can form a k[t] module out of it, where multiplication by t now acts by T.

now you use structure theorem on this k[t] module

[pastel cliff]

oh so is it that map on each index of the thingy?

[rustic crown]

right so you wanna understand the multiplication by t on your space V. so you decomposite it as a k[t] module and understand each summand nicely

each summand looks like k[t]/(a(t)) like you said

and this has an obvious choice for a basis

but yea people do mix up the order

some people like to do 1, t, t^2, ...

others do the reverse

t^n-1, ..., t^2, t, 1

[pastel cliff]

why doesnt it go up to t^n

[rustic crown]

because as a k-vector space k[t]/(a) has dimension deg a = n

[pastel cliff]

ohhh

[rustic crown]

since you started at 1 = t^0

you need to end at one before n

[pastel cliff]

ok hm

i need to phrase this all out myself

a k-linear operator is what we represent with a matrix right

[rustic crown]

it's useful because if you have a vector space V and two operators S and T on it... these operators are similar if and only if the corresponding k[x] modules are isomorphic

[pastel cliff]

ohhhhh wait ok so

given the k-vector space and an arbitrary k-linear map

wait no i lost it

grahhhh

[rustic crown]

right so you wanna write the matrix for "multiplication by t" map

[pastel cliff]

and that's a map from the k[t] module to itself right

[rustic crown]

yep

[pastel cliff]

so once we've expressed the k[t] module as the direct sum of those other guys, looking at how that endomorphism works on each summand gives us the characteristic polynomial for a given summand

then putting each companion matrix for those polynomials in one bigger matrix gives us RCF...?

[rustic crown]

yee

[pastel cliff]

:O

[rapid junco]

Do you guys think AI poses a threat towards Math research?

[rustic crown]

no

[fleet pelican]

Penrose argues that godel's theorem implies no

[rustic crown]

also go to some general channel

[rapid junco]

Because it can only spit out stuff that has already been discovered right?

[pastel cliff]

yeah #discussion

this is the det tutoring channel

[rustic crown]

[pastel cliff]

[sonic coral]

and mine

[pastel cliff]

he's the one tutoring us

[next obsidian]

[pastel cliff]

[rustic crown]

[pastel cliff]

this is the same as saying that S and T have the same RCF right

[rustic crown]

yep

so for example with that you can show that any matrix is similar to its transpose

[pastel cliff]

the last simple thing im a bit fuzzy on is how we get from the k-vector space to the k[t] module

do we just like

do it

[rustic crown]

univewsal pwopewties

an R-module M is a ring homomorphism R --> End_{Ab}(M)

so a k-vector space is map k --> End(V)

if you have a nice element T in End(V)

you can extend it to k[x] --> End(V) by sending x to T

T needs to be k-linear to be able to do this

because in k[x] scalars and x commute

[pastel cliff]

yeah im assuming that for the T and S from before too

[rustic crown]

that they're k-linear?

or if S and T commute?

[pastel cliff]

here yeah

that theyre k linear

[pastel cliff]

and k-linearity is all we need for this to be "nice"? since that means we can write it as a matrix?

[rustic crown]

you already know it's additive since it's an End of abelian groups

[pastel cliff]

thank you det

super helpful as always

[rotund aurora]

quite the contrary? like having computers proving the Riemann hypothesis or something wouldn't "ruin" research. Also, AI is also mathematically interesting

also yes #math-discussion

mb

[pastel cliff]

john carmack has a podcast with lex fridman where they talk about AGI, pretty interesting (i dont have access to those channels so this will be the last i say on it)

[rustic crown]

studying role is uwu

[rapid junco]

I meant like jobs

[pastel cliff]

.

[pastel cliff]

sometimes i wish it wasnt perma but then i remember i here to do me a learn

[rustic crown]

okie det also has studying role :3

[pastel cliff]

studying!

,iam dying

[cloud walrus]

Gave you the studying! selfrole.

[pastel cliff]

double study

[rotund aurora]

how did you get perma-study

[pastel cliff]

just ask a mod for it

i bothered roketto like a year ago and havent taken it off since

[rotund aurora]

[pastel cliff]

you have to ask a mod to take it off too though

and note that you still have access to obsidian and advanced lounge

[rotund aurora]

Does the inequality $||x|-|y||\leq |x-y|$ hold for normed vector spaces in general?

[cloud walrus]

Croqueta

[void cosmos]

yes

all u need is triangle inequality

[rotund aurora]

k

ah yeah right

thanks

[coral shale]

doesn the defn of norm involve triangle ineq

[rotund aurora]

yes

but like idk why I just didnt see immediately that you could derive it directly lol

[haughty shale]

Hi! If $\omega$ is $n$th root of unity, is $\sum_{i=1}^n \omega^{4i}$ $n$th root of unity as well?

[cloud walrus]

emphatic_wax

[rotund aurora]

n=2

thats like $\frac{\omega^{4n+4}-1}{\omega-1}$, no reason why that should be an nth root of unity

[cloud walrus]

Croqueta

[haughty shale]

Okay gotcha

[coral shale]

using i as an index

[haughty shale]

yes

is n=2 like for a counterexample?

[rotund aurora]

the second roots of unity are 1 and -1

like omega could be 1 lol

[haughty shale]

ohhh okay

got it then

thank you

[rotund aurora]

but thats true if omega is a 4th root of unity that is not one

and this is the only case (besides when n=1 and omega=1)

try figure it out

[haughty shale]

okay thank you!

[rotund aurora]

well its what I wrote

[charred crescent]

when we have a mapping f: G/N to G/H defined by f(xN) = xH
(and N,H normal in G, with N subset of H of course), is surjectivity obvious?

