groups-rings-fields

[serene dune]

also my head can never accept the fact that, same action in semidirect product yields different groups

[coral steeple]

I meant to reply sorry. Maybe when I know more topology I can revisit your proof

[grizzled spindle]

I’m getting something weird when I try to use this, is this not an action?

[kind temple]

what are you getting

this is definitely a group action

[grizzled spindle]

That any H is a normal subgroup somehow

[kind temple]

how do you get that?

[grizzled spindle]

Doing this I get that the kernel of phi is exactly H but I know I just messed up, idk what

[kind temple]

S_G/H is the set of all bijections G/H --> G/H?

[grizzled spindle]

Yes

[sacred mulch]

Saw this problem in Contemporary Abstract Algebra. I'm don't want help with proving it but I don't see how it's correct. I came up with this counterexample. The cayley table is:

$\begin{matrix}
_ & a & b & c\
a & a & b & c\
b & b & a & c \
c & c & c & a
\end{matrix}$

[cloud walrus]

Lukecell
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sacred mulch]

Where a is the identity element, and it's possible to create a subgroup {a, b}

[kind temple]

cb = c means that b is also the identity

[sacred mulch]

Does that mean it's impossible for a group to have an order < 3?

wait nvm im an idiot

[kind temple]

no. if you intend for a and b to be distinct, then this just isnt a group. otherwise, a and b are equal. there may be some relations which show that c is also the identity here. otherwise, this is just a group with two elements

also, sanity check, the order of any subgroup must divide the order of the whole group

2 does not divide 3, so {a,b} can't be a subgroup of this group (if it were a group)

[sacred mulch]

I think I get it, in this instance a=b since bc=ac (where c is the only other element) so it's basically a group of 2 elements

[kind temple]

i believe thats correct, the presentation you have here is Z/2Z

[delicate bloom]

a group multiplication table can't have repeats in any row or column

(why? 😮 )

[kind temple]

Take G = S_3 and H = {(), (12)}. H is not normal in G. put g = (13). m_g is not a bijection since m_g((12)(13)) = H, and (12)(13) = (132) is non-trivial

[kind temple]

somewhere in here you are assuming H is normal, implicitly.
i can't quite figure out where

[void cosmos]

this isn't a particular problem

but is there a way to generlize like

like i learnt about de rham cohom and theres both homotopy invairance and meyer viteoris

turns out there is too with simplicial homology

and like probably the proofs go the same ( iahvent gotten there yet )

is there like

a generlization for this

or is it jusut that cohomology and homology are "dual" ( in a way i still don't know ) and that's it

[quiet pelican]

It’s not well defined if H is not normal
Different choices of a within the same coset would give different results

[inner steppe]

I was watching Michel Penn's video on Quotient Ring with title: Abstract Algebra | Constructing a field of order 4.

I don't really understand the last step and he doesn't explain it either. He says that (x^2+x+1) is the ideal so it is like 0. Anyways how is that equality true. I am confused.

[rose ridge]

in the quotient domain youre essentially doing computations mod (x^2+x+1), which means that you treat x^2+x+1 to be equal to zero

in the same sense that 3 mod 3 is congruent to zero

so in Z_2[x]/(x^2+x+1), x^2 and x^2+x^2+x+1 represent the same element

as another example, we have x^3 = (x^2+x+1)(x-1) + 1 = 1 mod x^2+x+1

[inner steppe]

I am following but how come 1 and not x?

Anyways so I have like two mods? Excuse me if that sounds stupid but I have to check mod 2 and as well as the polynomial p(x) where Z_2/p(x) ?

[rose ridge]

thats right

[rose ridge]

kinda confused on why you think it should be x

[inner steppe]

The reason why I think that doesn't even sound logical to me. I'd rather you explain the right way.

[rose ridge]

err alright then, ill be more specific with the steps
x^3 = (x^3-1) + 1 = (x^2+x+1)(x-1) + 1 %= 0(x-1) + 1 = 1
where we used the congruence mod (x^2+x+1) at %

tried to use * instead of % but got italized lol

note that since this is all over Z_2, we may also replace all the - by +, but i didnt to make it more clear

[inner steppe]

Okay thanks, this was so beautifully explained! Hopefully I get it as I solve more questions!

[rose ridge]

glad i managed to get it across 👍

[opaque finch]

I have a small doubt

Is $\mathbb{Q}(2^{1/3},\omega)$ same as $\mathbb{Q}(2^{1/3},i3^{1/2})$ ??? $\omega$ here is cube root of unity.

[cloud walrus]

mycroftholmes1703

[opaque finch]

can someone verify if it is true

[south patrol]

Just look at the formula for omega

[opaque finch]

yes

$\omega = \frac{1}{2} + i\frac{\sqrt{3}}{2}$

[cloud walrus]

mycroftholmes1703

[opaque finch]

And that's why I feel it's true

[south patrol]

Yes it is true

sqrt{3}i and w can be obtained from one another by adding and multiplying by rationals

[inner steppe]

What is a primitive polynomial and what is a irreducible primitive polynomial? I really cannot understand the formal definition.

• An element f in F is called a primitive element in F if it generates the entire multiplicative group, i.e., (F \ {0}, ·) = <f>.
• A polynomial k(x) in F[x] is called a primitive irreducible polynomial if [x] is a primitive element in F[x] / (k(x)).

[coral spindle]

Can you elaborate on what in the definition is confusing you

[inner steppe]

Honestly now that I think about it, it is not confusing and I feel like I understand them but when I am solving a question I just get stuck for example this problem.

Let the polynomial $ k(x) = x^3 + 2x + 1 \in \mathbb{Z}_3[x] $ and
$ R = \mathbb{Z}_3[x] / (k(x)) $, the ring of equivalence classes in $ \mathbb{Z}_3[x] $ modulo
$ k(x) $. Determine whether $ [x] $ is a primitive element in $ R $?

Answer: Since $ k(x) $ is irreducible, $ R $ is a field with $ 3^3 = 27 $
elements. The multiplicative group has 26 elements. The possible orders for elements
in $ (R \setminus {0}, \cdot) $ are therefore 1, 2, 13, and 26. Let’s take powers of $ x $ and compute:

$$
[x^3] = [x + 2],
$$
$$
[x^6] = [(x^3)^2] = [(x + 2)^2] = [x^2 + x + 1],
$$
$$
[x^{12}] = [(x^2 + x + 1)^2] = [x^4 + 2x^3 + 3x^2 + 2x + 1] =
[x(x + 2) + 2(x + 2) + 2x + 1] = [x^2 + 2],
$$
$$
[x^{13}] = [x^{12} \cdot x] = [(x^2 + 2)x] = [x^3 + 2x] = [x + 2 + 2x] = [2].
$$
Thus,
$$
[x^{26}] = [(x^{13})^2] = [2^2] = [1].
$$
We see that $x$ does not have order 1, 2, or 13, but has order 26. Therefore, it generates
the entire multiplicative group and is primitive.

My question
What do they mean by: The possible orders for elements are therefore 1, 2, 13, and 26."

[coral spindle]

OK so you should ask the question at hand instead

[cloud walrus]

DatON1

[coral spindle]

Do you know the theorem of Lagrange

[inner steppe]

if G is a finite group and H is a subgroup of G then |H| | |G|?

[coral spindle]

Yes.

Set H to be a subgroup generated by x.

Let G, of course be R\{0}, which is a group under multiplication.

What do you get?

[inner steppe]

I am not sure

[coral spindle]

I really don't like it when people say they're not sure. When you say that, it seems to imply that you do have something in mind. So tell me.

If you just don't know, I would prefer you tell me that you just don't know

[inner steppe]

I don't understand the question. Are you asking what H is or what G is?

or both?

[knotty badger]

I think they’re asking you to describe H

[coral spindle]

I am setting G to be the group R\{0}

I am setting H to be the subgroup of G generated by x

Now what does the theorem of Lagrange tell you?

[inner steppe]

That the number of elements generated by x will divide then number of elements in G. Other than that I don't know.

[coral spindle]

OK

What is the number of elements in G?

Recalling, of course, that I am setting G to be R\{0}

[inner steppe]

26 elements

[coral spindle]

That's right!

Now you referred to this quantity given by the number of elements generated by x

Do you know another name for this quantity

[inner steppe]

order

[coral spindle]

(you do, but perhaps you are unaware of this connection)

Yes indeed

So the theorem of Lagrange now says that the order of x divides 26, agreed?

[inner steppe]

yes

[coral spindle]

So what can the order of x be? There are only so many divisors of 26

We cannot determine the order with certainty because we have not specified x, but we can look at options

[inner steppe]

1, 2, 13 and 26. Okay now I am following!

[coral spindle]

Yeah, so we're done.

Those are the only possible orders.

[elfin wraith]

Did not scroll up far enough to see the original question and thought you were just losing it trying to apply lagranges theorem to \mathbb{R}^x

[coral spindle]

not calling it \mathbb R^\times is fucked up

[elfin wraith]

I’m on my phone, I’m not putting in anymore backslashes than I have to

[inner steppe]

so the question states: Determine whether [x] is a primitive element in R ? I guess I have show that all elements in R can be generated using x but then I wonder what is the point of doing everything we did so far. How is finding out the order of possible subgroups being 1,2, 13,26 doing? Sorry but I still cannot see the connection.

