groups-rings-fields

[still dew]

Right

Very cool

Thanks for the help man

[rocky cloak]

Or maybe a better perspective. The characteristic polynomial measurers exactly when A - xI is invertible. So if A - xI is not invertible, then x is a root of the characteristic polynomial

[drifting mauve]

Hi everyone, I didn't really know whther to put this in discrete maths or here, it doesn't really seem like advanced maths, so let me know if I should move this. But I have this question:

List the elements of a subgroup $H$ of $\mathbb{Z}_{45}^*$ such that $H$ has order $8$.

The thing that gets me is that doesn't the order of a subgroup have to divide the order of the group by Lagrange's theorem? So what does this question ask for?

[cloud walrus]

theaveragejoe6029

[vivid tiger]

\totient(45) is probably a multiple of 8

3^2*5 so 3 (3-1) 1 (5-1) = 24

so note that 8 | 24

the 2-sylow subgroup will be an order 8 subgroup

[drifting mauve]

sorry, I don't know what that means.

could you explain why you chose 3^5*5?

[boreal inlet]

(Z_45)* is the group of units in Z_45

And you can prove that group of units is precisely the elements which are less than and coprime to 45 in this case

Which leads to the totient function, as it counts the numbers of how many elements less than n is coprime to n

This has a certain formula, which Xela has worked out, if you want to know why this comes, check out any elementary number theory book. It's a simple combinatorical argument

Which then leads us to know that the unit group has precisely 24 elements

And hence is a group of order 24

Now as 8 divides 24, there must be a subgroup of order 8.

Now by sylow's theorems, 24 = (2^3) * 3, hence, there's a 2-syloe subgroup of order 2^3 = 8

You have to find it

Also wait, actually, that subgroup is unique.

U_45 (that is, (Z_45)*, the unit group) is abelian, and hence any subgroup is normal.

Now a p-sylow subgroup is normal iff it is the only p-sylow subgroup.

So there exists exactly one subgroup of order 8, and it is normal, and of index 3

[vivid tiger]

okay so that part is just computing the totient.

totient is multiplicative (in the sense that \varphi(ab)=\varphi(a) \varphi(b) for coprime a,b), so to compute \varphi(n) we need only to multiply \varphi(p^n) for p|n. for p, note that we have that the only way to not be coprime is to have a power of p in you, so since there are p^n/p=p^{n-1} of those we have p^n-p^{n-1}=p^{n-1} (p - 1). Thus the order of the (Z/45Z)^× as given by my computation above is 24

oh. you also want to show the subgroup is normal?

we know that there are either 1 or 3 2-sylow subgroups.

(since 24=2^3*3)

unfortunately 1=3 mod 2 so we can't immediately rule out 3.

oh it's abelian

duhhhh

yes so of course it's normal

[vivid tiger]

how do you know there isn't an order 8 subgroup that isn't a 2-sylow subgroup?

oh because 8 is the largest power of 2 in 24

[boreal inlet]

Yes.

[boreal inlet]

U_45 isomorphic to C_6 x C_4

[drifting mauve]

oh duhh. my bad. I got mixed up thinking that I had to considder that it was ismorphic to Z_{+}^{44} which would give different factors lol

[boreal inlet]

nono lol that only happens if it's prine

[vivid tiger]

I know it's ableist just from definition.

[boreal inlet]

uhm do you mean abelian?

[drifting mauve]

Okay, I've caught up to what you're saying. But how do I then compute this subgroup explicitly?

The question doesn't necessarily need reasoning or working, although it would be interesting, I can write some code to compute it but I don't really know the logic.

[vivid tiger]

I know it can be done because Sylow constructs it (to the point that there's a polynomial time version of Sylow). I can't remember how it's done, though. Perhaps check the proof of Sylow.

[drifting mauve]

Ah, got the elements of the subgroup. @vivid tiger and @boreal inlet Thank you guys so much!!

🙏

[tribal niche]

can anyone give me a hint for the following?

If R is noetherian then any submodule of the R-module R^n is finitely generated

i've been trying to proceed by contradiction by supposing you can have a chain of submodules that don't stabilise

[delicate orchid]

sounds like the way to do it

I'm presuming you don't have the result that says M is noetherian <=> N < M is noetherian and M/N is noetherian?
Nor the result that says M is noetherian <=> every submodule is finitely generated?

[tribal niche]

no, i didn't know either of those

but i think i've got it, thanks anyway 🙂

[tough raven]

Is there any non-constant rational function p in any number of variables X1, X2, … over any field and a positive integer k such that p(X1^k, X2^k, …) = p(X1, X2, …)?

[delicate orchid]

does a degree argument work here

or do we be cheeky and do something like p(x_1, ..., x_n) = x_1/x_1 which technically isn't constant because it's undefined with x_1 = 0

[daring nova]

Or take a field for which X^k = X
Like k = p in Fp

[delicate orchid]

true!

they did say "any field" though, which makes this a lot more annoying - so following the Lefschetz principle we should probably work in char 0

[true ingot]

Can someone help with the following one?
We have $C_n = \mathbb{Z}_n$, the ring of integers modulo m.
And a bijection $f: C_n \to C_n$, $f(x) = ax+b$ (where $b \in C_n$ and $a$ an invertible element of $C_n$.
f, being a bijection is also a permutation on $C_n$. The exercise I am trying to solve is asking for the cycle structure of such a permutation.

[cloud walrus]

Faputa

[true ingot]

I have tried setting $x_{n+1} = a x_{n} + b$ and then solving to find an explicit formula for $x_n$ which I could then equate to a given $x_0$ to find the cycle length for it. But I am not sure how to do that or if it's the right approach (trying to get a formula for $x_n$ always amounts to having to invert something like $a-1$, which may not be invertible..

[cloud walrus]

Faputa

[rain grove]

What is the isomoprhism of field of quotients of the ring F[[X]] to the field of formal Laurent series F((X))?

[chilly ocean]

[rain grove]

F((X)) seems to only have a finite amount of coefficients by definition, whereas a quotient a/b where a,b is from F[[X]] appears that could have an infinite amount of non-zero coefficients...

[last spoke]

Can you give an example?

[tough raven]

I mean I want an example in one field, not an example for every field.

[tough raven]

That's never true as rational functions.

[tough raven]

As a rational function, it's equal to 1, so no 🤷.

[rain grove]

example of?

[tough raven]

That would work: you can solve for x_n (hint: try an exponential function of n plus a constant).

[tough raven]

Factor any fraction of power series into an integer power of X times a fraction of power series with constant coefficient.

Fact: a power series is invertible (with inverse a power series) iff its constant coefficient is invertible.
Using this, you can simplify the fraction of power series with non-zero constant coefficient to a single power series. It's now clear how to interpret the result as a Laurent series.

[delicate orchid]

funny that it's not true as functions, but is true as rational functions - a proper subset

[true ingot]

I have tried the following "standard" procedure for difference equations.
Homogenous part: $x_{n+1} = a x_n$ is solved by $x_n = c_0 a^n$.
I am having issue with the nonhomogenous part (the constant).
Setting, for example, $x_n = c$ and plugging in we get $c = a c + b$ or $c(1-a) = b$ now im not sure how to proceed since there is no guarantee that $(1-a)$ is invertible

[cloud walrus]

Faputa

[tough raven]

You need to add a constant as well.

[true ingot]

Can't I do it seperately since it's a linear equation?

[tough raven]

But constants are not solutions.

[true ingot]

The usual "solution is homogenous part + particular part"?

[tough raven]

This is a homogeneous equation already.

Its homogeneous part needs a constant.

The general solution will be
x_n = c (a^n + b),
where a, b depend on the recurrence and c on the initial condition (value of x_0).

[true ingot]

Are you saying the annsatz should be $x_n = c_0 a^n + c_1$ or something like that?

[cloud walrus]

Faputa

[tough raven]

Yep

[true ingot]

Why is this not equal to what I have attempted?

Wait I will try it.

I've tried it now and I get basically the same thing

$x_n = c_0 a^n + c_1$ is my ansatz.
Plugging in we get that $c_1(1-a) = b$, which I again don't know how to proceed since there is no guarantee (1-a) is invertible

[cloud walrus]

Faputa

[true ingot]

Or maybe (1-a) is always invertible for an invertible a? I have to check that

[tough raven]

I was incorrect. It is heterogeneous. Your homogeneous solution is correct.
However, for the non-homogeneous part, you need to find a particular solution ie an (x_n) such that x_n+1 - a x_n = b.

As I understand it, you tried the ansatz of ‘a constant sequence’ and as you noted, it didn't work. That just means you have to find a particular solution by some other means.

[tough raven]

No: a could be 1, after all.

[true ingot]

Perhaps if true if it is not 1, and the case a=1 can be checked seperately?

[tough raven]

oh.

You actually do need a - 1 to be invertible for that ansatz to work.

[true ingot]

I am thinking of a similar trick Ive seen before where 1/(1-a) can be expressed as kind of a geometric series 1+a+a^2+a^3+...

[tough raven]

I spoke hastily based on memories of doing these over fields 💀.

[tough raven]

In Z/nZ, there are other units which differ by 1 from a non-unit.

