groups-rings-fields

[torn warren]

I guess the author wants to have a wording correspondence between normal extension and normal subgroup when states the Galois main theory, so he renames normal=spliting+separable

[rocky cloak]

I mean the correspondence is still there. Normal subgroups of the Galois group correspond to normal intermediate extensions. It's just that any intermediate extension of a Galois extension is seperable

It just seems a lot more convenient to have two words for the two conditions you have to check. But maybe they make up a new word for normal extension...?

[torn warren]

this book didn't use Galois extension

[rocky cloak]

I mean they clearly use them, they just call them 'normal extensions' instead

[torn warren]

yes, but the author didn't use this word Galois extension, he only uses normal extension

[rocky cloak]

Right, but do they have a word for what one would usually call normal extension?

[torn warren]

normal extension is called normal extension

[rocky cloak]

I think we're talking past each other.

You posted a picture where a 'finite normal extension' was defined as the splitting field of a seperable polynomial. This is the usual definition of Galois extension.

I'm asking if they also give a name to having all it's minimal polynomials split. I.e. what is usually called a normal extension.

[torn warren]

oh, i see your points, no, they didn't give a name, they just call it splitting field

[rocky cloak]

Alright, fair enough. But then bottom line is that 'normal' means something different in these two books.

[torn warren]

yes, in Lang's, normal = splitting, in Fraleigh, normal = Galois

[white oxide]

i see, so would it be adequate to say that it's normal over F since it satisfies Nor 1, i.e. it's the splitting field of p(x) where p(x) = irr(a, F)

or would you opt for the traditional definition where every irreducible polynomial of F[x] which has a root in E splits into linear factors

[rocky cloak]

Both work, but just saying it's the splitting field of p should be good yeah

[white oxide]

thank god i'm sane

[torn warren]

this is false

p=x^3-2, E=Q(2^1/3)

[white oxide]

nah bro i'm saying that's one of the conditions for E to be a normal extension

By Lang's axioms

[torn warren]

you said what?

[cloud walrus]

ταυταυ

Is the function $\varphi : \mathbf Z / 5 \mathbf Z \to \mathbf Z / 3 \mathbf Z$ with :
\begin{enumerate}
\item $\varphi\,(0) = 0$ ;
\item $\varphi\,(1) = 1$ ;
\item $\varphi\,(2) = 2$ ;
\item $\varphi\,(3) = 1$ ;
\item $\varphi\,(4) = 2$,
\end{enumerate}
a morphism of $\mathbf Z / 5 \mathbf Z$ ?

[coral spindle]

No

[wooden rover]

hmm, i tried at least

(also consider the additition)

i don't see where is the mistake in there ?

oh

nvm

i saw one

[coral spindle]

For reasons I won't reveal, I can tell immediately it's not a morphism, but if you try each pair one by one you will quickly find one that doesn't obey the requirement to be a morphism.

[wooden rover]

4 + 4 = 3 but
2 + 2 = 1 ≠ 0

[coral spindle]

Yeah there you go

[dry pebble]

a

[wooden rover]

oh wait no, beacuse phi(3) = 1 that's not the mistake

[rocky cloak]

Have you considered that 1+2 = 3?

[wooden rover]

i figured yeah

[toxic zephyr]

i'm a lil rusty on non PIDs. in the ring R[x,y] the kernel of the evaluation map at (0,0) is just <x,y> isn't it?
i'd like to find the intersection of the kernels of the evaluation map at (0,0) and (1,1), but i'm not sure how to go about finding a generating set for the intersection of two ideals.
i know it's contained in <x,y,x-1,y-1>, but is it possible to know

1. the least number of elements necessary to generate (if i had to guess, i'd say 3)
2. a general method to find them
   i would conjecture that <x,y,x-y> might be a generating set, but i'm unsure nope x+y doesn't do it

would it be something like <x(x-1),y(y-1),x-y>?

[delicate orchid]

that looks more like it to me

you're trying to find all polynomials f with f(0,0) = f(1,1) = 0 essentially

[toxic zephyr]

yes

[delicate orchid]

only issue with that ideal is that x(y-1) and y(x-1) also work

[toxic zephyr]

oh no

single variable polynomials are so much easier haha

so <x(x-1),y(y-1),x-y,x(y-1),y(x-1)>?

[delicate orchid]

being a pid is so nice yeah

ok I'm still not convinced that's the whole thing - if R is algebraically closed I am convinced

[agile burrow]

If I is the kernel of evaluation at (0, 0) and J is the kernel of evaluation at (1, 1), then I and J are comaximal ideals (that is, I + J = R[x, y]). In a commutative ring, for comaxmal ideals we have I intersect J is equal to the product IJ. Then generators for IJ are given by pairwise products of generators for I and J

[delicate orchid]

did not know that result about comaximal ideals

oh wait no I remember seeing it like 3 years ago

so it's just <x(x-1), y(y-1), x(y-1), y(x-1)>

[toxic zephyr]

so using x-y is unecessary? makes sense since it's in <x,y>

[agile burrow]

That's right

[delicate orchid]

y(x-1)-x(y-1) = x-y

[toxic zephyr]

ah yes that too

[agile burrow]

The kernel of the evaluation map at (a, b) is the ideal (x - a, y - b) for any commutative ring R. If S is the quotient R[x, y] / (x - a, y - b) then we get an induced surjection S -> R since (x - a, y - b) is in the kernel of the evaluation. The induced map S -> R is injective as well because f(x, y) = f(a, b) in S. So if f(x, y) maps to 0 in R, then f(x, y) = f(a, b) = 0.

[delicate orchid]

oh yeah dw I convinced myself like 3 seconds after posting that

[agile burrow]

awesome

[delicate orchid]

I had some scare with zero divisors in my head

but then realised that it would still be the same ideal

[agile burrow]

I think I was mostly convincing myself

[delicate orchid]

that's fair

[toxic zephyr]

okay cool the ultimate goal of the problem is to get a basis of polynomials in R[x,y] such that p(0,0)=p(1,1)=0 and the gradient at (0,0) and (1,1) are any desired vector. in the 1d case, this was trivial as we just needed x(1-x)g(x) where g is a linear polynomial with the necessary values at x=0,1.
but this case is much more complicated as this ideal IJ is generated by 4 instead of just 1 element. so i'm not sure considering p=x(x-1)g1+y(y-1)g2+x(y-1)g3+y(x-1)g4 is going to really cut down the necessary computations.

thank you @delicate orchid @agile burrow for your help! (:
i definitely learned quite a few useful things from you guys

[delicate orchid]

you can kinda really quickly get a nice general formula out of this actually

for like, R[x_1, ..., x_n] and some random collection of (a_i1, ..., a_in)

ahhh no not quite, very sad

ker(2,2), ker(4,4) are not comaximal

[toxic zephyr]

well actually the gradients of these are pretty nice...

[delicate orchid]

but if the matrix (a_ij) has non-zero determinant then we should be able to use this idea for that collection

[toxic zephyr]

ah i tried to map p(x,y) to (grad p(0,0), grad p(1,1))
but the matrix is singular. not unexpected as the linear combination would only be a quadratic, and a cubic is what i should expect. but i think this is good actually. evaluating the gradient of each times a function g should yield something somewhat nice like the 1d case.

yeahhh bc product rule means evaluating at the points will just yield some constant times g(0,0) or g(1,1). okay no this is good

[tribal moss]

Hmm, that's not even a linear subspace, unless the desired gradient is zero.

(Oh, I may have misunderstood, I first read it as "a basis for the polynomals in R[x,y] such that ....")

[toxic zephyr]

yea more like finding $p_{00}(x,y)$ such that $p(0,0)=p(1,1)=0$ and
$$\nabla p_{01}(0,0)=(1,0),\quad \nabla p_{01}(1,1)=(0,0)$$
and then similar for $p_{ij}(x,y)$ where $i=0,1$, $j=1,2$

[cloud walrus]

eigentaylor

[toxic zephyr]

then that would form a nice basis which i could combine to get any desired grad values

and the nice thing about this generator basis is that $\grad (e_{ij}(x,y)g(x,y))(0,0)=g(0,0)(\grad e_{ij}(0,0))$ and similar for $(1,1)$

[cloud walrus]

eigentaylor

[toxic zephyr]

okay sorry this isn't #groups-rings-fields anymore imma hop out so Zaylar can get help

[delicate orchid]

Spec is covariant right

my idea for a proof is to look at the varieties generated by (F) and (G) and shove it through the nullstellenstatz but the arrows are going the wrong way

I need something contravarient

[tribal moss]

(I wish to just shout "Nullstellensatz + UFD", but I suspect you don't have those tools ...)

[delicate orchid]

ok so I'm not completely off

[tribal moss]

G is a polynomial that vanishes at all the common roots of (F), so by the Nullstellensatz some power of G is multiple of F, or in other words F divides a power of G. Since F is irreducible (and the polynomial ring is UFD), F must divide G itself.

[delicate orchid]

I don't see how nullstellenstatz implies that, unless I \subset J => \sqrt(J) \subset \sqrt(I) which I don't believe is true

[tribal moss]

The phrasing of the NST I'm invoking is

Let k be algebraically closed and I any ideal of k[x,y,...,z]. Then for every polynomial g that vanishes at all common zeroes of I, some power of g is in I.
with I being (F) in this case. But it's my own notes, so I may have swapped something around when I wrote it down ...

[delicate orchid]

nullstellenstatz to me is "the ideal of polynomials that vanish on a variety V(I) is the radical of I"

which is the same yeah

yeah I can follow it now

just wish I had the capacity to come up with it myself

[tribal moss]

Trust me, I'm feeling uncommonly accomplished right now for being able to figure out an application of it :-)

[delicate orchid]

I had the scent but I got distracted trying to figure out some way to reverse the monomorphism (F) included in (G) for some reason

[cedar vault]

Anyone?

