groups-rings-fields

[oblique river]

what book is this

[novel parrot]

dummit and foote

wdym about crossing lines?

[oblique river]

like, graphically

the set of points where v(u+1) = 0

[novel parrot]

oh

[oblique river]

it's the set of points where v = 0 or where u = -1

those are each two lines and they cross each other

[novel parrot]

whats the significance of crossing each other

[oblique river]

none

my point is just that this set doesnt look irreducible to me

becuase it has two irreducible components

[novel parrot]

oh ok

[oblique river]

what edition is this from btw? i'm trying to look for errata

but cant find it

hmmmm

[novel parrot]

edition 3 i think

[oblique river]

hmmm

yeah idk this is funky

cuz like that ideal does not look prime......

[novel parrot]

but

v(u+1) is an irreducible polynomial?

[oblique river]

no -- you've just reduced it haha

irreducible means "can't be written as a product of two non-units"

but you just wrote it as a product of two non-units

v and u+1

[novel parrot]

hm

how about the other ideal

[oblique river]

idk it looks complicated and it's late (for me)

[novel parrot]

ok

would it be a good idea to try and reduce the ideal first

[oblique river]

what do you mean reduce?

[novel parrot]

finding minimal generators with grobner basis

[oblique river]

oh i see

hmm does that help you prove an ideal is prime? (also i'm wondering now if the problem is supposed to say "radical" instead)

[novel parrot]

idk hah

im also wondering

if an ideal is generated by prime elements

will it be prime?

[oblique river]

if it's generated by a single prime element, yes

[novel parrot]

and not if by multiple?

[oblique river]

actually i guess multiple prime elements should still be good too

[novel parrot]

oh

[oblique river]

wait no sorry that's not true

[novel parrot]

yeah sounded too good to be true

lol

[oblique river]

the ideal (y - x^2, y + x^2 - 1) is not prime

(over C, lets say)

so just to pull back the curtain, im actually thinking of this geometrically

(f(x), g(x)) is the intersection of f(x) and g(x)

so the question is: "if f(x) and g(x) are prime, we know that their corresponding algebraic sets are irreducible. is the intersection of two irreducible sets still irreducible?"

[novel parrot]

hmm

[oblique river]

and the answer is no: Take the graphs of y = x^2 and y = 1 - x^2 and intersect them

you get two points

a two point set is not irreducible

that's how I came up with that example

[novel parrot]

ah

[oblique river]

the reason i tell you this is just to show you like

how thinking geometrically can actually help you answer purely algebraic questions

[novel parrot]

yeah .. that was nice

[oblique river]

also I found the official errata

this problem is a typo, here's what it hsould say

sry idk why the image looks like shit

it says "Show that I and J are radical ideals that are not prime. Conclude that I = I(V) and J = J(V) and that V and W are reducible algebraic sets"

I guess we've kind of already done part of that lol, we shoed that I isn't prime

skimming through the book I saw a section on "how to compute the radical of an ideal" so I guess you could do that to compute the radical of I and then show that's just equal to I again

similarly for J

also sorry but i need to head out now

it's 4:30am

and i'm a bad person who makes bad decisions

like staying up until 4:30am

when i have to be up at 10am tomorrow, fuck

[oblique river]

yeah i think this is really cool math :)

[novel parrot]

oh wow ok

thanks for the help

[oblique river]

np and gl

👍

[novel parrot]

[oblique river]

@weary terrace you can come back now haha

[weary terrace]

Thanks 🙂

I'll paste it again

[cloud walrus]

RiesZ

[weary terrace]

Ignore, I'm not sure it's true

[novel parrot]

how did they see that the grobner basis was (1)?

im just following the algorithm the usual way

but not getting 1

we first reorder it to (x^2 - y^2,xy,-xt + 1) yes?

im getting $(x^2 - y^2,xy,-xt+1,-y^3)$

[cloud walrus]

ActiveChapter

[novel parrot]

and this doesnt reduce to 1

<@&286206848099549185>

[weary terrace]

Okay, I managed to prove it:
Assume $g \in G$ s.t $\rho (g) = id$. Since any irrep $\lambda$ of $G$ appears in some tensor power of $\rho$, it means that $\lambda (g) = id$ for all irreps. But then it means that each irrep has multiplicity at least 1 in any other irrep. Contradiction.

