groups-rings-fields

[void cosmos]

imo

[rapid tinsel]

Okay how about this then: instead of an arbitrary integral domain, I know my elements live in a Dedekind domain

This somewhat limits the possible examples, but in general they're still not UFDs

I'm guessing I have to construct examples in the ring of integers of some number field with non-trivial or possibly large class number

Or the coordinate ring of some affine curve...

[celest furnace]

So real

This guy is still less agressive than that one guy who knows a ton but is always so angry when solving the problems lol

[delicate orchid]

that's me lol. Anyway he was just joking

[celest furnace]

Nah not you hold on

[delicate orchid]

I'm FURIOUS

[celest furnace]

Foghorn

This guy always comes with some crazy problem, drops 400 f bombs in two messages, and solves it and leaves

Frankly i wish i could be more like him

[delicate orchid]

miz is just a little goofy that's all...

[celest furnace]

Perhaps

[sonic coral]

Let F be a field and K/F an algebraic extension. Let α and β be two elements of F.Show that there is a polynomial m(x) ∈ F[x] with the following property: if f(x) ∈ F[x] satisfies
f(α) = f(β) = 0, then m(x) | f(x).

I know that if f(α) = 0, then the minimal polynomial of α divides f. Since the extension is algebraic, is that enough to say that such a minimal polynomial exists?

[celest furnace]

Yes. Algebraic just means there exists a polynomial with alpha as the root, so you can take the monic one with smallest degree and this will be the minimal polynomial

And such a polynomial exists since every nonempty subset of the naturals has a smallest element !

[spare isle]

I've seen the use of $\mathbb{Q}[\sqrt2]$ and $\mathbb{Q}(\sqrt2)$ used interchangeably. Do they both mean the polynomial ring with (concrete) coefficients $\sqrt2$?

[cloud walrus]

JJCUBER

[delicate orchid]

in this one specific case they happen to be the same but no, not generally

Q(\sqrt(2)) is the smallest field containing Q and sqrt(2)

Q[sqrt(2)] is as you described

[spare isle]

I see, do they have formal names?

(the first)

[delicate orchid]

Q(sqrt(2)) is a field extenstion

[spare isle]

thank you!

[chilly radish]

I think you mean alpha and beta in k?

You can take the least common multiple of the minimal polynomials

Think why this is the correct polynomial

(lcm in the polynomial ring)

[sonic coral]

the problem says F, but this may be a typo. i don’t really see the purpose of the beta in this problem. would alpha and beta lying in K change things?

my understanding is that since the extension is algebraic, i have some m_alpha(x) and m_beta(x), and from that i can construct a m_alpha,beta(x) that divides all other such polynomials that alpha and beta are a root of since it is minimal

[dull ginkgo]

my mom dropped me down the stairs as a child

[chilly radish]

How do you define m_alpha,beta?

Also the problem is just trivial if a,b are in F since then x-a, x-b are in F[x]

[sonic coral]

i was thinking the minimal polynomial that contains both alpha and beta as a root. this would be constructed like you said, taking the lcm of both minimal polynomials respectively

[chilly radish]

Yea, it's just the monic polynomial that generates the ideal (m_a,m_b)

[sonic coral]

what do you think isthe point of introducing a beta in this problem?

is it just to think about constructing a minimal polynomial that contains two things as roots

like this could have been stated and proved the same way with just alpha

[toxic zephyr]

i think so. it seems like this is basically constructing the minimal polynomial of two potentially distinct elements, alpha and beta.

i.e. that there exists minimal polynomials for not just one, but multiple elements

[sonic coral]

okay i see, thank you

[rapid tinsel]

I think this example works: in the ring Q[X, Y]/((X+1)^2 - YX^2), the elements x = X+1 and y = X satisfy x^2 = zy^2 for z = Y, but z is clearly not a square

I think it’s Dedekind cuz the polynomial gives a nonsingular affine curve

I still can’t think of an example in the number field case…

[rocky cloak]

Notice, if
x^2 = zy^2, then (x/y)^2 = z. So t^2 - z is reducible in the field of fractions.

So if your domain is integrally closed x/y must be in your domain. So there shouldn't be any examples coming from number fields.

[rapid tinsel]

I see! Thanks!

But I like how this distinguishes number fields from function fields lol

[rocky cloak]

I think you'll find that
YX - X - 1
squares to Y.

Any Dedekind domain is integrally closed. So there shouldn't be a difference whether you get your Dedekind domain as the ring of integers of a number field or the cordinate ring of a curve.

[scenic saffron]

For this proof

I dont understand the bottom line

Since its a subgroup, why does that mean g^2 is in it?

[tough raven]

g in H U gH

[scenic saffron]

Oh right i see

Right so why do we have if g^2 in gH

I understand the proof following that

Just confused on why we assume that now

[glad osprey]

It's just assumed for sake of contradiction. You have that g² is in H U gH, then you show that g² is not in gH, so it must be in H

[scenic saffron]

Oh right

[glad osprey]

But I don't understand why H U gH is a subgroup. Can you post the question with the definition of H?

[scenic saffron]

Oh thats just the question

But sure

[glad osprey]

Oh, it's just an assumption. I thought it was a known result or something

[rapid tinsel]

Oh crap

[sonic coral]

this was actually the case lol. i’m not sure how to approach it. i know the size of the multiplicative group but that’s about it

[rocky cloak]

Well, notice that
a^31 = 1 means that
a^124 = 1, so a^125 = a means that a is in F_125. Thus [F5(a) : F5] is either 3 or 1

Then you just need to conclude it's not 1

[sonic coral]

that’s neat, cool

[tough raven]

In general, you can deduce such degrees purely from the multiplicative order, since
a in F_p^n <=> a^{p^n-1} = 1
<=> ord(a) | p^n - 1
<=> p^n = 1 (mod ord(a)).
So [F_p(a) : F_p] is the smallest n such that p^n = 1 (mod ord(a)).

[dull ginkgo]

I take it that every isomorphism between a finite dimensional (rank) vector space (module) and its dual induces a symmetric bilinear form on it right?

[mighty kiln]

Why is it symmetric

[dull ginkgo]

Conjugation by the isomorphism?

Let me check

Need not be symmetric actually

[mighty kiln]

Nondegenerate bilinear form

[dull ginkgo]

Yeah

[rocky cloak]

A map
M -> Hom(M, R)
is the same as a map
M(x)M -> R
Which is exactly a bilinear form.

[dull ginkgo]

TENSOR HOM ADJUNCTION ECKS DEE

How did I not notice that

[frank cosmos]

my exercise is to show that if H is an index n subgroup of A_n, then the left action of An on An/H is isomorphic to the alternting permutation group of An/H. can i get a hint showing that if x \in An then its action on An/H is an even permutation?

i think it suffices to prove this when x is a 3-cycle

hmm i proved that the set of left actions is a subgroup of the symmetric group of (An/H) of index 2

hence is normal

ok so i just need to show that 1 3-cycle is in this image

then all of them are, hence the image is just A_n

wait, actually am i done? index 2 subgroup of S_n is A_n?

does this stream of arguments work

[dull ginkgo]

Quick question

By the alternating permutation group of A_n/H are you referring toA_[H : A_n]?

[rocky cloak]

An is indeed the only index 2 subgroup of Sn, so that should do it

[tough raven]

Technically, you don't need it, since it is proved by showing that Hom(M, Hom(N, R)) is the same as bilinear maps and then using universal property of the tensor product.
If you skip the last step, you get the equivalence without mentioning tensor products.

[tough raven]

For completeness:
Any injective map M -> Hom(M, R) gives a non-degenerate bilinear form. Forms for which this map is bijective are usually called perfect (at least for R = Z and M finite-rank free over R) to distinguish them from the non-degenerate ones.

[south patrol]

Real

[dull ginkgo]

I just realized this is how I defined symmetric bilinear modules in shit I’ve done before. I just forogot

Incredible mizalign L

[whole basalt]

So I just had a thought I'm curious about

The field ℤ₅(√2) contains square roots of all elements of ℤ₅.

Does that mean that any quadratic over ℤ₅ will split over ℤ₅(√2)?

Since I imagine the quadratic formula should work?

[rocky cloak]

That is correct. In fact for any finite field, if you adjoin the square root of any number that doesn't already have a square root, then every quadratic will now split.

(Note a finite field of characteristic 2 already has all square roots, but not every quadratic splits. The quadratic formula involves dividing by 2 after all)

[whole basalt]

Yeah I figured that would fail for characteristic 2, lol

Awesome, that'll be my last question on my (take-home) final XD

[celest furnace]

Wait.. F_2 has all square roots but it has a quadratic that doesn't split?

Oh like x^2 + x + 1

[whole basalt]

Yeah, x² + x + 1.

[celest furnace]

Woah that's weird

[rocky cloak]

Are you taking the exam or making the exam? (Or both ;))

[whole basalt]

Making the exam. I teach undergrad AA.

[celest furnace]

Niiice

[whole basalt]

It's meant to be a little lighter on the "prove something you've never seen" side and more on the "show that you understand what these concepts mean" side

It's been a tough semester

[rocky cloak]

Yeah, sounds like a good exercise

[rocky cloak]

For you or the students?

[whole basalt]

Both

[rocky cloak]

Feel ya

[whole basalt]

All my students in all my classes have been talking about how rough this semester has been

[celest furnace]

What do the averages tend to be on your exams?

[whole basalt]

But in my AA class I had 2 out of 10 people turn in my most recent Problem Set that was due Thursday

[celest furnace]

Took an exam in a non - math class today and the avg is going to be like 50% 😢

[whole basalt]

I don't do exams, I do problem sets, and they have the ability to revise their work based on feedback.

