groups-rings-fields

[south patrol]

Is this even correct as a proof

uhh

oh i should've been more careful

[delicate orchid]

have you done $\frac{1}{x_1+...+x_n} = \frac{1}{x_1}+...+\frac{1}{x_n}$ on the 2nd equality?

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

third*

FUCKING MOVE THE | UNDER THE FRAC RAAAAAAAAAGHHHHHHHHH

:thecosmoshumswithatunemostsweet:

[south patrol]

so equality holds iff Z(G) has index 4

ig

which holds for non-abelian order 8

makes sense

well

[delicate orchid]

I'm not seeing the third equality am I skill issu- yeah ok

[south patrol]

non-abelian order p^3

[delicate orchid]

[neuron activation]

[south patrol]

sorry

yes

[cloud walrus]

Süßkartoffel

[south patrol]

good point

sorry

it's

equality holds iff

[delicate orchid]

no no

it's fine

[south patrol]

1. Z(G) index 2

3. every conjugacy class size 1 or 2

which means p^3 doesn't work necessarily unless p = 2

[silver kettle]

hate algebra

[south patrol]

it's cause i left in an extra = by accident

[silver kettle]

hate finite group theory

[south patrol]

i'm bounding |g^G| >=2 for non-central elements

[silver kettle]

SIMPLE AS!

[delicate orchid]

there are many other channels on this server for you to enjoy then, buster

[south patrol]

Fair enuff

[delicate orchid]

I'm skill issuing again uhh

oh yeah

got it

[south patrol]

i wanna try the char theory now

ugh

[delicate orchid]

character theory has no uses other than the odd order theorem

[south patrol]

I cannot find a reference to this thing lol

for my diss

Sad!

[delicate orchid]

ur using the 5/8th theorem in ur diss? :letroll:

[south patrol]

lol unfortunately not

but did remind me about characters

tbf what i am doing rn is kinda chracter theoretic i guess

[delicate orchid]

oh lord here we go

[south patrol]

lol

[delicate orchid]

considering u said the word "etale" earlier I'm scared

oml Q_{2^n} has such a nice subgroup lattice

[summer path]

It's not symmetric :c

[delicate orchid]

the trivial group ruins everyone's life once more

[vivid tiger]

how?

looks symmetric to me

[delicate orchid]

the C_1...

https://tenor.com/view/mid-its-mid-ugly-nah-gif-19839131

[vivid tiger]

this is symmetric

oh

i see

[dusk whale]

I could use a hint on how to get started for this problem

The hint says use induction but im a little lost setting up a base case

[vivid tiger]

When a grid's misaligned with another behind it's a moiré!

[vivid tiger]

Well, Z/pZ is normal in Z/nZ with p dividing n.

Well actually you probably want that Z/(n/p)Z is normal in Z/nZ with factor group Z/pZ.

Then huzzah

[dusk whale]

How do we know this

[vivid tiger]

Normality? That's easy.

It's abelian.

The part you actually need to show is that Z/nZ / Z/mZ with m|n is in fact Z/(n/m)Z.

This is just due to uniqueness of cyclic groups of specified orders.

[dusk whale]

So Z/pZ is normal in Z/ nZ iff p divides n where p is a prime and thats something we know a priori?

[vivid tiger]

Definitely not

every subgroup of an abelian group is normal

[dusk whale]

Oh

I think I remember seeing that

[vivid tiger]

You can rederive it.

It's immediate from the definitions of normality.

ghg^{-1} is what?

in an abelian group, you'll always get h

[dusk whale]

I see

[dusk whale]

how come the m | n part is necessary?

[vivid tiger]

otherwise it's not a subgroup.

at the very least you have lagrange's theorem

yeah, I think that's true.

oh, obviously.

it's still just that generated by the generator

m * index = n

i retain my math license.

why do I feel like sometimes wrong

[crystal vale]

For 3, characteristic polynomial is det(tI -A ) =0 which are satisfied by A.

det( tI - A) = t^n so A^n =0, right ?

[dire siren]

that's correct
but the polynomial is det(tI-A), not "det(tI-A)=0"

[crystal vale]

I am not sure about characteristic polynomial, thank you

[white wraith]

I was reading a survey and there's 35.3% of people that think rings need to be commutative, is there some convention in another country that I don't know about??

[south patrol]

I imagine there is some ambiguity here - often people work in contexts where every ring is commutative

So it is an assumption rather than actually having commutative part of the definition

This survey was weird though lol I guess because it is largely undergrads

[dull marsh]

Oh lmao I thought the question was asking about the number of commutative rings and noncommutative rings

Only 35.3% of rings are commutative fr

[dull ginkgo]

commutative algebra mfers when they need to write out 2 extra words

[crystal vale]

Is it correct?

But I don't know how to show this set is uncountable

[rocky cloak]

This is correct.

To see that it's uncountable just notice that for any b in [-1, 1] there are c and d such that b^2 + c^2 + d^2 = 1.

[crystal vale]

If R is a Boolean ring without unity , can it be commutative, I think yes

[crystal vale]

Okay, thank you

[dire siren]

it's always commutative

[crystal vale]

So unity not necessary

[dire siren]

yes, not necessary

[white wraith]

I don't think so, you can show it with the idempotent property

[south patrol]

a+b = (a+b)^2 = a^2 + ab + ba + b^2 = a + b + ab + ba so ab = -ba

but uh

[white wraith]

Characteristic 2

[dire siren]

(x+x)^2=x+x

[white wraith]

-ba = ba

[south patrol]

exactly

uwu

no identity required

[white wraith]

Ending all my proofs with this from now on

[south patrol]

:)

I like "so we win", like on the Stacks project

[crystal vale]

Yes

[white wraith]

\renewcommand\qedsymbol{$so we win$}

[south patrol]

lol

[white wraith]

I'm testing that when I get home

[crystal vale]

So it also follows characteristic 2 without unity, right?

[white wraith]

Yeah

[white wraith]

Or in particular, people answering questions about things they don't know about

Like I didn't answer any applied questions because I haven't done it since I did intro to applied in my first year of uni 😆

[dire siren]

they completely messed up question 72

[crystal vale]

If R is a commutative ring with 1. Then is it true that if f(x) is an element of R[[X]] and all coefficients of f(x) are nilpotent in R. Then f(x) is nilpotent in R[[X]].

[dull ginkgo]

I think that’s for R[X]

Though my de facto proof for that is just proving the nilradical is additively closed, then just showing that if all the coefficients are nilpotent, then any polynomial is nilpotent because it absorbs multiplication

[crystal vale]

Yes, but I want to check on R[[X]]

[dull ginkgo]

Hmm

[crystal vale]

In the ring, the existence of the right unity implies the existence of left unity ?

[dire siren]

for R[x] there is a nice proof using the characterization of units:
f is nilpotent in R[x] => 1+f is a unit in R[x] => the free coefficient is a unit and the other ones are nilpotent

[crystal vale]

It can be true if we can replace 1 by any other unit element, right?

[dire siren]

yes

[crystal vale]

But for this , R must be commutative

It's given

[crystal vale]

How can I prove that if all coefficients are nilpotent then f is nilpotent

[dire siren]

oh, wait, I just realized the arrows are just "=>"

the other arrow goes by induction

[dull ginkgo]

The best way to show it imo is the nilradical is an ideal

[crystal vale]

Okay, thank you

[crystal vale]

This one ?

[dull ginkgo]

For the case of R[[X]]

[dull ginkgo]

This is a group theory exercise afaik

Or moreso module

I haven’t done much at all with R[[X]]

Though from what I’ve tried it’s like R[X] but “in reverse”

[rocky cloak]

Not necessarily no.

Like take
R = k[x1, x2, x3, ...] /(xixj, xi^i : i not equal to j)

Then
x1X + x2X^2 + x3X^3 + ...
is not nilpotent

[dire siren]

actually no need of induction
say the coefficients have nilpotence indexes k_1, k_2, k_3, ..., k_n
and raise f to the power k_1+k_2+k_3+...k_n... you'll get 0
(this is for R[x])

[dull ginkgo]

Is there a nice universal property for R[[X]]?

[rocky cloak]

It's the adic completion of R[X]. So I guess continuous maps of R-algebras are freely determined by maping X to a continuously nilpotent element.

[crystal vale]

Hint

[dull ginkgo]

Give me a moment, I’d have to find it

Correction, I am wrong

Misread the problem

[rocky cloak]

I don't think it's true, but don't have an example at hand

[crystal vale]

Let R be a non-zero ring such that the equation ax=b has a solution in R for all a,b belongs to R and a≠0. How can I show that R has unity?

[crystal vale]

Okay

[dull ginkgo]

This can be proven for semigroups

[dire siren]

it was asked recently

[dull ginkgo]

This IS an exercise in Jacobson

let me try to remember

[dire siren]

I found it, but apparently this has one extra condition: #groups-rings-fields message

[dull ginkgo]

I think it needs right unity

[crystal vale]

If it is commutative then it is true

[chilly ocean]

In what ring? Any?

