groups-rings-fields

[oblique river]

what you wrote is wrong

(assuming you meant a + bx + cx^2 + (x^3))

because x isn't an element of R

so you can't have an x in your expression for the image of p(x)

[languid meteor]

but isnt the a + bx + cx^2 part in Z[x]?

[oblique river]

yes

what I'm suggesting is just literally write down phi(a + bx + cx^2 + ...)

it's equal to a + b*phi(x) + c*phi(x^2) + d*phi(x^3) + ...

what do you know about phi(x)

@chilly ocean https://en.wikipedia.org/wiki/Schur–Zassenhaus_theorem is the theorem I was talking aboiut in general

[languid meteor]

I know that if phi(x)^n where n >= 3 it's going to equal zero because phi(x)^3 = 0 and phi(x)^n = phi(x)^n-3*phi(x)^3 = 0

[oblique river]

great

so

what do you know about a + b*phi(x) + c*phi(x^2) + d*phi(x^3) + e*phi(x^4) + ...

[languid meteor]

well phi(x^2) = phi(x)phi(x) = phi(x)^2

so a + bphi(x) + cphi(x^2) + dphi(x^3) + ephi(x^4) + ... = a + bphi(x) + cphi(x^2)

[oblique river]

why did the c go away

[languid meteor]

sorry just mistyped

[oblique river]

okay so no wyou're done

[languid meteor]

ah ok then that makes perfect sense, do I have to go into specific detail in regards to uniqueness?

[oblique river]

if a + bn + cn^2 = a' + b'n + c'n^2

then their difference is 0

i.e. it came from an element of (x^3)

i.e. it's a multiple of x^3

what does that tell you about it

(also gotta go now sry)

[languid meteor]

np thanks for the help! I was gonna say you can write a + bn + cn^2 = a' + b'n + c'n^2 + rx^3 or something

but ill work on it, im sure its not too hard to figure out

[languid meteor]

if anyone is still around in this channel im trying to figure out why we're able to take the constants outside of the ring hom in the equation: phi(a + bx + cx^2 + ...) = a + bphi(x) + cphi(x^2) + dphi(x^3) + ...

because a,b,c,d etc are also elements of Z[x] so surely I would have phi(a) + phi(b)phi(x) + ...

unless phi(a) = a somehow if a is an integer

but phi(a) has to be in Z[x]/(x^3)

[carmine fossil]

Since b is in Z

Just do phi(bx)=phi(x)+phi(x)+...phi(x) b times=bphi(x)(similarity for c,d..)

[languid meteor]

oh god of course, thank you

[leaden finch]

thanks lol @chilly ocean

[bronze trench]

Wassup my people! I have a question about modules and I'd love some help 🙂

Ok so here F is a field and E an extension, I have to show R is artinian as a left module over itself with its length being 2

ok so I can ignore the artinian part since a module has finite lenght iff it's artinian and noetherian so I only need to show a composition series of length 2 right?

Issue is, I think I found one of length 3...

[next obsidian]

It might be that you're forgetting how length is defined?

One of the terms is considered index 0

so if you have 3 things that's length 2

It should me like 0 = M_0 < M_1 < M_2 = M

this is a length 2 composition series

[chilly ocean]

Hello. Could you help me solve this problem? ""Find a ∈ R so that the list v=[v1, v2, v3] ^t is a base of R^3, where v1 = (a, 1, 1), v2= (1,a,1), v3 = (1,1,a)""

I initially thought a should be 1, but that can't be right, no?

t means transpose

a = 1 wouldn't give you a set of linearly independent vectors or a spanning set

[old lava]

1 obviously cannot be it

since all three vectors are the same

[chilly ocean]

also this belongs in #linear-algebra

[old lava]

cannot span a space of dim=3

[chilly ocean]

smh just twist the definitions a little and it works

[old lava]

I mean linear algebra is abstract algebra

[chilly ocean]

[old lava]

just say it's a module

but extended

to reals

or some shit

[chilly ocean]

did I post in the wrong channel?:)) Sorry if I did

[old lava]

ya, #linear-algebra is for the linear algebra questions

this channel is more for group/ring/module/field/galois

stuff

[chilly ocean]

okay, I'm moving there. Sorry! I thought we were being taught only abstract algebra this year

[old lava]

linear algebra is part of abstract algebra

it's just so big, and such a commonly asked about topic

it needs its own channel

[chilly ocean]

thanks. I understand now:D

[delicate bloom]

and since calculus is just derivatives and integrals which are linear operators, calculus is part of abstract algebra

[chilly ocean]

😮

[old lava]

dont take that seriously

[chilly ocean]

justify differential geometry being a subset of AA

:))

[delicate bloom]

lol

[chilly ocean]

inb4 algebraic tangent space

[sturdy marsh]

all math are words in some free group

[delicate bloom]

what you're talking about is actually a monoid, of which a free group is but a small area

[bronze trench]

Yes, I have a thing with 4 things... lol

[next obsidian]

O_o

[bronze trench]

I mean maybe I got it wrong

[next obsidian]

If the answer is 2 then... yes

[bronze trench]

consider these

0 E
0 F

and

0 E
0 0

pretend the matrices have the delimiters lol

so these are both left R modules, unless I messed the calculations up

[next obsidian]

I don't believe the first one is a submodule

[bronze trench]

but calling them A and B this makes
0<A<B<R

[old lava]

what you're thinking of is a subgroupoid, of which a monoid is but a small area

[next obsidian]

how do you multiply an element of E with one of F on the bottom right part of that matrix? won't it exit F

[bronze trench]

ah crap you're right

subfields aren't ideals

fields have no ideals

[next obsidian]

Right

[bronze trench]

fuck me

[next obsidian]

the submodules are exactly ideals

[bronze trench]

thanks

[next obsidian]

of that ring

[bronze trench]

yes yes I know

[next obsidian]

yee

[bronze trench]

I knew something was up because I've seen this before

which was confusing me even more lol

anyway thanks, this means F is irrelevant for this exercise, the following ones are the ones which will depend on F being a subfield

Also dumb question, if I quotient
E E
0 F
by
0 E
0 0
will I get
E 0
0 F

or is this too simple and dumb of me? xD

[bronze trench]

wait I'm confused again

E E
0 0 is a left ideal

even in the normal matrix ring

and
0 E
0 0 is too, so those 2 make a composition series of length 3?

what the hell is happening?

[next obsidian]

Oh wait sht

This isn't being considered as an E-module it's being considered as an R-module

so you act via matrix multiplication

gross

[sturdy marsh]

0

<

0 E

00

<

EE

00

<

R

does this not work?

@bronze trench

[bronze trench]

yes my point

[sturdy marsh]

yeah i dont think it's length 2

[bronze trench]

but the exercise says the length of the thing is 2

I'll ask the prof

[next obsidian]

I don't think
0 E
0 0
Is a submodule

Or...

uhhhhh

maybe

probably actually

[sturdy marsh]

found the analyst

[next obsidian]

I don't like matrices

also analyst

L O L

[bronze trench]

well I can't compute stuff so this may be wrong lol

but I'm pretty sure it's even a bilateral ideal, not just a left one

because I have to find all the ideals of that ring for the next exercise 😂

[drowsy oasis]

can permutations be defined on infinite sets?

