groups-rings-fields

[sullen tinsel]

Yea

[latent anvil]

you (Kirby) asked about the coprimality thing

[sullen tinsel]

Cause they said people who would answer the questions wouldn’t check those channels

[latent anvil]

They'll answer your question

But they wouldn't answer kirby's (usually)

I don't look at questions

[sullen tinsel]

Ahh I got you

[slate forum]

Alrighty

[latent anvil]

Also Kirby I don't remember how to do this off the top of my head, I'll think about it

You should probably repost your question in case someone else looks at the channel

[slate forum]

I did

Oh again?

[latent anvil]

Yeah

[slate forum]

How do I show for an algebraically closed field k, that if f and g are coprime in k[x,y], then they are coprime in k(x)[y]?

I need this for trying to prove the zero set of two coprime polynomials in k[x,y] is finite

[prime gale]

Thanks, HelixKirby. Now I can post here. Anyway, I must be missing something. Haven't tried writing anything down, but can't you just clear the denominators?

[latent anvil]

you can clear denominators but it's not clear how that helps

[slate forum]

Uhhh possibly

[prime gale]

That is, if you have a relation in k(x)[y], just multiply by denominator and you get relation in k[x,y]?

[latent anvil]

At least not to me

you get a relation but the relation has more terms

[slate forum]

I also have an issue

[latent anvil]

You get a p = f q and b p = g q where p, a, b in k[x, y] and q in k[x]

yeah?

[slate forum]

If gcd(f,g)=1 and it's still the case in k(x)[y]

Then af+bg=1

So I can plug in to get 1=0

[latent anvil]

I don't get what you mean?

Like, what are you plugging in

[slate forum]

In Z(f,g)

[latent anvil]

but that might be a point where a, b aren't defined

[slate forum]

Oh ok

[latent anvil]

does that make sense?

[slate forum]

Divide by zero

[latent anvil]

so I think you might be able to reduce to the case where f, g are irreducible

I need this for trying to prove the zero set of two coprime polynomials in k[x,y] is finite
@slate forum
You can do it for this question for sure

factor f and g into products of irreducibles

Write Z(f, g) as an intersection of unions and distribute

You get that Z(f, g) is a finite union of sets Z(f', g') where f', g', are distinct irreducibles

[prime gale]

Sorry. I was thinking of something weird. Anyway, for your original question, I think dimension counting works. Though I guess that is cheating.

[latent anvil]

I think you'll need something like this to prove bezout's lemma anyways haha

[slate forum]

Uhhh

That looks really complicated

[latent anvil]

At least I remember this exact lemma being used in the proof of bezout at some point

well it's not really helix

When is p in Z(f, g)?

If f factors as f1...fn and g factors as g1...gm

[slate forum]

It's a zero of one of f_i,g_j

[latent anvil]

Exactly

So it's in some Z(fi, gj)

that tells us Z(f, g) = union of all Z(fi, gj)

[slate forum]

Oh ok

[latent anvil]

And then wlog f, g are irreducible

Does that make sense?

[slate forum]

Ok

Finite union of finite sets?

[latent anvil]

Yup

[crimson arch]

So I realize this is trivial but I've never actually heard a solid answer for this: what is the proper way to actually say "X = {x : x satisfied P}" where X is a set, x is a element of that set, and P is a property that describes the elements of X
is it really just "x belongs to the set X as long as it satisfies P" .. idfk

[latent anvil]

@crimson arch not appropriate for this channel lol

[slate forum]

Oh that top thingy is irrelevant?

[latent anvil]

try #proofs-and-logic

[crimson arch]

literally abstract algebra related ....

[latent anvil]

How?

[slate forum]

Tangentially related

[latent anvil]

Anyways Kirby what top thing?

[crimson arch]

are you kidding me ..

[slate forum]

On your whiteboard

[latent anvil]

It's as algebra related as it is analysis or topology related

Oh lmao

That's totally unrelated

[slate forum]

Ok

[latent anvil]

It was commutative algebra I was doing earlier

[crimson arch]

or even linear algebra by that def

[slate forum]

That's why I was confused

[latent anvil]

yes

I agree

[crimson arch]

lol

[prime gale]

I must be missing something. Don't really remember definitions. f and g coprime means <f>+<g>=1? If so, then isn't it obvious?

[latent anvil]

no am that's not the case

That's coprimality of ideals

[slate forum]

Only in a gcd domain I think?

[prime gale]

<f> = ideal generated by f.

[latent anvil]

But in a UFD it usually means no common irreducible divisors

[slate forum]

Or like a pid?

[latent anvil]

In a PID these two definitions coincide

I think gcd domain suffices?

Like if (d) = (f, g) then d divides f and g

[slate forum]

For showing the gcd is still 1 though....

[latent anvil]

So d is a unit

well

The proof above works right?

[slate forum]

I think

[latent anvil]

Anyways I think you can reduce to the case of distinct irreducibles

The geometry side suggests it

[slate forum]

Well, I do need to show irreducibles have finitely many common roots then?

I can use Gauss' lemma there though

[latent anvil]

yeah I mean I showed that's equivalent

Oh can you?

hmm

Sorry why is that?

Oh right

You totally can

Yeah

Sorry I was being dumb

[slate forum]

Irreducible in k[x][y] then irred in k(x)[y] I think

[latent anvil]

okay so gauss is exactly the irreducible case

[prime gale]

I really must be missing something. f, g relatively prime means <1>=<f>+<g> in k[x,y]. Then trivially, <1>=<f>+<g> in k(x)[y], right?

[latent anvil]

right?

that's not what it means am

But in a UFD it usually means no common irreducible divisors
@latent anvil

[prime gale]

That's not what what means?

[latent anvil]

Consider f = x and g = y

these are coprime since they have no common divisors except units

but (x) + (y) = (x, y) is a proper ideal

[prime gale]

Yes. You are right. Googled definition of relatively prime ideals. Obviously it was wrong definition.

[latent anvil]

Yeah coprime ideals and coprime elements aren't the same unfortunately. Coprimality of elements is only really sensible in a UFD imo

maybe others would disagree though

Okay so helix we have a proof that Z(f, g) is finite right?

not saying to stop thinking about the original question

Just making sure we're on the same page

[slate forum]

I think so, factor f and g first

Then use how Z works on products

And intersections

Prove it for irreducibles

Via gauss' lemma I guess?

[latent anvil]

Actually I don't understand how the lemma you gave shows that Z(f, g) is finite when f g are irreducible

So gauss's lemma proves the question you originally asked

but I'm having trouble connecting that to the finite zero set

[slate forum]

Though that still doesn't explain why their gcd stays 1

[latent anvil]

Wait so if f, g are distinct irreducibles, why is Z(f, g) finite?

[prime gale]

Because k[x,y] is noetherian?

[latent anvil]

I'm sorry I'm not sure what you mean?

I don't see why it's immediate from noetherianness

[prime gale]

k[x,y]/(<f>+<g>) is noetherian zero dimensional. But I guess I am cheating again.

[latent anvil]

Oh I see why it's zero dimensional

But I am also cheating

[slate forum]

Reee no dimensions

[latent anvil]

dim k[x, y] = 2 so if p contains (f, g) we have a chain 0 < (f) < p and thus p is maximal

Is that what you were thinking of?

Sorry Kirby

I also feel like this is cheating lol

[slate forum]

Ok, look say f and g have a gcd in k(x)[y]

[latent anvil]

But yeah @prime gale is that how you were thinking of showing 0 dimensionality?

Is this for general f, g?

[slate forum]

Brb

Like hour or two

[latent anvil]

Sure, I'll think about this

[prime gale]

dim k[x,y] =2 and irreducible. k[x,y]/f is proper subset, thus must have lower dimension, -1, 0 or 1. k[x,y]/f is irreducible by assumption. k[x,y]/(<f>+<g>) is proper subset by assumption. Thus dimension must decrease again.

So actually, dimension = 0 or -1. But either way, finite.

[latent anvil]

I'm not sure what you mean by k[x, y]/f irreducible or proper subset but I think that's essentially the same argument, cool

[prime gale]

I am using krull dimension of spec(k[x,y]/f).

[latent anvil]

Oh you meant the specs were proper subsets

gotcha

[prime gale]

Yes. Should have said that, but since we were all talking geometry anyway, I was lazy.