[dull ginkgo]

[crystal vale]

Yes

[chilly ocean]

2x2 matrices with zero lower entries

[rocky cloak]

This seems like an overconvoluted way of saying that it
nr = 0, then nrs = nsr = 0 as well

[chilly ocean]

[1 0; 0 0] is a left identity but not an identity

[dull ginkgo]

lol it was mainly me just thinking it was neat

[crystal vale]

Okay, thank you

[crystal vale]

How can I prove that?

[dull ginkgo]

idk off the top of my head tbh

[crystal vale]

And if it is true then I can show that if R has no zero divisors and it is finite, then it is a division ring, right?

[dire siren]

that's always true

I mean the second part

[crystal vale]

Yeah but then I need unity

[dire siren]

yes

[crystal vale]

Centre of division ring is a field, because it is subring of division ring containing 1 and also it is commutative, right ?

[rocky cloak]

It's true that it's a field. But a commutative subring of a division ring doesn't have to be a field

[rocky cloak]

You don't need unity, in the sense that it follows that the ring has unity without you having to assume it.

[crystal vale]

And according to your subring definition 1 has to be in it or not?

[rocky cloak]

Not really relevant

[crystal vale]

Yes

[crystal vale]

Then?

[crystal vale]

It is not necessary that they have inverse closure, right?

[rocky cloak]

Personally, I use the definition that subrings have the same 1, but the statement is true regardless

Z is a subring of Q for example

[crystal vale]

Means centre has closed under inverse right?

[rocky cloak]

The center is yes

Which is how you would show it's a field

[crystal vale]

Yes

So if the subring of the division ring is closed under inverses and commutative then it is field , right?

[rocky cloak]

Yes, a commutative ring where every nonzero element has an inverse is by definition a field

[crystal vale]

Got it thank you

Is Z_2 × Z_2 Boolean ring?

Yes ?

[rocky cloak]

Maybe try writing out the definition and seeing if it holds

[crystal vale]

Yes it is hold

[rocky cloak]

Then there's your answer

[crystal vale]

So the Boolean ring does not need to be integral domain

[rocky cloak]

They very rarely are

[south patrol]

If e^2 = e in an integral domain, then uh

0 = e(1-e)

[crystal vale]

Yes got it

[rocky cloak]

There's Z/2, and maybe the trivial ring depending on your definition of integral domain

[crystal vale]

How it depends on integral domain definition, trivial ring means ab=0 for all a and b ?

[rocky cloak]

Trivial ring, means the ring with one element

Many people have 0 =/= 1 as part of the definition of integral domain

[crystal vale]

Okay, thank you

[chilly ocean]

Alternatively, define an integral domain as: if a product is 0 then one of the terms in the product is 0. Then the empty product on trivial ring is 0 despite none of the terms being 0 so it isnt a integral domain

[crystal vale]

Empty product?

[rocky cloak]

That's an interesting aproach

Seems to be leaning more towards obfuscation than elegance, but still

[dull marsh]

The empty product is usually defined to be the multiplicative identity

[rocky cloak]

It makes it so that the product over a disjoint union is the product of the products over each set

Which is nice

[crystal vale]

Okay

[chilly ocean]

This makes me wonder

Let R be a ring (unital or not idc, preferably commutative)

Can there be a subset of that ring that is closed for ternary products, but isn't a subring?

[crystal vale]

ternary products ?

[rocky cloak]

I guess take some element x where x^3 is 0, but x^2 is not. Then consider {0, x}

[chilly ocean]

nice move

[crystal vale]

In the ring of all real valued continuous function on [0,1] , the nilpotent element only 0 ?

[chilly ocean]

What have you tried?

[hidden wind]

help i'm stumbling over something really elementary 😭

[chilly ocean]

bet

[south patrol]

All maths is elementary

[crystal vale]

Let f be a nilpotent element then f^n(x) =0 for all x and for some positive integer n.
But it shows that f(x) =0 for all x, right?

[rocky cloak]

My dear Watson

[hidden wind]

namely, if S is a finite semigroup, in which the two laws of cancellation holds (i.e. if ab = ac then b = c, and similarily with a on the right), then S is a group

[rocky cloak]

On a finite set injective = surjective = bijective

[hidden wind]

i'm not sure how to go from this "ab = ac implies b = c", to "a has a multiplicative inverse"

[hidden wind]

whaaaaat

[rocky cloak]

Are you one of the lucky 10 000 today?

[hidden wind]

what do you mean

[chilly ocean]

if you have a function f: X->X and X is finite, then injective is equivalent to surjective

[rocky cloak]

https://xkcd.com/1053/

xkcd: Ten Thousand

Saying 'what kind of an idiot doesn't know about the Yellowstone supervolcano' is so much more boring than telling someone about the Yellowstone supervolcano for the first time.

[hidden wind]

ohh

[crystal vale]

Is it correct?

[chilly ocean]

Well, lets fix an x

you have that f^n(x)=0

what does that imply about f(x)?

[crystal vale]

From there we have f(x) =0 , right?

[chilly ocean]

why?

[crystal vale]

Because f(x) is the real number

[chilly ocean]

yeah

[crystal vale]

It follows that for all x, f(x)= 0, right?

[rain grove]

We can look at the set of square matrices over Complex numbers M_n(C) and over quaternions M_m(H) as vector spaces over Real numbers... is there a pair of natural numbers n,m so that these vector spaces are isomorphic?

[chilly ocean]

Do you know of any condition that is equivalent to two vector spaces being isomorphic?

[rain grove]

My answer is no, as M_m(H) ≅ R^(4 * m^2) and M_n(C) ≅ R^(2 * n^2)

but then that means n = sqrt(2) * m
is this ok?

[rain grove]

yeah, they are the same dimension

[chilly ocean]

sounds gud to me

[crystal vale]

Let R be a ring and Z be the centre of R, then M_n(Z) is never the centre of M_n(R) if n>=2.

So let's first for n>=3.
If I take A be strictly upper triangular matrices in M_n(Z) and B be the strictly upper triangular matrices in M_n(R). If AB= C and BA= D then C[1,3] = A[1,2] × B[2,3] and D[1,3]= B[1,2] × A[2,3] so if there are at least two elements, then I have distinct B[1,2]≠B[2,3] , right? But what if A[1,2] × B[2,3]= B[1,2] × A[2,3] ?

[dire siren]

I would suggest choosing matrices with only 0 and 1s, but very few 1s

for example: ||A = the matrix with 1 in the top right corner and 0 everywhere else|| and ||B = the matrix with 1 in the bottom right corner and 0 everywhere else||

[crystal vale]

But it is not given that R has unity

[dire siren]

@crystal vale I'm fixed on rings with unity; you can replace the unity with any element in Z

[crystal vale]

Yeah got it, thank you

[dire siren]

we should assume of course that R is not commutative (otherwise the problem is false)

[crystal vale]

What if R is commutative then Z= R , right?

Then how is the problem false?

[dire siren]

oh, nvm

I think I meant Z=/={0}

but I wonder if this is true though

[crystal vale]

But if Z={0} then maybe it is true

[dire siren]

I think we have to check this

because my proof works under the assumption Z=/={0}

[crystal vale]

Then M_n(Z) ={0}

[dire siren]

@crystal vale So if Z={0}, then for any element x we can find y such that xy =/= yx
so let's take a non-zero matrix A in M_n(R)
say the element on row i, column j is non-zero... we can denote it by a

now take an element b different from 0 and a

and consider the matrix
B = b on row j, column i and zero everywhere

I just hope I didn't mess up the multiplication, but I think AB is not equal to BA

yeah, the multiplication is messed up... have to search for another matrix B

[crystal vale]

But if Z={0} then M_n(Z) = {0} ?

[dire siren]

I know, but I have a feeling that fails to be true

@crystal vale I might be wrong, but my intuition comes from matrices with real entries... in this case the center consists of scalar multiples of the identity
so if we move to a general ring which might not have identity... my intuition tells me that it could happen for the center to only contain the zero matrix

[crystal vale]

Okay, thank you ❤️

[dire siren]

that's not a proof yet

[crystal vale]

Yeah I know, assume Z(R) ≠ {0}

[rocky cloak]

If ab = 0 for all a and b in R, then the same is true for Mn(R), so Mn(R) is commutative

I think things get pretty crazy in general if R isn't unital.

