groups-rings-fields

[cobalt heath]

I mean, the minimal polynomial divides x^n - 1, right

[boreal inlet]

Yes

It definitely should

[cobalt heath]

Now, what can we say about x^n - 1

[boreal inlet]

Hmm... it's monic, also p doesn't divide n

So the gcd of x^n -1 and it's derivative is 1

Does that say it is separable though..?

Oh wait I think it does.

Consider K[x]/(f), the residue algebra - (which is also a field)

The derivative f'(x) = nx^n-1 is a unit iff p does not divide n

Which also means we can generate 1 using f and f'

[tough raven]

The derivative is non-zero iff p does not divide n.*

[boreal inlet]

And also, that condition gurantees all roots are distinct

As f and f' do not share a root

[tough raven]

But in that case n x^{n-1} divides x^n so you can generate 1 = (something)(n x^{n-1}) - (x^n - 1).

[tough raven]

Oh, mb. Exactly this.

[boreal inlet]

Damn me can do math for once

https://cdn.discordapp.com/emojis/722022152328445985.png?quality=lossless&size=48

Ok wait if x^n - 1 itself is separable

Then f(x) must have phi(n) different roots

[tough raven]

Exactly; factors of a separable polynomial are separable.

[boreal inlet]

This is the minimal polynomial of the primitive nth root

[tough raven]

That's always true under your assumptions (char 0 or does not divide n).

[boreal inlet]

Yes

What I want to show is that f(x) is the nth cyclotomic polynomial

[tough raven]

That doesn't follow from x^n-1 being separable 🧐.

Unless you already showed something else earlier or something.

[boreal inlet]

My bad, I meant to say that we know rhe degree of f is exactly phi(n)

I'm trying to prove that it has to be the cyclotomic polynomial

[tough raven]

By assumption?

[boreal inlet]

Yes

[tough raven]

Got it.

[boreal inlet]

We do know that for zeta_n, the image under any K-algebra automorphism of K(zeta_n) is another root of f

[tough raven]

To put you on a different track: you know that f and Φ_n both have degree φ(n) (and monic, let's say). What would be a particularly convenient way to show that they are equal?

[boreal inlet]

Yes

Hmm..

If one divides other i g- OH

wait.

I think i might have got it

cyclotomic polynomial also has zeta_n as root

So the minimal polynomial must divide the cyclotomic polynomial

Both have same degree

so we done

Right

[tough raven]

Because?

[boreal inlet]

That's the definition no

Or am i tripping

Cyclotomic polynomial is product of (x - (zeta_n)^a) where zeta_n runs through Un and gcd(a,n) = 1

[tough raven]

Oh, sorry, I thought you were using a different definition for some reason.

[boreal inlet]

What are the other definitions which are useful to work with

[tough raven]

Π_{d|n} Φ_d = X^n - 1 as polynomials.

[boreal inlet]

Oh damn

Yeah i remember this now

[tough raven]

This (plus monic) determines the cyclotomic polynomials uniquely in Z[X] by a multiplicative version of Mobius inversion (although it's not a priori clear that they are polynomials rather than rational functions).

[boreal inlet]

Well uhm wait

That's true if I'm starting with Q i think

[tough raven]

In particular, AFAIK this is the way to define Φ_n if the characteristic may divide n.

(Although you just end up getting Φ_{p^k m} = Φ_m^{φ(p^k)} if char = p, n = p^k m, m coprime to p.)

[tough raven]

Anyway, with this definition, you're right — trivially any primitive n^th root of unity is a zero of Φ_n.

So f | Φ_n, which gives you 1 => 2, I think.

[boreal inlet]

nice

[versed charm]

1+1=

[boreal inlet]

Now only thing that remains is to show the automorphism group is isomorphic to the group of units

We do know the automorphism group simply permutes the roots of the cyclotomic polynomial

That means if we can determine where zeta_n is sent by one automorphism, i think we determine the whole map

It can only send zeta_n to another primitive root

That is, (zeta_n)^a for some a coprime to n

[tough raven]

Do you know the characterisation of K-homomorphisms from K(α) to L, for L another extension of K?

[boreal inlet]

The cardinality of it corresponds to the separable degree of the extension L/K

Afaik

[tough raven]

That's true.

But you can also kind of determine it as a set.

[boreal inlet]

Ohhhhhhhhh

Wait is it just simply Aut(K(alpha)/K) in this case

I think that happens if L/K is normal

[tough raven]

For every root β in L of the minimal polynomial of α over K, there is exactly one K-homomorphism mapping β to α.

You can use this here with L = K(α) to find the automorphisms.

(And α = ζ_n.)

[boreal inlet]

I think we did this but we didn't prove each such homomorphisms are unique

For every root there indeed is one K-algebra embedding

Oh mb that's trivial

[tough raven]

Any element of K(α) is a polynomial in α with coefficients in K (or rational function if α is transcendental over K), so the image of α determines a K-homomorphism entirely.

So that gives you uniqueness.

[boreal inlet]

Yes

Nice

Wait that gives the well-definedness of the map

Now we prove it's a group homomorphism with trivial kernel I think

Because finite sets and injective maps with same cardinality so it's isomorphism

Cardinality part I can say because the extension is already seperable

ab mod n -> Sigma_b • sigma_a i think

because like zeta_n goes to (zeta_n)^ab can be done by first sending it to (zeta_n)^a and then sending that image to (zeta_n)^a)^b

Last thing to show is 3 => 1 because that completes the equivalence

So we already have this isomorphism...

Oh i think that's trivial right

Because you already know all the primitive roots are of form (Zeta_n)^a for gcd(a,n) = 1

And the fact that sigma_a(zeta_n) = zeta_n^a

Is also a root of f

The automorphisms allow us to go back to each unique root and hence we can clearly claim deg f is at least totient of n.

Now as f is seperable

We are done

Oh no wait we aren't done

How do we know there isn't a root which is non-primitive

Let's pick a root which is non-primitive

[tough raven]

Btw you use here that sigma_b, which by definition maps zeta to zeta^b, also maps zeta^a to zeta^ab.
It's true but pointing it out.

[tough raven]

Now recall that f | Φ_n.

[cobalt heath]

Huh

I mean you do have to use what the roots of x^n - 1 is like

[languid trellis]

#1182793396776091708

[boreal inlet]

Yes

[boreal inlet]

Bruh

I am an idiot. Thanks for the help

[tough raven]

Not really. The problem is that when one proves f | Φ_n assuming something, one doesn't realise that it's true in general.

The picture in my mind is that algebraic conjugates of primitive n^th roots are n^th roots (because both are zeroes of Φ_n, which always has coefficients in K). Hence the primitive roots are partitioned into various equivalence classes, each of which is a set of conjugates, i.e. the roots of a single irreducible polynomial. Clearly then those ireducible polynomials form a factorisation of Φ_n (when char = 0 or does not divide n, so that Φ_n is separable).

With this picture, your equivalence is more or less clear: a minimal polynomial of a primitive n^th root is deg(φ(n)) iff it is Φ(n), iff all primitive n^th roots are (algebraic) conjugates. We can also see that the automorphisms are precisely ζ -> ζ^a for some a's (those for which ζ^a is a conjugate of a) in (Z/nZ)^×, and it can be checked as you have done that composition corresponds to multiplication. Thus Aut is a subgroup of (Z/nZ)^×, equal to the whole thing iff the other equivalent conditions hold.

[boreal inlet]

That's a very clean explanation

Thank you again

[rotund aurora]

anyone knows a reference to this? the topological closure of an infinite subgroup in a one-dimensional Lie group is open

I don't know about Lie groups, so may be easy to prove or standard

[rocky cloak]

Z inside R isn't open...

[rotund aurora]

mmh right

then idk what's this about, I'll show you where I read that

[rocky cloak]

If you add compact, then R/Z is the only 1D lie group I guess. Then you can basically just do case by case analysis of the subgroups of R/Z

[rotund aurora]

oh, right

yeah so this was in the context of elliptic curves over R, and I think their associated group is that

well actually it can also be (R/Z) x (Z/2Z)

[boreal inlet]

Wait isn't R/Z just the circle

Or am i tripping

Oh yes it is

[rotund aurora]

yes but the group isomorphism is not algebraic

[boreal inlet]

Wait uh

What do you mean by not algebraic

[rotund aurora]

it's not defined by rational functions, it's just an isomorphism of Lie groups

[boreal inlet]

ok got it

[rotund aurora]

elliptic curves over C are isomorphic to C/L where L is some lattice, so the result over R is kind of what you would expect

[dull ginkgo]

[dull ginkgo]

Assume the normalizer of a set S is the whole group, then <S> is a normal subgroup of G

trying to think of a way to prove this hmm

[rotund aurora]

conjugation is an automorphism

[dull ginkgo]

So the intersections of all groups containing S is fixed under conjugation for each g because all the groups containing S are just "permuted around"?

