groups-rings-fields

[rotund aurora]

So you start with a ring, call its idealization I(R). Now I(R) is a semiring, and you can idealize it again. The question is: are I(R) and I(I(R)) isomorphic as semirings?

xd idk

[delicate orchid]

(a)+(b) = (gcd(a,b)), (a)(b) = (lcm(a,b))
Ok now my mind is thinking categorically these look like products and coproducts but w/e

[barren sierra]

they are coproducts and products 👀

[rotund aurora]

(4)(2)=(8) no?

the lcm is the intersection, not product

[barren sierra]

have an arrow x -> y iff x | y

[left estuary]

Yeah I was just looking for extra questions

Makes sense I didn’t read that many questions

[delicate orchid]

Actually yeah the way to go is probably categorically but that’s way too much thinking

Train internet being slow

[delicate orchid]

Yeah whoopsie

(a)(b) = (ab) which is way nicer tbh

I found this http://www.quasigroups.eu/contents/download/2018/26_29.pdf which looks at the ideal semi ring of a semi ring @rotund aurora

[formal ermine]

my algebra exam is in a week and I don't feel ready

do I just do more exercises?

[elder wave]

probably

[woeful sage]

finish every serge lang exercise and you'll be ready

[formal ermine]

I'm scared about the ring and module part

groups and galois is ok

also construction using ruler and circle (or compass idk what it's called)

[agile burrow]

good luck on your exams

[formal ermine]

it's just algebra lol

as in

it's the only exam

[elder wave]

you'll be fine

[chilly radish]

Straightedge is compass

That one is easy

If number is constructible it's contained ina degree 2^n extension

If the extension is galois the converse is true

And you use the contrapositive to show some numbers are not constructible

[rotund aurora]

Question 1: Construct the 65537-gon

I mean I guess straigthedge and compass is simpler since they are towers of quadratic extensions, but not necessarily easier

[past temple]

is there an elegant way to prove that every element in the dihedral group of order 2n is of the form r^m, sr^m?

[boreal inlet]

I think that can be proved from the generator relations?

Dihedral group is generated by two elements r and s which are Permutations, with r being an n-cycle (1 2 ... n) and s being product of flip transpositions except the vertex on the flip axis

Now however way you combine these two, elements will always be of the form r^m or sr^m

[chilly radish]

I mean I think the main point here is proving Dn is really generated by these 2 elements

If you define it as the group of isonetries of an n gon

[boreal inlet]

That gives us two relations I think

r^n = e = s^2, rs = sr^-1

And uh

Was there anything more?

[rotund aurora]

these completelly specify the group

[formal ermine]

my course has yet to do it

we have 2 lectures left

I don't understand this part of the argument

[delicate orchid]

Well it ain’t gonna be C_49 or C_7^2

[formal ermine]

wdym

[delicate orchid]

Wdym wdym

[formal ermine]

s_7 is either 1 or 8

how do we know it's 1

[delicate orchid]

Yus

That part doesn’t deal with that

The next line does

If it was n = 8 the entire group would have to be C_7^8 (order argument) thus abelian thus trivially solvable

[formal ermine]

why would the entire group need to be C_7^8

also is that C_(7^8) or (C_7)^8

[delicate orchid]

7^8 > 56

And (C_7)^8 is what I just wrote

Yus

Because it’s a group of order 56 that contains 7 distinct groups of order 8, right?

Fuck

8 distinct groups of order 7

That’s the ticket

If there’s some semi-direct memes going on then I might cry

[formal ermine]

yes

[south patrol]

Prove it

[delicate orchid]

No

[formal ermine]

I don't get it

[delicate orchid]

Yeah me too

Invoke burnside pq and proceed to never think again

[formal ermine]

why are 56 - (8 * 6) = 8 elements left

like why are all of the sylow 7-subgroups trivially disjoint

[delicate orchid]

If they weren’t their intersection would be a subgroup of both - which is impossible, C_7 has no subgroups other than the trivial

Sorry I thought that was clear

[formal ermine]

intersection of groups is a group?

[delicate orchid]

Intersections of subgroups is another subgroup yes

LEMMA: prove this!!

[formal ermine]

ah I see

proof: trivial.

what if their intersection is C_7

ah wait

lol that was a stupid question

then they would be equal and no longer distinct

we have 8 distinct subgroups

right

[delicate orchid]

Yus

[rotund aurora]

When does a irreducible iff (a) maximal hold for rings?

[formal ermine]

in pids because irreducible is equivalent to being prime there

[rotund aurora]

yeah right, a irreducible implies (a) is maximal among principal ideals

[rustic crown]

non-zero prime

[rotund aurora]

but that's it, correct? In general, the implication a irreducible implies (a) maximal fails, even for, say, Dedekind domains

?

[elder wave]

UFDs even

☝️

[formal ermine]

how can -8x^3 + 9x^2 + 3 in Q[x] irreducible by eisenstein with p = 3

if we look at it with gauß we have it in Z[x]

ah wait nvm

lol

I thought gcd(8, 3) = 3

[next obsidian]

You want irreducible => prime and your ring to be dimension one

[elder wave]

wait ignore this

i had a brainlag

[chilly ocean]

I want to write ${\langle; x,y,z,t ;\vert; xy=yx,x^6=z^3,x^4=t^5 ;\rangle}$ as an amalgamated free product. Can I just separate ${\langle; x,y ;\vert; xy = yx ;\rangle, \langle z\rangle, \langle t\rangle}$, give the isomorphisms ${x^6 \mapsto z^3}, {x^4 \mapsto t^5}$ and call it a day?

[cloud walrus]

Gev

[chilly ocean]

no wait

this doesn't work

[lavish nexus]

Is this presentation really enough to determine the entire group

[next obsidian]

I mean any presentation determines the group

[formal ermine]

why isn't it true in ufds

[chilly ocean]

this should most likely turn out to be an amalgamated free product but I don't see how

feels like we have different amalgamations between pairs of groups

z^6 = t^15

WAIT

[elder wave]

what you said is true for UFDs but we also need non zero prime ideals to be maximal

and every UFD where that holds is a PID already

[formal ermine]

what does that mean

[elder wave]

what

[next obsidian]

it's not true in k[x,y]

your ideal won't be maximal

[chilly ocean]

$\langle x,y,z,t \vert xy=yx,x^6=z^3,x^4=t^5 \rangle \= (\langle x,y \vert xy = yx \rangle \ast_{\varphi} \langle z\rangle) \ast_{\psi} \langle t\rangle, \ \phi: x^6 \mapsto z^3, \psi: x^4 \mapsto t^5$

[cloud walrus]

Gev

[chilly ocean]

this one should work

otherwise I quit math

[formal ermine]

every field homomorphism is injective because the kernel is an ideal

and fields only have two ideals

it can't be the field itself because then it would send 1 to 0 which would contradict it being a homomorphism

[chilly ocean]

fields only have two fields

[formal ermine]

so it has to be {0}

meaning that it's injective

yes?

[formal ermine]

[elder wave]

yes

[formal ermine]

ok nice

[chilly ocean]

minor spelling mistake

[long nebula]

is every matrix being uniquely decomposable as the sum of a symmetric and an antisymmetric matrix (when char F ≠ 2) analogous to the tensor product being decomposable as the sum of a symmetric product and an exterior product?