[latent anvil]

well I was avoiding talking about Spec or dimension because it seems like Kirby is only dealing with basic stuff about affine varieties

[prime gale]

Yes. Cheating.

[latent anvil]

oh I think I see why it's finite?

nope I was wrong lol

Argument didn't work

[prime gale]

Anyway, seems like noetherianness is important. That's where finiteness comes from.

[latent anvil]

I see the argument using dimension, to be clear

I think noetherianness can be thought of in terms of irreducible components here

Like if you want to stick to the language of affine varieties with as little comm alg as possible

Then you say Z(f, g) has finitely irreducible components and you want to show each is a point

[prime gale]

I mean that a non noetherian zero dimensional scheme can be infinite.

[latent anvil]

This is the same as showing if p contains f and g then p is maximal, i.e. the quotient is zero dimensional

[prime gale]

But obviously in this case everything is noetherian.

[latent anvil]

Sure, I agree, I'm saying that the reason zero dimensional noetherian rings have finitely many primes is because noetherian rings have finitely many minimal primes

I.e. they have finitely many minimal primes containing any given ideal, i.e. an affine algebraic set has finitely many irreducible components

[prime gale]

I agree. My point is just that probably any argument will use fact that k[x,y] is noetherian.

[latent anvil]

Oh I see how to do it

Using the original lemma

How do I show for an algebraically closed field k, that if f and g are coprime in k[x,y], then they are coprime in k(x)[y]?
@slate forum

Suppose this is true. We know it in the case where f and g are irreducible by gauss's lemma

Then f, g are coprime in the PID k(x)[y], so there's a relation p f + q g = 1. Clearing denominators we get p'(x, y) f(x, y) + q'(x, y) g(x, y) = r(x) in k[x, y]

So there are only finitely many possible x coordinates for an element of Z(f, g)

You can do the same thing to get only finitely many possible y coordinates by saying they're coprime in k(y)[x]

I don't think this is secretly using noetherianness but it's definitely using stronger facts about the ring k[x, y]

[prime gale]

You are right.

[latent anvil]

I think I see how to reduce the general case to the irreducible case too

[prime gale]

I guess I still don't know what coprime means though.

[latent anvil]

f and g are coprime if whenever d divides f and d divides g, d is a unit

In a UFD this is equivalent to the statement that f and g have no common irreducible factors

in a PID it's equivalent to (f, g) = (1)

[prime gale]

I see. k(x)[y] is PID.

[latent anvil]

Exactly

[prime gale]

Reducing general to irreducible is trivial, right? No common components because f and g coprime. Finitely many components because noetherian.

[latent anvil]

you need to be careful about components no longer being irreducible in k(x)[y] but it's pretty easy

Wait by components do you mean irreducible factors?

Just making sure I understand you

[prime gale]

Isn't image of irreducible always irreducible?

Sorry--got it backwards. This is pullback.

Yes. By component, I mean irreducible compoent.

[latent anvil]

@slate forum we figured the argument out, I can give more details of you want

[prime gale]

"we" = "S. Hamrock".

[latent anvil]

haha thanks

basically factor f and g into irreducibles, wlog neither f nor g is in k[x], then the irreducible factorization in k(x)[y] is mostly the same

You take away all the factors that are already in k[x], this doesn't change coprimality in either ring

if some irreducible in k(x)[y] divides both, it has to divide some irreducible factor of f and some irreducible factor of g, and then you can use gauss's lemma

I guess you need to know that it f, g are irreducible polynomials in k[x, y] (not in k[x]) and p(x) f(x, y) = q(x) g(x, y) for some polynomials p, q, then f = c g for some c in k

this says unit multiples in k(x)[y] => unit multiples in k[x, y]

this is easy though, p f = q g implies g divides p if f, g are distinct irreducibles, but p is a polynomial in only so this is impossible

[slate forum]

Where am I factoring f and g into irreducibles?

[latent anvil]

Okay so

Factor in k[x,y]

Then essentially the same factorization works in k(x)[y]

Except some of the factors are now units

[slate forum]

Yeah

[latent anvil]

Now if they're not coprime they share a common irreducible factor

Up to units

[slate forum]

Mhmmm

[latent anvil]

So we get f'(x, y) p(x) = g'(x, y) q(x) were f', g' are irreducible factors of f, g in k[x, y]

since f, g are coprime and f' divides g'q, we get that f' divides q

Does that make sense?

[slate forum]

Uhhh

What is the common factor they share in k(x)[y]?

[latent anvil]

If they're not coprime there has to be one

[slate forum]

Sure

Call it d

[latent anvil]

Okay well choose an irreducible factor of d

That irreducible factor has to be a unit multiples of some fi and of some gj

[slate forum]

Why's that?

Or is that because those are f and g s factors?

Irreducible factors

[latent anvil]

yes

[slate forum]

Ok

[latent anvil]

Necessarily fi and gj aren't in k[x] also

Because otherwise the irreducible factor of d we chose was a unit (and thus not irreducible)

[slate forum]

Otherwise they become units

Ok

[latent anvil]

Yeah

So now we get that fi and gj are a unit multiple in k(x)[y]

So fi(x, y) p(x) = gj(x, y) q(x)

[slate forum]

Ok

[latent anvil]

now gj can't divide p because gj isn't in k[x]

So it's gotta divide fi

So fi(x, y) is a unit multiple of gj(x, y) in k[x, y]

So f and g share a common irreducible factor in k[x, y]

[slate forum]

Oh shit

I think I get it, their gcd isn't one now

[chilly ocean]

Oh look who it is, @slate forum

[slate forum]

It's f_i at least

[latent anvil]

yeah pretty much

oh I also proved that Z(f, g) is finite directly from this but you might have had that already

[slate forum]

I think so

Thank you!

I'm really bad at this stuff

[latent anvil]

It's mega ultra super hard imo

The thing I'm worst at in math is AG

Like by a mile

[slate forum]

What do you study?

Are you a prof?

[prime gale]

The bar is very high in AG.

[latent anvil]

no lol not at all

[slate forum]

Yeah I'm taking a year long course in it

[prime gale]

Too high for me.

[latent anvil]

I'll starting my 3rd year of undergrad

[slate forum]

Bro wtf

I'm a 3rd year grad student

[latent anvil]

I kind of failed an AG class last year. Not on paper but only because he gave everyone a 4.0 due to corona

I didn't take the 3rd quarter of the class

[prime gale]

S. Hamrock--what text?

[latent anvil]

hartshorne

also please call me shamrock

[prime gale]

Not your typical undergrad text...

[latent anvil]

The nick is just a joke

It was a grad course

I wasn't prepared

[slate forum]

I wish I was that smart

Jeez

I was taking algebra and analysis my 3rd year

[latent anvil]

I gotta say, I don't think "dropped a course because it was too hard" is at the top of my list of academic achievements :P

[slate forum]

If you say so

How do I get good at AG, is it intuition?

[latent anvil]

lol don't ask me

[prime gale]

Study.

At least that's what my teacher tells me. I never could get anywhere though.

[chilly ocean]

Don't sweat not being able to invent it on your own, for one thing.

AG's has a long history of hand-waving

[prime gale]

Trouble is all the papers seem to be write only papers.

[chilly ocean]

and and even worse history of French intellectualization.

(eg, divorcing things from their intuitive roots to get a really good axiomatic foundation)

[prime gale]

Well, I think the French guy fixed just about all of the hand waving.

[chilly ocean]

Yeah. It had the one problem, then Grotehndieck and friends brought it to the other extreme.

But you are trying to solve as homework over a long weekend what probably fell out as a natural consequence to trying to answer much more intuitive questions.

[slate forum]

I just feel bad looking for help sometimes

[chilly ocean]

I did too. I think most people do.

[slate forum]

I should be able to do it myself

should

[prime gale]

Think of how bad I feel giving incorrect answers...

[chilly ocean]

But the people who come up with these proofs aren't "trying to avoid help"

[latent anvil]

Eh

Pretty much everyone has to ask for help sometime

[chilly ocean]

A researcher is going to use every tool they can to get to an answer.

[latent anvil]

my perception of math research is that it's very collaborative

[chilly ocean]

(Doesn't the word "mathematics" mean "that which is taught"?)

[slate forum]

Sometimes I think I don't try long enough

Before looking for milkies

[chilly ocean]

Basically everyone feels that way.

[latent anvil]

that might be true, but also maybe not. How long do you usually spent before looking for help?

