groups-rings-fields

[latent anvil]

With "divisible by a prime" replaced with "element of a prime ideal"

Sorry let me back up

it's not really a theorem about Q[x], it's about Z[x]

But gauss's lemma lets us reduce to Z[x] a lot of the time

[chrome hinge]

But it concludes irreducibility in Q[x]

[latent anvil]

so generally if A is an integral domain and P is a prime ideal of A, and f = a x^d +.... + c is such that a is not in P, all the nonconstant coefficients are in P, and c isn't in P^2 then f is irreducible in A[x]

Er wait

I screwed up

Ah sorry, I misremembered

You're totally right

It could be divisible by some element of A

But it isn't divisible by any polynomial of strictly lower degree

In A[x]

Returning to Z and Q, this is sufficient to conclude irreducibility over Q but not quite enough over Z

[chrome hinge]

Indeed, the only way i know it could be used that way is with also gauss's lemma

[latent anvil]

I think it's actually easier to generalize it yourself than to have me explain

If you look at a proof of it, you'll see how you can replace primality of a number with a prime ideal

hm no I disagree with myself now

Maybe the ideal needs to be maximal

[chrome hinge]

Even then, does the hypothesis hold here?

[latent anvil]

I don't think it applies, sorry for the diversion

[chrome hinge]

LOL

No worries

[latent anvil]

I said this already!!

.

[chrome hinge]

I didnt notice

[latent anvil]

No false advertising on my part :P

[chrome hinge]

Its ok, no worries at all

actually thanks, it would be such a wonderful thing to have a generalized eisenstein's criterion

I'll think of it as soon as i have time to spare

[latent anvil]

Good luck!

[chrome hinge]

Ty

[celest brook]

@chilly ocean

[chrome hinge]

It was surprisingly straightforward from here...

[latent anvil]

yup! the gauss's lemma trick is really cool

[chrome hinge]

wow

I got the t power thing, but i wouldnt have thought that would be the general form of a factorization from there

f(x)(g(x) + t h(x)) this one

very nice approach

[celest brook]

How do I prove $<f> \subset I_{x}$

[cloud walrus]

Moosey

[celest brook]

I've proven the other way around

[proud bear]

please use \langle ...\rangle instead of <...>

[celest brook]

oh

sorry

:(

[proud bear]

nah it's ok lol. it's just that \langle...\rangle looks better

[red imp]

if one has an associative algebra $A=\text{span}(e_1,\dots,e_n)$ with inner product $\langle\cdot,\cdot\rangle$ then does $$\langle e_\alpha,e_\beta e_1\rangle =\langle e_\alpha,e_\beta\rangle$$ for all $\alpha,\beta$ imply that $e_1$ is the unit of the algebra?

[celest brook]

nvm

[latent anvil]

@red imp it's equivalent to x e1 = x for all x

If the algebra is assumed to have a unit already this implies e1 is the unit

I don't think it's sufficient to imply e1 is a unit on the left but I could be wrong

[red imp]

yeah thats what I briefly tried to show but I couldnt find a contradiction when I assumed that e1 wasnt the unit 😭

oh the algebra is commutative

I should have said that

[latent anvil]

Oh lol

Then uh

Yeah it's a unit on both sides

If it's a unit on one side

sorry wait so do you agree that x e1 = x for all x?

[red imp]

nope thats what I was trying to deduce

[latent anvil]

Ah okay

so it suffices to show x e1 = x when x is some eβ

Do you agree?

[red imp]

yep

[latent anvil]

Cool

so then to show eβ e1 = eβ it suffices to show <y, eβ e1> = <y, eβ> for all y

[red imp]

mmm does it?

I think thats the part that got me stuck

[latent anvil]

this is a standard trick in inner product spaces when you want to show two things are equal. Subtract one side from the other and take y = eβ e1 - eβ

so general lemma

If <x, y> = <z, y> for all y then x = z

In an inner product space

This is what you're stuck on, yeah?

[red imp]

wew yeah thanks that makes it easy then

[latent anvil]

sure

So <x-z, y> = 0

Because the inner product is linear in its first argument

Now set y = x-z

[red imp]

oh yeah does this come from the zero inner product lemma

[latent anvil]

I don't know what that is

Is it the special case when z = 0?

[red imp]

yeah it was proven to me in class, it says if <x,y>=0 for all y then x=0

[latent anvil]

Gotcha

Then yeah, just use that

After subtracting one side from the other

[red imp]

cool thanks!

[latent anvil]

[next obsidian]

yoy

[thorn delta]

Okay, i am struggling to compute the galois group of the polynomial x^4 + 4x + 4 at the moment. I'm having trouble even proving its irreducible (couldn't get anywhere trying direct methods or applying eisenstein's criterion to shifts), but setting that aside, the only thing I've been able to tell is that it only has complex roots. Therefore complex conjugation is a composition of disjoint transpositions, but that doesn't narrow down things much

any ideas?

Its resolvant cubic is x^3 - 16x - 16, which im confident is irreducible, but idk if I can prove it. This means that the degree of the field extension generated by its roots is either 3 or 6, so the galois group is either A4 or S4

[latent anvil]

If x^3 - 16x - 16 = 0 then x^3 = 16(x+1), which is impossible for x an integer because x and x+1 are coprime

This plus gauss's lemma should give irreducibility over Q

[rustic crown]

showing x^4+4x+4 irred is not hard... going mod 3 the polynomial is x^4+x+1 = (x-1)(x^3+x^2+x-1)
now x^3+x^2+x-1 is irreducible over F3 as it has no roots, just try 0,1,2.
so if x^4+4x+4 was reducible then one factor must have been linear which reduced to x-1 mod 3. but this would give a root of the polynomial and x^4+4x+4 has no integer roots, indeed a root has to divide 4. it cannot be +-1 as mod 2 gives a contradiction and it cannot be divisible by 2 as mod 8 gives a contra

[latent anvil]

The same argument proves x^4 + 4x + 4 doesn't have a root over Z, so if it factored it would factor as two quadratics

ah okay, looks like det got it first

I was going to look at the coefficients of a product of quadratics and deduce a contradiction

[rustic crown]

what's resolvent cubic?

and how does it help computing galois groups?

till now my only way of computing galois group is bashing 😦

[latent anvil]

Iirc it's a way to test whether the galois group is A4 or S4

[rustic crown]

I know the thing where you look at the discriminant being a perfect square to see if galois group is contained in A4

[thorn delta]

see this

and this

[latent anvil]

did my argument for why the resolvent is irreducible make sense?

[thorn delta]

I think so. I'm digesting det's proof atm

[latent anvil]

It does look pretty magical haha

[thorn delta]

x^4+4x+4 having a linear factor over F3 implies it has a linear factor over Q?

[latent anvil]

But factoring over Fp for small p is really useful when trying to figure out if a polynomial is irreducible

[rustic crown]

no it implies it doesn't factorize into two quadratics

both would be monic, so going mod 3, both will retain x^2 term

[latent anvil]

it's not about having a linear factor, it's about having an irreducible cubic factor

right?

if it factored as two quadratics, the factorization into irreducibles over F3 would be made up of quadratic or linear terms

[rustic crown]

yep

[latent anvil]

based on whether the quadratics remain irreducible

which violates unique factorization, bc we know it has an irreducible cubic factor

[rustic crown]

how did you compute the resolvent?

it feels like if i multiply things out then the coefficients would be symmetric functions in the roots, so we should be able to compute them from the coefficients of the polynomial

[thorn delta]

Re Shamrock: mmkay i agree. I'm going to eat and come back to this. I figured out the galois group btw: its S4. There is a theorem which says that if the polynomial factors over Fp, and the degree of the factors are n1, n2, ...., nk, then the galois group contains a permutation of cycle type n1, n2, ..., nk. In this case, the factorization over F3 tells us the galois group contains an odd permutation.

