groups-rings-fields

[regal tulip]

Hello everyone, I have a simple (I think?) question: given a group $G$, can I say something about its order, knowing that for some $g,h \in G$ we have that the discrete logarithm of $g$ w.r.t. $h$ is k, that is, $g = h^k$ and $k$ is the smallest positive integer for which this happens?

[cloud walrus]

Miyo

[regal tulip]

I'm working on elliptic curves and knowing this would be helpful

[lethal dune]

Is your group finite?

Also does the log always exist?

[regal tulip]

My group is finite yes, its order is between 30 and 54

And I know that there are $g$ and $h$ so that $g = h^23$

[cloud walrus]

Miyo

antichronos

[regal tulip]

Yes, sorry,

[formal ermine]

I don't

[lethal dune]

there's a lot things you can say about the group with this information

[chilly radish]

Do you know if h is not the identity?

[lethal dune]

for example, if you know g =e then there is an element of order 23 and hence order of your group is 23n for some n. using the bounds on the order you get n=2 and hence the subgroup generated by that element is normal and you can classify your group

[regal tulip]

Yeah I feel so, I was trying to conclude something about the existence of elements of certain orders, so that the order of the group must be a multiple of those, and maybe getting an answer knowing that the order is between 30 and 54

[lethal dune]

in case g≠e you can one element has order greater than 23 that forces group to be cyclic

[regal tulip]

It is not

[lethal dune]

I should have spoiler-ed that mb

[regal tulip]

Oh by the way
I know even it's an Abelian group

[lethal dune]

lol

[regal tulip]

It's the group of an elliptic curve if that can help, and I'm using Hasse theorem

[lethal dune]

I'm pretty sure your group is || C_23×C_2 or C_46||

unless I made some trivial mistake

[regal tulip]

Why tho? With some reasoning I tried to conclude that the order must be 46 but I'm not so sure it's right

[chilly radish]

Those r the same group

Cathode Ray Tube Theorem

[regal tulip]

Oh right so just C_46 using the structure theorem of fg abelian groups

[chilly radish]

Oh is that what you meant?

[formal ermine]

where can I find this?

[chilly radish]

I was making a joke on CRT

Chinese remainder theorem

[formal ermine]

,w is 23 prime

[cloud walrus]

Results provided by WolframAlpha

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Requested by projective#0

[formal ermine]

ohhhh

I thought 23 wasn't a prime

[chilly radish]

You don't need 23 to be he prime

It's odd so it's coprime to 2

[lethal dune]

h²³=g and o(g) | o(G) so 23 | o(G)

now use the bounds on o(G)

sorry my internet got cut off

[lethal dune]

lmao

[lethal dune]

it's divisible by 23

[formal ermine]

23 feels like it should be divisible by 7, change my mind

[regal tulip]

I know $Q = 23P$ (let me use additive notation). Let $N$ be the order of Q, then

$0 = NQ = 23NP$

So we have

$ord(P) > 23$
\vskip 0.2 cm
$ord(P)$ divides $23N$ for some $N>0$

And from this I think I can conclude that the order of $P$ and of the group is $46$ (alongside with the bound of the order of the group being between 30 and 54), but I'm not so sure

[lethal dune]

7π ≃ 23

[cloud walrus]

Miyo

[chilly radish]

That would give you that the order of P is \leq 23

But the order of Q is more than 23

Wait sorry

I got it the wrong way around my bad

I think you're right

[regal tulip]

But how do you conclude that 23|o(G)? I mean, if h was the identity I would agree

[chilly radish]

23|o(P)| |G|

[regal tulip]

Wait, if $h^{23} = g$ can I conclude that $23 | o(g)$?

[cloud walrus]

Miyo

[chilly radish]

Lagrange

[lethal dune]

say gⁿ=e then (h²³)ⁿ=e so h²³ⁿ=e

[regal tulip]

That is if you know a priori the group is cyclic

So if you know that such an n exists

[chilly radish]

The group is finitr

Every element has finitr order

No need for cyclicity

[lethal dune]

again Lagrange

[regal tulip]

...
Well well well, looks like someone needs to go back to first year of university

It might be me, who knows

Thank you, I got lost for nothing

[rocky cloak]

I mean it's true for C24 or any cyclic group relatively prime to 23 and bigger.

So you can't really conclude much

In fact h can have any order bigger than 23.

Do you perhaps have some more information about the relationship between P and Q?

For example if you know that h is not a power of g, then the order of g must be smaller than that of h, thus h must have order 23n for some n.

[regal tulip]

And in fact, big plot twist, I found an online calculator for elliptic curves that told me that the order of the point P is actually 39

[regal tulip]

I'll give the complete text at this point 😅

Suppose $E(GF(41))$ is an elliptic curve of equation $y^2 = x^3 + 2x + 1$ on $GF(41)$. We know that the discrete logarithm of $Q = (30, 40)$ w.r.t. $P = (0,1)$ is $23$. Find the order of $E(GF(41))$

[cloud walrus]

Miyo

[regal tulip]

There is a theorem (Hasse) which tells us that the order is between 30 and 54, and from that I don't think it matters it's a group of an elliptic curve (if not for the fact that it's abelian), it's just a problem of group theory

[lethal dune]

actually now that I think of it, it's false

look at C_5 and pick nontrivial h with g=h², same argument should give 2|5.

what we get instead is 23|G| is divisible by o(g)

[rocky cloak]

Have you tried computing 2Q?

[regal tulip]

No, why?

[rocky cloak]

Well if 2Q is the identity then P has order 46, so the group is cyclic of order 46

[hot lake]

Points of order 2 here must have y=0

[regal tulip]

Well the problem is that I found a calculator online for elliptic curves which tells that the order of P is actually 39

And I don't know how I am suppose to deduce that

[hot lake]

I counted 39 too

[regal tulip]

Also, this

[hot lake]

Not sure what we're supposed to do with the information they gave tbh

[regal tulip]

Wait, counted?

[rocky cloak]

Alright, then the group must be cyclic of order 39

[regal tulip]

Well
Yes, I agree
But normally I wouldn't be allowed to use online tools 😅

Other thing I found out (by brute force) is that there is no element of order two, so the order of the group must be odd

[loud thunder]

Let F be a field and let G be a finite group of order n. Show that there exists a non-trivial homomorphism from G to the multiplicative group of F. ------------------??

[chilly radish]

Shit you're right

[rocky cloak]

That's not true in general

[chilly radish]

We're dumb

[rocky cloak]

For example take F = GF(2) or take F = GF(3) and n odd. Or F = Q and n odd.

[regal tulip]

Well it really wasn't me in the end

[lethal dune]

fair

[stuck fiber]

So is this saying that $A^{f}$ is the image of $f$ or is it just all the elements of A which are mapped to A'?

[cloud walrus]

kenshin5334

[chilly ocean]

A^f is the image of f

it is a subset of A'

what you wrote is not only a subset of A, but actually all of A trivially

[stuck fiber]

Yeah I originally wrote that but wasn't sure if it was fully correct

Thank you

[barren sierra]

I hate that notation lol

f(A) is so much nicer

[south patrol]

Hm if $G$ is a finitely generated abelian group, why do we have $\mathrm{Hom}_{\mathbb Z}(\mathbb Z/p^{\infty}, G) \simeq 0$? Here $\mathbb Z/p^{\infty} = \mathrm{dirlim}_n \mathbb Z/p^n = \mathbb Z[1/p]/\mathbb Z$

Actually this is probably easy

[cloud walrus]

potato

[south patrol]

Oh, just use e.g. classification of finitely generated abelian groups lol

[next obsidian]

Yeh

[south patrol]

G = Z is easy, for G = Z/q^k (q not p) consider p-torsion in Z/q^k, for G = Z/p^k note every element of the domain is a p^k-th multiplication of smth else

nice

And I guess it trivially doesn't hold in general (identity map on Z/p^infty)

:)

[chilly radish]

Why would you necessarily have p-torsion, am I missing something

Oh wait sorry you mean every element of prüfer is p-torsion

I still don't understand your phrasing, but it's just the fact that you'd need the image to have p^k torsion elements for all k innit

[south patrol]

Which bit of phrasing sorry

[chilly radish]

The part about p \ne q

[south patrol]

Oh yeah I meant like any map Z/p^infty -> Z/q^k must be zero as any element in the image has order dividing both q^k and a power of p

[chilly radish]

Ah ok you're taking primary decomposition here

Alright

[south patrol]

ye for ease lol

[chilly radish]

Fair enough

[south patrol]

ye sure lol oop

[white oxide]

can i get a hint to show that if R is an integral domain then the set of all torsion elements of A is a submodule of A

i'm just struggling to show existence of inverses

i feel like it's so obvious

[chilly ocean]

inverses?