[coral shale]

n in h...

[rustic crown]

well... a preimage of xH is xN

[fleet pelican]

yes

[charred crescent]

it's like

well xH comes from xN so the x matches up

but how should i go about "rigorously" showing this

[fleet pelican]

take any element gH in G/H....

[haughty shale]

hey is this what the sum is equal to?

[charred crescent]

i currently have; "take any xH in G/H. we know f(xN) = xH, thus f is onto"

[fleet pelican]

exactly

[charred crescent]

okay thanks, ik it's simple but i always feel as if im doing this stuff incorrectly

[rotund aurora]

Yes, if omega not 1

Its a geometric sum

[fleet pelican]

dw, you'll do it incorrectly next time

[haughty shale]

okay thank you!

[charred crescent]

you're right

[coral shale]

I mean, this is not rigorous

as is.

you want to briefly justify choice of x doesn't matter

OR rather than saying take xH in G/H

say take x in G

[charred crescent]

okay that is fair

thank you

[pastel cliff]

ok i going back to tensor products now

what is 1) sstating

2. is pretty clear

is it like a weak basis?

[thorn delta]

its stating that simple tensors m\otimes n for m in M and n and N generate M \otimes N as an R-module

[pastel cliff]

isnt that a bit circular

[thorn delta]

i.e. if $f \in M \otimes_R N$ then $f = \sum_i r_i m_i \otimes n_i$ for some $r_i \in R$, $m_i \in M$, $n_i \in N$

how is that circular?

[cloud walrus]

kxrider

[thorn delta]

in other words, M \otimes_R N is the R-linear span of all the m_i \otimes n_i

[fervent rock]

is this okay to say? trying to express Z_140 as isomorphic to product of cyclic groups

[pastel cliff]

hmmm

ok simpler question then but what is a simple tensor

[fervent rock]

And similarly U(140)

[pastel cliff]

i know it's in what i sent

but differently i guess

[thorn delta]

do you know how the construction of the tensor product works?

[pastel cliff]

that's a bit later in my notes but i vaguely know yeah

[thorn delta]

well okay,

you have a canonical bilinear map M \times N -> M \otimes_R N

a simple tensor m \otimes n is just the image of (m,n) under this map

[pastel cliff]

yeah ive seen the diagram for this

[pastel cliff]

so 1) here is just saying that it's surjective..?

no

nvm surjectivity is weaker than image spanning i think

[thorn delta]

yea, but your instinct is correct

[sick meadow]

1. tells you that you can choose to think of the tensor product space as the space of formal linear combinations of simple tensor elements

[thorn delta]

when you construct the tensor product, youll see that this canonical map is just the composition
$M \times N \to F(M \times N) \xrightarrow \pi F(M \times N)/K = M \otimes_R N$, and you're essentially just asserting that $\pi$ is surjective

[cloud walrus]

kxrider

[lime badge]

Hi everyone, I've been self studying aluffi and am stuck on the following problem (1.20 from ch 7 of the book)

Let $p$ be a prime number and $\alpha = 2^{1/p}$. Let $g(x)$ in $\mathbb{Q}[x]$ be a non constant polynomial of degree less than $p$. I need to show α can be expressed as a polynomial in $g(\alpha)$ with rational coefficients.

I tried tinkering with a few examples like $p=3$ and $f(x) = x^2/2$ and $g(x) = x^2$, and I think the overall intuition is to construct $f$ such that $f(g(\alpha)) = \alpha$. You can get rid of the constant term in $g(\alpha)$ if you define $f_1(x) = x - g_0$ and then my intuition is to try to define other $f_i$ such that you reduce the power of all the other terms in $f(g(\alpha))$ by one with each iteration.

But the issue I have is generalizing my toy examples to arbitrary polynomials $g(x)$ (with degree less than $p$) and $f(x)$

[cloud walrus]

Eternal Way

[lime badge]

Maybe there's a simpler way of going about this, like thinking about Q(g(α))

[agile burrow]

Indeed, you should think about the tower of extensions Q(alpha) contains Q(g(alpha)) contains Q

[grand cliff]

This seems like it should be easy but I'm not sure really what to do

I know there'll be a nonempty normal subgroup consisting of the elements of order 2

But not really sure where to go

I was thinking maybe that different group elements will either include or not include the element of order 2, therefore splitting the group

[vivid tiger]

well, it sounds like you are almost done

[chilly ocean]

exact sequences of that form dont nec split

so you cant view G/<g> in G necessarily

i think that problem is tricky. maybe sylow?

[tribal moss]

If Sylow were available, there wouldn't be a need to point to Cauchy's theorem in the problem statement.

[chilly ocean]

first thing to notice is that n odd is necessary as A_4 has no index 2 subgroups

[tribal moss]

Here's one thing that seems to work: ||Let the group act on itself by left multiplication, and send each element to the parity of the permutation it creates.||

[grand cliff]

I don't see why the index of that subgroup would be 2 though?

[tribal moss]

It isn't necessarily -- for example if G is cyclic of order 2n, there is just one element of order 2, so the index of that subgroup is n.

[vivid tiger]

(this is why i crossed it out)

[grand cliff]

This sounds like the intended solution, since this question came up in a section on group actions

[lime badge]

Oh, but Q(alpha) has prime degree over Q, but such an extension has no subrings properly contained in Q(alpha) and properly containing Q. This implies Q(alpha) = Q(g(alpha)) right?

[agile burrow]

well we don't necessarily know that Q(g(alpha)) is properly contained in Q(alpha)

ah yeah, you've fixed it in the edit

Maybe it's worth justifying why g(alpha) isn't in Q, but otherwise you're essentially done

[lime badge]

Awesome, thanks!

