groups-rings-fields

[chilly ocean]

So...do i repost?

[tardy nimbus]

Assuming these are regular sets and not multisets I guess the statement here should be if S_|A| ~= S_|B| then is |A| = |B|. then it seems like it trivially follows (unless you want to say there exists cardinals Q1 and Q2 s.t. Q1 != Q2 yet S_Q1 ~= S_Q2 (which feels totally wrong but maybe theres a model of that))

[chilly ocean]

Regular sets

[chilly ocean]

S_A i believe is the standard notation for group of permutations on A

And how does it trivially follow

[tardy nimbus]

ok so in my notation i was indicating there is one symmetric group for every cardinal and indexed by that cardinal hence hte S_|A| notation as opposed to S_A. Then if S_|A| = S_|B| we have that |A| = |B| by our definition

is the notation S_A ever not a symmetric group on |A| elements? when you say "group of permutations on A" that could mean a lot of things actually

[chilly ocean]

But dosent that require this result

[tardy nimbus]

maybe i am just actually brushing this. Like why do I believe that if u,v are different cardinals that S_u != S_v

well suppose u < v w.log. then |S_u

|S_u| < |S_v|

is what i want to prove

[chilly ocean]

This follows i think by taking an injective map u to v

And for the permutations

You just replace them with the image to get an injection

In S_u and S_v

Isn't it right?

[tardy nimbus]

im trash at set theory but this seems fair to me. so every permutation in S_u will have an element in S_v via applying the map element wise to the individual parts of the permutation

and then there will be elements of S_v left over

now can we necesarily state there is a WHOLE cardinal more elements of S_v left oveR?

that feels like what needs to be argued (although i think a good formal logic person would be able to just one-line this somehow)

YOU showed |S_u| <= |S_v|

but we need |S_u| < |S_v|

[chilly ocean]

Do u happen to know one

[tardy nimbus]

I know one but he's a bit out there. Maybe math.se or mathoverflow would be the first place to go

not overflow

definitely just SE for now

[chilly ocean]

Okay maybe I'll post it on SE but I want to wait a bit more on this here...so I'll post this also in #foundations

Would that be a good idea

[tardy nimbus]

here you go: https://math.stackexchange.com/questions/3237056/symmetric-groups-of-sets-with-the-same-cardinality-are-isomorphic

this is your solution i think

no not really its only one way

its saying |X| = |Y| then Sym(X) = Sym(Y)

[chilly ocean]

Yes

I am asking the converse

[coral spindle]

For finite sets it is trivially true by cardinality. For infinite sets, I can't surely say, but there is a similar problem which iirc is independent of ZFC.

[chilly ocean]

Thank you for the response, This is resolved now in #foundations

[stuck cosmos]

Is the restriction of the spinor representation of an even dimensional complex Clifford algebra $\mathbb{C}l(2n)\cong$End$(\mathbb{C}^{2^{n}})$ to its even sub algebra $\mathbb{C}l^0(2n) \cong$End$(\mathbb{C}^{2^{n-1}})\oplus$End$(\mathbb{C}^{2^{n-1}})$ defined as:
\begin{align*}
\rho_0 : \mathbb{C}l^0(2n) &\longrightarrow \text{End}(\mathbb{C}^{2^n})
\end{align*}
or
\begin{align*}
\rho_0 : \mathbb{C}l^0(2n) &\longrightarrow \text{End}(\mathbb{C}^{2^{n-1}})\oplus \text{End}(\mathbb{C}^{2^{n-1}})
\end{align*}

[dull ginkgo]

Wouldn’t it hold for infinite sets due to the existence of a bijection and the functorial properties of the Symmetry group functor?

[cloud walrus]

SK2099

[dull ginkgo]

EVIL CHMONKEY: good latex so I won’t read it

[stuck cosmos]

Or the third option where it maps to End(C(2n-1))?

[south patrol]

Is it a functor?

[dull ginkgo]

I thought it was

Wait no

Was thinking of a different thing

[south patrol]

But you can clearly define it as a functor on injective maps

Which is enough

As in, wide subcategory of Set on injective maps

But yeah I mean you can use conjugation

Indeed this is a special case of the fact that isomorphic objects of a category have isomorphic automorphism groups

[crystal vale]

Suppose that M is a finitely generated module over a PID and M_t is its torsion part. If n = dim(M/M_t ), then any n+1 elements in M are linearly dependent.

Let {x_1,....,x_(n+1) } be the set of n+1 elements.
Because any n+1 elements in M/M_t is linear dependent, hence there exists a_i ≠0 such that a_1 x_1 +....+ a_ix_i +.... + a_(n+1)x_(n+1) in M_t and that n +1 elements lie in M_t, so there is r≠0 such that r(a_1 x_1 +....+ a_ix_i +.... + a_(n+1)x_(n+1) ) = 0.
Thus ra_i ≠0 therefore n+1 elements in M are linearly dependent.

Is it correct?

[terse crystal]

(1,0) and (0,1) are independent in Z oplus Z/2Z

[coral spindle]

That doesn’t follow. The power set functor is a functor yet, if I am recalling correctly, its “injectivity-by-cardinality” is independent of ZFC.

[crystal vale]

But what is the dimension of Z/Z_t, where Z_t is a torsion part of Z ?

[terse crystal]

? Weird question

[crystal vale]

Is it Z over Z/2Z ?

[terse crystal]

Direct sum of Z and Z/2Z

[crystal vale]

Oh

[terse crystal]

Over pid Z

[crystal vale]

Means Z× Z/2Z ? Because what the elements look like in Z + Z/2Z ? Sorry I have no idea

[terse crystal]

Coproduct

Or product , since it’s an abelian category

[crystal vale]

I don't know about the category

[terse crystal]

Then search direct sum

[south patrol]

Oh I think we both were talking about the easier direction lol

[crystal vale]

I know that definition, A = B + C where intersection of B and C is trivial then A is direct sum of B and C

[south patrol]

Tbh I'm unsure about the other way as the only obvious (obvious to me) obstruction to them being isomorphic is cardinality

[terse crystal]

Will this help?
$(1,0)$ and $(0,1)$ are independent in $\mathbb{Z} \oplus (\mathbb{Z}/2 \mathbb{Z})$ over PID $\mathbb{Z}$

Or I put an extra parentheses?

[cloud walrus]

Cogwheels of the mind

[crystal vale]

So the torsion part is { (0,0), (0,1) } ?

[terse crystal]

Torsion part of M=Z oplus (Z/2Z) is simply Z/2Z

So yeah

[crystal vale]

What will be the dimension of its quotient module ?

[terse crystal]

1

[crystal vale]

How ?

What is the generator?

[terse crystal]

Of its torsion free part?

[crystal vale]

Yes

[terse crystal]

(1,0)

[crystal vale]

But it does not contain (2,1)

[rocky cloak]

This depends on your definition of linearly independent

[terse crystal]

What contains (2,1)

I naturally assumed he meant that Σai xi=0 -> any i, ai=0 kind of independent

[crystal vale]

(2,1) is a torsion free element so it will be in torsion free part so how (1,0) generates (2,1) ?

[rocky cloak]

Well, then they're not linearly independent

[terse crystal]

I am dumb…

0(1,0)+2(0,1)=0…

[crystal vale]

Me too

[rocky cloak]

Anyway, this should be correct yes

[terse crystal]

Sorry, yeah you are correct

[crystal vale]

Okay, thank you

Finitely generated torsion modules over Z are a finite Abelian group.

I am wondering how it implies that the module is finite.

If M is a torsion module and generated by { x_1,....,x_n }. Then there is k_i ≠ 0 such that k_i x_i = 0 for all i.

So the total number of elements in the module is k_1 × k_2 ×...× k_n.

please correct me

[rocky cloak]

Could be less than that, but yeah it's at most that many elements. Hence finite

[south patrol]

Maybe I'm weird but I would add a little more detail as to why the total number of elements is less than that (but the details are easy)

Well, like without that this doesn't seem to be a proof

[dull ginkgo]

The only step here I’d assume is to prove the actual induction via considering submodules generated by less elements?

[dull ginkgo]

You can refine this a bit inductively

[south patrol]

Well that seems overly complicated i mean just like

[dull ginkgo]

Let M_n = <x_1… x_n> for n = 1 … N

[south patrol]

you can write every element in the form a_1 x_1 + ... + a_n x_n for 0 <= a_i < k_i

[dull ginkgo]

That too

[south patrol]

perhaps non-uniquely

[dull ginkgo]

If they are distinct then it’s at most that many

[south patrol]

another would be that there's a surjection Z^n -> M which factors through like NZ^n for some N >> 0 so it's finite

[long geyser]

If R is a domain and I do R[x] -> R by mapping x to 1 that works right? (x) the prime ideal

also, nothing comes to mind for (ii) here

[rocky cloak]

Yes, that works.