[chilly ocean]

This fucker.

was saying

"Don't just give me the answer"

when we were talking about it

[latent anvil]

wait what

[chilly ocean]

Obviously a good student.

[latent anvil]

?

[chilly ocean]

Hah

I invited Kirby to this server. We were discussing the problem. He was worried he was getting too much help by asking.

[latent anvil]

ahh gotcha

[chilly ocean]

(Lucky for him, fuck if I remember how the proof works)

[latent anvil]

Yeah my reaction was the same lol

"hmm I know I have seen this before..."

[slate forum]

Sometimes I just feel like I just learn the tricks

[chilly ocean]

Math is just tricks, theory, and bug collecting.

basically nothing else

[slate forum]

But then how did anyone learn anything?

[chilly ocean]

I'm telling you. They had more than a long weekend to work it out.

[slate forum]

Kind of a chicken and egg everythi

Scenario

[chilly ocean]

You have to think of it from a probabilistic framework.

No researcher goes out trying to prove some particular theorem. They go out trying to prove any theorem they can.

[slate forum]

For real?

I thought they had specific problems

[latent anvil]

no I don't think that's totally true?

[chilly ocean]

It's not totally true, of course.

[latent anvil]

like, at the very least people usually specialize to an area of math

I get the sentiment tho

If you end up proving something weaker than what you want that's still good

[chilly ocean]

But if you think of it as a process of discovery, then it makes more sense when you are having trouble.

You can't always decide on what you get to discover.

[latent anvil]

If you end up proving a vaguely related theorem that's good too

[slate forum]

I wish I knew what I liked

I dislike analysis

And differential equations

[latent anvil]

Omg what a mood

Except I don't dislike analysis haha

I just have no idea what I want to focus on

[slate forum]

So... I really like the stuff in homological algebra

But it's not really an active research field

[prime gale]

Plenty of research applies cohomology. Probably biggest hammer there is.

[latent anvil]

that's not really the same thing tho

like plenty of research applies calculus

You probably won't find a job researching calculus

[slate forum]

Like homology and cohomology of topological spaces was fun

[prime gale]

But if you like calculus, you can look into subjects that apply calculus.

[chilly ocean]

Don't worry. If you don't get a job, you just get to be like a normal person with a college education in 2020 😛

[slate forum]

Oh?

Indoctrinated?

[chilly ocean]

I always wonder how research works in math. It's too hard to get a broad idea of what's going on because it's all so damn technical.

[prime gale]

Of my class mates, about 95% did not get jobs. But those 95% are now VHNWI. The 5% are not.

[slate forum]

Vhnwi?

[prime gale]

In other words, don't worry about getting a job. You probably won't. But it'll all work out anyway.

[chilly ocean]

I kept my study to a hobby. It worked out nicely for me.

But I get the itch and then I just want to find someone to talk about sheaves. And no one knows wtf I'm talking about.

[scarlet estuary]

@slate forum "very high net worth individuals"

usually the cutoff i see is $5 milion net worth

[latent anvil]

oh lol I thought "very happily newly wed i(??)"

[prime gale]

Unless they are quasicoherent, I don't know WTF you are talking about.

[latent anvil]

Was very confused

[scarlet estuary]

very high newly wed Indianans

[latent anvil]

Good for them

[scarlet estuary]

its a surprisingly common life progression

[latent anvil]

gahhh why does dimension theory have to be so hard

[scarlet estuary]

the only real question is, waht comes first: the weed, the wife, or the move to indiana?

[latent anvil]

I'm trying to prove the like most baby statement

ht p + coht p = n

and it is f r u s t r a t i n g

[chilly ocean]

おｈ my~

[latent anvil]

matsumura has 2 exercises in this section and this problem has taken more time than all the other sections combined

[prime gale]

Miles Reid's translation?

[latent anvil]

yeah, I think so

[prime gale]

Scurve, Darb, Mole, Flem?

[latent anvil]

what?

[slate forum]

What's ht and coht?

[prime gale]

More or less codim and dim.

[slate forum]

Height and coheight?

What exactly is dimension theory?

I know dimension is like sup of chains of prime ideals

[latent anvil]

okay so

same idea

ht p is the sup of all lengths of chains contained in p

coht p is the sup of all lengths chains containing p

does that make sense?

[slate forum]

I guess?

[latent anvil]

Well the point is that this formula helps you calculate the dimension of an affine variety (hopefully)

[slate forum]

I know dimension 1 means primes after maximal

Are*

[latent anvil]

The coordinate ring of Z(p) is A/p, and dim Z(p) = dim(A/p) = coht p

similarly ht p = dim A_p

[slate forum]

Coordinate ring is intuitively functions that don't vanish on Z(p)?

[latent anvil]

Kind of

[prime gale]

Functions are considered equivalent if they differ by something that vanishes on Z(p).

[latent anvil]

so like, consider the variety Z(y^2 - x^3)

on this space the functions φ(x, y) = y^2 and ψ(x, y) = x^3 are equal

[slate forum]

Ok

And this is like stalks or germs or something?

[prime gale]

A_p is like germs.

[slate forum]

A_p is localization at p?

[prime gale]

Yes. That is, localizing at the set A\p.

[slate forum]

Ok

What's the coordinate ring for then?

[prime gale]

Sorry. I do not understand.

Figured out how to parse your sentence. Coordinate ring lets you apply Algebra to Geometry.

[slate forum]

Oh ok

[prime gale]

This is all intuitive nonsense though.

[slate forum]

Intuition is helpful

[latent anvil]

you should think of the coordinate ring of an affine variety like you think of the continuous/smooth functions on a topological/smooth manifold

[prime gale]

Without a coordinate ring, your space is very flexible--just a topological space. Once you add coordinate ring, your maps must respect the coordinate ring. Then your space becomes more rigid.

[bleak abyss]

Ooh we talking AG?

[latent anvil]

unfortunately

[prime gale]

Algebra question careened into AG.

[slate forum]

My question was AG

How do you intuitively see that a composition of two reflections is a rotation in a dihedral group?

[carmine fossil]

Composition of 2reflections is identity

[latent anvil]

that's not true

composition of the same two is an identity

[carmine fossil]

I think he meant composition of a reflection and rotation

[latent anvil]

@slate forum I'm not sure what a good way to do it would be. Maybe just write it out in matrices?

@carmine fossil why do you say that? I don't think that's true

Composition of a reflection and a rotation will have negative determinant

[slate forum]

Say rf composed with r^2f or something

Not the same reflection

[latent anvil]

yeah I understood what you meant

[carmine fossil]

Ok,like that

[slate forum]

But I don't want to use the relations

I want to see it visually

[latent anvil]

so maybe here's one way: think about a reflection in terms of rotating the line of reflection to the x axis, then reflecting, then rotating back

[prime gale]

Maybe imagine a kaleidoscope.

[carmine fossil]

Dummit and foote had a good way to do this

[latent anvil]

Does that geometrically make sense?

[carmine fossil]

Take a polygon (vertices are 1,2,3...)
Let r be rotation in clockwise direction and s reflection along the line of symmetry passing through 1.

If you apply r once on say a triangle whose vertices are numbered as (1,2,3),the vertices become (3,1,2)

[slate forum]

Yeah I guess that makes sense

[latent anvil]

sorry are you responding to me or Drake?

@slate forum

[slate forum]

Either

[latent anvil]

okay

so if you have two reflections, you can write them as $R_1^{-1} F R_1$ or $R_2^{-1} F R_2$ for rotations $R_1$, $R_2$ and $F$ reflection across the x axis. The product is then $R_1^{-1} F R_1 R_2^{-1} F R_2$. I claim that if $R$ is a rotation, then $FRF = R^{-1}$, so $R_1^{-1} F R_1 R_2^{-1} F R_2 = R_1^{-1}R_2R_1^{-1}R_2$.

Say $R$ is a rotation by angle $\theta$. Then for a point (complex number) $z$, we have $FRFz = \overline{e^{i \theta} \overline{z}} = e^{-i\theta} z = R^{-1} z$, so $FRF = R^{-1}$

[cloud walrus]

shamrock:

[onyx mirage]

The algebra text we got explicates that we can't generally view the polynomial ring R[X] as a subring of map(R,R) as polynomials aren't strictly defined by their values on R (for example X^p - X in F_p would be the 0 function, in spite of not being the 0 polynomial), but how does one then define the identity polynomial?

