groups-rings-fields

[prime sundial]

i was watching some lectures related to this, i think the point is to derive a bilinear form on a serfeit surface

or the matrix representation of one

but anyways, the exercise was a good one!

helped me process some of the polynomial stuff you were telling me about

[topaz solar]

Ok yep I see it

$\pdv{g} g^n = n+g \pdv{g} g^{n-1} = \frac{g^n-1}{g-1}$ ?

[cloud walrus]

Dragonslayer Sharp

[prime sundial]

i see what you're doing, im not 100% sure

[topaz solar]

Which is uhh $\sum^{n-1}_i g^i$

[cloud walrus]

Dragonslayer Sharp

[prime sundial]

so you did D(g)t(g^(n-1)) = n?

or well

with partials

[topaz solar]

Yeah I think

Well no that should be 1+wtv

[prime sundial]

hm im not quick enough to see it but im sure what you're doing is correct haha

[topaz solar]

Not n+wtv

[prime sundial]

ok yes

i agree then

[topaz solar]

So this is like x^n -> (x^n-1)/(x-1)

an interesting derivative

[prime sundial]

im excited to see what it brings, but the pages coming after all this are loaded with info

im going to give it one more good hour of reading and then im calling it for the day

[topaz solar]

Can’t exactly get the usual polynomial derivative out of this I think

Since that t(v) term

So this is definitely a more specialized thing

[topaz solar]

Btw relate this to the root of unity thing earlier

[distant summit]

(Sorry if interrupting)

What is the right answer here (no mark scheme unfortunately)?

First is equivalent to saying there is a solution to $ab-1=-(pq)k$ where $k$ is an integer. This is obviously the same as saying $ab+(pq)k=1$. Since $a$ is coprime with $pq$, there must be a solution to that equation.

Second, you can just do a table of $x$ and $x^3 \pmod 8$. The odd values of $x$ will return odd values of $x^3$, so this will be odd in the integer mod 8. Then $2^3=8\equiv 0$ and likewise for 4, from which it can be determined that the same holds for 6 as well. Hence there is no $x$ for which $(x^3-4)\equiv 0 \pmod 8$ has a solution.

Third, we have $d$ as the least integer of the form $ax+(a+6)y$. This can be re-written as $a(x+y)+6y=d$ or $ax'+6y=d$. So $\gcd(a,a+6)=\gcd(a, 6)$, meaning whatever $d$ is, it is necessary (but not sufficient) for it to be a divisor of 6. Hence the given set is correct.

Fourth, $p\mid a^5$ means $a^5=pk$ where $k$ is an integer. Since $p$ is not composite, we can't "distribute" it five ways, e.g. $a=p^{1/5}$, as this would not be an integer (or rational for that matter). So we actually have $p \mid a$, which of course does imply that $p^5 \mid a^5$.

So I would say that none of them are false.

[cloud walrus]

Douglas

[topaz solar]

Yeah

[distant summit]

Not that it's something I'm struggling with, but the division questions always make me think of prime factorisation, so I can see why NTists are so interested in primes

Like I can see why prime factorisation is useful

You can factorise numbers in various ways assuming there are prime factors with multiplicities greater than 1, but stuff like this makes me think prime factorisation is the "privileged" factorisation or an objectively more helpful one

[crystal turtle]

Wait until you learn that it generalized to a broad class of rings (Unique factorization domains)

Also yeah they're all true

[distant summit]

Got another one.

1. $g^d=e$ obviously and $g^{d^2}=g^{(d^2)}=(g^d)^d$, so the latter is $e^d=e$.

2. Prime factorisations of 15 and 21 are $3 \cdot 5$ and $3\cdot 7$. Only common factor is 3 so this must be the order. (I'm not sure on the reasoning here. Does this have something to do with $21\equiv 15 \pmod m$ having its least (non-trivial) solution when $m=3$?)

3. True by definition. Order is least that returns g to e, so all integers less than order must give something other than g.

4. $ord(g^n)=lcm(n,17)/n$. Since 17 is prime, the lcm is $17n$, giving $ord(g^n)=17$.

All appear to be true.

[cloud walrus]

Douglas

[distant summit]

also idk how to do \DeclareMathOperator in texit but that wld be good to know

[topaz solar]

I think so yes

2 I’m not sure your relating to the “least (non-trivial) solution” is right

When m > 3, then can that be true?

[distant summit]

this is a good point

thinking abt it

no

bc mod 3 "get rid of" the 3 in the prime factorisation of each

but for mod 6 there is no factor of 6

etc.

and after mod 7 then obvs 21≠15

so actually mod 2 and mod 3 are the only ones

and mod 1 ig but thats a bit weird to use

[hollow tartan]

I was watching a modular forms lecture by don zagier and I was excited to see it's a group action under invariants, I have a special love for group actions b/c it explains all of group theory , which also reminded me of mobius transformations! That's a remarkable combination of topics, it's so rich that I wasn't sure which section I should put this. Also this defined a moduli space of elliptic curves, which again I can somewhat understand. I can't imagine how many group actions are out there to discover, lastly how does one generate the moduli space of all types of smooth algebraic varieties? I suppose this a like classification of finite groups...

[molten viper]

A problem I just thought up, are there any non-trivial examples of rings in which every ideal is prime?

(the trivial examples being fields and the 0 ring)

[hollow tartan]

I thought the fact that it has prime ideals already answers the exercise by itself. I thought of something similar what property or invariant does the ring has to have in order to determine the number of primary ideals. I think this is similar to asking how many combination of divisors a number has?

[next obsidian]

Impossible for eg Noetherian local rings by Nakayama’s lemma since then I^2 = I for every ideal

[molten viper]

I'm not sure why that means it's impossible

[next obsidian]

This forces I = 0

By Nakayama

[molten viper]

gotcha

hmmmm

[next obsidian]

I feel like via localization this makes it impossible for all Noetherian rings but there’s one thing I’m not sure about

[molten viper]

dumb it down for me

[next obsidian]

Dumb what down

[molten viper]

Like, what's localization

[next obsidian]

Umm

Look it up I guess? I don’t really want to get into explaining localization in general

But you can turn any ring into a local one by inverting stuff

To make a prime a maximal ideal of the ring

And you can test if a module is 0 by seeing if it’s 0 after localizing at every prime (or even every maximal one)

So you can usually reduce a problem from an arbitrary ring to a local one by checking it “at every prime”

[molten viper]

interesting

so

Just off of a cursory thing

think

we need such a ring to have no zero disvisors

[next obsidian]

Yes

[coarse kestrel]

or at least the thing your localizing is not a zero divisor.

[hollow tartan]

This is how I see it, suppose every I is prime, but then it would always be forming another non prime ideal, assuming some extended module

[next obsidian]

@molten viper I don’t think you can understand the proof, but I’ll write it or regardless

I mean, I’m gonna use a fact

Which you probably don’t know

[molten viper]

(what are you proving?)

[next obsidian]

It’s impossible for a Noetherian ring

[molten viper]

ooooo

[next obsidian]

So I handled the Noetherian case, and your observation it’s a domain is important

[hollow tartan]

Exactly, because there's always another ring contained

[next obsidian]

The set Supp(I) is the set of primes p where I_p ≠ 0

[molten viper]

I feel like there's not many rings which are domains and not Noetherian

[next obsidian]

Supp(I) = V(Ann(I)), the primes containing the annihilator of I

[molten viper]

What's I_p refer to here?

[next obsidian]

Ann(I) must be 0 because there’s no zero divisors

The localization at p

So what you end up seeing is that I_p must be nonzero for every prime

But for any prime containing I, the proof for local rings shows that I_p is 0

So this is a contradiction unless either I = (1) or 0

So the only ideals are (1) and 0 so you’re a field or 0

[molten viper]

interesting

I believe you even if idk what exactly is going on

[hollow tartan]

I think you should look at my intuition first, and then see his proof. sometimes the geometric intuition is what gives you a sense of direction.

