groups-rings-fields

[frank cosmos]

i feel like this requires the condition of normality...

[tender wharf]

send original question

[dawn latch]

If * is a binary operation on set A, given a ∈ A and e is the identity element, does a * a⁻¹ equal e?

[tender wharf]

is A a group?

[dawn latch]

no sir, just an arbitrary set

let's say ℤ

[tender wharf]

unless you verify the existence of inverses, not necessarily

and Z is a group under + lol

[south patrol]

• could just be basically anything unless you have further assumptions

[dawn latch]

oh ok

[frank cosmos]

Suppose you have finite subgroups $H_1,\ldots,H_n$ whose elements commute with each other, such that $H_1 \dots H_n=G,$ and such that $H_{i+1} \cap (H_1 \cdot H_N)=e.$ Then, G is isomorphic to $H_1 \times \cdots \times H_n$

[dawn latch]

thanks

[cloud walrus]

Hello1

[frank cosmos]

Show using induction ^

[tender wharf]

they are necessarily normal

[frank cosmos]

oof ok

[tender wharf]

if they commute with each other the subgroups are normal

yeah

[frank cosmos]

also i dont even need induction then?

[tender wharf]

I haven't thought that far lol

[frank cosmos]

alr

oh wait

im stupid

if you have $x_1 \cdots x_n=e,$ then $(x_1 \cdots x_{n-1})x_{n}=e$ which implies that $x_{n}=e$

[cloud walrus]

Hello1

[frank cosmos]

now i want to use induction to show that x_{n-1}=e as well

but, i only have the condition of disjointness, H_1,...,H_{n-1} do not product to make G

so can i still use the idnuction hypothesis?

[median pawn]

where does the last bound on alpha_i come from?

[formal ermine]

yeah but how does that help us

[primal tusk]

what is the cancellation property?

[chilly ocean]

that things cancel

a + c = b + c implies a = b

[primal tusk]

ah okay thank u

[chilly ocean]

where a, b, c are elements of some set and + is some operation

you learned an example in elementary school

[lusty marlin]

Since m ∈ ℤ+, shouldn't it directly follow that:
|alpha_i|≤m|alpha_i|
≤m×max|alpha_i|=mB≤max{1,mB}

But then this seems too trivial to be explicitly stated, which probably means I've completely misunderstood what is going on here.

[chilly ocean]

I was thinking once, because in any semigroup we have its left/right translations

and for monoids for example, this is actually a representation

so we can consider semigroups which can be represented as left/right translations of themselves

and if you take any small category, then by adding an element to it, you can make it into a semigroup which has the above property

just send composition of morphisms to that additional element when it's undefined

I mention this because it reminded me of the property of being cancellative

it's recreational mathematics, I doubt anyone cares

[chilly ocean]

for a) does it suffice to say that $g = g^{-1} \implies gg = gg^{-1} \implies g^2 = e \implies |g| = 2$

[cloud walrus]

n!

[chilly ocean]

for the left-sided proof

[formal ermine]

yes

[chilly ocean]

and for the right-sided proof (since this is an iff statement) we have $g^2 = e \implies g^{-1} g^2 = g^{-1} e \implies g = g^{-1}$

[formal ermine]

and 2 is minimal because otherwise g would be e

[cloud walrus]

n!

[formal ermine]

yeah

[chilly ocean]

thanks

for b)

g^n = e

$g^n = e \implies g^{-n} g^{n} = g^{-n} e \implies (g^{-1})^n = e$

[cloud walrus]

n!

[chilly ocean]

right?

thanks

what about c)

I'm stuck on it

Suppose $G$ is finite with order $2n$, then $G$ contains an element of order $2$

[cloud walrus]

n!

[chilly ocean]

can I just use Lagrange's theorem?

because o(g) | o(G)

[rustic crown]

Two coprime things dividing the same number that means the product divides it. This let's you calculate all the degrees in the extension diamond.

[formal ermine]

this seems intuitively true but how do we prove it

[tender wharf]

lagrange's theorem doesn't state the existence of such an element

[chilly ocean]

yeah thats what i thought

[tender wharf]

of course, this is true by another theorem

[chilly ocean]

OH

the cyclic subgroup generated by an element

right?

[tender wharf]

hmm

not quite

what's your idea

[chilly ocean]

my idea is

ok nvm

it doesnt imply that the order is 2

my bad

[tender wharf]

nw

[chilly ocean]

hm

[tender wharf]

let me think of how to hint at this without giving too much away

[chilly ocean]

ok

[tender wharf]

2 is prime. now

are there theorems that come to mind

okay that's a lot cleaner

but you gave too much away

[chilly ocean]

Euclid's lemma?

[lusty marlin]

yeah sorry, should I delete it?

[tender wharf]

its too late so whatever

there's a rather famous french mathematician who has a lot of analysis things named after him

[lusty marlin]

from the fact that they're still asking questions, it may not be too late

[tender wharf]

there's a theorem he made

sure

[chilly ocean]

yeah delete it i havent read it properly

but thanks for trying to help @lusty marlin

[lusty marlin]

np

[tender wharf]

knightwatch ill let you take over alright

your solution is quite a bit cleaner than mine

also im popping off coincidentally to study some algebra

[lusty marlin]

nice

[tender wharf]

I was about to overkill it lmao

[chilly ocean]

so do you have any hints knightwatch

[lusty marlin]

hmm

the order of the group is even

[chilly ocean]

right

which can be written as 2n

[lusty marlin]

so there is an even number of elements

and a set having even cardinality means that one can...

partition it into pairs of elements

I've given away a bit too much now, it is just a step away.

[chilly ocean]

disjoint pairs of elements?

[lusty marlin]

pairs, the word partition implies disjoint. It was redundant

[chilly ocean]

so I should pair the elements with one another?

[lusty marlin]

Yes, but what should the rule for pairing be?

[chilly ocean]

good question

hm

pairing each element with its inverse?

[lusty marlin]

Ok, and what does that achieve?

