groups-rings-fields

[restive star]

nah cause it isn't necessarily a subgroup of S4

the proof assumes it divides 24

[hidden haven]

An automorphism of the splitting field of a polynomial can be viewed as a permutation of the roots of the polynomial

thats an injective homomorphism from Gal(splitting field/Q) to Sym(set of roots)

because if 2 automorphisms agree on the roots of f, and the roots of f generate the extension, then they agree everywhere

and a root can only go to another root

[restive star]

https://math.stackexchange.com/questions/315413/galois-group-as-a-subgroup-of-s-n I hope you're right lol. Here's where I found someone saying something different

[hidden haven]

The answer that says that f needs to be irreducible has comments correcting it

[restive star]

oh shit sorry lmao

that's why proof by math stack exchange prob isn't the best method

[hidden haven]

he says that if polynomial is not irreducible then it would have repeated roots (which you may have with irreducible f anyway) but that is not really a problem

[hidden haven]

lol

[restive star]

proof by "but it looks legit!"

[hidden haven]

is there are repeated roots you get galois group as a subgroup of S_k for a smaller k but that obviously embeds into S_n

in this case the only way you could have a repeated root is if r = s in the notation you were using, or any of them is 1

neither presents an issue

[restive star]

Ah great! Thanks for clearing that up

[hidden haven]

np

[vast quiver]

yup

you definitely know fields are integral domains

and for the other direction, you need to show that every nonzero element x has an inverse

you want to find its inverse

keep multiplying x with itself, eventually you gotta get 1

otherwise you contradict the ring being finite or that the ring is an integral domain

[inner acorn]

When it says the "set of pairs {i, j}"

I'm guessing it's implied {i, j} = {j, i}

Drawing up a rough solution, it would look something like this right?

[cloud walrus]

mudhi

[delicate bloom]

no because there is no x since it's being quotiented out

technically we probably should simplify it further but fractions are fine

-1 = 2 mod 3 so really it's just plain alpha^2

since Z_3 is a field you can always divide by nonzero elements to get 'fractions'

[hidden haven]

yes

and merosity is saying that -alpha^2/2 is equal to that, so your original thing is correct

and about fractions, they are just shorthand for inverses

1/2 is notation for 2^{-1}

and you know that 2^{-1} is 2 itself in Z/3Z

[languid meteor]

Hey could anyone point me in the right direction with this problem? If I let $\sigma$ be the image of $g$ in $S_n$ then $\sigma$ can be written as a product of disjoint cycles (which are commutative). However, im unable to show that I can write this as a product of specifically n/d, d-cycles. I'm trying to relate the order of $\sigma$ (which is d) to the fact that a d-cycle has an order d but im not really getting anywhere

[cloud walrus]

Saintscratch

[hidden haven]

By Cayley embedding do you mean some particular embedding, like something the book has defined?

Because if you take the embedding that is defined by the action of G on itself by left multiplication then it is false

[languid meteor]

I assume G is the subgroup of Sn that some group of order n is isomorphic to

I think we used the right multiplication definition to define the embedding

dont know if that changes much

[hidden haven]

Then it seems false, because that's a transitive action

Actually I think my reasoning is wrong

Nvm

[languid meteor]

the fact that it's a cayley embedding must be the trick, im trying to find some fact about the decomposition of a group element of order d into a product of elements and try imagine the images of those elements in G

[hidden haven]

Yeah I think I got it

So the image of an element g under the embedding tells you what multiplication by g looks like in G

Because it tells you how right multiplication by g permutes G

[carmine fossil]

Find an element g of order d

[hidden haven]

Pick an element, say identity, of G, then look at its orbit

[carmine fossil]

||You can divide the group into sets {Ga_1,Ga_2...},where G=<g>||

[hidden haven]

What would be the size of the orbit?

Buncho spoilers

[languid meteor]

1, there should only be g right?

or 2 if you include the identity actually

[hidden haven]

It would also contain g^2

Orbit means set of all elements where your element could go under the action of g

And by action of g we really mean action of subgroup generated by g

So it will be {1, g, g², ..., g^{d-1}} where d is the order of g

Do you agree with this?

[languid meteor]

looking at this definition, when you say I take the identity im assuming you mean fix x = e. Then O_e = G right?

[hidden haven]

It would be if you were considering the G-orbit

I'm taking the <g>-orbit

[languid meteor]

ah ha, gotcha

[hidden haven]

And the orbit just becomes <g> itself

[languid meteor]

right, that makes sense

[hidden haven]

So on this orbit, how is g permuting elements?

As in, can you see what the cycle decomposition is?

[languid meteor]

something like (1,2,3,...,d)

if I label the elements (n,1,2,3,...d-1)

[hidden haven]

Yeah, which is exactly why you wanna take <g> orbits

Now G is partitioned into <g> orbits, we've found one orbit of size d, can you say anything about the other orbits?

Try to relate other orbits to thee orbit of identity somehow

[daring ibex]

moldi

i have a question

[hidden haven]

sup

[daring ibex]

Doesn't this mean that A(p) is union of cyclic subgroups of order p^k where k \in N

[hidden haven]

I thought it'd be bs but actual algebra

[daring ibex]

But union of subgroups is a group iff they are contained in each other so wadda fuk

[daring ibex]

[hidden haven]

This is not true

Union of all subgroups of a group is always the group itself

[daring ibex]

like A(p) = set of all elements k such that p^n k = 0, and so p^n (k + k) = 0

[daring ibex]

hmm

[languid meteor]

If I fix s in G, then the <g>-orbit will be {s,sg,...sg^(d-1)} which is an orbit of size d. This works for any s in G so I have n orbits each of size d

[hidden haven]

Exactly

[daring ibex]

I mean

Hm

[hidden haven]

And on each orbit g is just acting like a d cycle

So you get the decomposition

[daring ibex]

hrummm

[daring ibex]

i feel like this is untrue

Like

[hidden haven]

Can you stop being a Minecraft villager for a sec

[daring ibex]

if you consider proper subgroups

[daring ibex]

[hidden haven]

Take a non cyclic group. Then every g in it is contained in <g> which is proper

[daring ibex]

oh yeah

fair

I'm being sully

[chilly ocean]

This is only true for 2 groups, not necessarily for more

[languid meteor]

are we saying that the image of g in Sn is defined by its action on the other elements of G? So that I can split g up into how it acts on each element of G, which happens to be a bunch of disjoint d-cycles

[daring ibex]

Ah i see

[hidden haven]

Yeah that's the Cayley embedding

[daring ibex]

That makes more sense then

[hidden haven]

Doesn't my reasoning for why it's false work for non cyclic 2 groups

[languid meteor]

hmm that makes a lot of sense, ill have to go think about that. Our lecturer showed us cayley embeddings without using the word orbits haha

thanks a million for all the help

[hidden haven]

Ah I see, np

[chilly ocean]

not sure what you mean. If you have 2 groups their union group iff contained. but if you have more their union can be group without contained i think

[hidden haven]

Oh shit

I read it as 2-groups

Lmao

[chilly ocean]

oh lol

[carmine fossil]

I don't remember if 14 is related

But w/e

[warm holly]

if we suppose that $|G:H|=2$, does the fact that $\forall g \in G$, $g^2\in H$ follows from : $G/H$ is 2-torsion so isomorphic to $Z/2Z$ and contains $eH$ and $gH$. But, as $gH$ is of order 2, then $g^2\in H$ from definition of quotient gpe? And as $gH=g'H \ \forall g' \in G$ the $g^2 \in H \ \forall g$

[cloud walrus]

Dаniil

[carmine fossil]

Yes

[warm holly]

thx .)

