groups-rings-fields

[chilly ocean]

As in - not in context of homology/chain complexes

Where groups are commutative

[tall jay]

Why is x^2 = -1 in Z_3[x]

[north sand]

it isn't tho

[tall jay]

Oh wait, nvm I read it wrong

How are we getting these elements for R/I?

[chilly ocean]

Because of division of polynomials

And uniqueness of it

[patent ocean]

?

[chilly ocean]

So the elements are just polynomials of degree < 2

w(x) = q(x)(x^2+1) + ax+b where q(x), a, b are uniquely determined by w(x)

[tall jay]

ahh

thank you

[chilly ocean]

Np

[patent ocean]

I want to find the smallest n such that a subgroup of Sn is isomorphic to some group and to find that subgroup. I'm trying to find this for C2XC2XC2. I know all the elements are of order 2 except the identity and that the order of the group is 8. How do I then find this subgroup and the smallest n?

Sorry if that was very unclear

[deep sky]

Could try thinking about it in terms of permutations + permutation matrices and how info is stored using entries in them?

[south patrol]

Ye

[south patrol]

So one wy to think about this is to consider how elements order 2 interact in Sn

[tough raven]

A map C_2×C_2×C_2 to a group G is the same as 3 commuting elements of order dividing 2 in G. This map is injective iff those 3 elements have order 2 and are pairwise distinct.
So you need to find the smallest n such that S_n has 3 distinct elements of order 2 which commute with each other.

[south patrol]

beat me to it aha

[tough raven]

Yes, this

[patent ocean]

how do you know it is injective iff the 3 elements have order 2 and pairwise distinct

I think I found the answer but it was more of a guess :/

[south patrol]

What did you get?

I mean I found it easy to guess but hard to prove you couldn't use lower n aha

[patent ocean]

So far I have {e,(12),(34),(56),(12)(34),(34)(56),(12)(56),?} I added a ? because i think i'm missing an element

since there should be 8

[south patrol]

Yeah just write that as <(12),(34),(56)>

you have three copies of C2 in S6 given by <(12)>,<(34)>,<(56)> and everything commutes nicely

You're missing (12)(34)(56)

[patent ocean]

oh right i forgot about triple tranpositions

thanks

[south patrol]

np

[patent ocean]

is this the reasoning why this is the lowest n?

[south patrol]

Well you can sort of adapt this reasoning to show it doesn't embed in S5

If you pick any three elements of order 2 in S5, they won't behave as you wish, essentially

Iirc I did this by an over the top method though lol

[patent ocean]

Damn now i'm confused

why do order 2 elements in S5 behave differently

oh nvm

not an even number

so can't get 3 disjoint transpositions

Is there anyway that incorporates the orbit-stabaliser theorem?

[tough raven]

In this case, you can write out the morphism on every element of C_2^3 and check this is the condition for nothing to be in the kernel (i.e. go to 1).

The idea is that C_2×C_2×C_2 has group presentation
‹a, b, c | a^2 = b^2 = c^2 = 1, ab=ba, ac=ca, bc=cb›
so morphisms are what I described them as.

[tough raven]

A map would be injective iff the kernel is trivial … and I can't think of any better way than this to show that is equivalent to a,b,c being distinct

Oh also abc shouldn't be 1

Otherwise you have a map to C_2×C_2 by taking the three non-trivial elements which are of order 2 and commute, but isn't injective because their product is 1

[south patrol]

You can consider by cases what happens when you have three elements order two

for example, if you're given three transpositions, then they must 'overlap' so you have (wlog) (12) and (23) and hence an element order 3 by taking their product

so no subgroup of S5 generated by three transpositions can be isomorphic to C2 x C2 x C2

[south patrol]

There is yeah

Let me think how I should phrase it.

[south patrol]

But yes so the slightly ott way which is that if G is a finite group and p prime and p^k the largest power of p dividing the order of G, then all subgroups cardinality p^k are conjugate (and hence isomorphic)

this is the second Sylow theorem

The proof is not very long but obviously nonexaminable lol

Anyway you can then say that oh, well S5 has a subgroup isomorphic to D8

and hence it can't have a subgroup isomorphic to C2 x C2 x C2 from the theorem I just quoted

[chilly ocean]

oh

[chilly ocean]

What does this mean more precisely. Are we postulating that there is a map from H to G such that the diagram commutes?

And that images of both injections have {e} as intersection

[hot lake]

yeah if there is a map from H to G such that the right composition is the identity of H

that both image intersect in {e} is a consequence of exactness

I think

[chilly ocean]

Yeah I think so too

Thanks!

[dull root]

If I have the polynomial $x^{({5^3} \cdot 3)} + x^{({5^3} \cdot 2)} + 2$ in the finite field $F_5$, is there a good way to see how to factor it in $F_5$? The $x^{5^3}$ looks suspicious and I know that $F_{5^3}$ is the splitting field of the polynomial $x^{5^3} - x$ in $F_5[x]$, but I don't know how to use that fact here

[cloud walrus]

MasakaBakana

[dull root]

I am hoping that this relates the the much easier polynomal to deal wtih $x^3 + x^2 + 2$ which is easy to see has no roots in F5 thus irreducible...

[cloud walrus]

MasakaBakana

[delicate orchid]

looks like a fermat's little theorem meme to me

x^p = x mod p

[dull root]

wait so that polynomail is just actually x^3 + x^2 + 2 right?

with fermat thm

[delicate orchid]

I'm not sure, due to the x^(5^3)
if it was (x^5)^3 then this is just x^3 but it's not

meaning I actually have to think

[dull root]

It was x^375 + x^250 + 2

so i mistyped

[delicate orchid]

ok so we can rewrite that as (x^5)^75+(x^5)^50+2 = x^75+x^50+2 = (x^5)^15+(x^5)^10+2 = x^15+x^10+2 = x^3+x^2+2
so yeah

sorry I had to do it in such a long winded way lol

[dull root]

Then, it is easy to ssee x^3 + x^2 + 2 is irreducible in F_5 since we can just plug in 0,1,2,3,4 and see it never equals 0, so the orginal polynomial is also irreducible

There was also an earlier question I had for finite fields, that I don't quite see: So If I am in the field F_p where p is prime, I see that F_p(\sqrt 2) = F_p^2 up to isomorphism, but how do I see that F_p(\sqrt 2, \sqrt 3) = F_p^2 too?

namely if I take any quadratic extension of F_p, that is F_p^2 up to isom

[delicate orchid]

I don't deal with field extensions sorry I'm awful at them

[hot lake]

that's not how any of it work

[dull root]

Wait why not? x^3 + x^2 + 2 is certainly irreducible right? since if it factors, it must have a root factor

[hot lake]

so you found that y^3 + y² + 2 was irreducible where y = x^5^3 ?

so Fp(x) must contain Fp(y) = F(p^3), and then x is a p^3-th root of some element of F(p^3)

and that is in F(p^3)

cuz x -> x^p is a bijection

so Fp(x) = Fp(y)

and so your degree 600 polynomial wasn't irreducible

oh wait

it's just (x³ + x²+2)^125

[dull root]

So I understand that since I work in mod 5, (x+y+z)^5 = x^5 + y^5 + z^5 right

[hot lake]

yeah the binomial coeffs are all 0

(a+b)^5 = a^5 + b^5

[dull root]

is that also true for (x+y+z)^5k that I can raise to powers termwise?

[hot lake]

just do that 3 times

[dull root]

o nvm, we needed that it was 5^3

so the key here really was that 2^5 = 2

[hot lake]

x^5 = x is only true for elements of F5

[dull root]

yes because it is cyclic of order 5

yea

[hot lake]

uh

[dull root]

so any x in F_5 we have x^5 = x

[hot lake]

who is cyclic

[dull root]

F_5 I thought of it was Mod 5

Z/5Z

[hot lake]

as an abelian group for + then ?

[dull root]

yea you're right, I am messing up between + group and multiplicative group. I need U_5 as a multiplicative group, so all elements have order 4. so x^5 = id x = x

[hot lake]

yeah

x^5 - x = x (x^4 - 1), 0 is a root of x and the rest of F5 is a root of x^4-1

[dull root]

Zef, do you know why F_p(\sqrt 2, \sqrt 3) = F_p^2? I am stuck on that for a long time. I def see that F_p(\sqrt 2) = F_p^2 and F_p(\sqrt 3 ) = F_p^2 but I don't see why adjoining both doesnt change it

[hot lake]

well if you say sqrt2 and sqrt3 are both in Fp²

[dull root]

where the equality is since there is only quadratic extension of F_p up to iso

[hot lake]

surely Fp(sqrt2, sqrt3) <= Fp² ?

so you just need to be sure that Fp(sqrt2,sqrt3) is not Fp ?

[dull root]

oh yea.... what am I thinking.. don't I have that immedaitely from what I wrote

So everything works in a more generally setting right? So if I am in F_p where p is prime. If I take an irreducible deg n poly, say g, over F_p, then F_p^n is the simple extension of F_p by a root of g

[hot lake]

yes

[dull root]

More, there is only one such F_p^n up to isomorphism right?

[hot lake]

yeah that's the surprising thing

every irreducible degree n polynomial generates isomorphic finite fields

[dull root]

Then if I were to take any deg n poly, say h \neq g$, then all the roots of h must also be contained in this F_p^n?

[hot lake]

it must be irreducible

[dull root]

yea, i meant irreducible too

That would mean that F_p^n completely splits all irreducible deg n polynomials over F_p?