[south patrol]

Okay short answer yes, but here is a more like general ting relating to rep theory, idk how much you know lol

More generally, if you have a G-representation V (a vector space with a linear group action basically) then you're often interested in subrepresentations (i.e. bits on which G acts more easily) and in particular direct sum decompositions

In the special case of G = Z/2 with generator g, say, the point is that we can (given assumption on char not 2) decompose it into bits where g acts as multiplication by 1 or -1, e.g. since g^2 = 1 and is thus diagonalisable for field not char 2

In fact you can get a p explicit formula for V^G (i.e. elements v of V with g.v = v for all g in G) and in the special cases you gave it spits out what you expect

[I should say of course that here the group actions are given by either g.A = A^T or g.(v \tensor w) = w \tensor v]

Oh, and just to throw a third example into the mix: decomposition of functions F -> F into sums of even/odd maps (which again has the char not 2 hypothesis ofc)

:)

@long nebula

[long nebula]

oooh interesting

[long nebula]

yea I was thinking about this haha

that is cool

I just told my friend about decomposition into even/odd functions, so I will also tell him that the other things are actually the same thing too

[south patrol]

Yeahh

[long nebula]

gotta love mathcord for giving me things to blow my friends away with

[south patrol]

In fact there's also more generally a fact that like

[long nebula]

oh and this is also like every harmonic function being able to be written as a sum of a holomorphic function and an anti-holomorphic function :o

[south patrol]

well basically usually your conditions on the characteristic of the field correspond to primes dividing the order of the group

[long nebula]

(just looked it up)

[south patrol]

Yeah nice

[long nebula]

that makes sense

[south patrol]

I'm trying to think of a simple analogous thing for a number higher than 2 lol

[long nebula]

Lol

this kinda reminds me of fourier analysis on finite groups but idk why

[south patrol]

Oh well in a sense this is related to like

quadratic formula only really working in characteristic not 2

and cubic in char not 3 and stuff

[long nebula]

hmm that makes sense

[south patrol]

cause the proofs of solvability I know basically use representations lol

[long nebula]

gotta learn me some rep theory then

[south patrol]

Well

Basically this is within Galois theory

But the proofs can be phrased/motivated representation-theoretically lol

[south patrol]

It's fun :)

[long nebula]

ty spud!!

[south patrol]

np owo

[pliant forge]

is there a quick way to check if R_{\Gamma}^4 \subset (\rho) when R_{\Gamma} is the two-sided ideal containing all non-trivial arrows in the quiver

💀

[rustic crown]

idk anything quicker than just computing it lol

and that's not so bad

R_{Gamma}^4 is generated by paths of length 4 right. notice that in the quotient by (rho), there are no non-trivial paths of length 3 starting at the vertex 1, you can't do alpha twice and can't do beta before gamma. if you start at vertex 2, there is no choice but to take the first step using gamma, but then taking 3 more steps will definitely kill it as now you're on vertex 1.

[pliant forge]

ok fair

[glossy crag]

If C is algebraically closed and i^2=-1 with i\notin F, is it automatically clear that \sqrt{-a}/\sqrt{a} belongs to F?

I had to think about this for a minute, came up with the following:

If x,y are the square roots of a and -a respectively and g is the nontrivial element of Gal(C/F), then g(x)=-x and g(y)=-y, since x and y both generate C/F and g\neq 1 (so g has to move both), so g(x/y)=x/y, so x/y\in F and -1=(x/y)^2.
Because I had to pause and think about this and it wasn't in the text (which is usually very thorough), I just wanted to check whether there wasn't an obvious explanation I might've missed.

[oblique river]

That’s right. In general, for any field K, if K(sqrt(a)) = K(sqrt(b)), then a/b is a square in K and your argument works just fine

You can also do it directly without galois theory. If sqrt(b) is in K(sqrt(a)) then write sqrt(b) = x + ysqrt(a), for x, y in K. Square both sides: b = (x^2 + ay^2) + 2xysqrt(a).

Since sqrt(a) not in K, we have 2xy = 0, so either x = 0 or y = 0. In the latter case we conclude sqrt(b) in K, contradiction. So we come to b/a = y^2

Hmm i guess tbis doesnt work in characteristic 2 haha

[delicate bloom]

everything's a square in char 2

[glossy crag]

What does he mean by "if there is 2-torsion, then the field has characteristic 0" (C again alg. closed)? I can see "if there is only 2-torsion, then characteristic 0 or 2", since char p & alg. closed => p-torsion.

[oblique river]

No?

For finite fields, yes

[glossy crag]

x not a square in F_2(x)

[delicate bloom]

oh lol my blunder

[oblique river]

But there are infinite fields in characteristic 2 which arent perfect

[glossy crag]

Oh wait, nvm, for some reason I thought the Frobenius endomorphism is torsion.

[glossy crag]

So what does he mean by this?

[oblique river]

Idk if im familir with this kind of fact :O

[glossy crag]

Damn, what now

I'll repost this in ANT, maybe someone will know it

[oblique river]

Can you post more context? Is this supposed to be a statement avout a general algebraically closed field? Or is this like, a conclusion of some sort

[glossy crag]

The context was Artin-Schreier theorem

Ah, I think it's this: proof of AS shows that |C:K|<\infty implies |C:K|=2 and char C=0 (I forgot about this last bit for a minute), so an alg. closed field of positive char. can't have finite degree subfields and hence torsion elements.

[oblique river]

Ok cool yes i recognized the artin-schreier context but i thought tbis was maybe a step in the proof as opposed to a conclusion

[rotund aurora]

What chapters are relevant for commutative ring theory in Lang?

So there's one chapter on noetherian rings

and you have basic rings and modules in the first part

is there anything else I should be aware of?

DF seems to contain more on these, idk if I'm crazy

[lament dawn]

One chapter for lang is a lot of content

It goes quick

You can just flip through and see

[rotund aurora]

What is this notation? I don't understand it

[rustic crown]

yea non standard notation

just means $\not\subseteq$

[cloud walrus]

det

[elder wave]

[rotund aurora]

🤦‍♂️

[elder wave]

Why

[rustic crown]

ig they want you to think of A > P as there is an element in A which is not in P

but meh

.<

[rotund aurora]

Does (i) iff (ii) require commutativity?

(i) implies (ii) holds without commutativity, but I think the problem of the direction (ii) to (i) is that (ab) need not be the same as (a)(b), as ideals

?

[elder wave]

Isn’t ii) usually the definition for prime ideals in a non commutative ring

[rotund aurora]

yeah but take (i) to mean ab in P implies a in P or b in P

[rustic crown]

whta if you consider R = M(2, k),

a = [1 0]    b = [0 0]
    [0 0]        [0 1]

R is a simple ring, so only two sided ideals are 0 and all of R. take P to be the zero ideal. a and b are not in P but their product is.

and it's easy to see it satisfies (iii)/(ii)

cause very few ideals

[novel plover]

What does it mean to say multiplying complex numbers and adding angles mod 2pi gives the same algebra?

[formal ermine]

what?

[novel plover]

I know, it’s weird right? I saw it in my book last night. Hang on, let me show you

[formal ermine]

if you go around 2pi on a circle you get to the same number again

because 2pi = 360°

which is just one full rotation

[novel plover]

That’s right

But how does that relate to complex numbers?

[formal ermine]

do you know the polar form

[novel plover]

Yeah a + bi is the same as |r|cos@ + isin@ right?

Which is just |r|e^i@

[formal ermine]

your notation is weird

lol but yeah

and sin and cos have a period of 2pi

[novel plover]

So 2pi is like the identity element? Cause if you add 2pi it doesnt actually change the angle

Lol Im sorry I have no theta on my phone keyboard

[formal ermine]

then write out theta

[formal ermine]

2pi = 0 in a way, yes

360° = 0°

[novel plover]

Okay, so 2pi in this case is in some sense equal to 1 when multiplying complex numbers

[formal ermine]

the isomorphism is then given by theta -> e^(i theta)

[novel plover]

So, let me see if I understand the essence of Abstract Algebra.

