groups-rings-fields

[sly crescent]

Hmm

[chilly ocean]

i.e. when there exists a bijective linear map f: g -> h such that f([x, y]) = [f(x), f(y)]

[frail perch]

Silly question, but I am learning about sl2 modules right now and in many places I see expression lile Y^n(v), where Y is a nilpotent matrix of rank 2. However wouldnt that mean Y^n(v)=0 for all n bigger than 2? Something doesnt seem to make sense for me

[chilly ocean]

maybe they mean a nilpotent matrix that also has rank 2? i've never seen "rank" used to describe the powers to which a nilpotent matrix vanishes

[frail perch]

This is the matrix

[chilly ocean]

ok then certainly Y^2 = 0

and this definitely does not have rank 2

[frail perch]

For example here serre writes those relations which dont make sense to me as e_n should be all trivial for higher n no?

(g is the lie algebra sl2)

[prisma shuttle]

anybody can give me a hint on this problem?

[broken stirrup]

I've already asked it once but still don't get quite get the difference

[sharp sonnet]

one of them is an element of Hom(A tensor B, A' tensor B') and the other an element of Hom(A, A') tensor Hom(B, B')

you can use the universal property of the latter to show that there is a natural homomorphism Hom(A, A') tensor Hom(B, B') -> Hom(A tensor B, A' tensor B')

[simple mulch]

hum interesting

[barren sierra]

How do I study for a comp exam

My prof said the final is going to be a mock comp exam

And idk how that's different to a normal final 💀

[chilly ocean]

do previous exams

[broken stirrup]

thank you I'll think about it

[prisma shuttle]

can someone explain this proof

i don't understand the final line

how does (a/s)(s/1) being in J imply that a is in I?

[chilly ocean]

I = h^{-1}(J)

[prisma shuttle]

coudl u elaborate

[chilly ocean]

h(a) is in J

[prisma shuttle]

ohhh i see

oops how did i miss tht

ok thx u so much

[broken stirrup]

Here it tries to prove that not every element of tensor product is a simple tensor. So e_1 and e_2 are basis vectors and it assumes the above sum is a simple tensor to get a contradiction. I didn't understand how we get that system of equations

[languid walrus]

the above shows that $$e_1 \otimes e_1 + e_2 \otimes e_2$$ is not simple

[cloud walrus]

adl

[chilly ocean]

Is every K a splitting field for F?(F<E<K)

[south patrol]

No

For starters any splitting field will be a finite extension

Besides, you seem to have omitted what E, K, F actually are

[chilly ocean]

F<E<K

[south patrol]

So are you asking if the definition is pointless?

lol

[chilly ocean]

@south patrol yes

[plush wasp]

what is meant by "F - isomorphic"

reference to : any two splitting fields of f over F is F-isomorphic

[next obsidian]

I assume it means that the isomorphism fixes F

You can consider it like a triangle, you have two inlcusions F -> L and F -> L'

an F-isomorphism should be a map L -> L' such that the triangle commutes

[chilly ocean]

What does a splitting extension over F mean? .. answer plz

[south patrol]

So suppose we have a polynomial p(x) in F[x]. A splitting extension of p(x) is a field extension K/F such that that i) p(x) splits in K ii) K is minimal over F with respect that property - that is, such that p does not split in any intermediate field L with F<=L<K

I guess here they've generalised it to there existing a finite set of polynomials, though really that's a splitting field of the product of those polynomials so it's a splitting field in this other sense anyway

[chilly ocean]

alpha=2^(1/4) here. Why is K a splitting extension of Q?

[thorn delta]

we need more information. Splitting field over Q of what polynomial(s)?

[chilly ocean]

Then why is K a normal extension of Q..

[uncut girder]

U just need to show it has conjugates of every element in the field

Enough to show conjugates of the basis exist on the field

[chilly ocean]

I still dont know what normal extension's definition is

even though i could find what my textbook says

[eager glade]

normal extension if 1, 2, or 3

1. every embedding of K into the algebraic closure induces an auto on K
2. K is a splitting filed of some family of non zero polys
3. each irreducible polys in F having at least 1 root in K has all roots in K

1 2 and 3 are equivalent and all define normal extensions

3 is also equivalent to say that eaxh irred poly in F having at least 1 root in K splits completely in K

[barren sierra]

Aight so like

I understand the words on the page

But have absolutely no intuition for what a split extension is

[chilly ocean]

see "splitting lemma" for some ways to think about split short exact sequences

[barren sierra]

Got it

Wait what is a short extension

[eager glade]

i feel that when i took alg top

[barren sierra]

yea

This class is way out of my league I'm too dumb to be doing this

Hopefully I can keep my grade after this final but prolly not

[lavish nexus]

as long as you’re above average

[long obsidian]

Does the degree of a polynomial divide the degree of it's splitting field?

Or any other type of relationship between these two

[barren sierra]

not good enough lol

[barren sierra]

would it not be the opposite? the degree of a splitting field will divide the degree of the polynomial

oh wait no no no

if $\Sigma / F$ is a splitting field of some polynomial $f$ of degree $n$ then we have that $[\Sigma / F : F] \leq n!$

[cloud walrus]

Spamakin🎷

[barren sierra]

that's from my notes

but I can't imagine it's hard to show that $[\Sigma / F : F] \mid n!$

[cloud walrus]

Spamakin🎷

[long obsidian]

Ahh okay thank you. There is that one

[next obsidian]

Hurb

@long obsidian yes it does

Or well, are you assuming that the polynomial is irreducible?

If so then yes, because if alpha is a root of f then you have the chain of extensions
splitting field/F(alpha)/F

The degree of F(alpha) over F is the degree of f so it follows that the degree of f divides the degree of the splitting field over F

And spamakon’s final thing is true, the degree of the splitting field divides n!

This is because you can obtain it by continually adjoining roots

[long obsidian]

I was assuming irreducible thank you

[eager glade]

IT IS

well not too hard but a pain in the behind

[next obsidian]

Not really?

Wait no!

You totally can’t

Or uh…

[barren sierra]

isn't it just

induction

[next obsidian]

I guess it’s a little harder

[barren sierra]

a little harder

but in my head it's induction

[eager glade]

yes but... STRONG induction which is so weird to need

[barren sierra]

strong induction is the best induction

[eager glade]

indeed

[barren sierra]

why tie n - 1 hands behind your back

[eager glade]

lol

[barren sierra]

(stole that phrase from a professor of mine lol)

but it's true

[eager glade]

im using that

[barren sierra]

I mean they're logically equivalent

so like

why not always use strong

[next obsidian]

I don’t ever write my inductive hypothesis anymore

So if it needs strong I’ll use it

[barren sierra]

I wish I could do that but graders be stickers

[next obsidian]

But often I only need the n-1 case

[barren sierra]

yea

[next obsidian]

I mean often I just say “so we are finished by induction”

But granted I’m not taking like

[eager glade]

true but i never HAD to use it u know

[next obsidian]

Intro algebra anymore

[barren sierra]

I mean idk even in this grad level algebra course I'm doing the grader is being a stickler for all all details, even trivial stuff like that

[next obsidian]

I'll be honest

I don't do algebra for a grade anymore

So I don't have a reason to make it that forma

[barren sierra]

I'm jealous

[eager glade]

it do be the good life

[next obsidian]

Yeah I mean also I’m at the point that if I ever have to write proofs for a grade again (which I will next year I guess) I just don’t care

If they don’t like that they can just cope

Idk after reading hard books that leave a lot of stuff out it became really hard to feel motivated to write stuff out in such painful detail lmao

[eager glade]

indeed but its always nice to see a slick ass proof

also thesis is a thing so dont get too rusty

[next obsidian]

Yeah but you don’t spell stuff out like that for a thesis either haha

I don’t think anyone’s thesis is actually saying.

We prove this by induction on n.

Base case n = 1: insert proof

Assume that we know the result for all k < n: insert proof

Thus by induction we are done.