Does this imply N_G(S) is a subset of N_G(<S>) for any subset S?

[delicate orchid]

gsg^-1 \in S
x \in <S>, x = s_1s_2...s_n => gxg^-1 = ... you finish it

[dull ginkgo]

I kind of want to use the intersection def

[delicate orchid]

ok I don't care

try that if you want but I think it's STUPID ah ah ah ah

[dull ginkgo]

Let $\sigma$ be an automorphism of $G$. I assume for subsets $A$ and $B$ that $A \subset B \Leftrightarrow \sigma(A) \subset \sigma(B)$

[cloud walrus]

A Fibrous Powder

[coral spindle]

Why would you assume something that is easily provable

[dull ginkgo]

because it's a bijection

[coral spindle]

[dull ginkgo]

brain fried

[coral spindle]

But yes this is essentially the key observation, it should be straightforward from here

[dull ginkgo]

Assume S is mapped to itself under an automorphism. Sp_G(S) is the set of subgroups of G containing S, then <S> is the intersection over SP_G(S). \sigma is a bijection from SP_G(S) onto itself, by the lattice automorphism induced on P(G). Thus the intersection is preserved by the automorphism and thus the automorphism fixes <S> in terms of images

[delicate orchid]

or you just put gg^-1 inbetween each s_i

also I don't know what P(G) is

[dull ginkgo]

power set of G

[delicate orchid]

ok I really hate this proof now wtf

ew

[barren sierra]

what is the original problem?

[delicate orchid]

_ _

[dull ginkgo]

But the automorphism part can be used as a lemma

[barren sierra]

oh bruh why are we talking about lattices...

[dull ginkgo]

also helps with the next problem

[delicate orchid]

let f be an automorphism that fixes S, x in <S>
f(x) = f(s_1...s_k) = f(s_1)...f(s_k) \in <S>

like

come on dude

[dull ginkgo]

[barren sierra]

why do we need automorphisms at all for this?

[delicate orchid]

his proof worked for general automorphisms

[barren sierra]

ah

[dull ginkgo]

It's useful for a problem coming up

[delicate orchid]

for the original problem, absolutely you just need inner autos

[coral spindle]

Let me guess

the next question is "a subgroup generated by a characteristic set is characteristic"

[delicate orchid]

no.

I won't let you guess

[coral spindle]

Too late binch

[dull ginkgo]

it's coming up a think yeah

[delicate orchid]

"characteristic set" wtf are these constructions nobody uses these

[dull ginkgo]

i saw it when skimming from what I remember when I first got the book

[coral spindle]

If sharp were here...

[delicate orchid]

logic is crankery

[coral spindle]

I can't deny that

[delicate orchid]

except the "true over char p for all but finite p <=> true over C" is nifty

[dull ginkgo]

this is a safe space

[coral spindle]

yeah but logicians care about "SNOREly rank" or "stab-ILL-ity" or something

[dull ginkgo]

Next one

Let $T$ be a subset of $G$, and $T_N = \bigcup_{g \in G}{gTg^{-1}}$, then $\langle T_N \rangle$ is the normal subgroup generated by $T$

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

So:

Assume $H$ is a normal subgroup between $T$ and $\langle T_N \rangle$. Therefore, by definition of the latter, it cannot contain $T_N$. Thus there is a $t \in T_N \setminus H$

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

It's obvious that for all $g \in G$ that $gtg^{-1} \in T_N \setminus H$

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

yeah I'm stuck

it meant to be H\T_N

Let me try again. Let $T$ be a subset of $G$. Then $T_N = \bigcup_{g \in G}{gTg^{-1}}$ is the \textbf{intersection of all conjugate-closed subsets containing T}. Now let $\langle T \rangle_N$ be the normal subgroup generated by $T$, i.e the intersection of all normal subgroups containing $T$. Likewise $\langle T_N \rangle$ is the intersection of all subgroups containing $T_N$ and is normal\ \

Therefore, we get $T_n \subseteq \langle T \rangle_N \subseteq \langle T_N \rangle$. I need to prove that there cannot be a normal subgroup strictly between $\langle T \rangle_N$ and $\langle T_N \rangle$

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

I realized I am an idiot chat

[chilly ocean]

wrong channel #❓how-to-get-help

[worldly lintel]

in z/2z, (0) is a maximal ideal?

[delicate orchid]

(0) is maximal in any field

[barren sierra]

^ it would be a good idea to prove that the only ideals in any fields are (0) and the field itself

[dull ginkgo]

If the set of units is the set of elements not in any nontrivial ideal then is it the antideal?

[summer path]

🐜ideal

[barren sierra]

idealn't

[vivid tiger]

https://ncatlab.org/nlab/show/anti-ideal+predicate

https://ncatlab.org/nlab/show/antisubalgebra

Instead of ideals (of rings), use antiideals. (Technically, these are antisubalgebras of the ring as a module over itself.) Again we can omit ≠-openness by strengthening the nullary condition. In detail, a subset A of X is a two-sided antiideal (or simply an antiideal in the commutative case) if p≠0 whenever p∈A, p∈A or q∈A whenever p+q∈A, and p∈A and q∈A whenever pq∈A.

[dull ginkgo]

Hey chat

so in general what's the procedure to show that a given Group is isomorphic to a given presentation

[barren sierra]

Projection in what context?

[dull ginkgo]

like I don't quite understand how to prove the full isomorphism if you can prove the relations hold in a group, but like, "no more further relations"

[tardy hedge]

@dull ginkgo bro u are mathing all day man

[dull ginkgo]

spring break homie

[tardy hedge]

Nice

I respect it. I should put more time in tbh

ESPECIALLY study without my phone on me. Im so much less productive with my phone next to me

[dull ginkgo]

Anyway like here's an example

This is a presentation I have found for SL(2,Z)

[somber sleet]

Hey guys I'm trying to prove that any primitive root of unite is a root of the nth cyclotomic polynomial, but I can't manage to conclude, I have the following:

Let $\xi \in \mathbb{C}$ be an nth primitive roots of unite, which means $z^n-1 = 0$ and $\forall 1 \leq m < n: z^m - 1 \neq 0$, then we know that $(z - \xi) | (z^n -1)$ and $(z - \xi) \nmid (z^m -1) \forall 1 \leq m < n $. We now assume that $(z - \xi) \nmid \Phi_n(z)$, from which it follows that $(z- \xi)\Phi_n(z) | z^n-1$ und $(z- \xi)\Phi_n(z) \nmid z^m-1 \forall 1 \leq m < n$.
From this we automatically follow that $deg((z-\xi)\Phi_n(z)) = deg(\Phi_n(z))+1 > deg (\Phi_n(z))$ .

Can now please somebody tell me where I get the contradiction from, I completely don't see it

[cloud walrus]

damn_guuurl

[rustic crown]

what's your definition of the n-th cyclotomic polynomial? the usual definition is prod (z - ξ) over primitive nth roots of unity ξ

[somber sleet]

nope it's the polynomial of biggest degree that devides z^n-1, and I guess this is the direct contradiction I was looking for

[rustic crown]

that's z^n - 1 tho

do you want it to be like coprime with z^m -1 for m | n and m < n?

[dull ginkgo]

Presentations suck

I have to prove that $G = \langle a, b, c | ab = ba, ac = ca, cb = abc \rangle$ is isomorphic to $V = (\mathbb{Z}^3,\cdot)$ where $\begin{bmatrix} n_1 \ m_1 \ k_1 \end{bmatrix} \begin{bmatrix} n_2 \ m_2 \ k_2 \end{bmatrix} = \begin{bmatrix} n_1 + n_2 + m_1 k_2 \ m_1 + m_2 \ k_1 + k_2 \end{bmatrix}$

[rustic crown]

and what have you tried? >.<

[dull ginkgo]

i know the generators

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

Just proving that they actually are generators in (Z^3, dot)

is gonna suck a bit

[rustic crown]

you have pretty nice relations, so it's not as bad

[dull ginkgo]

I have an idear

[dull ginkgo]

If $V = (\mathbb{Z}^3, \cdot)$, i think $Z(V) = \langle \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} \rangle$

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

I am an idiot!

