groups-rings-fields

[oblique river]

this problem is asing about multiplicative order, not additive order

yes that's fine

you just need the two diagonal elements to be nonzero

[chilly ocean]

wait but it's in Z so the determinant must equal +-1

[oblique river]

which you can phrase as "a \neq 0, c \neq 0" or "ac \neq 0" or just writing "a and c are nonzero"

[chilly ocean]

ac = +1 or -1

otherwise its not a unit group

[oblique river]

sure

you can also phrase it that way if you want

you could also phrase it as (ac)^2 = 1

but yhou really don't gain anything

by doing that

[chilly ocean]

i don't?

[oblique river]

no

[chilly ocean]

but if i say theyre just nonzero doesnt that leave ambiguity

[oblique river]

what ambiguity does it leave?

like, it's true that a and c are each both \pm 1

[chilly ocean]

because they can still equal some integers that make it so that the matrix doesnt have integer entries

[oblique river]

if A is a matrix in R, then the statement "A is in R* if a and c are nonzero" is true

oh wait

fuck

i'm so dumb

you are right

lmao

yeah sorry for saying you were wrong

I was thinking this was over a field

not over Z

[chilly ocean]

no worries i just thought i was going crazy for a second

[oblique river]

no

because you corrected me this means you get to steal my math degree

[chilly ocean]

depends

where did u do it

[oblique river]

congratulations, dr. isomorphism

[chilly ocean]

wtf ur a phd?

[oblique river]

yes

well

i was

until now

[chilly ocean]

LMAO

thank you so much @oblique river

[oblique river]

np :P

so yes you do need to specify that matrices in R* are those with a, c = \pm 1

[chilly ocean]

yessir I will do that

[oblique river]

you can still just approach the problem directly, by starting with a matrix A in R and calculating A^3

and seeing what happens if you try to set A^3 = Id

you can also approach the problem with more theory and using det's suggestion from above

to think in terms of characteristic polynomials

and eigenvalues

[chilly ocean]

i think det;s suggestion is beyond me

i have to do it the layman's way

[rustic crown]

it was basically this, but not doing the whole computation

if you have a upper triangular matrix A with diagonal entries a, c then the diagonal entries of A^3 are just a^3 and c^3

so we wanted a^3=c^3=1 which gives a = c = 1

[oblique river]

don't do the whole problem for them.....

[rustic crown]

.< oopsie

[wooden ember]

just putting this back here in case someone sees a mistake

Also had another question: if I have a semisimple lie algebra $L$, a Cartan subalgebra $H$, and a root $\alpha$, then letting $h_\alpha$ be the unique element of $H$ such that $\kappa(h_\alpha,h)=\alpha(h) ~\forall h\in H$ is there an easier way to find $h_\alpha$ than computing $[L_\alpha,L_{-\alpha}]$ and then checking $\kappa(h_\alpha,h)$ for some $h\in H$?

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

My main problem is the part where you have to choose h in H at the end, being smart about the choice can be annoying. Would be kinda nice if there was some quicker closed form based on already known values

ig it's not so bad though cause usually you have a pretty explicit basis of H so ig im just being whiny about it

[formal ermine]

how can the galois group of $\bQ(\xi)$ over $\bQ$ where $\xi$ is the $13$th root of unity be $\bZ_{13}^*$?

[formal ermine]

isn't the minimal polynomial $x^{12} + ... + x + 1$? so the degree would be 12

[south patrol]

yes and |Z_13^*| = 12

In fact, the minimal polynomial of a prim nth root is the product of (x-a) as a ranges over all primitive nth roots of unity and thus the extension formed by adding such a root is of degree phi(n) = |(Z/nZ)^x| in general

[formal ermine]

oh lmao

0 isn't in it

[south patrol]

indeed

[rotund aurora]

Is there a way to define the characteristic polynomial of a matrix without the determinant?

I think this deserves thought

[elder wave]

axler moment

[rotund aurora]

no seriously

what if, say, all algebraic closures were canonically isomorphic

[chilly ocean]

product of invariant factors

[rotund aurora]

I wouldn't have problem going to algebraic closures in order to define determinants lmao

[chilly ocean]

https://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain#Corollaries

The last invariant factor is the minimal polynomial, and the product of invariant factors is the characteristic polynomial.

this works over any field

[rotund aurora]

👍

I think Roman actually does that

haven't read the chapters on eigen things of Roman yet, will soon

For real, structure theorem of finitely generated modules over PIDs is one of my favourite theorems. Even simply the Jordan decomposition is one of my favourites

[formal ermine]

let $p$ be a prime and $L$ the splitting field of $x^p - 2 \in \bQ[x]$. how do I show that the galois group of $L/\bQ$ is isomorphic to [ \Set{\mtrx{a & b \ 0 & 1} | a,b \in \bZ_p, a \neq 0} ]

[rotund aurora]

I think what I did was first determine the galois group which is pretty simple

and then, you will see that this matrix group is also pretty simple and like you just write a bijection

I think you will need to invoke a primitive root

I don't remember all the details now

[formal ermine]

ah so I literally just construct an isomorphism

oke I'll try tomorrow

ty

[chilly ocean]

nvm got it

[vagrant zinc]

Why abstract algebra is so cool!

[lone violet]

how can i show that the trace of the trace is the trace in a module?

the first contention is obvious but im confused in the other contention

i take $x\in Tr_{A}(M)$ then $x = \sum im f$ with $f \in Hom(U,M)$ with $U \in A$

[cloud walrus]

alef0

[lone violet]

@slim stag

<@&286206848099549185>

[celest furnace]

Pinging helpers doesn’t do anything in these channels (for the most part) so don’t

[chilly ocean]

need help with a) ii)

[lament dawn]

do you know the definition of the order of an element of a group?

[chilly ocean]

yes

smallest positive integer $n$ such that $a^n = e$

[cloud walrus]

isomorphism

[lament dawn]

right. So you've done (i) I assume

[chilly ocean]

yes I defined $\langle g \rangle$ as ${g^n : n \in \mbb{Z}}$

[lament dawn]

(latex wants those angle brakets in math mode)

good then you pretty much have the answer

[cloud walrus]

isomorphism

[lament dawn]

you just need to show that the only elements you get are those specific n in the set there

[long geyser]

all that's left is to show g^n is in that set for all n

[lament dawn]

[chilly ocean]

how would I go about doing so?

[lament dawn]

in particular, you want to use this

what is g^m

[long geyser]

division algorithm

[chilly ocean]

g^m = e

[lament dawn]

and g^m+1?

there's many ways to do this

[chilly ocean]

g

oh

so it cycles

[lament dawn]

yep you get it

[chilly ocean]

can I similarly show that it cycles through all negative powers of g

[lament dawn]

to prove it you can use contradiction, induction, whatever basically

[chilly ocean]

like g^{n-1} = g^{-1}

no wait

yeah

[chilly ocean]

how would I use induction here?

