groups-rings-fields

[maiden ocean]

(locally free M implies exterior power is locally free)

[next obsidian]

[maiden ocean]

Can i do this in terms of Sym^n

That seems like itd be the easiest way

[next obsidian]

I think so

But you can also do it from tensor powers

I forget the specific relation

Just go to Wikipedia haha

[maiden ocean]

wikipedia is saur unhelpful its like 20 billion pages long

[next obsidian]

I think the basis I gave you is right tho

[maiden ocean]

Ok keith conrad has something online

defining it

In terms of Sym

Good

[next obsidian]

Oky so

I think you just quotient by the simple tensors containing a repeat element

[maiden ocean]

UGH wait never mind its not its just in terms of the simple tensors

Yeah

Ok

i guess ill just do this manually

[next obsidian]

And you could try to do it by quotient it by x (x) y + y (x) x but this only works in char ≠ 2

Do you have a copy of Aluffi?

It’s in Aluffi

[maiden ocean]

So we let Lambda^n(M) be nth tensor power of M quotiented out by tensors that repeat

[next obsidian]

Yeah

[maiden ocean]

We define the presheaf U -> Lambda^n(M(U))

[next obsidian]

I think that’s right

[maiden ocean]

We sheafify it

we again reduce to the free case

[next obsidian]

Yahhhhh

And the basis I gave you for exterior products was right

[maiden ocean]

many such cases

[next obsidian]

And so for this

I believe the size is actually r choose n

So in particular the r-th wedge product is dim 1

That’s a really important construction for diff geo and AG

[maiden ocean]

I see

[next obsidian]

In fact

I’ve seen it called the determinant

[maiden ocean]

Yeah kempf does that

[next obsidian]

That has a really good reason too, the determinant is related to this

But there’s this formul

For free modules M and N

You get to decompose the n-th wedge products of M (+) N

As a convolution of the wedge products of M and N

So it’s like

(+)_0^n wedge^k M (x) wedge^n-k N

But when n is the rank of M (+) N

Something remarkable happens where these are all 0 except when k = rk M

Then n - k = rk N

So the det(M (+) N) = det M (x) det N

This ends up coming up a lot more than you’d think lol

[maiden ocean]

I see

i skimmed and saw it used in reference to the sheaves of differentials stuff next chapter

so i assume ill be seeing it pretty soon

[next obsidian]

Yeah it is

I think it’s like

[maiden ocean]

saur yummy

[next obsidian]

Determinant of the differentials

Is the…

Normal sheaf…?

I forgor

[maiden ocean]

madlibs

[next obsidian]

It ends up being important tho

And mimics the story for diff geo

[maiden ocean]

wait

[next obsidian]

But idk that story so

[maiden ocean]

How do u actually see that x otimes y = y otimes x in the 2nd exterior power

actually

instead of answering that

let me not be dumb and try something

[next obsidian]

Take their sum

Oops

[maiden ocean]

Yeah]

[next obsidian]

This is why

In char 2

You have to mod out by the ideal of repeats

[maiden ocean]

wtf is a treat

Lol

[next obsidian]

Autocorrect

Outside of char 2 you can do this thing

[maiden ocean]

Oh sure but i meant like in the modding out by repeats defn

how do u see it

[next obsidian]

And show you can mod out just by the x (x) y = y (x) x relations

Oh

Take their sum

Then it’s

(x + y) (x) (y + x) = 0

So the two were equal

Or uh

[maiden ocean]

[next obsidian]

Difference…?

Wait

Lmfao

[maiden ocean]

Chmonkey

[dapper nebula]

gabe moment

[maiden ocean]

[next obsidian]

Uhhhh

[tribal moss]

Is (x) the same as wedge here?

[maiden ocean]

Yeah

[next obsidian]

Yeah

[maiden ocean]

Well its just normal tensor but like we're quotienting

[next obsidian]

We should I’ve switched to wedge

[maiden ocean]

into the wedge product

[next obsidian]

Yeah moth I forgot to say

For the wedge product

[maiden ocean]

I know

[cloud walrus]

Ultramothematics

[next obsidian]

Okay okay

Oh

I was right!

Okay take their difference

Then factor out a -1 lol

You end up with their difference ebeing

[cloud walrus]

Ultramothematics

[next obsidian]

-((x - y) wedge (x - y))

= 0

[maiden ocean]

Oh right

Yeah

whatever signs r overrated

[next obsidian]

[wise igloo]

what I find weird is how exponentiation is repeated multiplication (for the integers) yet exponentiation is the connection between addition and multiplication, as demonstrated by the fact that e^(x+y)=e^x*e^y

is there something more interesting at play here?

the fact that exponentiation is the "third" operation in the sequence of addition, multiplication, and exponentiation (again, for integers), but how it connects the first two

[barren sierra]

So I have a presentation

a = (1 2)
b = (1 2 3 4)

and my idea is now I want to show that all elements in S_4 can be written as a combination of 4 elements in some order

of which there are 4! orders

obvious 2 of those elements are going to be a and b

but what about the other 2

how can I determine that

[coral shale]

If I have a set of generators for an ideal of a polynomial ring of a field, the gcd of this set generates the ideal.

This is true?

[barren sierra]

I think so

that sounds familiar

[barren sierra]

Anyone got any idea with my question?

I have the set and relations <x, y | x^2, y^4, (xy)^3, (yx)^3>

[woven delta]

What do you mean "written as a combination of 4 elements in some order"

[barren sierra]

So like

[woven delta]

(abcd)?

[barren sierra]

I want to show that this set with these relations have 24 elements

[woven delta]

That form

Oh okay

[barren sierra]

24 = 4!

so intuitively maybe

I can show that every word in the group can be written as the product of 4 chosen elements in some order

there are 4! orders, done

I just need those words

playing with the relations hasn't led to anything fruitful

[woven delta]

What generators of S_4 are you choosing?

[barren sierra]

x = (1 2) y = (1 2 3 4)

[woven delta]

Okay so you want to show uniqueness I guess

ie that every order corresponds to a different permutation

[barren sierra]

yea

[woven delta]

And what are your 4 words?

[barren sierra]

that's what I'm trying to figure out

so presumably 2 of them are x and y

[woven delta]

I don't think this approach is that promising tbh

[barren sierra]

the other 2

¯_(ツ)_/¯

hm

ok then what approach could work?

because idk of any other approach

I'm using that approach since it worked for another problem

[woven delta]

Oh okay

That's interesting

[barren sierra]

tho that was for Q_8

a much simpler group

this has many more elements

for Q_8 I was able to show that every element took the form a^i b^j where 0 <= i < 4 and 0 <= j < 2 and a -> i and b -> j

so 8 elements

for my presentation

[woven delta]

Does permutations of the form (ab) generate S_n?