[celest furnace]

Thats awesome

[rocky cloak]

I'm holding exams in about 2 weeks

[whole basalt]

But even then a bunch of them are dead-set on getting everything right the first try

To the point of not turning the damn thing in

That and they have never worked together

Despite me telling them they could and should :V

[celest furnace]

Yeah that happens when the policy is a more relaxed than normal, can they turn it in later and get the same grade

[celest furnace]

Yeah 😭 I always have mixed feelings about policies like these because i have no friends in a class then im way less likely to work together with other people

[whole basalt]

Yeah. I've been trying to go off the "if you've demonstrated that you understand it, that's the most important thing" idea ... but I need to add more restrictions to that

[whole basalt]

True. But they all sit at tables of 2 or 3 (there's 10 students) and have been working in those groups in class all semester as well

[rocky cloak]

So they work together, just not on the problem sheets?

[celest furnace]

Yeah i feel that but you have to like someone a lot to go do homework with them on your own time vs in class

[whole basalt]

If I'm rambling let me know :V

Also, if anybody is up for looking at said take-home final and giving their opinion on how hard it is, please DM me

[rocky cloak]

I'd be up for it, but I guess it's hard to say how hard something is without knowing what you've taught them

[crystal vale]

I want an example where lcm does not exist. They state that 2(1+i √ 5) and 6 have not lcm in Z[i √ 5], and reason that if it exists say l, then N(l) =72 or 144 , where N(l) is the norm of l, and it is not possible.

But N(l)=144 it is possible when I let l = 12, right?

[rocky cloak]

12 is a common multiple, but 6(1+isqrt(5)) is also a common multiple. And neither of these are multiples of each other, so they can't be the least common multiple.

[crystal vale]

Okay got it, but why did they stated that N(l)=144 which is not possible?

[rocky cloak]

Well if the lcm exists, it's norm must be a common multiple of their norms. So if you look a little bit at what norms are possible, 72 and 144 should be the only options

[crystal vale]

Yes, and no element has norm 72 in that ring, but 12 has norm 144, but 6(1+i√5 ) is also common multiple and neither of these are multiple of each other, my question is how did you guess this 6(1+i√5) ?

[rocky cloak]

I'm not sure I did anything special. I'd argue it's easier to spot that 6(1+isq(5)) is a common multiple than 12.

I guess you can think that you'll have to use the weirdness of this ring somehow. Which comes exactly from that
6 = 2*3 = (1 + isq(5))(1 - isq(5)) doesn't have unique factorization.

So when you have both 2 and (1+isq(5)) it's natural to compare what happens when you multiply by 3 to what happens when you multiply by (1-isq(5))

[crystal vale]

Okay, thank you

[crystal vale]

In Dummit and foote, they state that Let R be unique factorisation Domain with field of fraction F and let p(x) in R[x]. If p(x) is reducible in F[x] then p(x) is reducible in R[x].

I think p must be a primitive polynomial.
If I let Z[x], then 5 is irreducible in Z[x] but 5 is not irreducible in Q[x], right ?

[hollow mica]

If L / K is a field extension, and x_1, …, x_n are elements of L, then how do I show that all subfields of K(x_1, … x_n) that contain K must contain one of the x_i ? (or potentially none of them if the subfield is just K)

[sonic coral]

this is what number theorist deserve for writing an algebra textbook

[hollow mica]

Wait is this even true

[south patrol]

No

This is very false in fact like

[delicate orchid]

K(x_1+x_2) moment

[south patrol]

Every finite extension of Q is of the form Q(a)

So your result would imply that you never have any intermediate fields

I guess this is particular striking if you take like Q(x)/Q so you have Q < ... < Q(x^4) < Q(x^2) < Q(x)

[hollow mica]

I wanted something like that to hold to prove this

[sonic coral]

what’s wrong with their proof?

[hollow mica]

Here splitting field is “minimal subfield over which f split”

[hollow mica]

Well…

[delicate orchid]

well you definitely can't expect it to split if all of the roots aren't in M, so you better adjoin them all!

[hollow mica]

Ok I see it I think: if G is a subfield of M containing K, and f splits over G, then if G is not equal to K f will have too many roots

[south patrol]

Well like

Any extension of K contained in L over which f splits must have the alpha_i and hence contain M

by definition of splitting

and f splits in M

[hollow mica]

“a polynomial can split in only one possible way”

[halcyon peak]

When it splits in M, you can view it as a element in L[x] which is a ufd, so the roots must be the alpha i's.

[hollow mica]

I don’t see how L[x] being a UFD implies that, but the way I thought about it above was

if a polynomials of degree n splits in two different ways, it will have more than n roots

in fact, in the two different splittings, the constant must be the same, and if a x-x_i ever appears in both splittings then the rest must be the same

so you only really need the number of roots argument when both your splittings are completely different (share no x-x_i)

[dawn saffron]

Guys I have something that is literally breaking my brain rn

Are angles real numbers?

[tribal moss]

Angle measures are real numbers, yes.

[dawn saffron]

But angles themselves are not right?

Bc they don't form a ring over multiplication

[coral spindle]

You can see them as the Abelian group – and yes, not ring – R/2piZ

[tribal moss]

I would say that an angle itself is a geometric shape, different from a number.

[dawn saffron]

But if they are a geometric shape do they exist in Euclidean space?

[tribal moss]

Yeah, perhaps it would be better to say that angle measures are congruence classes in R/2piZ, rather than literally real numbers ...

[dawn saffron]

Wait...

Could one consider angles a vector space?

Because they have scalar multiplication

[tribal moss]

It's generally most productive not to obsess overly about what things really are, so long as we agree about how they behave in the situations we use them in.

[dawn saffron]

And are an abelian group under addition

[dawn saffron]

Ik ik but I'm writing a thing defining angles so I wanna get this right

[coral spindle]

No.

[delicate orchid]

they're a module but not a vector space

[dawn saffron]

Okay that makes sense

Would someone mind proof reading my definition here for me

I am explaining a lot of stuff based on this definition of the set of angles which is something I don't see most texts use so I want to ensure its 100% accurate and clear

[dawn saffron]

Also could I phrase this as real numbers are a metric used to define angles?

[tribal moss]

I sort of withdrew that. To each real number there's a corresponding angle, but each angle has many real numbers that it corresponds to.

[dawn saffron]

That is true

But I think my phrasing still holds

Since it doesn't indicate the kind of map between the two

Maybe the phrasing: real numbers are a metric used to define angles modulo 2pi would be better

[tribal moss]

(At least if "angle" is a geometric thing, rather than some kind of synthetic concept that remembers whole revolutions. The latter is definitely useful sometimes, but doesn't lend itself to a fully geometric definition).

[dawn saffron]

Depends on how you define a triangle

I like to define it as the addition of three noncollinear vectors who's sum starts and ends at the same point

These vectors form three interior angles all non-reflexive

[delicate orchid]

no it doesn't

[dawn saffron]

Wdym?

[delicate orchid]

what do YOU mean?

[dawn saffron]

Idk lol I was just covering my ass in case my definition of a triangle was wrong

Mb

[delicate orchid]

a lot of "highschooler tries to formalise things in odd ways" vibes

[dawn saffron]

That is fair I'm an engineering major not a mathematician

Is there a better way to formalize it?

@delicate orchid ?

What about with vectors?

Thank you

[dawn saffron]

Still want to see you elaborate on this, otherwise a Discord classic to drop an insult like disappear

[delicate orchid]

I was getting my nuggets out of the oven

the better way to formalise it is to just define them as elements of R/2piZ as tropo said and leave it as that, rather than writing 2 pages

[coral spindle]

As tropo said??? Wtf my work is being stolen from me

[delicate orchid]

I ain't scrolling up buddy!

[coral spindle]

I am going to sue you Wew

[dawn saffron]

Most of those two pages aren't even on that though?

[delicate orchid]

I look forward to seeing you in court, we can get lunch after

[dawn saffron]

Only one setence?

[delicate orchid]

then why post them

[dawn saffron]

^

Context

[coral spindle]

I think we've said everything that needs to be

[dawn saffron]

No you guys were very helpful ty

I appreciate it

[delicate orchid]

I was not helpful. I was evil and mischievous.

[dawn saffron]

One last question sorry

Angles do obey scalar multiplication right?

[coral spindle]

What does that mean exactly

2pi radians is the same as 0 radians. So is 0.5 * that either pi or 0?

[tribal moss]

Well, only if you limit them to at most one revolution. Otherwise, is 0.1 · 10° supposed to be the same as 0.1 · 370°?

[coral spindle]

Seems to me it doesn't.

[dawn saffron]

I guess you could say they obey it modulo 2pi?

[delicate orchid]

with scalars in what ring is the question

[dawn saffron]

Real numbers

[delicate orchid]

R/2piZ is obviously a Z-module

[tribal moss]

Multiplication of reals modulo something is itself not well-defined, even before we start talking about angles.

[tribal moss]

But what is then the result of 0.1 · 10° which had better be the same as 0.1 · 370° if you're working modulo 360°?

[dawn saffron]

I see your point

How would one define angle dililation then?

[tribal moss]

If I need to, I'd say explicitly "measure the angle with a number in (-pi, pi]" or whatever makes sense in the situation. One just needs to accept that doesn't create a particularly nice algebraic structure.

[dawn saffron]

Aww...

I wish there was a way to well define the abelian group of angle along with the ability of dialation over the measure of all real numbers

[delicate orchid]

if you really want to multiply an angle I've thought of a way to do it but it's not pretty

[dawn saffron]

I can take it 🥺

[coral spindle]

No, multiplication mod 2pi is not well-defined.

R/2piZ is not a ring, like I said

[coral spindle]

wowee

[dawn saffron]

Scalar multiplication is already not a ring operation mb I should have specified this

[coral spindle]

If you are describing it mod 2pi, then it is in fact a ring operation.

[dawn saffron]

No because it isn't binary I thought?