[narrow marsh]

Some elements can be their own inv (which is guaranteed in this case)

[chilly ocean]

not sure, that's just my only thought

[median pawn]

noo wait

B is the max of the coefficients, not the roots

so the bound isn't trivial

[lusty marlin]

yes, thank you for that. a and α should never be used together

[median pawn]

ikr

[chilly ocean]

@lusty marlin I found a way to use Lagrange's theorem to solve this

would you like to see

[lusty marlin]

nah it's fine, I have seen this question in the past

[chilly ocean]

well thank you for the help regardless

appreciated

[lusty marlin]

did you get how pairing elements would give you the result?

[chilly ocean]

I did not

I understand it using L's theorem

[lusty marlin]

Ok, do you want to know?

[chilly ocean]

I do

my only question is why does p not divide 2

[lusty marlin]

So by partitioning the set into sets {a,b} where a and b are the inverse of each other, one finds that the identity element is put in a singleton set by the partition. But since the group has even order, there must exist another element that is in a singleton set, which means that it is its own inverse, which means that squaring it gives identity, which means that it has order 2.

[chilly ocean]

ah

that makes sense

[chilly ocean]

I'm still a bit confused about this

[tender wharf]

if p = 2 we are done

so suppose p > 2

[chilly ocean]

oh

gotcha

[lusty marlin]

Assume |α|≥1
|α^m|=|∑a_i× α^i|≤
∑|a_i×α^i| ≤ ∑|a_i|×|α|^(m-1)

Where the summations are from 0 to m-1
Divide by |α|^(m-1) throughout to get |α|≤∑|a_i|
Which is ≤m×max|a_i|

And if |α|≤1, the bound trivially follows.

[rustic crown]

prove what? sorry i just came back home

a | n, b | n and gcd(a, b) = 1 then ab | n?

more precisely, lcm(a, b) * gcd(a, b) = a * b

[rotund aurora]

So I was trying to prove that unique factorization of ideals in an integral domain $\implies$ Dedekind domain. In the strategy I followed I arrived at a point where I have to show that if p and q are (nonzero) primes such that $p\subseteq q$ then $p=q$. I'm not exactly sure how to conclude that (notice that we cannot use the fact that $p\subseteq q\iff q\mid p$ as ideals, because that's what we are trying to prove). One possible proof that it occured to me is to use Krull's intersection theorem, that would tell us that there exists an $n$ such that $q^n\subseteq p$, and since $p$ is prime, we would have $q\subseteq p$ and we would be done. But this feels overkill.

[cloud walrus]

Croqueta

[rotund aurora]

I'm not sure what's an easy way to show this

Localization also seems not helpful at all here

mmh actually not even Krull intersection gives that

q-p forms a multiplicative system actually

[formal ermine]

yeah

[rotund aurora]

you can just use the fundamental theorem of arithmetic

[rustic crown]

write ax + by = 1

then a(nx) + (ny)b = n

so ab | n

[tribal moss]

How do you get from the penultimate to the last line there?

[rustic crown]

b | nx and a | ny by assumption

so ab divides both a(nx) and (ny)b

[tribal moss]

Okay.

[rustic crown]

or tbh, just use prime factorization. you can assume that Z works nicely instead of proving everything from axioms of ZFC.

[formal ermine]

yet another application of bezout

thanks

[rotund aurora]

To prove the fundamental theorem of arithmetic, you would have to use Bezout anyway, so I guess its easier to just do Bezout

[rustic crown]

yea

wait so i wonder if there is a way to prove bezout from div alg

oh ofc

we can do the algorithmic proof by extended division algorithm

[rotund aurora]

yes

Euclid

[rustic crown]

i was confused if we still needed to know that N is well ordered or not

but was a weird confusion

[rotund aurora]

uhm I think you do need that tho

[rustic crown]

because ofc when i say division algo, we mean the valuation to N

like i was thinking of taking the smallest positive element in the set aZ+bZ

weird thought

nvm

ignore me

[formal ermine]

is the converse true too?

[rustic crown]

nah

[formal ermine]

sadge

[rustic crown]

[Q(sqrt(2), sqrt(3)):Q]=4

L = Q(sqrt2), K = Q, alpha = sqrt3

[rotund aurora]

If you have a ring extension $R\subseteq S$ and $M$ is an $S$-module, free as an $R$-module, is $M$ projective as an $S$-module?

[cloud walrus]

Croqueta

[rotund aurora]

M finitely generated over S

[rustic crown]

probably not

R = k, S = k[x], M = k[x]/(x^2)

for f.g./pid, projective => free

[rotund aurora]

where can I quickly learn some facts about projective modules

[rustic crown]

you only need to know 3 things on top of yoru head. the definition, direct summand of free, and can be used to compute derived functors

[somber sleet]

hey guys, why are the only primeideals contained in Z (0) and (p) for p a prime number?

[rustic crown]

do you know that Z is a pid?

[somber sleet]

I guess so

[rustic crown]

yee, so the prime ideal would look like (n) for some integer n

what conditions do you have on n, such that this is prime?

[somber sleet]

n is not a unit and for all x,y in the ring R, n|x or n|y

[rustic crown]

and that's one way to define what a prime number is

(this definition included 0 as a prime number)

but if you wanna show the other thing, which is usually called "irreducible" say n was non-zero and n = ab. from this you want to conclude exactly one of a or b is a unit.

[somber sleet]

Another question following, we actually defined a prime ideal of R a ring as a proper ideal p in R with: for all x,y in R: xy in p ->(x in p or y in p)

[rustic crown]

right

[somber sleet]

are both definitions equivalent?

[somber sleet]

like has the p mentioned here always the form (p), for a prime number?

[rustic crown]

an element p in R is a prime element if and only if the ideal generated (p) is a prime ideal

[rustic crown]

nah. there may be prime ideals which are not generated by a single element. the simplest example of such is for the ring R = Q[x,y] and the ideal P = (x, y)

it requires a little effort to see that you can't write P = (f)

[somber sleet]

but in Z all the prime ideals have the form (0) or (p) right?

[rustic crown]

yep!

because Z is a pid

every ideal is principal here

[somber sleet]

now it makes much more sense, thank you so much!!