Think about Z as a subring of Q for example

[long geyser]

omg

so true

[long geyser]

I need help with this theorem, this is a total mess

first of all, p is the irreducible poly of z over k

then, we map p to k'[x] using phi*, and find a root z' of phi*(p) over k', and then let p' be the minimal poly of z' over k'
I think there is a typo here where author use p' for two different things (first as image of p under the isomorphism and then as the minimal poly of z') but ignore that

I don't understand the rest of the proof, the straightforward generalization of A-3.87 he talks about is a theorem concerning two different roots of the same polynomial in the same field

in this case, we have two separate fields, two separate polynomials, how is there an isomorphism?

do we even know that p' = phi*(p)

[patent girder]

all Rotman is saying is that we know there is an iso k(z) -> k(z') [by that result 3.87 ii], but k(z') is isomorphic to k'(z')

[long geyser]

does k(z') even make sense? z' is an element of E' extending k'

[barren sierra]

I'm stuck here on (ii). I'm looking at a really ugly expression of the form

a [b1, c1]...[bn, cn] a^-1 [bn, cn]^-1 ... [b1, c1]^-1

and I'm not really sure where to go with showing such an element is of the form [a, [b, c]]

especially since the product of commutators may not be a commutator

NVM I think I solved it?

[patent girder]

right im not being precise: \psi gives psi* that induces the iso k[x]/(p(x)) -> k'[x]/(psi*(p(x))) or something like this

[long geyser]

yes, I get that there is an iso like that, which is why I asked if phi*(p) equals p' (minimal poly of z' over k')

there is definitely something wrong with this proof

if, instead of saying, let p' = irr(z', k'); he attempts to show that irr(z', k') = phi*(p) then everything makes sense

[patent girder]

i think the words "corresponding monic irred. polynomial" are saying that

[long geyser]

he says: let z' be a root of p' in E' (I assume here he is using p' to mean phi*(p))

and then he says: let p' = irr(z', k') be the minimal poly

if he wanted to say phi*(p) = irr(z', k'), it is not obvious at all

it's not obvious he meant that and it's even less obvious that this is true

[patent girder]

Well he defines p' after he uses it but it's like let z' be a root of p' which is the image of p under psi*

His clauses are a strangely placed, but i think that's what is being said

[long geyser]

yes, I'm saying he defines p' in two different ways

one as phi*(p) and one as irr(z', k')

[patent girder]

i think he gets that for free no? since z has irred polynomial p, then z' in k' obviously has p' as the its minimal polynomial. Take the definition either way, but it's gonna work out for free cause theyre isomorphic anyways

[long geyser]

no, this is not obvious a priori

because z is not in k and z' is not in k'

they live in the extensions

I'd be happy if you can show me why they'd be the same

I guess, suppose phi*(p) is not irr(z', k')

then irr(z', k') must divide phi*(p)

so inverse image under the isomorphism of irr(z', k') must divide p, but p is irreducible

I think this works

and fixes the proof

"fixes", it's just lacking important detail coupled with a highly confusing typo

[soft wyvern]

hi !, can someone help to prove that $\sqrt[3]{5}\notin\mathbb{Q}(\sqrt[3]{2})$

i found this but i think there a simply way to solve this https://math.stackexchange.com/questions/4843313/proving-sqrt35-notin-mathbbq-sqrt32-with-solvability-of-system?noredirect=1&lq=1

[cloud walrus]

benjazul

[celest furnace]

not sure what that did but you can probably just say if cube root 5 = a + bcube root of 2 + c 2^(2/3), then you can cube both sides and reach that cube root of 2 is rational somehow or something like that

That's probably what that link did though

[tribal moss]

It is.

[celest furnace]

Found a pretty cute argument

Notice that $a/\sqrt[3]{2} = a\sqrt[3]{4}/2$.

[cloud walrus]

Ultimate Chad

[celest furnace]

Thus if $\sqrt[3]{5} = a + b\sqrt[3]{2} + c\sqrt[3]{4}$ we would have $\sqrt[3]{5/2} = a + b\sqrt[3]{4}$ for some new $a,b$

[cloud walrus]

Ultimate Chad

[celest furnace]

Letting $p = 5/2$ and $q = 4$, we would have that $p^{1/3}- aq^{1/3} \in \mbb Q$. This says then that $(\sqrt[3]{2} - a\sqrt[3]{4})^3 = -qa^3 + 3a^2p^{1/3}q^{2/3} - 3ap^{2/3}q^{1/3} + p \in \mbb Q$. This in turn says that $ap^{1/3}q^{2/3} - p^{2/3}q^{1/3} \in \mbb Q$

[cloud walrus]

Ultimate Chad

[celest furnace]

Plugging in the values of $p,q$ would show that $a(5/2)^{1/3} 2^{4/3} - (5/2)^{2/3} 2^{2/3} = 2a \cdot 5^{1/3} - 5^{2/3} \in \mbb Q$. But this is impossible--$1, 5^{1/3}, 5^{2/3}$ form a basis for $\mbb Q(\sqrt[3]{5})$ over $\mbb Q$.

[cloud walrus]

Ultimate Chad

[celest furnace]

I leave it as an exercise to you to show that this argument actually depends on 5 and 2 being coprime. Hint: Why is b not 0? (I also switched a and b in the middle, sorry)

[south patrol]

Honestly like, my lecturer for Galois theory said this was a mess himself aha like this is not a pretty theorem

[soft wyvern]

thankss

[long geyser]

it's not just the theorem though, if you read what I typed further down the proof has a key mistake/omission

[soft wyvern]

other question, if $A\subseteq B$ then $[B:A]=1$?

[cloud walrus]

benjazul

[long geyser]

but yeah, the thm itself is messy

[south patrol]

hm i didn't read it sorry

[soft wyvern]

and then $A=B$

[cloud walrus]

benjazul

[celest furnace]

Yes that one is true

[south patrol]

What is the context

I mean you seem to be saying there can be no proper inclusions ever

[celest furnace]

B is probably a field extension of A here? In which case this is true

[delicate bloom]

see that x^3-2 has a solution in Q_5, but all valuations are integers, which means 1/3 can't appear. So in the larger field, Q(cbrt(2)) in Q_5 even doesn't have it.

[south patrol]

Chad lol

Wait I'm confused what you mean by the 1/3 etc

[delicate bloom]

v_5(cbrt(5)) = 1/3

[south patrol]

okay sure

[delicate bloom]

but v_5(x) of every x in Q_5 is an integer

yeah

[south patrol]

i thinki i am too used to like lol

|-|_p

But yes sure

[delicate bloom]

same I usually hate using the valuation direclty but here seemed more succinct

[south patrol]

Yes it is

[soft wyvern]

yes, is the notation for a field extension

[celest furnace]

usually you use F/K for a F a field extension of K

[south patrol]

Is using adics like that often useful lol

[celest furnace]

Not subset it doesn't really tell you taht much

[south patrol]

or is that more because it happens to be of the form x^p -a

[delicate bloom]

usually I'd go straight to the newton polygon

since that tells you the valuation and multiplicity of all the roots in the algebraic closure, and if the newton polygon is a single line doesn't hit any lattice points except the starting and ending, then it means it's irreducible - eisenstein's criteria is a special case of this

[south patrol]

Well I must admit idk about newton polygons

i probably know enough to learn about them now lol i should have a go soon, thank

[delicate bloom]

yeah plus it's more fun 😛 yup later lol

[south patrol]

i am free from exams soon enough 😭

[barren sierra]

how do I show B is not nilpotent for (ii)

I mean I don't even have a concise expression for [B, B] which would be the second term in the lower central series of B

[celest furnace]

I think you should be able to find one of that form and then you would find that [B^(1), B^(1)] = B^(1) or something like that

[barren sierra]

what is B^(1)?

[celest furnace]

[B,B]

[barren sierra]

oh right

[celest furnace]

This would then immediately imply that its not nilpotent right

[barren sierra]

if I could show that yes

but then a clean expression for what [B, B] eludes me

by clean expression, showing [B, B] = set of matrices such that ...

[celest furnace]

what have you done so far

it might also be all of B which would be nice

[tribal moss]

Definitely not all of B, since every commutator has determinant 1.

[celest furnace]

Ah ur right

[long geyser]

can anyone check to see if my proof is valid

by induction on deg(h). If deg(h) = 1, h = x - c for c in k forces p = x - c by irreducibility, proves the base case

now, let deg(h) = n. Since h and p share a root in some larger field, gcd(h, p) \neq 1. Since p is irreducible, gcd(h,p) = p. So, p divides h, i.e. h = pq where deg(q) = n - deg(p) < n
since every root of h is a root of p, every root of q must be a root of p and the inductive hypothesis applies to say q = p^m for some m and we are done

[celest furnace]

Spamakin you can get that ||[B,B] = N|| after you do all the calculations (or ask a calculator)

[tribal moss]

The calculations aren't all that bad if you notice that [kM,N] = [M,kN] = [M,N] for any nonzero constant k, so you can start by normalizing all the matrices to have d1=1.