[cloud walrus]

RiesZ

[frank fiber]

if M:K is a field extension and if M:L:K implies L=M,K, then M:K is simple?

[chilly ocean]

I think so? If it wasn't simple, then it has at least 2 elements alg independent over K, call them a,b, then there would be K(a,b):K(a):K?

and K(a,b)=/=K(a)

[hidden haven]

Ye, any element of K-M generates the extension

Because it generates an intermediate extension

[frank fiber]

thanks

[magic owl]

Long question time. Let $A_$ be the dual steenrod algebra at $p=2$, which is given by a polynomial algebra $P(\xi_1,\xi_2,...)$ such that $|\xi_i|=2^i-1$ and with coproduct $\Delta(\xi_n)=\sum_{i=1}^{n} \xi_{n-i}^{2^i}\otimes \xi_i$. I keep seeing the following claim. Suppose $a\in A_$ is in even degree and is such that $\Delta(a)=\sum_i b_i\otimes b'_i$ and $b_i$ and $b_i'$ are both in even degrees. Then $a\in X$ where $X$ is the subcoalgebra given by $P(\xi_1^2,\xi_2^2...)$. I think I found an easy counterexample, namely $\xi_1\xi_2$ with $\Delta(\xi_1\xi_2)=(1\otimes \xi_1)(\xi_1^2\otimes\xi_1+1\otimes\xi_2)=\xi_1^2\otimes \xi_1^2+1\otimes\xi_1\xi_2$. Then everything seems to be in even degrees as required, and I cannot tell if I am just being very dumb

[cloud walrus]

FpMaxJ/Fp+1MaxJ

[magic owl]

@latent anvil you like graded rings

Here is (one) source, where here P=X and you can ignore all but the last paragraph

(if you think of something pls ping i dont have notifs for this channel <3)

moment

[latent anvil]

I was only doing graded ring stuff bc chmonkey wanted help on a comm alg problem lol

Is this in a hopf algebra?

Oh there's more lol

Hmm

This is strange

I agree with your calculations

@magic owl yeah I agree with your confusion

I am now also confused. Your counterexample seems to work

[magic owl]

If you’d like the original source is Ravenel and the screenshot is from akhils blog

Maybe I’ll email Achilles

Akhil*

Akhilles

[next obsidian]

Chmonkey doesn’t understand what’s going on moment

We should separate whatever this is from algebra so I never don’t know what’s in this channel 🧠

[latent anvil]

lmao

[digital yoke]

I have a question

say, lambda is an eigenvalue of some linear transformation f with multiplicity of n

that doesn't mean that there will be n vectors in the basis for the generalised eigenspace, right?

(maybe this isn't the right place to ask this)

[hidden haven]

An eigenvector gives a one dimensional generalized eigenspace

So if you have n eogenvectors, they'd all separately generated 1 dim generalized eigenspaces

[digital yoke]

wait I'm confused, isn't a generalised eigenspace associated with an eigenvalue

and it can definitely be higher than one dimensional?

[hidden haven]

It is higher when you don't have an eigenvalue, only a generalised eigenvalue

Like if you look at the Jordan blocks

When you have an eigenvalue you just have a 1x1 Jordan block

But you get larger generalised eigenspaces when you only have generalised eigenvalues

@gritty sparrow I hope I'm not saying anything wrong here

[digital yoke]

I haven't gotten up to the jordan normal form yet so rip me

doing that soon though 😔

ok let me post an actual example of what I mean

so you have this transformation, and when you compute its characteristic polynomial, it's (x+2)^3 (x+1)

for lambda = 1, the eigenvectors are (2, 0, 1, 0) and (0, 2, 0, 1)

for lambda = 2, it's (1, 1, 1 ,1)

this is the next question

so for the generalised eigenspace where lambda = 1, is it necessary that there would be 3 vectors for its basis

because lambda = 1 has a multiplicity of 3

[dusty river]

oh I misunderstood what you meant completely

[digital yoke]

wait are you @hidden haven 's alt?

