groups-rings-fields

[round hull]

there is no multiset M, such that {1, 2} applied to M, gives M

no that's not true, we can always count each element in Z infinitely many times

[white oxide]

thought i shouldn't post it in the geometry channel or whatever since this is a section on field extensions, but when they mention obtaining new points by, say intersecting two distinct lines in Li(M), do they mean include all the points of each of the lines in Li(M)?

certainly it would be because then i couldn't think of any "new" points lol

[rocky cloak]

No, they mean include the intersection point. They're intersecting the lines

[glossy crag]

Aaaaayyy, fellow Lorenz enjoyer. Be prepared, it's a pretty tough book (part 2 in particular). Also I find it neat that he uses "F" for most theorems, it has a quaint archaic quality to it.

[karmic moat]

why does rank-nullity not hold for arbitrary commitative rings?

[peak root]

Consider the short exact sequence between k-vector space 0-->kerT--->V--->W--->0, it always splits because vector space over a field is always free (it has a basis). But if you work in R-modules, not every R module is projective for a general ring R, so the short exact sequence may not split.

[karmic moat]

oh true thanks

[rocky cloak]

I guess it depends what you mean by rank-nullity when not over a field, since there is no concept of dimension. But the first isomorphism theorem still holds.

[next obsidian]

Rank nullity holds for length

[tardy hedge]

As im learning more abstract algebra im noticing im looking at mathematical objects in a different way than i have before, did anyone else feel that?

We are doing stuff on polynomials and now i look at polynomials as just like a kind of number system , more removed from the “function” aspect of it

Its kind of interesting

Does anyone relate and does that continue as you learn more?

[glossy crag]

It's a perfectly natural paradigm shift that every one goes though. In fact, you don't need "abstract" algebra to encounter this, since already in analysis and linear algebra one often deals with spaces of functions.

[tardy hedge]

Yeah thats cool

[supple phoenix]

How do u do question number 1

[glossy crag]

A field homomorphism fixes 1 (by definition), think about what they entails for the prime field.

[tardy hedge]

Can we have polynomials that have the exponents be some weirder thing

Instead of just integers

What if the exponent was also any field or smth

[wraith cargo]

a polynomial is defined to have integer exponents but you can have exponents of elements by some weird things

[supple phoenix]

So we use the fact that the prime field is a subfield of K then we restrict the automorhism on it

Since the homo fix 1 we can use the other homo properties to show that the image of psi is the prime field

[crystal turtle]

The homo fixes 1 🏳️‍🌈

[sturdy mirage]

I'm asked in an exercise here to show that two... things?.. are isomorphic. But as what? As A-modules? A[X] modules? As rings? Is there information missing here, or should this just be obvious...

(here rings are commutative, and with 1)

[crystal turtle]

This should be as B modules. Since the point is that you're extending the scalars on A[x] to B, which turns out to give you B[x]

Iirc, at least. Not sure if they end up isomorphic as other structures too

[chilly ocean]

isomorphic as sets

[cobalt heath]

Maybe it could also be A-algebra isomorphism?

[sturdy mirage]

as A-modules they're isomorphic because A[x]⊗B = (⊕A)⊗B = ⊕ (A⊗B) = ⊕B = B[x], correct?

[next obsidian]

Yes

Sorry no, as B algebras even

[sturdy mirage]

so are you saying it's isomorphic in pretty much every way imaginable

[summer path]

chmonkey seems so much less tilted outside of AG channel

[rocky cloak]

They're isomorphic as A-modules, B-modules, rings, A-algebras, B-algebras. So take your pick.

Though my guess would be the exercise is asking about A algebras

[next obsidian]

[next obsidian]

I think B algebras

It’s the most natural one, and implies every other one

Do I seem tilted in AG channel?

[rocky cloak]

Doing AG in general is pretty tilted I guess

[chilly ocean]

hard to not be tilted when every other question is euclidean geometry

[summer path]

maybe i only remember the times chmonkey would call questions stupid

[tardy hedge]

The gcf of two polynomials are the same up to a constant multiple ?

So when talking about divisors o polynomials, those that differ by a constant multiple dont matter right?

[glossy crag]

Over a field, yes. Over a general integral domain, as long as they differ by a unit (which in the case of fields is any number \neq0). The relevant term is association/associates/up to associates.

[tardy hedge]

Ok cool

Rn we have been introduced to polynomials over a field and practicing division algorithm and finding gcf etc

So p(x) and c*p(x) are technically distinct divisors but they can be treated as the same ?

[glossy crag]

Yes. As literal elements of K[x] they're different, but all their properties related to divisibility are the same (whether they are divisors of some other polynomial, whether they're irreducible, etc.). That's why when dealing with polynomials over a field in matters of divisibility you can always assume the polynomial to be monic, i.e. have leading coefficient one (since you can always scale it appropriately if you wish to). Over e.g. Z you can't do this.

[tardy hedge]

Thank you

[white oxide]

Oh lol this was just an assigned reading

Since we’re going over Galois theory rn

[broken quartz]

Bedephull

[white oxide]

But I’ve heard good things about it and my prof keeps raving about it so I might continue reading

[white oxide]

It really is lol

Must complete the Lang quest tho 😤

[broken quartz]

Honestly I got kinda concerned when I sent that cuz it could be read as two different things

But glad u read it as beautiful lo

[white oxide]

Lol what were u going for

[broken quartz]

Beautiful

[white oxide]

Oh lmfao nice

[broken quartz]

also can be read as pedo 💀

[white oxide]

Yeah I thought that’s what you meant by the other thing but I wasn’t Tryna say it out loud 😭

[dim widget]

Over Z the analogy is multiplying by -1

[broken quartz]

😭😭

[glossy crag]

Well yeah, I just meant not every polynomial is associated to a monic one.

[glossy crag]

It's really a wild book (both volumes).

[white oxide]

How so if you don’t mind me asking lol

[glossy crag]

1. For starters it assumes absolutely nothing beyond a linear algebra course (although a German one, which in some cases are much more advanced than US ones) and by the end of the first book it's covered not only the basics of what is called "abstract algebra" in the US, but also most of what usually passes for a "Fields and Galois Theory" course and some rather advanced stuff too (including Lindemann-Weierstrass, transcendence degree, integrality, basics of AlgGeo, and so on).

2. It has a pretty interesting and somewhat unique outlook of having all of the material be subordinate to, motivated by, and developed through field theory.

3. The 2nd book goes 10x wilder, I've never seen so many advanced and disparate topics covered in a single book (excepting Bourbaki, but that's not really a textbook and more of a working reference).

[next obsidian]

What topics are in 2nd book

I wanna learn some deep field theory, but the only book I’ve seen cover the stuff I care about is Bourbaki Alg 2 lol

I found table of contents

[next obsidian]

It just does class field theory, lol

[glossy crag]

well not "just" though

[next obsidian]

I mean that it like

Just goes ahead and covers it

In an algebra book

Which is funny

[glossy crag]

yep, that's pretty wild ain't it

no ifs no buts no coconuts

[white oxide]

Holy shit

I’ll return to it then when I have enough algebra under my belt

Tbh tho Lang prolly covers all these things

But then again Lang is a reference book

[white oxide]

(trying to prove that i is constructible from {0, 1} = M) why exactly is it that we can say that -1 is constructible from M? as far as i'm aware, we're saying that it's obtained from an intersection of the unit circle with the x-axis on the left, but how can we say that since that's not a line constructible from 0, 1?

like i understand that the unit circle is constructible from 0, 1 sure

but by definition, don't we only include points that are intersected in lines and circles constructed from 0, 1, and in particular -1 is not contained in the intersection of any two distinct lines or points

like they say "intersecting R with the unit circle"

but R isn't a part of M

so how can we do that

oh wait

is it because we consider the real axis to be in Li(M)

oh i guess that makes sense

since it joins 0 and 1

well hm i'm not so sure about the complex axis tho

[rocky cloak]

-1 is the intersection between the line going through 0 and 1, and the circle center at 0 with radius 1

[white oxide]

ye ok that makes sense

why exactly would the complex axis be in Li(M) tho?

i can't really see what two distinct points it joins

oh

it's because we can consider it as a line intersecting the two distinct unit circles centered at -1 and 1?

wait no

because that would intersect at 0 which is not distinct

uhhh

where my algebraic geometrists at?

[buoyant echo]

hi all, when talking about the solvability by radicals of a polynomial, we only consider it over Q right, not R ?

[wraith oak]

I'm trying to determine how we define the equivalence relation here

Is this the correct relation?

[white oxide]

ye

[glossy crag]

You can consider it over any field, but usually it's Q, yes.

[wraith oak]

Does my proof make sense? I didn't make use of the hint given, so I have a feeling I'm misunderstanding something

[rustic crown]

nu, it asks you to show every ideal looks like "A x B" so you can't start the proof with "Let A x B be an ideal"

take an arbitrary ideal F in R x S and masage it until it looks like A x B for ideals A in R and B in S.

e.g. it's certainly false that any subgroup of G x H looks like G' x H'. you can have G = H and the subgroup is the diagonal, {(g, g) in G x G | g in G}

but in case of rings and ideals, situation is nicer

[wraith oak]

Alr, I'll try again

[white oxide]

can somebody explain to me what it means to "raise the perpendicular"?