$a = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}, b = \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix}, c = \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix}$
\ \
Then $\begin{bmatrix} n \ m \ k \end{bmatrix} = \begin{bmatrix} n \ 0 \ 0 \end{bmatrix} \begin{bmatrix} 0 \ 0 \ k \end{bmatrix} \begin{bmatrix} 0 \ m \ 0 \end{bmatrix} = a^n b^k c^m$

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

yippe that works

[dull ginkgo]

yay

Idfk what this top one means but I’ll just try to find the presentation on my own

I know product of intersecting two cycles is a three cycle

oh it's i neq j neq k neq i

okie

[vagrant zinc]

Dummint?

[vivid tiger]

just solve in general the problem of deciding if two words are equal smh.

[dull ginkgo]

Nope. Jacobsont

[hollow mica]

yea this is basic

(like the book is called basic algebra)

[dull ginkgo]

i love the irony of it and how many people ask about if it's basic college algebra (like before calc)

and I bullshit like it is just long winded and does basic precalc stuff

[hollow mica]

I just say it's what constitutes basic algebra in european gradeschool

[dull ginkgo]

lol

[dull ginkgo]

For the presentation of A_n

I wonder if I can use the problem before it and append a new relation to describe the subgroup

$x_1^3 = e$ implies $x_1$ is a 3-cycle \
$i > 1 \Rightarrow x_i^2 = e$ implies $x_i$ is a product of disjoint 2-cycles

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

the annoying part is that just the single x_1 is a 3-cycle

[dull ginkgo]

so all of these presentation proofs depend on finitivity but what if the fucking thing isn't finite

[cobalt heath]

Yeah, should have solved halting problem already, smh my head

[cobalt heath]

P sure you can derive this from how the elements commute

[dull ginkgo]

No but you need like inequalities and shit to prove the presentation is sufficient

[cobalt heath]

Ah

Well, for a group G

You can find its presentation by identifying kernel of F_n -> G, no?

[dull ginkgo]

Like wtf is this shit https://math.stackexchange.com/questions/152158/presentation-of-a-n-from-jacobsons-basic-algebra-i

[cobalt heath]

Yeah, that's analogous as what I propose

[dull ginkgo]

Like all the inequalities and shit

Exercise 1.11.2 - Jacobson Basic Algebra

[cobalt heath]

Ah, I guess element counting?

[dull ginkgo]

for what purpose?

[cobalt heath]

The question here is doing element counting to show that this surjective homomorphism is bihective

[dull ginkgo]

ah

[cobalt heath]

(Where surjectivity is.. well.. computation )

[tacit hemlock]

why does it NOT make sense to check for abelian for R,x

why is it -

[tender wharf]

weird

R\{0} under multiplication is a group

however, R under multiplication is not a group

[tacit hemlock]

ok thx

what's so special about matrices with det 1

[tender wharf]

they're called the unitary matrices

wait

[tacit hemlock]

is there some useful property of them that makes checking these group axioms trivial / obvious

[tender wharf]

sorry thats the special linear group

[tacit hemlock]

also det non zero

[tender wharf]

that's the general linear group

basically these are all the invertible matrices

[tender wharf]

well the special linear group is a subgroup of the general linear group

[tacit hemlock]

ok

very briefly, what's "cool" about them

why do they deserve their own special name

"general linear group"

[tender wharf]

honestly idk lol

[tacit hemlock]

Why is the set of transpositions/reflections not closed?

is it because no identity?

[tender wharf]

yea

RR = e but e isn't in there

[tacit hemlock]

can't you compose 3 reflections

to get the identity

[boreal inlet]

I mean identity isn't a reflection itself @tacit hemlock

That's why not closed

You have to manually add the reflection there

[boreal inlet]

That group can also be used to represent the set of linear automorphisms of the finite dimensional vector spaces over the field it's on.

[cloud walrus]

Riku

[boreal inlet]

I don't know about if this group "deserves" the name but that's one reason it might have this name

[tacit hemlock]

wtf is an automorphism

[boreal inlet]

... oh wait.

What is your background in algebra? Genuine question

[tacit hemlock]

3 weeks in

first course

[boreal inlet]

No i mean what are you guys reading

[tacit hemlock]

Artin's Alegbra

I guess Pinter's as reference

[boreal inlet]

Ok hmm have you studied Linear Algebra seperately before or this is the first time

[fading field]

real answer: giving standard names to things that come up often allows mathematicians to talk about these things and have readers understand what they’re talking about at a glance rather than having to redefine them over and over again

[boreal inlet]

^^

[tacit hemlock]

is an automorphism like

the generalisation of permutations

but to groups

[boreal inlet]

You're right to a certain extent but not exactly

Automorphism is a map from an algebraic structure to itself, which also preserves certain properties of the structure.

Basically a "homomorphism from one to itself, which is one-one and onto"

[tacit hemlock]

group automomorphism

ok so bijections but with groups

f: X --> X

[boreal inlet]

Not only bijections, it's also a group homomorphism

When I said Linear automorphism, I meant automorphism of a vector space

[tacit hemlock]

yah

[boreal inlet]

You'll get there when you read Matrices and Vector spaces

[fading field]

important: in order to be an automorphism you also need the inverse map to be a homomorphism (this is guaranteed for groups and finite dimensional vector spaces but is not true for any structure in general. you can ignore this point if you currently find it to be too pedantic)

[boreal inlet]

Wait, really?

[tacit hemlock]

so basically just an isomorphism to itself

[fading field]

yeah

[boreal inlet]

I doubt the thing fails for algebraic structures

Do you have an counterexample

[tacit hemlock]

is there any intuitive way or standard example of thinking about automorphisms

that i can get some intution of

i.e., for permutations f: X --> X

I can tihnk of some small set of integers

1,2,3,4

and some permutation could be just switching all elements around

2,3,1,4

etc

[fading field]

yeah the automorphisms of a set are precisely the permutations of that set

[boreal inlet]

For vector spaces it's easier

Like taking a vector in R^n, and we scale and rotate every possible vector in the space. Whatever you originally had now will morph into something bigger or rotate in some angle, but the shape stays the same

For a set, where there's no other underlying structure, all permutations make up the automorphisms

[boreal inlet]

.

If your automorphisms are "homeomorphisms" or "conformal maps in C" or something like that then maybe yes it's non-trivial

And we need to prove it

Actually for conformal maps between open connected subsets of complex numbers it's automatically true my bad

But yeah

Ok let me actually ask a question that I am trying for the longest time it's probably trivial but I'm missing it

So I have K = C(t), the field of fractions of the polynomial ring on one variable over C. I take f_n(x) = x^n - t to be a polynomial in K[x], and consider the splitting field of f_n to be L_n. When can we guarantee L_n is a cyclic extension, that is the automorphism group is cyclic?

What I know until now is

x^n - t is first of all irreducible over K. This is true because this polynomial is first primitive (it is monic) and hence is irreducible over C(t)[x] iff it is irreducible over C[t, x]. Now we see t \in (t) but not in the square ideal. By Eisenstein, it's irreducible in C[t][x] = C[t,x]. Hence we are done.

Now K is characteristic 0. Hence it's perfect and so, x^n - t is hence seperable.

So, L_n/K is a Galois extension as it's a splitting field of a seperable polynomial.

But what to do after that

What I think we might be able to do (but I don't see exactly how) is to find a certain automorphism of order n.

Because L_n/K is galois and hence seperable, so |Hom_K(L_n, L_n)| = n, the number of distinct roots of f_n

And L_n/K is normal so that's equal to the automorphism group Aut(L_n/K)

So the cardinality of the Galois group is n, hence one automorphism of order n is sufficient to show the group is cyclic.

My question is under what conditions we can find it

[vast quiver]

I could be very wrong here (i am still trying to get good at galois theory), but how do you know this? Can’t you have a polynomial of degree 5 with the full symmetric group as automorphism group?

I still think it’s true, but you do that by showing computing the degree of the extension L_n/K. I’m going to claim you only need one root of f(x)=x^n - t, call it a, to get the splitting field. Then, multiplying a by the n different nth roots of unity (which are in C), you get n different roots of f (plug them in and get 0), so indeed these are all the roots of f. So,

L_n = K[a]/(a^n - t),

so [L_n : K] = n. Next, this also gives us insight about what the automorphism might be to generate what we think should be a cyclic group. An automorphism is determined by where we send the generator a, and there are n options (so these choices exactly correspond to the n automorphisms). So, try sending a to zeta a, where zeta is a primitive nth root of unity.