[soft wyvern]

the context is: Let $F$ a field . Prove that if $[F(\alpha):F]$ is odd the $F(\alpha)=F(\alpha^2)$. Clearly $F(\alpha^2)\subseteq F(\alpha)$, but to prove the other inclusion, we use tower rule $[F(\alpha):F]=[F(\alpha):F(\alpha^2)][F(\alpha^2):F]$. As $\alpha$ is a root of $x^2-\alpha^2\in F(\alpha)[x]$, we get $[F(\alpha^2):F(\alpha)]\leq 2$, but how $[F(\alpha):F]$ is odd, so [F(\alpha^2):F(\alpha)]=1$ and the we get the equality $F(\alpha)=F(\alpha^2)$.

[celest furnace]

Ah i remember doing this exercise i found a pretty nice way to actually write alpha in terms of powers of alpha^2 and then there is just this amazing extremely short degree argument

[soft wyvern]

my doubt is the last line

when $[F(\alpha^2):F(\alpha)]=1$ the concludes the equality

[cloud walrus]

benjazul

[soft wyvern]

if the degree extension is equal to 1, i get the equality?

[celest furnace]

Yes. In general if W,V are vector spaces over F with W subset V, and they have the same dimension then they are equal

[long geyser]

did you mean to write [F(alpha) : F(alpha^2)]

[dull ginkgo]

ooo i kinda like this

Can’t have any “left over” chunks :3

[celest furnace]

Yeah ur right

[cloud walrus]

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

[soft wyvern]

yes

[long geyser]

at least in finite case degree 1 forces equal cardinality forces equality

[soft wyvern]

ahh ok

[long geyser]

(idk general case, but your case is finite so)

[soft wyvern]

thank you

[long geyser]

[dull ginkgo]

I am too lazy to use tower law, it uses degree and remainder theorem anyway (trollface)

[long geyser]

I'm confused abt this, when a question has two different parts almost always part 1 is used to prove part 2

but I think u can prove part 2 directly by the homomorphism k[x] -> k[x]/(f) x k[x]/(g), then use first iso thm

is there another way to prove part 2 using the stuff in 1?

[long geyser]

oh wait my bad about this, I don't know if you are working with finite fields @soft wyvern only that ur extensions are finite

this won't work, I'm guessing there's another way to show that

[dull ginkgo]

NOOOO DON’T BASTARDIZE MY BABY

[long geyser]

let K/k be a field extension and [K:k] = 1, so K = span_k(alpha) for some alpha \in K, notably you can write 1 = alpha * c for c \in k, so alpha ^-1 is in k so alpha is in k so k = K

works?

[dull ginkgo]

Wait which part are you trying to show

[long geyser]

uh, I was asking if there is a proof of (ii) involving what was shown in (i) instead of using first iso

(I'm hoping my first iso proof is correct)

[dull ginkgo]

idk how to do this at first glance without calling to classic CRT

Or just verbatim repeating its proof

[barren sierra]

ah yea I messed my algebra from earlier whoops

then I got the rest, thanks

[barren sierra]

I have tried many algebraic manipulations of this and I'm not sure how to show this >_>

[barren sierra]

Let $n_1 g \in N_1 H$ and $n_2 h \in N_2H$. Then I want to show $n_1 g n_2 h g^{-1} n_1^{-1} \in N_2 H$. I started with a right multiplication of $h^-1 h$ just to get an element of $H$ on the right hand side but manipulating $n_1 g n_2 h g^{-1} n_1^{-1} h^{-1}$ to be in $N_2$ hasn't born any fruit

[cloud walrus]

Spamakin🎷

[cobalt heath]

It is enough to show that both N1 and H normalizes N2H, do you see this? @barren sierra

[barren sierra]

ah yea that greatly simplifies it

thanks

[cobalt heath]

No problem

[crystal vale]

Okay, thank you

[crystal vale]

Should I go to Atiyah Commutative Algebra, I haven't covered module theory

[summer path]

a&m has stuff on modules so it's prob fine

[next obsidian]

Hi bubu :3

[summer path]

hi chuchu :3

[crystal vale]

If R is ring Z × Z then R-module R has a generator (1,1) ?

Because for any (a,b) we have (a,b)•(1,1) = (a,b)

[cobalt heath]

Yep

[dire siren]

I think you actually need part (i) in your proof for finding the image of that homomorphism. It is not that obvious without the theorem why it is surjective.

You can also consider the homomorphism h (mod fg) |--> (h mod f, h mod g) and conclude that it is an isomoprhism using part (i)

[winged void]

is there an easy way to visualize that ther is 24 rotational symmetry

of a cube

[rocky cloak]

There is a visualization where you look at how rotations of a cube permute the 4 diagonals.

Otherwise, what I find quite simple is that you have 6 faces, each of which could be on top, then you can rotate that face in one of 4 different ways. So 6*4 = 24 in total.

[winged void]

that is true i agree on that

but is there a way to determine it per axis

so for example the elements due to rotation on x-axis

is 4

because sometimes i need to use burnsides formule

[rocky cloak]

Sure, you rotate a cube either about a face or about a diagonal or an edge

So either about the x, y or z-axis or about one of the 4 diagonals, or about a pair of opposing edges. Excluding the identity, you have 3 rotations about each face and 2 rotations about each diagonal and 1 rotation about each pair of opposing edges. That makes 3*3 + 2*4 + 6 = 23. And the identity makes it 24

[winged void]

sure i agree on that

the first three are easy to see

but how did you see that you can rotate about a pair of opposing edges

this is for me so difficult to be seen

i used geogebra to explore it

[rocky cloak]

Well, if you just hold a cube in your hand it's obvious

[winged void]

lets say this is a cube in my hand

because for me its so not obviious to be fair

[rocky cloak]

Well, if you just rotate it 180, you see that it gets back to the right position

[winged void]

that is true but that is so difficult for me to see to be fair

[rocky cloak]

Like imagine the plane going through both edges. Obviously the cube is symmetric in that plane. So if you just turn it around, that's a symmetry

[winged void]

that is true

[rocky cloak]

But anyway, I would always recommend just picking up an actual cube

[winged void]

i did that

next week is my exam

and the professor recommend that everyone to bring a cube

):

i drew a plane right now but how are we going to rotater around the plane

now

[cobalt heath]

For the line connecting midpoint, maybe try "reflection across the line"

[rocky cloak]

Sure, you could also rotate in the plane. That would just give you a rotation about the edge facing you

[hidden wind]

based

[hidden wind]

if you fix the midpoint of this plane segment and draw a perpendicular line through it, that’ll be the axis of rotation, and the two points it intersects on the edges of the cube will be the fixed points of the rotation (though this rotation has no fixed vertices)

[molten viper]

Back to doing math, trying to work on learning fields + galois better. Does this argument look alright?

I've got a finite field F of characteristic p, and I want to prove that |F| = p^n for some n. So far I know that F_p (the finite field of p elements) is a subfield of F, so I can find the degree [F : F_p], which I call n, and I say the basis is {1, e_1, ..., e_n-1}. Then, every element in F can be written uniquely as a_0 + a_1e_1 + ... + a_n-1 e_n-1. Each a_i has p choices since they're in F_p. So we have p choices for n coefficients, so p^n total elements.

My plan is to just do like, all the exercises in the field + galois chapters of Dummit and Foote

[hasty sentinel]

is the trivial group considered to be a sylow p-subgroup for all other primes? as in |{e}| = 1 = 5^0 is a 3-subgroup of |S3|=2*3?

[coral spindle]

Yes

Wait hold on

you mean a 5-subgroup

not a 3-subgroup.

But indeed, it is a p-group with p=5 and it is a maximal such subgroup, so by definition it is the Sylow 5-subgroup of S_3.

[hasty sentinel]

in that example it could not by only a 2 or 3-subgroup as it would have to have order 2 or 3, right?

[coral spindle]

It is a 2-subgroup and a 3-subgroup, but it is not Sylow.

Because it is not maximal.

[hasty sentinel]

oh yea

not sylow

but still a p-subgroup

[coral spindle]

That's right

The Sylow 2- and 3-subgroups would indeed have to have orders 2 and 3 respectively

[hasty sentinel]

danke!

[coral spindle]

bitte

[crystal vale]

Let G be a finite group and | G |>=2. Then G has an element of order p.

It can be done by Cauchy theorem but they haven't introduced the subgroup yet.

So if we use induction then let this be true for n<| G|.

Let n = p_1....p_k be the order of G.
Then, let a in G if a has order | G | then G is cyclic and it has an element of order p_1.

Let a have order < | G |. Then by induction hypothesis there is an element of prime order in (a).

