groups-rings-fields

[rocky cloak]

Just have the image into B be an integral extension

[cobalt heath]

Ah, hmm

[mighty kiln]

It's an inclusion though

[cobalt heath]

I guess usually one needs to work elementeise for integral extension anyway

[mighty kiln]

^

[cobalt heath]

Sorry for not clarifying, I am specifically asking for when A -> B is not inclusion in general case.

[dire igloo]

Que :
Let G be a group of order 75 which has an element of order 25.
Then the number of elements of order 5 in G is

Ans :
75=(5)*(15) implies number of subgroup of order 5 is 1 because of sylow theorem. Therefore all elements of order 5 generates the same subgroup which is equal to number of generators of subgroup of order 5 and ie phi(5)=4

Answer is correct according to ans key but is my justification correct? If wrong please correct me

[dire igloo]

<@&286206848099549185>

[delicate orchid]

Looks fine to me boss. Except sylow’s theorem implies that the number of subgroups of order 25 is 1, not 5

[dire igloo]

No I wrote number of sylow subgroup of order 5 is 1

[delicate orchid]

Then that’s complete nonsense

I’m well aware that’s what you wrote. I’m correcting you

[boreal inlet]

And that subgroup is also normal

[delicate orchid]

I mean sure but that’s not particularly relevant

[boreal inlet]

.....oh right that's true

I think the order 5 elements all should be in this subgroup

[dire igloo]

Oh okay thanks !

Oh now I understood why it's wrong thanks wew lads tbh

[delicate orchid]

no problem Exuberance

[frank cosmos]

im trying to show that Z[i] is prinicpal

i want to show every ideal is generated by an element with minimum magnitude

say this is (a+bi)

then i need to show that for every (c+di) in the ideal, letting m denote a^2+b^2, m divides both (ac+bd) and (ad-bc)

if not, then im trying to use division algoroithm but getting stuck

[south patrol]

Good

Well I mean really you just need to show you have a euclidean function if you've seen that

[dull ginkgo]

I was about to ask if euclidean domains need to be integral and I realized that's an exceptionally stupid question

[south patrol]

I don't think it's exceptionally stupid lol, isn't it just part of the definition

[dull ginkgo]

the fact that it is part of the definition is aforementioned exceptional stupidity

[south patrol]

Well it's okay dw

[frank cosmos]

i just proved that if a1,...,an are nonzero elements in a pid, and (a1,a2,...,an)=d then d is a gcd. my proof didnt use the pid hypohtesis though. is it really necessary

[languid trellis]

We need it to be a pid for the ideal (a1, ..., an) to be principal for any choice of a1, ... an

[frank cosmos]

isnt that part of the assumption

(a1,a2.,,,.an)=(d)

[celest furnace]

Yeah looks like for this direction it’s not necessary 👍

The big one is there exists a generator which you need PID for

[frank cosmos]

lang moment

[languid trellis]

Hm. Yeah maybe

[dull ginkgo]

Well if we have a commutative ring, and two principal ideals (a) and (b), if their intersection is also principal, (c), then c is by definition the greatest common divisor of a and b

[dull ginkgo]

I just had a kind of weird observation,

[lapis latch]

the intersection is the lcm the sum is the gcd

[dull ginkgo]

oh shit thanks man

[cloud walrus]

THE TUBE

[dull ginkgo]

i am wrong ignore me lol

[south patrol]

there is actually one thing here which you may be interested in

if G is a group then R^G can naturally be given a coalgebra structure too

as in like, the multiplication G x G-> G induces maps R^G -> R^G (x)_R R^G if you apply R^(-) to everything

and this coalgebra structure is nicely compatible with the algebra structure and gives it the structure of a bialgebra lol

[dull ginkgo]

yeah I'm just trying to ponder

Namely if we have a commutative ring R and the fixed point subring R^G for |G| finite, like what can we siphon from this besides the fact that it's an integral extension

[south patrol]

oh lol i guess R^G here is a pain

i mean R^G as in functions G -> R

[dull ginkgo]

Yeah that's an interesting part too

[south patrol]

well if R is noetherian i believe R is finitely generated R^G-algebra

[dull ginkgo]

Algebraicly or like as a module

oh shit

[dull ginkgo]

I wonder if that is a byproduct of it being an integral extension

[south patrol]

No because I'm p sure this is false if R is not noetherian

maybe I am wrong though lol

[dull ginkgo]

interestingly this gives artin's lemma for the case when R is a field using Noether Normalization if you can prove the size of the generating set is |G|

[south patrol]

i think for example if you consider R = k[x_1,x_2,...] and let G act by x_i -> -x_i then R is probably not a finitely generated R^G algebra

[south patrol]

Well the problem is you don't know about the size of the generating set ig

Maybe it would work if you used the fact that k is a field and optimised the proof of every step but idk

[dull ginkgo]

Yeah

[south patrol]

i think artin's lemma needs some extra lin alg to make use of teh special vector space structures invoved

[dull ginkgo]

Yeah that's the thing

Artin's lemma implies Independence of Characters

I don't think it goes in the reverse direction though

no wait I'm a dumbass

automorphisms are characters from R^x to R^x

[hidden wind]

when i first had a peel at group theory some years ago i was so confused by group actions… now they’re like my favorite things in grouo theory

(i barely know any group theory)

[south patrol]

They are very nice

[round jay]

How can this direct sum in problem 5.17 be a ring if I does not contain the identity?

[delicate orchid]

Here's what I think is going on: this direct sum has to be as a direct sum of R-modules - if it was the coproduct of rings it would be a tensor product. Hence the identity given by the embedding of 1 \in R into the 0-graded part gives you the identity of the ring.

[hidden wind]

tfw the number of groups of order 2^(2^n)

[delicate orchid]

1, 2, 56092, quite a few

[hidden wind]

[dull ginkgo]

Hey chat so

we know that every transformation of a finite field is a polynomial

Is finding the polynomial of a given transformation the same as finding the discrete fourier transform of the function if we plug in (f(r^n) - f(0))r^-n where r is the multiplicative generator

[dull ginkgo]

Like let $\mathbb{F}q$ be a finite field of order $q$ and $f$ be a transformation on it. Assume a priori that $f(x) = \sum{n = 0}^{q - 1}{a_nx^n}$. Then immediately, $a_0 = f(0)$. Likewise: $\frac{f(x) - f(0)}{x} = \sum_{n = 0}^{q - 2}{a_{n + 1}x^n}$ for $x \neq 0$. Since $\mathbb{F}q^\times$ is cyclic of order $q - 1$, there is a multiplicative generator $r$ that is the primitive $(q-1)$-th root of unity \\

So thus, we can say $\frac{f(r^k) - f(0)}{r^k}= b_k = \sum_{n = 0}^{q - 2}{a_{n + 1} r^{nk}}$, so isn't solving for $a_{n+1}$ just solving for the discrete fourier transform of the $b_k$?

[cloud walrus]

THE TUBE

[old sphinx]

Anybody here

I asked for help on this to see if I got this correct or not

[chilly radish]

How did you conclude the answer for a? Because to me it seems like you didn't understand the problem

[old sphinx]

I was thinking it was the degree as no idea what the true answer might be

If you can help me out

[dull ginkgo]

file is called quiz

[chilly radish]

Well first of all, it says that it's looking for irreducible factors of degree 7

[dull ginkgo]

is this a take home quiz fam

[chilly radish]

Hmmm

@old sphinx is it

[old sphinx]

Practice quiz from my professor as not the actual quiz

[chilly radish]

Aight

[old sphinx]

Just need help with trying to figure out

[chilly radish]

Do you understand what this means

[old sphinx]

No not at al

just the irreducible

[dull ginkgo]

128 oh god it's just splitting it a bunch of times

wait no

x(x^127 - 1) oh good lor

PRIME

[old sphinx]

Yes that is what I thought for

[dull ginkgo]

we are working in char 2 obviously lmao

wait what finite field is that

[old sphinx]

No idea as just x^128

[dull ginkgo]

I assume it's F_2^7

[old sphinx]

Me too

[dull ginkgo]

Then that's LITERALLY the product of (x - a_i) for all a_i

[old sphinx]

(x^127-1)

[dull ginkgo]

tldr:
finite field under mult is cyclic of order q - 1, so 127

so each element solves x^(q - 1) = 1, i.e x^q - x = 0

so every element is a root

So finding the "degree 7 factors" confuses me

[chilly radish]

Surely this isn't the field you're working over

Otherwise it trivialises the question

[dull ginkgo]

because finding the septic divisors would be nCr(q, 7)

[old sphinx]

No idea what is meant by that

[dull ginkgo]

idk what you mean by order 7 factors tbh

Like wouldn't this polynomial split for like, any F_2^k for k <= 7

[old sphinx]

Probably

Looks right to me

[dull ginkgo]

like "degree 7" wtf does FAM MEAN

x^7 + x + 1 is irreducible mod 2 (obviously literally just check 0 and 1 lmao) so F_2[X]/(X^7 + X + 1) is iso to F_128

i am yapping to myself to understand the problem gimme a moment to condense it

[old sphinx]

Okay as I do not understand it myself

[dull ginkgo]

I take it alpha is THE primitive 127th root

I hate field theory

[old sphinx]

Makes sense

[dull ginkgo]

@chilly radish This is confusing because if alpha is the primitive 127th root, then wouldn't alpha^3 be in GF(128)?

so it's minimal poly would be (X-alpha^3)?