[latent anvil]

nice!

[thorn delta]

like this, det

[latent anvil]

wait, I'm confused

Won't the cycle type be 1,3?

Which is even?

[rustic crown]

oh okie

[thorn delta]

a 3 cycle is a product of two transpositions

am i being dumb

[latent anvil]

two is an even number

[golden pasture]

all symmetric polynomials can be expressed as elementary symmetric polynomials, and the proof is constructive

[rustic crown]

but 2 cycle is odd 🙂

[rustic crown]

yea that's what i thought

[latent anvil]

right, and a three cycle is even

[thorn delta]

oh wait yeh, for some reason I thought a 1 cycle was even. losing my mind...

[latent anvil]

hahaha

Wait...it is!

[rustic crown]

lol

[latent anvil]

[thorn delta]

oh yeah i guess it is 🤦‍♂️

okay, so the discriminant is an integer, so A4 it is

[rustic crown]

no

[latent anvil]

hahaha

whoops

[rustic crown]

discriminant perfect square => contained in A4

right

discriminant is 2^8 * 37

[latent anvil]

Yeah I think I showed that there's a 4 cycle?

Which would imply it's not contained in A4

[rustic crown]

discriminant is defined as product of (u_i - u_j)^2 over i < j

[latent anvil]

ah nvm I missed that the polynomial actually has to factor here

I showed it's irreducible over F5

[rustic crown]

this is a symmetric polynomial so discriminant of a monic integer polynomial is always an integer.

[thorn delta]

Im using this fact, det. Delta is the discriminant

[rustic crown]

oh from what i have heard people define Delta^2 to be the discriminant

[latent anvil]

wait

[thorn delta]

oh wait no, Delta^2 is the discriminant fuck

[latent anvil]

,w factor x^4 + 4 x + 4 over GF(23)

[cloud walrus]

Results provided by WolframAlpha

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

Requested by shamrock#0598

[latent anvil]

The galois group contains a transposition

[latent anvil]

Can you send me a reference for this btw?

It feels familiar but I don't think I've seen it in a book

This magic plus the resolvant thing solves it

But is a little unsatisfying

[thorn delta]

there is an extra condition that 23 should not divide the discriminant, which it doesn't. I originally heard this from my professor, but we did not prove it. Here is a stack exchange post i found about it

https://math.stackexchange.com/questions/836795/proof-of-dedekinds-theorem-on-the-galois-groups-of-rational-polynomials

[latent anvil]

ah okay

It sounds like there's a more direct way if you're looking at the discriminant anyway

it's been too long since I thought about this stuff

[golden pasture]

seems direct from splitting of primes on extnesions🤔

nt moment

add chebotarev for existence fun

[thorn delta]

I was focused on det's proof earlier, but I'm not sure I'm understanding this. Why wouldn't this same argument work with a reducible polynomial like, say, x^2 + 4x + 4?

ah yea that makes sense.

thanks!

[chilly ocean]

In the statement of the fundamental theorem of algebra, it says that p(x) is a polynomial over a field. I understand that that’s the same thing as finite integral domain. Why is the theorem not valid for infinite integral domains?

[carmine fossil]

What's the fundamental theorem of algebra over F?

afaik,FTA just says C is closed

[chilly ocean]

It says that a polynomial p(x) over a field F of degree n has at most n roots counting multiplicity

[rustic crown]

how many roots does x^2+1 have over Q?

[chilly ocean]

None

[rustic crown]

you just said 2

[carmine fossil]

*at most

[rustic crown]

oopsie

sorry

i'm dumb

[rustic crown]

afaik, this is one of Lagrange's theorem

[chilly ocean]

How does the statement change if it’s an infinite integral domain not a field?

[carmine fossil]

It doesn't?

Take a polynomial over R[x] where R is an integral domain

Let's say it has more than n roots

[chilly ocean]

Yea

[carmine fossil]

Now look at it as a polynomial over F[x]

Where F is field of fractions of R

[chilly ocean]

Now it has at most n roots correct?

[carmine fossil]

Yes

But,we have a contradiction

[carmine fossil]

.

[chilly ocean]

I can say It has more than n roots in R which is an integral domain and at most n in F then

[carmine fossil]

Well,If f=(x-a)p(x) in R[x] same factorisation works in F[x]

So,roots in R[x] will be roots in F[x]

[chilly ocean]

How can we prove that though?

And also it that’s true why didn’t they just use the word integral domain in the first place?

Maybe this theorem doesn’t apply to infinite integral domains which aren’t fields

Why is the question

[carmine fossil]

Do you know what a field of fractions is?

[chilly ocean]

I’m afraid not

No wait just googled it

[rustic crown]

(if you want, lets not bring some other ring to the picture... consider the non-zero polynomial of degree n, we'll do a simple induction. if n = 0 then its easy. for a general n, if the polynomial has no roots, then nothing to prove. otherwise says it has a root 'a'. Now you can write f = (x-a) * g. with g a non-zero polynomial of degree n-1 < n. Since we're working in an integral domain, only say (x-a)* g is 0 is when one of the factors is 0. so number of roots of f counted with multiplicity are number of roots of x-a counted with multiplicity + number of roots of g counted with multiplicity. But the latter is bounded by 1 + (n-1). Which completes the induction.)

[chilly ocean]

So it’s applicable also to infinite integral domains correct?

[carmine fossil]

Yes

[rustic crown]

yep

[chilly ocean]

Thanks so much

[vale idol]

hey guys, quick question, are T1, T2 and T3 all isomorphic to ℤ ?

[carmine fossil]

Are they all abelian groups?

[vale idol]

assuming i’ve constructed Tm correctly (which i believe i have) T1 and T3 should be but not T2, but i don’t know why

they’re quotient modules

sorry i should have mentioned, x and y are previously defined to be generators of ℤ

i have a module B=ℤ=<x>, a module F=ℤ=<y>, and a module Pm=ℤ(m,-4) and Tm is defined as B+F/Pm

so other than T0 i'm just struggling to see what the structure of Tm "looks like", since the relation mx=4y doesn't feel like it really does anything

[carmine fossil]

Are x and y integers?

[vale idol]

yes

[carmine fossil]

So does T_1 refer to the set ,{...(0,0),(4,1),(8,2)...}

[vale idol]

from what i've understood, the elements of Tm should be the classes of (b,a) with b∈ℤ and a∈ℤ4

to be honest i haven't really understood this T module

in my particular example i have A=ℤ4, B=ℤ, F=ℤ, P=4ℤ

and i have calculated that Ext(A,B)=ℤ4

and i'm trying to compute the short exact sequences associated to each element of Ext(A,B)

[chilly ocean]

@vale idol Is it a Ring?

