groups-rings-fields

[old lava]

to let you know that you're bad

[pine patio]

i didnt know a name existed for it

thanks

[old lava]

that's objectively false

joe

[chilly ocean]

Anyone knows of an example of a polynomial of degree > 3 that is irreducible, eisenstein criterion doesnt apply but theres some simple trick to prove its irreducible? (over Q or over some UFD)

[old lava]

x^2 + x + 1 in Z[x]. Can't use eisenstein, but x^2 + x + 1 is irreducible in (Z/2)[x], so it is irreducible in Z[x] as well

if for any ideal I of R, if the image of a polynomial in (R/I)[x] is irreducible, then the polynomial is irreducible in R[x] as well

that's the simple trick

[chilly ocean]

thats not of degree >3

Poros cannot count

that one can use root test so its izi

[fiery berry]

idk what exactly would be an easy trick, but x^4 -10x^2 + 1 is irreducible over Q but reducible mod every prime and eisenstein obvoiusly doesnt apply, however you have to put in some footwork to prove irreducibilty

at least i dont know of a fast trick

so this is probably not what you were searching for

i think that's a cool one nonetheless

[oblique river]

an easier example of that is x^4 + 1

"easier" in that the coefficients are smaller

but yes I agree that's very cool :)

[chilly ocean]

Is there a set of tricks that is complete in proving irreducibility?

[oblique river]

Carla, what does "easy" mean? I'd say it's really easy to prove that a degree 4 polynomial is irreducible over F_2 because we can just write down all the reducible polynomials of degree n inductively and then just see if our poly is on that list

nope, not really

[chilly ocean]

(you could give an algorithm, and this would be complete, but this feels sort of cheating)

At least, over say Z, or a finite field maybe

[oblique river]

for finite fields you can

just because there are only finitely many options

[old lava]

I mean you can use the same principle for a polynomial of deg > 3 . Like x^4+x^3+x^2+x+1 works too

[chilly ocean]

Do you mean there is no algorithm to prove irreducible over Z? But I think I recall det or someone gave one

[old lava]

x^4+x^3+x^2+x+1 is irreducible over Z/2Z, hence over Z

Cant use eisenstein for it

[oblique river]

I think carla's point is that it's not "easy" to see that that poly is irreducible over Z/2Z

for some definition of "easy"

[old lava]

True, I suppose so

I mean it doesn't have a root in Z/2Z and theres only one irreducible degree 2 polynomial in Z/2Z, and the poly i gave isnt a square of it

So it's irreducible

It's not terribly hard to prove either

[oblique river]

I agree

I think that carla should be more specific about what they are looking for exactly

also dinner time, good bye all

[quartz lake]

quick question, why isn't the least upper bound of 2 & 5 not {20, 30}?

and I guess going off of that, why isn't the least element in this hasse diagram {2,5}?

Both of the "correct" answers to these questions are listed as "they do not exist"

But the upper bounds of 2 and 5 are {20,30,60}, so how can the least upper bound not exist?

[oblique river]

the least upper bound must be a single element

it can't be a set

there are 3 upper bounds of {2, 5} but none of them is the least

the set {20, 30, 60} does not have a least element

[old lava]

also there's no least element because you can't compare 2 and 5

[quartz lake]

ahhh i see

thank you!

[mortal hill]

😩 the notation threw me off, I was thinking of (a0,a1,...,an) = (1) => some product of ai's is a factor of 1, and I was completely lost. Thanks for the help det and buncho!

[rustic crown]

[languid meteor]

@chilly ocean x^4-4x-8=0 in Q[x]. It has no roots in Q so you check if its possible for it to be the product of two degree 2 polynomials in Q[x] (x^2+ax+b)(x^2+cx+d). Comparing coefficients you eventually get to b^2 = -8 which has no solutions in Q

[rustic crown]

\sqrt{2} means a solution to x^2 - 2 = 0

i assume by Z_7 you mean F_7 or Z/7Z?

x^ = 2 = 9 and so x is 3 or -3 = 4

so \sqrt{2}*1 is either 3 or 4

what's the action of Z7 on Z[\sqrt{2}]?

first what do you even mean by Z[\sqrt{2}] do i look it as a subgroup of R?

i don't understand how you're doing this

maybe look it like this

an R-Mod M is an abelian group with some extra data

namely a ring homomorphism R --> End_Z(M)

oh isee

So notice that End_Z(Z/7Z) is isomorphic to Z/7Z as a ring

and giving this a Z[\sqrt{2}] module structure is same as telling how you multiply by this new element

So how do we give find a ring map
Z[sqrt(2)] --> Z/7Z

you must send sqrt(2) to either 3 or 4

but once you decide this. you have no other choice.

also first let me get this clear... do you want Z[sqrt(2)] as a Z/7Z-Mod or Z/7Z as a Z[sqrt(2)]-Mod

so Z/7Z is Z/7Z

how do you bring up Z[3] = Z = Z[4]?

[rustic crown]

this is a nice way to think about it

you have 2 such maps, in the first action of sqrt(2) is same as multiplying by 3

in the second action is multiplying by 4

does this make sense?

yep

it means the abelian group homomorphisms from Z/7Z to itself

you can check that this structure forms a ring under pointwise addition and composition

one way to think about this is like the ring structure on Z is not arbitrary at all

once you have Z as a group, you automatically see that this set has a natural ring structure

[golden pasture]

alternatively you know that
Z/7Z is a Z module in one way only since knowing how 1 acts on Z/7Z is enough to determine how everything in Z acts on Z/7Z

similarly, if we know how sqrt(2) acts on Z/7Z, we immediately know how Z[sqrt(2)] acts on Z/7Z

Furthermore it's enough to know sqrt(2)•1 since Z/7Z is generated (as an abelian group) by 1

but we also need (sqrt(2)sqrt(2))•1 = sqrt(2)•(sqrt(2)•1)

alternatively when you see tensor products it's really sayibg Z/7Z\otimes_Z Z[sqrt(2)]=(Z/7Z)^2 where the action of Z[sqrt(2)] on each component is your Z[3] and Z[4]

[steady axle]

If we have sylow p-subgroup P of some group G then is H intersection P a sylow p-subgroup of H (H is a subgroup of G)?