[lime badge]

And just to confirm, this is because g has degree less than p right?

[agile burrow]

yes, that's right

[fervent agate]

I'm trying to prove that if F is a finite field, then there exists a prime p such that pa = 0 for all a in F (where pa = a + a + ... + a).

If we suppose F has n elements, then certainly na = 0 due to F being a group with addition. My professors hint to us is to then consider that n1 = 0, and to leverage properties about fields and integral domains. But I really have no idea how to get started.

[agile burrow]

Well, what happens if n isn't prime?

[fervent agate]

Well the group (F, +) doesn't care whether n is prime. It's happy to be a finite abelian group regardless. So we must get some funkiness when looking at multiplication... Not sure what

[agile burrow]

Sure, but like

What explicitly does it mean if n isn't prime?

[fervent agate]

If n isn't prime, then the orders of nonzero elements w.r.t addition can be less than n

[tribal moss]

Is n the size of the field?

[fervent agate]

yes

[tribal moss]

I wouldn't start there. Instead I'd look at the smallest positive m such that m1 = 0.

(n1=0 is still relevant for that approach, I suppose, because it tells us that at least one such m exists).

[fervent agate]

If m1 = 0 then ma = 0 for any a, so the order of every element divides the order of 1. Now if I can just force m to be prime

[tribal moss]

What happens if m is not prime?

[fervent agate]

If m is not prime, then we could have a nonzero element $a$ whose order is less than $m$ but divides $m$. Let's say $k\cdot o(a) = m$. Then $\sum_{i=1}^k a = 0$. But this is just $a\sum_{i=1}^k 1 = 0$. However, $a$ is nonzero and $F$ is an integral domain, so $\sum_{i=1}^k 1 = 0$. This contradicts that the order of 1 is $m$.

[cloud walrus]

ecurtiss

[fervent agate]

Hence the order of 1 is prime, and in fact the order of every nonzero element is the same prime.

(order w.r.t addition)

[tribal moss]

Yes.

[fervent agate]

Thank you!

[white oxide]

are all finite integral domains isomorphic to Zp where p is prime?

[thorn delta]

no

there are finite fields of order p^n for any prime p and n > 0 for example

[white oxide]

oh ok i didnt know that

thanks!

[thorn delta]

npnp

[lethal dune]

But these are the only ones

[white oxide]

hmm

do you know what texts discuss this?

i believe you fasho but i just want to read up more on it

[lethal dune]

every finite integral domain is a field

and finite fields have cardinality pⁿ

proof is easy as R → R by x ↦ax is injective and being finite it's surjective. so there is one b s.t. ab=1=ba so every nz element is invertible

[formal ermine]

the p^n part is then just it being a Z/pZ vector space where p is its char, by simple counting you get that it has order p^n for some n > 0

uniqueness follows from it being the splitting field of x^p^n - x

[grand cliff]

Hey! I've been stuck on this book exercise for a while:

If G acts transitively on a set S and H <= G with index s.
If the induced action of H on S has l orbits then l <= s. (G can be infinite)

My intuition is that each coset essentially adds more actions and eventually when you add all the cosets together you get the original transitive action.

My expectation is that somehow we can prove that each orbit progressively gets eliminated by adding more cosets and (for some reason) can only be up to s orbits

[grand cliff]

Maybe it would be better if I try and find an injection from cosets to orbits

[thorn delta]

is H normal?

[grand cliff]

Sadly not

[thorn delta]

have you tried letting G act on the set of orbits of the action of H on S?

[grand cliff]

No, what is the intention of that

[thorn delta]

i thought i had an idea but realized i made a silly mistake

[thorn delta]

actually, i think my idea does work

[grand cliff]

Okay nice, I'm guessing the action of G on the orbits is the "typical" one

where you act on a representative or something

[thorn delta]

wait hold on, i may have made a mistake

[grand cliff]

lol

All good, what's your idea/

[rustic crown]

another cute argument is by saying if it had two different prime factors, then you have elements of (additive) order p and q. this would force char = 1 which is bad.

[thorn delta]

i don't think the action i had in mind is well defined, but my "idea" was to let G act on the set of orbits of the action of H on S, call it S', by g[x] = [gx]. This action (if it were well defined) is transitive.

Then you get a homomorphism G -> Sym(S'), where Sym(S') the symmetric group on S'. Since the action is transitive, this homomorphism is surjective. Let gH be a coset of H in G. Then
gH[x] = g[Hx] = g[x] = [gx]. This defines a surjective map G/H -> S(S') which would show s >= |S(S'| >= l.

[grand cliff]

Yeah, I'm currently trying to prove the well-defined-ness

Like, I get to this point

x^g is is g acting on x if that's not obvious

[thorn delta]

Since the action is transitive, this homomorphism is surjective
also, i think this is wrong actually

[grand cliff]

Last line is incomplete

I'm kinda stumped, though, doing an action on the orbits seems like the right idea. Just need to make a well defined action

does conjugation work?

[thorn delta]

wdym by conjugation?

[grand cliff]

apologies my brain is fried

for some reason I was thinking we were acting on a group for a second there lol

[thorn delta]

its okay mine too

[grand cliff]

act by g^-1 instead of g

and it's well defined I believe

[thorn delta]

i thought of that. i don't think it works

[grand cliff]

why not

wait nvm, I think you're right

[thorn delta]

so, define [x]g = [g^-1 x]. then we have to show if hx = y then [g^-1x] = [g^-1y] and we pretty much have the same problem

[grand cliff]

yeah true

WAIT

i got it I think

your action needs to multiply the other direction

so let:

[x]g = {xhg | h in H}

rather than what we were doing which was
[x]g = {xgh | h in H}

[thorn delta]

hm

is the action of H on S a left or right action?

i thought it was a left action

[grand cliff]

Is there anything wrong in these steps?

