groups-rings-fields

[bleak abyss]

Aka it has an inverse if x is coprime to 30

[keen sparrow]

@median pawn I would use binomial expansion to show that it is nilpotent but I’m not sure if there’s a simpler way

[chilly ocean]

yes

[median pawn]

That would be painful, considering that f(X) is an infinite sum

Moreover binomial expansion is only for finitely many terms?

[keen sparrow]

What I was thinking is $f(X) = \sum_{k = 1}^{n} a_{k}g_{k}$, then apply "n-nomial??" expansion because what matter is the coefficients

[cloud walrus]

Mike Desgrottes

[rustic crown]

i wasn't able to find much on the internet about this 😦

[keen sparrow]

aren't lie algebra vector spaces?

[rustic crown]

vector spaces with some additional structure

now the morphisms need to preserve this new structure which reduces the number of such

which later restricts what kernels can be, and so reduces the things by which you can quotient

[next obsidian]

Here’s what’s going on, Z_n is Noetherian. The nilradical of any ring is well… an ideal, so that the nilradical N is finitely generated, say by a1 through an. Now since there’s finitely many elements here, some M is large enough so that ai^M = 0 since each ai is nilpotent. Now, we see that for any f in N, that f^(Mn) = 0, since f = Sum biai, then you apply the binomial expansion and see every term has some ai being raised to a power >= M

There’s a much much much much much simpler answer in the specific case of Z_n though, there’s only finitely many numbers so you’re taking the max of a finite set

As for raising a power series F to the n, you do indeed use the binomial theorem

You want to look at the coefficients of F^n, each coefficient will be a finite sum of powers of things using a multinominal coefficient, the multinomial coefficient part is irrelevant though.

I can just say that if the coefficients are in Z_n, and are nilpotent then I think you f^(n^2) will always be 0, but I didn’t spend much time trying to justify it so it might be wrong, kind of doubt it tho

[dreamy fiber]

If K/F is a splitting field and separable extension (the requirement's probably not necessary), but let's say f(x) \in F[x] is a minimal polynomial for A \in K, and B is also a root of f(x), is f(x) also the minimal polynomial for B?

nvm if there's a smaller minimal polynomial it means f(x) is reducible

[delicate bloom]

Suppose $f(x)=\sum_{k \ge 0} a_kx^k$ with each coefficient nilpotent in $\bZ_n$. Then that means each $a_k$ is a multiple of $\mathrm{rad}(n)$, the square free radical of $n$. We'll write this as $a_k=\mathrm{rad}(n)*b_k$. This is nilpotent of order $N$, and is our upper bound for the order of nilpotency of each $a_k$, in other words $a_k^N=0$ for all $k$.

Now let's look at $$f(x)^N = \sum_{k\ge 0} \left(\sum_{i_1+\cdots+i_N=k} a_{i_1}\cdots a_{i_N}\right) x^k$$

Its coefficients are the convolution of the series with itself $N$ times. It is the sum of all products with indices satisfying $i_1+\cdots+i_N=k$, which is just a reflection of the simple fact that these terms from each of the $N$ series contributes to the monomial $x^{i_1}\cdots x^{i_N} = x^{i_1+\cdots+i_N}=x^k$ term.

$$= \sum_{k\ge 0} \left(\sum_{i_1+\cdots+i_N=k} (\mathrm{rad}(n)*b_{i_1})\cdots (\mathrm{rad}(n)*b_{i_N})\right) x^k$$

$$= \sum_{k\ge 0} \left(\sum_{i_1+\cdots+i_N=k} (\mathrm{rad}(n))^N(b_{i_1}\cdots b_{i_N})\right) x^k$$

$$= \sum_{k\ge 0}0x^k = 0$$

[cloud walrus]

Merosity

[median pawn]

I didn't understand what you did

Since there are only finitely many terms, I understand the existence of N anyway

[delicate bloom]

if $n=\prod_i p_i^{a_i}$ then I'm saying $N=\max_i (a_i)$

[cloud walrus]

Merosity

[delicate bloom]

$\mathrm{rad}(n) := \prod_i p_i$

[cloud walrus]

Merosity

[delicate bloom]

trying to fill in some definitions I left out I don't know if that helps

I'm not psychic through, tell me what else you're having trouble with

[median pawn]

thanks, it was just the definitions, let me go through the argument again now

[warm holly]

In general to count the number of morphisms between 2 finite groups G and H (G-> H) is it legit to count the elements of H which's order divides the order of G?

[hidden haven]

Only works for cyclic finite groups I think

[delicate orchid]

I can tell you that if both groups are finitely generated then all morphims are just maps between generators so if you count them that’ll be the number of morphisms - I think

Which reduces it to a combinatorics problem that I can’t do without paper

[hidden haven]

But not all generators can map to any generator

The generator for Z/2Z can't map to the generator of Z

Hence the order divisibility condition when you have cyclic groups

[delicate orchid]

Ah very true

[cloud walrus]

Hausdorff

[median pawn]

I'm thinking of Gaussian elimination, and maybe x' would be a product of elementary matrices?

[hidden haven]

This is just axiom of choice

[median pawn]

Wait what

[hidden haven]

Think of the matrices as linear maps

[wooden ember]

lol nice

[hidden haven]

For all set maps f, there is a g such that f = fgf is an equivalent way of stating AC categorically

[cloud walrus]

Hausdorff

[wooden ember]

this isnt just a set map though it's a linear map

[hidden haven]

You just gotta make sure that the g here is linear

[wooden ember]

so surely axiom of choice isnt enough

[hidden haven]

ye

You gotta extend a map on the basis linearly

[median pawn]

That's super interesting

[wooden ember]

ah fair enough

[median pawn]

Wait, where do I start?

[hidden haven]

x is a map K^n to K^n

Take a basis of the codomain

define x' on that basis such that xx'x = x

extend this to everything and you should be done

[wooden ember]

that's pretty neat

will keep it in mind in case something like this pops up on my lin alg exam lmao

[median pawn]

easier said than done

how do you know that we can define x' in such a way

[hidden haven]

Let b be a basis element

If there is some a such that x(a) = b

Then do something

If there is no such a

Then do anything

Ah you gotta be careful in what basis you choose too

Of the codomain

[median pawn]

then define x'(a) = b^{-1}?

