groups-rings-fields

[coral spindle]

I do not know about this

This is interesting

[south patrol]

Realising maybe I need to learn more about G-sets lol

[delicate orchid]

if you have X = G/H and some other K \leq G then m_X(K) := m(H, K) = |{gK | HgK = gK}|

which is obvious but it's so useful for computing stuff

like you just automatically know that if H isn't conjugate to a subgroup of K, then the mark is 0

which makes decomposing the product of G-sets a lot easier

[crystal vale]

Thank you I will look into it

[south patrol]

Specifically what I am doing is like global equivariant cohomology theories, so like uh combination of Mackey functor and cohomology theory ig

[mighty kiln]

Is that the same as the skeleton of G-Set

[south patrol]

And then a generalisation of characters for that

Now I am just trying to prove some lemmas which are apparently clear

[delicate orchid]

the underlying set of the burnside ring is all isomorphism classes of G-sets but I'm not sure how the morphisms would work

[delicate orchid]

which lemmas

[mighty kiln]

Oh wait is this like similar to Grothendieck groups on manifolds

[south patrol]

Uhhh I am reviewing a paper on this topic lol like by Hopkin-Kuhn-Ravenel

Like im showing given constructions r compatible w induction

[mighty kiln]

Where you put operations on class of some objects over something and then universal property into a group

[delicate orchid]

yeah all Grothendieck groups are the same in that respect

say guys, this K(-) functor seems rather important... perhaps we should invent some sort of theory to study it???

[delicate orchid]

sounds annoying. Just tensor with the group algebra omegalul?

[severe linden]

Does anyone have a hint to get me started on this one?

[mighty kiln]

Consider gH ∩ g'K

[delicate orchid]

and then hopefully it's clear why the first statement implies the second

[south patrol]

This is like induction of G-spaces rip

Well works the same way lol

But ye

[delicate orchid]

almost like taking representation rings is a Mackey functor or something

[south patrol]

I mean topological spaces etc hm

anway

i have proven it

[hidden haven]

No that's the underlying set of the Burnside rig

[hidden haven]

Lemma balls

[merry dawn]

what was the definition of a rational function $$R(x, f(x))$$?

[cloud walrus]

Sweet Tea 🧋

[merry dawn]

like, just a function whose numerator and denominator are linear combinations of the terms like x^n • f(x)^m where n,m are integers?

or smth different

[delicate orchid]

are you meaning to tell me that "-G/1" is actually just nonsense? No way....

[hidden haven]

Just as nonsense as "group"

[hidden haven]

Never seen this notation boss

[delicate orchid]

yeah

a rational function is just something in the field of fractions of ur polynomial ring in one variable

[glossy crag]

Got anything with some bite, chief? The contents are too basic.

[mighty kiln]

Rig

[delicate orchid]

it's an introduction. The first chapters are going to be simple

everyone has this complaint

[glossy crag]

Well I looked through the rest and it's very basic, I'd be fine with a text handling this in ~20 pages (except the stuff on induction and whatnot).

[delicate orchid]

out of spite I'm going to say Fulton-Harris

[glossy crag]

I leafed through it the other day. It's pretty tough, I agree, but what I really don't like is the haphazard nature of the presentation.

[delicate orchid]

abcdefghijklmnopqrstuvwxyz

[delicate orchid]

ok then Isaac's

[glossy crag]

What, am I wrong?

[delicate orchid]

no I just needed the alphabet for something

[teal vessel]

𐑶𐑬𐑫𐑜𐑖𐑗𐑙𐑘𐑡𐑔𐑭𐑷𐑵𐑱𐑳𐑓𐑞𐑤𐑥𐑒𐑢𐑣𐑠𐑪𐑨𐑦𐑩𐑧𐑐𐑯𐑑𐑮𐑕𐑛𐑾𐑲𐑴𐑰𐑚𐑝𐑟𐑼𐑸𐑹𐑺𐑻𐑽𐑿

if you're looking for the alphabet

(not in order, sorry)

[rocky cloak]

This hebrew or something?

[delicate orchid]

nah it's not hebrew - where's the aleph ykwim

[teal vessel]

it's ·𐑖𐑱𐑝𐑾𐑯, obviously.

[rocky cloak]

Shawian

[teal vessel]

it's an English alphabet, as opposed to the latin alphabet which we use for english all the time

[rocky cloak]

Might as well write in IPA at that point

[coral spindle]

If you haven’t heard of the Frobenius–Schur indicator and Clifford’s theorem, you have something to lean from James and Liebeck. Alternatively just read Isaacs.

[delicate orchid]

there's also a lot of great chapters which are nothing but computations, something I wish more books did

for example the one on GL(2, q) is a really nice introduction to parabolic induction

[teal vessel]

the whole point is to not have to do that. It's not phonetic, it's phonemic. The same spelling would be used regardless of if you pronounce your R's or distinguish between cot and caught, etc. a phonetic spelling would be unique to the individual, and make everything a nightmare, on top of having to memorize over 100 symbols and combinations. This is a very limited set, and it's arranged in groups so that you're really only dealing with a few symbol groups that then follow certain rules (𐑑 and 𐑛, for example, are t and d, since they make very similar sounds). But I digress.

[cloud solar]

The number of automorphisms of a finite abelian p group with p^n elements is product from i=1 to k of (p^(a_i)-p^(a_i-1)) where sum from i=1 to k of a_i = n ?

[delicate orchid]

so $A = \oplus_{i=1}^n C_{p^{k_i}}$ and $\text{Aut}(C^n_{p^{k_i}}) \cong GL_n(p^{k_i})$, so this is true for cyclic groups. My problem for it in general is take $C_2 \times C_4$ then $\text{Aut}(C_2 \times C_4) \cong D_8$ which is also a power of 2, so your formula wouldn't work

sorry not cyclic groups, I meant to say powers of cyclic groups

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

although tbh, I can also barely decipher what you're saying

[glossy crag]

I think I'll read Jacobson for now, then try Isaacs or Curtis/Reiner.

Man, why is Jacobson so heckin good and wholesome

[delicate orchid]

it's because the rep theory is the same with or without the Jacobson

if you know you know

[glossy crag]

I wish the man had written 20 more textbooks.

[hollow mica]

how do we get this line

this is a proof that shows coset product is well defined when your subgroup is normal

[delicate orchid]

bN = b'N <=> b^-1b' \in N

[hollow mica]

that's the line after it no?

[delicate orchid]

hence b^-1b' = n', so n(b^-1b') is in N

hence (ab)^-1(ab) in N

[delicate orchid]

it's also just obvious, multiply both sides of bN = b'N by b^-1, you get N = b^-1b'N

you don't need N to be normal, is the key part

[hollow mica]

I understand your first method

For the second method

[delicate orchid]

I only stated one thing

[hollow mica]

How does N = b^-1b'N help you when you want nb^-1b'

[delicate orchid]

because N = b^-1b'N if and only if b^-1b' is in N

[hollow mica]

ohh

nice

I see

gH = H iff g in H is something I forgot

[delicate orchid]

I agree that if they're going to be this annoyingly intricate with this proof then they should put an extra step in explaining this

[hollow mica]

proofwiki.org lol

am I being dumb

or is this a bunch of balogne

this is supposed to be another proof for above

[coral spindle]

It is a correct proof

[hollow mica]

oh coset product is the same as subset product for normal subgroups is a fact I didn't know

[coral spindle]

Yes this is the secret that they're not mentioning

They're using the symbol o differently to how one might otherwise

[hollow mica]

I still don't fully see it though

The first equality is justified by the fact I just mentioned

but then what the last equality shows is... the fact I just mentioned?

where's the a' b'

[barren sierra]

wdym

[hollow mica]

if we wanna show coset product is well defined, we need to show that for any a, a', b, b' with aN = a'N and bN = b'N that (ab)N = (a'b')N

that's the only phrasing of "coset product is well defined" that I know

[delicate orchid]

(aN)(bN) = aNbN = a(bNb^-1)bN = abNN = abN

[hollow mica]

That makes sense

[hollow mica]

Where is the connection to here tho

[delicate orchid]

they're just straight up equal as sets

and (a'N)(b'N) = (aN)(bN) follows from aN = a'N, bN = b'N immediately

[barren sierra]

If you really want to convince yourself

Write out the definition of aN as a set

And what the set (aN)(bN) means as a set

etc etc

Do the subset inclusions and all that jazz

[hollow mica]

This is saying that the product of aN and bN as subsets of the group (i.e. {xy | x in aN and y in bN}) is equal to the coset (ab)N .
Now if a'N = aN and b'N = bN then (a'b')N = (a'N) * (b'N) = (aN) * (bN) = (ab)N.

