groups-rings-fields

[winged void]

well this means that two equivalence classes are equal to each other

[dull ginkgo]

thus x ~ (-x) for each x, i.e n | x - (-x) = 2x

[winged void]

which mean if i pick an elemnt out $\overline{x}$ and another out $\overline{-x}$

[cloud walrus]

Mootje

[winged void]

this mean that n | a - b

[rocky cloak]

So let me write [] instead of overline:

2a is in [2a] agreed?
[2a] = [0], per assumption
Thus 2a is in [0]
Hence n divides 2a-0

[dull ginkgo]

*equivalence c l a s s$, i.e x ~ -x where x ~ y iff n | (x - y)

so n | (x - -x) for each x

there's your forall quantification

set x = 1

2 | n

n even

[winged void]

i agree about that but then my point here we picked like an element out 2a but that elemnt is arbitrair so thats why

[rocky cloak]

Sure

[dull ginkgo]

it's arbitrary because it holds for EVERY one in the equivalence class

[rocky cloak]

Pick whichever you want, as long as it's equal to 2a modulo n

[dull ginkgo]

really it's because it's an equivalence relation, that's honestly a bit more important here

maybe because it's equivalent

[winged void]

oh i see what you mean i get it

well you meant actually that 2a is in [2a]

has that form

i mean

its just like when saying z/3z

{1} = has the form of 3z + 1

for example [1] = [4] in z/3z this is equivalent to saying that 3z + 1 = 4+3z mod n but this is equivalen to 1 = 4 mod n

is this a good way of thinking about it ?

i mean here mod 3

actually

so this mean that 1 = 4 mod 3 which is true as well

[dull ginkgo]

Bruh wait is showing x = y (mod A) and (mod B) is equivalent to x = y (mod A \cap B) just because of x - y being in A \cap B

[coral spindle]

Yes

[winged void]

what do you mean

[toxic zephyr]

x=y mod A means x-y is in A
x=y mod B means x-y is in B
so x-y is in A cap B
so x=y mod (A cap B)

the implications go backwards too. so it's equivalent

[dull ginkgo]

The general form of Chinese Remainder Theorem was much harder to prove in full than I thought

[celest furnace]

I dont think so

Could be misremembering correctly but im pretty sure it still is just writing x as the nice linear combination from bezout basically

[dull ginkgo]

the ring theory proof

[winged void]

Sure I agree about that

But what has that to do with this

.

[coral steeple]

Hello! I am asked to find an isomorphism between $\mathbb Z_2[x]/(x^4+x^2+1)$ and $\left{\begin{bmatrix}a & b\& a\end{bmatrix}:a,b\in\mathbb F_4\right}$

[cloud walrus]

person2709505

[coral steeple]

I have identified the non-units of the matrix ring, and can use the fact that the iso must map non units to non units to narrow down the search space, but I am really quite stuck figuring out how to get something that isn't a 16-case piece wise function

Z2 means z mod 2. No p adics here.

One of my guesses was to take q,r obtained through division of a polynomial in the poly ring by x^2+x+1, and having q,r be the entries of the matrix in the matrix ring. No luck

[dull ginkgo]

@coral steeple maybe try Cayley-Hamilton. What is that characteristic polynomial?

[coral steeple]

I will chew on that, thanks...

As far as I can tell, Cayley-Hamilton applied to these matrices gives the fact that $\begin{bmatrix}a & b\& a\end{bmatrix}^2=\begin{bmatrix}a^2 & \& a^2\end{bmatrix}$

[cloud walrus]

person2709505

[gritty sparrow]

The matrix ring is isomorphic to the ring $\mathbb Z_4[t]/(t^2)$ by sending the matrix with entries a and b to a + tb. This latter ring is isomorphic to $\mathbb Z_2[y,t]/(y^2+y+1,t^2)$. Now this ring is isomorphic to the ring you want by sending $y \to x^2$ and $t \to x^2+x+1$.

[cloud walrus]

chmonkeynumber1fan

[coral steeple]

I'll think about that, thanks

[coral steeple]

How do we interpret the first isomorphism between the matrix ring and Z_4[t]/(t^2), seeing as the matrix entries are in F_4 while a+tb is supposed to be in Z_4?

[gritty sparrow]

Apologies, I meant to write F_4 in all the places I wrote Z_4

[coral steeple]

Ah got it thx

[coral steeple]

Should Z_2 be replaced by F_2 as well?

[gritty sparrow]

Those are the same anyhow

[coral steeple]

True, big brain moment for me

[gritty sparrow]

In any case, is the other isomorphism ok with you?

[coral steeple]

I'm still mulling over the polynomial ring in two variables. I've only read about them in passing

Does the isomorphism $\mathbb F_4[t]/(t^2)\to \mathbb F_2[y,t]/(y^2+y+1,t^2)$ exist because of the fact that $\mathbb F_4\cong \mathbb Z_2[y]/(y^2+y+1)$?

[cloud walrus]

person2709505

[coral steeple]

@gritty sparrow

[gritty sparrow]

Yeah

[cobalt heath]

Huh, how is F_4 isomorphic to an extension of 2-adic integers

[coral steeple]

^

[cobalt heath]

Ah, I see. I would advise to use F_2 instead

[coral steeple]

Fair enough. I'm not sure why we use Zm in class while the prof says the most consistent notation is Z/(m)

[cobalt heath]

Oh that's interesting

Maybe prof just succumbed to students' request?

[coral steeple]

Idk, he is very careful with notation. He insisted on writing the gcd of a,b as GCD(a,b) so that when the notation (a,b) appeared, the statement (GCD(a,b))=(a,b) would have meaning beyond notation

[coral steeple]

Well I think I understand how these functions work more or less now. Thanks again!

[gritty sparrow]

Np

[crystal vale]

If a ring R, for all a≠0, b there exists c such that a•c = b. I want to show that R has no zero divisor.

So if I let x ≠ 0 and y ≠ 0. So, for any b there exists a_1 such that x•(a_1) = b and for a_1 there exists a_2 such that y•(a_2) = a_1.

Hence for any b there exists a_2 such that (x•y)•(a_2) = b , so x•y can not be 0.

Is it correct?

[tough hedge]

why couldn't a1, a2 be 0?

@crystal vale

[crystal vale]

I think if it is then it does not matter

If a_1 is 0 then b will be 0 but I want to prove that for any b

[tough hedge]

if a2 is 0, then xy* a2 = 0

and you couldn't conclude anything about xy being nonzero

[crystal vale]

But when b ≠0 then ?

[tough hedge]

you tell me

what happens if xy * a2 != 0

[crystal vale]

Since if b≠0 then xy can not be 0 because if it is then b will be 0

[tough hedge]

yep

so you need a separate argument in the other case that b=0

[crystal vale]

Okay, but I don't understand why I consider this case

[tough hedge]

o i misread the statement

myb

but yes the argument looks good

we just force b to be nonzero as well

[crystal vale]

Okay, thank you

[delicate orchid]

Haar measure on a finite group

[rocky cloak]

If lim_n a^n = e
and your multiplication is continuous, then
a * lim_n a^n = a*e = a
a = lim_n a^n+1 = lim_n a^n = e

So this only happens if a=e

(or I guess if your topology is not Hausdorff then a sequence can have several limits)

[crystal vale]

If G is a group and H is subgroup of G. Then mapping from G -> G/H such that g-> gH.
Then H is the kernel of this mapping?

[coral spindle]

Yes. Note this only works when H is a normal subgroup.

[cobalt heath]

Can one describe normal groups with exact sequences?

[coral spindle]

Do you mean normal subgroup?

[rustic crown]

or if you define kernels of pointed sets

[coral spindle]

Yes, a normal subgroup is the image of the first map in an SES.

[coral spindle]

Kernels as equivalence relations is more cute, but alas...

[crystal vale]

But if I don't take H be normal then ?

[coral spindle]

Then the question doesn't make sense

Because G/H is not a group.

You cannot then talk of homomorphisms or kernels, unless you expand the definition as det alluded to.

[crystal vale]

But I don't want G/H as a group I just want to take as a set of left Cosets, then?

[coral spindle]

That's fine, but you cannot talk about the kernel of something that is not a homomorphism

G/H is not a group; it has no identity

[crystal vale]

So in that case kernel terminology is not good ?

[coral spindle]

Yes.

The closest comparable statement is that G/H is a G-set. And it is true that the stabiliser of the element H of G/H in G is H. If you don't know what this means, don't worry about it.

[crystal vale]

Means G acts on G/H ?

[coral spindle]

Yes.

[crystal vale]

Yes then the stabilizer of H is H

[coral spindle]

That is what I said, yes.

[rustic crown]

you could define the notion of a kernel as soon as your category has a zero object.