[white oxide]

or rather that if a is in T(A) then so is -a

sorry should've specified A is an R-module

[chilly ocean]

did you show that if a is in T(A) and r is in R then ra is in T(A)?

[white oxide]

yea

[chilly ocean]

-a = (-1)a

-1 meaning the additive inverse of 1 in the ring R

[white oxide]

yeah that makes sense for some reason my dumbass got caught up in thinking about Zp and the additive inverse of 1 lmao

[chilly ocean]

there are only 3 things you need to check when you're proving that something is a submodule. contains 0, is closed under addition, and is closed under scalar multiplication

you don't need to check the entire list of module axioms. some of them are immediate from these ones and the fact that your submodule (well, the set that you're proving is a submodule) is just a subset of a module

[white oxide]

oh i thought you had to show that it was also an additive subgroup, meaning that you had to show existence of additive inverses

or does one of those things that you have to check imply that

because an additive inverse -a of a need not be in a submodule right

[chilly ocean]

closure under scalar multiplication implies closure under taking additive inverses

[white oxide]

oh yea right

that makes sense

got it ty

[stuck fiber]

Is this statement really trivial or am I tripping

[rocky cloak]

Guess it depends on your definition, but if N_H is defined as the union of all subgroups where H is normal, then it's pretty trivial yes.

[stuck fiber]

Yeah in this case I think it should be the set of all k\in K such that kHk^{-1}=H

[rocky cloak]

Still pretty trivial

[stuck fiber]

Yeah

[stuck fiber]

So is $\lambda$ here just the restriction of $f_{*}$ to the Im(f)?

[cloud walrus]

kenshin5334

[rocky cloak]

corestriction, but yes.

[stuck fiber]

Ah is restriction only for domain?

[rocky cloak]

Yeah, that's the usual naming convention at least

corestriction might be used differently as well

But restriction usually refers to the domain

[stuck fiber]

Cool, I had worked out that it was surjective if I restricted the codomain to Im(f) so I wanted to make sure there wasn't some other homomorphosm I was missing. This proof of the first isomorphism theorem actually seems really cool

[rocky cloak]

Yeah, it's pretty cool. In some sense it feels "too obvious", but that's also a good thing in a way

[stuck fiber]

The initial idea seems a bit out of nowhere to me but once the requirements are setup (injectivity and f=) it seems pretty obvious

Then at the end I was looking at lambda like what's this supposed to be and I was like well if we just corestrict f_*

[solemn dew]

Guys can someone explain how the rubik's cube is a group?

And the operation is a move on it

I can't wrap my head around this because I can't see why it is associative

[hidden haven]

Permutations of the cube form a group much like permutations of any set form a group

The only difference here is that not all permutations are valid, so the group is a subgroup of the symmetric group rather than the entire symmetric group

It is associative because the group operation is composition of permutations

[worthy sonnet]

When is H a normal subgroup of G (G here is the group of dihedrals)? Is this when H is the stabilizer of one point of the n-gon?
Because I thought H(P), where P is one point of the n-gon would be H(P)={id, S1} therefore not being normal since reflections aren't commutative in general.
So when is H a normal subgroup? Is it when H is the rotation group of the n-group? I am confused.

[gusty thistle]

How can we talk about Q/R? R is certainly not an ideal of Q right?

Is this alternative notation for the quotient field ?

Wait do they mean Q/R as an abelian group maybe

[rustic crown]

as an R-module, yee

[gusty thistle]

ah yes

that makes sense thanks

[gusty thistle]

anyone have a hint on how to show uniqueness in a ?

Its equivalent to tha last thing u have to deduce in a right

[wraith cargo]

Maybe an unusual approach but induction on n?

[gusty thistle]

yeah good idea

its also a good way to get a hand on the problem i think

[lethal dune]

what I would suggest is show M ⊗ Rⁿ = Mⁿ by tge natural map m ⊗e_i goes to (0,0.., m('th), 0,..)

[gusty thistle]

oh thanks

ill try

[lethal dune]

It adds one more step but the result M ⊗Rⁿ=Mⁿ is worth it

[gusty thistle]

yeah def

[gusty thistle]

how do we know that this map is well defined

iknow u can do stuff like that for free modules but M ⊗ Rⁿ is not thee free module generated by M x {e_1, ... , e_n} right

[lethal dune]

show ∑ m_i ⊗r_i =0 then it's image is 0

you need to use the fact that 0⊗e_i=0

[gusty thistle]

ty

[solemn garden]

What is an example of an isomorphism of A8 to GL(4,F2)?

[delicate bloom]

maybe we can construct one by thinking of the group action on sets of 8 elements to vectors in (F_2)^4

[solemn garden]

There are 15 non-zero vectors there

[solemn garden]

what group action is that

[delicate bloom]

I think I worded that ambiguously

the alternating group on 8 elements has an obvious group action on {1,2,...,8} and GL(4,F_2) has an obvious group action on the vectors of (F_2)^4

so I'm picturing those sets that they have a group action on as having some kind of bijection between them

and then we can think of the isomorphism of groups that way by seeing how, for instance pairs of elements of {1,2,...,8} get moved around is the same as linear transformations on that vector space moving vectors around

[solemn garden]

So do the elements 1,2, .. 8 correspond to vectors or sets of vectors?

[gusty thistle]

anyone have a hint on how to approach this problem

[delicate bloom]

I think it's more like subsets of pairs of elements of {1,...,8} correspond to vectors, idk that's what we'd have to work out

[delicate orchid]

Split it up into an “if” and “only if”, the reverse direction is pretty clear

[gusty thistle]

yeah

[delicate bloom]

like which sets/vectors are fixed by the operations and try to associate these

maybe start simple thinking of specific cases, like the identity element or just permuting two pairs of elements

[solemn garden]

identity goes to identity

double transposition has to go to an element with order 2

[gusty thistle]

do you have a hint for => v = av' ? What property of vector spaces should i use?

[delicate bloom]

ok so let's say (12)(34) what subsets does this fix, and similarly what does the corresponding matrix look like along with its fixed vectors

[gusty thistle]

I feel like i dont rlly have a lead yet if that makes sense

[solemn garden]

(12)(34) fixes 5,6,7,8
what is the corresponding matrix?

[delicate bloom]

maybe it corresponds to the block matrix made up with a permutation matrix in the top 2x2 and bottom right is an identity 2x2

just a guess

[delicate bloom]

nah, what subsets does it fix, not elements

for instance {1,2} is fixed, {} and {1,...,8} are fixed

[delicate orchid]

Nothing immediately comes to mind but I’ll be something obvious like considering v’ (x) v - v (x) v’ = 0

[delicate bloom]

so {5} and {6} are fixed too like you're saying

[gusty thistle]

okay hmmm

thanks

[solemn garden]

and also {3,4}, {1,2,3,4}, and all subsets of {5,6,7,8}

[delicate bloom]

so I guess what's the simplest way to describe these

trying to set up a consistent way of looking at this to make any kind of pattern more obvious to us if possible

I'm thinking any unions of subsets of {{1,2}, {3,4}, {5}, {6}, {7}, {8}} I guess

[solemn garden]

oh

[delicate bloom]

at least that gives us one package, and it basically comes from (12)(34)(5)(6)(7)(8) as a permutation directly

[solemn garden]

then (123) would fix any unions of subsets of {{1,2,3},{4},{5},{6},{7},{8}}

[delicate bloom]

yeah, since each (....) is going to be internal to its own little corresponding subset

althought that won't be one of our elements

cause it's not in the alternating group

[solemn garden]

isn’t it even?

[gusty thistle]

i think ill just get back to this problem when i know more stuff about bases of vector spaces it feels p unaproachable rn and all the stuff i find online uses a basis

[delicate bloom]

oh you're right it is, my bad

I guess the next thing might be to try to work out a similar picture on the other side

[solemn garden]

so then find a matrix and see what subsets it fixes?

[delicate bloom]

yeah

I say pick the simple example of block diagonal with a 2x2 permutation matrix in the first block, then identity in second block

since that's at least the same order

so seems like a decent enough candidate

[solemn garden]

\begin{matrix}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 1 & 0
\end{matrix}

[cloud walrus]

even order group => solvable
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[delicate bloom]

yeah that works too

[solemn garden]

this has order 2 and fixes (1,0,0,0),(0,1,0,0) and does not fix (0,0,1,0) or (0,0,0,1)

so by linearity it fixes (0,0,0,0),(1,0,0,0),(0,1,0,0), and (1,1,0,0)

[delicate bloom]

I guess it fixes any linear combination of {[1 0 0 0], [0 1 0 0], [0 0 1 1]}

[solemn garden]

So that’s 8 elements it fixes

And it permutes the other 8

[delicate bloom]

interesting

[delicate bloom]

I guess back up here this fixed 2^6 elements

[solemn garden]

it fixed 2^6 subsets but not elements

[delicate bloom]

2^3 vs 2^6 this isn't some kind of complement, directly

I was thinking of them as elements in the power set, but we're on the same page I think

since those are subsets lol

[pallid oracle]

You should be able to do it by using the definition of tensor product. Which definition(s) of tensor product do you know?

