groups-rings-fields

[solar glacier]

then x-y = a1-a2 + b1-b2 where a1-a2 \in I same for bi's and J

thus x-y \in I+J

and for the ideal property

let r \in R

then by distributive property r(a+b)=ra+rb

where a \in I and b \in J

thus by idea proerty of I,J ra \in I and rb \in J

thus I+J is an ideal

[south patrol]

Yeah nice

[solar glacier]

thx

preparing for alg final

[rotund aurora]

What happens if instead of using the trace we use the determinant?

It will be something different, but det(\wedge ^k A) must also encode a lot of (interesting) information about A

[hard flame]

this look right to yall , problem is [G:H] < p |H| where H subroup of G then H is normal

[obtuse bear]

why do you say that O(gH) = [G:H]? That would be true if the action were G x G/H -> G/H, but your action is H x G/H -> G/H

to see that it's not true, take any g in H, then gH = H and O(gH) is just {H}

[white grotto]

if the gcd is defined for any PID, whats the gcd of 5,7 in R?

[rustic crown]

R, as in real numbers?

gcd is a thing which is defined up to units. gcd(a, b) is a common divisor such that every common divisor divides it. with the definition, you can see that it only makes sense to talk about gcd up to multiplication by an invertible element. for fields every non-zero element is a unit, so gcd of any two elements will be 1, unless both are 0 in which case gcd(0,0) = 0

[rotund aurora]

Is there a useful analog of this for multilinear forms?

[white grotto]

wdym up to multiplication by an invertible element

[barren sierra]

GCD is defined in terms of ideals right?

Errrr

[white grotto]

AH i got it

[barren sierra]

Lemme phrase that better

[white grotto]

Nvm

[rotund aurora]

in a PID it makes sense to do that yes, the gcd of a,b is the ideal (a)+(b)

[barren sierra]

Ah cool

[white grotto]

no elaborate plz

[barren sierra]

Yea that's just what I was about to say

[rotund aurora]

idk if you would say "the ideal", but that ideal has single generators, those generators are gcds.

[barren sierra]

Croqueta elaborated for me lol

[white grotto]

Ah yeah i get it

[rustic crown]

but there is a problem, this definition only works for PIDs

for general GCD-domains (basically ring where gcd makes sense) the ideal (a, b) might not even be principal

even for UFDs, that's not the case

[barren sierra]

Yea

[rustic crown]

3d matrices

idk lol

[rotund aurora]

I said useful

xd

[rustic crown]

[rotund aurora]

I guess thats all you can do lul. I guess this simply states that to evaluate f(v_1,..,v_n) you only need to know the coordinates of v_i and the image of the n-tuples of basis vectors.

[rustic crown]

yea >.<

[white grotto]

show that Z is noetherian

[lethal dune]

No

Hint Z is PID

[white grotto]

i want to use the property that Z is principal

so i am taking the Union of ideals since we have an increasing seq of ideals

and i know that this is generated by just one element

[south patrol]

Yes

[white grotto]

but i am not sure how to move from there

i want to show that there is a smallest one

[lethal dune]

Use one of the equivalent criteria for Nietherian

[south patrol]

What can you say about that element

you are nearly done

[white grotto]

gcd

[south patrol]

Well I mean

[lethal dune]

All ideals are finitely generated

[south patrol]

So you have a union of ideals

[white grotto]

yes

[south patrol]

and this generates the union

[white grotto]

yes

[south patrol]

So how does that relate to the original ideals

[white grotto]

all original ideals are included in it

i am not sure what else

Ah

my element is the the generator for the biggest ideal, the last one

[lethal dune]

You already have that all ideals are fg, which is equivalent to Noetherian

[white grotto]

how?

how is it equivalent

[lethal dune]

Otherwise you can always construct an ascending chain which does not stabilize

[white grotto]

ok i see the first implication, i am asking about the equivalence

[lethal dune]

It’s standard and given in all books

Other way is what you just did

Take the union of the AC and its fg so must stabilize after a point

[white grotto]

you mean the infinite union

Ah

[lethal dune]

Hm

[white grotto]

yeah i see

how about arthinian

how to show its not arthinian

[lethal dune]

You can construct a dc that doesn’t stabilize

Fun fact, all Artinian rings are Noetherian

[white grotto]

ouf

[white grotto]

how

in Z for example

Z, ill remove an ideal, lets say 100000

if i remove all ideals below that

[lethal dune]

|| (2) ⊃(2²) ⊃ (2³) ... ||

[white grotto]

i am bound to go to 0

[white grotto]

fucking hell :p

nice

[lethal dune]

lol

[white grotto]

i am so fucking bounded in thinking :/

[lethal dune]

also

Artinian Integral domains are fields

Z is not a field

[white grotto]

is there some characterization for artinian?

[lethal dune]

hint (x) ⊃(x²) ⊃ (x³) ...

[lethal dune]

dcc?

every set of ideals have a minimal element is another characterization

[white grotto]

i am not sure whats dcc, but i mean like the finitely generated ideals thingy

[lethal dune]

unfortunately no

[white grotto]

whats a minimal element? a minimum?

i know what a minimal polynomial is

[lethal dune]

minimal under inclusion,

[white grotto]

could you elaborate what this means then?

[lethal dune]

meaning there's no smaller ideal contained inside it

Have you read any standard book for these definitions? It's literally an equivalent definition

[white grotto]

i have not, this came up in an exam 2 years ago its not in my book

[lethal dune]

😑

😐

[white grotto]

i was doing old exams and we're asked to prove this

[lethal dune]

I see

just using the acc, dcc conditions I suppose

[south patrol]

Yup

[white grotto]

whats acc

[lethal dune]

ascending chain condition

[white grotto]

ah

[lethal dune]

the increasing chain stabilize thingy

[white grotto]

but wdym i want to understand this

[lethal dune]

you know what a partial ordering is?

[white grotto]

no

[lethal dune]

ok I'll phrase it differently then

[white grotto]

i think i do tho

ok

ah yeah yeah i do

its just a generalization of > and <

to include inclusion and such

[lethal dune]

it means that if you take a set of ideals of R then you must find one ideal among them which is "smallest" in the sense that there's no smaller ideal contained inside it which is also in your set.