[lament dawn]

assume g^k is in there, then multiply by g. you showed manually that g^m+1 is in there so you're done

[chilly ocean]

so base case n = 0

g^0 = e which is clearly in the set

Let $n = k$, assume $P(k)$ is true for some $k \in \mbb{Z}^+$, i.e. $g^k \in \langle g \rangle$

[cloud walrus]

isomorphism

[chilly ocean]

uhh not sure how to do the inductive step

[lament dawn]

Either way this will just get you the postiive powers

[long geyser]

not sure why you said n=k

[lament dawn]

you still need the negatives

also yeah n only appears once in that statement

[long geyser]

tbh you could double induction on the integers

[long geyser]

how you do this is by induction hypo, g^k is either 1, g, ... g^(m-1)

[lament dawn]

yes but I imagine just using the division algorithm is easier at that point tbh

[long geyser]

then g^k+1 is clearly g, g^2, ... or g^(m) = e

same set

[chilly ocean]

i dont know the divisin algorithm

is there any simple way to do this

[lament dawn]

lems just told you

[long geyser]

division algorithm is the simple way

one should learn about it before cyclic subgroups

in any proper algebra text, at least

[chilly ocean]

can I argue that we get every negative power of $g$ by leveraging $g^n = 1$

[cloud walrus]

isomorphism

[chilly ocean]

i.e. $g^n g^{-1} = g^{n-1} = g^{-1}$

and so on

[cloud walrus]

isomorphism

[chilly ocean]

$g^{n-2} = g^{-2}$

[cloud walrus]

isomorphism

[chilly ocean]

$g^{n-k} = g^{-k}$

[cloud walrus]

isomorphism

[long geyser]

yes that would be the same induction backwards...

[lament dawn]

you would need two inductions to do it without the division algorithm

the funny thing is that you are basically doing the division algorithm there

but that would work yes

[chilly ocean]

can you show me how it'd be done with the division algorithm

[lament dawn]

step one is learning the divison algorithm

[chilly ocean]

A division algorithm is an algorithm which, given two integers N and D, computes their quotient and/or remainder, the result of Euclidean division. Some are applied by hand, while others are employed by digital circuit designs and software.

[lament dawn]

I could explain it to you but you'd be better served just using your textbook

[chilly ocean]

is it like a = qb + r?

[lament dawn]

yes thats it

[lament dawn]

so first you show that the given set is a subset of <g>. This is clear by what you said earlier

[next obsidian]

I 🤬ing hate when people refer to this as the division algorithm

[lament dawn]

then you want the reverse inclusion

what do you call it chmonkey

[next obsidian]

This isn’t the algorithm, you’re just using that division with remainder exists

The algorithm tells you this exists, but this statement that p = aq + r isn’t an algorithm

[chilly ocean]

trivially {1, g, g^2, ..., g^n-1}$ is a subset of {g^n : n in Z}

[lament dawn]

yeah

[lament dawn]

to do this, use that any element $k\in G$ is of form $g^\ell$ the apply division algorithm to $\ell$

[lament dawn]

and do what you were doing here

[cloud walrus]

The Migillope

[chilly ocean]

$\ell = qb + r$

[cloud walrus]

isomorphism

[lament dawn]

yes but you get to choose $b$

[cloud walrus]

The Migillope

[lament dawn]

so choose it intelligently. What choice solves the problem for you

[chilly ocean]

m

$l = qm + r$

[cloud walrus]

isomorphism

[lament dawn]

then what

[chilly ocean]

$g^l = g^{qm + r} = (g^m)^q g^r$

[cloud walrus]

isomorphism

[chilly ocean]

$g^l = g^r$

[cloud walrus]

isomorphism

[lament dawn]

yep

and with what you know about r...

[chilly ocean]

$r$ is the remainder

[cloud walrus]

isomorphism

[lament dawn]

well yes but to be done with the proof what do you do

[chilly ocean]

uh

$0 \leq r < m$

[lament dawn]

strictly less than m but yes

[cloud walrus]

isomorphism

[chilly ocean]

is there anything else?

[lament dawn]

no, thats it

just putting it all together

[chilly ocean]

what is the implication of this

[lament dawn]

well, we were trying to show that $<g> \subseteq {1,g, \ldots, g^m}$

[cloud walrus]

The Migillope

[lament dawn]

and you showed that, given a generic element $k\in \left< g \right>$, $k= g^r$ such that $0 \leq r < m$

[cloud walrus]

The Migillope

[lament dawn]

so... that's it

$k = g^r \in {1,g,\ldots, g^{m-1}}$

[cloud walrus]

The Migillope

[lament dawn]

mind this typo, should end at $g^{m-1}$

[cloud walrus]

The Migillope

[lament dawn]

now, if you haven't proven the division algorithm in class yet then you should use the double induction argument

but that seems incredibly unlikely

[chilly ocean]

would it be strong induction then?

[lament dawn]

not as you've written it

weak induction is fine

you do need it twice though

[chilly ocean]

how come

[lament dawn]

because if we start at 0 and go up, we dont get the negative numbers

hm, well, alternatively, you could do the induction for the positive numbers, then show that the set you are given is closed under inversion

either way there is just a little more work after the first induction

[chilly ocean]

is there any way to do this without induction or the division algorithm

[lament dawn]

bro 😭

[chilly ocean]

i just don't understand the division algorithm

like why are we using it

[lament dawn]

because it solves the problem

do you not see how it does that?

you wrote out the proof yourself...

[sullen island]

hi, this is taken from the wikipedia page on the Schur-Weyl duality, and im having trouble understanding this statement

why $U_i \otimes W = \mathbb{C} \iff U_i \cong W$ ?

[cloud walrus]

xy

[tender wharf]

you should open your algebra textbook to the first chapter where it will likely be covered

[ruby sundial]

$u_iz_i \otimes_A w_jz_j = z_i/z_j u_i \otimes_A w_j = C \implies u_i \otimes_A w_j = 1? $

Not sure if this makes any sense at the moment

[wooden ember]

U_i and W are simple submodules of A so you can write them as A/their complements

and these are comaximal if U_i is not W

so that their tensor product is 0

[ruby sundial]

Does comaximal mean Ui+W=A?

[wooden ember]

their complements

U_i^c + W^c = A (M^c denotes the direct sum complement in A)

[ruby sundial]

do you have any references for this?

[wooden ember]

not in particular sorry

what part is unclear?

[ruby sundial]

I havent learned much about simple modules so I am curious what reference did you use to learn about them. I have only encountered them on rare occasions

[chilly ocean]

I got (1/5,1/5) + Z^2, (2/5,2/5) + Z^2, (3/5, 3/5) + Z^2, (4/5, 4/5) + Z^2

is that correct?

[wooden ember]

there are more

[chilly ocean]

such as

wait

wouldn't (1/5, 2/5) technically work too

and (1/5, 3/5)

and (2/5, 3/5)

and (0, 1/5) too

there's so many of them

<@&286206848099549185>

[wooden ember]

yes those examples work too

at this point it seems youve got the gist of what makes an element be of order 5 though, you should be able to enumerate them

[coral shale]

do you have a visualization for this group

not that it necessarily helps the question

[static needle]

i'd think of that as just 2d euclidean space mod 2d space with integer coords

idk

[wooden ember]

visualizations are great but in this case it wont help much, the algebraic formalism is more than enough

you can visualize R^2/Z^2 as a torus if that interests you

[coral shale]

Its just a form of intuition

[coral shale]

Sometimes intuition like this lets you immediately see your guess can't be right

[static needle]

oh yeah that

[wooden ember]

oh yeah definitely

[static needle]

wait how is R^1/Z^2 homeomorphic to T^2

what's the homeomorphism

[wooden ember]

R^2 not R^1

[static needle]

sorry i meant that

[coral shale]

draw a unit grid on 2d plane.
You identify all the unit squares.
In particular, taking just one of them, notice the left and right edges are 'glued' together. And same for top and bottom edge.

