groups-rings-fields

[rustic crown]

oh okay

[delicate orchid]

idk it's just

I can't parse it

[summer path]

a u w u b

[delicate orchid]

I have to draw out the entire lattice in my head

[rustic crown]

(same)

at least Q -----> Q(a) n Q(i) -----> Q(i)

[summer path]

those are some long arrows

[delicate orchid]

not long enough imo

[rustic crown]

lol

Q ---------------> Q(a) n Q(i) ---------------> Q(i)

[white oxide]

does i have degree 2 over Q(a) since it's just a root of x^2 + 1 in Q(a)

[delicate orchid]

yes

[rustic crown]

not "in Q(a)"

usually you only know x^2+1 is irred over Q, this has the same content as [Q(i):Q] = 2

[delicate orchid]

but it is also irreducible over Q(a) right

[rustic crown]

but now you also know x^2+1 is irred over Q(a)

[delicate orchid]

oh I see
right now I get the intuition

[rustic crown]

but yea i doesn't lie "in Q(a)"

[white oxide]

ok fine

with coefficients in Q(a)

[delicate orchid]

is that not the same thing?

[rustic crown]

i mean i is also a root of x^2+1 in C[x]

but that's not saying it's irreducible over C right

[tardy hedge]

Is there some sort of shortcut or is showing associativity really annoying here

Q12

Like i need to unravel all these set theory formulas

[rustic crown]

"see it"

[tardy hedge]

Poo

I am not smart

I dont see such things

[delicate orchid]

commutativity is obvs, then I imagine the rest is just De Morgan's bashing

[tardy hedge]

Yeah thats what i was trying

Just seems like a pain

[delicate orchid]

annoying!

[tardy hedge]

De morgans

Yeah

[rustic crown]

anyway, show that there is a bijection between P(I) and Maps(I, Z/2Z)

[tardy hedge]

Reminds me when i did symbolic logic course

From philosophy department

[rustic crown]

you're using this bijection to turn P(I) into a ring

[tardy hedge]

Damn thats some advanced shit

For me lol

[rustic crown]

where Maps(I, Z/2Z) is a ring under pointwise addition and multiplication

[delicate orchid]

wait what

oh this is for 14?

[rustic crown]

12 right?

[tardy hedge]

Im doing 12

[delicate orchid]

and 14 actually

[rustic crown]

A+B is the symmetric difference and A*B is intersection

[delicate orchid]

the first ring is boolean

so 14 applies backways

[tardy hedge]

Oh thats cool^

[rustic crown]

det no understand

[delicate orchid]

you can think of boolean rings like those maps you mentioned

I didn't see how we could see the ring in 12 as those maps, until I realised it was also boolean

my apologies my thought process is very scatterbrained constantly and doesn't come across very coherently sometimes

[rustic crown]

nah, but you need to first check axioms to show it a ring then only you can use 14 right

[rustic crown]

not really. a subset A of I has the same data as a function f : I --> Z/2Z which maps elements of A to 1 and rest to 0.

that's called the characterisitc/indicator function of A, denoted \chi_A or 1_A

[delicate orchid]

yeah, I didn't meant as a solution to 12 - it was more me understanding where your construction came from

[rustic crown]

oh interesting

[rustic crown]

another similar idea is to show "XOR-ing" is associative.
x in (A+B) iff (x in A) xor (x in B)
then, x in (A+B)+C iff (x in A) xor (x in B) xor (x in C) iff x in A+(B+C)

[white oxide]

wait wut

how did Lang get that the rational root must be 1 or -1

how does that make sense too since it's not a root of it

[rustic crown]

that's the rational root theorem

if p/q with (p, q) = 1 was a root, then p^3-pq^2+q^3=0

[white oxide]

damn i must have missed it

[rustic crown]

so p | q^3 implying p | 1 and similarly q | 1

[white oxide]

i see, thanks

[white oxide]

would the Galois group of $f(x) = x^3 - 10$ over $\mathbb{Q}$ be $\text{Gal}(\mathbb{Q}(\sqrt[3]{10}, \xi \sqrt[3]{10})/\mathbb{Q}) \simeq S_3$ where $\xi$ is a third root of unity?

[cloud walrus]

okeyokay

[vocal patrol]

how is this done?

[delicate orchid]

not the best notation I've ever seen

[vocal patrol]

this is in a field Z5[x] mod that polynomial

[vocal patrol]

its from a teaching assistant

Is it by polynomial division?

[delicate orchid]

then it's all just using properties of mod

like, the definition of what it means for x to be congruent to y mod z is that x = y+kz for some k in your ring

[vocal patrol]

could u run me through 2 or 3 examples rq?

[delicate orchid]

so when we mod out by z we can just say that x = x+kz because obviously x is congurent to itself

[vocal patrol]

i have Z5

[delicate orchid]

ok then x = y mod 5 if and only if x = y+5k for some k in Z

[vocal patrol]

mod the polynomial x^2+4x+1

and I wanna compute the powers of (x+3) mod that polynomial

so (x+3) mod that

(x+3)^2 mod that

and so on

[vocal patrol]

this would just stay the same i suppose

[delicate orchid]

there's a different polynomial in the image which is strange but ok

[vocal patrol]

yes that was for another exercise

[delicate orchid]

modding out by x^2+4x+1, so x^2+4x+1 = 0, so x^2 = -4x-1 = x+4
we can now use this to cancel down all higher powers of x into just linear stuff

[vocal patrol]

so this would then equal to x^2+6x+9 mod that polynomial no?

[delicate orchid]

yeah, but since you're in Z/5Z x^2+6x+9 = x^2+x+4 = (x+4)+x+4 = 2x+8 = 2x+3

[vocal patrol]

ah true

[vocal patrol]

how does one get from x^2+x+4 to (x+4)+x+4

[delicate orchid]

yes, btw

[delicate orchid]

we just calculated that because x^2 + 4x+ 1 = 0 in our new ring, that x^2 = -4x-1 = x+4

(x+3) is either going to be order 5 or order 25 so lets hope (x+3)^6 = x+3 lol

sorry, that's not true

[vocal patrol]

well im trying to prove it for Z5 star

that x+3 is a generator

and my idea was to show its of order 24

so I wanted to show trhat its not of order 1 2 3 4 6 8 12

[delicate orchid]

yeah, that'll work

[vocal patrol]

yea but im a bit confused still

doing the modular arithmetic with polynomials :(

[delicate orchid]

you have the relators 5 = 0 and x^2 = x+4

other than that it's exactly like you'd expect it to work

[vocal patrol]

well 5=0 i get

but why x square = x+4

[delicate orchid]

not typing it out again

[white oxide]

would the galois group over Q adjoin root 2 be the same then

[vocal patrol]

ahh got it tysm

[delicate orchid]

not sure about that one! Are the degrees the same

oh wait it's the cube root of 10

yes, then they would be the same

[white oxide]

cool thx

[delicate orchid]

there will be a nice way to do this hold on

multiplication by (x+3) is an endomorphism of the underlying ring - which is a degree two extenstion of Z/5Z, so we can write it as a 2x2 matrix in M_2(Z/5Z). Specifically:
(x+3)(ax+b) = ax^2+3ax+bx+3b = ax+4a+3ax+bx+3b = (4a+b)x+(4a+3b)
So we can represent mulitplication by x+3 via the matrix
(4 1)
(4 3)

check the order of this matrix in GL_n(Z/5Z) and boom

[vocal patrol]

okay thats too advanced for me sorry 😭

but ty ill look into that approach later

[delicate orchid]

$\begin{pmatrix} 4 & 1 \ 4 & 3 \end{pmatrix}\begin{pmatrix} a \ b \end{pmatrix} = \begin{pmatrix} 4a+b \ 4a+3b \end{pmatrix}$
this is exactly the same action that multiplication by $x+3$ has on $ax+b$

[cloud walrus]

Wew The Lads Tbh

[delicate orchid]

that's the idea, it's just a reframing

[vocal patrol]

uh huh

how do i extract the solution for x+3 to the power of 3 from that then for example

[delicate orchid]

who cares, you just want the order

[vocal patrol]

yea true

so

24

[delicate orchid]

the answer is uh very painfully though

yeah good point I should actually check that this has order 24

it does!

I'm not chatting shite for once

[vocal patrol]

thats crazy

well Im doing it by just calculating the first two now

and then just substituting the next ones with those

to simplify them

for (x+3)^4 i got x

[delicate orchid]

yeah

that's right

[delicate orchid]

matches up with the matrix as well, if you take this matrix and raise it to 4 you get
(1 1)
(4 0)
and x(ax+b) = ax^2+bx = ax+4a+bx = (a+b)x+4a, just incase you weren't convinced!