[next obsidian]

I don’t understand what you are saying

To be honest

[molten viper]

So, what happens if R is a PID

So, every ideal is prime, and is generated by one element

(I'mma call it p for obvious reasons)

[next obsidian]

A PID is Noetherian

So it can’t happen

[molten viper]

TIL

Ok

[next obsidian]

I mean Noetherian is fg ideals

PID is every ideal is principal

[molten viper]

oh duh

[hollow tartan]

All I'm saying is that if every ideal is prime, it would always be forming another module with a third prime ideal, forming another non prime ideal of the first two prime ideals

[next obsidian]

What?

[molten viper]

so, we need a ring which is a domain, and it has to have some infinitely generated ideal

yknow

My mind turns to the cyclotomic polynomials

those are irreducible in Z[x] right?

[hollow tartan]

it has equivalence classes up to irreducibility

[next obsidian]

Z[x] is Noetherian

[molten viper]

TIL!

[crystal turtle]

Hilbert's basis theorem gives that it's Noetherian iirc

[next obsidian]

To find non-Noetherian rings you have to construct huge rings

[crystal turtle]

(chapter like 6 or 7 in AM?)

[next obsidian]

It’s that deep?

That’s absurd

[molten viper]

I've heard of hilbert's basis theorem

[next obsidian]

It says a polynomial ring over a Noeth ring is Noeth

[molten viper]

If R is Noetherian than R[x] is too right?

yeah

and Z is noetherian

[hollow tartan]

I'm not sure if it was the daughter or the father who came up with that lemma lol

[molten viper]

For a ring to not be noetherian, it needs to not only be infinite, but very infinite, you say?

[next obsidian]

What lemma….

[hollow tartan]

it needs to be finite

[next obsidian]

In some sense yes

[hollow tartan]

R noetherian implies R[x] noetherian

[next obsidian]

That’s not a lemma it’s a theorem

And that’s Hilbert

Not Noether

There’s no daughter

[molten viper]

what about something weird like Q[\pi]?

[void cosmos]

construct a ring that doesnt have IBM now

[next obsidian]

That’s super Noetherian

[molten viper]

Makes sense yeah

[next obsidian]

It’s finitely generated over a Noetherian ring so Noetherian

[molten viper]

I was thinking it might be useful because transcendental

[next obsidian]

That’s isomorphic to Q[x]

Because pi is transcendental

[molten viper]

that makes sense

[next obsidian]

But the infiniteness isn’t about cardinality. You have Noetherian rings of arbitrary cardinality eg fields

[coarse kestrel]

the trivial example is always something like Q[x1, x2, x3,....]

[next obsidian]

The “hugeness” is because you usually make them by appending a shit ton of crap to a ring

Like in the above example

[molten viper]

hmmmmm

I feel like I need some particular example of a non-noetherian ring

[next obsidian]

They gave you one

Adjoin infinitely many variables to any ring

[molten viper]

Oh I misunderstoof

well, sadly that ring doesn't have our property 😔

[coarse kestrel]

https://math.stackexchange.com/questions/1820304/what-is-an-easy-example-of-non-noetherian-domain

[topaz solar]

Or infinite products I think, but that’s not an integral domain

[next obsidian]

I really don’t think you’ll construct an example

It would be so hard to construct one

If you want to find an answer you’d probably only have a shot at trying to prove it for non-Noetherian rings

But I have no idea how you’d approach it

[hollow tartan]

so don't tell me artinian rings wasn't proved by artin

[next obsidian]

Maybe via topology of Spec

How do you prove Artinian rings?

[molten viper]

Sadly I know nothing about topology

[next obsidian]

That’s like, a definition

I think Artin defined them tho

The father

[molten viper]

Yknow

crazy idea

What about something weird like Q[x1,x2....]/(x1^2,x2^2...)

[next obsidian]

That’s not a domain

[molten viper]

duh

yeah hm

[hollow tartan]

wouldn't that be a point?

[next obsidian]

Yes

A non-Noetherian one

[hollow tartan]

maybe trivially noetherian lol

[next obsidian]

It isn’t Noetherian lol

The maximal ideal is still not fg

I really don’t think you’ll make an example

[molten viper]

yeah probably not

[next obsidian]

It has to have the property that I^2 = I for all ideals

I don’t know a single example of a ring like this that isn’t 0 or a field

[molten viper]

hrmmm

[topaz solar]

Can you get some wacky Boolean ring thing?

[next obsidian]

It won’t be a domain

[molten viper]

well, at least we can cross polynomial anything off the list

[topaz solar]

Dang right

[molten viper]

Can you walk me through how Nakayama's implies this?

[next obsidian]

This isn’t Nakayama, Nakayama shows I = 0

The proof of that is this

[topaz solar]

I^2 is prime

[next obsidian]

The radical of I^2 is I

But I^2 is prime

So it is its own radical

[molten viper]

oh gotcha

[topaz solar]

Well there’s the 0 ring ig

But uh

[molten viper]

Well that's part of the trivial case

[next obsidian]

And a field

[hollow tartan]

I think it is actually an open problem for I^2=I, if R/I have nilpotent elements

[molten viper]

Fuck I've decided my thesis is not getting worked on

[topaz solar]

Well it’s a prime ideal so

[next obsidian]

My friend did some topology and said the ideals are totally ordered

So any such ring is a valuation ring

[molten viper]

wow literally scooping my problem smh

[topaz solar]

So it’s not discrete

[molten viper]

idk what a valuation ring is 💪

[next obsidian]

PQ < P and Q so PQ < P\cap Q

Right so

P < P\cap Q or Q < P\cap Q

And then P < Q or Q < P

For any ideals P,Q (using that they’re prime for some prime avoidance type stuff)

[next obsidian]

(This step)

[molten viper]

So, we have a stronger condition

basically

[next obsidian]

I mean basically

You’re looking for the most cursed ring ever

[molten viper]

R has to be both a valuation ring and non-noetherian

[next obsidian]

Yes and have I^2 = I for everything

[molten viper]

This tracks with my personality outside the math world

sounds monumentally cursed

[topaz solar]

Abyssal ring

[molten viper]

Shame I don't really have the breadth of knowledge to even begin to imagine what that might be like

[topaz solar]

What would the value group need to be?

If we can give any restrictions on that beyond “not Z”

[next obsidian]

Wait

I think this is impossible now

Let a have non-zero valuation

And we know (a) = (a^2)

Then v(a) = v(ba^2) = 2v(a) + v(b)

And then you’re screwed right?

[molten viper]

I'll take your word for it?

[next obsidian]

Everything in a valuation ring has positive valuation?

I think

Right?

[hollow tartan]

I understood from wiki that in simple words if u have (a,b)~(c,d) and you do the regular (ad+bc)/bd either that or its inverse is in the integral domain, because i don't get P < P\cap Q or Q < P\cap Q

[next obsidian]

It’s because PQ < P\cap Q and P\cap Q is prime

[topaz solar]

Well it’s not a field so I think you can say so?

[next obsidian]

So this forces P or Q to be contained in the intersection

[next obsidian]

Yeah like

It’s literally defined as

The things of non-negative

I just was wondering if I was smoking pack

[molten viper]

I wish I understood any of what was going on here

[next obsidian]

Or wait wait

My friend hella simplified it

(a) = (a^2)

[topaz solar]

Well, idk if you can have some negative things but uh

[next obsidian]

a = ba^2

Domain

So 1 = ab

a is a unit

Okay done

[topaz solar]

Based

[molten viper]

oh so the property actually implies R is a field?

[next obsidian]

Nah

I mean

Or a was 0

So 0 or a field

[topaz solar]

Ye

[molten viper]

wow

[topaz solar]

TFW the obvious answer

Why didn’t I see it

[molten viper]

so, for a principal ideal (p), we have (p)^2 = (p^2) yeah?