[white grotto]

why does this fail in Z

because we can always go smaller

ok

[lethal dune]

say (2), (2²) thingy

[white grotto]

so its kind of like the minimum property for intervals in R

[lethal dune]

if you pick (2ⁿ), you'll always get (2^(n+1)) contained inside it

so no minimal ideal

[white grotto]

yeah

is this to show that R[x] is not artinian?

[white grotto]

.

and can i say that an example of a non noetherian ring is the set of matrices Mn(R)?

or what other examples can i propose

[lethal dune]

they are noetherian over field k

even artinian

because of finite dimension-ness

[white grotto]

right xd

[lethal dune]

maybe you should get comfortable with the def first

[white grotto]

so Z[X]

[white grotto]

is any ideal of matrices finitely generated?

[white grotto]

i am, by trying to find examples

[lethal dune]

Z[x] is Noetherian ring btw

[white grotto]

AAA

give me an example of a non noetherian ring then

and ill live you alone xd

[lethal dune]

look up Hilbert basis theorem

[white grotto]

so i need infinite dimensions for that?

[lethal dune]

ok can't recall an easy example but here's one

ℤ [1/p]/ℤ

or easier is ℤ [x1, x2,...] constantly many variables

[white grotto]

okay

got it

thanks a lot

[wind steeple]

ℂ{z}

[lethal dune]

Can anyone have an example of when a quotient of an injective module is not injective.

I know it's not possible over PIDs in general

or even semisimple 😪

[south patrol]

the examples i can find online are where R is an infinite product of fields e.g. countable product of F_2

[lethal dune]

okay and why the quotient is not injective?

[south patrol]

wait hm okay the thing i found online was flawed lol

you might find this helpful https://math.stackexchange.com/questions/2864675/quotient-of-injective-module-which-is-not-injective

this is interesting to me too now hm

[white grotto]

a necessary and sufficent condition for this to be an integral domain

my guess is that there's only one element in P(E) other than E and the empty set

[chilly ocean]

there can't be 3 elements in P(E)

but your guess is still close

[white grotto]

there definitely should be no disjoint elements

[chilly ocean]

yes! (non-empty disjoint)

[white grotto]

and why finitude of E implies this ring is principal

[chilly ocean]

principal ideal ring?

[white grotto]

yeah

[chilly ocean]

let's see

Say I is an ideal of it

[white grotto]

yeah, so its the intersection of all elements of E with a selected set of elements

[chilly ocean]

yes so let's say X is an element of our ideal

what can you say about the subsets of X?

[white grotto]

so when E is finite, it suffices to take the union of all intersections with all elements

am i right?

[white grotto]

i am not sure what you mean

[white grotto]

that's an element of P(E) and will generate the ideal

but why does it break down when E is infinite

[chilly ocean]

So, if X is an element of I and X' subset of X then X' is in I right?

[white grotto]

yes

[lethal dune]

kind of got it

[chilly ocean]

but just being closed for subsets doesn't mean it's principal yet

you have to look at the other operation too

for example the subset containing {1}, {2}, Ø isn't generated by a single element

however {1} Δ {2} = {1,2}

and if you add that element then this thing will be an ideal and it's generated by {1,2}

more generally $A\cup B = (A\cap B) \Delta (A\Delta B)$

[cloud walrus]

Just A Dishwasher

[chilly ocean]

and you can extend this for any finite amount of things

So if P(E) is finite and I is an ideal then I is finite so it is closed for unions so you can take the union of all elements of I and that'll give you an element that generates I

so P(E) is a principal ideal ring in that case

[white grotto]

nice

and if P(E) is principal

why is E finite

[chilly ocean]

if P(E) is finite then E is finite

[white grotto]

yeah

[chilly ocean]

E being finite is equivalent to P(E) being finite

[white grotto]

yeah

i mean if we dont have P(E) finite

i am doing the other implication

P(E) principal iff E finite

[chilly ocean]

Okay, suppose P(E) is infinite

Let I be the set of all finite subsets of P(E)

prove that this is an ideal that is not principal

[white grotto]

god why is thinking this hard

[white grotto]

idk

i think of taking some kind of union

[chilly ocean]

Suppose it is principal. Then there'd be an element X that generates the ideal. Can you reach a contradiction from here?

Hint: ||An ideal here is closed for subsets so X would have to contain every finite set...||

[white grotto]

yeah so X would have to be infinite

whats the problem

[chilly ocean]

What was our ideal?

[white grotto]

all finite subsets

but there is infinite finite subsets

[chilly ocean]

if it was generated by X then X would belong to the ideal

but X is not finite

[white grotto]

so what we're saying is that the ideal generated by all finite subsets cannot be generated by other than an infinite subset

but the infinite subset cant be the generator for finite subsets, because it's not a finite subset

and finite subsets are closed under intersection?

[chilly ocean]

yeah

[white grotto]

lol

how did you think of it :(((((

i am failing my finals for sure

[chilly ocean]

alternatively, suppose X generates the ideal. Then X is finite. So you can find an element outside of X to create a bigger subset that is not contained in the ideal generated by X

[lethal dune]

Cool name

[chilly ocean]

well i've seen other situations where similar techniques are of use

[white grotto]

okay?

[chilly ocean]

the finite ones are principal because we proved ideals are closed for finite unions

so it's natural to try an example where we'd need an infinite union

[white grotto]

yes

very cool

thanks a lot

[coral shale]

uwu math girl gone sad

[chilly ocean]

She's still there

@chilly ocean

uwu

[hard flame]

But g not in H by construction cause g is in that and other cosets

Right?

[white grotto]

i want the diagramm that shows where each structure lies within the others, like rings PID, fields...

whats it called. i remember seeing it on wiki someday

[hard flame]

Oh wait I jsut need the number or orbits is bounded by the index

[elder wave]

do you mean this?