[indigo nymph]

do you know about C_i fields? see for example https://en.wikipedia.org/wiki/Quasi-algebraically_closed_field#Ck_fields

there is also a version for non-homogeneous polynomials, see https://en.wikipedia.org/wiki/Tsen_rank

[cedar vault]

I didn't know about these, also every finite field being C_1 is just the chevalley warning theorem right?

My bad, thats a special case of chevalley warning

Anyways, thanks

[torn warren]

why here it is the intersection? the intersection is Q ?

[prisma ibex]

I mean yes it's Q

[torn warren]

thank you

another question is, when we wrote F(alpha^1/n), does the alpha^1/n stands for the real root ONLY, not other complex branches?

[chilly radish]

Yes, that's why the intersection is Q

Well, heuristically

This requires proof

Which they provide right afterwards

[white oxide]

Is this Lang

[glass zealot]

given some prime number n/2 < p <= n. isn't the group generated by (1,...,p) not always a sylow p group of Sn?

or am i missing something

since p^1 is the largest exponent of p in n!

as p^2 is > n for n > 2, given that n/2 < p

[coral spindle]

since p^1 is the largest exponent of p in n!
Yes that should be right, assuming those strong conditions on the size of n

[glass zealot]

hmm

this seems too easy to be true

[delicate orchid]

well it is true

[torn warren]

yes

[white oxide]

you switched to the dark side i see

[torn warren]

the splitting field for x^4+4 over Q is Q(i+1, i)=Q(i+1)=Q(i), which has degree 2, correct?

[delicate bloom]

I'm thinking you'll need sqrt(2) and an eighth root of unity in there, which you don't currently have

[torn warren]

no, we don't need sqrt2

the four roots are i+1, i-1, -i+1, -i-1

[delicate bloom]

you're right, 1+i is basically sqrt(2) lol

not saying we actually have sqrt(2) here though, I didn't think past the knee jerk newton polygon having slope -1/2

that's pretty fun to factor now that I'm looking at it, once we see that x |-> ix is a symmetry and complex conjugates we are stuck with solutions of the form k or k(1+i) for k real

[karmic moat]

for k a field and R = k[x_1, x_2], what's Hom(k, R)?

i know that each element is either 0 or injective, but i'm not sure what i can say about the whole module

another way of viewing it is as Hom(R/<x,y> , R) but i'm still not sure what to do with that lol

[chilly ocean]

Let $(H, \star)$ be a semigroup.
How can i show, that $\
x_1 \star a=a \land a \star x_2=a \Rightarrow x_1=x_2$?

[cloud walrus]

Plazzi

[chilly ocean]

I tried to argue with associativity, but that doesn't work

a,x_1,x_2 are elements of H

[coral spindle]

Why would this be true?

Unless there's some combination of quantifiers I'm not seeing, this is just false.

In any case you need to clarify the statement with quantifiers

[delicate orchid]

Also as a rule of thumb please use words instead of logical connection symbols

I legitimately thought there was a wedge product here

[coral spindle]

You can also use the much more readable \implies instead of \Rightarrow. Spacing matters!

Ofc as wew says, words are best

[chilly ocean]

I have to prove that the semigroup $(H, \star)$ is a group, if the right and left translation $\star_a$ and${}_a \star$ are surjective. I have proved everything, except that $e_r \star a =a$ and $a \star e_l =a$ implies $e_r=e_l$

[cloud walrus]

Plazzi

[chilly ocean]

I should have mentioned the translation part

[coral spindle]

Context will be my literal end. Context will drive a spear through my torso and that's how it'll all end

[delicate orchid]

ntext will be my literal coend

[coral spindle]

Anyway just look at $e_r \star e_l$, that'll get you what you want.

[coral spindle]

die

dice

[cloud walrus]

Boyt

[delicate orchid]

MODS

[coral spindle]

just practicing my plurals ✨

did you know the plural of sheep is sheep

crazy

[delicate orchid]

What’s this mysterious translation operator gizmo

We speaking German? Basque?

[coral spindle]

Hayır

[delicate orchid]

No but fr

[coral spindle]

I've seen the left application of an operation be called left 'translation' before

Makes it feel like addition

I don't think it's that uncommon

[chilly ocean]

How?

[delicate orchid]

Oh is that what that means? That’s completely deranged beyond recognition

[coral spindle]

Pfft

[delicate orchid]

if ae_1 = a and e_2a = a for all a then e_2e_1 is…?

[coral spindle]

Group actions on geometric objects (varieites, manifolds) are also often called translation

V common

[delicate orchid]

That’s more understandable…. NOT!

[coral spindle]

Please let me use my pretty words

[delicate orchid]

It’s sandwich time chat

[chilly ocean]

But this is isn't the case

It also could be, that two different semi group elements a,b have the identity elements e_b_l and e_a_l, but e_b_l≠e_a_l

The definition of the left translation (in my course) is:
For all x in H, x *a is also in H

Anyway, i found the solution, but thank you for your help ❤️

[chilly ocean]

I had to show that e_la=a is true for every a in H

I had to proof this part

[rocky cloak]

So a map from R/(x, y) is the same as a map from R that maps x and y to 0. Which maps R to R map x and y to 0?

[silk horizon]

Can I ask a very basic Galois theory question? I'm new to the subjects and there's like one simple example I can't figure out

[summer path]

[silk horizon]

if we consider the splitting field of p(x) = x^4-2 which is Q( 2^1/4 , i ) with Galois group <r,s> = D4 where r is the element of order 4, r( 2^1/4 ) = i 2^1/4, and s is the element of order 2, s(i)=-i, what subfields correspond to the subgroups of order 2 <sr> and <sr^3>?

[silk horizon]

looks like this example comes up here often...

[torn warren]

write the number as the linear expansion by basis with undetermined coefficients, then act sr and the result is invariant, then you compare coefficients, done

[rocky cloak]

So the corresponding subfield is the one fixed by the group. In this case you just need to find the elements fixed by sr.

A useful trick can be to note that x + sr(x) is fixed by sr, and that if x is already fixed then this is just 2x. So if you pick a basis for your extension and then apply x+sr(x) to each, you get a spanning set for your fixed field

[silk horizon]

oh

thanks

I think I got it

||Q(2^1/4 - i 2^1/4) and Q(2^1/4 + i 2^1/4)|| I totally overlooked those options for some reaon

[karmic moat]

just the typical mod x and mod y?

[rocky cloak]

Wait are these suposed to be R-linear or k-linear maps?

[karmic moat]

R

yeah i shouldve been clear my bad

[rocky cloak]

alright, then an R-linear map is determined by where it maps 1

[karmic moat]

mhm

[rocky cloak]

and we must have f(x) = f(x*1) = x*f(1)

[karmic moat]

mhm

ok thank u i will revisit this in a few hours i just woke up and i am gonna sleep some more

thank u

[regal star]

Someone help me out... I am not sure how to get started . I guess I will have to use the second Isomorphism theorem somehow

[delicate orchid]

do you know about sylow subgroups yet, no worries if not it's just so I know if I can use certain words lol

[regal star]

No, Slylow groups are given in later chapters. I am following Dummit and Foote

[delicate orchid]

ok no problem - so you were right in that the 2nd iso theorem here is useful, if we assume we know that |P \cap N| = p^b can you show that this implies |PN/N| = p^(a-b)?

or vica versa, going |PN/N| = p^(a-b) => |P \cap N| = p^b

[regal star]

Yes. It is clear from 2nd iso theorem

Also I know p^b divides the order of P and N as a>=b here

[delicate orchid]

Ok I think the best way to show this is using indexes.
[G : P] = m which is coprime to p, so because [G : P] = [G :PN]*[PN : P] both [G : PN] and [PN : P] must be coprime to p. Now use the fact that |PN| = |P||N|/|P \cap N| (which follows from the second isomorphism theorem in case you can't see it) to relate [PN : P] to [N : P \cap N] and thus [N: P \cap N] is coprime to p

and then ||because you know [N : P \cap N] = |N|/|P \cap N| = p^bn/|P \cap N| is coprime to p, |P \cap N| = p^b||

[regal star]

Almost got it. But what if |P \cap N| = p^b n. Is this not possible because then it won't be a subgroup of P ?

[delicate orchid]

exactly

[regal star]

Okay got it. Thanks

[tardy hedge]

Can every field be constructed by some F[x] \ <p(x)>?

[delicate orchid]

gonna say yes on this as fields are commutative algebra over some base field F and then polynomial algebras F are free objects in F-CommAlg

[tardy hedge]

Interesting

[delicate orchid]

maybe not if your field isn't finitely generated as an algebra over F

hmmm actually no I'm starting to not buy this very much at all. It works for F any extenstion of Q by the primitive root theorem

[tardy hedge]

I saw the example of C isomorphic to R[x] \ <x^2+1> so i was wondering

I thought that was kinda neat

[delicate orchid]

ok yeah, you can't get the algebraic closure of any finite field like this

works for Q so it's good enough for me

[tardy hedge]

It made me think of C differently, in that its not necessarily some special entity, its just a field with certain properties that can be built in different ways

But somehow doing calculus from functions on C turns out to be good

What if u take some other field and try developing calculus for it?

[delicate orchid]

good luck doing that on any countable field

their topologies tend to be rather boring

[tardy hedge]

Is that cause you cant really define limits ?

[elder wave]

a lot of properties come from the topological similarity to R^2

so in that aspect C is very special

[delicate orchid]

yeah you kind of want your field to be a manifold

does H work actually?

timo you might know this

[elder wave]

work in what aspect?