Is it correct? But how can I use it here (a) as a subgroup of G because they haven't introduced subgroups yet

[long geyser]

ohh, you're right

I need to find, given b + (f) and c + (g), d such that d is b mod f and c mod g

and that is exactly what (i) proves

[rocky cloak]

This seems somewhat overcomplicated. Like you just take a of order n, and take p to be prime dividing n. Then a^(n/p) has order p. That's it

[crystal vale]

Yes, this one is better thank you

[chilly ocean]

The group of rigid motions of two solids are isomorphic if they are duals of each other, now if the group of rigid motions of two solids are isomorphic and they are not the same solid does it mean that they necessarily have to be duals ?

I think not , but i couldn't come up with a counterexample

[rocky cloak]

I mean you can cook up somewhat silly things like a rectangular prism and a slightly taller rectangular prism. If that counts

[chilly ocean]

Yeah that dosent

[rocky cloak]

You can also cook up lots of solids where the symmetry group is trivial.

[chilly ocean]

Hmm okay

This one's true ig

Surely this works

Like something highly asymetric

[rocky cloak]

Like a hand and a cube with a different shaped spike on each face

[chilly ocean]

Okay cool thanks a lot

[crystal vale]

Let G be a group and H be a unique subgroup of given order. Then H is a characteristic subgroup of G.

If f is an automorphism of G then f(H) is a subgroup of order | H | but H is a unique subgroup of a given order. Thus H = f(H).

Is it correct? But I think it implies that H is a normal so normal Subgroup is a characteristic subgroup, right. Is there any other way to prove it?

[coral spindle]

The first thing you wrote is correct.

But I think it implies that H is a normal so normal Subgroup is a characteristic subgroup
No.

[crystal vale]

For a finite normal subgroup?

[coral spindle]

The definition implies that every characteristic subgroup is normal, not the other way around.

[sonic coral]

normal subgroups may not pass the “the” test

[tribal moss]

Let G be your favorite nontrivial group and consider the direct product G×G. The subgroups {1}×G and G×{1} are normal (they are the kernels of the projection homomorphisms), but they're also related by an automorphism, so they're not characteristic.

[crystal vale]

Yes they are normal but what is meant by they are related by an automorphism, does I need to find the automorphism such that they are not invariant under it?

[tribal moss]

f(g,h) = (h,g) is an automorphism of G×G

It takes {1}×G to G×{1} and vice versa.

(Normal subgroups are invariant under inner automorphisms; characteristic subgroups are invariant under all automorphisms).

[warped shadow]

when we characterize the structure of so(n) the standard proof seems to be to take the constraint R^TR = I and differentiate both sides at zero which yields R = -R^T

however SO(n) has another constraint det[R] = 1

never mind

I figured it out

but maybe as a follow up question

if I decide not to differentiate at zero

I would have

$$(\dot{R}(t))^TR(t) + (R(t))^T\dot{R}(t) = 0 \implies \dot{R}(t) = -R(t)(\dot{R}(t))^TR(t)$$

[cloud walrus]

criver

[warped shadow]

which is still workable but the determinant condition looks pretty gnarly

$$Tr(R(t)^T\dot{R}(t)) = 0$$

[cloud walrus]

criver

[warped shadow]

any idea how someone would deal with the latter?

or is it automatically satisfied, similar to the case when I differentiate at zero

[rocky cloak]

SO(n) is a connected component of O(n), so any path that goes through the identity in O(n) is contained in SO(n)

So the condition is automatic

[warped shadow]

so this implies the trace condition below? Let me try so X^T R = -R^T X -> Tr(X^TR) = - Tr(R^TX) -> Tr(R^TX) = 0, ok indeed

[cloud walrus]

Justin

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[chilly radish]

To explicitly compute the cocycle, there's a couple of different ways. Firstly, you can compute them directly from the extension by choosing representatives (I can explain if you don't know how to do that). I'm not sure about a more general technique

You can try and find constraints on the cocycle using the cocycle condition, and the fact that in this case the group is Z/3, and in general m-torsion for Z/m (It might actually be Z/m, i'm not sure)

[rocky cloak]

In general if you have an extension
1 -> A -> G -> Q -> 1

If you let s: Q -> G be a section (just as a function not a homomorphism) then
f(p, q) = s(p) s(q) s(pq)^-1
should define a cocycle Q^2 -> A

[rocky cloak]

So for example if you just choose s(p) = p where you think of p as an integer between 0 and m, then f(p, q) would be the carry bit of p+q. Don't know if there's a particularly nice polynomial expression for that though

In the example you gave, it seems s(1) = 1, s(2) = 2, s(0) = 3 will recover your chosen cocycle. Don't know if that helps

[cloud walrus]

Justin

[chilly radish]

Why would it be split necessarily? The cocycle is defined exactly the way it ought to be. Like if you define the crossed product group (idk what you'd call it, basically multiplication is given by the cocycle), with underlying set AxQ with A written additively and Q multiplicatively, then you have

(a,b)(c,d)= (a+b(c) + f(b,d),bd)

Where f is a normalised cocycle. Now consider elements of the form (0,q), then
(0,q)(0,q')=(f(q,q'),qq')
Now if we multiply by (0,qq')^-1=(0,q'^-1q^-1) (Note that here we used the fact the cocycle is normalised, otherwise we can't properly define the inverse, and Ax{1} is not a subgroup, but every cocycle is Cohomologous to a normalised one)

we get

(0,q)(0,q')(0,qq')^-1= (f(q,q'),qq')(0,q'^-1q^-1)=(f(q,q'),1)

(Again we used normalisation). So we see that f(q,q') is equal to s(q)s(q')s(qq')^-1 where s(q)=(0,q)

To summarise I showed that the section construction allows you to construct any (normalised) cocycle by considering its 'crossed product' group, so that it needn't be split

i.e. we can get arbitrary cocycles by the same construction

s is not a 1-cochain

It's a map from Q to G, a 1-cochain goes from Q to A

That's the difference

I think that's why you're confused

Lmao no worries

Yea, since we are assuming this is an extension given by the cocycle, the map A->G given by a->(a,1) is injectivr

So it's uniquely identifiable with an element of A

Good luck

Happy to help

[limpid ferry]

Given a a\in G, G being a group, how should I prove that |a| = |a^-1|? Please help

[delicate orchid]

(a^-1)^n = (a^n)^-1

[frank cosmos]

i dont understand this part

where do we use that k[S]/m is a field

what is the root of f^sigma

how do you take the extension of sigma(k) with (possibly uncountably) many elements?

[cloud walrus]

Hello1

[frank cosmos]

is this just the intersection of all subfields of k[S]/m containing these roots?

[late marsh]

should I ask for help here or use a help channel? wasn't sure if this was for help or just for general discussion

for my ring theory hw

nvm I just made a dumb mistake

[ashen heron]

[crystal vale]

Got it, thank you

Let K be the characteristic subgroup of H and H is a normal in G. Then K is normal in G.

For any g in G, let the automorphism of H x->gxg^(-1), and K is a characteristic subgroup of G.

Thus gHg^(-1) = H implies that K is normal in G.

Is it correct?

[crystal vale]

Are two subgroups of the same order conjugates ?

[coral spindle]

No

You just saw an example

Infinitely many examples, in fact

[late marsh]

didn't want to if it was the wrong channel 🙃

[crystal vale]

Okay, thank you

[crystal vale]

Let G be a group and H be a subgroup of order n. Then intersection of all subgroups of order n is a normal subgroup.

Let M be the set of all subgroups of order n.

If m in M, that means m in all Subgroups of order n. Let m in K then m also in g^(-1)Kg it implies that gmg^(-1) in K. Hence, gmg^(-1) in M.

Is it correct ?

[daring nova]

You just proved K is normal, which is wrong. So that is very poorly phrased, as your initial claim "m is also in g^-1 K g" is wrong.
It's more like
let m in M, K a group of order n.
Then m in g^-1 K g so g m g^-1 is in K. Hence g m g^-1 is also in M

[crystal vale]

My mistake I didn't mention that K is a subgroup of order n.

[daring nova]

no that is obvious

[daring nova]

the problem is your proof doesn't work
"Let m in K then m also in g^(-1)Kg" is basically stating K is normal

[crystal vale]

But I assumed m in M then m in K. Yes that's my mistake I didn't write correctly

Not for all m in K

[crude flame]

How can I find the number of homomorphisms from symmetric group $S_n$ to itself . I have seen a method for $n \not =4,6$ , $n!+1+\text{number of elements of order two in } S_n$ . Is that correct ?

[cloud walrus]

Bilal ns

[delicate orchid]

Any endomorphism of S_n must have kernel equal to S_n, A_n, or 1 (for large enough n), If the kernel is 1 then this is just Aut(S_n) = S_n (again, for large enough n), and the morphisms with kernels A_n is just choosing an element of order 2, so yes that seems to be correct

For n = 6 you add in the exceptional outer automorphism, and for n = 4 you have kernels that can be isomorphic to the klein 4 group of elements of the form (xy)(ab)

[south patrol]

how do you see that Aut(S_n)=S_n (at least for n >> 1)?