[carmine fossil]

If H is normal yes

Not sure in the general case

[steady axle]

yea if H is normal then it's easy

[carmine fossil]

I have a feeling it's false

[rustic crown]

if the sylow p-subgroup isn't normal then its false

just take two different sylow p-subgroups

say P and Q
now clearly P intersection Q isn't a sylow p-subgroup of Q

[steady axle]

right! nice.

[rustic crown]

(just wondering what was your proof when H is normal?)

(do you show that [HP:P] = [H: H n P]?)

[chilly ocean]

Hi, question about a problem I'm trying to solve. I want to find the Galois group of the splitting field over Q for $x^4 - x^3 -3x+3$. I'm pretty sure that field is $L = Q(\sqrt[3]{3}, \omega_3)$, with $[L:Q]=6$, but I'm thinking there are only 2 automorphisms of this field? (that is identity on $\sqrt[3]{3}$ and $\omega_3 \mapsto \omega_3$ or $\omega_3 \mapsto \omega_3^2$). Where am I going wrong?

[cloud walrus]

Godel

[chilly ocean]

(If I am, but for sure this extension can't be Galois, right?)

(or maybe it can, but I don't see how I can get 6 automorphisms)

[rustic crown]

you're right about the splitting field as that polynomial is just (x^3-3)(x-1). So we only need to split x^3-3

Why do you say that any automorphism will be identity of cbrt(3)?

[chilly ocean]

hmm idk actually, for some reason I factored (x -cbrt3)(somethign else) but obv I cant do this in Q

[rustic crown]

yea the factors are (x - r)(x - rw)(x - rw^2) where r=cbrt(3) and w = omega_3

you can do this over L

[chilly ocean]

Hmm okay so I guess this is some S3 stuff right

[rustic crown]

yep your guess is right 😄

[chilly ocean]

I should check that out of 9 "possibilities" only 6 give rise to automorphism and then find the subgroups?

[rustic crown]

the thing is cbrt(3) satisfies the minimal polynomial x^3 - 3... so an automorphism in principal should be able to send any root to any other root

[rustic crown]

nah no need to do so much work for galois group of cubics. This case is easy, so we don't even need help of discriminant.

the automorphism could take r to either of r, rw or rw^2

[chilly ocean]

(actually the problem asks to find all the subfields, but I guess findiing out its S_3 first is easier)

[rustic crown]

yea you're right

[rustic crown]

do you see this?

[chilly ocean]

yeah I mean Ive seen the theorem that said roots need to map to roots

[rustic crown]

so the automorphism is determined by where it sends r and w to. we have 3 options for r and 2 options for w. This gives atmost 6 automorphism.

(we just need to say now that all 6 are actually automorphisms)

[chilly ocean]

Yep, aight thanks for the help, I'll try to finish from here, don't want you to solve it all for me:P

.

[rustic crown]

okie

[warm holly]

Can I ask a question on group theory exercice here or only in question rooms?

[next obsidian]

Here is good

[warm holly]

Alright 1 sec x)

[next obsidian]

Oh, not that question tho

Just kidding

It was a joke

🥺

[warm holly]

So I can?

[next obsidian]

It’s okay to ask that

Yes sorry haha

[warm holly]

Yo, if someone could help with the following exercice, I would really appreciate it.

Let $G_1,...,Gn$ be finite groups such that their orders are prime two by two . Show that $\xi:$$\prod_{i=1}^{n}\text{Aut}(Gi)\cong \text{Aut}\Big(\prod{i=1}^{n}G_i\Big)$

Here is what I've done so far:

Let $\phi_i \in \text{Aut}(G_i)$ for $i=1,...,n$ and $(g_1,...,gn) \in \prod_{i=1}^{n}G_i$. Define $\xi$ as the following:

$\xi(\phi_1,...,\phi_n)(g_1,...,g_n)=(\phi_1(g_1),...,\phi_n(g_n))$.

One can check that $\xi$ is a group morphism and that $\xi$ is injective. My problem now is how to prove that $\xi$ is surjective..

My idea was to define an injection in the other sens (so if I have injective morpisms in both directions I could conclude by Cantor Schroeder Bernstein), but I'm not sure how to do it properly. If someone is ready to help, I could develop more the other direction injection.

np)

[cursive temple]

doing it for n = 2 suffices right

[next obsidian]

Indeed

Because once you know it for n = 2 you can do induction

Write G_1 x ... x G_n as (G_1 x ... x G_n-1) x G_n

apply the case for two

[cloud walrus]

Dаniil

[next obsidian]

So Danil, do you know that in a finite group G, that the order of any g in G divides |G|?

[warm holly]

Yes of course

[next obsidian]

Okay, so this lets you break up an automorphism of G x H

Into an automorphism of G, and an automorphism of H when G,H have coprime order

The idea is as follows

Suppose phi is an automorphism

Then phi(g,e) is some (g’,e)

The reason is phi preserves order

But if phi(g,e) was instead some (g’,h), for h in H not the identity

Then |(g’,h)| = lcm(|g’|,|h|) can’t be equal to the order of (g,e)

Because the latter is |g|

This uses the fact that the order of g’ and h are coprime because
The order of g’ divides G, the order of h divides H, and the order of G and H are coprime

This implies the order of g’ and h are coprime

And then you can prove what I said

Similarly you can look at phi(e,h), and this sends this to some (e,h’)

So you can look at what phi does to G x {e} and {e} x H separately and this gives you an automorphism of G and H from phi

Then just notice xi sends that pair of automorphisms back to phi

[warm holly]

Oh alright I see the idea I will try, thank you very much

[next obsidian]

Np

Oops sorry

Autocorrect 😦

[sinful mirage]

are representations by definition surjective?

