groups-rings-fields

[next obsidian]

So just as a formula

f ≠ 0 such that there exists a g ≠ 0 with fg = 0

[upbeat swift]

how does this work?

I don't really understand

[next obsidian]

What do you mean?

[lavish nexus]

you must have two functions that are not everywhere zero
f and g
s.t. f(x)g(x) = 0 for every x in R

[next obsidian]

I’ll give an example of a zero divisor

[upbeat swift]

if a product of two elements that are both nonzero, how could they equal 0?

[next obsidian]

Let f(x) = 1 if x > 0, and 0 if x <= 0

Let g(x) = 0 if x > 0, and 1 if x <= 0

Then we see f(x)g(x) = 1•0 = 0 when x > 0

And f(x)g(x) = 0•1 = 0 when x <= 0

So fg = 0

[upbeat swift]

is this different from what you just said?

[next obsidian]

No

[upbeat swift]

or is it the same?

[next obsidian]

It’s the same

You have to recall that f and g are functions there

not real numbers

[upbeat swift]

true

[next obsidian]

We see the f I defined is not 0

Because it takes on 1 when x > 0

Also g is not 0 because it takes on 1 when x <= 0

But their product is 0 because it’s 0 at every point

[upbeat swift]

ohh

[next obsidian]

So let’s return to the case where f is non-zero at every point

If it were a zero divisor, there’s a nonzero function g such that fg = 0

But think about what happens if you plug in x?

[upbeat swift]

would it not be possible?

[next obsidian]

Well just write it out

f(x)g(x) = 0

But what did we assume about f?

[upbeat swift]

f is a zero divisor

[next obsidian]

The other assumption

[upbeat swift]

f is nonzero?

[next obsidian]

Not just that it’s nonzero

That it has no zeroes

Or is non-zero at all points

But this means something when we look at f(x)g(x) = 0

We know that f(x) ≠ 0

So we can conclude something!

[upbeat swift]

this is true if g(x) is 0 are every point

[next obsidian]

Right

In fact we are forced to conclude that

[upbeat swift]

at*

[next obsidian]

So we can conclude that g is 0

Do you see why we can conclude that g is 0?

It isn’t just that f(x)g(x) = 0 is true if g(x) = 0

f(x)g(x) = 0 implies that g(x) = 0 because we know f(x) ≠ 0

[upbeat swift]

so this contradicts the definition of divisor of 0?

[next obsidian]

Right

Because we need that g was nonzero

The definition of zero divisor was that f ≠ 0 such that there exists g ≠ 0 with fg = 0

We just showed that if fg = 0, then g = 0

[upbeat swift]

so there's no divisors of zero in F(R)?

[next obsidian]

Well not quite!

We assumed that f had no zeroes

That was crucial in the argument

So we can focus our attention to f in F(R) which have at least one zero

Does that make sense?

[upbeat swift]

yes

[next obsidian]

Okay so let’s suppose that f(x) = 0

For some specific x in R

Can you think of a non-zero function g such that fg = 0?

Using just that f(x) = 0

[upbeat swift]

g(x) = 1 could be an option?

[next obsidian]

Well that won’t necessarily work

Let me replace x with x_0

So all we know is that f(x_0) = 0 for a single x_0 in R

If you took g(x) = 1 for all x, then
f(x)g(x) = f(x)

This is certainly 0 at x_0

But it might not be 0 at other points in R right?

But you can fix that using g!

[upbeat swift]

so you'd have to change g?

[next obsidian]

Right

Here, I drew a specific function

The red line represents the function f

It’s 0 at 0, and 1 at every other point

Can you find a g that works for this specific f?

[upbeat swift]

g would be 1 at 0 and 0 at every other point

[next obsidian]

Right

So adapt this for more general f

Remember, we know that f(x_0) = 0

And that’s it

[upbeat swift]

for some x_1, then f(x_1) = 0

and 1 for the other x's?

[next obsidian]

Well…

I’m not sure what you mean

We’re assuming that f has a zero

So I said let x_0 be such that f(x_0) = 0

We want to produce a non-zero g such that fg = 0

[next obsidian]

And I’m saying you can adapt this g a little bit to make it work

[upbeat swift]

what do you mean by adapt?

[next obsidian]

Well like

Use the same idea you used there

It’s just that this g exactly as-is won’t work

Because f(x_0) = 0, not f(0)

But you can just use the exact same idea you used to make g when f(0) = 0

And make one that works for when f(x_0) = 0

[upbeat swift]

I'm still not sure by what you mean

[lavish nexus]

say you have f that is zero on some set U and nonzero anywhere else
find a nonzero g such that fg = 0

[next obsidian]

I just mean to let g(x_0) = 1 and g(x) = 0 for all other x

You can compute that fg = 0

Under the assumption that f(x_0) = 0

And all I did was adapt your g which was g(0) = 1 and g(x) = 0 for all other x

I just changed where the one non-zero point was

[upbeat swift]

oh, so that is what you mean by adapt?

[next obsidian]

Yeah

I mean it isn’t a rigorous term, I was just saying you can mess with the function very slightly and get what you need

[upbeat swift]

oh okay

thank you

[next obsidian]

[lavish nexus]

the notation

deliberately choosing the worst notations

[trim grove]

Given that $I$ and $J$ are ideals of $R$ where $R$ is a CRU such that $I+J=R$, what can we say about $I^5+J^7$? any hint?

[cloud walrus]

Solution

[lethal dune]

what's the connection between lie algebra and group? where does the lie bracket come from?

[upper pivot]

how much do you know about manifolds and vector fields?

[gritty sparrow]

Hint: if i+j=1, (i+j)^1000000=1. Use the binomial theorem to expand

[trim grove]

in question they are asking about is it maximal or not, prime or not?.

[gritty sparrow]

My hint shows exactly what it is, you can check maximality and primality from that

[lethal dune]

I'm currently reading manifolds

ISM specifically

[upper pivot]

right, so have you been introduced to tangent spaces, tangent bundles or vector fields? I will taylor the description depending on which part ur at

[lethal dune]

yes

[upper pivot]

alright nice

so i guess do you know a lie group is?

[lethal dune]

i just know the definition, I wouldn't say the motivation is clear

[upper pivot]

right so a lie group is a manifold with this additional structure of being a group

it turns out that this additional structure like, instantly allows you to do a lot on the manifold

[lethal dune]

yea, group object in Top category right?

[upper pivot]

in the category of smooth manifolds

[lethal dune]

oops yes

that was topological group

[upper pivot]

so one of the reasons is this lie algebra that we will talk about, you can translate problems on lie groups to lie algebras which are much easier to work with

but even before that there are some nice properties you get from just being a lie group

so for example

every closed subgroup of a lie group is an embedded sub-lie group

which is a really strong statement

[lethal dune]

[trim grove]

but how this will help me, can you please elaborate a little bit, i am confused ?

[upper pivot]

on general manifolds you dont have this relation between being closed and being an embedded submanifold right

one of us should thread btw

lol

[lethal dune]

idk about this one

[gritty sparrow]

This will show that 1 is in the ideal, which means that the ideal is the whole ring

[upper pivot]

oh well i dont wanna spend too much time talking about this so take what i said for granted

well think about it like this, you know how painful it is to show something is a smooth manifold

[lethal dune]

I have read lie algebras independently of manifolds, i.e. the def and examples nothing else

[upper pivot]

here if you can somehow realize something as a closed subgroup of something we know to be a lie group you instantly have that the thing is a smooth manifold

and this comes in handy

right so far im only talking about lie groups and manifolds in general right

[lethal dune]

ok that's useful

[upper pivot]

but yeah the idea is this additional group structure just gives you a lot more tools to work with

now here is one thing about the group structure

if you have a lie group G, then L_g: G->G, left translation by g (so h-> gh) are self-diffeomorphisms right

[lethal dune]

yes

[upper pivot]

yeah so this is going to turn out to be really useful

lets take a... tangent for a second

so for a general manifold M

you have the space of all vector fields X(M) on it right

[lethal dune]

we should really make this a thread fr

[upper pivot]

oh i was just saying that cause there was another discussion

right so we can also view X(M) as maps D: C^infty (M) -> C^infty(M) such that D is linear and satisfies the product rule right

you familiar with this? if not i can describe it a bit more

[trim grove]

okk 1 is in I+J, so that means there exists i and j such that i+j=1

but how this will help to check I^5+J^7?