[crystal vale]

I don't know about category

[crystal vale]

Got it, thank you

[cobalt heath]

Yeah then I wonder why no textbooks does this

[coral spindle]

Would you sincererely want to learn about SESs before the first isomorphism theorem?

[cobalt heath]

Maybe

[coral spindle]

It is such a trivial fact about SESs but would be so confusing to a first-time learner

[rustic crown]

G/H is not a group, but it still has a distinguished point, namely the left coset H. so you could still copy your old definition of kernel to define kernel of a map of pointed sets f : (X, x) --> (Y, y) as the pointed set ({a in X : f(a) = y}, x).
but yes, this doesn't tell you a lot of structure about the kernel, here it's just a pointed set.

[cobalt heath]

I recall being confused about purpose of first isomorphism theorem, it could be helpful to know stuff about SES

[delicate orchid]

Well of course the image of the inclusion of 1 is a normal subgroup :trollshiro:

[coral spindle]

Lol

Yes the second map I suppose

Wow the normal subgroups are exactly the kernels? If only there were some theorem that told us this

[delicate orchid]

If only

[crystal vale]

At this moment I don't understand it properly but thank you, is pointed set terminology?

[delicate orchid]

A set with a point

[crystal vale]

Single point ?

[delicate orchid]

It’s very Google-able but it’s just a set with a distinguished element

[crystal vale]

Okay, thank you

[cobalt heath]

Is G/H just a pointed set with a projection from G, or does it have more identifying properties?

[coral spindle]

As I said earlier, it's a G-set. It's a transitive G-set, in fact

Every transitive G-set is isomorphic to some G/H also

[delicate orchid]

Every G-set is isomorphic to a disjoint union of G/Hs

For a rather small class of possible Hs

[cobalt heath]

Hmm, G-set

[languid trellis]

g check

[coral spindle]

You have surely heard of group actions before

A G-set is a set on which G acts.

[cobalt heath]

Yeah, I mean it reminded me of.. ah better not say that

[delicate orchid]

well now you HAVE to

[coral spindle]

Funny sex word

Cmon spit it out

[delicate orchid]

If it’s “G-string” you are a massive prude

[coral spindle]

Not that one

[delicate orchid]

And I am very disappointed in this outcome

Anyway

[delicate orchid]

The real sex appeal is the basis of the burnside ring

[coral spindle]

What a mensch burnside was

[crystal vale]

Let R be a ring with commutative and P is prime ideal in R.
Then I want to show that P[x] is the prime ideal in R[x].

So if I let f(x) = a_n(x^n) +.....+a_0 and g(x) = b_m(x^m) +.....+ b_0.
Then if fg belongs to P[x] then a_0b_0 belongs to P.

But if I take b_0 does not belong to P then recursively show that a_k belongs to p .

But if I take both a_0 and b_0 belongs to P then at some stage I get a_ib_i belongs P. So again two cases arrive a_i belong or not. But at the end we get one of them f(x) or g(x) belongs to P[X].

Is it correct?

[coral spindle]

Looks like the right idea but there are plenty of details missing.

You can show this much more simply by just showing that the kernel of the map R[x] → (R/P)[x] is exactly P[x], which is much easier. Since this is a surjective map and (R/P)[x] is an integral domain, P[x] must be a prime ideal.

[delicate orchid]

Boring!

[coral spindle]

Call me a stick in the mud but to me there is nothing more boring than endless manipulation of polynomials

It's like what applied mathematicians call algebra lmao

[delicate orchid]

I thought u liked number theory :trollshiro:

[coral spindle]

I've been destroyed.......

[delicate orchid]

Actually I take it back. It’s really annoying to see by direct manipulation

It’s immediate that it’s an ideal but primality is poopy

[coral spindle]

[languid trellis]

common wew lads L

[coral spindle]

Dw wew, I stand with you in solidarity 😔 ✊ L takers unite

[delicate orchid]

It is not the fury of the river that wears the mountain, but its persistence

[languid trellis]

unironically kinda based

[crystal vale]

Oh, got it, thank you

[analog sedge]

I am trying to organize my thoughts. However, my thought process is often very abstract, making standard tools unsuitable for capturing the abstractness of my thoughts. For that reason, I want to start learning abstract algebra. Would this be a good idea?

What do you think?

[delicate orchid]

That’s a dreadful reason but everyone should learn abstract algebra so yes do it

[analog sedge]

What? Really?

[delicate orchid]

Yeah it’s fun

[coral spindle]

Abstract algebra doesn't help you organise your thoughts lol

It's a subject like any other. You may as well ask the same question but substitute English literature for abstract algebra.

[delicate orchid]

Except I think English lit would actually help you organise your thoughts better lol

Moreso than AA

[coral spindle]

Yeah, probably

If you want to learn it, start reading an introductory book and you can decide for yourself whether or not it interests you.

But if you are expecting some kind of system to understand abstract thought, you will be sorely disappointed.

[delicate orchid]

That is the realm of philosophy

[analog sedge]

Yeah... I'm not a normal person though.
I often do some vibe-based reasoning that I don't find a way to write down.
Well, let's put it that way...
Before learning equation, you can have a vibe of say... solving x+2=5.
But after learning equation, you have a solid understanding of how to solve it.
I'm just feeling maybe something I've been intuitively thinking lie inside the abstract algebra.

So, maybe I need to try out.

[delicate orchid]

If you want “vibe based” you should not do mathematics

[coral spindle]

This really has nothing to do with abstract algebra

The name isn't really helping. Most (pure) mathematicians just call it algebra, anyway.

I reiterate that you should just pick up a book to get a better idea, and you can decide if you are interested.

[languid trellis]

Terence tao told me that professional mathematicians are exclusively vibe based

[delicate orchid]

Yeah and he does analysis and crypto scams not mathematics

[languid trellis]

Crypto scams? Am I missing some lore?

[delicate orchid]

He was big into the web 3 shit iirc

[languid trellis]

Ah I have no clue what that is

[delicate orchid]

Good

[coral spindle]

I might also point out that doing intuitive reasoning does not make you "not a normal person." Being unable to formalise intuitive reasoning simply means you are early in your mathematical training.

I assure you that every mathematician uses their intuition. The difference is that they have trained their intuition to be accurate.

[delicate orchid]

And they’ve learnt the required language to write down their intuition formally

Which is what proofs are

[coral spindle]

If you have any specific questions about the content of abstract algebra, you can feel free to ask them here.

[sly rain]

Whats a nice not too tedious example to inspect a concrete group action on some set and answer some questions regarding basic group action stuff?

Like not conjugation, left multiplication

Maybe has a concrete example and good questions in mind

[coral spindle]

Well, the canonical example is any symmetric group S_n acting on {1, ..., n}

[delicate orchid]

^

G acting on itself as well

Any matrix group acting on k^n is a classic

[sly rain]

Can you formulate some additional questions like as an exercise

[delicate orchid]

Just start calculating orbits and stabilisers?

[coral spindle]

I'm not a textbook lol I don't have good exercises ready and waiting

[delicate orchid]

Prove the sylow theorems

[coral spindle]

Yeah ig the best thing would just be what Wew suggests

think about some orbits and stabilisers, see what you get for S_n acting on {1, ..., n}

Describe them

Maybe calculate their orders

[sly rain]

Soon,

Yeah fair answers, I was Just wondering If there are additional questions that you should ask yourself.

[coral spindle]

Idk maybe here's a wild example

The group of rotations about the origin of the plane acts on the set of lines (i.e., infinitely long straight lines) on the plane. Describe the orbits.

[delicate orchid]

Orbits of S^1 on C my beloved

Uhhh what else

A5 acting on a icosahedron

[coral spindle]

PGL_2(F) is sharply 2-transitive on P^2(F) right

This seems like a good and reasonable exercise to spring on a beginner :DDDDD

[delicate orchid]

How the fuck am I supposed to know

Yes it is

[coral spindle]

Tbh it did seem like the erudite knowledge you would know

LOL

[languid trellis]

the set of lines going through the origin, or are you describing something like V/W, where W is a 1-dim subspace?

or just all the lines

[coral spindle]

All the lines! All the cosets of 1d spaces on the plane.

[delicate orchid]

Same thing (I love semi simplicity)

[languid trellis]

my brain is struggling hard to visualise this space

[rocky cloak]

Show that if x and y are in the same orbit, then their stabilizers are conjugate.

[delicate orchid]

That’s a really good one

[coral spindle]

Don't try to visualise all the lines at once! Picture a single line (off the origin) and imagine rotating it. What does that look like?

There is a very simple geometric description of when two lines are in the same orbit.

[delicate orchid]

Almost like a homotopy

[coral spindle]

And to be clear, I mean rotation around the origin.