[coral spindle]

OK you haven't understood something

The order of the subgroup <x> is also the order of the element x.

We have shown that the order of the element x is one of those options

To be a primitive element, we require <x> = R\{0}. This is exactly to say that the order of x is 26.

We showed that the order of x is not any of the other options; it must be 26.

[somber sleet]

Guys I have a really dumb question. If I want to show that the identity is unique in a field. Is it wrong to show that if 1 and 1' are both the identiy if$x\neq 0$, then $1 x= 1' x$ implies that $1 = 1'$

[cloud walrus]

damn_guuurl

[somber sleet]

I feel like this is wrong and it's not enough to show uniqueness, but I odn't see why

[coral spindle]

You are overworking!

Hint: consider $1 \cdot 1'$

[cloud walrus]

Boytjie xoxoxoxoxo

[elfin wraith]

Technically you could do what you did because cancellation holds in fields, but the other proof is shorter and more general

I.e. works in any ring

[coral spindle]

You don't even need the rest of the ring structure tbh

[elfin wraith]

Fair

[somber sleet]

yes of course I know that proof, but my student all did the other thing

[inner steppe]

Yes okay I understand so all of that was to show tha tonly 26 was a possible. But then I don't understand what they were doing with the x's in the solution there. Sorry if I am bothering you but I really have spent a lot time going back and forth with this.

[somber sleet]

and I feel like that's not necessary at all

[coral spindle]

Elaborate on what is confusing you

[coral spindle]

OK so can you write out their logic in full

[knotty badger]

There’s sometimes a phenomenon where people are suspicious of a proof that feels too short

I feel like that may be what’s happening here

[coral spindle]

I'm inclined to agree

[knotty badger]

Like in maths you can totally have proofs which are one-liners

But irl you never really have a logical argument that’s just a single line, and that also works

[inner steppe]

For starters: "Let’s take powers of x and compute:
[x^3] = [x + 2]"
What they trying to show here?

[knotty badger]

There’s always some room for nuance

[coral spindle]

[x]^3 is not the identity, so the order of [x] is not 3.

[somber sleet]

they take an arbitrary $x$, then assume that there are the identites $1$ and $1'$ such that $1x = x$ and $1'x=x$. They then set it equal, devide by $x$ by assuming that $x\neq0$ and get that $1 = 1'$. But I feel like it is rally strange

[cloud walrus]

damn_guuurl

[coral spindle]

take an arbitrary x
This is faulty logic

They need to take a specific x, otherwise what they write is true for the empty set

[knotty badger]

Yeah so strictly they’ve shown that, for any x not equal to 0, you can deduce that 1 = 1’

[coral spindle]

Ofc, they can prove that their set is nonempty by appealing to the identity, but this is precisely to say that x = 1 and it's the same proof.

[somber sleet]

obviously in the field, which is not empty by assumption

[coral spindle]

See above

[elfin wraith]

Bit of a random question, just proved in the notes for my non comm class that:
if R is left noetherian and I is a proper ideal then there exist primes P_1,…,P_n containing I such that the product is in I and that min(I) is finite and non empty

Quite happy with this theorem and the proof, but afterwords we note that the noetherian property is really stronger than we need because we just need 2 sided ideals to satisfy the ACC, but that leads me to a somewhat interesting question, what would a ring which satisfys the ACC for all its 2 sided ideals but isn’t its self noetherian look like?

[knotty badger]

They are trying to show that the order of x is not 3

[elfin wraith]

Not sure if there’s a nice example or if someone could just cook one up, but I’m struggling to picture what that ring could be or a nice way to create one

[inner steppe]

but what is the result saying, I mean what [x^3] = [x+2]

[knotty badger]

It means the equivalence class of x^3 is the same as that of x + 2

In other words, x^3 ~ x + 2 for ~ the equivalence relation under consideration

[coral spindle]

Do you know what is meant by a quotient field? If I write R[x]/(x^2+1) do you know what this object is?

Just say if you don't know.

[inner steppe]

Not formally but if I understand correctly given the field R[x], we can get a quitient field R[x]/(x^2+1) if and only if x^2+1 is irreducible in R[x]?

[coral spindle]

So you don't know what it means, OK

You need to recap the formal definition of a quotient ring.

This [] notation is part of the formal definition of a quotient ring.

Idk what textbook/notes you're working from, but this should be in any introduction to abstract algebra and rings.

[inner steppe]

I cannot find it in my book and prof never mentioned it formally. He just showed examples and told us what it was.

[coral spindle]

OK so I assume "he" is your professor

The professor is assuming that you know this already.

[inner steppe]

I'm using Normal Biggs (Discrete math) edition 2 if you guys know the book and the chapter.

[coral spindle]

You will need to review this content

[elfin wraith]

This is a fantastic website and very happy to learn about it, doesn’t seem like I can easily search for that specific combination of traits though

Might ask the professor on Monday, just seems like a curious combination of traits to have, so I’m wondering if there’s any actually interesting rings like that or if they’d be sort of “synthetic” cooked up examples

[rocky cloak]

I guess simple rings would be the obvious example

[inner steppe]

any good resource, please don't say wikipedia, I don't understand at all what they are doing on wikipedia. But anyways what is wrong with my definition?

[rocky cloak]

So for example if you take V and countably infinite dimensional vector space take R = End(V) and I the ideal of endomorphisms with finite dimensional image, then R/I is non-noetherian, but the only two sided ideal is 0 and R/I.

And if I'm not wrong End(V) don't have any other ideals besides 0, I and itself

[knotty badger]

So R[x] is a ring here and not a field, but what you say is true in that R[x]/(x^2 + 1) is a field iff x^2 + 1 is irreducible

Remember that a ring is informally a structure where you can always add, subtract and multiply, and a field also allows division (by nonzero things)

[coral spindle]

Any introductory book on abstract algebra that includes rings will cover this. Maybe Fraleigh would suit you.

[elfin wraith]

Oh yeah that makes sense actually, that’s a nice one!

[inner steppe]

I was thinking of online resources so I can do that right away.

[coral spindle]

Pdfs are easily available online

Piracy is so terrible but who knows what happens when you google book titles and “pdf”

[rocky cloak]

And I guess you can cough up bigger examples by giving V dimension Aleph_n

[waxen moth]

What is the reason for the Zassenhaus lemma being also called the butterfly lemma?

[delicate bloom]

https://en.wikipedia.org/wiki/Zassenhaus_lemma

[waxen moth]

oh

thank you

[prisma ibex]

oh mein gott zie zassenhausenlemmatta

[waxen moth]

schmetterlingslemma

[slim kayak]

A great lemma. My brain almost instantly deletes all details any time I see and my only memory left is "haha funny butterfly diagam"

[waxen moth]

Learning this makes me feel everything except joy; it's pure agony.

cant even remember what a centralizer and a normalizer is. ive probably written down the definition at least 5 times today already

[slim kayak]

The algebra gods demand sacrifice, it is what it is. gl

[south patrol]

||Just blackbox it||

[keen badge]

Hallo,
Someone wants to talk about a cover of a finite group via cosets?

[tardy hedge]

wdym

For ideals a and b in Z, why is it true that (a+b)(a cap b) = ab?

[rocky cloak]

You can try to just prove inclusions in both directions

[keen badge]

$(a+b)(a\cap b)=a(a\cap b)+b(a\cap b)\subseteq ab$

[cloud walrus]

gingerGM

[keen badge]

Wdym wdim?

[south patrol]

This is also like an "ideal" version of gcd * lcm = product

[keen badge]

For example, take $\mathbb C_8={0,1,\cdots,7}$.
We can say that $\mathbb C_8={1,3,5,7}\cup{2,6}\cup{0,4}=\left(1+2\cdot\mathbb C_8\right)\cup\left(2+4\cdot\mathbb C_8\right)\cup\left(4\cdot\mathbb C_6\right)$.
$$a\cdot C_n = {a\cdot c \pmod n: c \in C_n}$$

[cloud walrus]

gingerGM

[keen badge]

This is a cover via cosets of a finite group

I want to prove that no matter what is your cover, and no matter what is your finite group, you always have at least 2 cosets with the same cardinality

[tardy hedge]

Thanks

I also understood more properly rn why that formula is true too

[rocky cloak]

So when you say cover, you mean that it needs to be disjoint?

[keen badge]

Yes, a partition one might say

@rocky cloak

[keen badge]

Someone wants to prove it

[tardy hedge]

They say a cap b = ab provided a+b = (1)

(a+b)(a cap b) subset ab, ab subset a cap b

I see how if a+b=(1) then it works ou

is this iff?

i guess the question is when does a cap b is a subset of ab

[hidden wind]

dunno how common this is but Stillwell literally uses the notation gcd(I,J) for the sum I + J of ideals

[delicate bloom]

I see (I,J) to mean the sum of ideals pretty often, which also lines up with using just plain (m,n) to mean gcd that's fairly common as well. I've never actually seen gcd written for ideals though.

[keen badge]

You all need to remember that if I and J are ideals of Z then there exist unqiue positive n.m such that I=nZ and J=mZ.
Thus ...

