groups-rings-fields

[rigid cave]

I know that $K_{m-1}(\alpha_m) = K_m$ by construction, but how about $K_m'=K'_{m-1}(\gamma)$?

[cloud walrus]

older sister

[rustic crown]

lol that's the definition of K'm

[rigid cave]

OHHHHHH

[rustic crown]

that's the first time K'm makes an appearance

[rigid cave]

Ohhhhh wow... I just can't read proofs... Thank you so much!

[rustic crown]

[rigid cave]

[celest brook]

so by definition, quotient group and normal subgroup are intertwined?

dunno why my prof didn't introduce them together then

[chrome hinge]

wow i just came in here with a question about field extensions lmao

[celest brook]

o

[chrome hinge]

You were asking first

I didnt get your question though

[celest brook]

i was mainly just seeking confirmation

i.e. if you have a normal subgroup, then there exists a quotient group<-> if you have a quotient group, then there exists normal subgroup

[chrome hinge]

hmm

its more like

The concept of quotient group has meaning when you quotient by a normal subgroup

[celest brook]

yes

[chrome hinge]

yep

[celest brook]

speaking of bijective correpsondence i just proved right cosets and left cosets have a bijective correspondence on HW :I

not fun

[chrome hinge]

you would not have commutativity in your classes if it werent a normal subgroup

[old lava]

I think you mean associativity

[chrome hinge]

yeah, im sorry

[old lava]

ya, if it's not a normal subgroup, then the left cosets aren't equal to the right cosets

so you can't have associativity

sadge

[chrome hinge]

Anyone who has studied Galois theory here? :)

Can i have a normal extension of Q generated by an element which i cant obtain from Q with just sum, product, inverses or sqrt?

I mean like $Q(2^{1/3})/Q$, that sort of extension

[cloud walrus]

Maikel

[chrome hinge]

(i know that one is not normal)

[chilly ocean]

yes

like Q(sqrt2,sqrt3) ?

I want to say yes but I can't think of an example

oh no sqrt ok

[chrome hinge]

yep thats the issue

no sqrt...

[chilly ocean]

Maybe a cubic with all real roots?

I think of 2^1/3 as not working because we need i to make the extension normal

[chrome hinge]

Yep that doesnt work because of exactly that

Hmm

I need it be of degree greater than 4

or maybe four

[chilly ocean]

Oh, did you prove it is not possible with degree 3?

[chrome hinge]

Nope

[chilly ocean]

2^1/4 also not normal

[chrome hinge]

Not normal neither obtained without sqrt

[chilly ocean]

oh yeah forgot its obatined by sqrt

[chrome hinge]

Ok so heres my problem: i want a polynomial in Q[x] of degree 4 with roots not obtained from Q with sqrt

Yeah its kind of tricky

So i was thinking it would be nice to find a normal extension generated by an element of that kind

If the extension is of degree 4 im done

[chilly ocean]

Maybe the 5th roots of unity? x4+x3+x2+x+1=0

One of them generates the rest

[chrome hinge]

if that has all real roots could be a great option

[chilly ocean]

Erm, they are not real

[chrome hinge]

because of the equivalence normal iff theres a polynomial that splits in the extension

Oh and also it would have to be roots not obtained with sqrt

at least one

idk

See? im quite lost with this problem haha

[lunar coyote]

does anyone know where i can find similar questions like these

[old lava]

any algebra textbook that covers ring theory I suppose

[lunar coyote]

any suggestions?

i'm still a beginner at this and would need a lot of help from solution manuals

[old lava]

that's probably gonna be an issue

you won't really find solution manuals (they just don't exist)

for any decent algebra tb

as far as I know

[lunar coyote]

ugh i'm stressing for my algebra assignment and upcoming exam and want to get better at proofs asap

been trying to revise this stuff but i'm getting no where

[chilly ocean]

I mean dummit foote has a lot of problems to which you can find solutions online, but you should be able to solve them on your own if you read the chapter

[next obsidian]

Isn’t there a quartic formula? Like, every quartic should be solvable by radicals so isn’t this impossible?

[lunar coyote]

will have a look

[molten silo]

Part e anyone

[next obsidian]

2 divides 20 but neither of the other things it factors as

Prove the last part of what I said

Then u r done

[left monolith]

What does the notation Z_15 mean here?

{0,1,2,...,14}?

[chilly ocean]

to be pedantic, {0 mod 15, 1 mod 15, ..., 14 mod 15}

taking a picture of your screen

[left monolith]

Yeah, they are classes, true

[left monolith]

It's not my picture

[chilly ocean]

[celest brook]

how do they know if the quotient group exists if we need to prove that H is normal???

oh wait sets

ffs

[chilly ocean]

[celest brook]

Its a set difference

not

wait

dshugig

hate

[chilly ocean]

[celest brook]

OK

ic

[chilly ocean]

Sometimes G/H is used to mean the set of (right?) cosets of H in G even if H is not normal

[celest brook]

yes

G and H are also sets

and \ is used for set difference

[old lava]

no, \ is

[celest brook]

i.e. the stuff that's in G that's not in H, i.e. gH

\

oops

used to latex

[chilly ocean]

//

\/

[celest brook]

but yeah, it's suposed to be set difference

I was so confused

but yeah it makes sense, since its index two it paritions into two cosets (disjoint)

[snow flint]

i use / for addition

[celest brook]

.

isn't / like division or something

i'm so tired why does my brain refuse to work

:(

but we can't take a quotient, because that's only true if it's a normal subgroup which we are trying to prove so it must be set difference in this context

if it's not i'm literally going to scream

[old lava]

thonk

[celest brook]

huh?

yes...

i haven't heard of pure quotient before

just

quotient group

what is a quotient

i know the cardinality of the set of cosets is index tho

so

maybe it's just G:H

isn't index |G:H|

oh wait

i am confused again

index is a number

cardinality

[chilly ocean]

If not normal then its a quotient that is not compatible with group structure

[celest brook]

ohhh

ok

yes

makes sense to me

so G\H still represents a set though

but it can be different sets in different contexts

i.e. it is used for set difference, and it's used for the quotient as you described

yes

i mean the set difference makes sense to me in the context

of that proof

i've been mulling this over for hours

it only gets harder

well not this one problem, just the homework in general

and the only group G/H can be ismorphic to is a group of order two? (Z/2Z for simplicity) coming from cardinality of index

yes

but Z/2Z is also isomorphic to any group of order 2

the wonders of bijections

yes

[chrome hinge]

But that would be on R right?

i mean C

so those roots might eventually not be able to be constructed with sum, product and sqrt

but require an extension that is not Q(sqrt(2)) - like

[next obsidian]

I think that it’s over Q no?

