groups-rings-fields

[sharp sonnet]

it's just that small groups are small

[paper flint]

I see.

Thanks for the help. I'll continue with my lectures now.

[sinful mirage]

does anyone have ideas what the remark wants to mean? what are the properties that have to be satisfied?

[chilly ocean]

I started studying group theory today and I love it. But I'm always struggling to remember which property it is that makes a group an Abelian group instead of a regular group. Is it associativity or commutativity?

[sinful mirage]

comm.

[smoky cypress]

commutativity

groups by definition are associative

[chilly ocean]

Ah, thanks! I love group theory so far. Looks super interesting. Really spices up the maths I did in my undergrad when I reframe it in group theory.

Also, I had another question. Are rings always a triple of a set, addition, and multiplication? Or can the operators be other things?

[smoky cypress]

?

There is no advantage of thinking of rings as a triplet consisting of a set and two operations

You should think of a ring as a set where you can add and multiply things

[chilly ocean]

I thought that was the definition of a ring? That's how it was introduced to me anyway.

[smoky cypress]

Well yes a ring is usually said to be a triplet (R,+,*)

But you should think of it as a set R where you can add elements and multiply elements

But formally it's a triplet (R,+,*)

[chilly ocean]

Can it only be (R, +, *)? Or can I use other operators and still have a ring?

[smoky cypress]

Yes

You can have three operations

and if you ignore the third operation and it is still a ring

[chilly ocean]

If a ring is a set and some operators, is a group a kind of ring?

Because that's a pair of a set and an operator, right?

Or do you have to have at least two operators to form a ring?

[charred glacier]

you have to have <S, +, *> with <S, +> being an abelian group, <S, *> a monoid, and * left and right distributive over +, those operations + and * can be rather abstract though, not necessarily some numeric addition and multiplication

[chilly ocean]

Monoid?

[charred glacier]

monoid is a group without the inverse axioms

[chilly ocean]

Oh, got you. They didn't mention those.

[chilly ocean]

Is totality the same as closure and invertibility the same as the inverse axiom?

[charred glacier]

yeah

[chilly ocean]

Cool. This is pretty interesting stuff

Are all things that are associative and nothing else called "semigroupoids"? Or is a semigroupoid a specific thing?

[charred glacier]

semigroupoids/categories/groupoids are higher level structures, also semigroupoids are pretty exotic i'd say, it is a (small) category without identity arrows, then categories and groupoids are studied in category theory, in basic abstract algebra you start with magmas (maybe implicitly), up to abelian groups, and combined structures like rings, fields and modules

[charred glacier]

https://en.wikipedia.org/wiki/Algebraic_structure list a bunch of other structures, building up groups as https://en.wikipedia.org/wiki/Universal_algebra would be something like

record AbelianGroup : Set₁ where -- is a proper class
  field
    -- magma
    G         : Set -- some set
    _•_       : G → G → G -- 2-ary operation
    -- semigroup
    assoc     : ∀ a b c → (a • b) • c ≡ a • (b • c) -- associativity path
    -- monoid
    e         : G -- 0-ary operation
    left-id   : ∀ a → e • a ≡ a -- etc.
    right-id  : ∀ a → a • e ≡ a
    -- group
    _⁻¹       : G → G -- 1-ary operation
    left-inv  : ∀ a → (a ⁻¹) • a ≡ e
    right-inv : ∀ a → a • (a ⁻¹) ≡ e
    -- abelian group
    comm      : ∀ a b → a • b ≡ b • a

but category-like things are more enriched, though similar

record Groupoid : Set₁ where
  field
    -- precategory
    G         : Set
    _⇒_       : G → G → Set
    _•_       : ∀ {a b c} → a ⇒ b → b ⇒ c → a ⇒ c
    -- semigroupoid
    assoc     : ∀ a b c d (f : a ⇒ b) (g : b ⇒ c) (h : c ⇒ d) → (f • g) • h ≡ f • (g • h)
    -- category
    id        : ∀ {a} → a ⇒ a
    left-id   : ∀ a b (f : a ⇒ b) → id {a} • f ≡ f
    right-id  : ∀ a b (f : a ⇒ b) → f • id {b} ≡ f
    -- groupoid
    _⁻¹       : ∀ {a b} → a ⇒ b → b ⇒ a
    left-inv  : ∀ a b (f : a ⇒ b) → (f ⁻¹) • f ≡ id {b}
    right-inv : ∀ a b (f : a ⇒ b) → f • (f ⁻¹) ≡ id {a}

then usually we want larger categories (in Set₂, etc.), as, for example, all groups (a proper class, not just set) and their induced homomorphisms (still sets) form a large category (in Set₂)

[chilly ocean]

whoa, is this a programming language

[charred glacier]

that is https://en.wikipedia.org/wiki/Agda_(programming_language), so at least there is no obvious mistakes 😂

[stone fulcrum]

In a ring (R,+,×), the operations don't have to be literal addition and literal multiplication

If that's what you were asking haha

[chilly ocean]

Oh, are the plus and times symbols just placeholders? And yes, that is what I was asking.

Are there conditions the operators must meet or can they be any operators at all?

[charred glacier]

you have to have <S, +, *> with <S, +> being an abelian group, <S, *> a monoid, and * left and right distributive over +, those operations + and * can be rather abstract though, not necessarily some numeric addition and multiplication
there are examples here https://en.wikipedia.org/wiki/Ring_(mathematics)#Basic_examples, e.g. matrix rings

[stone fulcrum]

Take all of the subsets of a set U. You can make a ring out of these with the symmetric difference (XOR) as an addition, and intersection as a multiplication

Makes a neat example that's very different than most

We call such a thing a boolean ring

[chilly ocean]

theorem provers takin' all our damn jobs /s

CompSci ftw lol

[toxic zephyr]

so like R^n is there something like Z^n? i know it wouldnt be a vector space but could it be considered a ring?

[chilly ocean]

yah a ring

[toxic zephyr]

okay cool

also does the second operation ("multiplication") have to be commutative?

[chilly ocean]

no

[slate forum]

Can someone tell if my gf's proof is correct?

[woven delta]

What is this a proof of?

Oh ok

[chilly ocean]

it is true that if m != n, and G is infinite, then <x^m> != <x^n>, but this is not true if G is finite

[slate forum]

I know, but G is infinite, this is her trying to prove it, I'm skeptical of the division happening

[next obsidian]

Okay sorry

What’s the statement?

it is true that if m != n, and G is infinite, then <x^m> != <x^n>, but this is not true if G is finite
@chilly ocean this isn’t even true. If m = n + k|x| then <x^m> = <x^n>

[thorn delta]

i lose her at the 1/p notin Z part. I don't think that is a contradiction or anything

[next obsidian]

This also doesn’t even require that. If m and n are coprime to |x| then the two subgroups are equal to <x>

[chilly ocean]

sure, they are not necessarily unequal

[next obsidian]

There’s even an earlier false statement

It’s that x^n ≠ x^im

[chilly ocean]

ah, true, <x^-1>=<x>

[next obsidian]

If r ≠ 0

Take |x| = 5, m = 2, n = 3

Then m = 1n + 1

But x^m = x^2 = (x^4n) = x^12 = x^2

Since |x| = 5

Err

Okay I mixed up the n and m

But it goes both ways

To be complete here

[chilly ocean]

i dont really follow what ur saying, i was just meaning <x^m>!=<x^n> doesn't follow from m!=n when G is finite

maybe i wrote it poorly

[next obsidian]

I’m referencinf the proof given

[thorn delta]

i thought we're assuming G = <x> is infinite

[next obsidian]

Not yours

[chilly ocean]

oh

[next obsidian]

We are???

FUCK

[chilly ocean]

lmao

[next obsidian]

Everything I said is moot

Then yes, these are different ffs AAAAAAA

If G = <x> (and infinite) then G is iso to Z

And then x^m = m in Z

And clearly <m> and <n> are different

Since they’re mZ and max resp.

I think the proof works but I’d say she should justify why x^n isn’t equal to x^im for any i

[thorn delta]

yea, I think the division algorithm is not quite the right approach for this tho

[next obsidian]

Actually I also don’t agree with 1/p not in Z

[thorn delta]

yep ^'

[next obsidian]

There’s no contradiction there

[next obsidian]

There’s a lot of stuff you can say about groups of order pqr

I believe you can show there has to be a normal sylow-r subgroup

Anyway

Also yes haha

Q = H/P for some subgroup of G called H

By the correspondence (fourth iso/lattice) isomorphism theorem

Then like Lagrange or whatever says |H| = |Q||P|

Aka H has order qr

You honestly

Only need like 2 of them mostly

The third one sometimes

(G/H/N/H) = G/N

And the last one you barely need

Lattice iso which is this one I referenced

Honestly is second most important IMO

When you hit ring theory and shit

Yes

Subgroups of G containing N bijection with subgroups of G/N

Respects normality

And meets and joins

So the subgroup lattice of G/N is the same as that you get for G which contain N

And the correspondence is really easy to remember

You just send a subgroup of G to its image in the quotient

And for a subgroup in the quotient you pullback it along the projection

Yes

Since it also is 1-1 for normality

For rings the correspondence is the same and respects primality

Legit all I’m saying is if pi:G -> G/N is the map it should be then the correspondexe goes
H < G goes to pi(H) and for an H’ < G/N it goes to pi^-1(H’)

[slate forum]

How do I make sense of a map
$\mathbb{P}^1\times \mathbb{P}^1 \rightarrow \mathbb{P}^1, (x,y)\mapsto x-y$?