Something else is needed to find the particular solution for a general b.

Actually

What if you solve the recurrence over Q?

The sequence will turn out to lie in Z.

You can then look at it modulo n.

[true ingot]

Let me try it

[low wyvern]

Quick fact check: If x,y \in I where I is an ideal of some ring, is it true to say x-y \in I? I believe so but I want to be sure.

[tough raven]

Yes.

[true ingot]

Hmm well I get $x_n = x_0 a^n + b/(1-a) (1-a^n)$ and I am not sure how to make sense of this now in $\mathbb{Z}_n$.
I would have gotten the same result by just iterating from $x_0$, getting $x_n = x_0 a^n + b*(1+a+a^2+..+a^{n-1})$ and using the formula for the sum of geometric series. However... This is pretty ugly and I am not sure how to get more information such as lengths of cycles from it

[cloud walrus]

Faputa

[true ingot]

I am starting to question whether trying to solve the recurrence was the correct way to approach this problem. Although it feels like it should be

[tribal niche]

how are you defining noetherian here for a module btw? is it just the thing about ascending submodules

[delicate orchid]

yur

[languid trellis]

what is meant by "its index in G is prime to p"?

[agile burrow]

The index is coprime to p, i.e. their greatest common divisor is 1

sometimes prime and coprime are used interchangeably in this context

[delicate orchid]

(G:1), or as I like to call it, |G|.

[languid trellis]

oh right I see. this follows from lagrange. thanks walt

[languid trellis]

yeah i was wondering about this. seems weird but oh well

[delicate orchid]

what follows from Lagrange

I'm scared!!!

[languid trellis]

[G:H] coprime to p

[delicate orchid]

that's just how numbers work

[agile burrow]

I guess the point is that (G : H) = |G| / |H|

[languid trellis]

yes

[agile burrow]

Is that Lagrange? I forgor 💀

[delicate orchid]

take n, divide by the largest power of some prime p in n, the result is coprime

oh

I guess that is a result

[languid trellis]

that is a result as much as FTOA is a result

no fun wew strikes again

[delicate orchid]

I just forgot you actually need to prove that

[languid trellis]

lol

[languid trellis]

p divides |G| by hypo. I can't see how we get to "if p does not divide |C| (centre of G) then p doesn't divide one of [G: C(yi)]"

[dire siren]

just assume the contrary

you get that RHS is not divisible by p, while LHS is

[languid trellis]

that makes sense, thanks

[boreal inlet]

Let zeta_m and zeta_n be some mth and nth primitive roots of unity over Q. How does the compositum Q(zeta_m)Q(zeta_n) look like?

[cloud solar]

What can i know about a finite unit ring with the property that x+y is a unit for every x unit and for every y non-unit

I got that every non unit element is nilpotent

And the set of nilpotents with + is a subgrupul of A,+

Where A is the ring

[south patrol]

If every non-unit element is nilpotent then there's a very strong thing you can say about prime/maximal ideals

wdym by "finite unit ring"?

A ring with finitely many units, or a finite ring with unity

[cloud solar]

Ring with unit

[cloud solar]

We have an unique prime ideal

[south patrol]

Yes exactly

That's very strong

[boreal inlet]

It is a particular type of ring then

[cloud solar]

But can we find the ring only with this information ?

[delicate orchid]

no there's more than one local ring

[cloud solar]

Yeah this is What i got in a problem: let A be a finite ring with unit such that for every f:A->A bijective function with f(0)=0, f is an automorphism of the group (A,+). If A is a field i got A iso with Z/2Z. Then i took the function f(x)=x when x is not a unit in the ring and f(x)=sigma(x) when x is a unit, where sigma is a permutation of the set U(A) with no fixed points (assuming the order of U(A) is at least 2). This is a bijection with f(0)=0 so by the hypothesis f is actually an automorphism of (A,+). so f(x+y)=f(x)+f(y) for every x and y in A. Now if we take y as a non unit and x as a unit we have f(x+y)=sigma(x)+y. If x+y is not a unit then x+y=y+sigma(x) so sigma(x)=x which is a contradiction because sigma is a permutation with no fixed points. So for every unit x and for every non unit y we have x+y unit. And now i have that every non unit is actually a nilpotent element. More, the set of nilpotent elements create a group under the addition.

[delicate orchid]

ok so is the problem to characterise A?

[cloud solar]

Yes

Actually to find A

[delicate orchid]

because every identity preserving permutation of the additive group being an automorphism is a FAR stronger condition than the ring being local

are these commutative rings

[cloud solar]

Nope. About that i have a question. Can we use CRT on a finite non commutative ring?

[delicate orchid]

also I can't see why Z/3Z fails your condition

[cloud solar]

I did not said that Z/3Z fails

[delicate orchid]

you said that if A is a field it's Z/2Z

it's implicit

[cloud solar]

Oops now i see i made a mistake

On the case when A field

[hollow mica]

how can I show that for any p > 0 there exists n so that x^n - 1 is divisible by x^2 + x - 1 in F/p ?

[delicate orchid]

I'm very surprised the exercise wants more out of you than just showing that such a ring is nilpotent

[dire siren]

some idea: if the characteristic is not 2, then just take f(x)=x for all x, except for 1 and 2, which will be swapped (i.e f(1)=2 and f(2)=1)
from f(1)+f(1)=f(2) you get 2+2=1, so the characteristics is 3

[cloud solar]

Haha thanks:))

[dire siren]

moreover, if char(A)=3, choose f(1)=a for some a∉{0,1} and f(2)=1 and the rest random so that you get a bijection

then a+a=1, so a is invertible, implying that your ring is a field

actually, since the characteristics is 3, a+a=1 gives a=-1

which means that the only element in A-{0,1} is -1

thus A=F_3

[cloud solar]

Wow

[dire siren]

and we are left with the case char(A)=2

[delicate orchid]

Z/2Z x Z/2Z works

do all exponent 2 groups work actually

[south patrol]

But not me. I don't cook, I don't clean

[delicate orchid]

potato this is groups rings fields

[hollow mica]

lol

[south patrol]

i know

[boreal inlet]

lmao

[delicate orchid]

nah there's annoyances

and by annoyances I mean "No wew lads it's obviously not true you fool. You charlatan"

[dire siren]

@cloud solar if char(A)=2 and (no need of characteristic) |A|>4 I think we can just violate the equality f(a)+f(1)=f(a+1) for some a different from 0 and 1

like saying f(a)=a, f(1)=1 and f(a+1)=b, where b is not a,1,0,a+1

therefore |A|<=4

[delicate orchid]

this guy is cracked

[hollow mica]

If x^n - 1 = p(x) * (x^2 + x - 1) then
x^n / (x^2 + x - 1) - 1 / (x^2 + x - 1) = p(x) inside (F/p)[[x]]
so we need (-F_{k + n} + F_k) to be eventually zero for p(x) to be a polynomial in (F/p)[x] (i.e. have finitely many nonzero coefficeints),
and since the fibonnaci numbers are always periodic we can just set n to be the period.

wallah

[delicate orchid]

oh wait nvm it's just the group theory argument I had in my head

[cloud solar]

So we have 5 rings

I think

[dire siren]

we have to inspect them, but I guess all of them work

[delicate orchid]

Aut(C_3) = S_2, Aut(C_2) = 1, Aut(C_2 x C_2) = S_3 and these are the only groups with Aut(G) = S_{|G|-1}

[dire siren]

oh Z/4Z doesn't work

[delicate orchid]

so our only possible rings have to be (additively) Z/2Z, Z/3Z, Z/2Z x Z/2Z, F_4

I think there's only one option for multiplication on the first two but not sure about Z/2Z x Z/2Z

[cloud solar]

F4 ?

[delicate orchid]

F_4 as well

I forgot!

wait no

wait yes

wait no

F_4 additviely is just Z/4Z

this is really a group theory question

[dire siren]

it's Z/2Z x Z/2Z additively

[delicate orchid]

ok if u say so

oh yeah cause it's a 2-dim F_2 vs

now do we need to think about monoids of order 3?

ok so any rings we're missing have to be char 2 or else they'd just be Z/4Z. So we can think about them as {0, 1, a, 1+a} as an F_2-algebra I guess, the only thing that isn't implicitly determined is a^2. So we can write them as quotients:
F_2[x]/(x^2) - which I think is Z/4Z
then F_2[x]/(x^2+x), which is Z/2Z x Z/2Z
then F_2[x]/(x^2+x+1) which is F_4
and a mysterious F_2[x]/(x^2+1), which we hadn't accounted for

so there is a strange fourth ring of order 4

[boreal inlet]

.

