groups-rings-fields

[celest furnace]

I tried something similar with Cayley-Hamilton theorem, and i got:
"If you wanted to pick some explicit theorem, say Cayley-Hamilton, and then summarize the history of it, who is Cayley, who is Hamilton, read the original papers and see what they actually did, who proved what etc. -- that might be quite interesting"

Don't really have anything to compare to either, since this class used to be taught by someone else who (as far as I can tell) didn't have as strict requirements

[summer path]

Hmm

[frigid lark]

Have you looked into Mason Stothers? A bit number theoretical

[summer path]

Perhaps you could look into some of galois' papers? Surely his papers have English translations

[lapis trail]

Super interesting guy.

Would be low hanging fruit

[summer path]

True but I'm not sure how often you'd find direct translations of original papers

[celest furnace]

Great ideas guys, ill send him an email about both... though I expect him to reject Galois if he rejected Noether

[lapis trail]

More analysis than algebra but Ramanujan has an interesting story

[frigid lark]

Why did he reject Noether?

[lapis trail]

Noether is a great choice imo

[celest furnace]

Well... not exactly reject but "How about doing the easier theorem https://en.wikipedia.org/wiki/Primary_decomposition in the special case of polynomial rings, having a look at the fascinating life of Lasker (the life of Noether is equally fascinating but there are many more books about Noether, Lasker is hardly known among mathematicians even though he was the World Chess Champion). "

Mason-Stothers seems good though because I could even go into the history of the real abc conjecture, and the guy who tried to prove it in 10,000 pages and a new theory, and so on

[prime sundial]

@rustic crown sorry to bother, i don't see how to do the inclusion ker phi into (f)

[rustic crown]

so you know e + f = 1 and ef = 0

(that's how you should picture idempotents, like e = (1, 0) and f = (0, 1))

[prime sundial]

ah that makes sense

[rustic crown]

and this is the prototypical example, as any (non-trivial) idempotent can be used to decompose the ring as a product of two smaller rings

[rustic crown]

anyway, so the second equation tells you that that (f) is contained in the kernel of phi. and the first tells you that if x is in the kernel, then fx = x, so x is in (f)

[prime sundial]

thanks a ton

i swear i used to be good at this

[summer path]

If you wanted, I think another potential topic is Kurt hensel via the resources listed here: https://math.stackexchange.com/questions/844756/history-of-p-adic-numbers

[prime sundial]

is there a name for the injection from Z[x] to Z_p[x]?

sorry

i meant the homomorphism from Z[x] to Z_p[x]

where coefficients are just reduced mod p

[celest furnace]

natural projection might be what ur thinking of

[prime sundial]

probably, thanks

[delicate bloom]

it would be an injection if Z_p were the p-adic integers lol

[prime sundial]

lol good to know

[delicate bloom]

🥲

(not reduction mod p though)

[prime sundial]

also, i'm showing a quartic p(x) in Z[x] is irreducible, and what i was shown is to assume p(x) = (x-a)q(x), deg(q) = 3
we showed earlier that in Z_2[x], p(x) reduced to a polynomial r(x)^2, deg(r) = 2. how is this a contradiction when i take the image under the projection? is it that we have a linear factor still even though we're not supposed to

there's also another step where we assume p(x) = q(x)r(x), each of degree 2, and do the same projecting to Z_3[x] (it should become x * s(x), deg(s) = 3), so we know it can't reduce in the end

also, an unrelated question to a different problem.
if i want to find a generator of (Z/pZ)*, does it suffice to find an element that is coprime to p-1? it makes sense to me but i haven't shown/tried to show it

i was thinking additively, never mind

[delicate bloom]

yeah, if it's linear in Z it'll be linear in Z/pZ

but there was no linear term in Z/pZ

for p=2 I mean

[prime sundial]

great thank you

sorry for the spam i'm studying for an exam and i'm very worried about how much i know abt the material

[pastel cliff]

mero

[delicate bloom]

you're welcome, don't worry about it, I think most people monitor these channels like sharks just waiting for questions so they can think about and help with math lol

[pastel cliff]

me last week (i most likely failed )

[prime sundial]

got a 75 on my first abstract 2 exam , trying to pick up the slack now so i can get an A

[pastel cliff]

literally same but i couldnt pick up the slack

[prime sundial]

though even if i do well on this exam, the final will be galois theory which im not sure i'll be able to do well on lol

[pastel cliff]

so im banking everything on the inal

[prime sundial]

ah good luck

you got this

[pastel cliff]

[pastel cliff]

no u

[prime sundial]

[prime sundial]

i wish this was me

procrastination a drug frfr

[pastel cliff]

me too im just gaslighting mysel into believing it until i do

[lapis trail]

I think you might be surprised

[prime sundial]

i've heard people say it makes stuff easier but i'm not banking on it

just doing my best and hoping the curve gifts me the A

[prime sundial]

that's more like it

ok i suck at this

i'll be back

[pastel cliff]

keep fighting

[cloud walrus]

maximo

maximo

[prime sundial]

i did this like 2 weeks ago and got x^2 + 2

and so did my classmates. were we all wrong the whole time

omg i reduced the wrong thing

great, extra practice for me

[frigid lark]

What are you trying to do here?

[prime sundial]

.

i was actually meant to reduce x^4 + x^2 + 1 though, and the answer was x + 2

[frigid lark]

Sorry, for some reason I thought you were trying to show that x^4+x^2+1 is irred in Q[x]

[prime sundial]

ah no worries

my messages here have been a mess so it's on me lol

[frigid lark]

are you only working over Z_3[x]?

well and a finite field of order 27

[prime sundial]

yeah

im reducing the polynomial in E = Z_3[x]/(p(x)), where p(x) = x^3 + x^2 + 2x + 1

man i hate algebra

i liked groups

[frigid lark]

What you wrote looks correct to me

[prime sundial]

this thing

[frigid lark]

ye

[prime sundial]

i'm glad that's right but i wasn't meant to reduce x^4 + 2x^2 + 1

it was supposed to be x^4 + x^2 + 1 lol

[frigid lark]

oh

lol

[prime sundial]

i figured it out though, thank you regardless

i do have a question though

in the field i mentioned above, E = Z_3[x]/(p(x)), what are the roots of p(x)

[frigid lark]

x

is one

[prime sundial]

i naively said everything is a root, but now i realize that's incorrect

[prime sundial]

right

[frigid lark]

The other ones are gonna be harder to get, or might just not be in E

Well you could brute force it, it's only 27 elements

[prime sundial]

there's most certainly a trick to it

because i doubt my prof wants us to brute force 27 elements lol
also he's a putnam nut so most of his questions resort to tricky solutions haha

pretty sure he's a two time putnam fellow

[frigid lark]

I think the group of automorphisms of F_27 over Z_3 is cyclic generated by phi, which sends a to a^3

[prime sundial]

i think there's another intended solution since this was before we dove into field automorphisms

but please continue

[frigid lark]

And these automorphisms should send roots of a polynomial in Z_3[x] to another root of that polynomial

so try x^3

[prime sundial]

x^3 is a root indeed from the frobenius homomorphism

[frigid lark]

And then x^9

And that should be it

[prime sundial]

yeah that makes sense

just don't know how to justify "that's all of them"

[frigid lark]

polynomial of degree 3

has at most 3 roots

[prime sundial]

ah right

thanks a ton

that's actually a really neat solution

i think i'm getting acclimated to these polynomial rings

[barren sierra]

get used to polynomials

they appear uh

everywhere

[prime sundial]

real

[barren sierra]

kinda annoying when you think about it but whatever

I'm reading some stuff on them rn

[prime sundial]

i wanted to do an alg. geometry independent study because of that

[barren sierra]

(they aren't annoying)

[prime sundial]

but idk if my schedule permits

[barren sierra]

yeaaaa alg geo is the big thing

[prime sundial]

i'd have to drop some pretty important courses to do so which sucks

[barren sierra]

I'm in alg geo rn and hate it

but that's prolly cause of my prof

not the material

I'm sure if I was actually being taught the material I'd like it

[prime sundial]

yeah i'd have a really great prof leading the independent study so i'm sure i would end up enjoying it

[barren sierra]

algebraic complexity theory (field of CS) uses a ton of polynomial stuff

and that stuff is cool

and what little algebraic combinatorics I've looked at seems to leverage alot of algebraic geometry as well

[prime sundial]

im not touching combinatorics for as long as my work permits

[barren sierra]

rip

[prime sundial]

i don't like it at all tbh

im hoping i don't see it anytime soon, but i know it's inevitable

[barren sierra]

fair enough but why

[prime sundial]

i don't like what's talked about. don't really care about the topics

though generating functions sound cool

[barren sierra]

they are

[white oxide]

how did they arrive at s = n?