[chilly radish]

It's always cyclic since K has roots of unity

If you fix a root a, it's generated by (e.g.) the automorphism a-> a*xi

Every root is of the form a*xi^k

Ah joseph already said this

[chilly radish]

You're right that what he said is not sufficient to conclude [L_n:K]=n. You need to do the extra step of showing L_n is simple and then it follows since it's the degree of the min poly of the generator

[wide brook]

Hello! Can someone help me with this exercise in Galois theory, I don’t know if I’ve done it correctly and I still can’t grasp exactly what it says….

[tacit hemlock]

why is this not closed under multiplication

oh

is it (a+ai)(a+ai)

= a^2 - a^2 = 0 = 0 + 0i

So why does it not have inverses?

Can't I take $\frac{1}{a+bi}$

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

rationalize to get $\frac{a-ib}{a^2+b^2}$

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

oh it's a^2-b^2

and I can choose b s.t. b^2 > a^2

[boreal inlet]

I completely forgot about that case, thanks to both of you for the explanation

[tacit hemlock]

wait doesn't $H=GL_2(\mathbb R)$ here?

[cloud walrus]

normalAtmosphericPa=101,325

[alpine plank]

Yes

[tacit hemlock]

what's a quick way to check for this?

I know that in any such cycle $\sigma$, I must have 1 foloowed by 4

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

so WLOG, take $(14a_1a_2...a_k)$ as our cycles, where $k$ is at most 4 (nothing else is fixed)

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

but then I still have 5x4x3x2 potential choices

[boreal inlet]

Think of sigmas as functions, it might be handy

[tacit hemlock]

i am though aren't I

with the cycles lol

is cycles not the best way to go about these?

consider listing all elements:

1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6

then map $1\mapsto 4$

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

in the second --> third row, I can map 1 --> 4 again

but that doesn't preserve 1 --> 4

thus, not closed

[alpine plank]

yes

[dull ginkgo]

Damn I would’ve just said there’s no inverses

[tacit hemlock]

what about this one

can I reason that this is equivalent to consider $S_5$ since it is clearly isomorphic to it

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

(i'm pretty sure at least)

b/c we're fixing one element

[boreal inlet]

Yeah just show the isomorphism

[tacit hemlock]

how

we're mapping cardinality 6 --> cardinality 5

im confused

[boreal inlet]

Not exactly

[tacit hemlock]

how is this not closed

...

unless $\sigma$ is a product of non-disjoint cycles

[cloud walrus]

normalAtmosphericPa=101,325

[boreal inlet]

The cardinality of S_5 and this is exactly the same, any counting argument shows that

[tacit hemlock]

oops

so on top of checking injectivity and surjectivity, I also need to check multiplicativity for the homomorphism

[boreal inlet]

What you do is, just send every such permutation to a corresponding permutations in S_5, after reenumeration of the elements being permuted

Basically send (a_1 a_2 a_3 4 a_5 a_ 6) to (a_1 a_2 a_3 a_5 a_6)

The latter being a permutation of 5 elements instead of 6

[tacit hemlock]

Also, for checking subgroups of S_n, does closure automatically mean that it's a subgroup?

i.e, I don't also need to check other groups axioms: associative

[boreal inlet]

I mean if it's not even a subset i don't think so

[boreal inlet]

You can still show it is a subgroup up to isomorphism, that is

[tacit hemlock]

here, they're only checking closure it seems

[boreal inlet]

Show an injective group homomorphism

That's the furthest you can do in that case

[boreal inlet]

That is because subgroup is first and foremost also a group

And group needs to have closure

If you don't even have that how can it even be a group

[tacit hemlock]

but if it's closed √ ==> do I still need to check associativity? ==> identity? ==> inverse?

[boreal inlet]

Absolutely yes

For group all of those conditions are necessary

[tacit hemlock]

wait, so why are odd cycles not closed

is it because we aren't assuming $\sigma$ is composed of disjoint cycles?

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

i.e., $\sigma=(123)(345)$

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

Which is equivalently $(12453)$

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

ahh but that's still odd

$(123)(23456)$

[cloud walrus]

normalAtmosphericPa=101,325

[dire siren]

(12345)*

[delicate orchid]

(12453) if u multiply them the right way around

[dire siren]

oh yeah, the weird convention that fg means gof

I don't know why people would prefer it

[delicate orchid]

I still don't know which way round gof goes

fg "f then g"

[dire siren]

idk, fog(x) always meant f(g(x)) to me, so g first and then f

[boreal inlet]

same

[chilly ocean]

i'm not really sure. Whenever i try to think about function composition all that is in my mind is fog

[dire siren]

L o L

[cobalt heath]

I always struggle with this directions

It almost seems like we have to read things in RTL style

[tacit hemlock]

they called it "star-compositoin" in class

for cycles

Left to right

to emulate multiplication

[hardy scaffold]

this is why I read manga, makes it easier.

[glad osprey]

You don't need to check associativity because if an operation is associative it is also associative when restricted to a subset. To show H is a subgroup it is enough to show that for a, b in H ab^(-1) is also in H. You can derive the existence of identity and inverses from this

[fading field]

gof is read "g after f"

[patent girder]

I'm trying to compute $\text{Gal}(\mathbb{Q}(\sqrt{1+\sqrt{2}}/\mathbb{Q}).$ So far, what I have done is considered the Galois closure, namely $E^=\mathbb{Q}(\alpha,\beta)$, where $\alpha = \sqrt{1+\sqrt{2}}, \beta = \sqrt{1-\sqrt{2}}.$ Then, I have that this is a Galois extension, so that my automorphisms send $\pm \alpha \to \pm \alpha, \pm \beta,$ which means $\text{Gal}(E^*/\mathbb{Q}) \cong D_8.$ Next, I consider the subgroup lattice of $D_8$, but I can't find an intermediate field for each subgroup, is that fine, or am I messing something up

[cloud walrus]

Zander

[chilly ocean]

what does Gal(L/K) mean when L/K is not Galois? automorphisms of L that fix K?

[coral spindle]

Yes, this is a standard extension of the definition

You just don't get the usual nice results and you have to keep that in mind

[patent girder]

Right, that's why I'm wondering if I'm allowed to do what I did; I got the Klein 4-group

[teal laurel]

So for 2.8 of Atiyah Macdonald $\mathfrak{m}$ is the maximal ideal of a local ideal $A$, and $M$ is a finitely generated $A$-module. $M/ \mathfrak{m} M$ is also a vector space, and the theorem says that the basis of $M/ \mathfrak{m}$ generates $M$
So my question is, is $\mathfrak{m}M =0$ here? Since $\mathfrak{m}M$ is the zero of $M/ \mathfrak{m}M$ and the way the basis generates 0 should be unique here so that there is only one element in $\mathfrak{m} M$ and it is $0$

[cloud walrus]

unixitary

[hardy scaffold]

No, mM is not necessarily zero. The point is that a lift of a basis for M/mM to M is a generating set.

For example just consider A as a module over itself, for instance C[[x]], formal power series in x.

[long obsidian]

Hey the nilradical ideal N of a ring R is the intersection of all NONZERO prime ideals right? Clearly it's only a subset of the zero ideal iff the ring is reduced.

[hardy scaffold]

the zero ideal is only prime if R is an integral domain

[long obsidian]

Oh so only integral domains have a generic point in their prime spectrum then?

[hardy scaffold]

yes

[teal laurel]

Oh ok thanks

[hardy scaffold]

though I guess if the nilradical is prime you could call that a generic point since it is still dense? (someone who knows AG please correct me)

[wraith cargo]

I think it's enough for Spec A to be irreducible

if it's also reduced then it's integral

more generally any closed irreducible subset of a scheme has it's own generic point

[rocky cloak]

The nilradical should be there intersection of all prime ideals. For example C[[x]] has only the primes (x) and 0, and the nilradical is 0.

[next obsidian]

No, it’s when there’s a minimum prime, which is then the nilradical

Example k[x]/(x^2) with nilradical (x)

[next obsidian]

And no, it’s all prime ideals

[noble lynx]

I was playing around with the frobenius map and I got a bit confused, if phi(x) = x^(2p) then phi seems to be ring homomorphism on Z/pZ -> Z/pZ, but (-1) gets mapped to 1, but I thought ring homomorphisms on fields must be injective?