[next obsidian]

yes

it's just a bijection of the set

to itself

This is what the symmetric group on a set is defined to be with the product given by composition

[drowsy oasis]

what maps to x_1 ?

sorry what i said doesn't make sense

[tawny pine]

@drowsy oasis what's x_1

[chrome hinge]

math SOS

[old lava]

damn galois reincarnated?

[chrome hinge]

How on earth can i find the gcd between 11 + 7i and 18 - i in Z[i] heeelp ive been HOURS on this

Maikel Galois is my name hehe

[sturdy marsh]

are you also a simp?

[drowsy oasis]

@tawny pine in this book it is just the first element of S^n or A(S)

[chrome hinge]

idk what you meant @sturdy marsh

[old lava]

your profile picture

he died simping

[chrome hinge]

Oh dang no im not

But please help me im in dispair

[sturdy marsh]

look at their norms

[tawny pine]

@drowsy oasis english?

[sturdy marsh]

what's 18^2 + 1

[chrome hinge]

325

[sturdy marsh]

okay

[chrome hinge]

Yee ive looking their norms for quite a bit

[sturdy marsh]

and 121 + 49 = 170

[chrome hinge]

Just cant conclude anything yet

[drowsy oasis]

@tawny pine ye its herstein

[chrome hinge]

Indeed

[tawny pine]

i mean translate what you said to english

[chrome hinge]

DAMN you gave me an idea!!!

Thank you so much

[sturdy marsh]

if p divides r, then N(p) divides N(r)

work with that

[chrome hinge]

Yeye thats exactly what i was gonna do ty

[drowsy oasis]

A(S) it says are the positive integer elemnt functions which map 1-1 onto itself and x_1 is same as what 1 will map to

actually it says S can have any amount of elements i am just considering when S is that case above

[tawny pine]

can you eli5 A(S)?

[drowsy oasis]

i was thinking you would have osme issue if what maps to the first element but i realized it doesnt matter you just scramble all N

[chrome hinge]

Seeking for the last non zero remainder?

Ok i'll give it a try too, ty. I think i got it with the other method though

DANG YES I KNEW THEY WERE COPRIMES

[drowsy oasis]

@tawny pine i think A(S) is called the cyclic permutation

[tawny pine]

nah. also ELI5

[old lava]

isn't A( . ) for alternating group

[drowsy oasis]

could be, i am just starting to learn about about this stuff

[thorn delta]

in some books A(S) is used to denote the symmetric group on S

[old lava]

groups with A_n or A([set]) are usually the alternating group

I see

I did not know that

[thorn delta]

because if you have something like a set S, then S_S would look weird i guess

[old lava]

ya ig, I thought the S notation was universal

for symmetric group

but anyways, you can in fact have an infinite symmetric group

which would be the symmetric group on all the natural numbers

[sturdy marsh]

how do you make sense of an infinite alternating group

if you're looking at permutations with finite support you could

but I don't see how it works in general

[old lava]

Schreier-Ulam Theorem makes some assertions about normal subgroups of the infinite symmetric group

although, I cannot say I have much experience with infinite groups

for the infinite alternating group, I have no idea

I don't know how you would really talk about even permutations

of an infinite group

[drowsy oasis]

i am thinking if you have say N just scramble N

[old lava]

that would be the infinite symmetric group yes

[drowsy oasis]

i guess you could even have uncountable permutation

if f is in S_n is f^(k.n) = i always ? where k is in N

[chrome hinge]

Hey just curious, at which stage of your math career are people in here in general?

[barren sierra]

I am a freshman in college rn. I'm not taking abstract algebra yet but some questions in my intro to proofs class are algebra so I ask them here

[next obsidian]

junior in college

[chrome hinge]

Nice nice

Chmonkey i wanted you to know that i put an observation in my algebra notes with your nickname since you made me notice that lmao

[next obsidian]

wait wut

[chrome hinge]

Remember the thing about formal series being invertible and polynomials not?

[next obsidian]

oh right

[chrome hinge]

Ok now i took note of that and referenced your nickname in my notes

[next obsidian]

I see. I'm glad I could help haha

[chrome hinge]

Thanks again 😂

[fading wadi]

can anyone help me go through this question?

I'm lagging behind the course so IDK what L_{BA} and L_{A} means

Also this problem I'm confused with:

Like what does it mean to be a special matrix

[fading wagon]

@fading wadi hmm I suspect it might be a matrix where only diagonal entries are nonzero where $a_{ii}$ divides $a_{(i+1)(i+1)}$?

[cloud walrus]

Element118

[fading wadi]

oh, what about $L_{BA} and L_{A}$

[cloud walrus]

LHC2012

[gritty latch]

find all elements of order of 6 in Z6xZ6

whats the easiest way to do problems like these

ik i can just list out the cases

but with groups with alot of elements its isnt very practical

[carmine fossil]

Order of a element (a,b) is lcm(order(a), order (b))

[gritty latch]

yea i know

but thats still just listing out cases

is there a better way for the general case

[carmine fossil]

List more efficiently

[languid meteor]

hey guys ive been jumping in and out of this question for some time now and I still find it really difficult to understand whats going on in the quotient ring. Im currently trying to find the units and zero divisors

so if a + bn + cn^2 is a unit then (a + bn + cn^2)(d + en + fn^2) = 1 for some d + en + fn^2 in R

but in the last part I showed that (a + bn + cn^2)(d + en + fn^2) = ad + (ae+bd)n + (af + be + cd)n^2 [Im not sure I proved this correctly I just figured I could multiply everything as usual]

so I have ad + (ae+bd)n + (af + be + cd)n^2 = 1

so ad = 1 and ae+bd = 0 and af+be+cd = 0?

but then I have infinite solutions

which feels wrong

so if anyone could point me in the right direction id really appreciate it

[chilly ocean]

a,b,c are given, so you have a unique solution

You realized, that (a+n * stuff) * (d + n * stuff) = 1 is the condition for (a+n * stuff) to be a unit

this implies ad = 1, so a is a unit

and in fact (a+n * stuff) is a unit iff a is a unit in Z

n is nilpotent, i.e. there is some expontent - in this case 3 - such that n^3 = 0

this means you can invert elements of the form 1 + n * stuff by a geometric series

(1 + n * t)^{-1} = 1 - n * t + (n * t)^2 and here the series ends since n^3 = 0

If you have a + n * stuff, and a is invertible, then a + n * stuff = a * (1 + n + t), so its inverse is
[a * (1 + n + t)] ^{-1}= a^{-1} * (1 - n * t + (n * t)^2)

[languid meteor]

we havent done anything with nilpotent elements so this argument isnt very familiar to me :/

[chilly ocean]

You don't need the term "nilpotent", just for background. It's the condition that n³ = 0

The geometric series enables you to do a tidy computation of the inverse

[languid meteor]

I guess the geometric series thing is the only thing tripping me up, so im trying to find the inverse element for (a + bn + cn^2) right? so where does this 1 + n*t term come from? I know that a = 1 or -1 but beyond that im not sure why I can drop the n^2 term when finding the inverse

[chilly ocean]

You have, that (a + bn + cn^2) invertible implies, a is a unit, that is +1 or -1, right?

