groups-rings-fields

[chilly ocean]

how is that subgroup

[rocky cloak]

Let V be this fixed 2d subspace.

It's fixed by the identity, and if f and g are such that f(V) = V and g(V) = V, then
f(g(V)) = f(V) = V

And
f^-1 f(V) = f^-1 V
V = f^-1 V

[chilly ocean]

if f fixes W and g fixes Y where W and Y are different 2d space , then fog fixes some 2d space how to show this

[rocky cloak]

It doesn't necessarily, that's like the whole point of the proof

[chilly ocean]

this confuses me about the subgroup , you ar also fixing 2d subspace

[rocky cloak]

The subspace spanned by the two first basis vectors is just one specific 2d subspace.

You can start with a different one if you want, but you can't just change willy nilly

[chilly ocean]

if matrix doesnt fix it , does it mean polynomial is irreducible

[rocky cloak]

No, you're supposed to apply 6.1

[sturdy spear]

Universial Property

[rocky cloak]

Take H the subgroup that fixes V, then gHg^-1 is the subgroup fixing g(V).

If every matrix fixed some 2d space Union gHg^-1 would be all of G, but it's not

[chilly ocean]

yes i got my solution i was thinking little wrong, trying to insert every fixed 2d subspaces inside same subgroup.

that solves it, i forgot to look at linear algebra part of this question, and conjugation together my bad in this question

This question is very nice, how it connects many things in one single piece,

[candid patrol]

WTH...

[tough raven]

But ℂ ≠ ℤ/pℤ... and I believe (1) does require algebraic closure of the field (or at least infiniteness, since after all, GL_n(F) is a finite group if F is finite).

[thorn jay]

oh no, is that mathfrak S for symmetric group I see

[tough raven]

Oh, I guess that's exactly the point.

[chilly ocean]

symbolic hate

[chilly ocean]

gordon james proud of him

[untold torrent]

Can anybody help me with this question?

[candid patrol]

Gx = Gy => x € Gy and y € Gx by taking g = e

$x \sim y$ iff $\exists g \in G, y = g \cdot x$

[cloud walrus]

UGOBEL

[languid linden]

Can somebody explain what is a Field and what f(...) ∈ F[...] means

[tardy hedge]

A field is a commutative ring where all elements ≠ 0 have multiplicative inverses

[vast stump]

this is saying take f to be a non-zero polynomial with coefficients in F

[languid linden]

oh I see

thanks

[alpine plank]

Do most people see the definition of a field before or after that of a ring? I learnt what a field is for the first time in my linear algebra class

[coral spindle]

Often before

[delicate orchid]

I once did S_4s subgroup lattice by hand but S_4 x C_2 is deranged

[chilly ocean]

you cant quantify sets

you cant quantify sets

goodbye

[ripe crest]

What is universal algebra and why isn't it as popular as category theory?

[thorn jay]

Omg

[thorn jay]

It is essentially because universal algebra is mainly based upon lattice theory. Lattice theory tried to be the foundations for math, but with the rise of bourbaki and category theory, which honestly are much better and cleaner in many aspects, it kinda got obsolete for that purpose

In logic, and some other places (like with quandles in algtop) universal algebra is still used a lot, because its great for that use, but in other cases category theory is simply better

[thorn jay]

Guess he lived up to his name

[thorn jay]

Next challenge: draw the full Post's lattice

[ripe crest]

I see, and it's great for those use cases because there is something about the lattice theory foundation that lends itself quite naturally to these areas?

[thorn jay]

In logic you very naturally deal with closed sets of things

Classes of models closed under satisfying certain types of axioms, sets of formulas closed under deduction etc

In algebra, too: subgroups generated by .., ideals generated by .., and so on

These are all examples of so-called closure operators, and lattices essentially abstract them and turn them into (finitary) algebraic structures

But closure operators also turn up in Galois theory, and hell, the Nullstellensatz too

[ripe crest]

Free x

I'm wondering then, what makes category theory better than universal algebra. When we talk about algebraic structures, they're closed wrt their operations, and when we talk about generating algebraic structures, like a free monoid, these are also closed

Closure seems to be a pretty fundamental property

[thorn jay]

Mhm, and lattice theory certainly has its place, but

1. Categories have a natural notion of "context", so different notions of equivalence, morphisms, etc.
2. Categories allow for much, much more than just algebraic structures themselves, so they are more useful more often
3. Lattices are special types of categories and things like closure become monads, Galois correspondences become adjoint functor pairs, and such

The real advantage closure operators and UA have is that they still work with sets and are logic focused, especially with polynomial terms and the like

[ripe crest]

I see, thanks for the explanation!

[thorn jay]

Anyone feel free to correct me

Im an expert in UA more than in category theory lol

[alpine plank]

How did you get into UA?

a slow descent into madness or something specific?

[thorn jay]

I had a project about Latin square-related stuff and needed a framework to study some algebraic structure that arised

(in particular, this is about pairs of MOLS, but I didn't end up going very far, only explored some special finite field-related case)

then it was indeed a descent further and further into the rabbit hole, and now I do UA for UA's sake, basically

[south patrol]

Tbh curious about quandles lol

[thorn jay]

just like a knot has a fundamental group, it also has a fundamental quandle

[south patrol]

Ah OK

[thorn jay]

and its a strong knot invariant; two knots are the same iff they have the same fundamental quandle

[south patrol]

Knot theory is the opposite site of topology to stuff I do tbh aha but nice

[thorn jay]

which is delightful
what is not delightful, is that quandles are not nice to compute 😢

[south patrol]

Oh wow lol

Lol somewhat unsurprising I suppose if it is that strong

But cool

[thorn jay]

yeah, the quandle axioms and Reidemeister moves correspond in some way, from what I could tell

anyways, as algebraic objects quandles are also simply very rich and intersting and have cool connections with groups with a load of functors and cool constructions

oh and the name obviously

auegh I'm reading this paper and their proofs have absolutely no elaboration

"then X, Y, Z"
I DONT SEE WHY IM A LITTLE SLOW, PLEASE EXPLAIN QWQ

ohhh okay I get it

[south patrol]

Ohnice

Quandale dingle

[thorn jay]

hi guys

quandale dingle here

[rocky cloak]

Is the fundamental quandle just the fundamental group, but viewed as a quandle?

[thorn jay]

no, because that would mean the the fundamental quandle isnt a faithful invariant

at least, I believe so

when looking at the conjugation quandle you actually lose information of the original group, I'm pretty sure

yeah, for example any abelian group will give you an isomorphic quandle, a projective one (where the binary operation just returns the first entry)

quandles also have a lot to do with colourings of knots

[rocky cloak]

From nlab

Every tame knot in ℝ^3 has a “fundamental quandle”. To define this, one can note that the fundamental group of the knot complement, or knot group, has a presentation (the Wirtinger presentation?) in which the relations only involve conjugation. So, this presentation can also be used as a presentation of a quandle.

So if I'm reading this right, the fundamental quandle is bigger because it's the free quandle modulo these relations, so if we add in the extra relations of being a group we would get the fundamental group

[thorn jay]

in a sense? for quandles you forget the group structure though

the book I use (Quandles and topological pairs) constructs it differently (from the ground up, rather than top down like nlab)

I'm not sure if they would be the same though, I only started learning them yesterday because I thought they sounded funny, lol

[rocky cloak]

Quandles always reminds me of
https://youtu.be/tTYzGEvLvMc?si=otFu4HLu5HAbN9OH

[tough raven]

the rise of Bourbaki and category theory

[thorn jay]

did you know there's absolutely zero mention of category theory in the book I used to learn UA? (Burris and Sankappanavar)

they had personal beef istg

[thorn jay]

oo that's pleasant, I like it

I can only not unhear "tell me quandle, quandle quandle" now

[south patrol]

Comedy shorts gaming here

[keen badge]

$$
\langle r,t: r^4=t^2=e \rangle \cong\mathbb{Z}_2\times\mathbb{Z}_4
$$

$$
\text{or}
$$

$$
\langle r,t: r^4=t^2=e \rangle \cong \text{D}_4
$$

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[thorn jay]

Neither, you get the free product of C_4 with C_2

you gotta add relations between r and t

[keen badge]

So right now, it is an infinite group?

[thorn jay]

ye

[keen badge]

cool thx

[thorn jay]

👍

[proper jolt]

is it necessary for the statement the intersection of two semisimple submodules is semisimple? or could you say that the intersection of a semisimple submodule and a submodule is semisimple or 0?

i think the latter is sufficient since the intersection is a submodule of both of the submodules and a submodule of a semisimple module is semisimple

[hidden wind]

yeah with the generators and relations as given here rt has infinite order

geometrically, this is the rotational symmetries of the truncated tetraapeirogonal tiling of the hyperbolic plane

this is a tiling of the hyperbolic plane by triangles with interior angles pi/2, pi/4 and 0, so two of the sides are parallell

the corner at the boundary of the disc here is an “ideal point at infinity” through which we may rotate the triangle but we’ll never turn around it (the order is infinite)

common notation for this group is $D(2,4, \infty)$ for the geometric picture or just $\mathbb Z/2 \ast \mathbb Z/4$ for the free product

[cloud walrus]

rødbet-jens

[tough raven]

Yes, exactly as you said. In fact, 0 should be considered semisimple, so you don't even need the "or".