[barren sierra]

What is the theorem

missing context here

[sonic coral]

it looks like eisenstein

[barren sierra]

oooo yea

[lethal dune]

such a bad proof

[barren sierra]

yea this is awful

[lethal dune]

reduce mod p

[white oxide]

damn ok

i'll look up another proof then

[barren sierra]

I don't think we can say $a_n = b_0 c_n = b_r c_s (\text{mod }p) \implies s = n$? idk

[white oxide]

im confused, did they just take x = x + 1? (underined in blue)

[cloud walrus]

Spamakin🎷

[barren sierra]

no

[cloud walrus]

Spamakin🎷

[white oxide]

ah i see thx

im confused, if f(x) is irreducible then are they saying that it can be only factored into a unit times some polynomial g(x)

[barren sierra]

yup

if you could factor it some other way

it wouldn't be irreducible would it

[white oxide]

ya

[summer path]

sometimes i feel like when im doing algebra problems, im just messing around with what is given, without much direction, until something i want pops out; is it normal or am i somehow not understanding the intuition behind what is going on

perhaps this is too general of a statement to make

[sonic coral]

i’m trying to show that (r+s)^2 = r^2 + s^2 for all r,s in a boolean ring

i’ve already shown that r+r = 0. i’ve got r^2 + rs+sr+s^2. can i rewrite the r’s in the cross terms as r^2 and factor it out to give s+s, or is that not okay since they’re being multiplied on different sides of s

[celest furnace]

Not allowed probsbly; since you are trying to show commutative by assuming commutative

[sonic coral]

that’s what i figured. where should i go next then?

[celest furnace]

Not quite sure hold on

[stone hamlet]

try using the idempotent property x^2 =x on (r+s)^2

[celest furnace]

Yep

[sonic coral]

so (r+s)^2 = r+s, i ended with rs+sr = 0 after using the boolean property on r^2 and s^2. since i’ve shown r+r = 0 for all r, i’m done right

since sr and rs are back in the ring

no wait they would have to be the same element for that to work

[celest furnace]

Well you got that rs = -sr

And you know 2x = 0

[sonic coral]

so then rs = sr, and i get commutativity

on accident, that was the next part

[prime sundial]

i'm given R is an integral domain with a subring F (a field), and we define the usual VS over F where addition is addition of elements in R, multiplication is multiplication by elements in F
i want to show that if dim_FR is finite, then R is a field

i don't know how to construct the multiplicative inverses from this alone

[frigid lark]

Given an element a in R, there is some polynomial f in F[x] that a is the root of (else dim_F(R) is not finite). Argue that there exists such an f in F[x] where f(0) is not 0, and then you should be on your way constructing an inverse for a

The reason why R has to be an ID is really subtle in this proof, but it is good to think about

[kind jacinth]

huh? how is it easy to determine the minimal polynomial

how did they get the polynomial

also shouldnt the basis of Q(sqrt(3) + sqrt(5)) be {1, sqrt(3), sqrt(5)}?

[lethal dune]

no

how would you represent √15 in terms of 1,√3 and √5?

also since it is deg 4 extension, cardinality of basis have to be 4

Also for small dimensions, it's not hard to determine the minimal polynomial by hand

[kind jacinth]

how did they get that the cardinality of [Q(sqrt(3) + sqrt(5):Q] = 4?
also Q(sqrt(3) + sqrt(5)) = {a + b(sqrt(3) + sqrt(5))} so why would 15 have to be represented?

[lethal dune]

try squaring and adjusting

because it's a ring so if √3 and √5 are there, so should be their product

[prime sundial]

thank you, i was really lost and didn’t think to use polynomials

[kind jacinth]

right..

[kind jacinth]

yeah i got the polynomial now thanks

[lethal dune]

don't forget to show it's irreducible

[kind jacinth]

for the second example what would the splitting field be then?

i thought u find splitting field by looking at the roots. So in first example u get sqrt(2) and u get 2i from x^2 + 4 thus Q(sqrt(2), i)

[lethal dune]

ℚ (³√3, ω)

[kind jacinth]

first one makes sense, why is it also omega?

[delicate orchid]

Cause you need to include the complex roots of x^3-3 too, which are of the form cbrt(3)omega with omega being a third root of unity

[kind jacinth]

ahh alright thanks!

[craggy lichen]

true or false? if g^p = 1 for every element g in group G (where p is some prime), then G is abelian

[lethal dune]

https://en.m.wikipedia.org/wiki/Tarski_monster_group

idk if it's abelian

it's not

it's simple

[south patrol]

I'm sure there are easier counterexamples than using this tho

[lethal dune]

I mean I had to

[south patrol]

How about a non abelian group of order p^3

Lmao

Maybe no

[craggy lichen]

yeah i tried coming up w p^3 counterexamples for a bit, can anyone think of one?

[south patrol]

Idk

I don't think that works

[lusty marlin]

How does one show that the order of all elements in such a group is p?

[lethal dune]

I was thinking about Frobenius group but didn't wanna calculate what's A³

[south patrol]

Ye idk

[lethal dune]

y^p ≠ e

[carmine fossil]

Ok not that

[lethal dune]

Nice

This can be generalised to any p^n n>2

[delicate orchid]

The heisenberg group over a finite field of order p has exponent p but is non abelian

Ah never mind should have scrolled down

[kind jacinth]

why is the answer 4? is it not isomorphic to Z1xZ3xZ2?

because my G = {(0,0,0) + <h>, (0,1,0) + <h>, (0,0,1)+<h>, (0,2,1) + <h>} right?

or are both correct?

[hot lake]

That's not a convenient way to write them, you are obfuscating the group law

Also that last one shouldn't be (0,2,1)+H

[plush gust]

can smn give a hint how i should go about proving this?

[south patrol]

Use the fact that C(a) is a subgroup

[kind jacinth]

if I am given GF(2^4) am i expected to come with the isomorphic thing myself?

how would i do that?

something to do with this?

[delicate orchid]

is GF(2^4) = F_16 in normal notation

yeah ok it is

[kind jacinth]

yes

[delicate orchid]

you're quotienting Z_2[x] by an ideal generated by an irreducible 4th degree polynomial, which gives you a field 4 times larger than Z_2, F_16

they're isomorphic almost by definition, which is why they don't bother constructing an isomorphism

[kind jacinth]

why is it irreducible to 4th degree? im assuming i need to do the splitting field of x^16-x right? (the theorem)

[delicate orchid]

try and factor it lol

[odd abyss]

Hello

(e1 ⊗ e3) : (e3 ⊗ e1) is like (e1 e1)(e3 e3) ?

[lethal dune]

[odd abyss]

What ?

[delicate orchid]

uhh what does ":" mean here

[odd abyss]

You do not know this mathematical symbol?

[delicate orchid]

I have literally never heard of this

[kind jacinth]

uh whats a way to factor this. i get x(x^15-1) and so i need to factor x^15-1) further

[delicate orchid]

I mean try to factor x^4+x+1 to show it's irreducible

[kind jacinth]

ohh this i agree is irreducible, but how would i get to x^4+x+1 in the first place given just that field F16

like how would i come up with the quotient myself is my question

[formal ermine]

tensorial product lol

[kind jacinth]

also how did they get thos egenerators

[delicate orchid]

you can take it as a given that the multiplicative group of a finite field is cyclic - I hope,
so call the generator alpha and then use the fact that you're quotienting out by (x^4+x+1) to determine all the possible powers of alpha - and thus in turn also determining all elements of F_16 as alpha is a generator

[kind jacinth]

could u show me an example of how one of them got generated?

[delicate orchid]

I literally can't other than just rewriting what they've written

the only one that's requires any thought is a^4 = a+1, which is true because we know a^4+a+1 = 0 (we've quotiented out by (x^4+x+1)), and so a^4 = -(a+1) = a+1 (we're in characteristic 2)

[kind jacinth]

what do u mean by the use the fact that you're quotiening out by x^4 + x + 1? how do i determine using that

i know that the factor contains the remainders

[delicate orchid]

it's literally written at the top dude

GF(16) is iso to Z_2[x]/(1+x+x^4)

[wraith cargo]

yeah but he's asking why that is

[delicate orchid]

I have also explained that

ok tbf I didn't mention you get a field because any ideal generated by an irreducible polynomial in F[x] with F a field is maximal but w/e

some handwaving did occour

[odd abyss]

what is the problem?