[old sphinx]

You guys are the experts in this anyways

[dull ginkgo]

i am DEFINITELY not

I just got to the finite field part of my textbook today and I wanted to do this for practice

[old sphinx]

Okay as makes sense

Good with practice then

[dull ginkgo]

mostly to get adjusted to notation also used

[old sphinx]

Okzy

[tribal moss]

Note that "irreducible" is not the same as "no roots", so you need to check for divisibility by polynomials up to degree 3.

[dull ginkgo]

oh shit brainfart yeah

[old sphinx]

What should I do then

[dull ginkgo]

tropo what exactly is alpha here, and is it asking for the minimal F_2 poly of alpha^3 in F_128

[tribal moss]

In fact every element of GF(128) other than 0 and 1 is a primitive 127th root of unity.

[dull ginkgo]

really?

[tribal moss]

alpha is a root of x^7+x+1 -- namely, the residue of x in F2[x]/<x^7+x+1>.

[tribal moss]

Yes, since the multiplicative group of a finite field is cyclic and 127 is prime.

[dull ginkgo]

huh guess I don't understand fields very well

Not every element generates that finite field though i thought

Like take 2 in Z/7Z

[tribal moss]

I think what you need to do for (21) is calculate 1, alpha, alpha^2, ..., alpha^6, alpha^7 in a standard basis over F2, and find a linear combination of them.

[tribal moss]

That is not immediately a parallel case because the number of nonzero elements in Z/7Z is not a prime.

[dull ginkgo]

hm

OH WAIT

YEAH

127 is prime

mersenne prime

[old sphinx]

127 is the answer

[tribal moss]

To what?

[old sphinx]

To the problem

[dull ginkgo]

Tropo by residue of x do you mean x mod (X^7 + X+ 1)

[tribal moss]

Which problem?

[tribal moss]

Yes.

[old sphinx]

20

[dull ginkgo]

alrighty,

[tribal moss]

127 irreducible factors of degree 7 would give a polynomial of degree at least 889, which I don't think sounds like x^128-x.

[dull ginkgo]

I still have trouble interpreting the F[alpha] ~= F[X]/(p(X)) thing sometimes

[old sphinx]

Wha6 should the answer be then

[dull ginkgo]

where alpha is a root

but I guess any choice of root gives a different iso?

algebraicly indestinguishable

[tribal moss]

We know the only subfield of GF(128) is GF(2), right?

[old sphinx]

Yes

[dull ginkgo]

that's because log_2(128) = 7 is prime right

[tribal moss]

Yes.

[dull ginkgo]

sorry jacobson doesn't cover finite fields very well thus far

I don't mean to be an inconvenience

[tribal moss]

This means that each of the 126 elements outside GF(2) generates the entire field, so it is the root of a minimal polynomial of degree 7.

[dull ginkgo]

Lets say we DIDN'T choose a prime-power of 2

like uh 2^8

so 256

PERFECT uh

1024

1023 is divisible by 3

[tribal moss]

(I think perhaps they're asking this for 2^7 because the question is somewhat easier there)

[dull ginkgo]

I'm just trying to understand what you would do

[old sphinx]

Yes as I do agred to that so 128 then

[cobalt heath]

What are we doing?

[old sphinx]

Question20

Figure it might be 128 then

[cobalt heath]

Oh, this is the factorization over finite field

[old sphinx]

2^7 is what I originally thought it might be but then told to do x(x^127-1)

Yes

[tribal moss]

That would make the degree at least 896.

[old sphinx]

Okay

[tribal moss]

But the degree is actually only 128, so you don't have room for all those irreducible factors you keep wanting!

[old sphinx]

Do agreed and I need to go to sleep but if you can figure it out then

[dull ginkgo]

do you understand this stuff well?

[cobalt heath]

Well, mildly

[dull ginkgo]

I'm kind of a bit mind melted right now and I'm trying to get a stronger understanding of this

[cobalt heath]

I think one can, at least, think of the maximal degree of the irreducible polynomial of this one.

[tribal moss]

Furthermore, irreucible polyomials of degree 2, 3, 4, 5, 6 cannot have any root in GF(128).

(Because such a root would generate a subfield of a size we have agreed doesn't exist).

[cobalt heath]

No way, it shouldn’t be that easy

[dull ginkgo]

Does this have to do with the degree of GF(128) over GF(2) being 7?

[tribal moss]

As we agreed to up here.

[cobalt heath]

This is boring, let’s talk about GF(64)

[tribal moss]

Since we know x^128-x has 128 different roots (namely all the elements), it cannot have any irreducible factors that don't contribute some of the roots.

[dull ginkgo]

So irreducible GF(p^a) polynomials of degree n with roots in GF(p^b) exist iff n | log_p(b) - log_p(a)

[tribal moss]

Correct.

[dull ginkgo]

i see

[tribal moss]

No, wait.

I'm confused by having both a and b in that statement.

[dull ginkgo]

I thought [GF(p^b) : GF(p^a)] = log_p(b) - log_p(a)

[cobalt heath]

I was wondering why binary coefficient would be important this whole time

Realized that it’s the one allowing reduction

[dull ginkgo]

by tower law if there is an intermediate subfield then it must divide that

and intermediate field F = GF(p^a)[x]/(p(x)) for irreducible p(x) of degree n must be an extension of degree n over GF(p^a)

thus n | [GF(p^b) : GF(p^a)] = log_p(b) - log_p(a)

okie I see now

[tribal moss]

Just tying this up: we know the irreducible factors of x^128-x must have degree 7 or 1, and they are all different since a multiple factor would mean we have fewer than 128 different roots.

[cobalt heath]

Hm. Does factors of x^128 - x over Q appear over GF2 as well?

[dull ginkgo]

If it factors over Z

Wait

[cobalt heath]

Wdym “factors over Z”

[dull ginkgo]

I was thinking, if it factors over the integers, we have the epimorphism into Z/2Z

so the factors would, well, remain factors

[cobalt heath]

Hmm I should be able to describe this e.g. respect to universal property

[dull ginkgo]

actually, that's a monic polynomial in Z, so wouldn't it need to factor over Z by rational root theorem

wait

i am thinking of gauss' lemma

[cobalt heath]

Yeah Z vs Q is not a problem here obv

[tribal moss]

Since every element is a root, we certainly need to include both x and x-1 among the factors. That leaves 126 degrees which can now only be covered by factors of degree 7, so there must be 126/7 = 18 of them.

[dull ginkgo]

The epimorphism Z onto Z/2Z extends to an epimorphism Z[X] onto Z/2Z[X]

[cobalt heath]

Does it remain to be epimorphism

[dull ginkgo]

Think about the kernel

X is outside of (2) lol

[tribal moss]

It is clearly surjective, so definitly epi.

[dull ginkgo]

So like, lets say we had like

some q = 2^p where 7 | 2^p - 1, i refuse to find the specific q

wait

[tribal moss]

p=3 perhaps?

[dull ginkgo]

yeah

I'm just parsing how this is all working here

Also thanks for your time man

[old sphinx]

Since every element is a root, we certainly need to include both x and x-1 among the factors. That leaves 126 degrees which can now only be covered by factors of degree 7, so there must be 126/7 = 18 of them.

Your right as thank you

[dull ginkgo]

So like, if we had like x^1024 - x

the irreducible factors must have a degree dividing 10, i.e 1, 2 or 5?

[tribal moss]

Or 10.

[dull ginkgo]

oh shit yeah

and GF(1024) is that polynomial's splitting field right?

[cobalt heath]

Over GF(2)

[tribal moss]

In fact, if we collect my arguments here together, I think what they add up to is that the irreducible factors of x^1024-x are exactly every irreducible element of F2[x] whose degree is 1, 2, 5, or 10.

[dull ginkgo]

hm

Can this allow you to compute the amount of irreducible polynomials of a specific degree in GF(2)?

[tribal moss]

So we can use arguments like this to count irreduible polynomials of various degrees without actually finding them.