[solemn garden]

the set of all (1,2,n) where 3 =< n =< 8 generates A_8. Does this help? I can’t think of 6 elements in GL(4, F2) which all have order 3 and are similar

[delicate bloom]

[1 0 0 0]
[0 0 0 1]
[0 1 0 0]
[0 0 1 0]

something like that probably

now you've basically got the 3x3 matrix and one 1x1 identity matrix in there

at least this thing is order 3, along with if you move the elements around and transpose I think you might get 6 out of it

[solemn garden]

So 4 ways to choose 3 elements to permute, and 2 directions to permute it in, which makes 8 of this type of matrix

[delicate bloom]

then magical algebra fairies take 2 away to make 6

[solemn garden]

\begin{matrix}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 1 & 1
\end{matrix}

[cloud walrus]

even order group => solvable
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[solemn garden]

this also has order 3

but there are 4*3=12 of it

[delicate bloom]

cool

well if we're finding these, maybe we can find matching sets back on the permutation side

err "sets" as in sets of elements of the group lol

[tribal moss]

It can't; it has determinant -1.

[solemn garden]

oh, these are called orbits, right?

[solemn garden]

then I made a mistake
edit: nvm

[pallid oracle]

-1 = 1 in characteristic 2

[tribal moss]

Ah sory, didn't catch we were in char 2.

[delicate bloom]

yeah

maybe we could start trying to bust out some stuff like orbit stabilizer theorem, but I was originally hoping it'd be more obvious

like we'd stumble on to some simple relation between the orbits and elements of (F_2)^4

[solemn garden]

so this fixes any linear combination of {[1 0 0 0], [0 1 0 0], [0 0 1 1]}, and the other 8 elements are split into 4 orbits of 2 elements each

[pallid oracle]

My idea is that some matrices of 0's and 1's with determinant 1 are permutations in S_4, and S_4 embeds into A_8 by sending permutation p to (p, p). But I don't know where to send the matrices which are not permutation matrices lol

[solemn garden]

don’t all of them have determinant 1 since it’s over F2?

[pallid oracle]

well some of them have determinant 0

[solemn garden]

In GL(4, F2)?

[delicate bloom]

0 matrix

for instance

oh

[pallid oracle]

well, I wasn't saying in GL(4, F2)

[delicate bloom]

not in GL lol

[pallid oracle]

I was just saying matrices, but yeah matrices with det 1 is the same as matrices in gln

in this case

[solemn garden]

2^6 subsets are fixed here, and that matrix which swaps two basis vectors fixes 2^12 subsets

[delicate bloom]

some random ideas tot ry to break down GL a bit

maybe we can think of it as 2x2 block matrix, and think of the 2x2 matrices in there, sort of breaks up the structure a bit easier maybe

and transposing gets us a bit of symmetry to argue with

I think there's gotta be some kind of clean bijection between the orbits

[pallid oracle]

one possible thing to do is find the conjugacy classes first and then decide where we send them but that requires doing the jordan normal form in F2

[delicate bloom]

that makes it feel obvious

[tribal moss]

Can we find a set of 8 things that GL(4,F2) acts on? Subspaces of dimension 1, 2, 3 don't work, perhaps a conjugacy class?

[solemn garden]

does this multiply the same as if you took the elements out of the matrices and multiplied them?

[chilly ocean]

could someone help me w algebraic topology

[delicate bloom]

yeah, basically

just have to be aware of it not commuting

[solemn garden]

A conjugacy class of permutations in A_8?

[tribal moss]

I was thinking a conjugacy class in GL(4,F2) itself.

[solemn garden]

(F2)^4 is abelian

[tribal moss]

Not sure how to go looking for ones of the right size, though.

[delicate bloom]

? A
[1 0 0 0]
[0 0 0 1]
[0 1 0 0]
[0 0 1 0]
? B
[0 0 0 1]
[1 0 0 0]
[0 1 0 0]
[0 0 1 0]
? AB-BA
[ 0 0 -1 1]
[-1 0 1 0]
[ 1 0 0 -1]
[ 0 0 0 0]

[solemn garden]

So the aim would be to try and find 8 conjugacy classes which GL(4, F2) swaps around?

[solemn garden]

I thought they were talking about the vector space, not the matrices, I realize now

[tribal moss]

What I was saying that if we can find one conjugacy class with 8 matrices in it, then GL acts on it by conjugation, which gives us a homomorphism into S8. If we're lucky, perhaps its image would then be just A8.

[solemn garden]

What if there aren’t any conjugacy classes with 8 matrices

[tribal moss]

Then my plan didn't work.

[delicate orchid]

I found a paper explicitly calculating the 2-modular representations of A_8 but I can’t access it :babyrage:

[tribal moss]

I'm brainstorming ideas here, not providing hints for a known soluton.

[solemn garden]

Ok, I understand

What are 2-modular representations

[pallid oracle]

do you know what representations are

[delicate orchid]

Exactly what you’re looking for lol, a group homomorphism from A_8 into GL(n, F_2)

I would get out paper and do the brauer character nonsense but I’m on holiday

[solemn garden]

How would you compute one

[solemn garden]

does this mean there is a method to use to find the representation?

[delicate orchid]

There’s a method to find all of modular representations at each prime yeah, you decompose the group algebra F_2A_8 into blocks and then look at the one that’s 4-dimensional (and in this case has a trivial defect group cause it’s a matrix algebra by assumption). But I’d suggest using more basic/less general machinery

I agree with the method of finding a set of order 8 that GL_4(2) acts on nicely

Is there a subgroup of index 8?

Any such subgroup will be isomorphic to A_7 by our assumption

[gusty thistle]

The one where u take the quotient of the free module over the Cartesian product and I know stuff about universal properties of the tensor product

[solemn garden]

Why

[delicate orchid]

Think about A_8 ig

[rocky cloak]

This only gets the projective representations right, or am I misunderstanding you?

[gusty thistle]

I am following dummit & Foote if u know it

[delicate orchid]

Oh true, good point

I forgot you don’t get all of them if ur field isn’t algebraically closed

So we could work in F_2 bar but that just seems really silly

[rocky cloak]

Still only gets you the projective ones I think

[delicate orchid]

Anyway

[rocky cloak]

A8 should be wild over characteristic 2 I think

So it should be hard to find an the modules

[delicate orchid]

It’s got enough powers of two in it for me to believe that

[pallid oracle]

aren't projective and regular space the same in F2?

[pallid oracle]

oh I see

[delicate orchid]

When we say projective we mean projective F_2A_8-modules

Can we use the fact that GL(n, 2) for n large is simple and then just cite the classification of finite simple groups

[solemn garden]

Even so, that doesn’t give the isomorphism

[pallid oracle]

I'm sure whoever wrote the classification included the fact that these two are isomorphic so there's not even a need to prove anything

just citation

[south patrol]

#point-set-topology

[delicate bloom]

I guess we just gotta roll up our sleeves and start working out the generators

[tribal moss]

https://en.wikipedia.org/wiki/List_of_finite_simple_groups notes:

no two finite simple groups have the same order, except that the group A8 = A3(2) and A2(4) both have order 20160 [... and another exception]
I can't quickly find a statement of what the An(m) notation means, but perhaps we're barking up a non-existant tree?

[pallid oracle]

Think about what the condition would need to be on the Cartesian product, forgetting about tensors.
Then to get tensors there's only the extra relation av⊗v' - v⊗av' = 0.

[pallid oracle]

proving that it equals A3(2) is indeed the current problem

An is SL_{n+1}

[tribal moss]

What is An(m) with two parameters?

[pallid oracle]

the m is probably the order of the field

[tribal moss]

I see.

[solemn garden]

The problem is to find an isomorphism(function ) between them

[tribal moss]

Ok, that checks out, SL(3,F4) does have the same size.

[rocky cloak]

Does it have 8 7-sylow groups? The would be my guess are least

[tribal moss]

Nice if true.

Unfortunately it is not so:
1 0 0 0
0 0 0 1
0 1 0 1
0 0 1 0
has order 7, and it has an eigenspace of dimension 1 which all its powers preserve. So has at least 15 conjugates (including itself) that are not in the same 7-Sylow subgroup.

[solemn garden]

a 7-cycle must map to it

[tribal moss]

And besides, there are 8·7!/7/6 = 960 7-Sylow subgroups in A8.