[delicate orchid]

tensorial is a very funny word

[kind jacinth]

Ok i got it now. thanks!

[wraith cargo]

btw this stems from writing x^16-x as the product of irreducible polynomials over F_2[x]
Then you can get three irreducible quartics and you can quotient F_2[x] by the ideal generated by any of them to get F_16
And the quotient will be unique up to isomorphism

[prime sundial]

sorry for the ping, would you be able to elaborate on why if a in R, then there is a polynomial f in F[x] such that f(a) = 0? it makes sense to me, just can't justify it

this is open to anyone else as well. if R is an integer domain, F a subring that is also a field, then if dim_F(R) is finite, why does that imply every a in R has a polynomial f in F[x] such that f(a) = 0

maybe i should say what the vector space is. it's addition from elements in R, multiplication by elements in F

[south patrol]

There are a couple of ways to do this basically like

One way is to note that there being a polynomial annihilating a (with coefficients in F) is equivalent to {1,a,a^2,...} being linearly dependent over F

[prime sundial]

sorry, i don’t see why that’s the case

oh

i think i see one of the directions

[south patrol]

Which one

[prime sundial]

any root of f is expressed as a linear combination of {1,a,…}

[south patrol]

Why?

Hm i'm confused by that lol

[prime sundial]

because that’s a basis for R over F

[south patrol]

What does it mean if {1,a,a^2,...} are linearly dependent over F

[prime sundial]

oh wow

yeah ok gotcha

[south patrol]

Noice

Okay

Now do you see why {1,a,a^2,...} is linearly dependent over F

[prime sundial]

if it’s a root for some polynomial?

[south patrol]

no like

Sorry okay

[south patrol]

So do you understand both directions of this now

[prime sundial]

i think so let me take a minute to think about it

ok i confused you saying dependent with independent

i understand that if the set is linearly dependent then there is some polynomial in F[x] with a as a root

and i guess the other direction is more of the same

ok what i don’t quite get is why {1,a,a^2,…} is necessarily dependent

[wraith cargo]

Assume there is no such polynomial, this then implies that for any polynomial of arbitrarily high degree in $F[x]$, we have that $p(a) \neq 0$. From this we can see that ${1,a,a^2,a^3,..., a^n}$ is linearly independent for any $n$ and hence $\dim_F(R) = \infty$.

[cloud walrus]

Irony Incarnate

[prime sundial]

ok i see that makes sense

so an easy way to see this is that
$${1,a,a^2,…,a^{\dim_FR}}$$
must be linearly dependent, so a is the root of some polynomial with coeffs in F

[cloud walrus]

maximo

[wraith cargo]

uh idk it's easy but sure

[prime sundial]

a set with more elements than a basis would permit

it makes sense to me haha

ok thank you potato and irony

i wish i could go back to spending time in #point-set-topology instead

[ripe basalt]

I have an action of the orthogonal group on the skew-symmetric matrices by $A^T X A$ where $A$ is orthogonal and $X$ is skew-symmetric. I know the result is skew-symmetric, however it looks like $G(HX)$ is different than $(GH)X$, am i doing something wrong?

[cloud walrus]

*-algebra

[ripe basalt]

$G(HX) = G^TH^TXHG$ but $(GH)X = (GH)^T X GH = H^TG^T X GH$

[cloud walrus]

*-algebra

[delicate orchid]

this action is just conjugation by A isn't it?

[ripe basalt]

yes

but look at what happens

it looks like incompatibility

what's up with that

[delicate orchid]

defining it backwards seems to work

G*(H*X) = G*(HXH^T) = G(HXH^T)G^T = (GH)X(GH)^T = (GH)*(X)

that is odd

[ripe basalt]

hm

is this one of those times where a left and right action are different

[wraith cargo]

yeah this is a right group action but ur writing it from the left

if you define it as $A \cdot X = AXA^T$ then it's a left group action

[cloud walrus]

Irony Incarnate

[ripe basalt]

man

who the hell uses right group actions

[wraith cargo]

In your case you can write $(X\cdot G)\cdot H = (G^TXG)\cdot H = H^T G^T X G H = (GH)^TXGH = X\cdot(GH)$

[cloud walrus]

Irony Incarnate

[ripe basalt]

yeah got it

ty

[prime sundial]

ok i think i got the idea for the proof. if someone could check i'd appreciate it: \
let $a\in R$ be nonzero. let $g(x) \in F[x]$ such that $g(0) \ne 0$. and let $f(x) = (x-a) \cdot g(x)$.\
then $f(0) = -a\cdot g(0)$, which is nonzero as we are in an ID and neither $-a$ nor $g(0)$ are $0$\
note that $g(0) \in F$, thus $1 = a \cdot -\frac{g(0)}{f(0)} \implies -\frac{g(0)}{f(0)} = a^{-1}$

ugh wait i flipped something here incorrectly
aaaa it's wrong. ok i've realized this doesn't work because (x-a) is not in F[x]

[void cosmos]

yo is 11 just saying that this thing is some exact functor?

11e

[agile burrow]

More specifically, it's saying that taking the torsion submodule is a left-exact functor

And (f) is asking you to show that it is not right-exact

[void cosmos]

what does left-exact mean

oh like

the exact seq is from the left

[agile burrow]

Yes

[void cosmos]

like 0-->A

but A-->0 is a right one?

[agile burrow]

It should preserve an extra term as well, but yes, that's the idea

[void cosmos]

cool af

yo 1 more trhing

problem c

let A be a Z_6-module

does this work as a counterexample?

ie 3 and 2 are torsion elements but not 3+2?

cuz their annihilators multiplied would be 0

so its not closed?

[agile burrow]

That sounds right

[void cosmos]

cool af

[agile burrow]

I've seen you ask about derived functors before, so I'll make a brief comment. If we have a short exact sequence 0 -> A -> B -> C -> 0 and a left exact functor F, then 0 -> F(A) -> F(B) -> F(C) is exact. You might ask if we can "measure" how far this is from being right exact, and that's what right derived functors R^i F let us do. They let us extend this to a long exact sequence 0 -> F(A) -> F(B) -> F(C) -> R^1 F(A) -> R^1 F(B) -> R^1 F(C) -> ...

[void cosmos]

okayyy

i guess exact functors are p important

so this sequence is exact iff F is also a right exact functor?

thats the point?

[agile burrow]

No, but if R^1 F(A) = 0, then the sequence is also right exact

[void cosmos]

oh so its like cohmology haha

like if its trivial then its gucci

okay so

what would R^1 Tor(A) be

like

or is that outside my sciope

[flint cave]

oh lol I just had to submit this same q in my last assignment

[void cosmos]

lucky you

i wish i would take this in class

[prisma ibex]

torsion has a nice interpretation in terms of left derived functors

[agile burrow]

Yeah, this specific example is a bit weird

[void cosmos]

which one

my question?

eh ig i will just leave it for later as its probably outside my scope

kinda weird tho that the problems are easier yet the definitions/intuition feel deep

the problms are literally definition unfolding

[prisma ibex]

I guess unfolding it should be relatively doable, it's just maybe outside of the current scope to see why this operation (right deriving torsion) is useful or why it shows up in nature

[silent oxide]

anything category theory

[void cosmos]

yea thats the main problem

why this is useful

gotcha

[void cosmos]

well ig ik what type of math i can only succeed at lmfao

[agile burrow]

Oh this specific example of taking the torsion submodule is really weird lol

[prisma ibex]

here's an important comment to make, the two main derived functors that you take in practice are Ext (right derived functors of Hom) and Tor (left derived functors of tensor product)

[silent oxide]

some like it some don't 🤷‍♂️

[prisma ibex]

if we want to understand this torsion example better, we should try to relate this functor to Hom or tensor

[void cosmos]

Hom is still 2 or 3 sections later

i will come back later , im productive today so ig will go there by today or tmrw

sadly i have a web programming lecture tmrw and wireless so hopoefully i can finish quickly

[prisma ibex]

and indeed it is related: if \mathcal{I} is the set of nonzero ideals in a domain R ordered by inclusion then you have like

$T(A)=\varinjlim_{I\in\mathcal{I}}\mathrm{Hom}_R(R/I,A)$

[cloud walrus]

nGroupoid

[prisma ibex]

so now if you want to right derive this torsion functor T, you just have to understand how right derived functors interact with direct limits, and then you can relate this to some Ext groups

[void cosmos]

oh god ir emember this limit thing

[prisma ibex]

colimit*

[void cosmos]

idk what that is i just know the regular "direct systeM"

[prisma ibex]

but yeah funnily enough this torsion functor is more related to Ext rather than Tor

[void cosmos]

like a pair of abelian groups and homomoprhisms such that something something

[prisma ibex]

but yeah e.g. this is how you get something called local cohomology that shows up a lot in algebraic geometry

$H^i_I(A)=\varinjlim_n\mathrm{Ext}^i_R(R/I^n,A)$

[cloud walrus]

nGroupoid

[prisma ibex]

(same thing but you're just looking at I-torsion rather than all torsion, and then right deriving this resulting I-torsion functor T_I)

[void cosmos]

looks like something out of a grothendeick documentary lmfaao

what does _R mean

derived?