[hot lake]

yeah

in fact

x^(p^n) - x

is the product of all those polynomials, plus the ones from the subfields

and every element of F(p^n) is a root of it

so it's the polynomial whose roots are elements of F(p^n)

[dull root]

yes, i start to remember that now

[hot lake]

so you can use that to count the number of irreducible polynomials of degree n if you feel inclined

[dull root]

btw, so far when I think of characteristic p fields, I have only dealt with finite fields. Are there fields of characteristic p which are infinte?

[hot lake]

yeah there are

Fp(x)

[dull root]

What do you mean by that? All polynomials in F_p[x]?

[hot lake]

no I mean the rational fractions with coefficients in Fp

[dull root]

hmm I have never seen that before, can you give me an example of a field element?

[hot lake]

quotients of polynomials

[dull root]

so lets say we are in F_5

[hot lake]

x^2 / (x^3+2)

[dull root]

oh i see what you mean then

over a field F of char p, we say an irreducible polynomial is separable if it splits into DISTINCT roots in SOME field ext of F right?

[tribal moss]

Yes.

[chilly ocean]

A nice problem: let Z(G) denote the center of G. Prove that if G/Z(G) is cyclic then G must be abelian.

[dull root]

Do we also have the following criterion to check that an irreducible polynomial, say f, IS seperable if f and its formal derivative f' in F[x] have gcd 1? F can be any field in general

[hot lake]

I think so yeah

[tribal moss]

Clearly if there's a double root, then it's also a root of the formal derivative.

[dull root]

and this poly needs to be irreducible right? we cant just check any poly and its formal derivative

[tribal moss]

On the other hand if there's a common factor of F and F', then they must have a common root in an extension field where F splits. This can't happen if the common root is only single in F.

[dull root]

I have another question. I see that in mod p, we have n^p = n... but is this true in any char p field that n^p = n?

[chilly radish]

If n is in F_p, then yes. If it's an element of an extension that is not in F_p then no, it cannot be fixed by frobenius

[dull root]

the forbenius hom sends x -> x^p. Then if I have a field K with prime field F_p, the forbenius hom fixing an element x of K iff x \in F_p right?

[chilly radish]

Yes

This is as we know that x^p=x for x \in F_p, but x^p-x can have at most p roots

Therefore these are the only elements fixed by it

[dull root]

If I have a char p field K, then any finite field ext of K also has char p right?

[lethal dune]

does 1_K also belong to the extended field?

[dull root]

yea it does, so it perserves characteristic

[pastel cliff]

what does the "formal" part of the ring of formal power series mean

referring to Z[[x]]

[chilly ocean]

you don't care about convergence

[dull root]

So I have the following situation: I know that if I have any quartic in Q[x] where the Galois group of the splitting field is D_4, then this quartic MUST be irreducible.

What I want now to show is that for such a irreducible quartic f with galois group D_4, the cubic resolvant of f is reducible. But I don't see how to use the Galois group to show this...

[hot lake]

if G acts on the root as D4, and if you know how the roots of the cubic resolvant are formed from the roots then you should know how G acts on them

[pastel cliff]

why do some sources require ring homomorphisms to preserve mult. id and others dont

[dull root]

So if I let $a_1,a_2,a_3,a_4$ be the roots of my quartic, then the cubic resolvant has 3 roots namely,

(a_1 + a_2)(a_3 + a_4),

(a_1 + a_3)(a_2 + a_4),

(a_1 + a_4)(a_2 + a_3)

if I am not wrong

Moreover, I know D_4 has an element of order 2, so wlog we have an automorph that goes a_1 \to a_2 -> a_1. This forces this order 2 automorph to also map a_3 -> a_4 -> a_3.

so to show our resolvant is reducible, we want to show it has a root in Q. Then it sufficeis to find an automorph that fixes one of these roots listed above, but this order 2 automorph certainly fixes one of those 3 roots listed above.

[dull root]

But doesn't this arguemnt show that all 3 roots are fixed, so this cubic only has rational roots/

[hot lake]

so it's the opposite of irreducible ?

I don't get how you decided that al of those roots were fixed by G

[dull root]

wait, nvm its just 1 root is guratneed fixed. This order 2 automorph must fix 1 of those roots. So the only thing we actually used was that there exists an order 2 eleemnt in D_4. Then if our Galois group was instead something like C_8 the cyclic group, the cubic resolvant would not be reducible right, since no automorphs have order 2

[hot lake]

... C8 isn't a subgroup of S4

and you have to look at all the elements of G, not just 1

[dull root]

yea mb. but I don't see why we needed to look at all the elements of G?

[hot lake]

because something is rational <=> it is fixed by ALL the elements of G

so if you want to conclude that a root of the resolvant is rational you need to show that all the elements of G fix it

[dull root]

I see.. but then there is an huge issue. In D_4 there is also an order 4 element... my original thought was that any order 2 eleemnt of D_4 fixes one of those three roots (not necessarily the same root), but i don't see that this order 4 eleemnt is doing.

[hot lake]

uh

you know what D4 is ?

as in you can write down some of its elements as permutations, right ?

then you can apply those to the roots of the cubic

btw it's irreducible <=> there is only one orbit under the action of G

which is what you want to prove

[marsh goblet]

I can only assume it’s because some sources don’t actually define rings with multiplicative identities
And they’d call a ring with M.I. A unital ring, or something like that

So if that’s probably why some sources wouldn’t mention that, because with their definition of a ring, there may not be a multiplicative identity to preserve

[dull root]

I see I think I understand now. Let me try to resummarize;

WLOG I will use 1,2,3,4 to label my 4 roots of this irreducible quartic. the three roots of the cubic resolvant is

1.(1 + 2)(3 + 4)
2.(1 + 3)(2 + 4)
3. (1 + 4)(2 + 3)

I listed all 8 elements of D_4 in cycle noatiton. There is no such perumation in D_4 that fixes 1 and 4, while switching 2 and 3, so we can not get via a group auto from root (1.) to root (2.) so the cubic resolvant is reducible, since the galois group is not acting transitvely on the 3 roots.

[hot lake]

there are other permutations than (23) that turn root 1 into root 2 ; but yeah it does look reducible

[dull root]

What other does that? We need to swap 1 and 3 don't we?

[hot lake]

they are like, 8 of them

24/3 = 8

but none of them would be in D4

1->2->4->3->1 for example turns root 1 into root 2

[dull root]

How did you see that none of them would be in D_4? I only see that the the one swapping 2,3 or swapping 1,4 not in D_4, but even when I listed all 8 permutations in D_4, it seems like a pain to check them 1 by 1

[hot lake]

because D4 changes "adjacent roots" into "adjacent roots"

if you see it as the usual group of isometries of a square

the pair of diagonals is invariant

because no other pairs of vertices have distance sqrt2

and so that pair of diagonals (on which G acts the same way as on (1+3)(2+4) in the usual numbering of vertices) is fixed by G

(we're talking about unordered pairs here)

and so the cubic resolvant should have exactly 1 root over Q and 1 degree 2 irreducible over Q factor

[dull root]

so 1,3 and 2,4 are the two diagonals of the square. an isometry must fix this distance, so if 1 moves then the position of 3 is already determined (to be on the diagonal of 1) similarly for 2,4. (1+3) is always fixed?

[hot lake]

no, 1+3 can be 1+3 or 2+4

[dull root]

ah yes, but there then 2+4 must be 1+3

since there are only 2 diagonals

[hot lake]

yeah

[dull root]

yea, i like that way of seeing it. It is clear geoemtrically

I also have this problem that seems to be ridiculously awful if we try to brute force:

1. Let α be a root of x^2 + ax + b
2. Let β a root of x^3 + px + q.

We want to find a polynomial with coefficients in Q(a, b, p, q) having α + β as a root.

My idea:

1.Let α1, α2 be the two roots of the quadratic, β1, β2, β3 be the three roots of the cubic. WLOG let α = α1, β = β1.

2.It is clear then the the following deg 6 polynomial (x- (α1 + β1))(x- (α1 + β2))(x- (α1 + β3))(x- (α2 + β1))(x- (α2 + β2))(x- (α2 + β3)) has a root at α + β since we basically took all the possible combinations of roots. We have to show the coefficients are where we want them to be though.

3. We can see that

1.β1 + β2 + β3 = 0.
2. β1 β2 + β1 β3 + β2 β3 = -p
3. β1 β2 β3 = -q.
4. α1+α2 = - a
5. α1α2 = b.

It is possible to brute force everything by expanding all six terms in the deg 6 polynomial we have above and maybe reducing with the 5 equations listed, but there is no way this is a good way to go about it. For example, even the constant term in the deg 6 polynomial is horrendous. There must be a better way to look at this problem...

[hot lake]

yeah it's not a nice thing to do by hand

[dull root]

I am hoping that alpha is in Q(a,b) but looking at the quadratic forumula, that depends on if \sqrt a^2 - 4b \in Q(a,b)

So this problem seems to have some tricks with the discriminant hidden...but idk. Morever, we have a depressed cubic and I know what such a discriminant is

[pastel cliff]

why is it interesting to consider when ideals are equal

[delicate orchid]

what

[pastel cliff]

is it interesting

in some lecture or another my professor said "it is interesting to consider when ideals are equal" gave a fact about (a) = (b) iff a=ub for some unit u and that was it

[delicate orchid]

oh I see

[pastel cliff]

t-minus 10 hours until my final

[delicate orchid]

t-minus 10 hours until I rbbrbrbbr

[pastel cliff]

rbrbbrbrbrbrbbrbrbrbb

[delicate orchid]

exactrly

anyway uh yeah something like this
(a,b) = (gcd(a, b)) yeah?

cool

[pastel cliff]

oh

i guess i can trust you on that

[delicate orchid]

blunder

[pastel cliff]

wew lads please

im fighting for dear life rn

[delicate orchid]

well what I said holds for Z so I'm gonna say it holds for all bezout domains
source: I made it up

[south patrol]

o ye this is kinda interesting

We had an exercise to find f,g in C(R) such that (f) = (g) but there's no unit u with f = ug

[dull root]

The n-th cyclotomic polynomial phi_n has integer coefficients and its root factors are the primitive n-th roots of unity.