[coral spindle]

Because you are describing an action of the integers mod 2pi on the integers mod 2pi.

Idk what you mean now. It is most certainly binary.

[delicate orchid]

instead of representing angles as elements of $\bR/2\pi\bZ$, instead represent them as matrices of the form $\begin{pmatrix} 1 & \text{cos}(\theta) \ 0 & \text{sin}(\theta) \end{pmatrix}$, then you can define $n\theta := \begin{pmatrix} 1 & \text{cos}(n\theta) \ 0 & \text{sin}(n\theta) \end{pmatrix}$, I haven't through through how addition works yet (most likely it just doesn't)

[cloud walrus]

Wew Lads Tbh

[dawn saffron]

My logic was:
Vector x scalar->vector
Angle x scalar->angle
so I guess binary was the wrong phrase and I should have said that it isn't within one contained ring

[dawn saffron]

You are a goat for this

[delicate orchid]

wait, just define them as rotation matrices, then everything works immediately

[coral spindle]

If only there were a well-known group this is isomorphic to

[delicate orchid]

replace $\begin{pmatrix} 1 & \text{cos}(\theta) \ 0 & \text{sin}(\theta) \end{pmatrix}$ with $\begin{pmatrix} \text{cos}(\theta) & \text{sin}(\theta) \ -\text{sin}(\theta)& \text{cos}(\theta) \end{pmatrix}$

[cloud walrus]

Wew Lads Tbh

[tribal moss]

What is $\frac12$ times $\begin{pmatrix}-1 & 0 \ 0 & -1\end{pmatrix}$?

[cloud walrus]

Troposphere

[dawn saffron]

Out of everything I think this may be the most relevant the more I think about it

All I need the reader to get is that angles act on points to perform rotations in Euclidean space

I went too deep

[delicate orchid]

annoying I can't pass this to the lie algebra

[tribal moss]

What a pity the exponential map is not injective ...

[dawn saffron]

I have done it boys

$\mathbb{A} = \mathbb{A}+2\pi\mathbb{Z}$

[cloud walrus]

mithrilsword1

[dawn saffron]

Where A is a angle measure

[delicate orchid]

start with a lie algebra rep of R, or as I like to call it - R, then pass to S^1 we did it reddit

[tribal moss]

Sure, as long as you don't want to multiply by non-integers.

[delicate orchid]

well if we wanted a Z-module structure then we didn't have to do too much to begin with

[tribal moss]

Yeah -- or in particular the complex unit circle, which is probably the most popular representation of angles in advanced work.
Unfortunately that's a perfectly good abelian group (aka Z-module), but not a vector space like Mithrilsword wants.

[delicate orchid]

U(n) reps, or as I like to call them, [BU(n), BU(n)]

[hidden wind]

bubu

[tribal moss]

Bun-bun.

[sonic coral]

How can i find an algebraic number α so that Q(α) = Q(√2,∛3)

[celest furnace]

Hint: Look at the proof of the primitive element theorem

[south patrol]

Heuristically sqrt(2) + 5^5^5 3^1/3 should work

[sonic coral]

from what i’ve seen i should just take a linear combination of the generators?

wait no, i need a linear combination that is not fixed by any of the Galois automorphisms?

[tribal moss]

That's essentially Potato's point. If you pick alpha = sqrt(2) + q·cbrt(3) for some q, and then write out the first six powers of alpha in the standard basis over Q, then the determinant of the resulting 6×6 matrix is a polynomial in q of degree no more than about 36. Choose a huge q and it's pretty much guaranteed that you won't hit one of the few roots of that polynomial.

[teal wind]

can someone help me with my work? These are my answers:

[last spoke]

whats the definition of normal that's given in your prior literature?

[pastel cliff]

how do you find the automorphism group of a graph

tryinig to do this one in particular

[tribal moss]

I don't think there's a good (that is, polynomial-time) known systematic algorithm; that would seem to involve deciding graph isomophism.

Think of this as a license to be ad-hoc and depend on insights/observatons specific to the particular graph.

[delicate orchid]

there are automorphisms that swap 1 and 5, 3 and 4, 2 and 3, 2 and 4. The trick I use to find automorphisms is to find pairs of edges that have the same neighbours. I'll leave it up to you to show that these generate the whole group

correct me if any of those automorphisms are wrong, it's 3am and I'm very tired

[pastel cliff]

wew lads tbh

still fucking insane to me that youre a mod lmfao

[tribal moss]

The particular graph there is the complete bipartite graph K_{2,3}.

[pastel cliff]

loosely speaking, those are just the permuatations that preserve adjacent vertices right

[delicate orchid]

Yeah

In general bipartite graphs you can permute either set of vertices however you want and it’ll give you a graph automorphism

So the overall thing should be isomorphic to S_n x S_m for Bi(n,m)

[tribal moss]

Except if n=m, in which case you can also interchange the groups.

[delicate orchid]

True!

[crystal vale]

There is a statement, any irreducible polynomial is primitive. Is the Non-zero constant irreducible element in Z[x] primitive?

[crystal vale]

To show the number of roots of a non-zero polynomial over a integral domain R is atmost it's degree.

Should I show first if there is root a then we can write f(x) =(x-a)q(x), q in R[x]? Is it true in R[x] ?

[chilly ocean]

yes that sounds like a good start

f(x) =(x-a)q(x) in R[X] iff f(a) = 0

[crystal vale]

But does the remainder theorem work in the non-Euclidean domain?

[chilly ocean]

you would have to work in R's field of fractions but in the end i think the q and r you get live in R[X]

but i think u can prove it without division algorithm

[lapis latch]

The division algorithm will work as long as the leading coefficient is a unit in the ring, so you can do it with x-a

[next obsidian]

How I WISH this was how it was taught to everyone

[crystal vale]

Should I work on its field of fraction or from it imply that if I have more than n roots ( polynomial degree is n) , say n+1, then p(x) =(x-a_1)×....×(x-a_(n+1) ) it's contradicts that p has degree n

[lapis latch]

Yeah you can work in the field of fractions if you want. Showing the degree contradiction is a good strategy

[crystal vale]

Without using the field of fraction, does it work because the leading coefficient is 1 of x-a, right?

[lapis latch]

Yeah, so you can generally factor roots like that in the ring itself

Being an integral domain is important in one of the steps

[crystal vale]

In degree contradiction?

[lapis latch]

Yeah that’s one, in integral domains you have the nice multiplicative rule for degrees. Also in factoring it as a product of roots, say f(x) has two distinct roots a,b you can then factor f(x)=(x-a)q(x) now in an integral domain you can say that b must be a root of q(x). That would generally fail, say f(x)=2x in Z/4Z

[crystal vale]

Okay thank you

[crystal vale]

In a commutative ring, the minimal ideal is always the principal ideal?

let a≠0 in minimal ideal then (a) is subset of I, so I = (a), right ?

[mighty kiln]

Minimal nonzero ideals are principal yes

(Although there can be multiple minimal ideals)

[crystal vale]

Any hint to show that Non-zero minimal prime ideal is principal ideal

[rocky cloak]

That would be kinda hard to show, seeing as it isn't true

Just look at k[x, y]/(x,y)^2 for example.

[mighty kiln]

In fact for Noetherian domains this is true iff UFD

[crystal vale]

Sorry I forgot to mention R is UFD

[rocky cloak]

Think about the prime factorization of an element contained in your ideal

[thorny knoll]

If $(G,e)$ is a divisible abelian ordered group, does $G$ have to be archimedian aswell? That is, for $e < a < b$, must there be $n \in \mathbb{N}$ with $a^n > b$?

[cloud walrus]

Literature-chan

[coral spindle]

No. The classic example of this comes from model theory, where one constructs the hyperreals

I think also one can just take Z^(N) with the infinite lexicographic order but I would have to double check.

[mighty kiln]

I think R^N works

[rocky cloak]

Or just R^2 even

[mighty kiln]

This is just infinitesimals in naive hyperreals

[crystal vale]

If I want to show that in Z[√5] where N(x +√5y)= | x^2 -5y^2 |.

x is irreducible if 4<= N(x) < 16.
Can I use here negation?
And I have shown that for non zero non unit N(x)>=4.

So if I let x= ab, where a and b are both non-unit then N(x) => 16.

[dire siren]

yeah, that's fine

[crystal vale]

Okay thank you

[crystal vale]

There is a statement that for a prime p in Z, the polynomial 1+x+x^2+...+x^p is irreducible in Z_2[x] only if p≠2. I think this statement is not correct, right?

[dire siren]

if p≠2 then the evaluation in 1 is p+1=0, so 1 is a root

maybe you mean Z_p[x] instead?

[crystal vale]

No

[dire siren]

then it's false

[crystal vale]

May be printing mistake

[dire siren]

it is irreducible only if p=2

[crystal vale]

What is meant by evaluation in 1 ?

[dire siren]

f(1), where f is that polynomial

[crystal vale]

Then f(1) = 1 , so how is 1 root?

[dire siren]

f(1)=p+1 and p+1=0 because you are in characteristic 2

[crystal vale]

But p≠2 so how characteristic 2 ?

[dire siren]

Z_2[x] has characteristic 2

[crystal vale]

Yes

[crystal vale]

I am talking about it you state that if p ≠2 then the evaluation in 1 is p+1=0, so 1 is a root

[dire siren]

I don't see the problem; as long as you work in Z_2[x], any even is equal to 0

even if it was Z_p[x], the statement is still false because f(-1)=0 for p≠2

[crystal vale]

Yes

Any hint to show Z[i]/(p) is isomorphic to field of p^2 elements, if p remains a prime in Z[i].

We have that p can not be written as sum of two numbers and if any element a + ib in (p) then both a and b are divisible by p. But how can I show it?