[south patrol]

there are ext groups for different ringies

[long obsidian]

Hey I saw something on Wikipedia that seems wrong. It said the Laurent polynomials in one variable are obtained by the polynomial ring in one variable by localizing by negative powers of the variable... That sounds wrong cause doesn't the multiplicative set S of the localization of R supposed to be a subset of R? (That is S is a subset of R).

So wouldn't the Laurent polynomials in one variable be obtained by localizing the ring of polynomials in one variable by the multiplicative set of powers of the formal variable.

Does this sound right?

[solar shore]

could someone provide a hint for proving that $\mathbb Z \oplus \mathbb Z$ is not cyclic?

[cloud walrus]

blanket

[solar shore]

im thinking of doing it by contradiction

[prisma ibex]

you're localizing at the multiplicative set of R[x] consisting of non-negative powers of x

[barren sierra]

what contradiction do you have in mind

[solar shore]

suppose that it is generated by some x, then come up with an element that it doesnt map to?

havent worked out the ideas yet but thats what im thinkin so far

[long obsidian]

Ahhh I must have missed the NON-negative part lol. Thank you

[solar shore]

like if x = (a, b), then it certainly cant map to (b, a)

[wraith cargo]

you should split it into cases where your generator has different a and b, and when it has a = b

but that works

[south patrol]

I remember seeing a really silly way to do this lol

[lethal dune]

I would just say Z not iso Z ⊕Z

[south patrol]

||End_Z(Z (+) Z) is isomorphisc to M2(Z), which is not abelian, whereas End_Z of a cyclic group is always abelian lol||

[elder wave]

Thats kinda funny

[lethal dune]

otherway might be showing that no matter what element you pick , you can always give an element that cannot be a multiple of the choosen element, basically the way you are trying rn

[south patrol]

indeed

[lethal dune]

lol

[south patrol]

Consider for example that ||if (a,b) is a generator then we can write (1,0) and (0,1) as multiples of (a,b)||

[solar shore]

ah yeah, that's what i have so far

uhhh

lemme type it out

[south patrol]

Cool beans

[lethal dune]

prolly the most concise

[south patrol]

I like how quickly you clicked on that lol

[solar shore]

lemme type it out and get back to you guys?

[lavish nexus]

I mean
there’s like only 2 elements that can generate (1,0)

so it’s a direct proof

[chilly ocean]

a cyclic group is either isomorphic to Z/nZ or Z. Then by cardinality, if Z^2 is cyclic it must be isomorphic to Z which it is not

as for instance they have different ranks

[prime sundial]

suppose F is a finite field. show F is not algebraically closed
can i just use x^{p^n} + 1

for p=/=2

oh it's way easier

nvm

[south patrol]

Ye 1 + product of le (x-a)

[prime sundial]

ye

[wraith cargo]

me when x^p + x + 1

[south patrol]

Is that the weird poly w like

[wraith cargo]

this reminded me lmao

[south patrol]

f(x-1) = f(x) or smth

[wraith cargo]

https://lovelylittlelemmas.rjprojects.net/algebraic-closure-of-the-field-of-two-elements/

my favorite result

[rotund aurora]

[wraith cargo]

I love how many ppl didn't get the joke in the comments lmao

[prime sundial]

omg i was so confused

[rotund aurora]

polynomials are not functions

[prime sundial]

everything is an arrow

[rotund aurora]

or not necessarily at least

[prime sundial]

you can't escape it

[void cosmos]

is c basically the same way u would find the span of a set?

remark c

proving that if F is an R-module with X being a basis then F is the direct sum of copies of R

or copies of R-modules each iso to R

so the proof is to say F is iso to sum(Rx) as x varies over X

and Rx is iso to R (each)

and the other waay around is to consider theta_x = (0,0,0,0,...,1_R_x,0,0...)

where htis is the x index

and this is your basis for F

[lapis trail]

When does irreducible in Z[x] imply irreducible in Q[x]?

[white oxide]

all kernels of ring homomorphisms are ideals, correct

[fleet pelican]

yes

[fleet pelican]

Always

[fleet pelican]

what about the field with one element tho

[void cosmos]

read about gauss's lemma

[lapis trail]

Ok thanks

[white oxide]

is Zp a simple additive group where p is a prime?

[fleet pelican]

what are you asking?

[white oxide]

are the integers modulo p where p is a prime under addition a simple group, meaning that it has no nontrivial proper subgroups

[fleet pelican]

yes

[fast stratus]

Yes, by lagrande’s only subgroup are trival groups

[white oxide]

oh yea i forgot about lagrange lmao

thx!

[broken stirrup]

Also, simple means having no non-trivial normal subgroups, there might be nontrivial subgroups

[white oxide]

oh right oops

if a is in M, would the set be trivial since that's essentially the same as (r + M)(0 + M) = 0 + M = M which is the identity of R/M

[formal ermine]

that is a weird proof

[white oxide]

fraleigh has lots of weird proofs

[formal ermine]

it's two sentences with correspondence

[lapis trail]

Is pi^(3k) for all k in N in Q(pi^3)?

I'm trying to show pi^2-1 is algebraic over Q(pi^3)

And it'll be helpful if I had pi^6 as a coefficient in some polynomial

[south patrol]

It's a ring, so yes

[lapis trail]

Ok I see how this field extension stuff works now

Is it a fact that we'd say that [Q(pi^3):Q] is infinity

[south patrol]

Yes

[lapis trail]

I see I see

[south patrol]

pi^3 doesn't satisfy a polynomial over Q

[lapis trail]

Therefore

[south patrol]

Though idk how to prove that even pi doesn't lol

[south patrol]

Well it means pi^3, pi^6, pi^9,... are all linearly independent

[lapis trail]

Yeah

And I was going tobsay that must mean that if x is transcendental over Q then [Q(x):Q] is infinity

[south patrol]

Yes

[lapis trail]

If y is algebraic over Q then is [Q(y):Q] the degree of the minimal polynomial?

[south patrol]

Yes

[lapis trail]

Aha

[south patrol]

Try to see why

[lapis trail]

It's starting to make sense

Ok I will thanks

[kind jacinth]

I think I solved this question using the help provided a while back but I would like some confirmation if I am right.
so theta(x*x) = 0. Thus,
the kernel of phi is the ideal generated by <x^2> in Z_2[x]. This is not a maximal ideal because it is contained in a larger ideal generated by <x>

[south patrol]

Why does theta(x^2)=0 imply it is generated by x^2

[kind jacinth]

because kernel contains{x^2, x^4, x^8, ...} if i do phi(x^4) = phi(x^2 * x^2) = phi(x^2) * phi(x^2) = 0 + 0 and the same holds for the rest

and it would also hold for phi(x^2 + x^2)

while if i do for example phi(x^2+x) this would not give 0.

[fast stratus]

A is commutative ring (non-unital) , and J is an ideal. How to prove the closure of addition… im thinking some trick with binomial theorem.. but couldn’t figure out

[south patrol]

Just because the kernel contains x^2 doesn't mean it is generated by it

[kind jacinth]

should i try other additive combinations?

phi(x^2+2) could work as well

same for phi(x^2+2x). then is it correct to say the generator is <x^2,2x,2>?