Since phi_n has integer coefficients, it makes sense to consider the mod p reduction of phi_n for any n, so let that reduction be denoted psi_n.

Then if I am working in the finite field F_p, how do I show that the splitting field of psi_(p^n - 1) over F_p is actually F_p^n.

To see how many roots Psi_(p^n - 1) has, we have to somehow count the number of coprime numbers to p^n - 1 mod p right?

This p^n - 1 seems to make it harder at first. If it was just p^n, then the number of numbers coprime is just p^n - p^(n-1)

but with the -1, p^n-1 might not even be prime power

[pastel cliff]

dumb question but in a ring R with subring S, a + S is as much a coset of S as aS right

[next obsidian]

aS totally is not a coset

At least not in a way that you’d want

[pastel cliff]

im thinking about first iso in rings

im looking back over the definition and i was just wondering why, for an ideal I we dont consider aIbI = abI

but im now realizing that the multiplicative part of a ring is obviously not a group so this would break

[lethal dune]

you need ideals for that, not a subring

[pastel cliff]

what i mean is like

when we went through the creation of a quotient ring in class, we defined the multiplicative part as (a+I)(b+I) = ab+I

the ring axioms hold for that definition of course but why dont we have to verify it for (aI)(bI) =abI

i could be missing something small/overthinking here

[lethal dune]

aI = I

the definition of ideal takes care of it

[pastel cliff]

i forgor 💀

[lethal dune]

when we quotient, the ideal we are quotienting with acts as the zero of the quotient ring

so you can read aI as a times zero which is zero so aI=I

what you are saying is, "do don't we have to verify zero times zero is zero"

[pastel cliff]

[grand sigil]

To check a given family of modules is a chain complex given a differential d do I simply check that d^2=0?

Do I need to check more?

[next obsidian]

No, this is the definition (or easily seen to be equivalent to)

[lilac trench]

I dont see the p(x) = p_1(x^p) for some p_1

[grand sigil]

If you take derivative of p_1(x^P) it is 0 in characteristic p

Infact Dp(x) is zero if and only if every term has a degree which is a factor of p in characteristic p.

Do you see why?

Also. I do not see how chain maps take boundaries to boundaries unless the chain maps are surjective?

[lilac trench]

right cause D(x^p) = px^(p-1) = 0

since characteristic p

tanks

[grand sigil]

So far I have that given a chain map un:Cn->Dn. u(Bn(C)) subset of Bn(D). I cant show that Bn(D) is a subset of u(Bn(C)). So far I take x in Bn(D). This gives me x = dn(y) for some y in Dn.

If un were surjective I feel like I would be able to show that x=dn(un(z)) for some z such that un(z)=y.

However un is not surjective so I am not allowed this

Or does chain map sending boundaries to boundaries not mean that the boundaries are equal

I misunderstood and now I understand

[untold basin]

Hello

I don't really understand what does "embedded" mean here

https://math.stackexchange.com/questions/3277160/classification-of-subgroups-of-multiplicative-group-of-complex-numbers

Can someone specify in detail what are those subgroups please ?

[chilly ocean]

Take $f(k) = \exp(2\pi ik/n)$ and note that this is a group homomorphism with kernel $n\mathbb{Z}$

[cloud walrus]

Blitz

[chilly ocean]

Embedded means it's isomorphic to a subgroup

I. e. there exists an injective homomorphism from Z/nZ into C*

[untold basin]

Thx I'm trying to convince myself

[chilly ocean]

Use the commutative property of chain maps

There is some element which maps into your element, use the chain map on it and commutativity then gives you what you want

[pastel cliff]

algebra final time

this exam has no idea the absolute brutalization that it's in for

https://tenor.com/view/goku-super-saiyan-super-saiyan2-super-saiyan2goku-goku-vegeta-gif-23177097

[delicate orchid]

gl boss

[hidden haven]

atb

[opal osprey]

Is there a specific terminology for (non trivial) groups whose all proper finite subgroups are trivial?

Like the integers

[tough raven]

The image of a under the quotient map is a+I, not aI. Indeed, aI need not be of the form x+I for some x, so it need not be in the quotient ring at all. So this is irrelevant to the ring operations on the quotient ring being well-defined.

[hidden haven]

Perhaps torsion free

At least for abelian grapes that works

Because any non 0 element in the group then generates an infinite group

Idk if we use the word torsion free for non abelian groups

But I mean that is equivalent to not having any elements of finite order

[delicate orchid]

we do

or at least I do

[hidden haven]

✓

[lethal dune]

non abelian torsion free group

[delicate orchid]

free group on two elements lol

[tough raven]

Hmm

[wooden ember]

the direct limit of a directed system of set inclusions is just the the union of these sets right?

[tribal moss]

Yeah.

[chilly ocean]

Groups of characteristic 0

[gritty sparrow]

Pedantic point, but it doesn’t work for Z/pZ, all of its proper finite subgroups are trivial

[hidden haven]

[chilly ocean]

do matrices have any identity elements for any multiplication operation

[carmine fossil]

Identity matrix

[junior harbor]

Should I learn abstract algebra by trying to find analog concepts in things I already know (i.e thinking of mapping as functions)

Or should I avoid that

[tribal moss]

"Mapping" and "function" are basically synonyms.

There are some people to prefer to reserve "function" for mapp whose values are numbers, but not everyone follows that convention.

[chilly ocean]

ok perfect, thx

[junior harbor]

But I mean in general, should I always try to link abstract algebra concepts to stuff from other fields of math?

[tribal moss]

It's good if you can make a connection, but there's not always one to find.

So by all means try to look for it, but not to the exclusion of everything else.

[junior harbor]

Gotcha, but at the start there's some pretty obvious ones like mappings and whatnot

But I reckon in the future (aka beyond page 6) there'll be way more stuff I can't connect

[white nymph]

i want to show that the ring of all functions from an infinite set X to Z mod 2Z is not noetherian (by direct counterexample). am i allowed to identify a function with an infinite binary tuple, or if X is not countable, is this not permissible?

also, im not opposed to a better way of approaching this

[devout crow]

I think the word tuple is generally reserved for finite or countable collections

another pov for these kind of things is to identify each function with a subset of X (in particular, the preimage of 1) so that the ring of functions X --> Z/2Z is identified with P(X) (considered as a boolean ring with symmetric difference as addition and intersection as multiplication)

[tribal moss]

Another approach: one way to make an ideal is to select a subset A of X and consider $$\mathcal{I}_A = { f \mid \forall x\in A: f(x)=0 }$$ These ideals are subsets of each other exactly when their As are supersets of each other, so then you just need an infinite decreasing sequence of subsets of X.

[cloud walrus]

Troposphere

[white nymph]

thanks. i appreciate you both

[coral shale]

The Galois group of a simple algebraic extension K(a) : K must be cyclic, C_d, where d divides the degree of the minimal polynomial of a in K, right?

[next obsidian]

No

If that were true every extension of Q has a cyclic Galois group, every extension in char 0 is simple

[coral shale]

where thinky go wrong, I will check

[uncut girder]

Yeah its FALSE

[turbid pond]

I have a field extension L:K and some element a in L, and the minimal polynomial of a has some but not all of its roots in L.
Does the galois group act transitively on this set of roots in L?
I know it is true when L:K is the splitting field and L contains all the roots

[kindred jay]

can i get some help on these questions please? for 8a, i know the answer is (x-1)(x^2 + x + 2) from some guess and check, but i'm not really sure how to proceed forward (also we never learned what an "evaluation map" is)

[grand sigil]

No

I got that part

I just assumed for chain map f, f(Bn(C))=Bn(D)

I didnt know they meant boundary elements

[coral shale]

Galois Theory

[tribal moss]

Evaluation map is for example the map that takes a polynomial f as input and returns the value of f(5). A different evaluation map might take a polynomial as input and return the value of f(7). There's one of them for each possible input you can give to the polynomials.

Guess-and-check is a perfectly good way to factor polynomials over finite fields.

For 8(2), do you know how to create a field with 9 elements?

[kindred jay]

i think i figured it out, but there's a key gap in my proof

im not exactly sure why (or whether) that isomorphism exists

[tribal moss]

Is that your proof or an answer key?

[kindred jay]

its my proof

there is unfortunately no answer key

i just don't understand whether the isomorphism F[x]/p(x) -> F[x]/q(x) x F[x]/r(x) exists when p(x) = q(x)r(x)

[tribal moss]

That ought to be what the CRT says.

[kindred jay]

i think i was a bit confused by the definition of a comaximal ideal

one definition said something like the product of two irreducible polynomials

but another said that for ideals I and J, they are comaximal if I + J = R or there is i in I and j in J such that i + j = 1

[kindred jay]

i don't really understand this part

[tribal moss]

Well, it's a theorem that two different (up to constant factors) irreducible polynomials will be comaximal according to the second definitions.

[kindred jay]

oh ok ty!

[tribal moss]

Or more pedantically: the principal ideals generated by two different irreducible polynomials will be.

Namely, you can compute their gcd using the Euclidean algorithm. Since they are irreducible the only common divisor they can have must be 1, so that's what comes out of the Euclidean algorithm. But then, just as in the integer case, we can extract p and q from the Euclidean algorithm such that pf+qg=1. Here p and q will be polynomials, but due to the definition of ideal, we have that pf in <f> and qg in <g>, so <f> and <g> are comaximal.