[cloud walrus]

HelixKirby:

[next obsidian]

That doesn’t seem like a map from P^1 x P^1 -> P^1

P^1 should include two variables

The image makes sense, but the domain doesn’t

[slate forum]

? it has x and y

I mean, x and y are really homogeneous coordinates or whatever

[next obsidian]

But that doesn’t make sense

P^1 itself should have two coordinates

So a map from P^1 x P^1 should involve 4 coordinates

[slate forum]

I guess he's representing it as taking x in P^1 and y in P^1

like they're elements of the space I guess?

[next obsidian]

Uhhhhh

[slate forum]

just not written explicitly

[next obsidian]

I mean...

I guess you c an say you send

[slate forum]

I think he wanted us to see this as the extended complex numbers

[next obsidian]

(f(x,y),g(z,w)) -> f(alpha,beta) - g(alpha,beta) or some shit

Honestly I don’t know what this means

[vestal snow]

How do I prove that every element of <x,y|x^2,y^3)> can be written as a finite product where y or y^-1 is the first term on the left?

Edited

I meant non-trivially

Here's the problem where its from

One of the first steps is showing that you can write every element as beginning with y or y^-1

[hot lake]

where does it say that you can write every element as beginning with y or y^-1 ?

[obsidian path]

isn't this just Lagrange's theorem?

[hot lake]

no

[carmine fossil]

Lagrange's theorem states there MIGHT be a subgroup of order d

It doesn't guarantee one

[obsidian path]

oh

if d = m then it's the trivial case, otherwise its either the trivial case or there exists subgroup of order d right? cause its cyclic it has a generator which means it has a subgroup

no?

[chilly ocean]

I mean if m = dk than you can check that <x^k> is a subgroup of order d

right?

[toxic zephyr]

so if i'm dealing with nxn square matrices with all integer entries that I am adding/subtracting and multiplying, would that form or be in a ring? if not what would you call it?

[chilly ocean]

yes

Matrix ring

[toxic zephyr]

interesting

could i call it isomorphic to something like $\bZ^{n^2}$?

[cloud walrus]

nix:

[chilly ocean]

no

multiplying matrices isnt commutative for example

[toxic zephyr]

ah good point

what could I say the columns or rows of the matrices are in then?

and sorry for the many questions but what would it mean to take a determinant in this matrix ring?

oh and what would be the notation for this matrix ring?

[chilly ocean]

lol sure I guess

[toxic zephyr]

and if i was include all the gaussian integers would I say Z[i]^n^2?

[next obsidian]

Call it M_n(R) for nxn matrices with entries in R

[mint gulch]

isn't this just Lagrange's theorem?
@obsidian path

No, Lagrange Theorem is the other way

If exist a subgruoup of order d, then d | m, is not necessary to be cyclic, to Lagrange Theorem Statement

[obsidian path]

I see

I'm still clueless though.. I sort of thought of a way to prove it but not sure of it yet

[mint gulch]

Hmm, think C_m as Z_m with sum, they're isomorphic anyways

[obsidian path]

oh I didn't see that

my bad

hm

I was heading towards that @chilly ocean but I wasn't confident on how to prove it

and I'm not sure what you mean Enigsis..

true

[mint gulch]

<x> is always a Subgroup of G

If x belongs to T

G

[obsidian path]

right

[mint gulch]

Well, every cyclic group have the same structure as Z_m

When m is the order of the cyclic group

With Z_m I mean Z/mZ

With sum

Adittion

[obsidian path]

I'm not sure I understand, Z6 has a subgroup Z3 but they're not the same structure right?

or am I missing the point

[mint gulch]

No, no, I say that if you have a cyclic group of orden m

Then, it has the same structure as Z_m

[obsidian path]

oh ok that makes sense

[mint gulch]

I think that can help to understand

But, you can do what @chilly ocean said anyways

To prove it

[obsidian path]

I'm trying to think of what you said but I think what it entails is beyond me

[mint gulch]

For example.

In Z6, 6 = 2x3, then, [2] in Z6.

If you add up three times you get

[2x3] = [6] = [0], you only need to prove that 3 is the least integer with that property.

If 1 ≤ k ≤ 2, then 2x1 ≤ 2k ≤ 2x2 = 4 ≠ 6

Then, you can't get 0

<x> is always a cyclic subgroup, you only need to figure out its order

[obsidian path]

so we wanna prove that k is the order of x?

[mint gulch]

In this case, is d

See that m = dq

If d = 1, then, the result is the neutral element {e}>

If d = m, the result is one of the generators of the group.

If 1 < d < m, then 1 < q < m

Try with <x^q>

You already know that

(x^q)^d = x^dq = x^m = e

You only need to prove that d is the least element such that (x^q)^k = e

Proving that is the least, you can know that <x^q> have d elements

[obsidian path]

true

[mint gulch]

The least in N -{0}

Because, (x^q)^0 = e

But that doesn't say much

[obsidian path]

right

[mint gulch]

Try Richard Dummit and David Foot Book

[obsidian path]

Alright, I'll write this down and think about it

thanks!

[cinder bone]

?

any hints on this 😮

[stone fulcrum]

Basically find the cycle

119 = 11 (mod 18)

[chilly ocean]

You need to reduce the exponent, not just 119

[cinder bone]

hmmmm

i mean i see how i could do it if it was 5^2020

since U18 is built around 5

[chilly ocean]

what do you mean "built around 5"

like generated by 5?

[dusk summit]

Can anyone help me with this theorem? I dont understand the part about the intersection being the identity element

i dont see how the formula forces the intersection of the two groups to be the identity.

Oh wait I think I kind of get it, although my reasoning doesnt come from the formula. Since m and m' are coprime, their prime factorizations share no common numbers. Thus taking exponents doesnt change the unique primes in their factorizations, and a power of m will never equal a power of m'. Thus no element can have a period that is both a power of m and m', except the identity element which I guess kind of has a period of 1

[uncut girder]

Hey all

I need help

The problem is to find all radical monomial ideals

[cinder bone]

@chilly ocean yeah

[chilly ocean]

It doesn't matter what the generator of U18 is, what matters is its order

Do you know about Euler phi function? (totient function)

[cinder bone]

no 😭

Wait

yes I have

[chilly ocean]

Do you know about Euler's Theorem (the one about totients)?

[cinder bone]

yeah ill take a look thanks

[cinder bone]

so basically 119^(2020) = 7 mod 18

so 7/18 gives the same remainder as 119^(2020)/18?

mmm i think i get it

[tacit saffron]

im trynna solve a problem that says "Let H and K be subgroups of a group G. Prove that the intersection H n K is a subgroup of both G and H." I know that HnK is a subgroup of G (e is in H, and e is in K, so e is in HnK, if g is in HnK, then g-1 is in H and g-1 is in K cuz they both subgroups, and if g,h is in HnK, then gh is in H and gh is in K cuz they both subgroups) but how do i know that HnK is a subgroup of of H

[chilly ocean]

there's not really anything to show

HnK is already a group, and trivially HnK is a subset of H, so HnK is a subgroup of H

"subgroup" is really "is a group" + "is a subset"

[tacit saffron]

oh swag

[tacit saffron]

i gots another problem where i let G be a group of order 25 and i have to prove that G has at least one subgroup of order 5, and that if it contains only one subgroup of order 5, then G is a cyclic group.
ik, let g be in G, other than the identity. because <g> is nontrivial, by Langrage theorm, it must have order 5 or 25, but idk where to go from there.

[chilly ocean]

if g has order 25, then g^5 has order 5

[tacit saffron]

so, if G has no element of order 5, then it must have g of order 25, but because |g^5| = 5, <g^5> is a subgroup of order 5 in G?

[chilly ocean]

yah, G always has an element of order 5

[tacit saffron]

and if there is a subgroup of order 5, H, suppose H is the only subgroup of order 5 in G, and if theres some x not in H, it has to have order 5 or 25 (cuz e is in H).

[chilly ocean]

mm i guess so, but that just follows again from lagrange right?

[tacit saffron]

trynna prove that second half

yeah |x| = 5 means that H = <x>, so x would have to be in H, so |x| = 25, so for same reason, G = <x> is cyclic

[chilly ocean]

er, i dont follow

|x|=5 and |x|=25?