[delicate bloom]

dare I say it - unique up to multiplication by a unit

[hidden wind]

pff

[hidden wind]

invalidated

[keen badge]

No one here wants to talk with me about cover of a finite group?
I really want to do smth

[slim kayak]

Cover by what? arbitrary subsets?

[keen badge]

@slim kayak

[slim kayak]

By disjoint shifted subgroups?..

[keen badge]

yes

[slim kayak]

Can you show that one always exists

[keen badge]

The correct term for "shifted subgroups" is "coset"

[keen badge]

One always exists

The trivial one

G is a "shifted subgroup"

and it covers itself

It is shifted by 0

[slim kayak]

And what are the two cosets there with the same cardinality?

[keen badge]

{0,4} and {2,6}

Their cordinality is 2

[slim kayak]

Thats from a different cover

[keen badge]

WDYM

Ohh

[slim kayak]

So you want to figure out if you can always refine a cover to one that has two cosets of equal size

[keen badge]

sorry

yes my bad

you are 100% correct

hmm

The statement says as follows:

For a non-trivial coset partition, there exist two subgroups with the same cordinality

You can also adjust the conjecture for infinite group - but I would like to stay on the finite part

Here is the adjustment:

Let ${\alpha_i G_i}_{i=1}^r$ be a non-trivial coset partition of a group $G$, where $G_1,\cdots,G_r$ are subgroups of finite indices. Then the indices $n_1=[G:G_1], \cdots , n_r=[G:G_r]$ cannot be distinct.

[cloud walrus]

gingerGM

[slim kayak]

What have you tried so far?

For the cardinalities to work out and the hypothesis to be false you need to able to find a strictly increasing sequence of divisors of n adding up to n. But such sequences are possible.

[inner steppe]

Let R = {c2α² + c1α + c0 | c2, c1, c0 ∈ Z₅}, where α ∉ Z₅ that satisfies (α + 3)(2α² + 1) = 2α + 4. We want to determine if (R, +, ×) is a field.

Solution: 0 = (α + 3)(2α² + 1) - (2α + 4) is congruent to 2α³ + α² + 4α + 4 modulo 5, so R is equal to Z₅[x] divided by the ideal generated by k(x), where k(x) = 2x³ + x² + 4x + 4.

This is not the full soution but I am confused, shouldn't the form of k(x) be something like: c2α² + c1α + c0 ??

[keen badge]

Correct, that means that if n is a number such that the sum of its divisors is less then n, then the conjecture is true for a group of order n.
We say that G is a HS group.

[slim kayak]

Oh, is this is an actual open conjecture?

[rocky cloak]

Why do you say that?

[inner steppe]

because of the definition of R being: "R = {c2α² + c1α + c0 | c2, c1, c0 ∈ Z₅}"

[rocky cloak]

Well if alpha satisfied
alpha^2 = -c1/c2 alpha - c0/c2
Then the alpha^2 term would be redundant

[inner steppe]

yes but then what is the definition R = {c2α² + c1α + c0 | c2, c1, c0 ∈ Z₅} telling us and more importantly how did they conclude k(x) = 2x³ + x² + 4x + 4.

[rocky cloak]

Like if alpha satisfies a cubic, that means alpha^3 can be expressed as
c2alpha^2 + c1alpha + c0

[rocky cloak]

How they concluded it was just by multiplying out (x+3)(2x^2 + 1) = 2x + 4

The definition is telling you which elements are in the ring

[inner steppe]

yes but what is special with it?

[rocky cloak]

Well it was given that alpha satisfies that equation

[inner steppe]

nm

lol

[inner steppe]

but what about R = {c2α² + c1α + c0 | c2, c1, c0 ∈ Z₅}? What is the point of this?

[rocky cloak]

Well, usually to make a ring, you might start with Z5, then adjoin an element alpha.

Then for this to still be a ring you would also need alpha^2, alpha^3, alpha^4, and so on.

But if alpha satisfies some cubic equation you can already write alpha^3 as a combination of 1, alpha and alpha^2. So you don't need to add that in explicitly

So then R does become a ring

[keen badge]

Yas

[slim kayak]

(i hate that emoji quickbar, ew)

So what was your plan on posting it here then? You dont seem terribly talkative about it

[inner steppe]

If a polynomial f(x) in F[x] has no roots in a field F, is it irreducible?
A: yes, always B: only sometimes C: no, never D: don't know

The correct answer should be A no?

[night tartan]

wait am i being dumb

surely it's false

wouldn't it be B tho?

cus something like ||x^2 + 1 is irreducible in R[x] and has no roots||

ahaha nw we all sometimes fail to read

[night tartan]

yeah ok so obviously considering something like F = C would not be fruitful cus every polynomial has a root in C

so let's try F = R and seeing what we can discover

can u think of a function in R[x] that has no roots? is it irreducible or not

[inner steppe]

is C a constant?

[night tartan]

sorry i meant $\mathbb{C}$

[cloud walrus]

LY

[night tartan]

i was just too lazy to write it out in latex lol

[inner steppe]

I personally can't come up on any right now but if I understand you correctly since you mentioned complex numbers. I am just gonna assume a function with a form of something like f(x) = x^2 + b ?

[night tartan]

well yeah if f(x) = x^2 + 1

does that have any real roots?

[inner steppe]

no

[night tartan]

is it irreducible?

[inner steppe]

yes

[night tartan]

what about i.e. f(x) = (x^2+1)^2?

is that irreducible? and does it have any real roots?

[inner steppe]

I don't think it has any real roots, but I don't know if is irreducible or not.

Ig not?

[delicate bloom]

what does reducible mean

[inner steppe]

that it can be factored

[delicate bloom]

cool, can you factor (x^2+1)^2

maybe it's confusing to see written that way, can you factor x^4 + 2x^2 + 1

[inner steppe]

lol I am dumb hahah

thx guys for the help

[delicate bloom]

you're welcome!

[hidden wind]

i wonder if there's a somewhat sensible linear order to put on A4

i would be quite surprised but

[topaz solar]

There’s some conditions for “right ordered groups”

I don’t remember what they are

But that’s some keywords of relevance

[hidden wind]

oooh thank you i'll def read up on that

Any left- or right-orderable group is torsion-free.

oh

[hidden wind]

G is right ordered if there is a total order ≤ on G which is right-invariant, that is, if a ≤ b, then also ac ≤ bc

if i may cite wikipedia

[golden turtle]

I'm confused because it says let R be the ring EndK^(inf) and then it wants me to show R iso to R + R as an R-module
so is R an R-module or a ring

[glad osprey]

R is a ring, but any ring can be considered a module over itself, where scalar multiplication is just ring multiplication. So I think that's what they want you to use

[slim kayak]

What the grothendieck looking fellow said

Ring multiplication defines a module operation

[golden turtle]

I am using d&f, I don't think it is defined what it means to be ismorphic as an R-module

[charred iris]

that and the notion of isomorphism they want is not ring isomorphism, but module isomorphism

ah

[golden turtle]

When I google that, it just looks the same as a group isomorphism

is that right

[charred iris]

basically

[slim kayak]

Yeah

[charred iris]

R-linear map that's a bijection?

[slim kayak]

Or, R-modules are "R-vector spaces"

[charred iris]

how bizarre that there's no definition otherwise

[golden turtle]

f:R->R+R such that f(x+y)=f(x)+f(y) and f(rx)=rx

bijective

Is the mapping obvious

I think maybe

f(phi)=(phi, phi)

[slim kayak]

You can pull out elements from your ring

So f(rx)=rf(x)

[golden turtle]

so

for f(x+y)

x and y are endomorphisms right

but for

f(rx)

is r an element of K^(inf) or an endomorphism

If we define f:R->R+R by f(phi)=(phi,phi) then it follows easily that f(x+y)=f(x)+f(y)

[slim kayak]

I meant like in general, rf(x) is the module operation of r on f(x).

r(a,b)=(ra,rb) for direct sums

And here the module operation is just multiplication

[golden turtle]

so f(rphi)=(rphi, rphi)=r(phi,phi) since phi is an endomorphism we can pull out the r

[slim kayak]

r is also an endorphism here

[golden turtle]

so this f is an R-module homomorphism and then I just need to show its bijective

[golden turtle]

oh

but the operation is multiplication still, not composition?

[slim kayak]

Same thing here

In the endorphism ring the ring multiplication is composition

[golden turtle]

so

f(rphi)=(rphi, rphi)=r(phi, phi)

[slim kayak]

Sure

[golden turtle]

but r is an endomorphism of K^(inf) so how can it take an input from R+R

[slim kayak]

Note that for R as a R-module over itself the map is determined by f(1) or f(id)

[charred iris]

We have a ring R with composition as the multiplication. We can also make it an R-module, where the scalar multiplication is the same as the ring multiplication so again composition. The module R+R has scalar multiplication defined coordinatewise (same as with vector spaces), so it's coordinatewise composition in this case

[golden turtle]

I understand this if we already know that f is a homomorphism

[golden turtle]

This makes sense

[slim kayak]

It is for addition

But like, just pull out phi as well

[golden turtle]

so f(rphi) = (rphi, rphi) by definition of f
and then we can do the scalar multiplication in R+R
r(phi, phi) which is like you said coordinatewise composition, is why this equals (rphi, rphi)

so I think I understand now why f is a homomorphism

I guess the last part I'm not getting is why r is an endomorphism

like we want f(rphi)=rf(phi) for it to be a module homomorphism

[slim kayak]

Ask yourself where r is from

[golden turtle]

in the definition is says it must come from R

so I see

R=endK^(inf)

so

r in R is an endomorphism

[slim kayak]

Yeah

[golden turtle]

I see thank you

[slim kayak]

The diagonal map won't work btw

[golden turtle]

What is the diagonal map, the one that I'm using?

