groups-rings-fields

[cloud walrus]

rando

[long swan]

it’s kind of poetic that this is the name of the blacklist role lmaooo

[trail hamlet]

Why ω(g, h) = [[g, h], [h, g^−1]] has nil(ω) = 5 ?
Why ω(f, g, h) = [[f, g], [g, h]] has nil(ω) = 4 ?

[lusty marlin]

What is the blacklist role?

The first part is fine but the second part is completely absurd. You have literally written that 2(1+√(-5))=6.
This makes me strongly suspect that chatgpt has generated this solution.

[swift current]

😆

[next obsidian]

Holyyyy

[next obsidian]

Yo I think I agree

Like look at the last part of it too

And nobody texs like that, those ; and the way the [ are formatted

[kind temple]

no way you get conflate arcf and 1/f

[next obsidian]

I would if I checked help channels a lot, I think

[kind temple]

😭

[next obsidian]

I could also check off mixed fractions if the hw I grade for calc 3 counted

[lusty marlin]

Yeah

[tardy hedge]

Hows the homework

Do you get a lot of sloppy stuff

[next obsidian]

Lol

[tardy hedge]

Well, do you?

[tardy hedge]

Mixed fraction 😂 😂

Thats actually very funny

[tardy hedge]

Dude i saw someone talking about french interval notation

[proper jolt]

hello i don't understand the explanation of this

for the line:

doesn't \phi have to be at least an R-algebra map for this to be true? i don't see how a abitary map could imply this

also,

[rocky cloak]

It just needs to be a ring homomorphism actually. But I would interpret the exercise as it being an R-algebra map.

Otherwise I don't know what they mean by "map".

[proper jolt]

can you explain why it works if it's a ring homomorphism i don't get that

or like just that line

how would a ringhomomorphism work with that line

cuz im thinking that \phi doesn't have to map the units of S^-1R injectively into W^-1R

or like the maps \phi_S and \phi_w and \phi don't have to form a commutative diagram for it to be true

so idk

[rocky cloak]

I just meant the first equation sorry.

But yeah units map to units, then you want it to be a map of Ralgebras for R to map to R

[proper jolt]

oh ok np thankyou

[glad osprey]

I've read that if A is a commutative ring with x in A, and x is idempotent, then A is the direct sum of Ax and Ax' where x' = 1 - x. But is Ax even a subring? I don't think it contains 1 unless x is a unit

[chilly ocean]

Ax is it self a new ring with x working as identity now so you are right

[tribal moss]

... which wouldn't usually be called a "subring" if the identities are different.

[chilly ocean]

yes

[tribal moss]

"direct sum" suggests that Ax and Ax' are considered as A-modules, rather than rings.

[glad osprey]

ah, I see I was misreading it, it didn't say subrings

[glad osprey]

The point is that coproducts do not exists for rings, right? But if we consider them as A-modules, aren't we losing some information about the multiplication? Tbh, I've never quite gotten used to the interplay between rings and modules. In this case, the structure is basically exactly the same, but we need to be in the category of A-modules to satisfy some universal property?

[thorn jay]

usage of rng for no identity supremacy

[trail hamlet]

Hi, I am looking at this paper : Breaking the cubic barrier in the Solovay-Kitaev algorithm

I do not understand the last expression (ccan_{SU(d)}(ω) ≥ nil(ω)) inside the attached screenshot.

[glad osprey]

btw, can't we say that A = Ax \times Ax' using just a normal product, forgetting about the direct sum?

(as rings I mean)

[tribal moss]

I mean, the property "every element of A is the sum of something from Ax and something from Ax' in exactly one way" is a meaningful statement no matter whether you call Ax and Ax' rings or modules.
The property just happens not to need the multiplicative structure to state.

[rocky cloak]

So A will be the product of Ax and Ax'. As rings, but that does not mean Ax is a subring.

It is also the internal direct sum as abelian groups / modules. Some authors have some confusing terminology in these cases

Also there are coproducts in the category of commutative rings. It's just not the direct sum (it's the tensor product)

[glad osprey]

aha, I think I get it thanks Tropo, jagr and Mishra

[tribal moss]

Product rings are probably a good example to have in mind. A simple way to make a ring with a nontrivial idempotent is to take a ring product A×B×....×C and look at elements where every component is either 1 or 0, but they are not all 0 or all 1 (that would be the 0 and 1 of the product ring, which is idempotent all right, but not "nontrivial").
The direct-sum property essentially says that all nontrivial idempotents can be understood in this way.
And at least for me it makes it more intuitive that 1-x will be an idempotent too -- that operation just interchanges 1 and 0 in each factor ring.

[glad osprey]

damn, that makes a lot of sense I was thinking about how weird it is that every ring containing a non-trivial idempotent is a product, but now I get it. Thanks

[tribal moss]

(And xx' = x(1-x) = x-x² = 0, which means that the cross terms vanish when we multiply out (ax+bx')(cx+dx') = acx+bdx', so A does actually have the multiplicative structure of a product ring too).

[marsh fulcrum]

Hello. I have come with a rather strange problem
How do I increase my problem solving abilities
I know this sounds very clichéd but please don't give the usual do more problems

Specifically in stuff like group theory ring theory I do problems by picking elements and chasing
Rarely can I use stuff like maps and compositions
I can do it but not so much
How do I incorporate that in my problem solving

And how to become stronger in real analysis problems

[chilly ocean]

by following your great teacher

[chilly ocean]

For real analysis i usually see common problems with solution to understand the pattern

[rocky cloak]

I think you come a long way by just writing down the given information, the wanted conclusion, the involved definitions and some relevant theorems.

Then you can start at both ends. From this I know this, does that help? If I could prove this the conclusion would follow, can I do that? Etc.

Groups/rings are pretty simple abstract structures, so usually you don't have that many tools available, meaning there are very few ways to do the wrong thing. Just doing whatever is possible often works.

[woeful sage]

yea writing down given information, and writing down the relevant definitions and theorems gets you half way to solving the problem most of the time, it's quite helpful

in both algebra and analysis

[chilly ocean]

i believe this helps but sometimes problems still remains unsolved then what we can do, i still have some unsolved problems and coudnt solve those to this day. do i have to be orthodox

[rocky cloak]

Well, some problems are harder than others. Sometimes you just don't have the tools to solve something, and then you just have to try again when you do.

Looking at simple examples or solving simplified problems is good for getting ideas.

[chilly ocean]

i coudnt solve that radical problem by @chilly ocean

[rocky cloak]

Idk which problem that is

[chilly ocean]

Yeah no response

[white oxide]

How does this proof look? $J(A) \subseteq N(A)$: Suppose for the sake of contradiction that there is $x \in J(A)$ with $x \notin N(A)$. Since $x \notin N(A)$, $(x) \not\subset N(A)$. By assumption, there is $ax \in (x)$ with $(ax)^2 = ax \neq 0$. Therefore, $xa(1 - xa) = 0$, so $1 - xa$ is not a unit. This contradicts the fact that $x \in J(A)$.
\
\newline
$N(A) \subseteq J(A)$: If $x \in N(A)$, then $x^n = 0 \in J(A)$ for some $n \in \mathbb{Z}^+$. Since $J(A)$ is prime, $x \in J(A)$, as desired.

[cloud walrus]

okeyokay

[rocky cloak]

What was the problem?

[chilly ocean]

I think this is the problem he is talking about, involving radicals

Although I solved 6 and 7 upto (i) part J I am unable to even start

yeah

[rocky cloak]

Yeah, this is pretty funky.

Not sure what "this" refers to, I guess all the preceding exercises...

[south patrol]

Curious

[white oxide]

lol wdym

[south patrol]

No I mean cool question (I remember doing it) but like I wonder when this would ever come up

[chilly ocean]

please ask for the hints from your prof, i guess you met him already 2 times but forgetting to talk about this.

[white oxide]

haha no clue

i love these types of questions though

[chilly ocean]

Just sending him mail fine

[rocky cloak]

Wait so what's the deal with part (i). In the for every F/E, find e in Q?

Is there some kind of typo? Are you supposed to find which F, E such an e exists? Should e be in E not Q?

[chilly ocean]

For any F/E we have to find e

[rocky cloak]

But that very obviously doesn't exist

[chilly ocean]

Why

[rocky cloak]

Because then every extension would be degree 2

[chilly ocean]

Okay yes

Assume it a typo

[rocky cloak]

Well sure, but what was it supposed to say

Maybe that is the thing we're supposed to use in j...

[chilly ocean]

Yes

But that I don't know how to use it

he also posted this

[rocky cloak]

Anyway, I had a though similar to this.

Figure out which automorphism fix either side. This determines the field they generate.

Now using composition series of the Galois group you can find a sequence of field extension each with degree 2. Then write them as F = E(sqrt(e)) for e in E.

Then the automorphism negating sqrt(e) can be used by
(x + s(x))/2 and (x - s(x))/2sqrt(e) reduces the problem to 2 equations in E.

Then you can recursively simplify until you have something with rational numbers

[chilly ocean]

Thanks

I will try it

[chilly ocean]

what do u mean by negating sqrt(e)

[rocky cloak]

Making it negative

sqrt(e) |-> -sqrt(e)

[chilly ocean]

how do we get 2 eqautions i didnt get that

[rocky cloak]

(x + s(x))/2 and (x - s(x))/2sqrt(e) are two different numbers

So if you have
x = y
And you apply these two on both sides, you get two equations

[chilly ocean]

s is automorphism?

[atomic mesa]

hi, I have that to prove something is a subgroup M of G I check it’s nonempty, closed under multiplication and inverses in 3 steps. I saw the last 2 steps gets merged into 1: let g and h be in M. Prove g^(-1).h is in M.
I claim this doesn’t work because if g^(-1) isn’t in M then the above statement proves neither condition … anyone see where I’m going wrong?

[next obsidian]

One thing you do is show that it’s nonempty

If you do that then take a g in M, and then use this property to see that g^-1g = e is in M

Now take h = e and you see that if g is in M, g^-1e = g^-1 is in M

[atomic mesa]

Oh that’s why I can exchange “M contains e” with “M is nonempty” in definition of subgroup?