[hidden haven]

When you try to prove xx'x = x at the end, you should be able to see what conditions you need

[hidden haven]

x' goes in the opposite direction from x

x is from V to W, x' should be W to V

I mean here V=W

But just think of them as distinct copies

[hidden haven]

This is just from the fact that we can compose them both ways

[median pawn]

oh the multiplication in this ring is composition, right

[hidden haven]

yep

[median pawn]

xx'x(a) = x(a) = b, xx'(b) = b, so define x'(b) = a

[hidden haven]

also b^-1 doesn't make sense here

yep

[median pawn]

why not? K is a field

[hidden haven]

You define x'(b) to be any preimage of b

[hidden haven]

b is a vector though

in K^n

[median pawn]

elementwise reciprocal was what i had in mind tbh

[hidden haven]

Are you taking coordinatewise inverse

[median pawn]

but nvm that's not needed

[hidden haven]

Some entries may be 0

[median pawn]

true

ok it is trivial that K^n has dimension n over K right

and so we are choosing a basis of size n

[cloud walrus]

Hausdorff

[median pawn]

I've gotten so far @hidden haven

[cloud walrus]

Hausdorff

[hidden haven]

Right so one thing to keep in mind is that not any basis would work, so you will have to come back later and turn this arbitrary one into a special one

[median pawn]

for all i basically

[hidden haven]

yep

and what if there is no such f_i

[median pawn]

exactly my question... can we just define x'(e_i) = anything under the sun

[hidden haven]

yep

[median pawn]

why so?

[hidden haven]

This gives you a function x' on the basis

[median pawn]

xx'x = x? why would this hold

[hidden haven]

because on that f_i

uh wait

[median pawn]

xx'x(e_i) = x(e_i) but there is no f_i such that x(f_i) = e_i

[hidden haven]

yeah because there is nothing to check lul

Like what we want to say here

Is that given any element g_i in the preimage of e_i

xx'x and x agree on g_i

And they do do

because e_i has no preimage

[hidden haven]

So yeah this is the claim

[median pawn]

hmmmm

[hidden haven]

And now that you have shown that this claim is possible to satisfy for any basis {e_i}

You have to pick a basis

such that applying this construction to that

gives you x' such that xx'x = x everywhere

So try to prove this equation everywhere for an arbitrary basis

And see where you get stuck

[median pawn]

ok go slow

[hidden haven]

And then just impose that condition on the basis

[median pawn]

right?

[cloud walrus]

Hausdorff

[hidden haven]

Ye

[median pawn]

but we need xx'x and x to agree on whole of K^n

[hidden haven]

yep

[median pawn]

how does that follow

[hidden haven]

When can you say that 2 maps agree on the whole space given that they agree on a subset S

Like generally in any algebraic structure

A trivial case is when S is the whole space

But is there anything more interesting

[median pawn]

when S generates the space

in the sense that, span(S)= space

[hidden haven]

yep

So that's what you want here

You might want to first consider the simpler case where x is surjective

[median pawn]

hmm alright

if x is surjective then we always have the g_i

[hidden haven]

ye but do they span

Oh ye btw this proof works in any number of dimensions and even if domain and codomain are different dimensions

So there probably is an easier proof

probably doable just algebraically

[cloud walrus]

Hausdorff

[median pawn]

indeed they do span

[hidden haven]

Right, this is using the fact that domain and codomain have the same finite dimension so a surjection is an isomorphism so a basis maps to a basis

Too easy

Shouldn't have given this case lmao

Because I don't see how to generalise this argument to when x is not surjective

[median pawn]

oops

if x is not surjective have we hit a dead end

[hidden haven]

With this strategy yes

[hidden haven]

but

I had a different thing in mind for when x is surjective

[median pawn]

ahh i see

btw the hint given for this problem was gaussian elimination lol

[hidden haven]

lol

I am TAing a data science course linear algebra and its applications this semester and I gotta learn about all those things and how to implement them in python

[median pawn]

damn that's painful i would never

[hidden haven]

But yeah since they are assuming square matrices of finite size it would make sense that there is some purely symbolic thing you could do

which I do not want to think about

[hidden haven]

I made a bad decision

[hidden haven]

Also how come this is a thing that you have just proved

When you have already done so much analysis

Did you not know what matrix multiplication was about this whole time

[chilly ocean]

[lethal dune]

[iron vessel]

Do you think Dummit and Foote is good for Galois theory?

Or perhaps is there a better book

[hidden haven]

I have never seen it recommended for Galois theory

[chilly ocean]

didn't you literally make the field??

[lethal dune]

don't

[hidden haven]

Probably for recommending more modern texts to others

😌

Try Milne's notes or Lang's book

Milne is good

[median pawn]

lmaooo so

i had an algebra exam today on which they asked both these questions

obvs i had proved it earlier but they made me do it again today

[hidden haven]

Ok nice 😌

[iron vessel]

My professor also was recommending Lang and Stewart. He also mentioned Duaady

[hidden haven]

Never heard of Stewart and Duaady

[median pawn]

i read that as daddy

[delicate orchid]

Actually mapping endomorphisms to matrices is an interesting problem imo

And requires some thinking

[hidden haven]

Ye I am not saying it's not

[delicate orchid]

I’d say you could do a fuck ton without actually proving it and just using the isomorphism implicitly

[hidden haven]

I was curious because Hausdorff asks grad level analysis problems

And I would assume that someone who does that would know this

[delicate orchid]

Yeah and I’m saying they might’ve like, intuitively known it, but never actually proven the connection

Especially if they’re an analysis enjoyer

[hidden haven]

perhaps

[delicate orchid]

Anyway that’s all I wanted to say

[steep basalt]

Hey, so I'm just trying to make sense of a solution to a question, say i have A_5 x C_2 acting on a dodecahedron. The question asked me to find the order of three subgroups of A_5 x C_2 that fix 1. a vertex 2. an edge 3. a face

The solution was basically "a dodecahedron has 20 vertices, 30 edges and 12 faces so the stabilisers have order 20, 30 and 12 respectively. A_5 x C_2 has order 60x2 = 120 so the size of the respective subsets will be 120/20 = 6, 120/30 = 4 and 120/12 = 10" I'm comfortable with the second sentence of the solution but i'm not comfortable with the first sentence, why does the order of the stabiliser for a vertex = the # of vertices???

[rustic crown]

their orbits have sizes 20, 30 and 12

and orbit stabilizer gives that stabilizers have size 6, 4, 10

[west violet]

hows judson compared to pinter for algebra(ideally for ppl with no algebra experience)?

[chilly radish]

I need to prove that for every odd prime p, there is an element a such that x^2-a is irreducible. Now, of course there are a lot of ways to prove this, but the hint specificlaly says to look at the squaring homomorphism on the group of units. I don't.understand what the purpose of this is. Of course I want to show that the kernel is nontrivial (probably by some order considerations)
but I don't see how to do that

[rustic crown]

squaring is a group hom from F_p* --> F_p* with image being precisely the non-zero squares mod p. we need to show that the map is not surjective, which is achieved by saying kernel of this is non-trivial.
kernel is the set of elements satisfying x^2 = 1, so just 1 and -1. (which are different as p is an odd prime)

but yea you can say this directly as well, since square of b and -b is the same thing, there can be at most (p-1)/2 non-zero squares.