[sonic coral]

clever way to do this, let alpha be (72)^1/6

the next part was to determine the minimal polynomial of alpha, so a sqrt(2) + d*cbrt(3) would’ve been hard to do that for

[dire siren]

(p) is a prime ideal in Z[i], which is a PID, so (p) is maximal, which implies that Z[i]/(p) is a field
it is therefore enough to find the number of elements in Z[i]/(p)

extra hint: ||you can find representatives||

[crystal vale]

Okay, thank you

[halcyon peak]

how to show S/I is flat here?

R is commutative with unit

[next obsidian]

wtf

Oh a flat S-module

Ummm, so my guess would be

If you look at testing flatness against the inclusion of an ideal J -> S

Then what you really need is that the multiplication map

J (x) I -> IJ is an isomorphism (injective)

I would then think that because I is basically a bunch of copies of R as a direct SUM which means it commutes with the tensor product OVER R

That you want to try and show that this kinda is true even for the tensor product over S

Or errrr

Hard to explain

Basically, IJ is gonna be given by like

Let J_i be the image of J under projection to the i-th coordinate

I believe IJ is then (+)^N J_i

By examination

And I believe you should be able to see that I (x)_S J should be the same thing

[vivid birch]

I'm having some trouble understanding the following: Let S be a semigroup with 0, and suppose that D = S \ {0} is a D-class such that D^2 is not {0}. Then D has at least one idempotent. Why is this true?

My teacher said that it's because of the Miller-Clifford theorem, but I don't get how this theorem can prove this

[halcyon peak]

oh thx, i forgot that examine tensoring J -> S is enough

J/IJ -> S/I is injective if we show IJ = I \cap J, and this is clear i think

[next obsidian]

Oh youre totally right, I don’t know why I was tensoring with I

Do you know how to show it isn’t projective?

It is incredibly restrictive for a quotient to be projective, and you can describe exactly what the ideal you are quotienting has to be, and it’s abundantly clear it isn’t satisfied in this situation

[sonic coral]

does this work?

[rocky cloak]

I feel like you're missing something about x^3 - 7 being irreducible over Q or something.

Right now it's not clear why the cube root of 7 couldn't be in K.

[sonic coral]

i mean K is a quadratic extension of Q so the cube root of 7 can’t lie in K

is that not a justified statement

i was thinking about trying to argue this using the fact that the degree of the splitting field of x^n - a is n*phi(n) or n/2 * phi(n), along with the tower law, but i’m not sure if that would work

[dire siren]

that's more or less what the problem asks to prove

[next obsidian]

you need to know that such a cube root generates a degree 3 extension of Q, which is equivalent to saying that x^3 - 7 is irreducible over Q as jagr was saying

you can definitely have a cube root of 8 inside of K for example

[sonic coral]

so the polynomial being irreducible over Q tells me that i need atleast a degree 3 extension to obtain a root of it

but for a cubic, being irreducible is equivalent to not having a root, so the polynomial can’t be reducible in K?

[gilded frost]

¿$(G / N)\times N=G$?

[cloud walrus]

cosín™

[chilly ocean]

not always

try some small examples

[south patrol]

Or there are funny examples. Like compare $\mathbf Z$ with $\mathbf Z/2 \times 2\mathbf Z$

[cloud walrus]

Crystalline Potato

[celest furnace]

If they have comprime order you can show that a certain short exact sequence splits so it is a semi direct product

But as stated this is false

[dull ginkgo]

that's burnside's right?

[celest furnace]

I don’t think so I think you can just define the inverse map in the obvious way and it works out ?

No

[crystal vale]

In the integral domain R, is this statement correct only ideals of R which are free as R-module are Non-zero principal ideals?

[next obsidian]

This is an if and only if actually

[long obsidian]

Hey just to make sure I can always write an ideal as an ideal sum J=\sum <f_a> of the principle ideals of all the generators right?

[next obsidian]

Yes

And it’s principal btw

[crystal vale]

I have shown if it is principal ideal then it is free but how can I show that If it is free then it is principal ideal?

[next obsidian]

Show that if M < N and both M and N are free, then the rank of M is <= rank of N

[halcyon peak]

yeah i figured it out first and stuck in showing it's flat

[long obsidian]

Is it natural to view the r:a->r(a) the radicalization of an ideal a in ring A as a projection since r^2 =r?

[hidden wind]

are there uncountable families of distinct (as in, non-isomorphic) groups ?

[cobalt heath]

Do you want any groups, or do you want some rules?

[hidden wind]

any

i’m guessing some sort of diagonal construction is possible

[paper flint]

Free group on a set of generators

And you have uncountably many sets of different cardinalities (and hence necessarily lead to non-isomorphic free groups when taken as generating sets)

[hidden wind]

ohh

[cobalt heath]

Ah, are cardinalities guaranteed to be uncountable?

I also thought of this but was unsuare about that

[paper flint]

Had to look this up actually, and it works assuming standard ZFC:

https://math.stackexchange.com/questions/8013/are-there-uncountably-infinite-orders-of-infinity#:~:text=In particular%2C for any cardinal,many infinite cardinals below it.

Alternatively, maybe upward Lowenheim-Skolem theorem does the trick

[hidden wind]

whaa i was not expecting model theory to enter into this

[paper flint]

This is one of the fundamental questions one can ask in, well, "model" theory

[cobalt heath]

I do wonder if there are uncountably many nonisomorphic groups with its cardinality at most uncountable.

Maybe working with subgroups help

[paper flint]

at most uncountable?

Do you wish to restrict to countably infinite groups, or did you mean to say at most the cardinality of the continuum?

[cobalt heath]

Ahhh, sorry. I mean cardinality of continuum.

[paper flint]

Morally it feels like there should be uncountably many (non-isomorphic) group structures on a set with cardinality continuum, but I don't have an argument

Interesting question in any case, might be worth looking up

[cobalt heath]

Yeah, let me think

For any set $S$ of primes, you can construct an abelian group $\oplus_{p \in S} \bZ / p \bZ$. By order argument, distinct set $S$ result in distinct isomorphism type of groups.

[cloud walrus]

Absta

[cobalt heath]

Since there are infinitely many primes, this gives P(Z) worth of distinct groups.

[paper flint]

How are you getting P(Z) worth of distinct groups?

Ah okay, I see what you mean

That works

[rocky cloak]

I think most of the typical families of groups become uncountable (or really a proper class) of you allow them to be infinite.

So the free groups, the symmetric group, alternating group, (projective) (special) linear group.

No uncountable cyclic group though.

[crystal vale]

In integral domain ring R, if a is right invertible then it will be left invertible also, right?

[delicate bloom]

suppose ar=e (r the right inverse of a, e the identity) then you can play around with ra-e and see what properties it has and what friction that forms with being an integral domain. ||ra(ra-e)=rara-ra=ra-ra=0 but ra != 0...||

[crystal vale]

Actually I need to mention here R is not commutative

[hidden wind]

i don’t see how that works :S for example the symmetric group on any two sets of the same cardinality will be isomorphic right? just due to the existence of a bijection… though i admit i’ve never really touched on any cardinalities greater than the continuum’s so it might just be that which confuses me

[delicate bloom]

yeah obviously otherwise the proof would be trivial lol

[crystal vale]

So if R is not commutative how ra ≠ 0 implies?

Wait

[delicate bloom]

😉

[rocky cloak]

That's right, a set is determined up to bijection by its cardinality. So the family of symmetric groups would naturally be thought of as indexed by Cardinals.

[crystal vale]

I used this argument
Let a≠0, then ab = 1, now let ba= y => a=ay.

a(y-1)=0 since R is integral domain so y=1, is it correct?

[hidden wind]

right so it really just boils down to having sufficiently many large cardinals

[rocky cloak]

Yes, and you have a proper class of them and can get as many as you like.

The Cardinals are indexed by the ordinals, and there are ordinals of arbitrary large cardinality.

[hidden wind]

whew

[delicate bloom]

yeah looks good to me, not too different to what I did

[crystal vale]

Yes but I am thinking about what if we remove integral domain property and add that there is no right zero divisor

[hidden wind]

have any book rec which gives an introduction to these things?

[crystal vale]

Then it will work, right?

[hidden wind]

also thanku very much for the explanations

[delicate bloom]

can you have right zero divisors without left zero divisors

[crystal vale]

Yes I think

[delicate bloom]

show me

[crystal vale]

Matrix

[delicate bloom]

AB=0 so B is a right zero divisor and A isn't a left zero divisor?

[crystal vale]

No I mean AB =0 but BA≠0

[rocky cloak]

I don't really have any specific recommendations. But a book on set theory I guess...

[hidden wind]

real

[delicate bloom]

sure, but what you wrote, is A not a left zero divisor?

you can't have only right zero divisors without left zero divisors, they're two sides of the same coin

[crystal vale]

Yes A will be left zero divisor but I thought the question is if A is left zero divisor then it will be right zero divisor

[delicate bloom]

this is what matters

[crystal vale]

I call a be right zero divisor if there exists r such that ar=0 where r≠0

[delicate bloom]

is this more or less restrictive than an integral domain

[delicate bloom]

are there any left zero divisors here

[crystal vale]

If a is right zero divisor then r will be left zero divisor

Means if there is no right zero divisor then there is no left zero divisor

[delicate bloom]

I'm just trying to understand if your alternative condition is really different than an integral domain or not, and if it's different how

[crystal vale]

No I don't think so

[delicate bloom]

cool

[hidden wind]

ok having thought about it for a second this is easily the prettiest of the constructions mentioned thanku

[crystal vale]

Thank you

[delicate bloom]

yup yw

[cobalt heath]

No problem!

[hidden wind]

especially since it does not rely on cardinalities larger than that of the continuum

[crystal vale]

If ab is a unit then a,b is not necessarily a unit.
Let R be the ring of Endomorphisms of all infinite sequence of R. (May be this is wrong interpretation)
Take a= (a_1,....,a_n,...) -> (a_2,.....,a_n,...) b= (b_1,.....)-> (0,b_1,.....,b_n,....)