I'm guessing when I do g^-1 on the right that doesn't work

[grand cliff]

right action I think

[thorn delta]

[x]g = {xhg | h in H}
I don't think is an orbit of the action of H on S. For h \neq h', gxh might not be in the same orbit as xh'g

[grand cliff]

Yeah I am concerned about that

we could do conjugation on H to get a different orbit, I'm not sure if that is closed though

Like:

Take H -> g^-1 H g

and then take the orbit of the element w.r.t. that

Does that make sense?

[thorn delta]

not sure, like [x]g^-1Hg = [xg^-1Hg]?

[grand cliff]

So the orbit of an element x under group H is the set { xh : h in H }

If H <= G,
Is the following also an orbit:

{ xg^-1hg : h in H }

Like, potentially this orbit of x under g^-1Hg is identical to the orbit of some y under H (this would need to be proven to show it's true)

Since the conjugates of a subgroup all kind of behave the same way

That's my intuition for why this action might work

[hot lake]

if N is the stabilizer of any element of S, S is pretty much G/N, and the orbits of the actions induced by H are the same as double cosets H\G/N. We have an obvious surjection H\G -> H\G/N

and uh according to wikipedia the sizes of H\G and G/H are equal

so H\G has size n and H\G/N has size l

the hard part is probably showing that |H\G| = |G/H|

I'm not sure how to prove it

[grand cliff]

somehow i am more confused 💀

[hot lake]

define H\G to be the set of right cosets of H

H\G = {{hg}, h in H}, g in G}

[grand cliff]

Yep okay

[hot lake]

if s is any element of S

you have a surjection G -> S given by g -> g.s

it's a surjection because the action of G on S is transitive

now I claim this surjection induces a surjection from H\G to the H-orbits of S

so I have to check that two group elements in the same right H-coset are sent to the same H-orbit

which is uh just unfolding definitions

hg.s = h.(g.s) is in the same H-orbit as g.s

so we get a map H\G to the set of H-orbits

and it's still surjective

[grand cliff]

Okay nice

[hot lake]

so the only thing we really have to show

is that the size of H\G is equal to n

and if we define the index to be the size of G/H, that's not immediately obvious

let N be the kernel of the action of G on G/H

it's a normal subgroup of G of finite index (because it's the kernel of a map into the group of permutations of (G/H), which is finite)

and hopefully it can make a clearer picture

then G/H and H\G are isomorphic to the cosets (G/N)/(H/N) and (H/N)\(G/N) between finite groups

[solar shore]

could someone explain why $a \in \langle 15 \rangle$?

[cloud walrus]

blanket

[hot lake]

so their size is [G:N] / [H:N] in both cases

[frigid lark]

So a must have order 1,2,5,10 else the order of (a,b), which is a multiple of the order of a, cannot be 10. Thus one of these hold, 150 | a, 150 | 2a, 150 | 5a, or 150 | 10a. Observe that if one of the first 3 hold, so does the last one. Now we know that 10a = 150b for some integer b, thus a must have factor 15, thus a is in the subgroup of Z/150Z generated by 15

[past path]

I'm having some difficulty with the proof that the sum and product of two algebraic integers are algebraic integers.

I proved that if u is an algebraic integer, that u + n and un are algebraic integers if n is an actual integer, but I'm not sure how to generalize, since the proof relied on the fact that I could construct another polynomial with integer coefficients.

[silent oxide]

One way to prove $\mathbb{A}$ is a ring is by proving $$\alpha \in \mathbb{A} \iff | \mathbb{Q}(\alpha):\mathbb{Q}| < \infty.$$

[cloud walrus]

cflau_

[past path]

It's not a trouble with the algebraic numbers

but the algebraic integers.

We can show that the algebraic numbers are in fact a field.

But the problem lies with the algebraic integers.

I'm not sure how to show that the sum/product of two algebraic integers is an algebraic integer rather than just a general algebraic number.

Intuitively, I know it to be true.

And rigorously, I've shown it from a number theoretic perspective, but since we're dealing with this from an algebraic perspective at the moment I'd like to try to wrap my head around it that way.

[delicate orchid]

number theory isn’t algebra
Very based

[past path]

Well

[delicate orchid]

What’s the definition you’re using for algebraic integers is it the polynomial one or the slightly stronger (but equivalent) matrix one

[silent oxide]

whoops, I think a way to prove algebraic integers is a ring is by considering the characteristic polynomial

[past path]

The way I proved it in number theory (and I can't remember it in full generality) is by using Sylvester matrices to construct resultant polynomials, which is not what we're dealing with right now in algebra.

Yeah, the polynomials were my first thought, but I wasn't sure how to show that the minimal polynomial for u+v is monic.

(and has integer coefficients)

If u and v are algebraic integers, they are the roots of some monic polynomials in Z[x].

I'm not sure how to show that the same is true for u+v and uv.

I proved it for u+n and un where n is an actual integer

because there is an explicit construction for those polynomials that keeps the integer coefficients.

[delicate orchid]

it’s easier if you think about them as satisfying the characteristic polynomial for a matrix

[past path]

I seem to recall the number theory proof being very constructive in a way that involved a lot more computations than we normally use in this class.

What do you mean about the characteristic polynomials for matrices?

[delicate orchid]

Then your algebraic integers are eigenvalues of those matrices

[past path]

It's been awhile since I've dealt much with those. We haven't actually looked at a matrix once in this class.