[next obsidian]

That’s not how that works
(x^p)^p = x^{p^2}

So F^2 sends -1 to -1 still

[noble lynx]

is x mapsto x^(2p) not a ring homomorphism? I was not trying to apply frobenius twice, I just figured the additive property would still work if p was replaced with 2p

[next obsidian]

No it is t

Is not

Compute some binomial coefficients and see what happens

(6C3) = 6!/3!3! = 5!/3! = 20

Which is not divisible by 3

[noble lynx]

that clears things up thanks

[spare isle]

My prof and book say that this is correct and consistent with permutation compositions; if you apply the permutation compositions left to right instead, then you also have to apply function composition out to in instead.

So if fg = f o g for you, then you would apply permutations right to left

i.e. (12)(13) = (132) if you use normal right-to-left application (as in you have fg(x) = (f o g)(x) = f(g(x)) ).

[stark helm]

If F is a finite field and we construct a map phi:F->F by saying phi(x)=x^p when x is in F. Then if I finish proving phi is one-to one, can I say because the field maps to itself( same cardinality) and it is finite, so it is on-to? the second question is if F has characteristic p but F is infinite, we can still prove one-to-one but are not able to prove phi is on-to?

[next obsidian]

Yes

To everything

[south patrol]

for the last bit you can still prove it is an auto if F is algebraic over F_p though ig

But yes

[stark helm]

I am somewhat confused about what is ig? and also, F is algebraic over F^p can prove the map is automorphism?

[south patrol]

"i guess"

[south patrol]

Well F is always algebraic over F^p

[stark helm]

in fact, I am considering x^p in F^p must exist x in F, so it should be on-to. But field is perfect iff characteristic is 0 or characteristic of p that F^p=F. so it seems that all field can be perfect field???

[south patrol]

i'm not really sure what your first sentence is saying

[stark helm]

I am considering an infinite field of characteristic p, then what will happen?

[stark helm]

modify F to be infinte field with characteristic p, and phi(x)=x^p.

[south patrol]

Sure I mean not all fields are perfect

It sounds like you're saying every element of F^p is a pth power (which is obvious)

and Idk where you're going from there

[stark helm]

my question is, if F is infinite field with characteristic p, and we still construct phi: F->F and write phi(x)=x^p. Then we can still prove phi is one-to-one here and on-to? if we can, it seems that F=F^P and this infinite field with characteristic p is perfect field, but it seems contradicts the theory

[south patrol]

No we can't prove it's onto

there's like the universal example of

F_p(t)

the Frobenius is not surjective

since t is not a pth power

[stark helm]

but it maps from field to itself, so can I consider the case Fp(t)->Fp(t)?

[south patrol]

yes

[stark helm]

and then say transcendental

like Zp(t)

[stark helm]

I think I understand it,if maps from Q(pi) to Q(pi) as an example, the map must not be surjective

[south patrol]

Well Q(pi) is char 0 so it is perfect

and the Frobenius isn't a homomorphism

[stark helm]

ok, then just saying Zp(t)

[south patrol]

That's what I said there yes

[delicate orchid]

r u telling me x -> x^0 isn't a homomorphism :letrollface:

[rotund aurora]

idk if this is trivial but given a topological group, can you embedd it inside a compact topological group?

mmmmh

[rustic crown]

embed an infinite discrete group?

[rotund aurora]

by definition of embedding I mean an injection G-->X where X is compact, the map should be continuous and the image should be dense in X and homeomorphic to G with the discrete topology

isn't it standard or I'm missing some subtlety?

[delicate orchid]

all of the notions (both of them ) of compactification for a topological group I know of are just a G-set that's also a nice topological space. I can't think of any that are groups themselves

but there must be some way to do this it would be too yucky without

[rotund aurora]

A nice example is with Z, you can embedd it in say Z_p for some p

I think the point here is that the quotients Z/p^kZ are all compact, so the limit is compact

[rustic crown]

this won't be discrete Z right

it would be Z with the p-adic metric, where 1, p, p^2, ... converges to 0

[rotund aurora]

ah you are right

ok so you can't even do it with Z?

[delicate orchid]

I've found a fairly nice construction https://en.wikipedia.org/wiki/Bohr_compactification

[rustic crown]

idk

[delicate orchid]

the only wrinkle is that you need a nice condition on the finite dimensional unitary representations for the "canonical map" to be injective

and god knows if the image is dense

[rotund aurora]

wow this Bohr guy was a footballer

also bro of Bohr

but this is universal among compact Hausdorff groups, so it's the best you could do right

[delicate orchid]

I suppose, if your hypothetical compact Hausdorff group existed with a nice continous injective homomorphism it would have to factor through this, giving you a continuous injective homomorphism into this space too

that explanation was more for myself than you

of course, this still leaves non-hausdorff mfs

[rotund aurora]

yeah I was wondering that too

But if Z is given the discrete topology you always have continuous maps to whatever compact Hausdorff group, so Z^Bohr maps to any compact Hausdorff group

[cobalt heath]

#1216202707220566137
is there a way to do this problem non-computationally?

[celest furnace]

Can anyone help me with this one

All i've been able to do so far is show that f(m) = r * m is injective for nonzero r (this uses torsion free but not PID at all...)

Thought maybe by knowing where r * x_{s+1} would go you might somehow be able to recover where x_{s+1} goes but this doesn't seem like its going to work (for example the above map fails for ZxZ -> 2Zx2Z for r=2)

[cobalt heath]

Are you working with embedding M inside the free module

[celest furnace]

Yes

[cobalt heath]

So, did you express r x_(s+1) in terms of x_1 .. x_s?

[celest furnace]

Yes

[cobalt heath]

Then one trick remains

It says “a free module of rank s” for a reason

Which means, you do not need to take x_1, .., x_s as the generators of the free module.

Alternatively, note that you can embed some rM into the free module for some r

[celest furnace]

Crazy clever. Thanks man

[cobalt heath]

No problem!

[short idol]

Has anyone here ever done any differential Galois theory? I'm stuck on this exercise (right from the first question, I can't see how to use Cauchy's existence theorem), so thanks in advance!

I also have a more specific question. When we look at automorphisms in their matrix form, how can we say that a matrix is in the Galois group? Are there any particular equivalent conditions? Because apart from saying that it sends fundamental matrices over fundamental matrices, I don't really see...

[chilly ocean]

what is L?

[short idol]

L is a Picard-Vessiot extension for the equation

[chilly ocean]

Also which Cauchy theorem do they mean

[short idol]

Cauchy existence thm say that for an ODE in an open U, for every disc in U we have the existence of a fundamental Matrix on U for the equation

[chilly ocean]

Wouldn't that mean that every solution of the ODE (over C) is of form Uv for some v, and therefore the set of those solutions is a subset of the ring of holomorphic (therefore also meromorphic) functions?

[tacit hemlock]

why is this part the case

i.e., you have one group generated by a single element, but another group not generated by a single element

why can they not be isomoprhic

[chilly ocean]

isomorphisms preserve every formula without parameters about a structure

in this case, consider the following formula

There exists a x such that every element of the group is of form x^n for some n

Let phi be an isomorphism from G to H where G is generated by a single element but H isn't

And let x be a generator for G

what can you say about phi(x)?

[short idol]

No because after i found explicitely non holomorphic solutions

[tacit hemlock]

no idea

is it supposed to be obvious

[chilly ocean]

everything is obvious after you know the answer

[tacit hemlock]

it is simply a rewriting

of x

by some element in H

[chilly ocean]

what properties do isomorphisms have?

[tacit hemlock]

bijective homomorphism

preserve order

[chilly ocean]

do you know if all the solutions meromorphic?

[chilly ocean]

phi(x^n) = phi(x)^n and phi is surjective

[short idol]

Not necessarily. If we have two independant meromorphic solution f and g then C(x,f,g) is a PV extension thus every solution is of the form Pf+Qg with P and Q rational

[chilly ocean]

isn't Pf+Qg meromorphic?

[short idol]

Yeah i'm stupid 🗿 sry

[chilly ocean]

its okay, me too

[short idol]

In fact i don't know if the answer is for exemple an argument on analytic continuation or if we can cover C by disk and then patch the solution (i dont know how) together to form a full solution

Like, if we cover C by disks and at each intersections of those (id don't know if my sentence is correct i'm don't really good in english) we can say that is a sort of pole

[chilly ocean]

the set of poles has to be discrete

[short idol]

It is in this case no ?