[next obsidian]

Yes

[molten viper]

that's kinda wild

So

glad we figured that out rather than me trying to work on my thesis lol

I think I've been kinda working hard on math stuff so I can take a break lol

[hollow tartan]

i guess that's worth a thesis

[void cosmos]

what about IBM tho

constructing a ring that doesnt have IBM is much harder i think

can someone give an example

[molten viper]

what's IBM

[void cosmos]

invariant basis number

[topaz solar]

I do not think so

[hollow tartan]

well i meant for galstaff since he wa swriting one

[south patrol]

lol

M standing for number

nice

[topaz solar]

This was not his thesis topic

[molten viper]

It's completely irrelevant to my situation lol

[void cosmos]

oh sorry

IBN

mb

[molten viper]

Considering every ring I'm worrying about is both Noetherian and Local

[next obsidian]

Non commutative rings do not exist

All rings have IBN

[south patrol]

well commutative => IBN holds

which i imagine is the case here

Exterior powers 😼

[molten viper]

so non-commutative => ~IBN

[hollow tartan]

non commutative rings are scary

[next obsidian]

I feel like

Left Noetherian should imply left IBN

[void cosmos]

this is true

[molten viper]

I guess very broadly the problem for my thesis is looking at the rank of multiplication maps on graded rings

[topaz solar]

Make the worst possible ring imaginable

[molten viper]

well, one specific map over a lot of graded rings

[void cosmos]

is a thesis like

a problem thats not been solved or

is it just exploring some subarea

or some "phoneomna " ( cant spell it )

[hollow tartan]

what do u mean by rank of multiplication maps?

[molten viper]

In this particular case I mean if we take a graded ring, the rank of a map f : R_a -> R_b is dim(f(R_a))

So like, for a simple example, say we have $R = k[x,y]/(x^3,y^3)$, and look at the map $f : R_1 \rightarrow R_3$, defined by $f(r) = rx^2$, then $rk(f) = 1$ (as the image of f is the subspace span(yx^2))

DAMMIT TEXIT

[cloud walrus]

Galstaff, Sorcerer of Light
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[molten viper]

whateverrrrrrr

The exercise for the reader if define g(r) = rx, find the rank of f: R_2 -> R_3

[topaz solar]

Undefined because f isn’t

[hollow tartan]

rk(f)=1 u mean rank(f)=1

[topaz solar]

Yes

[hollow tartan]

Isn't that obvious because x^2 is fixed?

[molten viper]

It's an example to illustrate the definition

[topaz solar]

So R_1 is monomial with 1 multiple of x or y?

[molten viper]

It's the space of all degree 1 polynomials in our ring

[topaz solar]

Where’s f(x) in span yx^2

[molten viper]

It's 0

because x^3 is in our quotient

[topaz solar]

Ah I’m blind

[molten viper]

I guess technically R_1 = span(x,y)

R_2 = span(x^2, xy, y^2), etc

[next obsidian]

Can’t you compute these via rank nullity?

[molten viper]

Yeah

Just giving examples

[next obsidian]

I’m guessing the actual problem is more than just like

“Give me the number” kekw

[molten viper]

lol yes

basically, we're assuming that we have some degree for which dimR_i = dimR_{i+1}

blegh

why foesn't that format right

anyways

[next obsidian]

In your problem?

[molten viper]

yeah

[next obsidian]

This is for all suitably large i yeah?

Not just once

[next obsidian]

You need to put a \ before the _

It broke because you wrote a _{

[molten viper]

It actually only needs to happen once, and it might be a good exercise for me to show that that's actually a max of the hilber function for R

[next obsidian]

Is it automatically going to persist past that point?

Or is it that if t happens once, it happens for all suitably large i

[molten viper]

Under the setup we're using it actually is 0 for suitably large i

[next obsidian]

Possibly larger than when it first stabilized?

[molten viper]

so the dimension goes up for a bit and then starts falling

[next obsidian]

Wait so is your claim that if dim R_i = dim R_{i+1}

Then that number is a max for the Hilbert function?

[hollow tartan]

I wonder if it gets more interesting if one asks when this happens $$rank(H^i)=rank(H_i)$$

[cloud walrus]

Invariants

[next obsidian]

What on god’s green earth are you saying

[next obsidian]

Cuz if so, I find that interesting

And want to ask what your assumptions on R are, like if R_0 is a field and R is fg as an algebra by R_1 or something

[molten viper]

It's not the main claim

[hollow tartan]

for homology and cohomology ranks

[next obsidian]

Of what?

[next obsidian]

Yeah but is this true?

[molten viper]

It might be under our setup

[next obsidian]

It seemed like you were saying it is, so I wanted to ask about a precise statement cuz I want to prove it

[molten viper]

I have to think on it for a bit

[next obsidian]

I see

If you can make a precise statement please let me know

[molten viper]

Prove it for yourself and don't tell me cause this might be a nice property to put in my thesis lel

[next obsidian]

I don’t know what I’m trying to prove

Because I sort of doubt the proposed claim I made, so if it was true it would be really surprising which is why I want to try and prove it

But unless someone’s telling me it is true, I would guess it isn’t

[topaz solar]

What exactly is the hilbert function

[next obsidian]

f(n) = dim R_n

[topaz solar]

Ah

[next obsidian]

Commutative algebra stuff basically will say that for all n >> 0, this is an integer polynomial

And the degree of said polynomial is the dimension of the ring

Or something like that, I might be misspeaking. The dimension thing is about the associated graded algebra

But the thing about the integer polynomial thing is true for any graded module I think

I never remember the statements, but when I need to know them I know where to look in Matsumura kekw

[molten viper]

Trying to remember my exact setup is fucking killing me rn

it's late

ok so, the simple version is:
let R = k[x1,...,xn]/I, where I is generated by all the squares and N (more on the value of N later) monomials of degree i + 1, then if H(R_i) = H(R_{i+1}), I guess the claim is that's the max of the hilbert function for R

I think

Like it feels obvious but I have to think about it

Probably applies for a much more general R too

[next obsidian]

Is that supposed to be the same i?

For the degree of monomials and the i in the index

[molten viper]

I believe so yes

[next obsidian]

Ah okay

[molten viper]

because I'm worrying about the map given by xL, where L is linear

It's kind of a tangential thing but I'll note it for later

Well

basically there is a specific value of N where that property holds

because we've got such a carefully chosen ring

but basically the only thing R_k can do for k > i is shrink

and the socle degree of R is conjecturally less than n^3/4

Conjectured based on some rough computation up to n = 1000

well, sorry, that's specifically if i = 3

I'm getting ahead of myself

@topaz solar you should have seen the gorgeous graph I made

[chilly ocean]

need help

[lusty marlin]

Try pairing elements of the group in a specific way

[delicate bloom]

what's left in the set when you look at G-t(G)?

I might be assuming where zzzzz got stuck was somewhere else than KnightWatch I just realized lol

what have you tried/what are you thinking @chilly ocean

[rocky cloak]

Fun fact you can generalize this to primes bigger than 2 like this:

Let X be the set of tupples (g1, ..., gp) such that g1*...*gp is the identity. And let Cp act on this by permuting cyclically. Show that the number of fixed points is equal to |X| = |G|^p-1 mod p. And that a fixed point is exactly an element of order p (or the identity)

[indigo ridge]

is it always the case that for the factor group G/H. that for a,b in G a-b is an element of the normal subgroup H

[coral spindle]

No

Why do you think this is true?

[rocky cloak]

Maybe what you're thinking of is that a and b map to the same element in the factor group iff a^-1 b is in H

[indigo ridge]

Its true in R/Z just wondering if it were true generally

[vale locust]

is it?

pi-e isn't an integer after all

[vale locust]

if this was true in general take b=0

then a-b = a is in H

hence H = G

it's true if they belong to the same coset however

that is, a+H = b+H

[formal ermine]

[vale locust]

sry but I'm not an engineering major anymore

[chilly ocean]

Key facts:

• $x$ has finite order $n$ and $n = 2k+1$. I am asked to prove that if $|x| = 2k + 1$ then $x^{i} \neq x^{-i}$ for all $i \in [0, n - 1]$.

Can anyone provide a hint?

[cloud walrus]

zzzzz

[formal ermine]

move x^-i to the other side

[chilly ocean]

ah

$x^{2i} \neq 1$

[cloud walrus]

zzzzz

[formal ermine]

yes

[chilly ocean]

can I do this with induction, then?

[formal ermine]

just use Lagrange

[chilly ocean]

how so

do you mean because 2i is always even?