The group is S^1

[languid trellis]

intuitively speaking, two lines should be in the same orbit if they have the same gradient and are the same distance away from the origin I think. Something similar to how, if we take a point z in C and consider all rotations, the orbit of the z is the circle |z|

[coral spindle]

If they have the same gradient?

But the gradient clearly changes when we rotate a line

[languid trellis]

Oh yeah you're right

[coral spindle]

You are close.

[languid trellis]

there should be some invariant though, it's clearly not all lines some distance away form the origin

[coral spindle]

Why not?

You say it's clear, why

[delicate orchid]

This is me when I rotate

Oh that’s the invariant

WHAT

[coral spindle]

Pride

be proud

🏳️‍🌈

[delicate orchid]

I see

[languid trellis]

We need to exclude reflections, as this picture demonstrates

[coral spindle]

The group doesn't include reflections in the first place

Like I said, it's the group of rotations about the origin

[delicate orchid]

Yeah which is why they can’t be in the same orbit.

[languid trellis]

Yes, but the point is that they can't thus be in the same orbit

[coral spindle]

Are you sure?

[languid trellis]

Wait.

[delicate orchid]

Yeah I’m pretty sure

[languid trellis]

we can generate D_n with any two "distinct" reflections

[coral spindle]

When you say distance from the origin, you mean the minimum distance, right? The perpendicular distance.

:)

There is a line from the origin to any other line, which forms a perpendicular.

[languid trellis]

I was going off of some "intuitive" understanding, but now I realise my intution wasn't quite clear enough to attack this problem

[delicate orchid]

That’s the invariant I saw

[coral spindle]

I think it's pretty clear once one spots that

[delicate orchid]

I think for a reflection you can use the fact that you’re not dealing with line segments to move one onto the other

[languid trellis]

Oh right, so the orbit is the set of all tangents to a circle a given distance away from the origin

[coral spindle]

Indeed. The orbits consists of lines of equal distance from the origin.

[delicate orchid]

Oh yeah? Watch this

[languid trellis]

What would we use to formalise this? How would we even describe "the set of all lines" or "all rotations"?

[delicate orchid]

All rotations is easy, it’s S^1 lolololol

SU(1) if you’re a nerd

[coral spindle]

Wym? The set of all lines is very easy to describe. You know this. Take all 1d subspaces V of R^2 and look at R^2/V. That's the set of lines parallel to V.

Similarly rotations, as Wew points out, are in bijection with the unit circle. The group structure on that is well-known, it's the same as complex multiplication.

[delicate orchid]

^nerd

[coral spindle]

Guilty as charged

The invariant is more interesting though. It takes some amount of work to show that there's a minimal-length representative for every coset of any V

[delicate orchid]

The representatives are parameterised by R with some quadratic form being minimalised

[languid trellis]

Maybe I'm just being a little pedantic but I'm curious on how we would formally describe "all the 1-dim affine subspaces of R^2". Like what operation do we do to go from R^2/V1 to all of them? are we just taking a big uncountable union?

[delicate orchid]

Quadratic forms on R are uhhh quadratics lol

And so have a unique minimum

Does that work?

[delicate orchid]

Yeah union the mfs

[coral spindle]

Pick a unit vector. Look at its span!

Every 1d subspace has a basis of one element. WLOG we can choose this to be unit. There will be two such generating vectors for each subspace, but it's not hard to choose one.

[languid trellis]

right

[delicate orchid]

Le double cover has arrived

[coral spindle]

So we have parametrised the set of all 1d subspaces in terms of (particular) unit vectors

[languid trellis]

I'm following

[delicate orchid]

Now just take all the quotients! Ah ah ah ah

[coral spindle]

Exactly. Then the set of their cosets is the set of all lines

[winged void]

I have a question regarding symmetric group: This is what my book says: Let $\sigma \in S_n$, then there are $n$ possibilities for $\sigma(1)$. For $\sigma(2)$, there are $n - 1$ possibilities
since $\sigma(1) \neq \sigma(2)$, as $\sigma$ is injective. For $\sigma(3)$, there are $n - 2$ possibilities, and so on. In total, there are
$n! = n \cdot (n - 1) \cdot \ldots \cdot 2 \cdot 1$ possibilities to choose $\sigma$.

A permutation can be represented as a matrix with two rows. In the top row, we have the elements ${1, \ldots, n}$ and below them their images:
[
\begin{matrix}
1 & \ldots & n \
\sigma(1) & \ldots & \sigma(n)
\end{matrix}
]
However, this representation is rather cumbersome, which is why permutations are usually represented in cycle notation. well i understand everything. However im confused about something well here when we say that $\sigma \in S_n$ i do not really get it what does that mean

[cloud walrus]

Mootje

[coral spindle]

Wym? It's saying sigma is an element of S_n

[delicate orchid]

Sigma is an element of Sn?

[coral spindle]

are you confused about what sigma is

[delicate orchid]

It’s standard set theory notation

[winged void]

because i thogght first that $\sigma(1) \in S_n $

[coral spindle]

No

[languid trellis]

Right. I understand this construction, but whatever reason it still feels a little bit fleeting. I think that just might be me though. thank you boss man

[coral spindle]

sigma(1) is in {1, ..., n}

[winged void]

yes but i do not get it

[coral spindle]

The elements of S_n are bijections (i.e. functions) {1, ..., n} → {1, ..., n}

So sigma is a such a thing.

[delicate orchid]

S_n is the group of permutations of {1,…,n} - bijections from- BOYTJIE STOP DOING THAT

[coral spindle]

TOO SLOW TEE HEE

Sorry I'll leave it to you

[winged void]

so the elements of S_n are function that are bijvectives

am i right about that

but then what is $\sigma(1)$ is it an elemen of a specific function

[cloud walrus]

Mootje

[winged void]

but its not an element of S_n

[delicate orchid]

It’s sigma evaluated on 1

[coral spindle]

[delicate orchid]

Calling that an “element” of a function, while rather based, is very misleading

[winged void]

and what is then the diffrence when we say $\sigma o \tau (a)$ or $\sigma o \tau$ they are just the same

[cloud walrus]

Mootje

[coral spindle]

sigma(1) is like sin(pi/2). It is the evaluation of a function

[winged void]

yes but is it not suppose to be also an element of S_n

[coral spindle]

No.

It is meant to be an element of {1, ..., n}

sigma is a function

sigma(1) is a number

[winged void]

then why the result of $\sigma o \tau$ is equivalent to $\sigma o \tau (a)$

[cloud walrus]

Mootje

[coral spindle]

No. $\sigma \circ \tau$ is the \emph{function} that sends some $a \in \set{1, \dots, n}$ to $\sigma(\tau(a))$. Whereas $\sigma \circ \tau(a)$ is frankly bad notation for $\sigma(\tau(a))$.

[cloud walrus]

Boytjie

[winged void]

thats what my book is saying

[coral spindle]

It is not.

You are almost certainly misinterpreting your book.

[winged void]

i see what you mean

[coral spindle]

It is clearer to write (sigma o tau)(a) instead of sigma o tau(a) in my opinion.

It leads to confusion like this

[winged void]

i see bcause one of the question that i was trying to solve was this

"Let $\sigma, \tau \in S_n$.
(a) Suppose that $a \in {1, 2, \ldots, n}$ is mapped to $a' \in {1, 2, \ldots, n}$ by $\tau$. Prove that $\sigma(a)$ is mapped to $\sigma(a')$ by $\sigma\tau\sigma^{-1}$."

[cloud walrus]

Mootje

[winged void]

like what do they mean here

what is the intuiton

behind it

[coral spindle]

There's no intuition to speak of, really

This is just using functions

Please be more specific about what's unclear here

[winged void]

well i mean what do they mean with the last sentence that sigma(a) is mapped to sigma(a')

using the function

[coral spindle]

When people say that a function f maps a to b they mean f(a) = b.

[winged void]

do they mean applying the function sigma tau sigma inverse

[coral spindle]

Of course.

[delicate orchid]

The number sigma(a) is mapped to the number sigma(a’)

Or like, really explicitly:

tau^-1(sigma(tau(sigma(a))) = sigma(a’)

[coral spindle]

Other conjugation!!

[delicate orchid]

RAAAAAGGGHHHH

[coral spindle]

Trivial typos

[winged void]

Well I thought first when we say something map to something

We mean the -> which gives domain and codomain

[coral spindle]

We also use that language in that case, yes.

[winged void]

But I thought we use the other pile

Like it’s a to

Pile

[coral spindle]

...pile?