[hot lake]

no

[golden pasture]

(in general they aren't surjective)

[chilly ocean]

is there an algorithm to calculate the Monic polynomial of a matrix?

[old lava]

the monic polynomial?

do you mean minimal polynomial?

or characteristic polynomial?

[chilly ocean]

well, idk, exercise sais monic

orthonormal polynomial?

@old lava what is each one?

maybe if u tell me how to get each one i can point it

the characteristic is $|A - I\lambda|$?

[cloud walrus]

nosequepasa

[chilly ocean]

cuz if so, i think not

[old lava]

that makes no sense

any polynomial with a leading coeff of 1

is a monic polynomial

"monic polynomial of a matrix"

is pretty meaningless

[chilly ocean]

of = associated to (?)

mmm

ill try to translate the best i can

Given an hermitian matrix of grade m, calculate the monic polynomial of grade n of the matrix

so idk

is this the minimal polynomial?

post the original

it isnt in english

so

unless u know spanish

🙂

"the monic polynomial" doesn't make much sense so it'd be better if you posted the original to let someone else translate

who knows maybe there's a fellow spanish speaker here

i found this

A nonzero polynomial is monic if its leading coefficient is 1.

do u know french? is weird cuz

if i search on the wiki "monic" on spanish

wiki has a few options to change language

well, english isnt there

but french is

or deutch

i know what a monic polynomial is, im asking you to post your question in the original language, because "monic polynomial of a matrix" is ambiguous and could refer to at least two things

is any of those 2 things related to orthonormal polynomials?

it is proving that this monic polynomials is exactly the orthonormal polynomials of grade n

and to prove it, she does this

$\langle P_n(t), t^{k}\rangle = 0$, for $0 <= k <= n-1$ and $\langle P_n(t), P_n(t)\rangle = 1$

[old lava]

\langle and \rangle

is what you're looking for

instead of < >

[chilly ocean]

ah

still, it is the dot product

inner*

so, does this clarify something?

xD less equal isnt lq? q.q

[old lava]

not really, no

there's 2 main (monic) polynomials associated with matrices, which are the minimal and characteristic polynomials

nothing really to do with sets of orthonormal polynomials

[chilly ocean]

But she is proving this

$\langle P_n(t), t^{k}\rangle = 0$, for $0 <= k <= n-1$ and $\langle P_n(t), P_n(t)\rangle = 1$

[cloud walrus]

nosequepasa

[warm holly]

I just have a question, probably a bit stupid.. If we consider $\phi \in$ Aut$(G \times {e})$, is it equivalent to consider $\phi \in$ Aut$(G)$?

[cloud walrus]

Dаniil

[warm holly]

I'm pretty sure that yes just I want to ensure myself)

[chilly ocean]

G×{e} is isomorphic to G so their automorphism groups are isomorphic as well

If you want to be super explicit

You should write done an isomorphism from Gx e to G

But yes I think you understand

[warm holly]

Alright, thank you very much for your answers!

[chilly ocean]

But just to be sure

Phi on the left is a function from Gxe

But you have an isomorphism to G

if f is automorphism on the left one then foI is automorphism on right where I is isomorpism and vice versa

What Carla said

what program you use to draw that?

Also. Is there an easy way to show if x²-(1+sqrt(3)) is irreducible over Q(sqrt(3))?

[warm holly]

I think it's word on Ipad

[rustic crown]

yep

notice that the root will also satisfy (x^2-1)^2-3 = x^4 - 2x^2-2

this is irreducible by eisenstein, so Q(root)/Q has degree 4

Hence Q(root)/Q(sqrt(3)) has degree 2

[chilly ocean]

oh ofc im so dumb

[chilly ocean]

GoodNotes on iPad

very good notes indeed

[chilly ocean]

@chilly ocean Name of app

[neat valley]

\newcommand{\notimplies}{\;\not\!\!\!\implies}

So given some normal subgroups $H,K$ of a group $G$, I know that \[H\cong K\notimplies G/H\cong G/K\] (for example, consider $G=\bZ$, $H=2\bZ$, and $K=3\bZ$). But is it true that \[G/H\cong G/K\implies H\cong K?\]

[cloud walrus]

Isaiah

[blissful ice]

This is also not true, unfortunately
You can find a counterexample in Z_2×Z_4
Then you take K={0}×Z_4 and H=Z_2×{0,2} =iso Z_2×Z_2
Notice they're not isomorphic!

However both groups are of order 4, so quotients will be of order 2 - the cyclic Z_2
@Isaiah#5685

I failed pinging for some reason. Thanks phone

@neat valley

[neat valley]

Damnit, thanks though!

[north widget]

quick question

if ur trying to find kernel of some ring homomorphism is this the correct line of thinking,
let f:A->B be homomorphism
ker(f)={a in A|f(a)=0_b}
so let a in A
f(a)=0 and then use implications that follow from that to find out what the kernel is?

[chilly ocean]

What implications?

Just need to find all a in A such that f(A)=0

This is the definition.