[languid meteor]

right

[chilly ocean]

For the converse assume, that a is +1 or -1

Then (a + bn + cn^2) = a(1+ba^{-1}n+ca^{-1}n²)

The second factor is of the form 1+nt for t = ba^{-1}+ca^{-1}n

[languid meteor]

ah ha, gotcha that makes sense

so then how does this translate to a geometric series exactly?

[chilly ocean]

Take a general geometric series in say s

1 + s + s² + s³ + ... = 1/(1-s)

So (1-s)(1 + s + s² + s³ + ...) = 1

Now s = -nt

The s³ term and higher vanish

giving (1+nt) (1 - nt + n²t²) = 1

[languid meteor]

ahhh ok that makes a lot of sense too, thank you! so my units will be 1 - nt + (nt)^2 and -1 + nt - (nt)^2?

[chilly ocean]

No, what we've just proven is that (a + bn + cn^2) is a unit if and only if a is a unit

So the units are 1 + bn + cn^2 and -1 + bn + cn^2 for all b,c

[languid meteor]

is this because the inverse we found a^{-1} * (1 - n * t + (n * t)^2) only depends on a?

[chilly ocean]

The inverse depends on all coefficients

We have just proven the following theorem: Let a+bx+cx² be an arbitrary element of Z[x]/(x³), that is a,b,c are in Z. The following are equivalent:
(1) a is a unit in Z, i.e. a = +1 or -1
(2) a+bx+cx² is a unit in Z[x]/(x³)

[languid meteor]

ahhhh I get it now, thanks so much for the help. that makes a lot more sense than the method I was running with 🥴 thanks again

[chilly ocean]

You are welcome

i was already typing something for the zero divisors, maybe i'll just past it

*paste

For the zero divisors: There are only two cases for an element a + bn + cn²: Either a = 0 or a != 0. In the second case it cant be a zero divisor, since (a + bn + cn²)(d+en+fn²) = 0 implies, that ad = 0, which implies, that a is zero in Z, unless d=0. But then aen + ... = 0, which implies that a is zero unless e=0. But then afn²=0, which implies that a is zero unless f=0. So a zero divisor must have a=0.

This implies, that any zero divisor is a multiple of n. Hence it is contained in the ideal generated by n.

Conversely every element of (n) is a zero divisor

[languid meteor]

(n) here is the ideal generated by phi(x) ?

[chilly ocean]

yes

i assumed you defined n = phi(x)

[languid meteor]

yeah we did

[chilly ocean]

n = eta

okay

[languid meteor]

awesome, thats really interesting actually thanks so much

this quotient ring has been kicking my ass

[chilly ocean]

yes this ring is funny. i just worked with a similar ring today 😄

[golden merlin]

How does a two sided inverse show that a mapping is bijective?

[carmine fossil]

Do you know that existence of a left inverse implies injectivity and right inverse implies surjectivity?

From those facts,your statement follows naturally

[golden merlin]

Why does the left inverse imply injectivity? (and same for right inverse with surjectivity)

Can you show me how

[carmine fossil]

Let the function f:A->B have a well defined left inverse g:B->A
Then g(f(x))=x,If f(a)=f(b)=c for a!=b,this would imply g(f(a))=a,g(f(b))=b,i.e., g(c)=a and g(c)=b and a!=b which implies g is not well defined,but by assumption g is well defined

So,f has to be injective

[golden merlin]

Ok, so that works for injectivity. What about surjectivity?

[carmine fossil]

Let f:A->B and right inverse be g:B->A,then f(g(x))=x if f wasn't surjective some values in B would be missed(i.e.,not in image of f) and fog wouldn't be identity on B for those values missed

[golden merlin]

And the identity mapping is surjective, right?

So that makes it impossible

[carmine fossil]

yes, fog:B->B is surjective

[golden merlin]

Thank you drake, you can now @ me if you ever want to.

[next obsidian]

This isn’t even algebra tho this is just for any function 😞

[golden merlin]

Shutup.

[next obsidian]

Dan

[old lava]

chmonkey spitting facts though

[golden pasture]

ya go to like #prealg-and-algebra or smt i actly dk what channel is appropriate

what even is the diff between #prealg-and-algebra and #precalculus

[next obsidian]

I think #proofs-and-logic is the right place for stuff like that

[golden merlin]

This came up while studying group actions.

[next obsidian]

Okay sure but it’s just a property of functions

I’m just saying you don’t need it to be a group homomorphism

Continuous function

Etc

This is true for anything that’s a function

[old lava]

ya, in my abstract algebra textbook, I think it was part of chapter 0, as prerequisite knowledge

before you start any abstract algebra

[next obsidian]

Aluffi: Chapter 0

F all of you who will sully that

[golden pasture]

best intro tbh

[next obsidian]

Finally someone with taste

[stone fulcrum]

I haven't read it because my brain is too small

[golden pasture]

really eliminates anyone who isnt a true meme

wait actly looking at Aluffi: Chapter 0

it doesnt seem that meme

[next obsidian]

The ideal is principal so the ideal being prime is equivalent to the generator being prime

[golden pasture]

check irreducibility

[next obsidian]

C[x] is a UFD so it suffices to check that it’s irreducible

Hint, C is algebraically closed so it 100% isn’t

You can see this directly too. Think of non-zero elements in the quotient whose product is 0

[next obsidian]

Yee

[maiden ocean]

can someone elaborate on what this means

is gradiation like additional structure on the ring?

in some way?

[next obsidian]

Do you know what a graded ring is?

Lol

If not just look it up. The prototypical example is a polynomial with the grading given by the degree of the polynomial

[maiden ocean]

i know what it is

im just confused as to how two graded rings can be isomorphic as rings but not as graded rings?

or is this something about tensors that makes it non-isomorphic : ?

[vital quail]

https://math.stackexchange.com/questions/1068252/graded-rings-and-isomorphisms

actually this https://math.stackexchange.com/questions/3706834/definition-of-isomorphism-of-graded-rings

[white nymph]

can someone help me figure out how to approach this instead of just aimlessly slapping elements together in an attempt to show set containment?

[sturdy marsh]

it probably will involve slapping

but not aimlessley

[uncut girder]

all my homies hate graded rings

fuck the artin rees lemma

[white nymph]

i have a hard time figuring out how to slap purposefully in situations like these

[sturdy marsh]

yes officer, this message right here

[white nymph]

i mean the strategies that come to mind are to insert the identity into places to make it work... or to rewrite/rearrange/solve for elements, but I feel like i dont know how to do it in a way where im actually making progress.