[proper jolt]

oh ok nice thankyou

[sly crescent]

That or the order-4 apeirogonal tiling or the infinite-order square tiling

[chilly ocean]

can anyone suggest some good excersises on groups and rings

[void cosmos]

google some exams u find on the internet

[keen badge]

Let $\phi:R_1\to R_2$ be a ring isomorphism, such that $R_1$ is a field. Does that mean that $\phi$ is also a field isomorphism?

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[thorn jay]

yes, because the restriction to the multiplicative group of R1 would be a group isomorphism, and therefore preserve inverses

in general, if u in R is a unit with inverse v, then for any (unital) ring homomorphism f : R -> S, f(v) is the unique multiplicative inverse of f(u), because multiplicative inverses are always unique in monoids

[keen badge]

So if I want to prove that $\phi:F\to R$ is a field isomorphism where $F$ is a field, and $R$ is a ring, I need to prove that:

1. $\phi(x+y)=\phi(x)+\phi(y)$

2. $\phi(x\cdot y)=\phi(x)\cdot\phi(y)$

3. $\forall r \in R \exists f\in F$ such that $\phi(f)=r$

4. $\phi$ injective because $F$ is a field

---

yes?

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[thorn jay]

basically

yeah

[keen badge]

thx

[thorn jay]

you only need that its a surjective ring hom

[keen badge]

yeah. I wasn't sure, so I thought the best thing to do is to ask someone 🙂

[thorn jay]

ofc!

[vivid kestrel]

can anyone give me an example of a "nontrivial" unit in a group ring KG (so not kg for k nonzero and g a group element)?

[delicate orchid]

1+g for any g in G

Oh UNJT

take it away jagr

[rocky cloak]

If G = {1, g, g^2, g^3} = C4 and K = F2, then
1 + g + g^2 should be a unit.

Inverse should be
1 + g^2 + g^3

[vivid kestrel]

interesting, may i ask how you came up with that

[chilly ocean]

any good book for artin wederburn theorem

[delicate orchid]

Or just 1+g in F_2[C_2] right

[rocky cloak]

Fp[C_(p^n)] = Fp[x]/(x^p^n) where x = 1-g.

And in that ring the units are exactly polynomial with constant coefficient 1. So I just picked one

So in particular F2C4 should have 8 units

[vivid kestrel]

oh cool, thanks!!

[rocky cloak]

I believe this is also the smallest example where it works

[delicate orchid]

Any book on representation theory should cover it. I learnt it from James-Liebeck personally

But there’s also not much to learn? It’s just one theorem so you can just Google the proof

[rocky cloak]

F4C2 also works, same size

[thorn jay]

terminology dropped

unjt

[delicate orchid]

Society if the group-ring group of units adjunction was an equivalence:

[thorn jay]

society if any monad was naturally isomorphic to the identity

[rocky cloak]

Or avoiding finite fields
QC2 = QxQ with the two idempotente
(1-g)/2 and (1+g)/2.

So the units are
p(1-g)/2 + q(1+g)/2 for nonzero p and q.

For example 3 + g should be a unit.

[delicate orchid]

Oh yeah that ones a classic

[vivid kestrel]

so let G be a finite group and K a field and KG the group ring and V a KG module (nonzero). Does V then have a non zero subset finite that is an FpG module?

[thorn jay]

if G is infinite then I believe KG doesn't

[vivid kestrel]

my bad, G is also finite

[thorn jay]

any KG-module is an FpG-module, so any cyclic submodule is a quotient of FpG hence finite as both G and Fp are finite

(assuming that Fp is the prime subfield of K)

[rocky cloak]

So I'm not sure I understand what you're asking, but if K has characteristic p, then FpG is a subring of KG, so V is an FpG module by restriction of scalars.

And it will have a finite submodule just by picking a cyclic one

[vivid kestrel]

ah that figures, thanks

[hidden wind]

wait how does the infinite-order square tiling work here?

is that a euclidean motion group? a quotient of the one for the hyperbolic tiling above?

[glad osprey]

I just proved this, but I didn't use the fact that m is prime, so I'm not sure it's correct.

Assume $E \neq F$. Then there exists $a \in E \setminus F$ such that $a^m = b$. Let $w$ be an n-th root of unity, and let $n = mk$ for some $k$. Then $a, aw^k, aw^{2k}, \dots, aw^{(m-1)k}$ are m distinct roots of $x^m - b$, all of which are in $F(a) \setminus F$, so $[F(a) : F] = m$, therefore $x^m - b$ is irreducible over F.

Is this correct? As far as I can see, this works regardless of whether m is prime

[cloud walrus]

sheddow

[tardy hedge]

You’re still doing galois theory?

[glad osprey]

hell yea 🔥

[tardy hedge]

For a class?

Or just studying it

[glad osprey]

exam in a couple of weeks

[tardy hedge]

Oh i see

Good luck! But also dang when did ur semester start?

Thats a long semester

[glad osprey]

Thanks! It started in january I think

[tardy hedge]

Same

[glad osprey]

I feel like the exam is way too early

[tardy hedge]

Ive been done for a couple weeks tho

Yea mine was early af actually

[glad osprey]

I have another class which started in january where the exam is in the beginning of june

[tardy hedge]

I gotta show some of the questions i had on it

we could take exam home

[glad osprey]

ah, nice

[tardy hedge]

Wtf

[glad osprey]

I bet its gonna be 95% category theory, 4% commutative algebra and 1% normal galois theory

[tardy hedge]

Lmaoo

I mean ok there was one category theory question thankfully hahaha

I didnt do that one

lol

[chilly ocean]

do you know there are 4 categories

[tardy hedge]

WTF

[thorn jay]

I thought there were 5

[thorn jay]

letss goooooooooooo

[chilly ocean]

can you nane ut

[thorn jay]

💀

[glad osprey]

I think there's only 2, there's C and C^op

[thorn jay]

Set
Set but spicy
Law / Eq (category of equational theories / lawvere theories)
Grp
R-Mod as one big category for all commutative rings R

[rocky cloak]

The fact that the roots are in F(a) doesn't mean it has degree m.

For example, say b doesn't have a square root and consider
x^4 - b^2 = (x^2 - b)(x^2 + b)

If you're looking for hints:

• ||Say the minimal polynomial of a has degree d, what's the degree of the minimal polynomial of w^k a?||
• ||What does this tell you about the factorization of x^m - b into irreducibles?||
• ||What's the relationship between d and m?||

[glad osprey]

I see, thanks! Love you

[chilly ocean]

the mth root of unity is also inside nth root of unity

can anyone suggest some good resource learn galois group of polynomials

[hidden wind]

those are the same

[glad osprey]

there are actually countably many categories, there's C, C^op, C^op^op, C^op^op^op, ...

[thorn jay]

just like a square has countably many rotational symmetries!
0, 90, 180, 270, 360, 450, ...

[delicate orchid]

so close - those are NUMBERS not symmetries!

[thorn jay]

GROUP ACTIONS ARE NOT REAL

WAKE UP

[delicate orchid]

numbers are group actions

[arctic trail]

category theory has more ops than gangsta rappers

[glad osprey]

the infinite group C_4 🔥

[thorn jay]

exactly

[thorn jay]

nuh uh

[arctic trail]

nuh uh

[delicate orchid]

they're literally constructed by repeated application of "+1", which is an action of Z on itself

[arctic trail]

"Hello I would like 'the thing that acts' apples"

[delicate orchid]

I win you lose night night don't let the bed bugs bite

[delicate orchid]

you gaining a number of apples seems like an action on a group to me

[arctic trail]

🤓

[delicate orchid]

brother is called "trivial lemma" nerd emojing me

[arctic trail]

if you're talking about finite cardinals might as well call them semigroup actions

[delicate orchid]

BUUURRRRRRPPPPPPPP

[thorn jay]

domain expansion: BUUURRRRRRPPPPPPPP

[arctic trail]

wait you're studying representation theory

I wasn't familiar with your game

[thorn jay]

the good ending

[arctic trail]

what type of problems? It is not a small topic

[delicate orchid]

Mackey functors

with the occasional Tambara functor if I'm feeling saucy

[arctic trail]

the things that generalize restriction and induction?

[delicate orchid]

yeah

[arctic trail]

not familiar with those functors

[delicate orchid]

Tambara has seperate induction/coinduction which is nifty

[delicate orchid]

I bet you can name one

[arctic trail]

wdym?

[delicate orchid]

like they're everywhere you've definitely seen one

[arctic trail]

I meant I'm not familiar with the abstraction that are the mackey functors / tambara

and the general theory

[delicate orchid]

there's numerous ways you can realise them and I think the direct definition of "pair of functors with associated morphissm" is the least enlightening by far

[arctic trail]

Are they adjoint?

[delicate orchid]

absolutely not

one is contravariant and one is covariant

[arctic trail]

ah lol

[delicate orchid]

if they were adjoint that would retroactively imply that cohomology and homology were adjoint

which would be WONDERFUL so obviously it's not true

[arctic trail]

yeah of course

[chilly ocean]

I was talking about categories in different context but

[delicate orchid]

I'm on the edge of my seat with this one captain

[arctic trail]

What's your opinion on abstract harmonic analysis?