I see

[hollow mica]

wait @coral spindle this is what threw me off

there's no "fact" that they're using here right

they're literally just looking at the subset product

[coral spindle]

yes

As it happens they are equal.

[hollow mica]

Yeah the "fact" I mentioned is true but it's because of what is proven, it's not actually used in the proof

I thought I fell into circular reasoning which caused my confusion

[indigo ridge]

the characterisitc of Z_n is (n-1) not n right?

[mighty kiln]

char(Z/nZ) = n

[south patrol]

n-1 is not 0 mod n (ok unless n =1)

[solemn kettle]

Is the radical of an ideal generated by polynomials over a field of degree n generated by polynomials of degree less or equal than n?

[south patrol]

(x^2 +1) is a radical ideal of R[x]

So not in general no

[solemn kettle]

Less or equal*

[south patrol]

Oh lol

[solemn kettle]

I was asked to find the possible degrees of generators of the ideals of five points in the affine space of 2 dimensions, I know that in general the five points can be written as the zero set of a set of polynomials of degree 5 (depending on their position can be 4 or 3)

I dont't know if that means that the possible degree is at most 5 because the ideal of the five points is actually the radical of the ideal generated by those polynomials of degree 5

[chilly ocean]

Let $n \in \mathbb{Z}_{\geq 1}$ be a fixed positive integer and let $G$ be an abelian group. Prove that $H = {g \in G: g^{n} = e}$ is a subgroup of $G$. Is the result still true if $G$ is not required to be abelian? Justify your answer.
\begin{proof}
The set $H$ will at the least contain the identity element of $G$. So $H$ is nonempty.
Let $a$ and $b$ be elements of $H$. Since $b$ is also an element of $G$, there exists an inverse $b^{-1}$ such that:
\begin{equation*}
\begin{split}
(ab^{-1})^{n}
& = a^{n}(b^{-1})^{n} \
& = a^{n}(b^{n})^{-1} \
& = ee^{-1} \
& = e.
\end{split}
\end{equation*}
Thus $ab^{-1}$ is an element of $H$, and we can conclude that $H$ is a subgroup. Note that if $G$ is not abelian, we cannot let $(b^{-1})^{n} = (b^{n})^{-1}$.
\end{proof}

[cloud walrus]

gian

[chilly ocean]

Is my reasoning for G needing to be abelian correct?

I'm leaning towards no but I don't know why...

[tough raven]

Almost correct, but in fact (b^-1)^n = (b^n)^-1 in any group G. There's a very related step that does use the abelianness of G.

[chilly ocean]

Hmmmm

Oh

Is it $(ab^{-1})^{n}
& = a^{n}(b^{-1})^{n}$

[cloud walrus]

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

[tough raven]

Yep!

[chilly ocean]

Cool!

Do you mind giving me an explanation as to why?

[tough raven]

Well, why should that be true in an abelian group?

(ab^-1)^n = a b^-1 a b^-1 … a b^-1, whereas
a^n (b^-1)^n = a a … a b^-1 b^-1 … b^-1.

[chilly ocean]

Oh

Duh lol

Thanks so much

One last question:

[cloud walrus]

gian

[chilly ocean]

Why do we disregard p|(a-1) but not p|(a+1)? Since 1<= a <= p-1, its obvious that a-1 will be less than p. But why is that not the case for p|(a+1)?

This might be more of a number theory question

[mighty kiln]

Cuz if a = p-1 then p divides a+1

Though p also divides a-1 if a = 1? And 1 is indeed its own inverse

This step is phrased weirdly actually

[celest furnace]

It’s wrong

[mighty kiln]

But anyway I think you can see what the argument is

[celest furnace]

1 is its own inverse

[chilly ocean]

Yes

1 is its own inverse

We know that $1 \leq a \leq (p-1)$ and $p \mid (a-1)$ Assume that $a = p-1$. Then $p \mid (p-2)$. This cannot be true.

However for $p \mid (a+1)$, if we assume that $a = p-1$, then $p \mid p$.

[cloud walrus]

gian

[chilly ocean]

Is this the correct line of reasoning as to why we disregard p|(a-1) but not p|(a+1)

I also assumed that 1 being its own inverse is trivial and has nothing to do with the solution at hand

[mighty kiln]

I don't think we "disregard" any of them

Both work

And you get a = 1, a = p-1 as solutions

For a² = 1

[chilly ocean]

I thought if b<a you cannot have a|b

[mighty kiln]

p | 0

[chilly ocean]

?

[mighty kiln]

p divides 0

[chilly ocean]

From where though

[mighty kiln]

If a = 1, then p | a-1

[chilly ocean]

Oh

Thats bad

My instinct is to change the inequality to $2 \leq a \leq p-1$

[cloud walrus]

gian

[chilly ocean]

This kinda circles back to what I was thinking where 1 being a solution is trivial

[celest furnace]

Basically you will find that Zp[x] is a field so x^2-1=0 has <= 2 solutions: you find the two, so there are no more

And you prove this by factoring x^2-1 as (x+1)(x-1) and clearly seeing the roots are +-1

[chilly ocean]

But this isnt Zp[x] this (Z/pZ)^x

The group containing the units of Z/pZ

under multiplication

[celest furnace]

Yes and in Z/p^x you are looking for solutions to x^2-1=0

[chilly ocean]

Ehhhhhhhh correct me if I'm wrong but I think theres a distinction to be made between solving $x^{2}-1=0$ vs $a^2 \equiv 1 \pmod{p}$

[cloud walrus]

gian

[celest furnace]

Well if you try to solve x^2-1 over Z/p (like coefficients/roots coming from that base field) then they are the same

[topaz solar]

not if your solutions are in the space of mod p things

[chilly ocean]

Hmm

So we shouldn't disregard p|(a-1) as a solution?

[lime junco]

wondering if yall an let me know if my proof is valid

or if im tripping rn

the question is

my proof is

[south patrol]

To me this seems a badly worded question

If they are both roots of f and both roots of g then it is easier than you write, because irreducibility means f and g are both (up to scaling) min polys of alpha and hence multiples of one another

And conversely if they mean to say alpha is a root of f and beta a root of g, then the statement is false

Because one of alpha and beta could already be in Q

[lime junco]

they mean thisi believe

[south patrol]

Sure. Then the problem is still weird lol

[lime junco]

why is this true? its easy enough to show that if f is irreducible over Q(a), then its a scalar multiple of the minimal polynomial of alpha

but why does g have to be?

[south patrol]

Well there is complete symmetry between f and g etc in that case

[lime junco]

wdym by complete symmetry

like you can interchange them in the statement of the theorem?

[crystal vale]

What if I take Z6[x] and take f(x)=(x-2)(x-3) has 3 roots ....

[dull marsh]

Is K an arbitrary ring?

[crystal vale]

I think so

[untold cloud]

Z/6 is not a field

Guess this is true only for field K

[dull marsh]

(x - 2)(x - 3) has 4 roots in Z6[x] btw

[jaunty rivet]

Correct me if i'm wrong, but aren't K and L isomorphic if either f or g are irreducible, implying the other is?

[crystal vale]

Oh yeah I just want more than 2 roots

[crystal vale]

Yes but there is no mention that k must be a field

[jaunty rivet]

I believe K has to be an integral domain

[dull marsh]

Is K nowhere mentioned before in that section?

[untold cloud]

I am not sure if integral domain is true. I am thinking the proof is like taking integral closure, then decompose f into linear polynomials. Probably need k to be ufd? If k is integral domain, will f not monic affect the existence of roots?

[crystal vale]

They don't introduce ufd yet

[jaunty rivet]

I'm gonna be honest, I have no idea, but I believe you

I mean clearly though anything weaker than integral domain will make that theorem not true

[south patrol]

Ye

[rocky cloak]

Moving from an integral domain to it's field of fractions doesn't remove any roots, and the statement is true for fields.

Being a UFD doesn't really play into it.

[crystal vale]

So is it for only an integral domain?

Hint

[viscid pewter]

essentially you need to show that t gets mapped to some linear function

so try and look for issues that would happen if t were mapped to a polynomial of higher or lower degree

[crystal vale]

Yeah I take t map to an_t^n + a_n-1_t^(n-1) + ........ + a1_t + a0 and then I want to show that these all an,.....,a2 are zero and a1 can't be zero because then t will be belong to F but t is variable over F

[dull marsh]

Sounds like overcomplicating

The only essential thing is the degree of the polynomial that t maps to

[crystal vale]

That means automorphism preserves the degree of polynomial

[dull marsh]

That's what you are trying to show, yes

[south patrol]

Think about surjectivity

[crystal vale]

Yes

[astral fractal]

I don't understand what this statement is trying to say or the signifiance https://gyazo.com/1f9f265b175a9f536c05954317282d37

[dull marsh]

What exactly is confusing you about it?