[dire siren]

oh, but R is commutative in that case

[rocky cloak]

But I guess maybe you could get away with a condition like, there is an element x in Z(R) and an element y such that xy is nonzero

Alternatively don't consider Mn(R), when R isn't unital 😛 . Like I can't really imagine a situation where that ring would come up

[dire siren]

we proved the case Z(R) ≠ {0}, but we are stuck with the case Z(R)={0}

I can only get that a matrix in Z(M_n(R)) would have 0 on the main diagonal

[rocky cloak]

If R is commutative then Z(R) = R, hence nonzero

[dire siren]

we assum R to be non-commutative

[crystal vale]

But in case Z(R) ≠ 0 , I get ab=0 what if a is zero divisor , how can I prove that there is another b such that ab≠0

[dire siren]

what I want to prove specifically is
if Z(R)={0}, then Z(M_n(R))={O}

[rocky cloak]

non-commutative typically means "not necessarily commutative", but if you insist on it being strictly noncommutative you could just append two variables that don't commute with each other

[dire siren]

sorry, I meant to say "not commutative" and I didn't realize

yeah, the thing is that I solved Notknow's problem under the assumption Z(R)=/=0 and I doubt it is true in the case Z(R)=0

[rocky cloak]

But I don't believe you solved it

[dire siren]

yes, you are right

[rocky cloak]

With Z(R) = 0 I can believe it though

That Z(MnR) = 0 I mean

[crystal vale]

Lie ring has no multiplicative identity element because it is anti- commutative, right?

[rocky cloak]

Corectomundo

[delicate orchid]

1x =-x1

[dull ginkgo]

char 2 motherfuckers be like

[crystal vale]

?

[rocky cloak]

Just trying out new ways to say yur

[dire siren]

@crystal vale So regarding the previous problem, you should ignore my approach for the first part because it is wrong. Sorry about that.
But on the other hand, the problem turns out to be false, as jagr2808 provided a counterexample above.

Did you find that problem in a book or did you come up with it?

[stark helm]

if f(x)=x^4-6x+1 and want to justify f(x) is irreducible over Q[x], is it correct to solve roots x= +- sqrt(3+2 sqrt(2)) or +- sqrt(3-2sqrt(2)) and write it into linear factor? then arguing that linear factor of four roots can not be combined into f=h*g in any ways when h, g in Q[x]?

[long obsidian]

Say f:A->B and a is an ideal in A. Is the extension $a^e=\cap_{I\supset f(A)} I$ I can't remember if this is the same as the ideal generated by the image

[cloud walrus]

HausdorffT1

[celest furnace]

Yes that works

You can let alpha be one particular root and say (x-alpha)(x-beta) doesn’t have coefficients in Q for all the other roots beta

[chilly radish]

||For any ring, one has

Z(M_n(R))=Z(R)I_n

So indeed in Z(R)=0 then Z(M_n(R))=M_n(Z)=0.

In the case Z(R)\neq 0 amd n>=2 one has, M_n(Z)\neq Z I_n since M_n(Z) has non scalar matrices||

[vivid birch]

is this the correct channel to discuss semigroup theory?

[dire siren]

@chilly radish I think that's true only for rings with unity

[chilly radish]

I think one could probably extend it further but yea I did assume unity

I think most reasonable textbooks do tho

[dire siren]

yeah, initially I also thought that the ring has unity, but the guy who asked told me it's not necessarily unital

[chilly radish]

Oh my bad

That's silly

I guess in that case one can just take a nonscalar matrix and show it doesn't commute with another matrix

I think you need to think of the case where Z(R) has only 2 elements tho

I think taking diag(a,0,..) for a in the center will work, you just have to choose a correct matrix (and disregard trivial edge cases)

[chilly radish]

Sorry not diag(a,0...) But rather something like
(0 a)
(0 0)
And the rest is zeroes

[rain grove]

h: G -> G' homomorphism of groups, let a be an element of a finite order of G. Show that the order of h(a) divides the order of a. Show that if h is injective then the orders are the same

what i did: a^n = 1
then: h(1) = h(a^n) = h(a)^n
multiply both sides with h(a): h(a) = h(a)^n
multiply both sides with h(a)^(-1) : 1 = h(a)^n
im stuck here

[dire siren]

@chilly radish it just works by adapting the standard proof that the center of M_n(K) consists of scalar matrices of the identity (K - field)
In that proof one takes the matrices E_ij, whose elements in row i, column j is 1, and the rest of the elements are 0

For our problem we should take the same matrices with some arbitrary element b instead of 1.

So what happens is that if A is in Z(M_n(R)), then E_ijA=AE_ij so we get ba_ji=a_jib, for any b, so a_ji is in Z(R), so it must be 0; hence A=0.

[dire siren]

On the other hand, due to the randomness of the ring, I'm not sure to what extent such an extension is possible
for example, jagr2808 mentioned above the ring in which ab=0 for any a,b. For this ring the center of M_n(R) is M_n(R) itself

[chilly radish]

This works in the case of the center being 0

But I don't think this will work for nonzero center

[south patrol]

Non-unital rings...

[chilly radish]

Oh huh I missed that

[south patrol]

did notknow ask it

[chilly radish]

Yea ok then it's not true in general

[dire siren]

yes, the proof I wrote is for the implication "Z(R)=0 => Z(M_n(R))=0"

[chilly radish]

Ye

[south patrol]

Yeah he did

Lol

[dire siren]

I actually mentioned it, but I deleted it by mistake; sorry

[south patrol]

Only person to use non-unital rings

[chilly radish]

No worries

I feel like the author sdid not intend for thos to be a question of nonunital rings

But idk

[dire siren]

I don't like rings without unity, but this was the request

[south patrol]

waiting for the scheduled debate on unital rings

tbf i do think using nonunital rings in an intro textbook is probably a pedagogical mistake lol

causes extra headaches

[chilly radish]

I mean, it's pretty much never given that a ring is without unity. Usually you'd specify in the beginning that all rings have unity and only say something when they don't

It's either that or the author was thinking of unital rings when writing the exercise

[rocky cloak]

How to tell if you're doing algebra or analysis:
Look at your rings
Do they have unit?
Yes -> you're doing algebra
No -> you're doing analysis
What rings? -> your title as mathematician is hereby revoked

[vivid tiger]

wat. analysis rings have units.

...i hope...

[rocky cloak]

Rings that often come up in analysis are compactly supported functions on Hausdorff spaces

(which don't have unit)

((unless your space is compact))

[long obsidian]

Hey if I have the injection Z->Z[i] apparently if I take the ideal 2Z and extend it through this homomorphism $(2Z)^e={(1+I)^2}$ why is this the case

[cloud walrus]

HausdorffT1

[wooden rain]

I know proofs 1 and 3 from this thread: https://math.stackexchange.com/a/331027/1058445, but I wonder if you could prove that algebraic numbers form an additive group only using linear algebra without resorting to multlinear mappings and the tensor product

[vivid tiger]

well, i mean, it's not that bad. the tensor product here can just be thought of as the kronecker product, and the facts about the eigenvalues shown

that is, take the companion matrix (matrix whose char poly is p)

[rocky cloak]

(1 + i)^2 = 2i = 2 * i

Since i is a unit, (2) = (2i)

[vivid tiger]

So, the point is that we need to show that the algebraic closure of Q indeed only has elements algebraic over Q?

Isn't that immediate?

Every element in the algebraic closure, by definition, is algebraic over Q.

By definition, every algebraic element is in the algebraic closure.

Admittedly, this needs the apparatus to prove the existence of an algebraic algebraically closed extension.

Which is definitely more complicated than the tensor product.

[wooden rain]

The goal is to prove the set forms an additive group. It might, potentially, happen that the sum of two algebraic numbers is not algebraic

[vivid tiger]

but we have the algebraic algebraically closed extension. algebraic extensions are such that every element is algebraic.

since elements algebraic over Q are in the algebraic closure, the sum of two algebraic elements must be algebraic.

this should work over any field.

[rocky cloak]

The answer that imho is very simple is the second answer here.

If a and b are algebraic, then F(a)/F is finite, and F(b, a)/F(a) is finite, thus F(a, b) is finite.

Thus a+b, a-b, ab, and a/b are all algebraic

[vivid tiger]

Oh, that's much better.

[rocky cloak]

The only two things you're using here is that F(a) = F[x]/(p(x)) where p is the minimal polynomial of a, and that you don't have infinitely many linearly independent elements in a finite dimensional vector space.

Both of which are pretty basic facts

This is less powerful though . Like it doesn't price that the algebraic integers form a ring, which the other argument does.

[wooden rain]

What does F(a) denote? Based on your message, it would be some subspace of polynomials of degree lesser than the degree of the minimal polynomial, but I've seen the notation F(x) to denote the field of fractions. How about F(a, b)?

[vivid tiger]

by other argument, you mean the companion matrix argument?

[vivid tiger]

F(a) = field generated by F and a = F[a] = space of polynomials in a

(for a algebraic)

this is because e.g. 1/sqrt(2) = poly in \sqrt(2)

[rocky cloak]

F(a) is the smallest field containing F and a.