[sturdy marsh]

it's just a matter of doing it

assume you have ab from the set on the right

a is in H and K'

so it is in H' and K'

so it is in the intersection

similarly the same is true for b

the intersection is a subgroup

[maiden ocean]

no kinkshaming

pty

perfect its time for gabe to explain to me

how can two rings be isomorphic as rings but not as graded rings

what does that even mean

so like

A = bigoplus A_i and B = bigoplus B_i

A is iso to B but A_i isnt iso to B_i?

hurb this has to be

up to permutation right

of the A_i

[sturdy marsh]

just shift the grading

shifting grading does wonders

[maiden ocean]

idk what that means im not an algebra coomer

can u do this example

[sturdy marsh]

you get all the line bundles on projective space by simply playing around with the grading

[sturdy marsh]

topologists care a lot about graded stuff

[maiden ocean]

exterior algebra

on one element

maybe i just dont understand the grading here

https://cdn.discordapp.com/emojis/526983414226747393.png?v=1

[sturdy marsh]

I think just do it on the vector space with that element as a basis

I think all he's saying is that alpha lives in degree 2n

for the cohomology ring case

but alpha lives in degree 1 for the exterior algebra

with the usual grading

z[a]/a^2 is the cohomology ring of S^2n

H^i(S^2n) is 0 for all i \neq 0 or 2n

alpha is the generator of H^2n

yeah and we're just taking the direct sum of all the H^i

which gives us a graded module

and then the cup product gives us a ring

which is iso to z[a]/a^2

as all higher cohomology groups vanish

the the usual ring iso induced by a mapsto a is not an iso of graded rings

it's not even a map of graded rings

[maiden ocean]

yes kinda

also i use they/them pronouns btw lol

[sturdy marsh]

the 'he' referred to hatcher

[maiden ocean]

oh lmfao

based

idk im still confused maybe brain just bad

[sturdy marsh]

the grading stuff is weird

[maiden ocean]

i guess wedge[alpha] = wedge^0(alpha) oplus wedge^1(alpha)

umm

idk wedge^0(alpha) is spanned by just 1 right?

oh wedge^0 is the underlying field right

[sturdy marsh]

yes

the exterior algebra is just the tensor algebra mod relations

if V is an F vect, then the tensor algebra is F + V + V^2 + V^3 + ....

those powers are tensor powers

and that is the grading

[maiden ocean]

okay i think i see it

[sturdy marsh]

alpha lives in the grade 1 part

[maiden ocean]

because alpha lies in V right

right

[sturdy marsh]

yes

the cohomology ring is Z + 0 + 0 + ...+0 + H^2n

[maiden ocean]

mhm

and the grading on Z/(alpha^2) is uh

[sturdy marsh]

alpha lives in the grade 2n part

[maiden ocean]

Oh right i see

[sturdy marsh]

it generates H^2n

[maiden ocean]

oh this makes sense

ohh so our isomorphisms ARENT up to permutation

like

[sturdy marsh]

yeah they need to preserve grading

[maiden ocean]

ok this makes sense

ty!

[sturdy marsh]

I was super confused by graded stuff too

[maiden ocean]

i think its weird because im used to thinking of oplus as like

the order doesnt matter

but it does here

cuz of the graded structure

[sturdy marsh]

right.

but the grading can be over a lot of things

not just Z

so 'order' is perhaps not the right way to look at it

[maiden ocean]

mhm other indexes though i havent seen them at all yet

are there any like relatively simple examples of when you dont want to index over Z

(or N)

[sturdy marsh]

yes

there's a cohomology theory which is graded over the representation ring of a group

the category we're looking at is nice top spaces with a nice group action

the word to look up is RO(G)-graded cohomology if you're interested

[maiden ocean]

hmm

ty

i will check this out later

[cloud walrus]

slimvesus

[light tusk]

There's the theorem [A:B][B:C] = [A:C] for extensions. Also, remember the definition is in terms of the degree of the minimal polynomial

[cloud walrus]

slimvesus

[chilly ocean]

Yes because over the obvious degree 3 extension it doesn't

yes, the real cube root of 10

say we embed everything into C

yes

yes, both are S3-extensions

because the Galois group is determined by its action on a generator

yes

3

why should it be 2?

X³-10 is irreducible over Q(sqrt(-3))

so the degree is 3

The point is rather, why isn't it 6?

In general, when you adjoin the root of an irreducible polynomial of degree 3 it might just have one root in the extension

but the point is, that Q(sqrt(-3)) already contains all three cube roots of unity

because they can be weirdly built from sqrt(-3) 😄

The splitting field L has degree 6 over Q(sqrt(2)), ok?

So the Galois group has 6 elements

Ok, then you should learn the following fact: Any group of order 6 is either isomorphic to Z/6Z or S3

So we only have these two possibilities

exactly

G acts on the roots of X³-10 and this defines an injection G -> S3, which must be a bijection

That's the standard argument

If you have a splitting field of a polynomial, then you always get an injection into a symmetric group because of that argument

In the other example adjoining cube root of 10 splits the polynomial completely, so we have an injection G -> S3 and G is isomorphic to Z/3Z

If zeta is an n-th root of unity then you get an extension Q(zeta)/Q of degree phi(n)

yes

(Z/nZ)*

yes

Gal(Q(zeta)/Q) -> S_phi(n)

but that doesn't really matter in this case, because S_phi(n) is really large

If E/F is a primitive extension yes

E/F is primitive, if there is some a in E, such that E = F(a)

So a has a minimal polynomial and its degree is the degree of the extension

and Gal(E/F) permutes the roots of this thing

You just have to make sure that the extension is primitive

And maybe one should mention that this works almost always

Because of the Primitive element theorem, it says every finite separable extension is primitive

and every finite extension of a perfect field is separable. And perfect fields include all fields of characteristic zero and all finite fields

😄

you are welcome

[marsh fractal]

maybe a silly question

is every group of odd order?

[delicate bloom]

that is silly

[marsh fractal]

or finite group

[delicate bloom]

you should sit in time out until you figure it out

[marsh fractal]

lol

[delicate bloom]

it's just too dead easy, how would you try to prove yourself wrong?

what's the easiest group you could think to construct that's even? try making a group of order 2

write down a multiplication table and try to fill it out

[marsh fractal]

yeah 1 and -1 with +

i guess i had the assumption that identity elements were unique in groups

[chilly ocean]

but that is true

[vital quail]

integers with addition modulo n

[marsh fractal]

indeed i have counterexamples

but can you explain why

[delicate bloom]

1 and -1 with + isn't a group

[marsh fractal]

i maybe am being silly it is late

[vital quail]

please dont use that term

also what do you mean why

why would you expect there to not be groups of even order

[delicate bloom]

1+(-1)=0 which means you have made an element out of the set

1 and -1 with * is a group though

[marsh fractal]

yeah sorry i mean with *

[vital quail]

also like take S_n for n >= 2

do you know what Sn is

it's the group of permutations on {1, ..., n}

[delicate bloom]

@vital quail he doesn't even know what the identity element is

[vital quail]

that is, bijections from that set to itself

lol true

anyways so how many permutations are there of {1, ..., n}

basically you have n choices for the first slot, n-1 choices for the 2nd slot, and etc

so n!