I think that's true

[barren sierra]

wdym

[woven delta]

Transpositions

[barren sierra]

I think you need 3 transpositions

and I have to use 2 elements only

[woven delta]

Well you just need to show you can generate all transpositions using those 2 elements

And then you're done

[barren sierra]

ohhhh

hm ok

wait no that gives me surjectivity

not injectivity

I have surjectivity

[woven delta]

What do you mean you need injectivity

[barren sierra]

that's my working so far

I need to show $\Tilde{\phi}}$ is injective

[cloud walrus]

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

[barren sierra]

last line is wrong

[woven delta]

You never actually show surjectivity there

[barren sierra]

should read $| F(S) / N \leq 4!|$

[cloud walrus]

Spamakin🎷

[barren sierra]

wait but I did show surjectivity

[woven delta]

Oh okay so you already know from previous stuff that they generate

[barren sierra]

yea

[woven delta]

Okay

[barren sierra]

that's the trivial part of the proof

[woven delta]

[barren sierra]

I need injectivity which is the pain in the ass part

[woven delta]

I see now

[tribal moss]

How sure are you that is actually is injective?

[barren sierra]

I don't yet

I have to show it though

that's the problem

Well I mean I know that I can because this presentation characterizes the whole group right?

[tribal moss]

So it's more of a hope that you have enough relations yet.

[barren sierra]

sure

a^2 = 1 cause a 2 cycle is it's own inverse

b^4 = 1 through similar logic

[tribal moss]

I agree that they are relations.

[barren sierra]

I should probably add both (ab)^3 and (ba)^3

[tribal moss]

I thought you already had those.

[barren sierra]

I had the first one

(ab)^3

[tribal moss]

Hm, so ab = (2 3 4) and therefore b^-1(ab)b = (1 2 3), so bbbabba is a transposition. But can we prove from your relations alone that (bbbabba)^2=e?

[woven delta]

I don't think there's anything in your relations to disallow arbitrarily long unreducible words

[barren sierra]

really

[woven delta]

Yeah, for example bbbabbbabbba...bbba

[tribal moss]

That one does reduce because bbba = (ab)^-1.

[woven delta]

Oh okay, how about using bba instead?

[tribal moss]

That looks more likely to be a problem, yes.

[barren sierra]

ok so I need more relations

ab^2 and b^2a are both 4 cycles

[tribal moss]

The brute-force solution would be to select one canonical word for each of the 24 known elements, and write down a relation for each entry in the group operation table, stating that xy=z for such-and-such particular words.

[barren sierra]

hm

so like

(1 3 2 4) = ab^2

so then add (ab^2)^4

?

[tribal moss]

Not quite what I'm describing.

But we could declare that abb is the "canonical word" for (1 3 2 4).
And furthermore bbabb is the canonical word for (3 4).
And bbabbb is the canonical word for (1 2 4)
Since (3 4)(1 2 3 4) = (1 2 4) we know that abbbbabb=bbabbb or equivalently abbbbabb(bbabbb)' = e i.e. abbbab'a'b'b'b' = e (where the prime means ^-1)

[wise igloo]

sorry for the late reply, but why does this happen?

[tribal moss]

There'd be one such relation for each of the 576 possible compositions.

[barren sierra]

hm ok that just seems really impractical lol

cause there would be like what, 12 relations?

more?

[tribal moss]

576, though some of them would end up being trivial.

[barren sierra]

oh yea 24^2

[tribal moss]

Yeah, it is more useful as an existence proof that you can always find enough relations in the finite case.

But it is at least constructive.

[barren sierra]

prof wants an explicit set of relations

but hm

tho I could just describe it

I think that'll be a last resort

[woven delta]

Okay here's an idea

So assuming your word starts with an a, you can (sort of) divide a word up into blocks of ab's abb's, and abbb's

[wise igloo]

oh

rip

[woven delta]

Anyway if you can just show how to reduce products of these blocks you'll have more reasonable relations

[barren sierra]

hm ok

I'll think about that later

[woven delta]

An interesting relation is abbabb=bbabba

Which tells you that in a reduced word you shouldn't have two abb blocks in a row

[barren sierra]

side note

fuck these problems

I'm dying cause it's so tedious

[woven delta]

Ugh that emote makes me feel gross

[barren sierra]

lol

but yea there's like no good pattern I can see

it's just

"lol do these computations go fuck yourself"

[woven delta]

I mean I kinda like this stuff personally

It's relaxing for me

[chilly ocean]

Same.

Abstract algebra is a fun subject.

[barren sierra]

it's fun not these problems IMO

[woven delta]

Spamakin which words do you want to represent the whole group?

Maybe knowing that would help

[barren sierra]

What do you mean?

[woven delta]

Like there should be 4! words in a and b which represent the group

[barren sierra]

isn't that part of what I'm supposed to deduce from this?

[woven delta]

Not necessarily

You can go about it the other way

Okay let's try to list out words

[barren sierra]

I guess a place to start is just combinations of a and b

not sure of a quick way to enumerate that

(), a, b, b^2, b^3 obviously

[prisma shuttle]

can someone explain why the purple part is true

i thought integral domain only implied that one of them was 0 not both

[barren sierra]

what do you mean one of them and not both

one of what

[prisma shuttle]

like ab = 0\implies a = 0 or b = 0

one of the constant terms

[next obsidian]

so like

assume for now that they aren't constant

if either of them had a constant term, then the product couldn't be x^n

I guess here's the easier way to say it

if the product of f(x) and g(x) in an integral domain is a monomial like x^n

then f(x) and g(x) would have to look like x^n-k and x^k

Okay forget what I said here, I'll just prove it easier lol

we know one of the constant terms is 0, WLOG a(x)-bar's constant term is 0

Assume b(x)-bar's constant term is non-zero

we can write b(x)-bar = xf(x) + b where b is non-zero (xf(x) represents the higher order terms)

and a(x)-bar = a^nx^n + ... + a_ix^i where a_i is non-zero and i > 0 (I'm just saying that the i is the lowest degree term)

when you multiply out a(x)-bar and b(x)-bar

you will end up with, in the lowest degree, a_ibx^i + higher degree terms

but you want this to actually be equal to x^n

so this is impossible since this isn't a homogeneous polynomial

[prisma shuttle]

why does this matter

[next obsidian]

because you want it to end up as x^n

and that's homogeneous

[prisma shuttle]

ohhhhh i see

ok thx so much

[next obsidian]

I complicated that even more lmfao

look at the bottom and top degree stuff

but whatever

if u understood, mission accomplished

[barren sierra]

How does Cayley's Theorem apply to this

or does it not at all

it seems like it would

[stone fulcrum]

Cayley's says that the quaternion group of order 8 is a subgroup of S8

[barren sierra]

well I'm talking about the first part

[stone fulcrum]

But no, that won't really apply here

[barren sierra]

yea

One thing I have realized is that because this minimal group is non trivial

that there can be no subgroups that have trivial intersection

oh maybe does it have something to do with transitive actions?