[chilly ocean]

there is literally a picture

in ur picture

[white oxide]

yeah but what would the original perpendicular be then

[chilly ocean]

i don't know if i'm equipped to answer that

fine. i don't want to answer that

[rustic crown]

you have a horizontal segment [-1, r] and you slowly draw a perpendicular segment from (0,0) until it hits the semi-circle ><
so you're slowly raising it ><

if you had a point outside a line, you would call it "dropping a perpendicular from the point", since the point lies on the line, ig nicer to say "erect/raise a perp"

[white oxide]

ohh okay i thought it wasn't as straightforward lol

i guess i was overthinking it

[white oxide]

oh yea that makes sense

thank you!

[karmic moat]

can i do it swiftly?

[wraith oak]

I don't really know how to proceed from what I got. Any advice?

[rustic crown]

okie, so since you want to show F looks like a product of two thingies, (that itself isn't true if you weren't working with rings) you need to identify these two thingies

a priori we do not know if F looks like A x B. whether they're ideals then would come later

you know for sure F contained in A x B where A is "all the elements of R that appear in the first coordinate of some element of F" and similary B. but you would need to argue that they indeed are equal.

[glossy crag]

@rustic crown can maaaaaaay I DM you about something?

[rustic crown]

yee >.<

[next obsidian]

No you can’t

[marsh quarry]

if i have a group multiplication that’s just a constant map to some element of the underlying set, does this form a group? clearly not because inverses aren’t unique

however im not sure which axiom this violates

in particular i don’t think uniqueness of inverses is an axiom, just a consequence of the axioms

[rustic crown]

violates identity axiom

there must be some e,
then g * e = fixed, for each g. but it should have been g.

[marsh quarry]

nvm im a moron LOL

thank you

[wraith oak]

I completed 1 but cannot find a surjective function for #2. Any hints?

[chilly ocean]

cannot find a surjective function for #2
well, you're supposed to find one which isn't!

to start trying to find counterexamples, think of some choices of S you could make which have reeeallly simple ideals

[wraith oak]

oh right, that was a typo, my bad

is addition commutative in an ideal?

[chilly ocean]

well addition is commutative in any ring!

[wraith oak]

oh right lol

brainfart moment

[chilly ocean]

(not to say ideals are rings, but rather that if all the elements of a ring commute under addition then the elements of its subsets should also do so)

[crystal turtle]

If you don't require identity ideals are rings

[chilly ocean]

all rings are commutative and with identity

[crystal turtle]

As they should be, ofc

[wraith oak]

How do you start out solving this?

[chilly ocean]

assume such a ring homomorphism exists and get a contradiction

[wraith oak]

This was all I managed to derive from my assumptions. Am I missing something or this is all I need?

[chilly ocean]

f(a) = f(a) is certainly true. i don't see how this is enough - there's no contradiction

my advice is to think about the f(1) = 1 condition

if we left that out of the definition of a ring homomorphism, then you could very easily produce one from Z/5Z to Z (send everything to 0), so it ought to be the problem

[wraith oak]

ohhhhh

okk I'll try that

[tardy hedge]

Wait basic question but can u even have a group homomorphism from Zn to Z?

[chilly ocean]

[tardy hedge]

My boy im a new algebra er

Besides f(x)=0

[chilly ocean]

homomorphisms on cyclic groups are determined by where they send generators

this may help

something something orders of elements

[tardy hedge]

Oh ok so a homomorphism of a cyclic group would need to be cyclic

So if u define function Z2 to Z by like having f(1)=2 fhat wouldnt work cuz 2 is not a generator of Z

[dim thicket]

I can help if you would like. I am not an expert but I can try my best as I am learning that as well.

[tardy hedge]

Nice

[dim thicket]

I would really like to understand math a bit better. Could you help me out? Learning high school math 9-12 I think would help me. I need to start with freshman math so, if you can let me know, thanks!

[coral steeple]

Is it a coincidence that $|SL_2(\mathbb{F}_p)|$ are factorial numbers for $p=2,3,5$?

[cloud walrus]

person2709505

[tribal moss]

The order of SL_2(Fp) is (p-1)p(p+1), which happens to coincide with (p+1)! for p=2 and p=3; for p=5 it is more of a coincidence that p+1 happens to give just the two factors that would otherwise be missing from 5!

[coral steeple]

Ah, I see

I was trying to recognise a formula from a table of the order, which is why I was sort of grasping at straws there

[tribal moss]

First count the order of GL2(Fp) which is (p²-1)(p²-p).
SL2(Fp) is now the kernel of a surjective homomorphism onto the multiplicative group of Fp, so Lagrange's theorem tells us how its order to relates to GL2(Fp).

[prisma ibex]

yeah these are due to some exceptional isomorphisms for small p

[coral steeple]

Oh true, I should have seen that. Thanks!

[coral steeple]

Is there anywhere I can read about this?

[tribal moss]

SL(2,2) is isomorphic to S3 (though not for any deep reason, there just aren't that many groups of low order to go around) so the factorial makes good sense there.
However, SL(2,3) is not isomorphic to S4 despite having the same order: there's an element of order 6 in SL(2,3) but not in S4.
SL(2,5) is not isomorphic to S5 either: it contains an element of order 20.

[coral steeple]

Cool!

[summer path]

If you're trying to learn high school math, this is the wrong channel, and moreover, this is the wrong section of channels

[torn warren]

The red part is the textbook, and the blue part is what I interpret, is this the right logic? the red part is due to 3rd isomorphism theorem, so I move the red part into the blue part.

without 3rd Isomorphism theorem, we can't guarantee G_i is normal in G_i+1

[rocky cloak]

Yeah, wheter you want to think of it as the third isomorphism theorem or the correspondence theorem, when you have

G > H > N with N normal in G, then H/N is normal in G/N iff H is normal in G.

[torn warren]

thank you!

[vagrant zinc]

Name of book?

[crystal turtle]

B-but it says algebra!

[tardy hedge]

Wut

[civic ferry]

i think we should be referring to highschool maths as "numeracy" tbh

would avoid all of these kinds of confusion

[coral spindle]

Too late

People now use the word algebra

That's not gonna change for a long time

[woven panther]

If I have a group G of order (2^n)*m and a Sylow 2-subgroup of G, must the Sylow 2-subgroup be of order 2^n or can it be any 2^i for 1 \leq i \leq n?

In my textbook the definitions are really confusing, Im pretty sure it’s the latter

[crystal turtle]

Sylow p-subgroups specifically have the of the largest power of p possible for their order

So here, 2^n

[woven panther]

Alright thank you

[crystal turtle]

A 2-subgroup would be the latter

[woven panther]

Got it

[dim widget]

Fine we can change undergrad algebra to “the theory of hyper complexes” as noether intended

[supple phoenix]

How to solve the 2nd part of the question

[dim widget]

You have to think of some examples of cubic extensions

I guess you’re in a Galois theory class? If so it might be helpful to think about what the galois group of an extension like Q(cuberoot(d)) is

[supple phoenix]

We didn't do Galois theory yet, we only did field extension

[supple phoenix]

Let's say I find cubic equation we some roots "t" then [Q(t):Q]=3, how to show that t can't be write as cuberoot(d)

[dim widget]

That’s not the same as what you’re trying to show

What you’re trying to show is that there is no possible generator t in the field Q(\alpha) which is a cuberoot

anyway I will give this hint: how does complex conjugation act on the cube roots of a number?

[supple phoenix]

if the discriminant <0 then we have one real roots and two conjugate roots, right?

[dim widget]

What is the discriminant of the equation X^3 - a

[supple phoenix]

-27a?

no 27a

[dim widget]

Okay, so in the case of a cubic equation which comes from extracting a cuberoot you know that two roots are always complex, right?

[supple phoenix]

oh so 27a^2

[dim widget]

-27a^2

It is negative

[supple phoenix]

I think so yes

[dim widget]

This is because any cuberoots of a number have ratio a third root of unity

[dim widget]

So can you find a cubic equation with all real roots?

[supple phoenix]

Yes, hmmm let's say x^3-3x+1

[dim widget]

Okay great! That works

[dim widget]

So can you show that this cubic extension cannot be written as a cube-root extension?

[supple phoenix]

hint? like should I find the roots explicitly then show that they can't be written as cubicroot(d) or there is a better approach?

[dim widget]

that would not prove the statement that you want to prove

[supple phoenix]

Sorry yeah u already said that
ok hmmm what if we try to show there is no isomorphim between Q(a) to Q(cubicroot(d))?

where a is the root of a cubic

[dim widget]

That is what you want to show yes

There’s a way to do this using the difference between real and complex numbers

Hint hint

[supple phoenix]

Okay so from our equation that has 3 real roots let's say a
then we can embed Q(a) into C, right? with the image has to be a subset of the reals?

[supple phoenix]

the reals have no imaginary part?