[marsh fractal]

what do you mean

[vital quail]

that means for n >= 2 |S_n| is even

[delicate bloom]

you just said a minute ago you had counter examples and said something weird about identity elements being unique

do you think -1 is an identity element in the group {1, -1} with *?

[marsh fractal]

no

[delicate bloom]

can you show me why it isn't an identity element

write down a counter argument showing how it can't be

[marsh fractal]

k lets not patronize lol

but yes 1*-1 = -1

[delicate bloom]

yeah looks good

[delicate bloom]

can you show some of these now

[marsh fractal]

yes the modulos were one

[delicate bloom]

seems like we should fix your misconceptions you have

[marsh fractal]

like Z/7Z

[delicate bloom]

ok good, show where the counterexample happens

[marsh fractal]

its of size 6

[delicate bloom]

ah you have to be careful

this is of size 7, it's by default we assume +

otherwise (Z/7Z)* means multiplication

Z/nZ has n elements, (Z/nZ)* has phi(n) elements

[marsh fractal]

but yea so i just thought the inverses for each element were unique, and if it was an inverse of itself then it is the identity element which is unique

and so you just have the identity element and a bunch of pairs

[delicate bloom]

can you show me an element that has multiple inverses

and what they are

just one example I mean

[marsh fractal]

yep no idea

i think i should just do this

not thinking properly

yes no

i am convinced they dont

[delicate bloom]

I thought you said you had examples though, what were they

or did you just work them out and find your mistake or something

[marsh fractal]

Yeah i should say Z/6Z

as a counterexample

[delicate bloom]

I don't think he ever claimed that, that's why he was asking to begin with I think

[marsh fractal]

yeah

erm

it is a group

all things hold

ok yes so you can have something that has itself as an inverse

and its not the identity element

so that was my confuzzle

[delicate bloom]

you can have groups where everything is its own inverse actually haha

arbitrarily large even, take nxn matrices with only 1 and -1 on the diagonal

[vital quail]

x = x^-1 holds for order 2 elements not just the identity @marsh fractal

[marsh fractal]

yip thats what i mean

so i just had a wrong assumption

and yes the idea sounded whack but now i dont have that as an assumption

great stuff ty

[delicate bloom]

cool cool

[marsh fractal]

getting patronized lmao

no i joke lol

indeed

[delicate bloom]

yeah I just was cutting through to the problem as directly as possible while trying to be clear

there's some kinda misconception somewhere so just had to say stuff kind of plainly, just how it goes

not patronizing haha

[marsh fractal]

no i get that lol

was just the one comment about me not knowing what the identity element is

but i really dont mind im not taken back or anything

i apprieciate the help and i understand where you are coming from

[delicate bloom]

no that was me telling archsys to stop overcomplicating it, he was bringing in a bunch of extra stuff with S_n when your question was if even order groups even exist

it was just totally off the mark for him to start bringing in these groups of order n! lol

[vital quail]

wdym 🤔 another good example is 2p is a factor of |PSL(n, p)| for certain n

[marsh fractal]

what would you say is the simplest proof of lagranges theorem

yeah lets say you can only use order, and the group axioms

is it possible

cardinality

yeah i guess thats what one might have to do lol

[vital quail]

@chilly ocean

[fading wadi]

I don't understand this statement, can someone explain?

the H at the bottom is the system of representation of mod H

but tbh i don't know what that means

[thorn delta]

Might depend on the text, but that means "join" for me

[fading wadi]

here's the context

[maiden ocean]

wtf mantra

[old lava]

lmao, I've seen proposition, theorem, corollary, lemma, fact

never mantra

[latent anvil]

I love it

[thorn delta]

V either means join (subgroup generated by union of sets) or disjoint union i think

[old lava]

is it supposed to mean G is the union of all cosets

[latent anvil]

I would understand it as disjoint union, I guess?

[old lava]

ya, I guess

that would make sense

[fading wadi]

I see, so what does the bottom H make the union do?

like the "left transversal" thing

[old lava]

nothing, it's just saying it goes through distinct cosets

of H

fancy notation

for example if x and y are in the same coset of H, it will not use both of them

[fading wadi]

I see, and i guess "index" denoted |H| just means how many cosets?

[old lava]

that's the size of the set

[fading wadi]

oh no i mean |H| where H is the left transversal

[old lava]

oh ya

[fading wadi]

also denoted [G:H]

[old lava]

that's how many cosets are in G

[fading wadi]

ok nice

another thing I'm wondering is why don't they use a normal union

why disjoint?

like why V but not U

[thorn delta]

you could write U. They are just making the point that the set of distinct cosets partitions the group

[fading wadi]

ah ok

thx

[old lava]

ya, in the tb I use, they just used two statements

1. the union of cosets is G
2. intersection of distinct cosets is empty set

I guess here, they combine the two

[fading wadi]

Also here, K, H,G are groups, I'm confused to argue for case when G is infinite

because i think haven't understand the concept good enough

K and H are subgroup of G

[glossy yoke]

show that the cardinalities are the same.

probably by establishing a bijection.

or something.

[old lava]

if G is infinite and K isn't infinite, then it's obvious ([G:K] is infinite, and at least one of [G:H] or [H:K] must be infinite)

if G and K are infinite, then it's a little fucky wucky

[fading wadi]

why is it obviou if k isn't infinite?

but Poros we still need to prove that the infinite equate though

[old lava]

ah ya, if you want the "correct" infinite, then you'd have to resort to cardinalities or something

I wouldn't be able to give a good enough answer for that

[fading wadi]

which I have no idea what to do, never constructed a group homomorphism without a hint before XD

[old lava]

you can search up the third isomorphism theorem

(I think it's called)

which establishes that result

but it's only for K and H being normal subgroups

[fading wadi]

yea this is not normal 😦

So here's the question again, would like help from anyone who knows how to deal with when G is infinite

K, H are both subgroups of G

[thorn delta]

Let $A$ be the set of cosets of $K$ in $G$, let $B$ be the set of cosets of $H$ in $G$, and let $C$ be the set of cosets of $K$ in $H$. This is equivalent to finding a bijection from $A$ to $B \times C$.

[cloud walrus]

kxrider

[thorn delta]

For simplicity, assume A,B,C are indexed by representatives corresponding to distinct cosets. You have for example that $\bigvee_{x \in C} xK = H$, and something similar to this for all the others.

[cloud walrus]

kxrider

[thorn delta]

as a hint, the goal is to show that all products of representatives from B and C make up the representatives from A.

(i may have worded a few things confusingly. i can clarify anything i u want)

[fading wadi]

yep thanks, im thinking about constructing a bijection

what you mean by representatives is for example {aK| a in G} is a member of A, so a representative would simply be a right?

or it could be ak, k in K

[thorn delta]

if aK is a coset, 'a' would be a representative. What I don't want is two representatives for the same coset. Only distinct ones. This is just to clarify that when we take the disjoint union V_{x in A} xK = G, we are assuming that we've removed any repeated terms of this union. Another way of thinking of [G : K] is the number of distinct representatives for the partition induced by the set of cosets of K in G.