[scarlet parcel]

I think they mean arrow

pilum means spear in latin 🪤

[winged void]

I mean arrow

Sorry

This

What I mean

$to$

[coral spindle]

A pile means something quite different in English, just so you know

[cloud walrus]

Mootje

[winged void]

$\to$

[cloud walrus]

Mootje

[delicate orchid]

I’d love to know the linguistics behind “arrow” and “pile” becoming synonyms

[scarlet parcel]

Pil means arrow in my language

In old Norse as well supposedly

I guess once you have "thin pointy thing" down in the language you're pretty much set

[delicate orchid]

I see, interesting

[winged void]

In my language as well 😅

[winged void]

But what I mean this one is for domain and codomain

And the other arrow

[scarlet parcel]

What is your language?

[winged void]

Is the one that giving a function

[winged void]

Dutch

[scarlet parcel]

I see

[winged void]

$\mapsto$

[cloud walrus]

Mootje

[winged void]

Well you are right sorry

[delicate orchid]

Debating what arrow we use is just being pedantic

[winged void]

I have confused two arrows the one to and the other one with each other

[coral spindle]

$f : A \to B$ means that the function $f$ takes elements of $A$ and outputs elements of $B$.

$f \colon a \mapsto b$ means that $f(a) = b$.

[cloud walrus]

Boytjie

[coral spindle]

The difference is very important, I think you'll agree.

[winged void]

I agree for sure

But maping is indeed the second one

You are right

I confused two different things with each other

[glad osprey]

Is there a nice reason why almost all substructures are preserved under intersection? Like subgroups, subspaces etc. I feel like it has something to do with how intersections are meets in the preorder of subsets, and a substructure is a particular kind of subset, but I can't think further than this

[coral spindle]

Yes it's actually quite nice

This has to do with universal algebra

One moment

I have to deal with something but I will explain momentarily

[crystal vale]

If I want an example of a commutative ring R with unity and its subring S which has same unity such that intersection of S and M is not maximal in S where M is maximal in R.

So if I let R be the polynomial overs set of all real numbers and S be the set of polynomials over integers and M be (x) then M is maximal in R. But intersection of M and S is not maximal in S, because it is polynomial generated by x over integers which is not maximal in Z[x].

Is it correct?

[coral spindle]

OK here is the story. The majority of algebraic structures work like so: you have some set, and you have some operations with various numbers of inputs. A distinguished element such as the identity can be seen as an operation with zero inputs. Such things are usually called "algebras" by universal algebraists, but this terminology is a bit confusing, so I will just call them structures.

Then we usually require our structures to have some properties. We like certain nice properties, which are of the form "Forall a_1, ..., a_n, we have (expression) = (expression)" where each of these expressions uses only the a_i and the operations. Such properties are known as identities.

We say that a kind of algebraic structure forms a "variety" if its axioms can be expressed only with some set of these identities. Groups form a variety, but one needs to include the inversion of a group as an operation in order to do so. Rings also form a variety. Fields to not form a variety – I might suggest you try proving that they cannot, under any set of operations, form a variety, as it requires a bit of creativity.

There are some facts about varieties that are easy to prove. This really comes from how nice identities are! Your observation that the intersection of two substructures is also a substructure (in the variety) is one of them. Another is that the Cartesian product of two structures is a structure in the variety. There is also a notion of quotients of structures but this gets more complicated.

@glad osprey forgot to ping

[crystal vale]

But if we take the intersection of S with prime ideal P it will be prime ideal in S. But if the unity of S and unity of R are not the same then does it change the result ?

[delicate orchid]

Oh look. The Grothendieck completion

[coral spindle]

So pretty

[glad osprey]

Thanks for the response! Is this equivalent to the varieties you are talking about?

[coral spindle]

No.

This is a completely different thing.

[coral steeple]

Wait, how do you know y mapsto x^2 and t mapsto x^2+x+1 is an isomorphism? I'm willing to accept it's a sort of change of variables (so a homomorphism), but why should that be injective and surjective?

Not by bashing out the multiplication and doing the long division by taking two arbitrary elements of the form (a+by)+(c+dy)t (I don't see how anyone could more readily see that that is an isomorphism than just finding an iso between the original two rings)

[glad osprey]

I think I get it now: for any variety S with subvarieties A and B, then A∩B inherit all the equational axioms, (like associativity etc.) from S, and for the existential axioms it is clear that if there exists an x in A and B then x is in A∩B. And closure is also trivially satisfied because anything that is in A and B is in A∩B. And apparently you can even rewrite the existential axioms as equational axioms? And algebraic structures that are not varieties have non-equational axioms, so those substructures might not be preserved under intersection?

[gritty sparrow]

I would just bash. You don’t have to do the multiplication as long as you accept it is a homomorphism (which is easy enough to prove)

[coral steeple]

How were you able to come up with this last isomorphism? (if you can remember)

[glad osprey]

@coral spindle sorry for pinging, just wondering if I understood this correctly

[coral spindle]

There are no existential axioms in a variety.

That is part of the definition

There are only identities.

Families of algebraic structures that aren't varieties may or may not have these properties, it's simply not guaranteed.

[cobalt heath]

Guess I found one sense where I might be better than him

[glad osprey]

How do you formulate the existence of an identity element in a group without existential quantification?

[coral spindle]

A distinguished element such as the identity can be seen as an operation with zero inputs.

[delicate orchid]

Zoinks

[rocky cloak]

Fun fact: the existence of inverses (meaning a * a^-1 * b = b) guarantees the existence of an identity, unless your "group" is the empty set.

[gritty sparrow]

I knew the identities t and y satisfy, so I looked for elements in the other ring that satisfied those relations

[glad osprey]

I see, thanks 👍 so id : {1} → G and inv : G → G is part of the definition of a group?

[coral spindle]

Yes, you need that to show it forms a variety.

[glad osprey]

and for a field, you could define unity : {1} → F and mult_inv : F -> F, but you couldn't make an identity out of it, because you need x = 0 OR x * mult_inv(x) = unity

[coral spindle]

mult_inv(0) is the issue, yes

[crystal vale]

If I want to show that if for any pair of ideals I and J with IJ be subset of P implies I is subset of P or J is subset of P then P is prime ideal in commutative ring R.

So if I let I = (x) and J =(y) then xy belongs to P and then x belongs to P or y belongs to P.

Is it correct?

[winged void]

im have solved the first deel of the question

However

im facing a problem with the second deal

part

i mean

[coral spindle]

OK

[winged void]

Let $\sigma, \tau \in S_n$.
\begin{enumerate}
\item[(a)] Suppose that $a \in {1, 2, \ldots, n}$ is mapped to $a' \in {1, 2, \ldots, n}$ by $\tau$. Prove that $\sigma(a)$ is mapped to $\sigma(a')$ by $\sigma\tau\sigma^{-1}$.
\item[(b)] Deduce from (a): if the cycle decomposition of $\tau$ is given by
[
\tau = (a_1 a_2 \ldots a_k)(b_1 b_2 \ldots b_{\ell}) \ldots (g_1 g_2 \ldots g_q)
]
then the cycle decomposition of $\sigma\tau\sigma^{-1}$ is given by
[
\sigma\tau\sigma^{-1} = \sigma(a_1) \sigma(a_2) \ldots \sigma(a_k) \cdot \sigma(b_1) \sigma(b_2) \ldots \sigma(b_{\ell}) \cdot \ldots \cdot \sigma(g_1) \sigma(g_2) \ldots \sigma(g_q)
]
\end{enumerate} here is the whole question

[cloud walrus]

Mootje

[winged void]

now im trying to solve for b

this is what i have until now

but im not getting that far

i tried to multiply both side with the inverse maybe i will get something but i did not

and the only thing that we know from a that t(a) = a'

and also that sigma tau sigma sigma(a) = sigma(a') more nothing

im not really sure if you suggest something

[night onyx]

I'd break it into 2 pieces:

1. explain why if you write pi = S_1 * S_2 * ... * S_k as the product of disjoint cycles, then (sigma * pi * sigma^{-1}) = (sigma * S_1 * sigma^{-1}) * (sigma * S_2 * sigma^{-1}) * ... * (sigma * S_k * sigma^{-1}). This means that pi and sigma * pi * sigma^{-1} have the same cycle structure

2. Because of 1, you only need to show the result for a single cycle. So now focus on showing that if pi = (a1 a2 ... an) is a cycle, then sigma * pi * sigma^{-1} = (sigma(a1) sigma(a2) ... sigma(an)). Basically this just comes down showing that if pi(a) = b then (sigma * pi * sigma^{-1})(sigma(a)) = sigma(b), which is exactly part 1.

if you really think about what the cycle structure notation represents, like writing pi = (a1 a2 ... an) means that pi(a1) = a2, pi(a2) = a3, ...., pi(an) = a1, then this questions is just applying part a) to part 1. of what I wrote above