I mean every real root is expressive just purely using real numbers

Since you have access to all square roots already

So there’s no point in considering eg sqrt(2) since that’s already a real number

[chrome hinge]

hmmmh

I might not be getting what youre saying actually

You mean every quartic polynomial can be solved in Q? Cause thats certainly not true :o

So i think im not getting it, sorry

[celest brook]

i did not expect cayley tables to be so useful in AA

oh sorry for interuption

[chrome hinge]

you mean the product tables for groups?

[celest brook]

ye

[chrome hinge]

Oh they certainly are

at low order tho

[celest brook]

yes

[chrome hinge]

at higher orders they become unhandy, at least in my experience

But to prove unicity modulo isomorphism at lower orders theyre great

[chrome hinge]

@next obsidian ping so it doesnt vanish

[candid pier]

Need help with this MCQ

[celest brook]

everything is the symmetric group

[chrome hinge]

Hmm its just the equivalence relation given by that partition right?

[candid pier]

Answer is 1

[chrome hinge]

yeah it would be the one induced by the partition

its kinda weird haha im curious what other equivalence relations could you define if theyre actually giving a partition of the whole set?

its already completely determined

a more interesting question would be: how many eq relations can you define in A?

[candid pier]

I got this question below wrong somehow . Shouldn't the answer be ** 2 ** ? because Two partition = { {a} , {b} } , { a , b }

[chilly ocean]

[chrome hinge]

i think youre right

[chilly ocean]

what's the definition of a partition?

[chrome hinge]

probably its that, yeah

seems like youre excluding the relation that relates every pair of elements

[candid pier]

hmm , maybe . it's confusing

@chilly ocean This is what my book says =

[celest brook]

visual proofs based

[chilly ocean]

under this definition, you are right, {a, b} has two partitions. namely, {{a}, {b}} and {{a, b}}

[candid pier]

hmm , maybe the google form i submitted the answer in is just glitched i guess .

[next obsidian]

I might just be completely off tbh haha

[vestal snow]

Suppose I have an epimorphism $g: B\longrightarrow C$, a map $d:B\longrightarrow D$, and a map $f:A\longrightarrow B$ such that $df=0$ and $gf=0$ and there is a monomorphism $\varphi:Ker(g)\longrightarrow Ker(d)$ such that $\varphi$ commutes with the kernel maps of $g$ and $d$ into $B$. How do I define a map from $C$ into $D$?

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

If we were working with modules, we would simply use that C = coker(f)=B/Im(f) and use that the fact that Im(f) is killed by d to get a map from C to D. How do I modify this argument to work in an abelian category?

I asked this on the category theory channel but I didn't get any replies

[next obsidian]

I don't see why f is important here

Like you could always let f be the 0 map from 0 to B

and it doesn't play a role at all further on

I don't see how C is the cokernel of f tbh with modules

Anyway, for an arbitrary abelian category you can simply use the fact that g is an epimorphism implies that C is equal to B/ker g, where this means the cokernel of ker g -> B which is... the coimage of g

The fact taht you have this phi implies that ker g -> B -> D = 0 so you get a map from coker(ker g) -> D, but that object is C

you just need to prove that epimorphism => C = im g

To see that this is true, we want to show C is the cokernel of the kernel of g, but in an abelian category this is the image of g which is the kernel of the cokernel of g

But coker g = 0 since g is an epi morphism

And thus...

I believe this argument works

[vestal snow]

I understand everything except this part

Why is the cokernel of the kernel of g equal to the coimage of g?

Is this an easy verification through the universal properties?

[vestal snow]

Wait I wanted to prove this lol

The original problem was to prove that if A -> B -> C -> 0 is exact, then coker(A -> B) = B -> C

I had proved that B -> C was an epimorphim

And I was trying to prove that B -> C satisfied the universal property of the cokernel of A -> B

So I was trying to construct a map from C to D

And obviously I cannot use the fact that C is the cokernel because that's what I was trying to prove

[oblique leaf]

was trying to prove that for |G|=108, G can’t be simple. I initially thought that since |G|=2^2*3^3, n_2 = 1, 3, 9, or 27 (because n_2 needs to divide 27 and also be odd). Since n_2 is the number of distinct Sylow 2-subgroups of G, I thought that n_2 can only be 1 since 2-subgroups have order 2^2=4 and there are at most 2 distinct groups of order 4. Since, n_2=1, then it follows that G has a normal 2-subgroup and hence can’t be simple. But this can’t be right because I feel like this reasoning is too simple. Can someone please tell me where the flaw is in this argument?