[white grotto]

yeah but like a diagram

with examples for each

this is... sad

[pastel cliff]

try and come up with examples

[elder wave]

why don't you try to come up with ecamples for yourself

[pastel cliff]

sniped

[elder wave]

i got sniped

[pastel cliff]

double sniped

[white grotto]

i want allllllllllllll examples

[elder wave]

[pastel cliff]

start with Z

where does that fit in

[white grotto]

its a ring

commutative

not a field

principal

and integral

[pastel cliff]

what other examples can you come up with based on Z

[white grotto]

after principal i shouldnt have specified integral

is what i just learned

[pastel cliff]

it's impliedd

[white grotto]

Z/nz

its also a ring

commutative

field if n is prime

not integral so ofc not principal

[pastel cliff]

ok ssure

anything else

[white grotto]

wait this deduction is false right

[pastel cliff]

yeah

but there's two more pretty informative examples

[white grotto]

i mean theres ofc Z[X]

[pastel cliff]

that's one

[white grotto]

and R[X]

[chilly ocean]

depends whether by principal you mean "principal ring" or "principal domain"

[white grotto]

i think principal ring, i ve never heard of principal domain

[pastel cliff]

oh

[white grotto]

principal ring is one which every ideal is generated by one element right?

[pastel cliff]

nvm then i thought you meant a PID

[chilly ocean]

principal domain is a principal ring that is also an integral domain

[pastel cliff]

this is a pid

[white grotto]

wait.... but principal doesnt imply integral?

[chilly ocean]

No

[pastel cliff]

im probably getting my terms mixed up i hadnt heard of principal rings

[chilly ocean]

pid also needs requirement to be an integral domain

while principal ideal ring doesn't

[pastel cliff]

when they said principal ring i assumed pid, ignore that then

[coral shale]

integral is a weak condition

[white grotto]

so Zn is a principle ring that isnt integral

?

[coral shale]

thats not possible

all PIDs are IDs

[white grotto]

so principle implies integral?

[pastel cliff]

ok see im not crazy

[white grotto]

i am getting mixed msgs

[coral shale]

like check the inclusions

[pastel cliff]

but they dont mean PID's

[coral shale]

and have a copy of it in your notes

its clear what is a what

[white grotto]

how do we move from this to integral

[pastel cliff]

sear it into your skull

[coral shale]

wait

what

whats a principal ring

[white grotto]

lol

stooooooooop

[coral shale]

when u say principal, you mean PID yes?

[pastel cliff]

they dont

that's what i had assumed to

but we're getting side tracked

[coral shale]

???

ok heres my list of rings i actually use

===

Rings
Integral Domains (ID) - no zero-divisors (ab = 0 implies a or b = 0)
Unique Factorisation Domains (UFD) - all factorisations are unique
Principal Ideal Domains (PID) - all ideals can be generated by one element
Euclidean Domains (ED) - euclidean algorithm exists
Fields

[pastel cliff]

same

[coral shale]

in order

[white grotto]

this is the only defniition of principal

[coral shale]

thats a principal ideal domain

[pastel cliff]

probably throw noetherian in there too

[coral shale]

iem a PID

[coral shale]

it doesnt live in the inclusion chain

[pastel cliff]

"it's not necessarily integral"

[white grotto]

ok so a PID is integral?

[pastel cliff]

yees

[white grotto]

how to prove that

[coral shale]

yes. check the inclusion hierarchy

The proof of inclusions are generally non trivial so just look them up

[white grotto]

so this deduction is actually correct

[chilly ocean]

A PRINCIPAL IDEAL RING IS NOT THE SAME THING AS A PID

[coral shale]

oh

[white grotto]

stop with the freakin def please

i want the thing where every ideal is generated by 1 element

[coral shale]

no i got something wrong

[white grotto]

name it SBHADVIDBHSKBFJKASFP

[chilly ocean]

Principal ideal ring

and it is not in those inclusions you saw

[coral shale]

so whats PIR vs PID difference

[white grotto]

SBHADVIDBHSKBFJKASFP is necessarily integral?

[chilly ocean]

r v d

[chilly ocean]

No

PID is a PIR that is also integral

[coral shale]

Oh.

Bruh.

[white grotto]

can you give me an example?

of a SBHADVIDBHSKBFJKASFP that isnt integral

[chilly ocean]

yes remember that P(E) thing

[white grotto]

yes

[chilly ocean]

for E finite with more than 1 element

[coral shale]

wait a second.

[chilly ocean]

that is principal ideal ring as we proved

[white grotto]

yes

[chilly ocean]

but it's not integral

[white grotto]

oh indeed

nice

[coral shale]

whats P(E)

and also

[white grotto]

Z/nZ is principal also but not integral?

if n is not prime

[coral shale]

is this context entirely in commutative rings with unity

[white grotto]

yes

[coral shale]

just checking

[white grotto]

for me at least

[chilly ocean]

Powerset of E with sum given by symmetric difference and multiplication given by intersection

[coral shale]

oki

[white grotto]

.

[chilly ocean]

for n not prime yes

[coral shale]

I think i agree

So if u have ideal (a, b) its equal to (gcd(a, b)) right

Z/nZ

[pastel cliff]

this is probably not the most enlightening discussion to understand that chain of inclusions tho

[coral shale]

yh, ive never heard of PIR before this

[pastel cliff]

^

not sure what the point of a PIR is beyond just existing

[coral shale]

The point of many of the inclusions is to characterize how much this thing behaves like the integers with different particular properties

[chilly ocean]

well not when you get to fields since integers is not a field

i'd think it's more like if you prove something is one of them then automatically you know it's all the lower ones

[white grotto]

ok so lets keep going

where were you taking me

[pastel cliff]

Z[x] and Z[sqrt{d}]

[white grotto]

Z[X] isnt PID

[pastel cliff]

there you go

[white grotto]

lol

tho it isnt obvious

[pastel cliff]

but it is noetherian

so that's an easy example of a noetherian ring that isn't a pid

[white grotto]

yeah thats very cool

Z[sqrt{d}] is p+qsqrt(d)?

its a ring

id guess its not principal tho idk why

[pastel cliff]

i like to use a and b instead of p,q, keep those for primes

[elder wave]

this is a common confusion tho

one of my profs uses PIR always

because he excludes rings with zero divisors from the definition completely

[pastel cliff]

depends

actually maybe it's always a PID i'm not sure

but that's not what the example is supposed to highlight anyways

[white grotto]

is it a PID? how can i prove that?