[rocky cloak]

Darn

[tribal moss]

To make a 7-Sylow subgroup, we can first select a dim-1 subspace to be the eigenspace (in one of 15 different ways), then select a 3-dimensional complement of that that our subgroup will fix (in one of 8 possible ways!), then select one of 7-Sylow subgroups of GL(3,F2) (in one of 8 possible ways!).
If only there were a way to lift out one of those 8's to be something that the whole GL(3,F2) can act on.

[tribal moss]

And in fact, the 960 Sylow 7-subgroups of A8 split into 8 families of 120 each (according to which position is outside the 7-cycles), and A8 acts on the families by conjugation!

[stuck fiber]

Everything except the underlined part makes sense I don't see the justification for why N is in the kernel

Oh wait

[tribal moss]

The kernel of f is one of the things you took an intersection of to find N.

[stuck fiber]

Yeah it slipped my mind that the kernel is a normal subgroup for a second

[barren sierra]

What is the Kronecker product of matrix groups?

I know what the Kronecker product of matrices are

but I can't find anything about this product in the context of groups of matricies

if it helps these are matrix groups over finite fields

I also have this

which I guess is good enough for my purposes but I'm just curious about the definition of this product outside of these central products

nvm I think I found it

[coral spindle]

Just a guess but it may be simply the action of G_1 x G_2 on the tensor product of the vector spaces on which they act

[barren sierra]

yea it is

[coral spindle]

epic

[barren sierra]

and the group is formed by just taking the kronecker product of each pair of elements in the constituent groups

[coral spindle]

as expected

[barren sierra]

that's what I expected also

but I was struggling to find it written down somewhere >_>

[vestal marten]

Is there a standard product operation for this ring? It's not immediately obvious to me whether we should define (fg)(x) to mean f(x)*g(x) or f(g(x)). I would have guessed composition, but the answer key seems to suggest multiplication

[delicate orchid]

Consider the fact it’s a ring

So multiplication has to distribute nicely

f(g(x)+g’(x)) isn’t always f(g(x))+f(g’(x))

[vestal marten]

I see, thank you!

[south patrol]

Yeah usually functions valued in a ring are given multiplication pointwise like this

[night onyx]

Composition isn't well defined in general since f(x) isn't necessarily in [0, 1] for f : [0,1] -> R continuous

[gusty thistle]

Oooooooooooooh

Wow

Good point

It does seem deceptively simple though

Because idk if it’s true for Modules

[tribal moss]

You also need v⊗(w+w') = v⊗w + v⊗w' and similarly to the other side, I think.

[gusty thistle]

And the subgroup generated by those relations might be way bigger right

[delicate orchid]

Good catch

[brazen girder]

Sorry, this might be an inappropriate channel to post this but I'm trying to understand this paper: https://arxiv.org/ftp/q-bio/papers/0510/0510029.pdf and its about galois fields... I was wondering if I learn enough linear algebra (which I'm currently in love with) and what's covered in a standard abstract algebra course (which is what I plan on covering next) would I have enough knowledge to read/decipher this paper in terms of the math required?

[summer path]

on a very quick skim, probably? but to learn as much algebra as you are saying will probably take some time

[charred crescent]

i have encountered something called the canonical epimorphsim from G to G/N (N normal in G) is this the same as the natural projection?

from context i believe it is, considering i believe i remember from category theory that an epimorphism is "surjective" (right cancellative)

[stuck fiber]

I don't know about epimorphism but that sounds like the canonical homomorphism f:G->G/N defined by f(x)=xN

[charred crescent]

yeah i just wanted to know if they were the same

but i think you're correct

[stuck fiber]

Yeah I think the canonical homomorphism and natural projection are just terms for the same thing as far as I can tell

Ah yeah it seems a canonical map is also called a natural map

[topaz solar]

Yeah should be, but also stating that it’s epic

[charred crescent]

what exactly is the different between epic and surj. ? it never really made sense to me other than the fact that epic means right-cancellative

i guess in this instance they are using epic to indicate surjectivity

[uneven fossil]

reading dummit and foote and i've gotten to the third isomorphism theorem but i'm a bit confused

[topaz solar]

Consider Z -> Q among ring homomorphisms

[charred crescent]

funny im on the exact same page

[topaz solar]

That should be epic

[uneven fossil]

in this H is a subgroup of K, not a normal subgroup, then how should K/H be defined to be a group?

[uneven fossil]

real

[charred crescent]

but not surjective necessarily ?

[topaz solar]

It’s clearly not

Since what hits 1/2

[charred crescent]

exactly

[topaz solar]

Pretty sure that’s epic anyway?

[charred crescent]

see this is what confused me, because i saw this exact example on the wiki, but i would think that because f: Z -> Q is an inclusion (Z subset of Q) then f is injective

so wouldn't it be a monomorphism

but the wiki said epic

so maybe if it's surjective then its epic but not vise-versa?

[charred crescent]

i think since they're both normal subgroups of G then H is normal in K a fortiori

[topaz solar]

Things can be mono and epic

[prime sundial]

gHg^-1 = H for all g in G, which includes g in K

[topaz solar]

It’s epic because it’s right cancellative or something

[stuck fiber]

I am also on this theorem but in Lang lol it's the second in his book (technically 3rd of 5)

[charred crescent]

wouldn't that be an isomorphism

[topaz solar]

Surjections aren’t really the same thing, but if you agree on every input you’d expect it to be right cancellation compatible

[topaz solar]

No

[charred crescent]

oh

[topaz solar]

Mono + epic doesn’t mean we have an inverse

[uneven fossil]

ah this makes sense thanks

[topaz solar]

Z -> Q should be mono too iirc

[prime sundial]

sounds right

[topaz solar]

It would be quite strange if it wasn’t

[uneven fossil]

for the definition of the feit-thompson theorem here, how would this be true say for |G| = 9?

[stuck fiber]

Well G would have to be simple and I don't think that's possible by Sylow's first theorem

[uneven fossil]

that really makes sense

😭😭

[prime sundial]

it's not by sylow's first i believe

ah never mind

[stuck fiber]

Isn't Sylow's first like if G is a group of order $p^{n}m$ where $p$ and $m$ are relatively prime then there exist sylow p subgroups of order $p^{i}$ for $i<n$ and any sylow p subgroup of order $p^{i}$ is normal to a group of order $p^{i+1}$ so we have that for 9 $|G|=3^{2}1$ so there exist sylow 3 subgroup which is normal in groups of order 9 so it's normal to $G$ and thus $G$ is not simple

[cloud walrus]

kenshin5334

[stuck fiber]

I haven't done Sylow's in a few months so I may have this a bit wrong

[prime sundial]

yeah it's basically if p^n divides |G| then there is a subgroup of that order

and here we can just take p^1

[stuck fiber]

We had a few homeworks like this I think where it was like use sylow's theorems to prove that $G$ is not a simple group

[cloud walrus]

kenshin5334

[prime sundial]

i thought p^n had to be maximal, but that's for sylow 3

[stuck fiber]

Ah

[prime sundial]

i hated the sylow theorems

finite group theory is difficult

[uneven fossil]

infinite group theory

[stuck fiber]

Composition series kinda destroyed me so I'm glad I'm about to get back to them

[dreamy plume]

Yuh

[solemn garden]

How do you construct a matrix fixing the specified subspaces?

[stuck fiber]

The isomorphism theorems are so cool

[stuck fiber]

I am ngl I do not know how to read diagrams like I see them in Lang but I have never been shown how to read one. I kind of get it like how the arrows represent mappings but I have no clue what it means when diagrams commute or what showing that is like

[tender wharf]

varphi is homomorphism

pi is projection

the dashed lines is the isomorphism asserted to exist

[solemn garden]

Why

[topaz solar]

diagram commutes means that the path you take to reach H doesn't matter

[tender wharf]

pi circ overline varphi is same as varphi

[topaz solar]

i.e. phi = \tilde phi pi

[tender wharf]

*overline varphi circ pi

theyre used in topology too

[stuck fiber]

Yeah I get that cause pi will be the map G -> G/ker defined by x->xK then over line phi will be xK->h Lang does it very nicely imo

The isomorphism theorem (G/K)(H/K) ≈ G\H felt super slick and nice was super fun to figure out like I was writing down all this stuff then it hit me that function is surjective with H/K as the kernel

[frigid lark]

Good thing Lang doesn't have too many diagrams

[stuck fiber]

Sorry I can't tell but is that a meme or like an actual thing cause it feels like a lot of diagrams

[frigid lark]

Also which Lang is this, undergraduate or postgraduate

[stuck fiber]

Gtm Lang

[delicate orchid]

If there’s not two on every page it’s not a lot

[frigid lark]

Yeah not a meme

[stuck fiber]

Oh nice lol

[frigid lark]

Like chapter 1/3 might have some diagrams, but ch 2, 4-6 barely have any

And I haven't read past that

[stuck fiber]

I'll reading 1-4 then 13-16 iirc

[frigid lark]

No dead french man theory

Sadge

Also if you've got time give ch5 a read

It's an amazing chapter

[stuck fiber]

I might check out then, I'm most going off what my professor told me to read for rep theory

[delicate bloom]

is there a nice formula for f(n), being the number of divisors of (x^n - 1)/(x-1) in Z[x] of the same form

maybe call this something, p_n(x)= (x^n - 1)/(x-1), if d|n then p_d | p_n, so if tau(n) is the number of divisors of the natural number n, then f(n) >= tau(n)

I think we can do better and say something like f(n) >= 2tau(n)-1 or maybe even write an exact formula for f in terms of tau

[frigid lark]

I have a feeling that you might get equality here, cause f(q) for some prime q is 1

[delicate bloom]

I think it starts to diverge for larger number of factors

[frigid lark]

we could try to use the fact that x^n - 1 is the product of all the d'th cyclotomics for d | n

[delicate bloom]

that might work with some inclusion/exclusion type of argument. I also edited my question earlier since I left it out

specifically I'm only looking for divisors of that form

[frigid lark]

of the form 1 + x + x^2 + ... + x^s?