Let M be nonempty, closed under multiplication and inverse.
nonempty implies e is in M. If not, then g^-1 is not in M which is a contradiction

[next obsidian]

No, I think you didn’t read what I wrote

I showed that nonempty + the condition that g,h in M => g^-1•h in M means M is a subgroup

I first showed e is in M, and then I showed how this gets you inverses

And then I guess to show that it’s closed under products you just take g = x^-1, y arbitrary and now you have xy in M if x,y are

[atomic mesa]

I missed (and tried to justify) why must g^-1g be in M but from writing it out I see take h=g ?

[next obsidian]

Yes

[worthy orchid]

Hello! I'm studying up for an abstract algebra exam, and I was wondering what are some topics specifics I should focus on. I studied ring theory, modules and field extensions this semester. Any pointers would be greatly appreciated!

[chilly ocean]

i expect irreducibility problems fron rings,

[chilly ocean]

will your prof give only problems ?

[worthy orchid]

Yeah, i don't think there will be any theorems to prove for the exam

Just simple proof problems for example proving a polynomial is irreducible in Q

[chilly ocean]

from field theory you can have problems on computing minimal polynomials and degree of field ext

and some cyclotomic stuff

[worthy orchid]

Oh ok, thank you!

[chilly ocean]

gauss lemma is quite imp as well

for exams

and most probably to proove or disprove two rings are isomorphic

asking about maximal and prime

ideals

proving disproving a ring ufd pid

[worthy orchid]

I see, alright I've got a much better study outline now

[chilly ocean]

for module tell me how much u covered

[chilly ocean]

some standard problems are in dummit foote see those

[worthy orchid]

The very basics, I think up to the isomorphism theorems

[chilly ocean]

most probably iso will come only then dummit foote is good for that also

[worthy orchid]

Ok

[chilly ocean]

ring iso theorems cant be avoided they will be used to solve some problems

[worthy orchid]

Understood

[chilly ocean]

for field, finite field and constuctions of finite fields

using irred poly

[worthy orchid]

Ok

[chilly ocean]

solve all the tutorial and class excersises or assignment

because sometimes we may be lo=ucky

[worthy orchid]

Yes, hopefully

[ivory ore]

assume $G$ is finite. $G' \le \Phi(G)$ $\Rightarrow$ $G$ is nilpotent.

[cloud walrus]

longboard kayak

[ivory ore]

we just have to show that all maximal subgroups are normal and we are done right?

[south patrol]

What is capital phi

[ivory ore]

frattini subgroup

here it is defined as intersection of all the maximal groups

[ivory ore]

so G/phiG is abelian

and also it generates G

[south patrol]

Isn't like a subgroup of a nilpotent group nilpotent

And then Phi(G) is nilpotent so G' is

[ivory ore]

oh yeah it is

[south patrol]

Which implies G is

[ivory ore]

is the reverse true too?

[south patrol]

As in reverse of your original statement?

I don't know tbh would have to think

I never learnt about Frattinis properly lol

[ivory ore]

oh yeah it was an iff statement

but no i asked about something that i misread from your response

[ivory ore]

how does this implication work?

[ivory ore]

i have been told that it has all the nongenerators and now i am being asked to proof some facts with nipotency and p-groups

[ivory ore]

i think from this we atmost say that G is solvable

extension of solvable groups i mean

over

[proud hedge]

Hey guys can you give an example of a ring with no 1 such that it has only 3 idempotent elements?

[coral spindle]

That's a really interesting question, I'll think about it. Where'd you get it?

[chilly ocean]

some matrix shit

[proud hedge]

it's part of a question that our proffessor gave us

[proud hedge]

i couldn't really find one

[coral spindle]

Oh can you show the full qn?

[proud hedge]

yeah let me translate it

[chilly ocean]

that is where we can easily try

[coral spindle]

Protip: if you can't say anything helpful, it's often better to simply say nothing

[proud hedge]

a) prove that there is no ring with identity such that it has only 3 idempotent elements.
b) show with an example that the identity condition is necessary.

[coral spindle]

Ah I see

[proud hedge]

i proved a

[coral spindle]

Right, using 1 - e

[proud hedge]

i have more interesting questions like this that i can send but i'll have to translate them

[coral spindle]

Well you're welcome to, I'm just thinking about this one at the moment

Aha I think I have one

Oh, wait, no that makes four

Just a quick clarification, we are counting 0 as an idempotent, right?

[proud hedge]

yup

[coral spindle]

Cool

Let me explain my idea

[ivory ore]

i was thinking of the ring decomposition in terms of idempotents

[proud hedge]

i haven't read about ring decompositins

[coral spindle]

I was thinking of the ring of upper triangular 3x3 matrices whose bottom-right entry is zero

That has four idempotents

I was trying to somehow take away the identity

But alas it didn't work

[ivory ore]

peirce decomposition, something i heard about 2 days ago

[coral spindle]

I don't think products of rngs will work here

[ivory ore]

sad

[proud hedge]

it's part of a competition question so it's a bit hard

[ivory ore]

cool problem nonetheless

[proud hedge]

yeah honestly all of them feel interesting and kinda unsolvable 😄

[coral spindle]

I have this silly feeling that there's a rng of size 3 that works

[proud hedge]

i also had idea but couldn't find a good one

w8 i have an idea!

[tribal moss]

If t is a nontrivial idempotent then 1-t is also idempotent (direct computation), and it's a different idempotent because t(1-t)=0 (direct computation again).

[coral spindle]

We're looking at nonunital rings

[tribal moss]

Ah, sorry.

[coral spindle]

It's alright

[ivory ore]

that was the first section

[ivory ore]

so we need to define ab and a+b

[proud hedge]

it's ok i used the t^-1 idea in the first section so your idea is new and helpful ❤️

[ivory ore]

in no way it would be non-commutative

[coral spindle]

Well if a, b are the nonzero elements then a+b = 0

[tribal moss]

And then ab = -aa = -a = b

[proud hedge]

guys

[coral spindle]

Darn

[tribal moss]

But with the same reasoning ab=a.

[proud hedge]

i found it

[coral spindle]

👀

Show us!!

[ivory ore]

then make everything 0

[coral spindle]

Nice reasoning btw tropo

[ivory ore]

oh well

[proud hedge]

2x2 matrices where the bottom row is zero and top row if from Z2

[ivory ore]

2a=2b=a+b=0

[coral spindle]

Oh that's ncie

[ivory ore]

did i break math here

[proud hedge]

it has exactly 3 idempotent elements

[coral spindle]

Wait, I'm not seeing them. What are the idempotents?

I can only see the 0 matrix

[ivory ore]

10
00

[coral spindle]

Oh you said Z2 not 2Z

My bad

[ivory ore]

guh

[chilly ocean]

thankyu sir

[proud hedge]

the zero element
1 0
0 0
and
1 1
0 0

[coral spindle]

Nicely done

Very nice indeed

[chilly ocean]

yeah i was working in Z

[coral spindle]

No you don't get to say this

If you haven't participated, you don't get to claim victory at the end.

[chilly ocean]

i never claimed

[proud hedge]

i have more questions if you guys are interested one of them is in rings and the rest are in groups

[coral spindle]

Sure! Sounds fun

[chilly ocean]

i said i was working in Z coudnt found in Z

did you read wrong

[tribal moss]

Why isnt (0 1 / 0 0) idempotent too?

[proud hedge]

no it becomes zero

[coral spindle]

It should square to zero right?

[ivory ore]

imagine if there were something like classification of rings

[proud hedge]

yup

[tribal moss]

Yeah braino for me.

[chilly ocean]

does there exist any commutative one with same?

Without identity 3 idempotent

[coral spindle]

There sort of is. For unital ones we can say a fair bit about the structure of a commutative ring of order n. But in general obviously it's hard

This is mostly because we have a classification of finite Abelain groups

I can't remember where I saw this...

[proud hedge]

Let R be a commutative ring such that it has n+1 zero divisors where n is a natural number.
prove R is finite and |R|<= (n+1)^2

[proud hedge]

i don't think so
i think i proved that it can't be commutative but i am not sure

[ivory ore]

wait this ring is not commutative?

i mean the 3 element one

confabulation

[proud hedge]

no it's not

i checked

[ivory ore]

just send me all of it

[proud hedge]

i'll send it don't worry

let's focus on this on i hadn't thought of it beforehand but it seems really hard

[ivory ore]

wait so what is ab and ba

[proud hedge]

i notice that if r (and element of R) is a zero divisor then for every r' of R, rr' is also a zero divisor

[proud hedge]

i thought u asked about my Z2 matrice example

[ivory ore]

oh yeah i did

sorry for not doing the calculation on my own

ab=b; ba=a

[chilly ocean]

this suggest i think then we can use orbit stabilizer on group R with multiplication and taking element thew zero divisior

for counting type prob orbit stabilizer is good

or burnside

[proud hedge]

i am sure we can prove it without orbit stabilizers

[rocky cloak]

Say 0, e and e' are the idempotents. Then ee' is also idempotent so equal to one of these 3.

Say ee' = 0, then e+e' is idempotent.

say ee' = e', then e-e' is idempotent.

[tribal moss]

In the first case e+e' might be 0. No wait, then my reasoning from before applies.

(Which could be said simpler: if e and -e are both idempotent, then -e = (-e)² = e² = e)

[rocky cloak]

There's a pretty elegant module theoretic argument here:

||Say x is a non-zero zero divisor||
||Consider the map R -> Rx given by multiplication by x||
||Let the kernel be K.||
||Note that K and Rx both consist of zero divisors so has at most n+1 elements||
||From the short exact sequence K -> R -> Rx you have |R| = |K|*|Rx| ||

[proud hedge]

shouldn't the map be a subjective homomorphism?

surjective*

i think it is not surjective

[rocky cloak]

Why do you think that?

[proud hedge]

because the map can't produce x or any other unital element

[rocky cloak]

I'm not sure what you're saying but the map is surjective by definition

Since the codomain is just the image of the map

[proud hedge]

no u r right i'm sorry

this is a very nice and sophisticated way of thinking about it u r really good at this man

[rocky cloak]

You can also make this into a noncommutative problem.