They’re probably saying n = 1 is obvious. some reduction to the n-1 case this we are done by induction

[eager glade]

LOL

what would pure mathematicians do without the words obvious and trivial

[next obsidian]

If you’re smart like me you say: “one can see that…”

[eager glade]

ah yes

[barren sierra]

"bruh just look at it"

[eager glade]

that makes the reader feel bad about themselves
BONUS POINTS

[next obsidian]

I like to think of it this way, I’m no longer lying

[eager glade]

bruv you dumb? just solve for x

[next obsidian]

One can see it

But it isn’t necessarily obvious, it’s usually just tedious

[eager glade]

if you really want to youll see it trust me

[barren sierra]

"source: trust me"

but yea idk I feel I write way too much for my HWs

but the second I write less and try to be more concise by omitting those (at least to me) obvious details

I get murdered by the grader

[vast osprey]

Hey, I am new here, but very interested in abstract algebra

I've been reading Dummit and Foote, what are your thoughts on that text?

[next obsidian]

It’s p good

[vast osprey]

I haven't had many issues with it other than it perhaps is a bit wordy

[next obsidian]

I find it kinda boring

Too verbose

[vast osprey]

for sure

I like the plethora of examples but it can be overwhelming

esp because I am trying to get through it before September

[next obsidian]

I’ve expressed it before, but there’s a few standard books for algebra, pick the one you give with the most

[next obsidian]

good luck

That book is too big to do in 4 months

[vast osprey]

I already killed the Group Theory section and done most of the exercises

I just need to get through chapter 15

to prep for an algebraic geometry course I'm auditing in the fall

Any tips for preparing for alg geo?

[next obsidian]

Should’ve skipped the group theory then

[vast osprey]

I breezed thru cause I knew most of it anyways

[next obsidian]

The only prerequisite to an intro AG class is an intro AG class

[vast osprey]

fack

[next obsidian]

Just try and get as cracked at commutative algebra as you can if you don’t want to learn AG beforehand

But the best prep is probably to start learning AG

[vast osprey]

is that possible wo/ foundational knowledge in ring or field theories?

[next obsidian]

No

Field theory is kinda tertiary

You need to know some of it, but you probably won’t really need Galois theory

But you should probably learn it in case your instructor does do something that uses it

[vast osprey]

yeah that's the plan

[next obsidian]

I mean D&F does some basic AG

[vast osprey]

yeah Chapter 15 is intro stuff

[next obsidian]

You really need to cover the sections on rings, homological algebra, and the commutative algebra parts

And I guess the field stuff, but I’d skip Galois theory if you don’t have time

[vast osprey]

okay cool cool

that's the plan as of now

[next obsidian]

Also fuck D&F

Every ring has a 1

Homomorphisms take 1 to 1

[vast osprey]

only have to get through a section every three days to make time

[next obsidian]

And something has to contain the 1 to be a subring

Don’t let them gaslight you into thinking an ideal is a subring

[vast osprey]

hahaha

[next obsidian]

This is going to be standard forAG and most contexts anyways

[vast osprey]

I did notice that D&F is p restrictive w their def of rings and subrings

[next obsidian]

It’s not restrictive

[vast osprey]

that is what my prof told me

[next obsidian]

They’re way too loose

In the AG setting every ring will have a 1 and be commutative

D&F allows rings to not have a 1, and it (fairly) deals with noncommutative rings

Because those are important for other reasons

The 1 thing is stupid tho

[barren sierra]

I fuck with the non-requirement of 1 and non-requirement of commutativity

[next obsidian]

I cast a curse upon your lineage

[barren sierra]

Imagine not thinking ideals are rings

Cringe

[vast osprey]

they don't have to be

[barren sierra]

You claim they're groups and yet the most important fact about them deals with multiplication

Curious

[south patrol]

ideals are submodules not subrings

😎

[dull root]

Notation question: If I have two groups A,B. Hom(A,B) denotes all homomoprhism from A to B right?

But is hom(A,B) a group too?

[south patrol]

homomorphisms needn't be invertible which obstructs defining multiplication by composition

[dull root]

So I have the following situtation: I have a finite abelian group G, and S^1 is the unit circle on C. Then Hom(G, S^1) is a group

why are these homomorphism invertible then?

o wait, the group operation on Hom(G,S^1) is defined differently than composition. It is defined like fg(x) = f(x)g(x)

[south patrol]

Yeah that'll be the important bit

[dull root]

so its not composition, but in general if I write Hom(G, S^1) this general notation just means the set of Hommorphism G to S^1 right?

[wraith ether]

I'm a bit confused on the different terms and notations surrounding groups so I'd like to clear up some things here if possible:
$\$ - Is a factor group the same as a quotient group?
$\$ - Are the following notations correct?
$\ Z = {\dots, -2, -1, 0, 1, 2, \dots }$
$\ 2Z = {\dots, -4, -2, 0, 2, 4, \dots}$
$\ GF(2) = Z/2Z = Z_2 = {\bar{0}, \bar{1}}\ with\ \bar{0} = {\dots, -2, 0, 2, \dots}\ and\ \bar{1} = {\dots, -1, 1, \dots}$

[cloud walrus]

S3BAS

[wraith ether]

And followup question, we know that we can use 5Z as a normal subgroup which generates the following cosets: 1 + 5Z, 2 + 5Z, 3 + 5Z, 4 + 5Z.

We know we can all 5Z the normal subgroup. The group of cosets can be called the quotient group, however, does that also include 5Z? Or is the quotient group only {1 + 5Z, 2 + 5Z, 3 + 5Z, 5 + 5Z}

[delicate orchid]

0+5Z = 5Z is a coset

So yes, 5Z is an element of Z/5Z

And I wouldn’t call 5Z the normal subgroup, it is a normal subgroup of Z

[wraith ether]

Okay, thank you, that makes sense

[delicate orchid]

All your notations seem correct as well

[wraith ether]

Okay great, that clears it up, and a factor group is the same thing as a quotient group?

[next obsidian]

Morrrrbbbiiiuuussssss

[delicate orchid]

Yeah they’re the same

[wraith ether]

Cool, that helps out, ty!

[hollow moth]

what is the correct interpretation of the ideal in a homomorphism context? more precisely, Ive read this proposition statin that no square matrix ring admits a non trivial ideal, but relating that with that theorem about the homomorphism between a dual space and the square matrix ring , I wonder if there are some implications that follow

and more generally, not related to that matrix-dual theorem, what's the interpretation of the ideal on a general homeomorphism space?

[next obsidian]

I think the word you’re looking for is homomorphism, a homeomorphism is a continuous map with a continuous inverse

[hollow moth]

yep, sorry

[chilly ocean]

Left ideal of a ring is a submodule of it when interpreted as a left module

[barren sierra]

why not both

[chilly ocean]

There's no reason to assume that subrings don't have the identity element

[kind jungle]

Let α be a zero of x^3 + 2x + 2 in an extension field of Z3. how can you find α^4 and α^5 in terms of 1,α,α^2

[chilly ocean]

If you divide w(x) = x^4 by x^3+2x+2, you get w(x) = q(x)(x^3+2x+2)+r(x) where r(x) is a polynomial of degree < 3

Now set x = alpha

Same with w(x) = x^5

[kind jungle]

dang, is that actually how you do it

[chilly ocean]

Yes, no matter how you write it, it's basically this

[kind jungle]

thank you!

[chilly ocean]

Np

[barren sierra]

is part (b) not just 121 and 17 are coprime so it's immediate?

[delicate orchid]

I mean, there are two groups of order 2057 that satisfy that property (that I can think of)

the wording of this question is very odd

[barren sierra]

yea lol

cause 121 = 11^2

[delicate orchid]

yeah and 17*11^2 is 2057

[barren sierra]

but either way

same result right?

[delicate orchid]

does it have to be for all groups of order 2057?

[barren sierra]

I presume so?