[tacit saffron]

hMMM no

with the assumption that x can't be in H, |x|= 5 don't work

so |x|=25

and then G=<x>

[chilly ocean]

ah yeah that works

[tacit saffron]

in a question im asked to "Make sketches showing the partitions of the half plane into left cosets of H in G and into right cosets of H in G."

what does this even mean

[chilly ocean]

what are H and G? (something related to the plane?)

[tacit saffron]

this is the problem

(not seeking an answer, just need help understanding what its looking for)

so by partitions of the half plane into left and right cosets, seems to be saying that in the matrices, (x,y) would be a point on the right half of the cartesian plane (because x>0)

[chilly ocean]

yah

[tacit saffron]

i don't see where i get right and left cosets from this

[chilly ocean]

so, if elements of G correspond to points in the half plane, and the cosets of H partition G, then in the version where we translate things over to the half plane, the cosets of H partition the half plane, yeah?

[tacit saffron]

yeah that makes sense

i guess my confusion from this comes from my confusion about left and right cosets

[chilly ocean]

so, i think this problem is two questions. one question is, "do this for left cosets" and the other is "do this for right cosets"

its not like the left and right cosets are simultaneously partitioning the half plane or smth

[tacit saffron]

this might be stoopid but i don't really understand difference between left and right cosets

[chilly ocean]

if G is abelian and H is a subgroup, then gH=Hg, so the left coset corresponding to H is the same as the right coset corresponding to H. in the non-abelian case, this is not the case though. but, even if G is non-abelian, we have gH=Hg as long as H is normal

i think the situation is basically, "if H is normal, then left cosets and right cosets are the same, but in general, they aren't, so you have to specify whether you're talking about left cosets or right cosets"

(well, if in the particular problem that you're working on, it is important that you're multiplying something on the left vs the right, then it matters that you're talking about left cosets i guess, but besides something like that)

[tacit saffron]

uhhhhh

[chilly ocean]

i guess i'm not sure what u mean "what is the difference between left and right cosets"

[tacit saffron]

kinda ya

ok that much i do know

somehow

ok yes that makes sense

now i just confused how i do that with matrices

[south storm]

Could someone please help me to understand the prove of that every finite group of order n is isomorphic to a permutation group on n symbols?

What I don’t understand is how showing that every finite group of order n is isomorphic to P is showing that it‘s isomorphic to a permutation group on n symbols. Since how I understood this. Is that P is only a subgroup of S_n.

Please ping me if someone answers

[chilly ocean]

i didnt read the pic but in general, groups are isomorphic only to subgroups of S_n, not to all of S_n

[south storm]

Oh, the question is probably then just phrased badly.

Thanks

[next obsidian]

If $A$ is a DVR why is $A[x]_{\mathfrak{m}_A[x]}$ also a DVR?

[cloud walrus]

Chmonkey:

[next obsidian]

Oh, I think I got it lol

If $A$ is a DVR then it's a Noeth local domain where $\mathfrak{m}A$ is principal. Then $A[x]{\mathfrak{m}_A[x]}$ is a Noeth Local domain since polynomial rings and localizations of Noetherain rings are Noeth, it's still a domain, since we localized at a prime it's local, and the maximal ideal is just the image of $\mathfrak{m}_A[x]$ which is principal

[cloud walrus]

Chmonkey:

[tribal pasture]

What are the conditions on the domain and codomain for an injection implies it is surjection and conversely. I know it holds true mappings between finite sets and finite dimensional linear isomorphisms? Any other?
My guess is that it would hold true for any morphism between a domain and codomain which are freely generated from finite sets

[queen vine]

The Morphism $f:\mathbb{Z}\to\mathbb{Z}, f(x)=2x$ is an injective Map $\mathbb{Z}$-modules which are free on one generator but it is not surjective

[cloud walrus]

leoli1:

[queen vine]

On the the other hand if we require surjectivity instead of injectivity it actually holds: If $M$ is a finite $R$-Module, and $f:M\to M$ surjective, then $f$ is an isomorphism.

[cloud walrus]

leoli1:

[queen vine]

Another example: $R[x]$ is a free $R$-Algebra on one generator but the map $R[x]\to R[x]$ given by $x\mapsto x^2$ is a injective but not surjective.

[cloud walrus]

leoli1:

[tribal pasture]

On the the other hand if we require surjectivity instead of injectivity it actually holds: If $M$ is a finite $R$-Module, and $f:M\to M$ surjective, then $f$ is an isomorphism.
Ah I see, I was thinking that the phenomenon is a generalisation of the fact that surjective maps between finite sets are bijective. So I guess what generalises is the part about surjectivity and not injectivity.

[dusk summit]

There 24 elements in A4

and we know that there are the same number ofeven and odd permutations

sorry of S4

4! ways of permuting 4 elements

I think for n>=2 that holds, not sure

[nova plank]

@vital marten

If your odd permutation is σ then σ = [σ(1 2)] (1 2) and σ(1 2) is even

Oh, I didn't see the original context, I just saw your last question

But the bijection f would be given by f(σ) = σ(1 2)

Yep, it's easy to show f is bijective.

[chilly ocean]

i think this was a nontrivial proof

https://en.wikipedia.org/wiki/Parity_of_a_permutation#Equivalence_of_the_two_definitions

this proves that the number of transpositions it can be decomposed to is a well defined number, by showing that it is always equal to the number of inversions in sigma

[chilly ocean]

what book are u using?

[hot lake]

yeah it's a very nonobvious fact

[next obsidian]

I find it hard to believe they didn’t prove the sign is well-defined

Perhaps it’s a consequence of some lemma proven earlier which wasn’t stated

Or if not and they really didn’t do anything to prove it, lol

Bad book

Oof haha

Okay book is saved

[chilly ocean]

they probably come from years of thinking lmao, when they were originally discovered

[next obsidian]

NGL

This stuff becomes obvious after a year

Usually

It’s kinda whack what 1 year of studying does to u

But also some of them are really clever and probably just smart guy who was doing this for a long time sat around for a bit and came up with the trick

Not even math maturity

But like, ___ subject maturity

You do a lot of algebra and you pick up tricks, get a good sense for how you might solve ___ type problems etc

At least not as good as a diff eq person

[prisma thunder]

That reminds me

Abstract algebra did not prepare me for the nonsense transcendental number theory has

[thorn delta]

hmm

[next obsidian]

Okay so hear me out, this idea is using higher level ideas

But I think you can find a subgroup iso to G/<x> that intersects it trivially

Coming from ideas from module theory

[thorn delta]

alright

[next obsidian]

From there it’s obvious that <x>G/<x> is G

By size considerations

I’m assuming you’ve learned internal direct products?

[thorn delta]

yes

[next obsidian]

Then you can apply that

Anyway, that’s gonna have to be your plan of attack

Honestly

Maybe the fact that it’s iso to G/<x> isn’t that necessary

But otherwise you’d have to like build up subgroups from the complement of <x>

So you take a y not in it

Look at <y>

And probably keep doing this and summing all those subgroups and that seems annoying

Anyway, that’s kinda all I can say tho 😬

Or if you know the classification of finite abelian groups...

Which I don’t imagine you have yet

[thorn delta]

i do, but theorem 2.1 comes before it, so i don't think i can use it lol

[next obsidian]

Ah

I mean if this is to give another proof

I don’t see why it would be bad unless u used 2.1 to prove the classification

It also isn’t even clear to me@how that solves the problem lol

But at least being able to write it down explicitly makes things easier

It at least let’s you find what such an x looks like explicitly I think as a tuple

[thorn delta]

yea, there is a chain of theorems that starts with 2.1 that proves the classification. 2.1 basically says that you can find integers m1 | m2 | .... | mn such that G is iso to the direct sum of cyclic groups Z_mi and some possibly trivial free abelian group.

[next obsidian]

Ah okay

Also if you know that abelian group is the same as Z-module and know the splitting lemma

If you let f be the inclusion of <x> into G

If you can find a section of that map then the splitting lemma implies it’s a summand

Aka a map G -> <x> so that going <x> -> G -> <x> is the identity

But this is kinda using needlessly sophisticated tools I think

[thorn delta]

huh, yea im not really familiar with any module theory. I've had this idea so far let $g \in G$. Then $g = nx + g_1 = kx + g_2$ for some $n,k \in \bZ$ and $g_1, g_2 \in < G\setminus {x}>$. Suppose $x$ has maximal order $c$. With this setup, the goal is to get $nx = kx$ and $g_1 = g_2$.