[delicate orchid]

with the standard topology on R^4

that's for YOU to figure out. Is it a manifold

and if it is a manifold, are the skewfield operations commutative

commutative

ffs

CONTINOUS/smoothhhh

getting my words backwards today

addition obviously is

[elder wave]

you don't really have a suitable carryover of smooth/holomorphic functions

but quaternionic manifolds are a thing

[delicate orchid]

[elder wave]

but the uuh something quaternions are a 3 manifold

i remember that exercise

unit quaternions

[delicate orchid]

oh yeah duh

SO(3)

I feel a little silly now lol

[elder wave]

idk i try to avoid quaternions whenever i can

they're relevant for symplectic stuff sometimes

[delicate orchid]

I like Q_8

that's about it

actually generalised quaternion groups in general are quite nice

there's a theorem about sylow 2-subgroups of SL(2,q) being generalised quaternions which is NEAT

[dim widget]

SU(2) ackshuallysfasldfjasdlfasy

or Spin(3)

[delicate orchid]

WHAT

BB-B-B-B--B-B-B--B-B-B-BUT IT'S THE WHOLESOME SPHEREINOOOOOO NOOOOOOOOOOO

[dim widget]

If SO(3) was a sphere so many engineers would lose their minds with bliss

[delicate orchid]

it is a sphere. Think about it.

but not too much

[dim widget]

the belt trick is a psyop

[chilly ocean]

Let $J \subseteq I$ be an arbitrary subset. Then there exists a canonical injective homomorphism $\vartheta_J: \prod_{\iota \in J} G_{\iota} \to \prod_{\iota \in I} G_{\iota}$. It maps an element $f$ to $f_J$ such that $f_J = f$ on $J$, and $f_{J}(\iota) = e_{\iota}$ for $\iota \notin J \$.
I don't understand the last part. If $\iota \notin J$, then it should not be possible to assign a value to $f(\iota)$ because this $\iota$ is not even present in our image set, or am i wrong?

[cloud walrus]

Plazzi

[delicate orchid]

what are these G_is precisely

because the use of f for an element of them makes me think they're sets of functions or something

this theta_J map looks a lot like the projection map for finite products

ok no, got it now

we're taking tuples $f = (f_i){i \in J}$ with $f_i \in G_i$ to tuples $f = (f_i){i \in I}$ by setting $\theta_J(f) = \left(\begin{cases} f_i & i \in J \ e_I & i \not \in J \end{cases}\right)_{i \in I}$

so it's just the natural embedding map

[cloud walrus]

Wew Mods Tbh

[delicate orchid]

there we go

[chilly ocean]

Oh sorry, i forgot to say. G are Just Groups

G_i

[delicate orchid]

yeah it turns out that it doesn't matter, this works for sets

[dim widget]

well it wouldn't work for sets, because sets don't have a distinguished element

but the G_i could be pointed sets and the same construction would make sense

[delicate orchid]

oh yeah lol, what I meant by that is this "projection of products" always works if you just collapse down all of the terms outside of the sub index set

[crystal turtle]

Zero object moment

[dim widget]

el momento de zero

[woven panther]

Is the isotropy group and the stabilizer the same thing?

[coral spindle]

It's relatively uncommon to see it called the isotropy group, but yes those mean the same thing

[chilly ocean]

I'm confused. What is theta(f)?

Oh, you Just defined it. I still don't get it , why we define f_J(i) when i is not in J

[delicate orchid]

because we want out new thing to be in the larger product

the most natural way to do this is to define a map that's the same as f on J, and then map everything outside of J to 0/the identity

[chilly ocean]

But why is it still injective?

[delicate orchid]

\theta_J(f) is definitely not injective

\theta_J is

[karmic moat]

im having a bit of a brain fart

given an exact sequence A -> B -> C -> 0, is applying Hom(D, -) left exact? i.e., we have 0 -> Hom(D, A) -> Hom(D, B) -> Hom(D, C)

i know it's true on short exact sequences, just not sure if it's true for an arbitrary right exact sequence

[south patrol]

Yes, it is left exact.

Well knowing it's true on SES is enough

[karmic moat]

okey donkey

[south patrol]

I think you are a bit confused though, because Hom(D,-) is a covariant functor

[karmic moat]

yeah i knew it was left exact on left-exact sequences, just forgot if it was true on right-exact

yeah

[south patrol]

Wll

Should beb 0 -> A -> B -> C in the hypothesis

[karmic moat]

yeah that's the issue

im trying to figure out if it holds for A -> B -> C -> 0

npnp

[south patrol]

but still your thing doesn't make sense

where have you got the 0 from the left from

This isn't what left exact means

Like, if a functor F satisfies what you want, then it turns every morphism into a monomorphism

[karmic moat]

ok so here's the context based on some class notes i took

[south patrol]

you may have confused it with Hom(-,D) or smth

th e summary here is handy

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

[karmic moat]

if we take a projective resolution of a module M, P_2 -> P_1 -> K -> 0, where K is image(P_2 -> P_1), then we have the thing 0 -> Hom(K, B) -> Hom(P_1, B) -> Hom(P_2, B)

[karmic moat]

also just for clarification this was not meant to be rude

just an animal pun

[south patrol]

yeah exactly, you've applied Hom(-,B)

lol ye dw

[karmic moat]

like okey dokey -> okey donkey, totally -> turtally, definitely -> dolphinitely, etc

[south patrol]

lol

[karmic moat]

yeah

im just trying to see if a similar thing holds for Hom(D, -), instead taking A -> B -> C -> 0

[south patrol]

Well it is left exact

But what you just said is not left exactness

[karmic moat]

yeah

[south patrol]

a functor F is left exact if it turns 0 -> A -> B -> C into 0 -> F(A) -> F(B) -> F(C) basically

but Hom(-,D) is a bit weird cause of contravariance

[karmic moat]

mhm

[south patrol]

Yeah you define on opposite category

[white oxide]

You made a mockery of my name.

[south patrol]

so an exact sequence there is the opposite

[karmic moat]

ok maybe it's better if i give full context lol

actually nevermind im gonna keep thinking about this on my own for a while

im gonna see how far i can get

[glossy crag]

Isn't this just constant 0? Or could the notation mean something else.

[white oxide]

is it true that if $[F(\alpha): F]$ is odd, then $[F(\alpha^2): F]$ is even?

[cloud walrus]

okeyokay

[delicate orchid]

yeah that uhh looks to be constant 0. Lol?

[wraith cargo]

no

Cuz

[coral spindle]

Probably a typo.

[wraith cargo]

[F(a) : F] = [F(a) : F(a^2)][F(a^2) : F]

[coral spindle]

I couldn't say what it's meant to be without more context

[barren sierra]

Isn't F(a) = F(a^2) if [F(a) : F] odd?

[white oxide]

ye this is what i got, i'm trying to prove that if E = F(a) where [F(a): F] is of odd degree then E = F(a^2)

[delicate orchid]

I buy what this guy is selling

[barren sierra]

Yes it is

[white oxide]

so we have F(a^2) contained in F(a)

[barren sierra]

Cute proof by contradiction

[white oxide]

and i'm assuming for contradiction that there exists some w in F(a)

that's not contained in F(a^2)

so I just wrote down [F(a): F] = [F(a): F(a^2)][F(a^2): F]

know that the lhs is odd

if i can show that one of the things on the right is even then that's my contradiction

cuz even times odd or even times even is even

[barren sierra]

bingo

[white oxide]

part of me just wants to square the irreducible polynomial for a over F and assert that it's irr(a^2, F) of even degree

but i'm very sure that's false

[wraith cargo]

hint: x^2-a^2

[barren sierra]

You can compute [F(a) : F(a^2)] without too much work using that ^^^

[white oxide]

oh damn that was the solution 😭

yeah that makes sense tho lol

are intermediate fields defined in terms of strict inclusion

[dim widget]

Not necessarily I guess

Depends on the context

[white oxide]

hmm ok

[coral spindle]

To put it another way

POST THE DAMN CONTEXT!!!!!!

[white oxide]

i always get yelled at by Boytjie for not posting context

next time i will POST THE DAMN CONTEXT!!!!!!

[coral spindle]

I am the context gremlin

But for real, this is the xy problem and I'm not the only one who's noticed this

[torn warren]

because $[E:F]=2k+1$ is odd, then you have $a$ as the root for the minimal irreducible polynomial $a^{2k+1}+c_{2k}a^{2k}+...+c_1a+c_0=0$, separate the odd and even terms $a(a^{2k}+...+c_1)+(c_{2k}a^{2k}+...+c_0)=0$, note that both coefficients $(a^{2k}+...+c_1)$ and $(c_{2k}a^{2k}+...+c_0)$ are in $F(a^2)$, and both of them are not zero, otherwise the order of the minimal irreducible polynomial for $a$ is less than $2k+1$, hence $a=\frac{c_{2k}a^{2k}+...+c_0}{a^{2k}+...+c_1}\in F(a^2)$, which implies $F(a)\subseteq F(a^2)$

[cloud walrus]

WT

[white oxide]

Lol nice

I did a less computational proof

[coral spindle]

Very overcomplicated

[torn warren]

how you did?

[white oxide]

Suppose that $E = F(\alpha)$, and let $p(x) = \text{irr}(\alpha, F)$ be of odd degree. Clearly $F(\alpha^2) \subseteq F(\alpha)$. If there exists $\omega \in F(\alpha)$ such that $\omega \notin F(\alpha^2)$, we may write [[F(\alpha): F] = [F(\alpha): F(\alpha^2)][F(\alpha^2): F]] Since $[F(\alpha): F(\alpha^2)]$ is even (we see that $\text{irr}(\alpha, F(\alpha^2)) = x^2 - \alpha^2$) this implies that $[F(\alpha): F]$ is even, which is a contradiction

[cloud walrus]

okeyokay

[torn warren]

you didn't use this sentence

[limber wharf]

They implicitly did

[torn warren]

contradiction is from the assumption that $[F(\alpha): F(\alpha^2)] >1$, so you get contradiction which implies $[F(\alpha): F(\alpha^2)] =1$

[cloud walrus]

WT

[torn warren]

I think the red line should be deleted, there is no use.