[rustic crown]

[rustic crown]

these things took me quite some time to digest, so don't worry too much if you're just learning. projectivity would make a lot more sense when you do homological algebra

[rotund aurora]

I was doing NT and some exercise asks me to show that something is projective. I have never done anything related to this lul

[rustic crown]

happens :p

[rotund aurora]

I think I can try tho

the fractional ideals are fintiely generated O-submodules of K, where K is the field of fractions of O

[rustic crown]

lol i was thinkign about something similar today

[rotund aurora]

and like these fractional ideals are free as Z-modules I think, unless I'm smoking

but idk if its related

[rustic crown]

yee that's true

ig what's more related is that each fractional ideal is invertible

[rotund aurora]

A is integrally closed in K, L/K a field extension and B the integral closure of A in L

K field of fractions of A

[rustic crown]

i can't think of a simpler proof lol, and using this feels like an overkill

hopefully i revisit some alg nt after reading more commie alg

[rustic crown]

(in a slightly different language, so fractional ideals correspond to line bundles and i know line bundles are vector bundles of rank 1, and vector bundles correspond to f.g. projective modules)

[rustic crown]

are you okay with using this?

i can tell you the proof i have in mind

[rotund aurora]

yes, at this point unique factorization is well set

and that the fractional ideals form a group etc

[rustic crown]

nice nice

[rotund aurora]

Im working through some examples, tho you can say the proof. I'll read it if I can't do anything

[rustic crown]

so wlog assume that a is an ideal in O

and find another ideal b such that ab is principal

you have a map,
a --> N

so get the map
ab --> bN

since left side is principal, this map looks like cO --> bN

bM --> bN is also surjective

so we sort of reduced the problem to the case where the ideal is principal

so as a module it looks like O

you have O --> N'

see if you can lift this to M'

and then convert back

that's the idea in my head

(phrased differently, if a is a fractional ideal and b is its inverse, then
you have an equivalence of categories O-Mod --> O-Mod given by tensoring over O with a and coming back by tensoring with b. so starting with a --> N tensoring with b would make this O --> b⊗N, and then you lift it here, O --> b⊗M, and now you go back by tensoring with a, to get a --> M)

maybe there is a much better proof >.<

chmonkey will enlighten us soon

[chilly ocean]

can I send my question from earlier? if you guys are done, that is

[rustic crown]

yep we done

[chilly ocean]

det this is the q I had from earlier that you helped me with

I couldn't really get anywhere, though

[rustic crown]

okie

so

the cool fact about D_n is that if r is the rotation of order n and f is any reflection of order 2

then (r^i)f are all reflections

in particular rf is a reflection

so rfrf = 1

[chilly ocean]

r^if?

[rustic crown]

which is same as frf = r^-1

[rustic crown]

oopsie

for i = 0, 1, 2, ..., n-1

[rustic crown]

which is further same as fr = r^-1 f

so you can push the r to left of f, at the price of inverting it

[chilly ocean]

rfr = f

[rustic crown]

see if you can picture this in your head

for a rigid motion of a regular n-gon, you just need to keep track of 2 consecutive vertices to determine the whole transformation

number the vertices 0, 1, ..., n-1

[chilly ocean]

but shouldn't we generalize this to order n

[rustic crown]

didn't get you

r has order n, no?

reflections are always order 2

if you do them twice, you do nothing

[chilly ocean]

oh

right

my apologies

didn't exactly internalize what you said

[chilly ocean]

but why do we care about i here

oh

I misunderstood

[rustic crown]

i was just saying how D_n looks like

[chilly ocean]

right

[rustic crown]

it has a subgroup Z/nZ generated by r

[chilly ocean]

should I try to look at D_3

and D_4

[rustic crown]

which are rotations

[rotund aurora]

The idea is pretty cool, but when you are writing bN you are really writing the tensor product no?

[rustic crown]

and rest half of the things are all reflections

[rustic crown]

i had tensor product in head, but idk i felt bN and b⊗N would be isomorphic. did i make an error?

[rotund aurora]

well if its not a tensor product then idk how you interpret bN

[rustic crown]

product of an ideal with a module

[rotund aurora]

oh right b is in O

[rustic crown]

IM = submodule generated by im

[chilly ocean]

oh I see

[rustic crown]

(yea that was teh entire point of forcing both a and b are ideals in O)

[chilly ocean]

aba is just a^{-1}

[rustic crown]

else them being fractional is fine if we look at tensor products

[chilly ocean]

so aaba = 1

so we have $a^{-1}b^{-1}(ab)^3$

[cloud walrus]

isomorphism

[rustic crown]

wait i think there is an error here

[chilly ocean]

how come

[rustic crown]

you should have abab = 1

not aaba

a was the rotation right?

[chilly ocean]

this is the solution

but like

I don't want to memorize this property in D_n

I want to be able to derive it

so I can solve problems without having to memorize certain properties of groups

oh

isn't the answer just a^3 b^3

why is it a^6b

[rustic crown]

usually people think of D_n through it's presentation... so won't actually call it like memorization

some people would define D_n = <r, f | r^n = f^2 = (rf)^2 = 1>

[chilly ocean]

this doesn't make sense to me, why is r^n = f^2 = (rf)^2 = 1

you mean r has order n

?

[rustic crown]

cause although it's nicer to define it has the rigid motions of a polygoan, making that precise takes some work which can be avoided

[rotund aurora]

I forgot to reply to this, the point of this is simply that O is free and the map is surjective. We just want f(h(1))=g(1), since f is surjective we can send h(1) to a preimage of g(1) and done

[chilly ocean]

oh

[rotund aurora]

Free --> Projective is easy

[chilly ocean]

for this we have $a^2 b a^{-1} b^{-1} a^3 b^3$ where $a$ is a rotation of order $n$ and $b$ is a reflection.