[chilly ocean]

Beutiful

[delicate orchid]

I don't go anywhere near that slop

all of my groups are finite

[arctic trail]

it's beautiful though

[delicate orchid]

or profinite

[arctic trail]

positive characteristic?

[delicate orchid]

sometimes

[arctic trail]

I mean

that's just finite with more steps I guess lmao

[delicate orchid]

precisely

[arctic trail]

and the fact that they always factor through a finite group is sad

but makes sense

[delicate orchid]

sad for some

also wdym by they factor through

[arctic trail]

if G is profinite and pi: G -> GL(V) is a continuous representation (V finite dimensional)
Does there not exist a normal N with finite index such that pi factors through G/N?

pretty sure this is the case if V is a complex space at the very least

not sure if continuous is required here (?)

[delicate orchid]

ah right so you're saying they're all lifts from finite groups

[arctic trail]

yeah

[arctic trail]

I think it is, but I'm not sure I haven't seen this in a bit

[arctic trail]

and they can of course be chosen as one of the groups in the profinite system

[glad osprey]

don't leave us hanging bro, what are the 4 categories?

[chilly ocean]

An Indian knows those

[delicate orchid]

yeah - thinking about it the other way, due to the compatability conditions in the limit you can take anything in R(G_i) and extend it to the profinite group by having everything else act trivially

which means that everything else is in the kernel, hence your extenstion is a lift from the quotient

very swag

[amber burrow]

kinda general question: how do you like problem solve in group theory>

ive been studying it myself

but i also take an actual real analysis + linear algebra class

like in real analysis, i draw it out

for linear algebra, idk it somehow works

but with group theory i kinda just sit there going through possible solution paths until i think of something that works

[delicate orchid]

yup

you just get better at picking which path to take

[thorn jay]

that's most of higher math, anyways

you can't visualise the stuff normally anymore, so you gotta rely more and more on intuition

I guess eventually you get commutative diagrams

[rocky cloak]

Is that different from what you were doing for real analysis or linalg?

[amber burrow]

now that i think about it, not really

most of the time i feel like it was pretty easy to figure out what to do just from the problem itself

[chilly ocean]

for me group theory is always harder than all other suject , some times i solve sometimes 90%

[rocky cloak]

I mean, going through the different possible solutions until something works sounds pretty easy.

I guess it might be timeconsuming if there's many things to try, but I don't feel like there usually is...

[delicate orchid]

it can also be an issue of identifying things to try

[glad osprey]

I like to think of proving things as a chess puzzle, where you have a bunch of possible moves, and you have to find the correct sequence that leads to mate. If you're experienced, you can immediately see the most promising candidate moves, and it's just a matter of calculation, whereas a newbie may not even know where to start looking

plus, if you're a newbie, you sometimes make illegal moves, both in chess and math

[ivory ore]

Give an example of an infinite group which has a composition series.

1 \to A \to G \to B \to 1

take A and G to be infinite and B to be C2

does that work?

not every A is gonna work for C2 has to be a normal subgroup in A too?

[tough raven]

\newcommand*{\GL}{\operatorname{GL}} It depends strongly on the base (topological) field. As you mentioned, this is true over $\bC$ (and IG over any topological subfield $K$ of $\bC$, since $\GL_n(K) \subseteq \GL_n(\bC)$ has the subspace topology(?)). It is false if e.g., $G = \bZ_p$, the base field is $\bQ_p$ and the representation is $a \mapsto \begin{psmallmatrix} 1 & a \ 0 & 1 \end{psmallmatrix}$.

[cloud walrus]

Raghuram

[thorn jay]

If A doesnt have a composition series then G doesnt have reason to have one either

[ivory ore]

right

[arctic trail]

luckily all my representations are either over C or the algebraic closures of Fp

[delicate orchid]

rip Q

[thorn jay]

I looked on stackoverflow and found this:
A5 \subset A6 \subset A7 \subset ...
The union of which is the group A_\infty, which you can prove to be simple

[arctic trail]

direct limit thingamabobs

[thorn jay]

I mean, i guess, but an infinite union is much nicer to work with

And you can use as overarching group S_\infty

[arctic trail]

wouldn't this result be true for any direct limit of simple groups?

[tough raven]

IG there are infinite simple groups (e.g., permutations of ℕ / permutations of ℕ fixing all but finitely many elements; PGL_2(k) for any algebraically closed field k according to Lemma 4.1 of https://www.ma.imperial.ac.uk/~buzzard/maths/research/notes/the_adjoint_functor_theorem.pdf or https://www.math.ucdavis.edu/~efuchs/250A/PSL2Rsimple.pdf), and those obviously have a composition series.

[thorn jay]

Hmm, yeah probably

[tough raven]

Is the topology discrete in the latter case?

[arctic trail]

cause you can just take the preimages of normal subgroups of the direct limit

[thorn jay]

Hmm

[arctic trail]

and get normal subgroups in the smaller ones (so either trivial or the whole group)

then the property of it being directed means that a non-trivial normal group must have full preimage on every part of the system

[thorn jay]

And so must be full in the original group, by construction

[arctic trail]

yeah

[thorn jay]

Damn, thats sweet

[rocky cloak]

This is an extension, not a composition series. But if your looking for an infinite group with a composition series you just need to find an infinite simple group I guess

[arctic trail]

PSL(2,C) is simple innit

[delicate orchid]

or is it 🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨🤨

[thorn jay]

I guess the composition factors of any infinite group with a composition series must include some infinite simple group, so we'd find an infinite simple group anyways, truly an <=> moment

[arctic trail]

I don't know

[delicate orchid]

it is dw

[rocky cloak]

I guess you could allow the defintion of composition series to be infinitely long

[thorn jay]

Transfinite series

[arctic trail]

transfinite induction on composition series

[rocky cloak]

like Z > 2Z > 4Z > 8Z > ...

[thorn jay]

Every automorphism tower terminates transfinitely

[arctic trail]

exists phi phi

[thorn jay]

Seeing
Z > 3Z > 6Z > 12Z > ...
This is a nice example of the Jordan-Hölder theorem not holding in the infinite case :0

[coral shale]

i conclude Z < 0

[sly crescent]

It’s the same group

[hidden wind]

ok but what is the infinite order square tiling?

[hidden wind]

bbbut that is a quotient!

[sly crescent]

Of what?

[hidden wind]

of this^

of the tiling

[sly crescent]

No it isn’t

[hidden wind]

one square consist of eight triangles

[sly crescent]

And?

[hidden wind]

doesn’t that make it a quotient

[sly crescent]

No

[hidden wind]

what’s the action of the order two generator here

[dim widget]

It's a subgroup not a quotient

[sly crescent]

It’s a point reflection about the midpoint(?) of an edge

[sly crescent]

Wait. Are we talking about the tilings themselves or their symmetry groups here?

[dim widget]

I was talking about the group of transformations of the tiling

[sly crescent]

Well then it’s both a subgroup and a quotient, because it’s the same group

[sly crescent]

Also, this is actually the dual of the truncated tetraapeirogonal tiling

[hidden wind]

i think i’m just confused by the action and which fundamental domain it admits

each of these have the same symmetry group but only the triangle is a fundamental domain

?

[sly crescent]

The squares here are not fundamental domains, if that helps

[sly crescent]

Yes

[cloud walrus]

longboard kayak

[ivory ore]

these would be equivalnet right?

the construction of both of these from smaller series as mentioned in Schreier Refinement Theorem is a bit wacky

[cloud walrus]

longboard kayak

[ivory ore]

is this okay ?

[void cosmos]

look at C[x,y]/I

[chilly ocean]

my solution claim $I=<x+y^2,y+x^2+2xy^2+y^4>=<x,y>$ because $(x+y^2)(x+y^2)-(y+x^2+2xy^2+y^4)=-y\in I$ wo we have $x+y^2-y^2\in I$ and $x\in I$ so $<x,y>=I$ so I is maximal.

[cloud walrus]

Curvature

[chilly ocean]

what was ur plan

[ivory ore]

altho i don't understand everything, it felt like a neat post

[thorn jay]

Just looking at the cardinalities shows that they must be composition series

So yes, by Jordan Holder

[stuck egret]

ah yes the

[thorn jay]

Says that any two subnormal series with simple factor groups must be equivalent

Insane theorem tbh

[ivory ore]

hmm, it was before introducing JH

[thorn jay]

Ohh im sorry lol

Just prove it quickly

[thorn jay]

(Also you can see that the factor groups are C2, C2, C2 and C3)

[ivory ore]

yeah now i can

[thorn jay]

Looks good

[ivory ore]

i did that on my own and in the exercise he asks to exhibit comp series for S4

[glad osprey]

These hints are great btw! The last part was easy, but I struggled a bit with showing that $x^m - b$ splits into irreducibles of equal degree. Here's my proof, I feel like it uses a bit too much machinery, but I learned a lot writing it: $\newline$

Let $a \in E \setminus F$ where $a^m = b$, and let $w$ be an m-th root of unity. Then $a, aw, aw^2, \dots, aw^{m-1}$ are $m$ distinct roots of $x^m - b$, so $F(a) = E$. Now let $f(x)$ be the minimal polynomial of $a$ over $F$, with degree $d$. Since $F \subset F(a)$ is a Galois extension, we have $|Gal(F(a) / F)| = d$. It's easy to see that $Gal(F(a) / F)$ acts freely on the roots of $x^m - b$, since any automorphism sends $a$ to some other root $aw^i$. $\newline$

Now let $aw^j$ be another root of $x^m - b$. Its minimal polynomial is given by $$\prod_{\gamma \in Gal(F(a) / F)} (x - \gamma(aw^j)$$, and since the Galois group acts freely on $aw^j$, this polynomial has degree $|Gal(F(a) / F)| = d$.