[white oxide]

aight bro 😭

it's just setting up the contradiction that the other one is not a subset of the other

i.e. F(a) is not a subset of F(a^2)

[torn warren]

but there is no connection with the red line, suppose you didn't write the red line, you still can write the equation

[limber wharf]

Yeah it is used in that $x^2 - \alpha^2$ is irred

[cloud walrus]

colejagdtiger

[limber wharf]

But then we don’t know that $[F(\alpha):F(\alpha^2)]$ is even

[cloud walrus]

colejagdtiger

[torn warren]

your assumption is that $[F(\alpha): F(\alpha^2)] >1$, and you get $[F(\alpha): F(\alpha^2)] =2$ which gives contradiction

[cloud walrus]

WT

[limber wharf]

I guess they could cut it out but it seems rather minor, the other way of saying things is really just saying the exact same thing but in different terms

[quaint tree]

I'm studying for an algebra midterm and so I'm doing a couple computational Sylow theory problems about "classify all groups of order whatever". Would anyone be willing to check my working real quick? I'd really appreciate it!

1. $|G|=138$. We factor $138=2\cdot3\cdot23$, and in particular we get $n_{23}\equiv1\mod23$ and $n_{23}|6$, so $n_{23}=1$. Thus $\mathbb{Z}{23}\trianglelefteq G$. So, $G\cong \mathbb{Z}{23}\rtimes H$ where $|H|=6$. This gives two possibilities: If $H\cong\mathbb{Z}{6}$ we have $G\cong\mathbb{Z}{23}\times\mathbb{Z}{6}$. This is because $|\text{Aut}(\mathbb{Z}{6})|=5$ is coprime to 23, so the only map $\mathbb{Z}{23}\to\text{Aut}(\mathbb{Z}{5})$ is trivial. In the other case, $H\cong S_{3}$, we likewise have $G=\mathbb{Z}{23}\times S{3}$, because $S_{3}\cong\text{Aut}(S_{3})$ has 6 elements which is also coprime to 23.

2. $|G|=140$. We factor $140=4\cdot5\cdot7$. This allows us to compute that $n_{5}\equiv1\mod5$ and $n_{5}|28$, so $n_{5}=1$. Likewise, $n_{7}\equiv1\mod7$ and $n_{7}|20$, so $n_{7}=1$. So we have normal subgroups $\mathbb{Z}{5}$ and $\mathbb{Z}{7}$. So $\mathbb{Z}{35}\trianglelefteq G$, and so $G=\mathbb{Z}{35}\rtimes H$ for some group $H$ of order 4. There are two possibilities; if $H=\mathbb{Z}{4}$ we need some map $\mathbb{Z}{35}\to\text{Aut}(\mathbb{Z}{4})\cong\mathbb{Z}{3}$, which must be trivial because 3 and 35 are coprime, so $G=\mathbb{Z}{35}\times\mathbb{Z}{4}$. If $H=\mathbb{Z}{2}\times\mathbb{Z}{2}$ then $\text{Aut}(H)\cong GL_{2}(\mathbb{Z}{2})$ has 6 elements, and 6 is also coprime to 35. So in this case $G=\mathbb{Z}{35}\times\mathbb{Z}{2}\times\mathbb{Z}{2}$.

So, the groups of order 138 up to isomorphism are $\mathbb{Z}{138}$ and $\mathbb{Z}{23}\times S_{3}$, and the groups of order 140 up to isomorphism are $\mathbb{Z}{140}$ and $\mathbb{Z}{35}\times\mathbb{Z}{2}\times\mathbb{Z}{2}$. \qedsymbol

[cloud walrus]

Abelian Grapes

[buoyant echo]

Hi all, I have a quick question. For quotient group, as far as I understand it is only defined for normal subgroup. If a subgroup is not normal, then a quotient group does not exist ?

[void cosmos]

yea the quotient does not become a group

[buoyant echo]

what do you mean it does not become a group?

what group property does it possibly violate ?

[void cosmos]

the operation itself wont be well-defined

[quaint tree]

You can certainly have a set of cosets, but it will not necessarily be a group.

Look at the dihedral group $D_{2n}=\langle r,s:r^{n}=s^{2}=1,rs=sr^{-1}\rangle$ for example. Look at what happens if you consider the subgroup $\langle rs\rangle\cong\mathbb{Z}_{2}$, and what happens to the quotient.

My keyboard is so janky and so I have to use an on-screen keyboard. I hate typing like this.

[cloud walrus]

Abelian Grapes

[buoyant echo]

that is because it is defined the multiplication of two cosets gN and hN to be (gh)N, if N is not normal , then (gN)(h*N) might not be well-defined, is that right?

[white oxide]

Yeah their product might not be well defined

[cobalt heath]

Hmm, so kernel can sometimes fail to be both-sided. I wonder why

Like, is it confusing me by kernels of different categories existing closely together?

[wild jasper]

both-sided what?

[wooden rover]

Is $\mathfrak G,(G)$ the group of all bijective endomorphisms of $G$ ?

[cloud walrus]

ταυταυ

[wooden rover]

best i could find was mathfrak lol

[delicate orchid]

“Bijective endomorphisms” are automorphisms

And it’s a bit of a strange notation but I could see it meaning that

[wooden rover]

yep my bad

[delicate orchid]

Mathfrak{S} being the symmetric group

[wooden rover]

okok

i just had forgotten the term automorphism haha

For so long I thought $\mathfrak S$ was a G okay I see now 💀

[cloud walrus]

ταυταυ

[delicate orchid]

Can’t say I blame you. I don’t get how it’s an S either

[chilly ocean]

nobody can read fraktur letters it's okay

i thought a mathfrak A in this banach algebras book was a U for the longest time

[delicate orchid]

It looks like a 4 it’s so bad

[daring nova]

You mean it isn't a weird sigma?

I don't know which one is worse

[south patrol]

Lol

[wooden rover]

Some letters make sense

Like B, E or Z eg

[delicate orchid]

Ok proved it

[dim widget]

what a king

[delicate orchid]

And I think I’ve spotted a pattern to determine for minimal k with a^k = 0 for a being a length l sum of basis vectors

[dim widget]

Oh, what is it?

[delicate orchid]

Where ^k means the obvious thing in the exterior algebra

[dim widget]

wait

[delicate orchid]

Like you wedge it with itself k times

[dim widget]

what is a here?

[delicate orchid]

Some element of the exterior algebra with homogeneous grading - I should have said that last part

It’s rather important

[dim widget]

but if a has odd degree then a \wedge a = 0 no matter what...

[delicate orchid]

Yeah that’s what I’m seeing, but if a is even it’s more interesting

Not very interesting but like

Interesting enough to keep me occupied on a boring train journey lol

[dim widget]

hmmm but what is the pattern then?

[delicate orchid]

Save me from doing Clifford theory of solvable groups over F_p bar

[delicate orchid]

Well when a is length 2 (the sum of two different basis vectors) and you wedge it with itself once you get a simple element

So I’m thinking it’s l = k but I’m checking other cases

[delicate orchid]

Ah no not so nice! If it’s length 3 you get a length 3 thingy out

So if a is of length l a^2 is of length l(l-1)/2. Cool.

Simple combinatorial argument shows that

[delicate orchid]

So in general if a is of length l and b is of length k, assuming k > l wlog. then a wedge b is of length T(k-1)-T(k-l-1) with T(x) = x(x-1)/2.

So now use this to compute the length of a^n

[tardy hedge]

In 4.2.11, are the repeated factors in question only those in R or does C count too

Like if we have a polynomial over R thats irreducible in R, can we use this proposition to show that it has repeated factors (that would then be factors in C)

[warm wyvern]

wait...

[delicate orchid]

Waiting

[warm urchin]

didnt expect while searching about frak font to end up on a rabbit hole related to its use in Nazi-germany

[wooden rover]

if it's bad then let's get rid of the letter Z as well

z = sj from now on

ʂ

wtf happened here

CLARIFICATIONS

Aut(G) = {automorphisms on G}

Int_h(g) = hgh^{-1}

Int(G) = {int_h | h in G}

[rocky cloak]

They show that the inner automorphism group is a normal subgroup of Aut(G)

[wooden rover]

yeah but the steps, idk how they did it

the g is applied from* what

[delicate orchid]

alpha is an automorphism

[wooden rover]

yeah

[delicate orchid]

So alpha^-1 is being applied to it

[wooden rover]

i know there's $\circ$ hidden

[cloud walrus]

ταυταυ

[wooden rover]

okok

[rocky cloak]

Which step is confusing you

[wooden rover]

third term

from 2nd to 3rd

[delicate orchid]

The first equality is definition of Int_h, second is using the fact that alpha is a homomorphism

[wooden rover]

OH

I se

i see

i realised wayyy too late

lol thanks

you've got a(h a^-1(g) h^1) and since it's a morphism you can decompose it into a(h) a(a^-1(g)) a(h^-1) = a(h) g a(h)^-1

also because alpha(h^-1) = alpha(h)^-1

[tardy hedge]

lol i was trying to see if x^4+x^2+1 was irreducible in R and I just happened to guess that it factored into (x^2+x+1)(x^2-x+1)

to be fair i knew that if it was reducible it would only have quadratic factors

but im wondering how i wouldve done this if i didnt just happen to guess it correctly

[daring nova]

it factors in C, where you see real roots, and complex roots coming in pairs, whose product is a real quadratic
Their product is guaranteed to be a product of real polynomials of degree 1 and 2

[untold basalt]

If $G=\mathbb{D}{n}$ is the dihedral group of order $2n$ and $n$ is even, then $G/Z(G)=\mathbb{D}{n/2}$ right?