[topaz solar]

What is the definition of order

[chilly ocean]

so it must divide 2k+1

[chilly ocean]

smallest positive integer n such that x^n = 1

[topaz solar]

Yes

How does 2i compare to 2k+1

[chilly ocean]

2i is even

2k+1 is odd

[topaz solar]

Ye

[chilly ocean]

how do i proceed

[topaz solar]

For any of your choices of i

Can it be divided by 2k+1?

[chilly ocean]

no

[topaz solar]

Then there you go

If the order of x is n, then x^m = 1 means n | m

(May want to show this)

Does this make sense?

[chilly ocean]

What kind of stuff do I need to know to generate an algebra using the Cayley-Dickson construction?

[rocky cloak]

I mean the construction doesn't use anything fancy. Might be useful to know what a ring/algebra is.

But ultimately not necessary

[chilly ocean]

A ring is just a set that has a multiplication and an addition on it, with inverses and an identity for addition and multiplication, right?

And an algebra is just a vector space with a multiplication operator on it, which has a different definition depending on the algebra (ex: Lie algebra vs Witt algebra)

[rocky cloak]

Yeah, exactly.

Basically all you do is you start with some algebra (typically the real numbers). Then you define a multiplication on R^2 (which will give you the complex numbers). Now you have a new algebra, so you define a multiplication on (R^2)^2 in the same way

And iterate

The way to define the multiplication is just a short formula, you can find it on Wikipedia for instance

[chilly ocean]

I kinda want to know what a 32-dimensional algebra looks like. Even if it is totally useless.

Incidentally, the Cayley-Dickson construction is why there are no 3-dimensional algebras, isn't it? Because you can't get a 3-dimensional vector space by doubling.

[rocky cloak]

Well, I'm not sure the fact that you can construct algebras of dimension 2^n explains much about why you can't for dimensions that are not powers of 2

But it's related of course

Typically you would prove that using some algebraic topology

[open sluice]

I’m sure Hamilton could explain if you asked really nicely

[rocky cloak]

I guess your real problem will be keeping track of tuples of 32 real numbers

[chilly ocean]

So that means when it says multiplication is defined by $(p, q)(r, s) = (pr - sq, sp - qr)$ then $p$, $q$, $r$, and $s$ are tuples with 4 real numbers each? At least, when defining multiplication on $\mathbb{H}$ or $\mathbb{O}$

[cloud walrus]

Dark Angel

[rocky cloak]

yeah, p,q,r and s are elements in the previous algebra

so in the constrcution of octonions they will be quaternions, which can each be written as a tuple of 4 real numbers (a + bi + cj + dk)

[chilly ocean]

I guess that means I need to know what a sedenion is and what to do with it.

[rocky cloak]

yeah, either that or unwrap the definition back down to something you can understand

will be long complicated expressions either way

[chilly ocean]

I wonder if anyone has actually bothered going past sedenions or if they just worked out they're useless and stopped.

[dim widget]

They almost certainly show up in physics somewhere

First google result

https://link.springer.com/article/10.1007/s12648-014-0562-y

[chilly ocean]

Someone said they were useless earlier LOL

[open sluice]

zero divisors are sad

[chilly ocean]

I didn't understand a single word of that abstract LOL

Gaining zero divisors is bad right? Because it means that $ab \neq 0$ if $a =0$ or $b=0$

[cloud walrus]

Dark Angel

[coarse kestrel]

that's backwards $ab=0$ but $a\neq 0$ and $b\neq 0$

[cloud walrus]

Zybikron

[chilly ocean]

Oh my bad

[rocky cloak]

yeah having zero-divisors means you cant divide by stuff, which I guess is a little sad 😢

[chilly ocean]

I wonder if it would make school students happier though because then they don't have to learn long division

[coral shale]

matrices are stoopid

[indigo ridge]

im reading the wiki article on quotient groups and i cant wrap my head around one concept i was hoping somebod could help walk througtht it it says "A quotient group or factor group is a mathematical group obtained by aggregating similar elements of a larger group using an equivalence relation** that preserves some of the group structure (the rest of the structure is "factored" out).**" what do they mean b factored out?

[wraith cargo]

it's just introducing terminology
You usually say you factor out a subgroup or something when you quotient out by it

[indigo ridge]

yeah but what does it mean lol

I know the formal definition of a quotient group

but Im tring to get some intuitive understanding

[coral shale]

see this #math-discussion message

[topaz solar]

Can you draw out the Cayley graph for some simple cases?

But literally speaking, it’s saying that you get an equivalence relation from it and a ~ b and x ~ y means ax ~ by kind of thing

[wraith cargo]

think of it like taking the group and pretending that the subgroup doesn't exist
i.e. you collapse the subgroup to the identity

[indigo ridge]

Ok I think I might be starting to understand now

[void cosmos]

quick question

[indigo ridge]

is it kind of like "mod-ing'?

[void cosmos]

the countabe union of algebraic sets may not bea lgebraic

wait is this taken

[summer path]

Yes, G/H is often said "G mod H"

[agile burrow]

This is true

[void cosmos]

yea ik its a problem

now

my proof was this

the point was that

ik k^1 is a pid

so the point for me was that any countable set of points will work

cuz for their union to be algebraic it require that all of these points be the zero set of only 1 polynomial which is not true

or not not true but just impossible

so just let k be a field

and define A_1 = {1} , A_2 = {2}..

and so on

these are all algebraic with their polynomaisl being x-a

given a

btu their union isnt?

is htis correct

or yea 2 may not be in k

take k = R or Q or something

[agile burrow]

or C

But yeah, that's the idea

[void cosmos]

but i have a very bad questino tho

[agile burrow]

Only polynomial in C[x] with infinitely many zeroes is the zero polynomial

[void cosmos]

why cant i consider this infinite product (x-a)

[agile burrow]

Because that isn't a polynomial

[void cosmos]

yea

got it

thank you

[chilly ocean]

What does it mean to "adjoin" something to a field when that thing isn't a number? Like, if I have $\mathbb{R}[i]$ then that means I have all the reals, and $\sqrt{-1}$, which is just $\mathbb{C}$. But what if I want to have $\mathbb{R}[x]$? What even is $x$?

[cloud walrus]

Dark Angel

[void cosmos]

R[i] would be the set of elements of the form a+bi

where a and b are in R

Q[sqrt(2)] would be elements of the form a+bsqrt(2)

where a and b are in Q

@chilly ocean

[chilly ocean]

Does it matter what the field is, are all elements of some field $\mathbb{F}[x]$ of the form $a + bx$?

[cloud walrus]

Dark Angel

[void cosmos]

yes

thats by definition

of the form a+bx where a and b are in F

now see how addition and multiplication is defined

and now u have a new field

an extension of the older field

as in F is F[x] where all the b elements are 0

are u done? cuz i have a question 😄

[formal ermine]

No

,, K[a] = \set{f(a) | f \in K[x]}

[night onyx]

For i and root(2) it is because i^2 and root(2)^2 are both in F, if you adjoin a transcendental element it won't be of that form

[formal ermine]

where K[x] is the ring of polynomials over K

[void cosmos]

yea mb

[chilly ocean]

My $\mathbb{F}[x]$ here isn't meant to be the ring of polynomials over $\mathbb{F}$, I just needed something to put in the square brackets.

[cloud walrus]

Dark Angel

[night onyx]

Yeah in general F[x] means the smallest ring containing F and x, where x is in some field extension of F. So like in the case of going from R to R[i], you're forming the smallest subring of C which contains both R and i. Technically this is all polynomials f(i) for f a real polynomial, but because i^2 = -1 you end up getting a space of dimension 2, or {a + bi | a,b in R}

[shrewd island]

hey all, just joined so not sure if it's the right place to ask, but:

I know the usual way to construct GF(2^n) is by GF(2^n) ~= GF(2)[X] / (P(X)) where P is an irreducible polynomial of degree n. do other ways work? esp for GF(2^(2^n))... is there a way to construct GF(2^(2^(n+1))) as ordered pairs of (a, b) in GF(2^(2^n))^2, for an inductive construction starting from GF(2)? if it is, obv it would be non-trivial to convert between this in the usual representation, however it doesn't matter for me

the main property i want is for one side of the pair to "keep to itself", i.e. for all a in GF(2^(2^n)), (a^k, 0) = (a, 0)^k

similar to the complex numbers

i keep thinking i've found a way to construct it this way, then it turns out to be wrong in some way... maybe it's just not possible?