[slim kayak]

Sending x to (x,x) is called the diagonal map

[golden turtle]

I see

and its probably not surjective

?

[slim kayak]

Take phi non-zero. What element maps to (0,phi) under f?

[golden turtle]

none

[slim kayak]

Yeah

[golden turtle]

atleast I finally get the question now though

thank you

[slim kayak]

My the R-module homorphism rules you are allowed to pull the "scalar" out of the function

(note the R-module axioms are just that of a vector space but without talking about fields)

But when thinking about R as it's own R-module, any r in R is just the scalar r acting on the "vector"/module element 1.

So f is completely determined by what f(1) is, since here f(r)= r f(1) holds

For f to be surjective there are two somewhat special elements in R^2 here you want to map to and then all others follow

[golden turtle]

right I was just typing like

so if we need f to be surjective something has to map onto both
(0, phi) and (phi, 0) for phi some nonzero endomorphism
and so we need
rf(1)=(0,phi)
r'f(1)=(phi,0)
for some r, r'

is that what you mean

[slim kayak]

Kind of, I mean two very particular ( - , -) here

[golden turtle]

two elements of EndK^(inf)

special ones

hmm

and its not (0,phi) (phi,0)

[slim kayak]

Again, keep in mind that phi are also in R

[golden turtle]

is one of them the identity?

oh

maybe we want to map to

(1,0) and (0,1)?

[slim kayak]

Ding ding ding

Yes

And f(1)=(g,h) for some endomorphisms g and h, this and some light work then gives you a very concrete problem.

[golden turtle]

So

f(1)=(g,h)

can we pick g and h invertible

so

hmnvm

[golden turtle]

doesn't this define our map though, as you said? So I just say f(1)=(g,h) and then I determine what maps to (1,0) and (0,1)?

[slim kayak]

Yes exactly

This gives you four equations

[golden turtle]

af(1)=(1,0)
bf(1)=(0,1)
f(1)=(g,h)

What is the fourth?

[slim kayak]

ag=1
ah=0
bg=0
bg=1 is what I was going for

[golden turtle]

I see

af(1)=(1,0)
bf(1)=(0,1)
=>
(ag,ah)=(1,0)
(bg, bh)=(0,1)
=>
ag=1
ah=0
bg=0
bh=1

[slim kayak]

Have fun working out the four endomorphisms involved here. You reduced the problem to just one about vector spaces

[golden turtle]

thank you sir

[amber burrow]

This is probably a really dumb question, and I’ll realize it in like 2 minutes, but a simple group can’t have a quotient right?

Like if G is the simple group, and H is a subgroup, G/H is never a group right?

Since simple groups do not have proper subsets

[void cosmos]

idk wdym by that

[void cosmos]

yeah

assuming H is a non-trivial subgroup

cuz H can be G and it's fine

or the trivial subgroup which is also fine

but if G/H were to be a group then H would be normal

[amber burrow]

Meant to say proper normal subgroups

[amber burrow]

Yeah now I see

I was thinking about SL(2,R)

[alpine island]

Can you do this constructively?

I have found an easy proof by contradiction with Burnside's lemma, but this doesn't give you a way to find an element with no fixed points.

[golden turtle]

that was not fun at all until I actually discovered the a,b,g,h and then it was very fun

[long obsidian]

Can anybody say why the subring k[xx,xy,yy] is not isomorphic to k[x,y]?

[delicate bloom]

probably how the x and y interact

[long obsidian]

Oh I guess since (xx)(yy)=(xy)(xy) its not a UFD while k[x,y] is a UFD

[delicate bloom]

yeah that looks good to me

[elfin prairie]

$\varepsilon : D_n \to {\pm 1}, \hspace{3mm} \varepsilon(\sigma) = \begin{cases} 1 & \text{if x is a rotation,} \ -1 & \text{if x is a flip} \end{cases}$

[cloud walrus]

James

[elfin prairie]

I want to prove that this is a homomorphism

how should I approach this?

From the definition of a homomorphism, it seems that I need to know how rotations and flips combine with each other.

And maybe provide some justification for say what a rotation combined with a flip gives out a (flip? not sure)

[elfin prairie]

my guess is that doing this and using the definition of a homomorphism would suffice

So my question is, how do I go about it? Should I just reference the Cayley table? (that feels like cheating lol)

[knotty badger]

I think it would be useful to use a presentation for D_n

Which I assume is the dihedral group of size 2n?

Presentations for a group (generators and relations) are particularly useful for defining homomorphisms out of groups

[night tartan]

one way to do this quite easily is to notice that D_n effectively represent linear transformations

so we could represent D_n as a group of matrices

now thinking of elements of D_n as a matrix, what is \varepsilon

[inner steppe]

can someone explain this to me?

f(x) = g(x)h(x) implies that exactly one of g(x) and h(x) belongs to the set of units (invertible elements) of F[x].```

When I tried to google I saw sites saying deg(h) or deg(g) is 0. What is the correlation between these two ways of expressing?

I get the deg(h)=0 or deg(g)=0 because we are saying that only time f is reducible is when f(x) = g(x)*c where c is a constant. Kinda like a prime but the other definition is the formal definition I'm supposed to understand but as of now it is not telling me anything.

[night tartan]

the definition u've given here is just the definition of irreducible in an integral domain

now for a field F, it turns out that g(x) is a unit iff g has degree 0

[inner steppe]

you mean g(x) is invertible if an only if it has a degree of 0?

[night tartan]

yeah

i mean if g(x) = constant c, then it's clearly a unit right?

just take h(x) = 1/c

if g(x) = say 1+x

then it has degree 1

and then degree analysis => h doesn't exist

[inner steppe]

there is no case at where that is true tho right? I am just asking additionally questions?

in any polynomial field F a polynomial is invertible iff it has deg = 0 always.

[night tartan]

yh

hence the alternative definition u saw

this makes sense cus intuitively, you can only do 1/polynomial (and still have it be a polynomial) if it's a constant

[inner steppe]

wait what? Are you saying that if we do 1/polynomial then it is no longer a polynomial so it is never possible

[night tartan]

sorry i don't understand ur query?

[inner steppe]

I didn't think what you said at first was clear so I was just asking if that is what you mean but upon rereading, it is clear exactly what you said. But I means that the only way a polynomial is invertible is if 1/p(x) is a polynomial but this is not a polynomial if unless p(x) is a constant which Ig is exactly what you said.

[serene dune]

$1 \to \mathbb{Z}n \hookrightarrow D{2n} \twoheadrightarrow \langle \pm 1 \rangle \to 1$

\ \

\langle \pm 1 \rangle \text{acts on} Aut(\mathbb{Z}_n) \text{such that} a \mapsto a^{-1}

[cloud walrus]

yeshua
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[serene dune]

does that work ?

[terse crystal]

Can you guys give me some examples of groups satisfying descending chain condition on normal subgroups? Strangely I only found Prufer group on google

(Groups satisfying that any descending (in terms of inclusion) chain of normal subgroups of it stabilize, or in other words having no infinitely long descending chain of normal subgroups)

[rocky cloak]

Group of bijections on N should be one

[terse crystal]

Oh yes only four normal subgroups

[rocky cloak]

And any finite group of course

[terse crystal]

Thank you. I actually want to disprove that “a mono automorphism of a group satisfying DCC on normal subgroups is always a isomorphism”, I tried prufer group but it turns out to be true on it

[rocky cloak]

It appears that for abelian groups the Prufer groups are essentially the only ones

https://math.stackexchange.com/a/698816/306319

[rocky cloak]

Sym(N) seems like it should be able to give you such examples

[terse crystal]

Oh , thank you

No idea how to construct a injective but not surjective homomorphism…

[rocky cloak]

Like consider the subgroup that fixes some finite set

[terse crystal]

I don’t get it, what homomorphism is derived from a subgroup…

[rocky cloak]

Well, what group is that subgroup isomorphic to....

[terse crystal]

Oh shit

[terse crystal]

Thanks a lot

And btw since it’s true for prufer group, so I can say that the statement is true given that the group is abelian right

[rocky cloak]

Yeah, the key thing is that Prufer group satisfies dcc on all subgroups.

As you might notice the subgroups are the things relevant when talking about surjectivity, as opposed to normal subgroups which is more relevant for injectivity

[terse crystal]

Thank you, I am going to think more about this

[rocky cloak]

It's a theorem that comes up a few times that a surjective endomorphism on a Noetherian module is an automorphism. And this is just the dual result to that.