[cloud walrus]

kxrider:
Compile Error! Click the reaction for details. (You may edit your message)

[next obsidian]

Yes this does it

Should be <x> instead of {x}

But yeah

[thorn delta]

< G \setminus {x}> is what i meant to write

[next obsidian]

Uhh

That won’t be right

That’s just G

You want G \setminus <x>

[thorn delta]

oh yea yea

[next obsidian]

Yeah I’m just not sure how to do this haha

Like honestly it’s probably doable

And I just spent too much time trying to prove it in a fancy way then just show it directly lol

I guess one caveat is

You want to let 0 <= n,k < |x|

Since really they just have to be congruent mod |x|

But you can force it such that’s it’s like that

[thorn delta]

right

[next obsidian]

This is a really really minor detail

[sturdy marsh]

the classification for finite abelian groups follows straight from the sylow theorems right?

it's just the product of all the sylows

[thorn delta]

we didn't use sylow theorems. Those are coming later for us

[sturdy marsh]

oh okay

[next obsidian]

Damn how tf?

You guys did that proof mad quick then

My favorite proof is to quote classification of fg modules over a PID...

[sturdy marsh]

I was talking about classification of finite abelian groups

not fin gen

the only proof for fin gen I know uses modules over PID

[thorn delta]

The relevant path of topics goes
direct and free products -> free groups -> free abelian groups -> finitely generated abelian groups (w/ classification)

[next obsidian]

Yeah my favorite proof literally does that

And specializes to the finite case

If you’re finite then the free part is 0 so all you have is torsion, which is of the form...

[vestal snow]

If $F$ is a finite field extension of $K(x)$ for some field $K$ and $A \in Div(F)$ has degree $1$ and $A \geq 0$, can we say that $A = P$ for some place $P$?

[cloud walrus]

The formerly edible banana:

[vestal snow]

It seems true to me but one of the theorems in my textbook has a very short and simple proof if this is true so wanted to make sure

[chilly ocean]

Could someone explain me why any a = a(1) and what it entails?

[sturdy marsh]

Im pretty sure they just mean a = 1 + ...+ 1 (a times)

Given any element g of a group G, what is the order of g^k in terms of the order of g?

[chilly ocean]

order of g divided by k?

[sturdy marsh]

Not quite

k doesnt always divide the order of g

[next obsidian]

I’m gonna assume you don’t yet know Lagrange’s theorem?

Or even possibly the weaker statement that |g| divides |G| for any g in G?

Because if you do this exercise has a very simple answer

[chilly ocean]

i know it but if I use it i have to prove the whole theorem because we havent learned it in class

[next obsidian]

I mean the proof isn’t too bad?

You can just do a coset decomposition

[sturdy marsh]

you dont need lagrange to figure out the order of g^k

[next obsidian]

I know

I’m just saying the proof becomes really easy for this exercise

[chilly ocean]

i havent learned coset and yeah probably i could prove it but i still want to understand how to do it without lagrange if possible

[next obsidian]

How did you prove Lagrange without cosets?

Anyway, to get a head start on the order of g^k

[chilly ocean]

we havent learned about lagrange

i just watched a socratica video

[next obsidian]

Note that g^n = e if and only if n is a multiple of |g|

This for example is enough to tell you that if k is coprime to |g| then g^k has order |g|

If you try to generalize this phenomenon you’ll get a general statement about the order of g^k

[sturdy marsh]

You could also try figuring out ord(gh) assuming gh=hg

[next obsidian]

I don’t think that there’s actually a general formula for that, I feel like I concluded a while ago you need the orders of g and h to be coprime

Even if the two commute

I assume you want to say it’s || ord(g)ord(h)/gcd(ord(g),ord(h))||

But if you take g = h = 2 in Z/4Z that’s not true

If there’s a different formula you know of that would be big news to me, and I really wished I’d known it before haha

[sturdy marsh]

sure there isnt a formula in general but you can definitely say something about the order

[next obsidian]

You can surely give an upper bound

Also I think I have a proof there isn’t a formula in general

Consider g = h = 2 in Z/8Z versus g = 2, h = 6 in Z/8Z

Both have the same orders in the same group but the order of their sum is different

So it would have to depend on things you can’t get just via their order

[sturdy marsh]

You can get ||ord(g)ord(h)/gcd(ord(g)ord(h))^2 \vert ord(gh)||

and ||ord(gh) \vert ord(g)ord(h)/gcd(ord(g),ord(h))||

[next obsidian]

That seems believable yeah, I know the second and imagine the former isn’t too hard to show either

[chilly ocean]

This for example is enough to tell you that if k is coprime to |g| then g^k has order |g|
@next obsidian can you explain pls im rly dumb with this class lol

[next obsidian]

Suppose k wasn’t a multiple

Derive that there’s a number n < |g| such that g^n = e

Oh wait

That’s for the wrong part

I think if g^k wasn’t order |g| you get that k and |g| aren’t coprime

[sturdy marsh]

Also I think I have a proof there isn’t a formula in general
@next obsidian Im pretty sure once you lose the commuting assumption you can get gh to be any order you want

[next obsidian]

Oh yeah for sure, I was assuming they did commute

[chilly ocean]

if g^k isnt order |g|, then wouldnt k and |g| be coprimes

[next obsidian]

If (g^k)^n = 1 (for n < |g|) then g^{kn} = 1 so kn is a multiple of |g|

Stare at that long enough and you get that k and |g| aren’t coprime

This says the lcm of k and |g| isn’t just k|g|

And that’s equivalent to saying k and |g| aren’t coprime

Via a formula relating lcm and gcd

[chilly ocean]

but (g^k)^n has order |g| right?

[next obsidian]

I’m arguing via the contrapositive

I’m assuming n = |g^k|

If the order isn’t |g| then n < |g|

And I got a contradiction since I and |g| aren’t coprime

[chilly ocean]

how do you know n < |g|

oh wait im so confused

[chilly ocean]

does the P in the formula mean any power of x?

[solemn rain]

p is the exponent

[chilly ocean]

can it be any exponent

like if I want to use that formula to prove something what can I assume the p to be in gcd(p,n)

[solemn rain]

yes

p is arbitrary no:?

[chilly ocean]

cuz like that proof they assume p is prime

[solemn rain]

read the theorem

[mint gulch]

Could someone explain me why any a = a(1) and what it entails?
@chilly ocean
By Lagrange Theorem, the order of every subgruoup divides p, p is prime, so, if is proper, then, the group have to be the trivial group.

Given a Subgroup H of G, think of gH, you can form a disjunt union

|G| = |g1H| + ... + |grH| = r|H|

Then |H| divides |G|

[chilly ocean]

@mint gulch thank you very much for the explanation but I cant use that theorem, did they use that in the proof?

read the theorem
@solemn rain p is the exponent of x no?

why do they assume it to be prime

and the proof i sent you is one i found online, i dont have access to thm 6.14

[mint gulch]

why do they assume it to be prime
@chilly ocean

They asumme prime to get (a,p) = 1

For every number in Zp

[chilly ocean]

but how can they do that

[mint gulch]

For example in Z4 (2,4) = 2

And Z4 have a Subgroup of order 2

<2>

but how can they do that
@chilly ocean

They suppose that z is primee

That's the assumption

[chilly ocean]

but in x^P P can be any power of x

doesnt have to be prime no?

[mint gulch]

I don't get what you mean

[chilly ocean]

https://cdn.discordapp.com/attachments/496784958430380033/762438365097754634/unknown.png

is this the theorem they use?

[mint gulch]

I don't know, you have the book

There it says theorem 6.14

[chilly ocean]

its a proof i found online for an exercise i have

i dont have this book

but its okay if you dont know

[steep hull]

Proof of other inequality: Consider the elements g^u and h^u where u=gcd(ord(g),ord(h)). These are elements of order s=ord(g)/u and t=ord(h)/u, meaning that the orders are relatively prime. Let r be ord(g^uh^u). Then g^(ur)h^(ur)=1, but then (g^u)^(rt)(h^(ut))^r=1. Hence (g^u)^(rt)=1 and so s|rt, implying s|r since gcd(s,t)=1. Similarly, t|r, and so st|r. But (g^uh^u)^st=1, and so r=st. The group generated by g^uh^u is a subgroup of that generated by gh, and so hence st=ord(g)ord(h)/gcd(ord(g),ord(h))^2|ord(gh), as desired.

@next obsidian

Ofc g and h have to commute or this doesn’t work

So I guess the takeaway is to build a power of gh with kth powers of g and h that have relatively prime orders and then use the lcm property (which holds given commuting and relatively prime order).

[dull shard]

Hiiiii guys

In context of this

How would you determine if a polynomial is "reducible"?

Is it just reducible if it has any zeros in Z_5?

Is this correct?

<@&286206848099549185>

[clever rampart]

No if it can be factored into non-trivial factors

It is possible that none of them are linear

Hmm, so if in (a) you found that their gcd was not one you already have your answer (at least in part).

[dull shard]

@clever rampart

Are you still here?

I need help with this

[bleak abyss]

Which part?

[mint gulch]

@dull shard ?

[dull shard]

d.

[mint gulch]

What do you have?