[wooden ember]

we can move this to #point-set-topology and alg top if you want but its a quotient

[static needle]

oh just identifying edges of a unti square. got it.

[next obsidian]

R^2/Z^2 = (R/Z)^2 = (S^1)^2

[static needle]

there it is

though i did not know you could extract the exponent from a q.g

[coral shale]

Something to prove

[static needle]

crud

what IS the identity of R^2/Z^2

[lavish spoke]

When something asks you to "find" you just need to describe really. You don't need to list all individually, you can say something like "the elements of order 5 are precisely the elements of the form [...] for [... parameters]" and that should be ok

There's 24 of them here which is why I say this

[kind temple]

think of this as asking which complex pairs (u,v) on the torus have the property
that (u,v)^5 = (u^5,v^5) = (1,1)
there are 24 distinct pairs, by choosing 5th roots of unity in each component, discarding (1,1) (of order 1)
in terms of your answer, this is represented by (k/5, l/5) + Z^2 for k,l = 0,1,2,3,4, discarding (0,0) + Z^2 (of order 1)

[coral shale]

what is the group op
and shape of elements in there

[chilly ocean]

thank you very much

so in total, there are 24 elements of order 5?

[kind temple]

((x,y) + Z^2) + ((u,v) + Z^2) = ((x,y)+(u,v)) + Z^2

[coral shale]

asking them

[wooden ember]

on wikipedia it says that the lie algebra corresponding to the orthogonal group is the algebra of skew-symmetric matrices but in our class we defined $\mathfrak{o}_n(\mathbb{C})$ as the algebra of matrices $A$ such that $A^TS=-SA$ where $S$ is the matrix with $1$s on the antidiagonal and $0$s everywhere else

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

is there a reason for this discrepancy?

so for us we're taking the orthogonal algebra with respect to the symmetric form induced by S while for them they're taking the symmetric form induced by the identity...

and the prof seems to have done the same for sp_2n(C) as well: instead of the Hamiltonian matrices we get the symplectic algebra associated to the form with 1s on the antidiagonal of the upper right quadrant and -1s on the lower left quadrant (while hamiltonian matrices have the identity on the upper right quadrant and negative identity on the lower left quadrant)

[wooden ember]

okay prof answered and I feel pretty dumb: the matrices are congruent so induce equivalent forms, and this choice was made so that positive roots would lie in the algebra of upper triangular matrices

[glacial prism]

Is there an easy way of showing that the $\mathbb Z$-module of locally constant function ${X \to \mathbb Z}$ of the topological space $X = {0} \cup {1/n \ \vert\ n \in \mathbb N_{>0}} \subseteq \mathbb R$ is not $\mathbb Z$-free? Notice that any such function $f$ has a neighborhood $U$ of $0$ such that $f\vert_U$ is constant -- but $X\setminus U$ must be finite since any neighborhood of $0$ contains an epsilon ball and thus at a certain point all $1/n$. So $f$ can only take finitely many values. I explicitly tried to derive a contradiction assuming I have a basis, but this didnt lead me anywhere but doing tedious calculations. I hoped there might be a pretty argument.

[cloud walrus]

MrMonday

[coral spindle]

Why do you believe that this isn't a free Z-module? I think it is, actually.

In fact I think a basis is quite easy to write down

Were you maybe hoping to show that it doesn't have a finite basis?

[rustic crown]

btw i was wondering if it was a discrete space, then the locally constant functions would be a direct product of Z, how does one see if an infinite direct product is free or not?

[coral spindle]

Oh boy it's super hard, det. I saw a proof at some point, let me see if I can get it

[glacial prism]

Because it should be the $0$-th Alexander-Spanier (so Cech) cohomology group $H_{AS}^0(X,\mathbb Z)$ and this should disagree with the $0$-th simplicial cohomology of $X$ (as far as I know this is one of the examples where Cech and Simplicial disagree, next to the Warsaw circle) which can be calculated via path components, so $H^0(X,Z) = H^0({0}) \oplus \bigoplus H^0({1/n}, \mathbb Z) = \mathbb Z\oplus \mathbb Z^{\oplus \mathbb N}$ which is free.

[cloud walrus]

MrMonday

[glacial prism]

Could be wrong tho!

I calculated the AS-cohomology myself, so there is that for example

[rustic crown]

i think the following is a basis
f_n = 1 on 1/n and 0 else
and f_0 = 1 constantly

[coral spindle]

This is the one I was thinking of also

[coral spindle]

I can't find the proof, but it was in Roman's homological algebra, and it showed that Z^N is not free

[rustic crown]

so weird

[glacial prism]

Oh maybe

Mhm

Then I have to check everything I know. This is weird.

Oh maybe thats fine!

I am unsure of the path component splitting thing. I revisited my source and this says that 0th singular cohomology if function X->Z constant of path components, so Z^X not Z^(oplus X)

which is not free as you said

And then everything is fine ;D

Ok thanks for your help, I will see what I can do with that!

[lethal dune]

ye it's called the Baire specker group, there's a "proof" given in MSE
https://math.stackexchange.com/questions/320444/why-isnt-an-infinite-direct-product-of-copies-of-bbb-z-a-free-module?noredirect=1&lq=1

[coral spindle]

Whew, nasty

Thank you for sharing the link

[hollow fjord]

Hi guys!, I have a soft question, can a ring be non-associative ? Yesterday i was in my thesis defense (undergrad), and one of the teachers told me that rings are always associative, that i commited a fatal mistake specifying that in the thesis. But i did that because one of the books i used as reference specifies that the rings to be worked on are associative. So i'm a little confused and i do not know rings that are not associative

[coral spindle]

Rings are always associative.

It's true that rings are not always commutative, though.

There are other algebraic structures (e.g. Lie algebras) that are not associative though

[hollow fjord]

Thank you

Perhaps the definition of ring changed at some point after the publication of this book:c

[wooden ember]

Are there shortcuts to showing that a given set in a Euclidian space is a root system? Trying to exploit linearity usually doesn’t help too much since you have to handle quotients of scalar products and it’s kind of annoying

If it has a nice form it’s doable

[sharp sonnet]

can you tell me the author? wondering if they are french

[wooden ember]

But if it’s a small dimension exceptional root system say, it can be a bit frustrating

Like showing G_2 is a root system for example

[hollow fjord]

Frank W. Anderson and Kent R. Fuller

[glossy crag]

If $A,B$ are $k$-algebras ($k$ a comm. ring), is there in general an isomorphism $A\otimes_kB[x]\cong(A\otimes_kB)[x]$? I worked out the case $L\otimes_K K[x]\cong L[x]$ when $L\supset K$ is a field extension.

[rustic crown]

yee

the tensor product is nice

[cloud walrus]

Ocean Man

[rustic crown]

it's also associative, so notice that A[x] = A ⊗ k[x]

A ⊗ B[x] = A ⊗ B ⊗ k[x] = (A ⊗B)[x]

(all ⊗ are over k)

[glossy crag]

Alright, that's neat. The case of L/K I had to do by hand with bases.