[terse crystal]

Oh

That ker f^n=ker f^(n+1)=…, take g=f^n, ker g=ker g^2 thing

[rocky cloak]

Yup

In similarly Im g = Im g^2 for the dual case

[inner steppe]

I have solved this using Möbuis function with the 3 if-statements but I was wondering if it was possible to use this theorem?

Can't I like split first summation like Sum[mu(d)] * Sum[1/d] but then the left side will always be 0. Anyways what is the point of this theorem anyways?

[signal knot]

https://github.com/priyanshupant/group-theory-python hey what do u guys think about this

i was just messing around in python but idk if this has any potential or sm

[warm ember]

this is mobius inversion

say you have F=f*1

then Fmu=f1mu=fe=f

e is the identity (you used sigma)

[warm ember]

if you multiply this by 18 you have

mu*id

but id=1*e

so this is e*e=e

[inner steppe]

What is F=f*1?

[warm ember]

e(18)=0

[warm ember]

do you know convolutions

[inner steppe]

Sorry my course is in another language so I have search for definition, I googled the definition of it and it says mathematical operation that allows the merging of two sets of functions.

is composite function classified as one?

then sure otherwise not really

[warm ember]

$(f*g)(n)=\sum_{d\mid n}f(d)g\left(\frac{n}{d}\right)$

[cloud walrus]

Kevin Yang

[warm ember]

im talking about dirichlet convolutions

for arithmetic functions

[inner steppe]

Ig the one to right? I don't really understand it tho.

[warm ember]

yes thats right

that follows from the fact that mu*1=e and associativity of convolutions

f * mu = g * 1 * mu= g * e=g

https://en.m.wikipedia.org/wiki/Dirichlet_convolution

[inner steppe]

Can I ask a stupid question? What is Möbuis inversion formula is saying? I mean when do we use it?

Almost every where I go including my book they are just proving the functions and not explaineing anything else besides that.

[warm ember]

a lot of arithmetic functions can be written as f * 1

using mobius inversion can find f

for example

Sigma=1 * id

(divisor sum)

d=1 * 1

(divisor counting)

also you have things like phi * 1=id

phi is euler totient

[warm ember]

sorry this is incorrect

[opaque finch]

Need assistance in a problem

Let $p(x)$ be an irreducible polynomial over a subfield $K$ of $\mathbb{C}$ and let $L$ be the splitting field of $p(x)$. Show that if the Galois group of $L$ is Abelian, then it's order equals the degree of $p(x)$

[cloud walrus]

mycroftholmes1703

[opaque finch]

I have tried to observe something from examples but failed

[quiet pelican]

Suppose some root θ of p(x) in L has non-trivial stabiliser under the action of the Galois group, and let σ(θ) = θ with σ non-identity. Let φ be a root of p(x) in L is such that σ(φ) =/= φ
Now the Galois group is transitive, so can you find an element that doesn’t commute with σ?

(Then after that, use orbit stabiliser to conclude)

[opaque finch]

hmmnn ok so

The last part if a bit unclear

if you could make it a more clearer please 😅

[quiet pelican]

The Galois group is transitive so you can find an ele that doesn’t commute with σ?

Hint: the element you’re looking for is some element such that τ(θ) = φ

[terse crystal]

I have two questions about this answer, “now let D be a divisible torsion artinian p-group, then its socle is a finite dimensional vector space over Z/pZ”, how to derive this? And why “therefore D=Z(p^infty)^n”?

[opaque finch]

yes so you mean that $\sigma$ is indeed a permutation of roots which fixes a specific root say $\theta$. Now an element that doesn't commute with $\sigma$ is any permutation that does not fix $\theta$

[cloud walrus]

mycroftholmes1703

[quiet pelican]

Not quite
It needs to move θ to an element that isn’t fixed by σ

[opaque finch]

hmmnn

[opaque finch]

Ok. So let's do it a bit clearly. Suppose $\alpha_1,\alpha_2,\dots,\alpha_n$ be roots of $p(x)$. Every element in the Galois group corresponds to a permutation.
Let $\sigma_1$ be a permutation such that it fixes a root $\alpha_i$ but is not identity. Therefore there exists a root $\alpha_j$ such that $\sigma_1$ does not fix $\alpha_j$. Then we take the permutation $\sigma_2$ such that $\sigma_2(\alpha_i) = \alpha_j$. This element does not commute with $\sigma_1$ since
$\sigma_2(\sigma_1(\alpha_i)) = \sigma_2(\alpha_i) = \alpha_j$ but $\sigma_1(\sigma_2(\alpha_i)) = \sigma_1(\alpha_j) \neq \alpha_j$. Is this what you mean @quiet pelican ?

[cloud walrus]

mycroftholmes1703

[quiet pelican]

Yes

[opaque finch]

cool

[quiet pelican]

You should say a permutation such that sigma2(ai) = aj though

[opaque finch]

yes

I should mention that

Now to conclude from here, since the Galois group is abelian and it acts transitively on the set of roots, and each element corresponds to a permutation then it must permute without fixing any roots. Which means that order of Galois group is the number of distinct automorphisms of the roots.

am I right ?

[chilly ocean]

ㅡㅡㅡ
Is there any example of a extension field that splitting but not separable

[wild jasper]

@chilly ocean $F_pX$ I think was the example given in our class

[cloud walrus]

mh_le

[dim widget]

by the way this is false

the powers of the norm are the unique maps N: F \to k such that N(a) = a^n if a \in k and N is multiplicative

so this is some power of the norm

[rotund aurora]

But why should it be multiplicative?

[rocky cloak]

Anything that is the splitting field of an irreducible non-seperable polynomial.

Like Fp(x)/Fp(x^p) for example

[rocky cloak]

So in general if a module is artinian, then the socle is a finite (nonzero) sum of simple modules.

From there it's just using lifting properties to show that a divisible group with socle (Z/p)^n is the sum of p-Prufer groups

[terse crystal]

Yes I forgot to reply that the first part was figured out, yeah the remaining part, why sum of p-Prüfer groups

Lifting properties?

[rocky cloak]

Like choose M = Z/p in the socle, then consider x such that x in M or 0 =/= px in M or 0 =/= p^2x in M ...etc

And show that this is a direct summand of your group

(this is slightly inaccurate as written)

But pick a generator for Z/p, then lift that.

and construct the prufer group as a summand

[terse crystal]

I am stuck at this step…

(Z(p^infty)^n is now a subgroup of D, I don’t know how to show that it’s a summand…

[rocky cloak]

Well, the socle is ( Z/p)^n. Pick a basis, then consider the submodule killed by p^2. Inductively lift to a sum of Prufer groups

[opaque finch]

Thanks for your hinta @quiet pelican I got the solution to it.

[terse crystal]

Submodule of D killed by p^2?

[rocky cloak]

Yes

And killed by p^3 etc

[terse crystal]

Not sure whether I got it, does this make sense? : D’=(Z(p^infty))^n, we want to show that D’=D, namely any element a killed by p^r in D, is an element killed by p^r in D’. p^2 a=0 for some a in D, pa is in (Z/pZ)^n, so pa=Σx_i e_i, where {e_i} is a basis of (Z/pZ)^n, then p(a-Σx_i e_i/p)=0 gives us a=Σx_i e_i/p +Σy_i e_i=Σ(x_i +py_i) (e_i)/p, so a is in elements of D’ killed by p^2?

Elements of D’ killed by p^2 is generated by {e_i/p}
Wait elements of D killed by p is in (Z/pZ)^n?

Oh nvm yes , so I guess this is fine?

[rotund aurora]

@dim widget The other day I could find this formula, I'm pretty sure it's correct. If you consider f(x)=x^3-x^2-2x+1 you get N(x, y, z)=-x^3-y^3-z^3+4(x^2y+y^2z+z^2x)-3(x^2z+y^2x+z^2y)+xyz. Your claim is that N(x, y, z) is symmetric, which as you see needn't be true (this will only happen when alpha_1 alpha_ 2^2+...=alpha_1alpha_3^2+..., maybe you can always choose a defining polynomial with this property, idk.)

Btw this is slightly imprecise, as idk which coefficient should be T_+ and which coefficient should be T_- lol

[unique holly]

I could use help with this problem: let G be a group of order p^n, where p is prime. Let H \neq G be a sugroup of G. Show there exists an x, x \not \in H such that x^{-1}Hx = H.

My understanding is that I need to let H be a proper subgroup of G and show that H is a proper subgroup of the normalizer of H in G. But I'm not sure how to go about this

[mighty totem]

I think this is in D&F 6.1

[warm ember]

G has a nontrivial center Z

if Z is not a subgroup of H then we are done

otherwise take G/Z and consider H/Z

solution: ||by induction, there exists xZ that is in normalizer of H/Z and x is not in H. then x is in normalizer of H||

[dim widget]

because if \sigma is a permutation of a set I and \alpha = \sum_i a_i \alpha_i, \beta = \sum_i b_i \alpha_i, then \sigma(\alpha\cdot \beta) = \sigma(\alpha) \cdot \sigma(\beta)

so \Prod_{\sigma} \sigma(\alpha) \cdot \Prod_{\sigma}( \sigma(\beta) = \Prod_{\sigma} \sigma(\alpha \cdot \beta)

[dim widget]

yeah it's correct

[cloud lynx]

hey which section is for commutative algebra?

advanced algebra?