[dim widget]

Yes in particular every embedding is totally real

No matter which root of the cubic you send a to it has to be real

[supple phoenix]

but if there is an isomorhpism between Q(a) and Q(cubicroot(d)) then two roots has to have a complex embeding?

[dim widget]

Why is that?

(Yes you are right)

[supple phoenix]

because if c is a complex root then Q(c) is a subset of C
like our equation is x^3-1=0 then Q(i) is subset of C but not R

[dim widget]

Er, basically

Anyway I think you get the idea

[supple phoenix]

Yeah thank so much

[dim widget]

The equation x^3 -1 is not irreducible btw so you have to be more careful

[supple phoenix]

yeah that's true, I was trying to create a quick example

Sorry one more question
how to find the degree of the extension of the last one over Q
I manage to do the first two and they were =

[dim widget]

What happens if you cube it?

[opal osprey]

I know nothing about modular representation theory, but I just stumbled upon a problem where I need to classify the representations of $\mathbb{Z}{9}$ and $\mathbb{Z}{3} \times \mathbb{Z}{3}$ in $V = \mathbb{F}{5}^{2}$.

Any ideas of how to do this?

[cloud walrus]

MisterSystem

[opal osprey]

Actually, just knowing at least one non trivial representation exists would be enough for me rn.

[dim widget]

Okay well there aren’t that many interesting representations in this case. Gl_2(5) has one element of order 3 up to conjugacy, and it’s just from (0 -1 | 1 -1)

So you can have both factors of Z_3 acting by this matrix

For Z/9 since there are no elements of order 9 the only representation up to isomorphism comes from sending the generator to this element

And it is not faithful

[dim widget]

Now you can subtract off some terms and square then simplify

[supple phoenix]

Okay I will do it

[dim widget]

I would subtract off 6 times the element that you started with to get rid of the cuberoot term, then subtract off 4, then multiply by the element that you started with

Etc

[supple phoenix]

By the element that I started with u mean 2^3/2

[dim widget]

No you have to subtract 6(cuberoot(2) + sqrt(2))

Anyway I think you can figure this out on your own

[supple phoenix]

Alright thank u for the hint

[glossy crag]

I don't get the last line. What it's saying basically is if r is maximal with Q_r\subset Q_m (this implies that Q_m=Q_r in fact), then the number of roots of unity in Q_m is r. I can see why this number is >= r, but I don't see the converse.

[wraith cargo]

use the maximality of r

[glossy crag]

Yeah I know, but how?

OK, I think I got it: if G is the torsion group of Q_m, then it's finite (because phi(n)->infty), hence cyclic (of order d, say), then Q_d<=Q_m and d<=r.

Still, I feel like there's some easy explanation I'm not seeing.

[wraith cargo]

if it's >r then there has to be some number larger than r such that mu_n is in Q(mu_m)

Remember that the group of roots of unity is finite cyclic
so there needs to be some generator of all these >r roots of unity
so r isn't maximal in this case

[glossy crag]

You're right, ignore my stupid reply.

What you're saying is literally a rewording of what I said above though, so it involves proving the torsion group is finite (which is perhaps not immediately obvious). What I was wondering about was if there wasn't some more trivial way of seeing it.

[dim widget]

You just said that this implies that Q_m = Q_r

Isn’t that enough?

[glossy crag]

Well yeah, but how do I know there are r roots of unity in Q_r.

E.g. there are 6 roots of unity in Q_3.

And before you say so, yes, there are [2,n] ROU in Q_n, the point of this proposition is to prove it.

[rotund aurora]

They showed that if x in Q(zeta_r) is a root of unity then x is an rth root of unity or a 2rth root of unity (the second case only possible when r odd)

[dim widget]

Well you know there cannot be roots of unity of order prime to 2r by degree counting, so then you are only allowed roots of unity or order dividing 2r… but all of these are just 2rth roots or unity. If r is even as I assume it is in your case then the 2 is not relevant, so we get that to roots of unity are just mu_r which only has r elements

[wraith cargo]

Isn't that trivial enough

Like

[glossy crag]

Well it's not complicated, you just have to know phi(n) goes to infinity, which is easy enough, I was just wondering if there wasn't a completely elementary argument that I might have missed.

[glossy crag]

If $G'$ is the abeliasation of a finite $G$ and $\widehat G$ is the character group, then the isomorphism $\widehat{\widehat{G}}\cong\widehat{\widehat{G'}}\cong G'$ (where we use the group isomorphisms $\operatorname{Hom}(G,A)\cong\operatorname{Hom}(G',A)$ and $\widehat{\widehat{A}}\cong A$ for $A$ finite abelian) is completely canonical and natural, right?

[cloud walrus]

Ocean Man

[torn warren]

dummit and foote

[rustic crown]

looks nice

[glossy crag]

If L/E/K is a finite extension and a\in K is a norm of L, is it a norm of E?
Edit: I think this works if L/K is Galois, write a=\prod_G gx, then you can write this product as \prod_{G/H}\prod_H r_ih_jx=\prod_{G/H} r_i(\prod_Hh_jx)=N_{E/K}(\prod_Hh_jx) (where E is the fixed field of H) and \prod_H h_jx belongs to E.

[silent bronze]

Hi everyone, small question

Is Z2 x Z2 isomorphic to S2?

Or, is there any abelian group isomorphic to Sn where n < the order of the abelian group?

[lusty marlin]

S_n is non-abelian for n≥3

Therefore it cannot be isomorphic to an abelian group

[coral spindle]

S2 is a group of order 2, so there's not even a bijection.

[lusty marlin]

No, they don't have the same order

[silent bronze]

Hold on let me rephrase

I wrote the question wrong

If G is an abelian group isomorphic to some SUBGROUP of Sn

Is there an example of G

Such that the order of G is greater than n

[coral spindle]

Yes

[silent bronze]

Which

[coral spindle]

The subgroup of S_5 generated by (1 2 3)(4 5) is of order 6

[silent bronze]

Right

Oh so it’s isomorphic to Z6?

[lusty marlin]

Find an element of order > n, that's all you need to do

[coral spindle]

Well it's cyclic, and of order 6 :)

[silent bronze]

Hmm

So Z6 is isomorphic to a subgroup of S5

That subgroup being generated by (1 2 3)(4 5)

How is that subgroup of order 6?

[coral spindle]

Because it has 6 elements

I'm not sure what the confusion is

[silent bronze]

It has 5 elements…?

[coral spindle]

It does not.

Are you forgetting the identity?

[silent bronze]

OH

The Sn group is really confusing to me tbh

I thought that Sn was a group of permutations on a set with n elements

[coral spindle]

It is.

[silent bronze]

So it contains all possible permutations of some set with n elements

Including the identity permutation

[coral spindle]

It does.

[silent bronze]

That’s all well and good

[coral spindle]

If it did not have the identity, it would not be a group at all.

[silent bronze]

But when you talk about subgroups of symmetric groups

Im not sure how you form those or even begin to think of an example of one

[coral spindle]

You can think about them in the way you would think about subgroups of any other group

The symmetric group is not special in that way

I already gave an example: the subgroup generated by (1 2 3)(4 5). Maybe you should try to write down all the elements of this subgroup to see how this works.

[silent bronze]

Okay I’ll try that out, I think that’s the issue, that I struggle to visualise what is meant by the subgroup generated by a permutation

[coral spindle]

Again there's nothing special about this being a permutation; this is the same as the subgroup of any group generated by any particular element

[silent bronze]

So you just keep taking powers till you reach the same element again as with any other group ?

[coral spindle]

As with any finite group, yes

For infinite groups this doesn't work (for the simplest example, see Z and the element 1)

[silent bronze]

Yes that makes sense because the elements have infinite order

Okay so I got to this point, and this is the identity

The order of a subgroup generated by this permutation is the lcm of the length of its orbits right ?

So that’s why this has order 6

[coral spindle]

Yes

[silent bronze]

Okay yea this makes sense, thank you

[karmic moat]

ana is my name and algebra is my game

[vagrant zinc]

yea

[steel pulsar]

Statement : Let G be the group of bijections of R into itself. Give a example of two elements of order 2 of G whose product is of infinite order

Correction : For a ∈ R, we set σ_a(x) = 2a − x. It is an involution. For a, b ∈ R, we have σ_aσ_b(x) = x + 2(b − a). It is a translation, of infinite order if b ≠ a.

I don't understand the question and the correction, can someone explain it to me please?

[tribal moss]

That "correction" looks more like an "answer".

[rocky cloak]

Maybe you can expand a little on which part you don't understand

[broken quartz]

Can someone tell me why the j-invariant has 1728 in the formula? My professor said he prefers it to have 1728 but it can be used without it too.

[steel pulsar]

1. What is the group of bijections of R into intself ?

2. What is "infinite order" ?

[tribal moss]

That's probably more #algebraic-geometry or at least #advanced-algebra.

[rocky cloak]

A bijection is a function that has an inverse. The bijections from R to R form a group under composition.

The order of an element g in a group is the smallest n such that g^n is the identity. If there is no such n, then the order is infinite

[steel pulsar]

OK thanks you
But even like that, I don't understand his answer

What do I need to know to understand his answer?