I have shown this

[next obsidian]

idk but H is necessarily cyclic of prime order

[barren sierra]

H or M?

you mean M right?

if you mean M, then M is also normal right but idk if that helps

[next obsidian]

oh wait

I misread the problem so that expalins a lot

but also it's still true that M is cyclic of prime order

and no, it does not mean M is normal

[barren sierra]

Hm I thought there was something about groups of prime order being normal

Obviously misremembering

[next obsidian]

what does that even mean

normality is a property of a subgroup

it isn't inherent to a group

[barren sierra]

I meant subgroup

My bad

[next obsidian]

this is not true

[barren sierra]

Yea

[next obsidian]

easy counterexamples exist in S_n

[barren sierra]

Yea

Like in S_5 we don't have <(1 2 3 4 5)> is normal

[next obsidian]

yeah

normal subgroups are unions of conjugacy classes

[barren sierra]

Ok but M is cyclic and prime order

[next obsidian]

and the conjugacy classes of symmetric groups are really easy to classify

yeah

I mean technically prime => cyclic haha

[barren sierra]

True

lol

[next obsidian]

Everyone still says cyclic of prime order tho

I mean you can write M as the intersection of all proper subgroups

yeah idk

[barren sierra]

Yea that's as far as I got

[hybrid island]

can someone check my work

[proud bear]

does this work? if we have an injection of G into S_{n-1} then G acts on some set of size n-1 where the intersection of all the stabilizers is 1. there must be at least one element with trivial stabilizer because otherwise M<=the intersection of all the stabilizers. but then the orbit of that element would have size n.

[gritty sparrow]

Looks good to me

[upbeat swift]

Describe all the distinct powers of α. How many are there? Note carefully the connection with addition of integers modulo s. Can someone help explain this?

[next obsidian]

They want you to say how many of the numbers alpha^n are distinct

Like if alpha = 2 and you’re doing stuff mod 4

there’s only 2

Well I guess 3

2^0,2^1,2^2 = 0

Cuz once you go past that, 2^k = 0

Maybe you’re in a group tho and then it’ll be a little different because you’ll just get stuck in a loop

[upbeat swift]

α is the cycle of length s, say α = (α1α2 …αs).

[next obsidian]

Idk what group this is

S_n for some n?

[upbeat swift]

A permutation

[next obsidian]

And this is just a single cycle?

[upbeat swift]

Mhm

[next obsidian]

Ggggggggg

I mean it’s easy to describe what the order is buuuut

It’s a bit more annoying to describe what the powers are

[upbeat swift]

Showing it is a whole other thing

Yeah

[next obsidian]

I mean I guess I would say like

If you think of this as a function

You can like inductively describe it

so like alpha^1 sends alpha_i to alpha_i+1

Then alpha^n sends uhhh

alpha_i to ummm

alpha^(n-1)_i+1 I think?

No this isn’t right

[upbeat swift]

I wrote these down

[next obsidian]

Yeah idk this sucks

Like

Was alpha a particular permutation?

Or literally just an arbitrary cycle

[upbeat swift]

I'm unsure, just some cycle given

[next obsidian]

Like

Is the cycle written down explicitly

Like is it (16392) or something

Or is it a generic cycle of length n

[upbeat swift]

Yes yes

[next obsidian]

Oh

[upbeat swift]

It's length

Length S

[next obsidian]

But

It’s any length s cycle

[upbeat swift]

Yes

[next obsidian]

Okay uhhh

Yeah idk it’s not too clear what the problem wants from you, at least to me

Sorry

There will be s distinct ones

And alpha^n = alpha^n+ks for any integer k

But beyond that i don’t really know what it wants

[upbeat swift]

It's all good

Idk what it wants either

[lethal dune]

A very basic question, the module generated by an element $x\in M$ is $Rx = { r\cdot x | r\in R }$ and also it is the smallest submodule of $M$ containing $x$. But if $R$ does not contain $1$ then how can $Rx$ contain $x$? like the first definition is wrong if $R$ is not unitial ring?

[next obsidian]

Modules over a non-unital ring omegalol

I don’t know if modules are defined over non-unital rings

[lethal dune]

[gritty sparrow]

I think I saw modules over nonunital rings in the context of banach algebras once

[lethal dune]

what about rings then?

[gritty sparrow]

In that case the first definition is indeed wrong and should be replaced with Rx+Zx where Z refers to the integers

[lethal dune]

oh then Rx+xR+Zx for general case right, (I hate the book)

[gritty sparrow]

I assumed we were talking about left modules here

Are you talking about bimodules or something?

[lethal dune]

no rings but that will work for modules

[gritty sparrow]

Ah yes a 2 sided ideal is indeed an (R,R) bimodule

And yeah you need Zx+Rx+xR+RxR terms in a general bimodule as well

What book are you reading btw?

[lethal dune]

introduction to rings and modules by C. Musili (reluctantly reading)

also nice pfp, Reisz was it? from H Dxd?

[gritty sparrow]

Rias

Yeah highschool dxd

[gritty sparrow]

Ok just took a look at this and he does indeed define exactly what you want for left modules in chapter 5. He doesn’t do bimodules in the book i think

[lethal dune]

actually I've seen the definition Rx in a different book so I got kinda confused because the author didn't assume R to be unitial or even commutative

[inner needle]

May I ask a question here?

[gritty sparrow]

Sure

[inner needle]

I can't seem to find an approach to this one. Even the forward direction.

It makes sense to me, and considering a few examples makes it much clearer, but what is actually happening here?

If G is cyclic and m is the order of G, then clearly, we only need to consider values of n less than m.