[vale idol]

yeah these are all modules of the ring ℤ

[chilly ocean]

Interesting

[languid meteor]

https://cdn.discordapp.com/attachments/540211747613704221/833762344945057802/Capture2.PNG

Am I correct in saying there's only one possible jordan normal form that corresponds to this characteristic and minimal polynomial pair?

[rustic crown]

yep

[languid meteor]

cool, thank you!

[static temple]

well not for 0

the field can be finite for eg

[rustic crown]

define the degree of 0 to be -infinity or infinity 😛 and change the theorem to polynomial of degree n has atmost |n| roots.

[static temple]

(well thats true even when not finite but if you say that its degree is + infinity at 0

no it doesn't work, except if you say that the degree of 0 for a field of size n, is n, then it works

but it is quite absurd I would say

[rustic crown]

0 has infinitely many roots whether or not the field is finite (if we count with multiplicity 😛 )

[static temple]

oh thats actually true!

[strong valve]

what are you smoking?

[static temple]

is he wrong?

except if we consider 0 to be a field...

[strong valve]

no i'm just messing wid him

[static temple]

ooh i see

thats adorable

[strong valve]

😛

[rustic crown]

.<

[chilly ocean]

i wonder if this proof that the set A of algebraic elements of a field K subfield of a algebraically closed field L form a field is valid, even tho it sucks cuz it uses axiom of choice without need i think.

Simply, make a well-ordering of all algebraic elements of K, and then by transfinite induction keep extending K by each element. In end we have a chain of subfields of L whose union consists precisely of the algebraic elements of K. The union of a chain of fields is a field, so should be done if i didn't make mistake.

How do you get a chain of subfields? (I don't see that the well ordering of the algebraic elements should play nice with the field properties of L)

[rustic crown]

i think the idea would be, for the smallest element define it to be K(smallest element) and for bigger elements, we have defined the thing for smaller elements i.e. we already have a chain of subfields, so take their union which should be a field and adjoin this new element.

But i think this still doesn't answer why we didn't add any non-algebraic elements.

[chilly ocean]

Ah I see, I was being dumb

[chilly ocean]

But we are extending by only the algebraic elements, so this should never introduce non algebraic ones right?

yea ^

[rustic crown]

so the question is K(a, b) if a and b are both algebraic over K, then why is a+b?

if you can answer the questions for a+b, a*b and 1/a if a is not 0, then you already proved it.

[chilly ocean]

oh yea then all i'm saying is pointless you can do directly from that (and have to do that i guess) then

Oh, wait yeah, I think I agree with det

[rustic crown]

[chilly ocean]

det honorable

i agree with det too, thanks for making me realize that what i said was complete bs

[rustic crown]

[chilly ocean]

I guess I feel sort of left out, the rest of the gang are all honorable now. But I will not waver, I stand here, based, and proud not to be honorable

[next obsidian]

Det is peepee color

[rustic crown]

[nova plank]

When you think you're part of the gang...

[next obsidian]

Sup Luna. I feel like I haven’t seen u in a while but ur in very active gang

[nova plank]

I haven't seen you in forever ever

Thought you would be Emeritus

[next obsidian]

What

[nova plank]

I thought you were inactive because I haven't seen you

[next obsidian]

Weird

I’ve had like... a couple thousand messages the past month

Like 3k?

[nova plank]

Oh, idk how to check that

But I've been here all the time too

[next obsidian]

On desktop client

I searched “from:username after:one month ago’s date”

I did this like... 5 days ago?

[nova plank]

I'll check out of curiosity

Anyways hii

[unique juniper]

Question about norm

[next obsidian]

Yes

[unique juniper]

if norm of a = norm bc will a=bc always?

[nova plank]

Just under 3k

[next obsidian]

No

[unique juniper]

i dont think i understand it

[next obsidian]

Lots of questions here

1: what are you considering a norm on? The real numbers?

2: what norm? Just any norm? The usual Euclidean one on R?

[unique juniper]

hm

[next obsidian]

But also, the answer is always no

Well

Okay, I can’t say for sure it’s always no but like

I’m pretty sure norm(-1) = 1

Or like the unit

[unique juniper]

yeah

[nova plank]

As an example, for the Euclidean norm, |2| = |(2)(-1)| but 2 ≠ -2

[next obsidian]

But I’m not confident in that

Yeah

So that’s what I was going to say

But again, I’m not sure in what context you mean a norm

If this is purely for R^n this probably isn’t an algebra question

Moreso analysis

But maybe you mean a norm in the sense of used for the Euclidean algorithm?

[unique juniper]

right, i mean norm so we can define a euclidean domain

[next obsidian]

Yee okay

So yeah, still no

[unique juniper]

i see yeah

[next obsidian]

Honestly the idea of it is just to give numerical values to the things in your ring

You can think of it as “size” but it doesn’t have to be related to size at all

And the properties of it just exist so that we can use these to make sense of dividing with remainder

Basically the whole point of the Euclidean algorithm is to let us divide stuff with remainder

[chilly ocean]

It's true on F2 as a vector space over itself

[unique juniper]

i see

[next obsidian]

And we want this to eventually end, so we can enforce that the remainder needs to always be “smaller”

Because the idea is if you always get smaller, eventually you can’t get smaller so you’re done

[nova plank]

You used bc, but ignoring a product, you are basically asking if |a|=|b| iff a=b, and I can't think of a single example where this is the case.

[next obsidian]

But in an arbitrary ring like, what does “smaller” mean

We don’t have a notion of it, so we instead make a norm which assigns to each element a number

And from there we can make sense of it

[unique juniper]

yea

but

[next obsidian]

Note: if your norm takes values in R tho what I said about eventually terminating isn’t true

[unique juniper]

what

if

[next obsidian]

If it’s only in N then it will

Because of well-ordering and blah blah descending induction

[unique juniper]

what if the norm(a) is a prime in integers

can we always show that a is irreducible?

[next obsidian]

There’s actually a few conventions for norms (in the sense of what they must satisfy)

I forget which combinations are used, so can you spell out what the ones you’re working with are?

[unique juniper]

um dont think we are using any

[next obsidian]

Eh?

I mean you surely have a definition of what a norm is right?

That’s all I mean

[unique juniper]

yes

oh

ohk

yeah

so

Norm : Ring -> Z+ U {0}

[next obsidian]

Minus 0?

Cant the norm take on the value 0?

(Probably like norm(a) = 0 iff a = 0?)

[unique juniper]

yeah youre right sorry lol

[next obsidian]

Okay yup

And then is that it?