[fading wadi]

@thorn delta sorry I still don't know how to construct this bijection, and i think I'm missing something

when I'm thinking about it all I can do is write down the definitions of A B and C, and i know what a representative means

[thorn delta]

ur good. Basically, by the definitions of $A,B,C$, you can write $\bigvee_{x \in A} xK = G$ and $\bigvee_{y \in B} yH = G$ and $\bigvee_{z \in C} zK = H$.

[cloud walrus]

kxrider

[fading wadi]

yea i did that

i could only intuitively think about making sense of it

and it's only vaguely, so i realized because {aH|a in G} is a coset in B then i could see a is a representative but then for this exact coset there's [h:k] more cosets in A, because ah1 and ah2 could be distinct representatives in A if h1 and h2 are not in the same coset of K, but since h1 and h2 are both in H, ah1 and ah2 are repeated/same representative in B

but 1. i don't know if this thinking is correct and 2. I cannot think of how to formally write this

[thorn delta]

yea, ur kind of on the right track. its kind of hard to do this if you haven't seen it before. Lets say you have zK a coset of K in H. If y is in B, then yzK is a coset of K in G.

This isn't really super obvious though unless you see that you can do this $$G = \bigvee_{y \in B} yH = \bigvee_{y \in B} y\left(\bigvee_{z \in C} zK\right) = \bigcup_{(y,z) \in B\times C} yz K $$

[cloud walrus]

kxrider

[thorn delta]

and since you have a union of cosets of K whose union is G, there are at least as many elements in BxC as there are in A

[fading wadi]

true, this makes alot of sense, thanks

[thorn delta]

you have to show that you have a disjoint union on the right hand side. That would complete the proof. Do u see why?

[fading wadi]

ok so my understanding is: To prove the disjoint union is to prove that there's same number of elements in A and in BxC

But then

I still need to create a function and prove injective or something

wait nevermind

if there is same no. of elements, then the mapping is naturally created

@thorn delta RHS is disjoint union because suppose x in G s.t. x in case 1: z1K and z2K or case 2: y1H or y2H where z1≠z2 and y1≠y2, then by definition of y1 and y2/z1 and z2 being representative, they belong to disjoint cosets

so x can't be in both of these cosets, so the intersection of any of these cosets are empty

and then I guess since V__{x in A}xK and V_{(y,z) in BxC}yzK share the same coset K, then the same number of element naturally produce a bijective mapping from A to BxC.

correct me if I'm wrong

[thorn delta]

ur argument sounds a bit terse, so im not 100 percent sure I understand it. Another way to it would be to assume that $y_1 z_1 K = y_2 z_2 K$. Then $y_1 z_1 K \subset y_1 H$ and $y_1 z_1 K = y_2 z_2 K \subset y_2 H$. Since $y_1 H$ and $y_2 H$ must be equal or disjoint, we conclude that $y_1 H = y_2 H$ so $y_1 = y_2$ (we are assuming $y_1$ and $y_2$ represent distinct representatives here ofc). And then now I think you can apply a similar argument to show that $z_1 = z_2$.

[cloud walrus]

kxrider

[fading wadi]

oh

[thorn delta]

um hmm there might be a subtle point here

[fading wadi]

yea i think i got it overall now thanks

this is hard..

im just not used to it

[thorn delta]

yea it is. the idea is kind of simple, but there are a lot of details and stuff

and np

[fading wadi]

@thorn delta sorry to bother again but man i cannot apply a similar argument to show z1 = z2 because it's on the inside..

[thorn delta]

its not actually a "similar" argument, but its a simpler argument. Basically, use the fact that you now know that y1 = y2

[fading wadi]

like why does z1yK = z2yK imply z1 = z2

i know z1K = z2K imply z1 = z2 because that's how each z is defined as representative of a coset of K in H, but with the yK I'm not sure anymore

[thorn delta]

so ur struggling with y1 = y2?

[fading wadi]

no, I understand y1 = y2 part

I'm just confused that why (z1)yK = (z2)yK imply z1 = z2

[thorn delta]

what is y?

[fading wadi]

some element in G i guess

[thorn delta]

Am I claiming somewhere that (z1)yK = (z2)yK implies z1 = z2?

for arbitrary y?

[fading wadi]

no, but how else do you know z1 = z2 after knowing y1 = y2

[carmine fossil]

I think he means z1K=z2K or smt

[thorn delta]

y1 z1 K = y2 z2 K implies that z1 K = y1^-1 y2 z2 K = z2 K

[fading wadi]

oh my god

im actually braindead

[carmine fossil]

No, It doesn't. z1K=z2K implies (z1)^-1(z2) is in K

[fading wadi]

what is going on -_-

[thorn delta]

nah drake, u need context lol

[fading wadi]

yea i understand now i can't believe my brain is so stuck rn

[thorn delta]

re drake: z1, z2 represent distinct representatives for the cosets of K in H

[fading wadi]

So the process is: assume y1z1K = y2z2K, then from the tex you wrote up there y1 = y2 can be concluded

then using inverse of y1 you get z1K = z2K, and since z1 and z2 are representatives, z1 = z2 must hold

in order for the equality to be true

[thorn delta]

yes. and this proves the map from (y,z) in BxC to yz in A is injective.

[fading wadi]

what about surjective then, when is that proven

[thorn delta]

since the union of the yzK over (y,z) in B x C is all of G, that means for each x in A, the coset xK is covered by one of the yzK

[fading wadi]

oh

ty

[thorn delta]

npnp

[leaden finch]

can someone check my work for this one

[thorn delta]

ye, good. You could even just provide a counterexample like (2/1)/2 is not equal to 2/(1/2)

[leaden finch]

oh okay

im having trouble on this one

ik homorphism is

f(a) +f(b) = f(a+b)

[golden pasture]

there is more

than jus f(a+b)=f(a)+f(b)

[leaden finch]

f(a) * f(b) = f(a*b)

[golden pasture]

m yea

some conventions may require f(1)=1

[leaden finch]

oh whys that

[golden pasture]

uh

cuz like rings we usually define them with units

so it is funny if the unit doesnt map to a unit

[leaden finch]

ohh

hmm will this be right

[golden pasture]

yes

[steel crane]

is the group of units in an integral domain always abelian?

the set of units in a ring forms a group but I was wondering if i can prove/disprove the above statement too

[thorn delta]

yes. this is just because the multiplication in an integral domain is commutative

the same is true for the group of units in a commutative ring

[sour plume]

I guess the question is only interesting for integral domains defined in noncommutative rings

Which seems a rare usage but wikipedia says it's allowed

But quaternions are a noncommutative integral domain, and every nonzero element in there is invertible, but, well, it's not commutative

[golden pasture]

uh dont we usually use
integral domain -> commutative domain

[steel crane]

yeah that is what i thought... units in an integral domain should be commutative

but i took the ring of integers as an example

couldn't formulate a proof so there could be a counterexample

[golden pasture]

the proof is direct from definition

[prisma ibex]

Integral domains are defined as commutative domains so this is immediate from the definition

[steel crane]

oh yeah makes sense... Thank you!!