Well

I take that back

I think we dealt with the quaternions as a group of matrices

[delicate orchid]

Have you seen the tensor product yet

[past path]

but we deal a lot more with the abstract side than the representation side.

Yes.

If it means anything to you, we're in Hungerford, chapter 5.

[delicate orchid]

Ok good, then you should be able to take A and B with eigenvalues u, v and construct a matrix with uv as an eigenvalue fairly easily

u+v is a bit more tricky

[past path]

Hold on a sec; I'm not sure how to do that.

[delicate orchid]

and then since it’s an eigenvalue it satisfies the characteristic polynomial

[past path]

We dealt with abstract tensor products.

I've dealt with eigenvalues before in the past

But we haven't looked at them in this course.

[delicate orchid]

Ok call this the krondecker product then

[past path]

I haven't heard of the krondecker product.

[delicate orchid]

Ngl this excuse is flimsily, you can do some research lol

[delicate orchid]

It’s not that complicated

[past path]

That's not really my point lol

I'm sure it wouldn't be that complicated to show that the ring of algebraic integers is in fact a ring

But

I think many ways of doing so may be outside the scope of what we're doing in the class at the moment.

Like I said, I've gone through a proof of it before, and even that the ring of integers of any field is integrally closed in its field of fractions, but that's also outside the scope of the class.

Right now we're in introductory graduate field theory.

[delicate orchid]

Outside the scope more like outside the cope… a proof is a proof!!

[past path]

I've seen some of this stuff in the past, but I'm not sure how to approach the proof regarding the algebraic integers using the tools currently at our disposal.

Well, I did get 1/4 on an assignment in another class for proving a complex analysis fact using calculus.

Honestly though

the grade isn't really important

I've already pretty much guaranteed full marks in this class.

I'm just trying to wrap my head around the proof of this fact from the field theory perspective.

Here's the remainder of the problem, for reference.

[delicate orchid]

Well I don’t know any field theory so I will wish you luck

[past path]

I appreciate the help!

Honestly, a lot of this has been mostly number theory, but I think going into explicit matrix representations of tensor products to construct matrices with appropriate eigenvalues might be a bit much when we haven't said the word eigenvalue this semester 😂

That said, there are a couple other places where I've been stuck this time around.

That's one; I'm not sure how to approach that problem at all (well, loosely, but not very well).

And this is the other one:

Currently, my cheeky solution to "I don't know why x^3 - ux - u has no roots in K(u)" is to say "It does not."

as if the rest of the contradiction had already been established lol

If there's anyone who can help with any of these 3 questions I would quite appreciate that.

[frigid lark]

I have an idea, but I just started reading about field extensions, so it may not be great. ||For the forwards direction, if E = K(a), then as F is algebraic, Irr(K,a,X) = f exists. As f has coefficients in K, sigma(f) = f. (If f = f_0 + f_1 X + ... + f_n X^n, sigma(f) = sigma(f_0) + ... + sigma(f_n) X^n). As sigma(a) is a root of sigma(f), it is also a root of f. Thus [E:K] = deg f = [K(sigma(a)) : K] = [sigma(E) : K], hence sigma is surjective, and thus an automorphism. I think you can do something similar for arbitrary finite extensions. I don't know how to argue for infinite extensions however. ||

[frigid lark]

For arbitrary extensions, ||if E is an infinite extension then for a in E let E' be the extension of K generated by all of the roots of Irr(K,a,X) = p(X) that lie in E. We observe that sigma sends a root of p(X) that lies in E, to another root of p(X) that lies in E. Thus sigma(E') is a subset of E'. But by a cardinality argument they must be the same field. Thus a lies in the image of sigma restricted to E', and thus a lies in the image of sigma, i.e. sigma is surjective and hence an auto ||

I think for the the other direction you could do contraposition.

Damn, no cute parrot emote on the server

[past path]

Thanks man! I'll read through it and see if I can make sense of it all.

[fervent agate]

I'm looking back over this, and I don't think it's correct. I showed that if one could find a nonzero element whose order is less than m but divides m, then m must be prime. But I don't know that such an element exists.

Alternatively, I could suppose that such an element does exist, and reach the conclusion that the order of every nonzero element is the same. But that doesn't tell me that that order is prime.

[fervent rock]

hi

Is this legal to say?

These are the facts im working off of

[delicate orchid]

I mean, yes the group of units of Z/5Z is Z_4 but I don't see how that relevant in factoring Z_140 into cyclic groups?

[fervent rock]

I dont know, maybe i am not understanding this

nothing makes sense

[south patrol]

Hm a question im doing asks me to show that if M is a simple A module and I a two sided ideal with I^n = 0 then IM =0

Ofc this also follows by Nakayama but we can just say like oh if IM is nonzero then it's M, so for all k we have I^k M = M, so in particular 0= I^n M = M, a contradiction

Because this is meant to be like 6 marks or smth lol

[lethal dune]

hmm lol

[south patrol]

lol what does that mean

[lethal dune]

maybe simple here is superfluous?

you need fg

[south patrol]

Indeed

I'm just repeating the question lol

[lethal dune]

maybe they want you to prove Nakayama before using

[south patrol]

Nakayama isn't in this course anyway lol

this is ye olde rep theory course

[lethal dune]

our grader once deduced 2marks for assuming Aut(S3)=S3

[south patrol]

lol

Tbf I didn't know that was the case lol

Well or at least I'd have to think about it a lil

[lethal dune]

lol even I don't see this immediately now, ig it was justified then

[south patrol]

lol

hm now i am just thinking back to previous weird grading moments lol

[lethal dune]

what's your worst grading experience

[delicate orchid]

A is commutative here right?