[formal ermine]

wdym

[rustic crown]

it's more or less like this only

once you know z generates L over F=L^H, then its minimal polynomial is f(x) = (x-z_1)...(x-z_n) where these z_i are distinct elements in the H-orbit of z. this f is clearly a polynomial in F[x] as if you apply some h in H to f(x), it will just permutes the factors, so the coefficients of f are in L^H. moreover this is irreducible as ifg(x) was a polynomial in F[x] with z as a root, then automatically all of h*z are also roots. and so |H| >= |Hz| = n = deg f = [L:F],
now an automorphism in Aut(L/F) is determined by where it sends z to, and we can send z to any conjugate of z, this means |Aut(L/F)| = n and finally, the elements of H are automorphisms of L/F so we get
n = |Aut(L/F)| >= |H|
and so
n = deg f = |H| = [L:F] = |Aut(L/F)|

there is a proof of this using some linear algebra stuff, but i'm not too fond of that... that one feels more like a trick to me than some nice theory

[glossy crag]

How do I prove the isomorphism for a general k and A though? To define a map in one direction is easy, but how do I define its inverse?

[rustic crown]

both maps should be easy to define

A⊗k[x] --> A[x] is defined by sending a⊗x^n --> ax^n and then extending by k-linearity

and for the other direction, start with the map A --> A⊗k[x] (which sends a to a⊗1) and extend it to A[x] by sending x --> 1⊗x

now just check both composites are identity

uwu?

(also k here doesn't need to be a field, just a commutative ring is fine)

[formal ermine]

what's a fast way to find the fixed field

like I'm working on

,, L = \bQ(e^\frac{2\pi i}{13})

[formal ermine]

and I'm trying to find $L^{\Set{\operatorname{Id}, \sigma^3, \sigma^6, \sigma^9}}$ for $\sigma : \xi \mapsto \xi^2$

[formal ermine]

I was thinking of just going through each sigma, looking at what they fix in a linear combination, combine the spans

but that seems really tedious

is there a better way?

[rustic crown]

will have to think, sigma^3 sends xi to xi^8
so i think xi+xi^8+xi^-1+xi^-8 will generate the thing

[formal ermine]

here's the permutation table btw

[rustic crown]

say f = sigma^3, then what i wrote is xi + f(xi) + f^2(xi) + f^3(xi)

[formal ermine]

so for $\sigma^3$ the span is $a + b(x^2 + x^3 + x^{10}) + c(x^4 + x^6 + x^7 + x^9)$

[formal ermine]

doing this for all 2 other elements seems really time consuming though

[wooden ember]

is there a decent way to go about this that doesnt just involve calling up a known algebra of type B_3? I want to believe there's some nice argument with the Weyl group... I can show there are at least 18 roots with a few root chain considerations

[prisma ibex]

you might also use the fact that any two roots of the same length are in the same Weyl group orbit. The short roots are in one orbit, the long roots are in another. Now use the class equation/orbit stabilizer

[wooden ember]

Thanks I’ll try that when I get the time

I didn’t think to look at root lengths specifically

Pretty smort

[prisma ibex]

you still have to compute the Weyl group stabilizer on the long/short orbits separately and you still need to know what the Weyl group is and how it's acting

but at least this lets you just pick one short and one long root to compute with

[formal ermine]

ok so I found the fixed fields of each individual elements to be $\bQ(x^2 + x^3 + x^{10}, x^4 + x^6 + x^7 + x^9), \bQ(x^2 + x^{11}, x^3 + x^{10}, x^4 + x^9, x^5 + x^8 + x^6 + x^7), \bQ(x^2 + x^{10} + x^{11} + x^3, x^4 + x^7 + x^9 + x^6)$ how do I "combine" them?

[formal ermine]

uh replace $x$ by $\xi$ in your mind

[formal ermine]

is it just the intersection

[formal ermine]

it looks like it to me

cuz like

if it's in the intersection

it's fixed by all elements

so by definition

[glossy crag]

I think the fixed field ought to be $\bQ(\zeta+\zeta^5+\zeta^8+\zeta^{12})$.

[cloud walrus]

Ocean Man

[formal ermine]

should be + x^11 in the first here

[formal ermine]

the fixed field is $\bQ(\zeta+\zeta^5+\zeta^8+\zeta^{12}, \zeta^2 + \zeta^3 + \zeta^{11} + \zeta^{10})$

[formal ermine]

someone on mse replied already

[chilly ocean]

multiposter!!!!!111

[formal ermine]

cope

[glossy crag]

I think the second element is redundant, the first generates the fixed field, but w/e.

[formal ermine]

how do you get the second from the first?

[glossy crag]

By applying $\sigma|_F$ (where $F$ is the sought after fixed field), that ought to be an element of $\operatorname{Gal}(F/\bQ)$.

[formal ermine]

what does sought and ought mean

[glossy crag]

gesucht und soll

[formal ermine]

just speak normal english

lol

[cloud walrus]

Ocean Man

[glossy crag]

Looking at the MSE answer, Anne already explained how you get the 2nd element.

[formal ermine]

"edited 1 min ago"

[glossy crag]

Right.

I think I can give a more conceptual answer why F should look like it does than she did. At least it feels more conceptual to me.

[formal ermine]

no I totally understand their answer

[tired horizon]

How do u factor a polynomial in Q

Like this one

[chilly ocean]

probably using part 1

[tired horizon]

Ik

They said if P(p/q) =0

Then p devide A0

[chilly ocean]

so write out your options and check if they work

[tired horizon]

And q divide An

[warm wyvern]

Bro...

[chilly ocean]

so write out all the p/q with p dividing -8 and q dividing 6 and see which ones, if any, are roots

[warm wyvern]

Anyhow

[chilly ocean]

once you actually have a root it should be straightforward

[warm wyvern]

Is the hint here true?

And can't i just pove the contrapositive?

Ello tterra btw

[tired horizon]

Do I have to try all options?

[chilly ocean]

up to you

again,

and see which ones, if any, are roots

once you actually have a root it should be straightforward

[tired horizon]

I found this somewhere

[formal ermine]

what does that have to do with it

[tired horizon]

Idk its confusing me cause I think the teacher said that there is a way to find the root without trying out all possibilities

[chilly ocean]

simply get a root on the first try

[tired horizon]

gcd(p,q)=1

So its 2/3

[agile burrow]

The hint here does work

You can try the contrapositive, but I think the suggested outline is the most straightforward

[south patrol]

Side-note, that's pretty cringe using d and e for the powers of the polynomials

[warm wyvern]

Are you sure?

[south patrol]

The hint they gave is the way suggested in atiyah macdonald too iirc

Or at least, I've seen that method used to prove this before

[agile burrow]

not really, i'm just saying that I did the way suggested in the hint and it wasn't too long

[south patrol]

Idk how else you could prove it anyway

Contrapositive sounds wack

[warm wyvern]

Assuming such a b doesnt exist, then f(x)g(x) can't be 0 (coz the multiplication of the terms with the highest degree can't be 0) which means f(x) isn't a left 0 devisor

[south patrol]

well idk how you'd do it by contraposition

[south patrol]

how does this follow

[agile burrow]

just because f(x) g(x) isn't 0 doesn't mean the leading coefficient isn't a zero divisor

[south patrol]

like it could be that a_d b = 0 but a_0b isn't 0

etc

[agile burrow]

potato have you worked through atiyah macdonald

[south patrol]

Hm I worked through a lot of the earlier chapters like last year but need to do more exercises for chap 4 onwards

I think my algebra/cat theory is much better now so hopefully will be easier

hbu

Got commalg this term as i think i said, you do too right?