[rocky cloak]

Yes

[cloud lynx]

Are u also active in advanced algebra?😍

[delicate bloom]

they would be if there were questions there... 🥹

[hidden wind]

sad

[cloud lynx]

[inner steppe]

x in Z77 is described by ([x]7, [x]11). Which of the following is invertible?
A: (2, 10), B: (0, 7), C: (1, 0)

I know that according to CRT if sgd(7,11) then there wil be an inverse. But how am I supposed to think here?

[quiet pelican]

It’ll be invertible iff both components are
Because the congruence xy = 1 mod 77 is equivalent to xy = 1 mod 7 and xy = 1 mod 11

[earnest hare]

Given some field F, F[x] is the ring of polynomials over an indeterminate x. F(x) is the field of fractions obtained from the integral domain F[x]

[tardy hedge]

Is this annihilator of principal ideal (x) for each x in the ring?

Ty

[wild jasper]

Say I have a field extension Q(alpha) of degree 8 and I would like to find the minimal polynomial of alpha = sqrt((2+sqrt(2))(3+sqrt(3)) in an effective way, what would be a way to do it?

[tardy hedge]

If for some x in ring, b1,b2 and a are ideals, why x(b1+b2) subset a implies xb1 subset a and xb2 subset a?

Nevermind got it. Use 0

[inner steppe]

If F is a finite field, then the characteristic of F is p = o+(1)

I know what characteristics mean, it when we take the mulitplicative identity 1 and add it until we get 0. The number of 1's we have to add then is our characteristics and it can only be 0 or a prime number.

But I don't understand the definition on my lecture slide, what do they mean by p = o+(1)?

[void cosmos]

look at the galois group action on alpha

[void cosmos]

what is o

wtf is this

[void cosmos]

btw u can do this manually easily

[wild jasper]

I've been trying for days 😦

[inner steppe]

Wait I think I get it, I just missed a little part of the slide where they mentioned that o+(1) is the order under addition. So I guess they are using g^n = 1 but with addition which I get is g^n = g*n = 0 so n in this case should be a prime?

[inner steppe]

does this sound right

[void cosmos]

i am not following but yes the characteristic of a field must be prime

[void cosmos]

well either i missed something or u did but it's fine

[inner steppe]

or 0?

[void cosmos]

yes or 0

[wild jasper]

can you show me any shortcuts to make the calculations easier?

[void cosmos]

i would probably just find out what the galois group is

this has order 8 and there are only 2 nonabelian groups of order 8 ig

[wild jasper]

this question is before the galois group is determined

[void cosmos]

yeah im telling u what iw ould do

[wild jasper]

ok

[void cosmos]

but yeah i think manually doing it sucks

[inner steppe]

Btw guys aren't these saying exactly the same thing:{x ∈ G ; x^d = 1}| and |{x ∈ G ; o(x) = d}|

The number of x in G whose order is d?

[coral spindle]

No

1^500 = 1, so is it true that o(1) = 500?

[thorn jay]

First one is all the elements where the order divides d

[inner steppe]

how?

[inner steppe]

isn't the order g^n = 1?

[coral spindle]

so 1^500 = 1

What's the order of 1

[inner steppe]

no but prolly mixing it up, let me double check

[coral spindle]

What do you mean no

It is absolutely true that if 1 is the identity of any group, then 1^500 = 1

[inner steppe]

no but the smalled n in g^n = 1 is the order of g

[coral spindle]

Yes there you go.

So you have the answer to your question

They're not the same thing.

[inner steppe]

how come "First one is all the elements where the order divides d"

[coral spindle]

Because it is a standard fact that x^d = 1 if and only if o(x) divides d.

You should try proving this.

[thorn jay]

Okay suppose o(g) = k
Suppose some d such that k divides d
Then d = n * k for some n
=> g^d = g^nk = (g^k)^n = 1^n = 1
Now suppose for some d, g^d = 1 and let r = d mod k
=> r = d + m * k < k for some integer m
=> g^r = g^{d + mk} = g^d * g^mk = 1 * 1^m = 1
But as r < k, r must be 0 by the definiton of order, hence k must divide d

Thus we've prove then g^d = 1 if and only if the order of g divides d

[inner steppe]

the funniest thing is that I remember proving this as an exercise 2 weeks ago. Why do I keep erasing things completely from memory.

[thorn jay]

Loll

Felt

[inner steppe]

Thanks for this tho btw. I really appreciate the help.

[thorn jay]

Ofc!

I hope it kinda made sense x3

[inner steppe]

it definitely did!!

[thorn jay]

Okii idk im used to trying to teach this to people with barely any understanding of the subject matter hehe

[inner steppe]

my problem is that I am gonna forget I ever did this in like a couple of days

[thorn jay]

Well it's basically number theory this

Modular arithmetic

Because working in the subgroup generated by g (which is always when u work with powers of it) is working in Z_o(g)

Would this belong here? Animation i made with motion canvas of S_nxS_n acting on a latin square

[gusty thistle]

Suppose I have a ring A and a maximal ideal m of A. How do you show that the dimension of (A/m) (x) A^n as a A/m-vector space is n? I know that (A/m) (x) A^n is isomorphic to (A/m)^n as an A-module but this is not enough to conclude that the dimension is n right?

Clearly the dimension can not be greater then n but im not so sure how to show that it can't be less either

[south patrol]

The isomorphism you mention is also an isomorphism of A/m-modules

So you're done

This is just because the A/m and A module structures are compatible "in the obvious way"

[gusty thistle]

ah yea

that makes sense

[gusty thistle]

thanks

[inner steppe]

Let (F,+,×) be a field with 2^6 (= 64) elements, ∣ 𝐹 ∣ = 64 ∣F∣=64. How many elements x∈F satisfy x^21 =1

If G is a field then it is also a group and for a group G where |G| = n, G is a cyclic group where if n|d n = |{x belongs to G ; x^1 = d}| = d

Am I correct this far that every field is also a group, then what remains to use this theorem is to find the order of G, now how do I do that?

[thorn jay]

26 (=64)
What

[inner steppe]

check again

wait

now

[thorn jay]

Lmao

Sorry lol i got very confused

[inner steppe]

Totally understandable lol

[thorn jay]

x^21 = 1 means that the order of x divides 21 in the multiplicative group F*
21 | 63 so the number is just the number of elements with order 21, order 7, order 3 and order 1

The multiplicative group of a finite field is cyclic so in this case isomorphic to Z_9 x Z_7

From here it follows that there are 6 elements if order 7, as it has to be of the form (0, n) and 7 is prime so all elements except 0 generate it.

Elements of order 3 are of the form (m, 0) as 3 and 7 are coprime.
3k = 0 mod 9 => 3k = 9n => k = 3n
So there are two elements which have order 3; (3, 0) and (6, 0)

Then there are 2 * 6 = 12 elements of order 21; (3, 1), (3, 2), ... , (6, 1), (6, 2), ...

Finally, 1 element of order 1

Adding these up yields:
6 + 2 + 12 + 1 = 21

I think

[thorn jay]

Does this seem good to you?

Oh ofc im stupid

21k = 63n
k = 3n
Which has 21 unique solutions

Lol

I thought it was suspicious already that it came out to be 21...

[wild jasper]

how can one tell if two elements in a finite field extension has the same minimal polynomial?

[thorn jay]

In an algebraic field extension?

[wild jasper]

yes

Finite extensions are algebraic

[thorn jay]

Oh sorry i didnt see the finite part x3

Seems like a pretty computational problem though

[wild jasper]

indeed

https://imgur.com/a/vPKZsq9 I'm trying to solve the second part of (b) without having to compute the minimal polynomial and check 7 other roots

[rocky cloak]

The existence of an automorphism taking one to the other would do it

[inner steppe]

sorry I was gone for a little, not entirely sure so far

[wild jasper]

indeed

[inner steppe]

0^20 = 0 != 1, and (F \ {0}, ·) is a cyclic group with 80 elements. Since 20 divides 80, x^20 = 1 holds according to the known theorem for exactly 20 elements in F.

This is the answe they provided in the solution but I've no idea what they are trying to say ehre

[thorn jay]

well the multiplicative group of F is isomorphic to Z_63
so x^21 = 1 translates to 21k = 0 mod 63
21k = 0 mod 63 <=> 21k = 63n <=> k = 3n
for n = 21, 3n = 63 = 0 mod 63
hence for all values n in { 0, 1, ..., 20 } you get a unique solution, thus there are 21 solutions

[rocky cloak]

Which you can apply to your example by using that the Galois group of E is an extension of
Gal(F/Q) and Gal(E/F)

[inner steppe]

I get the math part in this, but what is unique solution telling us?

What is " for all values n in { 0, 1, ..., 20 } you get a unique solution," saying?

wait actually

how come

21k = 0 mod 63 <=> 21k = 61n ?

[thorn jay]

it means that for all n=/=m in that set, 3n =/= 3m

[thorn jay]

definition of modulo

oh

63*

in my defence I am very tired

[inner steppe]

that makes 2 of us, no worries!!