[prisma shuttle]

also i don't really knwo wut u mean by guess-check here but like

there's a fool proof way to do it without guessing and checking

basically because \mathbb{Z}_3 is a UFD, hilbert's basis thm gives that Z_3[x] is a UFD

and thus if a polynomial of degree 3 or 2 can be factored in Z_3[x] that means it must have a linear facotr

aka. it must ha e a root

and so for the polynomila x^3+x+1 has one root which is 1

so that mean there must be some way to factor x-1 out of it

and we can "shift" the terms to our disposal

so we do x^3+x+1=x^3-2x+1=(x-1)(x^2+x-1)

and we can check that 1 is not a root of x^2+x-1

so thats the best we can do for factorization

and if u need anything about understand the idea of comaximality and stuff D&F ahs some good stuff about that

and its phrased really nicely too all the theorems about polynomials

in tha book

so thats def a good reference

[kindred jay]

yea that makes sense, thank you for the explanation!

my friend and i have been trying to solve these two for so long, and we cannot get anywhere lol

(the evaluation map question and another one)

[proud bear]

did you solve these

[kindred jay]

nah still stuck

[proud bear]

for the evaluation map one, show that the kernel of ev_\alpha is a maximal ideal/generated by an irreducible. then Q[x]/ker ev_\alpha will be a field. for the second one, i think you're supposed to show that varphi can't have trivial kernel

[kindred jay]

for the evaluation map does your explanation follow from the first isomorphism theorem?

[proud bear]

yeah basically

[kindred jay]

because like

ring / maximal ideal -> field

[proud bear]

yes

[kindred jay]

how does the second one work?

[proud bear]

the kernel will be a normal subgroup of A_n, but A_n is simple for n>4

so if you can show that the kernel cannot be just 1, then you have a contradiction

[kindred jay]

oh i see

like a proper subgroup?

cause the normal subgroup is either trivial or the entire group

[proud bear]

yeah

[kindred jay]

awesome thank you so much!

[dull ginkgo]

Let A:G -> Aut(G) for any group G, Does this mapping preserve the subgroup-inclusion order

[tribal moss]

Any homomorphism maps subgroups to subgroups and preserves the inclusion order, but not necessarily strictly.

[dull ginkgo]

Basically if F is a subgroup of G, is Aut(F) a subgroup of Aut(G)

[tribal moss]

Ah!

[dull ginkgo]

i mean Aut(G) acts on G, then is Aut(F) maybe a stabilizer of elements G\F

[tribal moss]

The elements of Aut(F) are not necessarily even in Aut(G).

[dull ginkgo]

let me rephrase, is Aut(F) isomorphic to the stabilizer of elements in G\F if Aut(G) acts on G

[tribal moss]

Definitely not. Take G=Z and F=2Z. There is only one automorphism of Z that fixes all of the odd numbers, but two automorphisms of 2Z.

[dull ginkgo]

I phrased this wrong, anyway I'm wondering if the whole automorphism thing with field extensions (which are like subgroup inclusions) hold for group

[tribal moss]

In general, there can be automorphisms of F that don't extend to automorphisms of G.

For example, we can let G be Z/4Z × Z/2Z, and F the subgroup generated by (2,0) and (0,1). F is isomorphic to V4 and therefore has an automorphism that swaps (2,0) and (0,1) -- but these elements are distinguishable in G, so there's no automorphism of G that interchanges them.

[dull ginkgo]

For any field F that extends G, then Aut(F\G) is essentially the stabilizer of the set of F when considering the action of Aut(G) on G right

Well can’t this also work for groups

[tribal moss]

You can certainly define an analogous concept for groups if you want.

[dull ginkgo]

Aut(F\G) must be a subgroup of Aut(G)

[tribal moss]

Yes.

It doesn't necessarily have much to do with Aut(F), though.

[dull ginkgo]

I just realized that, yeah

I mean that can probably be generalized to any set based algebraic structure with a notion of inclusion

wait also

if G is a subgroup of H, let Aut(G) act on G through a group action. Let Stab(H) be the stabilizer of H. If Inn(G) is a subgroup of Stab(H), then is H normal?

[dull ginkgo]

Also the fact that normal extensions and a group being normal in a supergroup are entirely unrelated bothers me

[wild solar]

Hi everybody, I wonder if anyone could give me some help on this problem....the question is to describe the multiplicative system R_s, and here is my attempt(I think it's wrong but I don't know where exactly did I do wrong, would be grateful if anyone could point my mistake out) 🙂

sorry I meant r1t^k1/r2t^k2=1 in the second last row

[patent crescent]

What does it mean when you adjoin multiple elements to a field?

[wild solar]

W8

I made some progress

[tribal moss]

In the simplest case you can adjoin them one by one in sequence.

[patent crescent]

for example I know Q(sqrt2) is the set of all p(sqrt2) where p is a polynomial in Q

[wild solar]

Could you please tell me does this looks right to you ;P

[patent crescent]

what does it mean to adjoin another element to that?

[wild solar]

sorry I don't understand the question

[tribal moss]

You can say Q(sqrt2,sqrt5) is the set of all p(sqrt5), where p is a polynomial with coefficients in Q(sqrt2).

[patent crescent]

ah ok, thank you

Is there another way to word that by talking about polynomials in Q?

I think that would be the same as the set of all p(a) where a is sqrt2, sqrt5, or sqrt2*sqrt5?

[tribal moss]

You can also adjoin them in one go by taking Q[X,Y]/<X²-2, Y²-5>.

[tribal moss]

That would mean you don't get the element sqrt2+sqrt5, for example.

You would need to say all p(sqrt2,sqrt5) where p is a rational polynomial in two variables.

[patent crescent]

Alright makes sense

[wild solar]

okay I'm still kind of lost.. DOes anyone mind taking a look on this?

[chilly ocean]

How would I prove that a group of order 56 has non trivial proper normal subgroup?

I tried using Sylow, but I couldn’t figure out how to prove that either the number 2-sylow or the 7-sylow was equal to 1.

I proved that the 7-sylow was abelian thru cauchy’s, but I don’t believe that that implies that it’s normal in the larger group.

[chilly ocean]

Third line is wrong. k_1 and k_2 should be switched

Also it's not the kind of description I or author expected I think

Consider the ring of Laurent polynomials Q[t, t^-1] = Q[x, y]/(xy = 1)

Show that your constructed ring is isomorphic to it

[chilly ocean]

Thanks!

[upper cape]

Could anyone help me understand what they mean by functoriality here?

[hot lake]

it means if there is a representation W and a morphism of representations phi : V -> W, then some square with the two isomorphisms, phi, and the map induced by phi commutes

[upper cape]

Ahhh of course, got it now, just showing they are the morphisms in a natural transformation

[wild solar]

ummm, how?

also I don't think we covered Laurent polynomial in the lecture yet

😦

[chilly ocean]

it's just a name

[chilly ocean]

You want to think of your ring as formal quotients r/s where s is in S and r is in R

if r = a_0+a_1t+...+a_nt^n then this can be written as (a_0+...+a_n t^n)/t^m = a_0/t^m + ... + a_n t^(n-m)

so take a map from Q[x, y] to R_S where any rational q gets send to qt/t, x gets sent to t^2/t and y gets sent to 1/t

prove that its a surjective homomorphism and its kernel is (xy-1) = {r(xy-1) : r in Q[x, y]}

consider r/s as notation for the equivalence class of the pair (r, s)

[sweet echo]

Small detail bugging me in the proof of Schur's Lemma for finite groups part (b). I understand the proof and can use the lemma etc, however, in the proof we use the fact F is algebraically closed to get an eigenvalue of $\theta$, but what if $\theta$ has more than one eigenvalue? The same thing can be done with any eigenvalue right? What am I missing here, seems to violate uniqueness?

[cloud walrus]

iCaird

[north sand]

well the proof shows that all eigenvalues are the same. basically it shows that "what if $\theta$ has more than one eigenvalue?" doesn't happen

[sweet echo]

ah yep immediately see it once you said that

[north sand]

I wondered literally the exact same thing the first time I saw it

[sweet echo]

haha great (or maybe confused) minds think alike

[coral shale]

#974750805490008114 help

[cloud walrus]

GUNILLA62

Blitz

[terse wadi]

I thought about but I don't see how I can use it

[cloud walrus]

Blitz

[chilly ocean]

so you can use rational numbers and trivial group for counter-example

[terse wadi]

Good point

I think that you are correct

[chilly ocean]

yeah, I must be

[cloud walrus]

Blitz

[chilly ocean]

as groups

[cloud walrus]

GUNILLA62

GUNILLA62

[chilly ocean]

ah

[terse wadi]

I didnt use that my group algebra iso is an algebra isomorphism. I think that this extra structure gives me what I want

[wild solar]

Hi, I'm extremely confused about the last sentence, "Also gcd(m,n)=1 so n|a_r" hmmmm.....BUT WHY??

Am I missing something trivial here

since you can't be sure the 1/m*(a_r-1+.....) is an integer how can you say that n|a_r?

[chilly ocean]

if you multiply both sides by a big enough power of m, then both sides become integers

the left side, which looks like m^s a_r for some s, is divisible by n then

but m, n are relatively prime

so m^s, n are relatively prime

[wild solar]

Ohhhhh...

[chilly ocean]

so in fact n divides a_r

[wild solar]

makes a lot of sense

Thankssss

[chilly ocean]

np

[vague granite]

Suppose we know that V4 is a subgroup of S4; is it sufficient to show V4 is a normal subroup by pointing out it's the conjugacy class of 2x2 cycles?