You have to prove that if A,B in S, then A - B in S and AB in S

That's because the operations +, •, have the associativity property, the distributive property, and the conmutative property of the addition by a)

Of course, remember to see that S is not empty

[vocal depot]

So I have a problem I'm working on and I was wondering if there might be someone who could help direct my line of thinking. I feel like I have a good idea of how to go once I get the initial slog done but it's that part I'm having issues with.

Let $p$ be a prime number. Determine how many subgroups of order $p$ there are in $\mathbb{Z}{p}\oplus\ldots\oplus\mathbb{Z}{p}$

[cloud walrus]

HisMajestytheSquid:

[vocal depot]

My first thought is to try and find the number of cases where I can have lcm$(|a_{1}|,|a_{2},\ldots ,|a_{n}|) = p$.

[cloud walrus]

HisMajestytheSquid:

[vocal depot]

But that seems like it could go on forever with the number of combinations of elements $a_{i}$ I could go through.

[cloud walrus]

HisMajestytheSquid:

[vocal depot]

So is there any way to narrow down the list so that I can make counting the number of elements a little more managable?

[mint gulch]

What is circ(+)

[vocal depot]

direct external product

[mint gulch]

Yes, but, with sum or product ?

And that's Z/pZ, right9

?

[vocal depot]

I'm not familiar with that notation but I believe the operation of the group is assumed to be addition.

The convention we have been using is that $\mathbb{Z}_{n}$ is addition of the integers modulo $n$.

[cloud walrus]

HisMajestytheSquid:

[mint gulch]

I think <x> with x ≠ 0 is a Subgroup of order p

For every x in that sum

The sum is conmutative, then the product (Or in this case, the addition) of two Subgroups is a group too

A Subgroup

[vocal depot]

So far that makes sense.

[mint gulch]

If G intersection H is {0, ..., 0}

Then G + H is a Subgroup of orden p*p

Order*

Because G + H is isomorphic to G circ(+) H

I know those facts with •, hahaha, it's weird to state them like that

Do you agree with those facts?

[vocal depot]

I can get behind it but is it really necessary to define an isomorphism here? It seems like there should be an easier way than that

[mint gulch]

No, you can prove that G + H have order p² in other ways

Then, (G + H) + T has order p³ if G + H intersection T is {(0,...,0)}

And so on

I think you can form all of the Subgroups like that

[vocal depot]

I guess I'm confused about how that's relevant. The first part of my task should be to count the number of elements with order p. And then use that count to break down how many cyclic subgroups of order p I have.

[mint gulch]

You can start with the following fact.

Calculate the number of Subgroups of order p of that Sum. Every element have order p (Except for 0), then, they're cyclic, hence, their intersection is always 0, then, to count the number, all you have to do is to sum k times

p-1, this is (p-1)k and add it one.

k(p-1) +1, you need to get p^n

k(p-1) + 1 = p^n

k(p-1) = p^n - 1

k(p-1) = (p-1)(p^{n-1} + p^{n-2} + ... + p + 1)

Then k is equal to the second polynomial

Every element has order p, because, every coordinate have order p, without count 0

now, from there, you can form groups of order p²

And so on

[vocal depot]

Hmm okay. Let me see if I can get anywhere with that and I'll come back if I have more questions.

[mint gulch]

Ok

[mint gulch]

If I know that f primitive in Z[x] can be factored uniquely up to unities

Can I prove that every element g in Z[x] not 0 can be factored,, just considering g/c(g) ?

Where c is the content of g

The gcd

Of the coefficients of g

[paper flint]

The instructor said that phi here is a group homomorphism from Z to an arbitrary group G, but it seemed to me that he implicitly assumed the group operation for G is ordinary multiplication. I'd like to know if phi is indeed a group homomorphism for any operation on an arbitrary group G.

[hot lake]

the dot in a^m . a^n should have been a star *

not sure what the (a * a^n) * (a * a^n) * ... * (a * a^n) thing is supposed to be doing there

[paper flint]

So the operation on G should've been ordinary multiplication on integers, right?

[nova plank]

No

[hot lake]

no the operation on G is whatever operation G has

[paper flint]

How would that lead to a^{m+n}, then?

[hot lake]

?

[paper flint]

Guess I'm a bit confused about homomorphisms

[hot lake]

a^n is just a notation for a * a * a * ... * a

[paper flint]

Oh, so the homomorphism on the right just takes operation of Z on the left and operation of G on the right?

[nova plank]

So if you write out m a's next to each, then another n a's, in total you have m+n a's

[hot lake]

yes homomorphisms have to be compatible with the operations on the respective groups

[paper flint]

👍

Makes sense now

[nova plank]

$\underbrace{\underbrace{aa\dots a} _m \underbrace{aa\dots * a}n}{m+n}$

Just practising my latex, and failing

[paper flint]

XD I understood.

[cloud walrus]

Lunasong:

[nova plank]

Yay

[paper flint]

Nice!

Thanks for the help.

[nova plank]

Yeah, so later on, you will likely drop the * notation completely and just write ab for a*b. So we use multiplication notation for any group operation, because it's shorter and because the same exponent rules etc apply

[paper flint]

I think I'm still a bit confused on the group homomorphism bit

The only bit I understand is that it's supposed to be a function from a group to another satisfying the constraint phi(ab)=phi(a)phi(b), but which group operations am I supposed to consider?

[hot lake]

the ones that make sense

[carmine fossil]

Isn't there only one group operation?

[hot lake]

there are two groups

so two group operations

[paper flint]

Say $\phi:G\to H$, then for $\phi$ to be a group homomorphism, $\phi(ab)=\phi(a)\phi(b)$, then\(1)Which operation is used to compute $ab$?\(2)Which group operation is used to compute $\phi(a)\phi(b)$?

[cloud walrus]

TedNowKaczynski:

[nova plank]

If you are saying, which operation does G have, it is arbitrary. So we are saying a^(m+n) = a^m * a^n FOR ANY group operation *. But φ(m) and φ(n) are elements of G, so the operation between them is the operaton from G. Whereas in φ(m+n) the + is the operation from Z.

[hot lake]

the operation to compute ab is the operation of the group that a and b belong to

[paper flint]

Ah, that answers my question.

Thanks a lot both of you.

[deft plume]

Yah I agree, it depends on the operation of G

The idea behind homomorphisms is operation preservation

[woven obsidian]

I'm doing an exercise on when two localisations S^-1A and T^-1A are equal

But I'm unsure what equality means here. It clearly cannot be an equality of sets if S is not equal to T

And I'm not sure if it's simply an isomorphism either

Perhaps the isomorphism should take a/1 to a/1?

[steady axle]

u should specify what are S and T

[woven obsidian]

Multiplicative subsets of A

[steady axle]

then the statement does not hold in general u could take one of them to be {1} for example

[cloud walrus]

Token:

Token:

[woven obsidian]

then the statement does not hold in general u could take one of them to be {1} for example
@steady axle There is no statement

I'm just wondering what does S^-1A=T^-1A mean

[knotty mason]

@elder condor what would be the generators and relations for D_8/<r^2>?

we know r and s generate D_8 and have certain relations

yeah

so we could replace the relation r^8 = 1 with say r^2 = 1, and we have the same s^2 = 1, and then the relation between them sr = r' s

well if r^2 = 1

then r = r'

so sr = r's can be rewritten

[knotty mason]

in the quotient group D_8/<r^2> we have r^2 = 1 or (r<r^2>)^2 = 1 if you want to explicitly write out the coset

[knotty mason]

rs = sr'

but r = r'

so rs = sr

[chilly canyon]

Hello ! I couldn't find anything on the internet, so perhaps it's no known result... Is there some kind of formula for either 1) the characteristic polynomial, or 2) the spectral radius of a Vandermonde matrix ?

[zealous mortar]

How can I simplify this composition of linearized polynomials in $F_q^m$? Any tips?

[cloud walrus]

jm:

[hot grotto]

Describe the maximal ideal of Q[x,y]corresponding to(√2,√2)and(−√2,−√2), i am not sure where to begin with this

shouldnt the ideal for this be (x+y, x^2+y^2-4)?

[prisma ibex]

Sounds about right to me...

Errr

x-y

[full holly]

For all symmetric groups Sn, treating [1, 2, 3, ..., n] as the identity permutation, does there exist an ordering $I$ of elements without repeats, such that if $s_{ij}$ s the position the ith permutation takes the jth element of the identity permutation, then either $s_{i, j} = s_{i \pm 1 \mod{n}, j}$ or $s_{i, j} = s_{i + ck \mod{n}, j}$ for some integer c, and for all integers k?

[cloud walrus]

Obscura:

[full holly]

In simple language, is there an ordering of permutations such that when visualized in the above manner, vertical blocks of color appear to be "periodic"?

[chilly ocean]

hmm

[full holly]

for n=3:

132
123
213
231
321
312

[hot grotto]

@prisma ibex why x-y? that makes it non zero? and should i include x^2-2 and y^2-2 ?