[cloud walrus]

Seagull

[coral spindle]

I believe that's right

[tardy hedge]

Oh yeah you can just factor it fully and then remultiply it ?

[daring nova]

that's the power of the splitting field

when studying polynomials in R, it never hurts to look at what happens in C

[buoyant echo]

hi all, ideal defined to make quotient ring with addition well-defined, is that right?

[coral spindle]

addition and multiplication

[south patrol]

ideals are definitely interesting enough objects in their own right. in fact i'm pretty sure they came about through the work of Dedekind to generalise prime factorisation and stuff

[buoyant echo]

we want R/I = {r + I, for any r in R}, right ?

[south patrol]

well that is essentially the definition as a set

[buoyant echo]

can you elaborate a bit more with some examples ?

[south patrol]

Well I mean that uh

Okay the examples may not make too much sense but basically if you have a number field $K$ (a finite extension of $\mathbb Q$, but think of examples like $\mathbb Q(sqrt{-2})$ or $\mathbb Q(i)$ or just $Q$), you can form this `ring of integers' $\mathcal O$ (for $\mathbb Q$ this is just $\mathbb Z$) and unfortunately $\mathcal O$ doesn't necessarily have unique factorisation. But if we think about ideals instead of numbers, then we actually do get unique factorisation. In the case of $\mathbb Z$, this is just cause all ideals are of the form $n\mathbb Z$ and so it's basically the same as regular factorisation

[cloud walrus]

Potato E-Girl

[south patrol]

and then this is often enough to do some stuff in number theory

Why is that opencry 😭

[white oxide]

nothing i just thought it was funny because i wouldn't understand that example if i was first learning about ideals/i don't understand it now

[south patrol]

well okay we can simplify this to like

If you want to generalise number theory slightly beyond Q, you have to use ideals if you want unique prime factorisation

and i'm p sure that's where they historically come from

"ideal numbers"

But the point anyway was that ideals aren't merely defined so there's a notion of quotient

[white oxide]

no no like i'm sure it's a great example i just thought it was funny for no particular reason

my b 😭

[south patrol]

Thank dw dw

Tbf it was good to simplify it so thanks xd

[broken quartz]

Let's assume $f(x, y)$ is an irreducible polynomial over a field $K$ with characteristic $p$. We want to show that $f\left(x^p, y^p\right)$ is also irreducible over $K$.

Firstly, let's consider the field extension $K^{\prime}=K(x, y)$. Since $f(x, y)$ is irreducible over $K$, it remains irreducible over the larger field $K^{\prime}$.

Now, let's evaluate $f\left(x^p, y^p\right)$ over the field $K^{\prime}$. Substitute $x^p$ for $x$ and $y^p$ for $y$ in the polynomial $f(x, y)$ :
$$
f\left(x^p, y^p\right)=f\left((x)^p,(y)^p\right)
$$

By using the Frobenius endomorphism in characteristic $p$, we can rewrite this as:
$$
f\left((x)^p,(y)^p\right)=f(x, y)^p
$$

Since $f(x, y)$ is irreducible over $K^{\prime}$, the polynomial $f(x, y)^p$ cannot be factored into nontrivial factors over $K^{\prime}$. Therefore, $f\left(x^p, y^p\right)$ is also irreducible over $K^{\prime}$, which means it remains irreducible over the smaller field $K$.

Hence, we've shown that if $f(x, y)$ is an irreducible polynomial over $K$, then $f\left(x^p, y^p\right)$ is also irreducible over $K$.

[cloud walrus]

.

[broken quartz]

Does my proof look alr?

[south patrol]

I think this statement is just wrong lol

x is clearly irreducible over K[x,y]

But x^p is not

In fact you have given an explicit factorisation of f(x^p, y^p) as f(x,y)^p

Unless there is some detail we are missin

[south patrol]

Also the first step doesn't really work and I'm not sure why you're doing it - if a polynomial is irreducible, it needn't remain irreducible over a larger field. Indeed this is why you can construct splitting fields

[broken quartz]

Yeah I realized I redid it

https://media.discordapp.net/attachments/811391852934725653/1174124487285932032/image.png?ex=656673a3&is=6553fea3&hm=cbbbad590377094483e5eded1e6c30b931df616612b966443af61b39131d4806&

The curve $C$ defined by the equation $X^2-Y^2=0$ over a field $K$ consists of two irreducible components in the affine plane.

The equation $X^2-Y^2=0$ can be factored as $(X-Y)(X+$ $Y)=0$.

Therefore, the curve $C$ consists of two irreducible components:
The component defined by $X-Y=0$, which is the line $Y=X$. The component defined by $X+Y=0$, which is the line $Y=-X$.

In characteristic 2, we need to be careful with the distinction between the two components since $X-Y=0$ and $X+Y=0$ coincide. In this case, both equations represent the same line, $Y=X$, so there is only one irreducible component in characteristic 2.

Therefore, in characteristic 2, the curve $C$ has a single irreducible component, which is the line $Y=X$.

[cloud walrus]

.

[broken quartz]

In my proof, do I have to show why in char 2 it’s just x - y

Or is what u showed fine

[south patrol]

Yeah in char 2 this line is just X = Y

(at least as points lol, there is more scheme-theoretic information in the former case ig)

[broken quartz]

Yup

[buoyant echo]

its hard to absorb the text, but r u saying given an ideal, every number will have unique factorization ?

[broken quartz]

https://media.discordapp.net/attachments/328208536029102081/1174128800905318502/image.png?ex=656677a8&is=655402a8&hm=dfdd87a8a85288b891d584c9dfbbd03bb6aadb968abb78de83e41016c786caa0&

For this question do I just plug in?

Kinda like this:

Let's consider the map $f: D \rightarrow C$ defined by $f(x, 0)=$ $\left(\frac{1-x^2}{1+x^2}, \frac{2 x}{1+x^2}\right)$.

Firstly, let's verify that the image of $f$ lies on the curve $C: X^2+$ $Y^2=1$

For any point $(x, 0)$ in $D$, where $x$ is not a root of $X^2+1$, let's calculate $f(x, 0)$ :
$$
f(x, 0)=\left(\frac{1-x^2}{1+x^2}, \frac{2 x}{1+x^2}\right)
$$

Now, let's verify that $f(x, 0)$ satisfies the equation of the curve $C$ :
$$
\begin{aligned}
& X^2+Y^2=1: \
& X^2+Y^2=\left(\frac{1-x^2}{1+x^2}\right)^2+\left(\frac{2 x}{1+x^2}\right)^2 \
& =\frac{\left(1-x^2\right)^2+4 x^2}{\left(1+x^2\right)^2} \
& =\frac{1-2 x^2+x^4+4 x^2}{\left(1+x^2\right)^2} \
& =\frac{1+2 x^2+x^4}{\left(1+x^2\right)^2} \
& =\frac{\left(1+x^2\right)^2}{\left(1+x^2\right)^2} \
& =1
\end{aligned}
$$

Therefore, for any point $(x, 0)$ in $D$, the map $f(x, 0)=$ $\left(\frac{1-x^2}{1+x^2}, \frac{2 x}{1+x^2}\right)$ maps it to a point on the curve $C$, verifying that $f$ indeed maps $D$ to $C$.

This shows that for each point $(x, 0)$ in $D$, the image under $f$ lies on the curve $C$, satisfying the equation $X^2+Y^2=1$, establishing $f$ as a map of curves from $D$ to $C$.

[cloud walrus]

.

[rotund aurora]

[broken quartz]

What happened

[white oxide]

bruh wtf

is this for an abstract algebra course

[broken quartz]

well it's a bunch of things combined into one

so ig u could say that

[white oxide]

like what subjects

[limber wharf]

Looks like classic AG

[broken quartz]

a few r potential
topics include elliptic curves, Chebotarev’s density theorem, Hilbert’s irreducibility theorem, and Dirichlet
series.

basically number theory

[broken quartz]

yea basically

[white oxide]

oh damn

[broken quartz]

any of u guys know if what i did is relatively correct? as in the approach

[limber wharf]

Yeah I’m pretty sure it’s correct

[broken quartz]

thx

[limber wharf]

Well I would double check that they don’t expect some other condition when they say “map of curves”

I’m assuming it is just map between the curves that is a rational function in each coordinate

Or something like that

[rotund aurora]

It's more interesting to ask when is that map a bijection

But you should include the point at infinity

Which corresponds to (-1,0)

[broken quartz]

yup

tdy i learned about j-invariant. does anyone know the significance of 1728 in the formula

my prof said it can be used without the 1728 but he prefers it with the 1728 cuz there's historical value. idk what he meant by that

[white oxide]

is this a permutation of the sigmas since the sigmas are all the distinct embeddings of E into E^a and i'm assuming that they form a group but i'm too lazy to check

btw E^a refers to the algebraic closure of E, not sure if that's standard notation or not but anyways that's what it is!

[topaz solar]

Yes because it’s all the distinct embeddings but no it’s not because it’s a group

What would your composition even be

But E -> K -> E^a is composing embeddings yes?

But we already listed all of them

[white oxide]

idk tbh!

but yeah that makes sense

[buoyant echo]

ok I did some more reading. I guess this is how ideals came into being

So initially, there are rings such that factorizations that are not unique, for example in Z[sqrt(-5)], 6 = 23 = (1 + sqrt(-5))(1 - -sqrt(-5)),

Then after that ideals came into being to study some special properties of rings, and then from there it allowed us to study some rings in which factorizations are unique, for example principal ideal domain is also a UFD.

Is this correct?