[vivid tiger]

(it's \sqrt(2)/2)

the only reason F(a) would be with polynomials lesser than the degree of the minimial polynomial would be because you can reduce higher degree things

[rocky cloak]

Yeah, I didn't read any further down

[vivid tiger]

e.g. reduct \sqrt(2)^2 to 2

[vivid tiger]

i didn't even click on the link

[wooden rain]

Yep, that's clear and very elegant. I like it

[vivid tiger]

stuff like this is why i'm liking algebra nowadays

the basic field theory and galois theory stuff is currently some of my favorite stuff

[rocky cloak]

"all proofs in algebra are trivial once you understand the definitions"

(if only it wasn't for those pesky definitions ey)

[vivid tiger]

false, that's DG and AT

part of why i like algebra is that some of these proofs still seem neat after understanding them!

[wooden rain]

I'm not sure I understand why $F(a) / F$ is finite to be honest. Couldn't we construct infinitely many elements of the form $\alpha_1 a + \alpha_2 a^2 + \ldots + \alpha_{d - 1} a^{d - 1}$ where $d$ is the degree of the minimal polynomial?

[vivid tiger]

like the proof in lang attributed to artin of the existence of an algebraically closed field extension

[cloud walrus]

Thingoln

[vivid tiger]

"finite" means "finite dimensional vector space"

it indeed is infinite as a set

it's just "a finite field extension"

[rocky cloak]

If the proof
Zorns lemma, qed.
?

[vivid tiger]

no!

i mean, that's part of it

but that's the trivial part

[rocky cloak]

Like you want to prove that the algebraic closure is it's own algebraic closure?

[vivid tiger]

the hard part is getting a field extension in which all polynomials of the previous field split. from there you can do zorn's lemma

at least, i think that was the hard part

[rocky cloak]

I think you mean the other way around

[vivid tiger]

the proof looked at the polynomials over the letters X_f where f is a polynomial in k[x]

[rocky cloak]

Getting a field where all polynomials split is just Zorns

[wooden rain]

Okay, I think I got it. F is a one-dimensional vector space over itself. If F(a) / F is finite, then F(a) is also finite because $\dim (F(a) / F) = \dim F(a) - \dim F$. We apply the same reasoning to $F(a, b) / F(a)$ and that yields that $F(a, b)$ must also be finite

[south patrol]

Huh lol

This is the opposite of what is the case imo

[vivid tiger]

and then, showed that the ideal generated by f(X_f) is not maximal

because if it was a finite linear combination of them would be 1

but since each one has a root, by plugging in separately you get 0+0+0...+0=0=1

so then you can take the quotient by this ideal and get a field

[south patrol]

Like dg doesn't have as much abstract nonsense immediately

[cloud walrus]

Thingoln

[rocky cloak]

That's a funky as proof

[south patrol]

Yeah

[vivid tiger]

it's wonderful!

[vivid tiger]

okay but immediately you get trivial proofs and hard definitions

e.g. generalized stokes

[south patrol]

Sure

[vivid tiger]

and the rest is abstract nonsense

[south patrol]

Lol kinda disagree w that but sure yes initial DG courses just seem a bunch of definitions

But then a first course in rings or smth will probably usually be similar

AT has stuff like excision which is not like that

[vivid tiger]

ehhh, i was perhaps wrong about AT

i feel that, say, PID=>UFD or structure of finitely generated abelian groups or galois theory is much more nontrivial than initial DG

maybe in a couple years I'll be like "obviously it's trivial consequences of the definitions, just like DG"

[wooden rain]

Can I have one more question to this? Why does the finiteness of F(a, b) imply a + b is algebraic? Is it because we could choose a finite number of non-zero elements of that vector space whose sum would be equal to the zero vector? In the case of F(a), I imagine we could start interpreting elements of this space as polynomials of the variable a and taking a linear combination that would equal zero would produce a polynomial of coefficients from F whose root would be a

[vivid tiger]

Well, if we have a finite extension E of K

take any element a of E.

Look at 1,a,a^2, a^3, ...

[wooden rain]

For what it's worth, we haven't covered the proof of the existence of an algebraic closure of an arbitrary field yet. But I know Zorn's lemma (if that's important for understanding this)

[vivid tiger]

they can't all be linearly independent.

Se ,eventually we have some relation like c0 * 1 + c1 a + c2 a^2 + ... cn a^n = 0

[rocky cloak]

So take an element x in F(a, b) (for example x = a+b)).

Then 1, x, x^2, x^3, ... must be linearly dependent, since it's infinite.

Hence there are coefficients f0, f1, ... such that

f0*1 + f1x + f2x^2 + ... fnx^n = 0

[vivid tiger]

And, well, what is that if not a polynomial with coefficients in K and variable being a?

[rocky cloak]

So x is the root of a polynomial

[wooden rain]

Oh, that makes sense

[vivid tiger]

The intuition: algebraic relationships are just polynomials.

so, if you have linear dependence, you have some algebraic equation.

[wooden rain]

Algebra really turns into the study of polynomials lol We're studying the properties of endomorphisms at uni now and they're ubiquitous, but it seems it never changes

[vivid tiger]

this realization is what made me realize why we care so much about polynomials and why we should expect them to be interesting.

[wooden rain]

That makes a lot of sense. The proof assumes that an algebraic extension of a field always exists, right? Like although previously we looked at F(a), F(b) and F(a, b) as vector spaces over F, here we assume that F(a, b) is a field

[vivid tiger]

"endomorphism" reads to me as "morphism to itself", which is much more general (e.g. an endomorphism in Top is a continuous map to itself)

[vivid tiger]

No, you are proving that every element is algebraic.

So, we know F(a) is algebraic since a is algebraic. It's finite as it's generated by an algebraic element.

Same for F(b).

Now, F(a,b) = (F(a))(b)

So, we still have finiteness.

Now, we can conclude F(a,b) is in fact algebraic (meaning that every element is algebraic)

Likewise this works for extensions that are finitely generated by algebraic elements.

[rocky cloak]

Well by assumption a and b (and F) all live inside some field. Then F(a, b) is just defined to be the smallest field containing them.

But you could also directly construct F(a, b) using their minimal polynomials

[wooden rain]

I got it. 🎉 Thank you both for the help, I really appreciate it

Out of those proofs, I personally find the tensor one the most intuitive. It makes it really easy to see why a + b and ab would be algebraic

[long obsidian]

Ooh okay thank you

[long obsidian]

If A is a ring such that x^n=x for some n for all x in A then any prime ideal is a maximal ideal.

Does this follow since the quotient kinda "inherits" the relation and so for nonzero y in A/p we have y^n=y and then y^{n-1} =1 by cancellation of A/p. Then it follows that y^{n-2} is the multiplicative inverse of y?

[south patrol]

Yes, great

Of course we should assume "for some n >= 2" but yes

this is always a nice kinda "ansatz"; to show any prime ideal is maximal given some assumption A, it often suffices to show that every domain satisfying A is a field

[crystal vale]

What counterexamples, Z={0} , this one?

[crystal vale]

From internet

[crystal vale]

This one?

[long obsidian]

If A is a Boolean algebra apparently any finitely generated ideal is principal. So for instance say <x,y> is an ideal of A. Then consider z=x+y+xy and then it follows xz=x(x+y(x+1))=x and similarly yz=y. And so <x,y>=z

Say by induction any ideal <x1,...,xk> is principal then say <y1,...yk+1>=<y1,...,yk>+<yk+1> and then by applying the inductive hypothesis <y1,...,yk+1>=<f>+<yk+1>=<e> where I guess you have to apply the inductive hypothesis twice.

Is there a non inductive proof of this? I hate inductive proofs. I saw something on line about Boolean algebras and the proof maybe being more clear from that perspective

[dull ginkgo]

what

[crystal vale]

Because we can write f(x) as the sum of nilpotent elements therefore f(x) is nilpotent in R[X], right?

[dull ginkgo]

Nilradical is an ideal in any commutative ring

[crystal vale]

In the commutative ring, right?

[dull ginkgo]

Yes

[crystal vale]

If R is a finite commutative ring, then if f(x) has all nilpotent coefficient then f(x) is nilpotent in R[[X]], is it true for R[[X]] ?

[dull ginkgo]

No idea for R[[X]]

https://www.ams.org/journals/proc/1971-027-03/S0002-9939-1971-0271100-6/S0002-9939-1971-0271100-6.pdf

It has to do with Noetherian-ness apparently

[crystal vale]

There is a question in which it takes R=Z_n and says that f(x) can be written as a sum of finitely many suitable nilpotent elements in Z_n[X], by collecting terms with the same coefficient.

[crystal vale]

Is it true that in M_n(Z_m), every element is a zero divisor or a unit, for all positive integers m and n.

I think it follows from that if a finite ring has 1 then if x is not a zero divisor then x is a unit.

[vagrant zinc]

Guys a quick question!
(Z_4,+)={0,1,2,3} and (Z_4^x,x)={1,3}
is this correct?

[crystal vale]

(Z_4^x,x)= {1,3} ?

What is Z_4^x?