Then ab=1 but ba ≠1, right?

[cobalt heath]

Yeah

Since ba changes first element to 0

[crystal vale]

Okay, thank you

[spare isle]

Would it be right to say that the identity hom is phi(x) = x while the trivial hom is phi(x) = e?

I feel like they are vague/up to interpretation (namely the former), but they seemed to be "assumed" in my class.

[rocky cloak]

Yes, those are fairly standard names for them

[spare isle]

thanks

and does the identity one come from the fact that the identity permutation is usually notated as Id(entity), which is itself a function (that maps each element to itself)?

I guess that's kind of backwards though; it would moreso be from the fact that the identity function is typically f(x) = x.

[hidden wind]

tfw tivial hom is different to identity hom is the same as trivial iso

[rocky cloak]

The identity homomorphism is the identity function and is also the identity with respect to composition.

The word 'identity' comes from it leaving things identically as they were.

So e is the identity because e*x = x, and Id is the identity homomorphism because Id(x) = x

[spare isle]

Yeah that was pretty much my intuition behind it. It was just weird since we never used either of those words as descriptors for homomorphisms (or functions in general, outside of the identity permutation) until the final the other day. I just went off of what seemingly made the most sense.

[amber wraith]

Enderton, Elements of Set Theory
Kunen, Foundations of Mathematics
both have pdfs available online

[hidden wind]

thanku thanku

[amber wraith]

also if you prefer videos, Antonio Montalbon has a full series of set theory lectures on Youtube, it's worth watching

[crystal vale]

I am not sure but can units be reducible in a ring?

[mighty kiln]

What's the defn of reducible element

An irreducible element in an ID is a non-unit non-zero non-[product of two non-units]

[crystal vale]

Yes

So the reducible element is a non-zero non-unit element which can be written as products of non-units ?

[tribal moss]

That would make most sense to me, but check the exact definition in your book (if you're following one).

[hidden wind]

where can i find enderton?

[crystal vale]

Okay, thank you

So units neither reducible nor irreducible

[tribal moss]

That would at least match "1 is neither composite nor prime".

[crystal vale]

Can I upload here?

[tribal moss]

No.

[crystal vale]

Okay, thank you

[rocky cloak]

Here I guess
https://docs.ufpr.br/~hoefel/ensino/CM304_CompleMat_PE3/livros/Enderton_Elements of set theory_(1977).pdf

[hidden wind]

whew wonderful

[amber wraith]

for me it's the first result googling "enderton set theory pdf"

[hidden wind]

ehehe

my google-fu has dulled

happens when my library has almost everything i need

[amber wraith]

lol I always just google "{book name} pdf" when I want to find something

[crystal vale]

I want to show that in an arbitrary ring, if 1-ab is left - invertible then 1-ba is also left invertible.

If I let c(1-ab)=1 then how can I guess the left inverse of 1-ba ?

[rocky cloak]

One way to think about this is to lend some intuition from geometric series.

If you imagine ab as some number < 1, then the inverse of 1 - ab would be
c = 1 + ab + (ab)^2 + ...

While the inverse of 1 - ba would be
1 + ba + (ba)^2 + ...
Which we can see equals 1 + bca.

Now you just need to check that the guess is correct.

[crystal vale]

This is nice

[chilly radish]

Another approach:
Note
c(1-ab)=1
Now if we multiply on the right by a and regroup we get
c(1-ab)a= ca(1-ba)=a
Now multiply by b on the left and subtract both sides from 1 to ger
1- bca(1-ba)=1- ba
So
(1+bca)(1-ba)=1

[dull ginkgo]

Lmao I had to do this by trial and error

[amber wraith]

the geometric series approach actually gets you to the answer quite nicely without having to guess or do (seemingly) weird algebraic manipulations

[crystal vale]

Yes, thank you

[crystal vale]

But I want to know how that intuition hit you ?

[amber wraith]

just seeing (1 - x)^{-1}

[rocky cloak]

Well, the inverse of 1 - x being a geometric series is something that comes up quite often.

I see we're interested in 1 - ab, that's something of the form 1 - x, maybe look at the geometric series. And boom it worked out

[amber wraith]

that probably triggered an alarm lol

[rocky cloak]

The fact that I've seen this proof before is also a big help

[amber wraith]

lol fun fact actually (not relevant to what you're talking about), in set theory a "large cardinal" is actually a thing, they're cardinals that are so large that they can't actually be proven (within ZFC) to exist

when I read "sufficiently many large cardinals" I was like

https://tenor.com/view/surprised-sorprendido-shaquille-oneal-gif-23222312

[hidden wind]

haha oops

[crystal vale]

Okay, thank you

[crystal vale]

Any hint to show any ring with identity of order p^2 is commutative?

And if I want a non-commutative ring of order p^2 without unity.

Consider R is a set of all 2 × 2 matrix over Z_p where all lower entries of matrix are 0.

Is it correct?

[lapis latch]

perhaps look at the center of the ring, and reason about what the quotient as a group looks like

[lapis latch]

mhh that wouldnt be closed under multiplication

[crystal vale]

how?

[lapis latch]

actually nvm it will be

[cobalt pilot]

Say M and N are two F-algebras. To prove that they are isomorphic, is it sufficient to find an F - vector space isomorphism that also preserves products?

[dull ginkgo]

THAT

oops caps

[cobalt pilot]

wait I’m not sure what you mean. F-algebras have more structure than vector spaces

[crystal vale]

Z(R) has at least two elements and Z(R) is also a subgroup of R so Z(R) has order p or p^2.

If it has order p then R/Z(R) as group of order p and it is cyclic group. So if a,b is notp in Z(R) then ab- ba in Z(R). But I don't know how it makes any contrary?

[dull ginkgo]

wait I misread

[rocky cloak]

That is indeed sufficient. Try proving it for yourself!

[cobalt pilot]

Thanks! I will

[rocky cloak]

Notice you haven't used that R/Z(R) is a cyclic group

[crystal vale]

I used abelian property may be I need to use generator property

Then R/Z(R) is cyclic additive group so any multiplication defined over it must be commutative

[lapis latch]

now try and move back to R, can you write the elements in a way that shows they commute?

[cedar gull]

Let Q(zeta_n) a cyclotomic field.
For all k < n such that gcd(k,n) = 1, let g_k the automorphism defined by g_k( zeta_n ) = (zeta_n)^k
How prove this result :

Suppose z in Q(zeta_n). Then z in Q <=> for all k such that gcd(k,n) = 1, then g_k(z) = z

?

[lapis latch]

What sort of results do you have to work with? Since Q(zeta_n) is a Galois extension an element is fixed by all the automorphisms iff it is in Q

[cedar gull]

Is there exist an elementary proof without knowing galois theory ?

[crystal vale]

Then any element x in R can be written as a^n + z_1 where a is non-zero fixed element and n is some integer, z_1 is in Z(R).

Then xy= (a^n + z_1)(a^s + z_2 ) = a^(n+s) + a^nz_2+ z_1a^s + z_1z_2 = a^(n+s) + z_2a^n +a^sz_1 + z_2z_1 = (a^s + z_2 )(a^n + z_1) = yx.

Is it correct?

[lapis latch]

Oops remember its the additive group so its an not a^n, but yeah

[crystal vale]

Yes, thank you ❤️

[cedar gull]

Are you talking about this ?
Let z in L
If for all g in Gal(L/K), g(z) = z then z in K ?

[lapis latch]

Yes

[cedar gull]

Is the proof of this complicated ?

[lapis latch]

Mhh not too bad, you assume z not in K then it has a degree > 1 minimal polynomial and you construct an automorphism that sends it to another root of said minimal polynomial. There’s a few preliminary results needed though

[rocky cloak]

One thing you can do is look at
f(x) = average of gk(x)
then f(x) = x iff x is fixed by each gk and the image of f is exactly those elements fixed by gk, and f is linear.

Since (zeta_n)^i is a basis for Q(zeta_n) you just have to prove that applying f to them gives you something rational.

[crystal vale]

If I want an example of a non - commutative ring of order p^3 with identity.

If I let R is set of all 2 × 2 matrix over Z_p where a_21( entry at A[21] ) is 0. Then R is a non-commutative ring with an identity of order p^3, right?

[rocky cloak]

That's right

[crystal vale]

Okay, thank you

[dire siren]

another approach: if the ring is Z_(p^2), we are done, so assume the ring is not Z_(p^2), so its characteristic is p
your ring contains 0,1,2,...,p-1
let a be different from these
prove that all ma+n with 0<=m,n<=p-1 are distinct
since the ring has p^2 elements, these are all the elements, and they clearly commute to each other

[crystal vale]

To show all ma+n are different with 0<= m,n <=p-1 are distinct.

If m_1 a + n_1 = m_2 a + n_2 and let m_1 ≠ n_1 and n_1 ≠ n_2 , then (m_1 - m_2)a = n_2-n_2 but since 0< m_1 - m_2 <p so it's inverse exists in mod p so a will be in Z•1 , I am not sure

[dire siren]

that's correct

[teal vessel]

given a quotient group G', is it the case that (g^n)'=(g')^n? I feel like this is overreaching, but based on the definitions of the cosets and such it seems this may be true.

using the prime notation instead of overbar because I'm lazy

[south patrol]

Yes

The quotient map G -> G' is a group homomorphism

[chilly radish]

Even before that this follows by how you define the quotient group

[teal vessel]

yeah, I'm just rusty

[hollow mica]

In a commutative ring with unit, can units have divisors other than other units?