[vocal patrol]

i see

no I was haha dw

[vocal patrol]

What would I get for 5x+3?

i.e. for the power of 6

or is it not simplifiable

[delicate orchid]

3

5 = 0 remember

[vocal patrol]

yea I see

so just 3

[delicate orchid]

we can just skip straight to 12 if you wanted btw

if 12 isn't the identity then we know that all the lower powers aren't as well

[vocal patrol]

well okay that is true

I shouldve thought of that haha

[delicate orchid]

it's a good exercise to do these calculations though! Get a feel for quotient rings

[vocal patrol]

yea!!

[white oxide]

lol one more verification, the Galois group over Q adjoin sqrt(-3) would also be the same right, since the splitting field for f(x) = x^3 - 10 over Q adjoin sqrt(-3) would be Q(sqrt(-3), sqrt3, and so f(x) irreducible over Q adjoin sqrt(-3) (it has no roots in it since the cube root of 10 is not contained in it) implies that we can look at the discriminant, whose square root is 30 times sqrt(-3) which is in Q(sqrt(-3))

what the hell is k^2

[final gulch]

that must mean that c is not already the square of something in k

[vocal patrol]

how would I do this?

looks like some type of binomial formula to me if i insert the hint

but I suppose I have to do some rewriting to get the actual one

[white oxide]

silly question, but if E/k is algebraic, how do we know there exists an embedding of E into k^a fixing k?

[coral spindle]

We do not

C/Q has no such thing.

[delicate orchid]

watch this

[coral spindle]

Oh

you don't mean k to the power a

you mean the algebraic closure of k

[vocal patrol]

can i get this to a square +2ab + b square somehow?

[coral spindle]

Man idk how to explain this better than simply saying this is basically the definition of the algebraic closure of k. Like, that's the property you want it to have. Idk how you're constructing it but it should have this property by the construction.

You could say that since k^a is by definition the direct limit (categorical colimit) of all the algebraic extensions of k, every algebraic extension embeds in it necessarily.

[chilly ocean]

help

<@&286206848099549185>

[coral spindle]

Picture this

Someone walks up to you while you're sitting on a bench

and they show you a book and say "help"

you wait for them to explain what their problem is

they don't do so

I think this is a funny image in my head.

imagine how confused you would be

[rocky cloak]

Do you know about the Jacobson radical?

Or perhaps easier, do you see what the units are in this ring?

[crystal turtle]

Perfect analogy Boytjie

[white oxide]

then they scream HELPERS!!!!! after 1 minute

[tardy hedge]

Some subsets of rings that form rings under the same operations have a different identity than the original ring

But a ring like Z has no subsets that are rings with the same operation

And there is no subset of Z that can have a different element other than 1 act as identity

So are these facts related somehow?

[coral spindle]

Rings that share the unit are called unital, recall.

[delicate orchid]

unimodular

[coral spindle]

Indeed Z has no unital subrings aside from itself

[tardy hedge]

I havent learned that defn before

[coral spindle]

This is because Z is the initial ring. If R is any ring, then there is a unique ring homomorphism f : Z -> R sending 1 to 1.

Now let's think about what this would mean in terms of Z itself

[delicate orchid]

I wonder if that's true of initial objects in other categories - they can't have any subobjects

[charred iris]

(some people include identity in just 'ring' so follow your course if this is a course)

[coral spindle]

If R were a unital subring of Z, then there is a map Z -> R -> Z which must be the identity map by uniqueness, so the image of the map R -> Z must be the whole of Z

Same thing for K-algebras over a field K, since K is initial. Dw if this is greek to you.

[cobalt heath]

Semisimple rings obv have trivial Jacobson radical, why is this so hard for me to prove

[coral spindle]

obviously
so hard
???

These things are contradictory.

[cobalt heath]

Well, it simply means I am bad at this

Ah right, I think I already proved it

Jacobson radical J is annihilator of simple left modules, and the entire semisimple ring is sum of simple modules, so J is annihilator of entire ring

[crystal turtle]

I don't see why this would be true, but I also can't think of an example offhand.

[delicate orchid]

yeah, likewise

like, we have a monomorphism going into the terminal object and we assume that it's not the identity map

[rocky cloak]

If you have a map f: X -> Ø, where Ø is an initial object, then because of the map Ø -> X, you have that f is split epi.
If f is both split epi and mono it's iso, so the only subobject of Ø is Ø itself.

[crystal turtle]

FUCK

[coral spindle]

Well any subobject of the initial object would factor through itself. So to write it out, if A is initial and f : B -> A is a subobject, we have a map g : A -> B which is a right left inv—dammit

[delicate orchid]

right ok, I didn't know the split epi result

I was considering chaining like, I -> X -> I -> X

and then this has to be I -> X

if X -> I -> X is the identity on X, I buy that it splits

but I can't see that atm

[cobalt heath]

Is 2Z submodule of Z as a Z-module

[rocky cloak]

So f: I -> X and g: X -> I, then
gf is identity on I, so gfg = g. If g is mono we can cancel it giving fg = identity

[rocky cloak]

Try checking the definition!

No, the initial object there is (0)

[crystal turtle]

Jagr just showed you can't have a subobject of initial, if that's what you're looking for

[crystal turtle]

Mfw abelian category

[cobalt heath]

Ah so intuition holds

[delicate orchid]

ahhh compose them the other way - clever

obvious in hindsight but still nifty

[chilly ocean]

bruh this is an intro to abstract algebra class

[chilly ocean]

wait how would units help

[cobalt heath]

(If gf = id, g is epi IIRC - if g is mono as well, clearly it is an iso I think, or idk)

[crystal turtle]

(that was the "if both split epi and mono then iso" thing jagr mentioned)

[cobalt heath]

Ahh, I see. Thanks

[celest cairn]

Might be a stupid question but what’s the difference between a polynomial who’s Galois group is isomorphic to C_2 and a one who’s Galois group is iso to S_2?
Can someone give an example?

[crystal turtle]

C_2 and S_2 are isomorphic, so... no difference?

[celest cairn]

Ah, so the Galois group of
x^2 - 2 is isomorphic to both S_2 and C_2?

[chilly ocean]

@rocky cloak if a and b are both odd then a/b is a unit. Idk how that's useful though

[dire siren]

It's a small secret from Commutative Algebra: A ring is local (i.e. it has a unique maximal ideal) if and only if the set of non-units is an ideal.
You only need the "<=" direction, which is easy to prove (hint: if an ideal contains a unit, then it is the whole ring, so it's not maximal).

[chilly ocean]

damn bro

looks like the professor has been keeping secrets

[delicate bloom]

Really this is nothing to do with Galois groups, just plain groups C_2 and S_2

[white oxide]

bro has a galois group as a bio

[chilly ocean]

abstract algebra studies the most random shit

[rotund aurora]

not true

[crystal turtle]

I mean, everything studied is motivated by genuine questions/problems people had

[chilly ocean]

name one theorem that makes sense to people outside of university math

[crystal turtle]

Remarkably, when you talk to people who don't study what you study, they don't know things they haven't studied

[chilly ocean]

no but you can still say interesting things about it

[maiden ocean]

There are interesting things to say

subjectively

[cobalt heath]

With group theory, you can talk about symmetry

Also there is this classical Galois theory which shows when polynomials have roots expressible with n-roots.

[round hull]

crystallography and cryptography

[ivory trail]

abel-ruffini (oh already mentioned)

[limber wharf]

Fermat

If you explained the idea of groups and symmetry then CFSG is pretty interesting

Some AT stuff is easy to explain

[rotund aurora]

To give another concrete example, ruler and compass constructions, and in particular, construction of n-gons

I think that the core ideas of any topic in math can be explained to any minimally smart person in a short amount of time (might not be 5 mins, tho)

[lethal dune]

Doesn’t have to make sense, it’s not about making sense to people

[low wyvern]

Guys, quick question (and rather stupid perhaps). I have to prove $$(x^{-1})^n = x^{-n}$$ . I am able to do it using a theorem thats proved the addition law for exponents and induction. Without this though, how would I prove it?

[cloud walrus]

KξRNΛL

[warm gate]

Why

[low wyvern]

why what?

[warm gate]

why do you hate induction

[low wyvern]

I don't but the exercise was asked before the theorem that proved addition of exponents. Now whilst it works in an exam, I wish to know a way that doesn't use that result.