[tribal moss]

Some group theory.

[steel pulsar]

Well I read his course and learn the definitions

[tribal moss]

What do you find is unclear about the answer?

[steel pulsar]

Why he specifies “It’s an involution”
“It’s a translation”

The elements of order 2 that he took are a and b

why did he define the function 2a-x precisely?

[tribal moss]

"Involution" is another way of saying it has order 2, which the problem explicitly asked for.

[rocky cloak]

An involution is a function of order 2. The function 2a-x is the reflection about the point a, so is a natural choice

[tribal moss]

Pointing out that the product is a translation seems mostly to be to help you form an intuition about what it does (and why it has infinite order).

[tribal moss]

No, a and b are just numbers; they're not even elements of the group. The elements of order 2 are sigma_a and sigma_b.

[steel pulsar]

2a-x is order 2 ? but there is no root ²?

[tribal moss]

What do you mean?

[crystal turtle]

It's a group under composition. When you compose it with itself you get the identity

That's what order 2 means here

[tribal moss]

sigma_a has order 2 because it is not the identity but sigma_a·sigma_a is -- that is, sigma_a(sigma_a(x)) = x for all x.

[steel pulsar]

Ok I understand

why σ_aσ_b(x) = x + 2(b − a). ?

[tribal moss]

The LHS means, by definition, sigma_a(sigma_b(x)). You can unfold the definitions of those two functions and then simplify.

[steel pulsar]

I'm going to do some research on my own...but I find it hard to understand

[tribal moss]

This is just high school algebra!

[steel pulsar]

Seriously ? 😢

[tribal moss]

Yes. Sorry if that came across as abrasive -- I meant it as "it looks like you're overthinking this".

[primal beacon]

If you define elliptic curves as $\mathbb{C}$ quotiented by a lattice $\mathbb{Z}+\tau\mathbb{Z}$ The j-invariant is a meromorphic, 1-periodic function of $\tau$

si it admits of fourier développement starting at $-1$ now 1728 is just the coefficient of this term

[cloud walrus]

rayane

rayane

[tribal moss]

That doesn't explain why it has been defined with that factor of 1728. A function that was 1/1728 as large would have the same properties you stated.

[primal beacon]

I should have said

The j invariant was « found » as a natural polynomial of eisenstein forms, these eisenstein forms being the coefficients of the differential equation that the weierstrass rho functions satisfit

satisfy

the point is, if you try to construct well defined functions on the set of lattices in C you end up with this coefficient naturally. I dont know about the « deeper » reasons for that

maybe you could say that the weierstrass rho function are the simplest examples of periodic functions on a lattice and the j invariant comes forms that ?

[low wyvern]

I am having trouble with raising a cycle to a power. So say I have some cycle f, how would I find f^99 or something?

[white oxide]

isn't $M \cup \bar{M} = M$ if $M = {0, 1}$?

[cloud walrus]

okeyokay

[tribal moss]

Well, you'd start by subtracting the length of the cycle from 99 until the power is less than the cycle length ...

[low wyvern]

Ye it's ok, I got it now but thanks 🙂 . Just divide 99/(order) then use the remainder

[tribal moss]

Yes, but it doesn't say that M = {0,1} -- just that 0 and 1 are elements of M, but there can be other elements of M too.

[white oxide]

oh, i thought they were going off of their previous definition that M = {0, 1}

cuz they only proved that the set of points constructible from {0, 1} is a subfield

of C

[tribal moss]

I don't see any such definition.

[white oxide]

the second pic here

oh wait

nvm

i'm dumb

you're right

containing the points

right

lol

[tribal moss]

That says "let M be a subset of C containing the points 0 and 1". It doesn't say M cannot contain points other than 0 and 1.

[white oxide]

ye

am i missing something here? z^2 is still a complex number that doesn't belong to Q?

[tribal moss]

The argument is that instead of writing Q(z), you get the same subfield of C as Q(sqrt(-3)).

[white oxide]

ohh okay

yeah i guess that makes sense

both fields contain the other element

[silent bronze]

I’m really lost on part c

For context, this is the first part of the question which I’m good with and I’ve done it

[tribal moss]

Note that the relation c b^-1 a^2 = 1 will allow you to express c in terms of a and b.

[silent bronze]

Yea I get that, I’m just not sure how I go from a^-2b and conclude that that’s 1

[tribal moss]

What?

[silent bronze]

This is what I have done so far

[tribal moss]

You should conclude that a^-2b is c, not that it is 1.

[silent bronze]

The idea is to say that c is the identity and that way we can get rid of it as a generator

Sorry it’s sideways

[tribal moss]

I don't see why "c is the identity" would be true.

[silent bronze]

So like, in one of the questions I covered in class, we had this question

When we solved the last part, there were two different answers

Like this

So the prof said that there’s no way A^5 can equal A^4 unless A=1 (the identity)

So I’m assuming that we have to do a similar thing if we want to express the group in terms of just a,b and d
Hence, we need to show that c=1

[rocky cloak]

The group being generated by a, b and d just means that every element can be described as some product of a, b and d and their inverses. There's no reason to think that this means c will be the identity.

[white oxide]

can somebody explain to me how w is a square root of z^2 - 4 (at least i think that's what they're implying?) because w^2 = zw - 1 which is not equal to z^2 - 4

[silent bronze]

@rocky cloak So if we just say that c can be expressed as a product of a and b and their inverses

Then it’s sufficient to say that c is not needed to generate the group ?

[rocky cloak]

Yeah, that's enough

[tribal moss]

This should be clear from any reasonable definition of what "generate the group" means.

[silent bronze]

Yea it does make sense

So we have this basically

[rocky cloak]

By the quadratic formula w = [ z ± sqrt(z^2 - 4) ] / 2

So E = K(sqrt(z^2 - 4))

[tribal moss]

Wait, what? Why?

[silent bronze]

Uhhhh

So I rearranged the equation to make c the subject

Oh wait

[tribal moss]

Yes, that much is good. But where do you suddenly get c²=1?

[silent bronze]

Nvm I see what I did wrong

Yep yep yep sorry ignore that

[white oxide]

ohh okay thanks

[silent bronze]

C^-1 is b^-1a^2

I’m also a little sus on my solution to (b), would you mind just having a look through it ?

I haven’t used the hint from the question, is that an issue ? Or is my method fine?

[white oxide]

the complex numbers are quadratically closed, but it's not true in general that any subfield of the complex numbers are quadratically closed right

oh wait yeah just take Q

lol

[steel pulsar]

Statemment : Let n be an odd integer > 3 and let G be the group of set bijections from Z/nZ to itself. Provide an example of two elements of order 2 in G whose product is of order n

Answer : For [a] ∈ Z/nZ, we define σ_[a] ([x]) = [2a − x]. This is an involution. If n = 2k + 1, then we have: σ_[a]σ_[a+k+1] = [x] + [1]. It is a translation of order n

I don't understand why :

1. We choose n = 2k + 1

2. σ[a]σ_[a+k+1] ([x]) = [x] + [1]

3. We put brackets everywhere in this exercise

[tribal moss]

That's not a choice -- n was a given odd number. Saying "if n=2k+1" is just a way of saying "let k be the number such that 2k+1 is the odd number you've been given".

[rocky cloak]

This exercise is a little weird, since it's not really important that n is odd, other then that they really like 2a-x.

You could just as well use f(x) = 1-x and g(x) = -x, to get f(g(x)) = x+1

[steel pulsar]

Please explain to me in more detail how to solve this type of exercise and the path to take in mind.
I have 5 like this and I would really like to understand

[wraith cargo]

dumbass question
But are the profinite integers abelian

[steel pulsar]

They specify that we are "let G be the group of bijections setists of Z/nZ"

and n> 3

[rocky cloak]

I don't think it involves that much cleverness. Like you just pick two involutions, then tweak the examples a bit until they're composition has order n.

[silent bronze]

:”) would anyone mind letting me know if I HAVE TO use the hint in the question to prove the statement by induction?

[silent bronze]

Like is the maths valid :””) plz

[white oxide]

can i get a hint for (i) => (ii) please

so I wrote \sqrt(a) = x + y\sqrt(b)

and i know i want to show that \sqrt(a) = \sqrt(b)c

for c in K

and i'm stuck

i have four relations that I can make use of:

$\sqrt{a} - \sqrt{b} = x + y\sqrt{b}, \sqrt{a} + \sqrt{b} = x + y\sqrt{b}, \frac{\sqrt{a}}{\sqrt{b}} = x + y\sqrt{b}$, and $\sqrt{a}\sqrt{b} = x + y\sqrt{b}$

[cloud walrus]

okeyokay

[white oxide]

all of them have led me nowhere

[woven panther]

Probably because your elements are residue classes

[white oxide]

anyone have a hint?

[white oxide]

that's tough

[steel pulsar]

an involution is not an element of a group that, when multiplied by itself, produces the identity of the group ?

[tribal moss]

I think "involution" is usually only used about functions that are their own inverses.