[cloud walrus]

sheddow

[rocky cloak]

So there is actually very little machinery needed.

What you need is that the degree of the minimal polynomial of a is the same as the degree of F(a)/F.

And that F(w^k a) = F(a).

Then you just observe that the factors of x^m - b are minimal polynomials of w^k a

[glad osprey]

Ah nice, that's much easier

Is it possible to generalize this result? For example if the Galois group of f(x) acts freely on the roots of f(x), is that enough to conclude that f(x) splits into irreducibles of equal degree?

[rocky cloak]

And I guess, following your proof, once you say G acts freely on the roots it follows that d divides m, so you're done

[thorn jay]

Is there some general method to determine when the commutator of two normal subgroups is equal to their intersection?

[rocky cloak]

Yes, the orbit of an element is just the other roots of its minimal polynomial, so if we're in the seperable case the size of the orbit is the degree of the minimal polynomial.

If the action is free then all orbits have the same size (the order of the group)

[glad osprey]

Nice, that's so cool I love that I can finally make use of group actions in Galois theory 🔥

[rocky cloak]

The group action being free seems like kind of a strong condition though

[chilly ocean]

what does he mean by explicitly

do i have to write the set

[rocky cloak]

I guess so

[chilly ocean]

is that C[X]* C[Y]*{ -- polynomial in t}

[rocky cloak]

You mean the map is surjective?

It might be... Would need to think about it

[chilly ocean]

i doubt it is surject there is something in third touple{t}

that can we have C[t]=C[X,Y]/<X-Y>

[rocky cloak]

I would guess the answer is something like
|| (a + bx + O(x^2), a + cy + O(y^2), a + (b+c)t + O(t^2))||

[chilly ocean]

HOW IS THAT

[rocky cloak]

Well, that certainly seems to be contained in the image at least

[delicate orchid]

let $f \in \bC[x,y]$ be given by the sum over all $a_{i,j}x^iy^j$, then the image under this map is given by $\left(\sum_{k} a_{k,0}x^k, \sum_{k} a_{0,k}y^k, \sum_k\sum_{i+j = k} a_{i,j}t^k\right)$ I think?

[cloud walrus]

☻ Wew Lads Tbh ☻

[delicate orchid]

so the image of this map is all of C[x] x C[y] followed by some strange graded-tensor subspace looking thing

if I'm correct on that actually being the image of course

whatever it is I'm ~53% sure that the image is iso to C[x, y]/(xy(x^2-xy), xy(y^2-xy))

[rocky cloak]

Can you get something of the form
(1, 0, ?)

[chilly ocean]

53% OR 69%

[chilly ocean]

1+y maybe

[rocky cloak]

That would give you
(1, 1+y, 1+t)

[rocky cloak]

Yeah, I feel like you can't get more than this

[delicate orchid]

Good point, even in my full answer it’s pretty clear that the constant term in all positions has to be a_0,0 lol

I also don’t know how to write it more explicitly

[delicate orchid]

It’s giving total complex

[chilly ocean]

if a+b=p then a and b are relatively prime integers

<a,b>=<d> so we have to show show d=1

[rocky cloak]

0 + p just entered chat

[chilly ocean]

yes p=ds either d=p or d=+-1

and yes a and b are positive given we also need that

d=p forces x+y=1 where a=xp,b=yp

[tropic obsidian]

soft question, but why do people call projective limits a type of gluing operation?

couldn't direct limits also be viewed as a gluing operation as in the union of opens can be re phrased as a direct limit

[chilly ocean]

it is identical some where for every n

[rocky cloak]

I think it's just about the context where these come up.

Like if you have two open sets and look at functions defined on them, then gluing functions that agree on the intersection gives you the elements of the pullback.

This is a useful idea that comes up a lot. And comes up in context where we think about gluing things together.

[tardy hedge]

fact so nice he said it twice

[lusty marlin]

Euler's theorem

[chilly ocean]

oh thanks

[chilly ocean]

how u applying it

[south patrol]

I guess consider a + b mod ab

Maybe my solution is a lil odd though lol

[chilly ocean]

quite even

[arctic trail]

?

solve:
a^n + b^m = 1 (mod a)
a^n + b^m = 1 (mod b)

I think this is much simpler

[chilly ocean]

yes so c.r.t right?

[arctic trail]

yes

take m = phi(a), n = phi(b)

[chilly ocean]

yes

[storm kiln]

I thought of expanding (a + b)^phi(ab) as a^phi(ab) + b^phi(ab).

Freshman's dream (a + b)^n = a^n + b^n :)

Also a + b is invertible mod ab, since gcd(a + b, ab) = 1

[chilly ocean]

clinging

[storm kiln]

?

[paper sonnet]

I dont think its freshmans dream

in this case it works because ab = 0 mod ab

[chilly ocean]

how many dreams are there , i heard about sophomore's dream

is the division algorithm(in smaller ring) is preserved in bigger ring(same quotient and remainder)

[ivory ore]

forbenius

[chilly ocean]

what

[paper sonnet]

the thing is that for freshman dream you need (a+b)^p where p is prime

and the extra terms vanish because p | n choose p, in this case the extra terms vanish because ab = 0 mod ab, just that d:

[glad osprey]

I know that for a Galois extension F <= E, the Galois connection sends normal subgroups of Gal(E / F) to normal intermediate fields. Does the quotient of Gal(E / F) by a normal subgroup correspond to anything in terms of fields?

[dim widget]

It's the galois group of that extension

[glad osprey]

Ah right, thanks

[glad osprey]

How can this be true? Take f(x) = x^2 + 1 and g(x) = (x - 1)f(x) over Q for example, they have the same Galois group, but only one of them is irreducible. What am I missing?

[dim widget]

The galois group of g is not transitive

because g has degree 3

being "transitive" is not an intrinsic property of the galois group, it's a property of the action of the galois group on the roots

the proof of the theorem is basically that the orbits of the galois group on the roots of f and the irreducible factors of f (up to scalar) are in canonical bijection

[glad osprey]

hmm, I'm not sure I understand I thought the Galois group of a polynomial is just the Galois group of its splitting field. f(x) and g(x) have the same splitting field, so don't they have the same Galois group?

[dim widget]

they do have the same galois group. But being "transitive" is not something that you can say about a group, only about a group action

S_2 acting in the usual way on a 2 point set is transitive, S_2 acting by permuting the first two elements of a 3 point set and fixing the 3rd point is not transitive.

[glad osprey]

I see I would argue the theorem is a bit sloppily stated in that case though. For context, here is the definition of a transitive permutation group in this book:

[dim widget]

Yeah but picking a map from H \to S_n is the same as picking a way of letting H act on n points

in the statement of the theorem you're right that they're implicitly thinking about S_{set of roots}

[glad osprey]

hmm, so when talking about the Galois group of a polynomial, it's clear that it acts on the roots, so blurring the line between the group and the group action is maybe not that unusual

[thorn jay]

Mathematicians are lazy

[glad osprey]

btw, can I reformulate the theorem to say that f(x) is irreducible iff the Galois group of f acts transitively on its roots?

[dim widget]

Yeah this is somehow where the theory started, by thinking of galois groups as subsets of all permutations of the roots and trying to solve equations using resolvents.

[dim widget]

This is a more canonical way of saying it, but you still have to rule out multiple roots

[glad osprey]

I see, thanks

[dim widget]

maybe: f is irreducible if f is separable and Gal(f) acts transitively on the set of roots.

A fun group theory exercise is to show that if f is irreducible at least on g \in Gal(f) doesn't fix any root.

[chilly ocean]

x^4-x+1 what is the galois group

[dim widget]

What have you tried?

[chilly ocean]

the discriminant formula is very big

i want to know sometricks to solve such problems

[tough raven]

||G ≠ ∪_g gHg^{-1}||?

[dim widget]

that's one way!

there's also a more general technique using a famous counting lemma

[thorn jay]

niche reference

lol

[chilly ocean]

Yeah we had some heavy discussions on this more than once

[thorn jay]

🔥

[chilly ocean]

That's the only reason to react

[thorn jay]

I am a theythem of few words

[next obsidian]

I thought you’re an any pronouns

[thorn jay]

I'm a shapeshifter without preference

rn feeling they/them but idrc

[tardy hedge]

there are a few nonbinary mathematicians here arent there

well i know at least one other

cuz they have it in their name lol

[thorn jay]

🔥

[chilly ocean]

What is any pronouns

[thorn jay]

tbh I don't really feel like a nonbinary, because I frankly dont care but tend to refer to myself differently from time to time

[dim widget]

One nice trick is to reduce modulo p and calculate what frobenius is

[thorn jay]

refer to me using any pronouns you want, idc 👍

[chilly ocean]

I actually want to know the meaning, I don't mean that bro

[dim widget]

but the most algorithmic way is to compute the resolvent cubic and the discriminant

[thorn jay]

that's what "any pronouns" means

[dim widget]

he/him/she/they/I/me/you/etc.