[delicate orchid]

this is where the wording really confuses me

[barren sierra]

but I mean still I thought if |P| and |Q| are relatively prime and G = P x Q

then Aut(G) = Aut(P) x Aut(Q)

like that's just a theorem that's known

(I think. It sounds familiar)

[delicate orchid]

oh yeah they just have to be relatively prime and not prime

very hazy in my head one sec lemme do some "researching" (googling)

[hazy flax]

Will an extension of a field $F$ of the form $F(a_1,a_2...,a_n)$ where $a_i$ are algebraic over $F$ have finitely many intermediate fields? I think yes but I'm not completely sure.

[cloud walrus]

PROnoob

[next obsidian]

Yeah

I think so…

This is just asking if a finite field extension has finitely many intermediary fields and this surely is true… right?

Wait I don’t think this is true anymore lol

[strong yacht]

Wow, surprisingly it isn't

[next obsidian]

Apparently this being true implies the extension is simple

[cursive temple]

Yeah its the primitive element theorem in that case

[next obsidian]

Are you claiming that like

Finitely many => separable?

[strong yacht]

https://www.mathcounterexamples.net/a-finite-extension-that-contains-infinitely-many-subfields/

[next obsidian]

Wtf

[barren sierra]

that's so weird

[sturdy marsh]

why is it weird

one direction is obvious

hmm the other direction is harder

no I was just being stupid

[barren sierra]

is there a quick way to do 2b

(I also don't get how it's related to 2a lol)

[sharp sonnet]

is euclidean algorithm not quick enough?

[barren sierra]

is there a better way?

[chilly ocean]

Does this need justification in said ring, though?

[sharp sonnet]

you need to know its an euclidean domain, yes

better way "just see it" i.e. ask sage

or if you can see the prime factorization, i cant

or you can do norm tricks

the norm of the gcd must divide the norm of the two numbers and also the norm of their sums

if that reduces the possibilities enough...

[chilly ocean]

https://math.stackexchange.com/questions/600424/prove-that-the-gaussian-integers-ring-is-a-euclidean-domain#600512

[strong yacht]

This is definitely the way to go

nvm

[steep scroll]

is this false?

[tribal moss]

Looks strange that it writes both (1 2 3)(4 5) without commas and (1, 2)(3, 4, 5) with commas. Are they different notations?

[steep scroll]

no its the same thing, my prof just wrote it different

[tribal moss]

It's a good exercise to explicitly write down a lambda that works.

[barren sierra]

Is this a mistake since it doesn't look like this minimal polynomial divides the characteristic polynomial (which is impossible)

[tribal moss]

Perhaps recall that $$\lambda(1;2;3)(4;5)\lambda^{-1} = (\lambda(1);\lambda(2);\lambda(3))(\lambda(4);\lambda(5))$$

[cloud walrus]

Troposphere

[steep scroll]

oh thank you for that hint! makes sense

[barren sierra]

also I'm not sure how Jordan canonical form works when the polynomials don't split. Would I not need to find the matrix over F(i) instead of F?

[tribal moss]

Recall that in F5 we have 2² = -1, so (x²+1) = (x+2)(x-2).

[barren sierra]

oh shit

nvm yea that'll do it

[prisma shuttle]

wait why is this true

and where does this come from

the problem is basically just asking u to prove that (123)(45) and (12)(345) are conjugates

does anyone know if there is any sort of fast way to do this rather than just listing all the terms

[tribal moss]

Consider what lambda(123)(45)lambda^-1 does to, for example lambda(2).
First, lambda^-1 maps it to 2.
Then (45) leaves it alone.
Then (123) maps it to 3.
Finally lambda maps it to lambda(3).

So lambda(123)(45)lambda^-1 maps lambda(2) to lambda(3) and must have "... lambda(2) lambda(3) ..." somewhere in its cycle decomposition.

In fact, whenever sigma maps a to b, then lambda sigma lambda^-1 will map lambda(a) to lambda(b).

So if you have a representation of a permutation that doesn't depend on any canonical ordering of the elements (such as a disjoint-cycles representation, or a product of transpositions), you can get a similar representation of that permutation conjugated by lambda, simply by applying lambda to every element number that appears in the original representation.

Furthermore, conjugation preserves the number and length of the disjoint cycles that make up the permutation.

And conversely, if you have two permutations that have the same cycle structure, they will be conjugate.

[obsidian sleet]

this is a good exercise to prove tbh

in general

[stable topaz]

Can someone tell me why sigma is an exponent here and what it means?

[thorn delta]

$p^\sigma = \sigma(p)$

[cloud walrus]

kxrider

[stable topaz]

Ah

[pastel cliff]

is the difference between R[x] and R[[x]] that the former is specifically finite polynomials

whereas the latter is not necessarily finite

[chilly ocean]

"polynomial" means a finite sum by definition

R[[x]] consists of (formal) power series

you could say that R[x] consists of those finite formal power series (just polynomials)

[pastel cliff]

ah okay understood

[barren sierra]

what does the 3rd condition tell us?

oh

there must be 1 jordan subblock of size 2 and 2 of size 1 corresponding to the eigenvalue 3??

or no there must be 2 jordan subblocks of size 2 I counted wrong in my head

[pastel cliff]

at the risk of sounding dumb

how should i be thinking about what x is in R[x]

[barren sierra]

a new symbol

at the most formal level that is all it is

it has some properties such that $x^m * x^n = x^{m + n}$ and other such things you know of

[cloud walrus]

Spamakin🎷

[barren sierra]

but it's not an element of $R$

[cloud walrus]

Spamakin🎷

[pastel cliff]

like necessarily not an element, or it doesnt have to be

[barren sierra]

I think necessarily

If you know what monoid and group rings are you can think of it like that

otherwise it's just a symbol with some rules that work the way you expect because we define it as such

[pastel cliff]

also is R[x,y] the same as R[y,x]

[barren sierra]

yes

[pastel cliff]

so R[x][y] also has to be the same as R[y][x]

[barren sierra]

same as R[y][x] same as R[x][y] (and really these 2 things are the formal definitions of R[x, y])

[pastel cliff]

so they really just imply a different way of writing elements ig?

like R[x][y] is emphasizing a linear combination of y with coefficients in R[x]

[barren sierra]

eh

I guess but no one really thinks of it that way

[pastel cliff]

prof went out of his way to point that out kek

oh well

[barren sierra]

I think I've maybe seen one proof where the problem really wanted you to think of it as a polynomial in y with coefficients in R[x]

but really everyone thinks of it as R[x, y]

cause R[x][y] and R[y][x] are (clearly) isomorphic

[pastel cliff]

[barren sierra]

or idk about clearly I think we just assume xy = yx

[prisma shuttle]

wow thx i never thought of the cycle notation thing like that

that's a cool way to think about it

[barren sierra]

yea that identity for conjugation of permutations is really powerful

also for part b of this is the answer dependent on the characteristic of F?

[thorn delta]

yea

[barren sierra]

ok cool

I mean the answer is just the characteristic right?

cause inductively sigma^2(f(x)) = f(x + 2) etc etc

[prisma shuttle]

yea ima pretty sure

[thorn delta]

for char F > 0 it is. Handle char F = 0 separately

[barren sierra]

oh wait yea char F = 0 then it's infinite cyclic group isomorphic to integers

I was specifically asking for finite characteristic mb

[prisma shuttle]

yea from the looks of it your work shoudl be correct

also

could someone take a quick look at the first two pages of this paper

https://arxiv.org/pdf/math/0103046.pdf

in the part on cycle lifting:

can someone explain why $f_{n+1}(X_i)\subseteq X_{i+1}$ instead of equal?

[cloud walrus]

JustKeepRunning

[prisma shuttle]

cuz i think they shoudl be equal

but ima not sure why they are using the weaker condition of subset instead

[barren sierra]

I don't get part c

aren't Cylcotomic polynomials irreducible?

so what factors are there?

[chilly ocean]

What's a difference between normal extension and splitting field...

[proud bear]

i guess they're only irreducible over Q

[barren sierra]

oh

yea that may be it

[thorn delta]

technically they are equivalent, but they are usually defined differently.
F/K is a normal extension if for f in K[x] irreducible over K such that f has a root in F, f splits in F.