👀

[dusty river]

yes

[digital yoke]

amazing

[dusty river]

Yeah if I recall correctly, multiplicity of the generalised eigenvalue will be the dimension of the generalised eigenspace

[digital yoke]

ah if that's the case

[dusty river]

no that's wrong xD

[digital yoke]

I've done some arithmetic error... oh thank god

I don't wanna go through the calculations again

[dusty river]

for the identity matrix, 1 occurs with multiplicity = size of matrix

but there are size of matrix many generalized eigenspaces

oh fuck I have actually forgotten a lot of this stuff, I might be misunderstanding the definition of a generalised eigenspace too I'll let someone else answer this lol

[digital yoke]

alright xd

thanks for helping anyway :D

[latent anvil]

@digital yoke yes, the algebraic multiplicity is the dimension of the generalized eigenspace

[digital yoke]

yup, I thought so

thanks a lot

[obsidian sleet]

oh god

jcf

[latent anvil]

Metal do you not like jcf?

it's good, actually

[obsidian sleet]

i love it its really pretty

but

please never make me compute it by hand ive done enough in my life

[latent anvil]

oh yeah

That is bad

[obsidian sleet]

ode moment

[latent anvil]

I can't recall ever computing the jcf

i would simply use wolfram alpha

[obsidian sleet]

this is the way

[latent anvil]

is there a faster way than like, computing the characteristic polynomial?

[obsidian sleet]

yes i would love to know any faster method that would be neat

[sturdy marsh]

the char poly is not sufficient tho

you need more info

computing jcf is kinda fun

like qual problem style jcf computations

[latent anvil]

blech

What are you doing next year metal?

er, this year I guess

I am curious about what classes you're taking

[sturdy marsh]

chromatic homotopy

[latent anvil]

> metal

[sturdy marsh]

metal will be doing chromatic homotopy

[latent anvil]

hmm

[sturdy marsh]

not me

[latent anvil]

hmmmmmmmmm

@obsidian sleet confirm or deny

[sturdy marsh]

that's what the prophecy says

#math-discussion message

[obsidian sleet]

uh no but i guess in time

but as for classes

i take first of two semesters of dummit and foote algebra

then one semester fourier analysis from stein shakarchi 2

then first of two semesters of intro real analysis from abbott

then some intro math methods for physics class

[sturdy marsh]

noice

[obsidian sleet]

on the side ive also signed up for like two tiny once a week things

one of them is like

i have to present a proof to the class

its like show and tell basically

everyone gets a proof from this random book of cool proofs and they get a day to show about it

[sturdy marsh]

tiny once a week things can end up being your highest workload class lmao

[obsidian sleet]

then the other one seems to just be a problem solving class where u get like a few problems to solve every week for fun

oh

yeah it is scary if it ends up like that i will be sure to drop it

[sturdy marsh]

this has happened to me several times

the amount of work a reading course ends up being does not depend on the number of units

[obsidian sleet]

i see

[wooden ember]

You entering second or third year?

[hidden haven]

second

[cloud walrus]

whzup

[chilly ocean]

I think there is a stack exchange or mathoverflow answer for this question

I think the answer was "it's a hard problem"

[wooden ember]

It’s not true in general

For some groups it is though I’m not familiar with the condition

[chilly ocean]

Or maybe not, there is a decent chance I'm misremembering

[terse crystal]

If and only if for each coset aN we can choose an element g(a) being commutative with any element from N such that those elements {g(a)} themselves form a group isomorphic to G/N

[cloud walrus]

whzup

[novel parrot]

i have a question about this

im confused on part 3

and the inverse of pi on an element of D-1R

how can it be disjoint from D

$\pi^{-1}{a/d} = (\pi^{-1}a/1 )(\pi^{-1}(1/d))$

[cloud walrus]

ActiveChapter

[terse crystal]

π isn’t surjective.

[novel parrot]

ik thats why im confused

on what it does on d

[terse crystal]

You can’t necessarily find an inverse image of any given element of D^-1R

I don’t know where you got lost. You just need to prove that ce=1 and ec=1

[novel parrot]

pi inverse on a prime ideal in D-1R is prime yeah?