[unique juniper]

yes

[next obsidian]

Nothing about linearity, multiplicativity

Okay then no

If you don’t enforce it to have more stuff, then for example

Norm(a) = 1 for all a ≠ 0, norm(0) = 0 works

But this won’t work in the sense of making the Euclidean algorithm work

[unique juniper]

yeah

[next obsidian]

So if you assume the norm makes the Euclidean algorithm work

Then let’s see if we can show norm(a) prime means a irreducible

I’m not sure tbh

Lol

I mean all we know is that we can write

a = bc + r for r having norm less than norm a

But we can take r = 0

So it always satisfies that

[unique juniper]

hm

[rustic crown]

polynomial degree >.<

[next obsidian]

I think you have to be clever to make that one work

Namely, I think you have to shift degrees by 1 to get an honest Euclidean function

But i think it works nonetheless as long as you just shift, like I bet you can define it so that 0 has degree -1 or like -infinity then basically shift everything over

[rustic crown]

The definition i usually see is map R\{0} --> Z_{>=0}

[next obsidian]

Yeah that one looks familiar as well

I honestly forget the specific nuances of this

Like... I think if you do that

You have to instead specify this

For any p, and q ≠ 0

we can write

p = aq + r where

Norm(r) < norm(p) or r = 0

[rustic crown]

yep

[next obsidian]

Or you can instead assert the norm of 0 is 0

And only 0 has this

Then you can simply say

Norm(r) < norm(p)

Or something like that

[rustic crown]

norm(q) >.<

[next obsidian]

Oh right

Lol

[rustic crown]

lol

[next obsidian]

Anyway, I think sometimes you require norm(a) <= norm(ab)

Blah blah blah

I am not a fan

Lol

[unique juniper]

this looks hard

so can we show that a is irreducible with this norm?

[rustic crown]

i remember seeing like you can define norm'(a) = min{norm(ar) | r in R\{0}} to get that property...

[next obsidian]

I don’t think so even with the condition I said

[unique juniper]

oh

[next obsidian]

If you have multiplativitiy

So that norm(ab) = norm(a)norm(b) I believe it’s true

[unique juniper]

like Norm(x)Norm(y) = Norm(xy) you mean?

i see

[rustic crown]

you can detect it if your norm is nice enough. For instance take R = Z[i] and define norm has N(a+bi) = a^2+b^2. Then N(xy) = N(x)N(y). Since this norm respects multiplication you'll be hapiper.

[next obsidian]

Since I think norm(a) = 1 iff it’s a unit

Yes

[unique juniper]

i see

[rustic crown]

but still you don't get the full story. 3 is irreducible in Z[i] but N(3) = 9 which isn't irreducible in Z. But you can definitely say that if N(x) is irreducible in Z then x is irreducible in R.

[unique juniper]

right

with multiplication of norms, i still dont see how its done

[chilly ocean]

Given this dot product, https://gyazo.com/d7985836b10a11aa9ac976abbb196f15 , with coeficients on C, how can i prove z^n is orthonormal?

[next obsidian]

Prove that norm(a) = 1 iff a is a unit

Then it follows by just writing x = ab, then applying multiplicativity

One of them (of a and b) has to be a unit since one has to have norm 1

Because well... prime

[chilly ocean]

?

[chilly ocean]

do you mean to say "prove that the set of functions of the form z^n for some natural number n is orthonormal?"

what are the a_n's and b_n's here?

[next obsidian]

So i think you might need to assume the norm actually is a Euclidean function

Basically, if you assume this

[unique juniper]

yea i was kinda struggling

[next obsidian]

Assume N(p) = 1

Write 1 = pq + r

Then r is 0 because N(r) < N(p)

So you just have 1 = pq

It’s true that given a multiplicative norm, a unit has norm 1

But the other direction requires the Euclidean algorithm I think

[unique juniper]

ah right

[next obsidian]

Oh lmfao this is actually obvious

If you take an intral domain

And take that stupid norm I did

N(a) = 0 if a is 0, and 1 otherwise

Point is, this is a multiplicative norm, but it’s not true that N(a) = 1 iff a is a unit

[unique juniper]

yikes im kinda confused

so if we have a euclidean domain

you were saying

and Norm(a) = prime

[next obsidian]

And it’s multiplicative

Hahaha

[unique juniper]

and multiplicative lol yes

[next obsidian]

Then a is irreducible

[unique juniper]

yes

ok

:D

ty

[next obsidian]

...

Hahahaha

Shamrock strikes again

[thorn delta]

how can I show that x^4 -7x^2 -3x + 1 is irreducible over Q?

[chilly ocean]

I guess you can easily show there are no linear factors, then write down this polynomial as a product of two quadratics. I'm not sure if there is an easier way

Well, that plus gauss lemma to assume it factors over Z

[past temple]

given that f is in F[x] for a field F, does g | f => g' | f' ?

where g', f' are the formal polynomial derivatives of g,f resp

[thorn delta]

wait, how do i know that there are no linear factors? Reducing modulo p, I get an irreducible cubic factor. Therefore if f factors in Z[x], then f has an irreducible cubic factor, and therefore a linear factor, right? This would be a contradiction

i absolutely suck w/ polynomials, so hopefully im not spouting gibberish

[chilly ocean]

You can show there cannot be a linear factors with rational root theorem

[old lava]

no

[chilly ocean]

I think the implications is backwards here, if f factors in Z[x] with a cubic factor, then it factors modulo p with a cubic factor

[past temple]

rip

[thorn delta]

hmm right, okay. Anyway, the rational root thing helps a little. Forgot about that

[chilly ocean]

That's probably one of my favorite theorems about polynomials

[thorn delta]

hmm honestly, im not 100% clear on what this tells us about f=x^4 -7x^2 -3x + 1.

by rational root theorem, there is no factorization into 1+3. So that leaves 2+2

[chilly ocean]

Ah, nice. I think I have seen this trick before, but totally forgot about it

[thorn delta]

yep, and this is a contradiction?

i see. But if we assume f factors as two irreducible quadratics, then reducing modulo 2 gives us a linear factor, which is also bad. Therefore f is irreducible?

so, the cubic factor is a problem since if f factors as two quadratics, then f can't have an irreducible cubic factor?

I think I get it. This technique is pretty big-brain ngl .

[next obsidian]

The actual secret here is

if you think a polynomial can't factor, test a few primes

if it holds for like... 3 of them, it probably just is true

this is a Chmonkey non-guaranteed, but morally guaranteed technique

also if you think something does factor it gives you a way to guess what it might factor as

[chilly ocean]

I was gonna type out a bullshit "theorem" about this chmonkey technique being asymptotically probabilistically correct, but I found myself spending too much time/spending too much effort for a meme

[thorn delta]

yea ive been learning mathematica just for this haha

[chilly ocean]

was trying to find the splitting field of g(x) = x^4+x^2+1 over Q and set u = \alpha^2 where \alpha is a root of g(x) and just tried to find the roots over Q closure and try to find the smallest adjoin of Q that I need to get all those roots
But I think the u = \alpha^2 -> u^2+u+1 = 0 to quadratic formula idea breaks down for some reason, none of the roots I got when looking at u^(1/2) and juggling the exponent in exp(i pi (..)) correctly actually worked when I tried yeeting them into wolfram

even though when I plug things on paper, these steps are reversible

and should get me back that what I got are actually roots

but I know I have to be wrong

Cause at the end there I can get extra roots from -u^(1/2) since -1 = e^(i pi) and that gives me 4 more "roots" that also work on paper but not on wolfram, but they're distinct and g(x) definitely cannot have 8 distinct roots since deg(g) = 4

wait

found my error lol

[next obsidian]

haha

[chilly ocean]

arithmetic error somewhere in there, no worries

[next obsidian]

for quartics like this you can totally do quadratic twice

so I'm glad you figured out what went wrong

[chilly ocean]

yeah it works now 😭

[thorn delta]

When is the splitting field of a polynomial a simple extension? I've noticed that for some specific polynomials like quadratics, you'll get nice relations between the roots like K(a+\sqrt b, a - \sqrt b) = K(a + \sqrt b) = K(\sqrt b), but this isn't always the case. e.g. when a polynomial in Q[x] has real and complex roots. Just wondering if there is some theory surrounding this.