[elder wave]

it's not

[white grotto]

what is it meant to highlight?

[elder wave]

it's not even a UFD

[pastel cliff]

it's a euclidean domain for certain values of d

[white grotto]

yeah i wouldve guessed so

[pastel cliff]

in which case it is a PID and in turn also a UFD

[pastel cliff]

what's the point of this

[white grotto]

nvm i see why, for it to be integral

[elder wave]

[white grotto]

is there some cool intuition for this

[elder wave]

Indeed

Why did you delete it

[pastel cliff]

timo still getting sniped

[chilly ocean]

https://math.stackexchange.com/a/2924138/1083256

[white grotto]

what does he mean by all numbers for 1 to n-1 are relatively prime to n and can be paired into pairs of multiplicative inverses

[coral shale]

n sounds like a prime

?

[chilly ocean]

if n is prime then every number from 1 to n-1 is relatively prime with n

[white grotto]

yes

[coral shale]

gcd(a, n) = 1

[white grotto]

and?

[coral shale]

if xy = 1

theyre pairs

the only thing to consider is x = y cases

[white grotto]

so if i take for example mod 5

2 would match with 3 and 4 with 4

[coral shale]

4 . 1 = -1

[white grotto]

can i always match them this way

4 with 4

[coral shale]

well just think what inverse means

and proof theyre unique

[white grotto]

something i multiply by to get 1

[coral shale]

and this is a field btw

[white grotto]

yes so each has an inverse

and its unique

so necessarily they get paired up

right?

ah so (p-1)! cong to -1 [p] is super intuitive

fuck

thanks

[coral shale]

just spend a moment convincing yourself yh.

also. x^2 = 1

can only have 2 solutions at most

is a key point

[white grotto]

why?

i mean 1 solution is 1 obv

and -1 ig?

[coral shale]

one way is the fundamental theorem of algebra

you need to show there arent more solutions

otherwise they dont all pair

[white grotto]

so its always 1 and -1

[coral shale]

unless 1 = -1

[white grotto]

nice

got it

[coral shale]

but u can manually check that case

[white grotto]

how to prove that a finite integral ring is a field

my prof did a seemingly complicated method

[coral shale]

finite integral ring?

[white grotto]

yeah, integral domain ring

[coral shale]

an integral domain with finite elements?

[white grotto]

yes

[coral shale]

hm

uh pigeonhole i think

maybe

yh i think so

try that idea and see if it gets u there

[white grotto]

wdym

xd

pigeon hole

whats that

[coral shale]

think thats an important proof technique

that you ideally shouldve come across by now. maybe another name

lookit up

[white grotto]

tell me the gist

[coral shale]

then again it's a name for something simple

[white grotto]

i study in french

[coral shale]

wiki has french

im lazy

[white grotto]

ah

got it

principe des tiroirs

how can it be useful here

actually i get it

very smart thanks

[south patrol]

wilson's theorem is cool

i think there are like a million proofs

My favourite is probably like $x^{p-1}-1 \in \mathbb F_p[x]$ factors as $\prod_{a \in \mathbb F_p^\times} (x+a)$ and then plug in $x = 0$

[white grotto]

find x so that x=2[19] and x=3[15]

[cloud walrus]

potato

[white grotto]

i should use the chinese remaindertheorem

not sure how tho

[white grotto]

what? xd

[south patrol]

lol

[white grotto]

whats Fp

[south patrol]

oh

[white grotto]

and how

and why

[south patrol]

well F_p = Z/pZ

but thought of as a field lol but i could've written it as Z/pZ

the point is that x^{p-1} and the other thing both have the same roots

because both of them vanish at 1,2,...,p-1

[white grotto]

ok.... so we have (p-1)!= -1[p]

[south patrol]

yup

[white grotto]

how do you want to prove that

[south patrol]

Well okay I'll add more detail to what I did then lol

[white grotto]

waaaaw

incredible observation

absolutely fabulous

i think i am good

[white grotto]

how bout this?

[cloud walrus]

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

[white grotto]

how do i use the crt

[south patrol]

but like if p is odd then that is what you want and for p = 1 it works fine

[white grotto]

19 and 15 are coprime

[south patrol]

actually tbf you don't really need the chinese remainder theorem here other than just to know a solution exists lol

yes

[white grotto]

so.... x=6[19*15]?

[south patrol]

So what I'd do is like uh so this is a general method basically like

write x = 3 + 15a for some a and then do that mod 19

so like 3 + 15a = 2 mod 19

go from there

[white grotto]

keep going

i have no idea what to do from here i already reached this

15a=-1mod19

[south patrol]

yes

or 4a = 1 mod 19

equivalently

now you just need to find an inverse for 4

and you can do that by guessing or something lol

well, or maybe you can see it

[white grotto]

how?

[south patrol]

15 = -4 mod 19

[white grotto]

ah yes

ok so 5 is an answer

[south patrol]

indeed

so we know that a = 5 mod 19

so what is the general solution?

[white grotto]

x=3+15(5+b19)

so x= 3+65[19*15]

[south patrol]

ye

[south patrol]

wait wdym

oh

sorry yeah wasn't sure what you meant by the []

[white grotto]

mod

[south patrol]

ye sure

[white grotto]

is there some intuitive way to make sense of this?

where phi is the euler function

[south patrol]

oh lol i couldn't see the 1

[coral spindle]

The Theorem of Lagrange

[south patrol]

Well I would basically take this as a statement that $\varphi(n) = |(\mathbb Z/n \mathbb Z)^\times|$ plus Lagrange

[coral spindle]

^

[south patrol]

Lol my latex today is not great

[coral spindle]

🫂

[white grotto]

i got what you want to write

xd

[south patrol]

lmfao

😢

[cloud walrus]

potato

[south patrol]

Also uhh

so I wanted to show that there's a finite group G which doesn't embed into GL_2(C)

Am I right in thinking that like

[chilly ocean]

this theorem is used so often that I wonder if it even needs a name

[south patrol]

I can take G = (Z/2)^n and then like if there were an embedding of G into GL_2(C), then the matrices in the image would be digonalisable + commute and hence be simultaneously diagonalisable

[white grotto]

so because the group of invertibles is exactly of cardinality phi(n) then a^ phi(n)=1 because every element's order divides phi(n)

[south patrol]

But there are only 4 diagonal matrices of order 2

So I can take n > 4 and be done?