[vagrant zinc]

I was able to solve the doubt, it is the multiplicative group, it is that god so many different notations that have these books.

[crystal vale]

Means groups of units of Z_n, right?

[vivid tiger]

Yep

Note the following neat theorem: any finite subgroup of the group of units of a field is cyclic.

Proof uhh lemme try and remember/rederive this

call it A

Look at the p-torsion A(p)

ah

since it's finite take the max order p^r

Point is that all the elements must be a root of the polynomial X^{p^r}-1

but there's only at most p^r of them

so since the element of max order generated a group of order p^r

we know that that must be all of them in A(p)

now since A = direct product of A(p)'s (since it's abelian), we are done since p^r is coprime to q^s and we can apply Chinese remainder theorem.

[stark helm]

How will you prove this question? I know we have formula to know three roots of p(x), but the hint let me to graph it? Is it really a proof if we just draw the graph of p(x)??

[prime sundial]

that hint makes me think they want you to use calculus, otherwise im not sure what they're getting at

[vivid tiger]

Well, think about what graphing it tells you.

That is, it's negative below some number, positive above some number, and if you look at the derivative you'll hopefully no problem there. thus there's exactly one root.

specifically: it's negative at x=0, positive at x=1, and the derivative is
3x^2+3
which is always positive

so there is exactly one root between 0 and 1.

(in R!)

[dire siren]

yes

[rocky cloak]

Personally I think the inductive proof is very clear, since it gives a completely explicit construction of how to find the generator of a finitely generated ideal.

But I suppose you could unravel it into a single formula. I.e. for an ideal (x1, x2, ..., xn) define z to be the sum of all monomials
x1 + x2 + ... xn + x1x2 + x1x3 +... + x1xn + x2x3 + ... + x1x2x3 + ...
Then xiz = xi, so the ideal is equal to (z).

[rocky cloak]

Or I guess a slightly easier way to describe this would be

z = 1 - prod_i (1 - xi)

[crystal vale]

This one?

[dusk whale]

I have a quick question

In dummit and foote it says G/K can be thought of as the set of fibers of a homomorphism from G to some group H. If every normal group N is the kernel of some homomorphism G to G/N, is it still useful to think of this codomain G/N as the set of fibres of a different homomorphism?

[delicate orchid]

well not every map is the canonical surjection G -> G/N

but the fibres would be the same anyway so I guess you might as well just think about them as fibres of the canonical map

[dusk whale]

whats a canonical map mean

[delicate orchid]

very vague and odd question I must admit

like yea, there's probably been some point in human history where thinking about the fibres of whatever map you have is more useful and immediate

like maybe if you're lifting representations in a topological way by using vector bundles over the classifying spaces of G and G/N? only example that comes to mind

[barren sierra]

If I ask you for a map G -> G/N what's the map that comes to mind?

[dusk whale]

g -> gN

[barren sierra]

That's the canonical map in this case

[mighty kiln]

Canonical means the "obvious" one

[dusk whale]

oh ok

[barren sierra]

canonical is used in that sense

The "obvious" / "natural" map

[dusk whale]

i guess i was confused like if we define a map G -> G/N, and G/N is the set of fibres of some homomorphism between groups, which we haven't defined yet a priori?

I should provide context that I saw this in a proof that any normal subgroup N is a kernel to a homomorphism

[dull ginkgo]

I mean that's also the "statement" of the first iso theorem for sets. The equivalence relation is x ~ y <=> f(x) = f(y)

[dusk whale]

Of course G/N is the set of fibres of the homomorphism pi, but we didn't know that apriori, so how is G/N "well defined" at the beginning?

if we think of it as set of fibres of a hom.

[ashen heron]

N is a normal subgroup

[barren sierra]

Have you not seen that G/N is a group iff N is normal in G?

If so, it's not a hard proof, worth proving it yourself

[dusk whale]

I have but is the definition that N is normal implying G/N is a group the same as G/N being the set of fibres of a homomorphism?

[ashen heron]

are you saying N is a normal subgroup so it is kernel of homomorphism phi: G to G/N. then, if we quotient G by this kernel N then G/N is the fibers of elements in G/N itself? as in elements in G/N are of the form phi^-1(x) for x in G/N?

H = G/N here and your K is N

[dull ginkgo]

Is this a proof of the existence of an element of order divisible by p if G is a finite abelian group, and prime p | |G|

Assume the converse, thus p does not divide |(x)| for any x in G*, thus p | |G/(x)|. The same converse applies to G/(x), so repeating this implies there are infinitely many factors of |G| that arent divisible by p, a contradiction

[south patrol]

Yes that works. You can also just say that like

[crystal vale]

How? If Converse is false then how does it imply the original statement is true?

[dull ginkgo]

I meant contrapositive

brain fried

[south patrol]

this is a fine use of converse tbf lol

It's converse in the usual english sense

Anyway

[crystal vale]

G* ?

[south patrol]

You can use that like

[dull ginkgo]

G\{e}

[south patrol]

G/(x) has an element of order divisible by p by induction

and then you can lift that to G

[dull ginkgo]

strong or standard induction?

[south patrol]

who cares

Uh i guess strong

idk I wouldn't worry about the type of induction beyond high school exams lol

[dull ginkgo]

i suppose

[south patrol]

But yeah

[dull ginkgo]

I am trying to understand Jacobson's proof of sylow 1 which is different from what I've seen before

[south patrol]

Oh interesting

How does he do it

The nicest proof I've seen is the non-inductive proof

[dull ginkgo]

He proves Cauchy's theorem

[south patrol]

have you seen that

doesn't use Cauchy lol

[dull ginkgo]

group action one?

[south patrol]

Ye

[dull ginkgo]

I think he leaves that one as an exercise actually

[south patrol]

Well almost all use group actions

but there's a nice one w/o induction which is particularly cute

on wikipedia lol

[dull ginkgo]

He uses Cauchy's theorem to show there's an element of order p in all abelian groups who's orders are divisible by p

[south patrol]

Sure

[dull ginkgo]

then uses two cases, where p divides the center, or not

[south patrol]

There's also a nice other proof of Cauchy for abelian groups which is popular in like elementary number theory courses lol

[delicate orchid]

we were supposed to case even in high school exams?

[dull ginkgo]

if p doesn't divide the center then by the conjugacy class eq then it cannot divide some conjugacy class order

[delicate orchid]

anyway just throw the structure theorem at it :letroll:

[south patrol]

Tbf no but people here seem to care sometimes lol

[delicate orchid]

so strange, unless there's some logic reason why they're different. Which there probably is

[dull ginkgo]

so it must divide the order of some centralizer

yada yada yada

[south patrol]

Wait nah i was wrong w the Cauchy thing lol dw

i meant Lagrange

[delicate orchid]

proof that if G is a p-group, then G has an element of order p. This is true for p-groups of order p, hence assume it is true for all p-groups of order less than G - then we can lift an element of order p from G/Z(G) as Z(G) \neq 1

problem, lagrangeheads?

[south patrol]

Gotem

[delicate orchid]

don't ask how I know Z(G) is non trivial please please please please please please

[dull ginkgo]

conjugacy class eq

[delicate orchid]

it's circular irregardless

[dull ginkgo]

how is it circular

nah wait

def of a p-group

[delicate orchid]

I don't want to think about this anymore. It was a quip

stop talking lalalalalalalalal I can't hear you

although it is funny that this implies the result for uhh all groups

cauchy's that's it

[dull ginkgo]

hur dur $|G| = |Z(G)| + \sum{[G : C_G(x_n)]}$ but $|Z(G)| = 1$ so |G| is coprime to $\sum{[G : C_G(x_n)]}$ but the summands all divide $|G|$

[cloud walrus]

Mizalign #1 simp

[dull ginkgo]

There's two actions on the set of p-subsets of G

conjugation, or left/right multiplication

I wonder if either can be used to prove sylow I

[delicate orchid]

I don't buy left/right multiplication being an action

but I don't know what a "p-subset" is

[dull ginkgo]

subsets of G of size p

so the group action maps a subset S to it's image by left/right multiplication by an element

[delicate orchid]

then there's nothing special about p being a prime

[dull ginkgo]

it's for sylow you dingus

[delicate orchid]

ok?

[dull ginkgo]

of course it has a generalization to any pos number

[delicate orchid]

no it doesn't?

unless your group is solvable

society if I didn't have to think about solvablity

[south patrol]

gotem

[cobalt heath]

p-subsets, not p-subgroups?