So if u is a unit, does there exists a non-unit r dividing u, i.e u = rs for some s? I know that s cannot be a unit, because then r = us^-1 is a unit. So the question reduces to: can a unit be written as the product of two non-units?

ok the answer is no

units are divisors of 1, so divisors of units are divisors of 1 (by transitivity of “divides by”) hence units

——————
Anyone know why in the definition of irreducible element, we specify that the element must be nonzero? To me it looks like even without specifying this 0 will still never fit the other criteria: 0 = 0 * 0, and 0 is a non-unit

Seems like a redundancy

[delicate orchid]

that is a good question, I didn't even realise that was convention. Maybe it's something stupid to do with the zero ring

ah wait, non-integral domains

[delicate orchid]

like if (0) is maximal then your ring is a field, so R/(0) is still a field. So it doesn't break the first thing I thought of to check

[hollow mica]

what does (0) being maximal have to do with 0 being irreducible? I know that an element p is irreducible iff (p) is maximal among principal ideals, but I don’t know if this is what you’re using

[next obsidian]

Well for example take a UFD. If 0 were irreducible then you can never be a UFD as 0 = 0•0 = 0•0•0 = …

[summer path]

0•0

[hollow mica]

I’m not asking why can’t we let 0 be irreducible

I’m saying that the “no non-unit divisors” part of the definition of irreducible element should already omit 0 as being irreducible

since 0 = 0*0 is and 0 is a non-unit

[next obsidian]

Ah

[rocky cloak]

For what it's worth nlab doesn't include "non-zero" in their definition

[next obsidian]

I guess it’s a 0-ring thing?

In the 0 ring 0 is a unit

But 0 shouldn’t be irreducible because then you’d need to consider (0) as a prime ideal or something…?

¯_(ツ)_/¯

I’m also unsure about whether or not the definition i learned specifically omitted 0

[hollow mica]

artin and wikipedia omit 0

I guess it’s not really that important

[rocky cloak]

I guess it's just sort of emphesizing that
0, unit, irreducible and reducible are the four distinct categories

[hollow mica]

Oh wait I overlooked that most people don’t call 0 a non-unit

Even though it technically is

[rocky cloak]

I think most people would call 0 a non-unit

But idk

[tribal moss]

Or at the very least wouldn't call it a unit.

[hollow mica]

“A UFD is a ring such that every nonzero non-unit element r can be written as…”

[rocky cloak]

This is implying that they do consider 0 a non-unit

[hollow mica]

Oh oops

different question:
In the definition of UFD there’s two parts, one part being a factorization into irreducibles exists and the other part being that the factorization is unique up to reordering and associates. Doesn’t a factorization into irreducibles (ignoring uniqueness) always exist, in a general ring?

[rocky cloak]

Not without some finiteness assumptions

Like consider k[x1, x2, ...] With the relation that
xi = (x[i+1])^2

Then x1 cannot be factored into irreducibles

[hollow mica]

Hm

[tribal niche]

another example is the the polynomials in Q[X] with integer constant coefficient

[hollow mica]

I guess my reasoning was: for a given element r, if it’s irreducible then stop, or else write it as a product of two non-units; now keep recursing this process on the individual factors. if this process never terminates then it means r can be written as the product of an arbitrary number of non-units

I guess that is where the finiteness assumption comes into play

[rocky cloak]

Yeah, so at least in a domain if (a) = (b) then a and b are associates.

If you keep factorization r into more and more terms you'll get an increasing sequence of ideals. If your ring is Noetherian, then this must stabilize, so you'll reach an irreducible.

So in a Noetherian integral domain any element is the product of irreducibles.

If you ditch the domain part you can have weird things like two elements both being multiples of each other.

Like in k[x, y]/(xy) you have
x = x(1 + y) = x(1+y)(1+y) = ...

[hollow mica]

Wow this is so cool

If you think of the factorization process I described as building an infinite binary tree starting from the root, then if the process doesn’t terminate, for every n > 0 there will exist an element at level/height n, and in particular there is an infinite path starting from the root that only moves down. it is then the elements along this path form the chain of increasing ideals

Wait random slightly related question: if for every positive integer n there exists a chain of increasing ideals of length n, does this imply the ring is not Noetherian?

[tribal niche]

no

though i don't have an example on hand

[rocky cloak]

The integers would be one example

Or any non-field domain, by simply looking at (x^n) < (x^n-1) < ... for some non-unit x

[crystal vale]

Let G be a group. And G' be the subgroup generated by all commutators of group G. I want to show that if G' <= H <= G then H is normal.

Then let any h in H and g in G. So h^(-1)ghg^(-1) in H it implies that ghg^(-1) in H. Hence H is normal in G.

Is it correct?

[mighty kiln]

Yes

[crystal vale]

Do you maintain some blog ?

[long obsidian]

Say A is a ring and a,b are ideals and r(a) denotes the radical of a. I wanna show r(r(a)+r(b)) is a subset of r(a+b)

Say x^n =c+d where c^p in a and d^q in b.

Is the trick to do something like (x^n)^{p*q}?

Cause x^{npq}=(c+d)^{p*q} I think from the binomial theorem each term will have either a c^p factor or a d^q factor and that might do it...

[mighty kiln]

My blog is my mathcord activity

[mighty kiln]

Yes

For commutative rings

x^(n(p+q)) probably

[cloud solar]

I have a question

If we have a finite ring

The sum of all elements is 0 iff the number of elements is odd right?

[tribal moss]

Yeah, since the number of elements that are their own negative must be a power of 2, so if there are an odd number of elements, zero must be the only such element.

(Being a finite abelian group is enough).

[cloud solar]

Ok but F4 has sum of elements 0 right?

[tribal moss]

Oh shoot. I read the question as "if", not "iff".

So "odd number of elements" => "they sum to 0", but you have a good counterexample for the other direction.

[cloud solar]

Ok thanks

[rocky cloak]

Notice that every element not of order 2 is cancelled by its inverse. So the sum of elements is just the sum of elements of order 2.

The elements of order 2 form a subgroup isomorphic to (Z/2)^n
If n=1, the sum is not 0. If n>1, then consider a nonzero element a. Then we can pick 2^n-1 representatives xi such that every element is equal to either xi or a+xi.

So the sum of elements equals
[sum xi] + [sum a+xi] = 2^n-1 a + 2[sum xi] = 0.

So the only abelian groups where the elements don't sum to 0, are those which have exactly one element of order 2. I.e
Z/2, Z/4, Z/8, ..., Z/2^n, ...
and products of these with odd groups.

[tribal moss]

The elements of order 2 or 1 form a subgroup isomorphic to (Z/2)^n

[delicate orchid]

The Omega-(2,1) subgroup….

[bold sparrow]

Could anyone reccomend Fraleighs book for advanced but not graduate level group/ring/field theory?

Or not-reccomend

I have to take a 2nd course in modern algebra later on and only the worst prof at my school teaches it so I was thinking about prepping myself

[summer path]

fraleigh is not very advanced i think

[glad osprey]

I'm using Fraleigh for my first course in algebra. I like it a lot, but if this is your second course it may not cover everything. We're covering almost every chapter except the two last ones

I don't think it covers modules for example

[summer path]

if you're doing a second course, I think rotman is nice

d&f is standard but then you're reading d&f >.<

[bold sparrow]

Hm okay, my intro course ended at like ring homomorphisms I think( we covered fields but not extensively)

And the next course goes much further

[celest furnace]

You don’t like D&F?

[south patrol]

I don't either

[celest furnace]

Maam may I inquire why?

[delicate orchid]

Just intuit the entire thing

[celest furnace]

Realest thing wew lads tbh has said to me in 3 years

[south patrol]

Tbh this is more just from them having slightly weird conventions about units in rings lol

but i think the book is fine lol

[lilac mango]

I was reading some lecture notes on Galois theory and there's an example where they calculate the Galois Group of x⁴+8x+12

And prove that it's A4

It's in spanish but anyway

So to prove that it's A4 they use that f(x) is separable and the discriminant of f(x) is a square in F

Hence it's a subgroup of A4

But then they use that f(x) decomposes as a product of polynomials of degree 1 and 3 working mod 5 and somehow conclude that the Galois group has a 3-cycle

Which I don't understand

[next obsidian]

Galois group acts transitively on roots of irreducibles

[lilac mango]

Are they implying that if you have f(x) a polynomial in Z[x] and:

1. You take its splitting field K1 over Q.
2. You reduce f(x) mod p and take its splitting field K2 over Z/pZ.

Then the Galois group of K2 is a subgroup of the Galois group of K1?

[next obsidian]

Oh reducing mod 5

Uhh

[lilac mango]

Yes

[next obsidian]

Uhhh

I feel like this is now some number theory stuff

Sorry bud

[lilac mango]

Np

[next obsidian]

Me no good at that

[lilac mango]

But it's pretty weird that they just say it like that

[orchid iron]

Hi i am learning about modular lattice and i found this comment of yours?

How are these two connected?

Could you please elaborate on this?
(I am assuming you are talking about fourth isomorphism theorem,by lattice isomorphism theorem)

[vivid tiger]

Sure! I might flub it though.

So we have

a, b

uhh lemme figurebout how to get meet and join chars

[vivid tiger]

.
a ∧ b
| |
a |
| b
| |
a ∨ b

that's an ascii drawing of the lattice

oh god okay the following drawing will be horrible to type lemme actual draw it

[dull ginkgo]

This is like trying to describe a song on discord through onomatopoeias

[vivid tiger]

yes lemme get my laptop

fuck it i need to moge ill get inages

[orchid iron]

No problem,take your time
Just ping me whenever you reply

[summer path]

I used in it my first course; it's fine but pretty dry to read

[round jay]

Is it true that if J/I is an ideal of R/I then J is an ideal of R?