[chilly radish]

this is actually what I did

oh wow

i've been working with abelian groups and modules so much recently that I forgot the kernel is 1

lmao

thanks

[delicate orchid]

"group mode" as like to call it

[chilly radish]

I just said the map is 2-to-1 so can't be surjective, which is what you said, but I thought that was too simple

[molten silo]

Anyone know how to do 3b(ii)

[hidden haven]

What could be the orbit of A?

A is normal iff the orbit of A is {A}

[molten silo]

i dont understamd

[hidden haven]

So you just have to eliminate all other options

[molten silo]

hmm

Not following

[hidden haven]

Is this fine?

This is the definition of normal

[molten silo]

yes

[hidden haven]

Normal subgroup is one that is fixed under the action of conjugation

Ok cool

[molten silo]

If the stabiliser(A) = G

[hidden haven]

Yeah

[molten silo]

i got that far

[hidden haven]

Equivalently, orb(A) = {A}

So what are the subgroups that could lie in the orbit of A?

Use part i

[molten silo]

Subgroups of the same cardinatlity

as A

Sorry K

[hidden haven]

Yep, so you only have 2 options

[hidden haven]

A in this case

K was only there in the previous part

[molten silo]

But this doesnt show that Orb(A) contains a single element

[hidden haven]

It doesn't

But it limits the possible values of Orb(A)

[molten silo]

But how does this show that A and B are normal

[hidden haven]

It doesn't directly

You have Orb(A) = 1 of 2 things

[molten silo]

yes

[hidden haven]

You have to eliminate the other thing using orbit-stabilizer

[molten silo]

ahhh

Because G has cardinality odd

So orb(A) can only contain 1 element A or B not both

I think

[hidden haven]

Yep

[molten silo]

thanks dude

[chilly ocean]

Am I plugging in correctly?

This is to show homomorphism btw I know I have to show iso

[tawny pine]

yes, ideally clean it a bit to show phi(ab)=phi(a)phi(b) in 1 line

[chilly ocean]

how would i do that?

i dont think im finished actually

they dont look the same

[tawny pine]

write a chain of equalities phi(ab)=...=...=phi(a)phi(b)

[chilly ocean]

idk how manipulate this

to make is look like phi(a)phi(b)

Simplify the expression in the last line of the first pic you sent

You will make the connection

Wait uh

Yeah

i cant

Wym

i can simpify further??

there's a g and g inverse right next to each other

oh shit

im blind

What happens when inverses of each other kiss 😘

All good 😩

got it

do you guys know how i would prove its isomorhpic?

thats simply proving its a homomorphism right?

And bijective

[tawny pine]

u also need to show its bijective, either by showing its surjective & injective or by providing an inverse function

[chilly ocean]

yeah if the homomorphism is bijective its an isomorphism

You can have homomorphisms that transform a group into one of its subgroups, so not every homomorphism on a group is iso :P

Yessir

my prof never taught us how to prove its iso

he just said if its bijective then it is

lol

should i start with injectivity?

lets x,y be in G?

Sure

f(x) = f(y) at the end right?

then its injective

[tawny pine]

in this case ||providing an inverse|| is somewhat faster

[chilly ocean]

whats the law again

if theres an inverse they bijective right?

[tawny pine]

yes

[chilly ocean]

A two sided inverse yee

hmmm okay

let see here

[tawny pine]

but showing injective+surjective is also valid, just maybe a bit slower

[chilly ocean]

how do i do the inverse method

its been a while

i just look at phi right?

[tawny pine]

explicitly write an inverse & show its actually an inverse

[chilly ocean]

for phi

okay

would it be

(g^-1) ( a^-1 )( g)

[tawny pine]

no

but close

[chilly ocean]

the goal is

to have my function and the original function to equal identity right?

if by "and" you mean composed together then yes

[tawny pine]

fog & gof must be identities for g to be an inverse of f

ie f(g(x))=x & g(f(x))=x

[chilly ocean]

its

yeah I should have clarified, when you said "identity" it would only be true if you meant "act as the identity function"

(g)(a^-1)(g^-1)

not meaning "a function that just spits out the identity as a constant"

ah

which I think you may be confusing

[tawny pine]

no, u were closer before

ok think of it like this

[chilly ocean]

how do ieven do this lol

[tawny pine]

take any element a of G

[chilly ocean]

okay

[tawny pine]

apply phi

we get gag^-1

[chilly ocean]

yes]'

[tawny pine]

ie left multiply by g, right multiply by g^-1

what can we do to gag^-1 to get a?

[chilly ocean]

lefft multiplay by g^-1

right multipl by g

[tawny pine]

yes so this is what our inverse should do

[chilly ocean]

so

thats my inverse??

[tawny pine]

ie $\vp\inv(a)=g\inv ag$

[cloud walrus]

RokettoJanpu

[tawny pine]

recall that the inverse acts as an undo button

if i apply phi on a then apply phi^-1 on the result i should get back a

thats the thought process we just went thru

[chilly ocean]

ohhh okayyyy so

[tawny pine]

now just verify this function actually is the inverse

[chilly ocean]

whaever value we get from phi

you run THAT thru phi inv

and it should get me a

[tawny pine]

yes

[chilly ocean]

gotcha

how do u verify its pretty clear

that it is

[tawny pine]

fog & gof must be identities for g to be an inverse of f
ie f(g(x))=x & g(f(x))=x

[chilly ocean]

I like that lmao

[tawny pine]

so we must show $\vp\inv(\vp(a))=a$ and $\vp(\vp\inv(a))=a$

[cloud walrus]

RokettoJanpu

[chilly ocean]

oh so i plug thm into each other

both ways

it worked

beautiful

this is the first thing in abstract algebra that made sense to me

I think the whole exercise (a,b,c,d) may be interesting to you

although it might be hard to see it from a birds eye view coming at it for the first time

imma be lost on those portions for a while haha

yeah

[tawny pine]

mapping a to gag^-1 is called conjugating by g, it comes up again later on

[chilly ocean]

what resources did you use to learn

abstract alg

its honestly a tough subject for me not going to lie

big tuition money bags

[tawny pine]

i didnt use any particular book

[chilly ocean]

my uni is ass

i cant focus on this class and operating systems at the same time

both require unlimited hours of work

[tawny pine]

book recs can vary a bit depending on if u need more/less exercise

[chilly ocean]

If algebra as a subject interests you, you will end up seeing these concepts again and again in different contexts and your intuition will continue to deepen

honestly whether you like it or not if you're studying math or something math related LOL

[tawny pine]

the usual recs on this server are in #book-recommendations or the list of stickers

wait nvm the only aa sticker we got is d&f

[chilly ocean]

i appreciate both of your help

roketto helped me lots when i took introduction to proofs too haha

[tawny pine]

ah then it helps to review proof stuff for these problems

[desert dome]

If K/E is a finite algebraic extension and is purely inseparable, how to show that [K:E]_sep = 1?