[chilly ocean]

idk that is cool tho, hmst

I think Galois groups answer similar questions

Oh?

nutty

[next obsidian]

cring

Hey Sham did you know that if an algebraic group acts on a variety, a lot of things are closed

Also I don't know anything about topological / lie groups

so I'm curious what is true there

[thorn delta]

(thanks sham)

[next obsidian]

and what isn't

ur welcome

Chmonkey is always helpful

Inverse limit of F_p[x]/(x^{p^n})

Idk

Probably not a field

Lmao

Uhhh

Maybe

Probably

As long as you’re okay to skip some ahit

Which I think you are

Yeah

I think this generates the same topology

So yeah

Lmao

Okay so it’s just powerseries over F_p

Maybe F_p(x)

Cringe

Why isn’t it perfect

Oof

Oh

Wait I meant

Double (())

Yeah

But only finitely negative terms

Cuz then you can’t multiply haha

Uhhh

I don’t see how

Oh I see what you’re saying

Ummm

I don’t see how you’d get that

You can?

I thought you’d get fucky exponents

Yeh

Right

Cring

Yeah idk lol

Okay tske the perfect closure of...

Wait is that a thing?

It must be

No way that isn’t a thing

That’s just morally wrong

Also dumb question but like

Why positive characteristic?

Okay so I thought this might be the case but

Was perfect just saying anything has a p-th root?

Inside the field

Ah

Well you can’t be not separable in char 0 right

[delicate hawk]

what is $\mathbb{Z}[i] \setminus \langle 1 + 2i \rangle$ isomorphic to?

[cloud walrus]

Fiwam

[delicate hawk]

isnt this just $a + bi$ such that $a$ is an integer, but $b$ can either be 0 or 1?

[cloud walrus]

Fiwam

[carmine fossil]

Yes

[delicate hawk]

is it isomorphic to $\mathbb{Z} \setminus (1^{2} + 2^{2}) \mathbb{Z}$?

[cloud walrus]

Fiwam

[delicate hawk]

which is Z/5Z

i thot we had Z[i] / (a + bi) is iso to Z / (a^{2} + b^{2})Z is a and b are copprime

no, i just think i saw it somewhere

[carmine fossil]

So,suppose you have 1+ai,you can do
(a+i)-i(1+2i)=a+2

So,You only have to look at Z/5Z

So,yea that's true

[delicate hawk]

its just weird becuz Z/5Z is a subset of Z which itself is a subset of this set

[next obsidian]

What

Z/5Z isnt a subset of Z

[rustic crown]

$$\frac{\mathbb{Z}[i]}{(1+2i)} \cong \frac{\mathbb{Z}[x]/(x^2+1)}{(1+2\overline{x})} \cong \frac{\mathbb{Z}[x]}{(x^2+1, 2x+1)}$$

[cloud walrus]

det

[delicate hawk]

chomsky isnt Z/5Z just the integers without 5, 10, 15, 20, ...?

[next obsidian]

No

You have to consider the operation

Without considering the operation, then you aren’t saying anything meaningful, you could say Z/5Z lives inside any set with at least 5 elements by taking 5 random elements and just artificially declaring one of them to be a generator

[carmine fossil]

Let ra+sb=1, map x+yi to x+yi-y(r(b+ai)+s(-a+bi)) mod (a^2+b^2)

[next obsidian]

No

Literally any set

[carmine fossil]

Kernel of that map will be (a+bi)

[next obsidian]

Tske 5 elements, label them a,b,c,d,e say that a^2 = b, a^3 = c,...

I mean you get a subset of that “isomorphic to Z/5Z”

If you mean to say the objects of Z/5Z live inside of Z... even then its not true

Since Z/5Z elements are (probably) defined as equivalence classes

[final pasture]

there's different ways to define Z/5Z

[next obsidian]

I mean you can define it however you want really

Like you could take it to be {0,1,2,3,4}

And then the operation is simply addition and then reducing mode 5

[final pasture]

and you'd still call it Z/5Z by convention

[celest brook]

Z/5Z is just isomorphic to subgroup of S_n

[nova plank]

Lol, cancel it by convention

Z/5Z = 1/5

[celest brook]

[rustic crown]

(1)/(5)

[nova plank]

Omg

That works

[celest brook]

(12)(34)(56)(78)

product of train positions

[red imp]

lol

[celest brook]

every finite group is just subgroup of S_n

yeah I saw that

[rustic crown]

lol

[thorn delta]

Suppose that $F$ is a field and $K \subset E$ are subfields and $[F:K] = [F:E] < \infty$. Then does $E = K$?

[cloud walrus]

kxrider

[rustic crown]

what can you say about [E:K]?

[hot lake]

$K \subset E \subset F$ ?

[cloud walrus]

Zef Klop 🍃 🌿 🌻

[thorn delta]

ah i see

[F:K] = [F:E][E:K] so [E:K]=1, nice

[rustic crown]

yep!

[cloud walrus]

squirtlespoof

[rustic crown]

i think you want is in the exponents as well.

Then notice that the field on the left is just Q(a) where a = e^{i pi /4} because other 3 are just powers of a.
then recall that a = (1+i)/sqrt(2)

So a definitely lives in Q(sqrt(2), i)

Conversely, i lives in Q(a) as a^2 = i and sqrt(2) also lives here as it is (1+i)/a = (1+a^2)/a

yep

why does x = a + bu

can it not be a+bu+cu^2 + (du^3 + eu^4 + fu^5)?

also at the bottom, you know that Q(u) is contained in the span of {1, sqrt(2), crbt(5),....} but how do you know every linear combination is possible?

like what information does u^2 and u^3 give you?

i can see that anything in Q(u) can be written as a linear combination of those 6 elements, but how do you get the reverse inclusion as well?

could you show me why this is an equality and not just Q(u) is a subset of that span

how does it take care of that?

but how will that show sqrt(2) in Q(u)?

[strong valve]

🤔

[rustic crown]

||if you're still struggling with that, try to cube both sides in u -sqrt(2) = cbrt(5), and then solve for sqrt(2)||

[chilly ocean]

I could use some help with a question about normalizors. I have an H = <(1 2 3)>. I want to find the normalizer in A_4. How would I go about doing that?

[carmine fossil]

A normaliser of H shouldn't consist of a cycle like (14)

In S_4,There are 6 normalisers

So,in A_4 there are atmost 6 normalisers

[hidden haven]

Try proving that a permutation is a normalizer for this iff it fixes 4

[carmine fossil]

(1,2),(2,3),(1,3),1,(1,2,3),(1,3,2) form the normaliser group if we were looking at S_4

[chilly ocean]

Wait why's that exactly?

[carmine fossil]

Because (14)(123)(14) is (423)

Not in the group

[chilly ocean]

Oh yhh

[carmine fossil]

This

[rustic crown]

(you can use sylow theorems to avoid any computations >.<, number of conjugates is index of the normalizer and is 1 mod 3. can't be 1 since that subgroup isn't normal, hence normalizer has index 4 which means its the subgroup itself.)