[dire siren]

isn't that a definition?

[delicate orchid]

It may seem like it, but no

[low wyvern]

no. The definition is $$x^{-n} = (x^n)^{-1}$$

[cloud walrus]

KξRNΛL

[low wyvern]

I was unable to prove it using this though without creating a circular argument.

[delicate orchid]

You know inverses are unique, so show that both of these things are inverses of x^n

[low wyvern]

I think that's how I tried it but ill give it another go, thanks

Can I post my written solution? (it is a bit messy)

here

uh

[delicate orchid]

Yur go ahead

[low wyvern]

I believe lines 3 and 4 don't work as I used the result itself I think

[delicate orchid]

$x^n(x^{-1})^n = (x…x)(x^{-1}…x^{-1}) = x…(xx^{-1})…x^{-1} = e$

[cloud walrus]

Wew The Lads Tbh

[delicate orchid]

You can do this inductively if you really want. And last time I got “erm aktually’d” by someone who demanded induction was needed

[low wyvern]

Um. The way I did it using induction though arrived at needing to add the exponents and that result is proved literally the theorem after this in the lec notes. So assuming I couldn't use it, I thought an argument like the above would be needed. Anyways ill use what you gave me and give it another go. Thanks 🙂

[delicate orchid]

Wdym “use”? That’s the full proof it’s an inverse lol

[low wyvern]

oh ye lol

nice one

That's easier than what I though lol. I hate algebra, so many strange identities to prove....

[low wyvern]

just out of interest with this you have used the fact that $$(x^{n})^{-1} = (x^{-1})^{n}$$ right? Is this just allowed because it is a definition. I think my main problem with this is that I wasn't sure what I could and couldn't use.

[cloud walrus]

KξRNΛL

[rocky cloak]

They didn't use that fact, that's the one thing they're proving

[low wyvern]

Ye, I see it now. Thanks.

[ebon pine]

If $N$ is a normal in $G$ and $a \in G$ has order $o(a)$, prove that the order $m$, of $Na$ in $G/N$ is a divisor of $o(a)$.

[cloud walrus]

numbpy

[ebon pine]

Since Na is a set, the order means the size of the set right?

In which case, shouldn't it be reversed like o(a) should be a divisor of m. For example consider G=S4, N=V (Klien 4 group) and consider a = (1 2)(3 4) which belongs to N also. Then ord(a) = 2 while Na = N has size 4 so ord(a) divides ord(Na)

Or am I missing something here?

[dire siren]

Na is an element of the group G/N, so the order of Na means its order in this group

so in other words, the order of Na is the smallest positive integer k with (Na)^k=N

[tardy hedge]

Heyy i remember doing that question

[delicate orchid]

no

if you'd like a hint, there's a fairly natural homomorphism from G to G/N

[crystal turtle]

Hint: ||show for a homomorphism φ:G --> H that the order of φ(g) divides the order of g||

And consider how that relate a la Wew

[ebon pine]

I know that but I'm not supposed to use it cause Herstein. What I was thinking was that since Na has order m so (Na)^m = Na^m = N = Ne so ord(a) divides m. Is something wrong with this reasoning?

[dire siren]

the conclusion is wrong

use the fact that, if in a group x^n=e, then ord(x) | n

[ebon pine]

That's exactly what I did a^m = e hence ord(a) divides m

[dire siren]

@ebon pine a^m is not equal to e

[crystal turtle]

Mfw people aren't allowed to use results to prove other results (because obviously mathematicians start from foundations to prove everything)

[delicate orchid]

wait... they don't?? That'll explain my slow progress on my PhD...

[dire siren]

@ebon pine you want to get m | ord(a)
so consider calculating (Na)^ord(a), refering to my previous hint

[ebon pine]

Ohhhh, I understand what mistake I was making. Thanks

Thanks @delicate orchid and @crystal turtle

[tardy hedge]

Im trying to show that for a ring where a^2 = a for all a, then the ring is commutative

I got that ab+ba = 0

But thats like ab = -ba so not exactly what i want?

[cursive spindle]

Start from a+a

See what you can conclude

[tardy hedge]

Ok ill try thanks

Ohh you get that every element is the same as its additive inverse

Nkce

Nice

[cursive spindle]

[delicate orchid]

an hour late but a really nice way to see that boolean rings are characteristic two is (-1)^2 = -1 = 1

because "one-element \overline{1}" is enough?

you ask some odd questions

oh right like why don't they use just the symbol 1?

[velvet steeple]

Well 1 = \overline 1 is shorter, so maybe with one-element, they don't even mean 1?

[delicate orchid]

I think it's to emphasise that these are equivalence classes of integers rather than just integers

until next week when we stop caring and just write Z/mZ as {0... m-1}

[velvet steeple]

Ah, alright

[tardy hedge]

Does Q3 just kind of directly follow from Q2?

Q3 says phi is onto and by Q2 phi must be one to one, ok so its an isomorphism

[delicate orchid]

yeah, you know they have to be injective - so if it's also surjective then obvs it's bijective

how did you do Q2 btw

[tardy hedge]

I havent finished/written it up yet but if it wasnt one to one then phi would send elements that have inverses to 0 which doesnt have an inverse, and homomorphism preserves inverses

[delicate orchid]

sure

[lusty marlin]

Do you know the first isomorphism theorem for rings?

[delicate orchid]

my favourite proof is that the kernel is an ideal

[lusty marlin]

And what the ideals of a field might look like?

[tardy hedge]

Ya it was covered in this section

Ideals not so much, we just did that last class and i need to read more about them

They cover ideals and factor rings in the following section in this book

@delicate orchid btw do u prefer rings or groups

[delicate orchid]

I'm a group theorist lol

[tardy hedge]

Oh cool

[delicate orchid]

rings make no sense they're strange and mysterious beasts

[tardy hedge]

I was more excited in the course when we were doing group theory

This stuff is still cool but i dont care as much for whatever reason

[delicate orchid]

rings are only studied so we can put modules over them

[tardy hedge]

Ive heard of modules but dk what they are

[delicate orchid]

this isn't true but I don't recognise number theory as legitimate mathematics

[delicate orchid]

vector spaces but instead of being over a field they're over a ring

[tardy hedge]

Oh cool

Wew what do you think about analysis?

[lusty marlin]

Why?

[delicate orchid]

oh yeah guys the uhhh "ring of integers" definitely depends on d mod 4. Yup.

I'm definitely not just making things up

[tardy hedge]

If phi was zero, image of phi is just the zero ring?

Q2

[delicate orchid]

sending things to 0 tends to map them into the zero ring yeah

[tardy hedge]

Lol

[delicate orchid]

judging by the stuff above these questions they really want you to do this

[tardy hedge]

Somehow use the homomorphism theorem ?

To me using that is like next level stuff lol

They didnt introduce ideals until next section

[tardy hedge]

For. Q3 i literally just said we know from q2 it must be one to one so if its also onto then its an isomorphism

I mean there is nothing else but that right?

[delicate orchid]

but... there's quotient rings above the question?

[tardy hedge]

This book does things weird

It introduces things without giving the definition

And then gives the defn later

This is the book by beachy and blair

No one mentions it so idek why my class uses this

[delicate orchid]

Hmm. The only logical option is to drop out and sue them for malpractice.

[chilly ocean]

Hello 👋 I hope you all having a great time 🌸
I have a question about lattices. If L is a lattice with a given diagram, how to find the congruence relations and draw ConL?

[teal vessel]

Is there a name for the extension to the fundamental theorem of arithmetic that shows that all members of Q⁺ have a unique product across the integral exponents of primes?

[errant shadow]

What is the key difference between the basis for free groups and the basis for vector spaces?

Like, why shouldn't we compare them even in terms of analogies, etc?

[earnest hare]

for free groups words are not commutative but for vector spaces addition is

[mighty kiln]

Free Abelian groups and vector spaces are similar, just one is over Z and one is over F

[tardy hedge]

Why is math addicting?

[broken quartz]

*meth

[stark helm]

Does anyone know how to determine if S4 has a subgroup of order 8?

[rocky cloak]

Do you know the sylow theorems?

[stark helm]

I don't know that

[rocky cloak]

Alright, maybe think of some groups that permute four things

[stark helm]

what does permute four things mean?

product of two 2-cycles?