That would get you a superset of the kernel, and if you know your implications are actually iff, then it would truly be the kernel

[north widget]

@chilly ocean kinda hard when there are lots of them

im just trying to use facts from its multiplication and addition

also

how do generators work for rings again?

is it just the same as groups + all elements when you multiply them?

yeah nvm its gonna sound confusing so ill stop

[blissful ice]

Generators of a ring also account for addition
That's a common theme, imo
If a set X generates some structure, it means you take elements from that x and merge them in every possible way the structure allows: for rings this means multiply arbitrarily and add

[scarlet estuary]

given a subset of a ring, the subring generated by that subset is the intersection of all subrings containing that subset

i.e. the smallest subring containing those elements

in other words, we can say a set generates a ring iff the only subring containing that set is the whole ring

so yes, generators factor in addition and multiplication

[north widget]

i wonder if it relates to closure or smnth

whenever i hear smallest blank containing blank i think of closure in topology

afaik subsets of groups dont have same type of property

but i guess they do

there is probably some relationship im not 100% aware of

[viscid pewter]

?

a set generates a group if the only subgroup containing that set is the whole group, right?

[chilly ocean]

I guess closure of a set A is generally like "the intersection of all supersets of A having such and such property"

[north widget]

ok

what is quickest way to check if a ring generator is an ideal?

can i just jump straight to absorption property?

im trying to show <x^2+x+1> in Q[x] is an ideal

im just thinking it is

oh

i dont remember what principal ideals are

its related though

[old lava]

it's by definition an ideal, that's how you denote ideals generated by finitely many elements

[north widget]

but why is it an ideal by definition?

cus like

[old lava]

(a_1, a_2, ..., a_n) is by definition the ideal generated by a_1, a_2, ..., a_n

[north widget]

x*(x^2+x+1) isnt in it

[old lava]

that's how you define it

[north widget]

generator

like

<x^2+x+1> is subring of Q[x]

it generates

[hidden haven]

Subring generated by

[north widget]

<x^2+x+1> is a set made by multiplying and adding x^2+x+1 by itself?

yes

[hidden haven]

Same as in subgroup notation

[north widget]

subring generated by

ye i gave a counter example

<x+1> should be an ideal

wait

when do polynomials in Q[x] generate ideals?

[hidden haven]

Would be my guess

[north widget]

they def do

right?

[hidden haven]

I don't see how you'd get absorption of multiplication just by closing it under multiplication

[north widget]

yeah wait

thats a good point

[hidden haven]

Hold on won't subring generated by an element always contain 1

[north widget]

but how do people quotient by polynomials?

?

[hidden haven]

But ideal won't unless it's the whole ring

[north widget]

clearly we have something wrong here

what is an ideal of Q[x] then

like just 1 example

polynomials though

yea

[hidden haven]

Oh this is a somewhat interesting point, you might be thinking how for vector spaces we quotient by subspaces and for groups we quotient by normal subgroups so we should quotient by subrings in rings, but that is wrong. In each case, we can only quotient by kernels, and kernels of ring maps are ideals rather than subrings

[north widget]

then why isnt x^2+x+1 one

yes

i just said it

subring with absorbtion

multiply x by x^2 +x +1

i dont think its there

[hidden haven]

When you generate an ideal, you add in all those things

Ideal generated by a set will by definition be an ideal

[north widget]

yes

it isnt the same as the subring?

ok

so i know this already wtf

wait

i know why im tripped out

[hidden haven]

Subrings are images of homomorphisms, ideals are kernels of homomorphisms

[north widget]

the question is asking find f(x) in Q[x] such that R isomorphic to Q[x]/<f(x)> but i guess the notation is not saying subring but ideal because thats the only way you could quotient

wow

lots of wasted time ig

[hidden haven]

Lol

[north widget]

is there no standard notation

or do you need to explicitly state what it is

im doubting () is the better notation

most used*

ok

[hidden haven]

() is the most used even though it's shit notation

[north widget]

google images

is why im doubting

nvm

u have no reason to lie

mine uses <>

[hidden haven]

Don't trust slimvesus

[north widget]

fraleigh

[delicate bloom]

it really doesn't matter lol if you see either notation you should just recognize it and move on with your life

[north widget]

lol what

[old lava]

why is it shit notation?

[opal osprey]

Suppose that I have a commutative ring with unity R such that C is a subring of R and R is a finite dimensional vector space over C

Then

If R has exactly one maximal ideal

I want to prove that this maximal ideal consists of exactly those nilpotent elements of R

I think somehow

That I can use Nakayama's lemma here

knowing that R is a finitely generated C-module

but idk really

any ideas?

what if I make the condition of R being commutative weaker and only consider R being a ring with unity?

[next obsidian]

Is C the complex numbers?

[opal osprey]

yup

[next obsidian]

and you're certain the result is true?

[opal osprey]

Yeah

It's one of the exercises

In Artin's book

[next obsidian]

Okay, well nilpotent elements of R is the intersection of all primes, so you need to show R has only one prime ideal

[opal osprey]

And I am kinda stuck

[next obsidian]

This follows by integral stuff

Do you know the definition of an integral extension?

and a few results about them?

[opal osprey]

nope

[next obsidian]

Do you mean no?

Ah okay

well so there's a result that for an integral extension that A < B, if p < B is prime and such that p\cap A is a maximal ideal of A, then p is maximal in B

in this case your ring R is an integral extension of C, but C is a field

so the point is any prime ideal when intersected with C becomes 0, which is maximal

so it was maximal to begin with

[opal osprey]

integral extensions are sort of generalized field extensions for rings?

[next obsidian]

so let me try to think if you can somehow use f.d. vector space

Yes

[opal osprey]

hm

[next obsidian]

it's a generalization of algebraic extensions

[opal osprey]

I will take a look into those then

looks fun

[next obsidian]

It's a bit of comm alg to get it

I bet there's a simpler proof

[opal osprey]

I was thinking about using stuff about field extensions at first

but R isn't a field

[next obsidian]

but you have to show R has only one prime

yeah, it won't be an ID in general either

if it is, then it's a field

by... integral extension stuff

haha

[opal osprey]

hmm

Do I really need to use these results?