[dull ginkgo]

y e s

[delicate orchid]

I'm still not following

the proof of sylow 1 relies on p being a prime

[south patrol]

this is painful given that p-subgroup means order a power of p lol

[delicate orchid]

we're ignoring this

[dull ginkgo]

wew I was just defining the group action to use it

[cobalt heath]

Err

[delicate orchid]

counter example is the fact that there is not an order 30 subgroup of A_5

[cobalt heath]

Well, I guess left-multiplication is bijective, at least that's kind of true I guess

[dull ginkgo]

We can define a group action of $G$ on $\mathcal{P}(G)$, and it stabilizes the partition of $\mathcal{P}(G)$ into subsets of the same cardinality

[cloud walrus]

Mizalign #1 simp

[dull ginkgo]

so you just consider it on sets of size $p$ for p dividing |G|

[cloud walrus]

Mizalign #1 simp

[delicate orchid]

ok

[dull ginkgo]

i'm literally going off the standard proof of sylow

[cobalt heath]

This does not seem to have relations with sylow imo

[delicate orchid]

jeebeoo beeboo

epblep pop op

[cobalt heath]

Other than, well.. using group actions

[delicate orchid]

anyway yes, keep going

[dull ginkgo]

I am looking at my textbook right now

[delicate orchid]

you are awful at expositing information (and I am awful at reading it)

[dull ginkgo]

true chat

[delicate orchid]

anyway chat

I see where you're going with this now

although I don't think you want size p

p^n would be better I think

cause then index of the size is coprime with |G|

[dull ginkgo]

ah yeah

[stark helm]

If I use geogebra and it gives me a graph of increasing function, then we get ngtaitve when x<=0, positvie when x>=1, and derivative is always positive between 0 and 1, so there is exactly one root, is that true?

[south patrol]

Source chat

[delicate orchid]

A_5 is simple, chat

[dull ginkgo]

no it's complicated :troll:

[uneven bobcat]

Being simple only means there's no normal subgroups. You can still have plenty of non-normal subgroups.

Although, you're correct in this case that there's no subgroups of order 30, because such a subgroup would be index 2 and all index 2 subgroups are normal. This extra step is important to mention though.

[south patrol]

Wew is aware dw

[dull ginkgo]

wew will kill you

run

[stark helm]

Consider the polynomial p(x) = x3 + 3x − 1 ∈ Q[x] and let E be the splitting field of p(x) over Q, If I want to find the size of Galois group Gal(E/Q), My arguement here: argue that because p(x) is seperable because it is in a field of char=0, and we have a real root and two complex roots of p(x) in E. Hence, |Gal(E/Q)|=[E:Q(alpha)][Q(alpha):Q], alpha is real root of p(x). Sicne we have three distinct roots of p(x), then |Gal(E/Q)|<=3!=6, but [E:Q(alpha)][Q(alpha):Q]>=3 because [Q(alpha):Q]=3 but Q(alpha) does not contain complex roots? Is it a true argument?

so |Gal(E/Q)|=6 at last

[rocky cloak]

Yes that works. Assuming you know the polynomial is irreducible anyway

[delicate orchid]

almost like that's exactly what I said

and is the exact reason the subgroup of order 30 counter example came to my mind

anyway

[uneven bobcat]

Exactly isn't quite accurate.

[uneven bobcat]

See here for the "exact" wording of your claim

[uneven bobcat]

and the "exact" wording of your explanation

[delicate orchid]

three (you)s :thecosmoshumswithatunemostsweet:

I now know how women feel when they're being "mansplained" to

[uneven bobcat]

glad I could be of service

[stark helm]

then if I want to find phi in Gal(E/Q). suppose three roots of p(x) is alpha( real roots), alpha1, alpha2(two complex roots). Then phi(a)=a if a is in Q. and phi(alpha)=alpha, phi(alpha)=alpha1, phi(alpha)=alpha2. phi(alpha)=alphaalpha1, phi(alpha)=alphaalpha2, I should miss one here, do you think this forms that I get is correct?

[rocky cloak]

The automorphisms are given by permuting the 3 roots.

alpha alpha1 is not a root, so you can't map alpha to that

[stark helm]

so basically alpha from alpha, alpha from alpha 1 and alpha from alpha 2 is correct right?

And what will other three automorphisms look like?

[rocky cloak]

So you have, the identity. You have the automorphisms that fix one of the roots and swaps the other two. Then you have the cycles alpha -> alpha1 -> alpha2 and alpha -> alpha2 -> alpha1

[rocky cloak]

Just saying "alpha from alpha 1" doesn't really determine an automorphism, since it doesn't say what happens with the other roots

[stark helm]

OK, then we have identity, alpha->alpha1->alpha2, alpha->alpha2->alpha1, but we have six automorphisms here, and it seems only three of them are listed?

it seems to me like S3?

[rocky cloak]

I listed 6 right here

[stark helm]

six? how to count them?

[rocky cloak]

1 identity.
3 automorphisms that swaps two roots
2 cycles that cycle all 3

1 + 3 + 2 = 6

[stark helm]

does three automorphism that swaps 2 roots refer to like alpha1->alpha2->alpha1?

alpha -> alpha1 -> alpha2, this one seems to me that cycle all 3 roots

[rocky cloak]

No, it refers to the ones that swap two of the roots

For example complex conjugation is one

[rocky cloak]

Yeah, so what you wrote there

Alpha1 goes to alpha2, alpha2 goes to alpha1 and alpha is just mapped to itself

[stark helm]

I think the stuff that I confused is how to express these three automorphisms. since we fix one roots, it seems to me that there is a cycle in S3 like ( alpha, alpha1), (alpha, alpha2), and (alpha1, alpha2). and how can we describe all three roots by using map?

[stark helm]

I mean how to say what happen to alpha 2 here using this map, in order to emphasize alpha 2 is fixed here?

[dull ginkgo]

I read over the sylow shit in jacobson and I don't know how to do even the first exercise

nice!

[terse crystal]

Bourbaki gave the best (constructive) proof for sylow. The proof in Jacobson is merely an induction one for existence.

As for where to find that proof given by bourbaki. Idk, it’s translated in many languages, the one I read is from a textbook in my language. So sadly I don’t know which books in English have that proof. Lang I guess? I haven’t checked.

[hollow mica]

I’m the proof, if we differentiate a constant polynomial (that is nonzero)

it has constant term zero

so “if some coefficient of our polynomial f is not zero, then the constant term of a suitable derivative will be nonzero too” seems wrong

[rocky cloak]

The way to say it is just to say "phi(alpha2) = alpha2" or "alpha2 is fixed" or "alpha2 maps to itself".

You don't need any hocus pocus, just say what you mean.

[delicate orchid]

yeah, you do have a point. But it's kind of easy to see that non-zero constants won't be in the kernel

[rocky cloak]

A function is it's own 0th derivative

[hollow mica]

If we allow 0th derivatives then the claim seems empty

[delicate orchid]

maybe the authors should write down what they actually mean by "a suitable derivative"

why don't people just write shit down

it baffles me

[rocky cloak]

Why?

[delicate orchid]

the claim seems empty for non-zero constant functions because it's REALLY obvious they don't get mapped to zero

[rocky cloak]

Just because something's obvious doesn't mean it's not true

[hollow tartan]

Is this it? https://download.tuxfamily.org/openmathdep/algebra_abstract/Algebra-Bourbaki.pdf Pg 103 on the pdf

[rocky cloak]

Like if a function is 0, then all it's derivatives (including the 0th derivative) vanishes at the origin.

That's what they're using, and that's the thing that's true.

[terse crystal]

Page 78 I think

[hollow tartan]

Yes, I said 103 because it's easier to find

[terse crystal]

I see

[hollow mica]

How’s this for a proof: suppose f is a nonzero polynomial that gets mapped to zero and is of degree n. there are more than n zeros. that’s a problem.

[celest furnace]

Yeah that’s what I was thinking

And this proof generalizes for any infinite field (and you can get rid of the continuous hypothesis)

But this is more algebraic and the proof u showed is more analytic which can be a cool observation

[ashen heron]

so you went back from ice spice to ultimate chad

[celest furnace]

Yeah I had to give my discord to someone IRL and I got embarrassed being the ultimate chad 💀

[terse crystal]

The symbols look too old. The version I read is like this:
|G|=(p^r)n, we don’t require that p doesn’t divide n btw. N is the number of subgroups of order p^r of G. Let X be the set of subsets of G having p^r elements. And we define an action of G on X: g•M =Mg^-1. Then X is union of orbits Tj. Each orbit Tj we choose an element Mj of it. Aj={g: Mj g^-1=Mj}, the fix subgroup of Mj. From Mj=MjAj we know Mj is union of some cosets , say nj many gAj. (nj=|Mj|/|Aj| we know |Aj| is a factor of p^r) nj>1 iff |Aj|=p^rj for some rj<r, iff |Tj|=[G:Aj]=0 mod pn. On the other hand, nj=1 iff |Aj|=p^r, iff |Tj|=n, iff Mj=gAj, in this case the orbit Tj contains a subgroup Mj g^-1=gAj g^-1. And each orbit contains at most one subgroup of order p^r. So one orbit Tj contains a subgroup iff |Aj|=p^r, iff |Tj|=n.
Finally C(np^r,p^r)=|X|=Σ|Tj|=Σ{|Tj|: Tj contains a subgroup}=nΣ{1: Tj contains a subgroup}=nN mod pn. This is true for any group of order (p^r)n, particularly, for cyclic group of order (p^r)n, so C(np^r,p^r)=n mod pn. Together we have nN=n mod pn, which is equivalent to N=1 mod p

[hollow tartan]

I'm sorry but what is Aj? is it the orbit of each Mj?