[chilly ocean]

Yes discriminant is not comfortable in exam for me

[chilly ocean]

Yeah man

[karmic cape]

I concur

[chilly ocean]

For cubic it's fine , using cardeno then finding discriminant and check if it is in base field or not for quartic things getting

[dim widget]

for this quartic in particular it's not that hard to calculate the discriminant

[chilly ocean]

Formulas getting bigger and bigger

[dim widget]

I wouldn't use the formula I would calculate the resolvent (which is the determinant of a 7x7 matrix which is mostly zeroes)

or you could use euclid's algorithm which is probably fastest

[chilly ocean]

Okay I have yet to see those things

[dim widget]

euclid's algorithm is long division

like when you have wanted to divide two numbers you have done eudlid's algorithm

[chilly ocean]

Can you give some references

[chilly ocean]

Yes I know euclid algorithm

[dim widget]

okay if you want to find the discriminant you would do long division between f(x) and f'(x) until you end up with a number as a remainder

that number is the discriminant

[thorn jay]

is this proof correct? it seems way too good to be true

(Proving that N is subnormal in G and H < H means N\cap H subnormal in H)

I guess it mirrors the corresponding fact for normal subgroups nicely

[tough raven]

Looks correct.

[thorn jay]

yippee

[barren sierra]

I cannot find a definition anywhere in this text (Sturmfels' Algorithms in Invariant Theory)

What does it mean for a Grobner basis to be triangularized?

Googling "triangularized Grobner basis" hasn't helped

[chilly ocean]

why

[dim widget]

It’s difficult for me to explain without introducing the concept of the “resultant” of two polynomials although I’m sure there’s an elementary way to explain this algorithm

[chilly ocean]

ops

[dim widget]

basically the discriminant of f is the resultant R(f, f’) and one can show that if f = pg + r then R(f, g) = sign * leading coeff(g)^(deg(f)) R(g, r)

then finally if g = A is constant then R(f, g) = A^{deg(f)}

So if p_n is the sequence of polynomials produced by the Euclidean algorithm then the discriminant depends only on the leading terms of the p_n and the final two remainders

This is a pretty fast algorithm to compute the discriminant if the coefficients of the corresponding polynomials are simple

[chilly ocean]

yeah many excersises having coefficint 0 in cubic and quadratic part

[dim widget]

Yeah

Well anyway this is O(deg(f)^2) which is as good an algorithm in general as you’re likely to find

And in this case it’s really easy since the Euclidean algorithm is pretty short for small degree polynomials

[crystal vale]

what is meant by conjugating element? is it mean if a = gbg^-1 then g is conjugating element?

[south patrol]

Yeah sure

[void cosmos]

yeah a is obtained from b by conjugating by g

[crystal vale]

okay

so it implies H is normal subgroup, right?

[void cosmos]

the "property" is saying that IF you have an element a such that a = gbg^-1 for some g in G and b in G

then you can pick this g to be in H.

[crystal vale]

yes it implies H is normal

[void cosmos]

yeah

[crystal vale]

i think we don't need H to be normal we can directly show G' \subset H

[crystal vale]

which normal abelian subgroup should i have to think ?

[rocky cloak]

Think about the normal subgroup generated by a

[crystal vale]

it is intersection of all normal subgroup containing a, right? i have to show it is abelian, does it have structure form?

[rocky cloak]

It does yeah. Maybe think about what kind of elements it would necessarily contain

[crystal vale]

conjugates of a must be in that subgroup

[chilly ocean]

to begin with, if a is in center we are done so we can assume it is not in center

[crystal vale]

can i say it is subset of the subgroup generated by conjugates of a?

and that group also normal subgroup

i think then it is equal to the subgroup generated by conjugates of a

and generators are commute, so subgroup is abelian

right?

[ivory ore]

well you can checl for the quotient since it's the claimed condition means it will be abelian

[chilly ocean]

at least post your solution

[ivory ore]

don't try to boss around

[chilly ocean]

you always comes in between says something and left .

[ivory ore]

i did what i did

[ivory ore]

^^^

[chilly ocean]

then dont post incomplete arguments , at least post something full or initiate something even if its wrong

so that everyone can buid it up

[ivory ore]

idc, don't try to boss around, for the 3rd time

yeah maybe mods can take care

[chilly ocean]

if i say something nicely thats how you react

[ivory ore]

mods can check deleted texts

[chilly ocean]

its fine

nothing sensitive is there

idc

[ivory ore]

calm down, take this sulliness somewhere else.

[delicate orchid]

fight fight fight fight

[elfin wraith]

I wonder why youre not a mod anymore wew

[crystal vale]

i know if G is finite then every proper subgroup contained in maximal subgroup, is it true when G is infinite group?

[novel umbra]

guys can someone help me

Prove that if G = H x K is cyclic, then H and K are cyclic.

[elfin wraith]

Nope https://en.wikipedia.org/wiki/Prüfer_group

[crystal vale]

you can see H and K as subgroups of G

[crystal vale]

thanks

[novel umbra]

I reached a point where if (a,b) is the generator of G with order nm with gcd(m.n)=1 then the order of a divides mn

but I think it's not a good idea to go there

[elfin wraith]

Do you know anything about subgroups of cyclic groups?

[thorn jay]

Literally 1984

[novel umbra]

Yes, these will be cyclic subgroups

[elfin wraith]

Use that fact

[novel umbra]

and that would be all?

[candid patrol]

H ~ H x {e_K} which is a subgroup of H x K which is cyclic

[elfin wraith]

Yeah pretty much

The other direction is the more interesting one

[novel umbra]

ohhh i see

I hadn't thought about that, to be honest.

[novel umbra]

The exercise only has one implication, I suppose, and it's because it would lack some hypothesis for it to be true, like the orders are coprime or something like that, right?

[elfin wraith]

Yeah thats the trick

C_2 x C_2 isnt cylic, so you need them to be coprime

Well that shows you need something else, not that they need to be coprime, but you get the idea lol

[thorn jay]

The world if cyclic extensions were always cyclic

[elfin wraith]

She Klein on my group till i cant cycle

[thorn jay]

🔥 🗣️

[tardy hedge]

Thats hot

Gettin heated in groups-rings-fields 🔥

[novel umbra]

🫡

[delicate orchid]

oh my science it's just like my heckin non-noetherian doodads

[chilly ocean]

Hi genius people , i have a doubt, $frac{\frac{R[X,Y]}{<X^2+Y^2-1>}}$ and $frac{R[X]}$ are isomorphic?

frac means fraction field

[delicate orchid]

I'm guessing you mean R[X, Y] on the top

[cloud walrus]

Akhi Mishra(Riemman Slayer)

[chilly ocean]

DONE

[elfin wraith]

Genuinley the only reason I know of this group

I know nothing about infinite group theory, but muh Artinian not Noetherian for modules

[crystal vale]

So this group has no maximal subgroup

[delicate orchid]

https://tenor.com/view/counter-point-soyjack-silly-goofy-book-gif-9352393372278751595

[elfin wraith]

Yeah

[chilly ocean]

i think of parametrization circle in rationa function

[crystal vale]

If H is a maximal subgroup then its conjugates are also the maximal subgroup, right?

[thorn jay]

It sends it to another maximal group, as conjugation is an automorphism

[stuck egret]

Olleh

[thorn jay]

But i dont think it would be the same

[stuck egret]

Hii

[thorn jay]

Yeah, take any nontrivial semidirect product C_p \rtimes G

[crystal vale]

So, is it true or not?

So, is it true or not?

[delicate orchid]

the property of being maximal is preserved under automorphisms, which includes conjugation. As enpeace said

if you want a proof. Assume H is maximal but gHg^-1 is contained in some K < G, then H is contained in g^-1Kg

which is a contradiction

[thorn jay]

Did you mean a maximal subgroup, or the maximal subgroup, or what

Because depending on what you meant the answer varies

[crystal vale]

Yes

[slate fulcrum]

Is H^2(G_F,F_sep^) the same as H^2_cont(G_F,F_sep^). Wiki says that Br(F) is isomorphic to H^2(G_F,F_sep^*) and I'm confused

Oh shoot wrong channel, ignore this

[worthy solar]

For Q(pi) why do the elements have this form? Is this just because it is basically Q(x) but you replace x with pi and then now you I guess have to consider inverses of these polynoimals with the variable being pi?

like i would think you have

Q(\pi) = {f(pi) | f(x) in Q[x])

as the set

$\mathbb{Q}(\pi) = {f(\pi) | f(x) \in \mathbb{Q}[x]}$

[cloud walrus]

Brandon7716

[worthy solar]

so in I guess appealing to this being a field extension then we need inverses of these f(pi) elemetns?