Okay it seems a little involved

[hidden wind]

an isomorphism between two permutation groups is just a bij

[south patrol]

I don't really know what you mean

it also has to be a homomorphism

[delicate orchid]

They were just asking if their answer is correct not my problem!

[hidden wind]

am i having a silly moment

i can’t imagine a bij Sn -> Sn which is not a hom

but i am also currently just sitting outside in the sun eating cake, listening to music and am not thinking very hard

[delicate orchid]

Fix everything and send (12) to (123)

Or send the identity to anything that isn’t the identity

There are (usually) n! Automorphisms of S_n not (n!)!

[rotund aurora]

Can I get a hint for showing that if A is a finite k-algebra (meaning finitely generated as a k-module), where k is a field, then Spec A has only finitely many elements?

[hidden wind]

and (123) to (12) ?

[hidden wind]

oh

[next obsidian]

Apply Noether Normalization

Or errrr

I guess this is sort of not useful actually

Okay, show that a finite k-algebra which is an integral domain is a field

[rotund aurora]

yes, I know this

[next obsidian]

Okay so minimal primes are maximal

And then…

[median pawn]

Can we find a positive number in 4Z + 1 with two distinct factorizations in 4Z + 1? That is, can we get non-negative integers x,y,a,b such that (x,y) is not equal to (a,b) and (b,a), but (4x+1)(4y+1) = (4a+1)(4b+1)?

[rotund aurora]

is A necessarily noetherian?

[next obsidian]

Yes

[rotund aurora]

this is just because ideals are k-vector subspaces right, and the dimension of A over k is finite

[next obsidian]

There’s many ways to prove this

That’s an insane way to prove it but sure

[rotund aurora]

wtdym insane

[next obsidian]

You only need k to be Noetherian

You can say

Apply Hilbert basis theorem

[rotund aurora]

well right

[delicate orchid]

This would contradict Z being a UFD, so no

[next obsidian]

Or say that A is a quotient of k^n

Which is thus Noetherian

[rotund aurora]

ok so Spec A is noetherian and thus has finitely many irreducible components, by a general topology fact. No?

[next obsidian]

Yeh

Or well

Sure

You should know how to prove this

[rocky cloak]

Notice that A is artinian, hence the product of finitely many local artinian rings

[next obsidian]

But yes as a result finitely many minimal primes

[next obsidian]

Hurb

[lusty marlin]

5×5=1×25

[next obsidian]

Part of proving that is showing an artin ring has finitely many primes lol

[median pawn]

makes sense

thanks!

[next obsidian]

Those are the same mod 4 tho

[median pawn]

but what about the UFD comment hmm

[next obsidian]

[lusty marlin]

And you can generalise it too, obviously

[rocky cloak]

Not the way I would prove it.

[next obsidian]

Oh this is literally 4Z + 1

[next obsidian]

How tf can you prove it to ithout that lol

[lusty marlin]

Doesn't matter though?

[median pawn]

yes quite literally

but also seems like trial and error

just looking at 4Z + 1, idk if this can be done or not

like is there a more fundamental reason this works?

[next obsidian]

But like

[median pawn]

and doesn't work for some mZ + n?

[next obsidian]

4Z + 1 isn’t a ring yeah?

Idk what you’re doing tbh haha

[median pawn]

correct, also we're only considering non-negative integers

[median pawn]

Time to make it rigorous. Let $S := {4k+1: k\in \Bbb Z_{\geqslant 0}}$. Show that $S$ has an element that can be factorized in two different ways, in $S$.

[cloud walrus]

hausdorff

[rocky cloak]

Consider a maximal ideal m, then consider a maximal ideal n with m\cap n strictly included in m. Continue inductively and get a decreasing sequence. By artinianness the radical is the intersection of finitely many maximal ideals. So by Chinese remainder theorem A modulo radical is a product of finitely many fields. By lifting idempotents modulo nilpotent ideals you get that A is a product of local rings.

[next obsidian]

Hmmmmmmmmmmmmmmmm

[south patrol]

Oh lol

[next obsidian]

I guess this combines a few steps I’d do

[coral shale]

number theory???

[south patrol]

I would prove it has finitely many primes differently too

Just like p_1 > p_1 p_2 > ... would be descending

[rocky cloak]

Notice the proof never uses that there are finitely many maximal ideals. You can do a similar proof for noncommutative rings, where you actually have infinitely many maximal ideals

[lusty marlin]

This boils down to the question of finding (a,b) and (x,y) that are not permutations of each other such that m(ab-xy)=n(a+b-x-y)

Not sure if this makes it easier to find a pattern though

[next obsidian]

Sorry I don’t see how you show the nilradical is intersection of maximal ideals

I see how you conclude that for the Jacobson radical

[south patrol]

Real

[lusty marlin]

For mℤ+n

[rocky cloak]

For artinian rings Jacobson = nilradical by Nakayamas lemma

[next obsidian]

Ummmm

How

Lmfao

[south patrol]

Can't you just show every prime is maximal anyway

Or do you want to avoid that

[next obsidian]

True this isn’t hard

[south patrol]

Tbf if you have that

[south patrol]

You may as well do this

[rocky cloak]

I'm actually not sure of a proof where it would help knowing there are finitely many primes apriori

[next obsidian]

You show that Artinian => Noetherian + dim 0

And then do CRT on the fact that J(R)^n = 0 for some n

So like to show it implies Noetherian you do what potato was saying

You can show that Artin rings have finitely many maximal ideals

And then you do this insane composition series by multiplying one prime and each quotient is a fg VS over a field

And then you conclude the ring is finite length

[rocky cloak]

Once you do CRT, then you already have that it's the product of local rings right. Like you don't need that there are finitely many primes

[rotund aurora]

Let A be a noetherian ring. For any proper ideal I there are only finitely many minimal primes over I. Suppose this fails for some ideal I. Let I be maximal with the property that there are infinitely many minimal primes above I. Without loss of generality, suppose I=(0) (consider A/I). A is not an integral domain, say xy=0 with (x), (y) proper ideals. Then there are finitely many minimal primes above (x) or (y), but minimal primes of A are minimal over (x) or (y)

[next obsidian]

You need to know that there’s finitely many so you can ensure that you can get down to 0, or to know that it’s a finite product. Admittedly Artinianness can also get you this, but then you’re basically proving both things at once

[rocky cloak]

I mean you already have artinianess, seems like a detour to not use that the ring of artinian at that point.

[next obsidian]

I mean the proof of showing finiteness does use that assumption

But you can prove that but separately and then later invoke it

[chilly ocean]

hello, I wanted to ask if anyone could explain how (1) $S_n$ is generated by (i), (ii) and (iii), and (2) the proofs of these statements. I just picked up abstract algebra after a while of not doing any math, so I'm revising, and I'm really rusty on permutation groups and such. would dearly appreciate some guidance here. thank you!

[cloud walrus]

45

[chilly ocean]

a thought just appeared - since the order of the permutation is equal to the lcm of the lengths of the permutations, doesn't that mean the order must be $lcm(2, 2, \hdots, 2)$, $n$ times? But the order of $S_n$ is $n!$ I'm a bit confused

so the order of (i), (ii) and (iii) will divide n!, right?

[cloud walrus]

45

[ashen heron]

S_n itself is not a permutation?

[chilly ocean]

it's the group of permutations yeah

[lusty marlin]

The proof is written here. What is the issue?

[lusty marlin]

When written as a product of disjoint cycles, the order of a permutation is the lcm of the orders of the cycles

[chilly ocean]

I don't understand how $(a, b) = (1, a)(1, b)(1, a)$ and also why that's relevant to proving (i)

[cloud walrus]

45

[chilly ocean]

yes, that's what I meant, sorry

[lusty marlin]

If that is what you meant, then what is the issue?

The cycles are not necessarily disjoint here

[chilly ocean]

oh, right, since they're not disjoint

[chilly ocean]

still uncler about this, though...

[lusty marlin]

It follows from the basic definitions. What do you understand by (a,b)?

[chilly ocean]

well I understand that $a \mapsto b$ and $b \mapsto a$

[cloud walrus]

45

[lusty marlin]

Good

Then you can compose the permutations on the right and check that you obtain (a,b)

[chilly ocean]

oh yeah

[lusty marlin]

1 maps to 1, a maps to a, and b maps to b

[chilly ocean]

let me draw it out, i'm more used to composition with two-line notation

[lusty marlin]

Now all you need is to understand why S_n is generated by transpositions, and then you're done

Because all statements follow then

[chilly ocean]

because all permutations can be expressed as products of transpositions?

[lusty marlin]

How? Can you prove it?

[chilly ocean]

well yes

$(x_1, x_2, \hdots, x_n) = (x_1, x_2) (x_2, x_3) \hdots, (x_{n-1}, x_n)$, no?

[cloud walrus]

45

[lusty marlin]

Yup

[chilly ocean]

thank you

[viral mountain]

Is $2x$ provably equal to $x + x$ in a field?