For the circle of the disk

[chilly ocean]

i don't think you can cover the plane with disks such that the intersections form a discrete subset

maybe #real-complex-analysis would be a better place to ask

[short idol]

I just don't know if even someone has worked with this theorem ahah

Ok the thm is in fact more precise. For every disk of radius R, D(0,R) in U, we have the existence of a fundamental matrix composed of holomorphic function on D(0,R)

[chilly ocean]

yeah im bad at complex analysis so im prob missing something obvious but i dont see how to construct a solution in the whole plane from that

[short idol]

I don't know if it's correct. But, If we have two disks suppose that they are centered at 0 for exemple of radius r and R st r<R. Then we have the existence of a solution on D(O,R) and on D(O,r). But the solution D(0,R) must come from a solution on D(0,r). Thus for any solution in D(0,r) we can extend it as much as we want in the complex plane ? Or an other idea is to cover C by disks wich intersect each other in an open subset of C. Then for each we have the existence of an holomorphic solution in this disk, It is then also defined in the other disk which intersects the first because it is defined on the intersection and so on we have the existence of an entire solution in C. (The Cauchy thm say that we have the existence of a fundamental matrix, thus we have in fact two independant entire solutions, f and g, thus C(x,f,g,f',g') is a PV extension and a subfield of the field of meromorphic function)

[chilly ocean]

yeah but if you have a solution on D(O,R) what guarantees there will be a solution on D(O,R') for (all) R'>R?

you can make them smaller but idk how to make them larger

[short idol]

I think the second argument is better

[chilly ocean]

why do we have an holomorphic solution in those disks?

[short idol]

By the Cauchy theorem

[short idol]

I think we have to see it in reverse for this one. Any solution defined on a disk D(0,R) comes from a solution defined on a D(0,R') with R'>R and therefore any solution defined on any disk can be extended on a disk of any size

[chilly ocean]

i dont see how that implies any solution can be extended to a disk of any size

its consistent with that that there are solutions on D(0,R) that can't be extended to a solution in D(0,R') for some R'>R

[short idol]

we just have to find two of them

The thm say that for any disk D(0,R) we have the existence of two solutions on it. Thus for D(0,R'), two solutions on it, come from two solution of D(0,R)

And we have the existence for any size of disk

[chilly ocean]

how do you rule out the possibility that there are solutions in every disk D(0,R) but not in the whole complex plane?

[short idol]

A function is defined on C if it is defined on every disk D(0,R) R>0

[chilly ocean]

we need a family of solutions f_R such that f_R and f_R' agree on D(0,min(R,R')) then

i don't see how to get that

[short idol]

Otherwise the argument with the cover is perhaps clearer, I don't know

[chilly ocean]

just to give an example of a similar example where this doesn't work

for every D(0,R) there are non-constant holomorphic functions bounded by 1 in D(0,R)

but there isn't one in the whole complex plane

[short idol]

We have it directly by the thm i think

for every disk D(0,R') we have the existence of two solutions f,g, then their restriction to D(0,R) are solutions in this open subset

f_R=f_R'|R

otherwise i have an other idea ahah

the problem is at infinity

If we make the change of variable x to 1/x, we have the existence of a solution near 0 by the cauchy theorem

Then this is a solution near infinity

[wide brook]

Hello! What do you think of this true or false question: if K is an extension of F of degree 30 then there exist a,b,c in K where K=F(a,b,c)?

I’m thinking that if a,b,c are linearly dependent then it’s always true

But I can’t think of a counterexample that would make it false assuming linear independence

[short idol]

If they are linearly independant, then, [K:F]=[K:F(a,b)][F(a,b):F]=[K:F(a,b)][F(a,b):F(a)][F(a):F]

no ?

[south patrol]

presumably it'll be to do with 30 = 2 * 3 * 5

so by the tower law, the problem is more or less equivalent to saying: if K/F is an extension of degree p, is K = F(a) for some a

and that is indeed the case

[wide brook]

With p being prime?

[south patrol]

Yeah

More rigorously, you can pick random a in K which isn't in F. Then F(a)/F is of degree > 1 and hence [K:F(a)] is only divisible by two primes (or fewer)

Then do the same game with K/F(a) instead (i.e. use induction)

:)

Does that make sense?

[wide brook]

Hmmm yeah I think I get it! So either [F(a):F] is going to be one of 2,5,3 or some product of them. And then [K:F(a)] is what remains, either a prime again or a product of the two remaining….

[south patrol]

Yeah exactly

Basically by induction on $n$ you can show that if $[K:F]$ is a product of $n$ distinct primes then $K = F(\alpha_1,\dots,\alpha_n)$ for some $\alpha_1,\dots,\alpha_n$

[cloud walrus]

Süßkartoffel

[wide brook]

Oh yeah that seems like a useful result! Thanks for the feedback!

[south patrol]

Np.

I had never thought about this lol so I am grateful

[stark helm]

in R=Z4[x], can you figure out a factorization x=f(x)*g(x) in which neither f(x) nor g(x) is a constant?

[cloud solar]

How to show the polynomial X^n+X+3 is irreductible over Q

[delicate orchid]

oh this sounds fun

[coral spindle]

It's easier, I think, just to find a two nonconstant units in R

[delicate orchid]

I've managed to get 2x

[stark helm]

2x+1 and 2x^2+x i think that works

[coral spindle]

There you go

(2x+1)(2x+1) = 1 :)

[wide brook]

Hello! Does this seem correct to you? I haven’t solved a similar exercise before and I couldn’t find something online

[teal lily]

What does the gothic font mean and how can i write it in latex

[summer path]

it's a mathfrak letter

$\mathfrak{A}$

[teal lily]

Thanks!

[cloud walrus]

:3c

[teal lily]

what does it mean do u know?

something about permutations

[summer path]

Probably alternating group?

[teal lily]

ok i will look it up thank you

[delicate orchid]

it's definitely the alternating group

[summer path]

mathfrak letter guessing game

fun times

[teal lily]

\

[celest furnace]

Try not to use GPT since people here don’t really trust it (for good reason)

It’s also against the rules

[teal lily]

oh my bad

[delicate orchid]

it's against the rules to use it to answer other people's questions

[summer path]

but also chatgpt is just pretty terrible for math

lol

[teal lily]

ye it really is

[delicate orchid]

I'm actually very surprised chatgpt can even produce mathfrak letters

[teal lily]

for definitions i find it ok tho

[summer path]

i think you're better off just googling things

[chilly ocean]

Some definitions are hard to find online

[celest furnace]

Ever used it for TikZ?

Coding is one thing it is great at

[chilly ocean]

Tho ig its probably hard for chatgpt to know them either

[delicate orchid]

no, I use quiver for tikzcd stuff

and I don't use tikz on it's own

[summer path]

https://q.uiver.app/

[chilly ocean]

i wonder if chatgpt knows what 1-complementable means or P-hard

[delicate orchid]

tbh I don't know what those mean

[chilly ocean]

i also didnt, and it wasnt easy to find what they mean using web search

[summer path]

something something complexity theory probably(?)

[chilly ocean]

P-hard yes

But its hard to find a reference defining it

[chilly ocean]

Oh i found one. Yesterday i had more trouble finding it for some reason https://cstheory.stackexchange.com/questions/8041/problems-outside-of-p-that-are-not-p-hard

Maybe cuz i was looking at PP-hard rather than P-hard

[long obsidian]

If I have a ring hom g:R->S and x not a zero divisor in R does it follow that g(x) is not a zero divisor?

[delicate orchid]

3 mod 3 is 0 not 1

if you mean over Z/2Z, then write f(x) = x^n+x+1 then use fermat's little theorem to reduce the x^n down

[rocky cloak]

Maybe try unpacking the definition a bit.

What does it mean for g(x) to be a zero divisors. It means that there is some y such that g(x)y = 0.

If there is no such y in R, can you always expect there not to be one in S?

[cloud solar]

I am idiot omg

[delicate orchid]

it's ok boss we all make silly mistakes

you were on the right track just with the wrong prime

[cloud solar]

Haha thx:)))

[delicate orchid]

there's a nice choice for S and g that make this very explicit, my hint is: ||matrix||

[rotund aurora]

how do you show this gives G^* the structure of a topological group

T is the circle group

inversion is clearly continuous

because x in P(K, U) iff 1/x in P(K, 1/U) where 1/U={1/u : u in U}, and inversion is continuous in T

but for multiplication, you would want to show that if f(x,y)=xy then f^{-1} P(K,U) is a product of open sets in G^*

mmh maybe it's better to try to show that every point in f^{-1} P(K, U) is an interior point, by finding some neighborhood that still makes it work, and I assume that's where you also have to use compactness of K

Maybe it's something along the lines:

• Suppose x,y in G^* are such that x(k)y(k) in U for ever k in K.
• Consider z in P(K, U'), then you would want to be able to find U' such that z(k)x(k)y(k) in U still. And I think you can do that because K is compact

well, xy just belongs to P(K, U). So say f in P(K, U) then you want to find U' such that if g in P(K, U') then fg in P(K, U)

unless I'm clowning

I forgot to say, you want U' to be a neighborhood of the identity

[hidden wind]

meow

oops sorry wrong server

[summer path]

you mean wrong channel

https://discordapp.com/channels/268882317391429632/1055183210444763205 (animal noises thread in #cats )

[rotund aurora]

maybe it's just true that if K is compact in an open U then you can find open V such that KV subseteq U

[teal lily]

what does it mean for a group G to be albainian? , all elements of G are from albainia?