If you have a set of polynomials X in K[x], then a splitting field over K for X is a field F such that each polynomial f in X splits in F. Also, F must be generated over K by the roots of polynomials in X. i.e. F should be minimal among other field extensions such that all polynomials in X split in F

Every normal extension F/K is a splitting field of the set of polynomials which have a root in F, and every splitting field over K is a normal extension of K

[chilly ocean]

I think normal extension can not be minimal field which splits f(x)
So splitting implies normal but vice versa I guess@thorn delta

@thorn delta then why are they equivalent ..

This sentence implies splitting and normal extensions are not equivalent imo

[hidden haven]

Sounds like there's some non standard definition here

[tough raven]

Isn't it just <=, not divisibility?

[broken stirrup]

I'm looking at my notes on Representation theory of finite groups and I can't understand this mapping. To be exact, what's that matrix [g_C[G]^*]?

[tough raven]

Maybe g acting on C[G] by left-multiplication, given concretely as a matrix using G as the basis of C[G]?

In which case g should just map to the permutation matrix corresponding to the permutation of G defined by (h -> gh) …

[chilly ocean]

Does the four-lemma (for modules) implies that the map is an isomorphism, or just epi/monomorphism?

In the exercise there's the former but online (and that's the version I could prove) is the latter

[hidden haven]

There are 2 versions

Dual to each other

[chilly ocean]

I know

There's one that says "blah blah" then this map is a monomorphism

In my book it says it's an isomorphism

[hidden haven]

Show

[hidden haven]

That be wrong

It'll be a monomorphism

[chilly ocean]

I've proved this is a monomorphism but isomorphism seems impossible

[hidden haven]

It is impossible

Easy counterexamples

0 → ℤ → ℤ → 0
Injects into
0 → ℚ → ℚ → 0

First map is epi, other 3 are mono, 3rd map isn't iso

[chilly ocean]

Yep, thanks

[hidden haven]

[broken stirrup]

yeah i think so basically a matrix representation of a linear transformation

[wraith ether]

Hi, I've got a question, knowing the following
\begin{enumerate}
\item the order of a generator is equal to the order of the cyclic group it generates, and
\item if we know $g \in G$, where $G = \langle g \rangle$ and $g$ is the generator of $G$, then we can define the order of $g$ to be the smallest positive integer $n$ such that $g^n = e$, and we write $|g|=n$ (assuming $G$ is finite).
\end{enumerate}

So, I was thinking this means that I can use that formula to determine the order of the generator, which in turns determines the order of the cyclic group, so I wanted to try it on an example.

Let $\mathbb{Z}^_{241}$ be the cyclic group where we pick the generator $g = 2$. Since this is a group under multiplication, we know $e = 1$. So if we insert $e$ and $g$ in the formula $g^n = e$, we get:\
\
\begin{multline}
\ 2^n = 1 \
n = log_2(1) = 0 \
\end{multline*}
\
But this makes no sense to me, since the order of the cyclic group $\mathbb{Z}^*_{241}$ is definitely not $0$ as it contains elements. Where am I going wrong with this reasoning?

[cloud walrus]

S3BAS

[wraith ether]

Ah 0 is excluded, we need the

smallest positive integer

[runic hemlock]

Let $L=K(\alpha)$ be a simple extension, and $M$ an intermediate extension. Let $M'\subset M$ be the field generated by the coefficients of the minimal polynomial of $\alpha$ over $M$. Any idea how to prove that $M'=M$?

[cloud walrus]

Porphyrion

[runic hemlock]

by "generated" i mean generated over K, obviously.

nevermind I solved

[prisma shuttle]

exactly

if u want look up orders modulo primes

u will find some useful info

[frosty umbra]

I have a question. If $N$ is submodule of $M$ such that $M \cong N \times P$ for some module $P$, does there exist a submodule $P'$ of $M$, such that $M = N \oplus P'$?

[cloud walrus]

SmeškoSnežak

[chilly ocean]

Take M = Z and N = 2Z, then M is isomorphic to N x 0 but if it were M = N (+) P' for some P', then P' can't contain element of N other than 0, but if x is in P', then 2x is in N and P', so 2x = 0, hence x = 0. It follows that M = N which is false. Summarizing: the answer is no

[frosty umbra]

Wow thank you, amazing example, thank you so much! Really appreciate it!

[wraith ether]

Thanks

[thin burrow]

Hello! A small thing to check: If we have M/L/K field extensions, do we have M/K separable iff M/L separable and L/K separable? I managed to prove it but I'm not sure whether it's correct.

[next obsidian]

This sounds correct to me

[tough raven]

What is Z_241*? Numbers modulo 241 which are coprime to 241 with multiplication as the binary operation?

In that case, do note that 2^n = 1 in that group does not imply n = log_2(1); that's only true when 2^n = 1 in Z (or Q, R, etc.). For example, mod 7 instead of 241, 2^3 = 1, and mod 241 we have 2^240 = 1 by Fermat's Little Theorem (241 is prime, right?).

[tough raven]

Although I realise you said you solved this, I'm interested in it.
Looking at L as M'(α), this seems equivalent to claiming any (“strictly”) intermediate extension M must have a smaller minimal polynomial for α.

… which holds because [M(α):M] = [L:M] < [L:M']. Nvm

[prisma shuttle]

i mean 241 is prime so "numbers modulo 241 which are coprime" is literally just all the residue classes from 0 to 240

[tough raven]

I was not sure 241 was prime when I wrote that part of the message

[prisma shuttle]

also idt log is defined in Z241

or at least not like conventiaonlly defined

idk how u would define the log function over a general ring

its a cool idea though

[tough raven]

I presume OP meant log in the real numbers.

[prisma shuttle]

well idt that would be well defined?

because the log function would have to be closed right

but log is not usually an element in Z_{241}

[tough raven]

I mean

[prisma shuttle]

and plus even if it was an "integer" in the intuitive sense it would not correspond to ar esidue class

like theres a difference between $2$ and $\overline{2}$

[tough raven]

n is not in Z_241 anyway. The order of a group element is always just a positive integer (or infinity).

[cloud walrus]

JustKeepRunning

[prisma shuttle]

maybe theres a way to do it

but atm i can't think of it

[young fiber]

couldn't you define log the same way

[cloud walrus]

giannis_money

[tough raven]

log_x(y) would have to be a solution for t in the equation x^t = y.
I'm pretty sure there's no ring where that has an integer solution for (say) all x and y, even upto a unit or something.
So you would have to come up with a notion of raising elements of the ring to non-integer powers. (non-(non-negative integer) powers for non-invertible elements).
At which point you'd have to answer the question of what the extended possibilities for t are actually going to be.

[young fiber]

it would be a non-negative integer for all elements in a cyclic groups right? if we take the log with base of the generator

[tough raven]

Then there's the problem of the equation ending up with multiple solutions, which happens in any Z/pZ by Fermat's Little Theorem, in C because e.g. e^z = e^(z+2πi), etc.

In short, not sure log is really a ring-theoretic notion

[prisma shuttle]

the thing is like log(n) is going to be a real number

which is not in $\mathbb{Z}_{241}$

[cloud walrus]

JustKeepRunning

[prisma shuttle]

so it would be a bit weird

ima sure there's still a way to somehow define it but it would not be as natural as defining log over the reals

[tough raven]

Firstly, that's not the ‘usual’ way, because when defining log in R, there's going to be a unique solution. Picking the smallest one doesn't have a precedent in R.
More importantly, what do you do when there's no such n?

[prisma shuttle]

u are actually defining the order of an element in a group/ring there

[tough raven]

This is really doesn't seem like a vital or even desirable property for log to have. If it was defined only modulo 241, we would need x^241 = x^log_x(x^241) = x^log_x(x^0) = x^0 = 1 for every x. That's not true.

[prisma shuttle]

thats different from log

[tough raven]

They wrote log_g(a) = smallest positive n such that g^n = a, not the identity.

[young fiber]

maybe a better definition is to say it's the inverse of the exponential

where you define the exponential with its taylor series

well.. doing exponents makes sense, but maybe not dividing by factorials

[steep scroll]

is there a trick to this

[next obsidian]

There’s only one group of order 7

So that means you just have it check one group has this property

[steep scroll]

Z_7?