[dull ginkgo]

I see

[dull ginkgo]

Would that be the number of embeddings from GF(2) into GF(2^k) for degree k?

[tribal moss]

There's always exactly one such injection (at least if it is to be a field or just ring homomorphism).

[dull ginkgo]

huh brb

I assume this is only a thing for finite fields?

[tribal moss]

Well, there's never more than one homomorphism from GF(p) to anything, because 0 always maps to 0 and 1 always maps to 1.

[cobalt heath]

I mean

[dull ginkgo]

Oh wait that’s why it’s a bimorphism

Wait no that’s

I should probably go to sleep I am brain soup

[cobalt heath]

Hmm, it somehow feels like there is often only one embedding from like

Q and GF(p)

[dull ginkgo]

Z maps into every ring

So if it’s a field

The generated subfield is Q or Z/pZ

I.e it’s <1>

so yeah what tropo said makes sense to me

[tribal moss]

Yes, since the images of 0 and 1 fix the images of all other elements of the prime field.

[dull ginkgo]

it’s generating set is {0,1} and if two ring morphisms are equal on a generating set they are equal, which is enforced by the ring morphism def I guess

Now for that next question

How to find the minimal polynomial of alpha^3 hm

[cobalt heath]

It all comes down to generators!

[tribal moss]

^

[dull ginkgo]

I am a bit dense so I’m trying to understand what you mean by this

[tribal moss]

Sorry, I need to sleep now.

[dull ginkgo]

No worries I greatly appreciate you helping me

[cobalt heath]

Seems like it is about computing polynomial whose root is alpha

But idk how you can compute alpha in the first place - what is it?

[dull ginkgo]

I'll come back to this later, I'm just going to babble in #1217259452621783154

[cobalt heath]

Ahh

[noble lynx]

if x^n - 1's splitting field over Q has galois group the group of units of Z/nZ then wouldnt this contradict that fact that Q(w) is degree n-1 where w is the nth root of unity?

[dull ginkgo]

the degree is the totient function, i.e the number of units of Z/nZ

[noble lynx]

is the minimal polynomial nto 1 + x + ... + x^{n-1}?

[dull ginkgo]

no

It's the cyclotomic polynomials

[dull ginkgo]

Think of Q[i]

it's minimal polynomial is x^2 + 1, not x^3 + x^2 + x + 1

it's a degree 2 extension because it's a quadratic

[noble lynx]

thanks makes sense

[dull ginkgo]

you're welcome

[noble lynx]

hmm, actually maybe I am a bit more confused

[noble lynx]

isnt this irreudicble

[dull ginkgo]

sup

[noble lynx]

and the primitive root is a root of that?

[dull ginkgo]

if n is prime

[noble lynx]

ohh

[dull ginkgo]

yeppers

Look up "Cyclotomic polynomials", that might help

[noble lynx]

is there like a rule like the eistenstein criterion for testing irreducibilitiy of these polynomials?

or do you just recognize its cyclotomic polynomial

[dull ginkgo]

The cyclotomics are always irreducible

[noble lynx]

right, but how would you show something like that?

[dull ginkgo]

It's by definition

The cyclotomic polynomials are the polynomials that divide x^n - 1, but not any prior x^m - 1

[noble lynx]

oh ic

[dull ginkgo]

actually there is another def that states them as irreducible

it's a bit annoying to show they're irreducible

[noble lynx]

but if I just expanded it out into \sum c_i x^i, it would be a bit annoying to recognize which was actually cyclotomic?

[dull ginkgo]

lmao in general it's hard to tell when a polynomial lies in a given family depending on how you define it

so if you were given a random ass polynomial, showing it's the nth cyclotomic sounds like, hell

[noble lynx]

yeah lemme just remember these

[dull ginkgo]

plus, their degree is the totient

so there are infinitely many cyclotomic polynomials of a given degree afaik

actually no

i am an idiot nvm

[noble lynx]

I have another question about galois groups tbh

like if my polynomial is irreducible, why isnt the galois group just S_n? What prevents me from permuting roots arbitrarily like that?

is it because minimal polynomials can change over intermediate extensions

[dawn quest]

This is supposed to be a proof that coset multiplication is well-defined for normal subgroups

I don't see how this shows that though

We need to show that aN = a'N and bN = b'N implies (ab)N = (a'b')N

[next obsidian]

You use bN = Nb is equivalent to N being normal

[dawn quest]

Here there's no a', b'

[next obsidian]

Yeah so this is showing that like, this equality holds as sets

[dawn quest]

What I see this proof showing is that if you multiply aN and bN as subsets (i.e. XY = {xy | x \in X, y \in Y}) you get the same thing as (ab)N

[next obsidian]

Yeah!

That’s the point

So if you have aN = a’N and bN = b’N

[dawn quest]

I don't see how this implies this

yea

[next obsidian]

Then you have abN is the set product aN•bN = a’N•b’N = a’b’N

[dawn quest]

oh

I see

[next obsidian]

You’ve shown that abN can be recovered from the product of aN and bN as sets, so if the sets are equal you get the same thing

[dawn quest]

Thanks

[next obsidian]

Swag

[dawn quest]

Also dyk if it's true that coset multiplication of ideals is not equal to their subset product?

I think I saw this somewhere

[dull ginkgo]

ring theory ideals?

[dawn quest]

Yes

[dull ginkgo]

like (x + I)(y + J)?

is that what you mean?

[dawn quest]

Yea

Like in Artin

The set of products ... isn't always a coset of I

[dawn quest]

Ok I guess I answered my own question

I have another one

Why do the following two facts imply that (left) cosets partition a group?
\begin{itemize}
\item The relation defined by $a \sim b \iff a^{-1} b \in H$ is an equivalence relation.
\item $a \sim b \iff aH = bH$
\end{itemize}
It is supposed to be "obvious", but I am at a blank.

[cloud walrus]

plexcty

[next obsidian]

Pretty sure it’s appealing to the fact that equivalence classes partition sets

[dawn quest]

Right, ~ induces a partition on G. But why does the second fact above imply that this partition is exactly the cosets of H?

I know how to answer this question, but my proof has a couple steps in between. My impression is that the second fact should make this proof very direct, but I don't know how.

[dull ginkgo]

there is exactly one element x for each pair (a,b) such that ax = b

and likewise one y such that ya = b

It has to do with this

[sonic coral]

a problem on my galois theory final today was to determine if cbrt(2+sqrt(5)) + cbrt(2 - sqrt(5)). was rational or not. i’m not really sure what approach he was expecting with this

i see that you can like call it x and cube everything but what approach would be expected on a galois theory final

[dawn quest]

True, but that's still not as obvious as I'd like

If I change the definition of $\sim$ to $a \sim b \iff a \in bH$, then after showing $\sim$ is an equivalence relation, it is \textbf{actually obvious} that
$aH = {b \in G ~|~ a \sim b}$ and then the conclusion follows.

[cloud walrus]

plexcty

[dawn quest]

This is what I was looking for, I guess I misremembered the right definition of ~ to make the proof as clean as possible

[rocky cloak]

The natural thing to do would be to embed it into a Galois extension, then check whether it's fixed by the Galois group

[sonic coral]

that’s what i was thinking but i wasn’t really sure how to go about it

[celest furnace]

This seems super long

[crystal vale]

Let G be a finite group, T an automorphism of G with the property that T(x) = x if and only if x=e. Prove that every g in G can be represented as g = x^(-1)T(x) for some x in G.

If I define the mapping x->x^(-1)T(x) then it is well defined and injective because T(x ) = x implies that x = e. And G is a finite set so mapping is one-one and onto. Thus for all y in G there exists x such that y= x^(-1)T(x).

Is it correct?

[hollow mica]

Looks good. I don't think you have to mention that the mapping you define is well-defined. There's no reason for it to not be

[dawn quest]

How can I show that for a subgroup $N \subset G$, if coset multiplication is well defined, i.e.
$$ aN = a'N \quad \land \quad bN = b'N \implies (ab)N = (a'b')N ,$$
then $N$ is normal?

[cloud walrus]

plexcty

[dawn quest]

Does this proof by contrapositive look good?
Suppose $aN = a'N$ and $bN = b'N$ but $(ab)N \neq (a'b')N$.
This means that
$$ (ab)^{-1} (a'b') = b^{-1} a^{-1} a'b' \notin N .$$
Writing this element as
$$ b^{-1} a^{-1} a'b' = (b^{-1} (a^{-1} a') b) (b^{-1} b'), $$
since $a^{-1}a' \in N$ and $b^{-1} b' \in N$, if it were the case that
$$ b^{-1} (a^{-1} a') b \in N ,$$
then
$$ b^{-1} (a^{-1} a') b) (b^{-1} b') \in N, $$
which is a contradiction. Thus
$$ b^{-1} (a^{-1} a') b \notin N ,$$
meaning $N$ is not normal.