[summer path]

surely you mean abelian

or you're trolling

[south patrol]

Yeah

[rotund aurora]

is it true that if U is open there exists V open such that UV subseteq U

like that should be from continuity of multiplication right

[south patrol]

what is UV

Oh

lol

[rotund aurora]

UV={uv such that u in U and v in V}

[south patrol]

Okay sure

[summer path]

UV = the stuff that comes out of the sun

[rotund aurora]

lol it wasn't that funny

[rustic crown]

det doesn't see this if U is like not compact

for compact U, we have U x G --> G and U x {e} maps into U, so by tube lemma U x V maps into U for some V nbhd of e

[rotund aurora]

the map f : G x G-->G defined by f(x,y)=xy is continuous, consider the inverse f^{-1} U which should be open, say U_1 x U_2 with U_1, U_2 open, then consider (U cap U_1) x U_2 ?

[rustic crown]

that doesn't show U U2 lies in U right

consider G = (R, +) and any open U

[rotund aurora]

you don't need U2 inside U

[rustic crown]

right

det wrote U*U2

[rustic crown]

even the tiniest V say (-eps, eps) will make U+V larger than U

[rotund aurora]

mmh true

ok yeah I wasn't paying attention to what the open sets in the product are

idk why I assumed they had to be rectangles of the form U1 x U2 with U1,U2 open which is untrue

[rustic crown]

anyway, det will go eep

[rotund aurora]

bump

[rotund aurora]

sleep well

[stark helm]

If R is a domain and the map a->a/1 is an isomorphism, then R is field. I am confused why we can get conclusion that R is a field?

[old hollow]

Wdym

They’re basically saying “if R and Frac(R) are isomorphic, then R is a field”

That makes intuitive sense right?

[stark helm]

So given the map: R-> Frac(R), it assumes that every a in R maps to every a/1 in Frac(R), where a/1 in frac(R) have the inverse, so a has the inverse in R. is that true?

[old hollow]

Well we know a/1 has an inverse in Frac(R), written as …

[stark helm]

And if we omit the condition: R->Frac(R), just saying R is domain and map:a->a/1, then R is not a field?

[stark helm]

yes, we know this, 1/a so I want to confirm if we omit the condition map: R->Frac(R)

[old hollow]

I’m confused by your question

The theorem says:

Hypothesis. R is a domain, and the map f:R —> Frac(R) where f(a) = a/1 is an isomorphism.
Conclusion. R is a field.

What condition are you trying to omit?

[stark helm]

trying to omit the map f:R->Frac(R), probably modify it as R->R' where R'={a/1|a in R}

[old hollow]

You’re inventing this new set R’. Are you claiming it’s a field? What are its operations?

[stark helm]

R' should be the subring of frac(R) I believe.

[old hollow]

Okay it’s a subset of Frac(R), sure, and I believe it’s a subring as well

Okay fine

So what is your new theorem?

Is this what you’re trying to claim?

Hypothesis. R is a domain, and the map f:R —> R’ where f(a) = a/1 is a ring isomorphism.
Conclusion. R is a field.

[stark helm]

this is the theorem I ask, I want to know if it is still true to argue R' is a field, or it is not

[stark helm]

it seems to me that if we don't say map R to a frac(R), then what will happen to R, still a field or not

just let R map to other set

[languid trellis]

Consider decomposing an even permutation into transpositions, then using what you know about sign and then go case-by-case

[old hollow]

We know R’ is not a field in general. Consider the domain Z, and now consider Z’ = {n/1 : n in Z} as a subring of Frac(Z). This structure Z’ is definitely not a field, since there are no multiplicative inverses.

[languid trellis]

im thinking, is there anything else that may be useful in the long lecture notes

and maybe try some easy examples

like the identity

(1 ..... n) (1 ,,,, n) ... (1... n) }n times

or try (123)

in fact decoming a transposition would be pretty useful

probably

post it here

[old hollow]

How do you express (12)(3) as a product of three-cycles

?

[languid trellis]

(12) is not even

[cloud walrus]

Steve Backshots

[old hollow]

Oh oh

[languid trellis]

you're being a bit inconsistent with notation

you mean (a_1 b_1 c_1) = (a_1 a_2 a_3) right

also, comparing images, a_1 should map to a_2 according to the lhs, but the far rhs shows that a_1 is mapped to itself right? a_1 -> a_n -> a_1

a_2 -> a_3 so that works

again a_3 -> a_3

wait

right

im trying to sanity check myself by finding the inverse image of a_2

hold on

a_2 doest get mapped to a_3

it gets mapped to itself

right

[noble lynx]

what is the standard proof that the Frobenius map generates all automorphisms on finite fields?

[cloud walrus]

Steve Backshots

[minor wraith]

what does commute refer to in commutative diagrams

normally, I think of 'commute' as meaning ab=ba

to refer to some binary operation

i.e., the ordering of the operands are interchangeable

what about $A \circ B = D = A \circ C$

[cloud walrus]

Kakaka

[minor wraith]

isn't that what is instead implied in the "commutative" diagram?

[cobalt heath]

Yeah I think this is (a bit of) conflation of terminology

You have a composition A -> C -> D and A -> B -> D

[minor wraith]

so they're using a second definition fo "commute" ?

i.e., equivalent compositions

Then are commutative diagrams only used for showing equivalent function compositions?

[cobalt heath]

Yes, mostly.

It is also good for illustrating what is going on, I think

(Though, it is quite abstract)

[minor wraith]

why the same name

[cobalt heath]

Fwiw,

[minor wraith]

so many close synonyms they could've chosen

[cobalt heath]

You can think of the "left multiplication by a" as an oprtation L_a

Then, commutativity means L_a \circ L_b = L_b \circ L_a.

[minor wraith]

ohh

and then for the square example

I can create a singleton set A={1}

and make D = {ab}

√

[chilly radish]

You can think of this as saying that the operation of"going right" in this diagram commutes with the operation of "going down"

Right then down or down then right you get the same map

[cobalt heath]

Yeah but usually for group commutativity, A = B = C = D = G

[coral steeple]

I'm trying to show that if I,J are relatively prime ideals of a ring R, then I^n, J^n are also relatively prime. I've mostly tried to write an arbitrary element x of the ring as a product i_1i_2+j_1j_2 given that x=i_1+j_1, but this has been pointless

R is not necessarily commutative

[cobalt heath]

Arbitrary element could be harder. You know one specific element is enough for this, right

[coral steeple]

I guess you mean 0?

[cobalt heath]

0 is always in any ideal

[coral steeple]

That is the only thing we know is in I^2+J^2

Yes

[cobalt heath]

So you cannot use it for coprimality

1, on the other hand..

[coral steeple]

What good is the fact that 1 is in I^2+J^2? The whole ring is not necessarily generated by 1

[cobalt heath]

Huh

Hm, do you recall what an ideal is?

[coral steeple]

A subset I of R so that 0 is in I, aI is in I for all a in R, and i+j is in I for i,j in I

(left ideals at least)

[cobalt heath]

a * 1 = a

[coral steeple]

Ah yes I am stupid

[cobalt heath]

Happens to best of us

[coral steeple]

The thing is I can't come up with any useful element of I^2+J^2

[cobalt heath]

Going to be quite technical, I think

[coral steeple]

If i+j=1, then (1+j)i+j(1-i)=1 but that's not in I^2+J^2 (and the most useful observation I had)

My other idea was something something binomial expansion of (i+j)^n, because that will be equal to 1, but it is again not in I^2+J^2 because of all the coefficients not equal to 1

[cobalt heath]

Still good to try, though.