Multiplying complex numbers might be hard. But if you realize that's the same thing as adding angles, it might make the calculations easier?

[coral shale]

no lol

[formal ermine]

the essence of abstract algebra? 😵‍💫

[novel plover]

Well, why care about Abstract Algebra? Why care about isomorphisms?

It must be a nice thing but I'm not sure exactly how because they give totally different answers

[formal ermine]

do you know what abstract algebra is

[coral shale]

u have barely touched on it

[novel plover]

I mean, I passed Abstract Algebra in school. I could do the problems without thinking too hard about what exactly I'm doing. That's the enlightenment I'm asking for. What really is the significance

[coral shale]

high school generally does not offer such a thing

even if it does its usually some elementary group theory

[rotund aurora]

This is almost like asking "why care about mathematics?"

[novel plover]

I mean, i^4 = 1 which is in some sense the same as pi/2 + pi/2 + pi/2 + pi/2 = 2pi , 1 = 2pi they're both identity elements

So I see there's some similar structure

[chilly ocean]

In algebra an isomorphism is analogous to an if and only if. A statement holds in an object iff the statement obtained by replacing the terms by their images along an isomorphism holds in the codomain of that isomorphism

[coral shale]

AA is the study of algebraic structures.

Impose a set of properties to define the structure and deduce more properties they must satisfy.

You havent seen the beginning of AA if you havent done the isomorphism theorems

The 1st iso thm especially, is the first 'big' result you will see

[novel plover]

I have seen it before, thanks

[coral shale]

Youre still talking very concretely here

It doesnt give me confidence you have done much AA

[novel plover]

Well you might have done a lot of AA but you can't exaplain simply what an isomorphism is I doubt how much you actually understand

[rotund aurora]

Your intuition is more or less right. What you are saying is that the additive group R/(2pi) is isomorphic to the complex unit circle under multiplication

[coral shale]

Discussing the implications of this would be enlightening if youre looking for the 'essence' imo

[chilly ocean]

So intuitively an isomorphism is basically a translator

[novel plover]

But like, why is it useful to have an isomorphism if not to write something complicated as something easier?

[rotund aurora]

and you can replace "unit cirlce" by any circle centered around the origin, of course. But the point with complex numbers is that you can write every complex number in polar form, so it is more interesting to look at the unit circle ofc

[coral shale]

An isomorphism in AA is analogous to equality in other areas

[formal ermine]

Lol

[chilly ocean]

Instead of proving the same result for every isomorphic structure, you can prove it for only 1 of them and then use isomorphisms to automatically prove a related one in any other isomorphic structure

[coral shale]

By defining it we can start talking about which structures are the same

[rotund aurora]

I mean, what I said doesn't provide anything new. It's just saying what you said but with more precise language

[formal ermine]

then what is equality equivalent to 🤓

[novel plover]

Ahhh, that makes sense

[chilly ocean]

For example, suppose you have two polynomials f,g and want to prove that af+bg=0 only if a=b=0. You can treat the space of polynomials as a vector space and use that isomorphism to prove this

[coral shale]

Another major result is cayleys theorem. All groups are (isomorphic to) permutation groups.

[chilly ocean]

yoneda moment

[novel plover]

Oh damn! Thanks for the example. That does sound really powerful

[coral shale]

So now, we can consider studying only permutation groups to obtain general results on groups

if im not mistaken rep theory exists because of this basically

[rotund aurora]

How is that related to representation theory, isn't representation theory just looking at groups as "subgroups" of general linear groups of some vector space (usually over C I guess) ?

[coral shale]

i havent taken rep theory so

[rotund aurora]

mmh I have read like the very basics once, but haven't gotten into it

[novel plover]

Is a Rubick’s Cube isomorphic to any groups?

[coral shale]

youre looking at actions on vec spaces right?

so like permutations?

thats why i thought of that

[rotund aurora]

But I mean, Cayley's theorem is nearly trivial, and the symmetric groups are absolute monsters (in the sense that they are so rich)

I think the point of rep theory is that linear algebra is "easy"

[coral shale]

The puzzle is related to a group structure, u can check on wiki what it is

its not trivial

semi direct product involved

iirc it can be split into some kind of product of 4 structures

orientation/permutation of edges/corners

[novel plover]

Honestly I just asked these questions to wonder how would I explain the importance, beauty and significance of Abstract Algebra to someone with no Math knowledge like my mom. So I’m sorry if my questions are not clear

[coral shale]

so interestingly you can decouple this in the group structure like how we would decouple it when thinking about the puzzle

[coral shale]

I think u might have to stick to analogies

I cant even explain the significance of math as a whole to my parents rlly

They are aware its 'important', but why, no clue

[rustic crown]

i can't explain it to me

i just say it's cute and forget about the question

[rotund aurora]

I don't think you can explain it tbh

[novel plover]

That makes sense though. If you wanna prove some algebraic property of a complicated set, you can prove it for an easier set if theyre isomorphic

[rustic crown]

another fun isomorphism is SO(3) = RP^3

[chilly ocean]

There is a group consisting of the all possible movements of a rubik's cube up to equivalence. As in, two moves are considered "equal" if they always have the same effect on the cube. This can be seen as another instance of isomorphism: moves that have always same effect on the cube are isomorphic. We just end up treating them as equal tho.

For example rotating a cube 4 times in same direction is "equal" or "isomorphic" to doing nothing

[novel plover]

Ah so twisting a face all the way round is the same as not twisting at all. Hahah. Okay

[chilly ocean]

When things are isomorphic i d say none is easier than the other. There is however a generalization of isomorphism called homomorphism, that does not preserve all the structure only some of it. By preserving less structure you might be able to reduce a question about a complicated structure to one about a simpler one

[rustic crown]

for example if you wanna rotate R^3 around some vector by an angle, this isomorphism helps you write down what it does it each point
you think of S^3 --> H as unit quaternions and R^3 --> H as the span of {i, j, k}. Then given a unit vector u and an angle q, define the unit quaternion Q = cos(q/2) + u * sin(q/2).
then the rotation around u with an angle q is given by conjugation v --> QvQ^-1

[novel plover]

Ohh. But it can make things easier to prove, like your example about af + bg = 0 => a = b = 0. Equally complicated structure though

[chilly ocean]

well its easier to prove if you already know the answer in the isomorphic object

but you could prove the same answer the same way in the other object, so in that sense its the same difficulty

[novel plover]

Hmm. Okay, it seems my feeling about Abstract Algebra making complicated things simpler is not quite correct

Unless you're talking about homeomorphisms

[chilly ocean]

homeomorphisms is topology, not algebra

i said homomorphisms and those are algebra

[coral shale]

it is in a way. You ignore all the 'noise' and look at only specific aspects of structure

R is a field, but here we dont care about it being complete

or stuff like that idk

[novel plover]

We don't really care if every Cauchy sequence converges

[chilly ocean]

completeness for ordered fields is stronger than Cauchy sequences converging actually

there are non (dedekind)-complete ordered fields where every cauchy sequence converge

the confusion is that the word complete has different meaning for ordered fields and for metric spaces

[novel plover]

Ordered fields just means there's some notion of an element being greater than/less than another element?