[cloud walrus]

plexcty

[crystal vale]

Okay, thank you

[languid trellis]

The criterion mentioned is eulers criterion, using this it's trivial to show that -1 is a quadratic residue, but I don't know how we get to showing p is not prime in Z/pZ. I would appreciate any hint 🙂

[south patrol]

Try to write Z[i]/p in a different way

hint ||Z[i] = Z[x]/(x^2+1)||

[languid trellis]

Z[i] = Z[x]/(x^2+1)

[dire siren]

@languid trellis after proving that -1 is q.r., you know that there is some integer a such that p | a^2+1
in other words p | (a+i)(a-i)
now assume that p is prime in Z[i] and get a contradiction

[languid trellis]

Then I should apply correspondence theorem in some way to be able to say something about Z[i]/p

Oh

[south patrol]

Well what I mean is you can write (Z[x]/(x^2+1))/p in another way

[languid trellis]

x^2 \equiv -1 (mod p)

So x^2 + 1 is congruent to 0 mod p

[south patrol]

Well

a^2 + 1 = 0 mod p if a is a sqrt of-1

(just being careful as this is a different sense of x to what you said above)

i guess filip and i may be doing different things lol

yeah Filip's is faster i think

I was meaning to suggest ||Z[i]/p =F_p[x]/(x^2 + 1)||

[cobalt heath]

Ooo ramification

[languid trellis]

Right, we just need to show that p doesn't divide a+i and a-i which holds because p divides neither a or i. i is irreducible and if p | a, then p | a^2, so p doesn't divide a^2 +1 due to even/odd considerations

[languid trellis]

Hmmmm

[south patrol]

it's a bit easier too since (a+i)/p = a/p + i/p [in Q(i)]

and that guy isn't in Z[i]

[languid trellis]

Ic

[cobalt heath]

Ack. It is not ramification behavior, what am I saying

[south patrol]

Well it's closely related

[languid trellis]

I think F_p/(x^2 + 1) is a field, so so is Z[i]/p, so all elements are a unit and so nothing is prime?

[south patrol]

It isn't a field

Do you know when k[x]/(p(x)) is afield for k a field

[languid trellis]

The ideal (p(x)) has to be maximal iirc

[south patrol]

Yes

there is an easier characterisation of that

(essentially as k[x] is a PID)

[languid trellis]

Oh but in F_p x^2 + 1 isn't maximal because (x^2 +1) is the 0 ideal? (x^2 cong -1 mod p)

[cobalt heath]

Uh

[languid trellis]

Am I tripping

[cobalt heath]

Do you have x^2 cong -1 for formal parameter x?

[languid trellis]

In this case I suspect prime ideals should correspond to maximal ideals

[cobalt heath]

Btw, (0) is a prime ideal.

[languid trellis]

my mind is broken rn

[cobalt heath]

Pity

[languid trellis]

Not sure what you mean by formal parameter but the context is that if p is a prime of the form 4n+1, then -1 is a quad residue mod p, and we're trying to get to p not being a prime in Z[i]

[cobalt heath]

Do you recall what “-1 is a quad residue mod p” means?

Basically, I have difficulty seeing where you deduced x^2 cong -1.

[languid trellis]

prime ideals are maximal when k[x] is a p.i.d because adjoining any element to a prime ideal (not in the ideal) means they are coprime and hence generate k[x] by bezout. maximal ideals in k[x] are prime because the ideal (d) is generated by the gcd of all elements in the ideal, so if the product ab is in the ideal, then d = gcd(a,b) is in the ideal, and so a = cd is in the ideal as is b = fd

[languid trellis]

-1 is a quad residue mod p means that (1) -1 is not cong to 0 mod p, (2) there exists an integer a such that a^2 equiv -1 (mod p)

[cobalt heath]

Yeah, so there exists an integer which satisfies the property.

[languid trellis]

yeah not all of them

my bad

[cobalt heath]

But when we say x in k[x], it is not a

[languid trellis]

so x^2 + 1 cong 0 is true when we apply the evaluation map

x -> a

but not in k[x]

@south patrol I need to go now, but I would appreciate it if you could flesh out your approach a bit more, and when I get a moment I'll come back and review it : )

[cobalt heath]

Yeah, I am sorry for pushing it

[hidden wind]

if H,K < G and HK = KH then also HK < G but i forget, what is the name for this sort of product of subgroups

?

[south patrol]

Cool: First note that $\mathbf Z[i]/p \simeq \mathbf F_p[x]/(x^2+1)$. Now $p$ is prime in $\mathbf Z[i]$ iff $\mathbf Z[i]/p$ is a domain, iff $\mathbf F_p[x]/(x^2+1)$ is a domain, iff $x^2 + 1$ is irreducible over $\mathbf F_p$, iff $x^2 + 1$ has no roots over $\mathbf F_p$ (since it's a quadratic!) iff $-1$ is not a quadratic residue

[cloud walrus]

Crystalline Potato

[sonic coral]

it was a 5 point t/f on a 200 point exam, 3 hours

[languid trellis]

Damn. What the fuck

I am majorly taken aback

I'll need to convince myself of that iso and 'p prime in Z[i] off Z[i]/p is a domain

[crystal vale]

@languid trellis the following proof is given in Herstein
If p is of form 4n+1 then there exists a^2 + b^2 = p.

|| Now if p is prime in Z[i], then p = (a + ib)(a - ib) divides one of them. Let p divides a + ib.

a+ib = p(c+id) by comparison p divides a and p divides b then p will be divide a-ib it implies that p^2 divides p, it is contradiction. ||

[south patrol]

The second is a super important fact

Well, okay you can divide itinto like

p is a prime element iff (p) is a prime ideal [that is easy]

and then if A is a commutative ring and I an ideal, A/I is a domain (resp field) iff I is prime (resp. maximal)

[south patrol]

this is very important

[languid trellis]

Proving a^2 + b^2 = p is the next exercise .-.

[languid trellis]

Yep this is just definition, if A/I is not a domain then there is some 0 divisor, I.e. (a + I)(b + I) = I which implies that ab is in the ideal, while a or b aren't which means I isn't prime

I remember proving the second part (maximality implies A/I is a field) in some wacko way with considering elements in R \ I but I'll have to remind myself later, I've got a lesson now

[hidden wind]

can someone at least verify my question makes sense?

HK being the set { hk : h in H, k in K }

[crystal vale]

Yes if HK = KH then HK is a subgroup of G

[hidden wind]

but does this product have a name?

[crystal vale]

I don't know about it

[next obsidian]

Product of subgroups

[mighty kiln]

A product of subgroups that is a subgroup

If one of them is normal and there's trivial intersection it's a semi-direct product

If both are normal and there's trivial intersection it's a direct product

[hidden wind]

excellent excellent thanku all

[next obsidian]

Okay but every subgroup is this with B = {e}

So it only really makes sense IMO to ask if the object AB has a name, with reference to the specific presentation and id just call that the product of A and B

Like, within G I guess

I dunno

Chmonkey

[rocky cloak]

It's called a Zappa-Szep product, and can be characterized externally (but it's a mess)

[hidden wind]

omg what an amazinh name

[next obsidian]

Nooooo you can’t say it has a real name, I look like I’m wrong then

[hidden wind]

wrong by not knowing a name

[plucky dome]

if p is prime then Z_(p^r) over addition has p^r-p^(r-1) generators but im struggling to see why

obviously you have divisors p^1,..p^(r-1) and then 2p^1,..,2p^r-1 up to (p-1)p^1,...,(p-1)p^(r-1)
shouldnt that be (p-1)^(r-1) divisors

[crystal vale]

Let 𝑅 be an integral domain containing C such that it is a finite-dimensional C-vector-space.

Then R not necessarily field , right?

Because if I take R is the set of all at most n degree polynomials over C.

[south patrol]

I believe R is automatically integral over C and hence C

Well like pick any r in R. Then C[r]/C is a finite extension

so it is trivial (as C is algebraically closed) and r is in C

Your example doesn't seem to have any multiplication on it

[crystal vale]

Yes

[round jay]

In exercise 6.4, why are we allowed to mod out a module by something which is not a subset of that module.