[void cosmos]

i think this is trivial but the point is

L intersec C is algebraic yeah but like

its like obvious geometrically that the intersection ( L intersec C) is F(X,Ax+B)

but idk is this like

like basically i think once i wrap my head around why L intersec C is this then i just use fundamental theorem and thats it

but why is the intersection htis

thi*

[wraith cargo]

this isn't what the hint is telling you to do tho?

it's like

telling you to write

[void cosmos]

it is

i just dont see why L intersec C is F(..)

[wraith cargo]

L \cap C = V(F) \cap V(Y-(ax+b))

now on the rhs you have that Y-(ax+b) = 0, i.e. Y = aX+b

so you can write that L \cap C = V(F(X,aX+B))

since you just subbed in that value for Y

you can imagine this as

you're considering all the points where F(X,Y) = 0 and Y-(aX+B) = 0

[void cosmos]

yea ^

this i the intuition

from like precalc

[wraith cargo]

now from the linear equation you can get that all the Y points that satisfy this are of the form aX+b

so your problem reduces to finding all the points where F(X, aX+b) = 0

and this basically solves ur problem

Fun fact this is a special case of a more general theorem called Bezout's theorem

[void cosmos]

yea will check this out

cool stuff

[south patrol]

says u

[void cosmos]

im having fun

im going to ask alot of this stuff here tho

not in ag chanl

[void cosmos]

would the set {(x,y)|y = sin(x)} be not algebraic as it intersects the line y=1 infinitely many times?

over A^2(k)

[dim widget]

For that and many other reasons

[topaz solar]

That sure doesn’t look like a polynomial to me

[void cosmos]

yeah got it

[open sluice]

nuclear lunch codes

[void cosmos]

its literally the proof

of the hardest problem in AG

which idk how to solve

[crystal turtle]

apparently that would be whether {(x,y) | y = sin(x)} is algebraic

[void cosmos]

the time it took me

to solve this

was a time of spirit stagnation

lmfao

(im copying your letter to ihes )

[next obsidian]

No shot this is an open question

[coral shale]

eli5 what algebraic means here anyone?

[next obsidian]

It forms an algebraic set

Like, you can cut it out via polynomial equations

[coral shale]

hmmcat

ic

[rocky cloak]

Really? I would have thought there was some obvious reason why this was not algebraic

That's crazy

[crystal turtle]

yeah I know, crazy how even the strongest methods cannot solve it yet

[coral shale]

collatz could help

[void cosmos]

If u put this problem into warm water and eait for 10 years

With the right amount of force it will rip open

and become solved

-inventor of ag

[rocky cloak]

So something like this should be possible I think.

If you start with GF(2), then you can adjoin a root of x^2 + x + 1, so you get something where (0, 1)^2 = (1, 1).

I think you can generalize this by having (0, 1)^2 be (a, 1) where a is the last element you just adjoined.

The recursive nature will make it a little complicated, but it will satisfy the property that (a, 0)(b, 0) = (ab, 0)

This basically comes down to the root of x^2 + x + a being primitive whenever a is primitive.

This is actually much easier to do in characteristic different from 2. I.e. you can construct GF(p^2^n) very easily by just adjoining square roots of the primitive root

(when p is not 2)

[shrewd island]

Sweet, thanks, I’ll see if it works out

[shrewd island]

The case I’m using this in relies on the representation being binary with the xor operation as addition 😦

Stuck with the pain of characteristic 2

Of course the polynomial representation also has the same property but I wanted to see if this other way works

[rocky cloak]

There are actually top of the line algorithms for this that are better than what I can cook up on the spot here

[shrewd island]

Yeah haha, it’s just for a toy little implementation so I wanted to try something new

Programming challenge where finite field dft is the best solution

[rocky cloak]

Fast Multiplication in Finite Fields GF(2N ) - Springer Link https://link.springer.com/content/pdf/10.1007/3-540-48059-5_12.pdf

Here's the first that showed up in my googling

[shrewd island]

Neat, thanks!

[rugged night]

i feel like im missing something here

[chilly radish]

If a^n is a unit, there is b such that a^nb=1

Rewrite this as a(a^n-1b)

[rugged night]

ah. thanks

thats easier than i thought

[lilac mango]

Is there a neat way to determine the maximum possible order for an element of $S_n$?

[cloud walrus]

tirib00

[patent girder]

Look into landaus function; we know upper bounds but that's it

[south patrol]

Interesting

[teal vessel]

so just thinking about it, the longest "path" in a cycle is to visit every position in that cycle, which makes sense. For sufficiently small n (less than 5) the longest paths are those which use up every position in the permutation, i.e. for S4, the maximum order is 4. However, once you get to 5, you can start splitting up the elements into sub-cycles that have coprime cycle orders, so the minimal order of such compound permutations is the product of the orders of the sub-cycles (notice for S5, the permutation (1 2)(3 4 5) has order 6, the product of 2 and 3). Given this, a guess for the lower bound of any "maximal" order would be the maximum of the set of LCM of the orders of partitions, i.e. being able to split up the elements into as many "collectively co-prime" sub-cycles as possible.

oh, wait, just looked up Landau's function, whelp. Good to know my reasoning process is on point ig.

[chilly ocean]

I've read a few proofs that $t$ can have no $p$th root in $F_p(t)$. But I have thought of a simple one: the degree function is additive over domains, $deg(fg)=deg(f)+deg(g)$. If $f^p=t$, then $deg(f^p)=p\cdot deg(f)=1$ so $deg(f)=\frac{1}{p}$ (contradiction).

[cloud walrus]

username0000

[chilly ocean]

Does this check out?

[coral shale]

deg doesnt sound like a legit argument

does it

[dim widget]

Yes it checks out! Though the proof of the degree function being additive is also a proof of this fact

[coral shale]

Oh Fp

[dim widget]

Why does it matter if it’s F_p

[coral shale]

was just thinking zn things

because me.

[dim widget]

Yeah over something which is not an integral domain the degree function would have to be defined differently

You would have to define it as the degree of the smallest unit coefficient of a Laurent polynomial

[teal vessel]

Suppose σ= (1 2 3 4), τ=(1 2 3) ∈ S4 and act on some object p. Shouldn't it be the case (because of the definition of group actions) that τ·(σ·p) = (τ o σ)·p ?

[crystal turtle]

Yes, by definition

Well, groups generally act on sets, but I assume you mean as an element of the set S_4 is acting on

[teal vessel]

huh, the question must be asking me to calculate it first via action-on-element then by composition of permutations first.

it's asking me to calculate both

[crystal turtle]

If you don't know a priori that something is a group action, then that makes sense to check, since you are basically checking that is is a group action

[teal vessel]

idk why that would be the case, since it seems obvious it's a group action. $p=p(x_1,x_2,x_3,x_4)$ is any polynomial in four variables such that each term contains non-negative powers of the variables, and an integer coefficient, and the action of any σ ϵ S_4 $σ·p = p(x_{σ(1)},x_{σ(2)},x_{σ(3)},x_{σ(4)})$

[cloud walrus]

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

[teal vessel]

eh, gets the point across. too lazy to fix

[crystal turtle]

well yeah I agree it should be one lol. But unless you're told you can assume it's a group action I think that's the point

Kinda like how you rigorously check the first few examples of groups you see

[teal vessel]

fair.

[crystal turtle]

(idk the full context, so this is just my guess)

[teal vessel]

guess I'll know if I got something wrong if my answers disagree

[glossy crag]

By ideal do you mean one or two-sided? Because if it's the latter, that's patently false.

[rugged night]

what do you mean

to be clear, the statement is in two parts: "a commutative, unital ring is a field iff it has no proper, non-trivial two-sided ideals" (not that two-sided needs to be specified if it's already assumed to be commutative)

and then: "a (not necessarily commutative) unital ring is a skew field iff it has no proper, non-trivial one-sided ideals"

is there something that's false here or is your issue with the unital part of it (which i guess i didnt mention)?