[rustic crown]

Sylow's second theorem says all sylow p-subgroups are conjugates and the third theorem says number of conjugates is 1 mod p and divides the size of group.

you're confusing between distinct and non-isomorphic

consider S3, it has 3 distinct sylow 2-subgroups, the ones generated by (12); (23) and (31).

but these are all isomorphic

[lunar coyote]

how do i show that something is an ideal?

say i have something like

{...,-2*n,-n,0,n,2n,...}

can i show it by showing the kernel

[chilly ocean]

closure under multiication by elements of the ring*

Q is additive subgroup of R and closed for multiplication but not ideal

[lunar coyote]

can i show an ideal like this?

or this is just completely wrong

[carmine fossil]

You are showing that's a subgroup

Not enough

[lunar coyote]

do i need to show it's under multiplication as well

[rustic crown]

elements = subsets

[lunar coyote]

hmm

[lunar coyote]

those two don't equate right

[rustic crown]

yep

[lunar coyote]

think i was just

trying to bring shit from linear algebra

to this

[hot lake]

I wouldn't even call it a proof that it is closed under addition

[chilly ocean]

thats why i prefer to call it absorbing property

[static temple]

should be a-b \in I

because you would like to see if their inverse are also in I

[lunar coyote]

is this still wrong?

[static temple]

then again you didn't use the fact that a and b are in the form nx

its because a = nx and b = ny that a-b=n*(x-y) is in this

[hot lake]

you didn't explain anything

[static temple]

You should fist begin with "Lets show that nZ is an ideal:"

[hot lake]

also you should stop mixing up = and $\in$

[cloud walrus]

Zef Klop 🍃 🌿 🌻

[static temple]

First we show that its a subgroup

then that its closed under multiplication by Z (or something like that, I should revise my english)

[lunar coyote]

no idea on how to show that

been struggling on this for days

[static temple]

well thats the def of an ideal xd

do you kknow whats a group/subgroup?

[lunar coyote]

not really because my shit teacher didn't explain anything

notes are dog shit so i'm left to self teach myself

sorry for ranting

[static temple]

no worries, my situation is a little like that too

But basically Z is a ring, because it is closed under multiplication and addition, and you have the distributivity

[lunar coyote]

i'm supposed to be able to do this stuff with no issues but my prerequisites were messed up because of covid

yeah

i know that much

like i know the definitions of an ideal

[static temple]

so if you take 34 and 23, you can add then multiply them and its still in Z...

[lunar coyote]

but i have no idea on how to manipulate things

[static temple]

ok no worries

so a ideal is a sort of smaller ring, which is not a ring

its smaller than Z

[lunar coyote]

basically a subset of Z

[static temple]

well a theorem says all the ideals of Z are in the form nZ

[lunar coyote]

think it's like

a,b is an element of R

[static temple]

but yeah a subset, which is closed under addition multiplication....

[lunar coyote]

so a+b is also an element of R

[static temple]

yeah pretty much, but its not enough

[lunar coyote]

and r,a is an elemtn of R

so r*a is also an element of r

[static temple]

yeah, a is in the ideal of course

[lunar coyote]

yeah so i know some of the definitions but it terms of application

[static temple]

so first you want to describe all the elements of nZ

[lunar coyote]

i cannot use them

[static temple]

no worries

step by step

give me an example of an element in nZ

[lunar coyote]

4z?

oh

i mena

rn

4n

[static temple]

4n, -4n .... 0... 2343n

[lunar coyote]

yeah accident

[static temple]

Yeah there are all in the form rn!

(where r is an integer)

so when you will manipulate nZ, you manipulate only numbers in the form r*n

where r is an integer

is that okay?

[lunar coyote]

yeah

[static temple]

lets first show that its a subgroup then!

first you have to show that its not empty

it is crucial

[lunar coyote]

but isn't that self explanatory

[static temple]

because if its empty, it can't be an ideal

[lunar coyote]

we can see it's a non empty set

[static temple]

you have to say it

for example, 0 is in nZ because 0 = n * 0

thats the first step

then subgroup, you take two elements a and b and show that a-b is in nZ. Thats the step to take to show that any subset of a group is indeed a subgroup

you have a proof that a subgroup is also a group but well, lets put that aside

[lunar coyote]

can i say

[static temple]

yes?

[lunar coyote]

-2*n - -n= -n

and -n is in nZ

or is that insufficient?

[static temple]

to show its not empty or ...

?

[lunar coyote]

to show a-b is in nZ holds

[static temple]

oh no no no, it should be for ALL elements in nZ

you took only 2 of them

but there is an infinity

so you have to show it for any elements a, b in nZ!

so lets take two a,b in nZ shall we? How would you describe a and b?

[lunar coyote]

uhh

can i say

[static temple]

yes?

[lunar coyote]

a = {...,-2a',-a',0,a',2a',...}

and same for b

[hot lake]

no

[static temple]

a is not a set ^^

[hot lake]

nZ = { nx, where x in Z }

[lunar coyote]

are we treating a as an element then

[static temple]

yes of course, a is in nZ!

not nZ!

[hot lake]

which means that y in nZ <=> there exists x in Z such that y=nx

and you have to use that definition of belonging to nZ

pretty much all the time

[static temple]

so basically, 4n is in nZ <=> there exit 4 in Z such that 4n = n*4, but here you replace 4n by y

because you consider any element

[hot lake]

4n is in Z because 4 is in Z and 4n = n*4

[lunar coyote]

am i supposed to know the definition for that

[hot lake]

yes

absolutely

[lunar coyote]

or it's just self explanatory

because i'm pretty sure thats not in my notes

[hot lake]

you are supposed to know how to use {n*x | x in Z}

[static temple]

there is nothing self explanatory, but yeah the definition of nZ is the set that contains elements of the form kn

(like Zef said)

[hot lake]

y is in {f(x) | x in S} <=> there exists x in S such that y = f(x)

and the other set-builder notation {x in S | phi(x)}

y is in {x in S | phi(x)} <=> y is in S and phi(y) is true

you will see sets described like that everywhere

[lunar coyote]

yep can confirm thats not in my notes

[hot lake]

all the time

you have to know how to understand them

[static temple]

there are a number of automatisms you have to grasp, but the set ones are essential

[hot lake]

so when you want to show that nZ is closed under addition

[static temple]

@hot lake do you have an introductory course paper in english for that? (maybe reading it first in a week should be better)

[hot lake]

you need to show "forall a in nZ and b in nZ, a+b is in nZ

and you should automatically translate that into

"forall a,b; if there exists x,y in Z such that a =nx and b=ny, then there exists z in Z such that a+b = nz"

maybe something like Velleman's how to prove it ?

never read it but people talked about it a lot some years ago

[static temple]

mmh, I was on something more "home made by lazy teachers but that do the job"

[hot lake]

idk , I remember my first real math lesson after high school was all the set theory basics

[static temple]

same more or less

but anyway @lunar coyote the notation {-4n,-3n...0, n,...} illustrate what contains nZ, but you should preferably view it like {kn | k in Z}

so x in nZ <=> x in {kn| k in Z} <=> there is a k such that x = kn

[lunar coyote]

are you guys finished with uni

or at masters/phd level now

[static temple]

work first!