[south patrol]

2 = 0

[kind jacinth]

thats what i was initially thinking. then what am i missing?

[south patrol]

You can show exactly what the kernel is like

The only things you need to check are 1, x , 1 + x

because we know that x^2 and hence x^3 etc are sent to 0

[kind jacinth]

would i do multiplication or addition? like phi(1+x) = phi(1)+phi(x) right?

[south patrol]

ye

[kind jacinth]

none of them give 0. so its from x^2 onwards that gives 0. Eitherway <x^2> is the wrong generator (only generates even numbers). I guess its <x^2,x>?

but no cause this includes x nvm

[kind jacinth]

oh... <x^2, x^3> should be it?

[sinful mirage]

@agile burrow I ended up writing stuff up now and I actually figured out more interesting things, perhaps it is also new for you

I think this is the clearest way to put what we discussed. I did not send proofs, because 1) we did parts of them, 2) the others are trivial. however, I can send any proof if you're interested

[sinful mirage]

new in the sense of remark 2.4 and prop 2.5

[wraith cargo]

If you raise x+y to a high enough power you'll have that x^k is in J and y^{n-k} is in J such that all the terms in the binomial expansion turn out to be in J

[craggy lichen]

whats F_p?

sorry

[celest furnace]

Z/pZ where p is a prime treated as a field

[white oxide]

why is the answer not a^6 + 20a^3b^3 + b^6? i expanded it out using the binomial theorem and cancelled out all coefficients that were multiples of 3

[celest furnace]

well in char 3 20 is 2

beacuse 20 = 18 + 2

(i.e., you got the same answer)

[white oxide]

Ohhh so it’s similar to computing mod 3?

[celest furnace]

yes

char should be that every a in the ring is so that 3a = 0

so just setting any multiples of 3 to be 0

[white oxide]

ah i see now

3(6) + 2 = 20

[celest furnace]

yep

[agile burrow]

this is neat stuff, very cool

thanks for sharing

Yeah, so from what I've seen we like for the cocycles to be normalized (that is, f(g, e) = f(e, g) = 1) because at least in the context of the correspondence between H^2(G, A) and extensions of G by A, these yield to normalized sections from G to the extension. It doesn't change the cohomology computation at all, which sort of makes sense because you can always normalize a section with an appropriate translation

[kind jacinth]

question: if i do long division and get (1/2)x + 1/4 as quotient and 7/4 as remainder what do i do?

its in Z_3 so i cant use fractions

do i just multiply them all by 4 and rewrite it as 2x+1 (quotient) and 7 (remainder)?

[tribal moss]

How did you get the fractions 1/2 and 1/4 in the first place?

(Assuming that Z_3 are the integers modulo 3, and not the 3-adic integers).

[kind jacinth]

long division 2x+1 as divisor and x^2+x+2 as dividend

trying to get 1 = q*f(x) +z*g(x) form

[tribal moss]

But 1/2 is not even an element of Z_3. You shouldn't be able to get that when you do the arithmetic parts of polynomial division.

If you want a multiple of 2x+1 that is quadratic with a first term of x^2, then that would be 2x(2x+1) = x²+2x.

[kind jacinth]

And how do i get it in the form 1=qf(x)+zg(x)?? Do i keep doing this?

[tribal moss]

Yes. (Assuming we mean the same thing by "this").

[kind jacinth]

This is my definition of this😂

Then i get 1= ...

[tribal moss]

(-x+2) = (-x+1)·2 + 1 is not true ...

[kind jacinth]

Woopsie.. i think i screwed up in g

[tribal moss]

Note that in Z_3 you have 2 = -1, so having both 2's and minuses around makes things look more complicated than they are.

[kind jacinth]

Thats good to know

Ill rewrite it

[glass vine]

Hey guys, I have a possibly stupid group theory question. Let G=(Z2)^n and H=Z3 and let x be such that <x>=H. I am interested in the wreath product H wr G. What is the order of this for a given natural number n? Also what else can we say about the structure of this group?

[tribal moss]

The wreath product is a semidirect product of Z3^(Z2^n) with Z2^n so computing the order ought to be straightforward. Pretty big.

[glass vine]

So is it just 3^(2^n) * 2^n?

[tribal moss]

Yes.

[glass vine]

Okay makes sense, sorry for the stupid question I guess

[kind jacinth]

Getting this but need my g out. Any idea?

[tribal moss]

I'll admit I don't understand what you're doing there. It doesn't look like any division procedure I'm familiar with.

[kind jacinth]

How would u do this?

Does the f=2xg + 2(x+1) make sense?

[tribal moss]

I'm a bit loath to just do what appears to be your homework here.

How does an ordinary integer long division in base 10 look for you?

[kind jacinth]

It has solutions which is x+2 but no explanations

[kind jacinth]

In general how id do it?

[kind jacinth]

This for example

[tribal moss]

That's a polynomial long division, but okay.

[kind jacinth]

A polynomial one yes

[tribal moss]

Okay, now I've actually read the original problem 😅. I thought you were just trying to do polynomial division over Z3.

[kind jacinth]

Ohh😂😂

Need to find the q and r in fq+gr such that it equals one

[tribal moss]

So you're trying to apply the extended Euclidean algorithm? Ngl, that one always confuses me unless I spend twenty minutes getting things straight.

[kind jacinth]

Yes im trying that

Im assuming i start with f(x) = x^2+x+2 and g(x)= 2x+1

[kind jacinth]

Then f(x) = q(x)*(2x+1) + r(x)

But how do i do long division Z3 on this for example?

[tribal moss]

Okay, from

f = 2xg + 2x+2
I would write
f = 2xg + g + 1 = (2x+1)g + 1

[prime sundial]

sorry to burst in, i have a short question
let F\subseteq E, E algebraically closed, and let K be the relative algebraic closure of F in E
suppose p(x) in K[x], then p(x) has a root, a, in E. how do we know a is algebraic over F? is it because E is an algebraic closure of F so in general E may not be the algebraic closure of F, right? since E may not be algebraic over F. so it must be for some other reason

[kind jacinth]

So i just rewrite to@not get fractions when doing long division?

[tribal moss]

Well in Z3, 1 divided by 2 is 2, so your first coefficient should be 2 rather than "½".

[tribal moss]

and then 1 = f - (2x+1)g which is the same as 1 = f + (1x+2)g

[kind jacinth]

How do u determine that 1/2 is 2 in Z3?

I can see it for -1 and 4 etc

[tribal moss]

Because 2·2 == 4 == 1 (mod 3)

[kind jacinth]

But nit fractions

[silent oxide]

2*2=4=1 mod 3

[tribal moss]

It's not a fraction -- it's a division.

[kind jacinth]

Why do u write 1/2 by 2*2?

[tribal moss]

I don't.

[kind jacinth]

Oh wait ur rewriting x^2 by 4x^2?

Cz this gives x^2

[tribal moss]

1/2 means the element that we can multiply by 2 and get 1. The element 2 is such an element because 2·2 == 1 (mod 3)

[kind jacinth]

Ohhh i see

So if it was Z4 1/3 would be 3 right?

[tribal moss]

Yes. (But beware Z_4 is not a field, so not all divisions are possible there).

[kind jacinth]

Ok i got 2x+1 as quotient and 1 as remaindr

[formal ermine]

Z4 is my favorite field!

[tribal moss]

Let L be the extension of F generated adjoining the coefficients of p one by one, and then a. Each of the extensions is finite, so the final L has finite dimension over F. Therefore, {1, a, a², a³, ...} must be linearly dependent over F, and a linear relation among them is a polynomial in F[x] that has a as a root.

[kind jacinth]

Ok got this!

[tribal moss]

So now we've done one step of the Euclid's algorithm, and already reached a remainder of 1.

[kind jacinth]

Ok and then f(x) - (2x+1)g = 1 i can rewrite it as f(x) + 2(2x+1)g whixh gives the inverse x+2

Cz thats just f(x) + (x+2)g

Thank u!!!

[tribal moss]

Yes.

[prime sundial]

so if p(x) = c_0 + c_1x + c_2x^2 + ... + c_nx^n
then by adjoining are we considering L = F(c_0, ..., c_n) and then L(a)?

[tribal moss]

I was saying L = F(c0, ..., cn)(a),but sure.

[prime sundial]

right yeah

ok that makes sense, thanks a ton. let me take it in a lil

[tribal moss]

(I'm adding a separately at the end because that makes it easier for me to convince myself that [L:F] is finite. It might not be strictly necessary).