[lethal dune]

yes I am familiar

[upper pivot]

alright nice

Now notice that under this identification

X(M) is not a ring right

the composition D circ D' of two derivations need not be a derivation unfortunately

[lethal dune]

ye

[upper pivot]

but its still atleast a vector space right

[lethal dune]

but they are module over C^inf (M) right

[upper pivot]

yeah

stronger than just a R-vector space but we will not be using that rn

[lethal dune]

okay

[upper pivot]

so DD' is not a derivation

but it turns out

DD'-D'D is a derivation

[gritty sparrow]

I+J=1

[lethal dune]

oh so - comes from C^inf(M)

[upper pivot]

like the rough reasoning is both of these are second order derivatives, and the 2nd order term cancels out when we do this and we now once again have a first order derivative

its easy enough to verify

just plug in f times g into it and see it satisfies product rule

[lethal dune]

ok I'll verify them

[upper pivot]

ill let u do that

[lethal dune]

so that turns X(M) into a lie algebra

nice

[upper pivot]

mhm

but heres like a problem right, X(M) in general is infinite dimensional

and understanding is structure is like

really hard

so unfortunately this doesnt end up giving us a lot of information

[lethal dune]

yea

[upper pivot]

but this is where the lie groups come in

so now instead of X(M), we will look at a subalgebra of it

we will look at the so called left-invariant vector fields

what are these? the condition is simple, so we have this differential d(L_g)_h : T_h(G) -> T_(gh) (G) right

where L_g was the left translation by g

we are gonna ask a left invariant vector field to have d(L_g)_h X_h = X_(gh)

edited for clearer notation

[lethal dune]

ye, it was confusing before

now it's clear

[upper pivot]

right so the idea is that if you left translate the vector field at point h you get it at point gh

[lethal dune]

what's X_h here? an element of tangent space at h?

[upper pivot]

so X is a vector field in X(M) (ok i used X twice here but to be fair in my head the X(M) X is math frak oops)

[lethal dune]

[upper pivot]

and X_h is like, the member of T_h(G) corrosponding to X

like X at point h basically

[lethal dune]

okay

a tangent vector ar h basically

[upper pivot]

sure

now you can verify if X,Y are two left invariant vector fields

so is [X,Y]

=XY-YX (just apply dL_g to it)

[lethal dune]

okay

[upper pivot]

so the upshot is we have found a sub lie algebra of this really complicated space X(M)

the left invariant vector fields right

now this turns out to be much much simpler

why? well notice that left invariant vector fields are determined by X_e (the vector field at identity)

because once you have that you can translate by h to get it at all other points right

(some verifications need to be made but hopefully the intuition is clear)

[lethal dune]

yeah

it's clear

[upper pivot]

right so we can identify this left invariant vector fields, lets call it Lie(G)

with T_e(G) right

[lethal dune]

yep

[upper pivot]

and guess what, T_e(G) is finite dimensional space

and its structure is generally not too hard to figure out

so Lie(G) is a finite dimensional space (of dimension n where n =dim(G))

and along with the lie brackets we defined, its what we call the groups lie algebra.

[lethal dune]

I see

[upper pivot]

and this is like, the main reason why lie groups are so overpowered

cause you often reduce problems on manifolds to problems on this finite lie algebra

[lethal dune]

now they indeed feel like overpowered

[upper pivot]

like some of the big theorems of lie theory like tell you how you can recover info about lie group from their lie algebras

so ok like we know how to go from the groups to the algebras

whenever you have a smooth homomorphism f:G->H

the induced map df: Lie(G)-> Lie(H) turns out to be a lie algebra homomorphism

(preserves the lie bracket)

[lethal dune]

[upper pivot]

now you can ask the questions, are all finite lie algebras from lie groups? are all maps Lie(G)-> Lie(H) coming from smooth homomorphisms

etc

[lethal dune]

alright thank you very much. I'll read this part again

[upper pivot]

mhm

[chilly ocean]

in mood?

what is lie group associated to C^infty(M)

also what is dual of lie algebra

actually i should probably look up if you don’t feel like telling me googleable things

[iron vessel]

Let $f \in \mathbb{F}_p[T]$ be an irreducible polynomial and $K = \mathbb{F}_p(T) $ the field of fractions of $\mathbb{F}_p[T]$. Prove that $X^p -f $ is irreducible in $K[X]$.

Can anyone give me a hint on how to solve this?

[cloud walrus]

Évariste Galois

[iron vessel]

What i was thinking was that if $\alpha \in K$ is a root, then $\alpha^p - f(\alpha) = 0 \implies \alpha - f(\alpha) = 0 \implies f(\alpha) = \alpha$. However i am not entirely sure how to reach a contradiction

[cloud walrus]

Évariste Galois

[coral shale]

i dont think you can just guess alpha is a root

it may reduce into non linears

[iron vessel]

I meant more " For the sake of contradiction let \alpha be a root"

ah true

[hot lake]

I don't think alpha is a root => alpha = f(alpha)

[iron vessel]

hmm ok, what do you suggest i should do

[coral shale]

Contradiction likely is the way though

assume it is reducible

[hot lake]

eisenstein criterion ?

[iron vessel]

how do i apply eisensteins to this polynomial?

[coral shale]

eisenstein isn't bidirectional?

[iron vessel]

I dont think so right?

[coral shale]

I was going to suggest to assume it is reducible. Try coming up with a factor

for X^p - f

Since you are in Fp, I feel like there are good guesses to try

[iron vessel]

Ok, can i assume that the degree of X^p - f is p?

[coral shale]

why

[hot lake]

1 is not 0 in K so yeah the degree of X^p - f is p

[coral shale]

oh

[iron vessel]

so if I have two polynomials, g and h, that are factors then deg(g)+deg(h) = p

[coral shale]

i think uh

[hot lake]

ah wait you meant the degree of a root ?

[coral shale]

f is a constant isnt it?

[hot lake]

but that's not really defined yet

[iron vessel]

$f \in \mathbb{F}_p [T]$

[cloud walrus]

Évariste Galois

[hot lake]

f is in K

so yeah

[coral shale]

f is a polynomial in F_p[T]
But it is a constant in F_p(T)[X]

don't mix that up

[iron vessel]

right

[coral shale]

so yes, degree p

X^p - f

X^p - gh
I would try guess a factor, maybe something works

[hot lake]

f is irreducible

[iron vessel]

Perhaps I can use that in another exercise i found that if $K$ is a field of $char p>0 $and $\alpha$ not a pth power then $X^{p^k} - \alpha$ is irreducible in K[x]?

[cloud walrus]

Évariste Galois

[coral shale]

contradiction

wait sorry

[hot lake]

then yeah that would be a corollary

[coral shale]

i wrote the contraction statement wrong

[hot lake]

or that other exercise

[iron vessel]

so i can apply that exercise to my current problem?

[hot lake]

K is a field of characteristic p and f is irreducible so it's not a pth power

[iron vessel]

Ok so it essentially just follows ?

from the other exercise i mean

[hot lake]

well yeah

[coral shale]

So if we do contradiction
Assume X^p - f(T) = P(X)Q(X)
Show f reducible

[iron vessel]

ok thanks guys 🙂

Damn galois theory is hard tho

Ok so what i did was:

Since $f$ is constant in $K[x]$, deg($x^p - f(T)$) = p = deg(A) + deg(B). Let A(x) = $X^m + \cdots + a_0$ and B(x) = $X^n + \cdots + b_0$. Where n+ m = p. Then, A(x)B(x) = $X^{n+m} + \cdots + a_0b_0 = X^p - f(T)$. Hence, $f(T) = a_0b_0 \in \mathbb{F}_p[T]$, which is a contradiction as $f$ was irreducible.