[rocky cloak]

Yes, in general Q(a) contains elements of the form
f(a)/g(a) with g(a) nonzero

[worthy solar]

i guess that seems reasonable

[worthy solar]

is there a clear difference in notation between say [] and ()
one problem i had was discussing

$\mathbb{Z}[\sqrt{d}]$ but i feel like it would be polynoimals of the the form $f(\sqrt{d})$ where $f(x) \in Z[x]$ but then like just enough to make it form a ring as opposed to a field

[cloud walrus]

Brandon7716

[worthy solar]

like

$Q[x]$ vs $Q(\sqrt{2})$

[cloud walrus]

Brandon7716

[rocky cloak]

In general R[a] is the smallest ring containing R and a and R(a) is the smallest field

[worthy solar]

i think you can write the latter as some polynomials but the higher order terms are already in Q (after degree 1)

[worthy solar]

okay got it

[glad osprey]

Equivalently, F[a] is the ring of polynomials in a. If a is algebraic then F[a] = F(a)

[sly crescent]

Cursed notation idea: you don’t need to specify R

So [a] is the same as Q[a] and (a) is the same as Z(a)

[white oxide]

What do they mean by a "new basis of F_q of the form..." is it just that this new basis is equal to the original basis of F_q?

[tough raven]

inb4 equivalence classes and ideals

[tough raven]

I think it means that B1 ∪ B2 ∪ B3 is also a basis of F_q, not necesarily the same as the one used to construct it.

[sly crescent]

(Assuming a is an element of a characteristic 0 ring)

[tough raven]

Actually wait it's really cursed that this isn't [a] = ℤ[a] and (a) = ℚ(a).

[sly crescent]

…I got the notation backwards

[tough raven]

DW I think this makes it more authentically cursed.

[sly crescent]

True

[chilly ocean]

Where $S =\frac{R[X,Y]}{\langle X^2+Y^2-1 \rangle}-{0}$ and $D=R[X]-{0}$.

Now in $S^{-1}{\frac{R[X,Y]}{ \langle X^2+Y^2-1 \rangle}}$ we have $x^2+y^2=1$ and $(x^2+y^2)^n=1$ so we have $\phi(x)^2+\phi(y)^2=1$ but left side is a polynomial in $x$ , either $\phi(x)=0 $ or $\phi(y)=0$ because if one of them is non zero then $1= (\phi(x)^2+\phi(y)^2)^n$ for all n, so we have $\ker(\phi)$ is non trivial . Hence no isomorphism. Please tell me if it is correct or not?

please verify anyone, i have deadline till 5:00 pm

[cloud walrus]

Akhi Mishra(Riemman Slayer)

[chilly ocean]

how to define norms in poly nomial rings

[delicate orchid]

use the class equation and do the case of 1, 2, and 3 classes separately

[chilly ocean]

having trouble in (3) case

[crystal vale]

i did this problem some days ago

[chilly ocean]

please send

[crystal vale]

what have you tried?

[chilly ocean]

1/|G|+1/m+1/n=1 this is what i got,

[crystal vale]

.

[chilly ocean]

thanks

[crystal vale]

But I would like to say don't go to the solution directly

[chilly ocean]

okay

can anyone verify my solution above

[glad osprey]

Is complex conjugation an automorphism of any normal extension of Q? It's an automorphism of C, and the restriction to an extension Q <= E has image E if the extension is normal. This implies that the Galois group of a polynomial with non-real roots has even order, right?

[rocky cloak]

It is yes, but there are normal extensions that are real. So in the case complex conjugation is just the identity.

A silly example could be just Q itself, but there are other examples

And yeah, if E is not real then G(E/Q) has even order, because it contains complex conjugation

[glad osprey]

yep, that makes sense still feels like a useful result, like you can look at x^3 - 2, notice that is irreducible and with non-real roots, then you can immediately conclude that the splitting field has degree 6

[rocky cloak]

Here you can also use the fact that the splitting field contains a square root of the discriminant. So if the discriminant doesn't have a square root in Q you'll get even degree

[glad osprey]

I haven't really learned about discriminants yet, but isn't the discriminant of x^3 - 2 equal to -108? Does that have a square root in Q(w, 2^(1/3))?

[rocky cloak]

-108 = -3 *6^2

And the third root of unity is
-1/2 ± sqrt(-3)/2

So yeah (2w+1)*6 would be a square root of -108

[glad osprey]

I see, thanks I found this too, which I think is related

[rocky cloak]

Yeah, more generally for an irreducible degree n polynomial you can think of the Galois group as a subgroup of Sn from how it permutes the roots, then it's contained in An if and only if the discriminant is a square in F.

[delicate orchid]

actually a useful proposition to remember

[rocky cloak]

For n = 3 the only options for Galois groups is A3 = Z/3 and S3

[glad osprey]

Interesting I think I see how this result is pretty powerful; for example if the discriminant is a square then the Galois group is contained in A_n, then if complex conjugation is a non-trivial automorphism then it must transpose an even number of pair of roots, so any polynomial where the discriminant is a square must have 4k non-real roots for some integer k, is that correct?

[chilly ocean]

can you share something similar for quartic

[glad osprey]

I'm not familiar with any specific result, but you can look at Chapter 4, quartic polynomials in Milne's notes: https://www.jmilne.org/math/CourseNotes/FT.pdf

[untold basalt]

why is p=1 mod 4?

[delicate orchid]

what are we proving

[untold basalt]

yeah sorry the screenshot sucks

Let $p$ be a prime and suppose $x^2 \equiv -1 \mod p$ for some integer $x$. Prove that $p \cong 1 \mod 4$.

[cloud walrus]

crazynakedgalois

[untold basalt]

btw I think I got it, hang on

[plain rock]

P has to be odd

[untold basalt]

yes sorry

I forgot

[plain rock]

Consider the the multiplicative group (Z_p)*

It has order (p-1), if any element has order 4, so 4|p-1

[untold basalt]

$x^4 \equiv 1 \mod p$, so $x$ has order $4$ and then I got it cus this means $4|p-1$

[cloud walrus]

crazynakedgalois

[untold basalt]

Yeah thanks

[barren sierra]

Gamma is a subgroup of GL(n, C). Why do we need that C is an infinite field here?

[rocky cloak]

You need the field to be infinite to identify polynomials and polynomial functions.

[barren sierra]

Ah yes

[coral spindle]

A field $\Bbbk$ is infinite iff any of the following hold:
\begin{itemize}
\item[(a)] The obvious map $\Bbbk[x] \to \Bbbk^\Bbbk$ is an injection. The ring on the right is the ring of functions.
\item[(b)] Any $\Bbbk$-vector space is not a finite union of proper subspaces.
\end{itemize}
I wonder if there is a way to show that (a) and (b) are equivalent without passing through the infinitude of the field?

[cloud walrus]

Boyt(ji=-k)e

[delicate orchid]

could be a silly trick where you embed any k-vector space into k^k via the dual space or something

although that only works for finite dim spaces

[coral spindle]

Ah well I guess I'd be happy with finite dim spaces

[rapid gyro]

Hi I try to prove Let $G$ be an infinite, indecomposable abelian group that is isomorphic to each of its nontrivial quotients. Prove that $G \cong \mathbb{Z}_{p^\infty}$ , any hint ?

[cloud walrus]

Algebraic Geometry Hater N.1mod4

[vivid kestrel]

how to prove that t^2 - 5 is irreducible over the field Q(X)

[rocky cloak]

You can just show that it doesn't have any roots

[vivid kestrel]

I thought that'd be tricky but by a theorem which name I forgot, any root would have to be in Q[X] and divide 5 which obviously isn't the case, is that correct

I feel I'm overthinking this

[south patrol]

This is a form of Gauss' lemma yeah

But since it is degree 2 it's easy just to check no roots

[vivid kestrel]

for Q[X] i see why that wouold be but not for Q(X)

[velvet hull]

oh you said Q(X) whoops

[chilly ocean]

is it true galois group x^4+ax^2+b is contained in D_4

[untold torrent]

I have a little doubt about interior point and the set of interior point where its differentiable so my doubt is that in a function there will be there can be many points which are interior of the function but it's not differentiable is this possible

D_f^°

[rocky cloak]

It's very easy actually.

You just take an element of Q(X), f(X)/g(X) and plug it into your polynomial. It gives

f(X)^2 = 5g(X)^2

Now just compare for example leading coefficients to see that there are no such elements

Your argument is also correct

[vivid kestrel]

ah thank you, that is the kind of argument i was looking for initially, much thanks

[chilly ocean]

How to show that X and 1-Y are irreducible in $\mathbb{R}[X,Y]/<X^2+Y^2-1>$, do we have to use some norm, or is it directly posiible to show this

[cloud walrus]

Curvature

[candid patrol]

Maybe I'm wrong, but you should start by showing that if G is a divisible group, then G ~ Q or G ~ Prüfer group

[distant summit]

What's a good method/intuition for these questions?

How do you know to conjugate A by B and C and not try any others?

Is it a general rule that if you have G=<a,b,c> that [a]={a,b,c}?

i.e. all the generators are in a single conjugacy class?

[thorn jay]

No, that is special to this case

[distant summit]

so how are you supposed to construct them?

bc if you go the definition and try to find M s.t. P M P^-1 = A for some P, then there's a lot of trial and error

surely there is a quicker way of finding conjugacy classes than doing it that way

[thorn jay]

a subset is a union of conjugacy classes whenever it is it closed under conjugation by generators, so you basically just start with a single element and conjugate it by your generators, and get a new set. Continue this until conjugating by the generators does not add any new elements

[coral spindle]

Generally speaking there is no quicker way.