I'm not sure there's complete consensus about whether the word also describe the identity itself.

[crystal turtle]

No I've seen it generally for groups

Or any unital binary operation really

[tribal moss]

Hmm, anyway, that difference is not relevant here, where the group explicitly consists of functions under composition.

[white oxide]

can somebody tell me how this would lead to a contradiction? How is K^n not the union of finitely many one-dimensional subspaces, namely those which are spanned by the standard basis vectors?

[alpine island]

The union of two subspaces is not a subspace

Consider in $\mathbb{R}^2$, $\text{span}{[1,0]}\cup\text{span}{[0,1]}$

[cloud walrus]

nHail

[alpine island]

visually, that would be the x axis and the y axis combined

but nothing between them

[white oxide]

ohh ok

[tribal moss]

Bringing projective spaces into it seems to make it look unnecessarily deep, though. Just note that if a is some fixed element of E\K, then 1+ka for k in K all lie in different one-dimensional subspaces.

[white oxide]

yeah i ignored that part

[white oxide]

hm ok i'll think about this

thanks

[steel pulsar]

If I Have f(x) and g(x) two functions, say that the translation is of infinite order, returns to showing that f(x) o g(x) ≠ x ?

[coral spindle]

As you've phrased this question, it doesn't make sense to me

Please try again

[steel pulsar]

you talk about me ?

[alpine island]

Yeah I don't get that either, what are f,g functions to and from?

[alpine island]

yes

[steel pulsar]

it's about my problem up

it's question for troposphere

[alpine island]

here

[tribal moss]

@coral spindle it refers to this problem, later generalized to Z/nZ instead of R.

[coral spindle]

The question frankly still makes no sense to me

[alpine island]

Is R a ring or the real numbers?

[white oxide]

are there no vector spaces $V$ with $V = \bigcup_{i = 1}^n S_i$ where each $S_i$ is a subspace of $V$ and $n > 1$?

[cloud walrus]

okeyokay

[coral spindle]

There are many such vector spaces

[white oxide]

oh wait yeah

[coral spindle]

E.g...... finite vector spaces

[white oxide]

just take V itself

[white oxide]

wait so what's the contradiction here

😭

[alpine island]

if S_i is every subspace, yes

[white oxide]

oh wait

what if each S_i is a 1-dimensional subspace of V

[coral spindle]

.

[white oxide]

right

wait so where's the contradiction in the hint they gave 😭

[coral spindle]

The field K is infinite!

[white oxide]

wdym

yeah i know that K has infinite elements

[alpine island]

this implies K is finite

but your question says K is infinite

[white oxide]

but like you could have a infinite vector space (in terms of cardinality) made up of finitely many subspaces right

[coral spindle]

No

[white oxide]

am i trippin

[coral spindle]

Provided dim V > 1

[white oxide]

LEMMA TIME

[alpine island]

even if you could, that doesn't hold for all. and this is a statement about all infinite fields K

wait

no that's wrong kinda

[crystal turtle]

lemma time

[tribal moss]

If V is a vector space over some field K and dim V >= 2, then there are at least |K| different one-dimensional subspaces of V.

[alpine island]

Okay, take K^2 as the vector space. then your family of subspaces can each be spanned by [1, a] for distinct a in K.

these are all linearly independant, and there is an infinite number of them.

for K^n, you can extend this to [1, a, 0, 0, 0, ...]

[wraith oak]

Does anyone know how I can write my proof more clearly? I get the problem conceptually, but I don't know how to say it clearly

[tender wharf]

a neat trick is to observe that for a+bi + I, if a or b > 1, you'd remove a bunch of multiples of 2's until it is 1

(i ignored negative numbers here but its not hard to fix)

so any a+bi + I is really just

(a mod2) + (b mod 2)i + I

[wraith oak]

Yeah that already makes sense to me

[tender wharf]

that's what I'd write at least

not much shorter but

[buoyant echo]

hi all, for a polynomial p(x) to be solvable by radicals, suppose p(x) is over field F and E is the splitting field of p(x) , then the galois group of E/F is solavble is necessary and sufficient, right?

[coral spindle]

necessary and sufficient for p(x) to be solvable by radicals, yes

[rocky cloak]

I think maybe you need F to be characteristic 0.

[coral steeple]

Are bases of F^n and elements of GL_n(F) not exactly the same thing?

[south patrol]

ordered bases, yes

If you don't want your bases ordered, then no

[coral steeple]

Ah, true

Thanks

[south patrol]

Np

[rocky cloak]

Like if F = F2(t) for example and p(x) = x^2 + x + t

Then the Galois group of p is C2, but the polynomial shouldn't be solvable by radicals

[coral spindle]

I imagine requiring the fields to be perfect would also be a solution

[rocky cloak]

Yeah, that's probably sufficient

[rocky cloak]

Hmm, wait
F2(t, t^1/2, t^1/4, ...) Should be a perfect field. But I'm not sure it changes the fact that x^2 + x + t is irreducible nor wheter it's solvable by radicals.

[steel pulsar]

a is congruent to b modulo n iff a-b is divisible by n
It is an equivalence relation, and we quotient Z by this relation
We have n - 0 = n is divisible by n so n is congruent to 0 modulo n ie n (bar) = 0 (bar) in Z/nZ

people I don't understand that, someone can help me ?

Why "n - 0"

[tribal moss]

What is the definition of "is congruent to"?

[steel pulsar]

By definition in Z/nZ, the class of n is equal to 0: n (bar) = 0 (bar)

congruence is that

this symbole

[tribal moss]

How is the "bar" symbol defined, then?

[tribal moss]

Wait, that is just notation, not a definition of what the notation means.

[steel pulsar]

I don't know, I want to give up

Group theory is too hard...

[tribal moss]

If you don't bother to figure out the definitions of what you're reading everything will be too hard.

[steel pulsar]

In fact I'll explain it to you from the beginning

[tribal moss]

Um, wait a minute.

[faint fractal]

idk, I feel the same about algebra but not for analysis 🧐

[tribal moss]

You explicitly quoted the definition.

[tribal moss]

a is congruent to b modulo n iff a-b is divisible by n

[steel pulsar]

I was asked this question : : We consider the translation $t_1 (x) = x + 1$ in the field of real numbers. Could you show that it is of infinite order on $\mathbb{R}$?

[cloud walrus]

Pg

[tribal moss]

So if you need to argue that "n is congruent to 0 modulo n", then the definition of that claim is whether n-0 is divisible by n.
That is why they subtract 0 from n!

[steel pulsar]

to say that the translation $t_1$ is of infinite order amounts to saying that
$$ \forall n \in \mathbb{N}, : t_1 \circ t_1 \cdots t_1 (x) \ne x $$

[cloud walrus]

Pg

[steel pulsar]

so we have : $$ \forall n \in \mathbb{N}, : t ^n _1(x) \ne x $$

[cloud walrus]

Pg

[buoyant echo]

hmm I see

[steel pulsar]

So it is of infinite order if we consider it as a bijection of $\mathbb{R}$ into $\mathbb{R}$.
On the other hand in the ring $\mathbb{Z}/n\mathbb{Z}$, $n=0$, so
$$ t ^n _1 (x) = x $$
So $t_1$ is of finite order (we can show that it is of order n) in the group of bijections of $\mathbb{Z}/n\mathbb{Z}$

[cloud walrus]

Pg

[steel pulsar]

And my problem was

"On the other hand in the ring Z / nZ, n = 0"

I didn't understand why n = 0

So I was told it's because By definition in Z/nZ, the class of n is equal to 0

But I still don't understand

[faint fractal]

Z/nZ is the group of integers mod n. Which just means you are taking the remainder of each integer when dividing by n. So if you were taking Z/5Z then 12 would have a remainder of 2 when dividing by 5. So would 17. But 5 would have a remainder of 0, which is why n = 0

[steel pulsar]

But here you take the example of Z / 5 Z

In my exercise, why do you apply this example?

In fact it doesn't matter what we take

Z / 5 Z

Z / 3 Z

Z / 2 Z

We would still have n= 0?

[faint fractal]

yes

[tribal moss]

In Z/5Z we have 5=0.
In Z/3Z we have 3=0.
In Z/2Z we have 2=0.

[faint fractal]

I solved the first part of part b, how do I show in the second part that the minimal polynomial has degree f?

[steel pulsar]

ok I think I understand

thank you!!

[teal vessel]

interesting question: do the solutions of the 8 queens problem constitute a group?

[faint fractal]

what would be the operation?

oh

[teal vessel]

probably matrix multiplication on the 8x8 matrices

[faint fractal]

hmm

[teal vessel]

can confirm, not a group

at least not represented as a matrix

[faint fractal]

I was thinking as like a subgroup of S_8

idk if that makes sense

[tribal moss]

That yields the same problem: the identity element of the group corresponds to all queens on the same diagonal, which is not a solution.

[faint fractal]

ah

I see

[tribal moss]

The 8 rooks problem does make a group.

[faint fractal]

😮

[tribal moss]

Namely exactly S_8.