[next obsidian]

No

[opal osprey]

This looks op as fuck

[next obsidian]

definitely not

I think there's a simpler proof

but I'm just airing out what I know to see if I can specialize to this case and come up with a simpler proof

I think you want to use linear algebra haha

which is kind of "duh"

Oh, I think I see it

Okay, so you want to do the thing I said about if you're an integral domain then you're a field

but you can get a simpler proof of that when it's a f.d. vector space

I can try to give you a hint, or I can just tell you waht to do haha

[opal osprey]

I've been thinking though this for a while by now lmao

And this is not even the main thing I have to prove

btw

[next obsidian]

OKay

I'll just go thorough it then

[opal osprey]

this is exercise 9.11 of Artin's book

[next obsidian]

Lemma 1: Let A be an integral domain which is a f.d. K-vector space, then A is a field

consider for any a in A nonzero the map A -> A given by multiplication by a

this is a K-linear map, and injective by A being an integral domain

[opal osprey]

On the chapter on algebraic geometry

[next obsidian]

by rank nullity it's also surjective

this implies there exists a b in A such that ba = 1

so a is invertible

thus A is a field

Now if you had C < R, and R a f.d. vector space, for any prime ideal P < R, consider R/P

this is an ID, and also f.d. over C

so R/P is a field, thus P is maximal

so any prime ideal of R is maximal, but there's only one maximal ideal

thus R has only one prime ideal, its maximal ideal, and this is equal to the nilradical

[opal osprey]

Oh dammnit

That's a nice result

[next obsidian]

It's more general

if A < B is an integral extension of integral domains, then A is a field iff B is

integral extension is like algebraic extension, except every element must satisfy a monic polynomial

A finite extension is in particular integral

anyway, that's not too important for your purposes at the moment

but it's a cool result

[opal osprey]

https://www.youtube.com/watch?v=EYAeBXVatS4&list=PL2xox8ncv81XaZWb9AupufVZegaSHmJwb&index=2

There's these commutative algebra lectures

I still have to watch lmao

[next obsidian]

hahaha

[opal osprey]

on ring extensions

[next obsidian]

integral extensions are pretty important for AG

[opal osprey]

atiyah also covers these on his book

[next obsidian]

yeah

I don't like A-M tho haha

[opal osprey]

But I still gotta finish the whole module theory part

before going into the deeper stuff

[next obsidian]

I did use its proof for going down for flat extensions tho

[opal osprey]

What's your favorite book on commutative algebra?

[next obsidian]

Matsumura

Commutative Ring Theory

[opal osprey]

I use Atiyah's because it's the only most people recommended me

[next obsidian]

This is a bit more advanced tho

I also like Reid's Undergraduate commutative algebra

I think A-M is fine if you can stand it

I just can't stand the book, the exercises kill my motivation to read it

Matsumura also covers a lot more material, but A-M is tiny so that's expected

[old lava]

what's wrong with AM exercises

[next obsidian]

I like to flip a page every couple days

not every couple weeks

[old lava]

kekw

so are the exercise parts really dense or something?

[next obsidian]

Yes

There's a lot

[old lava]

ah I see

[next obsidian]

and they can be pretty hard

It's like the Hartshorne of CA

[old lava]

time to do every A-M exercise for my next challenge

kekw

[next obsidian]

in that the exercises have all the material

I mean if you do every A-M exercise

you ought to go do AG

or ANT

use your CA knowledge for something

[opal osprey]

Anyway

[next obsidian]

lol

[chilly ocean]

Just skip the exercises

[opal osprey]

Thanks a lot chairmonkey

[next obsidian]

[opal osprey]

😳

[old lava]

ah yes, the electrical and computer engineering/algebraic geometry path

[next obsidian]

椅子猿

[old lava]

I definitely have the time for it

[next obsidian]

kekw

A prof at my school does computer vision stuff

using AG

[old lava]

sounds fun

[next obsidian]

maybe you'll become the Don of computer vision

[old lava]

I might do more algebra during my masters and all

on the side

who knows

[next obsidian]

kekw

okay, I go back to AG now

[old lava]

good luck

[north widget]

qq

is Q(sqrt(a)) isomorphic to Q(sqrt(b)) for a,b integers

isomorphic as fields

[chilly ocean]

sqrt(a) cannot map to anything, its image would have to satisfy x^2=a

[old lava]

if a = b then they are, sure

[north widget]

oh

[old lava]

otherwise, no

[north widget]

88 nice explanation

well i just gave a faulty explanation

damn i hate time constraints

[delicate bloom]

you're trying to cheat on a test or somethin?

[north widget]

no

if i was trying to cheat i would post questions?

[delicate bloom]

here's your question

[north widget]

also i wouldnt say i hate time constraints

why would i be obvious about it lol

[delicate bloom]

here's you claiming to have given the wrong answer because you didn't get their answer in time for the time limit of your quiz

[north widget]

homework exists.

[delicate bloom]

why are you being obvious about it? you're not explicitly denying it

[north widget]

usually students have homework due at deadlines like 11:59

also arent tests timed?

[chilly ocean]

[next obsidian]

Isn't the implication that it's already due

and they answered incorrectly and are asking what the real answer was because they were unsure?