[prisma ibex]

I mean (a,a) is in the zero locus of x-y

Not x+y

[hot grotto]

what is meant by that?

in terms of the ideal i dont see the connection

x+y is root2 + -root2 = 0 so isnt that in the locus?

[prisma ibex]

I mean you’re trying to find an ideal whose elements vanish at these points

[hot grotto]

yes

ok so i do include the terms like x^2-2 then?

[prisma ibex]

No no you have points of the form (a,a) and (-a,-a)

[hot grotto]

ok

[prisma ibex]

I mean sure the points you are interested in are in the zero loci of those too

You just might be adding redundant terms

[hot grotto]

ok so i have to include both then x+y and x-y or just x+y

oh wait

[prisma ibex]

if you include x+y you will get points like (a,-a) in the zero locus so no

[hot grotto]

its x-y as in (x1,y1) , (-x1,-y1) and im doing x1-y1 right?

per each term?

[prisma ibex]

yes

[hot grotto]

ok for some reason i was doing it from x on one ordered pair to the y on the other ordered pair

now i see yes

[dusk summit]

does anyone have a good resource for understanding the classification of finite abelian groups? I'm having a little trouble following the proof

[woven delta]

I think the proof in dummit and foote is pretty good. Basically if the group G is generated by a set {g_1, g_2, ..., g_n}, you can make a map phi from Z^n to G which maps the ith generator of Z^n to g_i. Then by the first isomorphism theorem Z^n/ker phi is isomorphic to G

Then you can show that subgroups of Z^n are free, so you can write it as Z^m for some m<= n

The map phi then turns into a map from Z^n to Z^m. You can turn this into a matrix

@dusk summit then the next step is proving that you can do something similar to gaussian elimination to matrices with entries in Z

I think Artin covers this at the end of the book

And dummit and foote covers it sometime in the middle

[dusk summit]

oh ok thanks

Hmm I was reading it for just finite groups, not finitely generated and the proof seemed to be a lot different (no matrix, no free groups, the book hasn't covered that yet)
is the proof very different between the 2?

Oh I checked out some background stuff in Dummit and Foote and I think It might make the proof a lot clearer for me through some stuff scattered throughout the sections

[woven delta]

I only know the more general proof

The matrix algorithm is very important in general, so it's probably worth reading the proof and learning the algorithm

[dusk summit]

I definitely will then, thank you for the help!

[slate forum]

Alright, so, say I have two irreducible quadratic polynomials $f,g$ in $k[x,y]$ for some algebraically closed field $k$, how can I show that if I look in $k(x)[y]$, I have $a(x,y)f+b(x,y)g = 1$, where both $a(x,y),b(x,y)\in k(x)[y]$, and I want to make sure that when I clear denominators, the right hand side will be a polynomial solely in $x$ and it has degree at most four.

[cloud walrus]

HelixKirby:

[slate forum]

Anyone?

[chilly ocean]

maybe, a coefficient p(x)/q(x) of a(x,y) must have p(x)= a constant because k is algebraically closed, so otherwise common factors of p and q would cancel, and q(x) cannot be degree >=3 (this is almost a penis) maybe because the highest degree of x in f is 2, or something like that

my algebra is ass though, so this could be way off

hm yah nvm i dont think this is going anywhere

[slate forum]

shoot

<@&681260374879633482> Anyone any good at AG?

[prime gale]

@slate forum I doubt it is true. For example, take f=g.

[slate forum]

irreducible and not equal

@prime gale

[kindred mist]

ok am I correct that direct summands of projective modules are projective? (I am pretty sure I proved it and I think my proof is correct and simple but I just wanted to be sure I didnt mix something up since I have heard "the direct summand of a free module is projective" but why even bother with that if literally "free" can be replaced with "projective" (I am saying because I get that free implies projective))

nevermind someone answered yes already

[prime gale]

@slate forum I think you can use the fact that k(x)[y] is a Euclidean Domain. Just use Euclid's algorithm to compute GCD and back up to get coefficients. Don't think it matters that k is algebraically closed.

If deg_y(f) = deg_y(g) = 2, then find q1 in k and r1 in k[x,y] with deg_y(r1)<=1 so g = f*q1+r1. Then find q2 in k[x,y] and q3,r2 in k[x] with f=r1*q2/q3+r2/q3 and some kind of bounds I haven't computed yet for deg_x of q3 and r2. Then r2/q3=f-(g-q1*f)*q2/q3.

[prime gale]

We can easily bound deg_x(q3) with 1. Also easy to bound deg_x(r2) with 3 (because deg_x(r1)<=2 and deg_x(q2)<=1), so deg_x(q3*r2)<=4.

[mint gulch]

Z[x] is countable? If so, I'm curious how to prove it, any hint?

[chilly ocean]

it's like the proof that Q is countable on steroids

[prime gale]

Is the number of polynomials with degree<=n and coefficients -n<=coeff<=n finite?

[mint gulch]

I think yes

[prime gale]

Does that give you a proof?

[mint gulch]

it's like the proof that Q is countable on steroids
@chilly ocean

But, that's because Q is a ordered pair of a countable set

[chilly ocean]

okay... think super steroids

[mint gulch]

Does that give you a proof?
@prime gale

I will think about it

[golden pasture]

eh

you know there are infinitely many primes

so

is there some really natural way to go from polynomials to integers via primes?

[prime gale]

Why all these weird complicated ways to prove it?

[golden pasture]

i mean like it's the most natural for me

cuz gives immediate bijection

by most natural probably means it was the first proof i saw probably

[vital quail]

@mint gulch easiest is to just prove AxB is countable if A, B are

and then induct

@golden pasture what do you even do with negative coefficients though lol

[prime gale]

@vital quail Does that really work? Doing what you say, you prove that set of polynomials of degree <=n is countable for all n. Need additional argument to show Z[x] is countable.

[vital quail]

transfinitely induct

[prime gale]

Huh?

[golden pasture]

true negative coefficients is a thing

then can just biject with rationals

[vital quail]

sure

[prime gale]

Positive rationals.

[prime gale]

@slate forum Computing super explicitly, I get this. First of all, assume deg_y(f) = deg_y(g) = 2. Maybe deg_y<2, but that is easier case. Dividing by y^2 coefficient, assume f=y^2+f1*y+f2 and g=g1*y+g2. I got rid of the y^2 term from g by subtracting f. f1,f2,g1,g2 in k[x] and deg(f1) = deg(g1) = 1, deg(f2)=deg(g2)=2. Degrees may be less, but that is easier case. Then g1^2*f-(y*g1+f1*g1-g2)*g = f2*g1^2-f1*g1*g2+g2^2. Notice that degree of RHS is 4.

[slate forum]

Oh jeez

[prime gale]

Jeez good or bad? It is like PG-13 level explicit. Maybe a bit too much.

[slate forum]

Why can you subtract f?

[prime gale]

You want to find a*f+b*g=1. If h=g-f, then (a+b)*f+b*h=1

The pair (a+b, b) has all the properties you wanted (a,b) to have.

Or vice versa--whatever.

[slate forum]

Uhhh

I don't know how you got that

[prime gale]

Are you still questioning that I can subtract f from g?

[slate forum]

I guess? And that a+b business

[prime gale]

Use this method to find a and b, so that a*f+b*(g-f)=1. Then (a+b)*f+b*g=1. And a+b and b have the properties you want.

[slate forum]

I guess?

Is it like doing a step in the euclidean algorithm

[prime gale]

Yes. You can think of it as a euclidean algorithm step.

[slate forum]

Ok I guess?

[prime gale]

tl;dr, just use euclidean algorithm.

[slate forum]

Maybe if I explicitly wrote out the division?

[prime gale]

You can. But I already did that.

[slate forum]

Oh...

These things are of the form ax^+bxy+cy^2+dx+ey+k right?

[prime gale]

Yes.

[slate forum]

So... a1x^2+b1xy+c1y^2+d1x+e1y+k1= quotient plus remainder

How do I actually divide these polynomials

Jeez

Say I look at xy-1 and y-x^2 for now

[prime gale]

You can just check that g1^2*f-(y*g1+f1*g1-g2)*g = f2*g1^2-f1*g1*g2+g2^2 is true. Then a+b=g1^2/RHS, b=(-y*g1-f1*g1+g2)/RHS.

Of course you still have to check those degenerate cases--but they are easier.

[slate forum]

What are fi,gi?

[prime gale]

f = y^2+f1*y+f2. g-f = g1*y+g2. Just read my solution.

[slate forum]

Ok

Where are the denominators in your solution?

[prime gale]

There aren't any. I end up with something in k[x] on RHS. If you want 1 on the RHS, then my RHS is your denominator.

[slate forum]

Ohh

How did you get that equation though?

Was it from EA?