[faint fractal]

https://en.wikipedia.org/wiki/Rook_polynomial

lol

[faint fractal]

tropo can you help me with my problem?

[tribal moss]

Not at this time of the night, I'm afraid.

[faint fractal]

unfortunate haha

[tribal moss]

I need to be fully awake and alert to even begin pretending I know cyclotomic stuff.

[white oxide]

sorry i can't help with this atm but is this from dummit and foote lol

[faint fractal]

yes haha

[white oxide]

ye recognized the font

[teal vessel]

I wonder if it forms a group under different representations. It would really depend on what we count the "identity" as

(representation in the informal sense, idkwtf representation theory is)

[tribal moss]

Well, right. If everything else fails, it's a finite set, so we could just arbitrarily assign elements of the appropriate cyclic group to each solution!

[faint fractal]

lmfao

so true

it's just Z/92Z 🧐

[teal vessel]

I suppose a more meaningful question would be this: what best exemplifies the structure of the 92 element set, given that it only has 12 unique (up to symmetry) solutions?

[faint fractal]

maybe you could try thinking of the 12 solutions as conjugacy classes, but still no operation would make sense

[tribal moss]

Making it into a group most likely isn't the best answer to that question. If we want to use group theory, a better starting point would be that the dihedral group of order 8 acts on the 92 elements such that they split into 12 orbits -- which means that not all orbits have size 8; some of the solutions must have internal symmetries.

[teal vessel]

solution (5 3 1 7 2 8 6 4) has deeper symmetries

[tribal moss]

Yeah, so its orbit only has size 4.

[teal vessel]

it has pi radian rotational symmetry

[tribal moss]

(It's alright, the kids are in bed so you can say 180°)

[teal vessel]

no has degree symbol

tau/2 symmetry

[faint fractal]

you play tf2? I saw your LFT med in your discord profile lmfao

[teal vessel]

used to. Haven't played competitively in years. I'll have to drop that lol

[faint fractal]

lmfaoo, I also played med for a few seasons

[steel pulsar]

When we say : Find a generator of the cyclic group (Z/13Z)*

Is there a method?

Or is it coincidence?

[tribal moss]

As far as I know, trial and error is what's usually recommended. Generally a fair fraction of the nonzero elements will be generators, so just random tries should succeed fast enough for practical purposes.

[steel pulsar]

But how to do the tests?

My teacher wrote this

correction

It can be verified by hand that
< 2 >= {1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7, 1, . . . } = (Z/13Z)*

We can speed things up by noting: the order of (Z/13Z)* is 12 = 2² x 3. The divisors
strict maximums are 4 and 6. An element n therefore generates the group if and only if n^4 =! 1 and
n^6 =! 1. We have 2^4 = 16 = 3 and 2^6 = 64 = 12. So 2 is of order 12

[tribal moss]

For cases as small as this you can just compute successive powers of the generator and see that they don't start to repeat until the 12th power.

For the "speeding things up", I don't think I can improve on that description.

[steel pulsar]

Why did he write <2>? why 2?

and why this number on bracket

{1, 2, 4? 8, 3, 6, 12 ...}

how do they find them

[tribal moss]

They are the powers of 2 modulo 13, starting with 2^0, 2^1, 2^2, 2^3, 2^3, ...

Since that only repeats at 2^12, this means that 2 is a generator.

[steel pulsar]

Ok so the fact that he chose the number 2 and not the number 3 or the number 4 or other number is just “chance”?

[celest furnace]

Hey guys. Im doing a pset on lower central series and I wasnt in lecture (it is VERY complicated). I wanted to know why the lower central series forms a chain. For starters, why is $[G^1, G] \leq G^1$?

[cloud walrus]

The Ultimate Chad

[celest furnace]

I tried showing that $z^{-1}[x, y]^{-1}z[x, y] \in G^1$ but i couldn't get anywhere

[cloud walrus]

The Ultimate Chad

[wraith cargo]

I mean this is by definition right
G^1 is the subgroup generated by all elements of the form xyx^-1y^-1

And the LHS clearly is of that form

[tribal moss]

It's common to start by checking 2, because writing out the list of powers is easier when you just have to double each number to get the next, compared to multiplying with a larger factor.

(The other generators in this case are 6, 7, and 11, which would take more metal arithmetic to create the powers of.)

[delicate bloom]

on top of that, it's not hard to pick out the other generators from that list, they're the ones where the exponent n on 2^n (I wrote them in red here) is relatively prime to 12:

[tribal moss]

(Which is how I found them, too!)

[celest furnace]

Thanks

[steel pulsar]

ok thank you very much for you help

[native fable]

Hello, Can I ask questions here, no one seems to be responding to me in the help room

I have been having a lot of trouble relating to constructing this group called Grothendieck

[chilly ocean]

just ask

[native fable]

Thank youu

I wanna ask about this specific choice of relation

Maybe I’m a bit old so I tend to ask Why a lot, and it puzzled me when this relation is defined and the material does not explain why

[tribal moss]

(a,b) intuitively represents the difference a-b, though that doesn't formally mean anything until after the group has been defined. But it still motivates the definition:

a-b = c-d iff a+d = c+b
as a matter of the abelian group structure we want to hold in the resulting group, and if we write the left-hand part of that as (a,b) ~ (c,d) instead, we get your definition.

A main example of this is how we can construct Z (and its addition) based on N and its addition.

[native fable]

I really appreciate when you help me understand the motivation behind this!

[woven panther]

If I have a finite group G of order k, is it true that for any g \in G, g^k=id?

[native fable]

I guess to have this relation we need to have cancellation property right?

[chilly ocean]

btw, help channels are for like, solve 2x = 8 kinda questions

generally high school or early university kind of stuff

[lime junco]

yes, becuase the order of all elements in the group has to divide the total size of teh group

[native fable]

omg i did not know cuz someone helped me with some difficult questions before

So I assume I would get lucky again

[tribal moss]

(I have to black out, anyone else feel free to jump in with further explanations).

[native fable]

Thank you for your comments!

I just found a very short explanation based on Troposphere comments, and apparently the relation is used to define a pair of natural number that can describe an integer

So for an interger like -1 we can have multiple pair of a-b that results in -1

[lime junco]

If youve talked about lagranges theorem - note that the size of any subgroup of G divides the size of G. Therefore the size of <g> for any g in G must divide the size of G, but this is the order of g.

[native fable]

Those pairs would belong to the same equivalence class, i wonder if I understood correctly

[native fable]

Guys I want to continue with the question about Grothendieck group

The whole proposition was given (S,+) commutative semigroup with cancellation property, there is a smallest abelian group containing S, in which S is Grothendieck group.

I was wondering about the “smallest” part, seems like it involves uniqueness. Is there an approach to prove it is unique?

[wraith oak]

Any advice on how to tackle this?

[topaz solar]

Well, try x -> sqrt p homomorphism

A good way to prove there’s an isomorphism is to make one

[wraith oak]

Oh damn I guess I was overthinking it

[topaz solar]

Yeah lmao

[random sail]

How do you prove that if f(sqrt p) = 0, x^2 - p divides f

[wraith oak]

it's the 1 am brainfart moments rip

[dire viper]

note that the min poly of sqrt{p} over Q is x^2-p

[torn warren]

Let $a\in Q$, If K is the splitting field of $f=x^n-a$ and $f$ is irreducible on Q, then $|Gal(K/Q)|=n\cdot \phi(n)$, is this correct?

[cloud walrus]

WT

[rustic crown]

not in general. n = 8, a = 2 should be a counter example. it holds if n and phi(n) are coprime. the splitting field is K = Q(a^1/n, zeta_n) as both a^(1/n) and a^(1/n) * zeta_n lie in there.
for a = 2, n = 8, you know zeta_n = (1+i)/sqrt(2), so Q(zeta_n) = Q(i, sqrt(2)). and certainly x^n - 2 isn't irreducible over it. so the degree of K/Q can't be n phi(n)

[torn warren]

Do you mean that revise it to $|Gal(K/Q)|=b\cdot \phi(n)$, where $b$ is the degree of the irreducible poly of $a^{1/n}$?

[cloud walrus]

WT

[rustic crown]

yea sure. but irreducible polynomial in the ring Q(zeta_n)[x]

and that follows clearly from the tower lemma for K / Q(zeta_n) / Q

[torn warren]

In what case the irr-poly of a^1\n are the same on Q[x] and on Q(zeta_n)[x]

[rustic crown]

one sufficient condition like i said is when n and phi(n) are coprime

these n are called cyclic numbers btw, any group of cardinality n would automatically be cyclic (i prolly don't know the best proof, but the one i read long ago wasn't super enlightening)

[torn warren]

Let $a\in Q$, If K is the splitting field of $f=x^n-a$ and $\gcd(n, \phi(n))=1$, then $|Gal(K/Q)|=n\cdot \phi(n)$, do you mean this? Does it need to assume $f$ is irreducible on Q?

[cloud walrus]

WT

[rustic crown]

it's not necessary tho. n = 4 doesn't satisfy that n and phi(n)=2 are coprime. but still for a = 2, the splitting field is Q(2^(1/4), i) / Q(2^1/4) / Q. so in this case the degree still manages to be n phi(n)

[rustic crown]

yep, and yee i want f to be irred/Q

[torn warren]

Is this sufficient condition proof difficult?