[rocky cloak]

Hint: ||for a linear transformation on a finite dimensional vector space injective = surjective = bijective||

[round jay]

That quotient module formula doesn't seem well formulated

[rocky cloak]

The start of the exercise specifically says to think of R^n-1 as a submodule of R^n by the described inclusion

[round jay]

I know, but "thinking of it as a submodule" doesn't make it a submodule

It feels handwavy

Unless you redefine the notation to mean what we want it to mean

[rocky cloak]

It just means we will from now use the symbol R^n-1 to refer to the described submodule

[round jay]

Gotcha

[rocky cloak]

I've never actually used the word myself except when answering the question of what it is called 😛

Though I guess in general Id use 'product of A and B' to mean the subgroup generated by {ab | a in A, b in B}, so I guess it's convenient to have a separate name for when that is a subgroup to begin with.

And I guess the name really belongs to the external construction anyway.

[quartz thistle]

can anyone help me with the proof of "every non zero element of Zn is a unit or a zero divisor"
i cant understand why all proofs have taken gcd of elements

[dull ginkgo]

Well, either it’s a zero divisor or a unit which is determined by if it’s coprime to n or not

Do you happen to know Bezout Lemma?

[quartz thistle]

nope

havent been taught this

i know euclids lemma

[dull ginkgo]

If we have a pair of numbers, a and b

There is a pair of numbers x and y (infinitely many, in fact) that satisfy:

ax + by = gcd(a,b)

[quartz thistle]

yes

i know this

didnt know the name of this lemma

[dull ginkgo]

Let’s say we are considering Z_n right

[quartz thistle]

if gcd(a,b)=1?

[dull ginkgo]

ax + by = 1

Taking that mod y

ax = 1 (mod y) and likewise
by = 1 (mod x)

I.e that’s the definition of a unit

[quartz thistle]

hows it a multiplicative inverse

can you break it down please😭

[dull ginkgo]

That’s the definition of a unit

There is an element b such that ab = 1 (mod n)

[quartz thistle]

oh right right

[quartz thistle]

understood

[dull ginkgo]

So yeah

But the other case is that there IS a common divisor

[quartz thistle]

1?

[dull ginkgo]

Yeah

[quartz thistle]

can you explain this
i just know this looking at solution

[dull ginkgo]

so let’s imagine that we have m, and the modulus is n

where d | m and d | n and d > 1

[quartz thistle]

d is the gcd

[dull ginkgo]

say ad = m, bd = n for sanity

[dull ginkgo]

need not be the gcd, can be any non-one divisor

But it helps to use gcd

[quartz thistle]

a and b be any arb integer?

[dull ginkgo]

No

Let’s make this a bit easier using Bezout again

So we have an + bm = gcd(n,m) = d right?

[quartz thistle]

yes yes

[dull ginkgo]

So

let’s say d * k = n

so n/d = k

if we multiply both sides by k:

we have:
ka * n + kb * m = n

Taking both sides mod n:

kb * m = 0 (mod n)

Thus it’s a zero divisor

So we have two cases

If m is coprime to the modulus, there is an x such that xm = 1 (mod n) so it’s a unit

Otherwise there is an x such that xm = 0 (mod n) so it’s a zero divisor

And notice that Bezout lemma gives us the other “halves”

So we can actually compute the x’s

[quartz thistle]

this helps a lot thank you so much

[dull ginkgo]

~~you can look up the “extended Euclidean algorithm” if you actually want to compute the x’s :3 ~~

[quartz thistle]

im rusty on basics, will understand this better after i study basics again

[dull ginkgo]

No worries

[quartz thistle]

gotchaaa

also

can i maybe dm you for other doubts in ring theory?

or ask here

[dull ginkgo]

Preferably ask here, a lot of people are more well versed than me

I’m currently on the rings section of an abstract alg textbook lol

[quartz thistle]

😭 this was so detailed i wasnt expecting anyone to break it down at this level

[quartz thistle]

yeah same

which book are you following

gallian?

[dull ginkgo]

Jacobson

[quartz thistle]

oh

have you done permutation groups?

[dull ginkgo]

Yes

Brb I need to shower lmao

[quartz thistle]

lol okay

[summer path]

Shower permutation group

[round jay]

Same. Have you looked at modules yet?

[dull ginkgo]

Not too much

Like I haven’t learned stuff like Nakayama’s lemma yet

Or like projective module stuff for homology

[crystal vale]

Any hint, in a group of G of order p^2 any normal subgroup of order p must lie in the centre of G. ( I don't want to use the result that G of order p^2 is an abelian group)

[dull ginkgo]

Every subgroup of G must either be trivial or have order p by Legrange’s theorem

Which is cyclic and thus abelian

?

[crystal vale]

Yes but how does that group lie in the centre of G follow?

[rocky cloak]

For any normal subgroup N < G, conjugation defines a map G -> Aut(N).

Think about what Aut(Cp) is and what such a map could be

[dull ginkgo]

[crystal vale]

Aut(Cp) ?

[rocky cloak]

Aut(N) meaning the group of automorphisms of N, and Cp being cyclic group of order p

[crystal vale]

Okay

[dull ginkgo]

How I’d handle this is

1. order is p^2, so every proper nontrivial subgroup has order p by Legrange theorem and is Cyclic

2. the center cannot be trivial, (this is a theorem, I forget the name, I can only remember the proof), so it must be all of G or is a cyclic subgroup of order p

3. you can prove that G/Z(G) is cyclic of order p in the latter case, and is thus abelian, so the whole group must be abelian

Wait shit

That just proves it’s Abelian lmao

That actually proves there is only two groups up to iso of order p^2 lmao

Z/p^2Z and Z/pZ^2

[crystal vale]

Yes by Conjugation classes we can prove that it has a non-trivial centre but I don't want to use that result

[dull ginkgo]

The rabid urge to use Sylow here

But that’s like taking out an ant with a bazooka

[south patrol]

Yeah

[dull ginkgo]

Immediately we have two cases: Z(G) = p^2 or Z(G) = p. The latter case implies Z(G) is the ONLY normal subgroup

[crystal vale]

But then any normal Subgroup lies in the center of G= G so it is correct but I don't want to use group action results here

[south patrol]

Well why not Z(G) = 1

[dull ginkgo]

can’t

[south patrol]

Showing the centre is non trivial is the hard part here

Yes but

Notknow wanted to avoid that

[delicate orchid]

ermmm sorry sweaty.... you're not allowed to be centreless...

[south patrol]

And that is the hard bit of this lol

Otherwise it follows very quickly

[dull ginkgo]

Stop being self-centered (Abelian)

[rocky cloak]

Yeah, I think proving that every p-group has non-trivial center is much harder than what notknow wants to prove

[south patrol]

Well it's pretty quick

but needs orb stab maybe

[crystal vale]

Without of that

[delicate orchid]

let G act on the subgroup of order p via conjugation, orbit stabiliser gets you a contradiction unless the size of the orbit is 1 => the subgroup is normal

[dull ginkgo]

so we want to show every normal subgroup of G of order p^2 must be contained in the center right without using Sylow

[quartz thistle]

how do we generalise the proof for For every prime p, Zp, the ring of integers modulo p is a field

[dull ginkgo]

we don’t those are the only non-rational prime fields

[delicate orchid]

quoitenting a ring by a maximal ideal is a field. This question is woefully incomplete if you're looking for any other answer

[crystal vale]

They don't introduce group action yet

[delicate orchid]

that's not my problem

[dull ginkgo]

Wait

[delicate orchid]

ok how about this

[dull ginkgo]

The intersection of Z(G) and H?

[south patrol]

How do you know Z(G) is non-trivial though mizalign

[dull ginkgo]

All subgroups must have order, 1, p, or p^2

[dull ginkgo]

I am trying to create a proof by contradiction

[delicate orchid]

ok no I'm just reproving orbit stabiliser

just use orb stab ffs

[quartz thistle]

How does it explain the theorem

[south patrol]

Writing Z_p gives me the ick

[celest furnace]

Certified p-adics moment🔥🔥

[south patrol]

Yes

[delicate orchid]

you need to be more specific with what you want from us. Do you want us to explain the construction of the finite fields that aren't quotients of Z?

[south patrol]

$W(\mathbb F_p)$

[cloud walrus]

Crystalline Potato

[dull ginkgo]

Consider G of order p^2. Then every subgroup of G must have order p^k for k = 0, 1, 2

Assume H is normal proper subgroup such that H cap Z(G) is a proper subgroup of H. Then clearly [H cap Z(G) : H] = p or p^2

[south patrol]

well [H: H \cap Z(G)] ig

usually

[quartz thistle]

this scares me😭

[south patrol]

maybe it don't matter

[dull ginkgo]

but the intersection must be in H and Z(G)

[delicate orchid]

https://tenor.com/view/erm-what-the-freak-puppy-stare-cute-reaction-gif-15215675160072457679

[dull ginkgo]

So consider Legrange’s theorem

We only have so many powers of p to work with here

[south patrol]

i don't see how you can get a contradiction from this if Z(G)=1 lol

[rocky cloak]

Sort of implicitly using group actions, but let x generate this normal subgroup of order p. And let y be an element in G.