[rustic crown]

(the last line isn't correct... gH isn't g'H for any g'... this is just saying all cosets are same so [G:H] = 1)

(if you mean g' = g^-1, then you shouldn't be saying for all g' in G? more like for all g in G, its correct but weird?)

[novel parrot]

Hi, i have some confusion with seperable polynomials, I have seen the fact that every polynomial over a field of characteristic 0 is separable. But (x-2)^2 does not have distinct roots and this contradicts the definition of separability?

[hidden haven]

Every irreducible polynomial over a characteristic 0 field is separable

[novel parrot]

Oh right! My bad. Thank you!

[rustic crown]

(there are some weird people who define a polynomial to be separable if all its irreducible factors have no repeated roots)

[chilly ocean]

then constant polynomials are separable

[old lava]

is that so wrong?

[rustic crown]

yes every polynomial over char 0 becomes separable

[old lava]

I thought it's this + no irreducible factor is repeated

[rustic crown]

i don't like it... (but my prof did )
for him x^p - a is separable over k(a^1/p) where a is not a pth power in k and char k = p

[rustic crown]

yea this is the usual definition

this is nice because it doesn't depend where the polynomial lives. and you have nice criterion gcd(f', f) = 1

[old lava]

ya, I like the gcd(f', f') = 1 criterion

it's nice

[chilly ocean]

it s very uh gucci

[rustic crown]

but you almost never use it lol, i only ever used it when f is irreducible

but there both definitions agree we just have f' = 0 to be an easier criterion

[old lava]

I only use it rarely when it's like obvious

often times when 0 is a root of the derivative or some shit

which works out nicely

[warm holly]

But as G/H is of order 2 then there are only 2 cosets: eH and gH but as we don't know the order of G, then for each g' in G/{e} gH=g'H, it is true, no?

[rustic crown]

if g' in H then g'H = eH

[sinful mirage]

on the space of functions on a group(which can be made into a vector space using pointwise multiplication and addition),we can introduce two structures:the convolution and the inner product

are they related somehow? or they are two different structures?

first one is the convolution,the second one is the inner product

[chilly ocean]

What kind of functions? groups dont have a pointwise addition and multiplication

[sinful mirage]

any functions defined on the group

let f:G->C, (f_1+f_2)(g):=f_1 (g)+_{complex numbers} f_2(g)

yes

G is finite

sloppy physicist not precising things

so G is finite,the above 2 products are defined and my question is whether there is a relation between the two structures

[sinful mirage]

this makes the space of functions on the group into a vector space, together with a multiplication (lambdaf)(g)=lambda_{complex numbers} f(g)

we can equip this vector space with a convoltion,i.e. first picture(which eats two functions and spits out a function)->this makes the vector space into an algebra

and we can put an inner product on the vector space, as the second picture above

is there any relation between the inne rproduct and the convolution?

or they are separate structures

yes i'm not sure if there is i was just curious cause in mathematics we usually put structures compatible with each other

(idk what compatibility in this case would mean)

yes,that makes the vector space into algebra

but the vector space a priori does not have an inner product

what my concern/doubt was, whether we need specify an inner product

or the convolution immediately induces one

[sinful mirage]

idk,that's my question if this even makes sense

or maybe this question is pointless(idk)

i'm new to abstract math(physicist here)

right that I see

that follows from the convolution

no?

the inner product is important for the space of class functions(orthogonality relations)

but idk if has relation to convolution at all

okay,so there is no relation between the inner product and convolution product

ohhh i see something

but it's ill defined

if we evaluate the convolution at identity,we get 1/|G| times the inner product

it's not exactly the same

i'll double check if the definition of inner product is correct

or def. of convolution

they don't seem to match,the definition seem to be right

now i'm confused how these two are compatible,if there's a factor difference

but there's some factor not working out

is that not a problem?

nothing

just curiosity,filling in extra interesting details in my lecture

[sinful mirage]

I am trying to prove associativity,but I fail,can someone help a bit please?
$((f_1 * f_2)f_3)(g)=\sum_{h} (f_1f_2)(h) f_3(h^{-1}g)=\sum_{h} \left( \sum_{h'} f_1(h') f2(h'^{-1} h) \right) f_3(h^{-1}g)=\sum_{h} \sum_{h'} (f_1(h') f_2(h'^{-1} h) ) f_{3}(h^{-1}g)$

[cloud walrus]

ProphetX

[sinful mirage]

while $(f_1*(f_2f_3))(g)=\sum_{h} f_{1}(h) (f_2f_3)(h^{-1}g)=\sum_{h} f_{1}(h) \sum{h'} f_{2}(h') f_{3}(h'^{-1} h^{-1}g)=\sum_{h} \sum_{h'} f_{1}(h) (f_{2}(h') f_3(h'^{-1} h^{-1}g))$

[cloud walrus]

ProphetX

[sinful mirage]

they do nott seem be equal

i tried renaming h->h'h,but it didn't work

[delicate hawk]

what is a basis of C[x] / <x^3> ?

every element in this ring is written in the form

a + bx + cx^2 such that a, b, c \in C

does that just make {a, b, c} a basis?

[chilly ocean]

Rather {1,x,x^2} I think.

[chilly ocean]

and because every element can be written in unique way*

[deft glade]

Wait going off of that question, Q[x]/<a,x> is just Q by that logic right? if a in Q is gonna be the basis?

[hidden haven]

If a is a non zero element of Q, that ideal would just be all of Q[x] so the quotient would be the trivial ring

[deft glade]

hmm ok that makes sense

[prisma ibex]

yes

[hidden haven]

prankd

[chilly ocean]

What is the free object in the category of commutative rings? (i think i know the answer just want to confirm)

[hidden haven]

Z[S]

Where S is the set

[chilly ocean]

as in Z-polynomials with variables indexed by S right

[hidden haven]

Yeah

[chilly ocean]

[chilly ocean]

what is R?

[rough lance]

oh the real numbers

[chilly ocean]

but yeah

<x> is a bigger ideal that contains <x²> and isn't the whole ring so <x²> is not maximal

[rough lance]

Oh I see, and that's because x^2 is reducible right?

[chilly ocean]

@chilly ocean Can I dm you?

you can

[chilly ocean]

since x² is reducible, it is the product of 2 non constant polynomials. since they are not constant, the ideal generated by either of them is not the whole ring, but its bigger ideal than original one bla bla hope you get the idea

[rough lance]

Yes, that makes sense. Thanks a bunch

[hidden haven]

Why are people deleting questions after they're answered

[green scroll]

very suspicious

[proud bear]

[sinful mirage]

why is $\delta_{g}(h)$ a basis on the vector space of functions on a finite group?

[cloud walrus]

ProphetX

[sinful mirage]

where $\delta_{g}(h)=g$ if $g=h$, 0 otherwise

[cloud walrus]

ProphetX

[sinful mirage]

how does this guarantee that the dimension of the vector space is the same as the dimension of the group?

i don't even see how $\delta_{g}(h)$ maps $G \to \mathbb{C}$

[cloud walrus]

ProphetX

[chilly ocean]

cuz you can write any function from G to C as linear combination of them

uniquely

ur basically evaluating at each point

[sinful mirage]

how do they map G->C?

what is $g$ in the def of $\delta_{g}(h)$?