[rustic crown]

nah, it super easy

well assuming you already know that degree of zeta_n over Q is phi(n)

[wraith oak]

I'm trying to make a bijective function from Q[sqrtp] to Q[x]/(x^2-p). Is it as straight forward as mapping "a" to the first sum highlighted in yellow, and "b" to the second sum highlighted in blue? Because to me I don't know how I would be able to show that's bijective and I feel like I'm missing something

[rustic crown]

the splitting field is K = Q(a^1/n, zeta_n)
it contains two interesting subfields Q(a^1/n) and Q(zeta_n) which have degrees n and phi(n) over Q. So the degree of the extension is divislbe by both and hence divisible by the product by coprimality. but the degree is easily seen to be bounded above by n phi(n) as well. therefore you have an equality.

[torn warren]

ah, that's smart!

[torn warren]

Is this equivalent to in what cases the Sn group has a subgroup with order n*phi(n) ?

[dim widget]

n and phi(n) are almost never coprime

but there are other ways to get the lower bound by n\phi(n)

[torn warren]

how to do that...

[dim widget]

For instance you could work out the automorphisms: there is an order n automorphism \nroot(a) \to \zeta_n \nroot(a) and the fixed field is Q(\zeta_n)

[torn warren]

yes, we need a n-cycle, which is equivalent to say the irr-poly over Q(\zeta_n) is the same as the irr-poly over Q, how to proceed next?

[wraith oak]

Is there a faster way to show that t^2+1 is irreducible in F, without plugging in all 27 elements?

[delicate bloom]

4 doesn't divide 26

[wraith oak]

What's going on

[dim widget]

The n cycle is totally explicit

You just multiply the nth root by a root of unity

[wraith oak]

What theorem makes use of this fact

[torn warren]

I mean n-cycle over Q(zeta_n)

[dim widget]

What do you mean? This automorphism is over Q(zeta_n)?

[delicate bloom]

sorry I'm occupied, I only meant to help the other person helping you

[torn warren]

yes

[dim widget]

Well it is… so doesn’t that prove what you wanted?

[torn warren]

but I couldn't prove there is a n-cycle over Q(zeta_n)

[dim widget]

Why not the one I wrote?

[torn warren]

the cycle might split, just as the counter-example given by @rustic crown

[dim widget]

What do you mean split?

[torn warren]

I mean one orbit splits into two orbits

[dim widget]

In this case you can see explicitly that this does not happen

[rocky cloak]

So using merositys hint, the multiplicative group of F is finite group. Note that if t^2 = -1, then t is an element of order 4. But the order of an element must divide the order of the group.

Another method could be to realize that F must contain F3, and that if it contains F3 and a root of t^2 + 1, then it must contain a subextension with 9 elements. But an extension of a field with 9 elements will have 9^k elements, so can't be 27

[torn warren]

but counter example here

[wooden rover]

||I want to ask a question but someone already is asking one :(||

[wraith oak]

oh damn, we haven't covered groups or field extensions yet. Should I still try to digest the info anyway?

[rocky cloak]

How are you constructing the field with 27 elements then?

Like are you just writing out a 27x27 multiplication table at random

[dim widget]

Yes if you start with a polynomial over Q you don’t always get an nth root extension, you could do the same with Q(cuberoot(-3)) etc

But if K is a field containing the nth roots of unity and a is not an dth power in K for d|n then K(nthroot(a)) is always cyclic of order n

So all of your questions come down to: this is true if and only if your a factors in a certain way in your cyclotomic field

[wraith oak]

I think this is what I need to use. I'm kind of lost though

[rocky cloak]

Yeah, so here you have exactly the thing I said about if t^2 + 1 has a root, then the field would have size 9^k

[torn warren]

Do you mean x^d=a, has no roots in K=Q(zeta_n)?

[dim widget]

Yep

[torn warren]

why it is x^d=a rather than x^n=a has no roots?

[dim widget]

Think about it, for instance if a is a square in K what is the degree of K(fourthroot(a))

[torn warren]

2

[dim widget]

Okay so we’re trying to rule that out

That’s why we don’t want any of the x^d - a to have roots in Q(zeta_n)

[wraith oak]

okok it's starting to make sense. Thank you

[dim widget]

This is implied if \phi(n) and n are coprime but I am giving you a necessary condition and not just a sufficient one

[native fable]

Can someone explain to me what approach the author is using to prove uniqueness?

[dim widget]

That it satisfies a universal property

They are not really giving a proof, they are describing a universal property that the group satisfies

Then you can use that to prove uniqueness

[native fable]

Oh that is so tricky I thought it was a proof from the line We have

[dim widget]

I can’t really tell what they’re saying

[native fable]

Is there a name to this property?

[dim widget]

It could be considered a proof

[dim widget]

Not really, there is some fancy name for this kind of property but it won’t help you

[native fable]

Oh I just want a name to look it up online

So I can understand why it gives uniqueness

[dim widget]

The point is that G(S) is initial for maps from S to abelian groups

So for any map of monoids S \to G there is a unique factorization S \to G(S) \to G

[torn warren]

but I don't get it, if for example, r is the root for x^2=a, then r^4 is not equal to a anymore, why let x^d=a for each divisor of n

[native fable]

Ok thank you for your explanation, let me take a minute to digest it

Sorry I’m not very well verse in abstract algebra, I barely know the concepts

[dim widget]

What

My point is that we want to demand that x^d - a doesn’t have a root in Q(zeta_n) for all d|n

[torn warren]

I guess you mean $x^d\notin K$ for each $d|n$

[cloud walrus]

WT

[dim widget]

That is another way of saying it yes

[torn warren]

because I am confused, in my op, I wrote f=x^n-a, so a is some fixed number in Q

[dim widget]

Yes and I am giving you necessary and sufficient conditions for that number to have splitting field with galois group Z/n \semi Z/(n-1), isn’t that what you want to know?

[torn warren]

but when you wrote x^d=a, you mean a is some changing number for each d, the a is different, but a is in Q(zeta_n)

[dim widget]

No

The a is staying the same and only d is changing

[torn warren]

how can this possible? if r^2=3, r^4 not equal 3

[dim widget]

This is true but what does this have to do with what I’m talking about?

At what point did I say that r^2 = r^4?

[dim widget]

So you start with a number a \in Q

[torn warren]

here you said a is not dth power in K

[dim widget]

No I said if a is not a dth power in Q(zeta_n) for all 1 \neq d|n then the splitting field of x^n - a has degree n\phi(n)

[torn warren]

is K=Q(zeta_n)?

[dim widget]

Yes or you could ask about the galois group of the extension K(a^{1/n}) more generally, then K would just be some field of characteristic prime to n containing the nth roots of unity

But sure for now let’s have K be Q(zeta_n)

[torn warren]

I still don't understand...

[native fable]

If I understand correctly then Fancy G is a collection of these Phi functions that maps a semigroup to abelian group, but can you help me explain why this Fancy G set is written as congruent to Normal G? Or is the condition for congruent the last 2 lines of the slide?

[torn warren]

the op is true if and only if x^n-a is irreducible in K=Q(zeta_n), right?

but why x^n-a is irreducible if and only if for each d|n, x^d=a has no roots over K?

@dim widget

maybe it can factorize as x^n-a=p(x) q(x) over K, and x^d-a doesn't divide neither p(x) or q(x)

[dim widget]

Maybe you should try to show this, but if r^d = a then x^n - a has a factor of x^{n/d} - r

[dim widget]

No there is just one phi function

Fancy G(S) is the group generated by \phi(S) inside of an arbitrary abelian group fancy G

[torn warren]

the remainder is not zero

[native fable]

My bad I understand it better now! What’s left is the congruent part and the last 2 lines with the appearance of Psi symbol

[dim widget]

You divided wrong…..

[dim widget]

X^d * X^t = X^{t + d} not X^{dt}

[torn warren]

oh, sorry..

[dim widget]

no worries, it’s a common mistake

[torn warren]

yes, this is correct, there is a factor...

[dim widget]

Since both G(S) and fancy G(S) are generated by S there is a natural map between the two groups which sends the S inside of G(S) to \phi(S) in fancy G(S)

That is the map \psi

[torn warren]

"x^n-a is irreducible if and only if for each d|n, x^d=a has no roots over K"

so we have shown the -> direction, because if x^d=a has some root r, then x^n-a has a factor, hence it is not irreducible, but what about the <- direction? even though x^d=a has no roots over K, but x^n-a may be factorized as p(x)q(x), just like x^4+4=(x^2+2x+2)(x^2-2x+2)

[native fable]

I assume this is related to Universal Property right? I found this picture on Wiki and understood it as
X in the picture is equivalent to S
F(A) is G
F(A’) is Fancy G
Direction of F(A) to F(A’) is psi
Is this correct?