[tribal moss]

Plus {e}, yes.
Beware that S4 has other subgroups that are also isomorphic to V4 but are not normal.

e.g. the subgroup generated by (12) and (34).

[dull ginkgo]

If H is a subfield of F, then when does the stabilizer of H of the group action of Aut(F) on F have the same order as the dimension of F as a vector space of H

I know this is sorta true for Galois extensions

[sinful mirage]

If I were to write out the rep, how would I do it? Rho:SU(2)->GL(C^2), ( a b -b* a * ) -> (a 0 0 a *)?

[hot lake]

maybe rho = identity ?

[sinful mirage]

wdym?

the elements are of the z 0 0 z^-1

[hot lake]

isn't SU(2) a subgroup of GL2(C)

[sinful mirage]

it is

[hot lake]

so the identity gives a 2 dimensional representation of SU(2)

[sinful mirage]

but this rep is that?

I don't see how

it's given by z 0 0 z^-1

[hot lake]

I'm saying V is C² and SU(2) acts on it by left matrix multiplication

that's the obvious 2 dimensional representation

[sinful mirage]

a rep is a map:SU(2)->GL(2,c)

I take a b -b* a*

I map it where by rho?

[hot lake]

the sentence about characters of representations seems to not be talking about V particularly

[sinful mirage]

to get this what they have here

[hot lake]

to a b -b* a*

[sinful mirage]

but how do I get z 0 0 z^-1 then?

[hot lake]

that's a different sentence that is not talking about V

it's saying that SU(2) contains some elements z 0 0 1/z

it's not trying to define a representation

[sinful mirage]

so if the rep is the identity map,then how is this irrep?

i'm confused

[hot lake]

V is irreducible because there is no proper subspace stable by SU(2)

[sinful mirage]

how do i see the character of this identity is the one that they give?

for me it's gonna be alpha+alpha*

a+a*

[hot lake]

z 0 0 1/z is not a character

it's an element of SU(2)

[sinful mirage]

i mean the character is z+z^-1

for them

for me it's alpha+apha*

[hot lake]

what even is alpha here

[sinful mirage]

a complex number

or sry

a+a*

an element of SU(2) is a b -b* a*

i map via id

trace=a+a*

[hot lake]

yes ?

[sinful mirage]

they have a+a^-1

[hot lake]

if the element of SU(2) is of the form z 0 0 1/z

[sinful mirage]

ah

[hot lake]

then the trace is z+1/z

[sinful mirage]

but why do they look at those

[hot lake]

because every element of SU(2) is conjugate to one of those

[sinful mirage]

that's right

ah the characters are constant on conjugacy classes

so it's enough to check them on conj classes

[hot lake]

yes

[sinful mirage]

and conj classes are of the form z 0 0 z^-1, all of them

for some given z

yeah

[hot lake]

so for this obvious 2 dimensional irreducible representation, the character on (z 0 0 1/z) is z+1/z

and that sums up all the data you need to determine the character

[chilly ocean]

If $$0\to \mathbb{Z}^{r_1}\oplus T_1\to \mathbb{Z}^{r_2}\oplus T_2\to\mathbb{Z}^{r_3}\oplus T_3\to 0$$ is exact with $T_i$ torsion, why does it follow that $r_2 = r_1+r_3$?

[cloud walrus]

Blitz

[next obsidian]

Tensor with Q

[sinful mirage]

is this ok?

[next obsidian]

Then apply rank nullity

[hot lake]

your rho isn't a group morphism

[sinful mirage]

ah

sry

obviously not i misswrote

[hot lake]

and I think that's 2 cos theta in the end

[chilly ocean]

I technically haven't "discovered" this yet, I'm trying to prove that the Grothendieck group of finitely generated abelian groups is isomorphic to Z

[next obsidian]

Localize at 0 then apply rank nullity

[sinful mirage]

now?

[hot lake]

yeah that's good

[sinful mirage]

how can I see though that they are indeed OK?

it should give that inner prod of this with itself is 1

[hot lake]

I guess by computing some integral ?

I'm not that much familiar with representation theory of lie groups

I don't know what the haar mesure is concretely there

[reef cradle]

It took me a whole day to get this result, I hope it helps you. If you already knew this result, do not hesitate to contact me!
.

[runic hemlock]

I'm having trouble proving this: Let $K$ be a finite extension of $\bQ$, and let $n>1$ be a positive integer. Then there exists an irreducible polynomial $f\in K[x]$ of degree $n$.

[cloud walrus]

Porphyrion

[runic hemlock]

Specifically, I have a homework problem to prove this for $K=\bQ(\sqrt{2}, \sqrt{7})$ and $n=7^2\cdot 11$, but I suspect it's true for the general case as well

[cloud walrus]

Porphyrion

[lethal dune]

how about xⁿ-a for some prime a?

you just need to argue such a prime exists

[runic hemlock]

yeah, do you have an idea how to prove such a prime exists?

[lethal dune]

if for all prime p, p^1/n is already an element of K then K can't be finite extension

[runic hemlock]

yeah, but irreducible is stronger than not having a root

obv you can pick the number a so that the poly has no roots

[lethal dune]

Eisenstein?

[runic hemlock]

then you need some sort of variant of Eisenstien for general number fields?

ig in the partical case K=Q(sqrt(2),sqrt(7)) you can somehow prove that the ring of integers is a UFD or smt?

[lethal dune]

[sinful mirage]

If V is self-dual and quaternionic, then why Sym^2 V can't have 1-dim summands?

[next obsidian]

@runic hemlock it suffices to get irreducible polynomials of degree p^k for any prime p, any k right?

If you have those, write n = p_1^k1•…•p_r^kr

Adjoin a root of an irr poly of degree p_i^ki for all i

The degree has to be exactly n

Using the primitive element theorem this extension is simple, generated by some alpha

Then alpha’s min poly is irr and degree n

[runic hemlock]

Ohhh cool argument

[next obsidian]

So you can reduce to prime power

And now it feels like maybe it holds because any of these fields are perfect or something?????

I don’t know haha

But at least it’s “simpler”

Does passing to a Galois closure of K feel like it would do anything?

IMO not really lol

It doesn’t seem tractable if you do that

[runic hemlock]

according to mse it's true but no proof or reference is given

[next obsidian]

Lol

I might’ve reduced it to solving the problem over Q, but I need a second to see if I’m correct

So here’s my thought @runic hemlock

Let j be the largest integer so that p^j divides [K:Q]

Let alpha be an element generating a degree p^(j+k) extension of Q

Then I think [K(alpha):K] = p^k

Whoops

[runic hemlock]

seems false

[next obsidian]

https://tenor.com/view/sad-baby-cry-tears-cute-gif-17855976

[runic hemlock]

intuitively I would guess that the min poly of alpha remains irreducible over K for "general" alpha

[next obsidian]

Hmm

Then pick a general alpha which generates a degree p^k extension of Q

My friend who’s way better at this than me says this

I think you're overthinking this. Sure you can reduce to prime powers, but at some point you will actually have to write down explicit extensions. So why not just write down an explicit extension from the beginning.

Suppose you want an extension of K of degree n. Find a prime p congruent to 1 modulo n[K:Q]. Then Gal(K(zeta_p)/K)=Gal(Q(zeta_p)/(K cap Q(zeta_p))) is a subgroup of Gal(Q(zeta_p)/Q) of index dividing [K:Q]. So it is cyclic of order divisible by n.

[cinder pelican]

The way I phrase this will probably be inaccurate, but a complex number is the sum of a real quantity and an imaginary quantity. Things multiplied by i cannot be combined with things not multiplied by i. This is also true for polynomials, where I can't add 1 to x any more than I can add x to x^2. This is also true for the geometric product which is a sum of an inner and outer product, which, again, cannot be combined.
Is there a term for objects which have this property of uncombinable-ness? Or something i could search to learn more about structures with this kind of property?

[tribal moss]

Perhaps you're looking for "direct sum"?

Or, if you're fancier than that, "graded ring/algebra"? The complex numbers are a Z/2Z-graded real algebra. And this language is also intrinsically useful for geometric products.

[cinder pelican]

Those seem like exactly what I was looking for, tyvm!
Are these phrases roughly correct?

• The complex numbers are a ring graded by 'i'
• Polynomials are a ring graded by polynomial degree 'x'
• Vectors are graded by dimension

[tribal moss]

The first sounds strange, since we're not usually speaking about C in terms of grading. We'd need to say something wordier lik "the complex numbers are graded by the distinction between real and purely imaginary numbers". In the language of graded rings we would say that numbers on the axes in the complex plane are "pure" elements of C, but it is so uncommon to treat the numbers on the imaginary axis as inherently special that there's no short common way to speak of "which of the axes this pure complex numbers lie on".

The last also sounds strange, mostly because it is rare to even speak of the direct sum of F^n for all n.

[hidden haven]

That is linear independence over integers

I don't think grading works here

[hidden haven]

Oh wait I didn't read this

[sinful mirage]

why is g_alpha aswell in W?(here V_n is the space of complex homogeneous polyonimals in 2 variables of degree n)

[next obsidian]

C isn’t a graded ring

Because i^2 = -1

The multiplication doesn’t respect the grading, and you can see this because the ideal of all relations i satisfies is generated by x^2 + 1, which isn’t a homogeneous polynomial

In effect all I’m saying is, C = R[x]/(x^2 + 1), and R[x] has a grading by degree and for this to descend to the quotient you need the ideal you’re modding out by to be a graded ideal i.e. generated by homogeneous elements

[hot lake]

C is a Z/2Z-graded real algebra

[tribal moss]

Say what? I say it is Z/2Z-graded so the product of two elements of grade 1 should have the base grade.