[prisma ibex]

a little backwards but yeah mostly correct

a lot of the historical motivation for introducing ideals was noticing that certain rings that show up in number theory like Z[\sqrt{-5}] don't have unique factorization of elements, but you do get unique factorization of ideals in so called Dedekind domains (of which rings like Z[\sqrt{-5}] are an example)

there is something called the class group which measures how badly Dedekind domains fail to be PIDs, in other words how badly unique factorization of elements fails

historically people were trying to prove Fermat's last theorem and were able to in cases where Z[\zeta_n] was a unique factorization domain, and later Kummer was able to prove more cases by working with "ideal numbers" even in the cases where unique factorization failed, and this led Dedekind to give a more modern definition of ideals

[stark helm]

How can I find a set of homomorphism from Z to Z? Does it mean I need to give an example of a homomorphism?

[coral spindle]

Wym "a" set of homomorphisms

Here's a set consisting only of homomorphisms from Z to Z: {}

If you need to find the set of all homomorphisms, yes you will need to describe all of them.

I would suggest you think carefully about what elements of Z generate Z in order to find the homomorphisms.

[karmic moat]

for a ring R and an ideal I, $S^{-1}(R/I) = S^{-1}R / S^{-1} I$. does this also hold true for ideals $J \subseteq I$?

[cloud walrus]

ana(functor)mono(morphism)

[karmic moat]

as R-modules*

[next obsidian]

Wdym

Just pretend I was J and then you have the statement for J

[rocky cloak]

Localizing is an exact functor, so for any submodule M < N you have

S^-(N/M) = S^-N / S^-M

[karmic moat]

sick thanks :)

oh wait sorry i meant

does a similar result hold for I/J

i forgot to mention that

but i guess J < I as a submodule

so it works i think

[rocky cloak]

Yes, ideals are examples of modules

[karmic moat]

yea

ok thank u

:)

[smoky ivy]

how do i show that the commutator group of the upper triangular matrices are the unipotent matrices?

[coral spindle]

Unipotent or upper unitriangular? There are going to be plenty of unipotent matrices that aren't upper triangular.

You can fairly easily argue that any commutator is going to be unipotent. As for showing unitriangularity, you should be aware of some nice generators for that subgroup, and iirc you can exhibit specific commutators that produce those generators.

(I am half remembering this so I apologise if I'm making a mistake here)

[rocky cloak]

Guess just explicitly show that elementary (unipotent) matrices are commutators. Then show that they generate all the upper triangular unipotent matrices

[smoky ivy]

im actually kinda struggling on showing that any commutator is unipotent

or more like

if you consider some commutator ABA^{-1}B^{-1}, you just need to consider the ii-th entry, but that seems so tedious with writing out the matrix multiplication

is there another way to argue that

[smoky ivy]

oh nvm, im stupid

[white oxide]

i'm a little bit confused, why does this argument not work if k is infinite?

[south patrol]

Well, do you know why the multiplicative group of a finite field is cyclic?

The proof uses properties of finite groups etc

But in fact the multipplicative group of an infinite field is never cyclic, idk if that's what you're interested in

[white oxide]

i kinda forgot the proof lol, doesn't it involve the fundamental theorem of abelian groups

[south patrol]

That's one nice way

But it only works for f.g. abelian groups

Which is why it doesn't generalise

[white oxide]

ohh

[south patrol]

If you want a silly example, note that uh

every cyclic group is at most countably infinite

so we run into obvious issues for fields of uncountable cardinality

[south patrol]

But yeah in fact you can prove this

[elder wave]

in fact

[white oxide]

oh because it's just Z if it's infinite which is countable right

[crystal turtle]

in fact

[white oxide]

what's the fact, i can't stand the suspense

[south patrol]

😭

do i say that a lot

[elder wave]

in fact that might be the case

[south patrol]

i've said it four times in the last 24 hours tbf lol

on here

[rocky cloak]

This you can also use to show that any finite subgroup of the multiplicative group of a field is cyclic

[white oxide]

oh right, i faintly remember that from the proof

tbh im just gonna review group theory once this sem is over lol

[limber wharf]

It does, however fun fact is that you can prove it with just very basic group theory and some clever tricks

One of my favourite exercises in the text algebra chapter 0 has you do so

Cause it really just rests on the existence of an element of maximal order that every other order divides

Which you can prove for abelian groups cause then the order of elements multiplied which have coprime order is their order multiplied

[rocky cloak]

Any obvious reason why it's never cyclic?

[south patrol]

Well i guess just more generally, (non-zero) divisible abelian groups cannot be f.g.

Okay I guess they needn't be divisible because of torsion but the idea still works i'm sure

[rocky cloak]

g^0 = 1

[south patrol]

wait what did i have in mind uhh

Okay yeah i'm not sure sorry

I did this a while back

Although tbf, maybe one way to see is it like

[rocky cloak]

I guess it's easy to see that Q^* isn't cyclic, and similar for transcendent extension of Fp.

And infinite algebraic extensions contain torsion elements of arbitrary high degree

[south patrol]

Your field K is an extension of a finite field F, of cardinality > 2 (say)

so K^x contains F^x as a subgroup

we can't have the former infinite cyclic and the latter finite cyclic

I think that works nicely?

[rocky cloak]

Yeah, that's a pretty simple argument. And as always there's an extra thing to check in characteristic 2

[south patrol]

Im assuming every infinite char 2 field contains F4 hopefully lol

But maybe I'm wrong

[rocky cloak]

F2(x) doesn't

[south patrol]

Rip

[rocky cloak]

So you would have to check that seperately

[coral spindle]

Well it either contains F_4 or it contains some transcendental element, right?

[south patrol]

Yeah okay

Yes

If you have anything algebraic ur fine

What even is the cardinality of F2(x)

[rocky cloak]

It's countable

[south patrol]

Is it countably infinite

Yeah

[rocky cloak]

But you should be able to show it's equal to Z^N using that F2[x] is a UFD

[broken quartz]

can someone give me hint on how to solve part b)?

also, just to verify for part a) im justing plugging in 1,2,3,4 in that eqn and see if i get 1 - e^-1

[stark helm]

How can I find the set Hom(Z, Z/2Z), is it correct to write phi(x)=1 when x is odd, phi(x)=0 when x is even?

[barren sierra]

Hint: Z is generated by 1, so how does this uniquely define homomorphisms from Z to Z/2Z?

[barren sierra]

That'll help you make this idea more precise

[stark helm]

I consider the Z/2Z implies the result can only be 0 or 1, and by preservation of operation, if x is odd, then odd number of 1 is still odd. If x is even, then even number of 1 will produce 0

[barren sierra]

Make this more precise

Again, use the hint

(you're also missing a morphism)

(which would be apparent if you used my hint)

You're right that Z/2Z only contains 0 and 1

Also are you trying to find the set or the group structure on Hom(Z, Z/2Z)? If the latter, my hint is even more important to make this clear

Does my hint make sense at the least?

[stark helm]

uniquely define means?

[barren sierra]

If you have a group homomorphism f:Z -> G, G some abelian group

if you know f(1), can you determine f(2) in terms of f(1)?

[stark helm]

by saying f(1+1)=f(1)+f(1)=2f(1)

[barren sierra]

bingo

hence, group homomorphisms from Z -> G, G some abelian group, are uniquely determined by where 1 gets sent

And like you said, Z/2Z only has 2 elements

Sooooooo can't be too much work

[stark helm]

Do you have any ideas about all homomorphism from a symmetric group like S5 to a Z/5Z? In my impression symmetric group must be a cycle, so I don't know why it can associate with the mod?

[dim widget]

There is no such homomorphism

Except for the trivial one

[broken quartz]

anyone iwlling to give me a hint for part b)

[stark helm]

You mean like from identity mapping to 0?

[south patrol]

a fancy way is that every element of A_5 is a commutator (i.e. of the form aba^-1 b^-1 for some elements a and b) and so it must go to 0 in Z/5

[south patrol]

constant mapping yes

[dim widget]

Not every element

[south patrol]

wdym by "symmetric group must be a cycle"?

[dim widget]

Every other element

[south patrol]

sorry yes

A5

But in fact this is besides the point anyway in a sense

The normal subgroups of S5 are just A5, 1 and S5

The kernel can't be 1, so it contains A5, so our map factors as S_5 -> S_5/A_5 -> Z/5

[stark helm]

like I mean S5 can contain (123), (12)(34) something like these cycles

[south patrol]

but then there are no non-trivial maps S_5/A_5 -> Z/5

[south patrol]

Sure, every permutation can be written as a product of disjoint cycles

(just see where each element goes and that gives you cycles)

[stark helm]

You mean every element has its inverse and it will goes to e?

[dim widget]

Are you in an algebra class?

[stark helm]

Yes

[dim widget]

You seem to be really confused about the difference between the presentation of a group and the group itself

Just because the description of S_5 as a set of permutations doesn’t look like it has anything to do with modular arithmetic doesn’t mean there can’t be maps between them

[stark helm]

Can you give me an example about what you are saying, I think f(sigma=(123))=2 something in this form can work, but besides constant mapping, i can not imagine f( sigma1*sigma2)=e because of operation preservation.

[south patrol]

not really sure what you mean

[south patrol]

This is kinda hard to read too

[dim widget]

(123) can’t map to 2 because they have different orders (3 vs 5)

[next obsidian]

It’s a little more than just them being different

[south patrol]

(also, it's a bit (very) non-standard to use e for the identity element of Z/5, please just say 0 (or 0 + 5Z if you are one of those people))

[faint fractal]

he just like me fr

[dim widget]

As I said before there are no non-trivial maps to Z/5 (all maps factor through the trivial groups) but to prove this you are going to have to review a lot of basic ideas from group theory

[limber wharf]

Note that the image of an element under a morphism must have order dividing the order of the original element, and every element of Z/5 has order 1 or 5

Then consider how you can write any element in S_5 as a product of transpositions (which have order 2)

[cloud walrus]

.