[astral fractal]

the range

And the notation of <+- 1>

[dull marsh]

The codomain is the multiplicative group {-1, 1}

But you can let it instead be Z/2Z as those are isomorphic

[astral fractal]

So the codomain is not supposed to be some generator presentation or something?

Since they use the "<" and ">"

[dull marsh]

No

I find it weird as well, writing {+-1} would suffice

[astral fractal]

Hmm ok, but what about the statement that every transposition is sent to -1?

[dull marsh]

It's a property that the homomorphism should satisfy

The one which exists according to the proposition

[astral fractal]

Right but is there a reason that all the transpositions must be mapped to -1?

Why not 1?

[dull marsh]

Because then everything will be sent to 1 and you will have a trivial homomorphism

(Everything will be sent to 1 because every permutation is a product of transpositions)

Also choosing every transposition to be sent to -1 gives a surjective homomorphism and therefore you will find a normal subgroup in Sn by looking into the kernel

That subgroup is called the alternating group and, as far as I know, plays an important role later on

[astral fractal]

So if epsilon's a homomorphism, then the condition that epsilon(sigma1 o sigma2) = epsilon(sigma 1) * epsilon(sigma 2) implies epsilon(sigma1 o sigma2) = 1? (sigma being permutations in S_n and o being composition)

and since epsilon(sigma) = -1 and <+- 1>'s group operation is multiplication

[dull marsh]

If by sigma, sigma1 and sigma2 you mean transpositions, then, yeah

[astral fractal]

i see

[astral fractal]

ah right

my statement would have meant that the compsoition of every transposition (permutations of Sn) would map to -1?

when it's really just every transposition

[dull marsh]

I don't think I understand what you mean

[astral fractal]

so

it says every transposition maps to -1

[astral fractal]

here i said sigma was a permutation and implied that epsilon(sigma) maps to -1

if my understanding's right then permutations can be compositions of transpositions, but the statement only says that transpositions can map to -1 and nothing of compositions

[dull marsh]

For the rest of the permutations you just use the homomorphism property

Write your permutation x as a product of transpositions s1s2...sn and you will have eps(x) = eps(s1...sn) = eps(s1)...eps(sn) = (-1)^n

So the only thing that matters in x is the number of transpositions that you can write it with

[astral fractal]

i see

[dull marsh]

A well-known fact is that the parity of that number is indeed well-defined

[astral fractal]

so transpositions must be -1 but permutations need not be -1 and can be 1 if n is even

[dull marsh]

Yes

[astral fractal]

i see that makes sense now

thanks for the help

[solemn garden]

I’m stuck on this question: Let R be a commutative ring with identity. Then let p(x) be a polynomial in R[x] which is a unit. Prove that the constant term of p(x) is a unit, and that all of its other coefficients are nilpotent.

I have already shown that the constant term must be a unit, but not sure how to proceed

I am allowed to use that a polynomial is nilpotent iff its coefficients are all nilpotent, if that helps

Can anyone give a hint?

[coral spindle]

Let q(x) be the inverse of p(x) = ax^n + (lower degree terms). Then ax^nq(x) + (lower terms)q(x) = 1. Think carefully about the coefficient of the highest power of x here, then the 2nd highest power of x, etc.

I am struggling to make the hint better than this, sorry.

You will have to use induction, that's about the best addition I can come up with.

[solemn garden]

Ok, I'll try that

[coral spindle]

Wait, I have a better hint

Let a_n be the highest coefficient in p(x). So you have already established that deg (a_n q(x)) <= deg q(x). What about deg( a_n^2 q(x) ) ? Continue with this pattern.

[solemn garden]

This looks like the proof to the theorem that if a polynomial p(x) is a zero-divisor in the ring of polynomials, then there is some constant c such that c*p(x)=0

[solemn garden]

but isn't a_n q(x) just 0

by your first message and comparing coefficients

[coral spindle]

Uh no I don't believe that's true

At least I don't see the argument

[solemn garden]

because a_n x^n q(x) = 0 by comparing coefficients, because all terms have degree greater than or equal to n

[coral spindle]

But the terms (lower terms)q(x) may have the same degree

You can only guarantee that the coefficient of x^(n + deg q(x)) is 0 immediately

[solemn garden]

Ohh

[coral spindle]

Perhaps I'm just forgetting something, but at least I do not follow your logic as you've stated it there.

[solemn garden]

No, I see my mistake

But I think deg (a_n q(x)) < deg q(x) since the leading term has to be 0, right?

[coral spindle]

Yes

It's also worth having an example to test, the simplest one I can come up with is (3x+1)(6x + 1) = 1 in (Z/9)[x], whence you can see what you claimed isn't true

[solemn garden]

Let a_n q(x) = a_n b_0 + … + a_n b_i x^i, where a_n b_i != 0. This b_i exists because b_0 isn’t a zero divisor, it is a unit. Then multiplying both sides of a_n x^n q(x) + (lower terms) q(x) = 1 by a_n, the result is a_n x^n (a_n q(x)) + (lower terms) (a_n q(x)) = a_n, so by comparing coefficients, a_n^2 b_i = 0, so that deg (a_n^2 q(x)) < deg (a_n q(x)).
Is this correct so far?

Then the same argument applies to lower the degree of a_n^r q(x) for each incrementing r, until a_n^r b_0 is 0. Then multiplying both sides by a_0 proves that a_n is nilpotent

[solemn garden]

Oh I just solved it

Because of this, the leading coefficient is nilpotent

So -a_n x^n is nilpotent

Then add that to the original polynomial, which is a unit

Since the sum of a unit and a nilpotent is a unit, and the result has degree n - 1, the inductive hypothesis can be applied to it, and that completes the induction

[solemn garden]

Thanks!

[fringe heath]

hey, for a question like this (1.2 ii) is it sufficient to show that if two elements were in fact gcds then they are in fact not associated and thus gcd doesn't exist?
and thus since 7 and 1+sqrt(-13) are irreducible AND not units, the quotient of them is not a unit and so gcd doesn't exist?
or am i missing something

[dire siren]

@fringe heath I think there is a problem in your reasoning, namely that you are assuming without proof that 7 and 1+sqrt(-13) are gcds

maybe use gcd(14,7+7sqrt(-13))=7gcd(2,1+sqrt(-13)) and part (i)

[fringe heath]

i see

oh i understand

wait nvm im not sure i understand

if we take 7gcd(2,1+sqrt(-13)) instead, how would i proceed?

oh i understand

i think

so 2 is irreducible, thus gcd(2, 1+sqrt(-13)) must be a unit
if we represent this by d, we are saying that 2d = (1+sqrt(-13)) which is equivalent to what i said before except without the assumption

that is, that they're not associate

thank you for the correction filip

[dire siren]

I'm not sure about the relation 2d = (1+sqrt(-13)), but once you found that gcd(2, 1+sqrt(-13)) is a unit, from the equality that I suggested it follows that 7 is a gcd for those numbers;
then it is enough to notice that 1+sqrt(-13) is a common divisor (no need to prove that it should be a gcd), so then 1+sqrt(-13) must divide 7, which is a contradiction

however, I'm not entirely sure if the equality gcd(7x,7y)=7gcd(x,y) can be considered obvious given that we just assume that the first gcd exists (more precissely, the equality should be read as: "if gcd(7x,7y) exists, then gcd(x,y) also exists and we have gcd(7x,7y)=7gcd(x,y)"

we can avoid this if we say something like: assume gcd(14,7+7sqrt(-13)) exists and denote it by D; then since 7 divides both numbers, we must have 7 | D, so D=7d, and then we can proceed as above

[fringe heath]

thank you thank you

[summer notch]

Given a group G and a subset S (not necessarily a subgroup), what does the group action from G to G/S (the set of left cosets of S) induced by left multiplication look like?

we mostly see this when S is some subgroup H (or especially some normal subgroup), but everything works when it's just some subset

although it seems like when S is not a left coset of a subgroup, the group action from G to G/S is just isomorphic to one from G to G

(note that of course when S is not a left coset of a subgroup, G/S will not be a partition of G; some left cosets will overlap)

[summer notch]

i guess it just depends entirely on whether there exists some nonidentity g in G such that gS=S

for example S={0,3,4,7,8,11} in G=Z/12Z, then the action from G to G/S is equiv to the one from G to G/{0,4,8}

[celest furnace]

What is the intuition behind $\R[x,y]/(x^2+y^2-1) \not \cong S^1$?

[cloud walrus]

rbmuk

[coral spindle]

Wym? It's not the circle, it is a set of functions from the circle into R. Why would it be the circle?

In what sense do you even mean isomorphic here. As topological spaces? You surely don't mean as vector spaces or rings

[celest furnace]

Set of functions from circle to R?