[broken stirrup]

it's definitely of an order 7

and yes

[delicate orchid]

is that statement even true lol

[broken stirrup]

how can i show that product(pointwise) of two irreducible characters is also a character

[delicate orchid]

I know of a proof for general groups but it's rather long winded

we should use the fact that G is abelian, so all of it's irreducible characters are linear (i.e. all the characters themselves are homomorphisms from G to C)

[tough raven]

A character is a homomorphism from a group to units of a field, right?

[delicate orchid]

now will showing that the pointwise product of two homomorphisms from G to C is another homomorphism be enough? I think so but I'll need to think

[tough raven]

Can't you just verify the definition of a homomorphism?

[delicate orchid]

no, linear characters are though

which is what we're using

[tough raven]

You should be able to multiply group homomorphisms pointwise and get a group homomorphism whenever the range is an abelian group.

[chilly ocean]

hello

can i get some help man?

[broken stirrup]

well I want to show that it's closed under multiplication

How do we know that product is also irreducible?

[tough raven]

Oh

[tough raven]

I thought it just had to be a character.

[chilly ocean]

CAn i get some help please?

[tough raven]

Sorry

[delicate orchid]

the product will also be a homomorphism from G -> C, which is another linear character- thus it's degree 1, therefore irreducible

if you know about the connection between products of characters and tensor products of FG-modules this becomes much clearer, we're tensoring two 1 dimensional FG-modules together when we take the product of their characters, so the resulting FG-module will be of dimension 1x1 = 1, and all 1 dimensional FG-modules are trivially irreducible

[tough raven]

Go ahead and ask your question

IG

[broken stirrup]

oh yes, G= \sum X(1)^2 but then each X(1)=1 so it's linear

but i didn't know that characters of degree 1 are irreducible

damn why is it so hard to find character theory related stuff online

[broken stirrup]

sorry i don't see why character of degree 1 must be irreducible

[stoic rose]

Because a one-dimensional vector space doesn't have any nontrivial subspaces

[delicate orchid]

bingo

a degree 1 character is irreducible if and only if it's related FG-module is irreducible, this module is 1 dimensional - so any subspace is either dimension 0 (trivial) or dimension 1 (the entire space), thus the space is irreducible, hence the character is irreducible

[broken stirrup]

oh yeah

thanks kings

[delicate orchid]

no worries, king

[pale jasper]

Could somebody help me to prove that $\mathbb{Z}[X]$ is not a Euclidean Domain?

[cloud walrus]

Scerball

[south patrol]

Show it is not a PID by finding an ideal which isn't generated by a single element

[prisma shuttle]

i think u asked something similar a couple of days ago

remember the chain thing

a euclidean domain is a type of PID

[pale jasper]

Thank you

Sorry if I asked something similar. I've got my ring theory exam tomorrow lol

[prisma shuttle]

its fine lol

i just wanted to remind u so like u could connect the ideas together

gl on your exam!

oh and if u want the ideal which cannot be generated by a single element an example of such an ideal is ||<2,x>||

[pale jasper]

Thanks :)

[barren sierra]

Not quite sure how to do this

I think the easy case is if the order is 4 distinct primes?

[chilly radish]

You can also do this differently. If Z[X] were euclidean, then X, being an irreducible element, would generate a prime ideal, but in an ED (In fact, in a PID) prime ideals are maximal, so Z[X]/(X) should be a field, but it isn't, it's isomorphic to Z, so Z[X] can't be euclidean (or a PID)

[barren sierra]

it's very possible

and a great exercise

I feel that

have you shown that -r = (-1) * r?

once you do that, it's quite easy

yes

1 is typical notation for multiplicative id

-1 is the additive inverse of 1

you showed -r = (-1) * r?

cool

multiplication is not always commutative

however multiplication by -1 and 1 is

well clearly 1 *r = r = r * 1 right?

you showed (-1) * r = -r right?

show -r = r * (-1)

so we have that now

you don't need that

you have (-r)^2 = (-1) * r * (-1) * r

then from there use what you've just shown

and associativity

kinda but (-r^2) is confusing notation

do you mean -(r^2)?

or even more concise
(-r)^2 = (-r)(-r) = -(r(-r)) = -(-(r^2)) = r^2

np

[sly crescent]

Is there a name for a group that cannot be constructed as a split extension of one nontrivial group by another?

[barren sierra]

What is this last part asking

"find a relation"

[chilly ocean]

it's probably referring to a group presentation for G

[barren sierra]

Oh

[barren sierra]

Not sure how to show ii

[barren sierra]

nvm

[broken stirrup]

what does [H,H] mean here?

[delicate orchid]

[H, H] is the group generated by the set of all commuters in H

[simple valley]

Just wanna double check: if a matrix M is such that M^z for z in Z is a finite group, then M is unitary (?)

Suffices to consider M = jordan block, then M^n = I means it's diagonal and the eigenvalue is on the unit circle in which case M* = M^-1

Wait maybe the properties of the similarity matrix are important

M is similar to a unitary but not unitarily equivalent?

[chilly radish]

Maybe a silly question, but given a homomorphism of abelian groups A-> B, can you lift it to a homomorphism from the free abelian group on the generators of A to B? This seems false but I haven't been able to come up with an example

[chilly ocean]

restrict it to the generators and apply the universal property

[chilly radish]

true TTerra I think that works

thanks!

[chilly ocean]

[steep scroll]

isn't this irreducible

[chilly ocean]

It has a root

[steep scroll]

i like cant factor anymore

this class would be easier without polynomials

thank u

[chilly ocean]

You're welcome

[hidden haven]

A is a quotient of the free abelian group on the generators, so you can just compose the quotient map and the given one

No lifting involved

[chilly radish]

also chrew...

I don't know why I didn't see that lmao

I was trying to figure out if every map of abelian groups descends from one of the free abelian group by as you said this is trivial by composing the quotient map

[odd flame]

Hey guys i dont understand this example

How is F_2 isomorphic to Z/2Z x Z/2Z

[delicate orchid]

It isn’t

[chilly ocean]

"the additive group of F_2[X}/(g) is isomorphic to ..."

[odd flame]

Sorry i mean as an additive group

[delicate orchid]

F_2[X]/(g) is not F_2, it’s F_4

The additive group of F_4 is iso to (Z/2Z)^2

[odd flame]

Yes, that im sorry i misspelt typing on my ipad

So thats the part i dont get

How is F_2[X]/g iso to Z/2Z x Z/2Z

[delicate orchid]

All of those listed elements are order 2 under addition, there are 4 of them, so they must form Z/2Z^2 as an additive group

Alternatively just compute the entire Cayley table if you want lol

[odd flame]

Alright, just curious, as a true/false question where i must give an explanation for why it is true or false, would that be sufficient to say

lol i hate computing those

in your opinion

[delicate orchid]

Yeah you could say what I said, maybe add in “there are only 2 groups of order 4, and one of them is Z/4Z which has an element of order 4, so this group cannot be isomorphic to Z/4Z”

And then the conclusion follows

[odd flame]

thanks!

okay so hope you dont mind, followup question

is F_2[x]/g not iso to Z/2Z x Z/2Z as a unitary group

unitary ring*

[tough raven]

Alternatively (and equivalently), take the composition of this map and the quotient map from the free abelian group on the generators of A to A.

[chilly ocean]

#groups-rings-fields message

[tough raven]

Actually is lift even the right word for this? A doesn't naturally map into the free abelian group.

[tough raven]

oops

[tough raven]

“It's a 2-dimensional vector space over F_2 and all of those are isomorphic (vetcor space structure; in particular additive structure) to Z/2Z × Z/2Z.”

is hopefully shorter.

[tough raven]

It's not, because
(i) g is irreducible; hence (because F_2[X] is a PID) prime; F_2[X]/(g) is an integral domain.
(ii) OTOH, A×B is not an integral domain whenever A, B both contain a non-zero element, because (a,0)×(0,b) = (0,0).