I will leave quite soon, I can't do much but I will do what I can in 7min

its not {ø} but ø

because else it would be the set containing ø xd

[lunar coyote]

true

what about the let a,b line

do you agree?

[static temple]

yeah it can work, n|a and n|b so n| (a+b)

[hot lake]

you should explain it more in low level detail

it's like you're trying to not having to do the proof

[static temple]

so a+b is in the form kn

[lunar coyote]

yessss

one step forward

[static temple]

but yeah like @hot lake said the proof is even easier XDD

its just factorisation by n

a = nk, b= nq so a+b = n*(q+k)

but okay lets say you have shown its a subgroup

now how do you do the second step?

[lunar coyote]

hmm

it's the same really

i think

let me try write it out

[static temple]

LOL NO NOT AGAIN

[lunar coyote]

😦

[hot lake]

I ?

[lunar coyote]

I?

[hot lake]

k in I ?

wtf is I ?

[lunar coyote]

meant to be the ideal

[hot lake]

that's nZ

[static temple]

just say, let a in nZ and k in Z, then a = n*q for an integer q, thus k.a= k.n.q = n.(kq) but kq is an integer so k.a is in nZ

you have to take an element of the ideal (nZ) and the ring (Z) and multiply them

and see if the product is in the ideal XD

[hot lake]

[static temple]

I really gotta go, good luck to both of you XDD

[lunar coyote]

well rip

my only help is gone

[static temple]

I gave you the answer! now try to understand it!

[hot lake]

what you wrote doesn't match the thing you need to show from the definition of ideal

[strong valve]

+1

what do you need to show btw?

[rustic crown]

[hot lake]

that nZ = {nx | x in Z} is an ideal of Z

[strong valve]

just do it using definitions

[lunar coyote]

i'm stupid so i can't

everyone has given up on me

[strong valve]

tell me the definition of ideal

that you know

[lunar coyote]

if a,b is in R

then a-b is also in R

if a is in R and r in in the ideal

then a*r is in R

[strong valve]

let's call ideal I for sanity purposes

and the ring R

[fossil lodge]

Let a is in ring R, and x is in the ideal I

then a*x must also be in I

@lunar coyote

[strong valve]

so if x,y in I then x-y is in I and for any x \in I and any r \in R, then rx is in I

you need to show nZ is closed in additive inverses

so directly show if x,y \in I then x = na , y = nb then x-y = n(a-b) \in I

[oblique leaf]

Just to confirm, in the second specification of the first Sylow thm, it is meant that H of order p^i is a normal subgroup of the “next” subgroup of G that has order p+i, specifically, correct?

[scarlet estuary]

order p^(i+1), not order p+i

but yes, thats the idea

if G has a subgroup with order p^i for i < m

then that subgroup is also a normal subgroup of some group of order p^(i+1)

[oblique leaf]

@scarlet estuary I'm wondering if the group of order p^{i+1} has to be specifically the subgroup of G of order p^{i+1}?

[rustic crown]

yea

[oblique leaf]

ok tysm!

[rustic crown]

To see this, try to show that [N_G(H):H] = [G:H] (mod p) and then take a subgroup of order p in N(H)/H

[oblique leaf]

ok, and in the case that G has multiple subgroups of order p^{i+1} is there a way of knowing which one H is a subgroup of?

[rustic crown]

like this gives an explicit construction of that subgroup

[oblique leaf]

ooooh okok

I'll try to prove it then thanks

[rustic crown]

using correspondance theorem will tell you that the subgroup of order p would look like H'/H where H' is a subgroup of N(H) of order |H|*p

since H' is a subgroup of N(H), you'll get H is normal in H'

[oblique leaf]

also quick question: this action has two orbits correct? orb([1 1]) that covers all of R^2 except the origin and the singleton orbit orb([0 0])?

[rustic crown]

i think it will have 4 orbits

corresponding to (0,0); (0,1); (1,0); (1,1)

right

[oblique leaf]

i thought those were the stabilizers?

cuz orbits of G are defined as orb_G(x)={g*x|g\in G} with * the action

[rustic crown]

okie so tell me in which orbit does [1 0] lie?

orb([0 0]) or orb([1 1])

[oblique leaf]

in orb([1 1])? because you can find an element in T, i.e. [1 0; 0 0] such that [1 0; 0 0] *[1 1]=[1 0]?

or is that wrong

[rustic crown]

you sure [1 0; 0 0] is in T?

[oblique leaf]

oh wait

d can't be zero

[rustic crown]

yep

[oblique leaf]

oops sorry

[rustic crown]

no need to apologize lol

[oblique leaf]

are there Abelian groups (with whatever binary operation) that are not closed under addition?

[scarlet estuary]

what do you mean by "addition"?

[oblique leaf]

"+"

[scarlet estuary]

sure but that doesnt make sense in general

do you mean like

"abelian groups that we happen to write with the symbols we often use for numbers"

[rustic crown]

how does your "+" relate to the binary operation?