Conjugacy classes are not easy to compute

[thorn jay]

Sadly 😔

[distant summit]

So here you would do

Conjugation of A by B: BAB=B(BC)=C where we have used B=B^-1 and AB=BC
Conjugation of A by C: CAC=C(CB)=B for similar reasons

Conjugation of A by AB: ABABA = ABABA = CAAAB = CAB = ABB = A

And similar stuff would hold for conjugation of A by AC

And so that exhausts the possible elements of [A]

And then you move onto AB

Is that all it is?

[thorn jay]

No, what I mean is that you conjugate A by B and A by C to get the set { A, B, C }. Then conjugate every element in this set by A, B and C and you'll notice you get the set { A, B, C } again, and then you're done

(at least I assume you get that set again, I haven't computed for myself)

[distant summit]

And then repeat for AB?

So [AB] consists of A AB A = BA and B AB B = BA and C AB C = CA AB = CB = BA

So [AB]=[AB, BA]=[AB, AC]

Since every element of G has now been placed in a conjugacy class, there is no more work to do

[candid patrol]

?

What is the order of A, B and C plz ?

[distant summit]

2

[thorn jay]

Yes, that's it

[distant summit]

Just checking I understand all of this...

G simple => Z(G) is trivial (either only identity commutes or everything commutes, the latter means G is abelian)

But Z(G) trivial =/=> G simple

Consider G=Z_4=<a>. This is abelian and a proper subgroup is H=<a^2>, so H is normal.

So Z(G) trivial (in this case G is abelian) does not imply G is simple, because we know G has a proper normal subgroup

Is that all correct?

[thorn jay]

There are abelian simple groups; cyclic of prime order

Your statement is correct (Z(G) trivial need not imply that G is simple), but your example does not demonstrate this fact (any abelian group has a full, hence nontrivial, center)

If you want an example of the reverse implication not holding, look at symmetric groups

Or do you mean with trivial "either the full group or trivial proper subgroup"

In which case this would be correct

[neat zodiac]

is this a mistake?

[distant summit]

Elaborate...

How is H=<a^2> as a subgroup of Z_4 not a counter example?

Also I'm using trivial to mean {e} or G, in the same way every group has two trivial subgroups, {e} and itself

[distant summit]

Yes

[thorn jay]

.

[thorn jay]

Yes, lol

[glad osprey]

Yeah, ab = bc doesn't imply b = c. Weird typo, I'm assuming they meant ab = ac implies b = c (ie. left cancellation)

[brave remnant]

so is {0} a field?

i conjecture it is

nvm figured it out

[rocky cloak]

Typically the field axioms include 0 and 1 being distinct. So no

[brave remnant]

i have another question

Q(sqrt2) injects into R as a field embedding

if Q to R is a field embedding then Q(sqrt2) to R is an extension

[rocky cloak]

Sure

[brave remnant]

if we consider the extension of (Q to R) to Q(i) to R there is no such extension

[rocky cloak]

That's right

[brave remnant]

then is there a theorem that states if Q to R then Q(x) is able to be extended when x is in R?

because i noticed sqrt2 in R but i not in R

[rocky cloak]

I mean, if x is in R, then Q(x) is also in R yes

I guess maybe you're asking about the converse?

[brave remnant]

so from F to E if we have an extension F(x) to E this extn only exists when E/F(x)?

[rocky cloak]

So you can sometimes embed Q(x) into R without x being in R

For example if you pick x to be a complex cube root of 2, then mapping x to the real cube root of 2 gives an embedding of Q(x) into R

[brave remnant]

i thought there was only one extension of Q to R to Q(cube root2) to R

[thorn jay]

X^3 - 2 can have 3 nonequivalent roots as it is of order 3

So there are 3 field extensions over Q, all adjoining a different root of X^3-2, which give 3 different embeddings of Q[x]/(x^3-2) = Q(cuberoot{2}) into C

[brave remnant]

yes

but thats to C

i want to R

[thorn jay]

Oh yeah, sorry, you take the canonical isomorphism to Q(cuberoot 2) composed with the inclusion Q(cuberoot 2) -> R

[brave remnant]

isomorphism from Q[x]/(x^3-2) to Q(cube root 2) then take Q(cube root 2) and include it into itself and send to R?

not sure

[thorn jay]

Ye

[brave remnant]

oh sorry you meant inclusion of Q(cuberoot2) into R

[thorn jay]

Yeah

[brave remnant]

then there is only one of those

[thorn jay]

No

I meant from Q(alpha) -> Q[x] / (x^3 - 2) -> Q(cube root 2) -> R

I should've specified better

[brave remnant]

okay so there are 3 alphas

[thorn jay]

Yeah

[brave remnant]

but each of those 3 alphas there is a unique extension of Q to R

[thorn jay]

But, as jagr said, only one of those is in R

[brave remnant]

this though?

doesnt that mean all 3 roots of x^3-2 we have an extension of Q(alpha) to R from Q to R?

or only does this work for the case when we take alpha to be the real root

[thorn jay]

No, extensions are inclusion mappings, not embeddings

[brave remnant]

no i meant extensions of a field embedding

is that what u are saying?

arent extensions of field embeddings also embeddings

[thorn jay]

What's an extension of an embedding?

[brave remnant]

a field embedding that is equal to a smaller embedding when domain restricted

[thorn jay]

Ah, right yes you're right, this gives 3 different extensions of the inclusion map into R, although the images are the same

[brave remnant]

so there is an extension of Q to R to Q(zeta_3 times cube root 2) to R

?

[thorn jay]

yes

[brave remnant]

where would you send zeta_3 times cube root 2

has to hit something in R

[rocky cloak]

You would send it to the real cube root of 2

[brave remnant]

i see

[vast tangle]

is gallian wrong here? the exercise was to show that if there is a surjective homomorphism from a finite group G to Z/10Z then G has normal subgroups with indexes 2 and 5

we previously showed that for a homomorphism phi from a finite group G to barG with H subgroup of barG that

shouldnt we have that $|\phi^{-1} ( \langle 2 \rangle ) | = 5|K|$ ?

[cloud walrus]

josemom2

[thorn jay]

Yeah, they made a typo

[vast tangle]

phew i thought i was going crazy

[thorn jay]

No worries

[rocky cloak]

I feel like this is kind of a convoluted proof anyway, and uses that G is finite which isn't necessary.

Like, just compose the map G -> Z/10 with the map Z/10 -> Z/5, that means you have a normal subgroup of index 5.

[vast tangle]

oh i am not aware of the fact that gives you this

what is it?

[rocky cloak]

The definition of index...

[vocal pebble]

first isomorphism theorem if you want

composing gives you a surjection G -> Z/5Z so its kernel has index 5

[vast tangle]

ah ok

[vocal pebble]

all of these proofs are almost the same, one could cite the correspondence theorem too

[vague canopy]

how can i show, if char(K) is not 2, then the Quaternion Ring is simple?

[tribal moss]

Quaternions over an arbitrary ring?

[distant summit]

Is it normal to use the notation $J=HG=G \ltimes H$?

Why does $H$ go on the left in the former, but on the right in the latter?

[cloud walrus]

Douglas

[vocal pebble]

HG =GH if H is normal

[distant summit]

which it is

[vocal pebble]

so you can say that J = GH and H is going on the right in both

[cloud walrus]

Asteroid

[rapid junco]

is it known wether or not O(n) is f.g.?

[vocal pebble]

O_2 is not finitely generated since it contains rotations by all angles

i think you can prove that for general n, O(n) is uncountable and so cannot be f.g

[rapid junco]

i see

okay yeah it contains a copy of R if you are smart

[vocal pebble]

it might be tempting to say that O_2 can be viewed as a subgroup of O_n and so O_n cannot be f.g. but it does not need to imply O_n is not f.g. but since O_2 is uncountable you are done anyways

[rapid junco]

thx

[rocky cloak]

I don't know if it's necessarily the best way, but the way I would attack it:

If you can show that the Jacobson radical is trivial, it must be semisimple.

Since the center is K the only possibilities for a 4d algebra are a division algebra or the 2x2 matrix algebra.

So the only tricky part is showing zero radical.

If you're working in the (a, b) Quaternion algebra and
x + iy + jz + kw
is an element, then multiplying by its conjugate gives
x^2 - ay^2 - bz^2 + abw^2.
If this is nonzero, then the element is a unit.

If it is zero, then we just want to show it's still not in the Jacobson radical.

Multiplying by i, j or k we may assume x is nonzero and dividing by it we may assume x=1.

Then (1 + iy + jz + kw)/2 is idempotent, so cannot be in the Jacobson radical.

[brave remnant]

can somebody help me with algebraic closure

$\overline{F}$

[cloud walrus]

redoftwored

[brave remnant]

suppose $F$ is a field then consider $\alpha$ transcendental over F then is $\overline{\overline{F}(\alpha)}=\overline{F(\alpha)}$?

[cloud walrus]

redoftwored

[brave remnant]

is this true in general? what if we let F'=$\overline{\overline{F}(\alpha)}$ and we have a nested thing

[cloud walrus]

redoftwored

[cloud walrus]

longboard kayak

[ivory ore]

sanity check

[cloud walrus]

longboard kayak

[ivory ore]

sounds?

[cloud walrus]

longboard kayak

[ivory ore]

a good way to generalise the non-transitivity of normal subgroups

but where is this classification of super-solvability is actually helpful?

[ivory ore]

ok, by the next exercise finite p-groups are supersolvable(quotient by the cyclic group = <"order p element" \in centre> and induct the process)
but can we get other examples?