[cloud walrus]

skeptician

[viral mountain]

nevermind. just realised that $2 \in \bR$, but $2$ may not necessarily be a member of the field under question.

[cloud walrus]

skeptician

[lime badge]

If "2" here denotes 1_F + 1_F in your field, then its fine (1_F is the identity in the field)

[viral mountain]

right

so $2x = x + x$ would be provably equal if $\bF = \bR$, correct?

[cloud walrus]

skeptician

[lime badge]

Sure, but something like this holds in every field that's not char 2

[viral mountain]

right right

[lime badge]

I'm assuming x here is just some element of your field?

[viral mountain]

yes

so in general, 2_F x = x + x is provably equal in any field

where 2_F = 1_F + 1_F

[lime badge]

Unless its characteristic 2

[viral mountain]

what's characteristic 2?

[lime badge]

a field where 1_F + 1_F = 0

[viral mountain]

ohh

got it

but even in that field, we could write 2x to represent x + x for any x in F, right?

where 2 is not necessarily 1_F + 1_F, just a notation of performing addition twice

[lime badge]

not really sure what you mean here since x + x = 0 in a char 2 field

x+ x = 1_F * x + 1_F * x = (1_F + 1_F) x = 0*x = 0

[delicate bloom]

as long as you're fine with writing 2=0 you're good

[viral mountain]

No I just mean that the notation $2x$ is defined to mean $x + x$ in any field right?

[cloud walrus]

skeptician

[viral mountain]

where the 2 is just a symbol

[delicate bloom]

that's what they proved here

[viral mountain]

oh yeah this makes sense then

[delicate bloom]

2x = (1+1)x=x+x if you've defined 2=1+1 then you're just using the distributive law

[viral mountain]

thanks folks!

[summer path]

waw mero back to being yellow

[lime badge]

This is somewhat related to my other question in the advanced algebra channel, but given a finitely presented subgroup S of a finitely presented group G, is there a way of finding out if some g in G is in S?

[barren sierra]

This is a generalization of the word problem in a group iirc

Which is computationally hard in general (in fact undecidable in general)

But if you look up "subgroup membership problem" you'll find the literature and such

[lime badge]

I'm using something like MAGMA to determine homomorphisms between two finitely presented groups. I suppose I can search for homomorphisms between the group generated by g (which in my case ends up being cyclic) and S and see if any correspond to the inclusion map right?

[barren sierra]

Maybe? I just don't know how you'd search for homomorphisms

[lime badge]

The software does that by finding mappings between the generators. But regardless, I just wanted to make sure I wasn't overlooking some slick group theoretic technique

[next obsidian]

Mellow

Yero

[summer path]

3 hours of eep + math =

[steel light]

How do you pick which exercises to do?

[sonic coral]

ones that look interesting or have some sense of generality

[steel light]

Same

[steel light]

Does this constitute a proof? I haven't done this in so long xd I didn't really justify my steps in between because it looked ugly formatting wise

But I think it's all justified

Oh

I guess maybe not that 0*(-1) = 0

[south patrol]

This should be an axiom

But yes this looks good

[steel light]

I'm referencing D&F, skipped past all teh goop theory stuff, these are the ring axioms given

[south patrol]

Ig 0*r then follows from distributivity

[steel light]

Yeah that's what I was thinking too

[south patrol]

(0+0).r

[steel light]

0a = (0+0)a = 0a + 0a
0 = 0a + 0a + (-0a)
0 = 0a

Onto the next :D

zz do I have to prove that (-1)u = u(-1) for this or can I just take it for granted

It shouldn't be hard to prove anyways but I'm ✨ lazy ✨

I guess it's already sort of done in the book so I'll just take it and move on

[tardy hedge]

This comment of Z-modules being the same as abelian groups

I'm a bit confused on the significance of it

[fading field]

it’s nice to be able to interpret things as different things

[tardy hedge]

Any module is itself an abelian group to begin with, but its saying that a Z-module, i.e, the abelian group with a ring action from Z acting on it, is still just an abelian group?

[fading field]

when you have a Z-module A the action of Z on A becomes explicit

[tardy hedge]

Also why when u have the action of Z only be na = a+a+a + ... , why is that the only possible action?

This whole thing is relying on the fact that based on module axioms the only action from Z we are allowed is that basic one right

[fading field]

it’s just stating that there is a natural correspondence between them, not that the only action of Z on any given module is repeated addition

[tardy hedge]

Ohh ok

but if u just say "Z-module" then its not explicit what the action is right

[fading field]

it’s sort of clear, also Z is initial in Ring, and since A is an abelian group we have that End(A) is a ring (iirc)

[tardy hedge]

Thats above my level rn

[rustic crown]

well you have two axioms for an action of a ring on an abelian group.

• 1*m = m
• (a+b)*m = am + bm

use these for a = b = 1

[steel light]

det

haven't seen you in a while

[rustic crown]

hewwo

[fading field]

well i just meant to say that it should end up being clear what the action is, n(a) = (1+..+1)a

=a+…+a

[tardy hedge]

any action of ring on abelian group needs this

?

[rustic crown]

yeep

[tardy hedge]

i only have looked at group actions so far so im not really sure the details for rings

Ok

[fading field]

group actions and ring actions are everything!

[tardy hedge]

yeah

THey are cool

In a vector space, why does av=0 imply either a = 0 or v = 0

because I saw the example that if ur talking about modules then this isnt the case

[south patrol]

if a is non-zero, then v = (1/a)(av) = 0

this fails for modules as a needn't be invertible if it's non-zero unless we're over a field (or division algebra)

[tardy hedge]

thx. and 0v= 0 because (0+0)v = 0v + 0v = 0v so 0v = 0 thingy

[tardy hedge]

I can see that if u define it to be na = a+a+a ... then those axioms are satisfied, but why do those axioms force it to be na=a+a+a ... ?

[rustic crown]

it true by induction

2a=(1+1)a is forced to be 1a+1a=a+a

then 3a=(2+1)a is forced to be 2a+a = (a+a)+a

then you just proved 0a = 0

and so 0 = 0a = (-n + n)a = (-n)a + na

so (-n)a is forced to be -(na)

[tardy hedge]

ohh right duh

Thanks

[tardy hedge]

So.. what about what this person said?

Arent u telling me that the only action Z has on abelian group is this one?

[slim kayak]

The trivial action is still an action, so there's that one as well

[tardy hedge]

oh yeah ok cool

trivial meaning just make na= 0 ?

[slim kayak]

na=a

[tardy hedge]

oh cause we want 1m=m

[slim kayak]

But for a Z-module you have stuff like (n+m)a=na+ma

[tardy hedge]

Yeah

[slim kayak]

which together with 1m=m forces it to be repeated addition basically

[tardy hedge]

I know group actions are kind of like another way of just seeing homomorphisms of the group

are ring actions similar? Im a bit confused on them

[slim kayak]

wdym?

[tardy hedge]

a group acting on S is the same as a homomorphism G -> permutation group of S

and vice versa

[barren sierra]

Is there a way to show Z[x] / (x^2 + 1) is a PID without resorting to the fact that Z[x] / (x^2 + 1) is a Euclidean domain?

[vapid vale]

what are you confused about?

[south patrol]

Well you could do it by some algebraic number theory lol

[tardy hedge]

a ring action on some set S is just a map RxS - > S that satisfies... what?

[south patrol]

But the method I know is way more complicated than just writing down the euclidean function

[south patrol]

Usually rings act on abelian groups

and then the axioms are what you would hope

[tardy hedge]

Why? Is it just cause thats whats most useful

[south patrol]

Well like

You have two operations (multiplication and addition)

And like idk how to make them interact unless you have extra structure on what you have

like obviously you want r.(s.x) = (rs).x and same for addition but like how can those interact for a set

For a group you can demand stuff like r.(a+b) = r.a + r.b and (r+s).a = r.s + s.a which is more interesting

[tardy hedge]

(r+s)+x = r+(s+x) is the addition one right

ok

[south patrol]

sure i mean that is associativity of addition

[south patrol]

Idk if a better way exists

i feel like often writing down a euclidean function is the best way to show something is a PID though

for what it's worth my tongue-in-cheek way was to show the ideal class group is trivial

and that follows immediately from the Minkowski bound lol

But then that is proven by like counting lattice points and stuff

which is similar in flavour to just, well, using the euclidean function in this case lol

[vapid vale]

well be careful because you only have one operation from R x S -> S; often this is just notated as multiplication or \cdot since right now youre using the same symbol for addition within the ring and the ring acting on S

[tardy hedge]

yeah

same thing can be said for r(sx)=(rs)x tho right

sx is ring action but rs is ring multiplication

[vapid vale]

yeah but the statement you made above specifically is true for ring multiplication (not addition); we want (r+s)x = rx + sx

[tardy hedge]

thats for a ring action on a group right

we were talking about what if u just had a ring acting on a set with no more structure

[vapid vale]

ah

[tardy hedge]

Yeah I guess I was wondering why do we usually care about rings acting on groups

[vapid vale]

yeah i mean the issue is that you can cleanly define group actions on a set S because with one group operation, we can imbue good structure on composition of elements in End(S), but i dont really know how to do it if you have two operations where you want structure to remain

i would not be surprised if theres a notion for it but i have not seen it

[steel light]

I'm still so confused by this problem lol

I don't understand how the center can contain the identity but the whole ring might not have one

If there's a 1 in Z(R) then that means 1r = r1 = r for all r in R, but 1 in Z(R) means 1 in R too no? So there exists a 1 in R such that 1r = r1 = r

[slim kayak]

"the identity" is usually used in rings with a 1, are you sure that in that section the definition of rings hasnt been slightly shifted?