If R has n left zero divisors and m right zero divisors it has at most n*m elements

[proud hedge]

can you explain how it is a homomorphism?
f(r1r2)=r1r2x

[glad osprey]

It's a module homomorphism, not a ring hom

[proud hedge]

oh! i guess there is so much i need to learn

[rocky cloak]

You only really need it to be a homomorphism of groups

All your using it for is to limit the size of R

[south patrol]

Hm interesting, I want to think about this lol

[proud hedge]

i couldn't understand that sorryt

[proud hedge]

tell me if you come up with anything

[chilly ocean]

if r is zero divisor then |R|= |orb(r)||stab(r)| as u said r' r is also zero divisor so $|R|\leq (n+1)|stab(r)|$ now $x\in stab(r)$ so $xr=r$ then $x^k r=r$ by piegonholes we must have $x r=x^j r$ where j is bigger than 1 so $x(1-x^{j-1})r=0$ and hence x is zero divisor so $|R|\leq (n+1)^2$

[proud hedge]

i don't know about orbits or stabilizers but i know you can prove without them

[cloud walrus]

Akhi Mishra(Riemman's Cat)

[chilly ocean]

i think i am still missing one thing

if x^{j-1} is 1 we have to take more higher power

[rocky cloak]

I think you need to pass to modules for this.

I don't think orbit stabilizer really works just for actions by monoids

[chilly ocean]

also 1 is also in stabilizer(c r u)

shit

if ring has unity

[rocky cloak]

Okay, I guess it works

[chilly ocean]

some gaps there

this works for ring without unity

i cant conclude if ring has unity but without unity it looks fine

i just messed up every thing without consider unity and withou unity

[chilly ocean]

how Rx has most N+1 elements

[rocky cloak]

The assumption was that there were n+1 zero-divisors

[chilly ocean]

but why do we need exact sequences here then

R/K is isomorphic to Rx and |R|=|K||Rx|

[rocky cloak]

Yup

[chilly ocean]

i see i have to go for module

isnt it possible without module

[tribal moss]

You can pass all the way to viewing them as abelian groups if "module" is not an abstraction you're comfortable with.

(Oh, Jagr said that already)

[swift current]

Just wondering if anyone could read through my proof of the following proposition: Let $R \neq 0$ be an integral domain such that $R[x]$ is a PID. Prove that $R$
is a field.

Proof:

Let (0\neq a\in R). We will show that (a) is a unit in (R). Consider the ideal
[
I = (a, x) \subseteq R[x].
]
Since (R[x]) is a PID, (I) is principal, say (I = (d(x))) for some (d(x)\in R[x]).

Because (d(x)) generates both (a) and (x), we have
[
d(x)\mid a \quad\text{and}\quad d(x)\mid x
]
in (R[x]). In particular, writing
[
a = d(x),f(x),
\quad
x = d(x),g(x)
]
for some (f(x),g(x)\in R[x]), the second equation shows (\deg(d)\le 1). Moreover, if (\deg(d)=1), then (d) is of the form
[
d(x)= ux+v,
]
with (u\neq0). But then (d\mid a\in R) forces (u x+v) to divide the constant polynomial (a), which can only happen if (u x+v) is a unit in (R[x]). Hence (d(x)) must itself be a unit in (R[x]). Thus
[
I = (d(x)) = R[x].
]
Therefore there exist polynomials (p(x),q(x)\in R[x]) such that
[
1 = p(x),a + q(x),x.
]
Now set (x=0). We get
[
1 = p(0),a + q(0)\cdot 0
\quad\Longrightarrow\quad
1 = p(0),a
]
in (R). Hence (a) has an inverse (p(0)) in (R). Since (a\in R\setminus{0}) was arbitrary, every nonzero element of (R) is a unit. Thus (R) is a field.\

[cloud walrus]

rando

[chilly ocean]

R[X] is pid so <X> is maximal idealand R is field.

[chilly ocean]

nice

[crystal vale]

If H is a subgroup of G with a finite index n, is it true that g^n \in H for all g \in G ?

I know if H is normal then it is true but what if H is not normal?

[swift current]

I'm not sure if this is correct or not, but if you consider $S_3$ to be the permutations of length $3$, then the alternating subgroup $A_3$ is the set of even permutations and has index $2$. I'm not sure if it is normal or not, but if you take an odd permutation, and multiply by itself, then odd * odd = odd?

[cloud walrus]

rando

[crystal vale]

A_3 is normal subgroup

And odd * odd = even

[sonic coral]

A_n is normal in S_n for all in since it has index 2 as you mentioned. the parity of the product of permutations follows the parity of the natural numbers under addition, so odd * odd = even.

[chilly ocean]

try as i didnt check take Gl_2(z/3z) and subgroup of upper triangular matrices with diagonals all 1 this subgroup is of index and take g=[1 0, 1 1] then g^16=g not in the subgroup

[chilly ocean]

i know u checked

[crystal vale]

Isn't g^2 = id?

[chilly ocean]

but u asked g^n

wt

[crystal vale]

But if g^2 = id, then g^ 16 = id

[chilly ocean]

now check i have changed Z/3Z

[crystal vale]

Then what is the index of that subgroup?

[chilly ocean]

NO i actually made typo

i meant Z/3Z

now see index is 16

[crystal vale]

Great

Thank you ❤️

How did you come up with this counterexample?

[chilly ocean]

i was lucky

[crystal vale]

[swift current]

Why does a common divisor of two elements in the guassian integers divide the greatest common divisor of the norms?

[chilly ocean]

the only non trivial proper subgroups of 2*2 invertible upper trinagular matrices are non zero diagonals matrices and uniptent matrices and scaler matrices right?

[prisma ibex]

Depends on which field or ring you’re working over

[chilly ocean]

Take field

Infinite char 0

[prisma ibex]

There is a difference between what happens in the algebraically closed setting and the non-algebraically closed setting

Say for C versus R

[chilly ocean]

take R

[prisma ibex]

Ah sorry I missed the upper triangular part of this, the examples I have in mind where something funny happens shouldn’t appear in this case

This should probably be all of them then

This is certainly false for GL_2(R) itself, one has a non split torus over R but not over C

[rocky cloak]

Consider S3 and the subgroup C2

[rocky cloak]

The norm of an element is a multiple of it

[swift current]

why is that?

[rocky cloak]

By definition I guess

N(a + bi) = (a+bi)(a-bi)

[ivory ore]

from this case just out of curiosity i wanna ask is it any special phenomenon when right and left cosets both follow a cyclic permutation but in a dual fashion: x_i H = H x_n-i where n is the index

and we can setup this action even when n is composite right?

[rocky cloak]

I don't understand what you mean. Like you're picking a set of simultanious representatives for left and right cosets? And then you induce a permutation on them and ask about the cycle type?

You always have eH = He, so that can't be part of any cycle

[ivory ore]

yeah right

i was think of two cycles

[rocky cloak]

And besides when H isn't normal, then you have left cosets that are not right cosets, so I don't see what
xi H = H xn-i should mean

Like maybe write out what you mean for S3 and I'll understand

[ivory ore]

i think those indices should be on the power

but yeah let me write it out explicitly

[rocky cloak]

Okay, but the sets with equals signs are not equal

[ivory ore]

oh shit

[rocky cloak]

If they were the group would be normal

[ivory ore]

i was under the impression that even if they are not same they are same upto labelling

[rocky cloak]

If xH = Hy, then xH = Hx

Because a coset is generated by any of its elements and xe = x

[ivory ore]

oh okay, the confusion in the mind was basically appearing from coomuatie and then normal subgroups and not commutative and then normal sugroups

so if the factor sets dont form a group they are never equal, right?

such a shame that these basics sometimes make me feel uneasy

[proven shadow]

Hi, I'd like some help with the following problem

Let {e} = H_0 ◁ ... ◁ H_{r-1} ◁ H_r = H be a normal series with abelian quotients. Meaning for all 1 <= i <= r, H_i / H_{i-1} is abelian.

Show that every refinement of this series also has abelian quotients.

[rocky cloak]

I guess a hint could be that a quotient or subgroup of an abelian group is abelian

[vestal sapphire]

Hey, I want to show that for all natural k there is only one subgroup of order k in Q/Z. I've managed to show that there exists such a subgroup but have no idea how to show it is the only one. Could someone help me?

[south patrol]

You can work out what all the elements of order k are

[swift tundra]

Does anyone know if there has been any work done in exploring whether, for any irreducible polynomial over an integral domain, there is always a way to apply Eisenstein?

My thought was that, for the cyclotomic polynomial, you can’t directly apply Eisenstein because all the coefficients are 1, but you can take a clever ring isomorphism f(x) maps to f(x+1) and apply Eisenstein to the image of this polynomial.

Now I’m curious if it is possible to prove, given any irreducible polynomial f(x), there always exists a (perhaps just injective?) ring homomorphism such that the image is Eisenstein irreducible (i.e. satisfies the Eisenstein condition ). Does anyone know if there are any papers on this? Maybe there are already counter examples?

I guess actually we would probably still want a ring isomorphism

[south patrol]

See https://math.stackexchange.com/q/3398787/932674 - the result is false if you only allow isos given by shifting the indeterminate

So I guess actually any isomorphism lol since there aren't many automorphisms of Z[t]

[keen badge]

Let $\phi:R[x]\to R[x]$ be a ring automorphism where $R$ is a ring(say commutative with unity).
If $f(x) \in R[x]$ is irreducible then so is $\phi(f(x))$?

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[keen badge]

(If f is reducible then phi(f(x)) is reducible, that's for sure because we can take phi(g)*phi(h) where f=g*h)

[coral spindle]

Right, this is all true, but since you're assuming that phi is an automorphism, it has an inverse, so you can just run it the other way.