[cloud walrus]

isomorphism

[chilly ocean]

and we want to leverage

bab = a^{-1}

[rustic crown]

right, and the best way is to focus on that ba^-1b^-1

because it gets rid of all the b's in the middle instantly

[chilly ocean]

(ba)^2 = 1

[rustic crown]

yee

[chilly ocean]

so.

thus.

oh

that's quite clever

(ba)^-2 = 1

thank you det

I got it from here

[rustic crown]

[chilly ocean]

should I just memorize this property that frfr = 1

[rotund aurora]

@rustic crown

This is from Clark btw

[rustic crown]

if you can quickly think why fr is also a reflection, then no need to. but you'll be used to it in no time

[rustic crown]

that's so slick lol

i need to think why that map splits though

[rotund aurora]

mmh but I'm confused, you just used that ab is principal (which you can easily get by inverting a and multiplying by a suitable principal ideal in O, and that's b). No?

[chilly ocean]

gotcha

thanks

again

[rustic crown]

yep

if you're comfortable with tensor products, then just use that language lol

i was trying to avoid it just in case

:p

[chilly ocean]

aba = b right

[rustic crown]

yep

cause b^2 = 1

[chilly ocean]

not abab = 1

[rustic crown]

same thing

[chilly ocean]

how come

[rustic crown]

ababb = b

aba = b

[chilly ocean]

o

i feel dumb

[rustic crown]

//patpat

[chilly ocean]

thanks for the 10th time today det

[rustic crown]

[rotund aurora]

So with Clark, given an ideal a you want to find an ideal b such that ab is principal and a,b coprime. You can do that by CRT: look at the ideal class of a^{-1}, by CRT you can pick b in O such that (b)a^{-1} is an ideal in O and coprime to a

Thanks @rustic crown

[rotund aurora]

Let $p$ be the minimal polynomial of $\zeta$, suppose it has degree $n$. How to prove that
[
\text{Tr}_{\bQ(\zeta)/\bQ}\left(\frac{\zeta^{n-1}}{p'(\zeta)}\right)=1
]
?

[cloud walrus]

Croqueta

[past path]

Is there a notation like $f:A\to B:a\mapsto b$ that explicitly means $f$ is a homomorphism as opposed to a general function?

[cloud walrus]

Amizar

[rotund aurora]

I dont think so, just say it

[past path]

I'm in the process of extending existing notation to include more algebraic structures, e.g. subrings and ideals etc.

And I'm making my best effort to stay along the same lines as existing notation.

[rotund aurora]

notation for what?

[past path]

For example, instead of saying $A$ is a subring of $B$, one could write (please hold while I take a screenshot, as the symbol here only exists in my own LaTeX package.)

[cloud walrus]

Amizar

[rotund aurora]

this is ugly

[past path]

It's rather low-res

[rotund aurora]

There's no point in doing what you are doing, in fact, things will become much less clear. Its easier to say "its a subring/group/algebra/field..." than to have distinct symbols for every algebraic structure

[past path]

I believe you're mistaken on every single thing you've said.

[rotund aurora]

rings fields and so on have been around for plenty of years

and most mathematicians didn't think there was a need to employ different symbols for each algebraic structure. Are you saying they were wrong?

[past path]

But this is not the place for that conversation. I'm not sure what is, but I'd rather not clog up #groups-rings-fields unless this is in fact the right channel for this conversation.

[rotund aurora]

mmh you are right

[past path]

Is there a better place to have this conversation/discuss notation?

[rotund aurora]

Probably #math-discussion

[formal ermine]

this channel is for conversations

[past path]

Oh, is this actually the right place?

[formal ermine]

Yes

as long as it's related to abstract algebra

[past path]

While on the topic, do you have any clue if there's a functional notation for homomorphism?

[formal ermine]

lol

notation creates ambiguity

words don't

[past path]

We have modifiers that imply injectivity, surjectivity, and a host of other properties

Then why have any notation at all?

Why not just say

one plus one equals two?

[rotund aurora]

ok so there's a proof of what I was looking in Lang. It didn't occur to me to look until now lmao. Lang is a king

[past path]

Why not just say "A is a subgroup of B that is not equal to B" instead of $A\lneq B$?

[coral shale]

You mean the other way round

words can create ambiguity

well and properly defined notation doesn't.

The issue here is convincing anyone that such notation is necessary

It probably doesn't serve much purpose and obfuscates the meaning

[formal ermine]

what definition of ambiguity are you using

[coral shale]

How are symbols ambiguous?

[cloud walrus]

Amizar

[coral shale]

You define them in such a way that a program could interpret it

[formal ermine]

they can mean different things depending on the context

[rotund aurora]

Notation should not be obscure, that's the point. Definitions should be introduced only when necessary.

[coral spindle]

This conversation is veering much further away from abstract algebra

[coral shale]

You're talking about overloading which is a separate issue.

[past path]

That's exactly the problem and why I'm defining non-ambiguous notation.

Exactly

You're making my point for me.

[coral shale]

Anyways - why not use =<

for groups, rings, fields

[formal ermine]

who says your notation is the standard

[coral shale]

whats wrong with the same symbol or all?

[past path]

The point is there is literally no existing notation.

[coral shale]

given context, there won't be confusion

imo

[past path]

I don't hate that, but it's still not fully satisfying imo.

[coral shale]

rather than inventing new notation

[past path]

We could go all the way to subsets

[rotund aurora]

What Boytjie said, but the point of notation is not just to make things unambiguous (in fact, I'd say that is not that relevant)

[past path]

Why distinguish between subsets and subgroups at all?

[coral shale]

In a way, subsets can cause confusion because sometimes you talk about the underlying set and want to use subset

Whereas in the case of fields and rings, you will not talk about the hidden group structure

and talk about one being a subgroup of the other when considering only one operation

[past path]

That's not always true.

[coral shale]

That rarely ever happens unless you make it explicit

[past path]

Particularly when it comes to rings and fields.

[coral shale]

No, not really

[past path]

You can certainly talk about a subring of a field.

For example, the integers contained within the rational numbers.

[coral shale]

Ok I agree with that

fields are rings

but fields and rings are not groups

so perhaps one more symbol only

[past path]

I agree that it's far less common to need to discuss the group structure of a ring

but

I would argue that there's potential for discussion nonetheless.