[cloud walrus]

ProphetX

[sinful mirage]

cause if g is a group element,then how is the image in C?

i'm confused about the notation

[chilly ocean]

cuz ur working with functions from G to C?

[sinful mirage]

in order to represent functions from G to C,my basis needs to be functions from G to C

or?

[chilly ocean]

yea?

I would think delta_g should map g to 1

oh yea

ofc

[sinful mirage]

and i can't see how this is a function to C

\delta_{g}(h)=1 if g=h,0 oterhwise?

so delta_{g}(h) not equal g

yikes

my prof wrote g

[chilly ocean]

Is this in a book?

im so used to delta notation i didnt even look at the definition lol

Oh

[sinful mirage]

lecture only

and if it maps to 1 or 0,how can we prove that any function can be written as a linear comb of these?

i.e. i can't see that the number of basis vectors would be equal to the number of elements of the group

[chilly ocean]

I think there are related things you do in homological algebra or something, where you consider the set of homomorphism into G from some other group, but presumably this is unrelated to what you are doing

[sinful mirage]

here we only have the functions on a group,i .e. $Fun_{g}:={f| f:G \mapsto \mathbb{C}}$ and this can be made into a vector space using pointwise product

[cloud walrus]

ProphetX

[sinful mirage]

and the claim is that this delta fns form a basis of this vector space

so I should be able to prove that $f(h)=\sum_{g \in G} \limits f'(g) \delta_{h}(g) \forall f in Fun_{g}$

[chilly ocean]

Is there any requirement as to how the functions interact with the group operation on G?

[sinful mirage]

the product between functions you mean?

or the vector space structure

[chilly ocean]

I mean like does f(xy) equal to, say f(x)+f(y)

[sinful mirage]

I don't think so,these can be any functions

need not be homomorphisms

[chilly ocean]

Yes (and assuming there are no requirements on f/weird complications) you should be able to quickly think what f'(g) should be

[sinful mirage]

f'(g) should be f(g)?

lol

[chilly ocean]

Yeah

[sinful mirage]

no

i'm a bit confused

what am i missing

ah i am missing f(h) on LHS,right?

[cloud walrus]

ProphetX

[chilly ocean]

Yeah, you also swapped g and h on the RHS

[sinful mirage]

right,my bad sorry

ok now it makes sense,this is a basis

thanks @chilly ocean @chilly ocean

there is also a claim that on class functions,i.e. specific functions satisfying f(hgh^-1)=f(g) forall g,h in G there is a basis provided by the delta functions "evaluated on conjugacy classes"

why do we need the "evaluated on conjugacy classes" part?

if delta function forms a basis on the space of all functions,in particular,it forms on the space of class functions(which are specific functions)

[chilly ocean]

sounds like a quotient

[maiden ocean]

whats the name for the notion that if a set carries two operations + and * with identity and the commution law holds then + = *

[hidden haven]

What's the commution law

[maiden ocean]

its eckmann hilton argument

but its (a + b) * (c + d) = (a * c) + (b * d)

i couldnt place it because i knew it wasnt eckmann-hilton duality

also hi john

[upper pivot]

hi moth

[final pasture]

well isn't that eckmann hilton argument, like you said ?

[maiden ocean]

yes i bullied brofibration into telling me the name

[final pasture]

kek

[upper pivot]

interesting property, what context were u using it or w/e

[maiden ocean]

showing that the operation on [\Sigma^k X, Y] is well defined

[upper pivot]

oh huh interesting

[maiden ocean]

regardless of which coordinate u chose

[final pasture]

[maiden ocean]

and also that pi_n is abelian for n >= 2

[upper pivot]

i can see that roughly lol, v cool tho

[maiden ocean]

loop space time

[upper pivot]

what book/resources are you using these days for topology, still hatcher?

[maiden ocean]

tom dieck

[upper pivot]

nice

[maiden ocean]

how have u been doing recently

also ive been using ur copy of AM to great effect

[upper pivot]

i got back into some math recently. also nice lol

i have been doing some prob memes

[maiden ocean]

[upper pivot]

this prob-methods book is like, actually the hardest thing i have done lol

[final pasture]

oh i'm planning on reading dieck soon

the book looks great

[maiden ocean]

have u done any AT before

like a first course

[final pasture]

no, was expecting to start AT with this book

[maiden ocean]

do a good chunk of hatcher first

dieck is hard and also not very visual

[upper pivot]

did u learn cat theory for dieck moth

[maiden ocean]

i dont think its very good for a first pass

i just winged it honestly john

[upper pivot]

fair lol

[maiden ocean]

it hasnt been a problem rly

[final pasture]

I tried hatcher and didn't like it tbh, too visual and not formal enough. Like hatcher is great for the visual intuition but since I was trying to formalize everything, reading hatcher was kinda hard

but some time passed since I first tried, maybe I should give it another shot, idk

[upper pivot]

eh having the visual as a backup is v good

[maiden ocean]

the visual intuition is important

[upper pivot]

rather than abstract nonsense kek

[maiden ocean]

if u want something less visual then u can do like bredon or something

but like

[final pasture]

No I like to have something visual, I just disliked the part "not formal enough"

[upper pivot]

thats fair, some stuff in hatcher was a bit sus

[hidden haven]

Same happened but that was when I was reading chapter 0

[maiden ocean]

i dont think people really benefit from their first exposure to covering theory being an equivalence of Cov_B and Tra_B

[upper pivot]

house with 3 room

[hidden haven]

Chapter 1 is formal enough I think

[final pasture]

oh I was also speaking of chap 0 lol, wasn't able to go any further kek

[maiden ocean]

chapter 0 is just preliminaries yea

i dont think the main book is informal

[upper pivot]

to be fair i think ch 0 is suppose to be super quick and cannot hence be formal

[maiden ocean]

just chatty

[hidden haven]

Bruh dieck defined a cover as universal if a certain functor was an equivalence of categories

[maiden ocean]

yea lmao

[hidden haven]

I had no idea what that meant intuitively at the time

[final pasture]

Then what do y'all recommend to start AT ? Like is hatcher really the best intro book ?

[upper pivot]

well be patient with it ig lol

[hidden haven]

Yeah Hatcher is really good I think

[upper pivot]

its p good

[maiden ocean]

it ends up working out in the end because for p: E -> B the associated functor A(p) is an equivalence of Aut(p)-Set and Cov_B iff the transport functor is or something

[upper pivot]

plus i can give u a series of lectures that v faithfully follow hatcher (look up lectures by pierre albin) helps fill in some more intuition

[hidden haven]

Yeah it probably does work out but I don't remember what the functor was lmao

[maiden ocean]

there r 2 functors

[final pasture]

will check it, thanks

[hidden haven]

I will read dieck after hatcher

Because I wanna see all the abstract nonsense too

[upper pivot]

tbh sounds fun, maybe ill join in lol

how fare are you moth

far

[maiden ocean]

one takes a cover p with right G-principal action on E and sends a set F with a G action on it to like

[hidden haven]

Oh yeah all that bs

[cloud walrus]

Moth In Shambles

[hidden haven]

I remember working through those definitions bit it was all so formal I forgot all of it

I did have some intuition for G principal actions so could probably decipher given enough time

#groups-rings-fields

[final pasture]