[winged void]

i will try that

[night onyx]

yeah it's a confusing looking question that will seem really easy once you think about what the cycle notation means

[winged void]

its really confusing

i hate permutation

this is what i did

im not so sure

I Find it really irritating question

[night onyx]

yeah, I would think about a simple example, like pi = (1 2 3)

don't try to prove the full result, just try thinking about an example first

like pi = (1 2 3) means that

pi(1) = 2
pi(2) = 3
pi(3) = 1

now using part (a) you know that

(sps^{-1}) (s(1)) = s(2)
(sps^{-1})(s(2)) = s(3)
(sps^{-1})(s(3)) = s(1)

[winged void]

Well I do not get what is here s2

[night onyx]

sorry i'm writing s instead of sigma

lol just shorter

and p instead of pi

I should have said that

[winged void]

Sure I get it

[night onyx]

okay, so now look at

(sps^{-1}) (s(1)) = s(2)
(sps^{-1})(s(2)) = s(3)
(sps^{-1})(s(3)) = s(1)

write this in cycle notation, like like for how pi = (1 2 3)

[winged void]

Well then the cycle notation will be (sps^{-1}) (s(1)) (sps^{-1}) (s(2)) (sps^{-1}) (s(3))

I guess

But not so sure

[night onyx]

no, remember than if

p (1) = 2
p (2) = 3
p (3) = 1

then we write p = (1 2 3) as a cycle

so if

(sps^{-1}) (s(1)) = s(2)
(sps^{-1})(s(2)) = s(3)
(sps^{-1})(s(3)) = s(1)

then we would write sps^{-1} as

...

[winged void]

Wait

[night onyx]

thats the question lol

[winged void]

I guess then s(2)s(3)s(1)

[night onyx]

almost

I mean yes lol

[winged void]

This is equivalent

[night onyx]

but also s(1) s(2) s(3)

[winged void]

To s(1)s(2)s(3)

[night onyx]

exactly

[winged void]

Sure

Agree but this works only if the cycle is a full cycle I mean like (1 2 3 )

But if it was like (1 3 2 ) it won’t work

Pi

[night onyx]

why not?

I mean it does work lol, or else the question would be wrong

and the question isn't wrong

[winged void]

Hahaha

Sure

But what has that do with our proof

[night onyx]

it's the same thing, like if

pi(1) = 3
pi(3) = 2
pi(2) = 1

then the same thing works, you have

(sps^{-1}) (s(1)) = s(3)
(sps^{-1})(s(3)) = s(2)
(sps^{-1})(s(2)) = s(1)

which is (s(1) s(3) s(2))

it works for any cycle like (1 3 2 5 6 8), whatever

but then you need to show it's also true for multiple cycles

[winged void]

Sure agree the intuition now is 100%

[night onyx]

like (1 3 2)(4 6) (5 9 7)

[winged void]

Oeps Well I mean every multiple cycle can be written as one cycle

So it’s true

[night onyx]

no, that's not true

you can't write disjoint cycles as a single cycle

[winged void]

That’s true

You have a point

[winged void]

Agree

[night onyx]

but to show it works for multiple cycles, just show that if you have something like

s (1 2 3) (4 5 6) * s^{-1}

then this is (s (1 2 3) s^{-1}) * (s (4 5 6) s^{-1})

each cycle in there is getting conjugated by s

[winged void]

Can we multiply that way

Because if that’s possible then we finish I guess

[night onyx]

yeah, just make sure you understand why you can multiply that way

like I said, not a hard question, just hard to wrap your head around the concepts and notations

[winged void]

To be fair I’m not so sure why it’s possible

[winged void]

True

And group theory is a difficult subject as well

[night onyx]

yep, it is

that was the first proof course I ever took in my first year of math, it was painful

[winged void]

Me too

Now I’m in my first year

[night onyx]

I was almost failing up until the very end, then it started to click, but it's hard stuff

[winged void]

But why multiplication that way is well good

[winged void]

Me too now I’m failing as well 😭

[night onyx]

remember that a permutation p is a function

[winged void]

True

[night onyx]

and multiplying permutations like that is function composition

[winged void]

But I mean we began with a composition

Am I right

s (1 2 3) (4 5 6) * s^{-1}

[night onyx]

ahh yes, here's another tricky part

(1 2 3) is a function

(4 5 6) is a function

[winged void]

Function I thought it’s a permutation

[night onyx]

s (1 2 3) (4 5 6) * s^{-1} is the composition of 4 functions

a permutation is a function

[winged void]

Sure

[night onyx]

it's a bijection from {1, ..., n} to itself

[winged void]

Sure

Composition of 4 functions how to go then to the composition of 3 functions multiply together

[night onyx]

like when I write (s (1 2 3) s^{-1}) * (s (4 5 6) s^{-1})

instead think of functions

like fgf^{-1}fhf^{-1}

like just composition all those functions

and since you have f^{-1} f in the middle, those cancel out, because of inverses

[winged void]

Oh I see super smart

Bro I hate that

I mean why are we even learning about permutations what is the intuition behind it why is it important

In math

[last spoke]

An exercise in Rotman asks to give an example where $H \lhd G$ and $G$ nonetheless has no subgroup isomorphic to $G/H$. An easy example is ${\pm 1} \lhd Q_8$ but $Q_8/{\pm 1} \cong V \not\subset Q_8$. What's an example with a different group? I couldn't think of any others

[winged void]

Thanks for the explanation I get it I will do some example like you did

But your trick is nice like try an example

And see what happen and prove

[cloud walrus]

Ryemo🌾

[last spoke]

@winged void (Alternating) Permutation groups are a large class of finite simple groups so they're interesting that way. Also every finite group is a subgroup of some permutation group, so that's a pretty big deal

[winged void]

I see

[south patrol]

For example subgroups of Z^n work nicely, because no (non-zero) element of Z^n has finite order

So for example 2Z as a subgroup of Z

[night onyx]

groups of bijections of sets are really important in math, and the symmetric group is the simplest example of that, it's the group of bijections of a finite set

[last spoke]

Oh of course

[night onyx]

but you can study other kinds of bijections too, like the group of permutations of an infinite set, or bijections which have some more structure, like continuous bijections, or linear bijections, or whatever kind of bijections

permutations are just the first, simplest form of that

[winged void]

Ah I see

[night onyx]

lol it's just weird and abstract at first

[winged void]

To be fair it’s still a difficult concept for me

[night onyx]

yeah don't worry, it's always tricky especially in your first proof based course, cause it's a lot of new ideas at the same time as learning proofs

[winged void]

Like half of the words you have said just like saying something that I do not understand that well

[night onyx]

well yeah, that's okay

[winged void]

I heard once that in linear algebra when we say that linear transformation is symmetric with another one

[last spoke]

some of the permutation groups are also the symmetry groups of important shapes

[winged void]

Has to do something with that

Or something

[night onyx]

no one expects you to know all terminology when you first start of course

[winged void]

True

[last spoke]

Like the rotational symmetries of the icosahedron are isomorphic to A_5

[winged void]

That’s too much for me @last spoke

Like my brain is already blowing due to that question

[last spoke]

who doesn't love spinning shapes in their head

[winged void]

I really love that for sure

I mean but why choosing a hard one

[last spoke]

if it's a tetrahedron? an icosahedron? guess what that's a permutation group

[winged void]

I mean triangle

[night onyx]

watch good youtube videos, having a visual/geometric intuition really helps when first learning group theory

[winged void]

Easy Nice shape

[night onyx]

because it's so abstract at first

[winged void]

Do you suggest any

[last spoke]

well actually the rotational symmetries of a triangle are also a permutation group

[night onyx]

only broadly, like 3blue1brown is good

I don't watch many so I don't know specific ones

[last spoke]

actually it's still tru if you include the reflections

[winged void]

3blue1brown does not have anything on algebra I guess

[last spoke]

all the different ways of rotating and reflecting a triangle rigidly is S_3

[winged void]

True

I guess because for rotation there are 3 elements

Not rotating

Rotating 60 degree

Rotating 120

I guess

And reflecting the same

[languid trellis]

#1217259452621783154

[coral spindle]

Indeed one has to look fairly far to find examples. This is the smallest possible one, if I'm not mistaken.

The family of dicyclic groups, hilariously notated Dic(4n), provide similar counterexamples. In fact Q_8 is Dic(8).

[last spoke]

the dic groups

interesting 🧐

[coral spindle]

There is a subgroup of Dic(4n) isomorphic to C_2, where the resulting quotient is isomorphic to Dih(2n). But there is no subgroup isomorphic to Dih(2n) in Dic(4n).