[celest furnace]

It’s dry??? 😭😭😭

[summer path]

lang is like a reference book

[celest furnace]

I don’t think I’ve seen a better book so maybe I need to look more

[celest furnace]

Yeah

[round jay]

Aluffi says here that if R is a ring, I is an ideal of R, and J is a subgroup of R containing I, then J/I is an ideal of R/I iff J is an ideal of R. The reverse direction is proven here, but the forward direction is not obvious to me.

[hardy island]

I know this question should be asked in #book-recommendations but I am asking it here, that is which book best suited for problem solving ?

[celest furnace]

If you know then go ask there instead and this question is way too vague to be answered in general

[barren sierra]

I answered in #advanced-algebra but tbh I should have answered in this channel. Don't spam a question in multiple channels

[rocky cloak]

I think by number theory magic, if O is the ring of integers of K1, and P is a prime lying over p. Then the decomposition group D(P) of automorphisms that leave P invariant surjects onto the Galois group of K2. So since the Galois group of K2 is C3, the original group must have an element of order 3.

[chilly radish]

This is basically a version of the Dedekind-Kummer theorem.

yea what jagr said, although it's a bit stronger, you can actually determine the cycle decomposition of the element by how the minimal polynomial decomposes mod p.

[vivid tiger]

woops i got distracted

sorry

now let [x,y] mean the interval from x to y

that is, {z : x\le z \le y}

now take [a ∧ b, b]

take x in that

now we map it to

[a, a∨b]

by mapping x to a ∨ x

likewise we can map the other way via doing b ∧ y

For a modular lattice, these two maps are inverses

(on the intervals)

so they are isomorphic

so now

in, say, groups

A \cap B = A ∧ B

AB = A ∨ B

if you interpret (not entirely sure how to formalize this, but I think it exists in the form of a theorem also called the lattice isomorphism theorem?) the interval as the quotient group

then you get the diamond isomorphism theorem

Groups form a modular lattice. So do Modules.

At least, I think that's right

Please check my connection between modular lattices and the isomorphism theorems before being confident about it

[languid trellis]

I've found two factorisations for 7 in D. $7 = (2+ \sqrt{3}i)(2- \sqrt{3}i)$ and $7 = (\frac{1}{2} +\frac{3}{2} \sqrt{3}i)(\frac{1}{2} -\frac{3}{2} \sqrt{3}i)$. If this is a euclidean domain, then by a previous theorem it is also a UFD, meaning something has gone wrong. Am I missing something? Does cancellation not hold in this ring for example?

[cloud walrus]

swifteeee

[languid trellis]

pls ping : )

[cobalt heath]

,w simplify (2 + sqrt(-3))/(1/2 - 3/2 sqrt(-3))

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by abstia

[cobalt heath]

@languid trellis ^ is a unit

[languid trellis]

How are we concluding that? Isn't the conclusion that 2 + sqrt3 i is in the ideal generated by 1/2 - 3/2 sqrt(3) i?

[cobalt heath]

You mean how we are concluding that (-1 + sqrt(-3))/2 is a unit?

[languid trellis]

Yep

[cobalt heath]

It's because this is a 3rd root of unity

[languid trellis]

Oh I didn't realise

[cobalt heath]

Its inverse is its complex conjugate, which also is in the ring

[languid trellis]

Right I gotcha

I wouldn't have caught that

So the factorisation is unique up to unit multiplier

[delicate bloom]

also it wasn't stated explicitly but Absta did it, it's probably a good idea to write sqrt(-3) and not sqrt(3)i since neither sqrt(3) nor i are actually distinct elements in this ring.

[cobalt heath]

Yep, always good to think of the unit - it can easily confuse you

[languid trellis]

Lol. Wacky ring

[dull ginkgo]

“Both in Z or both halves of odd integers” sounds like E8

[languid trellis]

I have no intuition on when Z[something] should be a ufd or not yet. Earlier on we showed that Z[sqrt(10)] isn't a ufd but Z[sqrt(2)] is

[dull ginkgo]

I’m pretty sure that is unsolved

[cobalt heath]

It's okay, I believe no one in the world has that intuition yet

[languid trellis]

Jacobson gives some criterion in some special cases

[dull ginkgo]

For negative sqrt it’s the Heegner numbers I think

https://en.m.wikipedia.org/wiki/Heegner_number

[languid trellis]

If q, p prime (q/p) = -1 (legendre symbol) then Z[sqrt(pq)] is not a ufd for example

[languid trellis]

There's 9 of them. What the fuck

OK back to studying thank you all

[cobalt heath]

So unexpected, right

[languid trellis]

It's hardly believable

[cobalt heath]

Yeah. For me, esp. last one being 163 and there's no more

[tribal moss]

Small-number accidents can keep happening for surprisingly large "small" numbers.

[rocky cloak]

This feels like it's closer to moonshine than law of small numbers, but whose to say

[dull ginkgo]

It’s original proof apparently used j function, complex multiplication and finding roots of a polynomial of degree 24

Which is extremely specific

Idk if the 24 is something completely unusual or

[tribal moss]

After all, what are sporadic groups other than a particularly grand set of small-number accidents?

[dull ginkgo]

wtf is a small number accident

[delicate bloom]

always potty train your small numbers

[dull ginkgo]

I wonder if the “integer or half integer” thing has to do with E8

[orchid iron]

Ok ok thank you

[hidden wind]

is there a standard symbol for “isomorphic to a subgroup of”

[vivid tiger]

half integer? did someone say root space? E8??

so half integralness comes from the definition of a root system. the motivation aka the connection with lie stuff comes from the rep theory of sl2C

as [H,X]=2X

[dull ginkgo]

But as just described in Jacobson that is a ring

[vivid tiger]

and [H,Y]=-2Y

and [X,Y]=H

[vivid tiger]

what is?

you mean the rep ring?

oh i was thinking of a thing

[dull ginkgo]

Is E8 not integers or half integer vectors who sum to even

This

[vivid tiger]

i wanted to say during a presentation that "irreducible decomposition is like the prime decomposition - in fact it's literally the factorization in the representation ring" but it doesn't look like they are the irreds in the rep ring?

because (assume complete reducibility, e.g. for semisimple lie algebra)

if we have V irreducible

that means V \neq V1 \oplus V2

it does not mean V \neq V1 \otimes V2

but the latter is the product in the rep ring

so instead irreds are like primitive idempotents

this can be made precise

as in

they literally correspond

and there's like algorithms using this to compute things like decompositions of the lie alg

but i don't know them yet

[rocky cloak]

I don't know whether it's related to E8 or not, but the more elementary reason is just that the quadratic formula has a division by 2 in it. So produces things of the form
x/2 + sqrt(y)/2

So when you look at the ring of integers of some quadratic field they will look like that.

[dull ginkgo]

Ah

[delicate orchid]

I was so excited to start expositing stuff until I saw “Lie algebra”…. The nice guy finishes last once more…

[languid trellis]

le nice guy

[lilac mango]

When I have a Galois extension and I know what its Galois group is, how do I find the fixed fields of the subgroups of the Galois group in general?

[wraith swan]

Hey, I got an exercise to show that G x H in the category of abelian groups is also the coproduct. But I am kind of confused when I have to use that all groups are abelian and that G x H is the product. And so I am kind of stuck, anyone got some hints? I have tried to write out some compositions of homomorphisms and sums of elements, but non gave me insides in the problem.

So I think I want to show that mu has to be defined mu(a, b) = j_G(a) + j_H(b), but I don't get how.

[tribal moss]

You're probably not supposed to work purely abstractly with the product. Instead you hopefully already know that in this category G×H is exactly the group whose elements are pairs of one element from g with one element from h, and the operation is (g1,h1) + (g2,h2) = (g1+g2, h1+h2).

[wraith swan]

Right, operations are piecewise

[tribal moss]

So based on that, there only "reasonable" thing to try in order to make it a coproduct to is to define i_G(g) = (g,0) and i_H(h) = (0,h).

[wraith swan]

And since mu is a homomorphism I can somehow show mu is also piecewise?

[lilac mango]

Say for example I have the cyclotomic extension Q(z_13). I know the Galois group is (Z/13Z)*, which is cyclic of order 12 and I know that the subgroups are cyclic for every divisor of 12, but how do I do to determine the subfields given by the Galois correspondence?

[tribal moss]

And then the two commuting triangles fix exactly what µ has to do on elements of the form (g,0) and (0,h), and all other values of µ then follow from being a homomorphism.

[wraith swan]

And where would you use that they are abelian? Since G x H is not the coproduct in Grp, right?

Or is that to show that mu is a homomorphism?

[tribal moss]

The crucial point is that Z (that is, the codomain of µ, not the integers) is abelian.
If Z were not abelian, we would get into trouble with
µ(g,0)µ(0,h) = µ((g,0)+(0,h)) = µ((g,h)) = µ((0,h)+(g,0)) = µ(0,h)µ(g,0)
so the whole thing wouldn't produce a homomorphism if j_G and j_H map to elements that don't commute.

[wraith swan]

Ok thank you, I think I get it

[rocky cloak]

The mapping of x to the avarage of g(x) for g in H defines a linear projection onto the fixed subfield by H.

So if you have an extension K(z) I'd start with the extension generated by the avarage of g(z), and see if that one has the right order. If not continue adjoining the avarage of g(z^2), and so on.

[lilac mango]

What do you mean by avarage?

[rocky cloak]

Like add them together and divide by the number of them

[rocky cloak]

The division is sort of unnecessary for your purposes. Just needed to make it into a linear projection.

[winged void]

Het Guys I have question How to find the number of morfisme c6 to S5. What is the intuition behind this and what to do

[south patrol]

Well here there is a specific thing

[lilac mango]

Oh ok I see that makes sense

[south patrol]

Note a morphism out of a cyclic group is determined by where you send a generator

That allows you to classify all maps from a cyclic group into any other group

[winged void]

Do you mean here by a generator the kernel

Or the generator of an element

[south patrol]

generator of the cyclic group

[winged void]

well that is an element that can generate the whole group sure

but then how can we know all the morfisme

[south patrol]

where can you send the generator?