[gritty sparrow]

Here is a hint: Let g be the generator for G. Show that the only elements with order n look like g^{km/n} where k is less than n. (Infact you can even say k is coprime to n but that is unnecessary for this direction). The converse is genuinely tricky I would be surprised if no hint was given

(Also note that if there is any element of order n, n has to divide m, so km/n makes sense)

[lethal dune]

you can write $|G|=p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}$ then decompose G into direct products of syllow pi groups

[lethal dune]

try showing the pi subgroup is unique

idk if that helps

[inner needle]

The book has not even covered direct products yet.

That is the next section. I don't mind coming back to this problem, but is there an elementary way to approach it?

[lethal dune]

I have none

[gritty sparrow]

Did you understand how the forward direction works atleast?

[lethal dune]

this works for fwd dir @inner needle

[inner needle]

And thanks for this. Puts into perspective what was really happening.

[inner needle]

Yes, I'm just going work it out myself.

[lethal dune]

@inner needle sidenote: that statement is also true even if G is not assumed to be abelian

[inner needle]

Yes, I did see this fact online. That's really cool, to be honest.

But I'm assuming the proof for that gets even harder.

[gritty sparrow]

Oh damn, I didn’t see that abelian part

[lethal dune]

not really

[inner needle]

Well, maybe difficult to approach in an elementary way?

[lethal dune]

ik a proof that uses syllow theorems

don't have anything else

[inner needle]

But that comes, much, much later. I have no idea about them or even the definitions they use.

[lethal dune]

ok wait i think I have something

using number theory you can show that for d < n the sum of number of elements with order d is < n

so that forces one element with order n

making it cyclic

[inner needle]

I don't get it, what is n here?

[lethal dune]

n is the order of |G|

[inner needle]

Oh, right. In the problem, n is used for something else.

[lethal dune]

idea is you show that there aren't enough element which can have order < n. so one element has to have order n

[inner needle]

Right! If I can prove that fact, the result does follow.

[gritty sparrow]

Here is an easier way to do it for abelian groups: first prove that given two elements of orders p and q, you can get an element of order lcm(p,q) then assume that the maximum order element is some d that divides m. If every other elements order divides d, we get a contradiction else use the thing i said above to get an element of order even larger than d which is again a contradiction

[gritty sparrow]

This is the hard proof i was referring to earlier, you first show that atmost phi(n) elements have order n and then use that fact that m= summ (phi(n)) for n|m

Where phi is euler’s totient formula

[next obsidian]

I think you should just classify finite abelian groups and then it follows from that pretty easily 🙂

[gritty sparrow]

True

[next obsidian]

I can’t produce a proof off the top of my head not using the classification, sad champ

Sometimes when you gain power you forget how to use it

[inner needle]

What is the contradiction to? The fact that d divides m?

[lethal dune]

shows there are more than m elements satisfying x^m=1

[inner needle]

But isn't d just m?

[lethal dune]

m = lcm(p, q)

[gritty sparrow]

There are two contradictions. The first one is the case where the orders of all elements divide d, the contradiction there is that all m elements will satisfy x^d=1 which is wrong

[gritty sparrow]

No m is the size of the group

[lethal dune]

oh, I assumed m means the maximal order

[inner needle]

So we are assuming d|m and d < m?

[gritty sparrow]

The second contradiction is that given an element x whose order ,t, does not divide d and y whose order is d, you can create an element whose order is lcm(d,t) which is strictly greater than d, which contradicts the fact that d is the maximal order

[gritty sparrow]

d|m will always happen since the order of an element will divide the size of the group

Yeah I assumed d<m though

[inner needle]

Oh right. So we are assuming d < m for contradiction

[gritty sparrow]

Yep

[inner needle]

Right! This is cool

Okay, I think I get it. I'll just write this proof down.

Wait, where have we used the fact that G is abelian?

[gritty sparrow]

The lemma that i didn’t prove about finding an element of order lcm(p,q)

That uses the fact that the group is abelian

[inner needle]

The getting an element of order lcm(p, q)?

Yeah, now that you remind me, I've already proved this.

[gritty sparrow]

Nice

[inner needle]

Okay, I think it all works out. The converse is rather tricky though, I would not have come up with such a proof myself.

[inner needle]

I don't think I've proved this though.

Seems similar to Lagrange theorem, which I've not done yet. But I'll try proving that regardless.

[gritty sparrow]

The size of a subgroup divides the size of the whole group and you can just look at the subgroup given by {1,x,x^2..,x^{d-1}}

[inner needle]

Yeah, I've not proved this yet.

[gritty sparrow]

Lagrange’s theorem?

[inner needle]

Yup

[gritty sparrow]

Oh you should definitely see lagrange’s theorem at least before doing any exercises

[inner needle]

It's actually there in the later sections

Not covered till this point

[gritty sparrow]

This book is very weirdly structured then

[lethal dune]

wait this exercise is before Lagrange's theorem?

[gritty sparrow]

There is no way you could have possibly proved this exercise

[inner needle]

Yes.

@gritty sparrow It's just one question xD

I don't think you can use that to judge the whole book, mistakes happen

[gritty sparrow]

Fair enough

[inner needle]

The rest of the problems are all reasonable, approachable with what has already been covered

It's just this one which I simply couldn't do

I think I understand why now

My favourite of the bunch was finding an equivalent criteria for <x^r> being a subset of <x^s> where the group is G = <x> of order n.

Turns out <x^r> = <x^k> where k = gcd(r, n). Also, k is the smallest positive power of x which lies in <x^r>.

[desert dome]

Hi I am trying to prove this problem using Artin's theorem learned in class. But I felt so weird about the proof. Did I understand this theorem wrong or Artin's theorem is just so much stronger than what I'm trying to prove? Thank you!

[inner needle]

I worked everything out. Thanks a ton, that was a fantastic proof. @gritty sparrow

Did we use the fact that d divides m, though?

I can't see where we used that, and if we didn't use it, we didn't use Lagrange's theorem.

[minor badger]

Could someone take a look and see why i cant get shit right? 🙂

[dapper nebula]

Your second last line is wrong

[minor badger]

yeah thats what i have suspected, i don't know how to continue after the third to last line

[cloud walrus]

det

[rustic crown]

@minor badger >.<

[fallow plume]

Quoting Aluffi's Chapter 0: \

one can define a field whose underlying group is $Z/pZ \times Z/pZ$, while the product ring is very far from being a field (why?).

[cloud walrus]

Thomas

[fallow plume]

Am I missing something obvious here?

Has an inverse, commutative, etc

[rustic crown]

product of two rings can never be an integral domain!