[winged void]

@night onyx thx alot for the intuition

this is last version of my proof

im not really sure if that is good proof

the first two pics are the intuition

the last one is the proof

[night onyx]

it's definitely the right idea and better, in proofs though you wouldn't want to include the part about the motivating intuition. Like the goal would be that the proof is just short/straightforward. That's part of writing a good proof, you figure out the intuition along the way, but you want the final proof to be short and direct

like if you write code, the final code you write should be simple/direct/clean even if along the way you're writing a lot of hacky stuff to figure out how to get it working

[winged void]

i see

but the last picture i did not include any intuition ideas

so is not that complete proof

[night onyx]

nono the last part is good

[winged void]

oh i was afraid

but you mean if im going to write a proof

i do not need to write intuition

intuition only for me

[night onyx]

yeah

[winged void]

now im working on another proof

also permutation

im going to try the intuition

example trick

i hope it will work

[crystal vale]

If a set has n elements then the total number of binary operations defined on that set is n^(n^2), right?

So if a set has 3 elements then the total number of binary operations is 729, right?

[coral spindle]

Yup

[winged void]

i tried to use the intuition trick but did not really find something intuitve and i got stuck again

maybe my intuition is incorrect

but i will show you what i did

[crystal vale]

I don't understand why it said that 19,683 binary operations on a set of 3 elements

[winged void]

Let $\tau = (1 2 3 \ldots n) \in S_n$ and $\sigma \in S_n$. Prove that $\sigma\tau = \tau\sigma$ if and only if $\sigma = \tau^i$ for some $i \in \mathbb{Z}$ with $0 \leq i < n$.

[cloud walrus]

Mootje

[winged void]

this is my intuition but i do not see something important from here

that i can grasp to prove the equivalence relation

[rocky cloak]

Think you need to check your math on that 729, because 3^3^2 = 19683

[night onyx]

so the first thing to notice is that if st = ts for permutations, then sts^{-1} = t

so you have t = (1 2 3 ... n) and they're saying s * (1 2 3 .... n) * s^{-1} = (1 2 3 ... n)

when you're working on intuition you want to actually make sense of why it's true in your head, often when you start doing proofs it's very mechanical, like it can feel like you're just manipulating symbols to try to get the answer, but really you should try to figure out "why" something is true before trying to prove it, and intuition is like really figuring out "why" it's true

[oak owl]

A non-zero element p of a domain R is called prime if p is not a unit and
whenever p | a · b for some a, b ∈ R, either p | a or p | b.
•An element a ∈ R is irreducible if is not zero or a unit, and if, whenever a = cb, either c or b is a unit
(b) [2 marks] Give an example of a prime element in a domain that is not an irreducible element. Justify your answer

[night onyx]

so like with this question, you have (1 2 3 .... n), and they're assuming that s * (1 2 3 ... n) * s^{-1} = (1 2 3 ... n)

[winged void]

well true

but this mean that s and s^{-1} should do anything to t

[night onyx]

but you know (from the exercise you did before) that s * (1 2 ... n) * s^{-1} = (s(1) s(2) .... s(n))

[winged void]

to have the good releation

[night onyx]

so you have that (1 2 3 4 .... n) = (s(1) s(2) s(3) s(4) ... s(n))

[winged void]

sure agree on that

sure until now i agree on what you are saying

[night onyx]

so now what does this really mean?

[winged void]

actually that is the prove

well this is a cycle

that can be written as

(s(1) s(2) s(3) .... s(n)) is equivalent to (s(1) s^2(1) s^3(1) ... )

and we can do the same thing for (1 2 3 4 ... n ) im not really sure if what im saying correct

but it makes sense to me

[winged void]

(:

[crystal vale]

Isn't 3^(3^2) ?

[rocky cloak]

Yes...

[night onyx]

so the question says that these two cycles are the same

[winged void]

sure agree on that

[night onyx]

okay, so the only difference between them is maybe that the one on the right is rotated

like you don't have to have that s(1) = 1, s(2) = 2, .....

you could have that s(1) = 3, s(2) = 4, s(3) = 5, ....

like you know how (1 2 3 4) = (2 3 4 1) = (3 4 1 2) ?

[winged void]

sure agree well they are all the same actually

[crystal vale]

Oh my bad, thank you

[night onyx]

exactly, you can think of each of them as just "rotating" by 1

it's the same cycle, just rotated

[winged void]

sure

agree

[night onyx]

so if the cycle on the right above is equal to the one on the left, it's just the first one, but rotated

[winged void]

sure but what has the power to do with rotation

[night onyx]

and if t is the one on the left, then t^k is rotating k times

that's the intuition (in my head anyways) t^k is like rotating t k times, and if the right one is the same as theh left one, it must be t^k for some k

[winged void]

sure agree with you

so actually this mean that composition here is rotating

something like that

i guess but its actually always the case

i guess that composition is rotating

[delicate orchid]

So like, (12345)^3 = (14253), you skip three places each time

(1234)^2 = (13)(24)

[winged void]

but i mean im also facing until now difficulty with intuition but when you explain it i see the intuition striaght a head how did you develop that

[winged void]

well its not always the same then

[wise helm]

Hello! This question is asking me to use the Fundamental Theorem of Group Homomorphisms to show that two groups are isomorphic to each other. I think I have shown that these two groups are isomorphic to each other (as there exists some function φ which forms a bijective group homomorphism).

I don't see how the Fundamental Theorem of Group Homomorphisms could be used to show that these two groups are isomorphic to each other. Any help would be appreciated.

[night onyx]

many years of doing math

[winged void]

i see but like do you have maybe some tips for me

or something

[delicate orchid]

For the record, I disagree with the notion of viewing powers of elements as rotations of that element

It applies to D_n and SO_n(k)

[night onyx]

yeah it's not always the case that rotations and powers correspond

[winged void]

well what is D_n and what is SO_n(k)

[delicate orchid]

Like good luck convincing me tensor powers are rotations of a space

Conjugates I’d buy being rotations. Cause conjugation is like “moving” an element somewhere else

[winged void]

One more question I’m not really sure because those were the only two question about permutations. Do you have like an extra question that I can do maybe it’s stupid question not so sure

[night onyx]

that's the question they were talking about, if s = (1 2 3 ... n) and tst^{-1} = s then t = s^k for some k, the intuition being that if (1 2 3 ... n) and (t1 t2 ... t^n) are the same, then one is the other "rotated k times", it was more of an analogy than any broad statement about powers being rotations

[crystal vale]

What is the fundamental theorem of the group, is it the first isomorphic theorem?

[wise helm]

Yes

[crystal vale]

I think Then there are two ways you can show mapping from (3) -> Z_4 and show that it is surjective mapping with kernel (12) otherwise take as mapping from (3) /(12) -> Z_4

[wise helm]

ah okay. I see now. thank you!

[crystal vale]

Like taking mapping from 3n -> nmod4 , does it work?

[wise helm]

<12> is the kernel of <3> when I have the function φ defined as (x/3) (mod4). I've shown that <3>/<12> is isomorphic to Z_4 under φ. So that's the same as showing <3>/ker(φ) is isomorphic to φ(<3>). φ(<3>) is Z_4.

it does. I just didn't see that it fit with the First Isomorphism Theorem.

[night onyx]

You could try proving that if two permutations have the same cycle structure they're conjugate. You already showed that if they're conjugate they have the same cycle structure, but you can also show the other implication too

[winged void]

I will try thanks a lot !

[random pasture]

i'm a little rusty with this, do I just have to prove reflexivity, symmetry and transitivity? And if so, how exactly does this support reflexivity?

[coral spindle]

Yes, to prove that E is an equivalence relation you prove that it satisfies the definition of equivalence relations.

How exactly is this reflexive? That's for you to prove.

[random pasture]

so is it valid to just say (a, b) E (a, b) => a+b<==>a+b satisfies reflexivity?

since it just maps to itself

or wait no it would be a-b to a-b since (a, b) basically represents a-b right

[coral spindle]

a+b<==>a+b
What does this mean

Do you mean a+b = a+b

The last thing you said makes no sense.

Use the definition.

The whole point of this exercise is that you cannot use -. Naturals don't support subtraction in all cases.

You are constructing the integers, so do not use the integers in the construction.