[coral spindle]

Yes, R[x, y] is (isomorphic to) the ring of polynomial functions from the plane into R

Similarly R[x, y]/(x^2 + y^2 - 1) is the ring of polynomial functions from the unit circle to R

This question confuses me, I think you're missing some large context that I really can't infer.

[celest furnace]

Yeah that’s what I was missing

Didn’t realize it would be functions

[coral spindle]

You didn't realise that R[x,y] consisted of polynomials?

I'm still very confused, but whatever

[celest furnace]

No I did but I thought quotienting out by f(x) Is the same as setting f to 0

So if you set x^2+y^2=1 maybe you just get the circle

But it was polynomials on the circle

[coral spindle]

Right...

[south patrol]

Note as well that there is a difference between polynomial functions and polynomials

fortunately as we are working over R, there is an isomorphism between the two but yeah

[south patrol]

But yes here is one way to view it is like

If we consider the ring of all functions on $S^1$, call that $A$, then the ring of polynomial functions is by definition the image of $\mathbb R[x,y]$ under the natural map $\mathbb R[x,y] \to A$. The kernel is those polynomials which vanish on the circle, and those are divisible by $x^2 + y^2 -1$

[cloud walrus]

søtpotet

[south patrol]

then first iso lol

[coral spindle]

Ig it's useful to remember that if k is an infinite field then there's no problem

We do get the occasional person here who thinks e.g. F_2[x] is a finite ring...

[celest furnace]

This helps a lot thanks

[south patrol]

Ye i just think like idk

They are sort of different concepts even if they end up being the same ig lol

[delicate orchid]

literally characteristic 2 sweaty.... how can it be infinite....

[south patrol]

Compute roots of unity in p-adics

[dull marsh]

{0, 1, x, 1+x} looks finite to me

[elfin horizon]

small silly question but none of the resources for my class clarify this: Are the elements of Z^x_n under multiplication the invertible elements in Z under multiplication, or the invertible elements in Z^x_n under multiplication

[coral spindle]

The invertible elements in Z under multiplication are just -1 and 1, so this would be a very uninteresting definition – the same set of two elements for all n.

Indeed when we talk about $(\bZ_n)^\times$ we are talking about the units in $\bZ_n$, the integers modulo $n$.

[cloud walrus]

Boytjie

[prisma ibex]

Z_n
ahh yes the n-adic integers

[coral spindle]

Yes well unfortunately algebra beginners use Z_n for Z/n, who am I to blow against the wind.

I would rather not confuse people who have never seen a quotient structure before

Fwiw I hate it too

[elfin horizon]

Oh yeah I just realised this would be v silly, Ty for the help ❤️

[crystal turtle]

One of my profs was trying to argue (to a senior UG/grad course) that you should write Z/pZ, Z_p, or Z_(p) , and that Z/p was not a good notation

[prisma ibex]

bruh

[coral spindle]

Surely Z/(p) at least was suggested?

[south patrol]

Weird

Z_p is the worst

And Z_(p) means something completely different (p-local integers)

Z/p is entirely unambiguous

[coral spindle]

Well I mean it could refer to the Z-module $\frac1p \bZ$

[cloud walrus]

Boytjie

[coral spindle]

But we'd write that in that case, and not Z/p

Woah I feel so awful

someone take away my tex licence

[mighty kiln]

What font is that

[south patrol]

yeah exactly

Looks palatino-ish but not quite hm

[coral spindle]

Some Palatino clone, I think it's newpx + newpxmath

[south patrol]

nice serifs

lol

nice

[coral spindle]

Nice one

👏

[south patrol]

second guessed myself

well look at my dissertation

:^)

[coral spindle]

Oh nice, probably exactly the same packages right

[south patrol]

yeah exactly

[coral spindle]

I love palatino's italic. Really a winner

[south patrol]

Fr

[coral spindle]

Anyway I could go on about fonts for too long

[south patrol]

Recently started enjoying learning about fonts lol

and appreciating them

[coral spindle]

It's honestly very rewarding. Lots of my friends now ask me about fonts lol

[south patrol]

i also really like lol all caps stuff but with bigger/small capitals like this

in palatino

this is uh

amsart document type

[barren sierra]

https://mathweb.ucsd.edu/~msharpe/mathsamples.pdf

here's a fun list of fonts to play with

[uneven jackal]

I don't know why but Z/n instead of Z/nZ always felt wrong for me

[barren sierra]

I agree

since nZ is common notation for an ideal

And yes (n) is alternative notation but then write Z / (n), not Z / n

[uneven jackal]

tbf it's not ambiguous but still it itches my brain for some reason lol

[barren sierra]

agreed

[crystal turtle]

I am a lazy bastard

[barren sierra]

[cobalt heath]

The most studious algebraist

[boreal inlet]

I am tasked with finding the intermediate subfields of the extension F_p^n/F_p which correspond to the subgroups of the corresponding Galois group.

What I did is that prove the extension is itself Galois, and also proved that Galois group is cyclic, generated by the Frobenius automorphism.

Now as this is a cyclic group of some order n, all the subgroups of this group are also cyclic, and the generator is precisely of the form (Frob)^d for d|n.

I showed the fixed fields of the subgroups are just F_p^d, and I claim those are the required subfields. Is this correct, or do I need to show something else also?

[cobalt heath]

Seems correct to me, that should make the field an intermediate subfield with corresoonsing galois group

[boreal inlet]

Thankss

Oh right, the correspondence is just H -> L^H

so this is fine

[cobalt heath]

Indeed!

[boreal inlet]

[cloud walrus]

Riku

[boreal inlet]

How do I show this?

If p was less than or equal to 3, that would be easy to show, as none of the elements in F_p are roots of this polynomial.

Oh right, I see that, if alpha is indeed a root of this polynomial, alpha + a is also

Doesn't that mean, this alpha is unique?

[rotund aurora]

for a in F_p, yes

[boreal inlet]

Yes

So, would I be right to assume that a polynomial of degree p in F_p[x] can have at most p roots?

[rotund aurora]

yes

[boreal inlet]

I'm asking because I have seen char p cases being really different than char 0

Ok thanks

So if it can have atmost p roots, this alpha has to be a unique number

[rotund aurora]

if the leading coefficient is not zero and the degree is n, there are at most n solutions always (in fact exactly n, with multiplicities, in an algebraic closure)

and this is true over any field

[rotund aurora]

no

[boreal inlet]

Oh I think I see what you mean

Take any root alpha and then the other roots can be expressed in this form

[rotund aurora]

Correct

so say alpha is a root of your polynomial. You know alpha not in F_p. Then, all the roots are alpha,alpha+1,alpha+2,...,alpha+p-1

[boreal inlet]

This proves the polynomial is seperable

I also want to show this is irreducible

Can we show that any finite product of (alpha + a_i) cannot be in F_p?

I think that is enough

....but that's also hard to show

[rotund aurora]

yeah, that would be enough

but it may be a bit hard

[boreal inlet]

Because if this is true, then there cannot be any polynomial in F_p[x] that divides the original

[celest furnace]

There is a weird trick

[rotund aurora]

don't spoil it pls

[celest furnace]

Have you determined it doesn’t have a root

I won’t

[boreal inlet]

In F_p, yes

[celest furnace]

Hint: think about ||UFD||

[rotund aurora]

and also, it's not weird

[celest furnace]

[rotund aurora]

mmh I'm not sure how that is useful lol

[celest furnace]

?

[cobalt heath]

How did you get here?

Ah, irreducibility

[rotund aurora]

You had the right idea here, but note that a polynomial has more coefficients other than the constant one

[rotund aurora]

I think I get what you meant, but Riku used that already I think, so felt out of place

[cobalt heath]

Perhaps one can use that if alpha is a root, then F_p (alpha) is the splitting field

Wait, that just works

Huh

Something is wrong

[boreal inlet]

ohhhh wait.

See the leading term we don't care about- oh no wait.

Let's say we are multiplying a quadratic polynomial whose roots are (alpha + a_i) and (alpha + a_j) with (x - (alpha + a_k)). The product will have the term x^2 with coefficients not in F_p.

So if we just show any product of form (x - alpha - a_i)(x - alpha - a_j) is not in F_p, we are done

oh. 😔

[rotund aurora]

it just doesn't follow, you could have the polynomial decompose as a product of irreducible polynomials of degrees >2

[cobalt heath]

Do you recall how each coefficient of polynomials is expressed in roots, right

[boreal inlet]

No i mean, if we don't have any degree 3 polynomial dividing this, then we are kind of done no? Let's say we decompose it in quadratics and linears. degree p is odd. So there has to be a linear factor somewhere. that can't work

[rotund aurora]

this doesn't follow btw. You could have (x^2-x+1)(x^2+x+1) (say over Q) then any root generates the splitting field and they are all distinct

[granite birch]

Is real analysis a prerequisite for this math?