[winged void]

well to an element that generates S5

i guess

or im not really sure

to be fair

[south patrol]

Well say f: C_6 -> S_5 and g is a generator of C_6

what can you say a bout f(g)

[wraith swan]

I just looked at it again and now it is so obvious... Thank you 🙂

[winged void]

well i mean f(g) must have the same order that it has in C_6

or am i saying something wrong now

[south patrol]

Not quite, but nearly

you can say that f(g)^6 = 1

do you see why?

[winged void]

not really to be fair

well i know but can we say that f(g)f(g) = f(g^2)

you mean that

[south patrol]

h=g

[winged void]

but 6 times

[south patrol]

exactly

in a slogan like

you can multiply before or after applying f

[winged void]

that is true

but sure what is the point of that

[south patrol]

Well it gives you a restriction on where you can send g

For example, you can't send g to (12345)

[winged void]

how did you know that

if i may ask

[craggy gyro]

How can I prove that if N is a normal s.g. where |N|=3 and isn't contained properly by the centralizer then there is a s.g. T of G s.t [G:T]=2 ?

[amber cradle]

What does the function f:F-> F given by x|->x^p-x look like when F is infinite of characteristic p?

Is it equivalently zero?

[south patrol]

No

[amber cradle]

Why not?

[south patrol]

Not even necessarily when F is finite

[amber cradle]

How

When F is characteristic p?

[south patrol]

Consider for example F_p(t)

[south patrol]

Well x^p - x has at most p roots

[craggy gyro]

It's not zero with infinite characteristic

[amber cradle]

But x^p \cong x mod p no?

[south patrol]

Mod p yes

But that's for x in Z or whatever

[amber cradle]

And if it’s characteristic p, isn’t that what we r doing

[south patrol]

Like that is just saying this holds in F_p

Doesn't say anything about other fields

Like F_p(t) as I suggested

t and t^p are different

[amber cradle]

This is the field of rational functions?

[south patrol]

Or any larger finite field

[amber cradle]

Over F_p

[south patrol]

Yes

Well rational polynomials ig, not functions

[amber cradle]

Rational functions refers to rational polynomials

Why is x^p not x?

[craggy gyro]

Take the field of real numbers

No exponent of x is x

Other than 1

[south patrol]

Well there can be a difference ig but yes

[south patrol]

We were over char p

[south patrol]

Well t is not t^p essentially by definition

[amber cradle]

I mean the function, not the formal expression

[craggy gyro]

I have tried constructing Z(G).N subgroup and tried to conclude something with second isomorphism and the order counting formula but couldn't proceed much

[south patrol]

Idk what you mean

You're talking about whether x = x^p for all x in F_p(t) right

And i'm saying t is not t^p

[amber cradle]

I see

I confused the function f:F_p(t)->F_p(t) given by x^p-x with the elements of F

So if x = t

f(x) = t^p - t

Which is not the zero polynomial and thus not equivalent to the zero function

Gotcha

[south patrol]

Yes

[dull ginkgo]

I'm stuck at showing independence of choice here

Assume $S$ is a multiplicative subset of domain $D$ without $0$, such that for each pair $(d,s) \in D \times S$, we have $sD \cap dS \neq \emptyset$. I need to show that if $s_1 d' = s_2 s', d_1 d' = d_2 s', s_1 d'' = s_2 s''$, then we have $d_1 d'' = d_2 s''$

[cloud walrus]

THE TUBE

[dull ginkgo]

Which is the independence

it's a bit annoying because idk how to use cancellativity here from the domain

i don't know how to go about this

[frank cosmos]

why is the product of 2 left ideals not nessecarily a left ideal

[south patrol]

It should be - do you mean the naive product like { ij | i in I, j in J}

[frank cosmos]

IJ=sums \sum xi*yi where xi in I and yi in J

ye

[south patrol]

i think there they mean the naive product i wrote down

Hm

Because otherwise it seems it should be a left ideal just cause like

a ( sum_i x_i y_I) = sum_i (ax_i) y_i

[frank cosmos]

ye

[south patrol]

and clearly closed under addition

[frank cosmos]

its a subgroup ye

maybe that writing is wrong

[real cradle]

whats about this discution ?

[dire siren]

@dull ginkgo horrible problem
so we have to prove that if bx=dy, ax=cy and bs=dt for some non-zero x,y,s,t, then as=ct

here is how to prove it: apply the right common multiple property for x and s, so there are u and v such that xu=sv=m
so bm=dyu and also bm=dtv
thus dyu=dtv, but we are in a domain, so this implies yu=tv

we have axu=cyu, which becomes asv=ctv, and since we are in a domain, this implies as=ct

[dull ginkgo]

I finally figured it out using basically the same technique as this

Holy balls that sucks

Thanks man,

[dull ginkgo]

Take $s'd^* = s''s^$, then we have $s_1d'd^ = s_2s'd^* = s_2s''s^* = s_1d''s^$, so $d'd^ = d''s^$ \ \
Thus $d_1d''s^ = d_1d'd^* = d_2s'd^* = d_2 s''s^* $ implying $d_1d'' = d_2s''$

[cloud walrus]

THE TUBE

[tough raven]

To add a little more information:

Your statement is true by Fermat's Little Theorem for x in Z/pZ = F_p i.e. the field with p elements, but not in any larger field.

In any field of characteristic p, the kernel of x^p - x is precisely F_p, because as a polynomial of degree p, it can have at most p roots, and F_p gives you p roots.

For example, in the field F_p^3 with p^3 elements, the kernel of x^p - x is F_p. (The kernel of x^p^3 - x is all of F_p^3 though.)

[cursive edge]

at the bottom, why would every element have at most order p^max(i,j)?

[crystal vale]

Every element has at most order lcm of p^i and p^j so what will be the lcm of p^i and p^j?

Let H be a normal subgroup of G. Then the product of left Cosets is left coset.

So let aH, bH be left Cosets so I want to prove that aHbH = abH.

So let ah_1bh_2 in aHbH. Then ah_1bh_2 = abh_3h_2 so aHbH is a subset of abH.

Now let abh in abH. Then abh = ah_3b so abH is a subset of aHbH.

Is it correct?

[cursive edge]

doesnt (x,y) have order lcm(order of x in Z_p^i, order of y in Z_p^j)?

how do we get that that's at most lcm(p^i, p^j) from that

[crystal vale]

Yes, but what will be the order of (1,1) ?

[rocky cloak]

Well, what could the order of an element in Z/p^i be?

[cursive edge]

lcm(p^i, p^j), so thats where we get p^max(i,j) from, but how do we know some element doesnt have a greater order than that of (1,1)?

since it seems like we're trying to argue every element has order \leq order of (1,1) = p^max(i,j)

[rocky cloak]

For any element in Z/p^i, it's order divides the order of 1

[winged void]

is it always the kernel of morphism is normal

?

is it always the case

[celest furnace]

Yes

And this can be taken as an equivalent definition

[winged void]

i see but what is the intuition behind that

idk if there is something to see why that is the case

just wondering

[celest furnace]

Basically anything * 0 is still just 0 right

So if you do phi(gxg^-1)=phi(g)phi(x)phi(g)^-1 = phi(g)phi(g^-1) = 1 for x in the kernel

[winged void]

but i assume you define here 0 as identity

right

or natural element

[celest furnace]

Yeah that was just intuition

[winged void]

i see

[celest furnace]

Like 0 + a - a is still 0

[winged void]

that is true

[celest furnace]

This would be for abelian groups where 0 is your neutral element

[winged void]

i see make sense

[winged void]

yes can someone check if the way that i determine the number of morphisms from C_6 to S_5 good is

[winged void]

i used your method

but not so sure if i used it good

and continued that way

[rocky cloak]

Not really sure what you've written here, but the main idea is that any map
f : Z/6 -> G
is uniquely determined by where 1 is mapped, because if f(1) = g, then f(n) = g^n.

So such maps are exactly described by elements g with g^6 = 1.

So you should try to count the elements in S5 with order dividing 6.

You might start by counting the elements with order exactly 6 (there should be ||20||)

[eager willow]

is this actually true? Translated poorly from Soviet article that gave no proof.

trying a proof by induction and can't even figure out a base case

[cobalt heath]

Seems like you cannot use induction if i goes infinitely.

[eager willow]

why not? I suspect it's possible to show that for fixed i, {x_i x_m}_m has finite support

[cobalt heath]

Hmm

[eager willow]

still struggling on that much though

[cobalt heath]

What do you mean by support here?

[eager willow]

set of j with nonzero coefficient in the terms

[cobalt heath]

Well, if x_i x_m = 2x_{min(n, m)} like the lemma claims, then it cannot have finite support.

[eager willow]

no fix i, then the support is 1, 2, ...., i as we range through m

[cobalt heath]

x_i x_{i+1} = 2x_i.

[eager willow]

yes, fix i, and the support is bounded by i

like S_i = {j | x_ix_m has nonzero x_j coeff for some m} is the set {1, ..., i}

but the functional hopefully lets us prove S_i is bounded to start an inductive argument

[cobalt heath]

Hmm, nonzero coeff?

Yeah I don't think that works, because you might have relations between x_i's.

[eager willow]

wym, if we assume the lemma, you understand the bound is i? the relations include that f annihilates x_n(x_m - x_m') whenever m and m' are both > n

[cobalt heath]

How will you define the coefficient when like

You can have c_1 x_i1 + .. + c_k x_ik = 0

Then, for any element of R, you can say it has nonzero x_{ij} coefficient.

[eager willow]

Ok I see what you mean now, in my wording it is not assumed that R is freely generated by the x_i