[random pasture]

Yeah a+b=a+b mb

[random pasture]

Oh i forgot it’s on N mb

[winged void]

i proved it

i guess but to be fair

with 0 intuition

this was my proof

im sure that its a correct proof

but without intuition

which is really sad actually

):

[night onyx]

oh that's not quite what I meant, I meant that if you have two permutations that have the same cycle structure, like (1 3 2 5)(4 6) and (1 2 3 4)(5 6) then those two are conjugate, so you can find some permutation s such that s * (1 3 2 5)(4 6) * s^{-1} = (1 2 3 4)(5 6)

[winged void]

Oh i will try that sorry i understood you wrong

However is my proof correct that I have written

[night onyx]

no worries, that's okay

[winged void]

It feels as if I’m playing actually

Because I just multiplied once left side and the other right side

And which are going to give the same result

[night onyx]

sometimes it's as simple as that, yeah

there's not always some deeper meaning to a problem, sometimes you just have questions that want you to apply formulas or do stuff like that

but in general if a problem is hard and it's not obvious how to prove it, the problem is that you don't actually understand what the problem is asking well enough

or maybe if you look at it different it'll make more sense

[winged void]

I see all those tips help really well

[night onyx]

and when you get to higher/more advanced math, actually understanding and having intuition behind things is way more important, a proof just shows that something is true lol, but we care about ideas and intuition in math, not just proving things

[winged void]

Well me as well I find intuition more important than the prove self

[night onyx]

that being said, you have to be able to do proofs in math, that's kindof like the basic grammar of math (after high school level), it's like you can't be a good writer if you don't know how to read and write or speak

or like in programming, you can't make interesting programs if you don't know the basic things about how to structure a program

[winged void]

And when we say two cycle have the same structure does that mean that they are isomorphic an each other

[south patrol]

Cycles can't be isomorphic

[winged void]

True agree

[night onyx]

no, it means that they have the same number and length of cycles

so like (1 3 2) (4 6) and (3 5 4)(6 8)

they have the same cycle structure, each has 2 cycles, length 3 and length 2

[winged void]

Ah I see

Thx a lot

[random pasture]

How exactly would I start proving - on N^2?

[delicate orchid]

Do you want part c or part d

They’re both trivial if you just write out what E actually means

And use basic arithmetic properties like a+b = b+a and~~ -(a+b) = -a-b~~

[random pasture]

okay yeah i wasn't sure if I was overthinking it or if there was some reason - couldn't directly be applied as a function since it's on N

[delicate orchid]

Ok you don’t even need the second one actually.

[random pasture]

oh yeah it's just (a,b) = (0,0) + (a,b), right

[coral spindle]

No

You haven't proved yet that -a + b = c is equivalent to b = a + c

Prove from the definitions that -(a,b) + (a,b) = (0,0)

You have everything in front of you!

[winged void]

i guess i have proof

i proved it with contradiction

but im not so sure

if its correct

but im like almost 90% sure that its correct

this is my proof

[summer path]

<@&268886789983436800> this is in multiple channels

Oh

[void cosmos]

this is the zodiac speaking

lmfao

[winged void]

lmfao im learing

[void cosmos]

no i meant your handwriting

[winged void]

oh i see

well thats my best

hahaha

[void cosmos]

no worries king just joking

[winged void]

do not worry!(

(:

[random pasture]

The proof for part a was relatively straightforward, but I'm not sure how to approach part b, or what exactly it means

[coral spindle]

Part b could be equally phrased as "show Q^N/I is an infinite set."

The approach is to be creative. Just try.

[random pasture]

wait just for clarification what exactly does Q^N/I consist of

is it the quotient set?

[coral spindle]

It is the quotient ring.

Are you unfamiliar with quotients of rings?

Det that kinda spoils the problem a bit...

I think the question expects the reader to work that out

[rustic crown]

i hope x496 didn't see anything ><

[random pasture]

nah I was confused because I didn't know how the quotient would come from a set rather than an equivalence relation

[coral spindle]

Are you clear on how that works or no?

[random pasture]

not really, how does the mapping work?

[coral spindle]

The mapping?

I'm not sure what you mean.

Anyway, quotient rings work like so:

If we have an ideal J of a ring R, the quotient R/J consists of the cosets r + J. It is the set {r + J | r in R}.

You already know plenty of these. Z/nZ, for example.

I expect that whoever set these questions also explained this previously, but if you don't recall, there is an underlying equivalence relation here. Two elements are equivalent if they differ by something in the ideal.

[random pasture]

i see

[coral spindle]

Is this clear? I expect you will have seen this before, otherwise these questions are not really well-pitched.

[random pasture]

so there must be infinite elements because there are (a_n)_n, (b_n)_n in Q^N/I that have a_n=b_n for n>m for some m in N, and all the sequences in each element just share that subsequence?

[coral spindle]

I don't follow the logic.

Maybe you could simply find infinitely many sequences explicitly, showing that they are distinct in the quotient.

[rustic crown]

how would you find infinitely many things in finite time

[random pasture]

lmao

[coral spindle]

As wew says, "watch this"

[rustic crown]

[random pasture]

okay wait i think the structure of my answer is off

[vivid tiger]

Ah.

||[(1,0,1,0,...)], [(1,0,0,1,0,0,...)], etc. The difference of two of them is not eventually 0||

For some reason it took me a bit to parse what was happening

[random pasture]

yeah that's what I was trying to say

[vivid tiger]

I think my firing on all cylindersness is starting to fade right now

[random pasture]

but also like

[vivid tiger]

||c: clearly not invertible. but add (1,0...)||

interesting.

wait

this is just like a not epsilon version of equivalence of cauchy sequences

[random pasture]

[0,0,-1,1,0...] is in the Ideal isn't it?

oh wait nvm

[vivid tiger]

So?

My elements differed in infinitely many places

A better example:

[random pasture]

it's repeating i got confused mb

[vivid tiger]

(1,3,5,...)

and (2,4,6,...)

[coral spindle]

Please don't just give away the answer...

[vivid tiger]

I did spoiler tags!

[coral spindle]

Yeah, for one solution.

[vivid tiger]

If he asks for questions about it

[vivid tiger]

oh woops my bad

I guess my instinct is that if the problem requires some thought on my part, I don't automatically spoiler even when I should

[south patrol]

The infinitely many would be much more interesting if they said uncountable

Oh lol for c) I was too silly

Thought it was N^N for whatever reason because I couldn't work out why the sequence wouldn't be invertible in Q^N

Lmao

[dull ginkgo]

Is this a valid proof of CRT

If $I_i$ is an arbitrary collection of ideals indexed by $i \in \Omega$, then $x \equiv y \mod I_i$ for each $i$ if and only if $x \equiv y \mod \bigcap I_i$, as $x \equiv y \mod I$ is equivalent to $x - y \in I$, so the prior statements are equivalent by $x - y$ being in each of the ideals. Therefore, the kernel of the map $\phi : R \rightarrow \prod_{i \in \Omega}{\frac{R}{I_i}}, \phi(x)_i = x \mod I_i$ is simply the intersection of all $I_i$, as it's equivalent to the set of elements congruent to $0$ mod every one of the ideals.

Now, if we assert that $\Omega = \mathbb{N}n$, and that the $I_n$ are all pairwise coprime ideals, we can show that $\phi$ is onto. First we need to show that $I_n$ is coprime to $\bigcap{k = 1}^{n - 1}{I_k}$. This can be shown by induction from the base case that $I_1$ is coprime to $I_2$ and through the following: \

Two ideals, $A$, $B$ are coprime if their ideal-sum is $R$. This is equivalent to there existing some $a \in A, b \in B$ such that $a + b = 1$. Now if $A$ is coprime to $C$, and $B$ is coprime to $C$, then there exists some $a \in A, b \in B, c_a \in C, c_b \in C$ such that $a + c_a = b + c_b = 1$. Thus: \
$1 = c_a + a = c_a + a(c_b + b) = (c_a + ac_b) + ab$. But $c_a + ac_b$ is in $C$ due to $C$ being an ideal, and $ab$ is in the product ideal $AB$, contained within $A \cap B$. Thus $C$ and $A \cap B$ are coprime. \

If we have two elements of $R$, $x$ and $y$, and two coprime ideals of $R$, $A$ and $B$, then we can find an element $x$ such that $z \equiv x (\mod A)$ and $z \equiv y (\mod B)$. By coprimality we have some $a \in A$ such that $1 - a \in B$. Then we have $z = x + a(y-x)$, as $x - z = -a(y - x) \in A$ and $y - z = (1-a)(y-x) \in B$. \

Thus if we have a sequence $x_n \in R$, assuming $z_{n-1} \equiv x_i \mod I_i$ for $i \in {1 ... n - 1}$, we can find a $z_{n}$ such that $z_n \equiv x_n \mod I_n$ and $z_n \cong z_{n-1} mod \bigcap_{k = 1}^{n-1}{I_k}$ as $I_n$ is coprime to $\bigcap_{k = 1}^{n - 1}{I_k}$. Because $z_n \cong z_{n-1} \mod \bigcap_{k = 1}^{n-1}{I_k}$, $z_n \equiv x_i \mod I_i$ for each $i$. \

Now, for each $x \in \prod_{i = 1}^{n}{\frac{R}{I_i}}$, we have the indexed equivalence classes $x_i + I_i$, and by the prior lemma and a choice function on those classes, we can find a $z \in R$ such that $z \equiv x_i \mod I_i$. Thus $\phi$ is onto. \

Therefore, by first isomorphism theorem, $\mathrm{Im}(\phi) = \prod_{i = 1}^{n}{\frac{R}{I_i}} \cong \frac{R}{\mathrm{Ker}(\phi)} = \frac{R}{ \bigcap_{i = 1}^{n}{I_i}}$

good luck reading it

[cloud walrus]

Fairy farts

[crystal vale]

(Q+, •) is not cyclic, because if it is generated by p/q then let s is prime number which does not divide p then s not belong to (p/q), right?