[gritty sparrow]

What definition are you using? If you are defining it as the degree of the separable closure of E in K, then since that is just E, you get the result from that. If you are defining it as the number of homomorphisms from K to the algebraic closure of E fixing E, notice that every element of K satisfies x^q-b as its minimal polynomial in E and this has a unique root in the algebraic closure, so if a homomorphism exists, it must be unique. But a homomorphism obviously must exist as we can just take the algebraic closure of K and that will be the algebraic closure of E

[desert dome]

thank you! We are using the number of homomorphisms.

[gritty sparrow]

Np

[barren sierra]

Stuck on showing that 2k + 1 | r

or is there a better way of showing the orders are equal

[thorn delta]

start with a = b^2. || raise both sides to the power of r ||

[barren sierra]

oh duh

thanks

[thorn delta]

npnp

[desert dome]

I am working on (2). I can show that for each a in K/k, the minimal polynomial of a over k is x^{p^n} - a^{p^n}. But I am not sure how to use that to show [K:k] = p^n.
I want to use if L/F/E, then [L:E] = [L:F][F:E]. If I write K/k as a tower of simple extension, then [k(a)/k] = p^n which means that for example, [k(a)(b)/k(a)] = 1. If so, it basically means that b in k(a). But I don't think that's true. Am I wrong?

[gritty sparrow]

The particular n we get for each element of K we choose can differ, so in that sense when you take some random a in K the extension k(a) need not be the whole field K. So there is still some room to choose b outside of this.

So in some sense the n in the second part is not necessarily going to be the n in the first part (and there is no single n in the first part anyway)

[desert dome]

i see. so should I aim to find an alpha such that k(alpha) = K?

[gritty sparrow]

No, such an alpha won’t necessarily exist. It seems easier to find a finite tower of extensions each of which has a power of p as its degree over the previous extension

[desert dome]

ok. I'll try to work on that

[desert dome]

btw, sorry I'm a little confused on this too. why x^q - b has a unique root in the algebraic closure?

[gritty sparrow]

let a be a root in the algebraic closure, so we have that a^q=b. (x-a)^q = x^q-a^q = x^q-b. So that means a is the only root of x^q-b

[desert dome]

gotcha. we can rewrite this way because characteristic = q. thank you!

[gritty sparrow]

Sorry i should have clarified, characteristic = p and q is p^m for some m (the rewriting in the case of p^m works in exactly the same way as the rewriting for just p)

[desert dome]

i see. thanks!

[gritty sparrow]

Np

[bitter spear]

This is a research task for us. I already searched but I couldn't find any answer on the internet. I don't know how to prove these statements because our professor haven't discussed it yet.

[chilly ocean]

hard to help you if you don't tell us what you do know

[chilly ocean]

why do they say "A is nonempty" it still true if A is empty

[bleak abyss]

Probably didn't want to think too hard about whether the empty map is a map or smth lol

[desert dome]

Suppose the characteristic of k is p. how to prove the characteristic of k(alpha), an algebraic extension, is also p? I know it's true but I don't know how to say formally
Like the identity is not changed, so it should remain the same?

[tribal moss]

The characteristic is all about which expressions of the form 1+1+1+...+1 evaluate to 0. In a field extension (algebraic or not), these additions all happen purely in the smaller field, so whether they are 0 or not cannot be influenced by what happens outside that field.

[desert dome]

I see. Thank you!

[small karma]

Can someone help me as to how is this part done? I know i can find it using 8 mod 65 then 8^2mod65 8^3mod65 and so on

But that methods seems wrong assuming it will make this question like 30 mins long

So what other method do i use

[rustic crown]

,rotate

[rustic crown]

ig you can quickly say that a^12 = 1 mod 65

[terse crystal]

It’s already told you what to do in detail, why don’t you just follow it…

[rustic crown]

and since a^16 = 1 mod 65, you get a^4 = 1 mod 65

[delicate orchid]

I think you really do just have to grind it out or use Euclid’s algorithm if you want specific orders of each element

[small karma]

Yeah i guess so

[terse crystal]

No elements of order 8 and not all elements are of order smaller than or equal to 2 so those 2 possible situations left… If it’s the latter case then there should be at least 7 elements of order 2 so the former one

[small karma]

I was hoping there will be a short way

Bcz these kinda questions are asked in my university exam previously

[delicate orchid]

Well as we’ve said you can eliminate a lot of the possibilities

[small karma]

So im guessing ill get one too

Yeah ill try it out rn

[delicate orchid]

So you only need to check a^2 and a^4 in this case

[small karma]

Yes

[delicate orchid]

I’m not seeing where this comes from though

[rustic crown]

oh 12 = lcm(phi(13), phi(5))

a^12 = 1 mod 13 and 5, so 1 mod 65

[small karma]

Quick question are u guys allowed to use calculater for this?

[delicate orchid]

Oh right it’s number theory

[small karma]

For a^2 or a^4

[rustic crown]

there are other ways to this this btw, but i'm guessing your book doesn't want that.
U(65) is isomorphic to U(13) x U(5) which is Z/3Z ⊕ Z/4Z ⊕ Z/4Z
now G is a subgroup of this of order 16

because U(p) is isomorphic to Z/(p-1)Z and we can then use CRT to break it further

[small karma]

Hmm I'll try it out and see

[small karma]

Yeah i cant use different method. My teacher is picky.

So ill have to stick with the book

[delicate orchid]

That’s not the way to teach abstract algebra at all

[small karma]

Yeah what can i do. Anything other then her methods is either googled or copied

[terse crystal]

Maybe her method can be shorten a little bit, x^4+1=0 mod 5 or mod 13 doesn’t have solution so the first and the second are ruled out, 8^2=-1 so an element of order 4, the fifth is ruled out, then you just calculate the square of those elements to see how many elements are of order 2, so 15 times of multiplication.

(If 65=5*13 divides a^16-1=(a^4-1)(a^4+1)(a^8+1) then 65 must divide (a^4-1) from my first two rows)

[lethal dune]

non commutative rings without unity are ugly

[wooden ember]

That is why we say they don’t exist

[lethal dune]

[delicate orchid]

I do quite like how they're a group and a semigroup though

[lethal dune]

also delete non commutative rings

[delicate orchid]

it's neater than a group and a magma

[rustic crown]

non-commutative rings are nice

but 1 is

[lethal dune]

[delicate orchid]

we must have unity!

for without unity there can be no (multiplicative) order!

[lethal dune]

Hence forth all rings are commutative and have unit

[rustic crown]

i like the ring of integral quaternions

[wooden ember]

Commutative rings are as bland as abelian groups

Change my mind

[delicate orchid]

ZQ_8?