[agile burrow]

we don't have an official text for comm alg so I'm thinking of working through it this semester

yeah, we're just following hochster's notes

[south patrol]

ah idk those notes

atiyah-macdonald is good but sometimes feels a bit old fashioned idk

well just like how they treat the more categorical stuff

but yeah it's cool

good problems

[warm wyvern]

Oh, hmm

[agile burrow]

yeah, it seems like we'll be doing a decent bit of categorical stuff since the professor started defining categories today so I'll see if I'm better off just using atiyah macdonald as a reference. at the very least I should be able to do the exercises

[south patrol]

mhm cool

[solar glacier]

have solutions verification question

here G_a denotes the stablizer ofc

[south patrol]

But yes sure this looks gud

It'd be a lil clearer to me if you used like some notation for the group action but maybe it's also standard to omit that

Just seems like you're going between sigma acting on a and sigma actually being a function taking in a or whatever but that is probs fine

Anyway, I'm pretty sure that if you want you can write this as just one chain of equivalences

Like, show that $x \cdot \sigma(a) = \sigma(a)$ iff $(\sigma^{-1} x \sigma) \cdot a = a$

[cloud walrus]

potato

[south patrol]

Since what you did was show that each direction holds separately

[solar glacier]

true

thanks!

i have an alg qual exam in three days!

[south patrol]

Gl gl!

[solar glacier]

Thanks 🙏

[static needle]

probably mult/addition

considering Z can form a group with + and R can form a group with both + and mult

[next obsidian]

R isn’t a group under multiplication you have to remove 0

[barren sierra]

they really should get around to fixing that

[static needle]

^

[gleaming fable]

Hello
Which one is the correct definition?
1)y∈f(A) ⇔ ∃x∈A and y=f(x)
2)y∈f(A) ⇔ ∃x∈A such that y=f(x)
Please reply as fast as possible if you can I got an exam after 4 hours 🥲🥲🥲

[formal ermine]

correct definition for what?

[gleaming fable]

For y∈f(A)

Some people use the first one and some use the second and I just got confused

[oblique river]

what do you think the difference between them is

[gleaming fable]

One uses "and" and the other doesn't
ik might seem obvious but my tutor says always that we need to use the CORRECT definition exactly in order to get it right

[oblique river]

i mean i can see that there are different words

but what do you think the actual difference in meaning is

[gleaming fable]

Oh

Umm like

And is that operation "٨"

[oblique river]

can you give an example of a set $A$, an element $y$, and a function $f$ such that $y \in f(A)$ according to one definition but not the other?

[gleaming fable]

Ok wait

[cloud walrus]

Buncho Bananas

[oblique river]

"such that" is not a logical operator

it's just natural language

my point is

they are the same thing

there is no difference

except that imo the one that uses "and" is more awkward

[gleaming fable]

For example if we need to prove that
f(A⋂B) ⊂ f(A) ⋂ f(B)
I assumed that y∈f(A⋂B)
I tried to use firstly this definition :
⇔∃x∈A⋂B ٨ y=f(x)
⇔∃x∈A ٨ x∈B ٨ y=f(x)
Now I got stuck because if you have
Q ٨ P ٨ R you can't write (Q٨R)٨(P٨R)
But in ther other definition where "such that" is not an operation I can do it as
Q such that R ٨ P such that R

Yet the first one works in other situations as well so that's just confusing

[gleaming fable]

Yes that's what I mean
Some use the operation "and"
And some use "such that"

Now idk which one is the definition to use

[oblique river]

(Q and P and R) implies Q, P, and R individually

in particular, (Q and P and R) implies (Q and R), and it implies (P and R)

so it also implies "(Q and R) and (P and R)"

[gleaming fable]

Wait what

Is that possible??

[oblique river]

what do you mean

"A and B" implies A

"A and B" also implies B

and if X implies A and X implies B, then X implies "A and B"

those are the only rules i'm using

[gleaming fable]

Wait I got confused..

Ohhh

Wait

Can I use this :

R implies R and R ?

[oblique river]

yes

[gleaming fable]

And replace it?

[oblique river]

that's true

again

if X implies A, and if X implies B, then X implies A and B

now set X = A = B = R

so "R implies (R and R)" is true

[gleaming fable]

Ohh

I got it

Thank youuuu

[chilly ocean]

for i) is $\rho = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \end{pmatrix}$

[cloud walrus]

isomorphism

[chilly ocean]

and is $\sigma = \begin{pmatrix} 1 & 3 \end{pmatrix} \begin{pmatrix} 2 \end{pmatrix} \begin{pmatrix} 4 & 5 \end{pmatrix}$

[cloud walrus]

isomorphism

[chilly ocean]

bc $1 \mapsto 3$, $2$ preserves its position, and $4$ maps to $5$

[cloud walrus]

isomorphism

[chilly ocean]

<@&286206848099549185>

[tender wharf]

@chilly ocean is there a question

[chilly ocean]

yes

[tender wharf]

what is it

[chilly ocean]

whether these permutations are correct

[tender wharf]

oh you're asking whether your answer is correct

[chilly ocean]

yeah

[tender wharf]

yes

technically (2) can be left out since it's fixed

[chilly ocean]

and is $\sigma \rho^2 = \begin{pmatrix} 1 & 4 & 2 & 5 & 3 \end{pmatrix}$

[cloud walrus]

isomorphism

[tender wharf]

idk probably

im sick => too lazy to compute it

[chilly ocean]

🥲

[tender wharf]

but im gonna assume you squared it correctly

[chilly ocean]

do you know how to do iii)

[tender wharf]

does your textbook compose permutations left ot right

or right to left

uh yeah

do you know what lagrange's theorem states

[chilly ocean]

yes

[tender wharf]

great

then it should be easy

[chilly ocean]

the cardinality of G equals the cardinality of H multiplied by the index of H in G

[tender wharf]

can D_5 have an element of order 7?

[chilly ocean]

which implies that the order of H divides the order of G

[chilly ocean]

uh

no

[tender wharf]

there you go

[chilly ocean]

that's not really

a mathematical argument

tho

[tender wharf]

well

why can it not?

[chilly ocean]

hmm

the order of the element divides the order of the group?

[tender wharf]

good

does 7 divide 10?

yeah obviously not

lmao

[chilly ocean]

is the order of the group 10

[tender wharf]

yeah

[chilly ocean]

why

[tender wharf]

the symmetries of a regular pentagon is D_5

that's what D_5 is

D_5 has 5 rotations and 5 reflections

[chilly ocean]

isn't the order of the group 7 lol

[tender wharf]

why would the order of the group be 7

[chilly ocean]

cuz

uh

[tender wharf]

you know what D_5 is right

[chilly ocean]

im dumb

[tender wharf]

lol

get some sleep my guy

[chilly ocean]

so because $7 \not\mid 10$

[cloud walrus]

isomorphism

[chilly ocean]

fuck

[tender wharf]

err

[chilly ocean]

why does it look so ugly

[chilly ocean]

how do u know

[tender wharf]

happens to me too

when im sleep deprived i miss the most trivial shit

yes 7 doesn't divide 10, which is the order of D_5, the group of symmetries of a regular pentagon

[chilly ocean]

so the main corollaries of lagrange's theorem is that order of a subgroup divides the order of a group and the order of an element divides the order of a group

right

[tender wharf]

yup

rather

the first thing you stated is the theorem

and the next one is a corollary

[chilly ocean]

and these are the implications

[tender wharf]

yup

[chilly ocean]

I see

[tender wharf]

now go to sleep

[chilly ocean]

thank you

uh

im in school

[tender wharf]

oh

well get some sleep when you can

[chilly ocean]

thank you will try

thanks for the help

[tender wharf]

no worries

[delicate orchid]

just for future reference, $7 \not | 10$ looks a bit better

[cloud walrus]

Wew Lads Tbh

[coarse forge]

\nmid gang (except you need the amssymb package)

[woven obsidian]

The map $GL_2(\mathbb{Z})\to GL_2(\mathbb{F}_p)$ is surjective right?