Suppose phi(f(x)) were reducible. Then phi^-1(phi(f(x)) = f(x) would be too

If you perhaps meant an endomorphism instead, rather than an automorphism, I think this is surely false, since for example phi : Z[x] -> Z[x] given by phi(x) = 2x should be an example

[keen badge]

yeah

[quartz wind]

did you mean automorphism here

[coral spindle]

No

OH

Lmao

I thought you replied to my other message

Yes indeed I did mean automorphism

Thanks

[keen badge]

wait

there is a theorem stating f(x) is irreducible iff f(x+alpha) is irreducible for all alpha.

Why can't I use this statement to prove that f(x) is irreducible iff f(x+alpha) is irreducible for SOME alpha

[coral spindle]

Well that is also true

Why do you think you can't prove this?

[keen badge]

oh lol

I thought it was false

[coral spindle]

It's true, unless I'm missing something hilariously obvious

Yeah, it's true by the automorphism observation

[keen badge]

thank you so much 🙂

[coral spindle]

No worries

[keen badge]

If I want to prove that f(x) is irreducible (say using Eisenstein's criterion) but a_0=1 then I would want to use what we discussed with some f(x+alpha).
Is there a point for this kind of questions?
For small deg(f), say deg(f)<4 we have that f is irreducible iff it has no roots. and if deg(f)>3 then it is just too much work to compute f(x+1).
Are there "useful" examples for this approach?

[coral spindle]

Sure there is. I think the best example is the polynomial f(x) = x^n + x^{n-1} + .... In this case, f(x+1)... ok jagr will explain I am with my friends lol

[rocky cloak]

I guess it's worth notice that you really only need to compute f(x+1) module the prime p

[keen badge]

But if the degree of the polynomial is 4 or 5 than the binomial has a lot of computations to calc.

[rocky cloak]

Right, but there's no need to calculate it is what I'm saying. Just determine if it's divisible by p

[keen badge]

I don't see how it is possible to compute it mod p before computing it regularly

[rocky cloak]

Well, cyclotomic polynomials are a good example I guess.

f(x) = 1 + x + ... + x^p-1 = (x^p - 1)/(x - 1)

Then
f(x + 1) = ((x+1)^p - 1) / x
freshman's dream tells us that (x+1)^p = x^p + 1^p, so
f(x+1) = x^p-1 modulo p

f(0+1) = f(1) = p, so f(x+1) is irreducible by Eisenstein, and we never calculated what it is

[keen badge]

ohh ok thx

[keen badge]

f(x+1)=((x+1)^p-1) / x =(x^p+1^p-1) / x=(x^p+1-1) / x=x^p / x=x^(p-1)
how'd you get x^p-1?

[rocky cloak]

I'm a fan of spaces
x^p-1 = x^(p-1) =/= x^p - 1

Besides if it was x^p - 1 it would not satisfy Eisenstein

[keen badge]

I thought the other way around.

how x^(p-1) gives us Eisenstein

[rocky cloak]

That's the criterion for Eisenstein. All nonleading coefficients should be divisible by p

[keen badge]

Oh, and the free variable(a_0) of f(x+1) is just f(1). ok cool

[keen badge]

If you like spaces, I think you'd like Typst.
(I mainly use it now instead of LaTeX)
https://typst.app/

[swift current]

what primes $p \in \mathbb{Z}$ remain irreducible in $\mathbb{Z}[i]$?

[cloud walrus]

rando

[velvet hull]

Exactly the primes where p = 3 mod 4

One direction is not hard, but showing that p = 1 mod 4 is never irreducible requires some subtlety

[chilly ocean]

coudnt get this i think i should watch pogo

[vivid kestrel]

Given a finite group G of order n and K the n-th cyclotomic field all character values of complex representations are sums of n-th roots of unity so contained in K. Is it then also true that every complex character can be obtained by a representation over K instead of C?

[rocky cloak]

This is true, but maybe not "then also true".

If all character values appear in F that does not mean every complex representation is realized over F. For example all characters of the Quaternion group are rational.

[vivid kestrel]

so the field containing all the character values doesn't suffice but the n-th cyclotomic field does? sorry not sure im parsing your message correctly

[rocky cloak]

Yes, that's what I'm saying

[vivid kestrel]

how does one see that the latter statement holds

[rocky cloak]

With lots of effort is the answer I think. theres a proof here if you can follow it.
https://groupprops.subwiki.org/wiki/Sufficiently_large_implies_splitting

Im thinking Boytjie and Wew probably knows about such things

[coral spindle]

This is unfortunately pretty tricky to show, I can't remmeber quite how the proof goes, but this is a celebrated theorem of Brauer

(indeed there are many such things)

You should think of the general idea as something like this: to split, we need End_kG(M) to be k for all simple modules. The argument in Schur's lemma for this to be the case requires the existence of a lot of eigenvalues. So at the very least if our field has a lot of irrationals, then Schur's lemma might hold anyway. The theorem of Brauer just tells you which eigenvalues we need.

[coral spindle]

Now the observation about character values is a little deeper

It is perhaps an unsurprising, but certainly a nontrivial fact, that there are representations that cannot be realised (i.e. given as a matrix) over the field that its character values generate.

The smallest example is the unique 2-dimensional irrep of Q_8. Its trace takes real values, but it cannot be realised over the reals.

In fact every character of Q_8 takes rational values.

So while it is not a trivial fact -- it's not a fact that we can just see abstractly -- it is indeed the case that a field generated by character values is insufficient for the realisation of those representations, hence the algebra would not split

[rocky cloak]

Hmm, is it known which division algebras appear in the decomposition of QG into matrix rings?

[coral spindle]

Great question

I don't know!

But I don't see any reason that there should be a restriction

It's easier if we look at local fields, since then the division algebras are parametrised by Br(Q_p) = Q/Z

[rocky cloak]

Well, they would need to be split by a cyclotomic field

[coral spindle]

Hm good point

[rocky cloak]

Doesn't sound like it should be true for every division algebra, but maybe it is....

[coral spindle]

Wait let me remind myself of something

OK so over Q_p, the Schur index always divides p-1. That is, the order of the division algebra in Br(Q_p) divides p-1. So in fact there is a very small subgroup of Br(Q_p) that can be achieved by group reps.

So ok, there's the answer. In fact it's a very small number. Whoops!

I am not clear on the details but it should be the case that the answer over Q looks like a tensor of things over Q_p

This is sort of funny to think about. I'm so fixated on the numbers that I haven't thought about what the algebras look like for so long

[rocky cloak]

So for Q8 for example, does the rational quaternions appear?

[coral spindle]

I believe so yes, they should be the rational quaternions

I'm pretty sure the calculations in R carry over to Q without a hitch

You can write down the unique 2d simple module of Q[i]Q_8, project down to get the unique 4d simple module of QQ_8, and then crunch the numbers

I think if you do so you get the obvious way to write down H as 4x4 real matrices, but it's been ages

[rocky cloak]

I guess there's also an obvious surjective ring homomorphism QQ8 to rational H

[coral spindle]

Ah true, good point

Not totally easy to see what the kernel is immediately

[rocky cloak]

Anyhow, so you have H, and you have all the cyclotomic guys.

Anything else that can appear?

[coral spindle]

There ought to be some generalised quaternion things

but frankly, I haven't checked

[rocky cloak]

Combinations of those maybe, like extending H by some odd roots of unity?

[tribal moss]

But there are plenty of odd roots of unity in H already.

[rocky cloak]

By H I mean rational H

[tribal moss]

Oh.

[rocky cloak]

Q[i, j, k]

[coral spindle]

If we are looking for algebras other than the rational quaternions, I think looking at representations of nonsplit extensions 1 -> C_a -> G -> C_b -> 1 would be a way to find examples

Non-Abelian ones of course

Specifically this is a way to get a whole load of nontrivial Schur indices for free. I think if we force b to be odd this would be fruitful

I will give these a look and maybe try to find an interesting example

This is something I should do anyway!

[tribal moss]

(Is this really for #groups-rings-fields rather than #advanced-algebra, btw?)

[coral spindle]

Haha true, I got sniped a bit

I'll leave it here for now in any case and maybe put an update in advanced

[tribal moss]

By the way, I think (-1+i+j+k)/2 ought to be a third root of unity.

[vivid kestrel]

i see, thanks jagr and boytje

[ivory ore]

If $(a, b) = 1$ and $K$ and $Q$ are abelian groups of orders $a$ and $b$, respectively, then there is only one (to isomorphism) abelian extension of $K$ by $Q$.

[cloud walrus]

longboard kayak

[ivory ore]

this can be done from the classification of abelian groups and chinese remainder theorem right ?

[tardy hedge]

in the tensor product why do we have the ar (x) b = a (x) rb condition?

[velvet hull]

the goal of the tesnor product is to convert bilinear maps into linear maps

we want r [a (x) b] = [ra (x) rb], because that's linearity, but bilinearity forces r [a (x) b] = [ra (x) b], and that's no good
see below

[tardy hedge]

oh because the constant "r" times a or times b should be the same result out of the bilinear map

not really sure what u mean by we want r [a (x) b] = [ra (x) rb], because that's linearity

i guess what does linearity vs bilinearity really mean?

[velvet hull]

well, bilinearity is not linearity

so none of the module theory (or linear algebra) applies to Bilin(VxW)

but then that's a problem, because the entirety of the theory is built off of using linear maps between modules to study the modules

so what that means is Bilin(VxW) becomes really annoying to study, because how does it even relate to other modules

[tardy hedge]

"we want r [a (x) b] = [ra (x) rb], because that's linearity" why is that linearity?

why not r(a(x)b) = (ra(x)b)

[rocky cloak]

If you think about the map
VxW -> V(x)W
that sends (a, b) to a(x)b
then r[a(x)b] = ra(x)rb would be saying the map is linear

But it is not, the map is bilinear (or at least R-balanced if R is not commutative)

[tardy hedge]

The domain is just cartesian product right so why write r(a(x)b))

[rocky cloak]

r(a(x)b) means the action of r on the element a(x)b

[tardy hedge]

Not really sure what im confused about atm lol

[tribal moss]

I think HChan typoed a negation too little here and that confused the following discussion; surely he meant "we dont want ..."

[velvet hull]

yeah I think that's what got me confused

[tardy hedge]

What does it mean for a map to be linear? A map with more than one input can be linear instead of bilinear?