[scarlet estuary]

or what

[oblique leaf]

I am trying to prove that endomorphisms of Abelian groups are rings with the pointwise addition and composition as multiplication

[scarlet estuary]

if so, consider a group of roots of unity.

[oblique leaf]

so right now I am trying to show that pointwise addition is an operation that takes End(A)xEnd(A)-->End(A)

[scarlet estuary]

well usually "addition" in that context is the binary operation of your abelian group

like your group is (G, +)

and the ring addition is the sum of homomorphisms

[oblique leaf]

oh so pointwise addition means that we're just taking whatever binary operation of the abelian group?

[scarlet estuary]

and ring multiplication is composition

typically yes

see https://en.wikipedia.org/wiki/Endomorphism_ring#Description

[oblique leaf]

oooh okok that makes more sense thank you!

[lunar coyote]

how do i show that the 0 element is in an ideal

[nova plank]

How can 0 be a subgroup?

But you are ignoring unity

Oh, nvm

I was thinking ring

Lol

Im ded

Joke fail

Bye

If only I read more Lax, I wouldn't fuck up like this

[rustic crown]

(what's Lax? 😶 )

oh okie

[junior moth]

{0} is non-empty
0 + 0 = 0
r.0 = 0 \forall r \in R

[chilly ocean]

as carla aptly put it lol

[lunar coyote]

i've already defined e_vx

[final pasture]

I'm confused, what are a and b supposed to be ?

[lunar coyote]

roots

[final pasture]

you're speaking about Ker(P) although you're looking for Ker(ev_x)

[lunar coyote]

hmm

should i change p(x)

to ev_x

[final pasture]

I think you're answering to a different question

'cause every step of your proof seems to not be related to calculatins the kernel of ev_x

[final pasture]

(I don't know what you are trying to prove though, I don't understand a lof of the things that you've written )

(like where does P(a+b) = P(a) + P(b) comes from)

[lunar coyote]

well i thought i had to show homomorphism for the question

well it asked for kernel of the homomorphism

[final pasture]

isn't p a polynomial ?

the homomorphism is ev_x

[molten silo]

why isnt z_10 is field?

[carmine fossil]

2*5=0

[molten silo]

i dont get it

[rustic crown]

what is z_10?

[rustic crown]

the ring with 10 elements?

[molten silo]

Its not even a ring?

[rustic crown]

if yes, then notice (the coset of) 2 is a non-zero element. Can you give me a multiplicative inverse?

[molten silo]

nevermind

[rustic crown]

if you mean the integers from 0 to 9 with addition and multiplication modulo 10, then yes its a ring. but not a field.

[molten silo]

why not a field though

[rustic crown]

if it were, then there should have been an n such that 2*n = 1 (mod 10)

[molten silo]

oh ok

so 2 is not a unit

[rustic crown]

but 2*n takes the values 0, 2, 4, 6, 8 modulo 10

yep 2 isn't a unit.

[molten silo]

How would you show this?

[final pasture]

so suppose 1+x isn't an irreducible polynomial

[old lava]

try to write it as a product of two non-units in Q[x]

[final pasture]

then you can write 1+x as QP, with Q and P nonunits polynomials

(nonunit = non constant and nonzero, in the case of Q[x])

what is the degree of a product of polynomials ?

[next obsidian]

Well this is an iff

If you want B -> C epi implies Coker = 0, simply show that 0 satisfies the universal property

First, clearly B -> C -> 0 is the 0 map

If you have C -> D such that B -> C -> D is 0, then C -> D = 0

Since if you took the 0 map from C -> D, the composition B -> C -0> D is equal to the original B -> C -> D

Now it becomes trivial that there is a unique map 0 -> D such that C -> 0 -> D is equal to C -> D

[vestal snow]

I was talking about showing coker(A -> B) = B -> C

Given that A -> B -> C -> 0 is exact

A -> B -> C is clearly the 0 map

Suppose A -> B -> D is 0

I want to find a map from C to D which commutes with the diagram

@next obsidian

We can show that g is epic

So the map, if it exists, must be unique

But showing existence has proven to be challenging for me...

@vestal snow Pinging so that I can look at it later

[next obsidian]

Oh lmfao

Well you can show that B -> C epi implies the cokernel is 0 which is all I used

This is how you want to approach it

[next obsidian]

This is the definition of coimage

In an abelian category the natural map coimage -> image is an isomorphism which is basically the first isomorphism theorem

[vestal snow]

I think my brain is too short deprived to do abstract nonsense right now

I'll try and make sense of it later

Thanks for the help though

[chilly ocean]

Given a field extenstion $K \subset L$, if deg a = p, deg b = q for $p>q$ prime and $b^p \not \in K$ is it true that $K(a,b) = K(ab)$?

[cloud walrus]

Godel

[molten silo]

Help

[chilly ocean]

just plug in 0 and 1, in advance check if the channels taken

Im probably overcomplicating something, I'm supposed to show that ab has degree pq, this was just my hypothesis which I couldn't prove as well, but might've solved the problem. Any hints?

[hot lake]

do you know if p and q are coprime ?

[chilly ocean]

well yes, both are prime

[rustic crown]

because [K(a, b):K] is divisible by both p and q

can we say the same for K(ab) tho

[chilly ocean]

hmm that's what I was writing out but I was missing something

Let me think, I just wasn't sure if we can say that about K(ab)

well K(ab) is in K(a,b) so I guess?

[rustic crown]

what about this

like you say K(ab) is contained in K(a, b)

[chilly ocean]

its a subfield right

[rustic crown]

so [K(ab):K] divides pq

so assume assume that [K(ab):K] < pq

then we'll verify the 3 other cases

[K(ab):K] = 1, but then b = k/a so must have minimal polynomial of same degree but p is not q so this is not possible

[chilly ocean]

ok then p and q

I'll try to finish

thx

[rustic crown]

wait, i'm still trying, i just thought this might work

c = ab, then if [K(c):K] = p, then notice that b lies in K(a,c) so q divides [K(a,c):K] can this happen?

ah no'

[chilly ocean]

I think if K(c):K = p then that would imply b^p in K which is contradiction

cuz minimal polynomial is of degree p therefore (ab)^p + something = 0

wait no nvm I dont know what im talking about

[rustic crown]

      K(a, c) = K(a,b)
      /   \
   q /     \ q
    /       \
  K(a)      K(c)
    \       /
   p \     / p
      \   /
        K

this is the picture

[chilly ocean]

yes

wait why [K(a,b): K(a)]=q?