With all the usual field properties

[chilly ocean]

yes + some properties about that order and relating it to the field operations

[chilly ocean]

Also, one example that you might have seen without realizing: to prove the sum of 2 odd numbers is even, you can consider the non-zero homomorphism from the integers to {0,1} with the operation 0+0=0, 1+0=0+1=1 and 1+1=0. It can be proven that any odd number gets mapped to 1 and even to 0. Therefore if a,b are odd then phi(a+b)=phi(a)+phi(b) = 1+1 = 0 so a+b is even, where phi is the homomorphism

[novel plover]

1+1=1

[chilly ocean]

what just happened .-.

[chilly ocean]

sorry i meant 1+1=0

[novel plover]

Ohhh

So odd numbers are 1, even numbers are 0. Then an even plus an even is, well, 0+0=0. I see

[chilly ocean]

yeah this is an example of a quotient

[rotund aurora]

Why can he conclude that? I don't get it. Also > is \not\subseteq

also this is very unclear, because I think he means an ideal maximal with respect to the property A cap M = empty

so like you Zorn up the set {A an ideal : A cap M = empty }

This is the discussion between the statement of the lemma and the proof

[lament dawn]

That's the worst notation ever

[rustic crown]

imagine using > for non-transitive stuff >.<

[rustic crown]

yea, have to do that

[rustic crown]

i have a feeling now that the person means $\supsetneq$

[cloud walrus]

det

[rotund aurora]

nu

he wants to prove P divides A or P divides B

ie, A subseteq P or B subseteq P, so he assumes these don't hold and tries to derive a contradiction

unless I'm wrong

[rustic crown]

if these don't hold and AB is contained in P
then consider (P+A) and (P+B)
these are strictly bigger ideals than P and you have
(P+A)(P+B) is contained in P

(so this proves (iii) => (ii))

[cloud walrus]

det

[rustic crown]

is there no index for notations?

[rotund aurora]

uhh ok that step was contained in the other proof lol (I didnt read the proof of the other lemma because notation annoying + very intuitive)

[rotund aurora]

But still, this is not correct, because you cannot conclude A cap M!= empty simply because P is maximal

[rotund aurora]

no

[rotund aurora]

but this proof is different than the one I posted before, its for this lemma ^

[rustic crown]

You can right? P is maximal which doesn't intersect M. So any strictly bigger ideal has to intersect with M

[rotund aurora]

But you dont know if A is strictly bigger

[rustic crown]

that why i thought A > P means that

[obsidian loom]

quick question

i said the only binary operation that would make this a group is the one where x * y = e, i.e. y is the inverse of x

So if that wasn't the case, then x * y not in G, so (G, *) not a group

is that enough to conclude that there's exactly one

[formal ermine]

this is equivalent to saying that there is only one group with 3 elements

have you had lagrange's theorem yet

[obsidian loom]

I've had 2 lectures so far lol

[agile burrow]

I think you should describe why you can't have x * y = x or x * y = y

[obsidian loom]

and its an introductory course, the last thing we'll do is Homomorphisms

so no isomorphism

[formal ermine]

oh

[rotund aurora]

what kind of course does that

[formal ermine]

yeah then what you're attempting is fine

[obsidian loom]

I prefer that then having to wait until second year to see groups

[formal ermine]

(fwiw every group with prime number size has only "one possible operation")

[rotund aurora]

I see.

Im also first year and its annoying as fuck, because they didn't let me take groups

but anyway, most of the stuff I learn is not following classes

[tender wharf]

what on earth

[cloud walrus]

sean

[void cosmos]

show that (25)(34) = (12345)^n(125)^m for some m,n integers

[south patrol]

well they don't even commute so that won't be the general form of an element in the subgroup generated by them (i'm also not really sure how that'd be helpful anyway lol)

Well, there is only one 60 element subgroup and you should know it contains (25)(34)

[lethal dune]

Can you say anything about the order of your subgroup ?

Oh already did

[south patrol]

Indeed

So now perhaps one thing to try is to show that this actually is A5

though there may be easier ways than that lol

[cloud walrus]

sean

[lethal dune]

Alternatively you can show it has all 3 cycles

That should be doable

[south patrol]

An is generated by the 3-cycles of Sn

So if a subgroup of Sn contains all 3-cycles, it contains An

[pliant forge]

Is there a "simple" example of a field F and an infinite dimensional F-algebra which has finite length as a module over itself?

[lethal dune]

so you want an field extension E/F that is infinite dimensional and E is finite length module over E itself?

try C/Q

[pliant forge]

how would the filtration be

[lethal dune]

C/C has dim 1

[pliant forge]

oh bruh

thx

ig the question was just hard to wrap around

[rotund aurora]

So he is saying that $A\not\subset U$ with $U$ maximal in $\mathcal X$ implies that $A$ only has finitely many isolated primes. This does not follow alone from maximality of $U$

[cloud walrus]

Croqueta

[rotund aurora]

Isolated primes of A are primes lying above A minimal with respect to the property of containing A

you only know that A+U has finitely many isolated primes, since U subset A+U

[glossy crag]

Isaacs
based

[rotund aurora]

I cannot decipher that shit tho

I'm missing something big

because he is using the same reasoning all the time

but its flawed, it requires ring theory

and I dont see the ring theory anywhere

I mean, this kind of conclusions do not follow from maximality alone

maybe Im being dumb

[rotund aurora]

okay the author is definitely using > to denote this notation

so annoying

[rustic crown]

at least that is transitive

[rotund aurora]

I'll be honest, when you wrote that, I read too quickly, and I read it as $\not\supseteq$, because of the crossed equality lmao, thats why I debated it. Should of listened 🤦‍♂️

[cloud walrus]

Croqueta

[rotund aurora]

He does it again here

It must be that, otherwise its absolute crankery

but wtf is this notation

[chilly ocean]

How to check whether these groups are isomorphic?
$\langle a,b,c,d \vert a^4=c^4=b^2=d^2=1,ab=ba^{-1},cd = dc^{-1}, a^2=c^2, b=d, a^2b=c^2d \rangle \
\langle a,b,c,d \vert a^4=c^4=b^2=d^2=1,ab=ba^{-1},cd = dc^{-1}, a^2=d, b=c^2, a^2b=c^2d \rangle$

[cloud walrus]

Gev

[chilly ocean]

I'm given two dihedral groups of order 8, and I need to find two generalized free products of them with K_4 amalgamated that are not isomorphic

I see 18 different ways of forming these products but I'm not sure if I can check the isomorphisms between any

[rustic crown]

[chilly ocean]

[bright coral]

So over the holidays my mind has been tourmented with trying to prove that the inner product space of trig functions span L^2(R) and now I don't even know if that is a sensible question to ask since trig functions arent even in L^2(R) but I wanna prove the fourier transform works for all functions in L^2(R). So I find the inner product space way of looking at Fourier series is quite elegant but I have no idea how to prove the trig functions of discrete frequencies span L^2[-pi,pi] and furthermore it occured to me that although in L^2[-pi,pi] the inner products of the sines and cosines of discrete frequency is normalisable, in L^2(R) when you inner product the same trig function you get something unnormalisable so trig functions arent even in L^2(R) so that makes it questionable whether you can apply linear algebra to it at all. I inquired about this in a physics discord and was told that trig functions indeed do not form a basis for L^2(R) since it is not in L^2(R) but there do exist other sets of countably infinite basis vectors that span L^2(R) so it still counts as a hilbert space. I've also been told that the trig functions do form something called a "continuous basis" for L^2(R) and it has something to do with what is called a "rigged hilbert space" and the person didn't elaborate because he wasn't familiar with it himself. Is it even sensible to ask how do you prove trig functions span L^2(R) if the trig functions arent even in L^2(R)? If it is a sensible question, how do you answer it? In any case how do you prove the fourier transform can decompose all functions in L^2(R) into a sum of sines and cosines.