Since the group is normal yxy^- = x^k for some k.

Notice that (yxy^-)^k = yx^k y^-, so y^2 x y^-2 = x^k^2, and similarly y^m x y^-m = x^k^m. Then think about what happens for m=p^2 and think about Fermats little theorem.

[crystal vale]

Aut(Cp) is isomorphic to U_p where U_p is the set of all units in Z_p. And its kernel is the centre of G, right?

[dull ginkgo]

Assume Z(G) = 1, then no element commutes with all G. The quotient group G/H must have order p and be cyclic, without a doubt because we asserted H is proper and nontrivial

[south patrol]

sure

[dull ginkgo]

I wonder…

Let’s rework this a bit

[rocky cloak]

Yes, or I guess the kernel is the centralizer of the subgroup, but that will also be the center of the group yes

[dull ginkgo]

H has to be cyclic if it’s nontrivial and proper

because it must have order p

Assume r is the generator

Would r not have to commute with everything and thus can’t have trivial center?

because G/H is cyclic, consider it’s generator s

Every element of G is of the form s^a * r^b

[delicate orchid]

reminds me of the proof that Inn(G) cyclic iff G

[dull ginkgo]

But the thing is

That H is normal

so s^a r^b = r^c s^a

sr^a = r^bs so
(srs^-1)^a = r^b

I’m trying to show that r must commute with s

Which if that happens, the center cannot be trivial, and (r) must lie in Z(G), so Z(G) must contain H which is generated by r

[crystal vale]

it implies that p divides k^(p^2) -1 , and Fermat little theorem says that a^p = a when a and p are relatively prime, right?

[dull ginkgo]

[rocky cloak]

Indeed

[crystal vale]

So I need to prove p divides k-1

[rocky cloak]

Yeah, or I guess you already did, since that's Fermats little

[dull ginkgo]

Full proof: assume H is proper, nontrivial, and normal in G of order p^2. Thus H must have order p.

Thus H is cyclic, generated by element h

G/H must therefore have order p^2/p = p and is also cyclic, so let it be generated by sH = Hs. Thus for each xH we have xH = (sH)^k = s^k H

We have that u = shs^-1 must be in H, and s cannot be in H. What’s left is to show shs^-1 = h

I am stuck here and pondering what I can do

[crystal vale]

It implies that k^p - 1 =0modp implies that k-1= 0 , right ?

[crystal vale]

G/N is isomorphic to Aut(N)?

[rocky cloak]

Not usually, no

[crystal vale]

Kernel is the centralizer of the N

[rocky cloak]

That's right, so G/C(N) will be a subgroup of Aut(N)

[crystal vale]

What will be the image of this mapping?

Should I need to find an image ?

[rocky cloak]

That completely depends on G and N

But in this particular case, G has order p^2 and Aut(N) has order p-1

So there are not many possible maps G -> Aut(N)

[crystal vale]

If C(N) can not be {e} and if C(N) has order p then Aut(N) has a subgroup of order p which is not possible. Thus C(N) = G, right?

[amber cradle]

We have p cosets of a group that has p^2 elements, how many elements does each coset have

[dull ginkgo]

p

[amber cradle]

You’re done then

[dull ginkgo]

Not quite

[amber cradle]

?

[dull ginkgo]

Yes both are cyclic (prime order p)

[amber cradle]

Order of each coset is the order of H

[dull ginkgo]

But it’s showing commutativity between their respective generators

[amber cradle]

What is the question you’re proving

I might’ve misread

[dull ginkgo]

This

[south patrol]

Just kinda funny how finding a solution that doesn't use that known fact seems to be taking longer than learning about group actions and providing a proof from that

[dull ginkgo]

Yeah lol

[dull ginkgo]

Wish I could just use conjugacy classes and fruit ninja that shit

[amber cradle]

I see

[dull ginkgo]

Sylow is way overkill

[amber cradle]

Can you use that a group of order p^2 is either isomorphic to the cyclic group or the direct product of order p cyclic groups?

[dull ginkgo]

Welp I give up I have more productive things to do and Jacobson made me not want to touch a group ever again

[dull ginkgo]

He wants to avoid that route or the one showing Z(G) cannot be trivial

[dull ginkgo]

Of which I autopiloted and was not what was desired

[celest furnace]

While we’re at it, I want the answer without the definition of the center either

[dull ginkgo]

Here sorry

[dull ginkgo]

Centralizer of G

[amber cradle]

Zentre

[south patrol]

Zentrum

[amber cradle]

What do you want to use?

Conjugation classes are like the most primitive thing in groups

Other than I guess primitive groups

[crystal vale]

They don't define conjugation classes yet

[amber cradle]

Wtf? How do you get to normal groups without conjugation

[crystal vale]

They do not define group action yet and this question included in Cayley Theorem section

[dull ginkgo]

OH

now I know the means of production

[amber cradle]

Conjugation classes aren’t necessarily from group actions

[dull ginkgo]

Inn(G)

It’s the action of Inn(G) on G

Is it not?

[dire siren]

@crystal vale Assuming I didn't make any mistake:
If the group is cyclic we are done. Let's assume it's not cyclic, so then the order of any nontrivial element is p.

Let H={1,x,x^2,...,x^(p-1)} be normal. Let y≠e be arbitrary. Then yH=Hy, so xy=yx^m, for some 0<m<=p-1.
It follows that y^(-1)xy = x^m.
By induction: y^(-k)xy^k = x^(m^k).

For k=p we get x=x^(m^p), so x^(m^p - 1) = e, so then p must divide m^p - 1. But m^p - 1 = m - 1 (mod p), so p | m - 1, but given the range of m, it follows that m=1. Thus xy=yx.

[dull ginkgo]

That is smart

[crystal vale]

Got it, thank you

Any hint, if a cyclic Subgroup T of G is normal in G, then show that every subgroup of T is normal in G.

[dim widget]

did you try writing out the condition that every cyclic subgroup is normal?

It follows pretty directly from that

[crystal vale]

Yes because the subgroup of the cyclic group is cyclic, right?

[dim widget]

no it doesn’t have anything to do with a subgroup of a cyclic group being cyclic

[crystal vale]

Wait I wrote the wrong question, sorry

I mean I need to prove that if T is a cyclic normal subgroup of G then every subgroup of T is normal in G

[delicate orchid]

hell, it's an action of G on G

[dull ginkgo]

QUOTIENT OUTS THE LAZY FUCKS

[delicate orchid]

https://media.discordapp.net/attachments/1104929486438420480/1225193473590820917/attachment.gif

[tribal moss]

Hint: Each subgroup of a cyclic group is determined by its order.

[round jay]

If we look at a ring R as an R module over itself, and it's a noetherian module, does this mean it's commutative as a ring?

[next obsidian]

Thanks Vegeta

[crystal vale]

But G is not finite

[rocky cloak]

It means that it's a Noetherian ring. Don't know if you meant to ask that, but it's unrelated to it being commutative.

[tribal moss]

Oh, right.

New attempt. Each subgroup of a cyclic group is determined by which multiples of a generator it contains, and those are the same multiples no matter which generator you pick.

[next obsidian]

It actually doesn’t matter as long as the cyclic subgroup is finite

The only issue is when the cyclic subgroup is infinite does Tropo’s comment not finish it

[rocky cloak]

Probably easiest to split into the two cases of T being finite and T being Z

[next obsidian]

Yeh

[tribal moss]

Yeah.

[old sphinx]

@tribal moss thanks for the help

[round jay]

Aren't noetherian rings required to be commutative? Thats what it says in aluffi

[next obsidian]

no

I think

[round jay]

Rip

[next obsidian]

there's a notion of left noetherian and right noetherian and I think noetherian is asking both at the same time

and I don't htink that implies commutative

but I'll keep it real, idk shit about noncomm rings

so maybe I'm wrong

[dull ginkgo]

I think it’s like a generalization of noetherian modules

[rocky cloak]

The nxn-matrix ring over a field is Noetherian (on both sides) for example

[next obsidian]

And that’s commutative

[rocky cloak]

If n=1 I guess

[dull ginkgo]

left or right modules depending on your

[next obsidian]

[rocky cloak]

There are noncommutative division rings, e.g
the quaternions, and they are very Noetherian as they have no nontrivial ideals.