[scarlet estuary]

excuse me, there is very clearly a computation there

[hidden haven]

imagine remembering sylow theorem statements

[scarlet estuary]

you had to compute 1+3

[strong valve]

please don't spam

[rustic crown]

(i thought you'll say how did i know the subgroup wasn't normal >.<)

[strong valve]

det stop doing Algebra

!!!

[hidden haven]

imagine reading anything after you use the word sylow

[rustic crown]

then teach me differential equations, i'm gonna do so bad in the quiz tomorrow

[carmine fossil]

PDEs?

[strong valve]

well there's a non-zero chance that quiz will contain lin alg

[rustic crown]

nah just ODE, need to read about rectification theorem and topological straightening

[strong valve]

english pls

[unique juniper]

$R = {a + b\sqrt{5}i | a,b\in \mathbb{Z}}$

[cloud walrus]

Yes

[unique juniper]

trying to show that $(1 + 2\sqrt{5}i)$ is prime

[cloud walrus]

Yes

[golden pasture]

I'd say the easiest is consider the norm function from R to Z

then you know what are the possible factors of (1+2sqrt(-5))

and show none of them is possible

[unique juniper]

i seee

with a norm (a + b\sqrt5 i ) = a^2 + 5b^2 and N(xy) = N(x)N(y)

[golden pasture]

yesh

[unique juniper]

i still cant seem to do it

norm cant be 3

[golden pasture]

so what's the norm of 1+2sqrt(-5)

[unique juniper]

21

[golden pasture]

ok ye

it can't be 3 right

[unique juniper]

yeah it cant

[golden pasture]

so that tells you you cannot factor into 3*7

[unique juniper]

yea

[golden pasture]

but there isn't other possible factorizations

[unique juniper]

yea

[golden pasture]

so you're done 1+2sqrt(-5) is an irreducible element

[unique juniper]

buttt

its irreducible sure

but we need to show its prime too!

[golden pasture]

(note that the ideal generated by 1+2sqrt(-5) is not actually prime i believe)

[unique juniper]

integral domain

[golden pasture]

Consider (2+sqrt(-5))(4+sqrt(-5)) = 3+6sqrt(-5)

that's cuz the ring you're considering is not actually a principal ideal domain, and not a unique factoring domain

here we found 2 different factorizations of 3+6sqrt(-5), (2+sqrt(-5))(4+sqrt(-5))=3(1+2sqrt(-5))

[unique juniper]

so you mean

show that (1+2sqrt5i) isnt a prime ideal

[golden pasture]

the ideal isn't prime cuz of this

although the element is irreducible

[unique juniper]

oo

ok

ty

[rapid meadow]

Hey everyone, does anybody have an idea about how to prove that in a PID every ideal is integral closed?

[hot lake]

integral closed ?

[wind steeple]

that means that if x is in Frac(A) (your ring) and is a root of a polynomial with coefficients in I (the ideal) then x is in I ig

[gentle pendant]

The poly should be monic. I think you can just mimic the proof of the rational root theorem.

[rapid meadow]

Here is the definition, I'm working in 4.15

I've been trying to look at the quotient of R by I and to show that the projection of the integral closure is zero, but it doesn't work

[vestal snow]

@next obsidian I was reading that argument from before and I had a question about it. You used that if g: B --> C is epic, then coker(ker(g)--> B) = B --> C

Why is this true?

[wind steeple]

this is bc Coker(Ker(g) -> B) = B/Ker(g), and you know that B/Ker(g) = Im(g) = C since it's epic

ig smth like that

[vestal snow]

What does B/Ker(g) mean here?

In an arbitrary abelian category

[wind steeple]

I see an abelian category as a full subcategory of some R-modules

so I can talk about quotient like that

btw B/Ker(g) is exactly the coker of Ker(g)->B in a category of R-modules

[vestal snow]

Okay so let's use this definition

So I should try and show that B/Ker(g) = Im(g)?

[wind steeple]

okay we can prove it directly ig but it's a bit tough

epic means that Coker(g) = 0

[vestal snow]

Yes

Wait

Yeah I agree

I proved that already

[wind steeple]

you want to prove that B->C is the coker of Ker(g)->B, let's prove that B->C satisfies the coker universal property

[vestal snow]

Yeah

We know that Ker(g) -> B -> C is 0

[wind steeple]

the composition Ker(g)->B->C is 0 then B->C factor through B->Coker(g)->C

this factors through B->Coker(g)->Im(g)->C

[vestal snow]

Why?

[wind steeple]

mh sorry let me think a little

okay, you have that Coker(Ker(g)->B) = Coim(g) by definition

[vestal snow]

Wait

[wind steeple]

so the statement above was a typo

I wanted to say that it factors through B->Coim(g)->C

[vestal snow]

Is this a simple verification through the universal properties?

[wind steeple]

Coim is by definition the Cokernel of the Kernel

what's your definition of Coim ?

[vestal snow]

The dual notion of this

https://en.wikipedia.org/wiki/Image_(category_theory)

[wind steeple]

lmao

okay

[vestal snow]

Let's use your defintion

[wind steeple]

ok let's admit that the coim is the cokernel of the kernel and the im is the kernel of the cokernel

[vestal snow]

Yeah

[wind steeple]

thus you have B->Coim(g)->C

which factor throught B->Coim(g)->Im(g)->C

[vestal snow]

Wait

So we have B -> C

Which is epic

[wind steeple]

yes

I haven't used the epic property yet

[vestal snow]

So why do we have the factorization B -> Coim(g) -> C?

[wind steeple]

because Ker(g)->B->C is zero

[vestal snow]

Okay

[wind steeple]

by universal property of the Coker, we have that B->C factors through Coker(Ker(g)->B)

[vestal snow]

That makes sense

[wind steeple]

which is Coim(g)

and Coim(g)->C->Coker(g) is zero because B->Coim(g)->C->Coker(g) = B->C->Coker(g) is zero and B->Coim(g) is epic (bc it's a coker)

[vestal snow]

Yeah

[wind steeple]

then Coim(g)->C factors through Coim(g)->Im(g)->C

this morphism Coim(g)->Im(g) is an isomorphism by the definition of abelian category

let's see now that Im(g)->C is an isomorphism.

[vestal snow]

Wait

[vestal snow]

Why is this true?

[wind steeple]

because Coim(g)->C->Coker(g) is zero

and because Im(g) = Ker(C->Coker(g))

[vestal snow]

Got it

[vestal snow]

What part of the definition?

[wind steeple]

for me the definition of an abelian category is an additive category with all kernels and cokernels

and such that the universal morphism Coim->Im is an isomorphism

[vestal snow]

So coimage and image are the same in abelian categories?

[wind steeple]

yeah that's litteraly the definition x)

[vestal snow]

How do we know that the morphism that we got is the same as the universal one?

[vestal snow]

This one

[wind steeple]

it's by definition the universal one

mh

it's more like is the name of "universal morphism"

for every morphism f : X->Y you have a factorisation X->Coim(f)->Im(f)->X

the factorisation is constructed as above

and Coim(f)->Im(f) is called the universal morphism

because it's the more natural you can construct

did you have an idea what's a abelian category is befora that ? What's your definition btw ?