[rocky cloak]

Many groups are naturally thought of as permuting things, like the symmetry groups of regular n-gons or cyclic groups or alternating groups or symmetric groups

Alternatively I guess you can just attack S4 directly, combining elements to try and make a group of order 8

[stark helm]

Here is what I have tried

[stark helm]

but It seems too complicated and I don't know why I even construct a group of order 7, while its elements are order of 2 or 4

[rocky cloak]

So it seems like you basically have it towards the end there. S4 has no subgroup of order 7 though...

[stark helm]

So I feel really confused why this kind of subgroup can be constructed

its element can have order of 2 and 4, which makes me feel more confused but can not find new elements in this group to make it more reasonable

[rocky cloak]

But you're looking for a group with 8 elements not 7, so there's no problem with not being able to construct one of order 7

At the end it looks like your considering the group generated by (12) and (1423)

That's a good idea

[stark helm]

Yes, I am working on that now

[rocky cloak]

You might recognize it as a group you know generated by a 4-cycle and a 2-cycle.

[stark helm]

it seems that ict can not work

[rocky cloak]

ict?

[stark helm]

it

[rocky cloak]

What do you mean?

[rocky cloak]

Why can't it work?

[stark helm]

I made a mistake, because I suddenly saw (12), (34) appear (but found that two 2cycles can't appear when I tried (12), (13)), that's my mistake

this should work now

[rocky cloak]

Indeed, and you might notice that this is the symmetry group of a square

Which you can see by for example labeling the corners 1, 4, 2 and 3

[stark helm]

Yes, it seems 1,4.2,3 and 1,2 as (1423)*(1423)

[stark helm]

but symmetry seems to be the same stuffs?

Another question, can someone check write down all abelian group of order 360 up to isomorphisms? I wrote Z2Z2Z2Z3Z3Z5, Z360, Z2Z4Z9Z5, Z2Z4Z3Z3Z5, Z2Z2Z2Z9Z5, Z8Z3Z3*Z5, is it a complete list? Or do I have to write down what does these groups isomorphic to?

[rocky cloak]

Looks pretty complete to me

[stark helm]

OK, really appreciate that!

[teal vessel]

uhhh.... because you can always generate G with the generating subset <G>, and therefore the order of G is always the upper bound for the generating subset?

this seems incredibly obvious

[next obsidian]

Congratulations

Now they can reference this fact if they need it, I guess

[teal vessel]

yeah, probably the fact that they're choosing to define "finitely generated groups" in the exercises rather than the main body of the chapter

"prove Z is finitely generated" there we go, now we actually have a (slightly more) nontrivial question

Z=<1>

also <-1> by symmetry, but who's counting?

[teal vessel]

ok but actually proving that Q is not finitely generated is quite beautiful tbh

[tardy hedge]

Nerd!!!!

[teal vessel]

absolutely.

[boreal inlet]

Try extending this for Integral domains which are not a field. Fun exercise

[cloud walrus]

Riku

[teal vessel]

haven't gotten there yet

[boreal inlet]

Oh my bad I thought you were doing modules

[rotund aurora]

Is there any way to prove there exist infinitely many nonabelian groups of order pq with p and q distinct primes, q fixed, without using that there exist infinitely many primes p with p=1 mod q ?

[teal vessel]

I'm doing generators and cyclic groups rn

[next obsidian]

Sorry what?

[boreal inlet]

Have fun in group theory it's cool

[next obsidian]

Oh, I see

Over all primes p,q

I mean, the answer is no

[rotund aurora]

ig you should fix q

[next obsidian]

You can classify all the groups of order pq and they’re all abelian unless p < q, p = 1 mod q

[rotund aurora]

The point would be to prove the particular case of Dirichlet's theorem

[next obsidian]

So if there existed infinitely many, then you get the existence of infinitely many such pairs of primes

[rotund aurora]

I'm talking about proofs

[next obsidian]

What?

[rotund aurora]

You didn't address the question, but the first message I sent was unclear

so let me rephrase

Fix an arbitrary prime q. Is there any way to prove that there exist infinitely many nonabelian groups of order pq with p some other prime but without using that there exist infinitely many primes p with p=1 mod q ?

so that as a corollary, you get that there are infinitely many primes p with p=1 mod q

I assume your answer is still no, but you didn't explain why, if you meant to

[tardy hedge]

No!! Ur supposed to be angry

Nerd and virgin!!

How about that

Math > Sex

Anyone?

[chilly ocean]

#algechill

[tardy hedge]

I will indeed take it back a notch

Im done with the tomfoolery dont worry

[tardy hedge]

But .. lowkey true ?

[kind temple]

they did not chill

[rotund aurora]

.

[teal vessel]

$N\wedge V = \bot \ N\vee V = \top$

[cloud walrus]

GoldenPhoenix

[tardy hedge]

Nerd and virgin is false? Nerd or virgin is true?

[teal vessel]

logically speaking, yes.

[tardy hedge]

Understood.

[teal vessel]

the determination of which variables have which values is left as an exercise for the reader.

[dim widget]

The existence of infinitely many is a lot easier than dirichlet’s theorem

[rotund aurora]

yes I know

[dim widget]

although you need some dirichlet style theorem

[white oxide]

can i have a hint for (1) => (2) please?

[errant shadow]

What would be an example for a normaliser of a subgroup H of G such that N(H) = H? Would simple groups satisfy this?

[crystal turtle]

Pick your favorite group G, and consider H=G as a subgroup of itself

[errant shadow]

Yea but if we are talking about proper subgroups?

[errant shadow]

Wait I think {id, (1 2)} in S_3should work? Is this correct?

[celest furnace]

Can anyone give me a small hint for this one

My geometric answer is just you are in the same orbit if you act the same in a different basis

My new conjecture is they are in the same orbit if they have the same eigenvalues through a quick calculation

[chilly radish]

Having the same eigenvalues isn't enough

Try to find a counterexample

[ivory trail]

is this ||stffgmoapid||

[chilly ocean]

What does GH mean?
Is it GH={gh}?

[rose prism]

what is the geometric picture you have in mind?

[chilly radish]

I mean in this case that's just ||Jordan normal form||

[celest furnace]

Yep I saw that

{{1,1},{0,1}} and I_2

[errant shadow]

How do you show that A_4 is not simple. .Like, I know there exists V subgp of A_4 such that V is isomorphic to Z/2z x Z/2z. But how do you reach there?

[rotund aurora]

Can't you show directly that there is no simple group of order 24?

[next obsidian]

What V is abstractly as a group doesn’t say anything about its normality

[next obsidian]

A_4 has 12 elements

[rotund aurora]

[chilly radish]

Also why would you do that instead of just showing klein is normal. If you know a generating set for A4 this is not hard

Hell just check directly it's only 8 elements

[errant shadow]

Oh yeah right

[next obsidian]

A_4 has 12 elements

[errant shadow]

Ahhhh that's easier

Thanks

[errant shadow]

How do you show GL2(Z/2Z) is isomorphic to S3?

[mighty kiln]

By finding an isomorphisjm

[errant shadow]

I couldn't find one

[mighty kiln]

maybe consider orders

Or you could consider the action on (Z/2Z)^2

[errant shadow]

Ykw, I have a silly doubt here. How is (0 0)(0 0) in Gl2(Z/2z)

[mighty kiln]

It's not

[errant shadow]

Okay yea right

[errant shadow]

Okay I'll try that

Thanks

[rocky cloak]

The Galois group acts transitively on the roots of irreducible polynomials

[dim widget]

can’t you show directly that there is no simple group of order 12?

[rotund aurora]

Yes you can

By doing some sylow things

Well, you can for groups of order p^2q

Order 24 is actually trickier

[rocky cloak]

||The number of 2-sylow subgroups is either 3 or 1. If it's 1 were done, if it's 3 then there's a map G -> S3. Since 24 > 6 this map must have a kernel.|| Badaboom

[rotund aurora]

Oh nice

But in that case, the 3-sylows and 2-sylows needn't be normal

You can do some argument for p^3q in general, but you get the special case p+1=q lol. But yeah, you get trivial inequalities

[dim widget]

they’re pretty much the same

A nice way that works for any p^a3 with p=2 is to note that if the group was simple you would have 3 p-sylow subgroups with no intersection, then there would be (p^a - 1)*3 + 1 elements of order dividing p, and so only 2 elements of order 3

standard trick

[coral spindle]

Because that is the definition.

[coral spindle]

Yes. Proceed by induction.

[velvet steeple]

Thanks

[velvet steeple]

Prove that $\overline a \in R$ has a multiplicative inverse in $\mathbb Z/m \mathbb Z$ $\iff$ $\gcd(a, m) = 1$.