[boreal inlet]

I mean, as far as I remember, (Q, +) is not even finitely generated

Let me ask this:

The definition of radical extension is also very similar

Where instead of the prime q_i, there is just some fixed natural number n

And any radical extension can be viewed as some n-radical extension

But how do I particularly find the q_i+1's for any general radical extension?

[crystal vale]

Yes

[cloud walrus]

Riku

[cobalt heath]

Maybe like

Q(2^1/4) is clearly a radical extension of Q, then

[boreal inlet]

Yes

[cobalt heath]

You can put Q(2^1/2) in between

So that (2^(1/4))^2 = 2^(1/2), which is the step between

Then you do it again to get to Q

[boreal inlet]

I mean like if Q is cyclic and is generated by p/q, how do you generate p/2q?

[boreal inlet]

Wait I mean that is cool and all but can we do this for any general one

[cobalt heath]

I mean, why not?

If a^n is in smaller field, you can factorize n and introduce intermediates appropriately.

[boreal inlet]

One thing we know is K looks something like

K = Q(a_1, a_2,.... a_r)

[cobalt heath]

What do you think would be an obstacle for introducing intermediate fields?

[boreal inlet]

The problem seems to be in the case if something like Q(a_1, a_2) occurs and you can't write it like Q(a_3) or something

[cobalt heath]

You can still write Q \subset Q(a_1) \subset Q(a_1, a_2) tho, and then do this stuff.

[rustic crown]

you make the filteration even finer

filter each successive K ⊆ K(a_i) into simpler radical extensions.

[crystal vale]

Here the set is a positive rational number under multiplication

[boreal inlet]

Oh

Wait

[boreal inlet]

Is (Q+, *) a subgroup of (Q - {0}, *)

[crystal vale]

Yes

[boreal inlet]

Then it should be cyclic

[crystal vale]

How?

[boreal inlet]

Every subgroup of a cyclic subgroup is cyclic right

Q is a field and hence the unjt group under multiplication is cyclic

[boreal inlet]

If this statement is true, your group is cyclic

[boreal inlet]

....i thought about it for some time but I might be missing how to do what you actually say here

The thing is I do think it's correct i just can't write it

[delicate orchid]

Q is not finite

[boreal inlet]

Oh wait

Wait.

[delicate orchid]

Assume you have a generator n/m with n containing prime factors p_1, … p_k - there will exist a prime q outside of these p_i

So how can you get to q/m

[crystal vale]

Yes

[boreal inlet]

.....lmao forget i said anything

My bad

[crystal vale]

How can we define a two sided prime ideal in a non-commutative ring without unity such that it will be consistent with commutative ring

[rustic crown]

if you had any field extension K ⊆ K(a) with a^n in K. factor n into primes and using this decomposition find a filtration of K ⊆ K(a) like absta suggested.

[boreal inlet]

Let's say n = pq for simplicity, where p and q are primes

Then a^n = (a^p)^q

[rustic crown]

yep

[boreal inlet]

Oh so uh

k \inclusion K(a^p) \inclusion K(a)

Or something like that?

[rustic crown]

yeep

[boreal inlet]

Wait bruhhh

Doesn't that

mean the question is trivial

[rustic crown]

yep

[boreal inlet]

💀

I hate myself sometimes

[rustic crown]

.<

https://tenor.com/view/mala-mishra-jha-pat-head-cute-sparkle-penguin-gif-16770818

[boreal inlet]

Ok so just to recap

We have this n-radical extension for now

[rustic crown]

(what's the n?)

[boreal inlet]

So for each step E_i+1 = E_i(a_i+1)

[boreal inlet]

It's just a common natural number

[rustic crown]

oh oki

[boreal inlet]

For which this happens, where a_(i+1)^n \in E_i

We now just

factorize n

[rustic crown]

yep

[boreal inlet]

And basically expand the chain

lmao

[rustic crown]

if p divides n, then you have a chain K ⊆ K(a^(n/p)) ⊆ K(a) and the first one is p-radical

inductively filter the second :p

[boreal inlet]

Ohh right right

I have to get the p-radical

Yeah so basically we always get primes for each step

Genius

Thank you guys

[rustic crown]

[delicate orchid]

[neuron activation] p-radical subgroups…

[boreal inlet]

Wait, subgroups?

[delicate orchid]

Yur

[boreal inlet]

How do you define this for subgroups again

[delicate orchid]

“Again” you’ve never seen this before

P is p-radical in G if the largest normal p-subgroup of N_G(P)/P is the trivial group

[boreal inlet]

....me bad at english

[delicate orchid]

It’s unrelated I’m just connecting words

[boreal inlet]

N_G(P) is the normalizer of P right

[rustic crown]

gwoup theowy

[delicate orchid]

Yur. It’s like saying that P is the largest normal p-subgroup in its normaliser

[boreal inlet]

Actually that's kinda rad tbh

Where do these appear naturally

[delicate orchid]

Radical almost

[rustic crown]

is P a p-subgroup?

[delicate orchid]

Yus

[delicate orchid]

They’re useful in getting information about G as a whole from just it’s Sylow subgroups

At least that’s why I care about them

[boreal inlet]

Damn

Good for determining finite groups

Especially the obscure ones

[delicate orchid]

You usually combine it with the condition that C_S(P) = Z(P) where S is a sylow subgroup of G containing P. This gives us the “local” behaviour we want, and being radical gives us the “global” behaviour that we want

I won’t go into more detail lol

[boreal inlet]

lol

[south patrol]

All groups are finite

[old zenith]

so apparently the free group of 2 elements is a subgroup of SU(2) right? how do you construct the subgroup

[delicate orchid]

Found this paper https://projecteuclid.org/journalArticle/Download?urlId=10.1307%2Fmmj%2F1028999969&isResultClick=False

Seems like you pick two real numbers n,m with a product greater than 4 and just put them in
(1 m). (1 0)
(0 1) and (n 1)

These aren’t unitary though… annoying!

Ok I’ve found a source but the url is literally too long to post in discord

https://www.sciencedirect.com/science/article/abs/pii/S0021869321001757

[cobalt heath]

Factorization is not easy tho

It could be difficult to think in terms of how things factor

[boreal inlet]

In specific cases yes

But we have an arbitrary integer n, and we just assume factorisation always exists

Because N is an unique factorisation monoid https://cdn.discordapp.com/emojis/722022152328445985.png?quality=lossless&size=48

[cobalt heath]

Does not mean it is easy to think of!

That said, I am struggling with the most basic rep theory..

Clearly for A = k[x1, .., xn], A/I is indecomposable if I contains all monomials of degree >= N

[boreal inlet]

That is true....

[cobalt heath]

Indeed, math is hard

[cobalt heath]

That said, I should have solved this already

Wonder if this has something to do with irrelevant ideal

[winged void]

@night onyx can you check the proof that i provided when you have time please (:

[crystal vale]

How the definition if aRb subset of I then belongs to I or b belongs to I implies that I is prime ideal is equivalent to prime ideal definition in commutative ring ?

[delicate orchid]

well in a commutative ring, aRb = abR and then it’s just the definition of a prime ideal

recall that (a) = aR

[crystal vale]

Yes but reverse ?

If R is commutative ring then ab belongs to I how implies that aRb be a subset of I ?

[delicate orchid]

ab is in I, so (ab) is a subset of I

[crystal vale]

Yes

[delicate orchid]

So (ab) = abR = aRb is a subset of I

[crystal vale]

Yes

If I want to show that the definition of prime ideal in a commutative ring implies the definition of non-commutative ring then how?

[delicate orchid]

I’ve just done that?

[crystal vale]

If I have ab in I, how can I show that aRb subset of I in non-commutative ring

[boreal inlet]

😔

[crystal vale]

Commutative ring?

[delicate orchid]

I’ve shown that, for a commutative ring, both definitions are equivalent

[crystal vale]

Yes

But for non-commutative ring?

[delicate orchid]

Is that not what you just asked?