That's how I read it at least

[delicate bloom]

sounds like you're implying that in your case it is due within some certain time limit of starting it, which could be any time lol

[north widget]

man

rename this server to forensics/mathematics

ig 4+ hour tests exist?

i think i asked question about homework around 6

[north widget]

curious question

if you have a group

and multiply each element together

what element do you end up with?

does it change for every group

i guess it does

:/

yea

i guess it depends

just thought of some exampls which r random

no

its gonna be whatever the 1 element would be

but it has an inverse

it still depends in each case

it wont have all their inverses always

it doesnt necessarily have a 1 element either

For finite abelian groups the product will be the product of all elements which are their own inverses

this false

groups

i said groups i thought

[sturdy marsh]

no

[north widget]

no?

z2

0+1

=1

but ur not adding it

multiplying it

ye

yes

no

forget i asked

question

why is notation for galios groups so similar to quotienting groups

besides the Gal part.

ig the notation for field extensions

[fresh vessel]

I may be incorrect about this, but I always thought of it specifying the field extension. So E/F is supposed to be read as E over F. So Gal(E/F) is supposed to suggest its the Galois group for the extension E over F. Feel free to correct me if thats incorrect though

[north widget]

sounds right

but field extension looks similar to quotienting

wondering why

[fresh vessel]

Its definitely not a quotient since you cant have non trivial ideals of a field

But I have no clue why the notation is so similar

[iron vessel]

Can anyone help me with a Group Theory question? I posted my question in #help-1

[solemn rain]

ask here

@iron vessel

[iron vessel]

Part C and D, they were explained to me in channel 1, but I am still not sure how to prove it.

[next obsidian]

the image of anything in S_3 has order dividing 6. So you can show the following, if x is in N, then psi(x) = psi(y)^6 for some other y. The hint tells you how to do this

this proves (c)

Oh whoops you said C and D were explained to you

[iron vessel]

It's C and D i need help with yeah

[next obsidian]

oh well what I said proves C

I'm not really sure how to approach D

[iron vessel]

thanks alot, it makes a little more sense now 🙂

[next obsidian]

Oh, well actually by C

you know a map G -> S_3 will factor through G/N

And so it suffices to find the number of maps from G/N -> S_3

and G/N = {-1,1} which is also just Z/2Z

so you can find how many maps Z/2Z -> S_3 exist, and this should be pretty easy

[north widget]

quick question

how do you know distinct elements for a given quotient ring?

so like say you have Q[x,y,z]/<xy^2-x-xz^2>

how would i even start?

[solemn rain]

think of long division

this is just asking how does the remainder woudl ook like

when diving a polynomial over (xy^2-x-z^2)

[north widget]

yes

[solemn rain]

but in the algebraic sense its just sets of cosets

[north widget]

i understand you are modding out by it

[solemn rain]

yea

[north widget]

but how would i go about finding each distinct element for a polynomial like that

im thinking of making a calculator that can do it

[solemn rain]

this is an infinite ring

[north widget]

yes

[solemn rain]

so i thinnk thta would be a bit hard

but u can just take random polynomials adn mod them out by the polynomial

[north widget]

i mean you can show what its isomorphic to

[solemn rain]

and then do out operations

to get other polynomials in the ring and etc

[north widget]

i heard that the standard method was something with division algorithim

and that you should only check what multiplication looks like for polynomials f with deg< degree of ideal generated by polynomial

[solemn rain]

well yea

modding is essentialy division algorithm ig haha

and by the division algorithm u always have the remainder being deg less than the thing ur dividing by

[north widget]

yeah

cus

[solemn rain]

when deg is ur norm

[north widget]

what is quotient relation again?

[solemn rain]

we say a = b mod H if a=bh for some h

[north widget]

a~b iff a-b = 0mod(polynomial)?

[solemn rain]

where H is a subgropu

yea this is for that specific case

what i wrote is for general groups

[north widget]

H being normal

[solemn rain]

H being normal implies we have a group structure

[north widget]

ideal*

[solemn rain]

normal , ideal whatever

this for groups this for rings

[north widget]

wait

is every ring isomorphic to some polynomial ring quotient?

like C[x1,x2,...]/(f(x))

just curious question

sound wrong though

[solemn rain]

i mean

try to think of first iso ig

if there exists a homo between a ring and some ring then yea

but

no

[north widget]

the answer is def no

[solemn rain]

there doesnt have to be a ring homo between two rings

take Z_p and Z_q

for p and q primes

[north widget]

yis

[solemn rain]

yea

[north widget]

wat about it

[solemn rain]

there cant be any homo

between those

[north widget]

well yea of course

but that wasnt question qq

[solemn rain]

yea haha

[north widget]

haha

[solemn rain]

damn i cant think of a ring not isomorpihic to a quotient of rings

idk tbh

[north widget]

i mean that wasnt question either

quotient of polynomial rings

because i think the only rings that wouldnt be isomorphic to a quotient of rings are ones where all of their homomorphisms have a trivial kernel

and

wait

i need to rephrase that

ok

im thinking if a ring exists where every surjective homorphism from another ring is such that its kernel is trivial then that ring isnt isomorphic to any quotient rings?

[solemn rain]

to any quotient rings of itself*

atleast

but ig

normal subgroups are just kernels

so yea

[north widget]

wait

isnt there a name for groups with no nontrivial normal subgroups

[solemn rain]

yes

simple

[north widget]

there is also a name for rings with no nontrivial ideals

or um

[solemn rain]

yes

simple

or

better

a field

[north widget]

nontrivial ideals mean maximal ideal = ring itself?

[solemn rain]

given the ring is commutative

[north widget]

but dont fields have nontrivial ideals?

[solemn rain]

no?

[north widget]

oh

i guess they dont

lol

how would i prove that though

wait

abosrption couldnt exist

[solemn rain]

try to think about it

[north widget]

absorbtion ?