[rocky cloak]

The product of two modules as again a module

So it makes sense to ask if a map from VxW is linear

[tardy hedge]

Ok

[rocky cloak]

Though I guess it's also fair to consider it a type error if you like.

Since when talking about bilinear maps on VxW the module structure doesn't come up

And indeed if R is noncommutative it doesn't have any module structure (since you're mixing right and left modules)

[tardy hedge]

Bilinear maps on VxW is like linear in each variable, separately, right?

[tardy hedge]

Because this means its linear in both slots simultaneously?

[rocky cloak]

Yeah

[tardy hedge]

Okk that makes a bit more sense i think

[swift tundra]

Hmmm, but they only account for shifts. I could be wrong but I think another valid isomorphism is flipping the polynomial (constant becomes leading coefficient, and so on). This gives a bit more flexibility.

[rocky cloak]

This is not an automorphism. You'll run into trouble when the constant coefficient is 0.

The only automorphism of Z[x] are given by mapping x to ±x + a

[tribal moss]

If the constant is zero then your polynomial is certainly not irreducible...

I think reversing the order is a valid automorphism of the multiplicative monoid of polynomials with nonzero constant term.

[swift tundra]

Oh wait yeah 🤦‍♂️

[swift tundra]

Interesting

I was thinking that f(x) is irreducible if f(1/x) is “irreducible “ meaning the numerator splits after some manipulation. And f(1/x) essentially has the effect of reversing the order of the coefficients

[tribal moss]

It doesn’t extend to a ring homomorphism (as far as I can tell), but you only need to preserve the multiplicative structure to fix the irreducibles anyway.

[swift tundra]

Yeah totally. This is interesting and something I should think about more. I guess we really do just want isomorphisms of this monoid.

[south patrol]

Boom

[swift tundra]

Yeah free R algebra on 1 generator I should have known lmao

[rocky cloak]

I guess if we go there, then the multiplicative monoid is the free commutative monoid on the irreducible polynomials, snce Z[x] is a ufd.

So there is an automorphism taking any irreducible to any other

[swift tundra]

Oh yeah that is quite interesting

[tribal moss]

Yeah, but for the purpose of concretely identifying irreducible polynomials, knowing some easy concrete automorphisms seems to be an improvement over the nonconstructive knowledge that they're out there somewhere ...

[swift tundra]

It seems like we lose too much structure by passing to the monoid, because then I can always send an irreducible polynomial to a p-Eisenstein polynomial assuming it exists. It would be nice if there was still some way to “respect” addition somehow. But I don’t know what such a notion would look like.

[tribal moss]

Hmm, how is that losing "too much" structure? Too much for which purpose?

[swift tundra]

I mean in this case we would still have to prove that the polynomial is irreducible before we can even claim that we have an isomorphism of monoids.

I still think that the morphism should respect the arithmetic structure of the polynomial somewhat. For example, mirroring the polynomial (assuming it is an isomorphism) still depends on the actual coefficients. But sending an irreducible polynomial to any other one doesn’t care about the coefficients

[tribal moss]

Oh, I've found your original question now. Yeah, allowing arbitrary monoid automorphisms trivializes the question of whether a given irreducible is Eisensteinable.

[rocky cloak]

I guess it's still a question of how much is needed for "Eisensteinabilization".

So far we know that the automorphisms f |-> f(x + a) are not enough. But what about all compositions of ring automorphisms and this funky flip the terms map.

What about all injective ring homomorphisms with this. Any other natural automorphism of the monoid?

Not that it really seems feasible to answer this question, but it's interesting

[swift tundra]

What is meant here by a “natural automorphism” of the monoid? Is this the idea of respecting additive structure, or do you mean the more categorical notion?

I definitely want to think about more, but yeah it doesn’t seem feasible to answer right now haha.

[rocky cloak]

I mean something when I look at it I can think "yeah that seems natural, I could have thought of that"

Curse mathematicians for ruining the English language I guess

[thorn jay]

If someone says a word that occurs in math it's like an instant neuron activation

https://tenor.com/view/neuron-activation-neuron-gif-1455356329209493514

[swift tundra]

Lol so true

[minor fulcrum]

I once had someone asked me if natural transformations are related to natural numbers

[thorn jay]

[mighty kiln]

They are; a natural number is an endo natural transformation of the forgetful functor Mon → Set

[thorn jay]

Because of course

[velvet hull]

True, though I doubt the natural numbers were name after that particular embedding

[thorn jay]

No it was obviously
Natural numbers -> that transformation -> natural transformation

[velvet hull]

Know what you’re right

[thorn jay]

Chronologically it makes more sense

[frigid shard]

In the sense that n corresponds to the nat whose components add the identity n times to each element of the monoid?

I wonder if this provides an alternative axiomatization of N btw as opposed to Peano

But surely there are endo-natural transformations other than these ones

[mighty kiln]

No, x → x^n

The forgetful functor Mon → Set is isomorphic to Mon(N, -)

So by Yoneda N = Mon(N, N) = [Mon, Set](F, F)

[frigid shard]

Oh crap the one I gave is not natural lol

[frigid shard]

Yeah

[mighty kiln]

You can replace Mon and N with any category of algebraic objects where "freely generated by one element" makes sense

Tao has a blog about the CRing and Z[x] case I thinkg

[frigid shard]

Ic

[swift tundra]

But this is not a canonical relationship …

[coral spindle]

Very cool

[tropic obsidian]

is the integral closure of a number field (finite extension of Q) always a dedekind domain? edit: every number ring is a dedekind domain so nevermind

[rocky cloak]

An integral domain is a Dedekind domain iff it is integrally closed of dimension 1

[vague canopy]

how can i show, that k[x,y]/(x*y-1) is not isomorphic to k[t]? k is a field.

i tried to show, that k[x,y]/(x*y-1) is not a PID, but i failed.

I also tried to compare the unit-groups. I got units(k) vs. units(k) x Z.
Is this the right way? Or are there fields k, such that there is an isomorphism between the groups?

[tribal moss]

k[x,x^-1] contains a unit u such that u-1 is neither 0 nor a unit.

[south patrol]

Lol nice

[obtuse merlin]

i need help

[frigid shard]

wrong channel

[obtuse merlin]

which channel then

[frigid shard]

in what context are you doing it? is it schoolwork? competition math?

[obtuse merlin]

like math ig

[frigid shard]

?

[obtuse merlin]

just random problems my froend sent me

so idk?

for fun to test what ever this is

though it looks easy

[frigid shard]

ic

if it's informal like that maybe try #math-discussion

[obtuse merlin]

alright thank you

[chilly ocean]

are you guyz inventing eisestinian ring

[vague canopy]

thank you for the tipp!

[chilly ocean]

but that is a pid

[vague canopy]

i already have the next question: How can i show, that a noetherian Ring only has finitely many minimal prime-ideals?

I found out: the intersection of all prime ideals are exactly the nilpotent Elements. (or sqrt(0)).
=> the intersection of all minimal prime ideals is sqrt(0) too. But how can i conclude, that it is a finite-intersection?

Sorry for the second question... But the homework is really hard this time xD

[rocky cloak]

So I don't know how much you know about commutative rings necessarily, but you can consider the union of all minimal primes and localize with respect to its complement.

Then you get a 0-dimensional Noetherian ring

[vague canopy]

thank you very much! I will try that.

[long obsidian]

An ideal J in C[x] is a vector subspace of C[x] right?

[chilly ocean]

can anyone clear the proof from line 2

nakayma lemma

[next obsidian]

If M is finitely generated it has a cyclic quotient:

Proof by induction on minimal number of generators, to do inductive step mod out by an element part of a minimal generating set.

So now M has a cyclic quotient which is isomorphic to R/I, so for a maximal m > I, the quotient R/m is a simple quotient of R/I, so now chain these together to get the simple W of the proof. From there proof is clear

[chilly ocean]

that I must be annhilator

[next obsidian]

Yeah

[chilly ocean]

so W is just R/m

[next obsidian]

All simple modules are of that form

[chilly ocean]

JW is 0 follows by simplicity of W

[next obsidian]

Yah

[chilly ocean]

if JW=W then WHATS THE PROB

[rocky cloak]

JW = JR/m < m/m = 0

(because J < m)

[chilly ocean]

oh forgot to use that

[swift tundra]

Do any of y’all know a good place to learn about monoid rings?

[chilly ocean]

Prove that sinx/x is irreducible

[chilly ocean]

i think you meant truncated poly of sin x/x

[coral spindle]

What are you looking to learn about them exactly?

[swift tundra]

Just some of the general theory. I’m particularly interested in whether this is a good path to explore irreducibility in polynomial rings.

[proven helm]

this is a bit stupid and simple, but when talking about a localisation, in the definition $(a,s)\sim (b,r) \Leftrightarrow \exists c \in S : c(ar-bs)=0$ why is the c necessary? couldn't it just be the equivalence relation as above without the c?

[cloud walrus]

aeiou

[dim widget]

if s \in S is a zero-divisor then we want (s, 1) \sim (0, 1) since we are supposed to be "inverting" elements of s and this should be impossible if s is a zero divisor. So this explains why the condition is necessary.

[proven helm]

ohhhhh yeah :P
thank you! ^^

[dim widget]

and this is kind of universally what this extra c is doing: we want to say (a, s) \sim (b, s') if as' - bs is a zero divisor with something in S

[proven helm]

got it, so the c is essentially a catch for zero divisor funkiness

[dim widget]

yep

obviously if S contains no zero divisors it is a vacuous condition

and you can ignore the c

[proven helm]

yeah, that was my issue :P
i need to get back into my abstract algebra mindset lol

[toxic zephyr]

if we have a field extn. K(zeta)/K, zeta a primitive nth root of unity, then is the galois group isomorphic to the mult. gp of Zn? prof said it was to a subgroup but wasnt tusre it's surjective. i feel like it probably is, but just looking to confirm. like if r is a unit in Zn^*, should sigma(zeta)=zeta^r be in the galois gp?