[lethal dune]

ok include only non commutative ones which has a unit

[rustic crown]

nah, the one i know is
{a +bi+cj+dk | a, b, c, d all in Z or all in Z+1/2}

[lethal dune]

[delicate orchid]

Z+1/2 wtf

there's probs a good reason for including that and now I want to know what it is

[rustic crown]

i think it's the same situation as in Z[sqrt(-3)] where you add the extra element (1+sqrt(-3))/2 to make it nicer

[wooden ember]

Was about to say the same

[rustic crown]

like 1/2(1+i+j+k) is a very nice invertible element

[delicate orchid]

what if you take i/2*j/2 though surely this is k/4 which isn't in the set

[rustic crown]

and you can use the structure of this ring to prove 4 square theorem

it's pretty much a euclidean domain but not commutative

[next obsidian]

Literally no

[small karma]

so i gotta decompose this right

?

[next obsidian]

Use the Chinese remainder theorem

[wooden ember]

I don’t care what you say you can’t change my mind

[next obsidian]

And just write every number as products of prime powers

Pick up a commutative algebra book

Hurb

[wooden ember]

Predictable

[small karma]

[rustic crown]

schemes are scary

[small karma]

me?

[next obsidian]

Is ur pfp wearing a Dino onesie? Also yes

[small karma]

yes

and yes

was just conforming

[next obsidian]

Confirming?

[next obsidian]

Scheeeeeemeeeeeee

[small karma]

yeah

wait

can i take the lcm of all the orders overall

like in edp

since its

and then use the fundamental abelian theorem

and make paritions easy

[next obsidian]

You don’t want to do it that way

[small karma]

why?

[next obsidian]

You can only break things up when they’re coprime and like already in “blocks”

Like if you had say Z_10 x Z_5

The lcm is 10

[small karma]

yes

[next obsidian]

But you should end up with Z_2 x (Z_5)^2

Unless I misunderstood what you meant

[small karma]

oh z5 x z5 x z2

[next obsidian]

Yeah

[small karma]

yeah correct

so for z60 x z30 x z15 x z5

i get

[next obsidian]

Just factor every number there as products of prime powers

And you’re done

You can then regroup them to make it look nicer if you like

Like here you will have four Z_5’s

So you can write (Z_5)^4

To make it look neater

[small karma]

(z5)^4 x (z3)^3 x (z2)^3

[rustic crown]

okie maybe a better way to write would be Z[u, i, j, k] = {au + bi + cj + dk | a, b, c, d in Z} where u = (1+i+j+k)/2.

also so i/2 isn't there because we want either all to be integers or all to be 1/2 + an integer in the old definition

[next obsidian]

So the last one is off

From Z_60 you get a Z_4

[delicate orchid]

ok I get you now

[next obsidian]

Z_4 ≠ (Z_2)^2

[small karma]

yeah true they are not relative prime 2 and 2

[next obsidian]

Exactly

So you should end up with Z_4 x Z_2

[delicate orchid]

chmonkey really has the not equals to symbol on their keyboard gosh darn

[next obsidian]

Nah

I’m on mobile

So I just longpress my = and it pops up

[delicate orchid]

wait what phone is that

[next obsidian]

iPhone

[delicate orchid]

huh doesn't work on mine

[next obsidian]

Tf?

[delicate orchid]

how... peculiar

[next obsidian]

What phone? I had it even on iPhone 7 years ago

Can’t you type accent marks like that too? Like ñ

[rustic crown]

is there a way to type unicode characters through keyboard without copy pasting stuff

[delicate orchid]

I can do accents and stuff

[rustic crown]

(on a laptop)

[next obsidian]

I think there’s a shortcut

[delicate orchid]

you got a numpad?

[next obsidian]

Like alt+ some numbers

[delicate orchid]

^

alt+numbers gives you uhhh some unicode and control+alt gives you the rest

╚ is alt+456 for example

[next obsidian]

Is that only if you input it via numpad?

[delicate orchid]

yup

[next obsidian]

So probably won’t work for laptop

Since they usually don’t have a numpad

[rustic crown]

⊕

OHHH

for gnome it's ctrl+shift+u then the code

[delicate orchid]

ah

[next obsidian]

Wtf is gnome

[delicate orchid]

it's like a little man with a big hat

[rustic crown]

(the Desktop Env i'm using >.<)

[next obsidian]

Wtf is DE

Wat

[delicate orchid]

DEez

[next obsidian]

Is that an OS

[rustic crown]

nah

[delicate orchid]

mfs really can't just install windows and be happy

[rustic crown]

if you just install linux you'll be thrown into a black ttyl screen

you need to install a desktop thing on top of it

but most of it is automated by popular distributions

[next obsidian]

I would assume the way to input Unicode is based on the Linux OS you use

Not the DE then

D:

[rustic crown]

the DE manages the keyboard shortcuts, so idk

[next obsidian]

Also Linux

Really?5

That’s whack

[rustic crown]

well it gives you the required control over everything if you think about it like that

okie now the question is how do i memorize unicode numbers

maybe there are other ways, i'll look

[delicate orchid]

ok I've been thinking

if you have an integral domain D with some homomorphism D -> Z then there's a homomorphism from Frac(D) to Q right?

[rustic crown]

i don't think so

[next obsidian]

No

[rustic crown]

Z[x] --> Z sending x to 0

[next obsidian]

It has to be injective

[delicate orchid]

ohhhh yes of course

[hidden haven]

laptops without numpads?

Aren't those just the tiny laptops

[delicate orchid]

or you could end up with 0s on the dominator in the homomorphism to Q right?

[next obsidian]

That’s a way to think of it, yeah

A different way to think of it is

You can obtain the map on the integral domains by restricting the map on fraction fields

And maps of fields are injective

[delicate orchid]

[next obsidian]

Existence of the map from the injectivity uses fractions tho, yeah

[delicate orchid]

ok yeah actually true

I didn't think about trying to see if it worked backwards I was only looking at inducing a homomorphism from D to frac(D)

[next obsidian]

there’s always a map there?

[delicate orchid]

no sorry, I'm inducing a homomorphism between D and Z to a homomorphism between frac(D) and Q

[next obsidian]

Ah sure

[delicate orchid]

mb

[next obsidian]

There’s nothing special about Z here

Taking fraction fields is a functor from domains with injective maps to fields

[delicate orchid]

yeah I can see it working for any 2 integral domains (with injective homomorphisms between them)

I was just working with mapping to Z because it makes it easier

[lethal dune]

oh

but then I'll need to remember the unicodes for the chars

unlike in my gboard where I can just do \oplus

⊕

[foggy merlin]

given some polynomial in the rationals (but if you know for general fields I'd be interested too), is there some algorithm that can construct a matrix over Q that has that polynomial as characteristic polynomial?

[lethal dune]

yes

one brute force technique is to just find the roots (if possible and put them in the diagonal)

otherwise you can construct the matrix replicating the rational canonical form

[foggy merlin]

but that doesn't work when the roots aren't rational though does it?

[lethal dune]

that is given $p(x) = x^n + a_{n-1}x^{n-1}+\cdots+a_0$ then the matrix will be given by ...