[glossy crag]

I thought you said "skewfield iff no two-sided ideals", which is false (there are plenty of rings like this that aren't skewfields).

[dim widget]

For instance matrices!

[dire python]

Are the elements of principal ideal ❬a❭⊆A polynomials expressions in a with zero constant coefficient? In other words, does ❬a❭ equal {f(a) | f∈A[x],f(0)=0}?

[dim widget]

Yes, but they’re also just the linear ones

With no constant term

[south patrol]

Oh lol i was confused by that description for a while

[dire python]

lol

gg

[dim widget]

Byyyyyyye potato’s comments

[dire python]

@dim widget that seems strange if we take A=𝐙[x] and a=x, since ❬x❭⊆𝐙[x] is not comprised of linear polynomials. What am I missing?

[formal ermine]

topos means ma where m in A

(a) = Aa

[mighty kiln]

Is it the same question?

[dire python]

Ah, never mind - I'd have to look at 𝐙[x][y]. My bad.

[dim widget]

I think you’re getting confused because now you took A to be a ring of polynomials

Yeah

[dire python]

One more question: suppose an ideal is generated by idempotent. How would you interpret this geometrically?

My mental image of an idempotent is an indicator function of a subset.

[rocky cloak]

For commutative rings having nontrivial idempotents is the same as decomposing into s product of rings. So geometrically of your ring has idempotents then your spectrum is a disjoint union of spectra

[south patrol]

If a commutative ring R has an idempotent e then R = Re x R(1-e), so that gives a bit of intuition in terms ofh ow it fits in the ring

lol

[dire python]

yes, so geometrically, an idempotent is like the indicator function of a clopen subset

[rocky cloak]

Exactly

[dire python]

Jolly good. Now what about an element whose power is an idempotent? How to think of it geometrically?

e.g fⁿ=e

[rocky cloak]

I guess that would be a bit like an indicator function times a root of unity

[dire python]

I hoped you'd say something sexier haha

[rocky cloak]

I mean you'll have f = fe + f(1-e), so f will be nilpotent on one component and root of unity on another

[void cosmos]

yo 1.15 is trivial right

cuz just consider the polynomial where the first n tuples are the polynomial for V and the rest m-n tuples are defined as the polynomial for W

correct?

[south patrol]

Sounds good, though not sure what you mean by tuples here

But yeah just collect the polynomials together lol

[void cosmos]

thank you

[void cosmos]

yo stupid question

to show that an element is in V(I) for some ideal I

its not enough to show that this element is the zero of some polynomial of I right

(given it may not be a pid)

i have to show that its a zero for all polynomials

is that correct

[chilly ocean]

moamen what is the definition of V(I)

[void cosmos]

V(I) is the set of points that vanish on I ( I being an ideal in k[x1,x2,...]

[chilly ocean]

meaning...

[void cosmos]

yea i need all polynomials

thank you

[void cosmos]

why do you always react like this to any of myq uestions

[chilly ocean]

a lot of your questions are answered by thinking for 30 seconds

usually just unravelling what a definition means

[summer path]

Sometimes thinking "aloud" or I guess in this case typing it out, does help with processing the problem and figuring out your own questions though

[open sluice]

you could always ask a duck 🦆

[crystal turtle]

quack

[chilly ocean]

[small bramble]

Dummit and Foote is simply omitting the "up to associates" bit in (2) right?

kinda weird how they bother to entail in (1) that the rep is unique up to order and sign but not the extra bit in (2) but w/e

for instance 3i is irreducible but not listed unless you assume it means to say up to associates

[crystal turtle]

correct, those are all up to association

And there are only 4 gaussian integers in each associate class (other than 0, which has 1 ofc)

[small bramble]

kool

[solemn garden]

,rotate

[rugged night]

isnt part b just like. 3rd grade set theory

[coral shale]

confused what 'show' is supposed to mean

that is just a definition

[glass vine]

Hey guys is there someone who likes representations online?

I am doing a set of problems regarding representations of dihedral groups. Most of them were concerning complex representations. Last one asks me to show that QD8 is isomorphic to Q×Q×Q×M2(Q) as Q-algebras

I get why this holds when you look at CD8 as a complex algebra. But idk how to do it for rationals

[dire python]

Let x∈A[X]/❬f❭ where f=x²+bx+c is a separable quadratic monic, i.e b²-4c∈A*. Suppose we go to the initial splitting algebra of f so that f=(x-ɑ₁)(x-ɑ₂). Then the characteristic polynomial of multiplication by 2x (as a linear operator on the quotient) equals (x-2ɑ₁)(x-2ɑ₂). Over A=𝐙/4𝐙, this characteristic polynomial is x²+2bx+4c, which has discriminant zero. How can I use this to deduce that 2x∈𝐙/4𝐙[X]/❬f❭ is not the root of any separable monic in 𝐙/4𝐙[X]?

[rocky cloak]

This just comes down to computing the irreducible representations, and the irreps of D8 are the same over Q and C.

[terse crystal]

Z/4Z is not a field, I am not sure this case you can still define those concepts, separate or whatever, on a ring. On a ring a polynomial can have more than the degree of it many roots, Z/15Z has 4 roots of x^2-x=0

[dire python]

@terse crystal everything is well defined. The trouble is that there are no minimal polynomials.

[celest cairn]

Mathematically, how would one determine how many degrees you rotate when multiplying by $e^{(2k\pi i)/n}$?

[cloud walrus]

Sapphire

[chilly ocean]

If you mean multiplying complex numbers, then it suffices to calculate the angle of the vector given by $e^{(2k\pi i)/n}$.

[cloud walrus]

username0000

[chilly ocean]

There's also the rotation matrices from linear algebra.

[chilly ocean]

Or I guess take a general complex number $a+bi$, and calculate the angle between a+bi and $(a+bi)\cdot e^{(2k\pi i)/n}$, using the angle formula from linear algebra.

[cloud walrus]

username0000

[celest cairn]

Thanks!

[celest cairn]

This a decent proof for my precalc problem?

[lethal dune]

minimal polynomials?

[indigo ridge]

Im trying to prove that: given a commutative ring R with $a, d\in R$ prove that $R$ is an integral domain if and only if with a not equal to the additive identity $0_R$ there exists at most one $b \in R$ such that $ab = d$

[cloud walrus]

jayz

[chilly ocean]

What happens if you have ac =d and ab = d?

[indigo ridge]

So the outline of my proof is as follows for the forward direction is as follows: I break it up into two cases, the first case is all the elements of R that are units (i.e all elements that have a multiplicative inverse) and the second case is those that don't.

Case 1: is easy to prove that there exists precisely one solution

Im having trouble proving the second case, where the elements are not units in other words there does not exist an x such that ax = 1

[chilly ocean]

There is a simpler proof that does not require two cases.

[indigo ridge]

Ok really what is it?

[chilly ocean]

If you consider this

What can you deduce if ac = ab?

[indigo ridge]

a(c - b) = 0 and since a is not 0 c - b = 0 and therefore c = b?

🤔

[chilly ocean]

correct

[indigo ridge]

but how does that prove the forward direction?

[chilly ocean]

That shows the uniqueness of b

Now you have to show if you have the uniqueness condition, then R is a domain.

[indigo ridge]

yeah but I need to show there exists atleast one b such that its true for each element of the commutative ring

[chilly ocean]

For the other direction, there's a useful choice of d.

[topaz solar]

Nah it just says at most 1

0 <= 1, so

[indigo ridge]

yeah but still I need to show that there exists at most one b such that for any two elements in R ab = d

we cant take ab= d for granted

[chilly ocean]

To show that there exists at most one, it suffices to show that if there is one, then there is at most one.

[indigo ridge]

but we didn't even demonstarte there exists a b such that ab = d

[chilly ocean]

" if there is one, then there is at most one "

So for the proof of that if-then statement, we assume that there is one, then show there is at most one

[topaz solar]

If there is 1, there is 1. If there is 0, there is less than 1

[indigo ridge]

if we cuppose that a is a unit then there exists an $x \in R$ such that $ax = 1$ then we know that $a(xd) = d$

[cloud walrus]

jayz

[indigo ridge]

that was my existence proof and then I showed it was unique

[topaz solar]

Why are you bringing up units

This is irrelevant

Stop

[indigo ridge]

then the second case would be all elements that are not units

[chilly ocean]

That's fine, but it will not help you prove the entire statement

[topaz solar]

You have shown that integral domain => there is at most one solution

Because ab - ac = a(b-c) = 0

[indigo ridge]

yeah.