[cloud walrus]

Évariste Galois

[hot lake]

f = 1 * f

also you are right now trying to reprove eisenstein's criterion

[iron vessel]

I have a few issues with this proof tho. First, can i assume $A(x) and B(x) have that form? Also as @hot lake just pointed out a_0 or b_0 can be units

[hot lake]

yeah if P is reducible it's A * B where the sum of the degrees is p

and the degrees are not 0

[iron vessel]

I dont understand what you mean, sorry

do you mean f or P?

[hot lake]

if x^p - f is reducible then it's A * B etc

that's correct

[iron vessel]

Ok, so that part is good

What if I suppose reducible, then (im not sure if this holds) by eisenstein, f(T)^2 | f(T) ?

that seems like a very flawed argument tbh

[hot lake]

yeah eisenstein works pretty well

[iron vessel]

So this is right?

[hot lake]

yes ?

[iron vessel]

Ah i wasnt aware that Eisensteins was bidirectional

[hot lake]

??

wha's eisenstein's criterion for you

[iron vessel]

Well i use it to show something is irreducible, but does it hold that for all reducible polynomials if the constant terms is divisible by a prime element then the polynomial is reducible?

[hot lake]

for all reducible polynomials, the polynomial is reducible

I am not sure what you are trying to say

[iron vessel]

Like, I know that if the constant term is divisible by P and is not divisible by P^2 then it is irreducible. Does the converse hold tho? i.e. if the polynomial is reducible then the constant term is divisible by both P and P^2 necessarily?

[hot lake]

that's not what eisenstein criterion says

[iron vessel]

there is more stuff to it

[hot lake]

yeah

[iron vessel]

yeah but my question is if in our case X^p - f(T) is reducible, then f(T)^2 necessarily divides f(T)?

[hot lake]

yes

A => B is logically equivalent to not B => not A

[iron vessel]

Ok, that was my question. Thank you!

[hot lake]

it's called the contrapositive, not the converse

[iron vessel]

alright, and what would the converse be? like A => not B?

[hot lake]

converse would be B => A

which is not always true

despite what my fingers tried to have me write for a second

[iron vessel]

lmao

but I understand now

[hot lake]

nice

[iron vessel]

Ill write out my argument and if you dont mind I'll ask for some feed back

[hot lake]

go ahead

[iron vessel]

Suppose $X^p - f(T)$ is reducible in $K[x]$. Then, by Eisensteins, it holds that $ \exists P$ a prime element such that $P|f(T)$ and $P^2|f(T)$. Since $f(T)$ is a prime element in $\mathbb{F}_p[T]$ its only divisors are 1 and $f(T)$. Since 1 is not a prime element, then $f(T)^2|f(T)$. A contradiction since f is not a unit in $\mathbb{F}_p[T]$.

[cloud walrus]

Évariste Galois

[iron vessel]

Am i missing anything? Or am i assuming something not true?

[hot lake]

ah no I wasn't thinking of exactly that kind of implication

[iron vessel]

damnit

Whats wrong with my argument?

[hot lake]

once you have stated that an is not in the prime ideal (f), that a1 ... a(n-1) = 0 are in (f), the criterion says that
if a0 is not in (f²) (and it is the case here) then P is irreducible, and the contrapositive of that last implication is that if P is reducible then a0 is in (f)² so f² divides f, but that's an unnecessarily roundabout way to go

[iron vessel]

Hmm, so what would you say I did that was unnecessary?

[hot lake]

you said "there exists P a prime element suc hthat" when all along we apply the criterion with f

so right from the start it's super strange

it's like you are saying

"if P is irreducible then that is provable with eisenstein's criterion from some mysterious ideal P"

which is not always the case

[iron vessel]

So instead of P i should just straight up use f(T)

?

[hot lake]

yes

[iron vessel]

Ok, i guess i could have done that but for some reason that seems overly simple and like a 2 line proof then

[hot lake]

or you say "there exists P" before "by eisenstein's criterion"

[iron vessel]

Im used to these exercises being half a page

[hot lake]

because P exists regardless of eisenstein's criterion

[iron vessel]

i see what you mean

[hot lake]

once we have it and have observed that the criterion will apply, only then you say "by eisenstein's criterion with P = (f) , ...."

[iron vessel]

Ok, i understand it much better now, for some reason I never thought i could have applied eisensteins that way.

thanks for your time by the way i really appreciate it

[hot lake]

idk if it's stated for anything else but Z[x] in your course though

[iron vessel]

what?

[hot lake]

the criterion

[iron vessel]

ah right, yeah we have done the general one too

[hot lake]

ah then it's fine

[iron vessel]

from D&F

[coral shale]

My Galois lecturer uses L : K to indicate a field extension L over K. But I see that the usual notation is L/K.
I wondered before if this could correspond to a quotient in any way and have been told it's just notation convention.
How about this though:

When we consider Gal(L/K) (or Gal(L:K) in the notation I see), the automorphisms of L which fix K, aren't we really taking Aut(L)/Aut(K)? ie. Gal(L/K) := Aut(L)/Aut(K)

[coral shale]

Hmm or maybe Aut(L)/Aut(K) doesn't even necessarily make sense now I think about it... I feel like there's surely some way to make it happen, though.

What I really want to quotient over is the cosets

{{f in Aut(L) : f extends g} : g in Aut(K)}

[woeful flint]

I know that there is a relation between irreducibility in $\mathbb{Q}[X]$ and irreducibility in $\mathbb{F}_p[X]$ where $p$ is a prime but I always forget what this relation is. Is it that if a polynomial is irreducible in $\mathbb{Q}[X]$ then it is irreducible modulo every prime. Is it the other way?

[cloud walrus]

Kraft Macaroni

[iron vessel]

Yeah check out Gauss' Lemma

Intuitively, if you have that you have a polynomial, f, reducible in $\mathbb{Z}$ then there is an element $\alpha \in \mathbb{Z}$ such that it is a root. Consequently, $\alpha \in \mathbb{Q}$ since it is the fraction field of $\mathbb{Z}$. So f is reducible in $\mathbb{Q}$ as well.

[cloud walrus]

Évariste Galois

[woeful flint]

Thanks @iron vessel

[iron vessel]

No worries 🙂

[woeful flint]

Remember to never duel for a woman, what good is that if you're dead

[iron vessel]

Lol

[woeful flint]

Take my advice

[iron vessel]

I will, dont worry

[woeful flint]

he didnt listen

[iron vessel]

Call an ambulance... but for me

[terse crystal]

Does it equal to Aut(L) by zorn’s lemma

[iron vessel]

To show that a field extension of the form K[x]/(f), with deg(f) = p and f irreducible, is an inseparable extension of degree p, does it suffice to show that f is inseparable with a root of multiplicity p?

[stoic rose]

Yes I think that's equivalent

[iron vessel]

Ok thanks! One more question, let $L_f = \dfrac{K[x]}{(f)}$. I must show that $L_f^p = K$.

[cloud walrus]

Évariste Galois

[iron vessel]

How should i approach this?

I think I should just use the frobenius automorphism applied to any element of L_f as K-basis

[stoic rose]

Hum, every element of L is of degree p, and I think that the only monic nonseparable polynomials of degree p are X^p-a

[iron vessel]

why is every element of L of degree p?

just because f is of deg p?

[stoic rose]

Because the degree of a subextension divide the degree of the full extension

And p is prime

[iron vessel]

ah right

so how would i get from that to L_f^p = K?