Can someone please check what I have for the ==> direction?

[delicate orchid]

oh you do know bezout's. Good

that makes my life easy

this proves both directions at once if you remove the first and last sentence

[velvet steeple]

Oh

Yeah

[velvet steeple]

True, Bézout's lemma is an iff

[south patrol]

I mean

[delicate orchid]

and there they go, back into the void

[south patrol]

Good luck picking a multiplicative inverse to 2 in the integers

[delicate orchid]

ring homomorphisms send units to units, so a^-1 only exists if a is a unit in Z aka if a = {1, -1} using this argument

[south patrol]

I assumed they were talking about complex numbers

Lol

And then scrolled up

[velvet steeple]

oh

[delicate orchid]

nah we're still at the stage where we care that elements of quotients are actually cosets

[south patrol]

Yeah fair lol

[velvet steeple]

^-1 makes sense only outside of the overline, true.

[delicate orchid]

no - it only makes sense inside of the overline

I feel like you had a perfectly good proof and are now just overthinking it and coming up with nonsense

[next obsidian]

Tfw $\overline{2^{-1}}$

[cloud walrus]

Chair MonkE-Girl

[delicate orchid]

this is equal to 0.5 if you knew anything about complex numbers (which you clearly don't) so perhaps you should take a look in the mirror (the real axis) before you start reflecting (conjugating) sweatheart

[velvet steeple]

We have $\overline a \cdot \overline{a}^{-1} = 1$, not $\overline a \cdot \overline{a^{-1}} = 1$, no?

[delicate orchid]

yur

I got confused

[velvet steeple]

Yeah, that's why I said it only makes sense outside of the overline, since inside, a^-1 wouldn't exist

[velvet steeple]

The last line I wrote is just wrong, isn't it?

We are in Z

[coral spindle]

Indeed

[delicate orchid]

I assumed everything after "let r = 0" was in the quotient but yeah I suppose you could make that a bit more explicit

[velvet steeple]

Well, but we can't stay in Z/mZ, the statement sa + rm = 1 is in Z

[delicate orchid]

did you miss the part where I said "after "let r = 0" was in the quotient"

cause setting r = 0 is taking the quotient mod r

[velvet steeple]

Oh

[delicate orchid]

anyway just write your little \overlines on everything to be safe

[velvet steeple]

I think it's clearer like this now

Maybe I should say that this also proves the other direction

[delicate orchid]

This is much better yeah

minor nitpick is that I'd say "by Bezout's lemma" rather than after

[velvet steeple]

Ah, alright

[delicate orchid]

apologies for being really nitpicky but it's important to get these things right early on

[velvet steeple]

Ofc, thanks. I'm from Germany, we often say "nach ..."

[delicate orchid]

yeah I had a feeling

it makes perfect sense in german - because it follows from bezout's lemma, so it comes after it!

and instead of saying r = 0 I'd be really explicit and write $\overline{sa+rm} = \overline{sa}+\overline{rm} = \overline{sa}+\overline{0}\overline{m} = \overline{sa} = \overline{1}$

[cloud walrus]

Wew The Lads Tbh

[delicate orchid]

just to make it absolutely clear that you know what's going on

[velvet steeple]

Alright, thanks

[delicate orchid]

yes

[south patrol]

nice

can you also use "laut" or is that wrong-sounding

i know a good amount of german but not really interacted with mathematics in the language

[velvet steeple]

You can, but I don't often read that. It doesn't sound wrong to say "laut den Ringaxiomen gilt ..."

Just CTRL+F-ed my linear algebra 1 lecture notes, and there is no occurence of "laut". They always say "nach" or just put the lemma in brackets

[south patrol]

Okay thank you

!

[delicate orchid]

!

[dull marsh]

In the context of groups with operators, is there any particular reason why operators aren't just called or identified with group homomorphisms?

[delicate orchid]

wdym a group with operators

is that the funny set action thing

cause if it is then yeah they're just group homomorphisms

[dull marsh]

I'm reading Waerden's algebra and he defines a group with operators to be a group G with a set of objects called operators such that if $\Theta$ is an operator and $a, b$ are elements of $G$, then the product $\Theta a$ is some element of $G$ and it also should hold that $\Theta(ab) = \Theta a \cdot \Theta b$

[cloud walrus]

A Lonely Bean

[dull marsh]

Is this the same as group actions?

[delicate orchid]

no because theta isn't a group

so how can it be a group action

if it is a group then this is just some semidirect product action on G

[dull marsh]

I don't really know what a group action is so

[next obsidian]

Isn’t theta literally just an endomorphism

[delicate orchid]

it is just an endomorphism

[ashen heron]

so this is a group homomorphism from a group to itself? is that called an endomorphism?

[next obsidian]

a group homomorphism G -> G

[dull marsh]

Yeah, he remarks that multiplying by an operator is just the same as an endomorphism

[rocky cloak]

I guess maybe to avoid such word salads as "homomorphism of group with homomorphisms", idk

[delicate orchid]

the reason why we do this chmonkey is that we'd like to talk about subgroups or subsets of G being "stable" under this set of endomorphisms

[dull marsh]

So I wondered if there's a reason for distinguishing between operators and endomorphisms

[delicate orchid]

and this is just the nicest set up to do so

[delicate orchid]

the elements of the set need not be actual endmorphisms of G - just identified with them

[dull marsh]

Maybe it's just old terminology

[delicate orchid]

that's pretty much the only difference

[delicate orchid]

but yes it is very old

[next obsidian]

Wait lmfao are you reading Van der Waerden

[dull marsh]

Yes

[next obsidian]

What pyramid did you recover it from?

[dull marsh]

None, my professor mentioned it as supplementary material

[next obsidian]

It was a joke

I’m just saying the book is ancient lmao

[delicate orchid]

ORIGINALLY PUBLISHED IN 1930

[dull marsh]

Well he's been sending the relevant pages from the pdf for the last two weeks

[next obsidian]

Is your professor a dinosaur or something?

[delicate orchid]

HOLY FUCK READ SOMETHING ELSE

ANYTHING ELSE

THE HUNGRY CATERPILLER

JUST ANYTHING

[next obsidian]

Van der Waerden was like, the first algebra text ever

That set it up the way all modern algebra books work now

[dull marsh]

Yeah, I know that it's old

[delicate orchid]

anyway I'm exaggerating for comedic effect

[dull marsh]

He's German so I guess he studied with that book in the past

[ashen heron]

no but is he a dinosaur

[dull marsh]

I can not confirm nor deny that information

[fair bay]

Hi, Im looking for examples of finite rings where there are elements so ab = 1 but ba does not equal 1

[delicate orchid]

there might be one in the space of non-square matrices over a finite field

I think if a non-square matrix is of maximal rank then it has a left or right inverse, I cannot find the original source I read that gave you a specific inverse

it wasn't too complicated though, something like A^T(A^TA)^-1?

[fair bay]

ah that makes sense

I forget non square matrices exist

[fair bay]

maybe I'm missing something how would that be a ring? since not all non-square matrices are multiplicable?

[rocky cloak]

So ab = 1 implies that multiplication by b is injective. On a finite set injective implies surjective, so there is a c such that bc = 1. Then Ill leave it as an exercise to you to show that a=c

[delicate orchid]

just an idea

we could then embed that non-square matrix into M_n(F) by shoving a bunch of zeros in the appropriate place

[fair bay]

if you embed into the space of square matrices dont you run into the same problem that rigth and left inverses are the same?

[delicate orchid]

jagr has given you an answer so I don't really care anymore

a far smarter answer

[rocky cloak]

I'd say the example of non-square matrices is pretty smart. It just doesn't work sadly

[delicate orchid]

tfw partial rings

or as I like to call them, additive categories

[fair bay]

what's the intuition for ab = 1 implying multiplication by b is injective? I get that it works but I wouldn't have noticed it myself

[rocky cloak]

Well being injective just means being left cancelable. And if you have a left inverse, then you're left cancelable

bx = by -> abx = aby -> x = y

[delicate orchid]

I was just discussing multiplication as a ring endomorphism the other day I'm rather annoyed I didn't see this

[fair bay]

thanks! I think I understand rings better now

[tardy hedge]

Im dying showing associativity for q12

For A+B

[mighty kiln]

If it helps you can draw Venn diagrams

[tardy hedge]

Ok

And then maybe by cases?