[paper flint]

This is not an appropriate question for this channel, try #advanced-analysis

[bright coral]

but cant the fourier coefficients be thought of as inner products? Thinking of fourier series in terms of an inner product space feels most natural to me.

[oblique river]

This is not an abstract algebra question

[bright coral]

i wouldve thought it was an abstract algrbra question because at least for fourier series you can think of periodic functions as abstract vectors whose basis are sine and cosine functions of discrete frequency under an inner product that is the definite integral of the product of two functions

[bleak abyss]

The content of what you're asking about is more the analysis behind it. Trig functions being in L^2 is a matter of convergence of integrals, for instance

[bright coral]

I see, thanks!

[bleak abyss]

You're not doing much with the algebra of inner products, and tbh in this setting there isn't much to do

[oblique river]

Just because you first learned about vector spaces in an algebra class doesnt make all questions which involve vector spaces algebra

[bleak abyss]

(Also for when you reask your question in analysis I'd recommend working on the readability)

[tulip cedar]

when it says something like "the group of automorphisms of L that fix K." what does fix K mean?

[rotund aurora]

fix pointwise

ie f(x)=x for all x in K

[formal ermine]

it's invariant under K

[rotund aurora]

Just a sanity check: Let $n=p_1^{v_1}\cdots p_k^{v_k}$, where the $p_i$ are distinct primes. Then $\sqrt{(n)}=(p_1\cdots p_k)$, correct?

[cloud walrus]

Croqueta

[rotund aurora]

also, technichally this question is nonsensical, because we are not provided a ring, but it is understood that the context is Z

[tulip cedar]

so its an invariant subgroup?

[formal ermine]

no

invariant like in linear algebra

just what croqueta said

[tulip cedar]

ohhh, wait i think i sort of understand

[formal ermine]

the reason we want it to be invariant under the base field is because then it fixes the coefficients of any polynomial over the base field

which then means that it basically permutates the roots that don't lie in the base field of the polynomials

(tl;dr roots get send to roots)

[rotund aurora]

👆

[tulip cedar]

it permutates them?

[formal ermine]

yes

that's why the galois group of a polynomial of degree n embeds into S_n

because the polynomial has n roots and so the galois group permutates those n roots

[tulip cedar]

oh wait i was thinking about it backwards, in regards to the base field

yeah that would make sense

in that context though not sure why the base field is needed

[formal ermine]

where does the polynomial live in

[tulip cedar]

oh, nice

[formal ermine]

if it lives in the polynomial ring over the extension field then well every root is in the field already

[tulip cedar]

so its just making life easier

[formal ermine]

it wouldn't make sense otherwise

Gal(K/K) is always trivial

[tulip cedar]

right

that definitely gives me a better understanding of it, i think

[south patrol]

Yeah

[formal ermine]

why do we need to specify T_x and T_y separately?

like wouldn't it suffice to just say Tx = 1 for all x

[coral shale]

Is there more context

but idk any rep theory so probs nvm

[formal ermine]

before that it talked about reps of S3 by looking at its generator <x, y, x^3, y^2, xyxy> and finding matrices R_x R_y such that R_x^3 = I blabla

[coral shale]

In that case, what do you mean by 'for all x'

for all x in what?

[formal ermine]

but here it says "every group" so idk if they're still referring to that x,y from before

in the group

a rep is a homomorphism G -> GL_n(C)

[coral shale]

x and y appear to have a specific meaning imo

but idk

generators?

x and y axis?

[formal ermine]

hmm

wait

in the next sentence it says

"we will see that every rep of S3 can be built from A, Sigma, and T"

A and Sigma were previously defined reps

so maybe they defined T like that for this?

[coral shale]

[oblique river]

My guess is that they are talking about reps of a specific group, like S_3, with generators x and y. For any group, to specify a representstion it suffices to specify what happens with the generators

[formal ermine]

yeah

[oblique river]

So, to specify the trivial rep, i will specift ehat it does to the generators: it sends them both to 1.

[formal ermine]

yeah

[oblique river]

yeah

Factor it

Galois theory over finite fields is “easy”. An irreducible polynomial of degree n has galois group Z/nZ, and for products of irreducible polys it’s cyclic of order lcm of degree of factors

Its cyclic permutations

The galois group is generated by frobenius

The subgroup is generated by an n-cycle

Oh yeah hahaha

Thats what makes finite fields “easy”

[formal ermine]

(does this hold even when n isn't prime?)

[oblique river]

Yes, it does

[oblique river]

Yes, thats the point

All galois groups of finite fields are cyclic

Because your poly could be thought of over F_2 itself — it has no roots in F_2 so it is either irreducible or factors into two quadratics over F_2.

The only irreducible quadratic over F_2 is x^2 + x + 1, so if it factors over F_2 it would have to be (x^2 + x + 1)^2

Which it’s not. So the galois group over F_2 is Z/4

And the splitting field is F_(16)

Which is a degree 2 extension of F_4

Hence it must factor into quadratics over F_4

[formal ermine]

how do we show this?

[oblique river]

By hand. The only finite fields are F_(p^n).

Prove that the galois group over F_p is generated by frobenius

[formal ermine]

I don't see how this would show the original statement

[oblique river]

Which part?

[formal ermine]

An irreducible polynomial of degree n has galois group Z/nZ

[oblique river]

Take an irreducible poly of degree n

Adjoin one root, that gives a degree n extension

[oblique river]

By this, that extension is galois and thus contains all the roots of f

And the galois group of F_(p^n)/F_p

Is cyclic of order n

[formal ermine]

why must the base field be Fp

why can't it be any Fp^k

[oblique river]

It can be

Then replace F_(p^n) with F_(p^(nk))

[formal ermine]

ah I see

why must the extension be galois?

I know that it's normal and simple

[oblique river]

Finite fields are perfect so all extrnsions are separable

Or just use the other definition, that the size of the automorphism group is equal to the defree of the extension

The order of frobenius is easy to calculate

[formal ermine]

how?

[next obsidian]

There’s lots of ways

But it follows from:
Classification of finite fields if you know it
Using the fact that there’s at most d d-th roots of unity

[oblique river]

What is the minumum d such that x^(p^d) = x for all x in F_(p^n)

If d is smaller than n, then you have p^n elements of a field which satisfy a polynomial of degree less than p^n

Which is bad news bears

[formal ermine]

ah so d must be n by euler

[oblique river]

Therefore frob has order n in the galois group of F_(p^n) / F_p

[formal ermine]

yeah

so it generates the entire galois group

[oblique river]

And since that galois group has size n

[formal ermine]

therefore the extension is separable

ok thanks

[oblique river]

No, you cant even talk about p not being prime here

Or else F_p wont be a field

Oh i see

[next obsidian]

Usually people use q if you want to include prime power I think

[oblique river]

I thought you wanted a general integer

Yes that is true for any finite field

[next obsidian]

I also was confused haha

Buncho what’s a source to learn about adeles

Cuz my class just blitzed through their definition

And it confusing

[oblique river]

I learned about adeles/ideles on the streets

[next obsidian]

Damn

Chicago streets are crazy mathematical

[oblique river]

Which is to say i just picked them up after sufficiently many random encounters haha

[next obsidian]

All i hear on the streets is “hiv/aids doesn’t exist”

[oblique river]

I think most class field theory books should have a good section on them

[next obsidian]

What’s a good class field theory book

[oblique river]

Lol uhhhhhhhh

[coral spindle]

You must live in adelaide

Ha

[next obsidian]

http://www.math.columbia.edu/~harris/website/courses1/mathematics-gr6657-spring-2023

Here’s my course syllabus. I’ve been told it’s “fast”

[coral spindle]

please laugh

[oblique river]

What do people use most often? Cassels frolich? Artin tate?