[rustic crown]

[K(a,b):K] is divisible by both p and q, so is at least pq.

[chilly ocean]

oh ok

[rustic crown]

but using the same minimal polynomial for b over K, for b over K(a) gives the bound that [K(a,b):K] <= pq

Now [K(a):K] = p so the other degree has to be q

[chilly ocean]

OH ok I see

[rustic crown]

okei atleast the other case can't be true, rigth... because the diagram will flip and p would be on top

      K(b,c) = K(a,b)
      /   \
   p /     \ p
    /       \
  K(b)      K(c)
    \       /
   q \     / q
      \   /
        K

this would be the picture

b can't have minimal polynomial of degree p over K(c)

this is just false

[chilly ocean]

oh ok yeah

cause we use the one over K

[rustic crown]

(since you already know a polynomial of degree q that works)

okei so only one case to verify... now i don't feel too bad 😛

i still haven't used that b^p is not in K... so i'll keep that in mind

[chilly ocean]

Sorry I have to go for like an hour, I appreciate the help, I will try to think about that last case on my own before looking here in case you solve it.

[rustic crown]

(i was about go as well... i have a homework due in 1 hour >.<)

[chilly ocean]

please don't bother with this problem then lol

[rustic crown]

nah its okay... i like doing this more 😛

[kindred mist]

Let $R$ be any nonzero ring (need not be unital) and let $R^2:= {\Sigma_{i=1}^nr_is_i\mid r_i,s_i\in R, n\in \mathbb{N}}$. Certainly $R^2\subset R$ Now let $x\in R$. then if $R$ is unital, $x = x1_R = \Sigma_{i=1}^1 r_is_i$, where $r_1:= x$, $s_1:= 1_R$. So $R^2 = R$ iff $R$ is unital right? My proff said its iff $R$ is commutative and unital so I feel like I'm missing something obvious but can't seem to figure out what it is.

[cloud walrus]

𝓒𝓸𝓾𝓷𝓽𝓪𝓫𝓵𝓮

[final pasture]

(you forgot to escape some of your brackets)

I don't understand how you get $\sum_{i}$ (sum on what btw ?) from $x1_R$

[cloud walrus]

Shika-Blyat

[kindred mist]

ah I forgot it kills them when I copy paste, 1 sec

[cloud walrus]

𝓒𝓸𝓾𝓷𝓽𝓪𝓫𝓵𝓮

[rustic crown]

ah submitted 16 minutes late 😛

[celest brook]

how do I prove the map is surjective?

[quaint ivy]

maybe use that (1,1) and (1,-1) are linearly independent in C^2

and thus a basis

[celest brook]

hmm

so essentially using the fact the direct sum of complex numbers is C^2

[next obsidian]

This is much easier than that

You’re looking at a + b and a - b

If you wanted to map to (c,d), set a to be the midpoint and then b to be half of c - d

[celest brook]

ohhh

very clever

just to clarify, are ideals the analogue of cosets in rings?

[old lava]

ideals are somewhat the analog of normal subgroups

[celest brook]

ohhh

[old lava]

(every ideal is in fact a normal subgroup)

[celest brook]

i was just confused because there's such things as left and right ideals

but looking further in my notes

if both of those hold, then an ideal

[old lava]

well it's hard to discuss an analog of that in groups, since groups don't have 2 operations

only 1

[celest brook]

which makes sense, since right cosets=left cosets imply normal subgroup

[old lava]

talking about "left normal" or "right normal"

doesn't make sense

[celest brook]

if it's analogous somewhat to normal subgroup

[old lava]

it's analogous in the sense you need those restrictions to take the quotient

don't think too deep

[celest brook]

I'm only looking at products too for ideals right

?

i'm asking because it will make this proof a lot easier

oh yeah

because it's a subgroup, I need only check for multplication

ok

oh wait ok

1. I need to prove it's a subgroup under addition. Easy, since it' essentially just like trivial group, right?

2.)0*any function=0 as well, so that also makes things easy too.

[chilly ocean]

For intuition, functions and lists are basically the same thing. the exercise is essentially the same as showing that {(x,0)} is a ideal of R²

[chilly ocean]

If h i doesn’t belong to Hj how can we use closure under multiplication

I'm not entirely sure what part you mean, but probably because of normality of Hj

Normality is what we have to prove

[old lava]

which part exactly do you find issue with

[chilly ocean]

I mean if Hi and Hj are disjoint, then how can we use closure under multiplication in the first 2 equations of the proof?

[old lava]

it says right there

they commute

[chilly ocean]

No that’s what we have to show

[old lava]

oh I see

we're given it's normal

read the sentence carefully

they're not proving normality

they're using it

[chilly ocean]

Ohhh got it yeah makes sense

[old lava]

the tb probably showed direct summands are normal

[chilly ocean]

So it’s normal which means Ha=aH so this follows

[old lava]

somewhere before

ya

[chilly ocean]

Thanks so much

[molten silo]

Are these irreducible

i think

no

[chilly ocean]

why

[molten silo]

By rational root test

only possible roots are 1,-1,3,-3 for (i)

these dont give zero

same for (ii)

[old lava]

you cannot use the rational root test for the second one

it is possible that the second polynomial is reducible (as a product of quadratics)

even if it has no roots

you should use eisenstein's criterion for (ii)

[molten silo]

why cant i use RRT

i see RRT only shows linear factors

[wind steeple]

for instance take (x²+x+1)²

this is reducible but has no roots

[molten silo]

i understand

god bless

Last one boys

no idea on how to go about doing this one

[chilly ocean]

I told you how before

[latent anvil]

What do you know about (ir)reducibility of quadratics and cubics?