(1, 0) * (0, 1) = (0, 0)

so it has zero divisors

[fallow plume]

Oo

Thanks so much

Can't believe I missed that

[rustic crown]

well it doesn't have inverses >.<

you can only find inverse of (a, b) when both a and b are non-zero

[fallow plume]

Haha yes I was only thinking on a surface level after you pointed that out

(also love the Eevee 🧡)

[rustic crown]

(sorry, a slight error. assuming both are non-zero rings >.<)

[stoic rose]

The zero ring is generally not considered an ingegral domain

[rustic crown]

right, but it will still be a terminal ring, so 0 x R = R

that looks so bad to my eyes >.<

[stoic rose]

Oh yeah true

[sinful mirage]

complete reducibility theorem relies on unitarity

is there any generalization?

i.e. finite dimensional representations of a Lie group being direct sum of irreducibles, if the rep isn't unitary

[stoic rose]

I think the usual theorem for finite groups generalizes in a straightforward way to compact groups

That is, you can always choose an inner product that makes the representation unitary

[sinful mirage]

yes

but I am interested in non-compact groups(physics)

Lorentz/Poincare

they admit finite dimensional non-unitary reps

[stoic rose]

Then I'm pretty sure complete reducibility isn't true in general

But maybe under some assumptions

[sinful mirage]

so for the unitary reps of the poincare/lorentz satisfy complete reducibility

but non-unitary reps not?

[minor badger]

omg i see it now, thanks ❤️

[barren sierra]

So I'm struggle with what I should be taking notes for in this class

So we just did the proof that alternating groups are simple

I normally don't write proofs but since the lecture covered the whole proof (and that took a large chunk of time) I did

should I be doing that for all proofs in this course and furthermore should I be expected to just be able to reproduce these results?

[delicate bloom]

I'd recommend going to your teacher's office hours and discussing this with them

[rustic crown]

taking notes is a good habit i'd say >.<

but yea that proof at first might seem pretty arbitrary and complicated. but eventually you'll notice what was the key ingredient that went into the proof. Also there are a couple of ways to prove it, my favorite one uses that A5 and A6 are simple and then later uses n = 6 case to prove it for any n > 6.
so technically with this proof, the hardest part is actually proving the simplicity of A5 and A6 which is a finite problem, so we don't really need to worry that much. (in the worst case, we could run a computer and ask it do solve it for us)

it usually takes me a couple of iterations of anything to really understand what went into the proof.

[chilly ocean]

What class is this for? O.O

[barren sierra]

Well this is more in general

I write theorems and stuff

But rarely the proofs

I read over the proofs and such, just never write them out

[rustic crown]

If the prof is faithfully following a book or is giving class notes, then it's probably fine to skip on taking notes, but i usually find it nice to have notes around so i can look at the clever ideas involved. >.<

[coral shale]

If x = -x for every x in a Field, does it have to be trivial?

[delicate bloom]

nope

[coral shale]

urgh.

[delicate bloom]

it's possible -1=1, in other words 1+1=0, AKA characteristic 2

[coral shale]

1, i mod 2 is an example right?

urgh maybe not

[delicate bloom]

I don't know what your notation 1, i mod 2 means

[coral shale]

sorry 😅

[delicate bloom]

the field with 2 elements, like regular mod 2 arithmetic is a field with characteristic 2 though

[coral shale]

Is that not the trivial field?

I meant that

[rustic crown]

if i is the sqrt(-1), then it's not going to give you a field
(1+i)^2 = 0 but 1+i isn't 0

[delicate bloom]

there is no "the trivial field"

[coral shale]

{0, 1} <-- I meant this

[delicate bloom]

it's the prime field of characteristic 2

but there are fields with characteristic 2 that aren't F_2

there's F_{2^k} for all k>=1 and F_2(X) as some examples

[desert dome]

Modern algebra

[coral shale]

Hmmm
So my thought process was this

If we have a vector space V over F

and V is a field

I believe the scalar multiplication function can be projected into a field homomorphism from F to V?

. : F x V -> V

We then define what I think is the homomorphism
f: F -> V

f(x) = x.1

[delicate bloom]

sure, what's the vector 1? looks suspicious

[coral shale]

V is a field.

[delicate bloom]

ok sure

[chilly ocean]

Cool. 👍

[coral shale]

I believe this is a field homomorphism

Which would mean it has to be injective.

If so, then suppose V is the field {0, 1}. F must be {0, 1}. I was checking if this is true, and got to x = -x in F.

[delicate bloom]

I don't understand why you're bringing a vector space into this if you just care about fields

[coral shale]

Well it's a thought exercise. I'm doing Galois Theory --- so Field Extensions naturally create a vector space.

Then I thought about the reverse process, if we had field V over F

[delicate bloom]

sure, I know that but I think I'm missing what your original question is

[coral shale]

uhhh

I guess it would be

'We have vector space V over F. Suppose V is the field {0, 1}. Prove/disprove F must be {0, 1}.'

[rustic crown]

yea your reasoning is correct

[coral shale]

This is the only bit I'm not 100% on rn

but pretty sure it has to be from the properties of vector space

😅 wondering if I was missing something obvious

[delicate bloom]

it has to be by closure, you know s in F and v in V makes sv in V, so we can write sv=u and then s=u*v^-1 so s is in V

[rustic crown]

we should probably make the distinction of the action of F on V vs the multiplication in V

[coral shale]

s.1 = uv^-1 is in V

[delicate bloom]

yeah s in F implies s in V so F is a subset of V

V has no proper subfields so F=V

[rustic crown]

wait isn't s * v a shorthand for m_s(v) where m_s is the function you get by currying F x V --> V

[delicate bloom]

yeah you're right

[coral shale]

I think Merosity's argument works up to isomorphism. Because every field F must contain {0, 1}.

Or am I misreading

[delicate bloom]

well what you said earlier s.1=s sort of seals the deal

s on the left is in F while s on the right is in V

[rustic crown]

i'm a little confused

[coral shale]

I don't think I agree/follow (how this seals the deal)

For starters, v has to be 1

s.1 = u is in V

===
An aside - all I have so far is
(a+a).1 = a.1 + a.1 = a.(1+1) = a.0 = 0
(a+a)((a+a)'.1) = ((a+a)(a+a)')1 = 1.1 = 1
therefore a+a=0 (the dash indicates multiplicative inverse which cannot exist for a+a)

[coral shale]

So I am suspicious of my previous argument

=====
(a.1)(b.1) = ???
(ab).1 = a.(b.1) = ???