[delicate bloom]

I haven't been clear I'm sorry, I made a mistake in describing my question

must be of the form $1+x^k+x^{2k}+\cdots+x^{mk}$

[cloud walrus]

merosity

[delicate orchid]

Vandermoode deterimant

[frigid lark]

ah how could I forget Vandermonde

[delicate bloom]

really just thinking about $$\frac{x^n-1}{x-1}=\frac{x^n-1}{x^d-1}\frac{x^d-1}{x-1}$$ factorizations

[cloud walrus]

merosity

[delicate bloom]

how will the vandermonde determinant help here

[delicate orchid]

It won’t

[delicate bloom]

oh what about it then haha

[frigid lark]

if m is one less than a prime p, this is the cyclotomic for p^{k+1}

[delicate bloom]

I'm only familiar with the vandermonde matrix as far as using it to find coefficients on a polynomial to fit through some points haha

[delicate orchid]

This situation is easy when d is prime, can we bootstrap our way up?

[delicate bloom]

yeah, I'd imagine we could do some kind of inclusion-exclusion formula or

I guess I was thinking we'd have some cleaner looking multiplicative function formula for it in terms of the mobius function

[delicate orchid]

Oh like a funky mobius transform

Yurrrr

[frigid lark]

it's morbin time

[delicate orchid]

Number 5 looks useful

[delicate bloom]

if we pretend these are numbers in base x, then we can write them like repeating 111...111 or 10101...101, or 100100...1001 etc

at least for counting them, it seems to be helpful

[delicate orchid]

Didn’t you just want to count them

[delicate bloom]

yeah

[delicate orchid]

Only problem is if mu(d) is -1

[delicate bloom]

for instance if it's n=12 we have like 111111111111=10101010101 * 11 = 1000001 * 111111 are sort of like two ways of factoring 12 into 2*6 I guess

[frigid lark]

we can expand on this idea, if n = p_1^r_1 ... p_s^r_s, then for each prime, there should be r such divisors

[delicate bloom]

or actually, I can just write it as, $$\frac{x^{12}-1}{x-1} = \frac{x^{12}-1}{x^6-1}\frac{x^6-1}{x-1} = \frac{x^{12}-1}{x^2-1}\frac{x^2-1}{x-1}$$

[cloud walrus]

merosity

[delicate bloom]

we're getting 4 instead of 2 distinct ways here, I'm thinking these kinds of examples break reasoning on primes like you're saying

but maybe I'm wrong about that

[frigid lark]

oh my prime reasoning is not all incompassing

it is just a lower bound

[delicate bloom]

yeah exactly

[frigid lark]

yeah

[delicate bloom]

I fell for this trap earlier myself, that's where my f(n) >= 2 tau(n)-1 sort of bound came from

at least, I think I came up with something like that, I last thought about this a year ago so going off memory here

maybe we can make some kind of recursive formula

[frigid lark]

I think we can identify these factors by two subgroups of mu_n, the n'th roots of unity

[delicate bloom]

I like the roots of unity idea

trying to think how to mold it together, cause I know there are going to be some cyclotomics that aren't of this form

but they'll be a product of cyclotomics at least, so I think that's good

[frigid lark]

basically the roots of unity that solve this equation will solve x^(m+1)k - 1, and the solutions of x^(m+1)k - 1 are the solutions of this equation, and k'th roots of unity

Well then we just have all the divisors of n, and then for all of those divisors we count their divisors

so something like $f(n) = \sum_{d | n} \sum_{r | d, r < d} 1$

[delicate bloom]

I think that ends up being the number of ways to write n=abc, up to reordering

at least that's multiplicative so whatever, just need to remember what it ends up being, it's like some kind of choose function on each prime

[frigid lark]

no not really

or maybe

I think order kind of matters here

maybe not

[delicate bloom]

I might be off on that description a bit

I'm not convinced that what you wrote down is the same as what f(n) actually is though, need to think about that

well, we could compute it for a special test case like n=12 or something and compare

call yours g(n) for now to make the distinction clear to me

[frigid lark]

wait, I might need to change mine a bit

gimme a second

[delicate bloom]

I believe $$g(p^k)=\binom{k+2}{2}$$ thinking in terms of stars and bars

[cloud walrus]

merosity

parrottea

[delicate bloom]

I think more generally in terms of the dirichlet convolution, with $u(n)=1$ we have $$u^{\star t}(n) = \prod_{p|n} \binom{v_p(n)+k}{k}$$

[cloud walrus]

merosity

[frigid lark]

if r = d we are referring to 1

okay I got g(12) = 8

[delicate bloom]

thinking about how this decomposition of factoring works out

[frigid lark]

well the d represents the equation x^d - 1

[delicate bloom]

I think I see

[frigid lark]

and then the r represents the equation x^r - 1, and we divide the 2

[delicate bloom]

for (x^n-1)/(x-1) we have for each d|n that x^d-1 will be in the numerator, and this will have x^k-1 in the denominator for each k|d

[frigid lark]

exactly

and if k is d, then we get 1 which you may or may not want to count

[delicate bloom]

hmm do we get other double-counting?

or is that the only case

I think that is but I don't wanna fool myself

[frigid lark]

I think it's only that case

cause if (x^d - 1)/(x^k - 1) = (x^d' - 1)(x^k' - 1), then their factors are d 'th roots of unity which are not k 'th roots of unity

hence the primitive d'th root of unity is a root of both equations, hence d' | d, similarly d | d'

Hence d = d'. We can argue by degree to get k = k'

If k = d, we can't argue by primitive roots of unity

[delicate bloom]

I need to think about that more, I thought I found a different way but it's incomplete

plugging in x=1 forces it to equal d/k = d'/k' which isn't quite enough unfortunately

but I think your way does it completely

[frigid lark]

mmm, if you find something wrong with it tell me

[delicate bloom]

yeah I see how your way works out now,I got distracted by playing with plugging in 1 making d/k=d'/k' but the degrees must also be equal d-k=d'-k' but then I decided to stop there

your formula can then be rewritten a little I think in terms of those known functions discussed above

[frigid lark]

well they aren't known to me, so good luck

[delicate bloom]

$$f(n)=u^{\star 3}(n)-\tau(n)$$

[cloud walrus]

merosity

[delicate bloom]

I think I might have left out something, let me see

all I'm really saying there is $$f(n)=\sum_{d|n}\left(-1+\sum_{k|d}1\right) = \sum_{d|n}-1 + \tau(d) = -\tau(n) +\sum_{d|n}\tau(d)$$

[cloud walrus]

merosity

[delicate bloom]

I think it technically needs a +1 when n=1

[frigid lark]

mmm

what was this for?

[delicate bloom]

staving off boredom

[frigid lark]

oh god

read a book

[delicate bloom]

lmao

[frigid lark]

well it did its job

[delicate bloom]

just one of those passing thoughts that seems simple enough but not quite lol

fun, thanks

[frigid lark]

alg

[median pawn]

SL(n,k) is certainly a group, but does it fit in the collection of examples above, i.e., is SL(n,k) = Aut(V) for a k-module V? For GL(n,k), just V = k^n works.