[hidden haven]

This is going in my Chmonkey gets owned compilation

[next obsidian]

Well this is a bit different than calling it a graded ring I think. Definitionally a graded ring is just a direct sum of abelian groups with an operation from A_n x A_m -> A_{n+m} which distributes

Like as an additive group you can’t combine the real and imaginary part, but once you start multiplying they get twisted

[tribal moss]

That sounds like it is specialized to a Z-graded (or N-graded) ring.

[next obsidian]

I mean you grade by ordered abelian groups

Err…

I guess so

Sure

I agree

It’s a superalgebra

Well… wait so I agree?

[tribal moss]

Yes, I think so.

[next obsidian]

This is weird idk I agree you should totally be able to say it’s Z/2Z graded

But Z/2Z isn’t an ordered group I think, so it’s a bit different than what I’ve seen

Because the operation doesn’t respect the order

[tribal moss]

AFAIK "superalgebra" means exactly Z/2Z graded, and you can use the language of grading with any monoid to index the grades.

[next obsidian]

Yeah superX is a Z/2Z graded X

Oh sure

Hmmmmmmmmmmmmmmmmm

[hidden haven]

[next obsidian]

You totally can grade by any monoid

I thought you needed an ordering on the thing you graded by, but I guess not

[tribal moss]

Ordering of the grades is a useful bonus if you have it, for sure.

[next obsidian]

If you don’t want a notion of ordering on the graded parts

It’s useful for like, proofs, to be able to take a minimal element wrt the grading

But you don’t need that

’s brain became bigger today

Or maybe more wrinkly

Same size, more surface area

[next obsidian]

[tribal moss]

I don't think we ought to require the index monoid to be a group, though. That would be awkward for R[X] or the tensor algebra, because we'd need to declare all of the negative grades to be trivial.

[next obsidian]

Yeah

I think monoid is the right setting

[delicate orchid]

I mean, indexing over N is pretty natural and that's a monoid

[next obsidian]

But I guess it’s equivalent since you can always groupify the monoid then just set the degrees which belong to new elements you introduced to be 0

Yeah

It’s certainly more unnatural though

[tribal moss]

Sure, but you get a lot of cruft if you groupify an arbitrary monoid.

[chilly ocean]

Wdym by grupify

[delicate orchid]

you can turn any monoid into a group by just appending a set of inverses for each element (and then quotienting so xx^-1 = e), I believe that is the method chmonkey was referring to

[chilly radish]

Indeed. My advisor for my research project studies G-graded Algebras and their identities for finite groups G, so you can definitely get away with not even having abelianity

[chilly ocean]

I guess so but more often than not you won't get the original monoid this way

[delicate orchid]

simplest example I can think of is groupifying N into Z by appending the negative numbers

and no? the result is a group, you won't get a monoid at all

[chilly radish]

Sure but that's a commutative monoid where the construction is much nicer

Grothendieck group

[chilly ocean]

It needs to be cancellative for it to be embeddable in such construction

[chilly radish]

You're correct

[chilly ocean]

But that's only equivalent if it's commutative

So you're not going to obtain the original datum of your monoid

[delicate orchid]

I don't know much about the non-commutative case I must admit

the way I think about it is with presentations of a monoid

and you just append loads of relators and duplicate the set of generators such that the duplicate set acts as inverses through the relators

this construction could get really complicated though that I admit

[sinful mirage]

does this hold? I thought they applied binomial,but coefficients are missing

https://www.math.toronto.edu/dburbani/work/quantumreptheory.pdf pg 5 bottom

[tribal moss]

Just having inverses of the existing monoid elements might not be enough -- you may also need to create things that are compositions of negated and non-negated monoid elements, which don't necessarily cancel out in nice way. I think for groupifying the most principled procedure is to create the free group generated by the underlying set of the monoid, and then quotient out relators that force compositions of the original monoid elements to give the expected results.

[delicate orchid]

I see, so it's almost like a backwards way of the way I had in my head, where you keep the number of appended generators small and add loads of relators instead. Interesting.

[tribal moss]

(This ought to produce a left-adjoint-left-inverse to the forgetful functor from groups to monoids).

[tribal moss]

Ah, after rereading what you wrote I think I understand it better. You're proposing to adjoin inverse elements in the monoid world while adding enough additional combinations to keep it a monoid, and then quotient out inverse combinations.

[delicate orchid]

yes that's correct

[tribal moss]

I think monoids have a good enough theory of quotients to make that work, but proving that everything ends up having inverses feels like it could take some footwork.

[prisma shuttle]

can someone explain how to get to the second line in this thing

where g=f^k for an arbitrary polynomila f

oh nvm i figrued it out

[broken stirrup]

I had an algebra exam the other day. There was a problem saying prove that every free module over a ring with identity is projective

and I said F = F (+) 0 and by characterization of projective modules F is a profjective module since there is a free module and submodule such that F is a direct summand

Do you think I'll get point for this genius solution? 🤡

[next obsidian]

…

[thorn delta]

a proof is a proof

[broken stirrup]

it didn't even say using definition or smth else so 😁

[next obsidian]

I just think this might be circular lol

[thorn delta]

i don't think it is. You can show from the definition that 0 -> A -> B -> P -> 0 with P projective splits. So take 0 -> ker f -> F -> P -> 0 (with F free)

[tough raven]

Not if you prove that without explicitly saying “a free module is projective”

[next obsidian]

Yeah like

You want to show a direct summand of a projective thing is projective

[tough raven]

No, show that a direct summand of a free module is projective without ever saying a free module is projective, but silently constructing the required map out of the free module during the proof

[next obsidian]

Yeah I mean

That’s literally just running the proof that a direct summand of a projective thing is projective lol

So I’m just saying to get that F is projective by being a summand of F, you literally have to just manually show you can lift maps along surjections

[broken stirrup]

so F=F + (0) doesn't imply that F is projective?

[next obsidian]

I mean it does

But to show that you would need to show that you can lift maps along surjections from F

Like my point is your proof literally will turn into “F is projective because it is a direct summand of F and that’s projective”

My guess is you’re gonna get very few points

[broken stirrup]

yeah i think so

[thorn delta]

i think if you had access to that characterization it was kind of a silly question to ask

[river nebula]

Anyone know of any example of a subring which is not an ideal and which doesn't contain unity?

[hidden haven]

If you talk about unital rings, then subrings are usually required to contain unity. Otherwise, you can look at even numbers in the polynomial ring ℤ[x]

[river nebula]

That's why I'm asking about subrings without unity

Oh okay

Yeah, thanks, that was easy, I should have thought of that

[next obsidian]

Even numbers are an ideal tho…

Oh in Z[x]

I won’t delete, I am not afraid

[hidden haven]

[river nebula]

I guess we can say that we can't find any such example in ring with dimension < 2 right?

[chilly ocean]

Take a non-unital subring generated by x^2 in k[x] where k field, dim k[x] = 1

Of polynomials which only contain monomials x^(2n) with n>0

Or even integers in Q, Q has Krull dimension 0

[river nebula]

Ah, yes of course, thanks a lot.

[iron vessel]

If i have $H \subset (\mathbb{Z}/n\mathbb{Z})^{\times}$ a subgroup and $L \subset \mathbb{Q}(\zeta_n)$ the intermediate field that corresponds to H under the map $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \to (\mathbb{Z}/n\mathbb{Z})^{\times}$, how do I show that $\eta_H=\sum \sigma(\zeta_n)$, for all $\sigma \in H$, is contained in $L$?

[cloud walrus]

Évariste Galois

[iron vessel]

So, I figured that $Gal(\mathbb{Q}(\zeta_n)/L) = H$. So, $H$ contains automorphisms that fix $L$. However, this would be simple if $\zeta_n$ were contained in $L$, which it isn't.

[cloud walrus]

Évariste Galois

[iron vessel]

Precisely, $[L:\mathbb{Q}]|[Gal(\mathbb{Q}(\zeta_n):\mathbb{Q}] \implies [L:\mathbb{Q}]|\varphi(n)$.

[cloud walrus]

Évariste Galois

[strong yacht]

What about the min poly of zeta_n over L?

[iron vessel]

Well it should divide $\Phi_n$

[cloud walrus]

Évariste Galois

[strong yacht]

And also, the sum of its roots are related to one of its coefficients

[iron vessel]

Yeah

[strong yacht]

That coefficient is in L

[iron vessel]

Is it for certain tho that the sum of its roots is a coefficient the min poly

Like I understand that for example for the minimum polynomial of zeta_n over Q, but how come that holds for the min poly of zeta_n overL?

With respect to the automorphisms in H

Is it because H is precisely the Galois group of Q(zeta_n) over L?

[strong yacht]

The roots of the min poly of $\zeta$ over $L$ in $\mathbb{Q}(\zeta)$ are precisely the orbit of $\zeta$ under the automorphisms in $Gal(\mathbb{Q}(\zeta)/L)$. Say the min poly is $x^m+ax^{m-1}+\dots + c$. Then over $\mathbb{Q}(\zeta_n)$ have the factorisation for this poly as $\prod (x-\sigma(\zeta))$ for $\sigma \in Gal(\mathbb{Q}(\zeta)/L)$, compare with the coefficient of $x^{m-1}$ and have that $\sum \sigma(\zeta) \in L$.

[cloud walrus]

Greenman

[iron vessel]

That makes sense

Thanks 🙂

[strong yacht]

No problem!!