[dull ginkgo]

nontrivial two sided ideals

[teal vessel]

I'm not sure exactly what it's asking for in the last part of this question. For reference, G-bar is D_16/<r^4>

[dull ginkgo]

[teal vessel]

the "isomorphism type" isn't something that's been described before. Am I just supposed to show the preimage of each element of H-bar?

preimage*

[dull ginkgo]

Oh I see

[rocky cloak]

"isomorphism type" really just means "which named group is this isomorphic to"

[dull ginkgo]

D_16 is the order 16 or

[teal vessel]

ah, so check if my intuition that it's D8 is correct?

[teal vessel]

it's order 16, yeah. The symmetry group on an octagon

[dull ginkgo]

Notation differs usually by a factor of 2

Anyway so

The quotient is basically just slapping an extra relation lmao

Wait D_16/r^4?

Doesn’t that just cut down on the order of r?

[teal vessel]

sr already has order of 2 in all dihedral groups

but yeah, it takes the order mod 4

[dull ginkgo]

Wrong presentation

There’s 2 lol

Yeah so you’re using

[teal vessel]

{s,r | s^2=r^8=1, sr=r^-1s}

[dull ginkgo]

(sr)^2, (s)^2, r^8 as representors right

oh

[teal vessel]

for D16

[dull ginkgo]

The last one I think is equivalent to (sr)^2 = e because s is involute

yeah

[teal vessel]

but yeah, I was just looking for clarification on the instruction. I have an understanding of what I'm looking for now

[dull ginkgo]

Idk what the isomorphism type is either tbh

I know the Klein 4 proof is rather easy through representation

Just rhe

[teal vessel]

yeah, a previous part already asked for the orders of the elements, so it's sufficient to show the orders all being 1 or 2

which is already shown

[dull ginkgo]

[teal vessel]

so name the archetypal group

[dull ginkgo]

Probably?

[teal vessel]

but I still have to look at the preimage of H-bar

[dull ginkgo]

What is G by default?

[teal vessel]

D_16

I got it, just tedium

[dull ginkgo]

Lmao is it proving that two elements involute elements who’s product is idempotent generate a dihedral group

Which sucks

[teal vessel]

I'm sure that makes sense, but I don't have those words in my vocabulary yet lol

idempotent ❌
involute ❌

[dull ginkgo]

Idempotent: some power of x is the identity
Involute: x^2 is the identity

[teal vessel]

I'll get there

eventually

[delicate bloom]

not quite

[teal vessel]

still only on chapter 3

[mighty kiln]

Idempotent is x² = x

[dull ginkgo]

AH nilpotent

brain fart

[mighty kiln]

I don't think it's that either

[dull ginkgo]

No?

[mighty kiln]

Nilpotent is x^n = 0 for some n

[dull ginkgo]

Oh yeah nilpotency in groups vs rings

[mighty kiln]

But notice how left side is multiplication while right side is additive identity

It's a ring

[teal vessel]

and then K4 is just H8/<-1> so that's easy to show

[dull ginkgo]

some resource I used before defined it for groups as x^n = e

For groups

I forget which

Rings it’s that some power is the annihilator

[mighty kiln]

Surely that would be finite order

[dull ginkgo]

Yeah I should use that

[mighty kiln]

And nx = 0 in a ring / Abelian group would be torsion

[dull ginkgo]

Torsion, that’s the word I was looking for

Thanks

[next obsidian]

Don’t get it in your testes. It isn’t fun I hear

[dull ginkgo]

yes, I know from experience

[next obsidian]

Real?

[dull ginkgo]

yes.

[next obsidian]

Rip in peace

Does it hurt?

Like is it like ouchie

Or is it like YOWZA WOWZA AAAAAAAAA

[dull ginkgo]

Agony, had to go to the hospital.

Also #groups-rings-fields

[next obsidian]

What?

[next obsidian]

Rip bruh. Is it just sudden like “WHAT THE”

[dull ginkgo]

yes

[next obsidian]

Bro that means it could happen to me next second

Every second I could be one second away from testicular torsion

[dull ginkgo]

“Oh boy I love not having unbearable pain”
SO(3):

[next obsidian]

[runic lava]

Hello, can anyone help me understand the Invariant factor decomposition method, if there is one, for Finitely generated abelian groups ?

[dim widget]

what don't you understand about it?

[old sphinx]

@dull ginkgo thank you

[dull ginkgo]

Welcome lol

[old sphinx]

Did you get the idea

[sterile gazelle]

is this for AP physics

[tawny dune]

no

[coral spindle]

Not that kind of field!

[hollow mica]

If Gal(K/F) = S_3, I want to show there exists a f ∈ F[x] such that K = splitting field of f.

First I note that S_3 has as a subgroup of index 3, so by the Galois correspondence there is an intermediate field L between K and F such that [L : F] = 3.

Now by the primitive element theorem we know there exists an r ∈ L such that F(r) = L.

Now apparently the minimal polynomial f of r in F should work.

But I have a couple unresolved questions

• why is the degree of f equal to 3?
• why is the splitting field of f equal to K?

[amber cradle]

I took group theory, and I feel very proficient in groups, but my major (bs math with specialization in economics) doesn’t let me take ring theory and field theory

How do I self study ring theory, module theory, and field theory over the summer so I feel as proficient in them as I do groups?

[celest furnace]

By putting a lot of time into it and reading a great book. I would recommend looking up an old class and following their exercises

[amber cradle]

I was referencing dummit and foote for this alg number theory reading program i’m doing, but I discovered Lang’s was far more motivated for ant. Should I stick with Lang?

[south patrol]

Well if it's finite Galois (which you sem to be assuming) this is always the case regardless of the group

[celest furnace]

I mean now this is book stuff but in general lang is probably way worse for learning

[amber cradle]

Oh lol

[celest furnace]

Book stuff goes in #book-recommendations

[south patrol]

Like you can always just ||take a set a_1,...,a_n of generators and multiply together their minimal polys||

[south patrol]

It's a general fact that if L = F(r) then [L:F] = n where n is the degree of the min poly of r. To see this, note that 1) you can write r^{n} as alinear comb of 1,...,r^{n-1}, and hence by induction you can see {1,...,r^{n -1}} span. and linear independence of this set is just asserting that the poly is minimal

Another way to see this is that if $L=F(r)$ then $L \simeq F[x]/m(x)$ where $m$ is the min poly of $r$

[cloud walrus]

Crystalline Potato

[south patrol]

and thelatter is well known to have the same bsis i mentioned

[hollow mica]

Sorry, I misstated my question. It should be there exists an irreducible cubic f ∈ F[x] such that ...

[south patrol]

Ah well that's more interesting, sure

Okay so we just need to see why the splitting field is k now

[hollow mica]

Yea

Wait before

[hollow mica]

you can write r^n as a linear combination of 1, r, ..., r^{n-1}

[south patrol]

Just from the min poly

[hollow mica]

Oh

[south patrol]

if $r^n + a_{n-1} r^{n-1} + \dots + a_0 =0$ for some $a_i \in F$

[hollow mica]

yea I see now

[cloud walrus]

Crystalline Potato

[south patrol]

ye cool

[hollow mica]

Lemme try to see why 1, ..., r^{n-1} spans L

every element of L is the division of two polynomials in r

[south patrol]

Well it's a polynomial in r

Although ig you have to prove that too at this point lol

[amber cradle]

The book recc says Lang is the goat, do u personally dislike it?

[hollow mica]

Maybe division algorithm to see this?

[south patrol]

But yeah tbh this is like field theory stuff i suggest you review, like kind of a prereq for galois theory imo

[celest furnace]

Which Lang? Lang Algebra or something else?

[south patrol]

Well you don't even need this i think

if you consider F[r] then it already contains r^-1

again from the min poly

good exercise

[hidden wind]

i can’t believe my luck as i got a physical copy of lang algebra for free

[amber cradle]

Lang Algebra

[south patrol]

i just wanna get a copy of Bourbaki algébre

well like all parts of it too

[hollow mica]

If I look at the minimal poly of r in L = F(r), and then multiply both sides by r^{-n}, I get a polynomial in r^{-1}

[hidden wind]

i have part two of their general topology

[south patrol]

Well kinda the inverse of what you want but yeah

[hollow mica]

Oh oops yea

[south patrol]

hint: a_0 is not 0

[hollow mica]

Look at the min poly of r^{-1}, say it is
a_m r^{-m} + ... + a_1 r^{-1} + a_0 = 0,
then after multiplying both sides by r^{m-1} I get
a_m r^{-1} + a_{m-1} + ... + a_0 r^{m-1} = 0,
or
r^{-1} = 1/a_m * (a_{m-1} + ... + a_0 r^{m-1})

Ok that works I think

[south patrol]

Sure

[hollow mica]

Because a_m, the leading coeff of the min poly of r^{-1}, is obviously nonzero

[south patrol]

leading coefficient is always 1

by definition

[hollow mica]

ye that too

oh

yea ok

[south patrol]

You can also note that if $r^n + a_{n-1} r^{n-1} + \dots + a_0 = 0$ then $-a_0 = r(r^{n-1} + a_{n-1} r^{n-2} + \dots + a_1)$

[cloud walrus]

Crystalline Potato

[south patrol]

dividing by -a_0 you see r is a unit

(and a_0 is nonzero by minimality of this poly)

[hollow mica]

we're in a field so everything is a unit?