Like, for example,

in module theory

One can confirm that an $R$-module $B$ is a submodule of some $R$-module $A$ if $B$ is an additive subgroup of $A$ and is closed under multiplication by elements of $R$.

[cloud walrus]

Amizar

[coral shale]

Even if you insist on introducing symbols for each, I would not use something completely different

[past path]

Anyway, I'm not actually here to make that case right now. I've already defined a good amount of notation and am still in the process of defining more.

Oh it's not completely different at all

In fact

[coral shale]

use =<_G, =<_R, =<_F maybe \frak for G, R, F

[past path]

it's extremely similar and follows intuitively from the symbols themselves.

I honestly don't hate that idea and might have done something similar if I'd spoken to you before I started.

[delicate orchid]

call a group G
…

[formal ermine]

wew did you ignore my ping

[past path]

The fact that subsets and subgroups already use distinct notation led me to consider different but similar approaches to the notation

[delicate orchid]

Probs I was out

[past path]

For A is a subring of B, I used

[formal ermine]

check your pings and answer pls

it's rep theory related

[coral shale]

i think thats just horrible to do in writing

[past path]

What do you mean?

[coral shale]

i wouldnt want to write this

[past path]

It's literally just $\leq$ with a circle

[delicate orchid]

Yus

[cloud walrus]

Amizar

[coral shale]

in the context of lecturing, i dont think this is easy to draw (well)

[barren sierra]

it's not standard notation but I have seen \leq also used for subring

never seen circle leq

[past path]

I mean, it's really not that difficult. I do it all the time on a whiteboard.

Or on paper.

[barren sierra]

If you really need to make the distinction between subgroups and subrings in the same context then I guess it's a fine symbol

[coral shale]

yh, I think its rare u need to

[past path]

I agree that it's uncommon, but one advantage is that it allows us to immediately consider A and B as rings as opposed to "Let A and B be rings such that A is a subring of B"

It condenses that whole thought down to 3 characters.

[formal ermine]

name on example where you need to distinguish them

[coral shale]

No it doesn't.

[formal ermine]

also you can just say B/A is a ring extension

[past path]

In what way does it not allow us to do exactly that?

[coral shale]

A =< B does not condense down to 'let A subgroup B'

without first defining at least one of them as a group

[past path]

You're right

However

The only reason it doesn't

Is because when subgroup notation was being defined

They used already existing notation that created ambiguity

If you say "Let $A\subseteq B$"

[cloud walrus]

Amizar

[past path]

Then it immediately contextualizes A and B as sets.

[coral shale]

Even so, its rare you would ever do this except where the context is very clear

what kind of sets are A, B? its usually going to be said

[past path]

The ring notation, as it does not conflict with any existing notation, would contextualize A and B immediately as rings.

Well sure, but that can be provided after the fact.

The whole point is that they're immediately known to be sets.

[coral spindle]

Can we move this conversation about notation, not algebra, elsewhere?

[past path]

I'd be happy to; do you have an elsewhere in mind?

I was told it belonged here.

[coral spindle]

Literally any discussion channel

#discussion, #serious-discussion, #math-discussion

This is not a channel to pontificate about notation.

[formal ermine]

a field extension over a finite extension field is always finite, no?

because if L/K where L is finite

then K = p^k and L = p^m with p^k | p^m

so p^m = (p^k)^l

[coral spindle]

Finally, some algebra

Yeah

[formal ermine]

ok epic

this seemed obvious to me but my intuition sometimes fails me

[rotund aurora]

I think it was the third time I asked about this lol. But in case you are wondering, this is due to this equation

where the alpha_i are the roots of f

it is trivial to see that the equation holds

0<=r<n ofc

[formal ermine]

lol why did I even ask this

a subset of a finite set cannot be non finite

🤦

[rustic crown]

reality check

hewwo walter

[agile burrow]

hi det

[rustic crown]

[agile burrow]

today i am reviewing hopf's theorems on group homology

and running away from complex analysis homework 🏃

[rustic crown]

hehe

[rancid quarry]

How would you guys go about proving this, assuming it’s an equilateral triangle and F stands for a vertical reflection and R stands for a 120 degree clockwise rotation

[rustic crown]

merosity talking about isometry

[delicate bloom]

I'd just go through and use the fact F^2=I and R^3=I to write them in terms of each other

lol

like multiply the first one on the left by R to remove R^2 an you have RFR=F, oh that's the 3rd one, etc just keep messing around

[rotund aurora]

Consider the linear forms L1(x,y)=x and L2(x,y)=y in R^2. The associated determinant is 1 and 3/2>1. So we should be able to find integers x,y such that |L1(x,y)|=|x|<1/2 ??

How would that example agree with what this theorem is saying?

(btw, it is understood that the integers m1,...,m_n should be zero)

[rancid quarry]

LOL thanks, I was overthinking it so hard

[rotund aurora]

how does this even make sense? The c_i can be permuted, so that is saying that we should find m1,...,m_n such that |L(m1,...,m_n)|<c_i for each i, with L fixed. But c_i can be arbitrarily small

thus {|L(m1,...,m_n)| : m1,...,m_n in Z} should attain arbitrarily small (positive) values, but in general this is not true at all

I mean that theorem is true ofc, I must be missing something lmao

[delicate bloom]

I think the det of those L_i is 0

idk didn't compute them now I'm probably wrong lol

L_1 has determinant -1/4 I think yeah

or wait

[rotund aurora]

$L_1(x,y)=a_{11}x+a_{12}y$ and $L_2(x,y)=a_{21}x+a_{22}y$. The matrix is
[
\begin{bmatrix}
a_{11}&a_{12}\ a_{21}&a_{22}
\end{bmatrix}
]

[cloud walrus]

Croqueta

[rotund aurora]

In my example I chose a11=1, a12=0, a21=0, a22=1

so the determinant should be 1

[delicate bloom]

maybe it's just not the origin of the lattice, so some 0 components are ok

[rotund aurora]

L is a linear form

L(0,...,0)=0 always

[delicate bloom]

yeah I know, I'm saying we exclude the origin

but (1,0) or (0,1) is fine

[rotund aurora]

btw I consulted some other text on this theorem, and the formulation is the same

(so Im just getting something wrong)

[delicate bloom]

yeah they say it there

nonzero vector

(1,0) is nonzero but has 0 components, that's ok

[rotund aurora]

ahh

lmao

I see

Thanks

[delicate bloom]

haha yeah cool, you're welcome

[rotund aurora]

So then, the tuple (m1,...,m_n) depends on i?