[stoic rose]

That gives you L^p contained in K

[iron vessel]

because what i thought initially was to show that [$L_f^p : K$] = 1

[cloud walrus]

Évariste Galois

[iron vessel]

I am not able to see what you mean

[stoic rose]

Every element of L\K is a root of some X^p-a, so L^p is contained in K

[iron vessel]

Ok i think i understand now

does the other inclusion follow from the fact that K is contained in L?

[stoic rose]

I think the other inclusion is false in general. Do you have any assumption on K?

[iron vessel]

yes, K = $\mathbb{F}_p (T)$ with f $\in \mathbb{F}_p[T]$

[cloud walrus]

Évariste Galois

[stoic rose]

Oh ok, yeah then then it's probably true

[iron vessel]

And does it follow from K being in L?

[stoic rose]

If you mean that K^p should be K then no that's false

You need the elements of L

[iron vessel]

No i mean, K being contained in L_f^p

[stoic rose]

How would it follow just from K being contained L then?

[iron vessel]

I was thinking that L is in L^p and since K in L then K in L^p

perhaps im mixing stuff up

[stoic rose]

L is definitely not in L^p

L^p is a strict subset of L

[iron vessel]

Sorry but could you explain why?

[stoic rose]

Well for example it's contained in K and K is strictly smaller than L

[iron vessel]

Im an idiot lol

that makes sense

what if i show that [L_f^p : K] = 1?

[stoic rose]

Hm, didn't thought of that but maybe it could work

Hum yeah actually, I'm pretty sure it works!

[iron vessel]

I did that before thinking that L_f was contained in L_f^p and thats why it didnt work

now it works by switching the two

damn thanks man you really helped 🙂

i have midterms soon and im grinding galois theory

[stoic rose]

You see how to prove that [L^p : K] = 1?

[iron vessel]

I did $[L_f : K] = [L_f : L_f^p] \times [L_f^p : K] \implies [L_f^p : K] = \dfrac{[L_f : K]}{[L_f : L_f^p]} =\dfrac{p}{p} = 1$

[cloud walrus]

Évariste Galois

[iron vessel]

Is this correct?

[stoic rose]

Hum no that doesn't work

[iron vessel]

how come?

[stoic rose]

A priori you don't know that K is in L^p that's what we're trying to prove, so the first equality makes no sense

[iron vessel]

hmm i see, but combined with your earlier argument that L^p is in K it works?

[stoic rose]

Hm no, then the degree [K : L^p] makes sense but not the other way around

[iron vessel]

Ugh im bad

So what is the way to do it?

[stoic rose]

You have to use properties of your particular K at some point because it's false for general K

[iron vessel]

I see

[stoic rose]

One way I see of doing it is by noting that L^p is a subextension of K/K^p which is strictly bigger than K^p. Then if you can prove that for you particular K you have [K : K^p] = p you're done

[iron vessel]

well I know that f is in F_p[T] and that L_f = K[x]/(X^p - f) with K=F_p(T).

This is the actual question.

[stoic rose]

Do you understand my sketch of a proof?

[lavish nexus]

what does the squiggly arrow on the second line mean?

[terse crystal]

It means The image of a is a_i

[coral shale]

Field Extension notation

[wraith obsidian]

This might be a stupid question, but Is there a group cohomological way of seeing whether H≤G is normal?

I know we can classify extensions N→G→G/N via some cohomology group (I forget which one) using equivalence classes of sections G/N→G, but in this context we do stuff after knowing that N is normal

Hm, perhaps this is really a question about the Normalizer N_G(H)

[chilly ocean]

thats a cool question

if you come to any progress lmk ill think about it and come back at end of day

[chilly radish]

I feel like i'm missing something very basic here. Ok, here's the setup: \\
We have a field, $F$, and we consider the free algebra of countable rank $F\langle X\rangle$ with $X = {x_1,x_2,\ldots,}$ (For the record, this is noncommutative and without unity). Consider now the free $F\langle X \rangle$ bimodule $R$ of countable rank with generators $r_1,r_2,\ldots$. Define the unique derivation $\delta:F\langle X \rangle \rightarrow R$ satisfying $\delta(x_i)=r_i$ for $i\geq 1$ (So this map is $F$-Linear and $\delta(fg)=\delta(f)g+f\delta(g)$). In particular, $\delta$ acts on monomials by
$$\delta(x_1x_2\cdots x_n) = \sum_{j=1}^n x_1x_2\cdots x_{j-1}r_j x_{j+1}\cdots x_n$$
If $U$ is a $2$-sided ideal of $F\langle X \rangle$, I'm trying to convince myself that that $\delta(f)\in UR$. In particular, i'm not seeing why this would even be true for some monomial in $U$, given the above formula for $\delta$.

[cloud walrus]

ShiN

[broken stirrup]

Hi everyone

[next obsidian]

Hello backpack

[broken stirrup]

We know that every ring is an R-module over itself. I want to find a simple example of finitely generated R-module such that it's not finitely generated as an abelian group

Would it work if I take Q over itself?

Q is not finitely generated as an abelian group

[next obsidian]

Q is not finitely generated over R as well

Oh wait

Yes

Q over itself works

[broken stirrup]

nice

it always feels nice to get validation :p

thanks

[chilly radish]

bump

[next obsidian]

Noncommutative free algebra, L

[fallow plume]

I'm a bit confused, (No proofs of the actual problems please 🧡)
but we're asked to prove:

and this seems reasonable, but it says that p_i is in Ass(M)

if M was a free module, shouldn't Ass(M) inherently be just (0)?

afterall if say M = Z^5, then clearly the only annhilator is 0 (which is indeed prime here since Z is an integer domain, so it is in Ass(M))

but then that doesn't make sense because there doesn't seem to be a way to make that question 5 true?

I'm assuming this should be isomorphic as modules as well

[next obsidian]

Well

Ass(M) isn’t actually always {0}

It includes in particular all minimal primes

[fallow plume]

If M is a free module*?

[next obsidian]

Oh sorry

M = R

[fallow plume]

gotcha

[next obsidian]

But let’s assume R is an integral domain

Then yes, Ass(R) = {0}

But then if M = R^n

You could do something like

[fallow plume]

oh right I forgot to include that part

[next obsidian]

R^n > R^n-1 > … > R^1

And then all the quotients are just R = R/0

So the theorem is still true

[fallow plume]

oh lol

i feel dumb now

[next obsidian]

No worries lol

[fallow plume]

Idk why but sometimes I think R/0 = 0

[next obsidian]

But does that clear up the confusion?

[fallow plume]

[next obsidian]

Hahahaha

[fallow plume]

yea

[next obsidian]

Awesome

[fallow plume]

thanks 🧡

[chilly radish]

nvm, this claim is most likely untrue in general and I was misinterpreting the author's vague writing

[toxic zephyr]

suppose i have a ring R .
can i say without proof that in any subset S of R, the additive identity of R (given it is in S) will act as an additive identity to all elements of S under the addition defined for R?
or do i have to explicitly prove that it does act as an additive identity to any arbitrary element of S?

im doing a proof showing there are additive inverses in the set S of 2x2 scalar matrices which is a subset of 2x2 real matrices. and im wondering if i can just say that the zero matrix is an additive identity in S.

[woven delta]

You can say that

[toxic zephyr]

great thank you 🙂

[tall jay]

Why is it true that if $a^{-1}b \in N$, then $(a^{-1}b)N =N$?

N is a subgroup of group G, and $a,b \in G$

[cloud walrus]

beeswax

[next obsidian]

This has nothing to do with a^-1b

You can replace that entire thing with g

In which case you just compute that the elements of gN agree with that of N

[tall jay]

So for any element g in G N, if I have gN, this will always give me the elements in N only?

[next obsidian]

No g has to be in N

[tall jay]

Oh, yes, N. My bad

Oh, right bc N is a subgroup and is closed

Lol oops

[trim grove]

How to use class equation to find the order of center of non abelain group of order 27 ?
it is easy to check that order of center is 3 , but how to show this using class equation . Any hint?