[dim widget]

Well for instance \zeta could already be contained in K?

then the galois group would be trivial

If you look at Q(\sqrt(5)) = K then K(\zeta_5) has galois group Z/2 and not Z/4 as you might expect

the ultimate point is that if K is galois then Gal(Q(\zeta_n) \cdot K/Q) is the set of elements in Z/n^* \times Gal(K/Q) such that they agree on the intersection of the two fields. The galois group over some Q \subset K' \subset K you can then calculate using basic galois theory, which tells you what's going to happen in general. But it's true that you just get some subgroup rather than all of Z/n^* in general.

[spice wagon]

Are we more generally using the definition of an action by a function $\varphi : G \times X \to X$ such that it satisfies certain properties, or rather having a homomorphism $\theta : G \to Sym(X)$?

[cloud walrus]

tm

tm

[candid patrol]

Second one

The morphism G --> S(X) is pretty usefull

[spice wagon]

ah rip i prefer the first one..

so $\theta(g)(x)=\varphi(g,x)$

[cloud walrus]

tm

[coral spindle]

These are so easy to convert between that I barely think of them as being different usually. However:

1. Often it's helpful to consider the image of G in Sym(X) so that is often the more helpful one in a lot of group-theoretic arguments.
2. When we are generalising the idea of a group action, for example for algebraic groups, or Lie groups, or profinite groups etc, we usually use the G x X -> X definition since this is more easy to translate category-theoretically, whereas Sym(X) often doesn't have a nice translation.

So the answer is "no"

:P

[spice wagon]

no for this ?

[coral spindle]

The answer is no, we don't prefer one over the other

[spice wagon]

ah ok lol mb

thanks

[coral shale]

Well I prefer the latter ! 😭

[spice wagon]

to show that $G$ operates faithfully on a set $X$, we need to show that with a single homomorphism of $G$ in $Sym(X)$, we obtain an injective homomorphism right ?

[cloud walrus]

tm

[spice wagon]

or all homomorphisms from G to Sym(X) must be injective

ah no i’m dumb, it’s the associated homomorphism that must be injective 🤦‍♂️

[coral spindle]

Well this is of course false for all groups except the trivial one, so this should be a good pointer

[spice wagon]

yeah i’ll keep that in mind 👍

[spice wagon]

how can we describe geometrically the fact that a group does not operate faithfully on a set?

[cloud walrus]

tm

[spice wagon]

but i have no geometric intuition 🥲

[tribal moss]

I don't know if it's "geometric" but "there is a group element other than the identity that leaves every point of X in place" seems to be a fairly intuitive notion.

[spice wagon]

so we can say that the points of X are out of "order" under the action of G?

[tribal moss]

Huh, that doesn't sound like a way I think "not faithful group action" could be described.

[dire bramble]

I think a good way to think about it is a faithful action gives an embedding of the acting group into the automorphisms of your object

I.e. G is "faithfully" represented as automorphisms

[spice wagon]

ahh okay yeah

[dire bramble]

Like, you can literally think of G as a subgroup of symmetries of X. That is a legitimate characterization

[spice wagon]

hmm ill need more training lol

[tribal moss]

A not faithful action is if there are distinctions between the elements of G that are invisible if all you can observe about them is how they act.

[arctic trail]

Me loving the permutation representations

[spice wagon]

so if it’s faithful, you can distinct the elements of G from their "form" and not only their actions ?

idk if i understood 😭

[tribal moss]

I'm not sure what "form" means here.

[spice wagon]

like their structure

idk how to say it

[tribal moss]

Perhaps another way to express it is to say we play a game where I think of a secret element of G, and you get to repeatedly ask me what the result is if I operate on such-and-such element of X.
If there's a way for you to figure out for certain what my secret group element is just by asking such questions (perhaps infinitely many of them), then the group action is faithful.

[arctic trail]

that seems undecidable

[tribal moss]

I'm not talking about computability here.

[arctic trail]

I know

[spice wagon]

hm okay but I can’t make the connection between knowing the element g through its actions and the fact that only the identity has to act trivially on any element of X for it to be a faithful action

[glad osprey]

If G acts faithfully on S, then different elements of G permute S in different ways. In other words, each element acts in a unique way. Does that make sense?

[spice wagon]

by permute you mean stabilize ?

or leaves invariant

smth like that

[tribal moss]

Suppose g and h are two group elements that both act in the same way. Then the product g^-1 h must be an element that acts by doing nothing, and g^-1 h is not e because I was assuming g and h are different elements.

[glad osprey]

No. Sym(S) is the set of permutations of S. A group action is a way to assign a permutation to each element of G in a way that respect the group operations

[tribal moss]

So if you have any two distinct elements that act the same way, then you also have a non-indentity element that does nothing.

[glad osprey]

G acting faithfully on S is equivalent to the group hom phi : G -> Sym(S) being injective, which in turn is equivalent to it having trivial kernel

[spice wagon]

its when the group acts faithfully?

[spice wagon]

yep but what is hom? 🥲

[glad osprey]

Homomorphism

[tribal moss]

When the group acts faithfully there are not two distinct elements that act the same.

[spice wagon]

ah ok yes

[spice wagon]

ok i think i got it

[rocky cloak]

A group action not being faithful just means there is some non-identity element that fixes everything.

In your example, scaling is a linear transformation that doesn't move any of the lines. So the action is not faithful

[spice wagon]

your analogy was very good thanks

yes i took $\lambda id \in GL(E)$ to show it acts trivially on every lines of E

[cloud walrus]

tm

[spice wagon]

i think I’m going to rest a bit rn lol thanks everyone for ur help

[limpid hemlock]

hi. what all do you need to know to start learning group theory? does it stand on its own without a wider knowledge of general abstract algebra?\

please @ me if / when i get an answer

[barren sierra]

Pretty much any decent abstract algebra textbook will start with group theory

I like Artin as a first abstract algebra text (although this is a counter example to what I just said, it starts with linear algebra which you can skip if you already know it)

I see Judson's text recommended alot as a first text

Dummit and Foote's textbook is also highly recommended, and very standard for most university courses to use as a textbook

[limpid hemlock]

Im trying to learn specifically group theory, but if algebra textbooks typically start with it, then sounds like im good to go

[barren sierra]

yea you should start there and if you learn more algebra you can later go into specific group theory texts and get recommendations (for which other people will be much more knowledgeable than me)

[limpid hemlock]

awesome

thank you for your answer

[tardy hedge]

That was nice

[crystal vale]

Yes

[vocal pebble]

regarding the prerequisites, not much so basically like what you said

[spice wagon]

when we say p-Sylow, we mean the p-subgroup of Sylow?

[candid patrol]

Yes, accepting p divides |G|

If |G| = \prod_{k=1}^rp_k^{m_k}$, then $G$ has a $p_k$-Sylow for each $1 \leq k \leq r$ of order $p_k^{m_k}$

[cloud walrus]

UGOBEL
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[woeful sage]

and for a particular p_k they are all nested within each other right?

like the subgroup of order p_k^{m_k - 1} is a subgroup of the subgroup of order p_k^{m_k} and so on

[candid patrol]

That's true, but it doesn't follow directly from Sylow. Your $G := p_k$-Sylow is, in particular, a $p$-group since it has order a power of $p$ ($|G| = p_k^{m_k}$). A classic exercise on $p$-groups is precisely to show that for $1 \leq s \leq m_k$, $G$ has a subgroup of order $p_k^{s}$, using the fact that the center of a $p$-group is nontrivial.

[cloud walrus]

UGOBEL

[rocky cloak]

If P < G is a subgroup of order p^k and p^k+1 divides |G|, then there is a subgroup P' of order p^k+1 containing P as a normal subgroup

Follows from Cauchy's theorem and the fact that
[N(P):P] = [G:P] mod p

[amber burrow]

do you think that the classification of finite simple groups would be a good thing to convey to martians to represent the human race?

[rocky cloak]

Not really

[tardy hedge]

Lol

random question and funny answer

[chilly ocean]

unexpected reaction

[slim kayak]

It is certainly an impressive personal achievment, maybe not the best to flex with

[amber burrow]

what would you flex on the aliens with?

[vocal pebble]

depend on how advanced they are, ig

[amber burrow]

suppose you have no idea

[thorn jay]

Amazing pfp

[sonic coral]

if we flexed that then they’d probably laugh and say that their spawn could solve our inverse galois problem

[long obsidian]

Can someone help me find the minimal x^a and y^b such that both are in the ideal J=<x^3y,xy^2>? My thought process is that this ideal corresponds to the union of the coordinate axes of the Cartesian plane. I think this has to do with the radical of J but I’m not sure

[trim wind]

They are never in the ideal

Every element of J is divisible by y, but x^a is not

[rocky cloak]

The radical of J is <xy> which geometrically looks like there union of the coordinate axis.

So you'd expect f^n to eventually be in J iff f vanishes on both axes. Which neither x or y by itself does

[long obsidian]

Thanks I was confused idk why I thought that lol

[tardy hedge]

Is As = 0 (localization at S) iff 0 in S? Or is it only if 0 in S

[valid fox]

iff

[tardy hedge]

If localization of a module is 0, does that mean union of all Ann(m) for m in M intersect S is nonzero?