[mighty kiln]

Ye

[tardy hedge]

Sounds much better than fking de morgans law

Lol

[stone fulcrum]

Note A+B is the XOR. Something is in A + B + C if it's in either 1, or all of the sets

[tardy hedge]

I was supposed to figure rhat out!!!

Jk thanks for help

[stone fulcrum]

I'm not sure if that helps with the algebra

[tribal moss]

A shortcut is to show that the structure is isomorphic to the set of functions I -> Z/2Z, with pointwise addition and multiplication. Since the latter is associative, your structure is too.

[tardy hedge]

Thats cool

I dont fully understand that yet but i would like to after i finish the question in the more elementary way

[tardy hedge]

I did pictures depicting what the set A+B looks like in 3 different cases, and it showed that something is in A+B if something is in A or B but not both

So i can use that to then easily show that (A+B)+C = A+(B+C) right?

Without algebra

[velvet steeple]

When doing calculations, is there any way to tell WolframAlpha that we're working in e.g. Z/7Z?

[summer path]

mod whatever is probably the closest up can get

[velvet steeple]

Oh

I'm doing a Lagrange polynomial interpolation and

Now adding all those up with appropriate factors would take pretty long, I guess

It seems WA doesn't understand saying mod 7 at the end of adding polynomials

[tribal moss]

Do the overbars mean reciprocals or just congruence classes?

[velvet steeple]

congruence classes

We can leave them out and mod by hand later, right?

[tribal moss]

So they're missing from L1 is just a typo? Yeah, one usually doesn't even write them for that reason.

[velvet steeple]

yep

[tribal moss]

I'd start by handling the denominators (by multiplying by the inverses of 5, 2, 6).

[velvet steeple]

Well, I did it by hand and I got $\overline{4.5}x^3 + \overline{2.4}x^2 + \overline{4.3}x + \overline 4$. I should check if that's correct

$\overline 4 L_1(x) + \overline 3L_2(x) + \overline 3L_3(x) + \overline 3 L_4(x)$ is what I need

[tribal moss]

Are those ... decimals?

[velvet steeple]

I only wrote them here to avoid frac...

[tribal moss]

But decimals don't even make sense in modular arithmetic. You can do addition, subtraction, and multiplication on raw integers and mod out later, but not division.

[delicate orchid]

this is one of the worst things I've read

[velvet steeple]

Well, but we end up with $\overline 4 x^3 + \overline 2 x^2 + \overline 4 x + \overline 4 + \frac{\overline 5 x^3 + \overline 4 x^2 + \overline 3 x}{\overline{10}}$.

[delicate orchid]

ok and you can cancel the last fraction

[velvet steeple]

Yeah, overline 3 in the denom

[delicate orchid]

yeah, and then you can cancel the fraction

[velvet steeple]

Oh

Making overline 12 out of the overline 5

[delicate orchid]

no you just multiply by 3^-1

which is 5

I think

[tribal moss]

Just multiply by 5 in the numerator and denominator.

[velvet steeple]

Oh

[tribal moss]

Since 10·5 = 50 == 1 (mod 7)

[delicate orchid]

you're in a field so there shouldn't be any fractions

(and tbh I don't like writing fractions for a general ring unless you're in a localisation)

[tribal moss]

Hmm ... Q is a field, lots of fractions there.

[velvet steeple]

Let's see if what I got is correct..

,w simplify 4(x^3 + 4x^2 + x + 1) + 3(x^3 - 2x + x)/5 + 3(x^3 - 6x^2 + 5x)/2 + 3(x^3 - 5x^2 + 4x)/6

Hm, nope

[tribal moss]

No, that assumes you're working in Q (or R).

[delicate bloom]

,w 4(x^3 + 4x^2 + x + 1) + 3(x^3 - 2x + x)/5 + 3(x^3 - 6x^2 + 5x)/2 + 3(x^3 - 5x^2 + 4x)/6 mod 7

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by merosity

[delicate bloom]

tries to solve for 0 for shame

[velvet steeple]

Shouldn't it be fine if we mod 7 by hand afterwards?

[delicate bloom]

yep

[tribal moss]

No, because there are divisions in there.

[delicate bloom]

as long as the stuff in the denominator isn't a multiple of 7

[tribal moss]

Okay, point.

I would still feel better about first replacing /5, /2, and /6 by *3, *4, and *6 first.

[velvet steeple]

[L_1(x) = x^3 + 4x^2 + x + 1] [L_2(x) = \overline 3x^3 - \overline 6 x + \overline 3 x] [L_3(x) = \overline 4 x^3 + \overline 4 x^2 + \overline 6 x] [L_4(x) = \overline 6x^3 + \overline 5x^2 + \overline 3x]

[delicate bloom]

I just wouldn't replace them if you're doing it manually with a calculator, just extra work

[velvet steeple]

What do you mean

replace the denominators?

[delicate bloom]

also back up back up

if you want to do this, there's a much easier way in Z/7Z

[velvet steeple]

What is it?

[delicate bloom]

g(x) = 1-x^{p-1} this is 1 for 0 and 0 otherwise

[velvet steeple]

But we'd need to prove that, no?

[delicate bloom]

fermat's little theorem?

[velvet steeple]

oh

[delicate orchid]

roots of unity adding to 1? this would never happen...

[tribal moss]

That's well and fine for Lagrange interpolation with 7 datapoints, but here it looks like there are only 4.

[delicate bloom]

it might not necessarily be minimal degree, but it's clean to get "input to output" you want

I guess it depends on what the goal is here

[open latch]

in the phrase "we can identify the field F with a subfield of K via the map..." what does identify means? Is it talking about isomorphism?

[tribal moss]

Yes, isomorphism plus "in the future we will not bother to write applications of that isomorphism explicitly -- figure out yourself where they should be inserted in the formulas for things to make sense".

[delicate orchid]

there is a subfield of K that is isomorphic to F

yeah

[open latch]

I see, thank you!

The book only proved that it is injective to K_f, but then we can restrict K_f to get a subfield that is surjective hence isomorphic

[tribal moss]

(Or alternatively we could say: suppose f: F -> K is an injective homomorphism, and assume wlog that F and K are disjoint as sets. Then extend f to a map (F union K\f(F)) -> K by making it do nothing to the elements of K\f(F). The extended f is a bijective map, and we can pull back the field structure of K back to F union K\f(F), which then literally becomes a field extension of F that's isomorphic to K.
In particular in situations where K is a field we have just constructed, we can now strip the name K away from what we made initially and instead use the name K about F union K\f(F), so we can say that F is actually a subfield of K.
But the details of doing it that way are a bit unwieldy, so people generally seem to prefer the invisible-isomorphism explanation).

[velvet steeple]

So I got \begin{align*}
L_1(x) &= \frac{(x - \overline 0)(x - \overline 4)(x - \overline 5)}{(- \overline 1)(- \overline 4)(- \overline 5)} = x^3 + \overline 4x^2 + x + \overline 1 \
L_2(x) &= \frac{(x - \overline 0)(x - \overline 4)(x - \overline 5)}{\overline 1(- \overline 3)(- \overline 4)} = \frac{x^3 - \overline 2 x + x}{\overline 5} = \overline 3x^3 - \overline 6 x^2 + \overline 3x \
L_3(x) &= \frac{(x - \overline 0)(x - \overline 1)(x - \overline 5)}{\overline 4 \cdot \overline 3 \cdot (- \overline 1)} = \frac{x^3 - \overline 6 x^2 + \overline 5 x}{\overline 2} = \overline 4 x^3 + \overline 4 x^2 + \overline 6 x \
L_4(x) &= \frac{(x - \overline 0)(x - \overline 1)(x - \overline 4)}{\overline 5 \cdot \overline 4 \cdot \overline 1} = \frac{x^3 - \overline 5 x^2 + \overline 4 x}{\overline 6} = \overline 6x^3 + \overline 5x^2 + \overline 3x
\end{align*}
And so \begin{align*}
f(x) &= \overline 4L_1(x) + \overline 3 L_2(x) + \overline 3L_3(x) + \overline 3L_4(x) \ &= \overline 4x^3 + \overline 2 x^2 + \overline 4 x + \overline 4 + \overline 2 x^3 + \overline 3 x^2 + \overline 2 x + \overline 5 x^3 + \overline 5 x^2 + \overline 4 x + \overline 4 x^3 + x^2 + \overline 2 x\
&= \overline 6 x^3 + \overline 4 x^2 + \overline 5 x + \overline 4
\end{align*}

Hopefully that's correct..

f(0) = 4, ok great

f(1) = overline(19) = 5

That's not correct....