[prime sundial]

ok this is all trying to build up to showing relative algebraic closures are algebraically closed.
i have a polynomial p(x) in K[x], K being the rel. alg. closure of F in E.
we know p(x) has a root in E, and just showed the root is in F as well, so the root must be in K? do i need to consider the irreducible factors of p(x), or are we done, since p(x) was arbitrary

oh i just realized K is a subset of E but F might not be a subset of K right

ok i might have overcomplicated

we take some p(x) in K[x]. then p(x) has a root in E since E is algebraically closed and K[x] is a subset of E[x]
then that root, a, is in F as well. now since a is E-algebraic over F, a must also be in K by the def. of relative algebraic closure. so the root a of p(x) in K[x] is in K

does that make sense

[tribal moss]

let K be the relative algebraic closure of F in E
Since any f in F is a root of the polynomial 1X-f which has coefficients in F, f is in K too.

[prime sundial]

oh ok yeah

that makes sense

[tribal moss]

The root of p is not necessarily itself in F, but we have shown it is also root of some polynomial in F[x]. That's enough to say that root is (by definition) in K.

[prime sundial]

ah alright yes

since p(x) was arbitrary, we can say p(x) = x - k for some k in K, and get that k is the root of a polynomial in F[x] as well right? so we get that K is an algebraic closure of F for free

[tribal moss]

No -- if you want to prove that K is algebraically closed, you don't get to choose which polynomials p you're dealing with. "Arbitrary" here means you need an argument that works for all of them.

[prime sundial]

well we've shown it's algebraically closed, so we'd just need to show K is an algebraic extension

so we just want to show every k in K is a root of some polynomial in F[x] right?

[tribal moss]

Yes, but that was how K was defined in the first place.

[balmy python]

how good is artin algebra to begin with abstract algebra

[prime sundial]

i'm sorry i'm not following.
we know K = { a in E | p(a) = 0 for some p(x) in F[x] }
we then showed for every p(x) in K[x], each root of p(x) is a root of some q(x) in F[x]
doesn't it suffice to take x - k in K[x] to show every k in K is a root of somy polynomial in F[x]

are you saying it's simpler than that, or that the logic there is wrong

[tribal moss]

Oh, right. But I was saying that K = { a in E | p(a) = 0 for some p(x) in F[x] } already tells you that such a p exists for every a in K.

[grand cliff]

https://math.stackexchange.com/questions/105433/does-every-set-have-a-group-structure

Just found this, thought the answer was surprising

[prime sundial]

ah right lol, so it was just simpler

thanks a lot for the help tropo!!

[south patrol]

Okay so quick q

I gotta show that if F is a field such that x^n - 1 splits for all n [not necessarily into distinct things - could be char p] and K = F(γ) is an extension w γ^p in F (for some p) but γ not in F, then x^p - γ^p is irreducible

Is the easiest way just to say that (if $\epsilon$ a prim pth root of unity) then $x^p - \gamma^p = \prod_{i=0}^{p-1}(x - \gamma \epsilon^i)}$, so any factor of $x^p - γ^p$ in $F[x]$ would have constant term $γ^{k}$ times something in $F$, so $\gamma^k \in F$ and so $p \mid k$ (since otherwise we could find $\gamma \in F$)

[cloud walrus]

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

[south patrol]

Because this feels slightly hacky lol maybe there is a more general way to look at this

[kind temple]

what is gamma allowed to be? just a little bit confused on that. if gamma is in F, then this claim is false

oh i just can’t read

that makes sense

[warm urchin]

What are some interesting examples of a Commutative (non-unitary) ring without any maximal ideal?

[fleet pelican]

https://arxiv.org/abs/1404.0135

[lethal dune]

bad notations but as you said say $f$ is a factor of $\prod (x- \gamma \xi_n^i)$ then it contains some but not all the $(x-\gamma \xi_n^i$. Now look at coefficient of $x$. It's some thing times $\gamma$ but that something is invertible giving us $\gamma \in F$

[lethal dune]

There's one, take (Q, +) and define product to be the 0 product, i.e. xy=0 for all x,y. Idk why would anyone study that

try arguing why it can't have a maximal ideal

[kind jacinth]

it needs to be mapped to (1 0
0 1 ) right? and index is size/

[warm urchin]

Yeah i saw that one , i also found a really cool example in a mse post
$$C_{0}(\Bbb{R})/C_{c}(\Bbb{R})$$

[cloud walrus]

Susilian

[warm urchin]

Maybe its not interesting but I'm curious to see why being non unitary stops us from finding maximal ideals

[lament dawn]

That is the identity on the rhs, yes. Assuming you mean 1 (mod 2)

[kind jacinth]

there arent infinite pos. right? since i guess index means size

[warm urchin]

A&M also mentions that since prime ideals are important (in comm alg), and maximal ideals are prime , then we guarantee there exists a sufficient supply of them , so that also justifies why we only deal with rings with 1

[south patrol]

oh sure like it's a polynomial in the roots of unity of too low a degree (or of the right degree but wrong other coefficients) so it doesn't vanish?

But yeah I guess that's very similar

[lethal dune]

yee can't claim that coeff of γ is not zero but your argument is fine

[lament dawn]

The index of a subgroup in a group is the number of cosets.

[lament dawn]

https://en.m.wikipedia.org/wiki/Index_of_a_subgroup

[kind jacinth]

ahh this is useful. I think I would need to apply first isomorphism theorem in this question right?

[pliant raptor]

Suppose I have n rectangles $R_1,\dots, R_n$, each with sides $a_i,b_i$, and thus with area $a_i \times b_i$, that fit together to form a rectangle of size $a \times b$. How can I show that $a \otimes b = \sum_{i=1}^{n} a_i \otimes b_i$ in $\mathbb{R} \otimes_{\mathbb{Q}} \mathbb{R}$?

[cloud walrus]

ImHackingXD

[lethal dune]

[pliant forge]

could i get examples of ideals generated by basis elements of the form b-b’ and b’’ of a K-algebra with multiplicative basis

[south patrol]

wdym "of the form b-b'" etc here

[lapis trail]

What's an example of an irreducible polynomial that isn't minimal?

[pliant forge]

basically the ideal is generated by basis elements of the K-algebra such that every element in the ideal can be written as sum_i a_i (b_i-b’_i) + sum_j a’_j b’’_j

[elder wave]

what do you mean by this

[south patrol]

What are the b' and b''

This is lacking context

[pliant forge]

b,b’,b’’ are all basis elements

a_i, a’_i elements in the field

[south patrol]

Idk like lol

What you mean basically

Well

Like what sort of examples you want lol

[pliant forge]

like a polynomial algebra or some stuff

like a simple example

[south patrol]

But like isn't this every ideal lol

since the b_i'' form a basis

[pliant forge]

it doesnt include all b’s

[south patrol]

Yeah okay this is just too vague lol

[pliant forge]

ye i think i needa sort my mind cuz thats messing with me

[lapis trail]

Nevermind

[south patrol]

an interesting q is like

find an irreducible polynomial f over some ring such that (f) isn't a maximal ideal

idk if that is what you meant or smth

[kind jacinth]

is the answer for this 6? Because SL2(Z)/ker(phi) is isomorphic to SL2(Z2)?

[wet zodiac]

how should i tackle 5.6

ive found nI where n is an arbitrary nonzero real number

[fleet pelican]

I think the hint is there. What commutes with an elementary matrix?

So I guess, you write out AE=EA for some elementary matrix E. As an explicit equation, this will introduce a single constraint, for every elementary matrix.

Then the question is, what is required to satisfy all such constraints simultaneously.

[kind jacinth]

its a true or false question but i dont really get what its saying

any hints on what i should be htinking about? is it like 1, 2 and 3 all get mapped to 2 in Z3? and since its a ring im assuming in my function i have 1*nr = 2 and 1+nr=2? and same for elements 2 and 3?

[lethal dune]

That’s correct

Also n=0 is also valid

[wet zodiac]

wait is that the only one?

also i dont think n=0 is valid because the matrix isnt invertible so therefore not in GLn(R)

[young fiber]

Choose B so you get useful information about A

[fleet pelican]

yeah.

it's surprisingly trivial.

[wet zodiac]

i was thinking of row/column operators

[fleet pelican]

or rather, the ring of matrices is highly noncommutative.