[tardy hedge]

Oh yeah cause 1

[wraith cargo]

I think it's enough to have a zero divisor in S

since then every element a/s is equivalent to 0/s' where s' is the zero divisor

[tardy hedge]

I think that just makes the natural map a -> a/1 not injective

[tardy hedge]

Oh yeah

[wraith cargo]

I think this condition isn't strong enough

What you need is that Ann(M) cap S != 0

since what could happens is that for example you have UAnn(m) cap S != 0
But that means only some elements are annihilated by elements of S
And there could still exist nonzero m/s that can't be equivalent to 0/s' because s'm can't be annihilated by anything in S

[spice wagon]

If $N \triangleleft G$, do we always have this exact sequence $1 \to N \to G \to G/N \to 1$?

ah

[tardy hedge]

Also the element that the zero divisor kills also needs to be in S right

need ss’ = 0 for s , s’ in S

[cloud walrus]

tm

[wraith cargo]

yes

[spice wagon]

okay thanks i wasn’t sure 👍

[tardy hedge]

If As = 0, it implies S must have zero divisors ? @wraith cargo

[wraith cargo]

I believe so yes

[tardy hedge]

I see why if S has zero divisors s and s’ with ss’ = 0, then As = 0

I dont get why A_s = 0 means S has nonzero s and s’ with ss’=0

I am comfortable with As = 0 iff 0 in S but now this zero divisor stuff has got me all mixed up lol

[tardy hedge]

Ok yea so im confused on this part

[spice wagon]

do sub-group lifts only appear in the case of semi-direct products?

or it can be defined in a general case

[delicate orchid]

You can lift many things from subgroups. Except hope and wonder

[spice wagon]

lol its because i think i don’t have the general definition 🥲

given $G$ a group and $N \triangleleft G$, we have a lift of $G/N$ if there exists a subgroup $H$ of $G$ such that $p_{|H}$ induces an isomorphism of $H$ onto $G/N$.

[cloud walrus]

tm

[elfin wraith]

This is true of all groups

Stick a monoid on there

[delicate orchid]

I think you're asking if N has a complement, which is indeed only true if your extenstion is split. If that's not what you're saying then I have no idea what this means

[spice wagon]

yeah it’s not what i’m saying but nvm i’ll continue with my definition

[tardy hedge]

Ik this is extremely uninteresting but …

[south patrol]

1/1 = 0/1 so there are s,t with 0 = st(1*1 - 0*1) by definition

I guess I assume here that s and t are nonzero, but like if you allow 0 to be in S then there is the counterexample of S = {0,1}

[wintry sluice]

Do notions of ON bases exist in modules? orthogonal projections?

[dim widget]

to define those things you need some kind of orthogonal or hermitian form

if you had some version of that you could make such a definition for a fairly general ring

[proven helm]

a bit of a stupid question here, i'm looking at ideals of polynomial rings (K[x,y] in particular here) and I've got to this process in a particular note:

$(x + x^2y + 3y^2 + x^4, y + 2y^3 + y^4 + 4x^7, x^2, xy, y^2) = (x, y, x^2, xy, y^2) = (x,y)$

are these valid as "processes" because, say in the case of $x + x^2y + 3y^2 + x^4$ "becoming" $x$, every other term there is a multiple of $x^2, xy, \mbox{ or } y^2$?

[cloud walrus]

aeiou

[proven helm]

i say "processes" and "becoming" because ik they'd already be equivalent and there's no real change going on under the hood

[velvet hull]

x + x^2y + 3y^2 + x^4 = x (mod x^2, y^2)

y + 2y^3 + y^4 + 4x^7 = y (mod x, y^2)

[proven helm]

yeah, but what i mean is like, that's reason enough to warrant "reducing" them inside the ideal?

[velvet hull]

well they are equivalent

[proven helm]

wait yeah of course i'm dumb

[proven helm]

thank you lmao

[south patrol]

The ideal (a1,...,an) is the smallest ideal containing a1,...,an and you can prove these manoeuvres from this fact

In particular a ideal contains a and b iff it contains a and a + b, so (a,b) = (a,a +b), and so on

[proven helm]

damn, so many more things make sense now. i was getting hung up on "proper reasoning" but it's just a matter of keeping it simple

at a base level i understand that ideals and R/I behave like modular arithmetic but the application part of my brain is still playing catch up :/

[spice wagon]

i have to prove that $Aut(Z/nZ) \cong (Z/nZ)^{*}$

[cloud walrus]

tm

[spice wagon]

do i need to construct a isomorphism between these 2 things ?🥲

bc i think i can’t use the first isomorphism theorem or things like that

[velvet hull]

oh, you can, but if you want to use the first isomorphism theorem you need some kind of group homomorphism to start with

[spice wagon]

I don’t see how I could form $(Z/nZ)^{*}$ using a quotient $G/Ker(\varphi)$

[cloud walrus]

tm

[violet spade]

Maybe just try seeing what an automorphism of (Z/nZ) looks like, and then try to guess what the isomorphism is

[velvet hull]

here's a hint: ||any homomorphism is uniquely determined by the image of a generating set||

[spice wagon]

$\varphi : \psi \mapsto \psi(\overline{1})$ with $\psi \in Aut(Z/nZ)$ ?

[cloud walrus]

tm

tm

[violet spade]

Okay but that's begging the question; what does psi look like?

[spice wagon]

$\psi(\overline{1})=\overline{k}$ with $gcd(k,n)=1$

[cloud walrus]

tm

[violet spade]

Yes that looks good

[violet spade]

Wait a sec.

[spice wagon]

and $\psi(\overline{m})=m \psi(\overline{1})=\overline{mk}$

[cloud walrus]

tm

[violet spade]

No that looks good indeed.

So we need the gcd property because it has to be a generator

Ofc

Yeah so now you're basically there already, do you see how you could construct an isomorphism now?

@spice wagon

[spice wagon]

yes

[spice wagon]

like that

[violet spade]

Yes that's it

So yeah. Now you only need to check it satisfies injectivity and surjectivity

[spice wagon]

yeah

[violet spade]

and of course, is a homomorphism, which is obvious

[spice wagon]

$\varphi(\psi_1)=\varphi(\psi_2) \iff \psi_1(\overline{1})=\psi_2(\overline{1}) \iff \psi_1(\overline{m})=\psi_2(\overline{m}) \ \forall \overline{m} \in Z/nZ$

[cloud walrus]

tm

[spice wagon]

$\psi_1=\psi_2$

[cloud walrus]

tm

[violet spade]

Yes because it was uniquely determined by the image of 1 this is valid.

[spice wagon]

yes

[violet spade]

Now only surjective which is not difficult either

Note that you could also first prove surjectivity of $\varphi$, and then prove that $\varphi$ must be injective because $|Aut(\mathbb{Z}/n\mathbb{Z})| = (\mathbb{Z}/n\mathbb{Z})^*$

[cloud walrus]

joel

[spice wagon]

i can do also injectivity -> surjectivity ?

bc they have same cardinal

[tribal moss]

Do you know they have the same cardinality, though?

(They do, but the proof of that fact that comes comes to mind for me is just to show from first principles that this mapping is a bijection).

[spice wagon]

no 😢

[spice wagon]

for the map u talk about $\varphi$?

[cloud walrus]

tm

[tribal moss]

Yes.

This is easy enough that there's a risk of overthinking it; you just need a few sentences to state that for every x in (Z/nZ)* theres an automorphism psi such that psi(1)=x.

[spice wagon]

for the surjectivity i take $\overline{k} \in (Z/nZ)^{*}$ and then we take $\psi(\overline{m})=\overline{mk}$

[cloud walrus]

tm

[tribal moss]

Yes.

[spice wagon]

okay 👌

but do we have other methods than just guess the isomorphism?

[tribal moss]

(You should probably have an argument that this phi is actually an automorphism of Z/nZ).

[tribal moss]

No.

There's no procedure to learn here.

[spice wagon]

ah

thanks for your help 🙏🏼

[tribal moss]

(At least not one where "guess the procedure" is any simpler than "guess the isomorphism").

[long obsidian]

Can someone confirm whether (c) can be done by saying (b) implies the ring quotient R/M is a field and hence M is maximal. Also no element of M is invertible so if M’ is a distinct maximal ideal from M then R\M’ cannot consist entirely of units like from part (b) because it would intersect M? Basically I think uniqueness in (c) can be proved by this contradiction

[storm kiln]

I think that works, another way to show that M is maximal is to say, if there is an ideal strictly containing M it must also contain a unit and would therefore be equal to R

[long obsidian]

Oh I see that is simple thank you

[dire wren]

why does d need to be squarefree? wouldnt d being positive integer suffice?

[dim widget]

It doesn’t need to be square free, but maybe for some later use of the question it does

[tardy hedge]

if m(x)n = 0 in M(x)N then the pair (m,n) is in the subgroup generated by those bilinear relations right

so shouldnt i be able to write (m,n) into a form like that?

[tardy hedge]

m(x)0 = 0 so (m,0) is in H, because (m,0) = -((m,0+0) - (m,0)-(m,0)) is in H? ik we're not supposed to really think about it in terms of reps and cosets in practice but just tryna to understand the basic stuff better

[delicate orchid]

tends to be how quotients work

so yes you can

[spice wagon]

i have to show that if $H \triangleleft G$ and $K \triangleleft G$, then $H \cap K \triangleleft G$. So let $t \in H \cap K$, $\forall g \in G, gtg^{-1} \in H$ because $H \triangleleft G$ and $\forall g \in G, gtg^{-1} \in K$ for same reason, then we can conclude $\forall g \in G, gtg^{-1} \in H \cap K$.

[cloud walrus]

tm

[spice wagon]

is it right ?

[dim widget]

seems good

[spice wagon]

okie thanks 👍

[candid patrol]

Of course it is right

[cloud walrus]

UGOBEL

[woeful sage]

Here is a nice exercise for you from D&F

[spice wagon]

😩

[spice wagon]

looks doable lol

lemme see

[spice wagon]

for b) i have $x^a=x^b \iff x^{a-b}=e \iff n | a-b \iff a=b \ \text{mod} \ n$ because the order of $x$ is $n$

[cloud walrus]

tm

[woeful sage]

yup

[spice wagon]

no idea for c) 😥

[dull marsh]

For c, consider the image of the generator

[dire wren]

how is this true?

[woeful sage]

w.r.t what operations?

pointwise addition and multiplication?

[dire wren]

that is what i am confused about

[dire wren]

in pointwise addition this doesnt have an additive identity?

[woeful sage]

I think it is pointwise addition and multiplication, based on f(x) \neq 0 condition

[dire wren]

but what will the additive identity be?

[woeful sage]

that's true

there is no additive identity if the operation is pointwise addition

[dire wren]

so what will it be?

[spice wagon]

by sigma?

[woeful sage]

under an arbitrary automorphism

[spice wagon]

ah

[tardy hedge]

Why not? it should be just f(x) = 0

Oh nvm

F nonzero

[woeful sage]

every homomorphism from a cyclic group is completely determined by where it takes the generator

[spice wagon]

yes i saw that earlier

by the generator 1 ?