[delicate orchid]

there is no way in hell compositum is a real word

oh my god it is

that one's going in the lexicon

[boreal inlet]

https://cdn.discordapp.com/emojis/756386972662235286.png?quality=lossless&size=48

[delicate orchid]

oh it's like a tensor product apparently

yeah ok it's the tensor product over a subfield in the intersection, in this case Q

so I imagine it's Q(zeta_{gcd(m,n)})

[boreal inlet]

It's the lcm, but I don't need that structure

[delicate orchid]

it's the lcm huh

[boreal inlet]

I have to prove it's the lcm thing

[delicate orchid]

so the lattices work backwards

makes sense actually, you'd expect the tensor product to be larger

[delicate orchid]

lets see if the standard proof for Z/nZ \otimes Z/mZ = Z/gcd(m,n)Z is useful

$x \otimes y \in \bQ(\zeta_m) \otimes_{\bQ} \bQ(\zeta_n)$, write this as [\left(\sum_{i=0}^{m-1} x_i\zeta_m^i\right)\otimes\left(\sum_{i=0}^{n-1}y_i\zeta_n^i\right)]
distrbutivity : [x\otimes y = \sum_{j=0}^{n-1}\left(\sum_{i=0}^{m-1} (x_i\zeta_m^i)\otimes(y_j \zeta_n^j)\right) = \sum_{i=0}^{m-1}\sum_{j=0}^{n-1} (x_i\zeta_m^i)\otimes(y_j\zeta_n^j)]
ok so lets just look at some $(a\zeta_m^i)\otimes(b\zeta_n^j)$ cause they generate the whole got dang thang. Since $a, b \in \bQ$ this is $ab(\zeta_m^i)\otimes(\zeta_n^j)$ so we just want that $((\zeta_m)\otimes(\zeta_n))^{\text{lcm}(m, n)} = 1_{\bQ(\zeta_m) \otimes_{\bQ} \bQ(\zeta_n)} = 1 \otimes 1$

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

which I think is true just by using the definition of multiplication when you tensor algebras together

nifty

so ur thing has a lcm(m,n)-root of unity

and I can't be bothered showing the rest hahahahaha

[boreal inlet]

I think that is the problem we needed to solve anyways

I'm not exactly sure how is this true

[delicate orchid]

I think it's another case where you just have to write the whole thing out and see that everything turns to 1s

actually no it's not that bad, you can directly move the power down like $(\zeta_m \otimes \zeta_n)^k = \zeta_m^k \otimes \zeta_n^k$

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

this should be true cause tensor products of COMMUTATIVE algebras is the coproduct

[boreal inlet]

There's a different way compositums are seen - that is, if M and N are fields, then NM is the field of fractions of the set of finite N-linear combinations on M (or vice versa)

[last spoke]

F2[x]/(x^2+1) is weird bc it doesn’t have cancellation (since x*(x+1) = x^2+x = 1+x but x doesn’t equal 1)

[delicate orchid]

huh. That is strange

[boreal inlet]

The problem is this looks fucking horrible to work with

[delicate orchid]

agree with you on that one

just tensor product.... tensors are nice...

[last spoke]

Idk how to make sense of it

[boreal inlet]

As a tensor product it looks million times nicer but it's still kinda pain to work with

[delicate orchid]

I'll google

[boreal inlet]

Wait what isn't it an integral domain

Oh wait it isn't

[delicate orchid]

yeah it's not surprising that it's not an integral domain

[boreal inlet]

x^2 + 1 is reducible over F_2[x]

[delicate orchid]

yur

anyway

it's this "L" ring apparently

[boreal inlet]

lmao

[delicate orchid]

stupid ring

[last spoke]

The dual numbers over Z2?

[languid trellis]

L

[delicate orchid]

I haven't heard of dual numbers in years

but wouldn't that be F_2[x]/(x^2)

aka Z/4Z

[south patrol]

2-adics

[delicate orchid]

this was actually a really fun exercise to do

giving a nice corrispondence between the commutative rings of order p^2 and irreducible quadratics over F_p

well, not irreducible

just all of them

[last spoke]

Wickerpedia says that’s the canonical matrix representation of the dual nums over a ring

[delicate orchid]

I must've gotten my initial classification wrong then

yeah I definitely did - how could you get Z/4Z from a quotient of a char 2 ring? duh

[last spoke]

F2[x]/x^2 has elems 0,1,x,x+1 with the same issue: x*(x+1) = x^2+x = x but x+1 doesn’t equal 1

So it looks like the same structure under a map A > A+1

[delicate orchid]

yeah I believe this

[dire siren]

apparently there are only four rings with p^2 elements

and they are all commutative

[vivid tiger]

well,

oh huh

prime p there are only two rings of order p up to isomorphism

[vivid tiger]

The purpose of this note is to give a complete classification of all finite rings of order p2 with p a prime. In particular, we show that up to isomorphism there are exactly 11 rings of order p2.

[dire siren]

I think those rings are not assumed to have unity

[dire siren]

also here, because there is only one ring with unity with p elements

[boreal inlet]

Finally found a better proof

[vivid tiger]

oh wow

(https://maa.org/sites/default/files/Classification_of_Finite-Fine04025.pdf)

[boreal inlet]

(zeta_m)^(m, n) * zeta_n is a primitive [m,n]th root. We are done

@delicate orchid

[vivid tiger]

well it's a group of order p so is cyclic. we also have that since 1≠0 the additive group generated by 1 is the whole thing

r(n1)=n(r1)=nr

err I mean

[delicate orchid]

Simply exquisite

[vivid tiger]

(n1)(m1)=nm1

[delicate orchid]

Any char n field contains Z/nZ, so for n = p prime we have that Z/pZ is the only ring of order p

[vivid tiger]

...how can this be possible? I thought it should be a field!

oh, (x+1)^2 = (x^2+1) so it isn't irreducible.

F2[X]/(x+1) is cancellative.

[delicate orchid]

F_2[x]/(x+1), or as I like to call it, F_2

[boreal inlet]

Actually nvm fuck this doesn't work for m = 4 n = 2

[delicate orchid]

simply diabolical

[boreal inlet]

😔

[delicate orchid]

yeah Q contains 2nd roots of unity

strange!

[boreal inlet]

man my professor is gonna kill me tomorrow morning

I haven't done a single problem correctly!

https://cdn.discordapp.com/emojis/1115546874728939630.png?quality=lossless&size=48

Ok can we like

let's say m and n are two natural numbers and uh

can we find a natural number k such that gcd(k,n) = 1 and zeta_k is an mth root of unity over Q

Because if this is true for any general m and n, we can always show that (zeta_m)^(m/k) * (zeta_n) is a primitive lcm(m,n)th root of unity over Q

[delicate orchid]

k would have to be a divisor of m

so yes, we can.

[boreal inlet]

Gcd doesn't work btw

because again same example

M = 4, n = 2

[delicate orchid]

2 is gonna be weird

cause Q(zeta_2) = Q

so I don't care about that for now

[delicate orchid]

ah no there's a slight annoyance

if M is the set of primes dividing m and likewise for N, n we need M \neq N

then we can find such a k by just picking a prime in M - M \cap N

[boreal inlet]

Oh yeah

Atleast one prime needs to be there which divides one but not the other

Otherwise this 2 4 shit repeats itself

[delicate orchid]

yur

[boreal inlet]

And we pick that exact prime

Or one of such multiple primes

Bruh

😔

This means I can't use this technique in general

wtf

Let's take Q(zeta_12)Q(zeta_18)

This should be Q(zeta_36)

what are the divisors of 12 again

1, 2, 3, 6, 12

[delicate orchid]

4 also

[boreal inlet]

Oop

Yeah

In anyways

<zeta_m> \cap <zeta_n> = <zeta_d> where d is gcd

Similarly, <zeta_m, zeta_n> = <zeta_m><zeta_n>

I think this is enough

The first one is true because LHS has cardinality d and it's a cyclic subgroup of <zeta_mn>

2nd one also is a cyclic subgroup of <zeta_mn>.

Now, all subgroups here are normal

So, <zeta_m><zeta_n> has cardinality mn/d = l, where l is lcm

Hence, there must be an element of order l, and that should be a primitive lth root of unity

Hence we can generate an lth primitive root of unity using zeta_m and zeta_n, and we are done

I honestly don't think the actual explicit construction for the lth root is trivial, group theory saved my ass

FINALLY.

[delicate orchid]

no idea what you're doing but I'm glad it's worked

[boreal inlet]

I showed Q(zeta_m)Q(zeta_n) \superset Q(zeta_l)

The other part is trivial

[delicate orchid]

we did it reddit...

[exotic verge]

anybody just start going through Aluffi chapter 0?

starting either later this week or next week

[grizzled niche]

i don't get Then (a^-1)^-1b^-1 in H part, could someone plz explain? ty

[delicate orchid]

a^-1H = b^-1H
H = ab^-1H <=> ab^-1 is in H

[grizzled niche]

oh multiplied both sides by a?

[teal wind]

Can anyone help me with these problems?

do my solutions to 3 and 4 work?

[tough raven]

Well, you can ask when is this congruent to x_0 modulo n.
x_k - x_0 = (a^k - 1) x_0 + b/(a-1) (a^k - 1) = (a^k - 1) (x_0 + b/(a-1)).
This is divisible by n iff a^k - 1 is divisible by n / gcd(n, x_0 + b/(a-1)).

[hidden wind]

hi, i’ve been meaning to

[tough raven]

OK that doesn't make sense, b/(a-1) does not necessarily exist in Z/nZ.

[tough raven]

x_k - x_0 = (a^k-1)/(a-1) (x_0 (a-1) + b)
This is divisible by n iff
a^k = 1 (mod (a-1)n / gcd( (a-1)n, x_0 (a-1) + b ) )

So you need the order of a modulo a certain factor of (a-1)n, which depends on x_0.