[rotund aurora]

No

[granite birch]

Thanks.

[rotund aurora]

No, this approach is doomed, really

[boreal inlet]

oh... sad

[boreal inlet]

For Quadratics yes

I think we can generalize it also

[celest furnace]

Could you not have it have a factor being an irreducible quintic?

[rotund aurora]

You would have to check that for every partition a1+a2+...+an of p there do not exist polynomials P1,...,Pn of degrees a1,...,an, respectively, such that your polynomial=P1(X)...Pn(X), but the degrees ai can be almost anything between p/2 and 1

[boreal inlet]

....oh my god.

[rotund aurora]

and primes behave generically wrt to partitions, so the fact that p is prime doesn't make the number of partitions be smaller or anything

it just rules out a particular kind of parittion LOL

namely, partitions of the type m+m+...+m for some m>1

[boreal inlet]

That is also possible...

[rotund aurora]

so if you don't know, try finding formulas for the coefficients of a polynomial in terms of the roots, this is something very useful. Maybe try to look for a coefficient which has a simple expression in terms of the roots. The constant coefficient is one as you pointed out earlier, but there does not seem to be an easy way to show that it doesn't belong to F_p

[boreal inlet]

Ohhhh

Sum of roots

That's the coefficient of x^p-1

Lesgo finally brain brained

ok so

uhm.

If I just add them up, it definitely isn't in F_p right

p • alpha

Not only that,

any combination as such should not be in F_p

[rotund aurora]

yeah so if you add, say, k roots, then you will get something like k·apha+beta where beta in F_p. So in particular, k·alpha in F_p. But this is only possible if k=0 mod p, because as you proved earlier alpha not in F_p

[boreal inlet]

......wait a minute i feel like I have seen this exact response from someone else here

I feel a crazy deja vu type feeling all of a sudden

In any ways, yes, this is correct

[rotund aurora]

[boreal inlet]

It probably was you telling someone else about this and i saw it before

Anyways

[boreal inlet]

This gives us a contradiction

[rotund aurora]

You have this https://en.wikipedia.org/wiki/Vieta's_formulas

this is pretty important

[boreal inlet]

Thank

[rotund aurora]

you don't need to remember exactly the formulas, with all the sign changes and whatever, might be confusing. But just be aware that you can express the coefficients in terms of the roots, and to derive such expressions, just multiply out (x-a1)(x-a2)....(x-an)

[boreal inlet]

yes that's what i did

[rotund aurora]

and look at the coefficients to see what you get

yeah, well done

[boreal inlet]

It inductively goes on

One question

p • alpha must be 0. In the algebraic closure.

Because otherwise there isn't even a root in the algebraic closure

Which is absurd

[rotund aurora]

the algebraic closure still has characteristic p

[boreal inlet]

Oh right

[cobalt heath]

Ah, my bad. Yea we need normality
EDIT: no, not even that is true

[rotund aurora]

if L/K is a field extension then the characteristic of L is the same as the characteristic of K (except perhaps if K contains just one element)

[cobalt heath]

Dunno why autocorrect decided to translate normality into locality..

[boreal inlet]

So I'm wondering how F_p[x]/(x^p - x + a) would look like

It is a field with p^p elements

[rotund aurora]

mh I guess the Galois group is Z/pZ

[rotund aurora]

this type of extensions are actually well studied, they are called Artin-Schreier extensions

[boreal inlet]

Oh, the automorphisms are generated by the t |-> t+1

[cobalt heath]

t ?

[rotund aurora]

right, that works

[boreal inlet]

I mean uh t is like a placeholder variable

[rotund aurora]

yeah, I understood don't worry

So, there is a converse to this

which is amazing

[boreal inlet]

Damn, actually?

[cobalt heath]

Ah, you still map 1 to 1 right

And only map alpha to alpha + 1

[boreal inlet]

Wait lmao isn't this just plain isomorphic to F_p^p

[sharp sonnet]

there arent too many fields with q elements for any given q 😄

[rotund aurora]

so if k has characteristic p and K/k is a cyclic extension of degree p (meaning it's Galois and the Galois group is Z/pZ) then K can be obtained from k by adjoining a root to k of a polynomial of the type X^p-X-a with a in k

[boreal inlet]

"Any two finite fields with p^n elements are isomorphic as F_p-algebras."

[rotund aurora]

You can call this Artin-Schreier theorem I guess

[cobalt heath]

Hmm I think there is a way to prove irreducibility using field extensions, tho not as simple;

[rotund aurora]

Irreducibility only makes sense if you specify a field with which the polynomial ought to be irreducible with respect to. So you are pretty much always considering field extensions when dealing with irreducibility

[rotund aurora]

You can look Lang page 290 @boreal inlet

What you would like to do is to classify all Galois extensions of a field with prescribed Galois group

over Q this is very difficult and the abelian case is known as class field theory

[cloud walrus]

Absta

[boreal inlet]

Thanks a lot

I'll check this out

[merry dawn]

how is $a^\frac{n}{m}$ defined, for $a \in \mathbb{R}$ and $\frac{n}{m} \in \mathbb{Q}$?

Is there any difference between
$$
\sqrt[m]{a^n} \quad \text{and} \quad \left(\sqrt[m]{a} ,, \right)^n
$$
?

[cloud walrus]

Sweet Tea 🧋

[merry dawn]

assuming that $a^\frac{1}{m}$ was defined constructively, by showing that $$s := \sup {t \in \mathbb{R} , | , t^m < a}$$ exists and is such a number that $s^m = a$

[cloud walrus]

Sweet Tea 🧋

[scarlet estuary]

the order is irrelevant* since x^yz = (x^y)^z = (x^z)^y for nonnegative y, z

* just normalize n/m to a nonnegative value by extracting a -1, and then taking the reciprocal at the end, to avoid sign issues

and yeah, a^n/m is just defined as (a^1/m)^n, or alternatively (a^n)^(1/m)

in order to make exponent rules work

again, the caveat is that, if n/m < 0, you instead take absolute values of n, m and then take a reciprocal

just to avoid headaches

[crystal vale]

Give me more hint

[dull marsh]

Let's say t gets mapped to a polynomial of degree m

What's the degree of the polynomial that a polynomial of degree n gets mapped to?

(Apply the homomorphism property)

[crystal vale]

mn?

[dull marsh]

Right

Now let's consider the case m >= 2

Can the linear polynomials be in the image of this mapping in that case?

[merry dawn]

thank you

[hidden wind]

this book (audrey terras) is rather odd, but ngl it’s grown massively on me haha

i was rather put off at first by how it gives very few proofs but leaves a lot to the reader, but since i first skimmed this book two years ago, i’ve grown rather fond of this approach 🙏

also, uhm:

[crystal vale]

No, got it

Thank you

[crystal vale]

I want to know what is the thinking process for these types of questions

[crystal vale]

In that case where he used the p is irreducible property and I think they assume k be p.i.d because how then they define gcd concept....if I am correct gcd not define on ufd ? Right
P is irreducible therefore gcd of p and f is 1 ?

[dull marsh]

Don't forget about the case m = 0 btw

Once you get a contradiction in both cases, m = 1 is forced

[elfin wraith]

I’m quite confused by this problem and I really don’t seem to be making any progress, I’ve to find all homomorphisms $$g:\mathbb{Z}/150\mathbb{Z} \to \mathbb{Z}/70\mathbb{Z}$$, and list all the values of $$g(1)$$

I know we start by noting a ring homomorphism is a group homomorphism and that $$150g(1)=g(0)=0$$ so it should just be a case of solving $$150a \equiv 0 \mod 70$$, and these are all the possible numbers that could be solutions, but I’m struggling to see which actually are solutions and why

[cloud walrus]

Nope

[elfin wraith]

Obviously g(a)=0 is always a solution but I’m not sure how to deal with the other multiples of 7

Also sorry for the slight latex gore but ya know, mobile

I’m also open to any methods to solve this using sage, as well as better understanding how to to it by hand

[crystal vale]

Yes when m=0 then it will be not hold because then t must be an element of F which is absurd I think

[crystal vale]

You should think about g(1)= a and then which element has a^2=a in Z_70

And a has order d which divides both 70 and 150

[elfin wraith]

Oh yeah that’s actually really simple lol

I don’t know why it didn’t occur to me to just solve a^2=a mod 70

Thank you

[balmy belfry]

The bears is funny

[stark helm]

can someone tell me why (12)(23)=(132)?

[dull marsh]

Statements that are not true cannot be explained

[stark helm]

then what will (12)(23) equal?