[lethal dune]

'A'

[delicate orchid]

what

[lethal dune]

'A'll rings are commutative

[delicate orchid]

actually no wait nvm you've specified it's two-sided irregardless

I can't see where that argument fails lol

[lethal dune]

yeah lol

checkmate

[south patrol]

Hm

Well definitely not the worst but a funny one was when the grader thought questions were out of 20 instead of 25 and like

One question went a little meh but was marked correct, but the other was perfect so he just gave it 20

[delicate orchid]

how the hell do you cock that up

[south patrol]

lol

Idk

[sinful mirage]

Is ker T=G for some reason?

Or not necessarily

[delicate orchid]

just scanning I see no reason why T needs to be trivial

[sinful mirage]

I was just wondering,because it is claimed in the book that E as a set is U(1) x G

[delicate orchid]

so

huh good point

[sinful mirage]

my guess was that E= (U,g) so that U is in the preimage of Tg under hat gamma (according to a SE hint)

so i was guessing T must be trivial

but I have no clue tbh

https://physics.stackexchange.com/questions/757607/projective-representations-and-cental-extensions-based-on-schottenlohers-book?noredirect=1#comment1696973_757607

[pastel cliff]

learning about tensor products rn, is it just my impression or is commutativity of that operation really powerful

[delicate orchid]

so E here is the particular subgroup of U(H)xG such that when you map the first coordinate through gamma you get the second coordinate mapped through T, so the composition gamma(S(U, g)) = gamma(u) = Tg = T(pi(U, g)) holds for all elements in E?

I'm struggling to se how this kinda isn't just true by definition of E

oh right the other square I can't just 5-lemma my way out of this one

yeah the other square commutes pretty obviously if the bottom map is U(1) -> U(H) given by lambda -> lambda*Id_H, I'm not too sure what you're confused about

it definitely doesn't need to be trivial

[sinful mirage]

I am trying to see why E=U(1) x G

do you see why is this true?

I am confused about this

it's not about commutativity

[delicate orchid]

right ok sorry I got distracted

firstly, E is a subgroup of U(H)xG - good start
it's a central extension so (U(1), 1) is in the centre of E - are there any elements outside of U(1) that commute with all of U(1)?

I must admit I don't actually know what these U(-) groups are

[sinful mirage]

unitary groups of a hilbret space

U(P) is the unitary group of the projective hilbert sapce

[sinful mirage]

agree

[delicate orchid]

cause if there aren't elements outside of U(1) (well, what I mean by that is elements in U(H) that aren't in U(1)) that don't commute with all of U(1), then this contradicts the fact that (U, 1) is in the centre of E for all U in U(1).

As we'd take some U* in U(H)-U(1) and then the inclusion (U*, 1) wouldn't commute with some (U, 1) in the inclusion of U(1), contradicting the centrality. So if there are NO elements in U(H)-U(1) that commute with all of U(1), then the group just has to be U(1)xG right?

the question is, are there no elements?

not sure

[past path]

I've solved one of my earlier problems; currently I'm having some trouble extending the proofs here in (b) and (c) to the question (d):

[south patrol]

Do you know that algebraic numbers form a field?

Otherwise this is much harder than a),b) and c)

[pastel cliff]

[lethal dune]

ignore, I thought you're talking about alg integers

c) you need to use the fact that algebraic integers form a ring

even for d

[pastel cliff]

how can we see Q and C as Zmodules

they are Z modules right?

[rotund aurora]

cuz they are rings that contain Z

like there's no additional structure, in fact its the opposite

[delicate orchid]

n(a/b) = (na)/b

n(x+yi) = nx+nyi

ta-da

and yes they're Z-modules are they're abelian groups under addition

[pastel cliff]

this means that Q \otimes Q should just be isomorphic to Q right?

and same for C \otimes C?

[past path]

Yes, we know that much.

So we know that if u and v are algebraic integers, then uv and u+v are algebraic numbers, but I don't know how to show that they're algebraic integers.

My intuition suggests trying to show that if u and v are the roots of monic polynomials in Z[x], then so are uv and u+v.

We know that they're roots of some polynomials in Z[x], just not necessarily that they're monic.

Of course, I know they do have to be monic, but I'm not sure how to argue that point.

[south patrol]

Honestly i've never tried to deduce the fact this forms a ring from the fact the alg numbers forms a field lol

How much stuff have you already done on alg numbers/integers? Like how did you prove that the alg numbers form a field

[past path]

We have that the set of all elements that are algebraic over a field form a field.

In general.

In particular, the algebraic numbers were (as far as this course is concerned) defined in the problem statement.

We're in introductory graduate Galois theory.

I've proven some of these things in a number theory course, but this class is sometimes particular about using content that we've covered.

For reference, if it means anything to you, we're currently on Hungerford chapter 5.

[pastel cliff]

i have this question

is this diagram enough for the first part

basically combining universal properties of product and tensor product

[rustic crown]

that diagram not correct >.<

f1 and f2 aren't bilinear, so they don't factor through the tensor product

[pastel cliff]

pain

is the idea right? though? in that i need some kinda diagram

[rustic crown]

depends on how much you've played with tensor products

if you've shown by now that it's a nice bifunctor and is right-exact then there is a much simpler diagram lol

[delicate orchid]

slightly related question that's just popped into my head, do functors R-mod -> R-mod preserve injections/surjections?

I know they do in set

[rustic crown]

which functor?