[chilly ocean]

why are presheafs contravariant?

[sinful mirage]

Why c_k=c_n implies A=id?

[iron vessel]

Let $H \subset (\mathbb{Z}/n\mathbb{Z})^{\times}$ a subgroup and $L \subset \mathbb{Q}(\zeta_n)$ the intermediate field that corresponds to H under the map $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \to (\mathbb{Z}/n\mathbb{Z})^{\times}$, how do I show that $L = \mathbb{Q}(\zeta + \zeta^{-1})$ when $H = {\pm 1 mod n}$?

[cloud walrus]

Évariste Galois

[iron vessel]

First off, by an earlier exercise I had found that $\alpha = \zeta + \zeta^{-1} \in L$; so $\mathbb{Q}(\alpha) \subset L$. Then, I tried a few things, for example, if $G = Gal(\mathbb{Q}(\zeta)/L) = {\pm 1 mod n}$, then $|G| = 2$; and consequently, $[L:\mathbb{Q}] = \dfrac{\varphi(n)}{2}$.

[cloud walrus]

Évariste Galois

[iron vessel]

However, I'm not sure how to conclude that $L \subset \mathbb{Q}(\alpha)$

[cloud walrus]

Évariste Galois

[broken stirrup]

I've done a) but got stuck at b)

I don't think that I really understand the problem

[shell agate]

You know that [Q(ζ):L]=2 and you know that L contains Q(a), so [Q(ζ):Q(a)]=[Q(ζ):L][L:Q(a)]. If you can prove [Q(ζ):Q(a)]=2, you're good to go.

That's what I'd try

And to do that, you just need to find a polynomial of degree 2 with coefficients in Q(a) and with ζ as a root

Since aζ=ζ²+1, that's not too hard

[iron vessel]

How could I possibly miss that?! Thanks a lot 🙂

[shell agate]

No problem

[broken stirrup]

okay solved this one

but now got stuck at this one

[fast stratus]

any recommendations on books about introductory algebraic number theory?

[tribal niche]

i'm stuck on this problem

[cloud walrus]

monkeman

[tribal moss]

Each coset in H1 of the intersection must be within a single coset in G of H2, so the index cannot be any larger than G:H2.

^ @tribal niche

Wait ... that gives an inequality in the wrong direction, doesn't it? Poop.

Instead, how about: Two different cosets in H1 of the intersection cannot lie in the same coset in G of H2.

Call the intersection K. Then for elements x and y of H2, the cosets xK and yK are different iff xy^-1 is not in H1. But that also means that xH2 and yH2 are different cosets in G.

[tribal niche]

so $(H_1:H_1\cap H_2)\leq (G:H_2)$?

[cloud walrus]

monkeman

[brazen frost]

So this is probably a fairly simple question but: I have two subspaces of a vector space over a finite field $B\subset A \subset \mathbb{F}^n_2$. I know a basis of $A$ and $B$ in terms of the standard basis of $\mathbb{F}^n_2$. Is it obvious how to compute a basis for the space $A\setminus B$ (ex. row reduction)? If this were something like the reals I would just compute orthogonal complements using the inner product, but there is no such structure here

[cloud walrus]

ConfusedTorus

[tribal moss]

Yeah.

First, A\B is not a subspace (it's not closed under addition). I assume you just want a basis of some direct complement of B in A.

[brazen frost]

Yeah

I forgot this when writing it out. I actually just want representatives (in F^n_2) for the generators of the coset A/B, but I came across some code that seemed to indicate it was a lot simpler in F_2

(for context, I'm actually computing representatives of generators of the first homology group with coeffs in Z_2 of a 3-complex and A is a ker, B is an image)

At least, searching directly for computing the homology generators in the literature got me nowhere 😅

[tribal moss]

Even over R, there's an easier way to create some direct complement of a subspace than by finding the orthogonal complement.

Suppose you have a basis for A, you want to extend it to a basis for all of F^n. (The new vectors in the basis will then span a direct complement).

Make each of your existing basis vectors into a row of a new matrix, and reduce it to RREF. Then add a standard basis vector corresponding to each of the non-pivot columns. The resulting matrix is obviously invertible, and if you now undo the row-operations that produced the RREF you will get a basis for the entire space that extends the orginal basis.

What's better, those row operations won't touch your new basis vectors, so they come out entirely unchanged.

[brazen frost]

oh, that does it

thanks!

[tribal moss]

And you don't need to reduce all the way to RREF, just enough to know where the pivot columns are.

[brazen frost]

RREF is easy enough, I've been computing SNFs xD

I was suspecting that this was something really simple, but I didn't really know where precisely to look (and not for a lack of trying 😢 )

[tribal moss]

Hmm, actually for your original problem there's an even easier way.

Take your basis for B as column vectors, concatenate it with your known basis for A so the A basis is to the right. Reduce to RREF. The (original content of the) pivot columns now correspond to a basis for B+A, and they include all of the B basis vectors. The selected A basis vectors span a direct complement.

[brazen frost]

btw do you happen to have a reference off-hand? Now that you say it, it's obvious, but I seem to be lacking some intuition for this

[tribal moss]

No sorry. I learned linear algebra from photocopied lecture notes that weren't even in English ...

[brazen frost]

fair enough

I scanned Roman for something like this but never found it

Turns out this is in 2.B of Axler's book

[urban ice]

Hello, the following problem has me stumped :

We have K = Z/pZ with p an odd prime

In the previous questions it was notably shown that every element of K can be written as the sum of two squares

This questions asks the following : for all $x\in K$, let $K_x = {(a,b)\in K^2, a^2+b^2=x}$. Show $|K_x| = |K_{-x}|$

[cloud walrus]

Syst3ms

[urban ice]

Now, this is easy if p ≡ 1 [4] because for ε²=-1, we have $(a,b) \mapsto (εa,εb)$ a bijection from $K_x$ to $K_{-x}$

[cloud walrus]

Syst3ms

[urban ice]

But I have no idea how to prove it if p ≡ 3 [4]

[barren sierra]

What is this notation p = 3 [4]

[tribal moss]

$p\equiv 3 \pmod 4$ I suppose.

[barren sierra]

do you mean p = 3 (mod 4)

[cloud walrus]

Troposphere

[barren sierra]

If so 4 is not an odd prime

And no a, b exist such that a^2 + b^2 = 3 (mod 4)

[tribal moss]

What.

[barren sierra]

Am I dumb

[tribal moss]

The modulus is an odd prime.

[urban ice]

i wouldn't say that, but that's not the question

[barren sierra]

Yea

[urban ice]

p = 3 (mod 4) but we're working in F_p

not F_4

[barren sierra]

But they said this

[dull ginkgo]

[urban ice]

where p defines this

[barren sierra]

Oh

I'm slow

Carry on

[tribal moss]

For p=3 we have 1 = 0²+1² = 0²+(-1)² = 1²+0² = (-1)²+0², whereas -1 = 1¹+1¹ = 1²+(-1)² = (-1)²+1² = (-1)²+(-1)².
There are four pairs for each indeed, but no obvious correspondence between them.

[urban ice]

Yeah, that's the issue

[urban ice]

One one I notice with an example is that not only does it look like |K_x| = |K_-x|, but it looks like the K_x all have the same number of elements as long as x≠0

[strong yacht]

Yeah, i think this is correct

The quadratic residues mod 11 are 1, 3, 4, 5, 9. Consider the equation (x+x^3+x^4+x^5+x^9)^2 mod (x^11-1). The coefficient of x^j counts the number of ways that j can be expressed as a sum of two quadratic residues of 11 (this is very related to your problem). In this case, that turns out to be 2 x + 3 x^2 + 2 x^3 + 2 x^4 + 2 x^5 + 3 x^6 + 3 x^7 + 3 x^8 + 2 x^9 + 3 x^10

So each quadratic residue of 11 can be expressed as the sum of QR's in 2 ways, and each non-QR can be expressed as the sum of QR's in 3 ways

[urban ice]

Wait, what is (x+x^3+x^4+x^5+x^9) mod (x^11-1) and how do you get the other polynomial

[strong yacht]

Sorry, square that poly

[urban ice]

right

That's weird, I got something different by doing it by hand

[strong yacht]

There are some extra solutions

What I counted does not distinguish between a^2 and (-a)^2

It also doesn't count using 0^2

[urban ice]

Also I whittled down the symmetries

ah, that's why

also i just realized i didn't properly whittle it down

[strong yacht]

So running with my mod 11 example for a little longer:

Every QR can be expressed as the sum of two QR's in 2 ways. This is really 8 ways once you account for a^2=(-a)^2. And you get 4 more from 0^2+a^2=a^2 etc.

So 12 ways

And every non-QR can be expressed as the sum of two QR's in 3 ways (which is really 3*4=12 ways when again, accounting for a^2=(-a)^2, and obviously non-QR's don't have solutions for a=0 or b=0)

And one final thing (i'm dealing with p=3 [4] since you dealt with p = 1 [4]), residues are anti-symmetric in the case p = 3 [4] meaning if a is a QR, then p-a = -a is NOT a QR.

This is motivation for the more general method which you can read about here: https://www.math.ttu.edu/~cmonico/research/qnr-charv5.pdf

[urban ice]

how does that computation work?