[lethal dune]

ok let me find the image on google, not in a mood to type it out myself

here

[foggy merlin]

oh wait that's it

[lethal dune]

[foggy merlin]

lol cool

I thought it'd be more complicated but I'm glad

[lethal dune]

(side note: that's also the minimal polynomial, and the matrix may not be unique upto similar)

[foggy merlin]

that's ok, I just wanted to get a matrix that had that property, didn't really matter how many there are

I just found it cool that with the help of tensor products one can explicitly construct polynomials that have the sum and the product of algebraic numbers a,b as roots, given we have their minimal polynomials (or just polynomials that have each as roots)

[dreamy fiber]

Let F be a field of characteristic 0, if A^n \in F for some natural number n, then let k be the smallest natural numbers such that A^k \in F, then is

x^k - A^k

The minimal polynomial for A in F[x]?

[lethal dune]

how many times will you edit the question

[dreamy fiber]

That’s the last time I swear 😂

[lethal dune]

ex $\omega ^ 3 = 1 \in \mbb{Q}$ but that's not the minimal poly

[dreamy fiber]

Okay to be fair, this was a pretty stupid question in hindsight

Thanks

[lethal dune]

(discoveries come from stupid question to begin with )

ok nvm

[dreamy fiber]

Thanks 😂 okay so I guess I will ask the larger question behind this

So I saw 2 different definitions of a radical extension and I wasn't sure if they are equivalent but here are the definitions.

1. $L/F$ is a radical extension if there's a sequence of field
   [F = E_0 \subset E_1 \subset ... \subset E_r = L]
   , where for every i, $E_{i+1}$ is the splitting field of $x^{n_i} - A_i$ over $E_i$, where $n_i$ is some natural number and $A_i \in E_i$.

2. $L/F$ is a radical extension if $L = F(x_1, ..., x_r)$ where for each $x_i$,
   [x_i^{n_i} \in F(x_1, ..., x_{i-1})]
   for some natural number $n_i$

I think I figured out 1 $\implies$ 2, but I am not sure if the other way is true?

[cloud walrus]

ScarletScorch

[next obsidian]

Well this is a little odd because in the first x is a variable, in the latter x are elements

[dreamy fiber]

sorry lol i can switch them up

[next obsidian]

But starting from (2), just let E_i = F(x1,…,xi)

And then you just get the statement of (1)

I think…

Is that the splitting field?

[dreamy fiber]

Well but is E_{i+1} necessarily the splitting field of the polynomial tho?

[next obsidian]

My brain says something something separable

Or something about the fact that a power of x lies in it that you’re okay

But

[dreamy fiber]

well like

[next obsidian]

I feel like

If you take n minimal for a

So like let a^n be in F

n minimal

Then a^i should all be linearly independent or something

But is this true

If it is, then F(a) is the splitting field

And then the proof of (2) => (1) works

This is true if a is like a root of unity over Q

[dreamy fiber]

yeah i guess that;s kinda why i asked the question earlier

problem is kind of like a general kummer extension

usually needs to like adjoin 2 elements

rather than 1

[next obsidian]

I wish I knew field theory

[dreamy fiber]

I wish I knew more linear algebra?

but yeah it would definitely work with a primeitiveroot of unity

[next obsidian]

what's the splitting field of cbrt(2) over Q?

isn't that degree 6?

it feels degree 6 to me

[dreamy fiber]

i think yes?

[lethal dune]

yes

[next obsidian]

then you have a counterexample

Q(cbrt(2)) is deg 3

but it isn't the splitting field of something or something like that

unless you have to change what element it is, but idk this seems fake af to me

[dreamy fiber]

yeah the galois group is S3

[next obsidian]

how could Q(cbrt(2)) be the splitting field of something over Q which is like an n-th power

This is a rigorous proof of that fact 🙂

"how could it"

Can you do something using Galois theory?

lol

[dreamy fiber]

well i don't think anything significiant splits in Q(cbrt(2))?

[next obsidian]

Like if it was a splitting field it would have to be of an element a which satisfies a^3 being in Q

but not a^2 or a

[dreamy fiber]

yeah

[next obsidian]

Idk like

Q(cbrt(2)) isn't normal over Q?

but if it was a splitting field it would be?

¯_(ツ)_/¯

[dreamy fiber]

it's not a galois extension

[next obsidian]

oh yeah that does it then

I think

which is equivalent to normality here cuz char 0

so everything is always separable

[dreamy fiber]

oh yeah then truly

[next obsidian]

I proved a field theory thing

or well, I said words and then other ppl proved it

[dreamy fiber]

well i guess the first definition of radical extension was pretty cursed

literally i have not seen this outside of my textbook yet

[next obsidian]

What I need to know is

why for a radical extension in char p you can assume the n is a power of p

or something idfk

I think I read a proof that used this once and I was like wat

[lethal dune]

[dreamy fiber]

Well in char p you like run into some Artin Schreier flummery

Which is like disgusting and gross

I think characteristic 0 is already enough for me

[median pawn]

hey i'm slowly getting into algebra, and i had proved it earlier lol don't make fun of me

[delicate orchid]

I was doing the opposite, I was defending you

[cloud walrus]

Hausdorff

[median pawn]

So I have an argument for this

[cloud walrus]

Hausdorff

Hausdorff

Hausdorff

[median pawn]

and since determinant is the product of eigenvalues we are done?

am I missing some justification? or is the argument fine?

[delicate orchid]

did you mean $A+iB$ has eigenvalues ${\lambda_{1, k}\pmi\lambda_{2,k}}$

[cloud walrus]

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

[median pawn]

Nope

I meant C

[delicate orchid]

I missed the ^2 in the original question whoops

[trim grove]

How to find the number of permutations that commute with a given permutation in Sn? Any hint

[wooden ember]

Consider how conjugation in Sn works

[trim grove]

How this will help me to solve this?

[next obsidian]

the issue is that "analysis enjoyer" is a slur

[small karma]

I'm ready for my exam tomorrow

[delicate orchid]

gl!

[wooden ember]

Two elements commute iff one fixes the other by conjugation

And conjugation in Sn is nicely characterized

[crimson canyon]

If $R_1$ and $R_2$ are two irreps of $\mathrm{SU(N)}$ and $T^a_{R_j}$ the generators in a given rep, is the following true? How?

$$\mathrm{Tr} (T^a_{R_1 \oplus R_2}T^b_{R_1 \oplus R_2}) =\mathrm{Tr} (T^a_{R_1}T^b_{R_1} + T^a_{R_2}T^b_{R_2})$$

[cloud walrus]

Siupa

[lyric spire]

There are 6 people, each person can either be friends or strangers with the others. Want to prove that there are either 3 mutual friends or 3 mutual enemies

why is this not a valid counter example?