[wooden ember]

You can use the class equation to show it isn’t trivial if you haven’t seen the proof for p groups in general, and then show that it can’t be of order 9 because then it would be normal and quotienting by it would give a cyclic group of order 3, which would give a contradiction (and not of order 27 since non abelian)

[south patrol]

If you need a hint to see why it would give a contradiction, ||prove the following fact: given a group G, if G/Z(G) is cyclic, then Z(G) = G||

[vocal mural]

I'm so confused with this formula

how to understand it?

???

what does it mean to do alpha to the power of beta??

[celest mantle]

read the purple sentence

[vocal mural]

the conjugate of a by b

what's that?

[celest mantle]

it's b^{-1} . a . b

[fallow plume]

I should've looked at the proof more. It's on our final so I can't look at it now ah well
I just started so time to see how bad this test is gonna be

[uneven island]

Hi guys, not sure if its the correct place to ask this question.. but its jacobson basic alg. vol 1..

I figured out its 2^(n^2) distinct binary relations.. but i got really stuck on counting all distinct equivalence relations..

Im talking about question 7 in the picture of course..

[chilly ocean]

Can I trust this diagram because it's completely the opposite of what I've learned

[maiden ocean]

Idk how it could be the opposite of what you learned

In what sense for you do rings "contain" groups

as a subclass of them

[quaint tree]

I mean technically venn diagrams are for sets and "the set of all groups" is too large to be a set but this otherwise seems to be correct.

[stoic rose]

I think it's a misleading diagram becuase rings and group don't have the same structure, I wouldn't have put both in the same Venn diagram

Like sure rings have an underlying additive group but they're not a subset of the set of all groups, since a given abelian group can have a bunch of different ring structures

("Set" of all groups doesn't technically exist of course but you should understand what I mean)

[fallow plume]

just define a group to be a ring with trivial multiplication since we're throwing out the need for an identity anyways

[tribal moss]

Rngs are Z-algebras, which are Z-modules with additional structure, and Z-modules are the same as abelian groups.

[stoic rose]

That would make the groups a "subset" of the rings, not the other way around

[stoic rose]

Yeah I know but as I said a Z-module can have a bunch of different ring structures

[tribal moss]

What? It is definitely not the case that every group is a ring.

[stoic rose]

If you don't require rings to be unital then every abelian group has the trivial multiplication a*b =0, I think that's what Thomas meant.

[chilly ocean]

Ah it makes sense to me now. I wasn't really focusing that every ring/field is a group with additional properties, though I've been studying them as groups with additional properties

[coral shale]

well you can say all rings are groups under addition

[chilly ocean]

Yes

[coral shale]

an intermediate set between the 2 would be abelian groups

[chilly ocean]

Wait a ring is an abelian group under addition, with multiplicative closure + commutative + distributive right

[coral shale]

For me, a ring is

• an abelian group under addition
• a commutative monoid under multiplication
• satisfies a distributive property
  ======================

non-commutative ring? rng?

[coral shale]

can't remember if I need to require 0 =/= 1, but I don't think it's uncommon to also write that

[coral shale]

oh nvm, u don't need to write that

suppose 0 = 1, we have 0 = a * 0 = a * 1 = a, so it has to be the trivial ring {0}

[stoic rose]

Yes but rings are not a subset of groups (even ignoring set-theoritic technicalities) since they have extra structure

[coral shale]

squares are rectangles with extra structure

You require they also have equal side lengths

[stoic rose]

That's not extra structure, that's extra properties

[coral shale]

If you're precise with what you're saying, you can say all rings are abelian groups

because they are - under addition

I can see your point about extra structure

but hmmm 🤔

[stoic rose]

Ignoring set-theoritic technicalities, you have a map from the set of all rings to the set of all abelian groups associating to a ring its underlying addtive group, but this map is not injective

So it's misleading to represent rings as a subset of abelian groups

[coral shale]

Yes, now that makes sense.

You can't take a subset of the abelian groups and say 'these are the rings'

That would be wrong.

[stoic rose]

Yeah exactly, for the question "is this abelian group a ring" to make sense, you need to specify what is the multiplication

It's not part of the data of the abelian group

[coral shale]

hmmm whats a good analogy

[stoic rose]

I guess an even simpler analog case would be asking which sets are groups

That makes no sense if you don't specify an operation

[coral shale]

I'm still big hmmm on an RL example lol, but a good one may not exist

or like an elementary example

[hidden haven]

Rings were never meant to be commutative

[coral shale]

moldi appears 👀

[hidden haven]

[coral shale]

can i have some thoughts on #954333596355788861

like whether what I was initially thinking about is even meaningful

[hidden haven]

[coral shale]

Aut(L)/Aut(K) = Gal(L/K) kindof idea

A way of thinking about the galois group, but I get this does not work for arbtirary L and K

Maybe notation is just notation and there is no deeper meaning to be had 🤷‍♂️

[stoic rose]

Idk, I have the impression that group theory is generally pretty much the first area of math where you encounter things with structure, so it's not easy to think of more elementary examples

[coral shale]

im thinking like saying this computer part

is like a type of computer

kindof analogy but I cant spit out a neat example

[stoic rose]

Maybe something along the lines "an adress is a street with extra structure".

[coral shale]

I would like to be able to explain this well if it came up, but I think the technical talk about structure is necessary

So nvm about an 'easy' analogy ig

Show $\bQ\qty(e^{i\frac{2\pi}{5}}) : \bQ(\sqrt{5})$ is a Normal extension

[coral shale]

(5th root of unity)

Does this suffice?

uhhh idk how to word it precisely

I can see Q[x]/(x^2-5) is 'inside' the bottom right thing

[regal carbon]

Hey, I'm currently referring to the book called "mathematical proof : transition to advanced mathematics".
Because my main target is to learn abstract algebra

The function theorems are difficult to go through. Please any suggestions?

[iron vessel]

How do I prove $\mathbb{Q}(\sqrt2 ,\sqrt[3]3) = \mathbb{Q}(\sqrt2 +\sqrt[3]3) = \mathbb{Q}(\sqrt2\sqrt[3]3)$?

[cloud walrus]

Évariste Galois

[iron vessel]

I tried using minimum polynomials but i end up with very long calculations and I doubt thats the correct way

[coral shale]

the 1st to 2nd makes me think of primitive element theorem

But you could also approach it without that

Let x = sqrt2, y = cbrt3
Consider equations formed by
x+y, (x+y)^2, (x+y)^3

And I think this gets you there.

[iron vessel]

I believe I may have cracked the first to the third

ok thanks, up until now i tried doing it ^6

so i wouldnt have any irrational terms but seems like i was wrong lol

[coral shale]

Well I chose 3

because uh

[iron vessel]

is there a particular reason?

[coral shale]

Basically, it will turn out to be 3 equations with 3 unknowns

So we are trying to show the middle thing

is contained in the left thing

[iron vessel]

yup

[coral shale]

agreed?

ok

x+y is a known

as are its square and cube

what isnt known is

x, y, y^2

x^2 and y^3 are also knowns

And this is precisely what I meant by '3 equations, 3 unknowns'

[iron vessel]

right, i must show that sqrt(2) and cbrt(3) are individually within the second field right?

[coral shale]

yes

There is another approach to this which is a specific case method of how to prove the primitive element theorem iirc

You consider ((x+y) - x)^3 and do stuff I think (can't remember the exact way).

[iron vessel]

ok thanks 🙂 also for the third field, i just did $(\sqrt2 \sqrt[3]3)^3 = 6\sqrt2 , so \sqrt2$ is in the third field, and consequently so is $\sqrt[3]3$. Is that good?

[cloud walrus]

Évariste Galois

[chilly ocean]

Can someone confirm this for me? Are tensor densities defined only for fields? Everywhere I look, tensor densities are discussed in the context of tensor fields.

[coral shale]

yh thats nice

[iron vessel]

Ok great

[uneven island]

Help please..