[cloud walrus]

AoiKunie

[woven obsidian]

I know that the map when you repace GL by SL is

[wooden ember]

That’s not a well defined map

Oh sorry over Z

[woven obsidian]

Just realised it isn't surjective

[wooden ember]

Im assuming that’s unimodular matrices?

But yeah won’t be surjective then

Cause matrix will remain unimodular

[woven obsidian]

yeah

[wooden ember]

In part d) does anyone get why the restriction of Delta to Inn(L) looks like an inner derivation? I’d need a result that would say something around the lines of Der(Inn(L)) = Inn(Inn(L)) but I don’t see why that’s a priori true

[somber sleet]

hey guys, if we have a diedergroup Dn=<r,s>, why is the commutator r^2? I don't see how you get it from rsr^-1s^-1

[lethal dune]

denoting ' as the inverse, note that s'=s so sr's'=sr's=ssr'' (generator reltn) =s²r''=r''=r, so we have r(sr's')=rr=r²

[somber sleet]

sr's=ssr'' why do you have this equality?

[lethal dune]

srs'=r'

(by definition?)

verify if you aren't sure

[somber sleet]

the problem is, that we didn't even define it

[somber sleet]

if I have this equality, then it's "trivial", like I see it

[glossy crag]

That makes sense, I was fixated on representing elements of A in terms of k. Thanks

[lethal dune]

you can verify it whatever way you have defined D_n

[coral spindle]

Here are defining properties of Dih(2n) which I personally find helpful for calculations. A group G is isomorphic to Dih(2n) if:
(1) it contains a normal, cyclic subgroup <r> of order n,
(2) it contains a cyclic subgroup <s> of order 2,
(3) srs' = r',
(4) and finally G = <r> <s> or equivalently G = <s> <r>.
I hope this helps.

Maybe also it would be better to write (3) as sr = r's.

[warm wyvern]

Can someone give me a hand with this?

I still don't see it

I get why a_d b_e=0

[cloud walrus]

DarQ

[warm wyvern]

And so on

But i don't see why all the terms must be 0

[chilly ocean]

use minimalist

minimality

autocorrect

the degree of a_d g is less than e

[warm wyvern]

OH

It's so obvious once you see it

[rustic crown]

i wonder why he says left-zero divisor if R is commutative

maybe that's why you didn't see it >.<

to say f * (a_d g) = 0 you need commutativity

[warm wyvern]

I just spent 15 minutes thinking "wait... doesn't R have to be commutative?"

Lmfao

[rustic crown]

[warm wyvern]

I need to remember checking out the errata

[rustic crown]

yea aluffi's first print has a lot of errors >.<

even the second one had quite a few

[warm wyvern]

I don't know where I can get the second one

[elder wave]

just assume every ring to be commutative

[rustic crown]

351

[warm wyvern]

Omg

[rustic crown]

second print has 98

[warm wyvern]

The holy library doesn't have it

[rustic crown]

it doesn't? 🙈

[warm wyvern]

Nope

None that I can see, at any rate

Soo

If anyone has a LEGAL version, please dm me :)

[crimson ibex]

Given X a finite set, why is a function g: X->X surjective if and only if g is injective?

[south patrol]

Not really algebra i guess

But intuitively, hopefully it is clear? If you want to fill up n boxes with n items, you can't put more than one item in each box

[chilly radish]

Not really abstract algebra, this is more #proofs-and-logic . To give a brief answer, if g is surjective, then suppose that X has n elements. If two elements were to be mapped to the same element (i.e. g is not surjective), how many elements will the image have?

if you want to prove this rigorously, induct on the cardinality

[crimson ibex]

I see thanks, sorry for the wrong channel posting.

[chilly radish]

you're all good

[solar glacier]

if G is a group and H is subgroup. Then H has index p in G iff it is the kernel of an onto homomorphism \phi:G \to Z/pZ

does this use the first iso theorem

supposing H has index p then

[formal ermine]

I swear you already asked this once like a month ago

could've been someone else though

[solar glacier]

nope not i lol

[formal ermine]

lemme find it

#groups-rings-fields message

[solar glacier]

sorry been doing so much algebra lately forgot i did this already

[formal ermine]

lol don't worry

[celest furnace]

Don’t know if you still need it but index p means G/H is iso to Z/p, and clearly H is the kernel of pi: G -> G/H, composing this with the isomorphism does the trick

For one direction

[slate mortar]

In the definition given by my teacher, an H-group is an homotopy associative H-space which has an homotopy inverse map but only along the right hand-side of the multiplication map. Is it also a "left inverse"? Or is it the case if and only if the space is contractible?

[wind steeple]

it implies it has a left inverse

like for groups

[slate mortar]

How?

[wind steeple]

like for a group

if a monoid has all right inverses then it has all left inverses which are the same as the right inverses

[slate mortar]

Oh and this translates in a commutating diagram proof I guess?

[slate mortar]

Anyway, thank you I didn't know about that

[wind steeple]

yes it should be the same for H-groups

[long obsidian]

Does someone have an easy example of a non simple tensor that they can explain to me why it's not simple? I saw an example of taking $e_1, e_2$ a basis of $R^2 $ euclidean space and considering $e_1 \otimes e_2 + e_2\otimes e_1 \in R^2 \otimes R^2$ as a non simple tensor but it wasn't clear to me why. They showed it was non simple by contradiction and showed that a system of equations was inconsistent but I can't tell the relevance

[cloud walrus]

fajitas

[south patrol]

I think that basically is the simplest example so I'm not sure what you want

[long obsidian]

My bad but can you explain why it's non simple? It seems like some thing about dependency but I'm not sure how to argue the non simplicity exactly

[cloud walrus]

potato

potato

[south patrol]

Is that clear?

[long obsidian]

Does the system of equations come about because $e_1 \otimes e_1 , e_1 \otimes e_2 , e_2\otimes e_1 , e_2 \otimes e_2$ are linearly independent in the quotient space of $R^2 \otimes R^2 $ so it must be that their coefficients are zero? But then no such coordinates of v and w could satisfy that said system of equations

[cloud walrus]

fajitas

[south patrol]

Well I wouldn't say the quotient space, not sure what u mean

But yes they are linearly independent in the tensor product

And so that's why I equated coefficients of the ei tensor ej in my solution

[formal ermine]

just for my own sanity

the roots of $x^p - 2 \in \bQ[x]$ are $\sqrt[p]{2}\zeta, \sqrt[p]{2}\zeta^2, \ldots, \sqrt[p]{2}\zeta^p$ where $\zeta$ is the $p$th root of unity

[formal ermine]

yes?

(ofc zeta^p = 1 fwiw)

[south patrol]

ye

well you should say "a primitive pth root of unity" (in some extension field) but ye

[delicate bloom]

if you don't say that we are obligated to pick zeta = 1

[formal ermine]

what does primitive mean here

[south patrol]

it means it's of order p in K^x

[formal ermine]

isn't there only one primitive pth root of unity

[south patrol]

No

Well like this all depends on what field we're using lol

But if there is one primitive pth root, then there are p-1

(if p is prime)

More generally there are phi(p)

This is just because cyclic groups of order n have phi(n) generators

[formal ermine]

ah

how can I imagine $\zeta^k$ visually?