[next obsidian]

you haven't drawn all the lines

you need to color every single line between each person, then show that there's always a triangle with all the same colors (in order to prove the fact)

also, this isn't really abstract algebra, I think this falls into Morse Theory

I can outline a proof in a different chat if you'd like

[terse crystal]

b commutes with a iff the image of any cycle of a under b is a cycle of b so n!/Π(j^λ_j) I think where a has λ_j j-cycles

[lyric spire]

Aha it's because one of the conditions is also that any given pair must be either friends or enemies, i missed that

[next obsidian]

yeah

[lyric spire]

and for sure, i'm going to try doing the proof myself and then maybe ask you if I cant figure something out?

[next obsidian]

sure

[terse crystal]

Looks like something like Tr(diag{A,B}diag{C,D})=Tr(diag{AC,BD})=Tr(AC)+Tr(BD)=Tr(AC+BD)

[crimson canyon]

Thank you, I was getting derailed by all the proprieties regarding Lie algebras and stuff, while it seems that it was a more general result

About really any matrix and about direct sums

It's hard sometimes to recognise which specificities can be ignored

[prisma shuttle]

can someone explain why this is true

like why for every polynomial there exists a multiplicative inverse?

i thought this was only true for power series

i don't think its true for polynomials but apparently d&f says it is

for reference this is on page 234 in the book

and i think that R is assumed to be a commutative ring with identity but idk why that implies R[x] is as well?

[terse crystal]

It didn’t say “for every polynomial there exists a multiplicative inverse” at all…

[prisma shuttle]

oh shoot idk wut i was thinking

i completely misunderstood it

nvm

[chilly ocean]

can someone explain to me what this means

how would i find the solutions?

[prisma shuttle]

basically an isomorphism would have to be injective and surjective, so because its injective, there can only be one solution (the kernel is trivial)

[prisma shuttle]

they are just the number relatively prime to 30, because this way there is a unique sol, otherwise, there are no sols or many sols

[chilly ocean]

so there would be only one solution in this case?

[terse crystal]

the ideal generated by 30/gcd(30,m)

[chilly ocean]

so many isomorphisms would there be?

1

?

and what does Z_30 mean

[terse crystal]

Don’t know what you’re asking, better if you can just give me your original question

[chilly ocean]

gokay

[terse crystal]

Oh I see

[chilly ocean]

the first part was easy

[terse crystal]

U(30)

=8

[chilly ocean]

U(30) ?

i dont think ive been taught that yet

[terse crystal]

Yeah, 8 isomorphisms

Sorry

[chilly ocean]

how did u get that so fast lol?

[terse crystal]

I meant |U(Z/30Z)|=φ(30)

Any homomorphism f, f maps 1 to k , it’s an isomorphism iff it’s surjective

And the order of k in Z/30Z is 30/(30,k)

[chilly ocean]

ohhh

[terse crystal]

That equals to 30 iff k and 30 are coprime

Order in terms of addition I mean

[chilly ocean]

so to find the number of isomorhpisms

i find all the coprime numbers to 30?

that are under 30

?

[terse crystal]

Yeah

[chilly ocean]

how did u get 8 so quick lol

u know em from the top of ur head?

[terse crystal]

There is a formula of Euler function

You can deduce that yourself using, like combinatorics

[chilly ocean]

hmm okay

[chilly ocean]

do u know waht this theorem is called

or link me to it pls

[terse crystal]

Just Euler function

If you want a number theory proof Cohen’s number theory has it

But just seems unnecessary, using combinatorics is more straight forward

[chilly ocean]

ahh okay

[terse crystal]

Inclusion–exclusion principle

[chilly ocean]

do we just assume when there is a cyclic group of order k

it is Z/km ?

[terse crystal]

You mean it is Z/kZ ?

Yeah there is only 1 cyclic group of order k up to isomorphism

[chilly ocean]

ohh okay

[terse crystal]

Sorry I misread it

It’s not a theorem

Just some calculation you do it yourself, not worthy of a theorem for it

[chilly ocean]

gotcha

one last question

then i thikn i fully understand

[chilly ocean]

i know the left side is order of it, but why do you set it equal to phi(30)

[terse crystal]

k is invertible in Z/30Z in terms of multiplication iff k is coprime to 30

[chilly ocean]

alrighty

so bascially i can use this method for all type of problems in this manner right?

if they are cyclic

and say the gcd between the two groups were 10

[terse crystal]

I don’t know I used any method… you have to ask a specific question…

[chilly ocean]

hmmm

[terse crystal]

And I don’t know what gcd of two groups mean

[chilly ocean]

it means that there are that many homomorphisms between the two finite groups

[terse crystal]

You mean how many homomorphisms from Z/mZ to Z/nZ?

[chilly ocean]

yeah

[terse crystal]

|{k: n|km}|

[tribal moss]

which is exactly the greatest common divisor of m and n.

[terse crystal]

Yeah I am dumb😂 , it’s (m,n)

There are φ(m) many injections when m|n, none otherwise. And φ(n) many surjections when n|m, none otherwise right?

[chilly ocean]

why is it so hard to understand what is given to my in an abstract algebra problem

like idek what b means

i know i have to show its associative inverse and identity

what is the actual thing im going to run these tests on?

[terse crystal]

ψ_g•ψ_h is defined to be ψ_gh

[tribal moss]

hmm, pretty sure I saw a long discussion about pretty much that question only a few days ago, but I wasn't in it myself ...

[terse crystal]

I mean the composition of ψ_g and ψ_h is exactly ψ_gh

[tribal moss]

(If you're talking about Par's problem, note that his variant uses phis rather than psis).

[terse crystal]

Bad at distinguishing them😂,this? φ

[chilly ocean]

its the one taht loosk like a y lol

[tribal moss]

Yeah. Just wanted to ward off unnecessary confusion.

[chilly ocean]

actually cogwheels for the problem we talked about before this

a lot of people are getting 1 for isomorphism

not 8

this is how my friend did his proof

[terse crystal]

? Mapping a to a^7 already give you another one other than identity

[chilly ocean]

wym

[chilly ocean]

that discussion was about part a for this problem

[tribal moss]

The mistake in your friend's proof seems to be that he the only way he can find a g^a such that g^(ka) = g^m is if ka=m. But we could also have ka=m+30 or ka=m+60, and so forth, and this freedom allows a lot of k's other than 1 to work too.

[terse crystal]

His last 7 rows make no sense at all😂😂

[chilly ocean]

ohhhh okay lmfao

[tribal moss]

Yeah, those 7 lines are pretty opaque, but I think I've at least gotten the gist of where it goes wrong, namely where he starts writing m/k for what I called a in my explanation.

[chilly ocean]

alrighty ill let him know

[terse crystal]

We can only give counterexamples to disprove something, can’t give examples to prove something. Also “any k” in the fourth row from below is completely nonsense…

[chilly ocean]

yeah i see it now

just wanted it to get checked

anyways for part b i assume i would compose the phi above with itself?

im kinda confused on that that notation means

[terse crystal]

Isomorphisms from G to itself is a group right?