[terse crystal]

Fix(Mj)

G acts on X, x from X, Fix(x)={g: g•x=x}

[hollow mica]

what does the sentence “But without proposing (2.4), …” mean

[void cosmos]

assuming you do not know it exists, shit gets hard

thats the point

[hollow mica]

wdym

[void cosmos]

it is saying that without this proposition it would be hard to see that the integer determined (gcd?) has this form

[hollow mica]

How does it follow that the gcd has this form though

2.6 is just one direction

it says the gcd satisfies these properties

[void cosmos]

im studying something rn so i can't help you with the details im sorry

[void cosmos]

probably the proof should show youw how

[hollow mica]

ok nw

[void cosmos]

b) c) are literally the definition of the gcd ig

[hollow mica]

OHH I see

only one integer can satisfy (b) and (c), since if d1 and d2 did then d1 | d2 and d2 | d1 so d1 = d2

and then our construction with the least prime power thing satisfies (b) and (c)

so it has to be the gcd

and so it satisfies (a) too

Nice

[stark helm]

I see that the order of this galois group is 4 by writing f(x)=x^2-sqrt(2) over Q(sqrt(2)) and x^2+sqrt(2) over Q(2^(1/4)). I feel very confused about what does say what each of its elements is doing to generator mean here?

[rustic crown]

galois group is the group of automorphisms, so it acts on the generators 2^1/4 and i. tell them what is sigma(2^1/4) and sigma(i) for each sigma in the galois group

[stark helm]

do you think the order is 4 here?

[stark helm]

and do you mean we need a map that considers 2^1/4, sqrt(2) and i? like the map from i to 2^(1/4) to sqrt(2)? since for this one i am not sure if it want me to find the elements of this galois group?

[rustic crown]

yee, size of the galois group is same as degree of the extension

[rustic crown]

nu
an element of the galois group is a field map φ : Q(2^1/4, i) --> Q(2^1/4, i) such that φ fixes Q(sqrt2)

[stark helm]

so we have the automorphism: sigma1: identity, sigma2: 2^(1/4)-> 2^(1/4) i sigma3: 2^1/4->-2^1/4 *i, sigma4: 2^1/4->i, do you think this is complete or any mistakes?

[rustic crown]

to specify an automorphism, you need to tell the image of both the generators, not just one

and don't forget the slogan "roots go to roots"

so in sigma4 you can't map 2^1/4 to i, as the image must also satisfy x^4 = 2

and in sigma2 you need to make sure that this automorphism is actually fixing the base field = Q(sqrt2). sigma2(sqrt2) = sigma2((2^1/4)^2) = (2^1/4 i)^2 = -sqrt2

so out of the ones you're written, only sigma1 is in the Galois group.

sigma2,3 would be in Gal group of Q(2^1/4, i) over Q once you also specify where i maps to

[stark helm]

I ignored this fact. So then sigma2: 2^1/4->-2^1/4. fix 2^1/4i sigma3: 2^1/4i->-2^1/4*i fix 2^1/4, sigma4: the combination of sigma2 and sigma3?

And one more question: does finding all automorphims answers this question( the diagram I post?)

[stark helm]

I think if I just tell them sigma(2^1/4)=-2^1/4 and sigma(i)=-i, then we have already answered it, does it what you mean?

[rustic crown]

yep

since sigma needs to fix Q(sqrt2), it must send 2^1/4 to another root of x^2 - sqrt2. Similarly, it must send i to another root of x^2+1

so there are at most 4 possibilities for such a sigma

and since we know there has to be exactly 4 automorphisms, it must be these ones

so yea sigma(2^1/4) = ± 2^1/4 and sigma(i) = ± i

[stark helm]

does A refer to all A in M_n*n? I am considering if I just let A=I, then let a0=-1, a1=1, and done. but is it true that all A can be linearly dependent in this polynomial?

[fleet citrus]

yes

[south patrol]

A was arbitrary and given

You cn't just choose what it is

[vivid tiger]

...Cayley Hamilton, QED?

[delicate orchid]

I feel like if they had cayley hamilton they wouldn't have been asked this exercise lol

[vivid tiger]

Well, look, you need to show that F[A] is algebraic, so you can show it's a finite dimensional extension. You can do this by showing that the powers are not all linearly independent.

Look at the Jordan decomposition, maybe? The problem is that just looking at powers still won't make the nilpotent part go away

oh wait

F[A] = F[D+N] ⊆ F[D,N] = F[D][N]

N is algebraic over F because N^n = 0 so you get the polynomial x^n.

D is algebraic because each element along the diagonal is

wait no

okay, why is a diagonal matrix algebraic?

[delicate orchid]

well if each element is algebraic can you just product their minimal polynomials together then shove the matrix through that?

[vivid tiger]

I also thought that but now I'm unsure.

[2,0]
[0,1]

product would be (x-2)(x-1)

Oh

Duh

[delicate orchid]

which is the characteristic polynomial lol

[vivid tiger]

Yeah, each indices gets zeroed by each part, but then the product is 0 because dude, they're diagonal matrices.

That was neat. Note that it generalizes to showing that for any finite dimensional matrix A over E with E an algebraic extension of k, then A is algebraic over k.

(that is, matrices over algebraic extensions of a base field are algebraic over the base field)

[delicate orchid]

also the first time I've ever seen the diagonalisable + nilpotent decomposition in the wild

[vivid tiger]

oh come on

i see it only more

and yet i still can't for the life of me remember how to compute the jordan decomposition after 5 minutes of relearning it

[dire siren]

the point of the hint is that M_n(F) is a finite dimensional F-vector space

so we can find some linearly dependent matrices in that sequence, and this gives a polynomial

[chilly radish]

Jordan and chavelley crying rn

[vivid tiger]

i thought of this but then somehow stopped thinking of it for some reason.

My thought was "algebraic => finite => algebraic, AHHH CIRCULAR" and "M_n isn't a field!" and dropped it instead of going "...it's still a fin dim vector space"

[stark helm]

D and N? probably because I don't know about jacobian composition?

[stark helm]

And also, why we know N^n=0? I am doubting if I miss too much knowledge here

[vivid tiger]

nilpotent

the point is that D is diagonal and N eventually dies a lonely death

[delicate orchid]

D is diagonalisable

[vivid tiger]

yes yes just diagonalize it

[stark helm]

I mean why it is nilpotent here?

[vivid tiger]

Jordan decomposition splits in diagonalizable and nilpotent

[delicate orchid]

it's Jordan-Chevalley decomposition, it's a standard linear algebra thing

[stark helm]

so if there is any n*n matrix A, then we all can splits A in a diagonalizable form? So all this kind of A is nilpotent?

[delicate orchid]

then you can write A as the sum of something nilpotent added to something diagonalisable

or "N" and "D" as Xela called them

[stark helm]

ok, so any matrix A can be the sum of nilpotent and diagonalizable?

[delicate orchid]

yes

[vivid tiger]

it's just like the spectral theorem

except now you account for
0 1
0 0

in general you can turn it into Jordan blocks, each with lambda on the diagonal and ones right above the.

representing the sum of a nilpotent part and a diagonalizable part

@delicate orchid wait so how do we get the polynomial that has D as a root when D is only merely diagonalizable?

i thought this would be obvious and it probably is but i'm not seeing it.

[delicate orchid]

f(XYX^-1) = Xf(Y)X^-1 for any two matrices X and Y

and polynomial f

[vivid tiger]

yes, but

Oh, 0

[delicate orchid]

exactly

[stark helm]

OK, so when we get every Ai can be written as Ni+Di, but how will it be related to linearly dependent matrices and do we first need to show tht F is finite vector space?

or we need finite->algebraic-> linearly dependent?

[vivid tiger]

everything's finite dimensional, you are given that

otherwise you should call the functional analysis emergency hotline

[stark helm]

OK, then F[A] can be argued as an finite extension of F right? so A is algebraic over F. but I don't see how A=N+D works here?

[vivid tiger]

so there are two proofs here

the easy one is realizing the matrices are a finite dim vector space

so of course the powers aren't linearly independent

the other way is to look at the powers of N+UDU^{-1}

since F[UDU^{-1} + N] \subseteq F[UDU^{-1}][N]

you can then just prove the latter is algebraic

so, note that F[UDU^{-1}] is fine as we can do the product of the minimal poly's of each element in the diagonal.