[boreal inlet]

You're welcome

[dense raven]

what is this saying

like what are "representatives for G-orbits"

[winter shore]

This may be a very stupid question but what is the relationship between being finitely presented and a concrete presentation <A | R>?

is it R is a finite number of relationships? I don't see the direct connection

[dense raven]

thats the definition yeah

R is finite

[winter shore]

my definition of finitely presented is we have an exact sequence Rn -> Rm -> M -> 0

[dense raven]

?

R? like the real numbers?

[winter shore]

oops overloading R, in the exact sequence it is just a ring

[south patrol]

Well <A | R> is usually used for groups

[chilly ocean]

Given an $\Bbb R$-algebra $\Bbb R^2$ with multiplication given as $$(x_1,y_1)(x_2,y_2)=(x_1x_2+y_1y_2, x_1y_2+x_2y_1).$$ Is there a way to determine the maximal ideals of this from knowing what the maximal ideals of $\Bbb R^2$ with the regular multiplication are or how should this be done?

[cloud walrus]

Emerton

[south patrol]

A presentation of an $R$-module $M$ is an exact sequence $R^I \to R^J \to M \to 0$ for some $I,J$. You think of $R^J \to M$ as specifying some generators, and then the kernel expresses relations.

[cloud walrus]

Süßkartoffel

[south patrol]

If you write out first iso theorem you'll see how it matches up to what you expect

[winter shore]

yes I got it up to there

oops meant to reply the last message

[dense raven]

im pretty sure there isnt much of a direct relationship between finitely presented R-modules and finitely presented groups

[winter shore]

Okay for simplicity let R=Z

[south patrol]

There is, like you have a surjection from a free group/module whose kernel is finitely generated

[winter shore]

from the relationships of my presentation <A|E> what can I say of the map R^I-> R^J sus mentioned?

[south patrol]

well like A specifies a map Z^J -> A

[winter shore]

yup

[south patrol]

and E is a set of generators of the kernel of that

which specifies a map Z^E -> Z^A

such that Z^E -> Z^A -> M is exact

[winter shore]

I get lost at this part

[south patrol]

Like uh say you have a and b in A and some relation e is that "a + 2b = 0"

then you send the generator e of Z^E to a + 2b in Z^A

[dense raven]

oh that makes sense lol

[south patrol]

(here, notationally, i'm viewing Z^S as formal linear combs of elements of S)

So like your map Z^E -> Z^A is formal, like you're sending the relation "a+2b=0" to a+2b lol

and Z^A -> M gives them "meaning"

[winter shore]

ohhhh yeah that makes sense

thanks

[dense raven]

this type of definition doesnt work for nonabelian groups right?

its just for modules and stuff

and abelian groups

[winter shore]

one last question, what about an exact sequence Z^D -> Z^E -> Z^A -> M -> 0? (I'm trying to get intuition of resolutions). What does the map Z^D -> Z^E mean? Or how to understand it?

[delicate orchid]

Identity this algebra with R[x]/(x^2-1) then correspondence theorem

[winter shore]

yeah it won't, because a relation may not generate a subgroup so the map F(E) -> F(A) sus described is not a group homomorphism

[chilly ocean]

Can I ask, how do you obtain R[x]/(x^2-1) from this?

[winter shore]

or rather, does not generate the subgroup it should

hmmmmmmm ok I'm doubting myself

[delicate orchid]

it's complex multiplication but with i^2 = 1 instead of -1

[winter shore]

Ok the map F(E) -> F(A) induced by mapping "e" to e is a group homomorphism, but it's image is not necessarily normal. So yeah it doesn't work. One could modify definitions to say exactness means ker f = normal closure of (Im g) and it'd work though

[dense raven]

its even worse for infinite groups lol

as in any sequence Fn -> Fm -> G is necessarily non-exact

[chilly ocean]

Does this actually simplify further as x^2-1 = (x+1)(x-1)?

[delicate orchid]

well that's how you'll be getting your maximal ideals

and uhh yeah you can chinese remainder theorem it

[lament bough]

are there moral reasons as to why a formal power series over a field is a unit iff the first element is not 0?

[coral spindle]

Bc it is a local ring with maximal ideal (x)

[south patrol]

Well one thing is that x is "topologically" nilpotent whilst non-zero constant polynomials are units

unit + nilpotent = unit

But ultimately i'd say it comes downt o like

1/(1+x) making sense if you allow for power series

There are other moral reasons in the sense that you are taking a line over the field and "zooming in" at zero

[lament bough]

makes sense

I'll ruminate on that

[languid trellis]

I'm trying to determine the order of uv, and i'm struggling. in particular, I'm struggling to discount the possibility that uv has some order greater than 2. (I have already justified to myself that uv=1 gives a contradiction, I want (uv)^2 = 1, because then we get uvu = v, exactly what I want to show that <u,v> is iso to D_2)

[coral spindle]

You’re gonna struggle with that! In certain cases, uv is the element of order n in Dih(2n)

What I mean to say is that it’s the rotation

[languid trellis]

Oh, so it's not the case that <u,v> is iso to D_2, but rather some D_2n

[coral spindle]

That’s right

[languid trellis]

hm

[coral spindle]

Try showing that D_n has this property, maybe that will help you see how to put the pieces togetger

[languid trellis]

that D_ n has the property that it is generated by two involutions

?

[coral spindle]

Yeah

[languid trellis]

Composing two reflections gives a rotation, r. My first idea was to compose this rotation with itself n times to get all the rotations in D_n, this this only works if |<r>| is coprime to n. I think that we should next consider compositions of this rotation with the reflections that we are given, (which will each result in reflections), and see if that's enough to generate D_n

Oh, and because these are involutions in D_n, they are reflections (for n geq 3). we know uv neq 1, because then we have u = v. So uv is a nontrivial (not equal to the identity) rotation

[wooden fulcrum]

hello

[vivid tiger]

oh hey it's this one again

miz had this argument but i got distracted and i didn't read it

[vivid tiger]

because once you do this you will be able to map u and v to them

err

we just have the coxeter group with two elements are n

that is, let n be such a that (uv)^n=1

we only need to show that there aren't any extra relations

[languid trellis]

chat what is a coxeter group

[vivid tiger]

oh sorry don't worry about it

point is that D_{2n} is <r,s : r^2=s^2=1 and (rs)^n=1>

[languid trellis]

yes I agree, that's exactly what we have with <u,v>

[vivid tiger]

that is, that the group generated by u and v in Aut is actually the free group on two generators with the relations u^2=v^2=(uv)^n

[vivid tiger]

how do we know that?

we know that those relations are satisfied.

[languid trellis]

u and v are distinct involutions, so u^2=v^2=1. uv has finite order as G is finite

[vivid tiger]

yes, so?

<u,v> is the group generated by u, v in Aut.

Just because u,v satisfy those three generating relations in D doesn't necessarily mean that there aren't extra ones that don't hold for D

D is a free group mod the normal closure of the subgroup generated by r^2,s^2,(rs)^n. We know that <u,v> in Aut is certainly not bigger, but what if it is strictly smaller?

[languid trellis]

what other relations could there possibly be?

[vivid tiger]

how do we know there aren't any!

for all we know the involutoons are special and thus have some relations D doesn't

after all, we came up with these

you could try doing an order thing

I think miz showed the order of <u,v> = order of D_{2n}

[languid trellis]

huh

[vivid tiger]

by induction, but, I don't know what he did

@unkempt stream enlighten us

wait no not you

@dull ginkgo

#1217259452621783154 message

So

so therefore (uv)^N-1 is vu

Every word in <u,v> is of the form uvuvuvuvuvuv.... or vuvuvuvu....

because involutions

okay so

combinatorics

(uv)^n=1

(vu)^n=1

[languid trellis]

supposing (uv)^N = 1 no>

[vivid tiger]

yes

not no

(uv)^n vu = vu

then (uv)^{n-1} = vu

so a string of form vuvuvu... is just (vu)^m optionally times v

and this, as we have established is for some l, (uv)^l optionally times v.

which would optional cancel the last v

giving us (uv)^{l-1} times u in the option case and (uv)^{l} otherwise

so every string looks like
(uv)^m Optional(v)

with m < n

so, there are 2n of these (namely: identity, u, uv, uvu, uvuv, uvuvu, ...)

QED

[languid trellis]

let me try explain this to myself

[languid trellis]

How are we getting from |<u,v>| = 2n to <u,v> iso to D_2n?

[vivid tiger]

Okay so firet

The group <u,v> is a subgroup of D_2n

err

Okay instead

The group <u,v> satisfies at least as many relations as D_2n (and it satisfies the relations of D_2n)

so, if we show that the generators u,v don't have any extra things glued to 0 besides the one that D_2n does, we will be done

more abstractly, <u,v> is a quotient of the free group on two generators by the normal closure of its relation, and so is D_2n

since <u,v> satisfies the relations of D_2n (namely uu=vv=(uv)^n=1)

all we need to show is that there are no extra relations.

by showing the groups have the same size, we will have shown that the relations are the same, because <u,v> doesn't have any extra

[languid trellis]

<u,v> having the relations u^2=v^2=(uv)^n =1 means that |<u,v>| \leq |D_2n|, so then showing that |<u,v>| = |D_2n| => these are the only relations on <u,v> (if we had another relation then <u,v> should be smaller) which them implies that they are isomorphic, Am I understanding that correctly?