I mean it's algebraic topology

[upper pivot]

lol

[maiden ocean]

and then the other is like you have a functor T: Cov_B -> [Pi(B), Set] that sends a cover p': E' -> B to the functor T(p') which sends points of b to their fiber and paths to the function F_b -> F_c given by like lifts and shit

lift ur path v to v_x at x in F_b and then T(v)(x) = v_x(1)

[hidden haven]

damn

so neat amirite

[maiden ocean]

and T is an equivalence iff you have a universal p: E -> B w/ right G-principal action on E simply connected

[hidden haven]

[maiden ocean]

its pretty elegant

its just

hilariously formal

[hidden haven]

yeah it probably is elegant

[maiden ocean]

the payoff is that classification falls out immediately

[hidden haven]

I just wanna have some proper intuition first

otherwise its a struggle to even remember the definitions

[maiden ocean]

ya

its just fun

[upper pivot]

feel like u need to spend quite a bit on each defination figuring out intuition

[maiden ocean]

mhm

also john im on chapter 4

elementary htpy theory

[upper pivot]

nice, ill try to catch up

gimme a few days

[maiden ocean]

u probably will im slow

well i finished ch 3 and then i didnt do anything for a while

cause i was doing AM

[upper pivot]

better than me quitting math for a few months and doing bio lol

[final pasture]

AM = ? @maiden ocean

[upper pivot]

atiyah macdonald

[final pasture]

imagine doing bio tho

[final pasture]

thanks

[daring ibex]

Sanity check, $\text{SL}(n,K) \hookrightarrow \text{Isom}(K^n)$ right

[cloud walrus]

mirzathecutiepie

[chilly ocean]

I don't think so, for example [[2, 0], [0, 1/2]] is determinant 1 in, say SL(2,R), but this is not an isometry

[daring ibex]

hmm fair

[chilly ocean]

I think the reverse is true though, isometries are determinant 1

[daring ibex]

I see i see, fair

[chilly ocean]

Or determinant -1?

[final pasture]

I think +-1 yeah

like if I'm not bullshiting [[-1,0], [0,1]] is an isometry

[hidden haven]

Is this the metric space isometry?

[final pasture]

yeah ?

[hidden haven]

What's the metric on K^n

[final pasture]

Well idk I assumed mirza had one, in the case of my example over R, I was thinking about the metric induced by the euclidean norm

[hidden haven]

[daring ibex]

euclidean normal

[waxen hollow]

How I can find the degree of the splitting field for p(x)=x^3+3x+1 over Q??

I know that the degree must divide 3! = 6, but dont know how to decide over 3 or 6

<@&286206848099549185> ?

[delicate bloom]

since it's a cubic it has at least one real root a, but since p'(x)=3x^2+3 > 0 it's always increasing and so has 2 complex roots b and b*

so now you can look at [Q(a,b):Q] = [Q(a,b):Q(a)][Q(a):Q] = 6

also, I'm assuming it's irreducible but you can prove it's irreducible with eisenstein looking at p(x+2)

@waxen hollow

I think I was a little terse, Q(a) is only a real extension so contains no complex roots, that's how we know it doesn't contain b and b*

[rustic crown]

(Rational root theorem is also pretty good here)

[delicate bloom]

oh nice, that is better

[rustic crown]

Also generally I go mod 3 first then decide the shift, like here going mod 3 gives x³+1 = (x+1)³

But I get sad if the constant is then divisible by 3² :p

[waxen hollow]

but how i'll know that b inst something like +-a+-ai?

[rustic crown]

It doesn't matter even if it is that

[waxen hollow]

I am confused, because I thought that in this situation we would have degree = 3

[rustic crown]

[Q(a, b) : Q(a)] must be ≤ 2

And can't be 1 as b isn't real

[rustic crown]

One simple test for irreducible cubics is to just calculate the discriminant. If that's a perfect square the galois group is A3, else its S3.

[waxen hollow]

understood

the discriminant is b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd?

[rustic crown]

Lol I don't remember but yea something like that

I just write the cubic on wolfram and it gives me the discriminant

[daring ibex]

Okay this one thing is driving me up the wall

here is the uni property

And in the existence proof it states this

is the $R_{G_1} \cup R_{G_2}$ supposed to be a disjoint union?

[cloud walrus]

mirzathecutiepie

[daring ibex]

I mean i think it is but

I might be straight up missing the point

[daring ibex]

I guess i didn't give enough info for it to be helpful

to anyone wanting to help

We are trying to construct a group G such that the uni property is satisfied, each of the \alpha_j are the same as in the uni property

And G = G_1 * G_2

or well, that's what we're trying to show lol

The issue here is if

i free product a group with itself then

$G_1 * G_1 = \langle {x_g | g \in G_1} \sqcup {x_g | g \in G_2} \mid {x_{\alpha_1(a)} x_{\alpha_2(a)} \mid a \in A} \cup R_{G_1}}$

[cloud walrus]

mirzathecutiepie

[daring ibex]

This allows for free mixing of each of the two disjoint groups

*sets

inside the free prod

like if we have

$x'_1 x_1$, we can reduce this if there is a relation between $x_1$ and $x_1$

[cloud walrus]

mirzathecutiepie

[daring ibex]

I thought this was contrary to what the point of the free prod was

or well what the idea behind it was

and plus doesn't this mean that there's no point to disjoint unioning the generators in the first place? If each x'_g behaves the exact same way as each x_g

Am i overthinking this too hard or is this like implied or sth

maybe the tagging has something to do with this, does that always force the union to be disjoint? I mean set theoretically definitely not, but maybe that's implied?

[carmine fossil]

I guess if the union was not disjoint,\beta_1 and \beta_2 may not be unique or smt

I don't see why else disjoint union might be useful here

[daring ibex]

The thing is one of the problems in the book requires this same thing, and if we assume a kind of nondisjointness then we get a completely out of the world answer

here

Consider $\mathbb{Z}/2 * \mathbb{Z}/2$

[cloud walrus]

mirzathecutiepie

[daring ibex]

By the given expression

$\mathbb{Z}/2 * \mathbb{Z}/2 = \langle x_0, x_1, x'_0, x'1 \mid R{\mathbb{Z}/2}\rangle = \langle x_1, x'1 \mid R{\mathbb{Z}/2}\rangle$

[cloud walrus]

mirzathecutiepie

[daring ibex]

And $R_{\mathbb{Z}/2} = {x_gx_h x_{gh}^{-1} \mid g,h \in \mathbb{Z}/2}$

[cloud walrus]

mirzathecutiepie

[daring ibex]

And so since $x'_1 \in \mathbb{Z}/2$ and $x_1 \in \mathbb{Z}/2$ we can do $x'1 x_1x{0}^{-1}$

[cloud walrus]

mirzathecutiepie

[daring ibex]

stuff like this

and so doing this we eventually reach the presentation $\langle x_1,x'_1 \mid x_1 x'_1 = x'_1 x_1, x_1^2, x'1^2\rangle$ which gives that $\mathbb{Z}/2 * \mathbb{Z}/2 \cong {e}$ but the problem was to show $\mathbb{Z}/2 * \mathbb{Z}/2 \cong D\infty$

[cloud walrus]

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

[daring ibex]

So kinda funky

[carmine fossil]

Differentiate the 2 Z/2 groups

[daring ibex]

Yeah that's the issue i think

there needs to be some form of distinction between these two guys

[carmine fossil]

It should be x_1 x_1 x_0 I think

Not x_1' x_1 x_0

[daring ibex]

But set theoretically i mean, here R_G_1 = R_G_2 so

[daring ibex]