(that stuff)[N] is fine because N satisfies an obvious polynomial, it's nilpotent

[stark helm]

so because matrices are finite dim vector space, then if we have a large enough degree of p(x)>dim(Mn*n), then powers of vectors are linear dependent

[stark helm]

by the way, is it true to say that dim(Mnn)=nn?

[vivid tiger]

also known as n^2

[delicate orchid]

well find n^2 K-linearly independent matrices

[stark helm]

I may just find n^2 basis by writting a_ij in each entries.

[celest furnace]

Yes.👍

[stark helm]

in this part, how will u show this one is algebriac instead of saying this field is finite extension?

[vivid tiger]

j did bere

let's take an example

2 0
0 1

let's do (D-2I)(D-1I)

now we get
0 0
0 -1
times
1 0
0 0

which equals the 0 matrix.

[reef trench]

Suppose a comm. ring R.

If I, J are distinct ideals, ie there exist elements i in I, j in J, s.t. i is not in J, and j is not in I, then does that imply that R/I and R/J cannot be isomorphic?

This seems like a natural conclusion to me, but at the same time, it also seems open to counter examples

[vivid tiger]

It doesn't seem true to me.

Imagine k[x,y]

take the ideal x

and the ideal y.

[reef trench]

Dammit, that ring

Ah

[vivid tiger]

the point being that it looks the same

[reef trench]

That's an interesting example

[vivid tiger]

like, do you see how i came up with that? perhaps the only difference between the parts is that there are 2 of them.

[reef trench]

bcs the isomorphism is swapping x's w y's lol

[vivid tiger]

yes, indeed

[reef trench]

Interesting

[next obsidian]

[reef trench]

I suppose this only holds in PIDs, then?

[vivid tiger]

what does?

[reef trench]

.

Oh shit

Maybe not

[vivid tiger]

...probably not?

[reef trench]

Okok

This is the question I had in mind

[vivid tiger]

not sure. poly ring over field in many vars may not be PID

[reef trench]

Mhm, i hear ya

[next obsidian]

You can factor y^2 - x^2 and can’t factor y^2 - x

[chilly radish]

They never are

[vivid tiger]

Ah, okay, that's what I thought.

[reef trench]

Yes but what exactly does that tell us?

[next obsidian]

🗿

[vivid tiger]

irreds

[chilly radish]

But I don't think your statement is true even in a PID

[vivid tiger]

prime ideals

[chilly radish]

Just take any 2 maximal ideals in C[x]

Both quotients will give you C

[reef trench]

Yeah, this is a thought I had, but maybe I am not fully understanding reducibility

[reef trench]

OMG

That example was literally 1 question ago

how could I forget

[chilly radish]

Lol

[vivid tiger]

So...

C[x]/m must be a field. How do we know it'll be C?

[chilly radish]

It's an algebraic extension of C

C is alg closed

So it has to be C

[next obsidian]

-1: you did not show that C[x] has more than a single distinct maximal ideal

[chilly radish]

I don't go to uoft

[reef trench]

Could you please elaborate

[chilly radish]

I don't have to show that

[vivid tiger]

Well, PID, so m is gen by a single element, so we have the minimal poly whose root this is an extension by.

[next obsidian]

No because you should figure it out

For your own remotest

[chilly radish]

But for posterity, maximal ideals are irreducible polynomials

[next obsidian]

Enrichment

[chilly radish]

Exercise: find 2 distinct monic irreducible polynomials in C[x]

[vivid tiger]

don't need monic

[chilly radish]

Yea, so it's a finite extension of C so it must be C

[vivid tiger]

irred in C is monic.

(over C)

[chilly radish]

Otherwise you need to specify nonassociate

[reef trench]

my god, it was that easy

[chilly radish]

Irreducibles are not defined uniquely unless you fix their leading coefficient

[reef trench]

One is prime, the other is not

Lol, ty yall

[vivid tiger]

Oh, right, I forgot what monic meant.

[reef trench]

Oh wait

Am I even right?

[vivid tiger]

I confused monic with degree 1.

[vivid tiger]

one ideal is. now make a statement about the quotients.

[cobalt heath]

Wdym prime

[chilly radish]

That's another way of using algebraic closeness to argue that

[vivid tiger]

prime ideal

[reef trench]

One is ID, the other is not

[cobalt heath]

Ah, thought we were talking about monic polynomials

[vivid tiger]

yes, yes, that's why I said "you don't need monic"

[vivid tiger]

not yes necessarily

[reef trench]

What?

[vivid tiger]

you can still have zero divisors if the prime ideal wasn't maximal

wait

nvm

[reef trench]

I think prime implies quotient is ID

[vivid tiger]

yup reproved that while trying to prove it false, dw, thanks.

[cobalt heath]

Algebra is confusing

[reef trench]

fr

I actlly prefer analysis

ts lowkey like taxonomy

[cobalt heath]

Woah you are fast

[reef trench]

no offense

[vivid tiger]

i was typing "well, imagine if there was xy=0 but xy not in the ideal" and then was like "oh"

[cobalt heath]

I cannot verify a property like this so fast

[crystal vale]

I am not sure about the cycle decomposition of S_n , what does that mean ?

Disjoint cycles ?

[celest furnace]

Most likely writing any permutation as a product of disjoint cycles

[crystal vale]

Okay

[dusk whale]

Just had a quick question on this

Im confused how the last part shows a contradiction if | H | = 6

[untold garnet]

The last part shows that H contains at least eight different elements, so |H| can’t be equal to 6?

[dusk whale]

Oh

[dusk whale]

how do we know there are 8 elements of order 3?

[untold garnet]

You have two rotations of order 3 per vertex of a tetrahedron.

[dusk whale]

Oh ok

Thanks

[crystal vale]

In Z_3 × Z_3, {( 0,0), (1,2), (2,1) } is a subgroup but not subring, right?

[toxic zephyr]

dumb question: so continuous functions on R form a ring right? and if f/g is not continuous (for f,g continuous), then can we infer that this happens if and only if g is zero somewhere?
and if so, would that imply that the strictly nonnegative and nonpositive functions make up the units of the ring?

[toxic zephyr]

under addition yeah it's a subgroup. you basically have {0,x,-x} where 2x=-x. so that's closed under addition and inverses.
but the product x^2 is not in the set so it's not a subring

[crystal vale]

Yes

[crystal vale]

And what's your operation?

[toxic zephyr]

regular addition and multiplication

[celest furnace]

How do you treat removable singularities? Like would (x+1)^2/(x+1) be considered continuous?

If not then yeah I think ur unit argument is correct

[prime sundial]

well this just wouldn't be included in the ring since the domain is different right

[celest furnace]

So what’s in the ring? How can you talk about quotients?

Feels like you have to assume the functions are never 0 to make this work

[prime sundial]

well not all functions in the ring have inverses

it's not a division ring

[celest furnace]

Sure

So the claim is f is invertible iff f is strictly positive or negative?

[prime sundial]

the units are also not the nonnegative/nonpositive, they must be strictly positive or negative

0 is exactly what you want to exclude

[prime sundial]

yes

for continuous that is

and so we are on the same page, by invertible we mean has a multiplicative inverse

[tacit hemlock]

guys

how many homomorphisms are there from $C_n\to D_n$?

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

for cyclic C_n, Dihedral D_n

since all homomorphisms are determined by where the generators are sent, I can consider all possible values for f(1) right?

since 1 is a generator of Z_n

but 1 is not necessarily the only generator of Z_n !!!

[crystal vale]

C_n is Z_n ?

[tacit hemlock]

isomorphic

so I can consider it instead right

all finite cyclics are isomorphic

[crystal vale]

Yes

[tacit hemlock]

ok

so 1 is a generator

additively

but so is any k such that gcd(k,n)=1 b/c by bezout's identity, we can construct 1

so suppose we have $b$ possible values for $f(1)$

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

doesn't that mean the number of homomorphisms is $\varphi(n)\cdot b$ instead of $b$?

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

where $\varphi(n)$ is euler's totient function, so the number of terms coprime to $n$

[cloud walrus]

normalAtmosphericPa=101,325

[rocky cloak]

It depends whether n is odd or even

[tacit hemlock]

why does it matter

[rocky cloak]

Because if n is even, then you can map the generator of Cn to elements of order 2 in Dn (which there are many of)

[tacit hemlock]

but c_n doesn't necessarily have a unique generator right

so do we have to consider where all the generators are sent?

[rocky cloak]

That doesn't matter. You just pick a generator then proceed

[tacit hemlock]

huh?

the image of ecah generator determines a homomorphism right

so if I map 1 --> b possible choices of images in D_n

that is b homomorphisms?

[rocky cloak]

Alright, so let's say g generates Cn. Then a homomorphism f:Cn -> Dn is uniquely determined by where g is mapped

[tacit hemlock]

but what if t also generates Cn

[rocky cloak]

That doesn't affect it

[tacit hemlock]

oh, because any other generator can be simply rewritten in terms of that generator

i.e. t = g^k for some k

[rocky cloak]

Yeah, that's right