[wind steeple]

wtf

I've never seen this definition before

I guess it's legit but for me it's not the standard one

[vestal snow]

Tbf, I was originally supposed to work with modules only

But I decided to do it in arbitrary abelian categories instead

Probably not the wisest decision since we hadn't talked about them in class yet

[wind steeple]

if you want to learn a bit about abelian category I suggest you to read the Grothendieck's Tohoku paper

Ig it's translated in english

[vestal snow]

I think I will learn a more standard definition when we do homological algebra

[wind steeple]

idk if mine is the standard one but that's the one I saw in particular in the Tohoku but in all other references I've read about homological algebra

btw to end the exercice you can see that since B->C is epic, you have Coker(g) = 0 and then Im(g)=C

(because Im(g) should be the Kernel of C->0)

thus Coim(g)->Im(g)->C is an isomorphism

you can as an exercice prove that the two definition of an abelian category are equivalent haha, maybe it's false tho

[vestal snow]

Thanks

[chilly ocean]

Everytime I see your name I read Hahn Banach at first.

[vestal snow]

What do you mean?

[chilly ocean]

[next obsidian]

This notation in an arbitrary abelian cat means the cokernel of ker g -> B

And the reason that coker(ker g -> B) = B -> C when g is epic is simply because you can prove that it’s true, but you need to do work

It comes down to the fact that c = im g

And then you use the fact that im = coim

[vestal snow]

Got it. Zak explained the main argument

I should probably hold off on doing this until I read a bit about abelian categories

Because in my mind I was using an ad hoc definition

ie kernels and cokernels exist

[next obsidian]

There’s a few equivalent ones

Are you able to access this?

[vestal snow]

Nope

[next obsidian]

https://www.overleaf.com/read/pfpwbtvvdvqh

Maybe this one

[vestal snow]

Yeah I can access it

[next obsidian]

Compile notes.tex

And find where I prove a lot of crap about abelian categories

I think I didn’t prove that 0 -> A -> B -> C is exact iff A is the kernel of B -> C since I did this in a class last year

But I prove a lot of general stuff about abelian categories

[vestal snow]

Cool I'll check it out

Thanks

[wind steeple]

btw you still can work on abelian category like if it was an R-modules category, it's simplier

[next obsidian]

No

It has to be small

Altho you can usually get around it by taking a subcategory containing only the objects you are using in a diagram

But I think in topology sometimes you can’t even do that trick to embed into R-mod

Also R won’t be commutative which is worthwhile noting

[wind steeple]

why it has to be small ?

[next obsidian]

¯_(ツ)_/¯

If you look into the statement of Freyd-Mitchell there’s size limitations

Also it doesn’t preserve injectives or projectives

[wind steeple]

yeah ok, in practise you'll have to take the full abelian subcategory generated by your objets and it works isn't it ?

[next obsidian]

Normally yes, but I think in topology sometimes you are working with more than a set’s worth of objects

[wind steeple]

yeah okay, but to manipulate ker and coker it seems like we can be in a category of modules

[next obsidian]

Yeah

[wind steeple]

I didn't know the theorem was for small categories

I believe it's not that annoying

[next obsidian]

It isn’t, but usually I think people trying to prove stuff in an abelian category are doing it to test themselves or just for the hell of it

Eg me writing the notes I put in this chat

So using Freyd-Mitchell sort of defeats the purpose

[rustic crown]

I remember seeing a simpler construction in Aluffi, so usually its easier to cook up map using universal properties and usually its easier to verify the commutativity using elements. The construction was pretty nice, and we could also verify exactness like if we were working in Set*.

[delicate hawk]

if we take 2 complex numbers a and b

with a gcd of theirs being c

do we have norm(c) divides both norm(a) and norm(b)

[chilly ocean]

Do you mean gaussian integers?

[delicate hawk]

yes

not necessarily, even stuff like $\mathbb{Z}[\sqrt{-3}]$

[cloud walrus]

Fiwam

[next obsidian]

what norm are you using

if your norm is multiplicative yes

and so if you're using the usual norm given by N(a + bi) = a^2 + b^2, then yes

[sinful mirage]

I am trying to prove this is a projector

I am trying to show $\pi(V^{}\otimes W) \in Hom(V,W)^{g}$ and $Hom(V,W)^{g} \in \pi(v^{} \otimes W)$. if this were true,its image would be exactly what we want. Now trying to prove the first one: Let $\phi \in V^{} \otimes W \implies \pi(\phi)=\frac{1}{|G|} \sum{g \in G} \rho{v_{i}^{}}^{*} (g)\otimes \rho{W}(g) (\phi)=\frac{1}{|G|}\sum{g \in G} \rho_{v_i}^{-1}(g) \otimes \rho(W)(g) (\phi)$

[cloud walrus]

ProphetX

[sinful mirage]

Now I am stuck, I do not know how to continute to show that this phi would be in $Hom(V,W)^{g}$(this is the space of intertwiners)

does anyone have hints?

[cloud walrus]

ProphetX

[tight otter]

I'm attempting to show that the polynomial $x^4-4x^2+2$ splits over $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ but I'm not sure if all the roots of the polynomial are actually contained in that field. I've found the roots to be $\pm\sqrt{2\pm\sqrt{2}}$ but I don't know how to show $\sqrt{2-\sqrt{2}}\in\mathbb{Q}(\sqrt{2+\sqrt{2}})$.

[cloud walrus]

panoramatopia

[sour warren]

Oh wait sorry

Well sqrt 2 is in q(2+sqrt(2))

Just multiplying 2+sqrt2 and 2-sqrt 2

Or rather sqrts of those

So then sqrt(2-sqrt(2)) is in the field

So basically I want to solve 0=(an)f+(a(n-1)x)f'+...+(x^n)f^(n)

I think it's probably not algebra question right

Oh btw @tight otter are you good

[tight otter]

yes thank you!

your explanation helps

still a bit on field extensions

@sour warren actually though you can't multiply $\sqrt{2+\sqrt{2}}$ and $\sqrt{2-\sqrt{2}}$ if you don't know that the latter is in the field, right?

[cloud walrus]

panoramatopia

[sour warren]

No

Multiply (2-sqrt2) and (2+sqrt2) first

[tight otter]

yeah, that gives you 2

ohhhhh

[sour warren]

:)

[tight otter]

you just DIVIDE them

because its a field

[sour warren]

There is probably some way with galois

But I know jack about that stuff

[tight otter]

well i'm trying to show its normal so i can show its like

a galois extension of Q

[thorn delta]

lol i did this exact same thing yesterday for part of a problem. You can show that sqrt(2) is in this field, and sqrt(2)/(sqrt(2+sqrt(2)) = sqrt(2-sqrt(2)) iirc

oh i guess someone already helped

[weary terrace]

Let $G$ be a finite group and $K_g={hgh^{-1} | h \in G}$ the conjugacy class associated with $g$. How do I prove that $G$ is a disjoint union $K_{g_1} \cup K_{g_2} \cup ... \cup K_{g_n}$ for some $g_1, g_2, ..., g_n \in G$ ?

[cloud walrus]

RiesZ

[chilly ocean]

consider the conjugation action, then apply the theorem that says the orbits partition the group

alternatively for a direct proof, first prove that conjugacy is an equivalence relation, then the set of equivalence classes will be your partition

@weary terrace

[rustic crown]

So you basically have to show that K_g and K_h either don't intersect, or they overlap completely.

[weary terrace]

@chilly ocean
Thanks. Is it correct to say that every equivalence relation on any set, partitions the set?