Ideally it would but we can indeed do x'_1 x_1 x_0 since with x'_1 we have 1 \in \mathbb{Z}/2

[carmine fossil]

Yea,It feels like that should be disjoint union

So you have x_1^2=e and (x_1)'^2=e

[daring ibex]

yeah

hrmm

[carmine fossil]

Instead of doing a disjoint union say G_2 is always a distinct group from G_1

[daring ibex]

tho even if it was disjoint union it wouldn't work out now that i think about it

[carmine fossil]

If G_2=G_1 identify G_2 with a isomorphic group

[daring ibex]

Yeah but it isn't tho

Like i guess yes we want distinction

how do we achieve that without doing sth really gimmicky tho

So this is what i was saying when i was asking whether this was implied

bcs set theoretically definitely R_G_1 = R_G_2 and we can do this, so i guess yeah we have to treat them seperately

And if we do that, then this is automatically a disjoint union i think

am i wrong

[carmine fossil]

I think that that is a kind of disjoint product

[daring ibex]

wdym

[carmine fossil]

If A and B have a common element, identity B with a isomorphic group

So,set theoretical this new isomorphic group has no common element with A

[daring ibex]

is it necessary that there is an isomorphic group that is not itself

like is it necessary if i have A \cong C where C \neq A and A \cap C = 0

[carmine fossil]

You know that given a set ,you can always find a set bijective to the original set

[daring ibex]

indeed

[carmine fossil]

Define a group operation on the new set so that it obeys the homomorphism rules

[daring ibex]

i see where this is going

okay sure

yeah makes sense

But then again this is sorta a gimmick no

Hrm

[reef agate]

Hi dear fellow numberbrainiacs I need some help in understanding a polynomial long division of two polynomials over a expanded Galoisfield GF(8). I don't know the result but I know the remainder and I am not able to recreate it. The division is: $5x^{6}+4x^{5}+3x^{4}+2x^{3}+x^{2} / x^{2}+3x+2 =$

[cloud walrus]

doodlejump007

[reef agate]

The suposed remainder is: 5x + 4

And my field generating polynomial is: $x^{3}+x+1$ from which I derived the multiplication table

[cloud walrus]

doodlejump007

[reef agate]

Can someone calculate this division so that this remainder results?

[hidden haven]

Isn't GF(8) the field with 8 elements? In which case you have characteristic 2 and both the polynomials become really simple (and you get remainder x which is the same as 5x+4)

[unique juniper]

anyone can help me out with part c?

i can guess theres only 2 cosets

but idk

[hidden haven]

Can you guess what the conjugate of a+b√D is?

[rustic crown]

use the last part where it says "In particular if K/F is Galois this is ..."

(K/F is indeed galois in that case if char F is not 2)

[unique juniper]

i dont actually understand that part

if K/F is galois, why are the representatives in Gal(K/F)

[hidden haven]

Because you're taking embeddings of K into the alg closure of F. The image of all such embeddings has to be the same as K is galois over F, so you can just view these embeddings as isomorphisms onto the image

And then prove that there's a 1-1 correspondence between these isomorphisms and the elements of the Galois group

[unique juniper]

i see

thank you

i understand

:D

but this wont work if F did have a characteristic of 2 tho

[hidden haven]

If characteristic is 2 then it seems a+b√D has no conjugates since it's minimal polynomial is non separable? In that case the norm should be itself but that just seems wrong so idk

[unique juniper]

oh

[hidden haven]

looking at the Wikipedia definition, the definition given in the problem seems wrong

only the case of Galois extensions seems correct

[rustic crown]

or equivalently you can say that the extension isn't separable so you can't even contain that in a galois extension. so the definition doesn't work

[hidden haven]

oh L isnt the algebraic closure

[rustic crown]

yea lol

[hidden haven]

but L isnt being used in the formula is it

[rustic crown]

you take product over cosets of H

H is the corresponding subgroup of Gal(L/K) so you are using L.

[hidden haven]

ah right lol

[rustic crown]

but i think you can define norm in this case as well... if the extension isn't separable. just muliply all the conjugates and raise the product to the inseparability degree.

this matches with the determinant definition

[hidden haven]

inseparability degree would be the same as the multiplicity of alpha in its min poly right?

[rustic crown]

what is alpha?

no primitive elements for inseparable stuff 😛

[hidden haven]

oh I was taking that to be the generator of the extension

yeah i was taking the subextension lol

but that would change norm

[rustic crown]

there are many definition... one i like the most is this.
K/F is finite define L = {elements in K separable over F}
then L/F is separable and K/L is purely inseparable

separable degree is [L:F] and inseparable degree is [K:L]

so that definition can be seen as a special case of property of norm

N_{K/F}(a) = N_{K/L}(N_{L/F}(a)) = (N_{L/F}(a))^[K:L]

purely inseparable means that every element satisfies a polynomial of the form x^(p^n) - b

[hidden haven]

oh epic

[rustic crown]

this was a long exercise in Aluffi lol

N_{K/F}(a) = N_{K/L}(N_{L/F}(a))
this property holds for any extension tower. Norm is just too cute.

[unique juniper]

this looks hard

[hidden haven]

lol the point is that the definition given in the problem doesnt work in characteristic 2

[reef agate]

Yes but the thing is I need to get the 5x + 4 as a result because the congruent over GF(2) is not enough. For context I'am programming a reed solomon decoder and 5 and 4 would be symbols later used in the decoding process.

[hidden haven]

because there is no choice of L

[unique juniper]

oohhhh

is a quadratic extension possible in a field of characteristic 2

[hidden haven]

oh I have no clue then but doesnt division over GF(8) only given an answer upto congruence mod 2?

[rustic crown]

yep

[unique juniper]

oh ok

[rustic crown]

GF(4)/GF(2)

[unique juniper]

oh

[rustic crown]

when did this happen? this new reacts system looks pretty nice.

[unique juniper]

yeah i noticed, i like it too!

[daring ibex]

oh that's not that bad as ppl were saying

ill update discord then

[hidden haven]

It look good

[chilly ocean]

how can I get GF(n)

[hidden haven]

I'm on discord beta

[rustic crown]

(btw i have been wanting to ask you something... which anime is your dp from? @.Yes)

[chilly ocean]

It's just an irl picture of a cat.

[hidden haven]

[rustic crown]

i was asking to @.Yes >.<

[unique juniper]

Koe no katachi

its very good :D

[rustic crown]

av, Yes

,av Yes

[cloud walrus]

Yes#0553's Avatar

Click here to view the image.

[hidden haven]

oof

[unique juniper]

i feel insecure now

[rustic crown]

lol

[daring ibex]

Okay beta

[rustic crown]

i only remember two characters from Koe no Katachi

😓

[unique juniper]

:(

[hidden haven]

I remember 3

MC's mom

[unique juniper]

she is the little sister, the coolest person imo

[rustic crown]

abstract algebra is so fun

[waxen hollow]

How I can find Gal(Fp^10/Fp)?

[rustic crown]

can you find one automorphism of Fp^10 fixing Fp?

except id

[waxen hollow]

Pls help x.x I don't even know how to miss

[waxen hollow]

I dont know

[rustic crown]

[hidden haven]

Do you know the cardinality of the group?

[rustic crown]

Have you seen that over char = p, (a+b)^p = a^p + b^p?

[chilly ocean]

||Frobenius?||

oh ur faster

[waxen hollow]

Yeah

[rustic crown]

so Frobenius is an automorphism right. what is its order in the group Gal(Fp^10/Fp)?