[steel light]

Yes, as far as I can tell D&F do not require rings to have identity or be commutative, and from what I've seen in other problems/definitions they specify when a ring does have identity

Here are the ring axioms they are using

2 and 3 are clearly optional right?

[south patrol]

for D&F yes

[slim kayak]

also, an element a can commute with all r and also be the identity for all other elements b but still not be the identity in certain situations

[south patrol]

subrings should contain 0 by definition though

[steel light]

Yeah so how do I prove that the center has an identity without assuming beforehand that R has an identity

[slim kayak]

1-a would then just be the zero divisor of a bunch of stuff

[steel light]

Yeah I proved that is has 0

[south patrol]

i would just assume R has an identity here

[steel light]

Can you give an example?

[south patrol]

since they usually do in ring theory anyway

[slim kayak]

sure, Z x Z and consider (0,1), it acts like the identity on {0} x Z

[south patrol]

indeed whenever you have an idempotent you can play this game

[slim kayak]

something something spec is disjoint union is vaguely floating around in my subconscious rn

at least something related to said game

[steel light]

I think I understand

[dull ginkgo]

It has an identity if R does but not necessarily otherwise

[steel light]

So if R doesn't have an identity then the center might not either, but it can?

[steel light]

but doesn't that contradict this

I think I got the first part down at least

[slim kayak]

rings without identity are an abominations, but regardless consider something like non square matrices

They do not have an identity element ( IA=AI=I just doesnt work for non-square matrices), but if you consider the subring of m x m matrices (assuming m < n) you suddenly get an identity

[steel light]

Great example

So even though a ring might not have an identity, its subrings can, but if we're talking about the center specifically then the center having an identity <=> the ring having an identity

Is that understanding correct?

[south patrol]

yeah

[slim kayak]

you remember what it was? please share 😭

[south patrol]

TFAE for a commutative ring A:

1. A has a non-trivial idempotent e
2. A = B x C for rings B,C
3. Spec A is disconnected

1 <=> 2 is p easy, 2 => 3 as Spec(B x C) = Spec B u Spec C and uhhh i always forget the other way lol

[next obsidian]

I feel like 3 is trash

[south patrol]

oh yeah it's okay

[next obsidian]

Like you should outright say how it’s disconnected

[south patrol]

agreed

[next obsidian]

In my opinion

[south patrol]

though

imo it's not obvious that being disconnected implies it decomposes as a sum of two spectra

I don't think there is a nice way to phrase this lol

i guess you can say moreover in this situation Spec A = Spec B u Spec C

etc

[next obsidian]

If you know something something sheaf

[south patrol]

But yeah the proofs are all important here

[next obsidian]

Then it’s trivial to get an idempotent

[south patrol]

heh

[next obsidian]

By gluing 0 and 1

🗿

[cobalt heath]

If we put together disconnected components, Is it really gluing?

[slim kayak]

As a snack, then its still part of the process

[south patrol]

Well if Spec A is disconnected then there are radical I,J such that V(I) and V(J) partition Spec A, then V(I+J) is empty so 1 = e + f and moreover V(IJ) = Spec A so IJ is nilpotent, so say (ef)^n = 0. but note (e^n,f^n) = A so we can just replace e by e^n etc to assume ef = 0 and 1 = e + f and then it's clear these are our desired orthogonal idempotents

actually dumb q but what is the easiest way to see that if 1 = (e,f) then 1 = (e^n, f^n) lol, i guess easiest to me is that otherwise (e^n,f^n) is contained in a prime p and then p contains e and f contradiction

i guess another is that radical of (e^n, f^n) contains (e,f)=A, and then a "power of 1" is in (e^n,f^n)

[cobalt heath]

I thonk this is nontrivial

[south patrol]

eh i mean ig it is trivial given i have just supplied two short arguments lol

[cobalt heath]

Hmmm, fair

[south patrol]

but yeah less trivial than expected perhaps

like idk how to just write down 1 as an A-linear combination of e^n and f^n

[slim kayak]

If e^n and f^n dont generate 1, then shouldnt they both lie in a maximal/prime ideal?

So by contradiction there must a linear combination of them producing 1. Idk about a constructive result either

nvm cant read apparently

[cobalt heath]

For e + f, it is relatively easy

(e + f)^(2n-1) has monomials where each is divisible by either e^n or f^n

[slim kayak]

oh right, we still have ef=0 to work with

[cobalt heath]

I mean we don't need ef = 0

[slim kayak]

yeah, silly me

[delicate bloom]

yeah, every element in Z/2Z is the same up to adding a multiple of 2, so you can just pick 0 and 1 as representatives for everything

[steel light]

The center of a ring is always commutative no? Since the center consists of elements that commute with every element of the ring, and every element of the center is an element of the ring

[slim kayak]

Yeah, if they commute with everything they also commute with each other

[steel light]

Oh

[slim kayak]

Oh?

[steel light]

So if this is right then I’m pretty much done with pt 2 LOL

[steel light]

Just didn’t expect the second problem to be so easy

[slim kayak]

not so fast

[steel light]

I just had to show that there always exists a multiplicative inverse in the center, making it a division ring, then I just showed that since it’s commutative that means it’s a field

[slim kayak]

Its an commutative subring of a division field. You have to make sure it contains inverses

oh well

No tricking you apparently

[steel light]

I did! Let a in R (division ring) with a nonzero, and let b in R. Then since a in center we have ar = ra <=> arb = rab <=> barb = brab, then use the fact that b is the multiplicative inverse of a to simplify the ba/ab products and you get rb = br, meaning b is in the center too

So every nonzero a in the center has a multiplicative inverse b (since R is a division ring) and then we showed that that b is in the center too, making it a division ring

Ya? :D

yayyy

I’m gonna take a break for now

I’ll have to see how long D&F takes to get to quotient rings

Since I think I remember that’s what I had the most trouble with when I tried to read AM

[steel light]

Maybe I can skip straight to it hehe

Or maybe that’ll just lead to the same result as last time

[slim kayak]

last time?

[steel light]

I tried to read AM without any ring theory background ^.^ besides just knowing ring axioms

It was a lot of fun but I didn’t get very far before I got stuck

It was my favorite math book though and made algebra look a lot more interesting than I used to think, and made me interested in AG too

[slim kayak]

AM?

wait not atiyah macdonald

[steel light]

Atiyah-Macdonald

Yes that AM :3

[slim kayak]

Saying that it makes algebra look interesting is one of the takes of all times

[steel light]

What

😭

[cobalt heath]

Woah

[slim kayak]

idk, for most people it looks a bit dry

[cobalt heath]

If you got interested after reading Atiyah-Macdonald, maybe it implies you should be algebraist, after all >.>

[next obsidian]

John would say that AM gave him depression and made him quit math for like a year and a half lol

[cobalt heath]

Lmao

[steel light]

Idk about that one :p

Analysis and topology are way more relevant to engineering

[slim kayak]

The spirit of bourbaki has haunted him personally

[next obsidian]

I like Bourbaki

[steel light]

Why?

[slim kayak]

Sure me too, in portions

[steel light]

What commalg book do you prefer Kerr?

[slim kayak]

Specifically commalg? Never touched any other than AM

[steel light]

And you didn’t like it? :sadge:

[slim kayak]

It was fine, arguably whenever I could throw some abstract nonsense via tensor products and adjunctions was actually kinda fun

Just kinda a terse and... well, work

[steel light]

Tensor products

[slim kayak]

yeah?