I guess this isn't necessarily true if V is a field (the 2 being equal), which is how f would fail to be a homomorpism

[hidden haven]

bro

Just call it multiplication by s

[rustic crown]

sorry lol i'm very sleepy right now

[hidden haven]

UGCT parts of the brain taking over while you drift off 🙈

[rustic crown]

so what i feel for that is V/F is a vector space, but this vector space structure doesn't exactly know what the multiplication on V looks like.
we probably can make (a.1) * (b.1) and (ab).1 different elements

• is the multiplication on V and . is the action of F on V

[coral shale]

Do we have a 'common' example where the 2 are different

I'm guessing not

since we usually use stuff like R, C, etc

[rustic crown]

that's what i'm trying to think, but brain no worky >.<

moldi help

[hidden haven]

Positive reals with exponentiation stuff 🙈

vector addition is multiplication

scalar multiplication is multiplication by exp(scalar)

magic 🙈

I think this works though I might have stated it wrong

[hidden haven]

Should be r.v = v^r

[coral shale]

so
(a.1) * (b.1) = (a+b).1 \neq (ab).1 in general

[hidden haven]

But real power

Let me just look it up lol

[hidden haven]

oh this works lol

[coral shale]

So its R+ over R.

[hidden haven]

Yes

[rustic crown]

what's the multiplication in R+?

[hidden haven]

r*v = v^r

[coral shale]

not commutative ?

[rustic crown]

not commutative

[hidden haven]

Why do we want that

[coral shale]

Uhhh we wanted V over F where V is a field

[hidden haven]

I didn't read the chat I just saw det mention currying so I was here to make fun of that

[coral shale]

and (a.1)(b.1) =/= (ab).1

sorry 😂

[hidden haven]

Oh I see

[delicate bloom]

maybe the vector space axioms get us stuck in the case of F_2 here

we need to look beyond to find counterexamples is my guess

like uhh, what's it saying if it were true, every algebra over a field is just the field itself?

[hidden haven]

Have ℚ act on itself but on the vector space version define the multiplication to be as if the positives are negatives and vice versa 🙈

I mean the multiplication of the vector space plays no role in any of this it seems

Wait does it

I didn't even bother reading the equations

What are . and * again

Actually don't bother lol

[rustic crown]

i think we can answer the question if we can find an abelian group such that you could define two non-isomorphic field structures on it.

[coral shale]

Is this non-trivial lol? surprised.

[rustic crown]

finite fields won't work i think

[rustic crown]

like ideally you would just say that the field structures and vector space structures are compatible with each other

okie i think i might have an answer
f : Q(sqrt(2)) --> Q(sqrt(3))
this is Q-linear and sends sqrt(2) to sqrt(3)
say its inverse function is f'

F = Q(sqrt(2)) and V = Q(sqrt(3))
V is also a field
now we want to pretend that V is the abelian group Q(sqrt(2)) and define the vector space structure.
so define a . v = f(a * f'(v))

I'm too sleepy to confirm if this forms a vector space or not, it feels it does so i'll stop at that

(a.1) = f(a * f'(1)) = f(a * 1) = f(a)

so (a.1) * (b.1) = f(a) * f(b)
but (ab).1 = f(ab)

i chose f to not preserve multiplication by taking non-iso fields for instance a = b = sqrt(2) then ab = 2 so f(ab) = 2 and f(a) * f(b) = 3

[coral shale]

That sounds right

[rustic crown]

i can sleep now

[coral shale]

Thank you for entertaining my wandering thoughts

haha

[barren sierra]

He's doing both

I take notes on concepts and examples and such

[coral shale]

Hmmm so there indeed is no field homomorphism from F to V.

2 = 1 + 1 = f(1) + f(1) = f(2) = f(rt2 rt2) = f(rt2)^2

[barren sierra]

So I have gotten a hint for this

I need to show that if G acts on any set of size n - 1, then the action cannot be faithful

But then this means that all the stabilizers of said action must be non-trivial

So does this amount to just using the fact that stabilizers are subgroups of G?

ok yea that's all it is

OK then my question is

how tf am I supposed to read that question and think about group actions

💀

how is that supposed to come to mind

[next obsidian]

embedding into S_{n-1} is equivalent to a faithful action on a set of size n-1

any map into S_n gives you a group action on a set of size n

if you know how the correspondence of acting on set of size n gives you a map to S_n, you can reverse that process

[olive mirage]

I think this is probably well known, but not to me. If I take k to be a finite field (F_p if you want to be super specific) then G=Aut(k(t)/k) is a finite group isomorophic to PGL_2(k). What is the fixed field of G in k(t)?

(by usual theorems it must have the form k(f(t)) for some rational funciton f(t)

(and f(t) has to have degree (q+1) * q * (q-1) )

[tribal moss]

If there's a non-abelian group involved always think about group actions no matter what else is happening.

[olive mirage]

If your group isn't acting on something, is it even really a group?

Actually, I think the answer to my question is something like f(t) = (t^(q^3)-t)/(t^(q^2)-t)^q

Which is not quite right, but I think the idea is you want one zero at every element that is in P^1(GF(q^3)) that isn't in P^1(GF(q)) and then you want to balance that out with a bunch of poles at elements of P^1(GF(q^2)) that aren't in P^1(GF(q))

(and hope the combinatorics works out)

[barren sierra]

Ok I thought I had it but I don't. How can I show that all the stabilizers are nontrivial? I know since we have a set G of size n acting on a set X of size n - 1 that for all x in X we must have that there are g_1 =/= g_2 in G where g_1 * x = y and g_2* x = y

can I do something like

g_1 * x = g_2 * x

let g_1^{-1} act on both sides

and get g_1^{-1} g_2 * x = x?

is that valid?

yea I think that works

[barren sierra]

Can someone help with part b? Since the proofs are basically the same, lets suppose I want to show n/d divides n'

I really don't know how the gcd of indices relates to the index of the intersection of subgroups

because it's not necessarily true that the order of H intersection K is gcd(m, n)

we have [G : H cap K] = [G : H][H : H cap K]

so [G : H cap K] = m * n'

and I guess similarly = n * m'

[barren sierra]

yea idk where to go from there

[chilly ocean]

What is a Lie group?

Also, what is a ring compared to a group?

[next obsidian]

A Lie group is a smooth manifold with a group operation that’s smooth, aka M x M -> M the product is smooth as is the inversion map M -> M

[chilly ocean]

Oh, wow.

[next obsidian]

A ring is an algebraic structure with 2 operations, addition and multiplication. Multiplication and addition distribute as they would for the integers

[chilly ocean]

Oh, so like a group + distributive laws?