[waxen hollow]

[rustic crown]

can we react on a react?

[golden pasture]

hmm

discord vuln hunting time

[rustic crown]

try to see the definition... how many times you do compose Frobenius with itself until you get the identity?
We're asking the smallest n such that x^(p^n) = x

[waxen hollow]

I'm not that familiar with this group ...

I just want to cry

But i'll see

[rustic crown]

pwease dun cri >.<

[waxen hollow]

Sorry, but I really dont know how to figure out the order

[rustic crown]

try looking at the group (F_p^10)\{0} under multiplication... what's its order? what does lagrange theorem say?

[waxen hollow]

when I try to check the definition I can only think of my mediocre existence

Well, the order of any finite subgroup of G divides G

[rustic crown]

yep, so if you take any element in (Fp^10)\{0} then its order should divide the the order of the group which is p^10 - 1

which means for any non-zero element x, x^(p^10 - 1) = 1

try to take it from here?

[waxen hollow]

Sure, I'll try

Thanks det

[rustic crown]

[chilly ocean]

[rustic crown]

are you Godel?

why use 2 accounts?

[chilly ocean]

No I'm Ledog

[rustic crown]

[chilly ocean]

Does anyone have any good intuition for fields of characteristic p?

[reef agate]

I thought so too. But maybe this is some kind of weird engineering math I have no clue 😂

[chilly ocean]

I understand algebra from the point of view of thinking of rings as rings of functions, and thinking of fields as being C

[reef agate]

I like examples to get an intuition for things lets take for example characteristic 2. You have the obvious Finite Field with two elements 0, 1 (basically binary) and then you have expansion fields with $2^m$ elements where m is a natural number. The expansion field 2^2 has the elements 0,1,2,3 and so on. The characteristic is often helpful since it can be used as a primitive element in the expansion fields with which you can then generate all the other elements in the expansion field.

[cloud walrus]

doodlejump007
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[daring ibex]

i mean they are just Z/pZ

Those are the most common finite fields i think

[reef agate]

And to do this expansion nosense your characteristic has to be prime. There really is not more to them

[hidden haven]

No

[daring ibex]

[waxen hollow]

Fp^10/Fp is the splitting field for x^p^10-x, so the order must be 10. 10=2*5, so all groups of order 10 is Z10 and D5. Since Galois group is abelian, Z10 must be Gal(F_p^10/F_p)

there's something wrong?

[daring ibex]

The only finite fields are those with char p, and hence that have orders p^k for k \in N, p prime

[hidden haven]

How do you know galois group of this is abelian?

[daring ibex]

The most common of these are like Z/pZ

[waxen hollow]

I think I've seen a demonstration of something like that

[sharp sonnet]

what does "most common" even mean

[hidden haven]

Tf is more common mean like occurrence in UG textbooks?

[hidden haven]

Well if you can assume that then it seems fine

[reef agate]

And characteristic 0 ist also interesting this occurs in no finite fields. Like the field of rational numbers.

[waxen hollow]

I saw a result like this but I think it is for Q, could you suggest any other way?

[hidden haven]

Galois groups over Q are also non abelian in general

So you know order of the group is 10

And you know one non identity automorphism

Which is the frobenius map

If you can figure out the order of the frobenius map, you will get a pretty good idea of what the group is

[chilly ocean]

If you are doing, say Fp^2, then you are also secretly doing Fp because Fp is inside Fp^2. So it must be the case that Fp is most common

[hidden haven]

So can you figure that out?

You already know it divides 10

[hidden haven]

Counterpoint they are all countably common

[daring ibex]

Most of the time it's literally just F_p

[sharp sonnet]

the norm 1 subgroup of F_p^2 is used a lot

[daring ibex]

i cee

[minor badger]

Im trying to understand the proof of Nakayama's lemma on wikipedia by using Cayley-Hamilton's theorem, but I'm struggeling, If someone could help me I would appreciate it

[golden pasture]

check atiyah macdonald ch2 if it helps

iirc they proved with cayley hamilton as well

[rustic crown]

i don't understand your argument for order = 10 and it being abelian...

[minor badger]

Their proof is a 1-liner, and its that line im struggeling with ;-;

[rustic crown]

i was hinting you torwards showing that order is 10 as the degree of the extension F_p^10/F_p is 10. And frobenius is the generator

[hidden haven]

If the Galois group is abelian then it follows, because as they said there are only 2 groups of order 10 and only 1 is abelian

But we don't know a priori if galois group is abelian

[rustic crown]

yea

and this method won't generalize as well... 😦

[hidden haven]

Yeah

[golden pasture]

which one specifically?

[cloud walrus]

bim

[waxen hollow]

Any example of a*b=c which a,b is transcendental and c is algebraic?

[minor badger]

$\frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}$

[waxen hollow]

nani?

[cloud walrus]

bim

[waxen hollow]

wow

thanks

[minor badger]

🙂

wait

[hidden haven]

1/root 2 is algebraic

[minor badger]

yeah i just realised XD

[hidden haven]

Any transcendental times its inverse

[waxen hollow]

[minor badger]

No, too much algebraic geometry for today. Time for the next algebra!

[cyan marten]

Given two distinct primes p < q, I am interested in exhibiting a nonabelian group G such that x^pq = 1 for all x in G. If q = 1 (mod p), then the nonabelian group of order pq works.

I thought of taking r = p - 1, and taking a semidirect product of the elementary abelian groups of order q^r and p

Is there a slicker approach?

Or, more importantly, does this work?

[chilly ocean]

I have a question about irreducible / reducible polynomials

If f(x) is irreducible how do we rewrite it as f(x) = g(x) h(x)

I mean, if f is irreducible then it cannot be written as product of two nonconsant polynomials.

How do we show that it is irreducible, it says either g(x) or h(x) must be a unit. But like there's infinite ways to write that.

I don't quite understand what you're asking, the definition of and irreducible polynomial is that if you have such product f = gh then g or h is a unit.

For an example, 5x^2 +25 = 5(x^2 + 5)

or 2.5(2x^2 +10)

It depends on the ring.

[rustic crown]

I think it works... the automorphism group of $\mathbb{F}_q^n$ is $GL_n(\mathbb{F}_q)$ which has order $(q^n - 1)(q^n - q) \cdots (q^n - q^{n-1})$ and for a a nice choice of $n$ we can guarantee this is divisible by $p$. So by Cauchy there is an element of order $p$ in $GL_n(\mathbb{F}_q)$, which would give us a Group homomorphism $\varphi:\mathbb{F}_p \to \text{Aut}(\mathbb{F}_q^n)$. Which could be used to get a semidirect product $\mathbb{F}q^n \rtimes\varphi \mathbb{F}_p$.

[cloud walrus]

det

[carmine fossil]

If q!=1 mod p,there is no nonabelian group

[rustic crown]

i still need to verify though that x^pq = 1 for every element in the group

[cyan marten]

No nonabelian group of order pq, sure. But there should be one of order p q^r yaing det's argument (and Fermat's theorem).

[rustic crown]

yea but that is easy so no need to worry

[carmine fossil]

mb

Misread the question

[cyan marten]

Yeah, that's where I am stuck. I think we can let C = <x> and consider its q-Sylow subgroup Q..