I take it as just "1" and not as "1 +pX" in spite of having same valued outputs mod p

I thought this to be the case, but then one of the homework questions asked me to find the multiplicative inverse of 1+2r in R[X]* with r in R[X] with R=Z / 4Z

[fair obsidian]

When asking about the identity polynomial do you mean which polynomial is the multiplicative identity?

(Instead of like the identity function. As you said polynomials are formal linear combinations, not just functions)

[onyx mirage]

I think I spotted the kicker

the coefficients are all members of Z/4Z, so (1+2r)(1+2r) would be equivalent to 1 in R[X]*, since 1+4r + 4r^2 == 1 + 0r + 0r^2 (mod 4)

[fair obsidian]

Yeah exactly. The polynomials 1 and 1 + pX are the same. In Z/pZ the coefficients are numbers mod p, so pX is just 0

And yeah that inverse looks good

[onyx mirage]

hype, thanks

bit rusty after the holidays 🤭

[fair obsidian]

No problem, you kinda solve it yourself though

[onyx mirage]

Would all multiplicative inverses still be unique in spite of there being potentially more representations for the same functions?

[fair obsidian]

Multiplicative inverses should always be unique yes, as you define them and can prove their uniqueness purely through the algebra

[prime gale]

Yes. You are talking about commutative ring, right?

[onyx mirage]

not necessarily commutative ones

For rings multiplication isn't necessarily a group operation is the thing, I'm not sure whether that causes anything to break down

[fair obsidian]

Ok it should still be true as long as you remember inverse means two-sided inverse in that case

[onyx mirage]

ah yeah

[fair obsidian]

True but the uniqueness can be proven using the single inverse, you don't need the inverses of any other elements

Actually, if you have a two-sided inverse x to an element y, I believe even a one sided inverse to y must be x

[onyx mirage]

I recall doing the proof for regular groups a year ago, and it'd lead to a contradiction of an element not being equal to itself (assuming it's not the identity) if the inverse wasn't unique

[fair obsidian]

Sounds about right

Although you can also do it directly I think

[onyx mirage]

I think so, another proof we had to do did involve right inverses

and I think the conclusion implied either unicity, or that it's also the left inverse, assuming partial versions of group axioms

[fair obsidian]

I'm not familiar with the term unicity

But yeah assuming that an element x has a left inverse y and right inverse z, you automatically get that they are equal:
y = y(xz) = (yx)z = z
and hence the inverse on any side must be y = z.

But having an inverse on a single side need not guarantee an inverse on the other side exists, or any uniqueness. Thinking about some counterexamples is good practice.

[onyx mirage]

uniqueness

[fair obsidian]

Ah ok

[onyx mirage]

I have to come up with a (potential) counter example of a unit in the same polynomial ring, not of the form 1+2r

[prime gale]

-1?

[onyx mirage]

that'd be r=-1

[prime gale]

I see. I thought your ring was Z/2[r].

Sorry--Z/4[r].

[onyx mirage]

It's (Z / 4Z)[X]

Anything without a constant term should suffice I think

[prime gale]

No way can you have a unit without constant term.

[onyx mirage]

or any polynomial with a coefficient that is not "divisible" by 2

I think you're most likely right, but I'm not sure

[fair obsidian]

And otherwise you have to prove all units are of the form 1 + 2r?

[onyx mirage]

Actually since 4 is not prime, any expression without a constant term can't possibly evaluate to 1 with "2" as input

[fair obsidian]

Yeah you definitely need a constant term

[onyx mirage]

And otherwise you have to prove all units are of the form 1 + 2r?
@fair obsidian yes

[fair obsidian]

Hmm ok

Since multiplying by -1 doesn't do anything that is basically the same as polynomials 1 + 2p for a nonconstant p not sure if this helps

[prime gale]

Think of the map Z/4[x] -> Z/2[x]. What are units in Z/2[x]?

[onyx mirage]

Would be like multiplying 2 binary numbers

Think only 1 can be a unit then?

[prime gale]

Does that tell you anything about units in Z/4[x]?

[onyx mirage]

in Z/2[X]

If something is a unit in Z/4[x] it necessarily has to be a unit in Z /2[x]

🤩 that's the answer I think

big thonk

Though I'm honestly not sure

[prime gale]

Your reasoning looks good to me.

[onyx mirage]

Let me take a bit of a break so I can feed

I have to say I really like algebra

It's a bit of a strange course, because the problems genuinely make you stray off so far from things actually covered in class or in the references, it's what I imagine being a mathematician would be like.

[prime gale]

Don't know. Never been a mathematician. Studied math for a little while, then flunked out.

[onyx mirage]

why so?

[prime gale]

Why did I study math? Inertia, I guess.

[onyx mirage]

and drop you mean

[prime gale]

Too stupid.

[onyx mirage]

It's definitely a tough course

SPJ got cucked out of following maths at cambridge, but have to say his contributions to computer science almost make that seem like a good thing

[prime gale]

Plenty of CS people are math flunkies--e.g. Knuth, Lamport.

[onyx mirage]

Only 1 and 3 in Z/4Z have multiplicative inverses

That means, any evaluation of any polynomial that has to have a multiplicative inverse, has to evaluate to either 1 or 3

which means the functions has to be of the form 1+2r where r is a polynomial in R[X]

[prime gale]

Well, that's true, but I don't follow your reasoning.

Your original argument seems simpler.

[onyx mirage]

It does

I'm essentially reducing the problem: If a polynomial has an inverse polynomial, that means it must evaluate to values that all have an inverse itself. Then we apply standard modular arithmetic knowledge: namely that if a polynomial evaluates to values with an inverse, it necessarily has to evaluate either to 1 or 3 (since they're coprime with 4) of which both classes can be written as 2k+1. The previous argument is very closely related to this

[prime gale]

But how does it follow that any polynomial that evaluates to 1 or 3 has form 1+2r?

[onyx mirage]

what I was missing in the previous argument is that units in Z/2Z and Z/4Z don't necessarily translate to units in Z/2[X] and Z/4[X], but they do if you're talking about evaluations in polynomials rather than polynomials themselves

[prime gale]

Are you wondering why the only unit in Z/2[x] is 1?

[onyx mirage]

No, rather that Z/4[x] doesn't have units that need not be in Z/2[x]

[prime gale]

Is the image of a unit under Z/4[x] -> Z/2[x] always a unit?

[onyx mirage]

It is

[prime gale]

What is preimage of 1 under that map?

[onyx mirage]

Anything of the form 1+2r

[lusty swallow]

hey im probably just a bit stupid but can someone point me to a definition of 'integral closure'? of a ring

having a bit of trouble with the googles due to the last brain cell taking a holiday

[prime gale]

Wikipedia? I guess you mean integral closure of ring in its field of fractions or something like that?

[lusty swallow]

yeah, im kinda confused for some reason. is it just the same thing as 'algebraic element' for fields?

[prime gale]

Well, the integral closure of a field in some bigger ring is an algebraic extension, if that's what you are asking.

Well, if it is a field anyway.

[lusty swallow]

right, i think i get it. think i just need some sleep. thank you!

[chilly ocean]

@solemn rain yo homeboy where did you learn group stuff from

[solemn rain]

onlyfans

[tribal pasture]

For the second part, is there any more nicer description of the Hom set except just R? Like I ask my TA and he said you have to give a "nice" description ):

[solemn rain]

was ur question to the TA

after u told him Hom(A,R) is iso to R?

or b4?

cuz i think thats as nice as it gets

[tribal pasture]

b4

cuz i think thats as nice as it gets
Oh okay thanks!

[prime gale]

@tribal pasture This is really nit picky, but I would say Hom(A,R) = R instead of ~=, because you have constructed a natural transformation from Hom(A,R) to R, not just shown the existence of an isomorphism. I doubt that is what your TA has in mind though.

[tribal pasture]

@prime gale I thought = meant an equality as elements? Is that not the case?

[prime gale]

Yes. It does. But at least in some subjects, we use = to indicate that things are not just isomorphic, but naturally isomorphic.

[tribal pasture]

Oh okay, I see! Thanks!

[prime gale]

In any case, ignoring notation, you have shown more than the fact that the two are isomorphic. You have shown they are naturally isomorphic.

[tribal pasture]

Also by natural transformation, you don't mean the category-theoretic definition of natural transformation right? Instead you mean the canonical right?