[fleet pelican]

that's the right approach.

[wet zodiac]

i thought there were more

i guess not

[fleet pelican]

.

[hollow mica]

the bilinearity of the tensor product directly corresponds to gluing together two rectangles with a common side length

idt the fact that you’re treating them as Q-vector spaces is relevant. they could also be real vector spaces and the same thing would be true

[pliant raptor]

So, you don't have to use any universal property arguments?

[prime sundial]

i have the following remark from my class:

a field F is algebraically closed iff for every extension E/F, the algebraic closure of F is F itself
is that correct?

[formal ermine]

algebraically closed means no finite extensions

[prime sundial]

is that so even for finite extensions? that seems wrong to me for some reason

[formal ermine]

take the algebraic closure of Q

it's not R nor C

[prime sundial]

but Q is not algebraically closed

[south patrol]

||non-trivial ones||

[prime sundial]

oh just read this again

[south patrol]

uh

this needs another line lol

[prime sundial]

that's all i wrote down haha

[south patrol]

algebraic closure of F in E in right

[prime sundial]

oh

yep

is that correct with that in mind?

i can't wrap my head around it

oh

ok it just hit me why

hm. so we're saying every polynomial has some root, so can we say every polynomial is of the form (x-f_1)(x-f_2)...(x-f_n) where f_i in F?

[formal ermine]

every polynomial splits in an algebraically closed field, yes

[prime sundial]

ok then it would make sense that any nontrivial extension introduces new elements that aren't roots of the polynomials in F[x]

[formal ermine]

yeah and thus non finite ones

as ones with just roots would be subsets of splitting fields

which are always finite

cuz it's a condition to be an algebraic element

[prime sundial]

ok that's actually pretty interesting

thanks for the help illuminator and potato

here's another question that i kind of half understand
suppose E/F is a field extension, a in E algebraic over F and of odd degree. prove F(a) = F(a^2 + 2023)

we basically said, let b = a^2 + 2023, then [F(a) : F(b)] <= 2, so [F(a) : F] = [F(a) : F(b)][F(b) : F] is odd, so [F(a) : F(b)] must be 1

ok writing it out helped me understand what was happening, sorry for the spam

[hollow mica]

Don't think so

[boreal inlet]

May I ask for help in the direction?

I only reached that order of H is of the form p^(alpha) * k where p does not divide k

[lethal dune]

which problem? 1 or 2

[boreal inlet]

can we show that all Sylow subgroups of G are sylow subgroups of H

[boreal inlet]

1

[lethal dune]

that's the goal

Are Pillow group of H also Pillow group of G?

[pliant raptor]

Sorry, didn't understand one thing: is it relevant or irrelevant?

[boreal inlet]

That should be

H is a subgroup of G

I presume you mean Sylow by Pillow

the opposite is not so obvious if it's true

[lethal dune]

P-syllow yes

remeber all pillow groups are conjugate and H is normal

just chase the definition

[boreal inlet]

oh wait so we can define an conjugacy action

whose kernel is the maximal normal subgroup?

you know the core subgroup

[pliant raptor]

Thinking about it again, I'm confused about why this is true, do you have any references I can look into?

[lethal dune]

you're overthinking

[boreal inlet]

🥲

[lethal dune]

gHg' ⊂ H

[boreal inlet]

yes

[lethal dune]

and for any 2 pillow group P,Q there is g s.t. gPg'=Q

can you see where this is going

[boreal inlet]

yes because they're conjugate

[lethal dune]

hint: || P ⊂ H ⟹ gPg' ⊂ gHg'=H ||

[boreal inlet]

hmm so if P is a sylow-p subgroup of H, gPg' = Q is a subgroup of gHg' = H hence a subgroup of G as H is normal

varying g can capture all the sylow p subgroups

but won't that say G contains all the sylow-p subgroups of H, and not the other way around?

[gilded osprey]

I am asked to determine if $A$ is an ideal in ring $R$. I am given that $R = Z_7[x]$ and $A$ is the set of polynomials with vanishing constant term.

Just a bit confused on the language, does that just mean that an element in $A$ is of the form
$a_1x+a_2x^2+\cdots+a_mx^m$
That is, the constant term is 0?

[cloud walrus]

JacobHofer

[south patrol]

ye

well

It means that A is exactly the elements of that form

[gilded osprey]

Okay, was just a bit confused on the wording of "vanishing constant term"

Also, A would be an ideal in this case correct? The informal argument being that the constant term of ra is r_0 * 0 = 0

[south patrol]

yup exactly

(also gotta check addition etc but that's obvious)

Or, use the fact that A is the kernel of a ring hom out of R

[hollow mica]

irrelevant

[hollow mica]

which part

[pliant raptor]

How does bilinearity correspond to gluing two rectangles with a common side length?

[hollow mica]

a ⨂ b + a ⨂ c = a ⨂ (b + c) is analogous to gluing together rectangles of size a x b and a x c along one of their common sides of length a

[sinful mirage]

@agile burrow remember we discussed that we can lift proj. reps to linear reps iff H^2{G,C^{x})=0. The direction that H^2=0 => we can lift is straightforward. how does the other direction work?

[agile burrow]

It's not a two way implication

[sinful mirage]

ah

[agile burrow]

If the cocycle is trivial then you should be able to lift to a linear rep

And if H^2 = 0 then certainly the cocycle is trivial, but the other direction need not hold

[sinful mirage]

the sketch for H^2=0 => linear is as follows I think: suppose H^2=0. Then, every cocycle is cohomologous to the trivial cocycle. But for the trivial cocycle, we immediately get that the lift is a linear representation. Now, every other lift must satisfy that it's a scalar multiple of the lift for the trivial cocycle. Thus, we are done

is this how ti should be argued?

in particular, we can simply cancel out the scalar(as it's nonzero)

[agile burrow]

Oh sure, that sounds right

Right because if you're cohomologous then you differ by a choice of scalar in the lift

Ok cool

[sinful mirage]

yes

[agile burrow]

This stuff is soooo cool

[south patrol]

Says u

Where did you learn this stuffs BTW lol

[sinful mirage]

in particular, this follows from this theorem

[sinful mirage]

yesss

[sinful mirage]

do you know a counterexample perhaps?

suppose H^2 neq 0. yet all projective representations are linear representations

[sinful mirage]

I am currently learning from a book called 'mathematical introduction to conformal field theory' from Martin Schottenloher

it's written for mathematicians mostly

[south patrol]

Oh nice

[agile burrow]

I think this comes back to the question of whether every cocycle is realized by some projective representation. I haven't thought about it much, but my intuition says the answer ought to be yes since the same idea holds for group extensions

[sinful mirage]

wait dym

do you mean the statement might be an iff?

[agile burrow]

Oh I misread your original message

Then yeah, I think it is an iff

Every projective lifts to a linear rep iff H^2 vanishes

[sinful mirage]

do you have an idea how to prove the fact that if proj lift to linear, then H^2 vanishes?

let us assume that the lift (T,\alpha). is a normal representation, i.e. a group homomorphism. In this case, the cocycle alpha associated to this is trivial. how to conclude that the whole cohomology group is trivial?
or is this not a two-way implication?

[agile burrow]

I mean my my thought is that every element of H^2 should correspond to a projective rep

So if there's a non trivial element of H^2, then there's a projective rep which doesn't lift to a linear rep

[sinful mirage]

ah I see

i'll try to think on how to prove this

so basically the proof reduces to showing that H^2(G,C^{x}) iso {proj reps}?

[agile burrow]

Yeah

[white oxide]

has anybody heard of this lol

[south patrol]

Lmao

[boreal inlet]

Help

💀

[hollow mica]

first page

[celest furnace]

☠️

[solar shore]

could someone give a hint as to why every abelian group of order 15 is cyclic?

[fleet pelican]

try contradiction, take an element, what order can it be? how many elements of that order can there be?

[solar shore]

by lagrange's theorem, an element can have order 3 or 5 (discounting 1 bcs i dont want identity, and 15 would mean its cyclic)

[sinful mirage]

how does one see this? This is my trial

it doesn't seem to hold unless lambda(g_1) lambda(g_2)=lambda(g_1 circ g_2)- and i don't see why this should hold

[fleet pelican]

can they all be order 3, or all be order 5? or do you need at least one of each?