[next obsidian]

This is kind of cruel lol

[next obsidian]

So assuming you’ve managed to show that # equivalence relations = # of partitions of the set you need to count the number of partitions of a set of size n

These are a combinatorial number called the Bell numbers, and there’s no closed formula for them

You can give a sort of recursive formula tho, I think

[delicate orchid]

yeah the recurrence formula isn't that bad

[next obsidian]

It’s like B_n+1 = Sum nCk B_k

The idea I think is you’re gonna isolate k-elements

Make that one block in the partition

Then you have to partition the remaining numbers

Or something like tha

Maybe that’s off lol

[coral shale]

bump (also if my way isn't good - I'm curious if there is an easier way to show this)

[next obsidian]

Honestly idk what you did but if it were me I’d just compute the minimal polynomial, find the roots, show they all live in the extension

Then this is a splitting field so it’s normal

[coral shale]

I suppose I did then

[next obsidian]

I see

[coral shale]

👌 makes a lot more sense the way you said it

[next obsidian]

[uneven island]

Yeah i felt that, i remember that i've seen that counting surjections is no easy job, but i noticed that the exercices that the book is proposing are really interesting.. so i just wanted to make sure to not miss out anything. Thanksss so much.

[lapis trail]

Ok I think I'm crazy. But $\mathbb{Q}(\sqrt{2})$ is NOT isomorphic to $\mathbb{Q}(1+\sqrt{2})$, or am I making a mistake?

[cloud walrus]

seth.delacroix

[lapis trail]

I defined f(a+b(2^(0.5)))=a+b(1+2^(0.5)) and multiplicativity is failing unless I'm smooth brained

[next obsidian]

Yeah they’re not

Notice that 2 has a square root in one, but not the other

For any supposed isomorphism between them, 2 has to go to 2 because 1 goes to 1, then additicity

Thus if they were isomorphic both would have to have a square root of 2

[lapis trail]

Ok great

And that makes a little sense when you think about how the minimal polynomials are different

[cloud walrus]

seth.delacroix

[lapis trail]

I think that is I adjoin root(1+root(2)) to Q I also have to adjoin powers of it

[coral shale]

this isn't true.

You will find that it fails the closure axioms

if you assume everything just looks like this

[delicate orchid]

you gotta get 1+sqrt(2) in there too dawggg

[coral shale]

Q(a) basically adjoins all 'polynomials in a' if you like (as well as the fractions in them - but there is a handy theorem...)

[lapis trail]

Ok thanks that makes a lot of sense

[lapis trail]

Holy shit I just understood

math is amazing

[chilly ocean]

I've been thinking about this, that a ring must also be a commutative semigroup not commutative monoid under Multiplication. Since a Monoid must have an identity element, but in formal definition of rings, it's not necessary for a ring to have a multiplicative identity element 🤔

[kind temple]

depends on your definition of ring. some definitions require it to have an identity.

[terse crystal]

If R doesn’t have 1, you can always embed it into some bigger ring that has 1. Like R’=Z×R, (m,a)(n,b)=(mn,ab+na+mb). Then R’ has a 1, which is (1,0). R is a subring of R’, a mapped into (0,a). If you want it to be of characteristic p then replace Z with Z/pZ

[chilly ocean]

Ooo

[chilly ocean]

Can we call a quasigroup an invertible magma, or is it always divisible magma

[chilly ocean]

what are your definitions of invertible magma and of divisible magma?

[chilly ocean]

According to my understanding, a divisible magma is a quasigroup where for every element x and y in the divisible magma (or Quasigroup) (M,.,/,\), there exist two unique elements l and r where:
lx = y => l = x/y [left quotient]
xr = y => r = x\y [right quotient]

An invertible magma (M,.,/,\) is an algebraic structure such that there exist a unary operation (-)^-1:M->M where:
(x.x^-1).y = y => y\y = 1
y.(x.x^-1) = y => y/y = 1
x.(x^-1.y) = y => y/x = x^-1.y
(x^-1.y).x = y => y\x = x^-1.y

Okay I'm confused now

[chilly ocean]

What is 1?

Identity

But wait won't that become a loop

magmas do not have an identity

(as in a distinguished one)

I'm literally confused now

I have a question
every nonzero element except to unit is zero divisors?

[chilly ocean]

nah

look at integers

nice

you just corrected them

[chilly ocean]

plz explain this...

[chilly ocean]

There can be elements of a ring that are neither zero divisors nor units

[chilly ocean]

Let L be the rationals and D be the integers

We know the integers are a subring of the rationals and that it has no zero divisors.

Also integers dont have units besides 1

since none of the elements are invertible

but that's not a proof..♡

[chilly ocean]

If your hypothesis is that every nonzero element in a ring is either a unit or a zero divisors

Then the counter example to disprove the hypothesis are the integers

[chilly ocean]

if D had a zero divisor, so would L, which would contradict L being a field ❤️

[chilly ocean]

if you want this property you need artinian rings

[chilly ocean]

isn't -1 a unit

Yes

I meant that implicitly

but there is also 2

which is nonzero

and isnt a unit

bro that poor guy barely knows what a unit is and you indoctrinating them about artinian rings 😂

[chilly ocean]

You should try and show that the additive inverse of a unit is a unit.

[chilly ocean]

Wait is invertible (unital) magma an invertible quasigroup?

[chilly ocean]

From definitions you gave I think so

unless there is a invertible magma where x*x^-1 isnt unique

for all x

Oh I see

That's why people prefer divisible magma définition for quasigroup I guess

[lethal dune]

is it convention to consider (0) to be a maximal ideal?

or prime

(in case R is a field/ id)

[sharp sonnet]

its maximal/prime if it satisfies the definition

i dont see how this is convention?

its just true according to the definition

[lethal dune]

is nonzero a part of the definition, that's what I'm asking

[sharp sonnet]

generally no

[lethal dune]

like ≠(1) is a part

[sharp sonnet]

you want R/I field/id iff I maximal/prime
i think this is good enough argument to not exclude the zero ideal in the definition

some other definitions or theorems will require "nonzero prime/maximal ideal" then but thats fine

(this is also the convention i am used to)

[lethal dune]

ok fine, tks

then is there a special reason to include ≠(1)?

also another question, when proving existence of a maximal ideal, why comm with 1 is assumed? the proof works fine even if it's non commutative and non unitial?

[sharp sonnet]

if you only care about commutative rings with 1, then you build your theory with those

[sharp sonnet]

mostly because this is "trivial" and would have to be excluded in many theorems for no benefit i think

[coral shale]

I think it's the same reason we don't want 1 to be prime in the naturals. You don't want unit ideals to be prime (note any ideal containing a unit is the whole ring, ie. (1))

[sharp sonnet]

you can come up with deeper reasons that ideals shouldnt include units due to number theoretic reasons maybe

[lethal dune]

true

[wind parrot]

You don't need commutativity, in that case you get existence of maximal left/right/two-sided ideals, but you need 1 to show that the union of a chain of proper ideals is proper (which is needed for applying Zorn)

[chilly ocean]

Is my notion of understanding these algebraic structures correct?

[sharp sonnet]

just look up the definitions?