[prime gale]

Yes. I mean both. When you say "canonically isomorphic" or "naturally isomorphic", you are really saying that there is a natural transformation, in the category sense. That is how you make the concept of "canonically isomorphic" rigorous.

[tribal pasture]

In that case, do you mind clarifying what are the ambient categories in which the natural transformation is being specified?

[prime gale]

Well, a natural transformation is a map from one functor to another. So I will describe the functors. First, you have category of Rings (commutative with unit). You have category of abelian groups. First functor is Hom(A,.), from Rings to Abelian groups. Second functor is R->R, the forgetful functor from Rings to Abelian groups. Your natural transformation, which takes a morphism, f, to f(X), takes the first functor to the second functor.

[tribal pasture]

Oh okay! Thank you very much. I understand now!

[prime gale]

Technically, you have to check that all the diagrams commute. But with experience, you don't actually do it.

Anyway, this is all very long winded, so in reality, people just say "natural map". This is just how you make it rigorous.

[tribal pasture]

So from going from your previous comment, whenever we talk about a canonical isomoprhism, we mean a natural transformation, or is it the case with any isomorphism?

[prime gale]

When we say canonical map, we mean natural transformation. When we say canonical isomorphism, we mean natural transformation that is invertible.

[tribal pasture]

But the same isn't true for any isomorphism, right?

[prime gale]

No. For example, it is easy to prove that any two one dimensional real vector spaces are isomorphic. But they are not canonically isomorphic.

Just look at fibers of a non trivial line bundle. If you had canonical isomorphisms, it would have to be trivial.

[tribal pasture]

Yeah, I get that part what I am asking is for an isomorphism to be a natural transformation, it must be canonical, right?

[prime gale]

I see. Yes--that is the definition of canonical.

[tribal pasture]

Because I was always told that the "canonical" was just the "intuitively" nice morphism. Kind of a shock that natural transformation formalises that idea xD. Thanks for the info!

[prime gale]

That is the way to learn it. You first get a good feeling of what a natural morphism is. But then you have to make it rigorous. If you started with the rigorous stuff, it would all seem really mysterious and unmotivated if you didn't already know what it is.

[tribal pasture]

True true, yeah, the worry was that the sense conveyed by "canonical" is that it is a subjective choice we make from a whole collection of maps. So maybe not, if it is formalised by the natural transformations.

[chrome hinge]

Im trying to prove that it doesnt exist a subgroup of $A_{4}$ that has grade 6

[cloud walrus]

Maikel:

[chrome hinge]

Any hint?

[prime gale]

Do you know what all the groups of order 6 are?

[chrome hinge]

Not really. I know $D_{4}$

[cloud walrus]

Maikel:

[chrome hinge]

What i was thinking is: such a subgroup would be normal, maybe that just cannot be in A_{4}

But alternate groups are simple for n>=5

[prime gale]

Well, it is probably pretty easy to see there are only two groups of order 6. Z/6 and D_6. Can you eliminate them as subgroups of A_4?

[chrome hinge]

You mean D_4 ?

[prime gale]

No. D_6, dihedral group of order 6.

(By "grade 6" you mean order 6? I assume you do--otherwise, I have no idea.)

[chrome hinge]

Yeah i meant order

I just realized there are two different notations dor dihedrals

D_3 as symmetries of the equilateral triangle (3 sides) and D_6 as the same group since those symmetries are 6

[prime gale]

By D_6, I mean symmetries of equilateral triangle.

[chrome hinge]

Oh, you're right

[prime gale]

Sometimes people call it D_3. But never D_4.

[chrome hinge]

I always meant D_3

sorry for that

[prime gale]

Okay. Do you accept that only groups of order 6 are Z/6 and D_3? Haven't tried to prove it, but I expect it is elementary.

[chrome hinge]

Thats an excelent starting point

Like, i can probably eliminate those as subgroups of A_4

I just didnt know there were only two

Thanks man, i'll give it a try :D

[latent anvil]

I think you can argue this at a higher level

As someone noted earlier, an order 6 subgroup of A4 would be normal

It's also gotta contain a 3 cycle and an element of order 2

hmm no the thing I was thinking didn't work

I think you could get it to work more easily then doing cases on whether the subgroup is S3 or C6 though

[prime gale]

Shamrock--I agree. It is trivial that a subgroup of order 6 in A4 is normal. Any subgroup exactly half the size of a group has to be normal. So that works.

[latent anvil]

i mean, I wasn't able to finish the proof from there

[chrome hinge]

Same

[latent anvil]

I was going to say the conjugacy classes are too big but they aren't

[chrome hinge]

I had already thought of that

[latent anvil]

yeah sorry maikel I was going to elaborate but realized I was wrong

[prime gale]

Well, it is pretty easy to eliminate Z/6 and D_6 as subgroups of A_4.

[chrome hinge]

No problem man, thanks for stepping in and tying to help :D

Yeah am, im going to try that type of argument

Thanks both!

[latent anvil]

oh I have a proof

The conjugacy classes of three cycles are
(123),(142),(134),(243)
and
(234),(143),(124),(132)

(132)(124) = (134) and (123)(142) = (234)

so if a subgroup contains one conjugacy class, it has to contain an element from the other

so any normal subgroup containing a three cycles has to contain all three cycles (and thus be the whole group)

[vestal snow]

Is there a slicker way to do this?

Or do I just compute the order for each of them separately

[prime gale]

Seems like pretty easy computation to do them one at a time.

[vestal snow]

Upper bound yes, but lower bound is gonna be annoying

[prime gale]

Have to be clear though--O(3) or SO(3)?

[vestal snow]

Don't know the difference?

[prime gale]

No--I mean pretty compute size of group of symmetries. Not provide bounds.

O(3) is rotations and reflections. SO(3) is only rotations.

[vestal snow]

O(3)

[prime gale]

Do you want help computing?

[vestal snow]

Sure

Which ones do you want to do?

[prime gale]

Okay. Lets start with tetrahedron. If it is sitting on table, one face is facing down. How many possibilities?

[chilly ocean]

(if it were me, I'd start with a cube, since I can hold a die in my hand and intuit it)

[vestal snow]

Oh I meant you can do one or two of the solids and I could do the rest

[chilly ocean]

tetrahedron would work too

[vestal snow]

and then we can share the solutions

[prime gale]

Okay. Start with cube. If it is sitting on table, one face facing down, how many possibilities?

[vestal snow]

24

Hold on

I think we need to use the algebraic definition of the group of symmetries

In that we need to prove that an isometry is completely determined by what it does to 2 adjacent points of the cube

[prime gale]

Okay. I computed them all. Okay. 24 is right.

[vestal snow]

Yeah the problem is that while its easy to see why 24 is correct, writing it down is

painful

[chilly ocean]

Isn't an isometry also decided entirely on which face and orientation a fixed face gets mapped to?

[prime gale]

Once you fix a face and fix rotation of that face, everything else is determined.

Then multiply by two because of reflections.

[chilly ocean]

right

[prime gale]

What do you get for the others?

[vestal snow]

I think this gave me an idea to do it quicker

Yeah I think I can do it now

Thanks guys

[prime gale]

Great!

Just making sure--24 is answer for tetrahedron, right? Not cube.

[vestal snow]

yes

[slate forum]

@latent anvil or @prime gale As for that little lemma, suppose $\gcd{f,g} = 1$ in $k[x,y]$, so let's break $f=f_1\dots f_n, g=g_1\dots g_m$, where each $f_i,g_j$ are irreducible elements of $k[x,y]$, clearly they are coprime. Suppose $\gcd{f,g}=d$ in $k(x)[y]$, where $d=\frac{d_1}{d_2}$ is not a unit, that is, $d_1$ is not a polynomial in just $x$. So, we have $f=\frac{d_1}{d_2}\frac{h_1}{h_2}, g=\frac{d_1}{d_2}\frac{k_1}{k_2}$, now clear denominators to get $fd_2h_2 = d_1h_1$ and $gd_2k_2=d_1k_1$. Now, every irreducible factor of $d_1$ is necessarily a unit multiple of at least one of both the $f_i$'s and $g_j$'s, and there must be at least one from each that is not a function of just $x$, as otherwise $d_1$ would be a unit. Say $f_i,g_j\mid d_1$. Now, we see that $fd_2h_2 = d_1h_1$, so $g_i\mid d_1$, so $g_i\mid fd_2h_2$. However, $g_j$ does not divide $d_2$ or $h_2$, since these consist only of polynomials in $x$, and since $g_j$ is irreducible, and we're in a UFD, we know that $g_j$ is prime and so $g_j \mid f$, so necessarily $g_j=f_r$ for some $r$, but this means that $\gcd{f,g}\neq 1$, which is a contradiction.