[topaz solar]

If there is at most one solution, how do you show that ab = 0 implies b = 0 when a is nonzero?

[indigo ridge]

because in an integral domain if ab = o then a = 0 or b = 0 since a is not zero b must be 0

[chilly ocean]

I'll be more explicit about the proof method to hopefully clear up any confusion.

To prove that there is at most one thing (<=1) satisfying some condition, you can do so in two cases: the case where there isn't anything satisfying the condition, and the case where there is something satisfying the condition. These two cases exhaust all possibilities.

The first case is trivial (if none exist, then that's less than 1). For the second case, you want to show that the one that exists is unique (less than 1).

[indigo ridge]

So case 1: we would say "Suppose there does not exist a b such that ab = d for any a,d"?

[chilly ocean]

It's more like "Fix a,d. Consider the case where there does not exist a b, and then the case where there does exist a b".

The cases are made after you fix a and d.

[indigo ridge]

I don't get it how are assuming the very thing we are trying to prove? The forward direction of the statement is "If R is an integral domain then for al $a,d \in R$ with a not $0_R$ there exists at most one $b \in R$ such that $ab = d$

[cloud walrus]

jayz

[chilly ocean]

I understand the confusion. It's an artifact of the way the cases are broken up, but to be precise we are not assuming what we are trying to prove. We are trying to prove that there exists zero elements or one element with the property. If you think about it in terms of sets, we are proving a set has at most one element and we consider two case: the set is empty and the set is not empty.

But assuming the set is not empty is not the same as assuming it has one element in it.

[open sluice]

at most one solution = either there are zero solutions or there is exactly one solution
if there are no solutions, then there is at most one solution and you are done
otherwise assume there are not no solutions, i.e. there is a solution
then you can show that if there is a second solution it must be the same as your original

[indigo ridge]

so we are not proving the existence of such a b, just that if there existed one it would have this property?

[open sluice]

yes

you will often hear people talk about “existence and uniqueness of solutions”

the existence part is showing that (number of solutions) >= 1

the uniqueness part is showing that (number of solutions) <= 1

these are often proved separately

[indigo ridge]

yeah I get that part I just assumed that we had actually show that such a b exists with this property

[open sluice]

you can have existence without uniqueness, and you can have uniqueness without existence

[indigo ridge]

of course yeah makes sense thanks @open sluice and username

[teal vessel]

weird thought: if uniqueness is strictly saying "for all x,y with property Z, x=y" (only one form of uniqueness statement among a couple), and it is then proven that there is no such x with property Z, then wouldn't it be the case, vacuously, that uniqueness would also fail? i.e. because the range of objects discussed is empty, we can arbitrarily assign properties to them (as that won't change the emptiness of the range)

[topaz solar]

No it’s vacuously true, since clearly it holds

[teal vessel]

yes, but its negation is also vacuously true, since there is no fact of the matter to be discussed

[topaz solar]

But that’s different from it being false

P -> Q and P -> -Q, because P was already false

But both implications are true

[teal vessel]

yes

[topaz solar]

Your P in this case being x\in Z and y\in Z

[teal vessel]

I'm saying that the statements "all X with Z are unique" and "there are X and Y with Z that are not identical" would both be true.

[topaz solar]

Yep

But that’s different from being false

[rocky cloak]

Which is why uniqueness is framed as a for all statement and not a "there does not exist" statement

[teal vessel]

I didn't say false, I said "fail" which was admittedly a poor choice of words, but it only implies falsehood in non-vacuous cases

[topaz solar]

Vacuous truths with sets are really kinda a variation of (x \in \emptyset) -> P(x)

It doesn’t fail, because it’s still true

It just fails to give you usable information

[teal vessel]

I guess the real question would be this, then: shouldn't uniqueness come after existence, generally, as it is meaningless without it? At the very least, existence would have to be assumed.

[topaz solar]

Well, nah?

Consider the aforementioned integral domain iff ab=d has a unique b for any d, nonzero a

Doesn’t have to exist, just be unique

[open sluice]

when you prove uniqueness you do assume existence

uniqueness is often easier to prove than existence tho

so in some proofs uniqueness comes first

[topaz solar]

Injective functions are things where you have uniqueness but not necessarily existence

Surjective is the opposite

[open sluice]

left and right inverses

[topaz solar]

These are very ubiquitous things

Or, if you’re working categorically, ig you have full and faithful kinda conditions

[indigo ridge]

I want to prove the backward direction of this statement. namely "if $a,b,d \in R$ with a $\neq 0_R$ then $R$ is an integral domain" since $R$ is already a commutative ring with the multiplicative identity $1_R$ not the additive identity $0_R$ all we now need to show is that if $ab = 0$ then $a = 0$ or $b = 0$.

[cloud walrus]

jayz

[indigo ridge]

We know that $a \cdot 0_R$ is always going to be $0_R$ (By Ring properties) so if we have $a, 0 \in R$ then we know that there is at most one solution b such that $ab = 0$ . my confusion is we have already assumed $a \neq 0$ is that an issue?

[cloud walrus]

jayz

[indigo ridge]

we can say $b = 0$ is a solution clearly but can we say it could also be a = 0?

[cloud walrus]

jayz

[topaz solar]

No because a is not zero?

That’s kinda how you chose it altogether?

[indigo ridge]

I know but I want to be able to say that in the end either $a = 0 or b = 0$ in my proof not that $b = 0$ is the solution

[cloud walrus]

jayz

[topaz solar]

Why not

you know the choice is unique, so ab=0 implies b=0 when a is not zero

If a = 0, then it still holds?

It’s commutative chief just swap a and b??

[indigo ridge]

if a = 0 then b could be any element of R

[topaz solar]

Yes, and?

So what

[indigo ridge]

idk 😭

sorry lol

[topaz solar]

Still fine for being an integral domain

[chilly ocean]

You can still say that.

Let $a,b\in R$ such that $ab=0$. Consider two cases: $a=0$ and $a\neq 0$.

If $a=0$, then we're done. If $a\neq 0$, then we know that $b$ is the unique element such that $ab=0$, but $a0=0$ so $b=0$.

Thus, if $ab=0$ then either $a=0$ or $b=0$.

[cloud walrus]

username0000

[indigo ridge]

Ok right makes perfect sense thanks once again

[formal ermine]

@indigo ridge what's your working

[indigo ridge]

so $x - (\sqrt{3}i + \sqrt{2}) = 0$

[cloud walrus]

jayz

[indigo ridge]

then we have $ (x - \sqrt{2})^2 =(\sqrt{3}i)^2$ $\Rightarrow$ $x^2 - 2x\sqrt{2} + 2 = 3(-1)$ $\Rightarrow$ $x^2 +5 = 2x\sqrt{2}$ \Rightarrow $(x^2 + 5)^2 = (2x\sqrt{2})^2$ \Rightarrow $x^4 + 18x^2 + 25$ my only question is how do i show minimality im a little uncertain because it is a complex number and there might be some subtlley im not aware of @formal ermine

[cloud walrus]

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

[formal ermine]

you swallowed a minus at the end

[indigo ridge]

where?

[formal ermine]

it's 2x^2

not 18x^2

[indigo ridge]

right its mean to be 10x^2 - 8x^2

[formal ermine]

ye

[indigo ridge]

so it should be $x^4 + 2x^2 +25$ then?

[cloud walrus]

jayz

[formal ermine]

yeah

you can show irreducibility with eisenstein

gotta substitute u = x + 1 first though

[indigo ridge]

right

wait what?!

substitution

[formal ermine]

f(x) is irreducible iff f(x - 1) is irreducible

[indigo ridge]

Ok hmm I

never seen that before

[formal ermine]

exercise: prove it

wait

I keep forgetting that 4 isn't a prime number

oops

yeah nvm I don't think eisenstein will work here

[indigo ridge]

wdym it doesnt have to be a prime power to show its irreducible

[formal ermine]

you could try reduction mod p

[formal ermine]

yeah but the p we choose in eisenstein has to be a prime number

[indigo ridge]

oh right oops

[formal ermine]

yeah

reduction mod 3 works

[indigo ridge]

Ok let me have a go at it.