I have just started revising rings and modules for an exam that I have in 2 days. Really appreciate you guys helping 🙂

[pallid wing]

Hi, I'm going through the abstract algebra book by Dan Saracino and he uses this notation to denote a special kind of equivalence relation

anyone know how to say it in words?

[pallid ember]

x is related to y iff the product xy^-1 is in H

that's it really

[pallid wing]

oh okay ty!

[steel crane]

How can a polynomial over a matrix ring be created?

Like i have the matrix, I am not sure about the process

[golden pasture]

you take the matrix ring R

then do

R[x]

a polynomial ring!

sage: R = MatrixSpace(GF(5),2,2)
sage: R[x]
Univariate Polynomial Ring in x over Full MatrixSpace of 2 by 2 dense matrices over Finite Field of size 5

[leaden finch]

can someone help me with isomorphism

im not sure if i did it right

[carmine fossil]

It's correct

[leaden finch]

oh okay

[carmine fossil]

Nvm,It's not surjective

The codomain is M2(R)

[chilly ocean]

tell us if any non-diagonal matrices are in the image of f

[leaden finch]

hmm is it because we dont have the ending result for M2(R)?

[golden pasture]

ok think about it right

surjective means for any element in M2(R), we can find a pre-image

so for the general element $\begin{pmatrix}a&b\c&d\end{pmatrix}$

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

a cute cat ٩(˃̶͈̀௰˂̶͈́)و
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[chilly ocean]

discord killed ur backslashes

[golden pasture]

ouch

the only time i am lazy

discord stabs me

thanks

can you find an element $x\in R$ such that $f(x)=\begin{pmatrix}a&b\c&d\end{pmatrix}$

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[leaden finch]

hmm ohh i dont think we can

[golden pasture]

ye

so easiest is just show a counterexample

[leaden finch]

okie

[light tusk]

try it out for a transpose

[old lava]

if you know semi direct products, you can show normalizer of a cyclic subgroup generated by a permutation in $S_n$ is isomorphic to $Z_n \rtimes \text{Aut}(Z_n)$

[cloud walrus]

PorosInMyAshe

[golden merlin]

Okay. Lemme get this straight. Does a coset have to be under a subgroup?

For a group G, can i have gG?

[viscid pewter]

you can take the coset of any set

[golden merlin]

because cosets are usually defined in terms of a subgroup H

[viscid pewter]

most used in like normal subgroups tho

[old lava]

you can take coset of any subgroup pretty sure, not any set, but only normal subgroups allow for quotient groups to be formed

[viscid pewter]

not even subgroups

just any set i thought

[old lava]

I thought it was for subgroups only

[golden merlin]

thats what i thought

[viscid pewter]

oh bugger

[old lava]

because some of the theorems/propositions only work

if it's a subgroup

[viscid pewter]

yeah wikipedia says subgroup nvm

[golden merlin]

so they HAVE to be of the form Hg instead of Gg (with H being a subgroup and G being a group)??

[old lava]

and ya, G is a subgroup of G, so why not

[golden merlin]

THATS RIGHT

[old lava]

it just results in the trivial quotient group, G/G isomorphic to 1

[viscid pewter]

gG is just G forever lol

[golden merlin]

:PPPPP

[old lava]

so it's meaningless to do

[viscid pewter]

like for all g

i mean it's not meaningless, it works

it just isn't too useful usually

[old lava]

meaningless as in they're usually not under consideration

1 and G are always normal subgroups of G

they're just almost never considered

for a lot of purposes

it gives no more information

[golden merlin]

Thanks

[old lava]

ah then just ignore what I said before lol

https://math.stackexchange.com/questions/1218392/normalizer-of-the-cyclic-group-in-s-n this post might be useful

I'm not sure what you know, so you can look at the top answer, and it has some good information

use what you can from it, to help you do it more intelligently than brute force

[old lava]

ah, last two points are related. Since $N_{S_n}(\langle (1, 2, ... n) \rangle) \cong Z_n \rtimes \text{Aut}(Z_n)$, and it's another fact that $\text{Aut}(Z_n)$ has order equal to the euler totient function of n

[cloud walrus]

PorosInMyAshe

[old lava]

is how they got there, I would assume

if you don't know semidirect products, I'd ignore those 2

[cloud walrus]

treemelia

[old lava]

it's true for any n cycle

in S_n

[white nymph]

looking for tips on part b.)

[chilly ocean]

||hahahah commutative ring more like c-ring more like cringe ||

[sturdy marsh]

(b) is a matter of just doing it

assume ab is in the intersection

what does the fact that Q is a prime ideal tell you?

[white nymph]

that a or b is in Q

i dont know how to show that a or b needs to be in R however

[sturdy marsh]

we assumed that

An ideal I of R is prime if for all a and b in R, ab in I implies a or b in I

[white nymph]

where did we assume that?

we assumed ab in Q \cap R so ab in Q and ab in R. Q prime makes a or b in Q, but we only have ab in R, how are we guaranteed a or b in R?

[sturdy marsh]

I did not write it down

a and b are in R

again, read the definition of prime ideal

[white nymph]

R is a subring of S, not a prime ideal

[sturdy marsh]

and think about what you're trying to prove

yes R is a subring

we're trying to prove some ideal of R is prime

[white nymph]

yeah idk man. still not seeing why we can say a and b are in R from ab being in R

[sturdy marsh]

ab in R doesnt imply and and b are in R

but we're looking at a and b in R to begin with

[white nymph]

unless im having a stroke, we're looking at the product ab in Q \cap R

and you also legit typed "a and b are in R" 8 mins ago

[sturdy marsh]

what do you need to do to check an ideal is prime

of a ring

[white nymph]

if ab in the ideal, nts that a or b in the ideal

[sturdy marsh]

what are a and b

[white nymph]

elements of S

[sturdy marsh]

no

Again, if I have a ring A, what is the definition of a prime ideal

[white nymph]

I is a prime ideal of A if it is an ideal of A and if it satisfies the property i typed

[sturdy marsh]

type the property again

with all quantifiers

[white nymph]

okay i see where im misunderstanding

a and b are assumed in R

[sturdy marsh]

yes

[white nymph]

thanks

I need to go back and brush up on my understanding of ideals

[steel saffron]

Question to double check a proof I did. In the group $\mathbb{F}^{\times}_{p}$, $p-1$ is the only element of order 2, correct?