[odd flame]

thanks!

[delicate orchid]

Unitary rings, or as I like to call them, rings

[tough raven]

ring rngs

[odd flame]

In part c i can only think of the two trivial idelas

Ideals*

How do i find more

[proud bear]

do you know the correspondence theorem? ideals of R/I are in bijection with ideals of R containing I?

[odd flame]

I dont know

[proud bear]

ok whatever. have you proven that R is a field?

[odd flame]

Not yet

Abstract algebra is my weakest topic

I apologize

[lethal dune]

no worries

is x²-x-1 irreducible in F3?

[chilly radish]

I would like a hint on this, I have no idea how to approach it: Let $p\geq 5$ be a prime, $m_{1},\ldots, m_{p-1}$ be even numbers, and let $n>0$ be even, then for sufficiently large odd $m$, the polynomial
$$f(x)=(x^2+m)\prod_{i=1}^{p-1}(x-m_{i})-\frac n 2$$
Is irreducible, with exactly $2$ complex roots

[cloud walrus]

ShiN

[odd flame]

yes it is irreducible

@lethal dune

so if i prove R is a field then the trivial ideals are the only ones i have to list right

[chilly radish]

Ok I think I might know how to show it is irreducible, but i'll have to think for a sec on how to show it only has 2 complex roots. My thought for irreducibility is showing that it's irreducible mod 2, and since it's monic this will imply irreducibility in general

[proud bear]

yes

[odd flame]

and

since i have proved that the quotient ring is irreducible

does that suffice?

[proud bear]

what do you mean by this

[odd flame]

I mean if i prove that $F_3[X]/(X^2-X-1) is irreducible does that show that it is a field?

[delicate orchid]

you prove the ideal (x^2-x-1) is maximal/x^2-x-1 is irreducible

[chilly radish]

nvm, apparently I miswrote and it's supposed to be $\frac 2 n$ not $\frac n 2$

[cloud walrus]

ShiN

[chilly radish]

so my reduction mod 2 strategy isn't gonna work because if I take common denominators the leading term will vanish mod 2

so back to square one

ok wait nvm turns out I misread even further

lemme reformulate

So the question is, $p\geq 5, m_1,\ldots,m_{p-2}$ even, $n>0$ even and $m$ sufficiently large odd number, then
$$f(x) = (x^2+n)\prod_{i=1}^{p-2}(x-m_i) - \frac 2 m$$
Is irreducible over $\mathbb Q$ with exactly 2 complex roots. This seems sus to me now tho, since $f$ is irreducible iff the polynomial gotten by multiplying out by $m$ and discarding the denominator is irreducible (because they only differ by a unit), but this new polynomial when reduced mod $2$ becomes $x^p$ which is obviously reducible, regardless of the size of $m$, so I must be missing something here

[cloud walrus]

ShiN

[chilly radish]

p prime ofc

[proud bear]

I guess just because its reducible mod 2 doesn't mean its reducible over Q. Like x^2+1 for example

[chilly radish]

right, you're right

the converse does not hold

I was thinking it was the contrapositive in my head but it's not

so I have no idea how to approach this

[cursive temple]

does this work for irreducibility: Pick $m = p^{p-1}$, and note then that for $g(x) = p^pf(x/p)$ we have

[g(x) = (x^2+p^2n)\prod_{i=1}^{p-2} (x-m_i p) - \frac{2p^p}{p^{p-1}} = (x^2+p^2n)\prod_{i=1}^{p-2} (x-m_i p) - 2p ]

and so is irreducible by eisenstein

[cloud walrus]

Pappa

[cursive temple]

oh wait im not sure if the constant term isnt divisible by p^2

the constant term is $p^p n \prod m_i - 2p = p(p^{p-1} n \prod m_i -2)$

[cloud walrus]

Pappa

[cursive temple]

which isnt divisible by p^2

[chilly radish]

I haven't checked out all the details but that looks righ

Right

How would I conclude there are only 2 complex roots? Do you.think another linear change of variables would work?

[tough raven]

Is it possible there's a typo and the numerator for /m is supposed to have an x in it? (specifically, could it be a monic polynomial in x)

[cursive temple]

i have no clue on how to approach the condition that a polynomial has two complex roots

[chilly radish]

No it shouldn't. At least this is how it was written in the lecture notes

Why do you think there's a mistake

[tough raven]

If so, this would be exactly like a procedure I saw for constructing irreducible polynomials over Q with a specified number of real and complex roots
And f would be irreducible for every m by arguments there
But IG it's only supposed to be irreducible for sufficiently large m here …

[tough raven]

Laziness (on my part), basically

[chilly radish]

I'm also not sure where we used the fact that p>=5

Lol

Honestly if you have that procedure that's good enough for me

I'm trying to construct a degree p polynomial with galois group S_p

[cursive temple]

the product from i=1 to p-2 might be bad for p = 3

[tough raven]

Just saying, I'm not sure how to construct that from what I described, but if you know how, 🤷

Consider a monic polynomial f with integer coefficients irreducible mod some prime p. Consider a monic polynomial g with integer coefficients with r real and s pairs of complex simple zeros, for any chosen r,s>=0.
Consider h_m := g + f/mp which has rational coefficients. Because
mp h_m = mpg + f \cong f (mod p);
mp h_m is irreducible mod p hence irreducible over Z, so h_m is irreducible over Q.

By a result called continuity of roots of polynomials (the only proof I know of which using Rolle's Theorem), if we draw disjoint discs in the complex plane around all the roots of g, and if m is sufficiently large so that the coefficients of (h_m - g) are sufficiently small, then h_m will have zeroes in these discs, with the number of zeroes in each disc equal to the multiplicity of that zero of g.

In particular, you can make the discs small enough that h_m has a pair of complex roots for ever pair for g; and because h has real coefficients, within each disc for the real roots of g (which are simple), the root of h also has to be real.

So h is irreducible, and has r real zeroes and s complex zeroes like g.

[chilly radish]

Ah, right

[tough raven]

You might need deg(f) <= deg(g) for this to work.

[broken stirrup]

I guess we are supposed to make use of it but don't know how to do it

[broken stirrup]

do we have [X_A,X_A] = X_1 [X,X] ?

[delicate orchid]

we basically just want to show that A is normal right?
A cannot be trivial as if chi(1) = |G:A| = |G| then chi(1)^2 > |G| which is a contradiction, and A is abelian by definition - so surely just showing it's normal is enough?

I think that lemma you've posted is the right thing to use btw

[broken stirrup]

but you didn't use it right?

[delicate orchid]

we still gotta show it's normal, lol

[broken stirrup]

sorry i thought you solved it in your head without referring to the lemma

[delicate orchid]

oh no lol

I thought I was onto something with the vanishing but then I read the line under the statement of the theorem

[broken stirrup]

or we could take the set that of elements that do not vanish under X and then consider the group generated by them. It will be normal since characters are class functions and will be subgroup of an abelian subgroup hence a normal abelian subgroup

[delicate orchid]

oh I do like that

very swag

[broken stirrup]

and will be contained in A for sure

since it vanishes outside A

[delicate orchid]

yeah A is a superset of the kernel

wait, G-A is a subset of the kernel, sorry lol

[broken stirrup]

so i guess we have to show that equality holds

then it implies that it vanishes

[delicate orchid]

oh

sure

[broken stirrup]

nice

and i can't do that

lol

[delicate orchid]

wait I can't see why this doesn't prove it

double nevermind, I can't see lol

[broken stirrup]

yeah to get to that point we have to show it vanishes outside A

[delicate orchid]

what a poopy situation

[broken stirrup]

so the generated group will be a subgroup of abelian group A and that will ensure that its not only normal but also abelian

[delicate orchid]

yeah and it's normal in G cause chi is a class function in G

[broken stirrup]

yeah character theory is like that sometimes

[delicate orchid]

most of the time

found this, could be useful

nevermind I didn't see the "index 2"

[broken stirrup]

maybe we could use definition of [X_H,X_H] to show that

[broken stirrup]

something like this

[delicate orchid]

$\frac{1}{|H|}\sum_{x \in H} \chi(x)^2$? Just asking cause I'm used to the notation specifying which group we're taking the inner product w.r.t

[cloud walrus]

Wew "Morbius" Tbh 🧛

[delicate orchid]

yeah I was just asking about which "G" we're talking about here

I presumed H cause it's the restriction

[broken stirrup]

A in our case

but doesn't matter that much

idk how do you pull out X(1)

[delicate orchid]

oh wait chi(1) is just |G : A| lol

it's already there

[tough raven]

Not necessarily; the inside A part plays the role of ensuring it's abelian.
You would just have to show it's nontrivial.
Possibly that's why you're asked to show there is some nontrivial abelian normal subgroup, instead of showing that A specifically has that property.