[molten silo]

did you

[chilly ocean]

Yes, when you asked right after I posted my question without even looking

[molten silo]

I dont know what you mean @latent anvil

[molten silo]

This

Are you saying substitute 0 or 1 to see if it gives a root

[dark lichen]

Should be in this case - if it is the product of smaller polynomials it must be the product of a degree two and degree one polynomial, so it is sufficient to test 0 or 1 to see if the function evaluates to zero (that would also give you the degree 1 root), if I'm remembering correctly (edited this because I forgot "root of degree zero" in my head just meant (x + 0), which is of degree 1)

[molten silo]

From this i believe its irreducible

[dark lichen]

Should be (since f(0) = 1 and f(1) = 1)

[molten silo]

thanks dude

[celest brook]

it seems obvious it should be something like f(x)=0 for all x in X, but i'm unsure

[chilly ocean]

Consider a function that is 1 everywhere except at x

[celest brook]

how would that help? Aren't I only looking at these functions evaluated at x?

[chilly ocean]

What is a principal ideal?

[celest brook]

an ideal which is generated via multplication through every element of the original ring

which are functions

[chilly ocean]

not really clear what you are saying

A principal ideal is an ideal that can be generated by 1 element

[celest brook]

yes

[chilly ocean]

How do the elements of Ix look like?

[celest brook]

they're all just 0

oh wait

f(1)=0, f(2)=0,...f(n)=0

n in X

[chilly ocean]

is that the only element of Ix?

[celest brook]

X is an arbitary set

yes

[chilly ocean]

[celest brook]

it's literally from definition of I_x

isn't it??

[chilly ocean]

kinda fancy notation for the trivial ideal wouldnt it be

If x=1 for example

what is I_1

[celest brook]

f(1)=0

set of functions

[chilly ocean]

I_1 is the set of functions from X to R such that f(1)=0

[celest brook]

yes

[chilly ocean]

if X={1,2} for example

is the function that is 0 everywhere the only one in I_1?

[celest brook]

no

[chilly ocean]

so why you said this

[celest brook]

look at what I said below

meant to correct

kinda weird when they don't put qualifies in the question too

like $\exists x \in X \mid f(x)=0$

[cloud walrus]

Moosey

[chilly ocean]

the x is fixed

[celest brook]

because some time it's implied, other times it's not

but I digress

[chilly ocean]

They mean to say. For x in X, let Ix be that

[celest brook]

so how does this help me in proving whether or not it's a principal ideal

because 0 function is an element of I_x

not the only element

oh wait

If I_x has a function that's not equal to zero at some y, then it won't generate I_x

ok

ic

[chilly ocean]

well any function that is only 0 at x generates it yea

[celest brook]

so basically we're proving that the trivial ideal is not equal to this ideal

since f(y) some y in X is possibly not 0

[chilly ocean]

no

[celest brook]

hmm

when we want to prove whether or not I is a principal ideal we want to prove I=<x> where x is the smallest ideal of R containing x x in R

[chilly ocean]

what no

you mean <x>the smallest ideal of R containing x ok

you want to prove that there exists a x such that I = <x>

if its commutative with identity yea

so that x be in it

[celest brook]

ok, so the smallest ideal of R that contains x is just the function where it's zero everywhere, right?

me neither

my profs own words

[chilly ocean]

ideals are not ring elements, they are sets of ring elements

[celest brook]

...how does a set generate

fijdsiofs

all of the members of S?

elements

head hurts

I should be a set of functions then

a subring

subgroup under addition

¯_(ツ)_/¯

ok. how come we can change f(input) input willy nilly

and that changes the ideal or makes it not principal

why is important to look at other elements in X besides x

yes...

my professor mentioned what carla mentioned earlier

about f(s)=1 for s not equal to x

i'm

confused

these are hints prof gave

yes

and those help prove that is a principal ideal?

how

OH WAIT

so y(x)f(x)

ok

they both have to be evaluated at same x

ok

yes

so in reality, it doesn't have to be 1, it could be any number, or we could have a myrad of conditions

but it just matters that at x, f(x)=0

oook

how come it matters that f not equal to zero elsewhere?

what if y was zero somewhere that f wasn't

yes

smallest ideal of R that contains a function f

ok wait

ok

hmm

how does y equalling 0 somewhere not break things?

oh wait

hmm

hmmmm

i wish my brain could understand this

it's a specific place where at f(x)=0

x in X, f(x)=0

hmm

i dunno how to describe it more

can you explain it?

f in f

(I,+)

hmm

but isn't it zero?

it's always zero

that's

in definition of it

wherever it ix

is x

yes

ok

hmm

at x it's zero

the function

a function

a fuction that maps X to R

reals

I_x is a set of functions that x, equal zero.

so I_x will always include the f=0 for all x in X

ok

so having y=0 would simply mean that the zero function is in the set I_x

for <f>

yes

so since y is any function, if f is equal to zero somewhere else other than x, how does this break?

what makes it 'not work'

in I_x we can have the zero function in it

okk

ic

f(s) is not guranteed to be zero, so with the work I did before

g(s)f(s) is not guranteed to be zero either

hmmm

ok so the <f> that is completely different that the fs in I_x

because sin(x) is in I_x

yes

ok

OOHH

OK! so it's basically showing what must be in and what's not in the ideal. We want things that are equal to zero at x.