[chilly ocean]

I was just checking here too, like some some sources say (for a Group) that it is an Inverse Semigroup + Identity. Other sources say that it's an Inverse Monoid, and when it is cancellative its a group

[sharp sonnet]

then you have to look up the definitions of those things

the definition of groups is very standard and it should be easy to verify

if you are unsure about a specific thing, supply your definitions and what you tried to show equality

[scarlet estuary]

the majority of this jargon is never used fwiw

[paper flint]

Semi-complete

[sharp sonnet]

recently this server learned that this diagram is wrong

[paper flint]

Oh

[scarlet estuary]

you can just google it for the 1 time every year it comes up

rather than learn a bunch of rare definitions, learn to do math

[carmine fossil]

Someone should edit that wikipedia page

[sharp sonnet]

top 10 questions science still cant answer: is a group an inverse semigroup with identity or a cancellative inverse monoid

[carmine fossil]

monad

[sharp sonnet]

maybe i already forgot what the problem is though

i dont think whoever mentioned that even provided a counterexample

just "not every xy is a z"

[coral shale]

not every associative quasigroup is an inverse semigroup
I was told

indeed, no example provided

[glacial moss]

Hello, I'm interested in understanding the gist of abstract algebra. So, I've started watching lectures. It would be great if someone could give an outline of which important insights about the structure there are to come. If you know something, please correct or add to the following understanding:

There are most insights about rings, when they are commutative. For rings it is possible to find ideals - which are like the "null elements" in subrings. You can divide out these ideals to get subrings, or even a field if you pick the maximal ideal. For rings which are an integral domain, you could also find a larger rings, which is a fraction field. There is a well-defined construction, but in a way it's like "throwing in all inverses" and knowing there will be no mathematical inconsistencies.
In field theory, there are unique finite fields of size p^k which are well understood(?) The most prominent field is the field of rational numbers (and it's subfields and extensions), but I do not consider those in this question, because it is the most complex part. If you extend the rational numbers to include all limits, then you get the real numbers. Those can be extended to the complex numbers, but there are not so many other interesting extensions or results about these (unless you increase the cardinality) (?)
And of course, there is vector algebra, which I do not consider for the moment.

Is this correct and what are some major results that would come next in ring or field theory?

[delicate orchid]

I wouldn't use the word "subring" to refer to qutoient rings

Z/nZ is not a subring of Z

[stoic rose]

Did I misread that sentence or did you wrote that there are "not so many interesting results about complex numbers"?

[delicate orchid]

given your current trajectory it sounds like you're either gonna go into algebraic number theory (rings) or galois theory (fields/groups), neither of which is my area of expertise

if you haven't seen a completion of a ring that's an important thing to learn before going any further in ring theory imo
and stuff like noetherian/artinian rings

[delicate bloom]

no you didn't, it's your duty to educate them

[glacial moss]

I mean in terms of field theory. Are there major results for the field of complex numbers in field theory?

[celest mantle]

i mean there is

it's a genuine thought though because in algebraic nt we don't deal THAT much with C but there are tons of results on it

[stoic rose]

Anyway your question is extremely broad so it's hard to answer but I can try to give some examples. If you want to study rings further there's commutative algebra, which is the study of commutative rings and their modules, which are a very important tool to study rings. Commutative algebra is also very closely related to algebraic geometry, which studies things like rings of polynomials and their quotient, and more generally studies commutative rings by thinking of them as rings of functions on some space. You mentionned the rationals and their extensions (things like Q[i], cyclotomic fields, etc.), the study of these particular fields is a topic of algebraic number theory, which studies number fields and particular rings in them, which are the analog of the ring of integers sitting in the field of rationals. Finally for more general field theory the big topic is Galois theory, which studies extensions of abstract field by stufying their group of automorphisms.

[glacial moss]

I know it's broad, but your answer helps to get an overview. Thanks

[stoic rose]

In field theory not really because field theory is concerned with the general theory of fields and not really specific fields. But otherwise the field of complex numbers is probably THE most important and best understood field in all of mathematics.

Though depanding of the specific area R or Q might do some competition for most important field

[glacial moss]

And C seems to be like the most "general" field in a sense?

[stoic rose]

What do you mean by that?

[glacial moss]

Hmm, not sure how to define that. But let's say you know some mechanism or physical relation satisfies the field axioms. Then some subset of C could represent that?

[stoic rose]

I'm not sure I understand what you mean by mechanism or physical relation. Do you mean like something in physics?

[delicate bloom]

when you said " If you extend the rational numbers to include all limits, then you get the real numbers." this wasn't entirely true, there are other ways to define distance between rational numbers which you can use to complete Q with by limits and you get entirely disjoint fields from R.

I am assuming you were coming from the angle of like, thinking Q is in C and since C is algebraically closed, that's the "end" or something

well, even simpler, no finite field or any of their algebraic closures are contained in C either, so C isn't "most general" in this sense either if that's what you mean, you can tell because they have different characteristic

[celest mantle]

well you can build C from algebraic closures of finite fields, so there is kind of an analogy here too

[stoic rose]

How can you, they have different characteristics

[celest mantle]

take a free ultrafilter on the set of primes, consider the equivalence relation induced by it, then take the infinite cartesian product \prod_{p \in P} \overline{F_p}/U and you have C

something along the lines of this should do the trick

you have most of the properties using Los's theorem

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwi_l8ya9NT2AhVCXBoKHfWZAJgQFnoECAUQAQ&url=https%3A%2F%2Fchoum.net%2F~chris%2Fcours_et_notes%2Fultra%2Fultra.pdf&usg=AOvVaw0FBvTZGIHVCpX9_AsUT2xs it includes of a proof of ax-grothendieck too if you want to read it

[glacial moss]

I meant when you observe objects (maybe just something in a board game or whatever) and you know that they satisfy the field axioms. Maybe I can phrase it differently: are most fields technically a subset of C and could be represented by a complex number for each object? and which fields cannot be represented like that?

[stoic rose]

Well all fields of characteristic p can't be embedded in C

[celest mantle]

for instance take the algebraic closure of the p-adic numbers (which are another completion of Q for a non trivial absolute value)

[stoic rose]

If you ignore the topology I think this can actually be embedded in C

[celest mantle]

really ?

how so

[stoic rose]

All algebraically closed fields of characteristic 0 and cardinality c are isomorphic to C

You need the axiom of choice of course

[celest mantle]

ah yeah, the classification of fields up to the cardinality of transcendence basis can give fucked up things, that's exactly why the isomorphism that i gave earlier is valid too actually

[stoic rose]

Of course, usually you care about the topology though, and in this case they are very different

[celest mantle]

yeah

[dull root]

So I have S_4 the symmetry group of 4 elements. It has order 8*3, so its sylow 2 subgroups has order 8. How do I see that these sylow 2 subgroups are isom to D_4,the dihedral group of the square. I am told this is trivial, but I am not seeing the intuitive picture

[tribal moss]

Do you doubt that there are subgroups of S_4 that are isomorphic to D_4?

Or is your problem to see that those are all the Sylow-2-subgroups?

[stoic rose]

All the Sylow 2-subgroups are conjugate, so if one is isomorphic to D4 then they all are

[delicate orchid]

would showing that D_4 has a natural embedding into S_4 be a valid proof here

[stoic rose]

Yes since then it's automatically a 2-Sylow since it has order 8

[tribal moss]

It would at least be part of a valid proof.

[delicate orchid]

yeah then you'd have to show that the other subgroups of order 8 do not have a natural embedding, I presume

[tribal moss]

Whether you can leave the subsequent appeals to general facts from Sylow theory implicit is a fuzzier matter.

[tribal moss]

As Radiateur pointed out, Sylow says that any two subgroups of the appropriate order are conjugate, so the rest will automatically have similar embeddngs.

[chilly ocean]

hi radiateur

[stoic rose]

Hello

[dull root]

I see that all sylow 2 groups od S_4 are isomorphic, but I don't see D_4 has a natural embedding into S_4

wait nvm, i am stupid, yea it is trivial

D_4 is just the group of symmetries of the square, so we can think of it as a subgroup of the permuation of its 4 vertices

[shell brook]

Suppose $f(x) = x^n + a_{n-1}x^{n-1} + \ldots + a_0 \in \bZ[x]$. If $r$ is rational and $x - r$ divides $f(x)$, then $r$ is an integer.

[cloud walrus]

jesse

[shell brook]

There is a SO post that says this follows from Gauss' lemma, but I don't see it?

I know there is another way to prove this

I'm not asking for that way, though. Just curious if this actually follows from Gauss

[stoic rose]

Gauss lemma tells you that if a monic polynomial factors over Q that factorization is actually over Z

[shell brook]

I am not familiar with this version

[stoic rose]

What version are you familiar with?