[thorn jay]

loll

[wild jasper]

Well the part where I show that E/Q is Galois is after this

but E/F is...

[inner steppe]

I am sorry but I still don't get where we are going with this

[thorn jay]

okay so we have 21k = 0 mod 63 <=> k = 3n right

that makes sense?

[rocky cloak]

Then I guess just compute the minimal polynomial

[wild jasper]

yeah... and check 7 other roots! 😄

[rocky cloak]

Or the listed roots are clearly the roots of a degree 8 polynomial, so all you need to do is show that the minimal polynomial of alpha has degree 8

Which is the same as showing the degree of E is 8

So you just need to show that F = Q(alpha^2) if you've already done (a)

[inner steppe]

yes

Ig

[thorn jay]

so we just go through all the n until we hit a number we've seen before

[wild jasper]

@rocky cloak ahha

[inner steppe]

sorry on a second thought k = 3n, not really sure where this comes from

[thorn jay]

so n=0 yields k=0
n=1 yields k=3
n=2 yields k=6
...
n=20 yields k=60
n=21 yields k=63 = 0 mod 63
n=22 yields k=66 = 3 mod 63
etc.
so there are only 21 unique values for k, hence 21 unique solutions to the equation

[thorn jay]

21k = 63n <=> k = 3n
division by 21 on both sides :3

[inner steppe]

then

[wild jasper]

@rocky cloak does this work? https://imgur.com/a/vE5wwTe

[rocky cloak]

Well, I'm not sure "is obvious" is the best argument. But otherwise yeah

[wild jasper]

thanks 🙂

[wild jasper]

actually it's not obvious that sqrt(2) and sqrt(3) are in Q(a) 😦

[rocky cloak]

Indeed, that's the main thing that needs a proof

[thorn jay]

Hm?

Then that

Yes

[inner steppe]

I am so lost because I am not even sure what we are trying to do sorry

Are you trying to show me hwo come we get this

0^20 = 0 != 1, and (F \ {0}, ·) is a cyclic group with 80 elements. Since 20 divides 80, x^20 = 1 holds according to the known theorem for exactly 20 elements in F.

Than I still have no clue where we are going with everything we did so far.

[thorn jay]

But you're trying to do x^21 = 1 right

Not x^20 = 1

[tardy hedge]

Bit confused on how to show v)

[thorn jay]

What is r here?

[tardy hedge]

radical of ideal

a and b are ideals

[tardy hedge]

Also kinda stuck on vi) for n > 1

[rocky cloak]

One inclusion is clear. For the other one, just start with x suck that x^k is in r(a)+r(b) then construct n such that x^n is in a+b

[tardy hedge]

x^k in r(a)+r(b) means x^k = y+z for y and z in ra rb but i wasnt sure on how to relate them

[rocky cloak]

So then you know there is an m such that y^m is in a and z^m is in b

See if you can use that

[inner steppe]

yes but that was the solution so that is hwy I was confused the whole time and still am

[thorn jay]

Oh lmao

Well x^n = 1 has n solutions

In finite fields

[inner steppe]

yes

[inner steppe]

but anyways is there any explanation to why they are solving it that way

[thorn jay]

Idk it's a very handwavey explanation of my proof

[inner steppe]

are you talking about this proof?

[thorn jay]

This proof

[inner steppe]

I finally see what you were tryna do there lol. Earlier I thought you were like doing similar things like them. Since I spent time understanding their solution I was not ready to receieve other ways until now. Thanks for the help!!

[thorn jay]

Hehe, i try, sorry for the confusion

[grave sedge]

(fun fact/rant) This is somewhat of an inefficient thing, although it is the one people usually do: you're basically repeating the steps you use to prove the radical is an ideal, instead of using the properties you already have.

A "better" proof would be to notice that r(a)+r(b) ⊂ r(a+b) (because r(a) ⊂ r(a+b) and r(b) ⊂ r(a+b)) and then take the radical of both sides

[tardy hedge]

Hmm ill think about that thanks

$(i,j) = (i,i+1)(i+1, i+2)...(j-2, j-1)(j-1, j)(j-2,j-1)...(i+1, i+2)(i, i+1)$

[cloud walrus]

kiand123

[tardy hedge]

is that true for permutations

looks messy and dumb but

[next obsidian]

This doesn’t finish it though you still just get an inclusion.

You need to do some argument to show you get everything, but for that you can just say that r(a + b) < r(r(a) + r(b))

[grave sedge]

The other one is more or less obvious (since a+b ⊂ r(a+b))

(anyways my point was that this specific result needs no element-wise manipulations assuming you have already done the stuff that is usually done before it, like r(r(a))=r(a), but for some reason it's very common to see proofs by picking elements)

[fading summit]

help! please

Let G be a finite group of order n, and let g ∈ G be such that G =<g> (a generator for the group). Let d ≥ 1 be a divisor of n.

1. Let x be an element of G. Prove that x has order d if and only if there exists an integer k such that gcd(d, k) = 1 and x = g^(nk/d).
2. Let x1, x2 ∈ G be two elements of order d. Prove that <x1>=<x2>. (Do not use the fact that there is a unique subgroup of G of order d – the statement in this problem was used in the proof of that theorem).

[brisk oak]

Hmmm maybe use iii) ? 🤔🤔

[thorn jay]

2 isnt true though? In the group Z_3xZ_3 one has elements (1, 0) and (0, 1) both of order 3

The respective subgroups generated only share (0, 0)

Oh im tired i didnt see G = <g>

Suppose x = g^m and o(x) = d
=> d * m = 0 mod n = k * n
m = k * n/d
Let gcd(d, k) = l and notice:
m * d/l = k * n/d * d / l = k/l * n
But l | k => k/l is an integer => m * d/l = 0 mod n
Hence by the definition of order gcd(d, k) = 1
The other direction is obvious

[fading summit]

the other direction would be this? x^d = (g^(nk/d))^d = g^(nk) = (g^n)^k = e^k = e

[thorn jay]

Mhm!

[fading summit]

thanks

[thorn jay]

x1 = g^{n/d}^k1
x2 = g^{n/d}^k2
consider then S = <g^{n/d}>
clearly |S| = d and x1, x2 in S
=> <x1>, <x2> < S
But |<x1>| = |<x2>| = |S| = d
=> <x1> = S = <x2>
Omfg that took me way too long

[fading summit]

but the problem says we're not allowed to use the fact that there is a unique subgroup of G of order d

[thorn jay]

Yeah

I know

Im using the fact that the cardinalities must be equal

And since everything is finite that must mean the sets are equal

[thorn jay]

Where in my proof do i use that?

[fading summit]

since <x1> and <x2> are both subgroups of S with order d, it must be true that <x1> = <x2>

[thorn jay]

S is also of order d

Hence you get
|S| = |<x1>|
Which means that <x1> = S as <x1> is a subgroup, and thus a subset, of S

[fading summit]

okay thanks!

[thorn jay]

Ofc!

[fading summit]

List all the possible orders of the elements of G := Z_30 × Z_24. For each number you list as a possible order, give an example of an element of G that has that order. Without listing all the 720 elements of the group, explain why the numbers on your list are the only possible orders.

i found out what the possible orders are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

but how do i find examples for each of them? is there a pattern that im not seeing?

[thorn jay]

well the orders are lcm(o(a), o(b)) where a in Z_30, b in Z_24, as o((a, b)) = lcm(o(a), o(b))

[fading summit]

oh wait i figured it out

i can just let one of the components of (a, b) be 0 and then use the multiple for the other one

like the order of (6, 0) is 5

or the order of (0, 4) is 6

[thorn jay]

that is one way yes

[golden turtle]

I am confused how the earlier part of this problem is related to the final sentence. Say I show that T is surjective if and only if det M is a unit in R. How does this connect to the case of mxn matrices when the determinant wont even be defined?

[thorn jay]

well if there was a surjection to R^m to R^n, then R^n would be isomorphic to some quotient of R^m.
But a quotient of R^m would induce a surjective endomorphism, which would need to be an automorphism, hence R^m would be isomorphic to R^n, which is impossible as any commutative ring has IBN and m=/=n

[golden turtle]

we have not covered IBN yet, is that required here?

[thorn jay]

eh you can show it directly

show that R^m ~ R^n => m = n

for R a commutative ring

[golden turtle]

Okay I will try that, thank you

[thorn jay]

good luck ^w^

[fading summit]

Let G be a group of order 8, and let a, b ∈ G. Assume that o(a) = 4 and b is not in <a>. If H is a subgroup of G that contains a and b, prove that G = H (in other words, G is generated by the two elements a and b, i.e. G = <a, b>).

im thinking:
|<a,b>| = |<a>| + |<b>| − |<a ∩ b>| = 4 + 4 - 0 = 8

[thorn jay]

by langrange's theorem: [G:<a>] = 8/4 = 2, and b not in <a>, thus
G = <a> U b<a>
G = <a> U b<a> < <a, b> < G
=> <a, b> = G

[fading summit]

huh? is it really that simple

[thorn jay]

mhm

cosets go hard

[fading summit]

could you elaborate it? im not following

also i havent learned what a coset is yet

[thorn jay]

well the index of <a> in G is 2, so it has two left cosets
b is not in <a>, so the other coset must be b<a> => G = <a> U b<a>
H = <a, b>, and notice that <a> U b<a> must be a subset of H
hence we get G < H < G => H = G

[thorn jay]

ah okay

so no langrange's theorem then

😔

[fading summit]

no i've learned lagrange's theorem, but not cosets

[thorn jay]

bruh

how have you proven it without cosets?