[terse crystal]

Yes

[obsidian sleet]

second

[novel parrot]

but i dont understand why no D can be in it

[terse crystal]

So you are proving for any prime ideal q of D^-1R, the intersection of c(q) and D is empty ?

[novel parrot]

yes i want to show that

[terse crystal]

Suppose that there exists a d from that intersection, then d/d’ belongs to q for some d’ from D

Which is a contradiction since d/d’ is invertible whose inverse is d’/d

[novel parrot]

about contraction ...

we cant always find inverses so how does it work

[terse crystal]

By definition any prime ideal isn’t the ring itself

So any prime ideal can’t contain any invertible element otherwise it would contain 1

[novel parrot]

yea

but how do we inverse an ideal in D-1R

[terse crystal]

?

[novel parrot]

i mean pi's inverse

$\pi^{-1}(J)$

[cloud walrus]

ActiveChapter

[terse crystal]

It’s not pi inverse q that you should talk about

But c(q)

Which = {x is from R: there exists d from D such that x/d is from q}

[novel parrot]

hmf

[terse crystal]

Wait they are the same...

[novel parrot]

how

[terse crystal]

x/d is from q then x/1 is from q

So it should be simpler(/more simple?) just if there exists a d from that intersection then d/1 is from q which is a contradiction since d/1 has an inverse 1/d

[novel parrot]

if x in c(P) and D

x = pi inverse on some a/d

ya?

[terse crystal]

If x in c(P) and D

Then x/1=π(x) belongs to P

which is a contradiction

Since x/1 has an inverse 1/x in D^-1R

[novel parrot]

yes

[novel parrot]

i was confused on this part but i think ive got i now

ty

[terse crystal]

NP

-I is an element of O(7)\SO(7) which is commutative with SO(7) I was trying to prove that I is the only element of O(8) that’s commutative with SO(8) but failed... it might work perhaps...?

Oh I got it

First any such matrix is commutative with diag {-I,I,I,I} diag {I,-I,I,I} diag {I,I,-I,I} diag {I,I,I,-I}

Where I is of order 2

Therefore such matrix has the form diag{A(x_1),A(x_2),A(x_3),A(x_4)}

Where A(x) means

(cos(x) sin(x)

-sin(x) cos(x))

And easy to prove that A(x_1)=A(x_2)=A(x_3)=A(x_4)=I

[novel parrot]

can i ask another question

extending P

is generating P a/1 over D-1R when for all a in P yes?

$r(a/1)$ a is in P and r in D-1R

[cloud walrus]

ActiveChapter

[terse crystal]

e(P)=P*D^-1R={x/d: x is from P and d is from D}

[novel parrot]

yea

[terse crystal]

Oh yes same

[novel parrot]

whats the point of proving the reverse inclusion

like taht

can i just not say, since P is prime, ab in P, a or b is in, so a/d will be in Q

since a/1 * 1/d

and 1 is in R

[terse crystal]

I don’t get it so there are 4 steps

1, any P prime of R, P is contained in ce(P).

2,ce(P) is contained in P.

3, any Q prime of D^-1R Q is contained in ec(Q).

4, ec(Q) is contained in Q.

Which step are you considering now?

[novel parrot]

the step we any Prime disjoint from D can be extended into a prime in D-1R

showing e(P) is prime on DR

this but the second half

[terse crystal]

If (x/d)(y/e) is contained in PD^-1R then xy/de = z/f for some z from P and f from D therefore there exists g from D such that xyfg =zdeg which is from P therefore xy is from P therefore x or y is from P therefore x/d or y/e is from PD^-1R

[novel parrot]

right ok

thank you

[terse crystal]

A(x_1)=A(x_2)=A(x_3)=A(x_4)=I because this matrix with its 2i-1 th row and 2i th row exchanged for all i equals this matrix with its 2i-1 th column and 2i th column exchanged for all i

This matrix M = (M_ij) where 1<=i,j<=4 then if i doesn’t equal j then M_ij=-M_ij

You can always make that happen since M is commutative with the above four matrices

The determinant should be 1 not -1

I made a little mistake but can be fixed. M=diag{A_1,...,A_4} where those A_j are I or -I but it’s easy to verify that M=I in the end