[formal ermine]

like if I have the unit circle

and $p$ regularly spaced points on it

[formal ermine]

then what does the operation $\zeta^m \cdot \zeta$ correspond to

[chilly ocean]

all elements of the unit circle are rotations

[formal ermine]

yeah but by what amount?

if it's just one left/right it wouldn't work

[ruby sundial]

e^ix->cos x + isin x ->cos x & -sin x \\sin x & cos x

if zeta^n=1 then zeta^k corresponds to a 2pi/k rotation

[formal ermine]

if zeta^m is the kth point (enumerated), then what point is zeta^(m + 1)?

[ruby sundial]

i wrote it wrong lol

what is zeta?

[formal ermine]

a primitive pth root of unity

[ruby sundial]

what form are you looking for?

we can just say its the zeta^m+1 point?

do we have any more information about m?

if you want a visual

imagine we have unit circle in complex plane

draw a line from 0 to 1

[formal ermine]

if we enumerate the points

and zeta^m is the kth point

then what point is zeta^(m + 1)?

[ruby sundial]

if we have that zeta is kth root of unity which means zeta^k=1 then imagine making copies and rotating the 01 line by 2pi/k

it depends ig?

if m=n-1

then zeta^m+1= 1

[formal ermine]

that's not my question

please

[ruby sundial]

it is?

it depends on what root of unity zeta is

[formal ermine]

I want a general form

[ruby sundial]

zeta^k= e^2i*pi k/n for zeta an n’th root of unity

[formal ermine]

that is not my question

please carefully read my question again

[ruby sundial]

no

i said it can be k+1’th or 1

it depends on which root of unity zeta is

i dont know why you arent satisfied

[formal ermine]

this does not answer my question

[ruby sundial]

can you elaborate on how?

you asked for two questions #groups-rings-fields message

#groups-rings-fields message

[formal ermine]

???

that's the same question

I give you the kth point, you multiply it by zeta, which point do you have now?

[ruby sundial]

i said it can be k+1’th or 1

[ruby sundial]

here

[formal ermine]

could've just said that at the beginning

but thanks

[ruby sundial]

[vagrant zinc]

Guys in this Cayley table, I have interpreted that the only table that fulfils for it to be a group is C, if a belongs to group G and that is the identity element then C would only fulfil

[tender wharf]

are you asking for your answer to be checked?

[vagrant zinc]

Well, they tell me to prove that it is a group, and the only one I see that fulfils the condition is the C group.

[tender wharf]

if that is so then you are correct

[ruby sundial]

to solve these quickly find the identity element by looking at each column and making sure that it acts like identity when looking down rows.
then check for inverses by picking seeing if the identity is in each row/column

[ruby sundial]

what about b?

[celest furnace]

if you know that there are only two groups of order 4, then you can just look for $\bZ/4$ and $\bZ/2 \oplus \bZ/2$ in there

[cloud walrus]

Geopchad

[ruby sundial]

good thing they dont ask for which group

[next obsidian]

If you’re looking at a cayley table

And asking if it defines a group

I don’t think you’d know how many groups of order 4 there are

This is probably like from week 1 of a group theory class

[ruby sundial]

oh yeah people are in school today

[vagrant zinc]

I think the multiplicative operation fails in the second row.

[ruby sundial]

why?

[celest furnace]

probably associativity would be my guess

[vagrant zinc]

Eha if I read that example, I understood it, I just need to wait for the exercises, I'm working on it :'D.

[ruby sundial]

hint it is a group

[vagrant zinc]

bb=a it could not be possible

[ruby sundial]

sure it can

[vagrant zinc]

So you tell me that the bb=a where would give us the identity element why ?

I don't understand

[ruby sundial]

well doesnt a^2=e in integers mod 2?

[vagrant zinc]

I didn't know that, where could I read about it?

[coral shale]

check klein 4 group

[ruby sundial]

do you know about modular arithmetic?

[coral shale]

thats not relevant to the original Q

[celest furnace]

by the way Aylou this is $\bZ/2 \oplus \bZ/2$

[cloud walrus]

Geopchad

[ruby sundial]

it isnt thats why i pinged the question they asked.

[coral shale]

apart from this how is C2 relevant

[vagrant zinc]

Yes of course, but I saw it in a numerical way eha not so deep, but yes.

[coral shale]

that aside they probably havent seen direct products

[ruby sundial]

ig the takeaway of the exercise is that you should think about the properties for a group

[coral shale]

The quickest way to approach the question is to know the groups of order 4

[vagrant zinc]

I've never seen that from Klein's group

[coral shale]

There are only 2

otherwise its a pain to prove axiom such as associativity

[celest furnace]

shuri this is what i was saying

[coral shale]

ik

[celest furnace]

then D monkey said that they prob don't know that

[ruby sundial]

in group tables, associativity and closed are usually implicitly there because you cannot determine order of operations being applied and closed if letters on the inside match letters on the border

[coral shale]

Then prove assoc by brute force 💀

[ruby sundial]

so you only need to check for identity and inverses which are given by checking this #groups-rings-fields message

[coral shale]

how can u see assoc by inspection?

[vagrant zinc]

OH mr
I have seen the example you gave me

[ruby sundial]

you dont see associativity because its implicit in how the table is supposed to work

[coral shale]

no it isnt?

is it?

U can construct a non associative table cant u?

[vagrant zinc]

The book gives it I had not seen that detail, I really am an idiot.}

[ruby sundial]

associativity is a law requiring three inputs, a group table only has multiplication in its definition

[coral shale]

???

I feel like uve misunderstood what assoc is

there only ever is one operation in a group

the assoc property refers to one op only

[ruby sundial]

a*(b*c)=(a*b)*c

the group table only tells about a*b

[coral shale]

no...

it tells u what bc is

hence what a(bc) is

[ruby sundial]

thats my point

[coral shale]

???????

[ruby sundial]

you can only know result of two elements at a time with a group table

[coral shale]

bruh

if u need to know what a(bc) is

compute bc

then compute a(bc)

[ruby sundial]

correct

[coral shale]

2 lookups

. a b c
a c a a
b b a a
c a a a

(ab)c = bc = a
a(bc) = aa = c

Not associative

===
Check this example.

[ruby sundial]

there is no identity

[coral shale]

Thats irrelevant

[ruby sundial]

no it isnt

[coral shale]

Ive constructed a table that is non assoc

contrary to your previous claim

[next obsidian]

[coral shale]

Anyways without me constructing an explicit example

[ruby sundial]

#groups-rings-fields message

[coral shale]

Algebraic structures that are groups without assoc axiom exist

[ruby sundial]

this always works though

[coral shale]

Hence their cayley tables will be non assoc ofc

[ruby sundial]

showing unique identity also shows associative ig

[coral shale]

It works only to show not a group

Your steps fail to prove it is a group

[coral shale]

no.

[ruby sundial]

it does

[coral shale]

If your claim is true, then a group need not specify it needs to be associative

this property would be redundant

[ruby sundial]

i disagree

[coral shale]

since identity => assoc according to you

[ruby sundial]

nah this is different

[vagrant zinc]

eha

[ruby sundial]

i agree now

[coral shale]

Just because almost all intro to algebra studies associative structures doesnt mean non-assoc ones dont exist. You just arent familiar with the examples (and neither am i)

[vagrant zinc]

Thanks!