[fading summit]

we didnt prove it in class yet. but we're still allowed to use it.

[thorn jay]

ah right..

of course

smg

smh*

hmm

well

I can quickly teach you cosets

they aren't that hard

if you want

[fading summit]

i was thinking we can let <a> = {e, a, a^2, a^3}, and let <b> = {b, ba, ba^2, ba^3}

so <a, b> = {e, a, a^2, a^3, b, ba, ba^2, ba^3}

so |<a, b>| = 8 and |G| = 8, so G = <a, b>

[thorn jay]

The <b> you defined is exactly the coset b<a>

[fading summit]

so the proof works?

[thorn jay]

Kind of akward notation

As <b> denotes the subgroup generated by b and <a, b> the subgroup generated by a and b

[golden turtle]

So I tried reasoning this out, and I'm not sure I understand completely. Are you saying, if there's a surjective module homomorphism f:R^m->R^n then by 1st iso R^m/ker(f) iso to R^n. But then I'm not quite getting what the "induced surjective endomorphism" is here. I think maybe you're saying something will show R^m iso to R^m/ker(f) iso to R^n, which is our contradiction. But Idk what the "something" is that shows that. Hopefully my question makes sense

[thorn jay]

Oh you're right i was kinda being dumb lol

[golden turtle]

So its nothing I was missing?

[thorn jay]

Nope

Wait

There is a natural projection of R^n onto R^m right?

[golden turtle]

for n<m?

yes

[thorn jay]

For n>m

[golden turtle]

is it to just kill the extra stuff?

[thorn jay]

Yup

[golden turtle]

ok

yes

[thorn jay]

h : (r_1, ..., r_m, ..., r_n) -> (r_1, ..., r_m)

[golden turtle]

yes

[thorn jay]

Then f : R^m -> R^n surjective

=> h * f, f * h surjective

But both are endomorphisms so both are bijective

[golden turtle]

and is that because h*f and f*h are R^m to R^m and R^n to R^n respectively, so the first part gives us that

[thorn jay]

Mhm

[golden turtle]

so this just gives two different isomorphisms from Rm to Rn

right

[thorn jay]

From R^m to itself

[golden turtle]

ah yea

[thorn jay]

And one from R^n

To itself

[golden turtle]

yes

so we want to get a contradiction from this right, we assumed that f exists as a surjection

[thorn jay]

The compositions of two surjections can only be an injection if both are injections

[golden turtle]

can't we let like, x=(0,0,..,0, r_(m+1), .. r_n) then f*h(x)=0 but this would contradict f*h being a bijection

[thorn jay]

Hence h must be an injection

Hence n = m

Wow thats elegant

[golden turtle]

would what I said also work as a contradiction?

[thorn jay]

Its the same argument essentially ^w^

Lets goo

That was fun

[golden turtle]

haha

I should've been able to think of that

thanks

[thorn jay]

Ofc!

And thanks for giving me a fun exercise too

Im gonna busy myself with the first part now

[golden turtle]

likewise

[golden turtle]

I think we have that ||Madj(M)=detMI|| and ||adj(M)M=detMI|| so if T is surjective <=> det M is a unit we have ||M^(-1)=adj(M)/detM|| exists which implies ||M invertible => T bijective => T isomorphism||

I put the spoilers incase you were still working on it

but if you're not, maybe let me know if that's fine approach 🫡

[thorn jay]

Hmm as it is almost 4 AM i will look at it tomorrow x3

[golden turtle]

okay no worries, ty

[vagrant zinc]

Hi guys, could someone tell me, if the demo outline is on the right track?

[gilded saffron]

I don't know Spanish but you have the right idea as far as $ker(p_1) = U_2$

[cloud walrus]

rays

[serene dune]

If $a^3 =a$ for all $a$ in a ring $R$, then $R$ is commutative.

[cloud walrus]

yeshua

[serene dune]

how often the converse is true btw ?

i was wandering if this is anyhow useful in commutative algebra or not

[delicate bloom]

pretty much never, I'd say this isn't useful, it's more just an exercise for some really restricted conditions forces commutativity.

[serene dune]

i was very exiceted nonetheless 😿

[delicate bloom]

it's a fun result, the fact that it doesn't hold for the integers basically means you can't have the integers as a subring which isn't a good sign

[serene dune]

in anyways, can u suggest me good problem set(s) for rings

[serene dune]

hmm thats a good note

[warm ember]

try romania national mo grade 12

there are some good ring problems in there

[void cosmos]

i would go through qualifying exams

from diff universities

algebra ones

surely u will find good (and should be deep ig?) problems

[serene dune]

https://blngcc.wordpress.com/wp-content/uploads/2008/11/rmc1997.pdf

this right?, i hope posting the link is approved by TOS

[warm ember]

hmm

i think rnmo is different from that

[serene dune]

alr

[serene dune]

any sheets or specific problems would be appreciated 🙏

[warm ember]

https://artofproblemsolving.com/community/c3365_romania_national_olympiad

the problems arent deep

and dont require theory

[void cosmos]

idk if i can remember lilke

a specific problem

give an example of a ufd that is not a pid

[serene dune]

well i was solving artin as well

[void cosmos]

like

[serene dune]

umm i think i have done those
XD

[void cosmos]

here is one that i got in an exam i got examined

show that the ideals of M_n(R) are of the form M_n(I) where I is an ideal of R

i did only like one and a half direction couldn't finish it

[serene dune]

yeah

i immediately thought bout that

[void cosmos]

okay i remembered one

[void cosmos]

similar to this

suppose we have a ring such that x^2=x for all x

[serene dune]

boolean ?

[void cosmos]

non zero ring that may not have an identityh

show that R is commutaitve

[warm ember]

boolean ring is commutative

[serene dune]

yeah i think i have proved that

thanks anyways

[void cosmos]

yeah ig the problems ik are like

the ones that involvle standard arguments or tricks that one sees in qualifying exams

ig for more exotic problems u can see what kevin yang sent

[serene dune]

alr

[warm ember]

Prove that a nontrivial finite ring is not a skew field if and only if the equation $ x^n+y^n=z^n $ has nontrivial solutions in this ring for any natural number $ n. $

ring here means noncommutative ring with unity, and nontrivial means x,y,z are nonzero

[serene dune]

yeah im looking forward to qualify some exams

ok i will return to this, hopefully soon, topology untill then

[rocky cloak]

||So if R is a skew field, then for n = |R|-1, x^n = 1.||

||So if x, y and z are all nonzero then 1+1=1 contradiction. ||

||If there exists x and y such that xy=yx=0, then x^n + y^n = (x+y)^n||

||You can argue that such x and y exists by looking at the radical or in the case the radical is 0, using the classification of semi-simple rings.||

[thorn jay]

determinant is defined in the usual recursive way?

[south patrol]

Is that the normal way lol

[rocky cloak]

Consider the surjective map
R^n ->> R^m -> R^n

is it an isomorphism?

(Where the first map is the usual projection)

[thorn jay]

We've already done the thing they asked about, now we're doing the main part of the question

[thorn jay]

Idfk i like the basis invariant way but you'd have to show they're equal lol

[rocky cloak]

Ah sorry, I missed some stuff

[south patrol]

Oh I mean the other main definition i know is just axiomatic

[thorn jay]

Its oki :3

[south patrol]

or you can use the Leibniz formula

I guess you have one in terms of like Laplace expansion

[thorn jay]

Yeah

Thats the one i learned

But they're all so annoying to work with istf, but i suppose we should be happy to have something computable saying so much about the transformation

[golden turtle]

Yes, I’ve realized this only works for 1 direction though

Not for T surjective => det M unit

[rocky cloak]

I guess the key idea is ||the adjugate matrix||

Not sure I know if a different way

[thorn jay]

You've proven det M unit => T surjective?

[golden turtle]

That only works for 1 direction

[golden turtle]

Yes

[thorn jay]

Clearly by the multiplicative property of determinants T iso => det M unit

[golden turtle]

We don’t have T iso though only surjective

For the other direction

[thorn jay]

Yeah yeah

But if we prove T surj => T iso we have
T surj => T iso => det M unit => T surj

[golden turtle]

Yes

[thorn jay]

Which implies
T surj <=> T iso <=> det M unit

[golden turtle]

Yes

[thorn jay]

So this isnt totally useless if we can prove T surj => T iso

Lol

[golden turtle]

Yes

I have done the other direction in a different way but I suppose it doesn’t matter how we get it

Using that

It’s free

[thorn jay]

Yup

[golden turtle]

Do you have an idea for surj => iso?

[thorn jay]

Not really

[golden turtle]

That’s fine, me either

[thorn jay]

I suppose we can steal the proof from linear algebra

Over vector spaces

:3

[golden turtle]

I tried something like that

Like if it’s surjective it maps to the basis