[delicate bloom]

nope

[rotund aurora]

ok I still dont get it I think 🤦‍♂️

What would be the (m1,m2) in the example L1(x,y)=x and L2(x,y)=y ?

[delicate bloom]

tell me your c_i

[rotund aurora]

1/Rayo's number

xd

1/2

[delicate bloom]

lol

ok and the other is

[rotund aurora]

and the other 3

[delicate bloom]

ok perfect then (0,1)

[rotund aurora]

ah ok

me dumb

[delicate bloom]

😛

[rotund aurora]

Thanks

[delicate bloom]

me too

you're welcome

I didn't really read your question but sort of going off what I remember about minkowski's theorem as intuition about how you have a symmetric convex shape and you can squeeze it past one lattice point in one direction, but not another, and caring only about getting a point that's not the origin

at least I figured it was like that

and like the reason we care is we can use that to bound the class number or something? lol idk

[rotund aurora]

you mean Minkowski's theorem or the linear forms thing?

[delicate bloom]

both?

[rotund aurora]

Minkowski's is apparently really useful

I have not seen linear forms thing in action

[delicate bloom]

like I'm thinking of the determinant as that volume of a convex shape, or related to it

[rotund aurora]

yeah, its along those lines

I mean, I think the case of dimension 2 is straightforward

[delicate bloom]

so if you're bigger than a fundamental cell you've necessarily grabbed one of the lattice points and that feeds into another thing about how the norm over the ideal norm is something I forget

I think it's basically the same argument as how you'd show ideals of Z[i] is a PID by assuming you have an element of smallest norm?

[rotund aurora]

https://www.math.cmu.edu/~ttkocz/teaching/1819/read-sem-notes.pdf In this it is proven the two and four square theorems with Minkowski (you still have to do some effort besides Minkowski tho). This is pretty cool

[delicate bloom]

I don't know, this is something I have to spend more time on eventually but haven't gotten around to it yet

[delicate bloom]

neat

[rotund aurora]

yeah the usual proof that Z[i] is a euclidean domain is essentially computing the volume of the fundamental mesh/domain/parallepiped

[rotund aurora]

They are still not trivial btw, and idk if three squares falls with Minkowski. I have actually never seen a proof of three squares

[delicate bloom]

I have but with the hasse-minkowski local-global theorem for quadratic forms, basically proving in Z_p for p != 2 is easy with a clever substitution, then p=2 case handled separately

[burnt sparrow]

Hey

Why can't (2, 2) be a thing?

[rich patrol]

Product should be reversed

[burnt sparrow]

Apparently the notion is additive

So like okay

Z2 = {0, 1} and Z3 = {0, 1, 2}

I get that

but how is

(0, 1) * (0, 2) = (0, 0) with this additive notion?

[rich patrol]

1 + 2 = 3 = 0 mod 3

[burnt sparrow]

gotcha

so is that how we do addition with the Zs?

with mod?

[lethal dune]

Elements are written in wrong order

It’s actually Z2xZ3

[burnt sparrow]

ahhhhhh

how do we write in proper order?

and what determines ordeR?

I feel silly, im sorry

@rich patrol

[rich patrol]

Do you know what Z_2 and Z_3 are

[burnt sparrow]

Z2 = {0, 1}, Z3 = {0, 1, 2}

Z2 is cyclic group of order 2

Z3 is cyclic group of order 3

[rich patrol]

See ryu's message

[rich patrol]

.

[burnt sparrow]

this is the question:

[rich patrol]

Wdym by this

[rich patrol]

Ok what about it

The solution is above?

[burnt sparrow]

lets just focus on this questiopn

how do I list the elements?

like I know what Z2 and Z3 are

but how do i figure out what the elements are

[rich patrol]

Okay so what is Z2 x Z3

[burnt sparrow]

Z6?

[rich patrol]

No

What does the x mean in groups

[burnt sparrow]

means an operation right?

ah no

it's the

direct product

[rich patrol]

Yes

So what is Z2 x Z3

[burnt sparrow]

{0, 1} + {0, 1, 2} = (0, 0) + (0, 1) + (0, 2) + (1, 0) + (1, 1) + (1, 2)

oh!

[rich patrol]

No

[burnt sparrow]

alright im confused af

[rich patrol]

Ok so what's the definition of the direct product

[burnt sparrow]

G x H is the group consisting of all the elements

x, y

where x in G

y in H

correct?

[rich patrol]

Ok, so what is Z2 x Z3

[burnt sparrow]

{(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)}

😄

[rich patrol]

Okay, and what is the group operation

[burnt sparrow]

group operation is additive

[rich patrol]

Tell me explicitly

[burnt sparrow]

well i know in Z3

4 = 1

because 4 = 1mod3

so wait i guess we're in mod3 now

not mod2?