[ruby sundial]

the examples are stupid pathological

[vagrant zinc]

Thanks for that tip 😄

[coral shale]

https://upload.wikimedia.org/wikipedia/commons/thumb/3/3f/Magma_to_group4.svg/820px-Magma_to_group4.svg.png

found it

[vagrant zinc]

THANKS ❤️!

[coral shale]

so what u want is...

a loop?

lol this is non trivial

[ruby sundial]

yeah these have to be bad looking

[coral shale]

brief scan, lie algebra might have examples

[coral shale]

Good to know theres a small finite example

[ruby sundial]

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

[agile burrow]

I'm reading that B3, the braid group on three strands, is the universal central extension of PSL(2, Z), but I was under the impression that a group needs to be perfect to have a universal central extension. Maybe I'm just missing something in the definitions

[formal ermine]

I wanna find the galois group of $x^p - 2$ for $p$ prime

[formal ermine]

I know that its order is $p(p - 1)$

[formal ermine]

so it's either $\bZ_{p(p - 1)}$ or $\bZ_p \rtimes \bZ_{p - 1}$

[formal ermine]

I know that a $\sigma$ sends $\zeta \mapsto \zeta^b$ for $1 \leq b \leq p$ and $\sqrt[p]{2} \mapsto \sqrt[p]{2}\zeta^a$ for $1 \leq a \leq p$

[formal ermine]

can I conclude that it's cyclic and therefore $\bZ_{p(p - 1)}$

[formal ermine]

or hmm

on second thought I'm not sure if it's actually cyclic

ah it doesn't even matter for my actual problem though

cuz the semidirect product one is just the cross product with a different operation, no?

[rustic crown]

this claim is false, for example if p = 5, then there are two abelian groups of order 20.

[rustic crown]

yep

[formal ermine]

hmm then how do I find the galois group

[rustic crown]

you know how to compose two automorphisms, that's more than enough to find the group up to isomorhpism

[formal ermine]

yeah but I don't wanna do so much work

[rustic crown]

is it that much work?

say t = 2^(1/p) and z = exp(2 pi i/p)

then a given automorphism is just a pair (a, b) like you wrote where t --> t*z^a and z --> z^b

the composite can be calculated directly

[simple valley]

Is this problem asking $\exists n, \forall W$ or $\forall W, \exists n$?

[cloud walrus]

mniip

[rustic crown]

(a1, b1)(a2,b2) should be the pair which sends
t --> t * z^a2 --> t * z^(a1 + b1a2) and
z --> z^b2 --> z^b1b2

which is (a1+b1a2, b1b2)

[delicate orchid]

I believe it’s there exists an n for all W

[simple valley]

This clarifies nothing

W are the irreps of G, do all W share the same n

[lethal dune]

Is this Serre?

I believe it's ∀W ∃n

[chilly ocean]

now we have two people claiming the exact opposite

[coral spindle]

One implies the other

Idk which one the question is specifying, but I'd certainly try to prove the weaker one

[lethal dune]

the wording implies the other way

ok idk anymore what the wording says

[simple valley]

The hint in the answers section seems to suggest the weaker version

[coral spindle]

What's the hint?

[lethal dune]

@coral spindle

[simple valley]

I don't see how this would let you relate the n for different phi

[lethal dune]

Are you asking why this gives you one n that works for all irreps?

I don't think that directly follows but we get the weaker version of the statement

[formal ermine]

@rustic crown I thought of a better proof

I was able to show that the galois group of x^n - a embeds into Zn rtimes Zn*

[lethal dune]

https://mathoverflow.net/a/192103
here's a second proof but long

[glossy crag]

How to show $\Phi_{p^{r}}(x)=\Phi_p(x^{p^{r-1}})$ is irreducible for all $r\geq1$? I know the proof that all $\Phi_n$ are irreducible, it's this special case I'm interested in. The hint suggests the substitution $x\mapsto x+1$ and Eisenstein, just like for $\Phi_p(x)$, but the substitution doesn't help me.

[cloud walrus]

Ocean Man

[runic hemlock]

it works in exactly the same way...

[hot lake]

If G = S2 and V = the nontrivial irreducible representation (which is faithful) then V^n alternates between the two irreducible representations so you can't have an n that works for both.

[glossy crag]

How so? For r=1 we have a simple expression for Q_p(x+1), namely (x+1)^{p}-1/x, that allows the direct application of Eisenstein. For r>1 the expression becomes the complicated (x+1)^{p^r}-1/(x+1)^{p^{r-1}}-1 and I personally don't see how to apply Eisenstein to this.

[runic hemlock]

a polynomial is eisenstein if it is a monomial mod p and the last coefficient is not zero mod p^2

in this case the last coeff is p

[glossy crag]

I am aware, I am saying that myself I don't see a good concrete expression for Q_{p^r}(x+1) as one would for Q_p(x+1).

[glossy crag]

I realise this, it's the other coefficients that bother me.

[runic hemlock]

do you know that (x+1)^{p^k}=x^{p^k}+1 mod p

[glossy crag]

Hm, I didn't think to reduce mod p from the outset, was trying to get a concrete expression for Q_{p^r} (as one does for r=1)...

This works, thanks.

[south patrol]

Moreover I'm fairly sure Q_{p^r}(x) = Q_p(x^s) for some s which gives another way to deduce it

[west sinew]

There seems to be two distinct ways in which the phrase "solvable by radicals" is used, and it is unclear to me how they relate. Can somebody explain?
i) A polynomial is solvable by radicals if its splitting field is a subfield of a radical extension.
ii) A polynomial is solvable by radicals if its solution has a closed form algebraic expression (did I say it right?).

[hot lake]

ii) should be if all its solutions have a closed form algebraic expression (obtained from rational numbers using field operations and nth roots)

the roots you have to take in ii) tell you what radical extension you use in i)

[solar glacier]

Does anyone else’s discord not work on PC or Mac I was tryna ask a question but it won’t start on my laptop 💻 😞

[tender wharf]

ah yes, abstract algebra question

press ctrl-r to reload discord on pc

[solar glacier]

I have mac

[formal ermine]

check your system's tray

[solar glacier]

What’s a tray lol

The control alt delete button ?

[tender wharf]

try cmd-r then

[solar glacier]

I’m gonna restart my laptop then see what happens

[solar glacier]

im trying to show that M is maximal iff R/M is a field using the 4th iso theorem. is this correct:

[cloud walrus]

MyMathYourMath

[solar glacier]

since no such I' exists

since the ideal I does not exist

[solar glacier]

whole RING* (which is ideal of itself) or trivial ideal

[elfin furnace]

yeah looks right

[solar glacier]

cool thx

[formal ermine]

what does "$\on{Gal}(\bL/\bK) \text{ acts transitively on } \Set{\alpha_1, \ldots, \alpha_n}$" mean

[formal ermine]

$\alpha_i$ are the roots of a separable and irreducible polynomial $f$, $\bL$ is the splitting field of $f$,

[rustic crown]

since Gal(L/K) are automorphisms of L, they clearly act on L. now roots of f will get mapped to roots of f under an automorphism, so you get that any such automorphism restricts to an action on {a1, ..., an}

you say Gal(L/K) acts transitively if this action is transitive