[next obsidian]

The addition forms an abelian group, so it has inverses, an identity (we call 0), and it’s commutative

Yeah

The multiplication only has to be associative plus distribute

[chilly ocean]

Oh, rings are all abelian?

What are rings useful for compared to groups and why are they called rings?

[next obsidian]

However: most sources also require the multiplication to have an identity denoted 1

But not inverses, so Z is a ring

But like, 2 doesn’t have a multiplicative inverse

[next obsidian]

Yeah the addition is

[chilly ocean]

Ah, cool.

[next obsidian]

Rings model a lot of things, commutative rings underly all of algebraic geometry

[chilly ocean]

Wow, this is so fascinating. I can't wait until my abstract algebra class coverts them.

[next obsidian]

You can define modules, it’s like a vector space over a ring, they’re very useful

They show up in algebraic number theory

[chilly ocean]

Right, I can see that.

[next obsidian]

And say, continuous functions into R form a ring

[chilly ocean]

I'm in linear algebra as well right now and it's pretty interesting.

[next obsidian]

Since you can add them and multiply just on the output

[chilly ocean]

So vector spaces over F can be seen as being over rings?

[next obsidian]

Yeah

[chilly ocean]

Because both R and C are rings?

[next obsidian]

A field is a special ring

Yup

A field is a ring with multiplicative identities

Also the multiplication has to commute

I forgot to mention but rings don’t need multiplication to commute

Thing of matrices

Those form a ring, but MN ≠ NM

[chilly ocean]

Oh, really?!?!

[next obsidian]

Yeah

[chilly ocean]

Oh, cool. it's just a special case.

[next obsidian]

Yeh

[chilly ocean]

But, there are no multiplicative inverses in general, right?

[next obsidian]

Right

[chilly ocean]

At least for vector fields.

[next obsidian]

It also doesn’t have to commute

Idk what a vector field is

[chilly ocean]

But in groups there must be inverses.

[chilly ocean]

Huh?

A field of vectors. 👀

[next obsidian]

We have vector spaces

And fields

[chilly ocean]

I meant vector space.

oops

[next obsidian]

I mean I know what a vector field is in like, manifolds

[chilly ocean]

But vector fields exist.

[next obsidian]

Well vector spaces have inverses

[chilly ocean]

In Calc III, for instance.

[next obsidian]

They don’t have a multiplication

Right yeah I know them in that setting

[chilly ocean]

I mean vectors in vector spaces.

[next obsidian]

I mean they have inverses

[chilly ocean]

v1 * v2 = 1?

[next obsidian]

You can’t multiply vectors

I just mean -v exists

[chilly ocean]

Yeah.

So there is no mult inverse.

[next obsidian]

Well sure

[chilly ocean]

Yeah, for addition...

[next obsidian]

There’s no multiplication lol

[delicate orchid]

More like there is no mult

[next obsidian]

So there couldn’t be a multiplicative inverse

[chilly ocean]

true, true

[barren sierra]

in general you can't multiply vectors

[chilly ocean]

What else is there in abstract algebra?

[barren sierra]

but in some cases you can

[chilly ocean]

What is a fuscian group?

[barren sierra]

like the vector space of matricies

you can multiply matricies

[chilly ocean]

Or 3-spaces.

[barren sierra]

yea

[next obsidian]

That’s disingenuous to say

You aren’t using their structure as a vector space there

As a vector space you can’t

[delicate orchid]

Other than rings, groups, modules, and fields it’s mainly just what are called “algebras” from my experience

[next obsidian]

Yeah

I mean there’s Lie algebras too

And you can get other weird stuff, but those are way more niche

[delicate orchid]

Which are a vector space with an additional vector multiplication rule

[barren sierra]

if you want an intro to Lie Groups and such

[next obsidian]

Semi groups, monoids, etc

[barren sierra]

Naive Lie Theory is a great text

[chilly ocean]

Wow, algebra is very interesting.

[chilly ocean]

Thanks!

[barren sierra]

it's and intro to Lie Theory with only linear algebra and calc as a prereq

[delicate orchid]

It’s all just magmas

[chilly ocean]

I will check it out. (literally 😉)

[delicate orchid]

Yeah there’s tens of subfields I’ve heard of and probably a hundred I haven’t

Really fantastic area of maths

[next obsidian]

COMMUTATIVE RINGS

THEY CALL ME CHMISTER COMMUTATIVE ALGEBRA

[delicate orchid]

*abelian rings

[chilly ocean]

what

How many?!?!

[delicate orchid]

Off of the top of my head

Galois theory, algebraic topology, algebraic geometry, Representation theory are the big boys that aren’t just studying a specific object like “group theory” or “ring theory” does

[chilly ocean]

Wow.

[delicate orchid]

For example, representation theory studies groups via vector spaces

Galois theory studies fields via groups

[chilly ocean]

There's so much.

Maybe I want to mainly do research in algebra. 👀

[delicate orchid]

Alg geom alone is an entire undergrad course

[chilly ocean]

I've been trying to find a main field to enter.

Like, analysis, algebra, geometry, topology, number theory, foundations, etc.

Or theoretical computer science.

etc.

[delicate orchid]

Ah there’s a funny thing there

The first 5 things there basically become the same thing after a while

[chilly ocean]

What do you mean?

[delicate orchid]

Algebraic number theory is a thing, I’ve already mentioned algebraic geo and algebraic topology

[chilly ocean]

Right.

[delicate orchid]

And number theory is applied ring theory

[chilly ocean]

I'm in an algebraic topology course right now.

[chilly ocean]

:o

[delicate orchid]

Wait as an undergrad?

[chilly ocean]

Yes.

It's a 400-level.

It's just an intro course.

[delicate orchid]

Lucky

[chilly ocean]

Math is one heck of a drug.

Maybe I don't even want to double major, if I can just keep taking more math courses endlessly. 👀

[barren sierra]

what major are you?

CS?

[delicate orchid]

I prefer LSD but group theory is good too

[chilly ocean]

Pure Math

[barren sierra]

oh

[chilly ocean]

Why?

[barren sierra]

wait then wdym double major

[chilly ocean]

I'm really ahead in math.

[barren sierra]

ah

[chilly ocean]

I'm running out of courses.

So to spread it out I was going to double major with either phil or cs.

[barren sierra]

yea I'm staying in math and doing the same thing (tho I'm doing alot of theoretical CS courses on the side)

you could take grad classes

[chilly ocean]

I have 12 in my course planner ,haha/.