[dull marsh]

(12)(23) = (123) assuming you have defined composition conventionally (from right to left)

[stark helm]

map from 2 to 3, 3 to 2 and 1 to 1 on the right side, then saying that 2 to 1

[balmy belfry]

I hate that notation

Why is it so confusing

[dull marsh]

From right to left?

[balmy belfry]

Bean do you know how to define even permutations without the weird () thing

[balmy belfry]

No ()

[dull marsh]

Ah

[balmy belfry]

Like (12) etc

I like the full map representation

[dull marsh]

You want me to explain the notation?

[balmy belfry]

Like the matrix

[balmy belfry]

Idk is it hard

[dull marsh]

No

[balmy belfry]

Oh ok

[dull marsh]

For each element you just look at the element standing right to it

And that's going to be its image

(If the element is on the far right, you look at the first element)

[balmy belfry]

Oh

[dull marsh]

So (1 2 3) is the same as 1 -> 2, 2 -> 3 and 3 -> 1

[balmy belfry]

(12) is 1->2

And 2->1

[dull marsh]

So the first permutation (2 3) maps 2 to 3 and then 3 is fixed by (1 2), meaning in the product we have 2 -> 3
Then (2 3) maps 3 to 2 and then 2 is mapped to 1 by (1 2), meaning in the product we have 3 -> 1
Obviously 1 must then be mapped to 2 so you get that the product is (1 2 3)

[dull marsh]

Yes

[slim kayak]

so true frogsie

[balmy belfry]

Yay

I’m so happy

Frogsie is such a sick nickname too

[dull marsh]

This is called cycle notation just in case

[balmy belfry]

Now I can finally stop avoiding alternating

Group

[hidden wind]

anyone know if rotman's introduction to the theory of groups is any good?

[static glen]

I used it to learn about free groups and presentations of groups, i did like it

[hidden wind]

wonderful

[coral spindle]

I haven't read that particular book, but Rotman is a good writer.

[cloud solar]

Let A be a ring with 2^n - 1 invertible elements and fewer non invertible elements. Prove A field

It is clear that A finite. So the elements are invertible or zero divisors.

Let d be a zero divisor the function f:U(A)->D (D is the set of zero divisors) f(x)=xd is not injective.

but f is surjective

So there is a in D with a=u1×d=u2×d with u1,u2 two distinct invertible elements

But again there is b in D with b=v1×a=v2×a with v1,v2 two distinct invertible elements. b=v1×u1×d=v2×u2×d.

Idk how to prove the only zero divisor is zero

I feel like it is an iteration to do

[hidden wind]

don't mind me posting this picture of camille jordan

[burnt matrix]

yo can I ask anyone here to take a look at a group theory question in my help channel

thanks

[silent delta]

is it true that in an integral domain if y does not divide x it also does not divide any integer powers of x?

[celest furnace]

No

4 doesn’t divide 2 for example

[silent delta]

wait lol I think I formulated my question wrong

suppose we have an integral domain, and that x, y are in the ring, and that x,y don't share common factors. Is it true then that x^n and y also don't share common factors?

[crystal vale]

Yes, If I can define the gcd concept on it then it is true

[silent delta]

would that require the ring to be euclidean though?

[crystal vale]

No

You can define the gcd concept on ufd

[silent delta]

so it would require the ring to be a UFD then?

[crystal vale]

I am not sure about it, I need gcd concept and there is a possibility that someone does not gcd concept for it

[silent delta]

i see

[formal ermine]

theres not really a concept of common factors unless ur in a ufd

[silent delta]

why not?

if x,y are both multiples of k then k is a common factor no?

[formal ermine]

Yeah thats true but they wont necessarily be unique i think

but then again its 6 am and i could be talking pure nonsense

ill try to think of an example later if its even true

[silent delta]

what do you mean by this?

[celest furnace]

There is something called a gcd domain

So maybe there it would be true

[formal ermine]

@silent delta Ok so I thought about this a bit

you wanna show that: there doesnt exist a v such that v | x and v | y => there doesnt exist a v such that v | x^n and v | y

right?

if yes then just take the contrapositive: exists v st v | x^n and v | y => exists v st v | x and v | y

and if you dont have prime decomposition (arent in a ufd) then this should be false im pretty sure

[rocky cloak]

The notion of common divisors makes sense, but there is no guarantee that there is a greatest common divisor.

For example in the ring Z[sqrt(-3)] if we consider 2(1+sqrt(-3)) and 4, then both 2 and (1+sqrt(-3)) are common factors. But neither are multiples of one another and no proper multiple of them is a common factor.

[cobalt heath]

Just curious, is it possible to formulate this geometrically?

Does it not work because they are not prime ideals

[rocky cloak]

Like algebrogeometrically?
I don't think so, it's very much a statement about elements in the ring, not the ideals.

[cobalt heath]

I see, I hoped there would be a way to visualize the common factors.

[crystal vale]

Oh , I don't know why I take the gcd concept, it does not need

[crystal vale]

I don't this is right channel for asking but,
I solved this question but I don't know what will be the proof writing for this question, if Subspace is finite then I can do but in-case when subspace is not finite.

[delicate orchid]

By assumption of the condition of y we know that x cannot be in M. If y is in H but not M then it has to be in the span of x, and thus x is in the span of y and therefore is in K

think this works

[crystal vale]

Okay thank you

[boreal inlet]

I am currently stumped with a seemingly simple question.
I have to show F_p^(nd)/F_p^d is a Galois Extension.

I was previously thinking of showing that the fixed field of the automorphism group is F_p^d, but I'm not sure how to do it here.

Do we take the Frobenius Homomorphism?

There's another option of showing that it is Normal and Seperable, but I believe that is significantly more strenous to show than anything else.

[coral spindle]

Are you familiar with the Galois group of F_p^n : F_p?

[boreal inlet]

I believe it's generated by the Frobenius endomorphism of F_p.

That is, a -> a^p

[coral spindle]

Right, so in particular it is Abelian, yes?

[boreal inlet]

Yes

[coral spindle]

So every subgroup is normal

And by the Galois correspondence, this means that every intermediate field produces a normal extension

So you have normality for free, right?

[boreal inlet]

Wait, is this true? I was only aware the correspondence exist between the fixed fields of the subgroups, and the subgroups.

Oh, nevermind. All the intermediate extensions are Galois.

My bad

[coral spindle]

The Galois correspondence is between subgroups of Gal(E/F) and subfields of E/F. There is no restriction on the subfields; they all occur as the fixed fields of some subgroup of Gal(E/F).

Is this clear at least? Are you happy with this?

[boreal inlet]

Ah, wait. I probably should say this - this is a convention I (consequently due to our course instructor) use: If the extension is not Galois, we do not denote the automorphism group as Gal. This might be irrevalent.

[boreal inlet]

This is correct, by definition of Galois extensions.

[coral spindle]

That's fine, I'm using this convention here already

So you know now that F_{p^{nm}} : F_{p^m} is a normal extension necessarily

[boreal inlet]

Yes.

[coral spindle]

So now you only need to prove separability

[boreal inlet]

Seperability should be straightforward right?

[coral spindle]

I believe so, the proof should come out easily due to finiteness but I'm having a mental block with it rn

[boreal inlet]

All intermediate subfields of a seperable extension should be seperable.

[coral spindle]

Ha well that also works

Well then, job done I reckon!

[boreal inlet]

Because finite number of roots, and if you descend down to the intermediate subfield, less number of roots exist, which is anyway a subset of the distinct roots available in the bigger extension.

[boreal inlet]

Thank you very much.

So I argue with normality.

If we do have to calculate the fixed field here, is there a way to do it?

[coral spindle]

Yes, it relies on knowing the Galois group of finite fields but after that it falls out quite immediately I think you'll agree

[coral spindle]

Fixed field of which group?

Well in any case

[boreal inlet]

The F_p^d algebra automorphism group of the extension

[coral spindle]

So Gal(E:F) where E,F are appropriate fields, ok

But this would just be F

I feel like I'm not understanding your question

[boreal inlet]

Hmm.. okay so I was trying to use the definition that was given to me at first:

L/K is a Galois extension if L^(Aut(L/K)) = K.

We later on proved that any Galois extension should be normal and seperable, which we definitely can use, I was just wondering if there's an way using the definition directly.

[coral spindle]

Oh OK I see.

See it this way. If F is a finite field and |F| = p^n then F\{0} is a cyclic group of order p^n-1.

Now the Galois group of F over the prime subfield is generated by the Frobenius map phi. Let's say we want to find the fixed points under phi^m

Pick some generator xi of F\{0}. Then the things fixed here – what are they? We're looking at the subgroup xi^k where k == kp^m mod p^n-1 (adjusted for clarity)

I've probably already given enough hints here, so I won't spell out all of the other details, but the point is that once you know the order of the fixed point group you also know the order of the field of fixed points: that group, with 0 included.