[delicate orchid]

just an arbitrary one lol

maybe the question isn't well phrased it was just a thought I had

[pastel cliff]

hmmm

[rustic crown]

yea dunno what you mean

[pastel cliff]

i'll be back later then det

confus

tired

[rustic crown]

,ti sebb

[cloud walrus]

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4.   sebb#2822 
5.   Sebbbbb#3562 
6.   sebbb (stμ₂dying#9923)
7.   sebbo#9220 
8.   Sebb#9575 
9.   SebbyLaw#2597 ```
Please type the number corresponding to your selection, or type `c` now to cancel.

[pastel cliff]

,ti

[cloud walrus]

The current time for stμ₂dying is 02:42 PM (EDT) on Fri, 31/03/2023.

[pastel cliff]

it's a reasonable time for once

[rustic crown]

did you not get a good sweep >.<

[cloud walrus]

Member selection timed out.

Mar 31, 2023, 6:43 PM

[pastel cliff]

been a long week

[rustic crown]

https://3.bp.blogspot.com/-8_y4EtOTyeg/WnqIsn11LXI/AAAAAAABFDo/XygoPm9_ifQzY8X5VUAxYFKWFbjlDpTjACKgBGAs/s1600/Omake%2BGif%2BAnime%2B-%2BSlow%2BStart%2B-%2BEpisode%2B5%2B-%2BKamuri%2BHead%2BPat%2BRecharges%2BEiko.gif

[pastel cliff]

[warm shoal]

yo

[rustic crown]

hewwo

[warm shoal]

What's going on with the server and channels

why do they have weird glitchy names

HACKERS?

[delicate orchid]

no way no way ABELIAN GRAPES

brbrbrbrbrbrbbrbrbrb

[sonic coral]

LOL

the channel names

[quaint tree]

Fuck yeah I love this new channel name!

[sonic coral]

look at all the other ones

[coral spindle]

suits it better

[trim sonnet]

realise the day

[coral spindle]

Oh cool det's a mod now

@rustic crown I'm sure you'll do great

[rustic crown]

[delicate orchid]

congrats!

[coral spindle]

I haven't been here in a while so sorry if this is old news

[delicate orchid]

Boyjie I don't know how to tell u this...

[rustic crown]

det only has the mod color >.<

[coral spindle]

Hahahahaha

[delicate orchid]

u have been april fooled

[rustic crown]

no

server got hacked

[coral spindle]

Ah yes hacked server

ofc

To make up for that...

I'll post something cool I think is true about monoids

[rustic crown]

https://tenor.com/view/byakuren-hijiri-touhou-sipping-juice-sips-juice-grape-juice-gif-15098400

[coral spindle]

Let's say you have a monoid $M$ and a submonoid $S$, and $M$ acts on some set $\Omega$. Then of course, the submonoid $S$ also acts on $\Omega$.

Now the question is (as has come up in some of my work) if we say that $M$ acts transitively on $\Omega$, when does it also hold that $S$ acts transitively too?

[cloud walrus]

Boytjie

[coral spindle]

Now this has come up in some stuff regarding Markov chains. We can define a Markov chain via an action of a monoid M on a set Omega, and a probability distribution on M.

Amazing channel name change 10/10

anyway

[rustic crown]

server is glitchy

[coral spindle]

so true king

but the point is, this probability distribution could be really badly adapted to the monoid

like, there could be a 100% chance of just choosing the identity for example

so we need to think about the submonoid generated by the support of the distribution

Anyway

It turns out that the condition seems to be that the transitive $M$-sets are the same as the transitive $S$-sets when $S$ contains any (two-sided) ideal of $M$

[cloud walrus]

Boytjie

[coral spindle]

It's super easy to show that this condition works, though I haven't given it any thought whether or not it's necessary

section 3 of this paper does this https://arxiv.org/abs/1401.4250

Prop 3.2 is the correct statement but the proof is totally wrong

[tepid viper]

Channel Topic (for specific): Homological Algebra

[primal tusk]

im having a hard time understanding this commutative grape problem

specifically the notation at the bottom right where the sigma is in the subscript

what is happening?

[coral shale]

sigma is a permutation

it maps elements in {1...4} to the same set

so sigma(1) is just a number.

[primal tusk]

oh true thank you

[raven pendant]

wo

[tender wharf]

If I have a rng of grapes (not necessarily commutative) such that x^3 = x, how do I show that 6x = 0?

Sorry let me rephrase

[coral shale]

nani is this lmao

[tender wharf]

🍇^3 = 🍇

[simple valley]

I don't think that's true because x=-1 in \bR satisfies this

[tender wharf]

hmm

interesting

I'm guessing the book made a typo

also whats the

fake moderator role

[coral shale]

<@&1091183017051050074> your authority is being questioned

[tender wharf]

can I eat abelian grapes or can I only eat nonabelian grapes

shuri why aren't you moderator

[lapis latch]

if you eat a red grape and then a blue grape is that the same as eating a blue grape and then a red grape?

[tender wharf]

hmm

I don't think eating is abelian sadly

for example, if you first eat a steak and then eat ice cream

you'll be fine

but if you ate ice cream and then steak

you'll die

[coral spindle]

If you turn them into wine first, they commute.

[lapis latch]

mix em up in a blender form the commutator

[pastel cliff]

what does this look like det

[rustic crown]

M1 (x) M2 --> M1 (x) N2 --> N1 (x) N2

And the other way to do this

[pastel cliff]

:O

i hadn't thought about that

[rustic crown]

By rightexactness all tiny arrows are surjective

So is the diagonal arrow then :3

[pastel cliff]

ok im on my phone but i'll try and type it up later

[tender wharf]

sebb please eat some abelian grapes

I mean not sebbb

what happened to sebbb

:(

[pastel cliff]

ok working on a different question now and maybe im being silly but

is that surjection A^n -> M always there?

oh i guess it is

after thinking about it like an extra minute lol

[south patrol]

there is a surj M x N -> M (x) N i guess

[pastel cliff]

mod potat

[rustic crown]

Nu whut. Not every tensor is elementary

[pastel cliff]

what's an example of a non-elementary one

[rustic crown]

Like literally anything :p

[hallow kiln]

like…any of them?