@rich patrol

how is (1, 0) + (1, 0) = (0, 0) in the operation table

1+2 = 2

2 = __ mod 3

[tender wharf]

recall your definition

which you literally sent

[tender wharf]

right here

[burnt sparrow]

a, b + a', b' = (a+a', b+b')

1, 0 + 1, 0 = (1+1, 0+0) = (2, 0)

[tender wharf]

thats why they asked you to state EXACTLY what the operation is

no, that is wrong

[burnt sparrow]

how

im so confused

[tender wharf]

look closely

very carefully at your definition of the operation

nevermind they didnt make it explicit

ok thats not your fault

let G and H be groups

Let * denote the operation of G

. denote the operation of H

then the operation of the direct product in G x H is

(g, h)(g', h') = (g * g', h. h')

is this clear? @burnt sparrow

[burnt sparrow]

ah yes

[tender wharf]

now try again

[tender wharf]

.

no wonder you were so confused

here the external direct product means the same thing as direct product

[burnt sparrow]

so like

[tender wharf]

"It is understood that each product g_i g_i' is performed with the operation of G_i"

[burnt sparrow]

(1, 0) (1, 0) are both in Z2 x Z3

[tender wharf]

yeah that part is correct so far

now perform the addition

[burnt sparrow]

and the 1+1 is from Z2

and the 0+0 is from Z3

[tender wharf]

yes

so how would you compute the addition

what book is this by the way

[burnt sparrow]

OH

1+1 = 2 = 0mod2

and 0+0 = 0 = 0mod3

THANK YOU ❤️

[tender wharf]

np

what book is it?

[white oxide]

how can a^h = a^k and simultaneously h > k? i know they're saying a^h = a^k for contradiction but it still seems like it wouldn't work

because h and k are positive

[tender wharf]

lets consider this

[white oxide]

because for instance, taking h = 5 and k = 2

[tender wharf]

consider Z_10

now let h=5

then 4h=2h

do you agree?

[white oxide]

yeah

[tender wharf]

there we go

of course without more context its not possible to see what you're talking about

here 2 is the order of 5

you're going to need to post

$\original$

[cloud walrus]

active mental mutilation liker

[white oxide]

what do you mean here

[tender wharf]

you know what an order is right

[white oxide]

i'm still a bit confused but i kind of get what you mean

yeah

the cardinality of a group correct?

but what does 5 refer to

Z5?

[tender wharf]

oh sorry case 1 is tackling the infinite case

yeah this is why I asked you to include the theorem statement

If G has infinite order and is generated by a this is of course true

for example $\bZ = \langle 1 \rangle$ under +

[cloud walrus]

active mental mutilation liker

[white oxide]

yeah i understand

but i'm more confused about the proof

logically it doesnt make sense

[tender wharf]

ok

[white oxide]

you can't have a^h and a^k and have h \neq k

[tender wharf]

of course not

thats a contradiction

they're showing you why it's absurd

it says that <a> generates an infinite group but somehow a has finite order

do you know what order of an element means

[white oxide]

why is there a need for it then

because if a^h = a^k

then a^ha^-k clearly equals e

i dont see a need for h > k

[white oxide]

no

[tender wharf]

ok

give me a few minutes

[white oxide]

actually wait

i see why they need it

to show that h - k is a positive integer and thus contradicts that for all positive integers m a^m \neq e

but it still doesnt make sense they contradict each other

anyways yeah take ur time

[tender wharf]

basically if for contradiction your a^m = e

then the group is not infinite

let's suppose m is the least integer such that that is the case

then given any a^k

we can write a^(qm+r)

= a^r * a^qm

and clearly a^qm = e

this group is not infinite

[white oxide]

okay i understand that

how does that apply here?

or more specifically where we're allowed to say a^h = a^k and say h > k

[tender wharf]

I just gave you an example

[trim grove]

Can Someone Please Help Me With First Line Of This Proof?

[trim grove]

,rotate

[rustic crown]

look at the class of g in H\G

since X is a set of representatives, Hg = Hx for a unique x

so g = hx for unique h in H and x in X

[trim grove]

Isn't X here is of form {Ha, .....}?

[rustic crown]

i think X is a subset of G such that X --> H\G given by x --> Hx is a bijection

for each coset you picked a representative

[trim grove]

Please wait i am confused now😅

[rustic crown]

yee im not going anywhere :p

[trim grove]

Does X have number of elements as much as G/H has numbers elements?

[rustic crown]

yep if they finite

[trim grove]

Okk

[rustic crown]

more precisely this bijection

(when they allowed to be infinite)

[trim grove]

So its like, if g in G that means it is somewhere in any coset lets say Hp, so g = hp?

[rustic crown]

yee

[trim grove]

That means g =hx form some x in X?

Because X is representive set here

Thankyou For The Help.

[rustic crown]

np :3

[trim grove]

How to show that Additive group Q is not a Free Z module?

[median pawn]

why is the text in blue true?

suppose E is a finite extension of F, and R is a subring of E containing F

R is an F-algebra, sure, but R may not be finite?

R is an integral domain since the cancellation law would hold (the cancellation law from E is inherited)

[lethal dune]

note that any such extension is algebraic, show there's an inverse

[median pawn]

ahhh idk what an algebraic extension is, yet

[lethal dune]

you can write the inverse in terms of a polynomial in F and α

[median pawn]

at least it's clear then that the result is not a corollary of lemma 1.23

[lethal dune]

ok so say α ∈R with F ⊂R ⊂E where E is finite ext, then R is a vector space over F

now for α ∈ R, 1, α, .. , α ⁿ cannot all be LI, since dimR ≤ dimE

[median pawn]

n = [E: F]?

[lethal dune]

so you get a equation of the form
∑ a_i α^i=0

[lethal dune]

yes

[median pawn]

ahh makes sense, now just choose the smallest j in 0 to n such that a_j is non-zero, take it to the other side and we have the inverse?

it is depending on alpha tho, but i guess that's fine

[lethal dune]

since α^(-1) ∈F, multiply by it and solve for α ^(-1) and show that it must also lie in R

so you get R has inverse of all nonzero elements

[median pawn]

i see, thanks

how does this result relate with the lemma tho?

doesn't seem obvious

[lethal dune]

it does, thing is there is a reltn

lol show it's an integral domain

lol that's much easier

[median pawn]

this is correct innit

but R may not be finite!

[lethal dune]

R ⊂ E, so R must be again integral and hence field by the lemma

[median pawn]

the lemma wants R to be "finite"

[lethal dune]

I think they forgot to mention that

otherwise the proof I gave works

[median pawn]

ahh i see, thanks

[lethal dune]

or finiteness is implied

idk

[median pawn]

so your proof works more generally, even if R is not finite?