Hope I’m on the right track to connect these ideas

[dim widget]

Er, the diagram has the right shape

But I think this diagram is about a different concept

[dim widget]

Ah i see

Okay yes I forgot that 4 was a special case

X^n - a is irreducible over any field F if and only if a is not b^d for any d|n and a is not -4*c^4 for some c in F if 4|n

[dim widget]

^so this is what you should prove

[native fable]

Ok, I guess I need to draw my own version. Do you mind if I ping you later?

[dim widget]

No that’s fine, I’m in bed sick so I’ll be around

[native fable]

Oh I wish you get well soon

Cannot believe that you are solving maths when you are sick

[wraith oak]

Does my ring homomorphism proof make sense, or did I do something wrong?

[dull marsh]

Looks good

[torn warren]

is there a textbook covering this part? is this part belong to number theory course?

[dim widget]

It should be in any textbook on Galois theory

This theorem isn’t just about Q it’s about an arbitrary field containing roots of unity

It is, for instance, in Lang’s algebra

[torn warren]

here the red line part, could you give an example, how does a factors in cyclotomic field?

[dim widget]

This was more a vague heuristic about x^d - a not having a root. (In the number field setting over Q(zeta_n) this means something about the factorization of the ideal (a) but this isn't the main focus.)

the "factorization" comment is just saying that you want to impose that x^d - a doesn't have a root

[torn warren]

According to this thm in Lang's book, the little k (in this thm) = Q(zeta_n) in my OP, so my OP is equivalent to:

In what case the extension Q(a^1/n, zeta_n) is a cyclic extension of Q(zeta_n) with degree n, right?

[dim widget]

Yep! And in Lang’s book you can find necessary and sufficient conditions for this to be true

[torn warren]

I find the thm you mentioned just now.

[torn warren]

but I don't find the necessary and sufficient condition...is it not thm 6.2?

[dim widget]

This is the necessary and sufficient condition for what you want.

Or rather you can extract it from this

[torn warren]

thm 9.1 gives the sufficient condition such that x^n-a is irreducible, but not gives the necessary condition?

[dim widget]

We already showed that the first part is necessary

[torn warren]

what is the first part?

[dim widget]

Also the second condition should only be necessary when you don’t have roots of unity in the field

[dim widget]

Prove that X^n - a is irreducible over a field containing the nth roots of unity if and only if a is not a dth power for any d|n

At this point it’s easier if you try to do this yourself rather than message me all day

If you are curious about when X^n - a is irreducible in general then this theorem gives the full conditions under which this happens.

good luck!

[torn warren]

yes, we have shown this hours ago, I forgot..

[torn warren]

which second condition do you refer to?

do you mean (ii) in thm 6.2?

[cloud walrus]

Seagull

Seagull

[untold basalt]

Apparently this is wrong

But where is my mistake?

[coral spindle]

You haven't elaborated on why surjectivity implies that

Your answer just needs more elaboration

[mighty kiln]

(correcting the last line,) you've only proven the center maps into the center

[coral spindle]

^ this is also true

e.g. you need to explain why it is true that phi(c)phi(g) = phi(g)phi(c) – you haven't said why.

I was not elaborating on what you were saying, I was elaborating on what I was saying.

I wish you wouldn't delete your messages, it just makes me look like I'm chatting to myself

anyway

[mighty kiln]

Sorry didn't wanna clutter chat

[coral spindle]

Seagull, it seems you have correct reasoning, but to be totally clear you definitely need to explain more carefully why what you say is true, as there are steps missing that could've been confused.

[untold basalt]

actually I should have written H invariant iff phi(H) \subseteq H, it is not necessary to have equality

Let $\varphi \in \text{Aut}(G)$ and $c \in Z(G)$. For all $g \in G$ we have $\varphi(g)\varphi(c)=\varphi(c)\varphi(g)$ (by applying $\varphi$ to both sides of $gc=cg$) but since $\varphi$ is surjective we have that for all $g \in G \quad g\varphi(c)=\varphi(c)g$.

[cloud walrus]

Seagull

[untold basalt]

oh crap I misread the definition of fully invariant

phi is just a homomorphism, not recessarillyan automorphism

I got it confused with characteristic, which was in my notes

[delicate orchid]

ah, abelian groups. Or as I like to call them, 1-Engel Z-algebras

[alpine island]

Oh yeah? what's the basis for (R, +) over Z?

nerd

[delicate orchid]

1 and 1.5 should do it

[alpine island]

Cool thanks

[upbeat lotus]

Is PID essential for conclusion? UFD is sufficient is a), what about b)?

[rocky cloak]

If B lies in the normalization of A, then the normalization of B equals the normalization of A. If B is a UFD, then it equals its own normalization.

[steel pulsar]

Can someone explain to me 1.why this group is finite and 2. why is not cyclical?

What is the difference between a finite cyclic group and a finite non-cyclic group?

[chilly ocean]

did you mean Z/2Z x Z/2Z (with an x, not a *)?

[steel pulsar]

yes

[chilly ocean]

you can see that it's finite by just writing out what it is as a set

[steel pulsar]

I don't see, I have a lot of difficulty in math

[chilly ocean]

find the definition of Z/2Z x Z/2Z

what does the notation A x B mean?

[coral spindle]

It's important that you write \times, not * here. The asterisk means something very very different in the context of group theory.

[steel pulsar]

It's not me, it's my teacher

[steel pulsar]

multiplication ?

[chilly ocean]

let me clarify the question

when A and B are sets, what is the definition of the set A x B?

just sets. we can't actually multiply anything

if you don't know it off the top of your head, that's fine, you can probably find it in your textbook, or on google by searching its name, "cartesian product"

[steel pulsar]

that ?

@chilly ocean

[coral spindle]

Yes.

[steel pulsar]

How do I apply it to Z / 2Z x Z /2Z?

[coral spindle]

You should be able to do this on your own now

List all elements and count them.

[steel pulsar]

but it's not the same..

[coral spindle]

OK

[steel pulsar]

?

well no matter, thank you

[upbeat lotus]

Does it mean: yes, PID is not required?

[wraith oak]

In (2) I'm confused why if t^2+1 is reducible then it implies F has 9 elements

[spice whale]

when does it say that you have to prove that

all you have to prove is that t²+1 is irreducible in F[t]

where |F| = 27

[wraith oak]

This is what I was told earlier when I asked about it

[hidden haven]

Not that it will have 9 elements, but it will have a subfield with 9 elements

I assume

Because the latter need not be the case

[spice whale]

but like

[wraith oak]

oh yeah that's prob what he meant

[spice whale]

F_27 has a subfield with 9 elements right

[hidden haven]

Nope

[spice whale]

oh

huh

[wraith oak]

Am I supposed to use this theorem or something else? Because otherwise I don't know what else to use to show it's irreducible

[hidden haven]

You can

[wraith oak]

So then how would the theorem help me conclude that t^2+1 is irreducible over F?

[rocky cloak]

Yeah, that's how it seems to me

[steel pulsar]

In this theorem they say: "Let G_1 and G_2 be two cyclic groups of respective orders m and n. (gcd(m,n) = 1 <=> (G_1 x G_2 is cyclic))

Here they say that: The group Z / 2Z x Z / 2Z is ​​not cyclic: (1, 0), (1, 1) and (0,1) are all three of order 2.

saying that gcd(2, 2) = 2 and that (1, 0), (1, 1) and (0,1) are of orders two is it the same thing?
Or are these two different ways of showing that the group is not cyclical?

[coral spindle]

These are two different ways of arguing that the group is not cyclic, yes.

[steel pulsar]

And from what order is a group cyclical?

because they say that order 2 means not cyclic

[coral shale]

[coral spindle]

Can you remember what it means for a group to be cyclic?

Explain what it means in your own words.

[steel pulsar]

I would first like to understand why they say order 2...

[coral spindle]

I will explain!

But because I am not going to simply give you the answer, I want you to work with me.

Now, let's try again: do you remember what it means for a group to be cyclic?

[steel pulsar]

A group G is cyclic if ∃ a∈G such that ∀g∈G,∃ k∈N such that g=a^k

[coral spindle]

Great, that's true. So we call this element $a$ a generator.

Now here's an important question:

what is the connection between the size of $G$, and the order of $a$?

[cloud walrus]

Boyt

[steel pulsar]

the order of a group is the size of the group, no ?

[coral spindle]

The order of $a$ is equal to the size of the group $G$, yes!

[cloud walrus]

Boyt

[coral spindle]

Now put this information together:

1. How many elements of G = Z/2Z x Z/2Z are there?
2. If G were cyclic, what order would a generator have?

[steel pulsar]

G have 2 elements : 0 and 1

[coral spindle]

No, this is false.

I am not saying G = Z/2Z.

I am saying G = Z/2Z x Z/2Z.

Try again

[steel pulsar]

2 elements x 2 elements = 4 elements ?

[coral spindle]

Yes, G has 4 elements.

Can you answer part 2?

[hidden haven]

If f(x) factors in F_27, then there's a homomorphism
F_3[x] → F_27 that you can construct on which you can use the first isomorphism theorem to get an embedding F_9 → F_27, which is not possible.

[steel pulsar]

If we have