Neukirch?

[next obsidian]

Maybe cassels frolich?

Idk

I know a lot of ppl use Neukirch too

Yes

[oblique river]

Well

The relative frobenius is x -> x^(p^n)

If F_p^n is your base field

[next obsidian]

Wut isn’t that identity?

Ah yeah

[oblique river]

That is what generates the galois group

[next obsidian]

Yee

You can see that like, once you know it for Gal over F_p

Using some Galois theory theorem yeah?

[oblique river]

Yeah

[next obsidian]

Yeah yeah

[oblique river]

Ok im gonna go take a nap

Gl all

[next obsidian]

Only one subgroup of the right size

Gn buncho

[rotund aurora]

maybe the answer is disappointing, but I think you just list all irreducible polynomials of degree 2 and check divisibility

First you check if it has roots, if it has none, then the only possibility is that it has quadratic factors

maybe I'm saying complete bullshit and there's a smart way

uh btw I thought you were asking how to factorise it over F4

ah ok

yeah ok my bad, didn't read carefully

[south patrol]

For the first bit, just look at the explicit formula for the map Gal(K/Q) -> (Z/nZ)*

[south patrol]

Indeed

For the second part like think of another way of writing K cap R

Which relates to the previous bit

Yes

The index of K cap R in K is hence given by what

Ye

Np enjoy your walk, can I join

Calm

[sonic coral]

Am i on the right track with these? for Lemma 1, i think i needed to assume that the subgroup is nontrivial as well, but the Lemma wasnt written that way so maybe not. For Lemma 3, im not sure what is special or even how to show that the intersection is contained in H, i feel like it would also be contained in K too, which would be pretty easy to prove.

[formal ermine]

why is <a> = {a, e}? what if a has order 3?

[sonic coral]

then wouldnt it be <a> = {a, a^2 , e}

[formal ermine]

yes

but you're saying that it's always <a> = {a, e}

[sonic coral]

so it needs to be up to n-1 where n is the order of G?

[formal ermine]

no

just say that <a> cannot be {e} and therefore has to be G

as <a> is a subgroup of G

it's what you were probably thinking when you wrote that

[sonic coral]

so we do need the nontrivial assumption then

[formal ermine]

yes

[warm urchin]

Have to note that the order of <a> divides order of G

[formal ermine]

btw use \langle and \rangle in latex

to make to look better

[formal ermine]

?

[warm urchin]

Did i say something wrong?

[formal ermine]

no, what you said is factually correct, but it doesn't make sense in the context

[sonic coral]

yooo thats sick

[warm urchin]

He said order has to be up to n-1 and im just saying that the order of <a> divides G generally, sorry if its unrelated

[sonic coral]

this is just lagranges theorem

[warm urchin]

Yes

[formal ermine]

they said that it should be { e, a, a^2, ..., a^(n - 1) }

not group order n - 1

[warm urchin]

Oh i see , apologies

Missunderstood

[formal ermine]

yeah I can see where the confusion comes from lol

[sonic coral]

and this only works because we know every group has a cyclic subgroup right

[formal ermine]

huh?

<a> is cyclic by definition of a cyclic group

[sonic coral]

okay yeah

[rotund aurora]

Consider Z and localize at all the primes except 2. The ring thus obtained is essentially Q except that 2 is not invertible. (2) is still prime and is in fact the only prime and this is still a PID. In particular, there are just finitely many minimal primes.

Why is this not a counter example?

[tired sierra]

ñ

[south patrol]

I don't see the issue

Well okay so maybe the thing is

(0) is the unique minimal prime in the ring you gave

Not (2)

So indeed the union of minimal primes is just (0), coinciding with what the question says

[dawn latch]

how do i prove that the set A = {n² | n ∈ ℤ+} is closed under addition/multiplication?

[next obsidian]

It isn’t?

[dawn latch]

oh ok?

[next obsidian]

I mean you can show it’s closed under multiplication super easily but it isn’t closed under addition

[dawn latch]

for multiplication, do i let a, b ∈ A, then define a = (k)² and b = (j)² for some k, j ∈ ℤ?

i think i got it now

thanks

[white oxide]

suppose i try to show that <nZ, +> is isomorphic to <Z, +>. I construct the following function f: <Z, +> --> <nZ, +>; for m, n in Z, f(m) = mn. is it possible to show that this function is injective? for assume that f(a) = f(b), so an = bn. The obvious way would be to divide both sides by n, but we are under no assumption that n is nonzero. is my function wrong therefore? or was there a mistake in the problem

[south patrol]

Well if n is 0 then the original thing doesn't hold

Since nZ is just the set {0} which ain't iso to Z

[white oxide]

so we just say that is nonzero?

that makes sense

but it's a bit annoying that they didn't include that in the problem

[toxic zephyr]

my professor said that metabelian groups are similar to semidirect products in a way (certain kinds, at least), but i didn't really understand what he meant. would anyone here be willing to explain (if they know)?

[oblique river]

semidirect p[roducts of abelian groups are metabelian

[trail stump]

let G be a group, and let <a> & <b> be two cyclic subgroups generated by element a & b, such that there is no element c∈G that can generate <a> or <b>. Question: can <a> & <b> have a non-trivial intersection.

my intuition is no, you can’t. So I try to prove by contradiction: assuming there is some intersection, and say a^m=b^n. assuming m and n are coprime, if not, we can always just find one that generate a^m and b^n so their exponent are coprime and prove that there is some combination of a & b that can generate <a> and <b> but i am not very familiar and good with number theory so i am stuck in the end.

[tender wharf]

hold on

if no c in G exists that generate <a> and <b> then how are <a> and <b> subgroups of G

[trail stump]

well i mean no c that is not a or b

obviously a generates <a> and b generate <b>

but no c like c^n=a or b

[tender wharf]

I'm assuming G is finite here

[trail stump]

that probably would simplify the problem but i think it won't matter maybe?

[tender wharf]

hmm I think that doesn't matter

I would consider cosets actually

one nice property of cosets is that they constitute a partition on G

yeah sorry I should've led with that

[formal ermine]

you can take c = a^k for some k coprime to the order of a

so the order of a has to be prime

similar with b

they cannot have a non trivial intersection because prime order groups have no proper subgroups

[trail stump]

But what if c=a^k and k divide the order of a

[formal ermine]

then it doesn't generate the entirety of <a>

take |a| = 8 and c = a^2, then you only get e, a^2, a^4, a^6

[trail stump]

But I am not asking the intersection to generate the entirety of <a> or <b>

Just non trivial

[formal ermine]

I don't see how this is relevant

I'm arguing that the order of a has to be prime if there's no c that generates its cyclic group

[trail stump]

i guess it was unclear to say "such that there is no element c∈G that can generate <a> or <b>" but i mean there is no element in G that is also not in <a> or <b> that generate the cyclic groups

[formal ermine]

that's not possible

if c generates either then it has to be contained in either

[formal ermine]

this is wrong btw lol

my brain is in C^inf mode rn

[trail stump]

yeah my first thought is in order to prove by contradiction, the order of a and b has to not be prime

[formal ermine]

your question is flawed I think

what are the restraints on c?

[trail stump]

c cannot be in <a> and <b>

[tender wharf]

okay can we maybe see the original question verbatim

[trail stump]

but somehow c generates a and b

or either one

[formal ermine]

👆

[trail stump]

so i am eliminating that

[tender wharf]

if not tip is to try cosets

i have no idea if thats what you want

[trail stump]

why do i feel like this guy is a bot - -

[formal ermine]

if c generates <a> then c has to be in <a>

[trail stump]

not necessarily

<3> generate 3z but 1 generate 3

[formal ermine]

1 doesn't generate 3Z

[trail stump]

1 does

i mean

its bigger i know

[formal ermine]

you mean that <c> is a supset of <a>?