[distant summit]

Can phi_g be as general as any automorphism of H, or does it have to be an inner automorphism (i.e. conjugation)?

Also how is Aut(H) actually being defined here? Because earlier we assume conjugation is of elements of G by elements of G

But here we're saying that conjugation of h' in H by g in G is an automorphism

And in general G and H are disjoint except for the identity

[vocal pebble]

It seems like this definition doesnt mention the fact that the map g \mapsto \phi_g must be a homomorphism. So no, phi_g cant just be any automorphism

[distant summit]

And the consequence of specifying that would be?

That it must be inner?

[distant summit]

This is thinking about it incorrectly I think

The reason is because G is a subgroup of J=HG=G \ltimes H

And since H is a normal subgroup it is invariant under conjugation by any element of J, in particular elements of G

[vocal pebble]

Well you need that map to be a homomorphism for the multiplication to define a group operation (see group axioms, associativity )

[delicate orchid]

So the one case I can think of off of the top of my head is that solvable groups are not necessarily monomial (all irreducible representations are induced from linear reps of subgroups), but super-solvable groups are.

[vocal pebble]

Usually (and in this case) Aut(H) is the group of isomorphisms from H to itself(this includes the inner automorphisms Inn(H))

[delicate orchid]

They’re using the same notation for different things. Phi_g is not (a priori) conjugation by g

They should really write that phi is a map from G to Aut(H) and then phi_g(h) is just phi(g)(h) if they want to use the given notation for conjugation

[ivory ore]

hmm, so is this somehow connected to the group acting on the minimal normal subgroups and then if it has got a vector space structure or not!

thinking this of in terms of equations in a bit out of my scope for me atm, tho i think it will not be hard

[solemn garden]

How do you show that, for all positive integers n except n = 4, (x-1) ... (x-n)+1 is irreducible in Z[x]? I know that if it's -1, then you can use the signs of the leading coefficients to show that it can't be the negative of a square, but here you can't do that with the constant terms, since it's an open problem

[chilly ocean]

Every proper subgroup of invertible upper triangular matrices of order n , is of finite index?

[echo vine]

Why are isometries in 2D classified into Reflections, Translations, rotations and glide reflections? Its just a composition of a translation and a reflection, why does it get its own category?

Like why isn't a glide rotation included for example? Or a rotation reflection

[vocal pebble]

this fails for the subgroup of upper triangular matrices of order 2 generated by $\begin{pmatrix}2 & 0\ 0 & 1\end{pmatrix}$, for example, since it is countable

[cloud walrus]

Asteroid

[chilly ocean]

is the index of this subgroup not finite

how to show that

[vocal pebble]

this is real upper triangular matrices right?

[chilly ocean]

yes

[vocal pebble]

so the parent group is uncountably infinte

[chilly ocean]

fine

[vocal pebble]

while the subgroup generated by that matrix is countably infinite

[delicate orchid]

the humble trivial subgroup:

[vocal pebble]

i should also point out that the trivial subgroup is another counterexample

sniped

[chilly ocean]

forget to mention non trivial

[vocal pebble]

anyways, .

[delicate orchid]

I knew you meant non-trivial. I just wanted to be annoying

[chilly ocean]

thanks

[vocal pebble]

because glide reflections dont fall into any of the aforementioned categories. a glide rotation is just a rotation about another point, and a rotation reflection is another reflection about a different line

[chilly ocean]

what if the case is complex

[rocky cloak]

A glide rotation is just a rotation, and a rotation reflection is just a glide reflection (or possibly just a reflection)

[vocal pebble]

the same subgroup is still a counter

[rocky cloak]

So to answer the question, they are classified as such because every isometry falls into one of those 4 categories.

[ivory ore]

(4.36) Prove that there is no nonabelian simple group of order less than 60.

without solvability and burnside

[ivory ore]

to prove that i have to workout syllow for every composition right???

single prime powers are easy to deal

there are lots of p^a q^b, but would be systematic

and same for pqr

[delicate orchid]

I’d rather not

[ivory ore]

rotman uses that to then state

umm, algebra is tautology

[delicate orchid]

Well you can still just show that all groups smaller than 60 are solvable

Directly,

Although I only recommend that for orders where you can’t use sylow

[ivory ore]

what would i be using ?

[delicate orchid]

Find an appropriate chain of subnormal groups?

[ivory ore]

no idea what that is but thanks for the input, would try to figure out

[rocky cloak]

I mean if you cover p^a q^b and pqr, then 60 is the smallest number not on that form

[ivory ore]

yeah

30 and 42 would be other of those pqr but easy to find out a normal subgroup

so for the other composite

just look at the big prime and a bit of calculation

no one but algebraists: this group is not complicated enough to be simple

also in the abelians it has to be the cyclic below 60

for simplicity

[rocky cloak]

All abelian simple groups are cyclic yeah

They're just Cp

[ivory ore]

also, out of the context, there are some cool composite order of pq for which there exist only a single group Cpq

15,35 and so on and so forth right?

[rocky cloak]

Yeah, should be iff p is not 1 modulo q (p>q)

[ivory ore]

yeah, alr

[distant sleet]

There are results that tell you if you have a group, preferably of infinite cardinal, then it can be decomposed into a direct sum or smth else?

[vocal pebble]

There is the fundamental theorem for finitely generated abelian groups, invariant factor theorem(generalisation of aforementioned)

Surely there are more general results

[distant sleet]

I will check about it. We are just working with finite case but we will consider compact then locally compact cases. So just wondering how to deal with non finite case

Thanks

[vocal pebble]

Im not sure what you are referring to, but i wouldnt suspect that such a general question can get you very helpful answers

[rocky cloak]

If the group is Noetherian (or artinian), then it is a finite direct sum of indecomposable groups.

But for example Z^N (the infinite direct product of Z with itself) cannot be written as a direct sum of indecomposables. It can be written as a direct product though (obviously)

[distant sleet]

Working with unitary finite reflection group. Then we will remove finiteness and add compactness. Its involving group invariance with complex analysis

[distant sleet]

Interesting, at some point we thought Noetherian stuff could appear

[chilly ocean]

: There does not exist a proper subgroup H of Invertible upper tringular matrix such that $G=\bigcup_{g\in G}g^{-1}Hg$

[cloud walrus]

Akhi Mishra(Riemman Slayer)

[late matrix]

is this a question?

this is true for any group G and subgroup H

[chilly ocean]

yes

[chilly ocean]

proper subgroup

[late matrix]

actually no isnt it just for finite G

yeah

[rocky cloak]

Think you need H to have finite index for it to be true in general

[chilly ocean]

thats why i asked that does thre exist a proper subgroup of non finite index of G

[rocky cloak]

There are lots of those

[late matrix]

ah

[chilly ocean]

i think to solve this we need some linear algebra and jordan canonical form

[rocky cloak]

That would make sense

[chilly ocean]

seeing this as C[X] module?

[coral spindle]

Might be helpful to observe that the group U of upper unitriangular matrices is nilpotent. Pretty sure you can argue then that any counterexample group H must contain U

[chilly ocean]

so we should proove any non trivial subgroup contains U

[coral spindle]

Any nontrivial subgroup whose conjugates cover B, yes

That's my bet

I haven't checked it

[solemn garden]

How do you show that, for all positive integers n except n = 4, (x-1) ... (x-n)+1 is irreducible in Z[x]? I know that if it's -1, then you can use the signs of the leading coefficients to show that it can't be the negative of a square, but here you can't do that with the constant terms, since it's an open problem to identify all n! + 1 which are also squares

I know that if the polynomial has odd degree or if n is not a multiple of 4 (so its triangular number is odd) then it can’t be a square, but this covers only 1/4 of all cases

[chilly ocean]

its a square of some polynomial

its remaining to show that coefficients are in Z for n=4

[solemn garden]

I have shown that it is reducible when n = 4, just not sure how to show that it is irreducible in all other cases

[chilly ocean]

this question has been discussed here

[solemn garden]

And I’m stuck on showing that, when n ≠ 4, this polynomial is never a square

[solemn garden]

Oh, I couldn’t find it when I used the search function, could you link to it?

[rocky cloak]

The answer exists here at least
https://math.stackexchange.com/a/1691272/306319

[chilly ocean]

its walways square though but not in Z

[solemn garden]

this answers the question for (x-1) ... (x-n)-1 but not (x-1) ... (x-n)+1

[chilly ocean]

go little up from here

from here at least

[rocky cloak]

I see, so you're only missing showing that it's not the square of an irreducible

[solemn garden]

Yep

[chilly ocean]

https://math.stackexchange.com/questions/503765/prove-that-the-polynomial-x-1x-2-cdotsx-n-1-n-ge1-n-ne4-is-i?rq=1

this answers it

[solemn garden]

Thanks!

why does a - b divide f(a) - f(b) for integer polynomials?

[chilly ocean]

This

[solemn garden]

oh, I see

Thanks!

[cloud walrus]

longboard kayak

longboard kayak

[ivory ore]

Gn-1 is abelian which is easy to see

do i have to now show that it is indeed normal in G?

[ivory ore]

or there is some other way to arrive at this

oh well those are higher commutator groups and indeed characteristic

[vague canopy]

thank you!

[coral spindle]

Yeah this is how I would show that fact