[vivid tiger]

yup

[languid trellis]

using the presentation of D_2n made this way easier. thanks for the help xela

[vivid tiger]

and thanks to mizalign for actually solving it

[languid trellis]

lul

[vivid tiger]

this is interesting to me because it gives me a reason to care about D_{2n}

likewise dynkin gives me a reason to care about coxeter.

[teal wind]

Can someone help me with these problems?

So far, I have this, but not sure if I'm on right track:

[delicate orchid]

1 is fine

3 is fine

all good

[languid trellis]

still funny to me

[dull ginkgo]

If we had an extra relation, than some uvuvuvuv…. LESS THAN (uv)^n would be the identity, but that would either imply (uv)^m*u = e, or (uv)^m = e. This violates the idea that n is the order of (uv) unless u = v

Of which the order is 2

[languid trellis]

u,v distinct by hypo

[delicate orchid]

D_n expert

[dull ginkgo]

Yeah

Took me a while but the assumption of the order is critical

it’s the whole “minimal n such that” part because more relations would contradict its minimality

Which that explanation makes sense to me more than just plugging and chugging for the order

Actually let me formalize this bastard

Let $u$, $v$ be involute elements in group $G$ such that $uv$ has finite order $n$, then $\langle u, v \rangle \cong D_{2n}$

[cloud walrus]

Request a new nickname

[dull ginkgo]

Every element in $\langle u, v \rangle$ is uniquely of the form $(uv)^m u$ or $(uv)^m$ for some $0 \leq m < n$.
This can be shown via induction by appending $u$ or $v$ to words of those forms and see they also are of either of those forms through $(uv)^m = (vu)^{n - m}$ = (a bit lengthy and repetetive, so I'll exclude it here).

We can show $|\langle u, v \rangle| = 2n$, because an element of the form $(uv)^m$ cannot be the identity as it'd violate the minimality of the order being $n$. Since $u \neq e$, $m \neq 0$. Therefore if $(uv)^m u = e \Rightarrow (uv)^m = u \Rightarrow (uv)^{m-1} = v \Rightarrow (uv)^{2m - 1} = uv \Rightarrow (uv)^{m} = e$, causing the same contradiction.

We can show there's a map from $D_{2n} = \langle r, s \mid r^n = (sr)^2 = s^2 = e \rangle$ to $\langle u, v \rangle$ where $r \mapsto uv$ and $s \mapsto u$, but we also know the domain also has order $2n$, thus it is an isomorphism

[cloud walrus]

Request a new nickname

[chilly ocean]

How do you prove that R^2 with multiplication given by (a,b)(c,d) = (ac, ad + bc) is isomorphic to R[x]/(x^2)? I was thinking about defining a R-module homomorphism f : R[x] -> A with f(p) = p(1,0), and then using the first isomorphism theorem, but looks like this wont work.

[coral spindle]

You should use the first isomorphism theorem. I’m not going to say anything more because I can’t possibly do so without ruining the whole thing

[chilly ocean]

How should one define the map f : R[x] -> A? The identity on A is (1,0) and 1 on R[x] so we need to have f(1) = (1,0)?

[rocky cloak]

Remember you want x^2 to map to 0, so x should map to something that squares to 0

[chilly ocean]

Riiiight its actually only x we need to take care of due to linearity right?

[rocky cloak]

Yes, that's right

[chilly ocean]

For any (a,b) in A we have that (a,b)^2 = (a^2, 2ab) and this is zero if a = 0. So any element (0,a) squares to zero in A.

So might as well map x to (0,1)?

The hard part now is then probably to show that ker(f) = (x^2)

[dire siren]

there is also the approach in which you guess the isomorphism

if you look at how two polynomials in R[x]/(x^2) multiply, you might get a good guess on what the isomorphism should be

[chilly ocean]

Any two polynomials in R[x]/(x^2) multiply as (ax + b)(cx + d) = acx^2 + adx + bcx + bd = (ad + bc)x + bd, but I'm bad at guessing

[coral spindle]

Now look at multiplication in A

[chilly ocean]

I know they are similar, but where should I send something like f(ax + b)? Maybe f(ax + b) = (b,a), in this way f((ax + b)(cx + d)) = f(ax+b)f(cx+d) = (bd, ad + bc) which is quite similar to multiplication in A.

[dire siren]

it's not just similar; it's identical, but you miscalculated the first component

my bad, you didn't miscalculate; I just had something similar in my head

your map works just fine

in my head it was f(a+bx)=(a,b)... the same thing

[chilly ocean]

Oh okay

[hollow mica]

How do we get the subgroup of G/(a) with order p^(alpha-1)

[cobalt heath]

Hmm, it says "By induction"

[hollow mica]

base case is obvious

induction step idk

[cobalt heath]

Maybe check up on Induction hypothesis?

[hollow mica]

Suppose for every group G with v_p(|G|) = n and element x of order p we have that there exists a subgroup of G/(x) of order p^(n-1).
Then if v_p(G’) = n+1 we have for any element x of order p that G’/(x) has order p^n, so if y is an element of G’/(x) of order p then (G’/(x))/(y) has a subgroup of order p^n

That’s not exactly what we want though

[cobalt heath]

Well

Can you identify the variable induction is applied over?

[astral apex]

how do i show that for any nilpotent $G$, nontrivial $N \lhd G$, that $N \cap Z(G) \neq {e}$?

[cloud walrus]

mwago

[hollow mica]

I’m trying to induct on n, which is alpha in the problem

[cobalt heath]

Yeah so

Let's say, we apply induction on Sylow's first theorem over alpha.

Taking the exact statement as P(alpha)

Then base case is P(0), and induction step is P(alpha - 1) => P(alpha).

Can you identify P(alpha - 1)?

[hollow mica]

wait in the proof they just use induction for a lemma, not the whole proof

[cobalt heath]

Lemma?

[hollow mica]

the existence of the subgroup of G/(a)

[dull ginkgo]

It’s strong induction

[cobalt heath]

Well, doesn't seem so to me

[dull ginkgo]

I didn’t know it’s possible to use a worse proof of Sylow than Jacobson but here we are

[cobalt heath]

It is a bit of abuse of statements, but it seems like it is inducting over the whole theorem

Which makes sense, you want stronger induction hypothesis.

[dull ginkgo]

It’s inducting over the existence of the Sylow-p subgroup for smaller orders, no?

[hollow mica]

Ok i don’t know what’s going on in the proof lol

[dull ginkgo]

Okay homie so our assumption is that every group of order less than |G| abide by the hypothesis

[cobalt heath]

Largest p-power factor of |G/<a>| is p^(alpha - 1)

[vivid tiger]

Lang probably has a worse one.

[dull ginkgo]

I.e there’s a subgroup of order p^k

[hollow mica]

yea

[cobalt heath]

I do think it is an induction over alpha

[dull ginkgo]

So, we have our conjugacy class equation: $|G| = |Z(G)| + \sum{[G : C(x_i)]$

[cloud walrus]

Request a new nickname
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dull ginkgo]

The size of the conjugacy class is the index of some centralizer of an element in G

So we have two cases

[hollow mica]

Ohh

[dull ginkgo]

Assume p DOESN’T divide |Z(G)|

Then it must NOT divide one of those centralizer indices

[hollow mica]

the induction part is literally the statement of the induction hypothesis

sorry

[dull ginkgo]

Okay do you get it now

[hollow mica]

yea

[dull ginkgo]

Sick

[hollow mica]

ok what

I like this proof

[cobalt heath]

Huh

[hollow mica]

how can it be made better

[dull ginkgo]

group actions

[hollow mica]

bruh are you talking about the first proof on the wikipedia

[cobalt heath]

At least they could have made the statement less confusing

[hollow mica]

which uses some shit called cummers theorem

[dull ginkgo]

Does Wikipedia use it

[cobalt heath]

By beginning with "induction over alpha" etc.

[dull ginkgo]

I don’t think you need the full strength of Krummer’s theorem

[hollow mica]

yea but once i got it made sense

[dull ginkgo]

Actually I think Krummer’s theorem can indirectly be proven as a corollary

[hollow mica]

Hm I’m try to go back and understand rn

[dull ginkgo]

Though it’s easier just to shortcut it with Krummer’s theorem there’s a stronger proof

Jacobson makes it an exercise to prove it that route

You know shit’s fucked when the exercise is easier than reading the goddamn section

[cobalt heath]

Wth

[cobalt heath]

Is Jacobson so bad

[dull ginkgo]

I implore you to try it out

The Krummer’s part is below it

I admittedly like guided proofs like this

[cobalt heath]

Stab A

That would hurt a lot for A

[chilly ocean]

Guided proofs are great

[dull ginkgo]

This proof of Sylow is a bit longer but just kind of working it out made it click for me

But it required a lot of partitioning lmao

• Power set gets partitioned into sets of the subsets of the same cardinality
• p^k-sets get partitioned into cosets of its stabilizer
• family of p^k sets get partitioned into sets with stabilizers of the same order
• each of the equivalence classes gets partitioned into orbits under the action

[tough raven]

bump

Any ideas?