[solar shore]

phi(3) = 2, so 1 + 2k can be 15, but phi(5) = 4, so 1 + 4k cannot be 15

so they could all be order 3

[hollow mica]

If g is of order 3 then what is the order of g^2 ?

[fleet pelican]

3

[hollow mica]

so elements of order three come in pairs

[fleet pelican]

this doesnt work, because gh also has order 3

and you can show these are distinct and come as triple of 6 elements

so you need 1+6k

[hollow mica]

oh hmm yea I didn't consider products between them

[fleet pelican]

hmm maybe thats too complicated

[hollow mica]

yeah it gets yucky now

[fleet pelican]

im not actually sure that works

[solar shore]

also sadly, i cant use sylow theorems or cauchy's theorem on abelian groups

:/

i have access to lagrange's theorem

[fleet pelican]

there should be a nice way to force ana element of order 5

possibly taking quotient

[solar shore]

we havent gone into factor groups neither :/

[fleet pelican]

ok so

yeah

gh

consider subgroup

g^ih^j

clearly a subgroup

order 9

contradiction

therefore there is only g order 3, no h order 3

take element that isn't order 3, product is order 15

[solar shore]

oo okay

got it

[fleet pelican]

ya a bit tricky. but i think the way the problem is set up suggests you should consider products, order of products, and subgroups of products

in general, order pq should set off alarm bells in your head

only slightly more complex than order p

[hollow mica]

so this shows at most one element is of order 3 in the group, but why does the conclusion follow

[fleet pelican]

at most two.

g, g^2

[hollow mica]

oh right

[fleet pelican]

what part isn't clear?

[hollow mica]

why g order 3, h order 5 -> gh order 15 is a contradiction/finishes

[fleet pelican]

abelian

(gh)^n=(g^n)(h^n)

[hollow mica]

oh if you get one element of order 15 then you win

okay nice

[fleet pelican]

(gh)^n=1 implies g^n=1 and h^n=1

because we already know g^-1 has order 3 as does h^-1 have order 5

[hollow mica]

You can just use this instead right

and it doesn't use the fact that G is abelian

[fleet pelican]

element of order 15 is a proof

[hollow mica]

like the result still holds if the group isn't abelian

[fleet pelican]

hmm i think to avoid using abelian you need sylow

but you can prove abelian if it isn't given

order 15 implies abelian

pq is abelian for p,q (need not be distinct) prime

[hollow mica]

this is a proof of |G| = 15 -> G is cyclic:

Proceed by contradiction so assume G is not cyclic.

Suppose G has an element of order 3; then we proved (without abelianity) that the only other element of order 3 is g^2. Now let k be an element of order 5. gk has order 15, contradiction.

Now suppose G is only elements of order 5. <Finish proof>

everything is sound up until the end right?

[fleet pelican]

yes

[hollow mica]

Finish:
Let g be an element of the group. Then g, g^2, g^3, g^4 are distinct and of order 5. Now let h be another element (not included in that list) of the group. Then g^i h^j generates a subgroup of order 25, contradiction.

[fleet pelican]

seems good

[hollow mica]

yeah sylow was just a really cool guy

[fleet pelican]

im not gonna dignify that with a response

[prime sundial]

was he not

can we always divide by the content of a polynomial in Z[x] to get a primitive (and so apply Gauss's lemma)?

i can't see why not, but i'm also not sure why the property of being primitive matters if we can just do that

[prime sundial]

also, by the isomorphism extension property,\
is the extension of \emph{any} homomorphism $$\sigma : F\to \overline{F}$$$F$ a field, $\overline{F}$ an algebraic closure of it to any other algebraic closure $\tilde{F}$ $$\tau : \tilde{F} \to \overline{F}$$ an isomorphism? or do we need some more conditions

[cloud walrus]

maximo

[prime sundial]

another question (sorry for the wall of text)\
an exercise asked to show $x^2 - 3$ is irreducible over $\mathbb{Q}(\sqrt[3]{2})$\
we concluded that $$[\mathbb{Q}(\sqrt[3]{2}, \sqrt{3}) : \mathbb{Q}] = 6, [\mathbb{Q}(\sqrt[3]{2}, \sqrt{3}) : \mathbb{Q}(\sqrt{2})] = 3$$how does this tell us that $x^2 - 3$ is irreducible

[cloud walrus]

maximo

[frigid lark]

If it was reducible, sqrt(3) is the root of a degree one polynomial in Q(cuberoot(2)), i.e. sqrt(3) is in Q(cuberoot(2)), thus [Q(cuberoot (2),sqrt(3)):Q] = [Q(cuberoot(2)):Q] = 3

[prime sundial]

i see that makes sense thanks

[vagrant zinc]

It is really important to learn the division algorithm in this topic of abstract algebra.

[summer path]

i need to show that every finite order element of the free product of groups $G * H$ is conjugate to a finite order element of $G$ or $H$

[cloud walrus]

Tubular Cat

[summer path]

i've been stuck on this one for a while, could someone give a hint for it?

proving it directly seems like it'd be hard so i tried looking at the contrapositive, but im not really sure where to go with that either
though actually i think maybe it wasn't so bad going by contrapositive, or i screwed it up

[wraith plume]

does anyone know a concise statement of which groups have linear representations?

i mean faithful representations of course, they all have representations :)

[frigid lark]

I think contrapositive is the correct way to go

[broken stirrup]

You can use the fact that if w = g_1...g_n is a word in normal form then w has infinite order if syllables, g_1 and g_n belong to different groups

[summer path]

yeah i ended up getting it via contrapositive

and using that fact

thanks though

[broken stirrup]

let g=g_1 g_2....g_n be a word in normal form and consider g^(g_n)^-1. And conjugation is an inner automorphism, so it also has finite order

[broken stirrup]

Nice

[broken stirrup]

How do we know that such S exists?

[coral shale]

whats S^F

conjugate?

[broken stirrup]

Normal closure of S

[coral shale]

I think that follows from the defn of pi?

you choose the relations that define the presentation

These relations involve a set of free generators

wait maybe I misthought

[agile burrow]

Can't you just take S to be the kernel itself

[coral shale]

Whats the defn of 'a set of free generators'

Oh ok nvm, I misread. Y has nothing to do with S.

[broken stirrup]

Oh yeah sure

Thanks

[agile burrow]

Happy to help

[lapis trail]

For field F is the field extension F(c+d)=F(c,d) when c and d are LI over F?

Nevermind

[kind jacinth]

i get that 9 and 27 are correct

but why is infinity correct?

[formal ermine]

take the algebraic closure of any finite char 3 field

[kind jacinth]

whats one finite field of char 3 thats algebraically closed? I know C is one example of an algebraically closed field but thats infinite

was thinking of taking {0, 1, i} but i also need 1+i so it wont be char 3

[formal ermine]

what does that have to do with what I said

take the algebraic closure of Z/3Z

it's char 3 and not finite

[coral shale]

order of a field is its size yes?

fields with infinite orders of finite characteristic exist.

[broken stirrup]

Check out the field of rational functions in indeterminate x over Z/3Z i.e. F_3[T]

[coral shale]

@formal ermine if something is incorrect with what I have said, please correct me.

[formal ermine]

shuri you're missing the context

you're just restating the question they had

[coral shale]

I am replying directly to this query

[formal ermine]

"it is correct because it is correct"

[coral shale]

If you pay attention to their response to your answer, they are not aware of it, clearly, so stating that does not help either unless you explain what on earth the "algebraic closure" of such is

They didnt seem to think that was an explicit example.

[coral shale]

As implied by what illum wrote, the algebraic closure of any char 3 field is infinite (or char p, hence any char, for that matter)

[coral shale]

Im not sure you have the right idea of what characteristic is from what you wrote here

You start with F3 = {0, 1, 2} and add stuff to it
All char 3 fields must contain this subfield

[chilly radish]

No need to go that far, just append an indeterminate

Consider the field of rational functions with coefficients in a field of order 3

[kind jacinth]

ohhh i see. yeah i did not hve the right idea

alright thanks! I will revisit some terminology

[chilly radish]

Also proving the alg closure of a finite field is infinite is not totally trivial

[formal ermine]

it is a one liner

[chilly radish]

Ok

That doesn't matter

It's nontrivial to someone who's asking a question at this level

[formal ermine]

imo it is trivial. it follows immediately from the definition

but whatever

[coral shale]

Its a good exercise for them to try, for sure. Within reach straight from defns, just need to come up with the right idea.