[solemn rain]

and suppose that F indeed has an ideal

[north widget]

how do you spell that

absorption

[solemn rain]

remember every element has an multiplicative inverse

[north widget]

! it turns into a p

[solemn rain]

lmao

why

does it turn into a p

[north widget]

yeah it cant exist because the field contains each inverse

[solemn rain]

hahaah

[solemn rain]

so yea no ideals

no nontrivial*

[north widget]

so any a*b also in the field

oh ok

[solemn rain]

do u know why?

[north widget]

oh wait

i think we found the answer

yeah i know why

[solemn rain]

okaay

[north widget]

fields cant be isomorphic to a quotient ring

[solemn rain]

well yea

haha

[north widget]

right?

[solemn rain]

not nly quotients but also products

R x S is never a field for R and S rings

its a consequence of what u just proved

do u see why

[north widget]

ya

[solemn rain]

but

what u said depends on how u look at it

like for example

think of GL_n(F)/SL_n(F)

[rustic crown]

Wait whattt

[solemn rain]

this is isomorphic to a field actually

[north widget]

i dont know special linear

[solemn rain]

matrices taht have det 1

[north widget]

general linear are invertible right?

[solemn rain]

yea

do u see the natural mapo

that comes to mind?

[north widget]

no

[solemn rain]

and now do u see that this quotient is iso to a field

[north widget]

what field though

[solemn rain]

F* itself

[north widget]

oh really

[rustic crown]

That's not a field... It's the multiplicative group of nonzero elements of the field

[north widget]

F*

[solemn rain]

by the determinant map

[rustic crown]

If not char=2

[solemn rain]

yea sorry given char

[north widget]

so

what if we say perfect fields

[solemn rain]

mb

[north widget]

wait not char 2

[hidden haven]

F[x]/(x)

[rustic crown]

ngl idu anything what's happening

[north widget]

so then what about char 2

[hidden haven]

It's isomorphic to F

[north widget]

oh um

det

our quest

[solemn rain]

.

should be to what ur saying

[north widget]

are there rings not isomorphic to a quotient ring

and then what are they

[hidden haven]

Any ring R would be isomorphic to R[x]/(x)

[solemn rain]

sad ending

[north widget]

ok

that is true i see

man

wait

not yet

(x) still counts as trivial

in my book

[hidden haven]

Bruh

[solemn rain]

what book is this

haha

[hidden haven]

[north widget]

my book

[hidden haven]

Trivial means (0) or (1)

[solemn rain]

C = R[x]/(x^2+1) haha

[north widget]

oh yeah good point lol

fine

trivial or (x)

[rustic crown]

This isn't correct, consider Q[x, y]/(y-x²).
When you mod out y is same as x² and "degree" Increased.

[north widget]

wait det

can you help me

is there a general formula for finding distinct elements of a quotient ring

so that you can find out what its isomorphic to

[solemn rain]

yea maybe i meant "same variable degree"

mb

again

[north widget]

like what would Q[x,y]/(y-x^2) be isomorphic to

[solemn rain]

thats not easy to answer ig

[north widget]

or i just havent learned divison

[rustic crown]

This one is easy... y is like pretty redundant.

[solemn rain]

maybe u need to look like it as

fields of fractions stuff

but in general i dont think its easy

[rustic crown]

So that's isomorphic to Q[x]

[north widget]

how do yo uknow

off top of head

[rustic crown]

I can see I'm replacing y by x² everywhere so y doesn't do anything which x has already done

[north widget]

oh

good point

wait

then what about Q[x,y,z]/(yx^2-x-xz^2)

thats a little harder

no?

[rustic crown]

If you wanna be formal, consider the map Q[x, y] --> Q[x] sending x to x and y to x², clearly surjective and kernel is (y-x²)

[north widget]

oh

that might be a method

nvm

[hidden haven]

What do you mean by "what is it isomorphic to"? Like do you want it to be a specific kind of ring, or just something "simpler"?

[north widget]

something that isnt a quotient

so simplier i guess, even though it might not be necessarily

[hidden haven]

I see

[north widget]

is it naive to think that quotient/products are easy to evaluate if you have method to do it?

[solemn rain]

trace back by first iso

[hidden haven]

But every ring is isomorphic to a quotient

[hidden haven]

So that isn't really well defined

[north widget]

to a not quotient ring

:/

like

[hidden haven]

Lol edited

Like what if I just gave you the quotient construction but instead of using equivalence classes I just used some random things as elements

[north widget]

um

example

[hidden haven]

Instead of Z/2Z which is a quotient, I could say R = {x,y} where x+x = y+y = x, x+y = y, etc

Then would you say R is not a quotient

Because I described R as not a quotient

But it's isomorphic to Z/2Z

[north widget]

yeah

um

im sorry for makig you waste some time

but

i meant to say polynomial rings

being quotiented

[hidden haven]

Yeah np lol

[north widget]

and yea R = {x,y} counts

as long as there is a ring representation

then its fine

but im wondering if its always possible to reduce a polyomial ring quotient into something that isnt a polynomial ring quotient

and if it is, what are the methods

for me the way i see it, quotient polynomial rings just give you different types of multiplication

[hidden haven]

But R[x]/(f(x)) always has the ring representation that you get by having one element corresponding to each coset of (f(x)) and defining addition, multiplication accordingly

ie the quotient construction itself

But you don't use equivalence classes

[north widget]

wait

one element corresponding it each of what did you mean to say

one element being some random variable?

yeah

[hidden haven]

To*

[north widget]

how do you find the distinct cosets

thats my question ig

i understand how to find multiplication for the most part

even though im slightly lost when its multivariabled polynomial rings

[hidden haven]

Writing down ideals is pretty hard, from there finding cosets would be easy

[north widget]

also this is more for infinite rings

[hidden haven]

Like we say ideals generated by ... because it's very difficult to write ideals in set builder form

[north widget]

is it

i never tried lol

someone today pressed me to try and explain what the ideal was

and i couldnt tell you honestly speaking

like ideal generated by (x^2+x+1)

where do i start

[hidden haven]

Ideal generated by a set S of elements of R is the smallest ideal containing S

[north widget]

what are principal ideals again?