[prime gale]

I understand what you are going through. I went through the same thing. We teach math literacy by teaching stuff like Algebra. So in the beginning, you have to struggle with literacy and Algebra at the same time.

Yes. I just used EA. But if you want to prove your statement, you don't need all of that. You can just use the result.

[slate forum]

I guess, but I like to know where these magic formulas come from

I don't like that when we write proofs we leave out the scratch

Though, it would get very messy

[prime gale]

You can read my post before the explicit one. I give outline of proof without computation, just assuming computation will work out. Then I got bored, so I did the computation.

[slate forum]

Thanks

What do you mean by literacy?

[prime gale]

Just ease of understanding the non Algebra arguments. For example, understanding that we can divide by coefficient of y^2 and assume it is one, and that we can subtract f from g and assume g has no y^2 term. Eventually stuff like that is just automatic.

[slate forum]

Oh?

When you say algebra, you mean modern algebra not college algebra right?

[prime gale]

Don't know what modern algebra is or what college algebra is. I just mean Algebra.

[slate forum]

Is it like knowing what to pick for delta in analysis?

College algebra is like functions, trig, pre calc

[prime gale]

What I mean by literacy is all the math reasoning stuff that is more basic than any subject, like Algebra or Analysis.

[slate forum]

I did well in my proofs class, so there's that I guess?

Though this is graduate school math

I'm a 3rd year

PhD student

The imposter syndrome is real

[prime gale]

I never had a proofs class--but I think you are right--probably proofs class teaches stuff like this. I just took a bunch of math classes and learned the language as I went. Kind of like learning to swim by being thrown into river. That's why I can understand the struggle.

[slate forum]

That's a scary way to learn math I think

Probably is a good way though

[prime gale]

Have you taken qualifier tests?

I have lots of friends who went to graduate school in math.

[full holly]

For all symmetric groups Sn, treating [1, 2, 3, ..., n] as the identity permutation, does there exist an ordering $I$ of elements without repeats, such that if $s_{i, \hspace{1mm} j}$ is the position the ith permutation takes the jth element of the identity permutation, then either $s_{i, \hspace{1 mm}j} = s_{i \pm 1 \mod{n!},\hspace{1 mm} j}$ or $s_{i, j} = s_{i + ck \mod{n!}, \hspace{1 mm} j}$ for some integer $c > 1$, and for all integers k?
@full holly

Can anyone help?

[prime gale]

Please proof read your statement and fix the errors. I don't think anyone can understand it as is.

Do you mean that s_{ij} = s_i(j)? i in I, j in [1...n].

[full holly]

In simple language, is there an ordering of the permutation, such that when arranged as below (vertically), vertical blocks of color appear to be "periodic"?

Do you mean that s_{ij} = s_i(j)? i in I, j in [1...n].
@prime gale
Yes

[prime gale]

Do you really want s_{i+-1 mod n}? That's weird. Not "mod n!"?

[full holly]

Oops, yeah

[prime gale]

Well, if I interpret your statement literally, it is trivial. Just take c=0.

[cloud walrus]

Obscura:

[prime gale]

Well, just take c=n!.

I am sorry--I am not trying to be cute. It is just I am struggling to understand what you have written.

If I interpret your statement right, it is true even if you ignore the c,k stuff. You can order your permutations so that each permutation differs from previous permutation by a transposition (easy to prove by induction, unless I made a mistake). So even ignoring your second condition, you can make an ordering that satisfies your first condition.

Well, I guess I have to be careful about where it wraps around n!. But you can probably make it work out.

[slate forum]

I high passed all my quals @prime gale

[prime gale]

Congratulations! Lots of my friends couldn't pass them.

[slate forum]

I feel like I got lucky

Anyway, for euclidean domains, if the polynomials are coprime, then the EA terminates when I get a unit right?

[prime gale]

Or the GCD.

Sorry--you said coprime. Yes.

But in this case, since degree is two, it terminates right away.

[slate forum]

The bottom part here, did I do this right?

[prime gale]

That is even more explicit than my PG-13 rated stuff. That is like R-rated. It looks about right in general. I'm not going to check every little thing.

[slate forum]

Ok, so, my r is a unit here, so, is that my gcd?

[prime gale]

I guess so.

[slate forum]

Seems weird, but my denominator is degree 3<=4, so that's nice I guess?

[prime gale]

I wouldn't worry about that. You probably didn't start out with random polynomials. If you start with random polynomials, your denominator will be degree 4.

[slate forum]

I don't think the algorithm will finish in one step if I use two quadratics in y

[prime gale]

Then use two steps.

[slate forum]

I know, but it's gross

[prime gale]

That's why I just subtracted f from g. Less gross than applying EA twice.

[slate forum]

I suppose

So.. assume they're both monic

[prime gale]

Working through concrete examples is good.

[slate forum]

Then the first step is kind of already done

The second step you divide a quadratic by a linear term

So you'll get a constant

Well "constant"

[prime gale]

Yes. Constant = element of k[x].

[slate forum]

Ok... jeez, maybe I can just claim without proof that the common denominator will be at most degree 4

[prime gale]

Probably. It is pretty easy.

What are you trying to prove overall?

If your intuition told you that denominator will have at most degree 4 before actually computing, you have excelling intuition. I can't tell without computing.

[full holly]

If I interpret your statement right, it is true even if you ignore the c,k stuff. You can order your permutations so that each permutation differs from previous permutation by a transposition (easy to prove by induction, unless I made a mistake). So even ignoring your second condition, you can make an ordering that satisfies your first condition.
@prime gale

How would this work? The first condition says that colors occur in blocks; if we ignore the second condition this is just saying that you can order all permutations without repeats where each color occurs as a contiguous block in each column (using the image I posted to simplify language): For S4, the first column has 6 consecutive red blocks, the next has 6 consecutive yellow, etc. The next column must also have six consecutive red blocks, but there's no way of placing it without there being at least 3 red blocks for any given color in the first column. This breaks the no-repeats condition, since there can only be 2 red blocks in the second row for any given color in the first row.

[prime gale]

I see your point. You are right. I need consecutive permutations to be disjoint. But that probably isn't possible.

[full holly]

Here's the n=3 positive example:

132
123
213
231
321
312

[prime gale]

I am just ignoring the stuff you say about colors and blocks. I can't understand any of it.

[full holly]

The question comes from the image. The poor wording of the more formal description comes from me trying to capture the visual characteristics of the problem. Again, stated simply, the question is whether or not permutations can be arranged stacked on top of each other, such that sequences in the same column occur as periodically spaced blocks.

[prime gale]

Yeah--I can't understand that at all.

[slate forum]

@prime gale the question is showing that two different irreducible conics intersect at at most 4 points

That's why I knew to look for four

Because I remembered how to show two different irreducible plane curves intersect in at most finitely many points

Which by the way, was definitely not intuitive at all

[full holly]

Here's a non example for n=3, in the first row second column s_{1,2} = 2, but s_{5,2} = 2, so c=4, but 5+4 mod 6 = 3 and s_{3,2} = 1, so the second condition fails.

123
132
213
231
321
312

[prime gale]

I think your proof works, though I might be taking too much for granted. You move around f and g and show that after moving them around, the intersection projects onto x axis as <= 4 points. But you have actually shown the intersection can project onto any subspace as <= 4 points. Only set that projects onto any subspace as <= 4 points is <= 4 points.

[slate forum]

Moving them around?

[prime gale]

That's what I call it. Moving g to g+a*f, etc. using EA. Quadrics move around, but intersections stay the same.

Well, I haven't thought about details, but it seems like your approach would work.

[slate forum]

Okie dokie, what do you know about jectivity of morphisms and their associates pullbacks?

[prime gale]

Don't even know what those words mean.

[slate forum]

Given $f:X\rightarrow Y$ a morphism of affine varieties and $f^*:A(Y)\rightarrow A(X)$ the associated $k$-algebra mapping on regular functions, what are the relationships between the injectivity/surjectivity of each

[cloud walrus]

HelixKirby:

[slate forum]

Any of those words mean anything?

[prime gale]

I understand question now. I have to think about it.

Well, if f is injective, f*:A(Y) -> A(X) is surjective. If f is surjective, then f*:A[Y]->A[X] is injective, but maybe A(Y)->A(X) isn't--I have to think about it.

[slate forum]

Yeah, I was able to prove if $f$ is surjective, then $f^*$ is injective

[cloud walrus]

HelixKirby:

[slate forum]

I know how the converse can go wrong, but it's not really clear how to make an example

If the image of $f$ is not dense, then $f$ is not surjective, and $f^*$ is not injective, for if I take any function that vanishes on all of the image of $f$ but not the whole of $Y$, then the pullback will be zero

[cloud walrus]

HelixKirby:

[prime gale]

Yes--f^* is injective for varieties because they are irreducible.

I agree. if image of f is not dense, then f^* is not injective.

[slate forum]

yes, how do I make this happen?