[pastel cliff]

ooh i thought you were saying like

not every element of a tensor product is elementary

[hallow kiln]

correct.

[pastel cliff]

i'll see myself out

[hallow kiln]

most are not, by definition.

[rustic crown]

V x V --> V (x) V where V is a dim 3 vector space

[rustic crown]

Oh oopsie

[pastel cliff]

i kid

[pastel cliff]

going back here for a moment, the surjective map onto a finitely generated R module exists because if we have M being finitely generated in R x M, then we can take n-many copies of R where id maps into each element in the generating set of M right

that's very weirdly phrased

ig im asking: the n in R^n -> M depends on the size of the generating set right

[rustic crown]

Yee

[pastel cliff]

so to show that the tensor product of finitely generated modules is also fin. gen.

this is the question

[crystal lodge]

abelian grapes?

[pastel cliff]

dont tell me det hold ont

[opal crypt]

The easiest way to prove this is probably by using the universal property

[rustic crown]

(Although idk what R x M is doing here)

[rustic crown]

lol I'm on phone, too hard to type here so I forget to read what's happening up :p

[pastel cliff]

ok we have R^n -> M1 and R^m -> M2, using the natural(?) map R^n x R^m -> R^n \otimes R^m, we get an induced map to M1 \otimes M2 right

then surjectivity of all that means it must be finitely generated i think

is natural a good/bad word to use there? idk what categorical implications it has if any

[rustic crown]

wdym "using the natural map"?

[pastel cliff]

hm yeah that's a bad word

i mean \varphi in the picture above

[rustic crown]

nu natural is fine

i was not sure about the usage of "using" lol

like what's happening is you have map M1 --> N1 and M2 --> N2 then you get a induced map between teh products M1 x M2 --> N1 x N2 which you compose with the map N1 x N2 --> N1 ⊗_R N2 and this composite is bilinear so has to factor through M1 ⊗_R M2

[rustic crown]

if you can define the map without making choices specific to particular objects, it would be natural, so dw about that

[pastel cliff]

do you think a diagram would make this clearer?

i'll do one anyways

[cloud walrus]

det

[rustic crown]

yea true... but i think it might be good to spend a little time with getting used to the fact that if you have a set of elements m_i then the span of them is exactly the image of R^{⊕I} --> M

if you have done some field theory, this argument is pretty much same as showing the tower law for degree

[pastel cliff]

oh god

soon

[rustic crown]

if {m_i} is a generating set of M and {n_j} a generator set of N, then {m_i⊗n_j} is one for M ⊗N

[pastel cliff]

it'll be on my exam lol

my last few lectures have just been like

https://tenor.com/view/halloqueen-milk-spongebob-patrick-gif-6136953

[rustic crown]

whut does milk on carpet mean?

[pastel cliff]

yeah

[rustic crown]

[barren sierra]

[rustic crown]

it means "yeah"?

[pastel cliff]

it means head empty no thoughts

(i havent followed the lectures very well and am behind)

[cloud walrus]

Parrot Tea

[rustic crown]

we doign algebra >.<

[frigid lark]

I have epsilons in my algebra book

wdym

[rustic crown]

like evaluation map?

[frigid lark]

No, for abc conjecture

[rustic crown]

or k[e]/(e^2)

[tender wharf]

that's just straight up true

obviously.

[frigid lark]

[rustic crown]

oh i think they meant 1-e for e<0

[frigid lark]

mmmm

Yeah, that's why there's no proof for it

Used the wrong epsilon

[frigid lark]

Is this for an undergrad course (like bachelors)?

[pastel cliff]

no it's a grad class

[rustic crown]

tensow pwoduct

[pastel cliff]

potato gave me this proof a little while ago

im a little confused about w in the first part though

in this equality, we go from lambda being a single element of k to the identity matrix times lambda right

and we're fine w this bc they act in the same way

ok nvm i think i see it

[vagrant zinc]

Does anyone know what happened to the discord channel?
I get all weird symbols and I can't differentiate the channels.

[solemn garden]

What channel are you trying to find?

[lethal dune]

det mod or april fools prank

[pastel cliff]

det mod

[summer path]

well one is called "Moderator" and the other is called "Moderators" >.<

[white oxide]

how exactly is i[D] a subdomain of F? of course i can check, but i'm curious as to where the "of course" came from

i know that if phi is a homomorphism from a group G to a group G' then phi[G] is a subgroup

but that only applies to group homomorphisms

ah well i suppose it is kinda trivial

nvm then

i still don't get how this "proves the following theorem"

how exactly are those elements of D

i is a map from D to F

therefore i(a) and i(b) are in F

i can't be the only one who thinks this is ass notation

is it because D is isomorphic to a subring of F?

so that subring is essentially D??

[solar shore]

ohhh okay thank you so much! that clears it up a lot

i've been trying to do some more problems where we count the number of elements and cyclic subgroups, so that insight actually helps out a ton

[frigid lark]

Elements in F are equivalence classes of tuples of elements in D, i.e. each f in F is of the form (a,b) where a,b in D

But once you get past the mental gymnastics to conclude that low quality materials are basically the same high quality materials isomorphism is basically the same as equals, you will be set to start your successful construction business algebra experience