[south patrol]

A unit in F[r]

[hollow mica]

oh

[south patrol]

I'm showing r^-1 is in F[r]

[hollow mica]

ok

ok I see that makes sense

so so far we've shown that the spanning set of {1, r, ..., r^{n-1}} contains r^n and r^{-1}

what we want to show is it is all of L = F(r)

Ok and by induction the spanning set contains any power of r (positive or negative)

[south patrol]

Indeed

Though I would say "the span of". A spanning set of a vector space should be a set whose span is the whole thing

[hollow mica]

Sorry yea

A little bit closer to showing every p(r) / q(r) is in the span of the r^i over F

Wtf they just take this fact for granted

[cobalt heath]

Lool

[delicate orchid]

As they should

[cobalt heath]

Just take inverse in F[[x]] and see how it reduces to F[r]

[delicate orchid]

I mean… yeah?

[cobalt heath]

Doubt this would actually work but >.>

What would be the easiest way to show that it is a field, hmm

[celest furnace]

Yeah I mean ask anyone here an entire 0 people will tell you that is good for self learning of all things 🤣

[cobalt heath]

If I is an ideal of F[x]/J, then there shall be an ideal I’ of F[x] containing J.
If J is maximal ideal of F[x], I’ should be F[x] or J.
I guess this tells about F[a] being a field.

[chilly ocean]

given a unitary ring $R$, a subring $S$ and a subset $Z \subseteq R$, is the smallest subring containing both $S$ and $Z$ the ring $\left{ P(z) : z \in Z \land P \in S[X] \right}$?

[cloud walrus]

Matias

[chilly ocean]

is that correct? polynomials with coefficients in S evaluated in all elements of Z?

[crystal vale]

Any hint, if G is a group of order pq, p< q, p and q are prime numbers and p does not divide q-1 then G is abelian.

I know Z(G), centre of G can be {e} or G in that case, how can I eliminate Z(G) = {e} case?

[chilly ocean]

use Sylow theorems to fully characterize groups of order pq, commutativity will follow

[next obsidian]

I feel like this is above the pay grade of the tools bro can use

[crystal vale]

They don't introduce Sylow yet

[tough raven]

Note that the denominator must not be divisible by the minimal polynomial of r, call it p. From here, it suffices to show that any polynomial f not divisible by p has a multiplicative inverse modulo p.
By the Euclidean algorithm, you can find a gcd of f and p and write it as a linear combination xf + yp, where x, y are polynomials.

Because p is irreducible, the gcd is either p or 1 (up to scaling by a constant polynomial). Because f is not divisible by p, it can't be p. Thus we can write 1 = xf + yp for some polynomials x, y. But then x is the inverse of f modulo p.

(Abstractly, we're reproving that the quotient by an irreducible element in a PID (or inspecting the proof above, more generally Bézout domain IG?) is a field i.e. that (p) is a maximal ideal if p is irreducible.)

In short

[tough raven]

.

[rocky cloak]

You'll have that Gal(K/L) is a subgroup of order 2, the splitting field corresponds to the core of this group, and S3 has no normal subgroup of order 2.

[wraith swan]

I got an exercise to show that fibred coproducts exist in Ab. Now I get that it cannot be a subset of the direct product/sum as then it wouldn't be possible to equalise elements. (Just as in Set)
Am I right to look at a subset of the free product but with extra relations dictated by the fibres(?) ?

Hm and all morphisms are homo, so I guess that limits it a lot. Let's come up with an example

[delicate orchid]

a pushout of groups is an amalgamated free product, Ab is a full subcategory so the pushout is also an amalgamated free product

[tribal moss]

How would that work in the case of a pushout of 0->G and 0->H?

[delicate orchid]

it would just be the free product of G and H

which makes sense, 0 is initial so there's no extra data added when you go from the coproduct of G and H to that pushout

[tribal moss]

But the free product isn't even abelian.

[rocky cloak]

It would be the abelianization of the amalgaamatef product. Not all subcategories are closed under colimits

[wraith swan]

Ok, I should look into that, sorry didn't see your message earlier.

[wraith swan]

Wait so the direct sum is also a free product? I guess... But the abelian groups make it special that the direct sum is isomorphic to the direct product, is there something similar for the pushout? Pullbacks are just subsets/groups? Of the product right?

[cobalt heath]

Does colimits change when you switch to subcategories? Under which conditions do you have preservation of colimits (or limits)?

[rocky cloak]

So the question just comes down to when the inclusion preserves (co)limits. Which would happen for example if the inclusion has an adjoint.

For example abelianization is left adjoint to the inclusion of abelian groups into groups, so the inclusion preserves limits.

And this also comes with the added benefit that abelianization preserves colimits, so a colimit of abelian groups is just the abelianization of their colimit in Grp.

You have exactly the same situation for the inclusion of sheaves into presheaves

[cobalt heath]

Hmm, I don't quite get how inclusion comes into play here

[rocky cloak]

Well, I mean that's just what it would mean for the limit in a subcategory to agree with the one in the larger category.

Do you get the same result of you first include your subcategory into the bigger one, then take the limit vs do you take the limit first

[quartz thistle]

is Zn[i] an integral domain?

[rocky cloak]

Depends on n

[quartz thistle]

has to be prime to be an integral domain?

[rocky cloak]

Yes, but that's not enough

[quartz thistle]

what else

[cobalt heath]

Ah, so inclusion here is the embedding of subcategory into the category? I thought the inclusion you were talking about was between objects of a category.

[rocky cloak]

Before I tell you the answer, maybe play around with Z/2[i], Z/3[i] and Z/5[i]

[tribal moss]

This also depends on what you mean by A[i] in the first place when A is a ring that is not a subring of C.
Usually it will be understood as meaning A[X]/<X²+1>, in which case it always adjoins a new element even if A already contains an element that squares to -1.

[quartz thistle]

i made the cayleys table of Z/3[i], ab is only 0 when either of them is 0

[wraith swan]

Sorry, what is this [i] notation? I came a cross it but it wasn't explained

[quartz thistle]

a+bi

complex number

[rocky cloak]

That's right, an what about Z/2

[quartz thistle]

doesnt A[x] generally denote a set of polynomials

[quartz thistle]

should be the same right

[tribal moss]

Yes, and then you quotient out the principal ideal generated by the polynomial X²+1.

[quartz thistle]

i havent studied ideals yet

[tribal moss]

Hmm, then I'm not sure you have the ingredients to even interpret the notation "Zn[i]" in a principled way.

[tribal moss]

Notations of the form "A[B]" generally mean something like "the smallest ring extension of A that also contains B".
This needs to be made precise in different ways according to which kind of things A and (especially) B are.

[quartz thistle]

im confused why Zn[i] is int domain when n is prime but Z[i] can be a zero divisor when there is no modulo

[wraith swan]

Ok, thank you. Haven't yet read about rings

[quartz thistle]

in my textbook set of polynomials has been written as Z[x]

[rocky cloak]

Well what is (1 + i)^2

[tribal moss]

Yes, that's the most common use of the notation. Conceptually, we can understand Z[x] as "the smallest ring of functions that contains all the integers (each viewed as a constant function), and also contains the identity function x".

[quartz thistle]

2i

[rocky cloak]

An you were working in Z/2[i]...

[quartz thistle]

does it also make a field?

[tribal moss]

No, because we don't require that everything has reciprocals.

If you want "the smallest field that contains such-and-such", the usual notation is A(B) instead of A[B].

[quartz thistle]

what if it is a constant

[quartz thistle]

understood

[rocky cloak]

Anyway, the answer is that n must be a prime congruent to 3 modulo 4. But I don't think you quite have the tools to see why that is yet. Would need to at least know a little bit about either prime ideals of finite fields.

[quartz thistle]

what is prime congruent to 3 modulo 4

also im getting the cayleys table wrong for Z2[i]

[tribal moss]

A more understandable criterion on this level would be that if Zp already contains something that squares to -1, then Zp[i] is not an integral domain.

Namely if a²=-1, then (a-i)(a+i) = a²-i² = (-1)-(-1) = 0.

In Z/2Z, 1 and -1 are the same, so 1 itself "squares to -1".

[quartz thistle]

can you give a bit easy example