[prime gale]

How to get an example of a map whose image isn't dense?

[slate forum]

yeah...

[prime gale]

Just take k[x]->k x->0?

[slate forum]

Like, I'm trying to find a counter-example

[prime gale]

Are you looking for counterexample of converse of "if f is surjective then f* is injective"? If so, just take (line minus origin) -> (line).

[slate forum]

f^* is injective here?

[prime gale]

Yes.

If two functions are equal everywhere except origin, then they are equal at origin. After all, they are continuous.

[slate forum]

Oh, ok

So, if $f^*(g)=0$, then $g\circ f = 0$, and this means $g$ is zero everywhere except at the origin, but by continuity, we must have that $g(0)=0$, so $g=0$?

[cloud walrus]

HelixKirby:

[prime gale]

Well, I wouldn't use the word continuity, because it only works for complex or real. But it is pretty obvious for the rings also.

Well, I guess you just want 1 example. So use real line minus origin -> real line. Then continuity works.

[slate forum]

It's obvious for rings?

How?

I don't have a notion of continuity though

[prime gale]

A(line) = k[x]. A(line-origin) = k[x]_x.

If you just need one example, you know that real polynomials are continuous, so that works.

[slate forum]

ok

By the way, stupid quick question, are the Zariski and usual topology on $\mathbb{R}^n$ and $\mathbb{C}^n$ comparable?

[cloud walrus]

HelixKirby:

[slate forum]

Like, is one finer or coarser than the other?

[prime gale]

Yes. Usual topology is finer than Zariski.

You can prove this using fact polynomials are continuous.

[slate forum]

Ok

So every zariski closed set is closed?

[prime gale]

Yes. Zariski closed set is preimage of {0} of polynomials. Since polynomials are continuous, preimage of {0} is closed in normal topology.

[slate forum]

ok, cool

[prime gale]

I have been way too addicted to news sites the last couple of days.

[slate forum]

The Zariski topology is weird though, huh, pretty much never Hausdorff

[eager bobcat]

Is the kernel of $\mathbb{Z}_n$ = n\mathbb{Z}$?

[mint gulch]

The kernel of what morphism?

[eager bobcat]

under modular ddition

[mint gulch]

$\mathbb{Z}\to \mathbb{Z}/n\mathbb{Z}$ defined by

$z \to z \pmod n$

[cloud walrus]

Enigsis:

[mint gulch]

$z \in \operatorname{Ker} f \iff f(z) = 0 \iff z \pmod n = 0 \iff n \mid z \iff z \in n\mathbb{Z}$

[cloud walrus]

Enigsis:

[mint gulch]

There you go

[toxic zephyr]

is this channel free?

[mint gulch]

Yes, I think so

[toxic zephyr]

okay just making sure. if ratsella still has a question i can delete mine or something

[mint gulch]

@eager bobcat

[toxic zephyr]

is it possible for a vector space to also be a field?
for example, the set of real numbers over standard addition and scalar multiplication over the field of the real numbers is a vector space, right? but since the scalars of the field are just the same elements of the set, the scalar multiplication ends up being just regular multiplication of the elements. so if i can then show that every nonzero element happens to also have a multiplicative (scalar?) inverse, then can i say it is also a field?
a linear algebra student asked if "the set of all real numbers with the standard operations of addition and multiplication" formed a vector space, and i guess i was just wondering if that sufficiently described a field as well.

sorry if that doesnt make any sense, im not extremely familiar with abstract algebra or solid on the proper definitions

well lets say i have a set of elements S which forms a field F under two defined operations (addition and multiplication). i guess my question is if that set S over the field F with the same definitions of addition and multiplications forms a vector space. i guess another way to say it is 'can you call a field F a vector space over itself?'

wow thats cool

thank you @open torrent 🙂

[mint gulch]

is it possible for a vector space to also be a field?
for example, the set of real numbers over standard addition and scalar multiplication over the field of the real numbers is a vector space, right? but since the scalars of the field are just the same elements of the set, the scalar multiplication ends up being just regular multiplication of the elements. so if i can then show that every nonzero element happens to also have a multiplicative (scalar?) inverse, then can i say it is also a field?
a linear algebra student asked if "the set of all real numbers with the standard operations of addition and multiplication" formed a vector space, and i guess i was just wondering if that sufficiently described a field as well.
@toxic zephyr

Yes, every Field F, is a F - Vector Space

If S is a subfield of F, F is a S - Vector Space, like C being a R - Vector Space or R being a Q - Vector Space

[golden pasture]

any algebraic extension of a field is a field as well

[uncut cove]

How do i get rid of the 0?

[golden pasture]

uhh as in?

what is $R$, $N$ and $\bot$ here

[cloud walrus]

ariana:

[woven delta]

I'm assuming row space, nullity, and orthogonal complement

[prisma ibex]

Yup that’s right

[glossy yoke]

am I confused or do those two spaces live in different dimensional R^blahs?

like $R(A)$ lives in $\mathbb{R}^n$, and $(N(A^t))^\bot$ lives in $\mathbb{R}^m$

[cloud walrus]

ball:

[woven delta]

A is a map from R^m to R^n. A^t is a map from R^n to R^m

R(A) is in R^n and N(A^t) is in R^n

[mint gulch]

Why is necessary to R be a integral domain for construct the Field of Fractions?

I did the construction, but, I think I didn't use that property

[oblique river]

if xy = 0 then 1/x * 1/y = 1/xy = ???

is basically the issue

[mint gulch]

I see, you're right

I have to say that in the definition of the addition and Product

Thank you

[oblique river]

np!

[vestal snow]

Why does $n-n'$ disappearing on $A_{K(x)} (-P_\infty$) imply that $n-n'$ is $0$?

[cloud walrus]

The formerly edible banana:

[next obsidian]

I did the construction, but, I think I didn't use that property
@mint gulch As a note, if R isn't an integral domain there's still something called the "total ring of fractions" where you just localize at all non-zero divisors. This is just putting in all inverses you possibly can basically, and you might run into it sometime. It isn't a field... unless nothing's a zero divisor

[mint gulch]

I see it, but the definition have to be different, right?

Not the same equivalence classes

Or just define the product when bd ≠ 0

[next obsidian]

Sorry

did you define the field of fractions via localization?

[mint gulch]

No

[next obsidian]

like did you cover what it means to make S^-1R

oh

[mint gulch]

Via equivalence classes

[next obsidian]

Yeah

Okay here I'll crash course localization really quickly

[mint gulch]

It's the same procedure to construct Q from Z

[next obsidian]

Call a subset S of R a multiplicative set if when s,t in S then st in S

you can make something called the localization of R at S denoted S^{-1}R as fractions r/s where r in R and s in S

and multiplication and addition are done as you expect

and you quotient out by the following

r/s = r'/t iff there is a u in S such that u(rt - sr') = 0

a fact about this is that S^{-1}R = 0 if and only if there exists s,t in S such that st = 0

so if you localize it away from all zero divisors this never happens

If R is an integral domain then S = R\{0} is a multiplicative set

in which case S^-1R is the field of fractions of R

you'll notice that the condition "there exists u in S such that u(rt - sr') = 0" actually is equivalent to rt - sr' = 0

since there aren't any zero divisors and u cannot be 0

so you just get that r/s = r'/t if and only if tr = sr'

[mint gulch]

Hmm, Do I have to take S integral domain?

[next obsidian]

S is not a ring

S is a subset of R which is closed under multiplication

Oh one thing I forgot to mention is S must contain 1

[mint gulch]

And S contains 0?

[next obsidian]

No

if S contains 0 then S^-1R = 0

since then everything is equal to 0

[mint gulch]

Then $0 \notin S$

[cloud walrus]

Enigsis:

[next obsidian]

take u = 0

I mean you want that to be true

you can include 0

but then you kill everything

so it's not very interesting

[mint gulch]

Yes, I see

[next obsidian]

but since S is closed under products

you can't include a pair of zero divisors

since if xy = 0 and x,y in S

then xy = 0 in S

[mint gulch]

And do you have to take a maximal S?

[next obsidian]

No

any multiplicative set works

[mint gulch]

Yes, but, to get the most big set

[next obsidian]

for the total ring of fractions

[mint gulch]

S^{-1}R

[next obsidian]

set S = {set of non-zero divisors}

then S^-1R is the total ring of fractions

in the case R is an integral domain then S = R\{0} and the total ring of fractions is the field of fractions

[mint gulch]

set S = {set of non-zero divisors}
@next obsidian

You could include here a zero divisor, but not the other zero divisor

[next obsidian]

you can but there's no canoncial way to do it

as in you have to make arbitrary choices

you'd also need to include all products of that zero divisor with elements in S

[mint gulch]

Hmm, it's messy then

[next obsidian]

right

so you just ignore zero-divisors

[mint gulch]

But, maybe it could be do it Xad

Lol

I don't know

Thank you