Maybe it was just meant as an additional comment and I'm digging too much into it

[solar vessel]

it's an additional comment

[median pawn]

Ah okay xD

[gusty thistle]

dummit & foote 😻

i remeber i also tapped out at the third isomorphsim theorem during my first algebra course lol

[lime badge]

I have a question about a problem I'm doing from Aluffi's book (problem 7.2 in chapter 7). We have the polynomial $P_n(x) \coloneqq (x-t_1)\cdots(x-t_n) \in \mathbb{Z}[t_1,\ldots,t_n][x]$. I need to show I can obtain any polynomial of degree n in $R[x]$, where $R$ is an integral domain, by specifying $t_1, \ldots, t_n$ in possibly a larger ring

[cloud walrus]

eternalway

[lime badge]

So clearly $\mathbb{Z}$ being initial in the category Ring will play a role in this proof, but I'm not sure how to proceed with specifying $t_1,\ldots,t_n$

[cloud walrus]

eternalway

[topaz solar]

well, lets say you wanted to specify them to be, uhhhh

i(t_n)

[south patrol]

Well you need to define a map from one ring to the other and there is only one choice (up to automorphism)

[topaz solar]

Z -> R is very helpful, you happen to know a map Z[x] -> R[x]

[lime badge]

Right, but one of my issues is that $P_n(x)$ is monic, but polynomials in $R[x]$ are not necessarily monic

[cloud walrus]

eternalway

[topaz solar]

I claim I can uniquely define a map Z[t1,...,tn][x] -> R[x] by specifying it's values on the t_n and having it be a homomorphism

since, after all, the Z[x] part is kinda already handled by Z being Z

[lime badge]

Sure, but the problem is then proving surjectivity, right?

[topaz solar]

well, we don't need our single map to be surjective

because the claim is we can get it by choosing (t_n)

[lime badge]

Ah ok, so there's an error in this problem as indicated here. Aluffi meant to write monic polynomial, not every polynomial in R[x]: https://www.math.fsu.edu/~aluffi/algebraerrata.2016/Errata.html

[topaz solar]

oh that, I missed the monic part of your worries

yeah you're not getting past that lol

[lime badge]

No worries. My other worry was that wouldn't such an assignment only get you the polynomials that split in R[x]?

I guess this is where the part of the problem statement about potentially specifying t1,...,tn in a larger ring comes into play?

[topaz solar]

yeah, kinda hard to split things that don't split

[lime badge]

Ok, so putting this all together, we can just consider f(x) \in R[x] over its splitting field and specify t_1,...,t_n by mapping these to the roots of f(x) in its splitting field in the homomorphism you indicated here

[topaz solar]

well ig that's not so much a map to R[x] oop

but yeah that should work I'd think

[lime badge]

Sorry, if F is the splitting field of f(x), this would be a map to F[x]

[topaz solar]

yeah whoops, but still sends P_n(x) to your desired polynomial

[lime badge]

Awesome, thanks a lot for your help!

[topaz solar]

Algebraic closure shenanigans also works but it's a bit of a sledgehammer, might be handy to have every such map (to the same R[x]) to the same codomain and all

I think

[lime badge]

Right yeah

[untold turret]

Let $f:K\to\Omega$ be a homomorphism of fields were $\Omega$ is algebraically closed. Assume that $\alpha$ is an element of some algebraic extension field $L$ of $K$. Then there exists a homomorphism of fields $\tilde{f}:K[\alpha]\to\Omega$ such that $f(z)=\tilde{f}(z)$ for all $z\in K$.

[cloud walrus]

nrs

[topaz solar]

took me a moment to realize the problem being asked

[untold turret]

why should I believe this is true? i.e., what would lead us to believe this/what's the intuition?

[lime badge]

Yeah, at first I thought it was asking me to come up with formulas that relate the elementary symmetric functions to the coefficients of our polynomial

[topaz solar]

I meant I thought it was about getting a map Z->R wtv, too eepy

more defining a map than the actual question

[topaz solar]

omega is algebraically closed and since alpha is algebraic, what about the image of the map K[x]->Omega[x] for a polynomial for alpha?

what's the roots

this satisfactory? @untold turret

[untold turret]

hm, right such a map has p(alpha) in the kernel

i.e. the ideal generated by it is the kernel

[topaz solar]

my idea was

a algebraic -> there's min poly p(x) where p(a)=0 -> f'(p) poly on the other side by K[x] -> Omega[x] then -> f'(p) has roots because alg closed -> f^(a) = a root for f'(p)?

then that should get you all of K[a]->Omega

[untold turret]

oh i see! yeah that works, thank you

[topaz solar]

the particular root doesn't matter since min poly -> it's only gonna be things not already in f[K]

but it'll have all the right algebra

something something actual argument Galois whatever

[oblique matrix]

If F is a field, consider the space of coordinates F^2 = {(x,y): x, y in F}. I want to prove that F^2 is an incidence geometry with points being elements and lines being subsets of F^2 (S) such that S can be expressed as {(x,y) \in F^2: mx+ny+c = 0 (with at least one of m or n nonzero)} . To do this, one has to verify the axiom that for any two points, there exists a unique line containing the two points. The existence part is trivial, but for some reason I'm having a lot of difficulty showing uniqueness.

[lime badge]

I have another silly question. Let $K$ be a field and consider the field of rational functions $F = K(t_1,\ldots,t_n)$ in the indeterminates $t_1,\ldots,t_n$. I'm supposed to show that the elementary symmetric functions $s_1, \ldots, s_n$ in the variables $t_1,\ldots,t_n$ are algebraically independent over $K$.

But isn't a simple argument for this that if they're not algebraically independent, then the transcendence degree of $F$ is less than n. But $t_1,\ldots,t_n$ form a transcendence basis of degree n over $K$, so you get a contradiction.

I feel like something is wrong with this argument though

[cloud walrus]

eternalway

[rocky cloak]

I guess it's not so simple to prove that the transcendence degree of F and K(s1, ..., sn) is the same. Though maybe there's a simple Galois theory argument for this.

[lime badge]

Well, you know transcendence degree is additive in towers of extensions. Galois theory shows that F is Galois over K(s1,...,sn) so F is algebraic over the latter. Thus this extension has transcendence degree 0 and the claim follows from the fact that F has transcendence degree n over K

But for proving that the elementary symmetric functions are algebraic, does my previous argument work (in addition to what I showed above)?

[rocky cloak]

Not sure what your previous argument refers to, but yeah if you've shown that the extension is Galois, then I think that should work fine.

[lime badge]

My previous argument for showing that the elementary symmetric functions are algebraically independent

[lime badge]

Actually NVM, this argument also establishes that

Thanks!

[untold turret]

$\mathbb{F}_p(t)$, what does this usually denote?

[cloud walrus]

nrs

[delicate orchid]

The field formed by extending F_p by t

[untold turret]

F_p being Z/pZ?

[delicate orchid]

Yes

[wraith cargo]

field of rational functions with coefficients in F_p and variable t

[untold turret]

ty all!

[summer path]

Note that this is different from k[x] which is the polynomial ring and k[[x]] which is the ring of formal power series

[delicate orchid]

Yur

It’s even different from k[x, x^-1] woahhhhwhhaha

[summer path]

And I guess k((x)) which is ring of formal Laurent series

[untold turret]

A \subseteq B are rings. do we have: B integral over A and q in B is prime ideal lying over prime ideal p in A implies B/q is integral over A/p?

[chilly ocean]

monic polynomials over A give monic polynomials over A/p

[untold turret]

ty!

[lime badge]

Just a quick question. Let $k$ be a field and suppose $f(x)$ is an irreducible separable polynomial of prime degree. Let $\alpha$ is a root in the algebraic closure of k, and suppose we can express another root of f(x), call it $\beta$, as a polynomial in $\alpha$ with coeffs in k (say $\beta = g(\alpha)$ for $g(x) \in k[x]$). I'm supposed to show that I can express all roots as polynomials in $\alpha$ and that the Galois group of $f(x)$ is $\mathbb{Z}/p\mathbb{Z}$

[cloud walrus]

eternalway

[lime badge]

I've already proved that the Galois group contains an element of order p that cycles through all the roots of f(x) (since the Galois group is a transitive subgroup of the symmetric group). I'm just having trouble showing that all other roots can be expressed as polynomials in $\alpha$. For instance, if $\sigma$ is a p-cycle, then $\sigma(\beta) = \sigma(g(\alpha)) = g(\sigma(\alpha))$, but this isn't a polynomial in $\alpha$

[cloud walrus]

eternalway

[white oxide]

Okay, here's what I got in the finite case: Let $E$ be a finite extension of $F$ so that $E = F(\alpha_1, \dots, \alpha_n)$. Suppose $\alpha_1, \dots, \alpha_n$ are separable over $F$. Now $F(\alpha_1)$ is separable over $F$ by the definition of $\alpha_i$ being separable over $F$. Now suppose that $F(\alpha_1, \dots, \alpha_i)$ is separable over $F(\alpha_1, \dots \alpha_{i - 1})$ for $i = 1, \dots, n - 1$. Observe that irr($\alpha_n, F(\alpha_1, \dots, \alpha_{n - 1}$)) divides irr$(\alpha_n , F)$, so it must have all zeros of multiplicity one. Hence $F(\alpha_1, \dots, \alpha_n)$ is separable over $F(\alpha_1, \dots, \alpha_{n - 1})$ and $E$ is separable over $F$.