[tough raven]

Note this clarification.

[last spoke]

It depends on what you mean by those functions being equal. As formal objects we might distinguish them anyways, but as @daring nova pointed out it's easy to find examples in finite fields. In Fp all elements satisfy a^p = a, so for any polynomial (and consequently any rational function over Fp) P(x1,...,xn) in Fp[x1,...,xn] we get the same evaluation pointwise under the map xj > xj^p

[tough raven]

I mean equality in the transcendental extension generated by the indeterminates, i.e. in the field of fractions of the polynomial ring generated by the indeterminates, i.e., ‘formally’.

[still dew]

I have had no progress yet on this

My attempts were looking at the kernels of those maps and trying to determine sth

Also I haven't been able to make use of the fact that their commutators generate them

How's that important

[fiery eagle]

Hello 👋

I am trying to understand this statement. Specifically the example that (Z,+) is generated by 1.

I can't convince myself this is true, because 1+1+1...+1 would never generate a negative number, so it cant generate all of Z. Am I miss interpreting the phrase "generate"?

[celest furnace]

Recheck the definition of a cyclic group. n in Z also includes negative n right? So you would also get (-n) * 1 as well

[fiery eagle]

I was getting lost in the sauce of definitions, you are right. I see now how we can get to generating Z. Thanks so much.

[toxic zephyr]

so we have this result from linear algbera (T is a linear transformation). isn't this a more general fact about homomorphisms, though? like, isn't this kind of a direct application of the fundamental homomorphism theorem?

[lament bough]

yeah its just a consequence of T being a homomorphism of the underlying groups

[drifting mason]

I have a very basic question. I'm reading about quotient groups on Wikipedia. I do not understand the first equality underlined red. Why is (ab)N=a(bN)? Nowhere have they defined this associativity to hold for cosets. Grateful for any explanation.

[lament bough]

it holds for cosets because it holds for elements

[south patrol]

yes, if you think about what each means it's clear

[lament bough]

in general, if you have subsets A, B, C of G (not necessarily subgroups) , then (AB)C = A(BC)

[south patrol]

like (ab)N is the set of elements (ab)n

[drifting mason]

ok 👍 so on the one hand we have {ab x n: n in N} and on the other hand {a x c: c in bN}. So I guess we can say that all elements in bN are of the form b x n and thus {a x c: c in bN}={a x b x n: n in N}

[lament bough]

exactly

[obsidian wadi]

do you guys have any good resources on abelion sets and intro to this topic

(videos esp)

[chilly ocean]

what are abelion sets?

[delicate orchid]

It might be another word for an abelian variety, I’ve heard algebraic varieties being called algebraic sets before

In which case, #algebraic-geometry will be more useful than here

[obsidian wadi]

oh i see, i thought it would be here bc we are doing sets and fields and stuff

[chilly ocean]

If I have something like the quotient k[x,y,z]/(y-x^2,z), since z = 0 in this ring can I say that this is isomorphic to k[x,y]/(y-x^2) or do I need some extra conditions to be able to say so?

[drifting mason]

I'm studying the group $\mathbb R/\mathbb Z$. It is said that $\mathbb{R}/\mathbb{Z}$ is in $1-1$ correspondence with the set $[0, 1)$. The group structure carries over to $[0, 1)$ and it is addition modulo $1$. \

As far as I understand, addition modulo $1$ is the positive fractional part of a real number. But since it is called addition modulo $1$, I wonder if it is somehow related to modular arithmetic? Is there any connection?

[cloud walrus]

Philip

[south patrol]

I mean, it is mod 1 instead of mod n, I don't know what more you want

[drifting mason]

well, the modular arithmetic I'm familiar with is; let n be a positive integer, and x,y integers, then x=y (mod n) if n|(x-y). Can we just swap x,y to be real numbers?

[last spoke]

R is a field so divisibility isn’t interesting

[dire siren]

you are correct; you can use the third isomorphism theorem to justify it

[last spoke]

It’s related to “regular” modular arithmetic in the sense that in both cases you have numbers equated when they differ by a certain amount

[drifting mason]

ok, so what does addition modulo 1 mean? I don't think I've seen a clear definition of this. At least I haven't been able to find an article about this on Wikipedia.

[delicate orchid]

if you have a quotient group G/H people sometimes say "modulo H" to refer to the inherited operation

I find this is more common with rings

Z as an abelian group is generated by 1, so they're just shortening it by only referring to the generator

[drifting mason]

ok 👍

[crystal vale]

I am not sure about notation R-linear combination when R is a ring

[delicate orchid]

it just means things that look like $\sum r_ix_i$ with $r_i \in R$

[cloud walrus]

Wew Lads Tbh

[crystal vale]

Okay, thank you

[chilly ocean]

how do I verify these compositions of all the possible combinations of rotations and flips?

My guess is that D_n is a subgroup contained within the group S_n

hmm does that help?

so certain permutations are allowed but not all without deforming the polygon

this subgroup is closed under the operation of composition

that would mean that it wouldn't be anything other than rotation and flip

[delicate orchid]

rotations and reflections have different cycle types after you embed them into S_n, you can use that to prove this

or consider the representation using 2x2 matrices - rotations have determinant 1 and reflections have determinant -1

[chilly ocean]

I am not familiar with this, sorry

[delicate orchid]

or take the subgroup generated by the smallest rotation and quotient D_n by it, you'll get two cosets - one of rotations and one of reflections

that's probably what they want

[chilly ocean]

the book haven't even introduced these topics yet

subgroups is in the next section

I am confused

I knew the concept beforehand hence I was able to think of it

[languid trellis]

An idea might be to use matrices

[delicate orchid]

_ _

[languid trellis]

When you Bury the relevant idea in a wall of text I am not going to see it boss

[delicate orchid]

skill issue

[languid trellis]

OK I am being hyperbolic I just didn't read what you said tbh

[delicate orchid]

wait maybe I should read the god damn image

use that homomorphism they've just defined

it's very straightforward

[chilly ocean]

I just want to verify those composition operations

they are closed

now how do I know which certain ones are which?

I can see it through visualization but that's not rigorous

[languid trellis]

Alex do you know what a matrix is and how to use one

[delicate orchid]

let phi be the homomorphism D_n -> {1, -1} as defined in your image (example 2.26)
let r,r' be rotations and f, f' flips
then phi(rr') = phi(r)phi(r') = 1, so rr' is a rotation
phi(rf) = phi(r)phi(f) = -1, so rf is a flip
and so on

[chilly ocean]

Yeah I am vaguely familiar with linear transformations

[languid trellis]

Yeah just read this

I mean this is the same thing as taking determinants of the appropriate matrices but wew is so right (as per)

[delicate orchid]

yeah, same thing as mapping to S_n and taking signs as well

[chilly ocean]

were you trying to construct a map? (sorry I am a bit new to algebra)

[delicate orchid]

who are you talking to

[chilly ocean]

swiftee

[languid trellis]

This is assuming what we're trying to prove?. Yes this si the final step ignore me

[delicate orchid]

both me and swiftee were coming up with different ways to get the map you've already been given

[chilly ocean]

Ah

[delicate orchid]

because I see a big block of text and my adhd activates my fight or flight so I didn't actually read it for like 5 minutes

[chilly ocean]

thought so

me adhd too 🫂

it seems like everyone has it these days

[delicate orchid]

nah I just think that mathematicians are more likely to be neurodivergent

[languid trellis]

Chat am I a mathematician

[chilly ocean]

[delicate orchid]

not yet

[languid trellis]

You're being nicer than usual

[delicate orchid]

maybe

[chilly ocean]

wait we have to verify the fact that it is a homomorphism and you are using the fact that it is a homomorphism?

isn't this circular?

[delicate orchid]

you're told it's a homomorphism

oh ok right yeah I really cannot be fuckin bothered?

it's the canonical surjection D_n -> D_n/<r>, which is a homomorphism as <r> is normal

it's the composition of D_n -> S_n followed by the sign homomorphism

define the action of D_n on the plane through how it transforms the n-gon, then it's the determinant which is a homomorphism

[chilly ocean]

hmm

[delicate orchid]

maybe actually post exercise 2.22?

@languid trellis did u know that the sign of a permutation is it's determinant as viewed as a F_1-module automorphism?

[last spoke]

What’s an F1-module

[chilly ocean]

...

[last spoke]

Something with cars?

[delicate orchid]

do you have a presentation of D_n

[rocky cloak]

F1 the mysterious "field with one element"

[chilly ocean]

yes but it's rather short

[last spoke]

F1 stuff sounds enticing but I can’t make sense of any of it yet in my learning journey

[delicate orchid]

ok then if we're defining a flip as just anything that isn't a power of a rotation then you should be able to do this from the presentation quickly

[rocky cloak]

An F1 module is just a pointed set, but a linear map between F1 modules is a map of pointed sets, that satisfies the first isomorphism theorem

I.e. if you mod out the kernel the map becomes injective

[delicate orchid]

writing the presentation as <r, f | r^n = f^2 = 1, frf = r^-1> we have (fr^n)(fr^m) = r^(m-n) which is a rotation, the other relations follow in the same manner

[rocky cloak]

They are somewhat useful for studying things over arbitrary fields, as "tensoring" F1 modules with arbitrary fields always gives you a vector space. So it's like describing linear maps that plays nice with a specific basis. Like permutation matrices for example.