[dull marsh]

The simplest fix seems to be just include the zero function

[spice wagon]

for example

[woeful sage]

right yeah

[dull marsh]

1 is not a generator unless n = 1

[spice wagon]

🤨

$Z/nZ=<\overline{1}>$

[cloud walrus]

tm

[woeful sage]

Here $Z_n = \langle x \rangle$ such that $|x| = n$

[thorn jay]

Addition "from the perspective of" f(x) = 1 perhaps

[cloud walrus]

Neamesis

[dull marsh]

The problem uses multiplicative notation for cyclic groups instead of additive, so let's stick to that

[thorn jay]

No clue how multiplication would work then though

[dire wren]

hm

[tardy hedge]

What does that mean

[thorn jay]

Nvm this would still require the zero map as additive inverse of f(x) = 2

[tardy hedge]

Maybe that question just badly written

[dire wren]

its an example

[spice wagon]

in the exercise $Z_n=Z/nZ$ no ?

[cloud walrus]

tm

[tardy hedge]

Maybe the example is badly written

[dull marsh]

.

[dire wren]

hm

[thorn jay]

Suppose you have an abelian group (G, +) with some element a ∈ G. Then we can define
x +' y = x - a + y
Then (G, +') is again an abelian group with identity element a

(G doesn't have to abelian but it makes notation easier)

[woeful sage]

They are isomorphic

but not equal, no

Z_n is written multiplicatively

Z/nZ is written additively

[spice wagon]

yes ik

[woeful sage]

you yourself used multiplicative notation just now when you did part b)

[spice wagon]

ah fck i thought, thats why i said 1 is a generator 😦

[woeful sage]

it is okay

[spice wagon]

let $\phi \in Aut(Z_n)$, then for $y \in Z_n, \ y=x^k$ with $k \in Z$. So $\phi(y)=\phi(x^k)=\phi(x)^k$

[cloud walrus]

tm

[spice wagon]

because x is a generator of Zn

[dire wren]

how would i approach proving this?

[spice wagon]

let $f : G \to H$ an homomorphism and $N \triangleleft H$, show that $f^{-1}(N) \triangleleft G$. My idea is to take $b \in f^{-1}(N)$, which is equivalent to $f(b) \in N$ but $N$ is normal in $H$ so $\forall h \in H, hf(b)h^{-1} \in N$ now i would like to take $g \in G$ such that $h=f(g)$ but am i allowed to do this ?

[cloud walrus]

tm

[wraith cargo]

this isn't totally trivial and you have to use Zorn's lemma

what you do is you construct the field by hand

[tardy hedge]

A/I as an A-module has Ann(A/I) = I right

Also an A-module being not faithful means there is some a in A so that am = 0 for ALL m right

So A -> End(M) is not injective

[glad osprey]

You haven't defined what h is, so I don't know what it means to take g in G such that h = f(g). You have your quantifiers a bit messed up I think, you don't pick a g in G, you have to show that gbg^-1 is in f^-1(N) for all b in f^-1(N) and g in G

[spice wagon]

h is in H

[tardy hedge]

And A is always faithful as an A-module?

[spice wagon]

but yeah it isn’t the cleanly written proof it’s only my ideas

[glad osprey]

that doesn't mean much, like a specific h? Or any h? Of course it's true that f(g) is in H, whether you can find a g that maps to a specific h depends on whether f is surjective. But that's not what you wanna do

[spice wagon]

no its for any h i wrote it

[glad osprey]

you should start with an arbitrary g in G, you have started kind of in the wrong direction. Like, of course it's true that f(g) = h for some h in H, that's the definition of a function. But you don't start with picking an h

you should write down the definition of what it means for f^-1(N) to be normal in G, with quantifiers and all

[spice wagon]

oki thanks i’ll write it cleanly

[dire wren]

I understand how to prove a minimal feild exists if i can prove a feild exists, but how do i show that a feild exists?

[wraith cargo]

You basically have to construct it by hand

[dire wren]

More detail?

I was thinking of adding elements for roots of polynomial with coefficients of already existing elements, but idk how to prove that this will end.

[rustic crown]

that's where the zorn/trans-finite induction comes in

[wraith cargo]

yeah that's basically the idea
You keep repeating this procedure indefinitely and use zorn to show there's a maximal element of this tower

[dire wren]

Oh, so you are saying something like "take union of all feilds you get after finite iterations"?

Oh, thanks alot

[rustic crown]

that is the idea, but it can be hard to implement

one problem is the lack of uniqueness of lifting of maps. if K ⊆ K_1 ⊆ K_2 is a tower, then there are many ways to embed K_1 into K_2

because of this it's hard to get a nice directed system of such K_i

[rocky cloak]

I like the proof where you take the polynomial ring with one variable per polynomial. Then quotient out by a maximal ideal containing f(x_f)

[jade mason]

This is the construction we did

[rustic crown]

that is det's favorite proof as well

[rustic crown]

and you can't just keep extending this tower, because each time you add a new root of a polynomial, you're also possibly enlarging the set of all polynomials that still do not have a root.

the proof jagr is talking about constructs a bigger field L containing K in one-shot with the property that any (non-constant) polynomial in K[x] has a root in L.

once you prove this, getting a directed system is easy, start with K_0 = K, and inductively construct K_{i+1} from K_i by using that proposition. the "union" of these K_i gives you an algebraically closed.

it may not be algebraic over K, but that you already know how to fix

[rocky cloak]

L will already be algebraically closed, but I guess it's easier to prove that this iterated construction works

[rustic crown]

what's the idea that you have? is noticing that L/K is algebraic as it is generated by a set of algebraic elements, and then bring a polynomial down from L to K by some norm/galois stuff?

[rocky cloak]

Let's see. For any polynomial with coefficients in L it's splitting field is finite over K, since you can pass through the field generated by the coefficients.

Now if K is perfect then the extension is generated by a single element, and since it's minimal polynomial has a root in L, the whole splitting field is contained in L.

It should also be true when K is not perfect, but I have to think a bit

[rustic crown]

me was thinking this:
any polynomial f in L[x] already lies in F[x] for some finite extension F/K. Then we look at the norm(f) = det(multiplication by f on F[x] over K[x]) and argue directly that f | norm(f)

[rocky cloak]

So for any element y algebraic over K there is an r with y^p^r seperable. Then there is a z such that K(z) is the splitting field of y^p^r, that means K(z^p^-r) is the splitting field of y.

[rocky cloak]

Like constructing an irreducible polynomial g over K[x] such that f divides g should be easy, but you can't guarantee that the root of g you're adjoining is a root of f

[rustic crown]

ah sad forgot about that, in my head i assumed this g splits completely >.<

[rocky cloak]

Yeah, so the heavy lifting comes from the primitive element theorem, which allows you to find gs that split completely

[rustic crown]

right nice

[toxic zephyr]

does this mean f' is zero everywhere? like x^p-a over a field of char p which has derivative zero everywhere. or that f' is zero nowhere like x^3+x?

[dim widget]

this is terrible wording

[toxic zephyr]

ikr 😭

[dim widget]

It should be that f’ \neq 0 on V(f)

[toxic zephyr]

V(f)?

[dim widget]

Of course f’ has to have some zeroes

[dim widget]

The zero set of f

[toxic zephyr]

oh I see. right that's basically saying no repeated roots

[dim widget]

Yeah I would guess that’s all they’re trying to say

[toxic zephyr]

f(u)=0 implies f'(u)≠0 I guess?

[dim widget]

yeah idk I can’t understand the statement really

[tribal moss]

Without context I would read it as "... iff f' is not the zero polynomial".

[dim widget]

Ahh okay f is irreducible

I missed that

[toxic zephyr]

but whatever no repeated roots is the ultimate point. we want to split polynomials with distinct linear factors for things to be nice is the point of a separable extension right?

[dim widget]

Then it just means that it is not zero

[toxic zephyr]

oh wait is that what it's supposed to be

[dim widget]

I think the point is that if f is irreducible then if f’ vanishes at a root of f then f divides f’ so f’ is literally zero

[toxic zephyr]

so if the derivative just isn't the zero polynomial we're Gucci?
tbh it feels like non separable field extensions are kind of pathological. like K(t)(u) for u^p=t in K(t) or whatever is the standard one right

[dim widget]

Yeah they all have to do with frobenius

[toxic zephyr]

(in a field of char p>0)
then it's like x^p-t=x^p-u^p=(x-u)^p
derivative is zero everywhere. x^p-t is irreducible and we see u is a repeated root

[dim widget]

yep

[toxic zephyr]

Frobenius how? haven't heard this

[tribal moss]

It's not merely that the derivative is "zero everywhere"; it is literally the zero polynomial.

[south patrol]

Indeed common misconception that seems to persist

[toxic zephyr]

and these aren't necessarily the same? is it like with a finite field you can have a product of x-a for all a in the field?

or is it something else...?

[south patrol]

It is almost immediate from the lemma here that inseparable irreducible polynomials over K (of char p > 0) are all of the form g = f(p^n) for an irreducible polynomial f. If the Frobenius is surjective on K, then in particular each coefficient is a pth power, so you can write g as a pth power (contradicting irreducibility), so this is impossible

Meanwhile if the Frobenius isn't surjective, then some a in K not a pth power and so x^p - a is an irreducible inseparable poly / K

[tribal moss]

Example: For odd prime p, over F_p, the polynomial 2x^(p+1) - x^2 has the derivative 2x^p-2x, which is zero everywhere but is not the zero polynomial.

[cinder bluff]

(Fix a finite field F. By pigeonhole, there exist [at least] two distinct polynomials f(x) and g(x) over F such that f(c) = g(c) for all c in F. [Why?] Now consider f(x) - g(x).)

[south patrol]

Yeah and also just like your phrasing seems to imply you are thinking of polynomials as functions

[toxic zephyr]

ohhhh this is why separable iff Frobenius map is surjective? because then it specifically gives you an inseparable irreducible?