[broken quartz]

im not sure what to do after this lol

[white oxide]

how exactly does $[k(\alpha + c\beta): k] \geq n$ imply that $k(\alpha, \beta) = k(\alpha + c\beta)$?

[cloud walrus]

okeyokay

[broken quartz]

is this ivan niven's intro to numb theory

[white oxide]

No lol this is Lang

[broken quartz]

ah i thought i recognized the font

[white oxide]

oh nah

having a brain fart, is the kernel nontrivial since q is sent to 0

okay yeah

[broken quartz]

yeah

[white oxide]

lol thanks

how exactly are these two paragraphs distinct? if splitting fields are unique up to isomorphism and we just showed that the splitting field of f(X) = X^p^n - X is the set of roots of f(X) in the first paragraph, what's the point of the second paragraph?

well i guess it's showing that it's a field and not a group

but like

the first paragraph shows that the splitting field for f(X) over Z/pZ is the set of roots of f(X), which must be unique up to isomorphism

[next obsidian]

@white oxide honestly idk what two paragraphs you mean but

Something that is being shown after the underlined thing is that the splitting field is the set of roots of the polynomial

It’s not true usually that a splitting field consists only of roots

See eg C as a splitting field of x^2 + 1 over R

[white oxide]

ye but didn't they show that (in the blue brackets) F is a splitting field for f, and every element is a root of X^p^n - X already

and since splitting fields are unique up to isomorphism it follows

[next obsidian]

Idk what F is

[white oxide]

a finite field of order q where q = p^n

[next obsidian]

You don’t know it exists

Until the latter part

The proof goes

Given a finite field of order q, it would be a splitting field of F, so finite fields of order q are unique

Now here is an example of a finite field of order q

[white oxide]

huh okay

i think i see

thank

s

[safe sedge]

Helo can someone tell me if a = 0000 0000 0000 0010 is considered a primitive element corresponing to x of GF(2^16) where the ireducible polynomial is 1,0,0,0,1,0,0,0,0,0,0,0,0,1,0,1,1 = 210013(in octal)

Just needing one alfa like the definition in GF(2^16) to implement some algorithm

[white oxide]

how exactly does this imply that d \geq n?

[topaz solar]

So, 1 <= d <= n since this is the order of phi ye?

[white oxide]

ye i got that

[topaz solar]

p^d <= p^n

So if d < n we have p^d < p^n because finite arithmetic

But q=p^n right?

[white oxide]

yea

oh

ohh ok i see so every element of F_p^n is an root of X^p^d - X, so if d < n then we wouldn't have every element being a root of the polynomial right

hence d \geq n

[topaz solar]

Ye

[white oxide]

cool

thanks sharp

[safe sedge]

help 😢

[white oxide]

aren't the homomorphisms just in bijective correspondence with the permutation of the 9th roots of unity which are roots of X^6 + X^3 + 1

[topaz solar]

Well, pick x a root, then sigma(x)^6 + sigma(x)^3 + 1 = 0

But if you can reach any other root from \alpha you gotta make sure those respective polynomials hold too ofc

[topaz solar]

Would you happen to know which roots of unity that includes?

[white oxide]

i'm not entirely sure, right now i'm just brute forcing

because i know that yea the homomorphisms have to send alpha to a conjugate

and that it divides X^9 - 1, so in particular all zeros of f(x) are going to be zeros of X^9 - 1 which are 9th roots of unity

[white oxide]

idk maybe the primitive ones or some shit lol

[topaz solar]

Well if x is a primitive, then you know it should map to a primitive, that sort of thing

[white oxide]

oh right generator maps to generator

[topaz solar]

That’s what I meant by “respective polynomial”

[white oxide]

huh okay

so it would be nice it alpha turned out to be primitive.....

nice and convenient...

[topaz solar]

Well, alpha sure isn’t 1

And it’ll be primitive when it’s not actually a 3rd root of unity or something right?

[white oxide]

where'd you get this

[topaz solar]

Well, primitive means it generates yeah?

9/3 = 3

[white oxide]

i kinda see?

[topaz solar]

If it’s a 3rd root it ain’t generate it

Since it’s order is 3

Not 9

[topaz solar]

Cyclic group

[frigid lark]

Question 5a from chapter 4

Also use the fact that there are 6 primitive 9'th roots of unity

[white oxide]

Damn bro memorized the question number

[broken quartz]

😭 can sum1 pls give me a hint. actually ik kind of what to do but idk how to apply it or prove the limt

oops i deleted the question

part b)

[tribal moss]

It is primitive iff its order is 2^16 - 1 = 65535.
Since the prime factorization of 65535 = 3·5·17·257, compute x^(65535/d) for d in {3,5,17,257}. If you get something different from 1 in each case, the order must be 65535, and x is a primitive element.

[broken quartz]

Anyone?

[buoyant echo]

Hi all, can anyone shed me some lights that the polynomial (x^3 - 2) is irreducible over Q imply sqrt[3]{2} is inconstructible?

[daring nova]

You mean a number that can be found as a length given a ruler and a compass ?

If so, iirc (from a 12 page pdf I read like 6 months ago) a number is constructible iff it belongs to an extension of Q that is a power of 2

https://artofproblemsolving.com/wiki/index.php/Constructible_number

i.e. its minimal polynomial's degree is a power of 2

[daring nova]

That's not the case for cbrt(2), whose minimal polynomial is obviously cubic

[buoyant echo]

yes

[buoyant echo]

so Q(cbrt(2)) is not a field extension of Q of dimension 2^k right?

[daring nova]

No, because it's 3 dimensional, Q, Q cbrt(2), Q cbrt(2)^3

Hence it's not 2^k dimensional

[grizzled crane]

How we do find the number of homomorphisms from G to G' ?

[coral spindle]

By thinking carefully about the structure of the groups and trying to find ways of enumerating them.

Without more information about the groups, this question cannot be answered. Ask the actual question at hand.

[grizzled crane]

from Z to 2Z, is there an isomorphism other than f(n)=2n?

[coral spindle]

There are many.

Oh, isomorphism.

There are indeed more.

Hint: the generator of Z must be sent to a generator of 2Z.

[grizzled crane]

ie 1 maps to 2 and -1 maps to -2

[coral spindle]

And what's the other possibility?

N.b. 1 mapping to 2 implies -1 maps to -2; you need only specify that 1 maps to 2.

[grizzled crane]

f(n)=n+1, here 1 maps to 2, but -1 maps to 0

[coral spindle]

And that's a homomorphism?

You sure about that?

[grizzled crane]

mb, it's not.

let me think

[coral spindle]

Think about this. What are generators of 2Z?

[grizzled crane]

2 and -2

[coral spindle]

So if we're going to have an isomorphism, 1 must be sent to a generator of 2Z. So what can the isomorphism do?

[grizzled crane]

we require f(1)=2 and f(-1)=-2

[coral spindle]

No.

[coral spindle]

You've forgotten this point again

Try putting this information together:

• A generator of Z – let's choose 1 – must be sent to a generator of 2Z.
• The generators of 2Z are 2 and -2
• A fact I have not told you yet: a homomorphism is determined entirely by where it sends generators.

No, that's incorrect.

I've laid out the work you need to do so I'm going to leave it to you now.

[grizzled crane]

sure

[coral spindle]

You don't have to delete your messages

It just makes it look like I'm talking to myself

[grizzled crane]

i just realized it was incorrect

@coral spindle just to make sure, will get a isomorphism which is a explicit function like f(n)=2n

[coral spindle]

You will.

[grizzled crane]

@coral spindle

tbh i can only think f(n)=2n

i don't know much expression having the property f(a+b)=f(a)+f(b)

i can only think of n(a+b)=na+nb

[tardy hedge]

how to characterize what structure is lost in a homomorphism?

[coral spindle]

The kernel

[glacial island]

Hi, I'm in twelfth grade and have 7 hours of math every week. Since I'm in my last year, we need to do a big research competence.

We need to research the applications of Conway-groups in other fields of mathematics, but me and my team can't seem to comprehend what those groups actually are. Most of the articles we read explain it with words we've never seen in our life and we're beginning to think this is way to advanced for us. Could someone explain what those groups are or send some links to articles that explain it clearly?

[coral spindle]

That is an ambitious project...

I will be very clear in saying that you will likely be unable to understand what the groups are in time. Perhaps if you had a couple of years you would.

You might have to simply look around the subject to see what people have said.

I'm not aware of any articles but I'll have a look around

[grizzled crane]

f(n)=-2n will work right ?

just flipping where the generators

[peak root]

Maybe you could try watching 3blue1brown's videos on groups, try searching groups in his channel.

[coral spindle]

3b1b's video on the monster is a pretty good start

[grizzled crane]

can you tell your function?

[coral spindle]

I think that was his 'huge favourite number' video or sth like that

[peak root]

His videos usually target a general audience, so I suppose that is a start

[coral spindle]

You found it.

[grizzled crane]

is that all ?

[coral spindle]

What do you think?

[coral spindle]

Remember what I said up here.

[glacial island]

do you coincidentally have some other things of john conway in mind which are a little easier for us to research?

[coral spindle]

I don't think researching other things john conway has done will be particularly helpful.

He was a very prolific mathematician.

[grizzled crane]

Answer is no,?

[rocky cloak]

the definition of surreal numbers and how it explains values of hackenbush/other games

thats pretty aproachable I think

[coral spindle]

Oh, unless you're just looking for a different topic. Yeah as jagr alluded to, his game theory is great

[rocky cloak]

Winning ways for your mathematical plays is a nice book(s)

[coral spindle]

If you can find a copy :,)

[rocky cloak]

Are they hard to come by?