[trail stump]

yeah

[formal ermine]

can we see the original question please

[tender wharf]

am I a bot?

[formal ermine]

yeah this isn't what generate means lol

[trail stump]

let G be a group, and let <a> & <b> be two cyclic subgroups generated by element a & b, such that there is no element c∈G that can generate the elements {a} or {b}. Question: can <a> & <b> have a non-trivial intersection.

[formal ermine]

send a screenshot

[tender wharf]

take a picture

[trail stump]

ok yeah the original question is flawed

[tender wharf]

did you rephrase or translate it by any chance

[trail stump]

no

[tender wharf]

please

[trail stump]

ok i changed the cyclic group <a> <b> to the element a and b

[tender wharf]

it has become even more confusing

please could you send an image

[formal ermine]

Lol

[formal ermine]

that is just a screenshot of this message

[tender wharf]

is this supposed to be funny?

I meant of the original problem that your prof gave you or something

[trail stump]

oh i was just working on some random theorem and i just came up with this

completely irrelevant trust me

[formal ermine]

yeah well the question is flawed to begin with

[trail stump]

is it?

[formal ermine]

if you mean it like this in the current state then take any a,b in Z and c > max{a,b}

their intersection will be their lcm

[trail stump]

but 1 generate a though if a is in Z

[formal ermine]

that's why I said c > max{a,b}

yada yada a,b not negative

who cares

[trail stump]

if c is greater that a, how does c generate a?

[formal ermine]

that's the point

it doesn't

[trail stump]

right

so it's not a counterexample

[formal ermine]

the answer is yes: it can have a non trivial intersection

[trail stump]

right

i agree, only if c cannot generate a or b

and if its not a counterexample, you have to prove it

[formal ermine]

I'm not sure what you're saying

[trail stump]

the example you give me is exactly that i want to prove it

like i know tons of examples in Z that are exactly what this statement says

[formal ermine]

it can have a trivial intersection though

[trail stump]

so i give up find any counter example in Z

[trail stump]

for sure

but i am interested in a non trivial one

[formal ermine]

?

there are cases when it has a trivial and when it has a non trivial one

I'm not sure what you're trying to prove

[tender wharf]

is this needed to prove your original question or just as like

[trail stump]

if it has any nontrivial ones, then there must be a c in G that generate a or b, in the case of examples in Z, 3 generate 3z, 5 generate 5z, then obvious 1 generate 3 and 5

[tender wharf]

"a fun thing I thought of"

[trail stump]

a fun thing lol

[formal ermine]

this is yet another different question to your original one

you keep rephrasing it

[trail stump]

i dont think i am

i only changed from the cyclic group <a> to the element a

which is still huge i know

[formal ermine]

please write down your question

[trail stump]

ok i think this is my final version of it

the last bit of the statement i think i have said different things but they all mean the same

[tender wharf]

,rccw

[trail stump]

gracias

so i was trying to prove by contradiction and think there is some combination of a&b so it generates a & b

[tender wharf]

ok so you are saying if a and b are not subsets of some other cyclic subgroup of G

then a and b have nontrivial intersection

(assume I said cyclic subgroups on a and b)

[trail stump]

a and b are elements of G

[tender wharf]

subset sounds really

suspicious

now there's an iff

okay

come back when you have a clear theorem statement

and write it with TeX

[trail stump]

If $G$ is a group, $a$ and $b$ are elements of the group and they generate the respective cyclic group, $<a>$ and $<b>$ , then they have non trivial intersection only if there exist an element $c$ in $G$ such that $<c>$ is a superset of $<a>$ or $<b>$.

[cloud walrus]

endless

[tender wharf]

so now its exist?

[formal ermine]

https://math.stackexchange.com/q/4628128/943349

[tender wharf]

its already downvoted

[trail stump]

holy shit how??!!!

O o

[trail stump]

apparently I am also confused lol

[tender wharf]

I am confused by what you are writing

your theorem statement has changed multiple times

are you sure this is your final theorem statement

[trail stump]

yeah i am 100% sure

i mean i changed a few times

but the idea is the same

but there were some "flaws" in the languages

[tender wharf]

these flaws you speak of make it impossible for us to understand

you need to communicate your idea with precise language

anyway

[obtuse bear]

if i understand the question right, the result seems false

[tender wharf]

yup

consider internal direct product

[obtuse bear]

say G = S4, and a = (1234), b = (3214). Then a^2 = b^2 = (13)(24), but there's no cyclic group in S4 with higher order than 4

[tender wharf]

from here if they have prime orders (like what illum said like about an hour ago)

follows very fast

^ should mention they have to be distinct prime orders

[obtuse bear]

i mean unless <c> doesn't have to be a proper superset, in which case it's stupidly true regardless of the intersection of <a> and <b>, just take c = a or c = b

[tender wharf]

the language OP used certainly seems to suggest not a proper superset

@trail stump

[trail stump]

oh wow it's as easy as that, thank you so much! really appreciate it!!!

[somber sleet]

I know it's in german, but this is just the construction of polynomial in more variables. Can somebody explain me how the jota function works? I somehow don't get it

[rustic crown]

i think looking at this a little more abstractly could be helpful.

the iota map sends an element of the ring to the constant polynomial

that has the coefficient a for x1^0x2^0 ... and all other coefficients are zero

[rustic crown]

instead of worrying with these tuples, you could abstractly construct the monoid ring. So say R is a (commutative) ring and M any monoid, then you could construct the ring R[M].

the elements of R[M] are formal finite sums of r_i . m_i. and you define multiplication by forcing the distributive law to hold and by defining (r.m) * (s.n) = (r*s) . (m*n)

to get the usual polynomials, you could take M = N, the monoid of natural numbers.

and actually instead of thinking of it as N, use it's isomorphic copy M = x^N, i.e. the monoid {1, x, x^2, ....} with the usual multiplication

once you abstractly understand this, polynomials over multiple variables are quite easy to construct. the monoid would be replaced with N x N x ... x N (or again its isomorphic copy M = {x_1^i_1 ... x_n^i_n, each i_k in N})

(if you wanna think of this more abstractly, x^N is the free abelian monoid on one generator x, and the other guy is free abelian monoid on n generators x1, ..., x_n)

(you could also play this game with non-commutative monoids, for example if M = <x, y> is the free monoid on 2 generators, i.e. finite length words made of x and y, then R[M] is the free R-algebra on x and y. these are called non-commutative polynomials)

i'll stop now :p

[rotund aurora]

thanks

[elder wave]

To be honest I’m not sure if referring to Monoids is the best idea here det

[rustic crown]

.<

i just feel doing all the tuples thing concretely can be weird

but, yea, if this the monoid ring looks too abstract, then you may ignore what i wrote >.<

[median pawn]

let F be a field and consider the ring homomorphism from Z (integers) to F, mapping n to n.1

n.1 = 1 + 1 ... + 1 (n times)

if the kernel of this map is non-zero, then we can find a smallest positive number p such that p.1 = 0. we can show that p is prime

so clearly pZ is contained in the kernel, but why is the reverse inclusion true? i'm trying to show that the kernel is generated by p

[rustic crown]

say something is in the kernel and use division algo with that and p

(you can also do this by using that ideals in Z are always principal. so the kernel actually looks like nZ for some n)

[median pawn]

oh yes neat

x = pq + r, for some 0\le r < p. then, x.1 = pq.1 + r.1, so r.1 = 0. this contradicts the minimality of p

why is this true? if char F = 0, then F contains Q, otherwise if char F = p (prime), then F contains F_p

but it's not clear from this that there are no other prime fields contained in F as subfields

right?