[coral steeple]

Unless I've misunderstood, it's a dead end because in Z we have 1=3*5-2*7 which is in (5)+(7), but (3*5-2*7)^2=9*25-2*3*5*2*7+4*49 , which one would be excused for believing is not in (25)+(49)

[cobalt heath]

So square does not work; how about cubic?

[coral steeple]

absolutely bizarre that that works

[next obsidian]

Relatively prime is equivalent to the statement that no maximal (prime) ideal contains both. Show that the maximal (primes) containing I^n and I are the same

Okay you said not necessarily commutative, so I’m not sure anymore, but probably this works using maximal 2-sides ideals

[coral steeple]

I realise the book probably intended for R to be commutative

[next obsidian]

What book

[coral steeple]

Lang's Undergraduate Algebra

[next obsidian]

Okay well in that case prove what I said

But I suspect it works without that assumption too

[coral steeple]

Yeah I think I see it both ways now. Thanks @cobalt heath @next obsidian

[summer path]

Hi chmuwu

[next obsidian]

Hi tububuwu

[vivid tiger]

Something cool, but basic:

1. Note that for finite groups, by Lagrange's theorem if a subgroup has size > 1/2 then it must be the whole group. (There are also generalizations using haar measures)
2. Suppose you have a group G and a way to test if elements are in a subgroup H, and want to know if H=G. Here's a probabilistic test: test n randomly chosen elements, and if they're all in H then H=G with probability at most 2^{-n}.

[rocky cloak]

Don't know if it's "the standard proof". But GF(q^n)/GF(q) is a degree n Galois extension, and the Frobenius map has order n. Hence it generates the entire Galois group.

[chilly radish]

These elementary probability results about groups are very cute. I like the 5/8ths theorem as well

[sonic coral]

can’t you stricten the bound to sqrt(n) for the first bullet point?

[dull marsh]

n can have divisors larger than sqrt(n)

3 divides 6 and 3 > sqrt(6)

[chilly ocean]

Do you mean probability at least 1-2^{-n}?

[sonic coral]

ah yeah we need like a factor of 2 or something

or we’ll actually if we go up to sqrt(n) we know all the divisors since they come in pairs, but i see the point now

[noble lynx]

is there any way to prove without galois theory?

[short idol]

Hey, i have a question about this exercise (question 19) find a generator of the polynomial relation. I just have to find one relation ? Because i found one but it seems weird to me

And after i found lambda by evaluating at 0

[amber burrow]

This is a chapter 1.2 on dihedral group in dummit and Foote - they don’t specify what r and s are so I’m kinda confused

Like if s is a symmetry why does it have order 2?

[short idol]

symettry is of order 2

s^2=id

and r is the rotation

[summer path]

s is the reflection element

So you can kind of think of it as flipping twice gets you back to where you started

[sly inlet]

https://en.wikipedia.org/wiki/Sylow_theorems#Fusion_results:~:text=1)2.-,Since%20q%C2%A0%3D%C2%A0pnm%C2%A0%2B%C2%A01,-%2C%20the%20order%20of how do they get the highlighted part

[dull ginkgo]

proceeds to not provide a proof of something not intuitive at all

i get the finite case

I'm just trying to explicitly describe the bijection

$\coprod_{g \in G}{X^g} = \coprod_{x \in X}{\mathrm{Stab}(x)}$

[cloud walrus]

Mizlang

[dull ginkgo]

a c t u a l l y

X is the disjoint union of

orbits

okay I get it lol

[delicate orchid]

yeah, I don't get it either
p = 1 mod q => q is a divisor of p-1, completely contradicting it being larger than p

[rocky cloak]

I mean the question is fundamentally a Galois theory question, but if you just unpack some definitions you don't have to use the word 'Galois'.

Another thing you can do, is use that the multiplicative group of GF(q^n) is cyclic, and let x be a generator. Then GF(q)(x) = GF(q^n) so x has a minimal polynomial of degree n, and any automorphism is uniquely determined by where x is mapped. Thus there can be at most n automorphisms, so powers of Frobenius is everything.

[dull ginkgo]

how the fuck do I even interpret this problem

Jacobson understand the exercise challenge (impossible)

[delicate orchid]

this is just counting the number of elements in a S_n conjugacy class I HOPE

so can be done entirely combinatorially

[chilly radish]

the point is that the elements r and s are just that, symbols, but their relation with each other is what represents the dihedral group

[hollow mica]

does the definition change if we instead require “at least one of a^n and b to be in I for some natural n”

I feel like it shouldn’t, but I don’t see why

[amber burrow]

Ok yeah I got it - didn’t realize s was a symmetry and that r was a rotation

And basically their point is that you can’t perform any number of rotations to cause a symmetry

[delicate orchid]

no that's not it at all

s is a reflection

every element of D_n is a symmetry

[amber burrow]

Yeah - by symmetry I meant it line folding over a line

Reflection is a better word for it

[dull ginkgo]

lol I understand the problem now only after finding the solution combinatorially and realizing what it asked for

Jacobson: understanding the problem is the problem and the solution is the solution

[vivid tiger]

haha yeah.

[rocky cloak]

The context is commutative rings right, so that would just be relabeling which one you call a and which you call b

[delicate orchid]

oh right that's what the difference is

it was like playing I spy

[mighty kiln]

Left-primary ideal

[dull ginkgo]

An idiot (mizalign)'s odyssey through jacobson

[rotund aurora]

If $\phi,\psi\c A\to B$ are isomorphisms of rings (commutative and with unity) are such that $\sqrt{(\phi(a))}=\sqrt{(\psi(a))}$ for all $a\in A$ does it follow that $\phi=\psi$?

[cloud walrus]

croqueta3385

[vivid tiger]

This is an interesting question. Can you ping me when you figure it out?

[rotund aurora]

sure

mmh

It's false I think

like consider k[x]/(x^2), then you have two isomorphisms given by sending x to x and another one by sending x to -x

the only radical ideal is in fact (x) and (1), so it works

@vivid tiger

[vivid tiger]

(-x)=(x). radical((-x))=radical((x))=(x). Yeah?

Alright, I see.

[rotund aurora]

ye

[vivid tiger]

Thanks!

[hollow mica]

oh you're right I thought it was for rings without assuming commutativity

[stark helm]

if R is a ring and I=(x) is the principal ideal in R[x] generated by x, then prove R[x]/I isomorphic to R. I am considering a ring homomorphism phi: from R[x] to R given by f(x)->f(alpha) where alpha is a real numbers. Then use the first isomorphism theorem to argue that R[x]/ker(phi) isomorphism to im(phi). The question is do I have to argue why im(phi)=R? if it is necessary, I can argue that im(phi) is a set containing analpha^n+...a1alpha+a0, and these elements are real numbers, is that adequate?

[delicate orchid]

it's pretty obvious that evaluating the constant functions on alpha gives you all of R

so the map is surjective

you'd need to pick a specific alpha for the kernel to be correct though

[stark helm]

then if I modify it as f(x) to f(1), will it be correct?

[stark helm]

then I can just argue that im(phi)=R, since it is just a constant functions on 1

[delicate orchid]

x is not in the kernel of this map

[stark helm]

then how will you construct the ring map here?

[delicate orchid]

f -> f(0)

pretty much the only way to send f(x) = x to 0

[stark helm]

then I can write it as f(x) to a0 right?

I want to prove the correspondence theorem of a ring

[stark helm]

The second question: I argue the isomorphism phi: R/I->image(phi) given by r+I->phi(r). but I feel confused why we can have the isomorphism between R/I and S/J given by r+I->phi(r)+J as well ?

[boreal inlet]

Do you know the existence of the natural map from R to R/I?

[crystal vale]

You can prove that the mapping from R-> S/J has kernel I and it's surjective mapping

[next obsidian]

R -> S -> S/J

Surjective as a composition of surjective maps, kernel is I

Then apply 1st iso

Riku more like miku

[boreal inlet]

Mbmb let me think

[next obsidian]

Wut

[boreal inlet]

What I was proposing as alternative is not correct

[next obsidian]

It works

You have to use uniqueness

[boreal inlet]

What is the map S -> S/J

[next obsidian]

I thought you’re saying you induce maps R/I -> S/J and backwards

And then show they’re inverse

[boreal inlet]

My plan was R/I -> S by universal prop, uniquely determined

The go to S/J from S somehow

[next obsidian]

You don’t get a map R/I -> S

[boreal inlet]

There is a map which is like x -> x mod J

[next obsidian]

No such map exists

[boreal inlet]

Why not

[next obsidian]

None that’s canonical at least

Because that comes from a map R -> S with kernel containing I

You are given nothing of the sort