And then I imagine in general you can take (Z/2)^{2^n + 1} and it doesn't embed in GL_n(C), by the same argument

Okay yeah sk this is basically saying any finite abelian subgroup of GL_n(C) can be conjugated to live in the maximal torus lol

And so we just need a finite ab group which cannot be embedded into U(1)^n

And then that is calm

[glossy crag]

Let A=Q[x,y], p=(x^2-y^3), is the following a valid proof that A/p is not integrally closed?

x^2-y^3 is irreducible, since considered as a monic polynomial in x over Q[y] it has no roots => A/p integral domain. If x',y' are the images of x,y in A/p, then x'/y' satisfies the equation t^2-y'=0 and is integral over A/p. It does not however belong to A/p, since if x'/y'=f\in A/p, then y'^3=y'^2f^2 => f=0 (since y' != 0 in A/p).

[trail stump]

can a group contain more than one maximal normal subgroup?

[glossy crag]

pZ is a maximal normal subgroup of Z for every prime.

[trail stump]

wait isn't Z doesn't have a maximal normal subgroup?

[south patrol]

ocean man has just given you countably many

idk like uh

any subgroup of Z is of the form nZ for some n

and if nZ properly contains pZ then n|p

[glossy crag]

Anyone?

[trail stump]

ok i think i get it, pz is not Zp

[white grotto]

why does this make sense, for the euler function if m,n are coprime

[coral spindle]

Wym why does it make sense

It is true

It's not a particularly obvious fact

[south patrol]

if $m,n$ coprime then $\mathbb Z/m\mathbb Z \cong \mathbb Z/m\mathbb Z \times \mathbb Z/n\mathbb Z$

[white grotto]

like suppose first number a= 2.3.5=30, and b=7 , so phi(m) are the numbers that are coprime with each divisor

i dont want that

[cloud walrus]

potato

[south patrol]

now take units

[white grotto]

too abstract for something intuitive

[solar glacier]

there we go ^

[coral spindle]

Yeah it does follow from this statement of the CRT

[coral spindle]

Yeah you're not gonna get something intuitive.

[white grotto]

it is intuitive

[next obsidian]

Does someone have a simple and intuitive

[south patrol]

Or maybe i should say: you share a factor with mn iff you share a factor with m or n

because you can work with primes ig

[coral spindle]

prime powers* tbh

[south patrol]

well i mean working with primes is okay right?

if two numbers aren't coprime, then a prime divides them both so that enough

[coral spindle]

Nah you need to deal with the prime power case separately

[white grotto]

yes

we can work with primes

[south patrol]

okay perhaps idk i basically view \phi(n) as order of (Z/nZ)^X

[coral spindle]

^ this is just unequivocally the best way of seeing this imo

[south patrol]

so we are basically asking why |(Z/mnZ)^x| = |(Z/m)^x| |(Z/n)^x|

and the most illuminating way is probably just to show that we have a straight up iso on the level of the original groups

to me at least

[white grotto]

how is iso equi to this

[white grotto]

.

[south patrol]

well not equivalent

[coral spindle]

If two rings are isomorphic, so are their groups of units

and the group of units of R x S is R^x x S^x

done.

[white grotto]

okay, still why would this be true |(Z/mnZ)^x| = |(Z/m)^x| |(Z/n)^x|

[coral spindle]

Bro

[south patrol]

$\mathbb Z/mn \simeq \mathbb Z/m \times \mathbb Z/n$, take units, $(\mathbb Z/mn)^\times \simeq (\mathbb Z/m)^\times \times (\mathbb Z/n)^\times$, take cardinalities

[coral spindle]

Z/mnZ is isomorphic to Z/mZ x Z/nZ, by CRT

[cloud walrus]

potato

[white grotto]

ah

right

ok got it

[coral spindle]

This is what I said earlier and you decried as being too abstract

[south patrol]

the more abstract the cooler

jk

well not jk

[white grotto]

lol, it still is but ill take it

since now i understand it better

[south patrol]

But like idk maybe there is a difference between like

[white grotto]

AND I am DONE with this yeeey

[south patrol]

smth which is simple algebraically vs simple if you are learning nt for the first time

[white grotto]

now to the 40 exercises lol

[south patrol]

like different aesthetic concerns lol

[white grotto]

:((((

[south patrol]

bruh

i am revise rep theory

[trail stump]

so is it correct to say that any group can be a normal subgroup of some bigger group?

[south patrol]

yes but fairly trivially

[coral spindle]

Yes; in fact this is trivial. If N is some group, then it is a normal subgroup of N x G for any group G.

[trail stump]

ok

i realized i just asked a stupid question but thx regardless

[coral spindle]

If you want an example where the larger subgroup acts nontrivially on G, the holomorph of a group is a good example, and is a larger group in almost all cases.

[chilly ocean]

Fun thing about the holomorph of a group G: if you see it as a split short exact sequence, it is the terminal object on the category of all such sequences with kernel G

[coral spindle]

That's a really cool fact!

[coral shale]

one day i shall understand the cat

one day

[trail stump]

man why you do this to me, i thought i am understanding something few minutes ago

[chilly ocean]

So you could say that the holomorph of G classifies semidirect products with kernel G

Why is the nilradical of a local ring trivial?

[chilly ocean]

I'm just saying it to people who might understand it. If you don't understand something it's okay, maybe one day you can study it later if you want

[coral spindle]

Yeah. It's definitely the "most generic" semidirect product

I suppose in some ways this is obvious

[trail stump]

ok... 🥲

[chilly ocean]

intersect all the primes maybe

at least as a first step to approaching it

[chilly ocean]

Is this for commutative rings or rings in general?