IF we have a function that's not equal to zero at x generating, then we will get functions that slip into the ideal that are not supposed to be there

i.e. if we have <sin(x)>

g(x)=(x+1)^2

I_0

then then the generator will incorrectly assume that at 0, (x+1)^{2} is equal to 0

[cloud walrus]

slimvesus

[celest brook]

s=pi

let s=pi

:)

so basically my gist was it splits functions into disjoint sets. Ones that equal zero at x, and ones that don't

sin(s)=0 at multiple s

I thought you said that couldn't happen?

@chilly ocean

ok

ok...

[cloud walrus]

slimvesus

slimvesus

[celest brook]

...ugh

it's different everywhere

but basically one is a subset of another

$I_x \subset <f> $ and $<f> \subset I_x$

[cloud walrus]

Moosey

[celest brook]

no

a set that is within another set

[cloud walrus]

slimvesus

slimvesus

[celest brook]

ok.

[cloud walrus]

slimvesus

[celest brook]

I know that this is true

I just haven't used it in awhile...

or uh

didn't know how to apply it

[chrome hinge]

Can i have some help with a field theory problem? Ive been stuck in it for a while

p prime, K=Fp(t) and f = x^p - x + t \in K[x]. I need to show that f is irreducible

[chilly ocean]

Ah, I hate it when they make the field that the coefficients come from all fucky and weird

[latent anvil]

I think gauss's lemma is useful here

[chrome hinge]

Same... i got a bit lost

[latent anvil]

I haven't thought out a complete solution

[chrome hinge]

I thought of using it shamrock

[latent anvil]

Hmm

[chilly ocean]

,w gauss's lemma field theory

[latent anvil]

this is not the gauss's lemma we're talking about

[chrome hinge]

So you say i should check irreducibility just in Fp[t][x] ??

[latent anvil]

Yup

[chrome hinge]

Ok

[latent anvil]

That's my intuition

[chilly ocean]

Ah, that theorem

[chrome hinge]

Thats what i was already trying

[latent anvil]

So suppose it factored

look at the x degrees, maybe?

Only one of them can have an x term, right?

[chrome hinge]

But were supposing it factors in what... a linear term and something else?

[latent anvil]

What do you mean?

[chrome hinge]

Or a general factorization

[latent anvil]

Just any factorization

[chrome hinge]

Oh, ok

[latent anvil]

If both have an x term, you'll end up with an x term in the product of degree > 1

right?

[chrome hinge]

Yes, indeed

So you say it must be zero

[chilly ocean]

Ah, wait a second

[latent anvil]

That seems like a big jump maikel

But I think you can take it from here

if it factors as f(t) (g(t) + xh(t))

Yeah 8da?

[chilly ocean]

Ah sorry, I was just thinking to myself. I just like eisensteins criterion and I recalled that weird trick problem where you use eisensteins after shifting via x->x+1, and I was wondering if this trick works here too

[chrome hinge]

is it? by expanding the factorization and equal with the original polynomial dont you obtain that?

[latent anvil]

I don't think it does

[chilly ocean]

But also, I thought we are factoring in x, not t?

[latent anvil]

Sorry, what was "it"?

I thought you meant one of the factors was 0

Did you mean the coefficient of the term with x degree > 1?

[chrome hinge]

the statement that said it was a big jump

[latent anvil]

Because I'd so I agree

[chrome hinge]

Oh no

[latent anvil]

No I meant what was "it" here?

[chrome hinge]

I must have explained myself bad

[chilly ocean]

But what about the shift x->x+t?

[chrome hinge]

Sorry haha

[chilly ocean]

Wait fucj

[chrome hinge]

I meant the coefficient of the degree>1 term should be zero

unless the degree were exactly p

[latent anvil]

You end up with (x+t)^p - (x+t) + t = x^p - x + t

Which is what you started with

right?

[chilly ocean]

Nvm yeah

[latent anvil]

Oh no, this is very embarrassing for me. I mixed up x and t

I meant look at the t degree haha

[chrome hinge]

Ooh

[latent anvil]

It has to factor as f(x)(g(x) + th(x))

right?

same sort of logic of looking at degrees

[chrome hinge]

Recall that t is not the variable of the polynomial we're working with

[latent anvil]

This is wrong. You end up with x^p - x + t^p. Doesn't seem easier though

[molten silo]

How would one do this?

[latent anvil]

What do you mean? We're working in Fp[x, t]. x and t are on the same level

We're just trying to show it's an irreducible element of this integral domain

[old lava]

(6) is contained in (2), so (2) + (6) is just (2)

[molten silo]

so i was right

nice

[chrome hinge]

Yeah i got confused but F[t][x] = F[x,t]

[latent anvil]

Yup

[chrome hinge]

I wasnt thinking it in that way...

[latent anvil]

Yeah, it's a weird trick

But that's why gauss's lemma is so powerful here

So do you think you can finish from the factorization x^p - x + t = f(x)(g(x) + t h(x))?

[chrome hinge]

I mean i probably can conclude that factorization cant be, but what about other factorizations?

[latent anvil]

Sorry, I got ahead of myself

Suppose it factors at all

Look at the t degrees of either term

We know that both can't be > 1

so one must be 1 and one must be 0

does that logic make sense?

[chrome hinge]

Yes it does

[latent anvil]

And that means one of the factors looks like f(x) and the other looks like g(x) + t h(x)

This is writing things as elements of Fp[x][t] essentially

[chrome hinge]

Ok so

Now i should have to show any of the factors is a unity

[latent anvil]

yup

(the term is just "unit" in english, not unity)

[chilly ocean]

Man, tbh eisensteins is the only theorem I know about irreducible, when it doesn't work, I go in a corner and cry

[chrome hinge]

+1

[chrome hinge]

Oh, thanks. Im not a native english speaker :)

Ok, let me try it from here

Thanks :D

[latent anvil]

hang on a sec...

Does Eisenstein work once we get down to Fp[x, t]?

Nah I don't see it

[chrome hinge]

I know the theorem in Z and Q

Does it have a generalization? :o

[latent anvil]

Ah okay it works for A[x] whenever A is an integral domain