[indigo ridge]

So once I reduce the polynomial therefore our poly is irreducible?

and we are reducing f(x-1) right?

[formal ermine]

nonono

reduce the original f mod 3

[indigo ridge]

Ok gotcha

[indigo ridge]

but what happens if we reduce it mod 3?

we show it's irreducible in mod 3 hence irreducible

[formal ermine]

yes

[indigo ridge]

Ok I see thanks seschsundersechgizmilli

[formal ermine]

fuck I'm way too tired for this

I didn't check whether it's actually irreducible mod 3 lol

it has no roots mod 3

but it can still split into deg 2 * deg 2 polys

[indigo ridge]

if it has no roots then its irreducible though so we good tho?

[formal ermine]

(x^2+1)(x^2+1) in Q

[indigo ridge]

Oh right

[indigo ridge]

division algo?

its tedious tho

and I don't wanna type it down in latex lol

[formal ermine]

yeah pretty sure it's irreducible

,wolf solve {au=1,av+bu=0,aw+bv+cu=-1,bw+cv=0,cw=1} over the integers

[cloud walrus]

Results provided by WolframAlpha

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

Requested by projective

[indigo ridge]

yeah theres an online irreducible polynomial calc that tells you if a given poly is irreducible believe it or not lol

theres a calculator for everything it seems

[formal ermine]

for mod 3 or the original one?

[indigo ridge]

the originial one

[formal ermine]

yeah wolframalpha can do that too

you can even ask wolfram to compute minimal polynomials

[indigo ridge]

oh lol

[formal ermine]

I'm more concerned about the mod 3 one

[indigo ridge]

wait so but if its not irreducible in mod 3 that doesn';t neccesarily imply it isnt irreducible thoug right?

I only went through this in passing so my memory is off

[formal ermine]

yeah

[indigo ridge]

rational root test?

the possible roots are +/- 1,5, 25 since the leading term is 1

[granite topaz]

It is, no?

= (x^2+1)^2 + 24 = 1 mod 3

oh, nvm, I better go to sleep

I'm too tired for this 😄

[celest cairn]

Stupid question but zeta_2 isn’t a primitive root of unity right?

[delicate bloom]

it is

-1

[south patrol]

zeta_n is also like notation for the "first" primitive nth root

[celest cairn]

Ah alright, thanks guys 🙂

[indigo ridge]

\begin{align}
\phi_{1} &= n_{20} + n_{02}\
\phi_{2} &= (n_{20} - n_{02})^2 + 4n_{11}^2\
\phi_{3} &= n_{30} - n_{12}^2 + 3n_{21} - n_{03}^2\
\phi_{4} &= n_{30} + n_{12}^2 + 3n_{21} + n_{03}^2\
\phi_{5} &= (n_{30} - 3n_{12})(n_{30} + n_{12})[(n_{30} + n_{12}^2) -3(n_{30} + n_{12})^2] \notag\
&\quad+ (3n_{12} - n_{03})(n_{21} + n_{03})[3(n_{30} + n_ {12})^2 - (n_{21} + n_{03})^2]\
\phi_{6} &= (n_{20} - n_{02})[(n_{30} + n_{12}^2) - (n_{21} + n_{03})^2] \notag\
&\quad+ 4n_{11}(n_{30} + n_{12})(n_{21} + n_{03})\
\phi_{7} &= (3n_{21} - n_{03})(n_{30} + n_{12})[(n_{30} + n_{12})^2 - 3(n_{21} + n_{03})^2] \notag\
&\quad+ (3n_{12} - n_{30})(n_{21} + n_{03})[3(n_{30} + n_{12})^2 - (n_{21} + n_{03}^2)]
\end{align}

[cloud walrus]

jayz

[crystal turtle]

okay?

[summer path]

Surely some context will be given.. right?

[indigo ridge]

Im being asked a true or false questio on if let R be a ring and a and b is in R then a divides ba in R. this feels like a trick question because it depends on how we define divisibility right?

i know what divisbility is in the context of the integers or eveb F[x] but Ive never seen it generalized before

[indigo ridge]

no 😎

[indigo ridge]

it was meant for #latex-help my bad lol

[south patrol]

There is only like one way to define it right lol

Well okay non-commutativity

But in commutative rings at least it's exactly what you expect i.e. x|y iff there's z such that y = xz

[crystal turtle]

but also who cares about divisibility in a non-commutative ring anyways

[chilly radish]

I do

[wicked harbor]

This is a definition in the notes for a topic I'm taking... By this map g, are they meaning a group action of G on E?

[chilly ocean]

G is a subgroup of Aut(E) so its elements are certain kinds of functions on E

i.e. if g is in G, then g is also an automorphism of E

that's how the g you see in this definition is a mapping

[dreamy plume]

no they mean g is automorphism on E meaning g is a bijection and a homomorphism on E.

Usually automorphisms have the property that they move elements around sorta like a permutation while preserving algebraic structure like the field structure.

Fixed field of a subgroup of automorphisms are the elements of the field that dont get permuted after applying automorphisms

[wicked harbor]

Oh, so it's the set of all elements of E for which the automorphisms in G restrict to the identity?

[ivory glacier]

Yes

[south patrol]

The largest such set

[chilly ocean]

What would be a good way to show $t\in K(t)$ Is transcendental over $K$?

[cloud walrus]

username0000

[glossy crag]

If it were algebraic, K(t) would be finite-dimensional over K, but it isn't, because K[t] isn't.

[chilly ocean]

Ahhh that's satisfying and slick, thank you.

[glossy crag]

You could probably also do it by hand, but it'd be cumbersome imo.

[chilly ocean]

What do you mean?

[glossy crag]

Assume that t is algebraic and derive some contradiction from its minimal equation. I don't have a pen and paper atm, so idk.

Actually what I was saying was dumb, lol

t is transcendental kind of by definition of K[t] and K(t).

A (formal) polynomial that is equal to 0 is tautologically the zero polynomial.

[chilly ocean]

I'm taking K(t) to be defined as the field of fractions of K[t], but yea that's the proof one usually sees

[glossy crag]

When I said K[t] isn't finite-dimensional, that's basically a masked version of this.

[glossy crag]

For a moment I confused what I was trying to say with the fact that a polynomial that vanishes everywhere isn't necessarily 0, hence my making it more complicated. It's literally just the definition of K[t].

Anyway, sorry for the confusion.

[somber sleet]

question for the intutition, cause I don't remember anymore:
If I have a polynomial ring $R[X]$ and I take $f \in R[X]$, what is contained in $R[X]/(f)$

[cloud walrus]

damn_guuurl

[somber sleet]

Is this the new abstract algebra channel?

[glossy crag]

The channel was split in two for elementary and advanced topics.

[somber sleet]

where is the line between them? Just so that I know where to ask haha

[glossy crag]

Idk what you mean by "what is contained in", by definition R[X]/(f) is the set of equivalence classes of a relation on R[X], it's literally a set of subsets.

[glossy crag]

It's kind of murky, but I'd keep most of the questions here unless it's advanced fields/groups or commalg. You can read the descriptions of either channel to get an idea.

[somber sleet]

I know this, since this is the definition of a factor ring, but my question is, intuitively, which classes of polynomials are contained there?

[glossy crag]

My point is that the question is moot, since every polynomial is trivially "contained" there as g+(f), you have a map R[X]->R[X]/(f).

The intuition is that you're collapsing (f) in R[X] to 0.

So if g and h are polynomials such that e.g. g=h+f^2, in the factor ring they are equal.

[formal ermine]

my intuition about it is either you remove all deg >= f polynomials (assuming f is monic or its leading coefficient is a unit) or it's just like you set x = a root of f

[chilly ocean]

In other words, it's the polynomials in R[x] modulo a multiple of f.