[steel light]

I thought it was the perfect balance between exposition and math

[steel light]

I had trouble understanding them, that’s all

[slim kayak]

in terms of math/exposition AM might violate the field axioms

[steel light]

LOL

What do you mean 😛

[slim kayak]

Aproaching via universal properties and what not is definitely the way to go, but it definitely helps having had some actual experience with them

[steel light]

I have a bit of experience working with tensors but not in the abstract

[slim kayak]

Central simple algebra stuff might be interesting in that regard, has some number theory applications in class field theory I believe too if that is decent motivation

[steel light]

Definitely not

NT is so big zzz to me

Well actually both ANTs are pretty cool but they’re still not that interesting to me

I skimmed Silverman once and it looked cool though

[slim kayak]

idk I find them to be more interesting as time goes on

[steel light]

We’ll see if I get there

[slim kayak]

okay thats mostly due to arithmetic geometry and having an use for the cool AG stuff but eh

[steel light]

My summer’s so busy this year I doubt I’ll be able to keep studying math

[steel light]

Hehe yeah I believe that’s what got me interested too

[cobalt heath]

Does it have even less exposition than Hartshorne

[slim kayak]

Well, at least the exposition in hartshorne was more memorable

But both are glorified exercise books honestly

[cobalt heath]

Yep

[quartz wind]

is this proof good enough? do i need to include that a1,...,ak is disjoint? is there a more formal way of approaching it?

[crystal turtle]

I really liked Atiyah Macdonald

[fading field]

if you take a1...ak not disjoint, then the statement is no longer true

[quartz wind]

right i just wanted to ask if the disjointness was implicit in the term "cycle decomposition"

[fading field]

yes it is

if you are worried then you can just write "cycle decomposition" as "decomposition into disjoint cycles"

not much more writing for a decent bit more clarity

it doesn't really matter though, as long as you and the reader both know what "cycle decomposition" means

[quartz wind]

i see ty

is the rest of the proof good though? i really didnt know how to approach it formally

[next obsidian]

When you have a bamboo bike that you threw away, call that cycle decomposition

[crystal vale]

Disjoint cycles commutes

[fading field]

if you want to be more formal, you can let $\sigma = a_1...a_k$ and then compute $\sigma^n = (a_1)^n...(a_k)^n$, and then the fact that $\sigma^{n_1} = (a_1)^{n_1}...(a_k)^{n_1} = e(a_2)^{n_1}...(a_k)^{n_1}$. Also, you probably shouldn't say that the decomposition of $\sigma^{mn_1}$ contains $a_1$, since that is not true. The right wording here is probably "$\sigma^{mn_1}$ eliminates $a_1$ for all $m \in \mathbb{N}$

[cloud walrus]

smay

[fading field]

@quartz wind

[next obsidian]

What the sigma

[quartz wind]

okay i see thank you!

[steel light]

So if I'm understanding right, I need to show that there exists some positive integer m such that any element of the residue class of (ab)^m is congruent to 0 mod n?

[cobalt heath]

Yeah

[steel light]

I think I sort of see it

Vaguely

[cobalt heath]

Can you describe that,

[steel light]

If you take (ab)^k then it's divisible by a^k so it should have remainder 0 mod n or something something, since n = a^k*b?? But there's an issue with the b term I think?

I haven't done this in so long sorry if I'm slow

[cobalt heath]

It's not slow at all

You are basically relearning these in the new framework of ring theory

Can you identify exactly when is (ab)^k = 0 in Z/nZ

[steel light]

Let me think about it

[next obsidian]

You can’t say exactly when, but you can give a lower bound

[cobalt heath]

I meant to say condition of c when c = 0 in Z/nZ

For c in Z

[steel light]

I know that (ab)^k cong 0 <=> a^k*b divides (0-(ab)^k) <=> -(ab)^k = a^k*bp for some integer p...I have no clue lol

I guess I can ignore the negative sign and just absorb it into p

[fading field]

upper bound

[next obsidian]

I’m saying you can give k for

Oh true

Idk it’s weird

[fading field]

i got so confused when you said lower

lol

[next obsidian]

I wanted to say you can give k for when j >= k means it’s 0

[steel light]

[fading field]

we know what you meant

[steel light]

what is j

the exponent

[fading field]

for all j > k (ab)^j = 0

[steel light]

yeah

mmm

[cobalt heath]

Its all becuz my abooze of notation

[next obsidian]

No it’s cuz I wasn’t specific

[fading field]

anyways, if you have x = 0 mod n then it means you can factor n out of x

[steel light]

Always...?

I'm so bad at modular arithmetic

[fading field]

if n = 12 then 6 is not 0 mod n

[fading field]

even though n = 2^2 * 3

[steel light]

@.@

[fading field]

actually you should try doing it for n = 12

that will serve as a good guide

[steel light]

So you fix n?

I thought you'd be given a, b, k

[fading field]

yeah you fix n

well it's saying that if you fix n, you automatically get an a and b and k

so what i am saying is you should try it for a,b,k = 2,3,2

[steel light]

Okay...

[fading field]

and also for a,b,k = 2,3,3

then that will be slightly illuminating i think

[steel light]

I don't actually know how to compute residue classes when the divisor is bigger than the number...or with negative numbers...💀

[fading field]

that's okay, because you know that if it's not 0 you have to keep goin

g

[steel light]

Oh

Is the lower bound just n then

Or well you said upper bound

[steel light]

I was thinking this still

[steel light]

No sorry not n

[fading field]

well try it out, you can actually get a better upper bound

[steel light]

Okay

[fading field]

certainly (2*3)^12 is 0 mod 12, but we can just do a bit better

[cobalt heath]

How many times should you multiply hour by 6 to get to 0 a clock

[fading field]

lmfao

[cobalt heath]

Sorry, this is mod 24

[steel light]

LOL

[vapid vale]

😭

[steel light]

I feel so stupid

[steel light]

thinking

[fading field]

okay, so try counting up, first, is (2*3)^1 mod 12 = 0?

and if the answer is no, keep going

and see how far you get

[steel light]

No, it's 6

[fading field]

right

so what about 2?

[steel light]

Yeah 2 and everything above is 0

But I don't really understand why

[fading field]

okay so let's look at the case of m = 2, what is (2*3)^2?

we get 36 of course

[steel light]

Yes

[fading field]

but now when we factor out 12, we can do it either knowing that 12*3 = 36, or we can do it looking at the prime factorization of 12 and 36

and seeing that the prime factorization of 12 is 2^23 and 36 is 2^23^2

ignore discord formatting, but is this clear?

[steel light]

hey 36 is (ab)^k

[cobalt heath]

Ye

[steel light]

that's all I noticed lol

um

[fading field]

okay but what is the prime factorization of 36

[cobalt heath]

Oh it would be good if you connect to prime factorization then classification of ideals in Z

[steel light]

2^2*3^2

[fading field]

great yes

okay so let's move to the case where n = 24

[steel light]

[fading field]

or a,b,k = 2,3,3

[steel light]

Okay

[fading field]

so do the same thing as before

[steel light]

Oh

Wait

[steel light]

I'm gonna guess that since both of these were to the power 2 and 2 was the lower bound, then this will have 3 as a lower bound?

[fading field]

yeah

[steel light]

Idk let me see lol

Okay

[fading field]

oh okay sorry i didn't mean to spoil it

i thought you were just asking

[steel light]

Nw

Same thing really

[fading field]

okay so now we can make guesses at the general problem

where n = a^k b

[steel light]

I'm sorry but isn't this what I was saying here?

Except now that I think about it there's no issue with the b term since b^k/b is still an integer

[fading field]

okay yeah, you have to say this about b. You can formalize this argument a bit more

and then you're done

[steel light]

So it'll be congruent to 0 when you take the kth power

So I had it from the start? O.o

Wtf?

LOL

I'm so shooketh

[fading field]

lol

[vapid vale]

🙂

[steel light]

Well okay then

Let me try writing something up

Since [ab] is a residue class do I just pick a representative? Or how am I supposed to work with this?

An element of [ab] x is such that x = ab (mod n) right? So it makes sense to take some x in [ab] and raise that to the kth power and then consider x^k = (ab)^k (mod n)? Is that how modular arithmetic works? 😭

[vapid vale]

in Z/nZ the residue class of ab is just an element

you can follow their notation and use \overline{ab}

[steel light]

why wouldn't the proof just be like one line then

basically exactly what I said

[vapid vale]

its not a long proof

[steel light]

https://tenor.com/view/well-that-was-easy-lesley-ann-brandt-mazikeen-lucifer-not-that-hard-gif-17380032

[fading field]

yes, you can pick a representative because the projection Z -> Z/n is a ring hom

so taking a computation in Z/n is the same as taking some representative and then taking the projection after

[steel light]

Goodness Microsoft Word looks so ugly with overline (don't flame me for not using a TeX editor 😭)

[steel light]

I'm just gonna do this I think

[fading field]

why are you using MS word

[steel light]

because learning how to format papers with tex is too much work for me

[fading field]

??????????? lol

[steel light]

It's always setting up the paper that I struggle with, I can write math okay but idk every time I die when I try to like make a header and whatnot lol

[fading field]

you can just use overleaf

they make all that shit for you

[sonic coral]

i learned overleaf two semesters ago and just got a file from someone and reuse all the formatting for every assignment ect

and just use all their shortcuts and what not to type the math

[fading field]

MS word is way more work than TeX as well

[sonic coral]

it’s not as overwhelming as you think it may be

i was in the same boat you are now, at one point