[rustic crown]

char of F_p is p

so that can't possibly be inside Q right?

[median pawn]

agreed, but still need to argue that F_p and F_q aren't in F

for p \ne q, p,q primes

[rustic crown]

okie so what i mean is if E --> F is a map of fields, then this forces that both have same characteristic

in particular an inclusion of fields

[south patrol]

if a and b are chosen such that pa + qb =1 then...

[median pawn]

tbh we can just take the intersection of all subfields of F

that'll be a subfield too, and it doesn't have a smaller subfield contained in it

[rustic crown]

but i don't think that would answer this.

[south patrol]

It's also clear that like

Well okay nvm it'd be the same as I said lol

But like if F_q is a subfield then the kernel of Z -> F contains (q)

and so you can't get another prime

[median pawn]

yep cool! thank u

[stiff karma]

how can we "cancel" a^j the cancellation laws only work if S in itself is a group right?

[next obsidian]

You’re using the operation in G where cancellation holds

It doesn’t matter if S is a group or not because you’re using an operation which has cancellation

[stiff karma]

oh right yes, my bad, thanks.

[next obsidian]

No need to apologize

[formal ermine]

ok so for the proof that $\tilde{v} = \frac{1}{|G|} \sum_{g \in G} gv$ is $G-$invariant, I don't quite see where we need the $\frac{1}{|G|}$ term in the front

[formal ermine]

I know that like when v is already invariant then we need it for obvious reasons

[south patrol]

You don't need it

[formal ermine]

ah

so it's just like

that when v is already invariant

we just get v back?

[south patrol]

Yeah indeed, this gives you a projection V -> V^G

[coral spindle]

We don't need it.

[formal ermine]

is every invariant vector of the form [ c \sum_{g \in G} gv ] for some constant $c$?

[formal ermine]

I wouldn't think so

but not sure

[formal ermine]

as in like

they all lie on a line

yeah would make sense actually

[coral spindle]

Suppose v is invariant. Then gv = v. So v = 1/|G| sum gv.

[formal ermine]

cuz it's like eigenvectors with eigenvalue 1 no?

[formal ermine]

yeah

ok nice

this was simpler than I thought lmao

thanks

so like I know that every operator on a finite dimensional nonzero complex vector space blabla has an eigenvalue

this is "needed" for this (is it?)

but I don't see where exactly it's being used

[coral spindle]

It's not being used anywhere here yet.

[formal ermine]

yet 👀

[coral spindle]

Algebraic closure is relevant to rep theory.

[formal ermine]

hm

but then like why would this argument fail when we just say if we can construct some rep to a real space and get some operator then it has an eigenvector with eigenvalue 1 by this computation

is it because our matrix is invertible

I remember that invertible <=> 0 isn't an eigenvalue

but do we have something about the existence of other eigenvalues

[coral spindle]

This operator can only have eigenvalues 1 and 0, since it is a projection.

[formal ermine]

why must it be a projection

[coral spindle]

Because it is idempotent.

[formal ermine]

why is it idempotent

[coral spindle]

I suggest you try to prove it is idempotent from the observations we made above.

[formal ermine]

I don't see how it is idempotent because we would need rho(g) rho(g) = rho(g), because rho(g) is invertible we'd have rho(g) = Id?

[coral spindle]

We are talking about the operator defined here.

[formal ermine]

yeah I can see why that is idempotent

because it's already invariant we just get |G| v

but we have the cancellation term at the beginning

so just v again

[formal ermine]

I don't see the connection of that to this question

[coral spindle]

Then I don't understand what you're asking.

[formal ermine]

we construct a rep to some real space. we can then use the computation from above to construct an eigenvector with eigenvalue 1 for some matrix in the space. yet not every matrix in a real space has eigenvalues?

[coral spindle]

We cannot use the computation above to construct an eigenvector for every matrix.

It is true that the action of every group element has an eigenvector, provided that the space is nontrivial of course.

However, this cannot be used to show that fact.

The reason is very simple: there is no guarantee that this does not produce the 0 vector.

[formal ermine]

ahhh

this was what I was looking for

thanks

[formal ermine]

what's your question

[chilly ocean]

we get 4 roots of x^4+1

but i can make two of them from the other two

so will the degree be 3?

[formal ermine]

no

[chilly ocean]

okay so the roots are

(1/sqrt2)(1+i)

[formal ermine]

we can do this simpler: the roots are i, i zeta, i zeta^2, i zeta^3 for zeta a primitive 4th root of unity

[chilly ocean]

1 is not the root

is it?

[formal ermine]

I read x^4 - 1

3 is still not the solution though lol

[chilly ocean]

yes

i think it should be two

[formal ermine]

no

[chilly ocean]

one zeta and then powers of it

[formal ermine]

you forgot i

[chilly ocean]

then what it is

[formal ermine]

instead of trying to find a basis we can use the fact that [K(a) : K] = degree of min poly of a

[chilly ocean]

yes

[formal ermine]

then we can just use the tower law [Q(zeta, i) : Q] = [Q(zeta, i) : Q(zeta)] * [Q(zeta) : Q]

[chilly ocean]

okay

so you mean that

by zeta and i we can make all roots

[formal ermine]

ah wait my brain is in C^inf mode right now i isn't a root

[chilly ocean]

yes that is true

[formal ermine]

i^4 = 1

[chilly ocean]

uhmm

okayy

roots will be (1/sqrt2)(+-1+-i)

[formal ermine]

the 8th cyclotomic polynomial is x^4 + 1

[chilly ocean]

cyclotomic?

[formal ermine]

polynomial that has all of the nth primitive roots of unity

it's irreducible in Q

so our roots will be generated by some primitive root of unity

[formal ermine]

and because it's irreducible we can use this

[chilly ocean]

yes

[formal ermine]

but this approach is good too

just look at the roots then manually find the splitting field

[chilly ocean]

yes but will Q[(1/sqrt2)(1+i)] contain all the roots

[formal ermine]

no, that's not the splitting field

also, [] stands for ring, () for field

[chilly ocean]

okay

[oblique river]

Some authors use [] here, too. The ring generated by any algebraic number over Q is automaticallt a field

And that is the splitting field…

[chilly ocean]

is it then

[glossy crag]

If A<=B<=C are division rings, then the degree formula holds the same way it does for fields, right? Or am I taking crazy pills?

[chilly ocean]

how can say that (1/sqrt2)(1-i) is in the field

[oblique river]

It’s the inverse of (1/sqrt2)(1+i)

[chilly ocean]

yes

[oblique river]

yes

[chilly ocean]

yes

okay then it's fine cus other ones are additive inverse and multiplicative inverse of additive inverse

then the degree is 2

[oblique river]

No 😭

Where did you get 2 from…

[chilly ocean]

1 and zeta

two basis elements

[oblique river]

What about zeta^2?

[chilly ocean]

[qzeta:Q]

hmm

[oblique river]

You cant write the multiplicatice inverse of zeta

[chilly ocean]

it should be of the form a+b zeta

[oblique river]

As a linear combination of 1 and zeta

[chilly ocean]

so we'll need to write them separatly

[oblique river]

Do you know that x^4 + 1 is irreducible?

[chilly ocean]

yes

over Q

[oblique river]

There’s your answer then

A root of an irred poly of degree d generates an extension of degree d