I tried long division and it gave a different result

[strong yacht]

This is from Mathematica

And it's done by just squaring that expression, and replacing wherever you see x^11 with 1

[urban ice]

I see

That's all fine and dandy, but at my level we don't even mention the term "quadratic residue" explicitly

And this is an oral exam question, so surely there has to be a more elementary proof

[strong yacht]

You know, that is very possible

When I saw your question, I googled 'additive properties of quadratic residues' and just did some reading on that, there could be a better way

Interesting question though

[urban ice]

The worst thing is, there is a question afterwards, and it looks even worse

[strong yacht]

This is the essential result from the paper I linked, I merely did a demonstration and how it links to your problem for mod 11

[urban ice]

Let $n=2m+1$, $X_n = {(z_i){1\le i\le n} \in K^n| \sum{i=1}^n z_i^2 = 1 }$ and $Y_m = {(x_1,\ldots,x_m,y_1,\ldots,y_m,z) \in K^n | \sum_{i=1}^n x_iy_i + (-1)^m z² = 1}$.

For n=3,5,7, link the cardinalities of $X_n$ and $Y_m$. Assume what was done these three cases is also true for any odd integer n. Deduce the cardinality of $X_n$.

[cloud walrus]

Syst3ms

[urban ice]

Yeah, no clue

Although, this problem is made significantly harder by the fact that there is no examiner to clue you in and to discuss with, as is the case during the actual exam

[strong yacht]

That sounds really difficult, I am not really familiar with oral exams (I'm an undergrad in the UK)

[urban ice]

the systems are different, but more or less the same for me

I'm two years out of HS, it's a French thing where it's two very intense years of preparation for engineering and research schools' entrance exams

[strong yacht]

The fuck? This is prep to go to uni?

that's insane

[urban ice]

No

Not uni

Uni runs in parallel to this

Most of the schools we are prepared for are engineering schools

But I'm aiming for a prestigious and research-oriented public school

[strong yacht]

Oh ok I see, when you said you were two years out of HS I didn't realise you meant you were at uni

[urban ice]

This specific problem is from one of the most prestigious engineering schools in France, so it's really on the hard end of the spectrum

[urban ice]

As i said, it's not uni, it's something else, but let's not fret the details

[strong yacht]

Ahhhh ok I got it, it's like the in-between process that you guys have in france

[urban ice]

sorta?

[strong yacht]

Well, best of luck with your preparations! I am very glad I don't have to take entrance tests anymore (no more maths exams for me after these finals)

[urban ice]

Thanks

I'm already done with the written exams

[long obsidian]

I cannot find this definition for the life of me.

There is some definition of an "algebra" relating to vector spaces. I think a vector space with a bilinar map is called an algebra but I'm not sure. Does anybody know what I'm talking about and can help me recall the definition?

[barren sierra]

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

@long obsidian ?

[long obsidian]

Yes! That's it thank you

[wooden ember]

Yeah in case you don’t know about prepa (that’s how we call it) it’s meant to prepare you for engineering schools or sometimes even just math/physics courses but you get there over prepared, I’ve always found the system a bit dumb. Like there are quite a few people in my year who did prepa and they’re way ahead of the rest, but not by enough for them to go to second year either, so it’s kind of that awkward middle where they just sorta wasted a year

[next obsidian]

Narwhal you’re French?????

[tribal moss]

Here is one thing I've thought of for @urban ice's problem. If we have a nonzero pair (a,b), then { (na,nb) | n in F_p \ {0} } produces either every nonzero square two times or every nonzero non-square two times. So perhaps we should aim to prove that there are as many of these cycles that produce squares as cycles that produce non-squares. The cycles are exactly generated by (1,a) for every a in F_p, plus one cycle generated by (0,1). So if we can show that exactly (p+1)/4 of the (p+1)/2 elements of the form a^2+1 are non-squares, then we can get through.

[astral galleon]

Dumb question does the symmetric group form a vector space ?

[chilly ocean]

what operations do you want?

[astral galleon]

I imagine no since the the S_n in general isn't abliean 🙂

[astral galleon]

everything that covers axioms of a vector space

[delicate orchid]

I can make it one if I try super hard

[thorn delta]

can you though? S_n is finite, so it would be a vector space over a finite field. but the cardinality of vector spaces over finite fields are p^n for a prime p

[delicate orchid]

Why would it be a vector space over a finite field

It would be finite dimensional

Not over a finite field

[thorn delta]

because S_n is finite

[astral galleon]

Isn't the requirement for a group to be a vector space that it has to be abliean ?

[delicate orchid]

The specific construction I had in mind was the CS_n-algebra

[thorn delta]

err hm maybe the field could be infinite

[next obsidian]

Any finite vector space has cardinality a prime power

[astral galleon]

So you can make S_n a vector space ? But what work would you have to do

[next obsidian]

There’s no way to make S_n a vector space unless n = 1 or 2

If you’re a vector space over F_p^n your size is a power of p^n

Because you just take a basis

[astral galleon]

ahh figures @next obsidian 👍 was at a talk when the guy started talking about lie groups this was the only counterexample that came to mind

[next obsidian]

When n = 1 it’s trivial, for n =2 it’s just F_2 even as a group

Like, S_2 = Z/2Z

[delicate orchid]

Oh wait you want to preserve the group operation as addition?

Then no, you can’t

[next obsidian]

It doesn’t matter

Like even if you want to define a new operation

As a set, there’s no way to put any structure on it to make it a vector space because of size considerations

[delicate orchid]

I simply don’t believe you, take the regular complex representation of S_n

[next obsidian]

That’s not on the set S_n

As a set S_n has cardinakity n!

[delicate orchid]

This can be formulated as acting on a vector space with a basis set of S_n

[next obsidian]

Any finite vector space is isomorphic to (F_p)^n for some n

As a basis set, S_n

Now you’re freely generating more stuff

[delicate orchid]

Sure

I did say “try super hard”

[next obsidian]

That’s not turning S_n into a vector space, you’re embedding it into a bigger vector space

[astral galleon]

Wait when you take the represetnation of S_n each of the elements essentially become linear transformations

[delicate orchid]

The main problem here is that the initial question is vaguely posed

[next obsidian]

I guess

[astral galleon]

Wait I thought what I asked for was clear

[next obsidian]

But if that’s the case then you can turn any set into a vector space by just letting them be basis elements for a vector space over any field lol

I don’t think that’s really interesting

[delicate orchid]

In my mind both me and chmonkey are right we’re just answering different questions

[next obsidian]

Yeah

[thorn delta]

its not that its not clear. its just kind of a weird thing to want to do lol

[astral galleon]

Oh heres the context for Directed Reading Program there an undergraduate student talking about lie groups so I was trying to come up with some non-examples and realized that S_n didn't quite fit the bill to be a lie group

[next obsidian]

I mean I think it is an issue of clarity. “Turn” isn’t rigorous like

You can “turn” Z into Q by adding fractions

[delicate orchid]

Yeah

[next obsidian]

But maybe you want to “turn” Z/nZ into a ring by introducing multiplication

Those are kinda different things

But both reasonably use the word turn

[delicate orchid]

I interpreted your question as more of the former where as chmonkey answered the latter

[next obsidian]

Yeah

[astral galleon]

But wait @delicate orchid it seems you tried to use rep.theory to make S_n behave as a vector space ?

[delicate orchid]

I brute forced it using rep theory

I defined a load of formal sums that basically forgot about the structure of S_n

But was still technically a vector space

[astral galleon]

but that fails the elements only would behave as vector spaces

[delicate orchid]

Perhaps the regular representation was a bit overkill, you could chose any faithful representation and get a much smaller space

[astral galleon]

not the entire the group tho

[delicate orchid]

I don’t follow

[astral galleon]

with represetnation theory don't you beasically write the elements of that group as matrices in general you are using that vector space to study the behavior of that group

[delicate orchid]

No no I was saying that that representation acts on the desired vector space

It was a long winded construction, apologises

[astral galleon]

So wait why would you want regular representations acting on S_n again ?

[delicate orchid]

Are you familiar with the group algebra? What I did was I basically constructed that and then ignored the vector-multiplication lol

Anyway, chmonkey’s answer is the more interesting one and in hindsight is probably closer to what you wanted your original question to be

[astral galleon]

so you just took linear combinations of the elements in g basically ?

[delicate orchid]

Yeah

[astral galleon]

But wait hold having the representations of a group when that group is not a vector space that is just imitation tho

[delicate orchid]

You specified no boundaries in your original question

[astral galleon]

oh true 😔

[delicate orchid]

So naturally I brute forced it

[astral galleon]

So you made s_n "act" like a vector space via representation theory when it's not

[delicate orchid]

Yup

[astral galleon]

Shame I didn't get to do representation theory as for my math final project 😢

[chilly ocean]

S_n is absolutely a lie group.

give it the discrete topology

and the obvious atlas

[astral galleon]

👀

@chilly ocean I thought lie groups at least form how the guy defined it in his slides had to be a vector space

But now looking up the defintion, A Lie group G is an abstract group and a smooth n-
dimensional manifold so that multiplication G × G → G : (a, b) → ab and
inverse G → G : a → a−1 are smooth.

oof looks like I made an oversight

[chilly ocean]

maybe you mixed up lie group and lie algebra

the latter of which is a vector space plus a commutator

[astral galleon]

ahh kk makes sense

So s_n would not be a lie algebra but it's a lie group. intutively I can see this since for example taking s_3 is the symetries of a triangle in euclidean space is basically a manifold

[astral galleon]

Yeah I think so I don't have his slides on hand but I rembered some of the stuff he said during the talk

[chilly ocean]

S_3 is not a triangle and a triangle is not a manifold

S_3 is a lie group for much more boring reasons

[astral galleon]

but why does it form a lie group ?

[chilly ocean]

give it the discrete topology, and it becomes a 0-dimensional manifold. such manifolds have unique smooth structures, and you can check that the group operations are smooth

your atlas is: for x in S_n, the map {x} -> R^0. clearly everything is smooth