[chilly ocean]

commutative rings

[woven obsidian]

Is it even true?

[chilly ocean]

i don't see why it should be

[lethal dune]

ya

[chilly ocean]

don't feel like thinking of counterexamples though

someone can think for me

[lethal dune]

like x is nilpotent then it's already inside the prime ideal P, localizing it doesn't send it to 0

[woven obsidian]

k[X]/(X^2) ?

[chilly ocean]

I also meant to include that it's a domain,

[woven obsidian]

A domain doesn't have zero divisors

[chilly ocean]

How did you define nilradicals?

[chilly ocean]

radical of 0

[chilly ocean]

oh of course

[woven obsidian]

Think there's an error when you say the quotient is not in A

You want to show it's not in A/P

[coral spindle]

People aren't not answering because they don't see your question; they're not answering because they don't know.

[glossy crag]

That's what I meant, just a typo.

[coral shale]

better to repost if it gets bumped away, it happens

[glossy crag]

Should I now or later?

[woven obsidian]

Then I guess only the last step is slightly confusing

[coral shale]

well like now. instead of posting that meme

but i think ppl have got u

[woven obsidian]

Isn't the result then f^2 = y' in A/p?

[glossy crag]

Ah shit you're right, i fucked up with the arithmetic.

[woven obsidian]

Still, should be possible to write down an argument that says that y' is not a square in A modulo p

[white grotto]

is there some characterization for similar matrices in terms of characteristic polynomials or eigen values

[south patrol]

unfortunately not

the closest is probably in terms of jordan normal forms

[white grotto]

only when its the same roots and same dimension of eigenspace for each root right?

ah no not even then

nvm

[south patrol]

hm i don't think even that is correct cause you care about how the generalised eigenspaces behave

[white grotto]

i want to show that if A^3 +A=0 then A is similar to this

[south patrol]

so you could have like three jordna blocks of size 2 vs two jordan blocks of size 3

[white grotto]

then theres no way :((

[south patrol]

i mean that isn't true in general because of stuff like A = 0

[white grotto]

Yeah lol thats true

[south patrol]

and probably more non-trivial examples

[white grotto]

this is an exam

[south patrol]

which field is this over

[white grotto]

imagine pointing out in the middle of the exam the mistake in the question lol

R

[south patrol]

hm okay

[south patrol]

so i think for 3x3 stuff this works

because like

there are only a few ways a matrix can split into jordan blocks

[white grotto]

if each factor is simple, no double roots

that means they're similar right?

in C they're both similar to the same diagonal matrix

in R.... well i am not sure

[south patrol]

actually so lol

you can sledgehammer a bit

[white grotto]

actually so means i am right?

[south patrol]

if matrices are conjugate over C, they are conjugate over R

[white grotto]

Ah nice xd

then i am done right?

[south patrol]

so yes, since in C all of them are conjugate to the same matrix, they are all conjugate over C

and hence all conjugate over R lol

gg

[south patrol]

and this isn't actually too bad to show

[white grotto]

thanks ❤️

[woven obsidian]

Two matrices are similar if they have the same rational canonical form in general

[white grotto]

whats that

[south patrol]

like you can put a matrix in the form diag(A_1,..., A_n) for some block matrices A_k where A_k is the companion matrix of some polynomial f_k and f1 | f2 | ... | f_n

but uh basically the important thing is that this is a canonical form that exists without assumptions on the field

and the min poly of A_k is f_k

so the rational canonical form of a matrix satisfying A^3 + A is very heavily constrained

[woven obsidian]

Isn't it the characteristic polynomial?

[south patrol]

min poly = characteristic poly for companion matrices

[woven obsidian]

Ah

[white grotto]

what does that mean :p

companion matrix isnt for cyclic functions?

[zenith trellis]

Yeah it's equivalent

The equivalence is not straightforward though

[white grotto]

what does it have to do here with anything?

this is very new to me so i need a bit of a thourough explanantion

the thing potato said

i just know that a companion matrix is a matrix for some cyclic f

[zenith trellis]

what about, given an endomorphism f over a vector space E, you construct a family of subspaces of E on which the restriction of f is cyclic ?

You can do that algorithmically

[white grotto]

but if the restriction on subspaces is cyclic doesnt it mean f is cyclic?

[zenith trellis]

nah. hum, I could try to tell what it does but I don't really know the words in English for that

[white grotto]

french?

or just imagination?

[chilly ocean]

nuclear bomb: do you have RCF at your disposal?

[zenith trellis]

yeah

[white grotto]

j etudie en francais

je vais comprendre

[white grotto]

whats that

in full name

[zenith trellis]

Bon. Si V est un sous espace de E stable par f sur lequel f est cyclique, alors tu peux réduire f sur une base de E adaptée à une décomposition du type E = V + S où S est un supplémentaire de V dans E. Regarde alors le bloc en haut à gauche de la matrice que tu obtiens pour f.

[white grotto]

ok so not on all subspaces, just some of them

right?

[zenith trellis]

What does that stand for ?

[white grotto]

f is cyclic only on some of the subspaces

[white grotto]

is this f1 divides f2 fivides f3?

[zenith trellis]

I just thought, what about you diagonalize A ?

A^3 + A = 0 gives a straightforward diagonalization over C

[white grotto]

yeah... we were talking about another method

[zenith trellis]

oh ok

[white grotto]

this thing

i really really want to understand this

can you plz either explain it fully, or guide me somehow?

[zenith trellis]

Well, I have only 2 years of studies past high school and I'm not in the university. That does not ring a bell to me.

[white grotto]

tu fais prepa?

[zenith trellis]

j'ai fini

chui en 1ère année d'école d'ingé

[white grotto]

quelle ecole

une bonne ecole?

[zenith trellis]

l'X

[white grotto]

ah fucking nice

my bro is there :l

i hope i can make it too

je suis spe

[zenith trellis]

Bruh

MP ?