[boreal inlet]

Wait, I do not really understand this line. I get the subgroups are cyclic, and all the generators should be of form (x_i)^k where k dividea p^(n-1).

[coral spindle]

What's x_i

Oh

I'm saying xi, the greek letter xi

lol

[boreal inlet]

https://cdn.discordapp.com/emojis/1026532993923293184.png?quality=lossless&size=48

My bad

[coral spindle]

The subgroup consists of those elements xi^k where k satisfies what I said

Like I'm literally just applying phi^m to an arbitrary element - xi^k – and seeing where it lands

[boreal inlet]

Oh right.

[coral spindle]

Maybe I should tex this

[boreal inlet]

it takes k to (k)^(p^m)

[coral spindle]

No that's not quite right

[cloud walrus]

Boytjie

Boytjie

[boreal inlet]

Oh correct, it's the product.

I meant the index k is raised to.

Yes, got it.

[coral spindle]

So to just hammer home the point a bit

The subfields of a finite field are uniquely determined by their order

So if you know the order, you know the field, in some sense at least.

[boreal inlet]

That makes sense.

Thanks for indulging me.

[coral spindle]

My friend, this was an indulgence for me. I have boring work to do!!!

[boreal inlet]

Understandable

I find math really interesting but university is sure trying hard to kill my interest.

[dim widget]

since the pth power map is surjective, if X satisfies some inseparable polynomial then there exists some Y such that Y^{p^n} = X and Y satisfies a separable polynomial. But then X is contained in the extension F(Y) so X also has separable minimal polynomial

[boreal inlet]

By inseperable you mean not seperable, right?

[dim widget]

yes

so the proof doesn’t come out of finiteness, it comes out of the surjectivity of Frobenius

but finiteness is an easy special case

[boreal inlet]

Wait, is this not enough?

[hidden wind]

how so?

[boreal inlet]

Mostly unreasonable deadlines and bad instructors.

[hidden wind]

oof

i just realized i don’t think i’ve had any bad instructors so far

ah wait i did in some physics course i took but that doesn’t count

the maths’ where it’s at, mwahaha

[boreal inlet]

The deadline thing is probably due to the fact that I'm very slow compared to the people who are fast thinkers, and my university batch is full of them.

[hidden wind]

i too have had a couple of, like, administrative problems

[boreal inlet]

The one who is teaching Field Theory to us is good though. He is an Algebraic Geometer and has taken Real Analysis and Multivariable Calculus for us before

[boreal inlet]

Those suck

[boreal inlet]

That's really good. We only have like only one course a semester who teaches well and stuff. Sad.

Currently our Measure Theory course is a total mess.

[warm wagon]

do you at least have good bibliography listed for the course?

cuz a good book can make a huge difference

[boreal inlet]

Not actually, because for some reason he makes his own notes. I am following Folland and Bass.

[warm wagon]

oof

[boreal inlet]

Saddest part is, his notations and terminology is too different from what you would find in books, which is making it even hard for me to understand a lot of things. I also lack intuition when it comes to these kind of things so certain definitions do not make sense to me, and when asked I get vague answers in return.

Trying to work my way through it so I don't butcher my grades, it's already bad. Can't risk ending my math major life here already, it's India.

[warm wagon]

you might get some luck asking one of your better instructors about books on the topic

[tacit ginkgo]

Does this mean that G is a group and i need to show G x G is or something else

[boreal inlet]

I'll probably have to. Is it funny that the three good ones I know all work in Algebra?

[boreal inlet]

G is a group should be enough.

When you say group, you're mentioning the operation.

Without it, G is just a set.

The first part is merely proving that the G x G -> G map is indeed a binary operation.

[tacit ginkgo]

so thats just showing G x G has closure?

[boreal inlet]

Yes.

[tacit ginkgo]

is this right so far for b

(ignore the identity part)

[warm wagon]

might be something to do with algebra guys being different. Back in my first year on uni, we had those interactive blackboards for the first time ever in class. And the very first week, somehow, the algebra guy (was well over 60yo) was already using it as if it had been teaching with them for decades.
Meanwhile, most "digital natives" students and many younger professors had every issue imaginable

[boreal inlet]

nonono. Wait. You have to prove G is a group. You can't assume that it is already.

Use the definition of the operation *. How is it defined?

[tacit ginkgo]

ahh im dumb i think i get it now

ty

[warm wagon]

it is a binary operation by definition. It takes two arguments.
You have to prove it has the properties required for it to form a group if equipped to the set G

[boreal inlet]

Seriously, they inspire me. The Field Theory instructor I told you about has probably read everything he needs to know in his life thoroughly. He never boasts about it but one day he slipped and told me about how he took extra years in his PhD to solve majority of Hartshorne by himself.

[boreal inlet]

Wait, I thought the closure is a property needed for the binary operation.

I was wrong.

[warm wagon]

no, and it needs not be closed. For example, substraction is not closed in N

[boreal inlet]

Thanks, I learn new things everyday

[warm wagon]

and that's why we have Z

and then division is not closed in Z.
and that's why we have Q

[boreal inlet]

Then- wait. R.

[warm wagon]

and sqrt is not closed in R. And that's why we have C

[boreal inlet]

Oh right, yeah, Irrationals and Transcedentals

[tacit ginkgo]

Why does some places online say it needs to be closed then

and my old teachers

[warm wagon]

"what" needs to be closed, and "for what"

[tacit ginkgo]

😳

[tacit ginkgo]

a group under a binary operation

[warm wagon]

that's different

[tacit ginkgo]

oh

[warm wagon]

A group is a set equipped with a binary operation, with the binary operation requiring some properties

[boreal inlet]

wait

How is subtraction in N a map of form N x N -> N

it is not right?

[warm wagon]

On a group, the binary operation needs:
associativity
identity element
inverse for every element

[warm wagon]

only for pairs (a, b) with a>=b

[boreal inlet]

oh goodness. Yeah.

[elfin wraith]

Controversial >=

[warm wagon]

if a<b, you have NxN -> Z

[coral spindle]

a >= b + min(N)

[warm wagon]

I use the N*, Z*, Q*, R* and C* being each except zero, so N is not an exception

0 not being in N makes it a lot more annoying

[elfin wraith]

I have literally no opinion either way and use whichever convention the textbook I’m using uses to be honest

[boreal inlet]

Same here.

[elfin wraith]

That or whichever happens to be more convenient for that exercise

[warm wagon]

Well, if 0 is not in N, addition in N has no identity element :)

[boreal inlet]

If we don't need N to be a monoid I think it is fine

[warm wagon]

i always liked this pic

[elfin wraith]

Same, algebra has too many strangely named structures

[coral spindle]

It should be mentioned that the vast majority of algebraists will only ever think about the things on the outer right edge of that square

[warm wagon]

for most stuff, you only care about groups, rings and fields tbh.

it's almost as if the name of the channel was that for a reason

[rocky cloak]

Maybe monoids if you're really out there

[boreal inlet]

That is, the major content of discussion of this channel.

[coral spindle]

There are a fair number of semigroup theorists, and monoids are so close to semigroups, and the rest of us study groups with some structure or another.

[dull marsh]

Magma theorists are my favourite

[coral spindle]

I have never met anyone who has described themselves as a magma theorist lol

Neither have I even seen a research group page, or even a personal page of one

You may as well say "I study functions. Just functions"

[elfin wraith]

Surely no one does lol, closure can’t be that interesting to look at

At least, only closure anyway

[rocky cloak]

Associative quasigroup theory is where it's at

I just love the empty set

[coral spindle]

The semigroup theorists at my institution seem to care a fair bit about inverse semigroups and the various affronts to god they call its generalisations

I kid. They do cool combinatorics with them. I just don't really care lmao ecksdee

Wtf is an Ehresmann semigroup don't tell me

[hidden wind]

i’m not sure if i understand this correctly

which words are mapped to the identity?

[boreal inlet]

What do they mean by x_g again

Is it just a letter? I think it is probably that. So we start with a set with same cardinality as the group, hence the bijection. The Free group can be generated from taking that alphabet, hence the bijection naturally extends to an group homomorphism, and rest works

I don't think we need to know what kernel is here. It can even be trivial, and then G itself is a free group. Otherwise, no

The quotient need not be free.

This looks similar to how modules can be represented in similar ways. Every module is a quotient of a free module.

[hidden wind]

ye just an abstract letter

actually my problem here is probably not with this in particular and rather just a confusion with quotients in general, i’ll take some steps back

[boreal inlet]

This proof is based on First Isomorphisn theorem. You may revisit that

[dim widget]

It depends on the group

We can think about an example

[hidden wind]

intuitively the idea is that phi "evaluates" the word as an expression in the group, right?