[open latch]

if you don't do anything with the elements of K\f(F) then how can it be injective? there will be multiple elements that map to the same element in K

[velvet steeple]

How do I check where the mistake is now

[tribal moss]

By "not do anything" I mean "return each of those elements unchanged". So two elements of K\f(F) don't map to the same element in the union, and something from F and something from K\f(F) cannot map to the same element either.

[open latch]

Oh then this explains

Instead of doing a restriction on the codomain, you extended the domain, making it bijective, right?

[tribal moss]

Yes.

[open latch]

Thank you for your explanations!

[velvet steeple]

[tribal moss]

So one of your base polynomials must give the wrong result at 1, or else you've botched the arithmetic of combining them.

Fortunately that's something you can test by direct computation in each step of your working.

[white oxide]

why does dividing by |z_1|^2 give you a quadratic equation?

[chilly ocean]

did you multiply out first?

[white oxide]

yeah i wrote down this long ass expression, but i'm hoping that i don't have to say something like z0 = a + bi, z1 = c + di, a = x + yi, etc and just write everything that way

unless that's the only way to verify this

if that's the case then i'll just believe him

[chilly ocean]

just staring at the left-hand side, without expanding, what is the coefficient of t^2?

[white oxide]

z_1 * z_1 conjugate right

[chilly ocean]

that's |z_1|^2

[white oxide]

ohh ok got it, subtract both sides and divide by it and yea you get ur result

cool thanks!

[chilly ocean]

Hello a good book for polynomial problems?

[south patrol]

That is a very broad question

What do you mean by polynomial problems

[tardy hedge]

solve x^2-4 =0

Good luck!

[delicate orchid]

Ok solved it

[glossy crag]

R((t)) (formal Laurent series with finite principal part) make sense over any ring, right.

[delicate orchid]

Yus R is arbitrary

[glossy crag]

And its units are the series that start with a unit of R, right (just like with R[[t]]).

[tardy hedge]

Good job

[delicate orchid]

there may be more considering t itself has an inverse now

haven't thought too much about it

[celest furnace]

This might be better suited for #competition-math

[toxic cedar]

I have to say group theory is much more interesting than I thought and offers insights into modeling computer science concepts

[sweet linden]

dead chat

xd

[dull marsh]

Is this even true? (divisor means subgroup and "in like manner" means using the second isomorphism theorem)

[mighty kiln]

Yes

[dull marsh]

I've started with $R_n T_n / T_n \cong R_n / (R_n \cap T_n)$ where $R_n$ is the subgroup of rotations and $T_n$ is the subgroup of translations, the right hand side obviously is just $R_n$, but I can't see why/how $R_nT_n$ would be equal to the entire group, so should I have a different setup?

[cloud walrus]

A Lonely Bean

[mighty kiln]

Can you find a description of any Euclidean motion in terms of a translation + rotation about origin?

[dull marsh]

Not with my geometric intuition apparently

In R^2, for simplicity, how would reflection around y-axis look like in that form?

[mighty kiln]

Oh hm

perhaps they meant R being rotations and reflections preserving a point

[dull marsh]

By they you mean me (since the author didn't name any of the subgroups) or you mean that the author probably meant to say that the factor group is isomorphic to the subgroup of rotations and reflections?

[mighty kiln]

The latter

Or they defined Euclidean motions to be positively oriented

[dull marsh]

Alright, I'll revisit wherever the definition is, thanks

[rocky cloak]

Yeah, the quotient of the euclidean group by translations should be the orthogonal group (rotations and reflections about a point)
And the quotient of the special euclidean group (orientation persevering maps) should be the special orthogonal group (rotations).

So I guess take your pick for which one of those you want to prove.

[dull marsh]

Can't find the definition in the pdf so I guess I will go with the special Euclidean group

[mossy lintel]

I need a hint for this exercise, I just don't know where to start:(

[rocky cloak]

I guess the first thing to consider is are you able to do it when p is on the form x^n - a?

[mossy lintel]

I think for x^n - a we can construct a chain by first adjoining one of the n-th primitive root, then adjoining one root from x^n-a.

then the result field will be the splitting field for p over F, therefore it's normal.

[rocky cloak]

Indeed, so for things to stay normal you want to make sure you just always add all the roots whenever you add a root of a polynomial. That way you become the splitting field of some polynomials

Other than that you just adjoin nth-roots of things corresponding to how p is solvable by radicals

[mossy lintel]

I'm still kinda confused, we can construct a chain where F_{i} is normal to F_{i-1} with this argument, but i don't quite understand how F_{k} will be the splitting field for some polynomial over F?

[untold basalt]

basis of splitting field

[rocky cloak]

So you want a little more than that. Whenever you add a root you should add all the other roots over F. It's not enough to just have Fi be normal over F_i-1

So for example if we have something like
sqrt(1 + sqrt(2)), then the we construct
F1 = Q < Q(sqrt(2)) < F2(sqrt(1 + sqrt(2)) < F3(sqrt(1-sqrt(2))

[languid trellis]

In the proof of this result, why are we allowed to "multiply inclusions". Can someone explain to me more precisely what's going on in the proof of (1) -> (2) please

Also when we say gHg^-1 = H, this is a set equality above all, so what are we doing when we multiply it? Are we just multiplying each element in each group in turn?

[coral spindle]

The definition is $gHg^{-1} = \build{ghg^{-1}}{h \in H}$.

[cloud walrus]

Boyt

[coral spindle]

In fact, that is explained on the very first line of the proof.

Now as for the proof (i) => (ii), the assumption is for all g, so we can apply it to g^-1 too.

[languid trellis]

I understand, I just still don't get why we're allowed to "multiply this inclusion on the left"

[coral spindle]

By the definition

[languid trellis]

so we multiply each element in the set {ghg^-1, h \in H}

[coral spindle]

Yes

You know that $g^{-1}hg$ is in $H$, so let's call this $x = g^{-1}hg$. So $gxg^{-1} = h$. We described this generically so this shows $H \subseteq gHg^{-1}$.

[cloud walrus]

Boyt

[languid trellis]

I see

That makes sense now

I was being a little pedantic I suppose by wondering where artin defined scalar multiplication of sets lol

[coral spindle]

But he defined it at the start of the proof

Whether or not this respects set inclusion was not proved.

[languid trellis]

hm

conjugation is a group automorphism would be another way to see 1 \implies 2 no?

Because certainly the image of H under the automorphism is H because H is normal, and we know there are an equal amount of elements in the preimage H and image H as an automorphism is a perfectly good iso of sets.

[delicate orchid]

no - normal subgroups are not automatically characteristic (closed under automorphisms)

[languid trellis]

I have never felt more humbled than asking anything on this server lol, but I appreciate you both, thanks wew and boyt

[tardy hedge]

5.2.11

Not sure how and why Z/ker being integral domain implies ker is 0 or nZ

Ohh wait

Is that not the implication? Is it just that ker(phi) for a homomorphism out of Z must be always either 0 or nZ?

[south patrol]

Z/nZ is an integral domain iff it's a field

except n=0 lol

I think they should've mentioned this lol like if Z/nZ is an integral domain, then either n =0, or n is a positive prime

[south patrol]

oh i mean

[tardy hedge]

I know that part

[south patrol]

any ideal of Z is of the form nZ

for some n

[tardy hedge]

Im just stuck on why ker phi is 0 or nZ

[south patrol]

(possibly 0)

it is a PID

[tardy hedge]

Ideals etc are introduced next section

[south patrol]

Oop

[tardy hedge]

But additive subgroups of Z must be nZ

So it just follows from that eh?

[south patrol]

Oh i just read what you're trying to prove lol

[south patrol]

yeah

(an ideal of Z is exactly an additive subgroup anyway so yeah)

[tardy hedge]

Yeah i think i get it now. I think i was initially confused cause i thought somehow Z/ker being an integral domain is what implied ker = 0 or ker = nZ

[delicate orchid]

that is quite the lol

[tardy hedge]

But that wasnt the case right?

[south patrol]

Yeah nah i mean that already holds just cause of what you know about subgroups

[tardy hedge]

Yeah

[delicate orchid]

introducing quotient rings without the objects required to make the notion rigorous is an interesting choice

[tardy hedge]

Yeah the book does that sort of thing the entirr way through

I actually dont mind it

Introduces the idea before the real rhing

[south patrol]

yeah saying "fundamental homomorphism theorem for rings" lol

but yeah ig quotient rings are easy after quotient groups have been covered

[tardy hedge]

This is also basically saying any integral domain is isomorphic to Zn for n prime?