[chilly ocean]

Yes but like you could have f(x) = 25 x^2 + 50x + 25 = 5 (5x^2 + 10x + 25)

@chilly ocean For example in Z there are only 2 units, so there aren't 'infinitely many' factorizations

And like its not irreducible but I wrote it to match the irreducible definition

You need to state the ring you're working with.

[rustic crown]

you have the explicit representation... so just use that

(a, b)^p = (something, b^p) = (something, 1)
(a, b)^pq = (something^q, 1) = (1, 1)

[chilly ocean]

I guess my question is saying it can't be rewritten as f(x) = g(x) h(x) is enough to say its irreducible?

[cyan marten]

Neat, thanks!

[rustic crown]

[chilly ocean]

What definition of irreducible are you using?

Every polynomial f can be rewritten as f=gh if you don't assume anything on g and h

Exactly, so a polynomial is irreducible if in any such product either g or h is a unit.

im just confused about how to show something is irreducible, so lets say i need to show x^2 +1 is irreducible over Z_3

Well over Z_3 it's easy, you can just see it doesn't have any roots by checking all 3 elements.

If it was reducible then it would be of the form (x-a)(x-b) for some a and b in the ring.

x^2 + 1 = 1 (x^2 + 1) and one is a unit so its irreducible?

[hidden haven]

not enough, you need to check all possible factorisations

[chilly ocean]

No, in the definition there is implies g or h is a unit for any factorisation.

[final pasture]

assume x² + 1 = gh

show that either g or h is a unit

suppose it's not the case

then g is atleast of degree 1

but it can't be of deg 2, because then h would need to be of deg 0 so that the product is of deg 2, and a deg 0 polynomial is an unit in Q (or in any field)

so g and h are of deg 1

yeah can't be deg 2 typo srt

sry*

so write h = a(x-x0) and g = b(x-x1)

you get x² + 1 = ab(x-x0)(x-x1)

now if you want to check it's irreducible over Z3, you need to prove this is absurd. How so ? Show that x² + 1 doesn't have any root in Z3 but ab(x-x0)(x-x1) has x0 and x1 as roots

[hidden haven]

[chilly ocean]

i see thank you!

I have never seen the proof for it, my professor just says polynomials are irreducible without much justification but in the exam we are going to have to show a polynomial is irreducible

[final pasture]

Well, take example on your prof during the exam

just say it's irreducible without much justification

[chilly ocean]

showing that a polynomial has roots is using a reducibility test and not just the definition. but in case he gives a degree 4 polynomial or something then i cant use the test lol

[final pasture]

yeah, depending on the polynomial it gets harder

there's no general method, just "a bag of tricks" to know

[chilly ocean]

ya makes sense

lots of different cases

ya but i think i got this! thank you guys 💗

Yeah, in finite rings you can do it manually like above, it's not as easy for infinite ones.

i hope for a finite ring, or a deg 2 polynomial then i can just use the quadratic formula lol

[chilly ocean]

A lie algebra is a vector space with a lie bracket, pretty much that means we require
[x,y]+[y,x]=0
[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0

Do we get any interesting structure or use out of extending this to higher cycles?

[x,[y,[z,w]]] + ... =0 for example

[maiden ocean]

is this actually true when A is not an integral domain or when S contains 0?

im pretty sure the proof they describe should not give you an equation of integral dependence for bt

it gives you like

[cloud walrus]

Moth In Shambles

[maiden ocean]

so if S does not contain zero and A is an integral domain then i think this works out

but otherwise

[oblique river]

I'm not sure I follow your point

if A isn't a domain or S contains 0, what that means is that "more things in S^(-1)A are identified with each other than what you might expect at first"

so what could go wrong in a proof is if you are trying to prove that something isn't equal to 0 but it might accidentally be anyway. Here, we're trying to show that something equals 0, but if more things equal zero than we expect, that's not a problem

I'm not seeing where you got your u from

In S^(-1)A, it's always true that s*(a/s) = a/1

[unique juniper]

my brain is hurting

[chilly ocean]

you should see a doctor, Yes

[oblique river]

proof: to show that a/b = c/d we need to find some u in S such that u(ad - bc) = 0. In this case, that means we need to find some u in S such that u(as - as) = 0. take u = 1.

@maiden ocean

oh wait as soon as I ping... let me re-think this.......

maybe I do see your point?

[maiden ocean]

no exactly like

we're trying to show that if x/s is integral then x/s is in S^{-1}C which is equivalent to saying that x is in C, or x is integral over A

[oblique river]

you're right, I was thinking about the other direction

[maiden ocean]

ya

[oblique river]

so i think we're still going to be fine, essentially i want to include the u in the definition of t

one sec

okay yeah so I'm gonna write this out for a quadratic just so I don't have to deal with indices. suppose we have (b/s)^2 + (a1/s1)(b/s) + (a2/s2) = 0

combining this into a single fraction we get (b^2 s1 s2 + b a1 s2 s + a2 s1 s^2)/(s^2 s1 s2) = 0

i.e. (b^2 s1 s2 + b a1 s2 s + a2 s1 s^2)u = 0 for some u in S

multiply this by (u s1 s2) and distribute everything

(b u s1 s2)^2 + (b u s1 s2)(a1 s2 s u) + (a2 s1^2 s2 s^2 u^2) = 0

which shows that (b u s1 s2) is integral over A and therefore is in C

now let t = (u s1 s2) which is in S, and then b/s = bt/st in S^(-1)C

[maiden ocean]

oh i think i see

because we can use the extra powers of u^n as our coefficients

[oblique river]

yeah

so I think AM is wrong here because they ignore this point

but i think the proposition is still true

so in general you would multiply by (u s1 s2 ... sn)^(n-1)

so that you could get a leading term of (b u s1 s2 ... sn)^n

[maiden ocean]

yeah

you definitely need to think about the u here

but it works

[oblique river]

so really the proof structure should be like:

b/s integral over S^(-1)A; coefficients have denominators s_i
=> (b s1 s2 ... sn u) integral over A for some u in S. call t = (s1 s2 ... sn u)
=> bt in C
=> b/s = bt/st in S^(-1)C

[maiden ocean]

mhm

[oblique river]

:D

good catch

sorry about my confusion at first

time for lunch now!

[maiden ocean]

bye

[unique juniper]

why is it necessary for it to be in an algebraic closure? i dont get this part

also

i know that when H is normal, every embedding of E into an algebraic closure of F is just an automorphism of E, but not sure exactly why from this G/H is seen to be isomorphic to Gal(E/F)

[cloud walrus]

PhysWiz

[cursive prairie]

Sorry about the formatting😩 Here's the question directly

Here's the solution I found. I understand the part bout 7C4 but why 6 & why do we have 3! ways instead of 4! (which makes more sense to me)

[past temple]

i'm not understanding the last line of this proof

"This contradiction shows that |G| = 1"

shouldn't it be that |G| = 2?

[next obsidian]

They’re saying as a result of the contradiction

We can conclude that in actuality |G| = 1

[old lava]

|G| = 2 is the contradition, which would imply |G| = 1

[next obsidian]

Not that the contradiction was |G| = 1

Back off Poros, this is my help

[past temple]

how does the contradiction show that |G| = 1

and if |G| = 1

then 1 = 2[N:C]

so [N:C] = 1/2 ?

i'm confused

shouldn't the contradiction show that |G| = 2?