[dim widget]

If G is the group Z/2 x Z/2 then there is a natural map from the free group on two generators sending a to (1,0) and b to (0,1)

[dim widget]

Yep, sending each letter to one of a set of generators

[dim widget]

Here the kernel is generated by aba^-1b^-1, a^2, b^2

[hidden wind]

right

thank you

[dim widget]

basically the point is that every group has some generators

[boreal inlet]

....So. if a polynomial f in F_p[x] has a root alpha in the algebraic closure, how do we prove that alpha^(p^n) for any n \in N is also a root?

[hidden wind]

like... itself

[dim widget]

at worst you can take as a set of generators every element of the group

[dim widget]

Yep

[dim widget]

well just apply frobenius to f(r)

[boreal inlet]

Ah man how did I miss it. Thank you.

The images are all roots of f.

[dim widget]

the hard part is

showing that this gives every root

showing that these give some roots is trivial once you know that this is a field automorphism

[boreal inlet]

I don't have to show it gives every root, just the roots of this form. But yes, it does give every root.

I think we do this by showing the automorphism group here is also cyclic and generated by the Frobenius map itself?

[dim widget]

that would do it!

[boreal inlet]

We have already proved a result regarding the correspondence of possible K-algebra embeddings and the algebraic elements of an extension.

But I think we don't need that for this one

[dim widget]

you just need to show that F_p^n stisfies x^(p^n) - x

[hidden wind]

actually, now i'm curious, is there any particularily interesting you could say about a group which has only itself as a generator?

[boreal inlet]

It is free

Oh wait, that might be false

nvm

[hidden wind]

would that just be the "freest" group? i dunno what the technical term is

[hidden wind]

this sounds intuitively right

[boreal inlet]

Yes

[dim widget]

this is impossible unless it’s the trivial group

if you mean the only generating set is the set G

[hidden wind]

ye

[dim widget]

yeah I mean, at the very least every element generates a cyclic subgroup

so you don’t need the other elements in that subgroup

[hidden wind]

ohhh that is very neat

[hidden wind]

i am loving it, he is indeed a great writer

A valuable technique in studying a group is to represent it in terms of something familiar and concrete. After all, an abstract group is a cloud; it is a capital letter G.

[boreal inlet]

Actually wait.

I am stumped.

Let's say this polynomial has degree d >= 1.

I have proved all alpha^(p^n) with n \in N are roots.

How do I show that

alpha^(p^r) = alpha^(p^s) iff r congruent s mod d?

What confuses me is that they call the roots to be equal.

[dim widget]

you use that \alpha lies in F_p^d

[boreal inlet]

I am aware of one result like this:

|Hom(K(alpha), cl(K))| = no. of. distinct roots of f in cl(K)

Which is also equal to the seperable degree of [K(alpha) : K]

But this does not help

[boreal inlet]

Wait a second. I get intuitively why it should be correct, but I'm having trouble showing it

Oh. Bruh.

F_p^d /F_p is an field extension with minimal polynomial of degree d.

Of course it would be there.

I was trying to do this like

assuming r = kd + s for some k,

alpha^(p^r) = alpha^(p^(kd+s)) = alpha^(p^(kd) * p^s) = (alpha^(p^kd))^p^s

Now the p^kd thing gets reduced to just p^k because alpha is in F_p^d

What about the opposite direction

[dim widget]

what opposite direction?

It seems like all of these things you’re asking about just fall out of the statement that F_p^d/F_p is galois with galois group generated by the pth power map

[boreal inlet]

It asked that the relation is true if and only if r cong s mod d.

Never mind though, some simple calculations showed it.

[boreal inlet]

Yes, it did

Thank you.

[worn spindle]

so i know Z/3Z means 'mod 3', but what does just Z/3 mean?

i saw it here https://www.math.columbia.edu/~khovanov/modAlgSpring2017/homework/solutions6.pdf

/nick ping when responding

[hidden wind]

just a different notation for the same thing

[worn spindle]

tyy

my hw asked me this, does this mean i need to write out 16 or 24 matrices??

[hidden wind]

you could, but it’s easier to work with cycle notation

[dull marsh]

I believe you don't need matrices for cycle notation

Well unless you do count it as a matrix

I guess they are matrices nvm

[hidden wind]

?

they are equivalent (that is, isomorphic), but the matrices also carry around lots of information that is not really interesting

[dull marsh]

(1 2 3), this is no different from a 1x3 matrix

[worn spindle]

sure but i would still need to write out like 24 cycles right??

[dull marsh]

Even more

Not every element of S4 is a cycle

[worn spindle]

wait but 4! = 24 how would there be more?

[hidden wind]

fake, sure you could write down a 1x3 matrix (1 2 3) but that would not be a permutation

[worn spindle]

oh i remember someone saying sth abt 2-cycles in class?

[dull marsh]

I think you are confusing permutations with cycles

[dull marsh]

That's the point of a notation

You write something (in this case, a matrix) and it is interpreted in a particular way (in this case, as a permutation)

[worn spindle]

ill do both since im a beginner

[hidden wind]

??????

[worn spindle]

why the question marks?

[hidden wind]

it is important to distinguish notation and meaning, that it is written just like a matrix does not make it a matrix

[hidden wind]

i'd say this is unhelpful, since the interesting part is the structure that relates these elements, and the structure on matrixes (typically that is matrix multiplication) is different from the structure on permutation (which is function composition)

[dull marsh]

What should I call it then? (The thing that I am writing, not the permutation)

[hidden wind]

there is an isomorphism of permutations with matrices, but that is different

[hidden wind]

cycle notation

[dull marsh]

Alright

[worn spindle]

can you elaborate on this

[dull marsh]

It asks you to write each permutation using its cycle decomposition and some elements of S4 have more than 1 cycle in their decomposition

[hidden wind]

e.g. (1 2) in S3 corresponds to [0 1 0; 1 0 0; 0 0 1] in GL3

[worn spindle]

ok sure but there would only be 24 permuations to write out

[dull marsh]

Right

The one in bottom left seems wrong

Ah wait I read that 3 as 2 nvm

It's correct

[worn spindle]

so this is all of them?

or am i missing one?

[dull marsh]

Yes

[stark helm]

How to justify part(b), is it restricted to the ring Z6 or some other field(this part makes me feel pretty confused)

[delicate orchid]

it wouldn't make much sense if it wasn't in Z_6

[stark helm]

Then in Z6, the only way that I can come up with is x(x+1), is there any methods to figure out other form?

[delicate orchid]

well if you have four roots, you should be able to factor it like any other polynomial using those roots

[stark helm]

OK, because a will be roots of f(x) if and only if x-a divides f(x)

[delicate orchid]

yur

[stark helm]

so we have four ways right

[delicate orchid]

we have 2 choose 4

maybe?

I'm distracted atm gimme a min

[stark helm]

Wait, I might make mistakes, the theorem I proposed only apply to a field

Z6 is not a field

[delicate orchid]

ok right I'm not distracted anymore

(x+a)(x+b) = x^2+(a+b)x+ab
ab = 0
a+b = 1

[ashen heron]

finite fields cannot be ordered because if K is an ordered field then the map N -> K, n |-> n.1 is an injection since n.1 = m.1 implies (n-m).1 = 0 but 1 > 0 in an ordered field which means m = n.

is this proof sketch to show that finite fields cannot be ordered correct?

[delicate orchid]

a = 0, b = 1
a = 3, b = 4

there we go

[stark helm]

The other question is, is it true that division algorithm can be applied only in PID or Euclid domain?

[delicate orchid]

I don't see how this shows it's not ordered.
Assume we have an ordering, then a < b => a+c < b+c, but if we chose c such that b+c = 0, then a+c isn't 0, and is thus larger than b+c. Contradiction

[ashen heron]

im trying to show the contrapositive that if it is ordered then it must be infinite

[delicate orchid]

this is the definition of a euclidean domain

and PID doesn't imply euclidean so you can't do it on all PIDs

[stark helm]

oK, I got it. Really appreciate that

[delicate orchid]

ah ok

ah I see now, very clever

[ashen heron]

i think i got this too

[boreal inlet]

Let's say we have a field K of characteristic p, and L/K is an algebraic extension, with a in L.

I have already proved that a is seperable iff K(a) = K(a^p).

I'm trying to show there exists an r >= 0, such that a^(p^r) is seperable.

So, I have to show for a certain integer r, K(a^(p^r)) = K(a^(p^(r+1))).

How do I go about this?

Only thing that I see is, this is a descending chain of subfields:

K(a) \superset K(a^p) \superset .... K(a^(p^r)) \subset...

I thought of using the tower formula but I'm not sure how it helps.

[south patrol]

Yes that's great

Note this is a descending sequence of K-vector spaces

Can you see how that means you can conclude