[white grotto]

yes

[zenith trellis]

Me too

[white grotto]

atten mais t es admis non?

[zenith trellis]

J'étais en MP

[white grotto]

ouais xd et t as gagne

[zenith trellis]

Kinda jokish that you go there if your bro did too

[white grotto]

yeah. I feel like the universe is against me tho

[zenith trellis]

mmh t'es à ginette ou llg ?

[white grotto]

it wont let me do this

[white grotto]

filiere international, liban

[zenith trellis]

ok

you have my respect

[white grotto]

thanks

you too xd

[chilly ocean]

rational canonical form

[zenith trellis]

Ah yeah the reduction over subspaces where the function is cyclic

[white grotto]

how can i understand this better?

[chilly ocean]

decomposes any matrix into companion matrices of factors of its characteristic polynomial

any matrix, not just one whose characteristic polynomial splits like jordan form

[white grotto]

so we take each factor, and see the companion matrix that we can get out of it?

[zenith trellis]

the proof seems straightforward for the existence of this reduction, but the uniqueness is not seemingly

first time i come across that too tbh

[white grotto]

how can it be unique? doesnt it depend on the vector i choose as a starting point?

[zenith trellis]

yeah but actually i've just read an additional condition

the characteristic polynomials of each companion block f1, f2 ... fk must satisfy f1 | f2 | ... | fk

[white grotto]

what does that mean

[zenith trellis]

f1 divides f2, etc...

[white grotto]

ik

i mean why

its a weird thing no?

[zenith trellis]

idk

if it's true i don't care if it's weird

[white grotto]

ok how can it be useful here?

A^3+A=0

[chilly ocean]

Does the correspondence in the correspondence theorem for ideals preserve maximality?

I guess this is obvious if you think about the lattices.

[zenith trellis]

well, the RCF of that matrix is going to be what ? if you take v an eigenvector associated to i and u associated to -i, then A(u+v) = i(u-v), A²(u+v) = -(u+v) so basically the RCF of A over C is given by a reduction over the basis x, u+v, i(u-v) with Ax = 0

and that gives you the matrix on the right

I've never manipulated RCF (and I don't think I ever will) though

[proud bear]

No. Consider the inclusion Z->Q. (0) is maximal in Q but its preimage isn't maximal in Z

[chilly ocean]

hmmm

[chilly ocean]

I meant in the context of quotients

[proud bear]

Ah oof

[chilly ocean]

i.e. if the projection of a maximal ideal is a maximal ideal

[proud bear]

Then yes

I think

[zenith trellis]

Oh well this is not the rcf

but still solves the problem

[white grotto]

how does f1|f2| f3

ah

xd

[zenith trellis]

anyways. rcf seems to be a super powerful tool and i know it's not taught in mp so just don't use it

[white grotto]

yes xd

[zenith trellis]

might help if you want to get into ulm though

[white grotto]

ulm?

[zenith trellis]

ens de paris

[white grotto]

ah

[zenith trellis]

(harder to get than l'x)

[white grotto]

non, X ca me va

xd

[zenith trellis]

(much much harder)

[white grotto]

ik lol

[chilly ocean]

Why is (xy-2) contained in (x-2,y-1) in R[x,y] where R is the reals?

[chilly ocean]

it's all an easy consequence of a (maybe) hard theorem: decomposition of finitely generated modules over PIDs

[zenith trellis]

stop abbreviations :/

pid ?

[chilly ocean]

principal ideal domain

[solar glacier]

^

[chilly ocean]

integral domain and every ideal is generated by one element

[zenith trellis]

ok. i've never studied ring theory in depth so not gonna help x)

and i don't think i will have the opportunity to

[solar glacier]

PID is a well known abbreviation for principal ideal domain

[chilly ocean]

all you need to know is that k[x] is a principal ideal domain when k is a field

[zenith trellis]

If you speak english*

ok i got it

[coral shale]

(x-2)(y-1) + 2(y-1) + (x-2)?

[solar glacier]

its a ring in which every ideal is principal i.e., generated by a single element

[zenith trellis]

ik ik

[chilly ocean]

thanks mymathyourmath

but you're getting sniped pretty hard

[solar glacier]

wdym by thAT

[chilly ocean]

not just ring, it also needs to be an integral domain

[solar glacier]

true

[zenith trellis]

Well i'm not gonna comeback on this channel. Far too high for me.

[white grotto]

wait i forgot, quand une matric n est pas diagonalisable.... intuitivement

C'est quand ca se decompose en block comme la somme d une matrice diag et matrice nilp oui?

seulement

[chilly ocean]

the intuition behind non-diagonalizability is not having enough eigenvectors

[white grotto]

so its non diagonalizable ,it can be written in blocks as the sum of a non 0 nilpotent and a diagonizable matrix right?

[zenith trellis]

yeah, just trigonalize it

[white grotto]

okay great

[zenith trellis]

and separate the diagonal and the rest of the triangle matrix

[white grotto]

yes

[zenith trellis]

(i mean if the characteristic polynomial is split)

[white grotto]

yes over C

[zenith trellis]

for example

in general, on an algebraic closure

[chilly ocean]

this applies to any matrix whose characteristic polynomial splits though, so what does this say about diagonalizability?

you need to be a bit more specific than just "triangularize"

[zenith trellis]

is that the right name in english though ?

[chilly ocean]

it is

[zenith trellis]

I just answered to him saying that any matrix can be written as the sum of a diagonal and a nilpotent triangle (in a certain basis). that's all.

[chilly ocean]

and i answered saying this has nothing to do with diagonalization of the matrix itself unless you specify what you mean

[white grotto]

wdym

[chilly ocean]

okay, there you go

[white grotto]

this is what i was asking it just glitched in my head

[chilly ocean]

any matrix whose characteristic polynomial splits, regardless of diagonalizability, can be triangularized in a basis. so you need to be more specific about the triangularization if you want to reflect anything about diagonalizability

their edit "in a certain basis" is meant to reflect that

[zenith trellis]

MMh, I think he did not really ask for a link with diagonalizability here

[chilly ocean]

[zenith trellis]

I'm not sure though

[zenith trellis]

indeed

but wasnt that a separate question ?

anyways

kinda childish of mine to argue over that

sincerely. I apologize for doing that.