[cloud walrus]

HelixKirby:

[latent anvil]

the bit about gj = fr is wrong

they might end up being unit multiples of one another

[prime gale]

I am not so smart--you have d_1 divides f* d_2*h_2. d_1 doesn't divide d_2. but why can't d_1 divide h_2 instead of f?

[slate forum]

That's fine, though, as the gcd still isn't 1

right/

?

[latent anvil]

yeah I think this mostly works

[prime gale]

Maybe I am interpreting your statement wrong. It looks to me like you are saying each factor of d_1 is one of the f_i.

[slate forum]

Each irreducible factor is one of the f_i or g_j

[prime gale]

I see. I misinterpreted your statement.

I would get rid of the word "both" to make it more clear.

[latent anvil]

Actually I'm confused h part of your proof. how do you get that "every irreducible factor of d1 is necessarily a unit multiple of at least from each of the fis and gjs"?

I agree this is the case for those irreducible factors which aren't elements of k[x]

Because d2 h2 is in k[x] and so are its irreducible factors

[slate forum]

hmm?

the irreducible factors of d_1 better be some f_i or g_j

or unit multiples of them

[latent anvil]

why can't they be irreducible factors of d_2 or h_2?

[slate forum]

uhh

Say I wrote it in simplest form

so that can't happen

[latent anvil]

That works for d2, but what about h2?

[slate forum]

iunno

[latent anvil]

(or k2)

[prime gale]

Shamrock's way of saying it is more clear to me.

[latent anvil]

You need to handle the irreducible factors in k[x] separately basically

basically like, take a gcd of f and g in k(x)[y]

We can multiply by units and it's still a gcd

Multiply by a unit to clear denominators and get it in k[x, y]

Then divide out by all irreducible factors in k[x]

then your proof goes through and you don't even need to consider d2

[slate forum]

divide out? Like cancel?

[latent anvil]

yeah pretty much

like if your gcd is x(yx+1)y^2

You can multiply by the unit 1/x and still get a gcd of f and g

but now it's got no irreducible factors in k[x]

[slate forum]

ok

[vocal depot]

I have a really, really, dumbass question for anyone who would be willing to humor me.

[prime gale]

Here is a question I came up with. Someone asked me why does a ring have two binary operations? Why not three? So I thought--does it really have two? Is one of them a consequence of the other? For example, given the additive group of integers, and chosen multiplicative identity 1, how many multiplicative structures are possible? How about for Q, R or a finite group? Or, given multiplicative structure and zero, how many additive structures are compatible? All of these are easy to answer in a couple of minutes.

[latent anvil]

@vocal depot what is it?

@prime gale ||C vs R have the same additive group, as do all number fields of the same degree, right?||

[vocal depot]

I'm in the middle of a proof where I'm using induction.

[prime gale]

Shamrock--you have answered one of the cases.

[latent anvil]

Oh I missed the other question

[vocal depot]

I'm at the point where I have assumed $\bigcup\limits_{i=2}^{k} \langle a^{i}\rangle = \langle a^{2}\rangle\cup\langle a^{3}\rangle\cup\ldots\langle a^{k}\rangle$ for some integer $k\in\mathbb{Z}$ is true.

Wow that failed horribly.

[fair obsidian]

@prime gale || if I recall correctly a finite abelian group completely determines it's compatible multiplications up to isomorphism iff it is cyclic||

[cloud walrus]

HisMajestytheSquid:

[vocal depot]

In the iduction step I am showing that $\bigcup\limits_{i=2}^{k+1} \langle a^{i}\rangle = \langle a^{2}\rangle\cup\langle a^{3}\rangle\cup\ldots\langle a^{k+1}\rangle$

[cloud walrus]

HisMajestytheSquid:

[vocal depot]

My question is whether or not it is valid or legal by proof standards to say everything up to K is what I had up to the previous step and then leave $\langle a\rangle$ as the portion that's left over. In other words write it as $\bigcup\limits_{i=2}^{k} \langle a^{i}\rangle\cup\langle a\rangle$.

[cloud walrus]

HisMajestytheSquid:

[latent anvil]

I'm a little confused, isn't the equality you're showing definitional?

[vocal depot]

Meaning?

[prime gale]

Well, such a definition wouldn't work for infinite unions. So probably there is another definition.

[latent anvil]

I don't understand what there is to prove about $\bigcup\limits_{i=2}^{k} \langle a^{i}\rangle = \langle a^{2}\rangle\cup\langle a^{3}\rangle\cup\ldots\langle a^{k}\rangle$

[cloud walrus]

shamrock:

[latent anvil]

How have you defined both sides?

[vocal depot]

It would probably be helpful to have the whole problem but ultimately I am trying to show that a finite cyclic group cannot be represented by a union of its proper subgroups.

And I'm sort of at a loss so I am attempting to prove it by contradiction using induction.

[prime gale]

Can you just say a generator can't be in any proper subgroup?

[next obsidian]

^^

[vocal depot]

I suppose that would be a more concise way to do that. But I'm a little inexperienced so I'm not exactly sure how to show that mathematically.

[next obsidian]

let g be a generator for a cyclic group of order n

then |g| = n

any subgroup containing g has g, g^2, ..., g^{n-1}

[prime gale]

What's wrong with "A generator can't be in any proper subgroup."?

[next obsidian]

all those are distinct

so the subgroup has at least n things in it, i.e. is G

[vocal depot]

So there would have to be something missing for it to be a proper subgroup, correct?

[latent anvil]

yes

[vocal depot]

Can I just pick any arbitrary element in the set the generator produces or is there a specific one I'm looking for?

[fair obsidian]

You should argue that if g is a generator, any proper subgroup can't contain g itself

[vocal depot]

That's the one I was thinking, my alternative was g^{n-1}.

[fair obsidian]

Yeah either works, but the point is that they are generators

Meaning any subgroup containing them must contain all of G, and hence can't be a proper subgroup

[vocal depot]

That makes sense. Thanks for humoring me, folks. This was really helpful.

[fair obsidian]

Glad I could help

@prime gale I believe Z and Q also have unique multiplicative structures, and possibly groups isomorphic to subrings of Q? But I would have to think about those for a bit

And I haven't been able to think of any other infinite groups whose addition completely determines the multiplication

[prime gale]

@fair obsidian I agree about Z and Q. Haven't thought of all those other cases you mention.

[vestal snow]

Quick question, for solids, the generators of the group of symmetries would be rotation in xy plane, rotation in xz plane, and reflection along the x axis right?

[prime gale]

Not enough dimensions. O(3) is three dimensional, right? You only have two.

[vestal snow]

O(2) is 2 dimensional, yet we only need one rotation?

[prime gale]

O(2) is one dimensional.

That's pretty funny. I think you generate a dense subset with your guys. But not everything.

[vestal snow]

So does it have 4 generators?

Oh wait hold on

I meant the symmetries of a fixed solid

Like a cube

[prime gale]

No--I take it all back. I think your set does generate.

[vestal snow]

Oh okay cool

[prime gale]

Actually, every element of O(3) is (rotation about y axis) * (rotation about x axis) * (rotation about y axis) * (reflection^{0 or 1}).

Here is a cute problem. Show O(n) is generated by rotations about n-1 axises and a reflection. But it can't be generated by fewer than rotations about n/2 axises and a reflection. That's a pretty big gap. I am sure people know the exact number.

(Axis = codimension 2 subspace of R^n).

[chrome hinge]

Hey @prime gale am thanks for giving me the useful fact about order 6 groups. I was wondering, is there a general procedure to know how many groups of a determined order there are?

At least in small orders

[prime gale]

Well, for prime p, there is one. For 2*p, there are two. Probably stuff known about stuff like p * q. And some other special cases like that. Don't really remember.

Easy to look up anyway.

In general though, it is one of the hardest solved problems in mathematics.

[chrome hinge]

Yeah i found like a group-wiki on google

Oh i knew prime order groups were cyclic. Now that i think about it, cyclic groups of the same order are always isomorphic?

[smoky cypress]

Yeah

[chrome hinge]

Niceee

Im just starting group theory its really fun

[smoky cypress]

Actually you can show that all groups of order 15 are cyclic

[prime gale]

Group theory gets really weird really fast.