[barren sierra]

however my grad abstract alg HW is murdering me rn

[chilly ocean]

My first is next semester.

But the problem is that I can only take up to 2 or 3 per semester, I think.

[barren sierra]

I am mega stuck

start with 1 lol

they're hard

at least for me

[chilly ocean]

If I take more graduate courses than undergraduate courses I will lose financial aid.

Hence the double major as filler.

[chilly ocean]

At first, yeah.

[barren sierra]

could just take lighter semesters

[chilly ocean]

I want to take MATH 536, which is abstract algebra.

big boi abstract algebra

[chilly ocean]

I guess I could just do 5 courses each.

But I like to grind.

[delicate orchid]

5

[chilly ocean]

Yes.

[barren sierra]

"just 5"

bruh

[delicate orchid]

I’m doing 2 courses a term this year

[chilly ocean]

I'm taking 6/7 this semester.

[chilly ocean]

wtf

[barren sierra]

I'm doing 3 and dying (tho I have other stuff outside of courses)

[chilly ocean]

I'm taking four math courses right now.

One CS, one art history, one fake class for learning assistants and I'm a learning assistant.

[chilly ocean]

Only three?

[barren sierra]

fake class for learning assistants

[delicate orchid]

I’m post grad I’ve got a thesis to do

[barren sierra]

like a TA training seminar?

[delicate orchid]

although last year I was undergrad and only did 3 per term

[chilly ocean]

LA

Yes.

[barren sierra]

same thing

[chilly ocean]

Yeah.

[barren sierra]

lol pay attention to that

[chilly ocean]

That.

[barren sierra]

I've had enough shit LAs

[chilly ocean]

ah lol

[barren sierra]

and it's not good for students

I'm an LA for two courses

hence why I'm only doing 3 courses

[chilly ocean]

Ah.

[chilly ocean]

Then it does make sense.

Maybe I'll just do the math major and take a bunch of independent studies credits.

OH

I can learn graduate-level stuff from those.

As undergrad credits 😎

@barren sierra: Would it make more sense to graduate early?

If I just do a math major, I can do the accelerated M.S. program and leave with a B.S. and M.S. in Mathematics by the end of my junior year.

[barren sierra]

I have no idea lol

[chilly ocean]

:/

😭

[desert dome]

Could anyone help take a look at this one

[rustic crown]

I don't understand that argument fully.

[desert dome]

Yeah I agree it is not very accurate. It should be |Aut(E/F)| = [E:F] instead

[rustic crown]

for the first one, say E/F is galois with galois group G having size n.
then it's easy to see that F is a subfield of E^G, because by definition the galois group will fix elements of F pointwise.
It's not completely obvious why F = E^G

that where we need Artin's theorem as you call it

[desert dome]

Then I can't apply Artin's theorem in either direction?

because in G = Aut(E/F), E^G might be more than F? but in that case, does it mean there is only one thing in Aut(E/F), the identity map

[rustic crown]

you probably have to use it somewhere

[desert dome]

I guess I can apply for the second case

[rustic crown]

we still need to see how the galois hypothesis is entering the picture

[desert dome]

emmm I see. For general Aut(E/F), the fixed field might not be F

but it is true for galois extension

[rustic crown]

yep, and it's one of the equivalent definitions for galois extensions

so what def are you using?

[desert dome]

galois extension is separable and normal

[rustic crown]

(like ideally i would use the primitive element theorem without thinking, are we allowed to use it or is that a no?)

[desert dome]

I don't recall we learned this theorem

[rustic crown]

oh it says any finite separable extension is actually simple

so it's much easier to think about the automorphisms groups as we only need to worry about 1 element

[desert dome]

no I didn't learn that...

[rustic crown]

anyway, the second part should be more or less direct from Artin's theorem. i'll have to think a little for the first one if we're not allowed to use that.

[desert dome]

Thanks! that's very helpful. I'll think about the first case more

[rustic crown]

say G = Aut(E/F) then like we just said F is a subfield of E^G
so we have a tower F --> E^G --> E
taking degrees [E:F] = [E:E^G] * [E^G : F]
by the theorem and hypothesis, [E:F] = [E:E^G] = |G|
so [E^G:F] = 1, showing E^G = F.
the theorem also says E/E^G = E/F is galois

[desert dome]

I see. tysm!!

👍

[barren sierra]

Reposting this again

stuck on 1b

ngl no idea what to do

I did this for 1a but I can't think of anything similar

my only thought with the divides relation is that I need to show some sort of thing with cosets

something like [A : C] = [A : B][B : C]

and then I can work from there

but playing with that hasn't gotten me anywhere

[rustic crown]

yep, that's true i think

so you'd get [G : H n K] = mn'=nm'

divide by d and use that (m/d, n/d) = 1

(m/d) * n' = (n/d) * m'

[rustic crown]

let me know if you need some help in proving this. a simple hint would be to define the map A/C --> A/B by sending aC to aB.
since the groups may not be normal, these are just sets and set functions. show it is well defined, surjective and analyse the fiber over each aB, show it's bijective to B/C.

[barren sierra]

no I have already shown that

I just didn't know how to use that >_>

[rustic crown]

oh >.<

[barren sierra]

and I already got that but the m and n' together made me thing it was the wrong track >__>

[barren sierra]

OH from here is it just
(n / d) divides both sides so (n / d) must divide n'?

and then vice versa?

[rustic crown]

yep!

[rustic crown]

draw some diagrams >.< like a helpful one here would be

[cloud walrus]

det

[rustic crown]

it makes it easier to see all the details at a glance >.<

[next obsidian]

Hey det y r u such a qt?

[rustic crown]

[barren sierra]

yea I just never divided both sides by d

which in hindsight

is obvious to do

but I never did it >_>

[dusty sapphire]

What is Automorphism of $\mathbb{F}_{p^n}$?

[cloud walrus]

BLツ

BLツ

[hidden haven]

Yes

[dusty sapphire]

ok

[languid meteor]

can anyone tell me why the first sentence of this proof is true? Its been a while since ive done field extensions, why does every element take this form? I thought you expressed everything in the field in terms of a basis 1, z, z^2 etc

[hidden haven]

What you're suggesting would just be f(z_1)

But that may not be a field if z_1 is transcendental

[languid meteor]

ah ok, so what is this field k(z1,...,zn)?

im adjoining z1,...,zn right

but why does the adjunction k(z1) give every element the form f(z1)/g(z1)

is this some really basic property of field extensions that ive forgotten about