[cloud walrus]

okeyokay

[white oxide]

lol i like ur about me too

can I get a hint to show that if E is an algebraic extension of a perfect field F, then E is perfect? i'm letting alpha in K and trying to show that irr(alpha, E) has all zeros of multiplicity one, and i don't really know how to use the fact that F is perfect because F being perfect doesn't imply that K is separable over F (since K may not be a finite extension of F)

also don't know how to use the fact that E is an algebraic extension of F, in the case that alpha in K but alpha not in E

[wraith cargo]

so since |Gal(E/F)| = [E:F] is you have a galois extension of prime degree the galois groups has order p so it's cyclic of the form Z/pZ (any group of prime order is cyclic)

[lime badge]

Yeah, that part is easy. Its the other part I was confused about. Doesn't it just immediately follow from the extension being simple?

[wraith cargo]

can you assume every separable extension is simple
I believe that is a very non trivial theorem (but yes that would make it follow immediately)

[lime badge]

Its proved by this point in the book, so I guess we can

Thanks!

[white oxide]

anyone?

[hidden haven]

Use the fact that given an embedding of F into F closure, and a finite extension K of F, [K:F] = number of extensions of the embedding to homomorphisms K → F closure iff K/F is separable

Scratch that, this is simpler. Alpha is algebraic over E with some minimal polynomial f(x). It's also algebraic over F with a minimal polynomial g(x). Both have coefficients in E so f(x) | g(x).

@white oxide

[white oxide]

oh okay thanks i'm doing some reading rn but when i return to the problem i'll look at the hint

thank u

[frigid lark]

Follow what moldilocks said, but in general if you want to deal with infinite algebraic extensions you can usually consider it on an element by element basis and find a finite subextension

[white oxide]

what's a subextension?

[frigid lark]

Intermediate field

If $K\supset F \supset k$, then we call F/k a subextension of K/k

[cloud walrus]

parrottea

[white oxide]

oh ok got it i'll keep that in mind

thanks

[noble hedge]

The minimal polynomial for something like $\sqrt[5] 3$ is the same in $\mathbb Q$ as it is in $\mathbb Q(i)$ right?

[cloud walrus]

strobilanthes

[noble hedge]

The big irreducibility tests I can think of are basically just over integers and are ported over to rationals via Gauss' lemma. Intuition tells me adjoining i doesn't add irrationals, so it can't lessen the degree of the minimal polynomial for one, but I'm not quite sure if there's a nice rigorous argument for that or not.

[frigid lark]

Yep one is a degree 2 extension the other is a degree 3

2 and 3 are coprime

[south patrol]

well degree 5 right

but i assume you just misread

[frigid lark]

Oh yeah

[noble hedge]

So what about if it were like \sqrt[8]{3} then?

[frigid lark]

I blind

[south patrol]

Same thing

I mean like

[noble hedge]

8 isn't coprime with 2 tho

[south patrol]

There are more general ways to argue

[frigid lark]

But x^n - a has a bit of nuance

And if 2|n it's annoying to deal with

[noble hedge]

So I've noticed

"adjoining i doesn't add irrationals, so it can't lower the degree of one" is about the only argument I can think of and that's just voiced intuition

[south patrol]

Why? Just use Eisenstein

since a =3 here

[frigid lark]

Over Q yeah sure

But over arbitrary fields

[noble hedge]

Eisenstein extending over Q(i) would require generalizing Gauss' lemma to Z(i) and Q(i), which seems like overkill

Well, Gauss' lemma taking irreducibility in Z(i) to Q(i)

[south patrol]

Oh i mean the result follows easiliy by another method using Eisenstein for Q

[frigid lark]

Well in this case you argue that Q(3^1/8) is a subset of reals

[south patrol]

Like note that if $\alpha = 3^{1/n}$ for some $n \ge 2$ then by the Tower law and Eisenstein $2[\mathbb Q(\alpha,i): \mathbb Q(i)] = n[\mathbb Q(\alpha,i): \mathbb Q(\alpha)]$

[cloud walrus]

potato

[south patrol]

Clearly the last term is 2n just by what Parrot is about to say lol

like Q(alpha) is real

So [Q(alpha,i): Q(i)] = n

[formal ermine]

couldn't we also use that L(alpha) : K(alpha) = L : K if alpha lives in the algebraic closure of K and L(alpha) : K and L : K are coprime

[noble hedge]

That's a neat result I haven't seen before

[formal ermine]

wait is that true

[noble hedge]

it'd be super convenient if it was

[formal ermine]

I know that it's true if L(alpha) : K and L : K are coprime

[frigid lark]

Nah this is false

[formal ermine]

oop

[noble hedge]

unfortunate

[formal ermine]

only in this case

it's true

[noble hedge]

But yeah I see what you're saying here, thanks!

[frigid lark]

Let L be Q(sqrt2), K be Q and alpha be Q(2^1/4)

[formal ermine]

yeah

need coprimeness

[south patrol]

np

ye holds nicely for coprime situation

there are probably funnier ways to solve this though in general hm

Well, like 3 is prime in Z[i] so you can apply Eisenstein i guess

[wraith cargo]

badass

we need more cool math names

[frigid lark]

ghost component

degeneracy in kummer extensions

afaic algebraists have a good naming sense

Potato where were you when I was struggling with Witt Vectors

[south patrol]

I mean idk much about Witt vectors either tbh

I wanna get more familiar with them for a project i'm doing though lol

[frigid lark]

oh alg

[south patrol]

wdym oop

[frigid lark]

oop?

[south patrol]

lol like wdym by alg there

[formal ermine]

what are witt vectors

[frigid lark]

pain in ass

[frigid lark]

not analysis or topology?

[frigid lark]

if you know some Galois stuff, the end of ch6 of Lang is good

[south patrol]

Uhh so in the case we often care about the point is that out of a perfect field k of characteristic p > 0 we can deform to a characteristic 0 ring W(k) which has a lift of Frobenius i.e. map φ: W(k) -> W(k) with φ(x) = x^p mod p.W(k)

which is a DVR

This assignment W(-) is also a functor from like commutative rings to comm. rings

[south patrol]

Noice

thanks

I didn't know Galois theory would really come up

[frigid lark]

https://web.stanford.edu/~dkim04/blog/witt-vectors/#3-construction-of-addition-and-multiplication

I'll also link this

[south patrol]

I read the Bourbaki treatment of it and it didn't use Galois theory lol

[frigid lark]

it's used in Galois theory

[south patrol]

Oh okay sure this is doing an application

That's cool thank you

Because I stumbled across them when looking at Artin-Schreier extensions but never actually looked at the application to that situation

[south patrol]

I basically don't do any analysis lol

But yeah I mean i'm interested in algebraic topology / homotopy theory mostly and this came up for smth in those topics

Namely Lubin-Tate rings

[frigid lark]

oh, I thought you meant the cool names

[south patrol]

oh okay lol

[frigid lark]

and I was joking

[south patrol]

lol fair

[wraith cargo]

there's some completion way to look at Witt vectors

[wraith cargo]

here's some mumbo jumbo for you to feed on

[frigid lark]

too bad I have zeta topology knowledge, where zeta is the primitive second root of unity

[formal ermine]

hurry up I'm trying to sleep fr fr

[frigid lark]

wdym its 10:36 am

[wraith cargo]

take a complete local ring R (with maximal ideal m) with perfect residue field K of characteristic p
Cohen showed in like the 40's that then there exists a (unique up to isomorphism) complete local integral domain W(K) whose maximal ideal is generated by p and whose residue field is K
This is called the ring of Witt vectors
Concretely what Cohen showed is that you can write the completion of R wrt m in the form W(K)[[x_1, ..., x_n]]/I for some ideal I of the power series ring

[formal ermine]

it's 2 am

[wraith cargo]

same

[formal ermine]

I told my dad to wake me up at 8 am

[wraith cargo]

I'm depressed so I'm valid in staying up late 👍

[formal ermine]

because my exam is on tuesday at 10 am

[wraith cargo]

exam
middle of summer

[frigid lark]

do you guys have trimestors or something?