[shell brook]

I know "If f, g are primitive, then so is fg", and "A non-constant integer polynomial is irreducible in Z iff irreducible in Q and primitive in Z"

For me primitivity is defined only for integer polynomials

(gcd of coeffs is 1)

[stoic rose]

It's not hard to prove my version from your first statement

[shell brook]

Okay

[stoic rose]

Actually if you look at the proof of your second statement it probably does it by proving my version

[shell brook]

Okay and so it follows from this because you get a factorization f(x) = (x-r)h(x) which is over Q, and so r must be in Z?

Okay nice

I will prove the version you gave me assuming the Gauss I know then

thx

[south patrol]

Jesse do you know the form of Gauss's lemma too where it's just saying content is multiplicative (where content can be taken of any element of Q[x])

[shell brook]

Like Con(fg) = Con(f)Con(g)?

[south patrol]

ye

[shell brook]

I haven't seen that stated as Gauss's lemma but it's used in the proof of the first Gauss I stated

[south patrol]

lol ok

Like if f,g are primitive so is fg follows immediately from this but it's a useful strengthening I'd apply to this problem

Basically you can show cont(x-r) = 1 through this

Which means r is integer (given f in Q[x], f in Z[x] iff content is integer)

[shell brook]

The proof of this is like... if f monic over Z[x] and reducible in Q[x], then f(x) = h(x)q(x), and since f monic we have Con(f) = Con(hq) = 1 => Con(h) = Con(q), and from Con(f) = 1 iff f in Z[x], we have h,q in Z[x]?

[south patrol]

how did you get con(h) = con(q) from that without more working

[stoic rose]

I don't know this thing about contents

[shell brook]

Con(hq) = Con(h)Con(q) = 1 and Con is integer so both are 1?

[south patrol]

content needn't be integer, no

Ok maybe i should explain content in the way I mentioned above

[shell brook]

Oh

Yeah I suppose it doesn't need to be lmfao

my bad

[south patrol]

Given f in Q[x], for any n in Z such that nf is in Z[x], we define cont(f) = 1/n cont(nf), where cont(nf) is defined in the usual way

i.e. GCD of coefficients (over Z)

[shell brook]

Right

Ahh

Thats cute

[south patrol]

And then the thing is that cont(fg) = cont(f)cont(g) still holds over Q[x]

i mean you literally just multiply through to get it all in Z[x] and then divide again
Also given f in Q[x], cont(f) is integer iff f is in Z[x]

[shell brook]

Yeah

[south patrol]

Anyway, so for the factorisation thing for example

Let's say f monic and f = gh for some g,h in Q[x]

1 = cont(g)cont(h)

Wait lemme check what I'm doing to ensure I don't make a mistake kek

[shell brook]

$$Con(g) = \frac{1}{Con(h)} = \frac{1}{(1/n)*Con(nh)} = \frac{n}{Con(nh)} = 1$$ and so $$n = Con(nh) = Con(n)Con(h) = nCon(h)$$ and so $Con(h) = 1$?

I made some cringe mistake here probably

ok Texy down

[cloud walrus]

jesse

[shell brook]

yeah something like that

Cause n = nCon(h) => Con(h) = 1

[south patrol]

Sorry, I had to go for a second

[south patrol]

But basically I think what this is meant to mean is essentially that it corresponds to a factorisation over Z

Now suppose yes f monic and f =gh for some g,h in Q[x]

Then 1 = cont(g)cont(h).

So cont(g) = p/q and cont(h) = q/p for some q,p in Z[x] where (q,p) = (1)

So in fact, (q/p)g and (p/q)h are in Z[x], since both have content 1

And so f factorises as their product

:)

[shell brook]

I see

Right I get it

okay thats nice

[south patrol]

Yeah it's nice, this form of content is nice

And it works for the thing you said with x-r, have a go :)

[shell brook]

Right yeah it follows right away

thx sm

[south patrol]

Np

And yeah this is exactly how I'd prove the thing about irreducible polys over Z being primitive and irreducible over Q

I mean we just proved that aha

since primitivity is immediate

[chilly ocean]

I suspect if my working is correct or not

[south patrol]

Sure, in summary the idea is essentially that phi(a^-1) and phi(a)^-1 are both inverses of phi(a) and inverses are unique

[chilly ocean]

Thank you 👍

[south patrol]

Np

[next obsidian]

I think it's weird that any dim 2 ring is catenary, and this implies that any dim 0 or 1 Noetherian ring is universally catenary

To do what?

Oh

[broken stirrup]

oh sorry

my bad

[next obsidian]

I am sorry, I do not know how to help you here

I assume it’s to show that R is not Noetherian

[next obsidian]

[broken stirrup]

exactly

[next obsidian]

Well the idea is that the reason it’s not finitely generated but Q is is that you have access to less things to multiply into generators

My idea maybe is like umm…

The ideal generated by all polynomials with prime constant term?

No

Bad

Uhhh

I have no clue

[broken stirrup]

is Q finitely generated?

[next obsidian]

Over Z?

No

[broken stirrup]

if we take its ideal of polynomials with zero constant term, wouldn't it be non finitely generated R-module?

[next obsidian]

¯_(ツ)_/¯

[broken stirrup]

we get copies of Q

[next obsidian]

That sounds reasonable

Like look at

x/p

Over all primes p

Yeah

I see

[broken stirrup]

and Q is not finitely generated as underlying group

[next obsidian]

Well it’s a bit more complicated

Because the generation is over R

But you’re right

Like the idea is that the generators for the linear stuff have to be just like

ax for a in Q

And then for these to give you elements say x/p

The thing you multiply needs to be a constant

Which comes from Z

Anyway, whatever, I agree with your solution

Very nice

[broken stirrup]

lol

thanks

[next obsidian]

[broken stirrup]

and thank you for your time

[next obsidian]

I didn’t do much tho

[broken stirrup]

i still appreciate

[dull root]

If I have extension K \subset K_1 \subset K_2 where K_2 is a normal extension of K_1 and K_1 is a normal ext of K, is K_2 necessarily a normal ext of K too?

You mean if a \in K_2, then its minimal polynomial over K divides its minimal polynomial over K_1?

yes

[south patrol]

And then that should give you it

wait sorry i completely misread

[dull root]

So the mimimal polynomial of a over K_1 splits over K_2.

[south patrol]

Or was it correct oops

[coral shale]

a polynomial in K2[x]

is also a polynomial in K1[x]

[dull root]

you meant a polynomial in K1[x] is a polynomial in K2[x] right?

[coral shale]

yeah the other way round

why not use latex since ur halfway there

[south patrol]

Ok sorry yes this is just not true

e.g. Q, Q(sqrt(2), Q(2^1/4)

[coral shale]

$K \subset K_1 \subset K_2$

[coral shale]

K2 : K1 and K1 : K are normal

you wonder if K2 : K is also normal

Notice polynomials in K[x] are also polynomials in K1[x]

[south patrol]

Q(2^1/4) not normal extension of Q because the min poly is x^4 - 2 which doesn't split completely since it has imaginary roots

[dull root]

yes, I think so, its mimimal poly is (x^2 + sqrt 2)(x^2 - \sqrt 2)

so it splits over Q(rt 2)

[coral shale]

Are we saying contradictory things or something

[dull root]

but not Q

[south patrol]

Yup

Oh shuri i'm saying the original statement like doesn't hold

i.e. composition of normal extensions not normal necessarily

I just thought about it wrong initially and said wrong stuff kek

[coral shale]

uhhhhhhhhhh

ah nvm you're right

godamn

wait

wait wait wait

what?

I claim all polynomials in K[x] either have split over K2 or have no roots in K2

Because all polynomials in K[x] are also polynomials in K1[x]

[south patrol]

Yeah but then e.g. if an irreducible polynomial has a root in K2 it will split over K1[x]

but not over K[x] necessarily