[chilly ocean]

Can a Noetherian ring have infinitely many prime ideals?

Z

is noetherian and has infinitely many prime ideals

[chilly ocean]

Oh duh because being Noetherian is equivalent to f.g. ideals and Z is a PID.

[formal ermine]

how do I show that for any matrix $A \in \mathrm{M}_n(\bC)$ there exists a $g \in \mathrm{GL}_n$ such that $gAg\inv$ is in JCF using the classification of finitely generated modules over pids?

[cloud walrus]

i believe in mathemagic

[formal ermine]

so we have $\bC^n \cong C[X]/((X - \lambda_1)^{k_1}) \oplus \ldots \oplus C[X]/((X - \lambda_r)^{k_r})$ as $\bC[X]$-modules

[cloud walrus]

i believe in mathemagic

[formal ermine]

for $\lambda_1, \ldots, \lambda_r \in \bC$ and $k_1, \ldots, k_r \in \bZ_{\geq 0}$

[cloud walrus]

i believe in mathemagic

[formal ermine]

(tterra btw I was able to figure out the other thing with the basis for the matrix)

[chilly ocean]

(how did you prove it?)

[formal ermine]

I just tried out random shit, got to 1, x - lambda, ... then reversed that set and got the correct solution

[chilly ocean]

from dummit and foote

[formal ermine]

yeah

[chilly ocean]

all you need to do now is apply that basis result to each summand here

you have C^n = V_1 ⊕ ... ⊕ V_r, and for each V_i you can pick a basis with respect to which the operator is a jordan block

What is the vanishing locus of (x-a,y-b) over K[x,y] ? Is it always (a,b)?

taking the union of all of the bases gives a basis of C^n with respect to which the operator is a block diagonal matrix with the blocks, as you can guess, jordan blocks

so you found a basis in which the operator is in jordan form

QED

[formal ermine]

pretty sure it should just be the single point (a, b)

[chilly ocean]

did you try proving it?

[chilly ocean]

So this is clear to me if (x-a,y-b)=( (x-a)(y-b))

But I'm too stupid to know if that's always true

[formal ermine]

gAg\inv is like a change of basis? (do I ask this in #linear-algebra rather than here?)

[chilly ocean]

yeah exactly

that's right

so you're done

[chilly ocean]

Is this always true?

no

(x, y) is not (xy)

[chilly ocean]

this, on the other hand, is true. did you try to prove it?

it should be a very "follow your nose" kind of thing

by that i mean you shouldn't be overthinking it

"should" is a dangerous word but i really think this should be straightforward if you just write out the definitions and, say, carefully prove each side is contained in the other

what have you tried?

[delicate bloom]

you can even go in desmos and graph (x-a)(y-b)=0 and the point (a,b) and compare the graphs (although you should be able to do this by hand if you can't do it in your mind)

[chilly ocean]

So any element in (x-a,y-b) is of the form p(x,y)(x-a)+q(x,y)(y-b), but I'm not sure how to go from there

what happens if you plug in x = a and y = b

Then that's 0

so (a, b) is contained in the vanishing locus of (x - a, y - b)

right?

I was going to try and find a polynomial that doesn't vanish when (x,y) is not (a,b)

or, if (c, d) is in the vanishing locus of (x - a, y - b), can you conclude that (c, d) = (a, b)?

in particular, x - a and y - b vanish at (c, d)...

[chilly ocean]

works too

[chilly ocean]

Oh duh, x-a and y-b are in there so they must vanish on (c,d)

and what does that mean?

so (c,d)=(a,b)

thanks a bunch

[chilly ocean]

to do this, if x is not a then take x - a, and if x is equal to a take y - b

[chilly ocean]

I understand, yea, that makes sense

[formal ermine]

stupid question but: "to which the operator" what operator?

[chilly ocean]

the one you're trying to find the jordan normal form of

well

you started with a matrix

[formal ermine]

ah ok got it

[chilly ocean]

same thing

[formal ermine]

I was confused with operator

yuh

thanks

[chilly ocean]

matrices are identifiable with their left-multiplication maps on vectors

[left estuary]

Can a range of numbers be considered a group?

[thorn delta]

the integers are a group under addition, so yea

[coral spindle]

I am once again asking you to read a book

[coral shale]

The definition of a group is precise

and not some vague concept

[coral spindle]

Fwiw also, there is a group of any cardinality, so any set "could be considered a group" with the correct additional structure

[thorn delta]

but like, the interval [0,1] would not be a group under the usual addition of reals because, for example, 1 + 1 is not in [0,1]

[viscid pewter]

false!

there is no group with cardinality 0 😏

[coral spindle]

Ha good catch

[left estuary]

So a group has to contain its operations?

[rotund aurora]

first page of an algebra book will answer the question

[thorn delta]

group is a set together with a single operation

[viscid pewter]

a group is precisely a set of elements together with a single operation

[left estuary]

I do better by asking many questions to come to my own understanding

[thorn delta]

satisfying some properties

[viscid pewter]

ok but you also waste a lot of time

i mean this stuff is just absolutely basic

[thorn delta]

why not just let the guy do what he wants

[coral spindle]

Nobody is going to teach you an intro algebra class in discord

[viscid pewter]

wastes ppl's time?

[thorn delta]

its anyone's free will to answer/refuse to answer any q in this channel

[left estuary]

But people can answer questions true likely no single person will teach me but that’s why this community exists

[rotund aurora]

I think it is good (for them) to redirect to algebra books or Wikipedia when this kind of questions are asked

[viscid pewter]

ok but like other ppl might have serious questions

[chilly ocean]

And it's also important for you to gain the ability to learn for yourself

[left estuary]

Then they can ask in help right

[elder wave]

[left estuary]

I literally had to rediscover the law of signs in order to understand how it works

[thorn delta]

i agree, but you tried that, and they still want to ask questions here, so i don't see why yall are pushing back so hard