And then you have some fun things like sign being the determinant, and Sn being GLn

[lament bough]

i hate hate hate the D_n exercise, its just a bunch of handwaving because "well its obvious there are no other symmetries" and its in the beginning of every first algebra course before you are able to prove the facts about planes and polygons rigorously

[last spoke]

@rocky cloak hmm interesting. Thank you!!

[delicate orchid]

agreed. It's dog water of the highest order

and I ain't talking like, a fine wolfhound

this shit is a pug

[delicate orchid]

these are the things I care about

you also have things like the burnside ring being the F_1-representation ring

and now I've just had a very scary thought about trying to do modular rep theory over F_1

like how much carries over

[languid trellis]

F_1 ? More like nonsense

Also that message made me feel really good about my algebra knowledge

[delicate orchid]

whyyyy

[languid trellis]

Because I only understood it vaguely

[delicate orchid]

oh it was sarcastic, in which case that's a good thing

F_1 is brainrot

[sly crescent]

F_6, on the other hand…

[languid trellis]

I thought you were talking about the field with one element which is why I said f1 is nonsense xd

[delicate orchid]

we are

[languid trellis]

I think there are some respectable mathematicians looking at F_1 though

I don't know anything precise, just will o wisps from reading other chats

[rocky cloak]

I guess that would just be studying G-sets, which is pretty well understood I guess... Idk

[delicate orchid]

yeah, I'd say they're completely solved tbh

what I want to do is go backwards, see how much block theory and brauer theory we can throw at it and how much of it will stick

this isn't a serious research suggestion I'm just posting words on the internet

[rocky cloak]

Words on the web

My favourite past time

[delicate orchid]

stealing this for my blog name if I ever make one

you will not be credited and I will be seeing you in court

[languid trellis]

'Words on the webs' sounds like tje name of a blog which has 3 posts from 7 years ago because the owner got bored and moved onto a different fleeting interest in his life

[delicate orchid]

exactly

[short tide]

Hi, recently started some galois theory and im doing field extension stuff rn, I just wanted to ask about what elements of fields of the form say Q[sqrt2] or like Q[a,b] in general. Ik elements of the first one are of the form a+bsqrt2 but how do I see this in general, like for eg how do we know elements of Q[cube root of 2] are of the form a+b times cube root of 2 + c times cube root of 2 squared. Notes dont rlly clarify this notation

[delicate orchid]

they're quotients of polynomial rings, so just treat them like polynomials

one moment please!

,w minimal polynomial of 2^(1/3)

[cloud walrus]

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[delicate orchid]

so Q(cbrt(2)) is Q[x]/(x^3-2), which has elements that look like a+bx+cx^2

so, passing that through the isomorphism, we have elements of Q(cbrt(2)) looking like a+b*cbrt(x)+c*cbrt(2)^2

[short tide]

Srry ig im trying to see how theyre essentially quotient polynomial rings cuz I have Q[a] defined as the smallest field containing both Q and a but I just dont see the link

[rocky cloak]

Intuitively, to get the smallest field you start with Q and a and do combinations of addition, subtraction, multiplication and division.

The first 3 are exactly described by evaluating polynomials in a (a polynomial is just a big sum of things multiplied together after all).

So the only question is where division fits in. But if a is the root of a polynomial, for example a^3 - 2 = 0. Notice how

a*a^2 = 2, so 1/a = a^2 / 2

So dividing by a is just multiplying by a^2 / 2. So division is already covered.

This brushes a few details under the rug, but it's the main idea.

[dire siren]

Q[a] is not a good notation because this normally means the smallest ring that contains Q and a, while Q(a) is the smallest field that contains Q and a
they are equal if and only if a is algebraic over Q

to see how the elements of Q[a] (in the sense of the definition I mentioned) look like, first notice, as jagr mentioned, that it must contain any p(a), with p in Q[x]
then convince yourself that these elements really form a ring, so by the minimality of the ring, you get Q[a]={p(a) : p in Q[x]}

now the elements of Q(a) are ratios of elements of Q[a]... so p(a)/q(a) with p,q in Q[x] and q(a)=/=0; if m is the minimal polynomial of a, then m and q must be coprime, so by Bezout there are u,v in Q[x] such that mu+qv=1, so q(a)v(a)=1, so p(a)/q(a)=p(a)v(a), which is a polynomial in Q[x] evaluated in a... this proves that Q(a)=Q[a]=polynomials in Q[x] evaluated in a

finally, by Euclidean division p=cm+r, with c,r in Q[x] and deg(r)<deg(m); and we have p(a)=r(a), so it turns out that evaluating polynomials from Q[x] in a is actually the same as evaluating polynomials from Q[x] of degree <deg(m) in a

[vivid tiger]

Q[a] is fine since a is a letter for algebraic things only

[south patrol]

Well it's a bit weird since usually people would still write Q(a)

[chilly radish]

I think i'm going insane

[summer path]

me too

[vivid tiger]

i already am

join me in insanity

it's better this way

[chilly radish]

I'm trying to prove a submodule of a semisimple module is semisimple, but I also want to prove something stronger. Suppose M' is a submodule of M. I proved that if
$$M = \bigoplus_{i\in I}M_i$$
Is the simple decomposition, then there is $I'\subseteq I$ such that
$$M = M' \oplus \bigoplus_{i\in I'}M'$$
Now I want to prove that $M' = \bigoplus_{i\in I-I'}M_i$. I want to start by proving every simple module $M_i$ for $i\in I-I'$ is inside $M'$.\\
I know that the intersection of $M_i$ with $\bigoplus_{j\in I}M_j$ is $0$ by independence, but I don't know how to rule out that $M\cap M_i=0$. I want to show that this implies $M_i=0$ but I am not able to show that

[cloud walrus]

ShiN 11R34

[vivid tiger]

I feel that that isn't true

Oh, no, now I think it is.

[chilly radish]

It is definitely true

the quotient that is isomorphic to M' is a direct sum of M_i for i\in I-I'

wait I think I know how

so call the sum over I' $M''$, now for $m\in M_i$ decompose $m=m'+m''$ for $m'\in M',m''\in M''$. Now since $M'$ and $M''$ by contradiction intersect trivially with $M_i$, then since we know they are semisimple, the elements $m',m''$ have a simple decomposition that doesn't have $M_i$ as a factor, but then everything is equal to $0$ by independence

[cloud walrus]

ShiN 11R34

[chilly radish]

this feels a bit....not nice

Cuz we first prove that M' is semisimple and only then do we prove the decomp

[chilly radish]

I think maybe I can prove the opposite inclusion first

Hmm

I'm starting to think this might not be true

Might only be up to Isomorphism

[south patrol]

Real

Hm

[chilly radish]

Which would make sense and I will try to proce

I think you can take a 2 dim vector space as an example

Cboose one basis

Then choose a different one with a common vector

But you can prove it's isomorphic through the quotient map I think

Because semisimple module decomps are only up to iso

Yea this example works

Ok I will try to prove this up to iso

[still dew]

I was able to do 1 I think the mapping is matrix maps to a+ib but I have no ideas for a map for bit 2

I think after knowing 2 , 3 is just exploiting the isomorphism and also the fact that this poly has roots in C

[south patrol]

Correct for 1

[south patrol]

Also correct

Can you find an invertible map taking A to an element of S? That's what I'd try

[still dew]

Like multiply some invertible matrix ?

Isomorphisms are meant to be invertible maps I don't understand what you mean by finding an invertible map

[celest furnace]

I don’t think that would be a ring homomorphism unfortunately

[still dew]

If you could pls elaborate more

[still dew]

Yeah right

[celest furnace]

You are most likely going to be looking for a different basis of R^2 and conjugating elements of S by that change of basis matrix

[still dew]

Ohkay so I need to find a change of basis so that A gets in the form of element of S

[celest furnace]

I think so

[still dew]

Okay so for getting around isn't it fair that I show that the jordan form of this is same as some other matrix in S

[celest furnace]

Sure

I think i might've found the basis and its less complicated than you would think so you could always just try to guess it as well

But yeah jordan form is definitely a less sketchy way of solving it lol

[still dew]

I am bad at guessing

[celest furnace]

Great idea actually

For later reference if you care (SPOILER ABOUT THE BASIS, ONLY LOOK IF YOU ARE READY!): ||I think the new ordered basis should be {e_1+e_2, e_2}||

[still dew]

Oh cool how did you find this

For the jordan forms idea the matrix in S has char poly (x-a+ib)(x-a-ib) so it's just diagonal I think

[celest furnace]

Yeah it doesnt work

[still dew]

And I just check the roots of char poly of A

[celest furnace]

I was thinking this is an issue because we need to keep the basis being real?