Yes

[terse crystal]

By choose any “permutation matrix” (permutations that aren’t the identity) of order 4 and replace 1 with I of order 2 and replace 0 with 0 matrix of order 2 if you know what I mean

So step 1, M_ij =0 when i doesn’t equal j , M=diag{A(x_1),A(x_2),A(x_3),A(x_4)}

Where A(x) means

(cos(x) sin(x)

-sin(x) cos(x))

Step 2 all those A(x_i) are either I or -I

Step 3 they can only be I not -I by considering “permutations 2-block matrices of order 4”

Actually just choose one permutation of 4 elements that doesn’t have fixed point will do the job

You mean there is a mistake?

In step 3 for example,MT=TM

Where T=

(0 I 0 0

0 0 I 0

0 0 0 I

I 0 0 0)

Therefore all those blocks can only be I not -I

So What’s the problem ? Step 1 I used that M is commutative with diag {-I,I,I,I} diag {I,-I,I,I} diag {I,I,-I,I} diag {I,I,I,-I}

Step 2 I used that M is commutative with

(0 1 0 0 0 0 0 0

1 0 0 0 0 0 0 0

0 0 0 1 0 0 0 0

0 0 1 0 0 0 0 0

0 0 0 0 0 1 0 0

0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 1

0 0 0 0 0 0 1 0)

Step 3 I used that M is commutative with

(0 I 0 0

0 0 I 0

0 0 0 I

I 0 0 0)

(This is a block matrix)

Oh so M = I or - I but both are contained in SO(8)

Yes,Matrices that are commutative with SO(8) = {I ,-I}

Both are contained in SO(8)

So you can’t find a matrix from O(8)\SO(8) being commutative with SO(8)

And my bad, yes step 3 can only show that all those blocks on the diagonal are equal

So M =I or -I

Oh M =diag{A_1,...,A_4} those A_i aren’t necessarily of that form but they were proved to be I or -I in step 2 anyway

( can’t be

(0 b

b 0)

Where b^2=1

Since those blocks are commutative with

(1/sqrt2. 1/sqrt2

-1/sqrt2 1/sqrt2)

[fickle pewter]

Hellooo everyone. Morning.-.

I have a question. It's about annihilator

If I have a ring Z5xZ5. What's ann of that ring?

Eh I mean Z5x{0}, not Z5xZ5

Is it only {0}x{0} ?? Or {0}xZ5?

[mild laurel]

I don't think it makes sense to ask about the annihilator of a ring. Usually you ask about the annihilator of a (subset of) a module over a ring

[fickle pewter]

Ohhh

So ann is for the subset?

Ann of Z5? Not ann of ring Z5?

[dusty river]

What is your definition of ann?

I am guessing it is defined as an operation on abelian groups?

[fickle pewter]

Ann (a) = {s | s.a = 0, a is an element of a ring R}

[dusty river]

you said Ann(a) but also quantified over a inside the set

did you mean "s is an element of R"

[fickle pewter]

Wait, lemme check

Ohh yes

S is an element of R

[dusty river]

a is an element of a module over R right?

So here you are viewing Z5 as a module over itself probably or maybe as a module over Z

[fickle pewter]

Actually I have no clue about what module is

[dusty river]

oh I see

a and s are both in R?

[fickle pewter]

Yaps

[dusty river]

is Ann(Z5) defined?

You only defined ann of a single element so far

[fickle pewter]

No. I have to find it

Ann of Z5xZ5

[dusty river]

ok so ann of a subset S of a ring R is the set of all elements that annihilate all of S

ie Ann(S) = {r | rS = 0, r in R}

[fickle pewter]

Exactlyy

[dusty river]

just replaced a with S in the previous definition and changed element multiplication to set multiplication

So where are you getting stuck with applying this definition?

You have to find all elements of Z5xZ5 which "act like 0 with respect to multiplication"

ie make everything 0 on multiplication

[fickle pewter]

Thhis one?

[dusty river]

well (0,1) is an element of {0} x Z5. Is (0,1) multiplied by anything equal to 0?

[fickle pewter]

no

but if I use Z5x0, so it will equal to 0,0

isn't it?