[smoky cypress]

Like all math

[prime gale]

Well, some subjects are more intuitive than others. Group theory isn't one of the intuitive ones.

[smoky cypress]

Here's an interesting fact that generalizes the fact that all groups of order 15 are cyclic

[chrome hinge]

Nice fact

[smoky cypress]

and yeah

like am said

this fact isn't very intuitive whatsoever

[chrome hinge]

Quite specific

[prime gale]

Didn't know that.

[smoky cypress]

the assumption q≠1 (mod p) tells you that you should probably use sylow theorems

Oh yeah also sylow theorems: they're useful and beautiful, but they aren't really intuitive

[prime gale]

I never properly studied elementary group theory.

[smoky cypress]

I don't think the fact I sent above is something that's usually presented in an elementary group theory course

[prime gale]

I studied one of the most intuitive subjects. It is so intuitive that for centuries, it was wrong.

[chrome hinge]

Probably not lol

Which one?

[carmine fossil]

Group theory?

[prime gale]

Group theory hasn't even been around for centuries.

[carmine fossil]

Number theory?

[prime gale]

So it couldn't have been wrong for centuries. Don't think it was wrong for any time at all, in fact. I studied AG before I flunked out. AG was wrong until about mid twentieth century.

[carmine fossil]

So,what is the intuitive one?

[prime gale]

AG is super intuitive.

[chrome hinge]

Sorry, AG?

[prime gale]

Algebraic Geometry.

[chrome hinge]

Ooh

Yeah ive heard few of that

[bleak abyss]

The Italians' incorrect theorems :0

(Jk I'm guessing it's more a moral schemes > varieties comment)

[prime gale]

Mostly the results were right, but the proofs were wrong. Then they ended up with some contradictions and things got bad. Then in the twentieth century, some guys fixed it up entirely.

[latent anvil]

ugh I was gonna type gabe premoment but he was too fast

[bleak abyss]

Oh you actually meant the Italians pulling a Weibel lol

I made that as a joke

I should refer to all incorrect stuff as pulling a Weibel tbh

[sullen tinsel]

quick question y'all, its not necessarily a problem but how do you express doing an operation a certain amount of times. For example, if I want to perform (4*5) 5 times. Or a better example would be how to calculate a percent x amount of times. i:e (10 x 1.15) and do that 3 times.

would that be to the ^x amount of times?

[bleak abyss]

I think you meant to have this in #prealg-and-algebra or something but idk I'd either use words or define a function

Like f(x) = 5x, f^3(4)

[sullen tinsel]

@bleak abyss is that f to the power of 3(4)?

[latent anvil]

it means "apply f 3 times to 4"

[bleak abyss]

f^2(x) = f(f(x))

Oh I get it, yeah to the power

[cloud walrus]

Daminark:

[bleak abyss]

You can use latex here

[sullen tinsel]

So lets say I wanted to figure out how many times i need to get a 15% return of $10 to reach $1000.

[cloud walrus]

Taziamoma:

[bleak abyss]

Also let me re-emphasize you want #prealg-and-algebra

[sullen tinsel]

i haven't been in school for so long i don't know if its anything 🤣 i just know it's some sort of algebra. But I'll post it there and see what tonight

[bleak abyss]

Yeah this is more like, late undergrad to early grad level stuff

Group theory, ring theory, Galois theory, etc

[sullen tinsel]

ahhhh makes sense

i thought it was for things that didn't quit fit in other algebra

[bleak abyss]

Nah it's a lot more advanced

In general the "Advanced Mathematics" channels aren't meant for people still doing high school or early division undergrad math

[prime gale]

Though I feel like I was lying when I typed ",iam adv". I am a total beginner.

[sullen tinsel]

ah i'm about to graduate college and I derived some formulas and every now and then i'll test it out by having people solve a question and depending on how easy it was for people to solve it, it'll tell me if i should've done more research before deriving a "new" formula that may have already existed

[bleak abyss]

I mean advanced here doesn't mean you have to be a world-renowned mathematician like myself, it's just a matter of

Do you have reason to be talking in these channels?

[sullen tinsel]

🤣 "like myself". You mean this one specifically or advanced in general?

[scarlet estuary]

no one has a reason to talk about math on discord

[latent anvil]

hmmmm

[scarlet estuary]

dont kid yourself

[latent anvil]

Someone appears to be impersonating me

@mods

what a sham

[scarlet estuary]

brb making a twin fantasy cover art but with bearded frogs

[bleak abyss]

Oh I'm one of the best mathematicians in the world

[next obsidian]

True

[scarlet estuary]

Toads of de Nile

[next obsidian]

Dami is almost 1/3rd of a mathematician I am

[bleak abyss]

Only Serre comes close

Oh guys so funny thing

In my AG class at one point my prof was like

Hey everyone here knows who Grothendieck is right?

[prime gale]

I want to delay brain rot as long as I can. Since I flunked out and became a hikikomori, if I am not careful, I will succumb to brain rot. Thinking about this stuff hopefully will delay the rot.

[bleak abyss]

And I was like hmm he was Serre's collaborator?

[next obsidian]

lol\

[bleak abyss]

And I feel like he had this momentary what the fuck before he realized I was joking

He was like "Well he's much more than that... oh it's a joke, that's funny"

[next obsidian]

You should've said "hey that's that guy who made homological algebra right?"

[bleak abyss]

"Oh he's Deligne's advisor!!! That guy yeah"

[next obsidian]

I think he wrote something in some tohoku journal?

[latent anvil]

He was an anti Vietnam War activist iirc?

Not sure why it came up in your AG class

something something indoctrination of ths youth

[next obsidian]

[latent anvil]

lmao

[bleak abyss]

Honestly the worse version of this is Atiyah

[latent anvil]

[prime gale]

He drew children's pictures.

[latent anvil]

Wasn't atiyah a singer?

[bleak abyss]

"Oh didn't he have that false proof of Feit-Thompson"

Oh Shamrock you win

That's it

Math is over

[latent anvil]

He had some kind of theorem from group theory about indices?

[scarlet estuary]

i had someone hit me with

"Who's Lawvere? Oh wait, he was Bunge's advisor, wasn't he?"

who the fuck knows marta bunge without knowing lawvere

and this wasnt a joke, as far as i could tell

100% serious

[bleak abyss]

Not even sure who Bunge is lol

Or Lawvere

[scarlet estuary]

bunge is some random category theorist that has done some good work and all (especially as she was a female mathematician in a time where the academic world was not very friendly to women) but overall not super notable

lawvere basically made half of category theory

the modern-ish definition of the topos is due to lawvere

(he took grothendieck's bad definition and made it not bad)

lawvere is also why Dalhousie of all fucking places had a prestigious category theory rapport

in the 70s

it is this project that led to the development of most of our understanding of enriched categories, especially in topological contexts

i think he's doing some weird categorical physics now

still chugging along at 83

[prime gale]

QTFT? TQFT?

[tiny pagoda]

TQFT functor

?

[scarlet estuary]

nah synthetic diff geo

[tiny pagoda]

Ahh

[tribal pasture]

lawvere basically made half of category theory
Isnt he also the dude who categorified Hegelian logic?

[vestal snow]

Say you are given a solid in three dimensions. Can one obtain the rotation along the xy plane using some combination of rotations along the xz and yz plane?

Can you get roatation by 90 degrees in xy as some combination of rotation in 90 degrees in yz and rotation by 90 degrees in zx

[steady axle]

@oblique river I was going over the proof once again and I found that I do not in fact understand the proof well enough yet.

$\overline \sigma \smile \chi \in \hat H^{-1} (G, Q/Z)$ but $\delta (\overline \sigma ) \smile \delta ^{-1} (\chi) \in \hat H^{-1}(G, I_G \otimes Hom_Z(I_G, Q/Z)) $ so what exactly do we mean by first the equality there? There is natural map, as you mentioned, $\hat H^{-1}(G, I_G \otimes Hom_Z(I_G, Q/Z)) \rightarrow \hat H^{-1} (G, Q/Z)$ and if we say that it's the image under this natural map that is equal to $\overline \sigma \smile \chi $ then we have to prove that right?