[lethal dune]

yea

[median pawn]

yeah im not sure either

[lethal dune]

in general for any k ⊂ R ⊂ cl(k) is a field, where cl(k) is the algebraic closure of k, whatever that means

you can replace cl(k) by any any finite extensions of k, which is the given statement

[median pawn]

i see, thanks!

[lethal dune]

oh btw

[rustic crown]

ryu

hiii

[lethal dune]

just noticed, the finite here doesn't mean finite cardinality

hii det

it means the dimension is finite

@median pawn

[rustic crown]

yea finite usually means f.g. as a module

and finite type means f.g. as an algebra

[median pawn]

oooo

[lethal dune]

[eternal prism]

A subring is a subspace so it's a finite dimensional vector space.

[untold basin]

I passed my Algebra exam
Thanks everyone ❤

[next obsidian]

Swag

[formal ermine]

ok so

some galois group

for x^4 - 5 in Q[x] I got Z2 x Z2

our roots are frt(5), zetafrt(5), -frt(5), -zetafrt(5)

so our splitting field will have degree 4

because 1, zetafrt, frt, sqrt will be a basis

we can list some automorphisms:

id : frt -> frt, zeta -> zeta

sigma : frt -> -frt, zeta -> zeta

tau: frt -> frt, zeta -> -zeta

obviously this is not cyclic

so we have Z2 x Z2

now for x^3 - 8 in Q[x]

roots will be 2, 2zeta, 2zeta^2

so degree 2, therefore Z2

now x^3 - 5 in Q(isqrt(2))[x]

it has no roots in the field, and is separable

I looked at its determinant: Delta(x^3 - 5) = -549

this is not a square in our field so we will have S3 as a galois group

are these correct?

[formal ermine]

{ 1, zetafrt, frt, sqrt } is a basis

because zeta^2 = -1

so (zetafrt)^2 = -sqrt

[hot lake]

what is sqrt * frt

[formal ermine]

good point

so 5

wait

yeah

5

so Z5

[hot lake]

nonono

what's your basis now ?

[formal ermine]

{ 1, zetafrt, frt, sqrt, 3/4th root }

[hot lake]

what's zeta frt * frt

[formal ermine]

{ 1, zetafrt, frt, sqrt, 3/4th root, zetasqrt }

[hot lake]

continue multiplying things

[formal ermine]

{ 1, zetafrt, frt, sqrt, 3/4th root, zetasqrt, zeta }

because sqrt * zetasqrt

still missing something

can't be 7

ah, zeta 3/4th root

this could work now

[hot lake]

yeah it would be 8

[formal ermine]

yeah ok

there are a lot of groups of order 8

[hot lake]

you're missing some automorphisms

[formal ermine]

yeah I see now

x^4 - 5 is the min poly of frt

for zeta we have x^2 + 1

this would've been easier with the tower law, yikes

ok I think I found them all

let me take a picture

,rccw

[formal ermine]

it seems to be generated by 3 elements

I would've guessed Z_2^3 but I'm not sure

[formal ermine]

w is zeta here btw lol

I can't write a zeta by hand

[hot lake]

no

are you sure it's commutative

r² doesn't seem to be id

[formal ermine]

yikes

you're right

tau has order 2, sigma has order 2, gamma has order 4

[formal ermine]

in an exam, would I just say "the galois group is { id, gamma, tau, blabla }"

or do I like have to find its name

[formal ermine]

next guess: Z4 x Z2

[delicate orchid]

we sure this is a group of order 8? Like, positive fr fr?

[formal ermine]

yes

100%

[delicate orchid]

might suggest just writing out the entire cayley table

nah nah

check to see what else has order 4

because if there's more than 2 things with order 4 we can eliminate D_8, and if it's non-abelian it has to be Q_8

[formal ermine]

gamma sigma has order 4

[delicate orchid]

are the two elements of order 4 in the same cyclic subgroup

i.e. are they both powers of some other element

if not, Q_8

if yes, D_8

[formal ermine]

it's Q_8 then

[delicate orchid]

it smelt like Q_8 i must admit

[formal ermine]

ok thanks wew

[delicate orchid]

my arcane knowledge of small p-groups comes in clutch

[formal ermine]

for this I am 99% sure

[formal ermine]

not sure about this

whether -549 is actually not a square in Q(isqrt(2))

I'm 70% sure there

[delicate orchid]

how can it be a square when it's a negative :mock2:

[formal ermine]

we have isqrt(2)

but 549 is not divisible by 2

[delicate orchid]

well that's convinced me

[formal ermine]

ok epic

now another one

I wanna show that Q(fifthroot(13)/Q has no intermediate fields

it obviously has degree 5

because the minimal polynomial is x^5 - 13

so Z5

Z5 has no proper subgroups, because prime order

so there are no intermediate fields by galois?

[rustic crown]

(ext not galois)

[formal ermine]

oh no

yeah I see

not sure what to do then

[rustic crown]

but doesn't matter

argue on degree instead lol

if E is an intermediate field, what is [E:Q]?

[formal ermine]

|G/Gal(E/Q)| I think

or was it Q(fifthroot(13))/E

[rustic crown]

Aut(Q(13^(1/5))/Q) = 1

[delicate orchid]

lol

[formal ermine]

ah I see

it's separable because char 0

so our extension will also have degree 1

thus no intermediate fields

[rustic crown]

[formal ermine]

wait

lol

2

[rustic crown]

you just said it had degree 5

[formal ermine]

not 1

wait

I'm really confused

uh

let me think

intermediate field degrees have to divide the degree of the entire extension

but that's 5

so it's either 1 or 5

meaning that there are no proper intermediate fields

ok I was thinking way too complicated at the beginning

thank you det

[rustic crown]

uwu

[silk bramble]

what's the most common symbol for the matrix form of a linear map T from basis A to basis B ?

$[T]_{A \to B}$ ?

[cloud walrus]

Delerik

[rotund aurora]

I have seen $[T]_A^B$ too

[cloud walrus]

Croqueta

[rotund aurora]

I think yours is actually pretty nice