(assuming you have shown it's inside A; IDK anything about that)

[broken stirrup]

what?

[broken stirrup]

i was talking about equality on the lemma

[tough raven]

Oh

[broken stirrup]

lol it happened once

we misunderstood eachother now it's the second time it happens lol

[delicate orchid]

bit in blue is chi(1) right?

[broken stirrup]

yes

i feel like a BIG embarrasment is coming

[delicate orchid]

I'm just gonna note here that chi_H has to be reducible or G is iso to A and the problem is trivial, might come back up later who knows

oh another thing I've remembered is that the induced character chi^A is 0 on G-A
frobenius reciprocity moment?

[broken stirrup]

sorry 😦

i haven't studied it yet

[delicate orchid]

that's alright I'll try and think of a way to do it without

tbh frobenius reciprocity is probably overkill

ok so what happens if chi(g) is non-zero for some g in G-H, what fails?

like I know the inequality is strict now but what else

[broken stirrup]

well i couldn't find the answer to your question but one thing to note is that since X is irreducible [X,X]=1

so [X_H,X_H] is less than |G:H|

if chi(g) =/= 0

hmm what does this imply?

[delicate orchid]

maybe since X_H is reducible we could try decomposing it into irreducibles? Because I have a theorem that states "the coefficients of the constituents of chi_H squared <= [G:H]"

with again, equality iff chi_H is 0 on G-H

could be something

and since A is abelian all of the irreducibles are linear?

[broken stirrup]

yes

they are

it just boils down to show that [X_H,X_H]=X(1)

[delicate orchid]

yeah that's what I'm trying desperately to do

[broken stirrup]

same

[delicate orchid]

chi(1) = chi_H(1) <= [chi_H, chi_H] <= [G:H] = chi(1)
is this true?

and if so does it imply equality

yeah it definitely implies equality actually

I swear chi(1) <= [chi, chi] if the underlying group is abelian is a theorem somewhere

[broken stirrup]

yes it implies it but tbh i haven't seen this result

so i didn't really get how you got those inequalities

but seems legit

:d

[delicate orchid]

it's from decomposing chi into irreducibles I think

so, letting $\chi_i$ be irreducible characters
$\chi(1) = \sum_{i=1}^r n_i\chi_i(1) = \sum_{i=1}^r n_i$
hmm so what's [chi, chi]

[cloud walrus]

Wew "Morbius" Tbh 🧛

[delicate orchid]

oh yeah it's $\sum_{i=1}^r n_i^2$ cause irreducibles are orthogonal

[cloud walrus]

Wew "Morbius" Tbh 🧛

[delicate orchid]

yeah there we go

so we have our equality

and then everything you said follows

[broken stirrup]

hmm i see

[delicate orchid]

I can go into more detail if you want

[broken stirrup]

thanks man

i can't thank you enough

[delicate orchid]

that was a brutal exercise lol

[broken stirrup]

you just spent like half an hour over a stupid problem of mine :d

[delicate orchid]

I need more character theory practice lol

it's fine

[broken stirrup]

oh i have more in store 😄

im working on chapter 2

from M.Isaacs

and they are usually hard

like this one

[delicate orchid]

wait this is chapter 2

[broken stirrup]

yes

[delicate orchid]

wtf

[broken stirrup]

i swear i dont like that textbook at all. Our instructor says this is one of the best, classical

but it's going too fast

[delicate orchid]

I should check this book out

seems spooky

[broken stirrup]

and Isaacs explains like reader has experience in the field especially in chapter 1, chapter 2 was considerably better

anyway maybe im just bad at it lol so im just making excuses

anyway thank you sm

[delicate orchid]

I mean I wrote my dissertation on character theory and this stuff is still wacky zany bananas so don't blame yourself

[patent ocean]

Can I get a clue for Question 4

I tried to show M2 and M3 are in the orbit of another 3x3 matrix but wasn't really getting anywhere

[odd flame]

Lofi hip hop

Nice

Sorry to hijack this thread again but can i get a hint for 2

2b

I dont understand this one at all

[patent ocean]

nvm I found out how to do a) by just taking the rre of M3

[junior stump]

Weird. Surely if you multiply X^2 by X you get X^3, which is the same as 1, so all elements?

[patent ocean]

Can someone give me a clue on how to do Stab(M2)=AStab(M3)A-1

[junior stump]

If you can find an invertible matrix such that A*M3= M2 you can prove that A works. Then think about row operations @patent ocean

[patent ocean]

I've found such a matrix through eros

but how does that help with the stablisers

[junior stump]

stab(M3) = {P | PxM3 = M3} agreed

[patent ocean]

yes

[junior stump]

Ok so what happens when you apply APA^-1 to M2

[patent ocean]

um is it a change of basis matrix or something

[junior stump]

Probably but you don't need to know any interpretation of it

[patent ocean]

hmm

ig the matrix is in the orbit of M3

[junior stump]

So we want to show that the matrix APA^-1 is a member of stab(M2) agreed

[patent ocean]

yeah

so it must be invariant when applied to M2

[junior stump]

correct

[patent ocean]

but how do you show that

[junior stump]

So we need to show that APA^-1M2 = M2

and we know that AM3=M2

What is A^-1*M2?

[patent ocean]

P^-1*M3?

[junior stump]

A is independent of P

So we can only express A^-1 * M2 in terms of M2,M3, and A

[patent ocean]

Oh its M3 then

[junior stump]

ok so then APA^-1M2 = AP(A^-1M2) = APM3

can you guess the next step

[patent ocean]

take (AP)^-1?

actually no

[junior stump]

Remember that P is a member of stab(M3) that's the only thing we know

so we have to use that info somehow

[patent ocean]

PM3=M3

[junior stump]

yeah

So APM3 = A(PM3) = AM3

[south patrol]

there's a theorem in the notes about stabilisers being conjugate under these conditions btw T

[junior stump]

which is what

yeah i mean this proof doesn't use anything about matrices so instantly generalizes to general groups.

All that's required is the existence of a matrix A, which I guess is the point of them being in same orbit

[patent ocean]

AM3=M2

[junior stump]

yeah, so if we zoom out we've shown that APA^-1M2 = AP(A^-1M2) = APM3 = AM2 = M2

So we've shown that APA^-1 is in stab(M2) in otherwords since it fixes M2

[patent ocean]

oh yeah im so dumb

[junior stump]

which is one direction

[patent ocean]

the whole point was to show APA^-1 is in Stab(M2)

[junior stump]

lol

[patent ocean]

so now we need to go the other way?

for double inclusion i think

[junior stump]

Yeah so show that if it's in P is in stab(M2) then there is some L in stab(M3) such that P = ALA^-1

You can pretty much redo the first proof though after rearranging to solve for L

[south patrol]

the version I've seen just changes if to iff everywhere essentially

If x, y lie in the same orbit and we write y = gx then hy = y iff hgx = gx iff g^-1 hg x = x

[junior stump]

cool

[south patrol]

And so Stab(x) = g Stab(y) g^-1

[junior stump]

i always forget all this orbit stuff

Isn't there some theorem about all stabilizers being the same size

does that just use thius

if they are all in the same orbit

[south patrol]

Well not all are the sane size

Oh if in same orbit either this or orbit stabiliser

Although I suppose this is better if we care about size as its a bit sharper and quicker

[junior stump]

Then you basically get a billion proofs of the form "Heres an action i cooked up, and all stabilizers are same size and divide group so cool result"

[south patrol]

class equation style stuff is cute

[patent ocean]

Thanks for the help @junior stump . I'm tired and dumb

[patent ocean]

where does the very last bit come from

[south patrol]

Sorry typo

Uhh

There we go

[patent ocean]

ahh yeah thanks

[south patrol]

Np

And in exams write dots ig lol

I'm lazy from module stuffs

[patent ocean]

yh

[odd flame]

So would it be R then

I dont get it

What would be the units in $F_3[X]/(X^3+1)$

[cloud walrus]

horridharry96

[patent ocean]

could I get a clue just to start the last bit of 4? idk why my brain is just dead

I'm trying a contradiction

[odd flame]

Can anyone just formulate what 3a is asking

Im not even asking for a solution just like formulate the task i have to do lol

[chilly ocean]

you have to show that it's an ideal

[odd flame]

Its very late here and my brain has become mush

I dont understand that ideal

Its a function of the sqrt of 5

?