[chilly ocean]

yes

well, the equivalence classes do

[weary terrace]

I tried to but it got really messed up really fast

[weary terrace]

Thx!

[chilly ocean]

and also conversely, every partition gives rise to an equivalence relation

so that's nice i guess

[weary terrace]

Only if you have the time and patience, I'd really appreciate if you could write a concise proof for what you offered

[weary terrace]

Yeah, well I guess it's the more elegant way

[chilly ocean]

It is not too hard

[rustic crown]

all 3 ways are pretty much the same.

[chilly ocean]

Basically just following the obvious path

[weary terrace]

I assumed $g\in K_{g_1}$ and $g\in K_{g_2}$

[cloud walrus]

RiesZ

[rustic crown]

Then try to show that K_g1 = K_g

[weary terrace]

then for some $h_1, h_2 \in G$, I wrote $h_1g_1h_1^{-1}=h_2g_2h_2^{-1}$

[cloud walrus]

RiesZ

[rustic crown]

okie looks good

[weary terrace]

Here it kinda got ugly

oh wait

so we get $g_2=h_2^{-1}h_1g_1h_1^{-1}h_2$

[rustic crown]

not quite

[cloud walrus]

RiesZ

[rustic crown]

yep!

[weary terrace]

oh nice, I guess I had a typo in my notes too

lol ok, sorry

[rustic crown]

no need to apologize

[weary terrace]

❤️

[delicate hawk]

is $\mathbb{Z}[\sqrt{-D}]$ always a PID?

[cloud walrus]

Fiwam

[rustic crown]

nope. D=5 is not even a UFD

[delicate hawk]

is it a pid for D = 2

[rustic crown]

yep

its Euclidean iirc

to see that, say alpha and beta are two elements with beta not 0. we want to find a quotient such that the remainder is smol.

look at the complex number alpha/beta, i wanna pick my quotient very close to this, but that q should be an element of the ring. if you draw the points of Z[sqrt(-2)] you'll see the plane tiled with a bunch of rectangles. Now given an arbitrary complex number, it will lie in some rectangle, so its distance from one of the vertices would be less or equal to half the diagonal which in this case is, sqrt(1+2)/2

|alpha/beta - q| <= sqrt(3)/2

Try to see what bound on the remainder you get using this.

[chilly ocean]

.

Why is deg g=0 in the second case?

[strong valve]

@rustic crown stop

[chilly ocean]

Cuz (g(x))=F[x], so if g(x) was not a constant, then (g(x)) wouldn't contain constants

[strong valve]

g*f = 1 for some f and hence g is a unit

[chilly ocean]

So if g wasn’t a constant then F[x] wouldn’t have any polynomials of the form constant multiplied by some polynomial correct?

[strong valve]

deg(gf) = deg(g) + deg(f) >= 1 if g is not constant

assuming non zero g and f

[chilly ocean]

Ohh got it thanks cos every f is a unit cos F is a field

[strong valve]

btw why do you love chores?

[chilly ocean]

Bruh this isn’t my real username someone changed it

[rustic crown]

that hedgehog is very cute

[chilly ocean]

Ikr

I have another picture if u wanna see

[unique juniper]

$R$ is an integral domain. If we had a map $\varphi : R \rightarrow \mathbb{Z}^+$ such that a = bq + r $\varphi(b) > \varphi(r)$ and that $\varphi(ub) = \varphi(b) $ where u is a unit. how can we show that $\varphi(ab) \geq \varphi(b)$

[cloud walrus]

Yes

[unique juniper]

i cant seem to do this

[chilly ocean]

Something is weird, because eg what if I pick a=0?

[unique juniper]

yeah should be non zero

[strong valve]

what is the statement exactly?

is it given a and b we can find (q,r) such that a = bq + r satisfying varphi(b) > varphi(r)?

[unique juniper]

yes

[chilly ocean]

If a is a unit, Ψ(ab)=Ψ(b). If a is not a unit, consider S that is not a unit. By the division algorithm, S=abq+r where 0<=r<ab. Now we can use the statement Ψ(b)>Ψ(r)

I’m pretty sure thats wrong but at least I tried

[rustic crown]

that's almost correct.

[chilly ocean]

Wow wtf >:)

[rustic crown]

say there was a counterexample (ab, b), replace ab with the multiple bc of b with c not 0 such that phi(bc) is the smallest.
so phi(bc) <= phi(ab) < phi(b)

now bc is not 0, so using division algorithm
b = (bc)*q + r.
r can't be 0 otherwise c is a unit and phi(bc) < phi(b)

Now r must be divisible by b, so r = bd and phi(bd) < phi(bc)
bd is a multiple of b with a smaller norm than bc.

this is a contradiction!

[unique juniper]

why is phi(bc) <= phi(ab) < phi(b)

[strong valve]

c is defined such

[unique juniper]

i mean

phi(ab) < phi(b)

[strong valve]

for the sake of contra

[unique juniper]

ok

i wouldnt have come up with that myself

[strong valve]

same

lol

[rustic crown]

just look at i love chores idea, we got r with phi(r) < phi(ab)

[strong valve]

i mean you have N so it has to do smoething with minimal element

[rustic crown]

so given a multiple of b, you can find a multiple with a smaller norm

but you shouldn't be able to do this indefinitely

i just phrased the argument in terms of well ordering principle instead of induction

[strong valve]

amirite @rustic crown

[rustic crown]

Hai!

[strong valve]

so you agree N = Z+

nice

[rustic crown]

not nice

[chilly ocean]

What are all ur names if I may ask?

[rustic crown]

i'm det

[strong valve]

mine's yaggur

[chilly ocean]

Bruh I thought det was short for determinant

[rustic crown]

(lol it is 😛 )

[strong valve]

lol

it's for determination

[unique juniper]

Im Yes

[strong valve]

@unique juniper

[unique juniper]

Yes?

[rustic crown]

xD

[chilly ocean]

I’m arryan but ppl call me arry

[strong valve]

I'll call you array

[chilly ocean]

Ooh nice

[final pasture]

[paper flint]

[paper flint]

I'll call you gandu

[tepid thunder]

Hey, I have been asked to show that the statement "The Union of 2 ideals of a ring R is again an ideal of R" is false

but I am not sure on how to go about this

[chilly ocean]

Pick a ring, try to find 2 ideals whose union is not an ideal. If you can't, try picking another ring

A reasonable choice for R: maybe Z, or Z mod n for some n

[nova plank]

And think about properties of ideals

[tepid thunder]

closure under addition and strong closure under multiplication no?

[nova plank]

You know if you add two elements from the same ideal, you get another element of the ideal. But if you add two elements from different ideals, what then?

The multiplication won't break because the original ideals are closed under multiplication with any element in R

[strong valve]

please don't swear

[rustic crown]

yes swearing = bad

what does sully mean?

make dirty i see

google

[hidden haven]

See you called yourself yaggur then started being annoying

[chilly ocean]

Yay I finally finished abstract algebra (almost)

[hidden haven]

Abstract algebra as in some book or the subject as a whole?

[strong valve]

yes sometimes i am yadetgu

[chilly ocean]

Not rlly tho I don’t remember many things

[chilly ocean]

A book

Joseph gallian

[hidden haven]

I see

[strong valve]

nice