[old lava]

you help too much in this channel

[next obsidian]

Oh wait, |G| should be 2 I think I think I was getting ahead of myself

[past temple]

yeah that's what i thought

i was assuming the proof went something like

"Suppose BWOC that [N:C] > 1"

which is equivalent to assuming that |G| > 2

if a G is a p-group with order > 2, then we should be able to construct a maximal subgroup of index 2 in G

[next obsidian]

Yeah, then you’re supposed to get a degree 2 extension of C I think

[past temple]

which then leads to the contradiction

[next obsidian]

Yeah

[past temple]

so is it a typo in the text

[next obsidian]

I think so yeah

I mean there’s even a typo earlier when it says T the fixed field of M

I think that should be fixed field of P lol

[past temple]

yeahh i noticed that too

tyty

[past temple]

does the isomorphism extension theorem guarantee that if you have an irreducible polynomial f in F[x]

and K is the splitting field of f

then for each pair of roots a, b of f, there exists a sigma in Gal(K/F)

mapping a->b

is this true?

[hidden haven]

Isnt that the statement of the theorem?

[past temple]

the way the theorem is stated in my textbook is much more convoluted and confusing

[hidden haven]

Yeah you can ensure what you are saying from this

because you can take F(a) and F'(b), and an isomorphism of F and F' extends to this with a -> b, and then you apply the theorem on this

[past temple]

what is F' in this case?

isn't it just F?

or am i dumb lol

[hidden haven]

oh right nvm

Yeah F(a) and F(b)

[past temple]

ah okay

[daring ibex]

right so

If we have $\mathfrak{a} + \mathfrak{b} = R$ then they claim that $R(\mathfrak{a} \cap \mathfrak{b}) \subseteq \mathfrak{a}\mathfrak{b}$ implies $ \mathfrak{a} \cap \mathfrak{b} \subseteq \mathfrak{a}\mathfrak{b}$ but how this this possible? I mean, the defn of an ideal $c$ is that $R\mathfrak{c} \subseteq \mathfrak{c}$

So we can't just apply the defn

oh wtf

that k is so ugly

[next obsidian]

wtf is R

[cloud walrus]

mirzathecutiepie

[next obsidian]

Oh

[oblique river]

R has 1

[daring ibex]

chomsky moment

[daring ibex]

yeah

[oblique river]

Therefore Rc = c

[daring ibex]

wait

[next obsidian]

RI = I

for any ideal

Mirza moment

[oblique river]

Thank you chm

For literally

Repeating

What i just said

[next obsidian]

oh

hurvb

[daring ibex]

chomsky moment

[next obsidian]

c is an ideal

[daring ibex]

Okay wait the defn given was Rc \subseteq c in general for c being an ideal, where did the inclusion c \subseteq Rc come from

[oblique river]

Because R contains 1

[daring ibex]

hm

[oblique river]

Why is everyone repeatinf me

[old lava]

why is everyone repeatinf me

[oblique river]

Thank you poros

[daring ibex]

i guess that works

[oblique river]

x = 1*x for any x in c

[daring ibex]

Yeah i got it i think

[oblique river]

Therefore x is contained in Rc

And therefore c is contained in Rc

[daring ibex]

Yeah

i got it loll

Hrm

okay makes sense

thanks

[green stone]

the classification of the sharply k-transitive groups

[hidden haven]

Yeah

If mn = 0 then m or n = 0 where you interpret a natural number k as the element of the field 1+...+1 (k times)

So characteristic can't be mn where m and n are >1

[prisma ibex]

this is true of fields in general

(+ characteristic 0)

[chilly ocean]

yo quick question what degree can splitting field have for a polynomial of degree n? I know it's at most n!, but can we say more, for example, what degree can't it be?

[rustic crown]

it has to divide n!

[chilly ocean]

aight that's what I thought

[rustic crown]

(definitely true when the polynomial is separable)

[chilly ocean]

I'll try to show it, can't be that hard

[rustic crown]

ye

[old lava]

the galois group embeds into S_n, so ez reasoning why degree must divide n!

[chilly ocean]

ye

[limpid edge]

"The degree of an extension is 1 if and only if the two fields are equal. In this case, the extension is a trivial extension."

What if you had the field {b*i \mid b \in \mathbb{Q}} and where i is the imaginary unit

isn't this a field extension of the rationals with one basis element i ?

[chilly ocean]

Is this a field?

[limpid edge]

yes

wait

associativity of addition and multiplication ✅

commutativity too ✅

[chilly ocean]

bi*bi = -b^2 isnt of the form ci for some c in Q

[limpid edge]

oh closure

I see

I think I just thought that

since a + bi was a field extensions of Q

bi should be as well

[limpid edge]

so this doesn't seem as obvious then

[chilly ocean]

it does

if you think about degree of extension as degree of minimal polynomial

In case of finite extenstions

[limpid edge]

ok I have to read up on the connection of fields and polynomials to understand that first

i won't get far in algebraic number theory without knowing that

[vast quiver]

I mean another thing wrong with your original example is that Qi does not contain Q. I think an easy ish way to see that the sentence is true is to view things as vector spaces. If something is a subspace of something else, and they have the same dimension, then they’re equal

[limpid edge]

so the definition of field extension, say A extends B, is that A is a field on the same operations as B

does this imply B is contained in A

or is that an extra thing in the definition

[chilly ocean]

field extension is a pair B \subset A of fields

such that everything works

[limpid edge]

ok so this is true then

but not the definition

[chilly ocean]

same operation?

[limpid edge]

oh wait

like * and +

sub+anything in abstract algebra usually means subset

i forgot about this

[old lava]

sub+anything usually means subset + some of the structure is maintained

[rustic crown]

and generally that means the inclusion map is a "nice" map

[chilly ocean]

yo are you an inclusion map?

[rustic crown]

idts?

I'm a very nice map though

[chilly ocean]

generally

it's just a subset of the original structure that is closed for all the (restricted) original operations

[rustic crown]

(that is generally not enough... take Ring for instance Z/6Z is a ring, consider the subset {0, 2, 4}... under the same operations. 4 magically acts like a new identity and this is isomorphic to Z/3Z)

[chilly ocean]

it's not closed for the "multiplicative identity" operation

[rustic crown]

what is identity operation 😶

[chilly ocean]

it's a nullary operation

[old lava]

subrng moment

[rustic crown]

ah well

[chilly ocean]

if you consider all the operations it works

[chilly ocean]

f: R⁰ -> R where f(empty) = 1 where 1 is identity of R

[old lava]

I mean if you necessitate the identity existing, then {0, 2, 4} is indeed a subset closed under the operations but isn't a subring, "multiplicative identity existing" isn't really an operation

[chilly ocean]

yes but det was trying to give an example of an algebraic structure where it fails

[old lava]

it does fail there, the operations succeed

[chilly ocean]

in rngs it's not a problem

[old lava]

but it aint a subring

[chilly ocean]

in rings it is

[old lava]

because "identity existing" isn't an operation

it is another condition

[chilly ocean]

no

[old lava]

but not really an operation

[chilly ocean]

it is a nullary operation

for a simpler example of a monoid. it's a set S with 2 operations: some associative binary operation * and a nullary operation 1

you might want to check this https://en.wikipedia.org/wiki/Universal_algebra @old lava

@chilly ocean Can I dm you?

yea and you don't need to ask anyone can dm me

i wish more people were told that for example obtaining the element 1 in a ring is a ring operation. i think its much prettier and you don't have to worry about different homomorphism definitions

[rustic crown]

yea pretty neat

[limpid edge]

what does the notation Z/nZ actually mean

like I know it is {0, 1, 2, ..., n - 1}