[chilly ocean]

square root of -5

[odd flame]

Right

[chilly ocean]

the ideal consists of polynomials with rational coefficients which send sqrt(-5) to 0

you must prove that this actually is an ideal

[odd flame]

So i need to show that the subset is an additive subgroup

And that if $s \in I$ and $r \in R$ then $rs$ and $sr$ are in I

[cloud walrus]

horridharry96

[chilly ocean]

yes

(rs and sr are the same thing)

[odd flame]

any clues on how i can show that this is an ideal

[chilly ocean]

by the definition

[odd flame]

im struggling to see how its an additive subgroup

[chilly ocean]

why

what have you tried to do

[odd flame]

so the subset is of polynomials in Q that when the coefficient is sqrt(-5) it sends it to zero

is that correct

[chilly ocean]

when x is sqrt(-5), you get zero.

[odd flame]

oh okay

yea ofc that makes sense when i look at it

[chilly ocean]

polynomials p(x), coefficients in Q, such that p(sqrt(-5)) = 0. this means you substitute sqrt(-5) for x

[odd flame]

so if we add two polynomials from that subset and substitute in sqrt(-5) it will equal zero

[chilly ocean]

this is what you must show

[white nymph]

can i also ask a question or is it bad form to interrupt?

[odd flame]

id say go ahead

Ive been interrupting all day

[white nymph]

fair.

given a vector space V that is the direct sum of cyclic F[x] modules where i know the annhilators of these cyclic modules, is this enough information to determine the invariant factors (and then the elementary divisors)?

like once i factor the annhilators, are these factors the elementray divisors? and if so, why? i can then push them together to find the invariant factors pretty easily i guess

for reference, this is from stuff regarding f.g. modules over PID/rational canonical form/jordan canonical form

[vast quiver]

yep!

first step: if M is a cyclic F[x] module with annihilator ideal (f), then M is isomorphic to F[x]/(f). To see this, we have a surjection F[x] -> M by sending 1 in F[x] to the generator of M (since M is cyclic it has one generator), and then by first iso thm, we get F[x]/ker is isomorphic to M. The kernel is exactly the annihilator of M.

Therefore, if we have V written as a direct sum of cyclic F[x]-modules and we know all of the annihilator ideals, we really have V written as the direct sum of modules F[x]/(f_i).

Now, we want to get invariant factor/elementary divisor form. To get elementary divisor, write each f_i as a product of irreducible polynomials, and use chinese remainder theorem to split it up. To get invariant factors, whenever you see F[x]/(f) and F[x]/(g) with f and g in your direct sum decomposition with f and g coprime, use chinese remainder theorem to combine them.

[west violet]

So...

how did we get from left to right in the picture?

[chilly ocean]

associativity and h_2 h_2^{-1} = e

[west violet]

associativity?

oh i see they just added terms

then used associativity, since those terms cancel out

[neat valley]

What does the group $(\bQ/\bZ)/n(\bQ/\bZ)$ look like exactly? Does it have any "canonical" isomorphism class?

[cloud walrus]

@neat valley

[next obsidian]

I think this is 0

[neat valley]

Yep, it is

[next obsidian]

n(Q/Z) should just be Q/Z

[neat valley]

found this:

https://pub.math.leidenuniv.nl/~smitbde/edu/lcft_2011/Solutions3.pdf

thanks!

[next obsidian]

Monka

[white nymph]

@vast quiver thanks much

[vast quiver]

yeah np

[chilly ocean]

Explain plz

[frail zealot]

the squares of the fermat primes that don't divide n are the primes that divide n

[tough raven]

In the prime factorisation of n the odd primes (which have to be Fermat primes as stated) can only have exponent 1.

2 can have any exponent though.

[odd flame]

In part 2b, how are there any elements

[hot lake]

are you saying X² isn't in the ideal generated by X² ?

[patent ocean]

can i get a hint for the very last bit of Q4)

I tried to do a contradiction

[hot lake]

I would just compute the stabilizers

[simple mulch]

So, I need to give a lecture on rings. Could anyone give some hints on what must be introduced in it?

[chilly ocean]

@simple mulch grab a book, look how it's presented

Pick topics

[simple mulch]

I don't think you understood what I was asking. Let me reformulate. I don't know much about rings, thus I am asking for some results that might be "more" important in the subject

If there is any such thing

[chilly ocean]

results important in ring theory

[chilly ocean]

More important for what

Also, you've asked a vague question so expect such answers

[simple mulch]

:(((((

[marsh goblet]

speaking of rings
i'm having a bit of trouble understanding the definition for two sided principal ideals
why is RaR defined as the set of all sums of {r_i a r'_i}
and not just the set of those elements themselves -- why do we need to sum them

[simple mulch]

I will just study everything in the chapter lol
Thanks

[chilly ocean]

Because we want it to be closed under addition

It's supposed to be an ideal and since we're working with rings here, we expect ideals to not be just about multiplication

[marsh goblet]

Yep, sure
So in that case, what does i range over — I can’t seem to find an answer anywhere. Just because if i can be 1; that is, there is no summation happening, then you just have that set of products which breaks the closed under addition rule

[chilly ocean]

Take elements of the form rar where r is any element of R

Then take all the sums of such elements

Finite sums

So elements are of the form

$\sum_{i=1}^n r_iar_i$

[cloud walrus]

Blitz

[chilly ocean]

With $r_i\in R$

[cloud walrus]

Blitz

[marsh goblet]

and n can be any number? (but less than the number of things in the ring i assume)

[chilly ocean]

Any natural number

[marsh goblet]

ah ok so the same terms can appear more than once in this sum

[chilly ocean]

Hmm... I guess?

[wooden ember]

for someone who reads french: i dont understand why we need to start talking about this finite set S: doesn't finding a single value of c for which deg m_a,K(gamma) = 1 suffice??

[runic hemlock]

yes, but that's not possible

what is possible, is to find all elements c that dont work, and to prove that there are only finitely many of them (the set S)

[wooden ember]

so it's just that finding an element c isnt a simple proof method, we just show that it exists by showing only finitely many dont work

i see

[patent ocean]

How would I do this?

Apart from the identity element

[hot lake]

write X M1 = M1 and solve for X

[patent ocean]

doesnt it just give the identity or am i bugging

[hot lake]

M1 isn't invertible

[patent ocean]

oh m1 doesnt have an inverse

so I have to just matrix multiply a random X and try to find its' coefficients?

[hot lake]

yeah the stabilizer of M1 is the set of matrices X in GL3(R) such that X M1 = M1

[patent ocean]

oh right multiplying it out you can see that it is independent of one of the columns

thanks

[south patrol]

Lol I remember doing this exact problem a lot T

We went over it w our prof in the sunshine with an outdoor blackboard

[patent ocean]

Oh lol nice

was this last year?

[chilly ocean]

Where are exact sequences of groups used

group extensions

e.g. central extensions

popped up in lie theory a few times

[chilly ocean]

Ah! Does it arise in topology?