[still dew]

Oh right hmmmmm

,w x(x-1) + 12=0

[cloud walrus]

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Requested by nads5404

[still dew]

They are similar in M2(C)

[celest furnace]

Yes

Oh my god dude

Do u wanna invoke a crazy result about Frobenius Canonical Form

[still dew]

Umm idk that

[celest furnace]

Rational canonical form?

[still dew]

Yea

I have heard of that

[celest furnace]

Youve heard of that one ?

They are the same thing !

[still dew]

Didn't know that name

Okay but still how it helps

[celest furnace]

(2)

Reference is Dummit & Foote Page 477

I think this might be a Tad bit overkill but that's part of the fun right!

[still dew]

Sure

Thank you

For the help

[celest furnace]

You're welcome! This problem was really cool lol

Got to use a lot of stuff i recently learned about which is very nice

[still dew]

Glad to hear you might be interested in looking at a problem I posted previously which was left unattended it's from the same source

[still dew]

.

[coral steeple]

Are there polynomials over non integral domains with more roots than the degree?

[agile burrow]

Consider f(x) = 2x in Z/4Z[x].

[coral steeple]

Ah yes

[next obsidian]

What. How

[coral steeple]

Oh wait yeah that polynomial is just 0

Lol

[hollow mica]

Wdym

[coral steeple]

Nevermind my brain somehow thought that said Z/2Z for a second

[hollow mica]

i think chamonke is joking

[celest furnace]

Sir does this mean that integral domain isn't real?

[agile burrow]

What

[vivid tiger]

not an ID

[next obsidian]

What. How

[vivid tiger]

4 is prime.

[next obsidian]

https://m.youtube.com/watch?v=JHiX50KOKP0

[vivid tiger]

looks, 4's only factors are 1 and 4.

another way to see this is that 4 is odd and >2 and so is prime.

[tough raven]

Reminds me of the Ace Attorney meme lol.

[coral steeple]

Is divisibility antisymmetric (up to associates) only in integral domains?

[tough raven]

I doubt that it fails to be antisymmetric in every non-integral domain, but as far as I know, there is no large class (or at least, a class basic enough that everyone is taught about it) of non-integral domains for which it's true.

Hmmm.

Suppose a | b and b | a, then a = ax for some x. If x is a unit then a and b are associate (assuming commutative rings here), so the question is to analyse to what extent that can fail.

[coral steeple]

Sorry yes I did mean to ask whether there are non integral domains where divisibility does not have this property

[tough raven]

If (0:a) = {x | ax = 0}, then ax = a precisely for x in 1 + (0:a). That consists solely of units iff (0:a) lies in the Jacobson radical, so that's one sufficient condition for divisibility to be antisymmetric up to associates: for every a, the ‘measure of non-cancellativity’ (0:a) should lie within the Jacobson radical, if not 0.

[tough raven]

Oh, then no.
If b = ax and a = by then a = axy, b = bxy.

If a is non-zero then this implies that xy = 1 => x, y are units => a, b are associate. Similarly if b is non-zero. And of course if a, b are both zero then they're associate anyway.

[coral steeple]

How do we know xy=1?

It's not uniqueness of 1 because we do not necessarily have rxy=r for all r (just a,b)

[tough raven]

a1 = axy

a is cancellative, if non-zero

[coral steeple]

Is that true if the ring is not an integral domain?

I know it holds in IDs: if a=axy then a(xy-1)=0, so a=0 or xy=1

[hollow mica]

How do we define R_m R_n as the group operation in R_m and R_n can be different?

or here is R_m viewed as the subgroup inside R

composed of elements (0, 0, ..., r, 0, 0, ...), where the r is in the (m+1)th place from the left, and r is an element of R_m

well hmm in that case isn't the product of an element in R_m and an element in R_n just (0, 0, ...) ?

[mighty kiln]

Take C[x, y]/(x²y), then y divides y+xy divides y

[coral steeple]

Oof. Who'd have thought. Thanks!

[tiny jolt]

Remember that R is a ring

It gets its addition operation from the R_n additive groups

but all the multiplication needs to do is distibute with addition

A good example is k[x] is graded as k (+) kx (+) kx^2 (+) ...

think about what multiplication does there

[hollow mica]

oh so, the R_i have to be subgroups of R?

Because then you can multiply elements so R_i R_j makes sense

[tiny jolt]

We're using the group structure of the R_i for addition, not multiplication

but yeah R_i are subgroups of the ring R

[hollow mica]

But R_m R_n \subset R_{m+n} means that for x in R_m and y in R_n that their product (as in multiplication in the ring) is in R_{m+n} ?

[tiny jolt]

yeah

[hollow mica]

here what did you mean

cuz like yea in a ring multiplication distributes over addition

but is that all you were saying

[tiny jolt]

Was more trying to say that while addition is component wise, the multiplication isn't necessarily

[hollow mica]

i see

[still dew]

,rotate

[still dew]

The question as we can see has two parts I am not able to get through a single part

Why is this an integral domain

I was trying to do some maps like y=t and x=t

If it wasn't an integral domain then wouldn't this Q[t]/<2t^2-1> not be an integral domain

But 2t^2-1 is irreducible in Q[t]

Does this work

Also how does one find the isomorphism for the second field of fractions part

[prisma ibex]

show that x^2+y^2-1 is a prime element of Q[x,y]

[tough raven]

Do you know that Q[X, Y] is a UFD?

[prisma ibex]

rationally parameterize the conic x^2+y^2=1, this is standard stuff with pythagorean triples

[still dew]

Not standard stuff for me 😭

[still dew]

Yes isn't it more than a UFD tho

[still dew]

Ok it's just they took t=tanx/2 or sth

[tough raven]

In this case, the useful thing to know is that it's a UFD, because then
X² + Y² - 1 is irreducible => (X² + Y² - 1) is prime => your ring is an integral domain.

[still dew]

How to know a multivariate polynomial is irreducible

[prisma ibex]

you can use Gauss' lemma to prove that x^2+y^2-1 is a prime

or you can just prove it by hand

[still dew]

Yes pls tell me more

[prisma ibex]

Gauss' lemma tells you it's enough to show that x^2+y^2-1 is irreducible

and this is easy to show

[still dew]

Okay why is this easy to show ;-; how do we work with multivariate polynomial

[prisma ibex]

[still dew]

I bash taking (ax+by+c)(dx+ey+f)?

[prisma ibex]

work in (Q[x])[y], assume that y^2+(x^2-1)=(a(x)y+b(x))(c(x)y+d(x)) for some polynomials a,b,c,d and then derive a contradiction

[still dew]

Okay

[prisma ibex]

you should end up with something like A^2=-1 as your contradiction where d(x) is Ax+B up to a constant

[tough raven]

I suppose this is a neat way to show that it is an integral domain as well: the map X |--> (1 - t²)/(1 + t²), Y |--> 2t/(1 + t²) extends to a ring homomorphism from Q[X, Y]/(X² + Y² - 1) to Q(t).
If one shows that this in injective and the image contains t, then this embeds the ring as an intermediate integral domain between Q[t] and Q(t), whose field of fractions must therefore be Q(t).

[still dew]

Why is injectivity necessary

[tough raven]

Otherwise you would be only embedding a quotient of Q[X, Y]/(X² + Y² - 1).

[still dew]

Okay.....

[tough raven]

You would get that that is an integral domain with field of fractions Q(t), but that doesn't tell you the same about the original ring.

[still dew]

Right 👍

So I needed to make such a parameterisation that was injective

Okay but isn't showing this as injective very difficult

[tough raven]

I'm not sure.

Maybe this isn't the best way to go about it.

[still dew]

Okay

[still dew]

So this is my mapping from field of fractions of Q[x,y]/<...> To Q(t)?

x-> 1-t^2/1+t^2, y-> ....?

[still dew]

Okay thanks @prisma ibex and @tough raven

[fresh gate]

sorry for the ping

[still dew]

No ideas so far

[vivid tiger]

=>: bezout.

err

clearly this works for n=2

[still dew]

Yeah so q1,q2,
..qn exist such that p1q1+p2q2+...pnqn=1?

[vivid tiger]

n=3:

a b c
m n 0
0 l/m 1

det is na - mb + lc which you can make 1 by bezout.

[still dew]

Okay

So there is some inductive construction

[vivid tiger]

Not exactly sure how it goes, but you can probably figure it out.

<=: you have \sum x_i p_i = 1

but...

but...

R over F is a PID.

we did need to know that to even do bezout

so then done

as then gcd must be 1

[still dew]

We are assuming determinant is 1?

[vivid tiger]

that indeed is what <= is

so for => only question left is what the matrices look like in general.

[south patrol]

Interesting question

[still dew]

Also what is l/m here

Is this allowed

[vivid tiger]

l divided by m

oh shit

halfway through i realized we were working over a ring and not a field and then i forgot to go back

[south patrol]

sorry so have you two proved either direction? i could try helping too

[vivid tiger]

<= is just bezout

(easily)

that is, det = 1 => gcd top row is 1

[still dew]

Yes

The other side is non trivial

[south patrol]

for <= you can just take determinants i think lol

yeah

[still dew]

I don't understand