[coral spindle]

Very very pricey. I don't think my uni library has copies lol

[grizzled crane]

My guess is as you mentioned for any isomorphism, it should map generator to generator, so given 2 generators i have only 2 ways to do so in a bijective fashion

[glacial island]

we originally planned to research his game of life, but when my teacher saw the word "game" we weren't allowed to, because "we're not supposed to explain games". I don't even think she even knew what is was

[coral spindle]

Sigh

You could look at that famous knot theory argument of his

his follow-then-swallow argument

There's a video of him explaining it online

I think that has significant approachability and merit generally

[rocky cloak]

Might even be worth trying to explain to your teacher that game theory is unrelated to video games. though I guess picking fights with teachers isnt always a good idea

[glacial island]

yup, especially with her

[coral spindle]

This is getting off topic so I won't say more, but here is the video of the argument: https://youtu.be/lwWeRMmXIoU?si=-uLfExCR72IJsLQu

@glacial island

[round hull]

sanity check

[void cosmos]

failed

[round hull]

if A is not an integral domain, then modules over A cannot be torsion-free

[next obsidian]

False

[void cosmos]

no

[next obsidian]

Torsion is only defined against non-zero divisors

[void cosmos]

all modules of R are torsion free does not imply R is an integral domain

[next obsidian]

Any ring is torsion free over itself

[void cosmos]

this is true only

for free modules

[void cosmos]

this.

or no i mean if R is an integral domain then the free modules are torsion free

[next obsidian]

Free modules are always torsion free

[coral spindle]

Probably talking past each other here, perhaps there's confusion between torsion = Z-torsion = finite order, and torsion as informed by the ring.

E.g. the Z/6Z-module Z/6Z is not torsion via the second definition :)

[void cosmos]

yeah

[void cosmos]

true

[round hull]

i did mean torsion informed by the ring

[round hull]

ok i remembered the definition wrong then

[buoyant echo]

hi all, for the problem of impossibility of trisecting an angel, for the case cos3x = 4cos^3x - 3cosx , if we use x = 20, we eventually get an equation y^3 - 3y - 1 = 0, this equation does not have rational roots, why does not having rational roots imply the trisection is not possible?

[coral spindle]

It doesn't.

[buoyant echo]

the proof says if we can show the equation above doesnt have rational roots, then we can say construction of 20 degree is not possible, or am I missing something ?

[coral spindle]

Considering we can construct a great many irrational numbers in Euclidean geometry, this is insufficient.

[rocky cloak]

So a cubic polynomial is irreducible iff it doesn't have a root, because of it was reducible it would be a product of a degree 2 and a degree 1 polynomial, and degree 1 polynomials have roots.

So y^3 - 3y - 1 is an irreducible polynomial that has 2cos(20) as a root. Thus Q(cos(20)) is a degree 3 extension. But extensions by constructable numbers always has degree 2^n. Hence it is not possible to construct cos(20) with compass and ruler.

Since we can construct a 60 degree angle, if it was possible to trisect angles we would be able to construct cos(20), but since we can't construct cos(20) it must be that we can't trisect angles.

[tardy hedge]

Poo

Anyone agree

[crystal turtle]

[summer path]

if this is your attempt at shitposting, you're in the wrong channel

[untold basalt]

How can I find an automorphism of Z[x] that maps these two polynomials into one another?

[glossy crag]

Eisenstein is not directly applicable lol, what prime would you use?

[crystal turtle]

Brainrot

[glossy crag]

I'm not sure why you want to map these polynomials onto one another, since the exercise (as far as I can tell) is asking you to prove they're irreducible via Eisenstein. Now in order to apply the criterion to either of them you probably need some kind of linear substitution, i.e. x->x+a for some a\in Z, e.g. how it's shown cyclotomic polynomials are irreducible via Eisenstein with the x->x+1 substitution. Try +-1, +-2, it'll probably work.

[crystal turtle]

Ignore me and listen to Ocean Man

[untold basalt]

allright thanks

[karmic moat]

i luv algebra

[dim widget]

This is debatable

[next obsidian]

[dim widget]

People have different definitions of torsion, a very reasonable one has non domains having torsion

This would then not agree with Tor but would agree with some other uses of the word torsion in the literature

[south patrol]

which other meanings do you have in mind?

[dim widget]

I guess I would say I-torsion means that M[I] != 0

[rocky cloak]

An element m in M being I-torsion if Rm = R/I seems reasonable

[dim widget]

What Jagr said but in the setting of the entire module

[south patrol]

sure

torsion means the connection has this property that

jk

[wraith oak]

The way I showed that (0_r, 0_s) is in I x J, does it make sense?

[next obsidian]

Yes :)

[round hull]

(the hint gives away the entire answer but) just making sure, we only need o to be dedekind to make sure that o_p is always principal, right

also i get the idea is to break M down into each prime, but how do we come up with the "There exists c_p in o \ p which pushes the image into F"

everything fits together really nicely but i'm not enlightened

[round hull]

i guess it's a general strategy for maps that go into bigger things than what you want

ideally we want g_p(M) to be in F, which would just work immediately as a candidate for g. so we can put some constants to make that happen (since f.g.), which will necessarily be in o \ p, and since prime ideals are cofinal this generates the unit ideal

(?)

it's like a messed up direct sum

[tardy hedge]

I dont understand the part ablut “so the primitive nth roots of unity are the generators of this subgroup”

[white oxide]

well by definition a primitive nth root of unity z is a nth root of unity such that n is the smallest integer such that z^n = 1

can you see how that relates to cyclic groups and generators?

[topaz solar]

Look at the unit circle in C

[tardy hedge]

Oh cause the group is order n and the order of z is n

So it gonna pass thru every other element

[white oxide]

ye

[tardy hedge]

Ya prob some pictures with the angles r smth right

Splitting up unit circle into pizza

[stable thistle]

there's an algebraic structure i'm considering, but i'm not sure what's the best way to represent it so that it's closest to actual things that people do mathematics on...

it's basically a group in that it supports all of the group axioms, but it has an extra structure on it

it's a group supplemented with "magnitude" or "numeracy" of its elements...

i think of it as multisets of elements of a certain group, and there are operations between these multisets that correspond to multiplying every element of the multiset by a certain element, but only a certain multiplicity of times so that only n of each element is affected

there's also a hangup that the identity element behaves a bit differently, it has an uncounted multiplicity, or an infinite multiplicity, such that the amount of elements never runs out

so here's an example, using the group C3
we have the multiset {a, a, a^2, a^2, a^2} and we apply the operation a with magnitude 3 to it, and get {a, a, a, a^2, a^2}

or let's take the group of integers under addition and the multiset where an integer's multiplicity is itself {1, 2, 2, 3, 3, 3, 4, 4, 4, 4...}, and apply the operation +1 with a multiplicity of 1, which... oh lol that's the same as the identity map, um, let's say it has multiplicity 2, which i think just switches the multiplicities of 1 and 2

uh anyways, there's got to be a better way to represent this, and it seems like such a simple object that i doubt i'm the first to imagine it... what's a better way to represent this to do group theory on it?

[topaz solar]

I don’t get why +1 would be the identity element

Also can you give a precise definition

[topaz solar]

Are you sure you did the C3 example right?

[stable thistle]

it's just an artefact of the infinite set

[round hull]

like the +1 brings 0 to 1

[stable thistle]

yeah

[round hull]

oh i get it finally

[topaz solar]

There isn’t a 0 though?

[round hull]

it's there but infinitely many

[stable thistle]

there is

yeah

the identity element has infinite multiplicity

[topaz solar]

That’s weird

[stable thistle]

it's like, the null element

[topaz solar]

Yeah still odd

[stable thistle]

if you "lose" some of an element it "actually" becomes the identity

[round hull]

does this satisfy the group axioms btw

[topaz solar]

Reminds me of convolution-y but scuffed

[stable thistle]

yeah

[topaz solar]

I doubt it should since the multiplicity of the applied action or such

[stable thistle]

oh no i don't think the multisets are elements of a group

but the operations have to be an application of a group element

i guess it's possible that it could reduce to a single group but i would be somewhat surprised

[round hull]

do you take the underlying group to be abelian

[stable thistle]

no

if a multiset has elements m and g and you apply the operation corresponding to g you evaluate m * g and g * g

[round hull]

what aspect are you trying to study?

[stable thistle]

it's technically describing the behavior of a thing i made up while half-asleep but the algebraic structure is interesting on its own, what i was first thinking to explore though was the analogues of easy group theorems like the relationship of groups and their subgroups in terms of what they do in the multiset group thing and similar questions

[round hull]

it's not associative: ({1, 1} applied to {1}) applied to {1, 2, 2} = {1, 1, 2} applied to {1, 2, 2} = {1, 1, 2, 2, 3, 3, 3, 4}, while {1, 1} applied to ({1} applied to {1, 2, 2}) = {1, 1} applied to {1, 2, 3, 3} = {1, 1, 2, 3, 4, 4} (this is in Z)

i don't think it's cancellative either

[stable thistle]

ooh right

[round hull]

actually i think it is cancellative it's not cancellative by your earlier example

[stable thistle]

shouldn't +1 and +0 having the same effect mean it isn't a group though? since a group can't have distinct identity elements? or do you not have to consider them different

[round hull]

it's not a group because it's not associative

also +1 only does nothing to {1, 2, 2, 3, 3, 3, 4, ... }

an identity element needs to do nothing to everything in the set

[stable thistle]

right duh

+1 doing nothing in that case is pretty interesting though, i'm sure there's other unusual behavior with infinite multisets

[round hull]

some actions don't have fixed multisets at all

like {1, 2} (or {n, 2n}) in Z

[stable thistle]

what do you mean?