[cloud walrus]

bert:

[steady axle]

for reference - https://discordapp.com/channels/268882317391429632/496784958430380033/752600803964878858

I think somehow I should use uniqueness of cup products but I cant exactly write down how

[oblique river]

@steady axle the image of the cup product of two classes in H^i(G, A) and H^j(G, B) lies in H^(i+j)(G, A \otimes B). If there is a map from A \otimes B to C then you can define the cup product as a cocycle taking values in C using that map

whereever the equality sigma cup chi = delta(sigma) cup delta^(-1)(chi) was proven

I would guess this gets discussed there

[steady axle]

@steady axle the image of the cup product of two classes in H^i(G, A) and H^j(G, B) lies in H^(i+j)(G, A \otimes B). If there is a map from A \otimes B to C then you can define the cup product as a cocycle taking values in C using that map
ya this map is clear to me. i dont think the equality was proven before but i'll check. I think I should understand the whole thing once I understand this equality.

Is the equality in the sense I think it is?

[oblique river]

hmm I'm not sure exactly what you mean by that, but I would think that the answer would come somewhere before if there was ever a discussion about delta and cup products

[steady axle]

i think i will have to read appendix on 'computations of cup products' from Serre's local fields
Serre does it by considering 0->I_G -> Z[G] -> Z ->0 tensor it with Q/Z and then observe that for this new exact sequence delta : H^{-1} -> H^0 is isom. So he shows that image under delta of both sigma cup chi and chi of sigma is same

btw the screenshot I uploaded was from notes on cft by a prof named Sharifi from ucla https://www.math.ucla.edu/~sharifi/algnum.pdf

[leaden finch]

Can someone help me

i got stuck at this part for gcd(a'b)=d

[opal vale]

In question 1, I think orbit will be positive line parallel to x axis and passing through (x0,y0).

Am I right?

[oblique river]

what do you mean by "positive line"?

I think you're on the right track but I'm just not sure what you mean exactly by "positive line"

if (x0, y0) = (1,1) could you write it in set notation? what about if (x0, y0) = (-1, 3)?

@opal vale

I'm going to bed now but I think you're on the right track, just try to make "positive line" more clear and make sure it's clear what the difference is between the case when x0 > 0 and x0 < 0 (or x0 = 0, in which case "positive line" doesn't really make sense)

[opal vale]

If (x_0,y_0)=(1,1) , then i mean to say by " positive line", line passing through (1,1) parallel to x axis, and lie on 1st quadrant.

@oblique river

${(\lambda,1):\lambda >0}$

[cloud walrus]

Manas:

[deft plume]

@leaden finch only thing I could think of is Bézouts identity

If d = gcd(a,b) then ax + by = d

Then maybe apply the fact that b|c

And if a|b and b|c, there’s a proposition I remember stating this implies a|c

So from bezouts identity, we can take the product with c from both sides:

cd = c(ax + by) = cax + cby

And I think with the givens, you might be able to take it from there

[oblique river]

it's actually simpler than that. if a divides b, what is the gcd of a and b?

[solemn rain]

suppose GL_n(F) is isomorphic to GL_m(K). Then the fields F,K are isomorphic and m=n.

a theorem i saw on fb

is there an easy proof for this?

should i try it my self or does it involve heavy stuff

i really hope it doesnt involve heavy stuff as i want to try it

i promise i will think for it for atleast 10 mins

[carmine fossil]

Try using something like all invertible matrices are product of elementary matrices

[solemn rain]

so it is doable

for someone at my lvl?

[carmine fossil]

Don't know

[solemn rain]

can u take a stab at it

[carmine fossil]

Try showing the only generators with order 2 are the swapping matrices and the scaling one by -1(except for char 2)

Seems promising

[bleak abyss]

The center of GL_n(F) is F^{\times}

Not sure if this is relevant but it might be

[opal vale]

@oblique river am i right?

[oblique river]

yes for (x0, y0) = (1,1)

what if x0 is negative?

or what if x0 is 0?

I'm not sure if "positive line" makes sense for those cases

[opal vale]

Then can you give some hint

@Buncho Bananas

[oblique river]

I don't think you need a hint yet

I asked you before -- if (x0, y0) = (-1, 3), what will the orbit look like

write it out as a set

or draw a picture

[opal vale]

@oblique river
I think it will be ${(-\lambda,3): \lambda >0}$

[cloud walrus]

Manas:

[oblique river]

great, that is correct

I hope you can see now why I took issue with the phrase "positive line"

because if you draw that out graphically, it kind of looks like the negative half of a line

unless you define "positive line through a negative point" to be this

all I was saying was that I think you need to be more precise than just saying "positive line".

[opal vale]

Ok

[prime gale]

@solemn rain Do you mean that GL(n,F) and GL(m,K) are isomorphic as plain groups? If you take n=1, m=1, then it is the statement that the multiplicative groups of F and K are isomorphic. There are examples of non isomorphic fields with isomorphic multiplicative groups.

[prime gale]

So I don't think someone at your level will be able to prove it.

[stoic rose]

Since it is false as stated I don't think anyone will be able to prove it 😛

[prime gale]

I think plenty of freshmen will be able to prove it.

[stoic rose]

I meant a valid proof

[chrome hinge]

Hi there

Im trying to prove that (Q,+) isnt finitely generated

Could anyone help me a bit with the intuition to why that statement is true?

I'm not seeing it rn

[prime gale]

Well, if you have a finite subset of Q, multiply all the denominators together. No way will you get something with a bigger denominator.

I mean that you won't be able to generate stuff with a bigger denominator that the product of all those denominators.

[chrome hinge]

Hmm but what if the generator of the finite subset, lets call it S, is infinite?

I mean, it is actually infinite right?

[prime gale]

Huh? I guess that is way too advanced for me.

[chrome hinge]

Every non trivial subgroup of Q is infinite, am i wrong?

[prime gale]

That is true.

[chrome hinge]

Just consider any element of the subgroup. Its generator is infinite

Nice

[prime gale]

You mean that any element has infinite order? That's fine.

[chrome hinge]

Therefore any subgroup

[prime gale]

Anyway, finitely generated means that a finite set, call it S, generates. If s={a_i/b_i} then 1/((product b_i)+1) is not in subgroup generated by S.

[chrome hinge]

Hmm strange

The definition i have is that the generated by S is the intersection of all subgroups containing S

i.e the smallest subgroup containing S

[prime gale]

I agree with your definition of subgroup generated by S.

[chrome hinge]

Great

[prime gale]

Let T={z/(product b_i) | z in Z}. T is a subgroup containing S. But T does not contain 1/((product b_i)+1).

[chrome hinge]

Then finitely generated means there is a finite set S such that the subgroup generated by S is G

Ooh i get it

[prime gale]

Not claiming S generates T though.

[chrome hinge]

So you found a strict subgroup of Q which contains S

Right, the generated by S is contained in T

[prime gale]

Sounds like you understand.

[chrome hinge]

Are you laughing at me?

[prime gale]

I am sorry if I came across as laughing at you. I am not.

[chrome hinge]

😂 alright

So thanks

[prime gale]

Everybody at every level struggles to understand stuff. If you aren't struggling, you aren't studying the right stuff. So everybody knows what it is like to struggle to understand stuff.

[chrome hinge]

I had been thinking i should find a finite subgroup containing S

I had been thinking i should find a finite subgroup containing S
@chrome hinge Didnt think it could take a weaker thing to prove

Everybody at every level struggles to understand stuff. If you aren't struggling, you aren't studying the right stuff. So everybody knows what it is like to struggle to understand stuff.
@prime gale Indeed. Thanks

I'll write a proof now

[solemn rain]

So I don't think someone at your level will be able to prove it.
@prime gale does it use higher stuff?

what would the prove use

l;ike prereqs

[prime gale]

I think it is false. That's why someone at your level won't be able to prove it. Though I have proven plenty of false things.

[solemn rain]

can u give a counterxample for me?

a basic 1

[latent anvil]

What's the claim @solemn rain?

[solemn rain]

suppose GL_n(F) is isomorphic to GL_m(K). Then the fields F,K are isomorphic and m=n.
[6:27 PM]
a theorem i saw on fb

@latent anvil

[latent anvil]

this doesn't feel super false to me

it is weird though

Oh no I see the issue am is pointing out

the n = m = 1 case says that if the multiplicative groups of two fields are isomorphic, those fields are isomorphic

Which does seem false

[solemn rain]

is GL(F) the multiplicative group of F

why

[latent anvil]

it is when n = 1

[solemn rain]

yea whats n=1

like 1x1 matrices