Nvm sorry

[coral spindle]

No! Absolutely not no!

The characteristic of a ring does not mean that it is Z/nZ

[tardy hedge]

Damn bruh

[coral spindle]

It means that it contains a copy of Z/nZ in particular

There are infinite fields of characteristic p.

So clearly these cannot be isomorphic to Z/pZ

[tardy hedge]

Ya i said nvm cuz i noticed the isomorphism was phi(Z)

Not with all of D

[untold basalt]

How can I prove that $x^4 + 2$ is the minimal polynomial of $i\sqrt[4]{2}$ in $(\mathbb{Q}(\sqrt[4]{2}))[x]$?

[cloud walrus]

Seagull

[rustic crown]

you can't

(since it's actually x^2+sqrt2)

[untold basalt]

sorry, X^4-2

Does that change things?

[empty rose]

well that's also not x^2+sqrt2 so still no

[untold basalt]

hold on yeah it's 2 the degree

Ok so since the degree is two once I found it it's obvious it's minimal

[mossy lintel]

So you want a little more than that.

[delicate orchid]

oh yeah? watch this

[uneven island]

Hi, isn't it okay to say that a subgroup of a group is a subset that is a group as a definition ?

[mighty kiln]

A subset that is a group under the same operation (restricted to the subset)

[uneven island]

yes

that's what i said i guess

i asked this because i've seen many textbooks define a subgroup by a subset that verifies the group properties. but they never shorten it by just saying that it's a subset that is a group

[crystal turtle]

Because it needs to be a group with the restriction of that operation, that part is very important

[uneven island]

hmm, so it's okay to say it is a subset that it is a group under the same operation ?

[south storm]

A subgroup of a group $(G,\cdot,^{-1},1)$ is a set $H\subseteq G$ s.t. $(H,\cdot|_{H^2}^H,^{-1}|_H^H,1)$ makes sense and is a group.

[cloud walrus]

27182818284tropy

[south storm]

This is one way of formulating it

Where |_A^B denotes birestriction

[uneven island]

oh now i think i see, the problem is with the birestriction!

[south storm]

Indeed, one is to verify closure under the operations, which is exactly the meaningfulness of the birestriction and containment of the neutral element.

[uneven island]

nice, thanks!

[south storm]

This is the situation for any algebraic structure more generally

[uneven island]

thanks all for clarifying : )))))

[untold basalt]

why exactly is it that if u is trascendental on R[x] then R(u) is isomorphic to the field of quotients of R[x]? (R ring)

[coral spindle]

This is not relevant to the subject of the channel.

Please read channel descriptions before posting.

[coral spindle]

(Someone deleted their post)

[tribal moss]

Hmm, what does "u is transcendental on R[x]" mean? Should it have been "u is an element of some field that extends R, and is transcendental over R"? If it doesn't, then what exactly does "R(u)" mean?

[primal tusk]

Not trying to interrupt but what shape represents the symmetry group of order 4? I can't seem to figure it out I've found 1-3 and 5,6

and also how is a Dihedral group different than a symmetrical group?

[delicate orchid]

Canonically, 4 points

Non-canonically, the cube without reflections

I’m impressed you managed to find 5 and 6 considering I’m not even sure what the lowest dimension polytope would be off of the top of my head

Oh wait duh, 4 and 5 dimensional

[primal tusk]

maybe i am not doing the same problem

i said 5 is a cube with different length sides for all dimensions

and what i mean is every possible symmetry

S1 = point

S2 = line

S3 = rectangle

S4 = ??

[rocky cloak]

The tetrahedron has symmetry group S4 (if you include reflections)

[delicate orchid]

Well yeah you’re describing the action of S_5 on the set of basis vectors of R^n

[delicate orchid]

No? That’s C_2xC_2

[rocky cloak]

And the rotational symmetries of a cube are also S4

[primal tusk]

rectangle has two reflections and one rotation i thought

[delicate orchid]

Which isn’t 6, it’s 4

The symmetric group should be thought of as rearranging n points

Without structure

[coral spindle]

1. it has more than one rotation 2) you seem to have forgotten about the identity element.

[primal tusk]

ohhh true

[delicate orchid]

No it really does only have one rotation, by 180

[primal tusk]

but true i was just trying to see if there was a pattern when using shapes to represent

[coral spindle]

Oh wait I do see how you could say it has just one rotation. The other one is the identity

[rocky cloak]

Rotation by 0 degrees 😉

[primal tusk]

wait so i dont have a shape for S3 now

[delicate orchid]

S_n canonically acts on the basis of R^n by permutations so you can take the cube in n dimensional space for example

[rocky cloak]

Triangle is one

[coral spindle]

Do you mean S_3 or C_3

[delicate orchid]

The triangle

[primal tusk]

a triangle is S6 i thought

[coral spindle]

No

S_3 has 3! = 6 elements

[delicate orchid]

How on earth could a shape with 3 sides have 720 symmetries

[coral spindle]

Are you sure you don't mean C_3

If you don't know, just say so

[primal tusk]

i dont know im like just learning this stuff now

[coral spindle]

OK

[primal tusk]

sorry for the confusion

[coral spindle]

S_3 is the symmetric group on 3 elements. It consists of all permutations of three elements.

C_3 is the cyclic group of 3 elements

Typo, let me correct that

|S_n| = n!, but |C_n| = n

C_3 is the set of rotational symmetries of the triangle. If we only think of rotations, this is what we get.

S_3 is the set of all symmetries of the triangle, but this pattern doesn't continue for higher S_n.

If you want a description of S_n as some set of functions, like I said it is all permutations of n elements.

Clear?

[tardy hedge]

Any integral domain has a subring isomorphic to Zn?

[coral spindle]

No, e.g. Z.

[primal tusk]

The way of describing them makes sense but I swear you can flip the triangle to make symmetries

[coral spindle]

Yes, which is where we get the other elements in S_3.

[primal tusk]

ohh

okay i see now

[tardy hedge]

Oh. Any integral domain with nonzero characteristic has a subring isomorphic to Zn

Better?

[coral spindle]

Yes

[tardy hedge]

Thanks

[coral spindle]

But this holds for any ring.

Not just integral domains.

[tardy hedge]

Ah

[coral spindle]

In fact if char R = n then there is a copy of Z/nZ in R.

This even works for n = 0 :)

[tardy hedge]

Coool thanks

[coral spindle]

Oh the benifits of notation...

[primal tusk]

okay but on the basis of a point having 1 symmetry, line has 2 symmetries, what are these groups called?

C(n) ?

[coral spindle]

S_1, S_2?

But these are isomorphic to C_1, C_2

So

Bleh

[tribal moss]

How does the sequence "point", "line" continue for you?

[coral spindle]

There's only one group of order any prime.

[primal tusk]

well initially i thought rectangle had 3 symmetries but it has 4, and a rectangluar box has 5, a triangle has 6

so its how can you pick up a shape and move it and put it in a hole with that shape for 0-2D

and 3D too i think

[tribal moss]

I'm trying to understand your question up here, and just those two examples don't help me understand which sequence of groups it is you want a name for.

[rocky cloak]

Now I'm a little curious.
We have
S1 symmetry group of point
S2, line
S3, triangle
S4, tetrahedron / rotational symmetry of cube
S5, icosahedron
but what about S6, is there some nice geometric object with S6 as symmetry group?

[primal tusk]

i was thinking about using a mirror to duplicate the points. So you hold a mirror in front of a point and make a line. If you hold a mirror in front of the line you make a square or rectangle depending on the distance of the mirror

[tribal moss]

A rectangular box (with three different side lengths) has eight distance-preserving symmetries, not 5.

[primal tusk]

and i wanted to see if there was a pattern

[rocky cloak]

I guess by the classification of finite subgroups of SO(3) it can't be a 3dimensional object....

[primal tusk]

Triangles in a way kind of messed up what I was thinking about but I was curious about the concept of "using a mirror" to make symmetries

[primal tusk]

three 180 rotations and 3 flips and what else?

[tribal moss]

You can flip or not flip each of the three dimensions independently.

[rocky cloak]

So you are sort of close to a pattern here, that Sn is the symmetry group of the n-1 dimensional simplex.

That is, S1 is the symmetry group of a point, S2 is the symmetry group of a line segment, S3 is the symmetry group of a triangle, S4 is the symmetry group of a tetrahedron, and S5 is the symmetry group of some 4dimensional generalization of a triangle.