[chilly ocean]

yeah

[terse crystal]

under composition, Inn(G) is a subgroup of this

[pastel cliff]

class syllabus finally dropped

anything stick out to anyone?

i would've thought groups would've come first

[delicate orchid]

$\phi_g(x) = gxg^{-1}, ; \phi_h(x) = hxh^{-1}$ and so on, you have to prove that the set of all functions of this form is a group under composition

[pastel cliff]

there's also another exam later that's not everything

https://tenor.com/view/epic-embed-fail-ryan-gosling-cereal-embed-failure-laugh-at-this-user-gif-20627924

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

ty lol

[chilly ocean]

so compsoing phi(h) into phig

i would i just check if associativity inverse and identity exist?

[delicate orchid]

and closure

[pastel cliff]

look at my syllabus and give me excuses to complain

im so excited to actually take this class tho

[delicate orchid]

I've got my own classes to stress about

[pastel cliff]

lol my semester starts next week im chillin still

catch me pulling 72 hour marathons 2 months from now

[delicate orchid]

I got an exam in 5 days and another in 7

I've ran out of past papers and revision material as per usual

[pastel cliff]

what class

#chill ?

[delicate orchid]

sure

[chilly ocean]

Like this?

[delicate orchid]

yup!

[chilly ocean]

then i show the 4 properties

okay

[barren sierra]

why does the notation in D&F suck 💀

D_2n rather than D_n
Q_8 for quaternions
why

[next obsidian]

D_2n thing is up to convention because it describes the number of elements

Q_8 is cuz it has 8 elements, there's a generalized quaternion group which has other numbers

[delicate orchid]

Q_8 is a universal standard lol

[prisma shuttle]

hey guys

can someone explain how associativity of multiplication of group rings follows from the associativity of the group it is based on?

the pruple part in d&f

isn't it because of associtiavity in the ring not the group?

[terse crystal]

both

[prisma shuttle]

oh so like its because of associativity in both of them?

[terse crystal]

yeah

[prisma shuttle]

i mean ig that is pretty obvious but like idk the book only says group so i got a bit confused

[prisma shuttle]

bruh i literally just spent an hour not knowing why the stuff in the book was true because they used this weird unconventioanl notation 😦

[trim grove]

Didn't this formula will number of elements conjugate with given cycle( idk i am confused now)

[terse crystal]

I was dumb, actually it should be

b commutes with a iff b maps any j cycle of a to a j cycle of a

[lethal dune]

u can also do it by using orbit stabilizer

[terse crystal]

Therefore in total Πλ_j!j^λ_j

[trim grove]

I tried i also get the formula , but didn't get the motivation behind that , how they get there

[lethal dune]

it's basic combinatorics

[trim grove]

And this is what over head projector for me

[terse crystal]

Yeah b looks like permutation of cycles of a

Therefore Π λ_j ! Then there are j ways to map a j-cycle to another j-cycle

[lethal dune]

like I could explain but I'm low on energy, just woke up

[terse crystal]

So Πλ_j!j^λ_j

[upper cape]

Anyone know what Irr() stands for?

[gritty sparrow]

minimal polynomial of alpha in K (it is also the only monic irreducible polynomial that alpha satisfies in k)

[upper cape]

thanks!

[small karma]

@rustic crown hey u know mathematica?

[rustic crown]

nope 😦

[small karma]

oh okieeee

[rustic crown]

.<

[next obsidian]

Wait what the, when did you become chmonkeynumber1fan >_<

[hidden haven]

.<

[rustic crown]

.<

[lethal dune]

[delicate orchid]

O_o

[trim grove]

[lethal dune]

[next obsidian]

Pls

Is this about subrings with a different 1?

Cuz this is trash, it isn’t considered a subring if you require 1 to exist because it breaks too many things

[hidden haven]

also 4 in Z/6Z is the unit for 2Z/6Z

[next obsidian]

Any ring with unity with a non-trivial idempotent will have these sorts of “subrings”

[hidden haven]

They never said subring

subrngs 🙏

[lethal dune]

like every author seems to have their own def of subrings

FFS

[lethal dune]

have seen one definition where the author requires to have unit to be subrings,

like 2Z is not subring of Z

[hidden haven]

Yes that is the legit definition

[lethal dune]

legitimacy is relative

[hidden haven]

[lethal dune]

I would consider 2Z to be a subring

[next obsidian]

Wrong

Bad

[hidden haven]

It is the most common and perhaps the most reasonable one

[next obsidian]

The inclusion map isn’t a map of rings

That makes it not a subring

[lethal dune]

(like if you don't bring some shit out from category theory and show it's wrong def)

[next obsidian]

It’s the wrong def because theorems no longer work lol

It should only be a subring if you don’t require rings to have unity

If you do, the “sub” structure should respect that

And I think you can formalize this in like a universal algebra way

Or model theoretic

[hidden haven]

(LIkE if YoU dOn'T uSE MatH)

[next obsidian]

Like 1 is a 0-ary function

And we require them to be the same sort of deal

[lethal dune]

ye that's the definition I would go with but like... why do ppl can't just formalize a subring

[next obsidian]

why do ppl can’t just formalize a subring

[lethal dune]

[next obsidian]

Does this book require rings to have a unit?

[sharp sonnet]

this is the correct thing

[lethal dune]

no

[next obsidian]

That’s why this definition is like it is

[sharp sonnet]

you could in the same way argue "why do we care about homomorphisms"

the unit is part of the data of a ring

so maps (and substructures) have to respect that data

[next obsidian]

But now the problem is just one step deeper because YOU SHOULD HAVE A 1

[lethal dune]

this book even considers 0 to be zd in Z, (but not in {0})

[next obsidian]

Wtf is zd

[lethal dune]

zero divisor

[next obsidian]

Hurb

[sharp sonnet]

i mean that is fine?

[lethal dune]

don't you require the element to be non zero itself?

[sharp sonnet]

you might have to change some theorems to mention nonzero

but thats not that bad

[lethal dune]

this book does not even require integral domains to have 1 or be commutative, FFS i'm deleting this book rn

[hidden haven]

what book

[lethal dune]

Introduction to Rings And Modules by C. Musili

this is the book our prof is following

[hidden haven]

literally never heard of this

[lethal dune]

yeah reason is clear

[hidden haven]

but sounds like a book written by an italian dictator

[lethal dune]

[sharp sonnet]

i TAd some linear algebra class that started with monoids

and it had theorems like "if an element in a monoid has an inverse, it is unique"

there are probably reasons to keep the theory general (at first)..

[hidden haven]

If an element in a monoid has an inverse, that element is unique

[next obsidian]

I helped someone with their French math hw

It said to show that a finite associative regular magma is a group

[sharp sonnet]

bourbaki moment

[hidden haven]

What's a regular magma

oh no