[formal ermine]

and I usually stay up till like 4-6 and get up at 4 pm

[wraith cargo]

completions as in the algebraic kind

[formal ermine]

the semester ended last week

[wraith cargo]

the inverse limit kind

the m-adic kind

[frigid lark]

oh alright

[formal ermine]

🫂

[frigid lark]

this smells like a cycle

[formal ermine]

ok but like

who cares

[wraith cargo]

want to die
math
happy :3 :3 :3 cozy math!!!!!!!! fun!!!!!!
stop doing math
sad
want to die

[frigid lark]

there's an obvious solution here

[wraith cargo]

idk if you're familiar but completions are kind fucked and Cohen's theorems are huge because they actually give you a description of what completions looks like

[formal ermine]

studying for my exam forced me to learn so much math that I despise that I temporarily lost fun

[wraith cargo]

aren't you in HS

[formal ermine]

uhhhmmmm 👉👈

yes but I am an idiot sandwich

[wraith cargo]

dw I finished HS last year
The strat is to not give a shit

and do math

[frigid lark]

last YEAR

damn, people on this server are cracked

[wraith cargo]

I'm (supposed to be) a first year UG
I kinda dropped out lol

[formal ermine]

🫱🍞😔🍞🫲

[formal ermine]

I still have 2 more years go go

fml

[wraith cargo]

I moved half way around the world to end up in fucking San Diego

[formal ermine]

LOL

[wraith cargo]

I spent 6 months at UCSD just to end up with severe chronic depression and now I'm moving back home to a local uni

50k wasted well

[formal ermine]

us moment

[wraith cargo]

at least I had a cool algebra prof

[frigid lark]

🫂

[wraith cargo]

bruh ur doing alg geo in like 2nd year HS??
I was an idiot back then

[formal ermine]

i LOVE my nt prof

I'm an idiot too bruh

[frigid lark]

nt prof? in HS?

[wraith cargo]

bitch where do you go to HS that you have a NT class

sounds like france

[frigid lark]

where do you go to HS that you have a prof?

[formal ermine]

I take coueses at uni but idk I can continue doing that

because my grades are terrible

[wraith cargo]

fucking french HSers doing topology in HS

[formal ermine]

and my teachers don't like me

[formal ermine]

GERMANY

[frigid lark]

wouldn't French HS's be doing Cat theory?

[formal ermine]

wait what kind of completion are we talking about

[wraith cargo]

I'm 4h drive south of you bruh

[formal ermine]

wya

[wraith cargo]

m-adic

[formal ermine]

italy?

[wraith cargo]

croatia

[formal ermine]

eh?

[wraith cargo]

the completion I'm talking abt is

[formal ermine]

colim R/m^iR or something lol?

[wraith cargo]

$\hat{R}_\mathfrak{m} \cong \varprojlim R/\mathfrak{m}^i$

[cloud walrus]

ironyincarnate

[frigid lark]

is colim inverse limit?

[formal ermine]

fucking nailed it

[wraith cargo]

colim is direct limit

[formal ermine]

istg

[wraith cargo]

limit is inverse limit

[formal ermine]

I always mix up lim and colim

I meant inverse limit

[frigid lark]

but the arrow is pointing to the left

[wraith cargo]

yeah because in the diagram you're going AWAY from the limit

[formal ermine]

or whatever has the limit to the left

[wraith cargo]

and for the direct limit you're going TOWARDS the colimit

that's how I always remember it 😎

[formal ermine]

you should visit me

then we can play smash together

[wraith cargo]

just don't tell me you like live in the northern most point in germany or something lol

[formal ermine]

look up bochum

anyway I should really sleep now

[frigid lark]

ban projective for 5 hours

[spare hatch]

h7

[formal ermine]

this server is the only thing keeping me sane

[frigid lark]

and awake

[formal ermine]

true true

otherwise I'd be on reddit right now

mindlessly scrolling

[frigid lark]

okay

under no circumstances ban projective for 5 hours

[wraith cargo]

okay feels like the ultimate burn

[formal ermine]

it's better than being left on read

I went on a date with my crush

now I texted a bit with her

and after my response to her response

she left me on read

two days in a row

I guess I'm not interesting enough

[wraith cargo]

people usually have a life

so they don't text 24/7

I learned that the hard way

[formal ermine]

impossible

[formal ermine]

yeah but I mean like

I asked something

she ignored it

the next day we text a bit again

she still hasn't answered it

same thing happens that day too

gn algebra channel

[wraith cargo]

good night

[frigid lark]

So is this just Lang being Lang, or is there some other nuance I'm missing

[frigid lark]

gn

[wraith cargo]

oohhh profinite groups
go on

I am familiar with them

[frigid lark]

cause here it's an inverse limit

and the arrow goes to the left

[wraith cargo]

yes

that's how it goes

inverse - left
direct - right

[frigid lark]

but here it's a direct limit and the arrow is going to the left

or am I missing something

[wraith cargo]

it's an inverse limit

completions are defined using inverse limits

think like

p-adics

[frigid lark]

but then what about this?

[wraith cargo]

proj was wrong

[frigid lark]

oh, okay

[wraith cargo]

it's the limit of the diagram

not the colimit

[frigid lark]

alright

makes sense

[wraith cargo]

I need to learn more about profinite groups

they are interesting

say parrot ur reading Lang rn right?

[frigid lark]

mm

[wraith cargo]

how long has that affair been going on lol

[frigid lark]

7-8 months

[wraith cargo]

I hope ur wife doesn't find out
but anyways what are ur plans afterwards

alg geo?
ANT?
Just more algebra?

[frigid lark]

Probably gonna have to do some topology

[wraith cargo]

wow

going soft eh

[frigid lark]

cause I already feel the lack of topology hurting me

[wraith cargo]

like

point set?

[frigid lark]

yeah, maybe alg as well to motivate hom alg

but idk, Lang is gonna take a while

but yeah more Alg sounds like what I want to do

[wraith cargo]

ngl I've been kinda winging my point set knowledge for a while now
realistically the most I've read is what I needed to know for Folland's analysis book

I needed to read that to know some stuff abt radon measures

And like Haar measure

[frigid lark]

maybe I could do my uni's topology course and kinda wing it from there

but my uni teaches baby stuff so, maybe not

[wraith cargo]

go to Folland and read ch. 4 it contains literally everything you'll ever need to know

[open sluice]

when you want to learn X but it requires you to learn Y first

[open sluice]

better yet, when X requires you to first learn Y, Z, and W

[wraith cargo]

what if you skip to K

[open sluice]

working on it

[tender wharf]

to learn X you need Y, to learn Y you need Z but to learn Z you need X :sofalse:

[wraith cargo]

cycle of trauma

[timid oriole]

Hello, just checking, "trivial" here means "the trivial subgroup" right?

[warm wyvern]

yea

so it only contains the identity

[timid oriole]

ok great thanks!

i was hoping the author wasn't just insulting me for not immediately knowing what the center was

[long nebula]

hahahaha

[ivory trail]

i guess I can visualize a sequence of objects at 0, 1/2, 3/4, ... with the limit at 1. then you draw arrows between consecutive objects pointing the same way as the arrow in the notation

this is kinda useful actually

before I just pictured the sequence with a floating (co)limit object above for the universal property

and then I usually don't have to remember which is the inverse or direct limit

but I can just recall the p-adics/completion of a ring and stalks of a sheaf

if I think too much about this I'll mess my recall up

[rocky cloak]

Limit is left exact, arrow to the left. Colimit is right exact, arrow to the right.

[wraith cargo]

Direct limit is exact so uh

[rocky cloak]

Direct limit yes, colimit no

Also, direct limits are not exact in arbitrary abelian categories.

[wraith cargo]

R-mod my beloved

[lethal dune]

You need filtered, arbitrary DL may not be exact

[rocky cloak]

I wonder if there is a nice categorical explanation for why direct limits are exact in R-mod. Something about the forgetful functor preserving direct limits perhaps. Though I guess that just pushes the question...

What is the thing that makes R-mod AB5...?

[rocky cloak]

Are you one of those people who use "direct limit" to refer to colimits?

[lethal dune]

Good point, it depends on the interpretation.

As much as I hate the terminology, I do mean colimit

What’s AB5 🙄

I don’t enumerate my axioms

[rocky cloak]

Filtered colimits are exact

[wraith cargo]

aren't direct limits a type of filtered colimits?

[rocky cloak]

Usually yes, but terminology may vary

[wraith cargo]

Why does the Ext_R^n(R,R) disappear here?

[rocky cloak]

R is projective

[wraith cargo]

thanks
I knew it was flat and I remembered that like if a modules is flat and something else then it was projective

I think the condition is that every flat, finitely presented modules is projective

[rocky cloak]

That is true, but seems like kind of a circular way of deducing that free modules are projective.

[wraith cargo]

ohhh
true fuck
Idk sometimes I just pull stuff out of my ass

I'm reading abt injective stuff and I know R isn't necessarily injective so my brain made the genius leap that R might not always be projective as well

even tho earlier I knew that it is but eh
brain fart

[rocky cloak]

Happens all the time

[south patrol]

Pain

[frigid lark]

What, are you going to say that if M is a free module, there exists some module N that $M \oplus N$ is free

[cloud walrus]

parrottea

[frigid lark]

Preposterous