[cloud walrus]

Riemann-Zeta1

[latent anvil]

Not always! If p = 2 there are no elements of order 2

[steel saffron]

Well, for p>2, that is

[latent anvil]

but in the case p ≠ 2 this is correct, since the equation x^2 = 1 has two solutions, and one of them is 1

[steel saffron]

Great, thanks, just wanted to make sure that was right

[carmine fossil]

That's simply orbit counting

There are 6 conjugates of a=(1234) but in the group {gag^-1,ga^2g^-1,ga^3g^-1,1} there are 2 conjugates of a

Take {(1423),(12)(34),(1324),1} as an example, if (1234) maps to (1423) you get the same set as when (1234) maps to (1324)

So, There will be 6/2 unique sets

In the language of orbits and stabilisers,you are considering the action of G/(C(a)) on <a>(g.<a>=g<a>g^-1)(The number of elements in kernel of this is exactly the number of elements of same cycle type in <a>)

@chilly ocean

[tacit saffron]

im being asked to find the minimal polynomial for sqrt(3)+sqrt(5) over fields Q(sqrt5) Q(sqrt15) Q(sqrt10) and it make me confusion

so i need to find the monic polynomial of smallest possible degree where sqrt(3)+sqrt(5) is the root

i dont' get what the difference between the fields are

like how does that affect the way i find my min poly

[carmine fossil]

For example, p(x)=(x-√5-√3)(x-√5+√3) is in Q(√5) but not in Q(√10)

p(x) is minimal in Q(√5) btw

[tacit saffron]

im confused how you arrived at your p(x)

[carmine fossil]

Well,I guessed that

Because (x-a)(x+a)=x^2-a^2

[tacit saffron]

so do i just like come across these

[tacit saffron]

nvm

i kinda figured out

[leaden finch]

can someone help me with latex 😦

[paper flint]

Try #discussion

[pliant raptor]

Could someone explain me what is the rule for one permutation inducing the other permutation of partitions? I am completely lost

[vital quail]

@pliant raptor for starters what does A become after (1 2 3 4)

basically the induced permutation is what happens when you apply the permutation on the inner parts of A, B, C

so A goes to {{2, 3}, {4, 1}} right

now can you tell me what happens to the rest

[keen jackal]

sorry, i tried to answer but i did it in another channel because I didn't know how to access this channel earlier

[vital quail]

ah yeah you just do ,iam adv in #bots

[keen jackal]

im trying to do this problem now actually now lol and it's kind of scary that i can barely remember any of the math i did at uni before i dropped out, i had to pull out old notes and remind myself what a homomorphism was

[vital quail]

nah, that's just normal brain stuff

the point is that once you learn it and store it away, you don't have to actually relearn it the next time, just refresh

[keen jackal]

for a normal uni maths course what year would you say this is?

this is year 1?

[vital quail]

not really

in the US did you know only 15% of students take calc in hs

[thorn delta]

year 1 or 2 for europe or year 3ish for U.S.

[vital quail]

US math major curriculum would on average be like real analysis 2nd year, and algebra 3rd or something

[keen jackal]

i feel like this is the kind of problem i might have got maybe 2-4 weeks into the maths course

of year 1

[vital quail]

i mean certainly they could introduce this topic as part of an intro course and not via an entire algebra course

[keen jackal]

as part of group theory

[vital quail]

but no it wouldn't be common to take algebra 1st year

in the US

[keen jackal]

i see

this is like one of those things where it's really intuitive, like you just know it's true but im struggling to think of how to prove it

[vital quail]

prove what?

[keen jackal]

that it's a homomorphism

[pliant raptor]

I see, thank you very much to all of you!

[keen jackal]

Actually can I just say that since every permutation can be written as a product of transpositions, I only need to check that theta {(1,2) (1,3)} and {(1,2) (3,4)} = theta (1,2) theta (1,3) and theta (1,2) theta (3,4) respectively

by symmetry I don't have to check everything

[bronze trench]

Hey people!

I need a little help

one of the questions is: consider E=F an infinite field. Show R has an infinite amount of left minimal ideals

tbh I don't know where to start, maybe I just need a little push

[light tusk]

$\begin{pmatrix}a & b \ 0 & c \end{pmatrix}\begin{pmatrix}x & y \ 0 & 0 \end{pmatrix} = \begin{pmatrix}ax & ay \ 0 & 0\end{pmatrix}$
Looks like an extra degree of freedom there

[cloud walrus]

Apopheniac

[light tusk]

@bronze trench $I_y = R\cdot \begin{pmatrix} 1 & y \ 0 & 0 \end{pmatrix} $

[cloud walrus]

Apopheniac

[bronze trench]

oh that makes sense, I was looking at it the wrong way. In class we did an example like
Q Q
0 Z

and the ideals had to do with that Z and I was trying to make it work with fields

thanks for the help!

[light tusk]

yeppers 🙂

[bronze trench]

oh you just got me out of the hole I was digging myself into, I was totally on the wrong path

thanks 😄

[leaden finch]

when we have subgroups

do we usually prove for associativity?

[carmine fossil]

Associativity always works since the subgroup is a subset of the group

So,no

[leaden finch]

oh okay, so for all subgroups we just prove for closure, inverse and identity?

[carmine fossil]

Yes

[leaden finch]

okie ty

[leaden finch]

can someone check my work for this one for well define

[broken crag]

End sentences with a period.

If you're proving a claim, state the claim clearly, state that you're then proving the claim

Be consistent on notation at the last line

e.g. 2(n) and 2w on the last line

Either both have parantheses or neither do

Other than that syntactical/structure stuff the proof looks fine to me

Also the s needs an apostrophe

(Last line should end in a period, too)

@leaden finch

[leaden finch]

oh okay thank you

[broken crag]

yA I know it's a real pain in the rear doing all these grammar things

I hate it, but it's been drilled into me

[leaden finch]

yeah same lool

[paper flint]

https://kconrad.math.uconn.edu/blurbs/proofs/writingtips.pdf

Worth checking out.

[leaden finch]

oooo nicee

[old lava]

To prove a subset is a sungroup you can in fact just prove that a subset

1. Is not empty
2. If x and y are in the subset, so is xy^{-1}

It's usually much faster do just just this check (1 is almost always trivially obvious since identity is usually in the subset) than to check all three properties separately

[golden pasture]

does these actly matter in papers

[devout crow]

does grammar matter in papers? ye

[old lava]

does grammar matter in papers? ye

being clear in your communication always matters, if I spelled out every word phonetically in english, it would still be readable, but extremely unclear. like you can get away with worse grammer informally (because people will probably understand you, and you want to be fast), but if you have time, you should always have the best grammar you can

[pallid wing]

spent embarrassingly long on this; would be nice if someone could push me towards the right direction- i don't necessarily want the answer, thanks :)

[kindred jolt]

follows from Cauchy's theorem

@pallid wing

[pallid wing]

thanks 👍

[light tusk]

Does this look legal?

[astral galleon]

Is the quotient ring isomorphic to its ideal ?

@light tusk what book is that form

[light tusk]

Hall's "Lie Groups, Lie Algebras, and Representations", exercises 9-10 of chapter 1

[astral galleon]

Ahh I see @light tusk care to answer my question

[light tusk]

In general $R/I$ is not isomorphic to $I$. Take $R=k[x,y,z], I=(x)$. $R/I = k[y,z]$

[cloud walrus]

Apopheniac

[golden pasture]

it doesnt quite make sense to ask for isomorphisms

the ideal isnt even a ring

[smoky cypress]

Ideals are rngs

[light tusk]

an iso I would assume implies module iso in this case

[leaden finch]

can someone check my work if im on the right track