[hidden haven]

Smallest meaning intersection of all ideals containing S

[daring ibex]

ideals generated by single element

sniped

[hidden haven]

Notice that intersection of any family of ideals is also an ideal

[north widget]

but isnt (x^2+x+1) generated by 1 element?

[hidden haven]

It is

[daring ibex]

It is a principle ideal

[north widget]

so whats its set builder?

[daring ibex]

ideal generated = {k(x)(x^2+x+1) | k(x) \in R[x]}

[hidden haven]

Set of all polynomials of the form p(x)(x²+x+1) where p(x) is another polynomial

[north widget]

thats cheating mirza

[daring ibex]

it's literally

the definition

[north widget]

oh its not cheating now

ok i guess so

[hidden haven]

Yeah you use generators to write it in set builder

[north widget]

thats obvious then if you can write it like that

oh shit

[daring ibex]

well yes they are obvious

[hidden haven]

It's hard to describe it without generators is my point

[north widget]

i didnt even begin to think of non principal ideals

[hidden haven]

Which is what you seem to want here

Like given an element, finding which coset it lies in

Is a harder problem than finding out if it's in the ideal

Which itself can be pretty hard

[north widget]

finding if its in ideal is just seeing if it divides

[hidden haven]

For principal ideals yes

[north widget]

holy smokes

[hidden haven]

For non principal it's way harder

[north widget]

i never thought of how you would quotient over nonprincipal ideals

jesus christ

nah

this isnt concrete dawg

gimmie something concrete

and ill tell you if ur right or not

[hidden haven]

Yeah this isn't concrete lol

[north widget]

apparent if you are the savior himself

no

i want to know what is concrete though

not by you

just in general

im thinking most things i learned in algebra are pretty crazy once you look deeper and arent always intuitive

[daring ibex]

[hidden haven]

Concrete means you have forgetful functor to Set

[daring ibex]

Been trying to do that for a while

[hidden haven]

Don't ask me for a definition of forgetful functor tho

[daring ibex]

Functor that forgets bro

[hidden haven]

Bro

[daring ibex]

Alzheimer functor bro

It's forgotten it's family

[hidden haven]

Wasn't asking seriously lol

[north widget]

,w alzheimer functor

[cloud walrus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by Marlin Flier#7333

[daring ibex]

lmao

chad

[north widget]

alzheimer functor site:https://ncatlab.org/nlab/

[hidden haven]

Bro no

[daring ibex]

Going through a rough time functor

[hidden haven]

Don't remind me of games on graphs

[daring ibex]

lol

how did that test go

i forgot to ask

[hidden haven]

Shit

[north widget]

wait

what test

[daring ibex]

damn

[hidden haven]

Prof is shit

[north widget]

game?

[daring ibex]

[hidden haven]

I hate him

[north widget]

oldest story second to the bible

[hidden haven]

Games on graphs, part of a course

[north widget]

league of legends?

go?

chess?

[hidden haven]

He basically makes shit up when he doesn't have an answer, and when he makes mistakes while grading

[north widget]

connect4?

[hidden haven]

Bro I wish

It's literally the shittiest game you can think of

[north widget]

connect 4 is a solved game right?

[hidden haven]

Like imagine playing the most boring game ever

But forever

[solemn rain]

i am literally logging in to league now

[north widget]

:ban:

[solemn rain]

what a coincidence i check this chat

[north widget]

,ban @solemn rain

[solemn rain]

anyways

[cloud walrus]

This may only be done by a moderator!

May 8, 2021, 6:09 AM

[solemn rain]

how do u get that good at math @hidden haven

[daring ibex]

moldilocks didn't get good at math

[north widget]

mirza

[daring ibex]

where did you get that idea

[north widget]

can you help me

[hidden haven]

Subconcious takeover successful

[daring ibex]

[north widget]

i want exercises with semi direct product

because i dont have intuitive understanding

[daring ibex]

Dummite and foote has them probs

[hidden haven]

Mirza pls

[solemn rain]

check df

[north widget]

like how tf is R -> AutN a homomorphism

[daring ibex]

Like for most things in algebra

you're not gonna have intuition ig

At least that's how it is for me

[north widget]

then who the fuck is making these things up with no intuition?

are algebraist actual nerds?

[daring ibex]

It's built up as machinery

[hidden haven]

I'm really good at math, I have the advanced role in the mathematics discord server

[daring ibex]

for use in other fields

you have to look at the context of those fields to get an idea

[north widget]

id expect it to be found in other fields first and then math decides to formalize it

[solemn rain]

for algebra its just like

book worming really

and brute forcing questions

like what u did

gj

[north widget]

but i have no idea what the other fields are

[daring ibex]

no i mean fields in math

[north widget]

book warming!!

[hidden haven]

You do develop intuition for it

[daring ibex]

like topology or analysis

I mean yes

Of course

[hidden haven]

Not visual intuition

[north widget]

oh

[daring ibex]

I'm saying it's not the same kind of intuition

as usual

yeah

it's not visual

[hidden haven]

But intuition about kind of stuff will be true

[solemn rain]

no intuition

u only get mathematical intuition

having the sense

[daring ibex]

But sometimes u do have an idea

[solemn rain]

but no actual intuition

[daring ibex]

"yeah this is probs how this'll be done"

[north widget]

there are other ways to develop no visual intuition through anecdotes, but mathematicans are bad at stories WTF!

[daring ibex]

"this has something to do with this that and the other"

[hidden haven]

[daring ibex]

tho actually