[dusty river]

Yeah, but then you would be finding ann(Z5 x 0)

instead of ann(Z5xZ5)

oh wait that was a typo

[fickle pewter]

OHHH

I see

[dusty river]

wait was it a typo or not

[fickle pewter]

uhum

omg it makes me confused. I've studied this for over 6 months lol

oh I think I know the answer. a is element of Z5xZ5, so ann a is (0,0)

it won't cause any trouble then

[dusty river]

ann(0,0) is Z5 x Z5

ann(Z5 x 0) = 0 x Z5 as you said

[fickle pewter]

so I was right, I just mislead the main concept lol

oke thank you so much for your help

[dusty river]

[fickle pewter]

I appreciate it a lot

[golden pasture]

is there a name for the construction where you have a left $G$-set $L$ and a right $G$-set $R$ and you take the product by quotienting the setwise product $L\times R$ by $(lg,r)=(l,gr)$.

Feels like a natural thing to do oso cuz it appears when looking at covering spaces but idk how to put it in a bigger picture (some categorical picture or something)

[cloud walrus]

ari 亲

[golden pasture]

istg seen it before in rep theory but i forgot the name

[wooden ember]

This may be completely outside the box cause I hadn’t understood the concept completely but isn’t this somewhat related to the central product?

[golden pasture]

hmm i feel it is somewhat related as well

[golden pasture]

edit: (gl,r)=(l,rg) because im directionally impaired as usual

[chilly ocean]

yo I came across this induction proof at the end ( https://sriasat.wordpress.com/2014/10/27/transcendence-bases-an-application-of-zorns-lemma/ )and I'm not sure I follow 1 thing - when he says "Since P(1) is true..." doesn't he assume independence of t_m over k(t1, ... ,tm_1)? I don't see how can you conclude this with P(1).

[mild laurel]

@golden pasture I think this is an induction functor. I think the rep theory thing you're thinking about is induced representations maybe

[golden pasture]

ahh sounds about right yea

thanks!

[mild laurel]

@chilly ocean I feel like it's a typo and they meant to write Since P(m-1) is true

[chilly ocean]

wait how does that make sense then? @mild laurel

I think this is just wrong, P(m-1) doesn't imply p_i(t1,..., t_(m-1))=0

[mild laurel]

@chilly ocean ah you're right my bad. P(1) is the correct thing. I think you're correct that this is wrong

I think the argument is overly complicated anyways. For any m, you can make the same argument as their argument for P(1) and say that t_1, ... , t_m must be contained in some S in C and so must be algebraically independent

[chilly ocean]

hmm ye I see, but how do we know that all t's are in some S in C?

we take the "biggest" S and since chain all others are in it? (Like, a chain is of the form S1 \susbet S2 \subset ... and we take the furthest S ?)

[mild laurel]

I mean, we can't assume there's a biggest S cause that's what we're trying to show

But without loss of generality, if you assume t_1 is in S_1, t_2 is in S_2 and so on, and assume that S_1 \subset S_2 \subset ... \subset S_m, then S_m will just contain all of the t's

[chilly ocean]

ok yeeeeeeeeee thats what I meant, ty I agree

imma message this guy maybe he will appreciate it or just ignore me

[sturdy marsh]

Theyre called balanced products

[golden pasture]

ahh cool thanks!

[untold cloud]

Hi, guys, R is a ring, then given a\in R, aR=Ra is an ideal. But for principle ideal I = aR, do we know that aR=Ra?

[next obsidian]

Actually idk,m

Can’t you have a left-invertible element which isn’t right-invertible?

If so then aR will be everything but Ra won’t be

[untold cloud]

sorry, i am not quite understand, do you mean 1\notin aR and 1\in Ra? So we dont have aR=Ra.

[next obsidian]

Yeah

[untold cloud]

ok, thank you hah

[next obsidian]

Oh hurb

Yeah I got it backwards but whatever

[chilly ocean]

Do not even know left from right, we must be kindergartners

[weary terrace]

Hi, I'm trying to prove that all irreducible representations of a group G cannot return Id for g \in G which is not the identity element of G.
My approach is rather cumbersome for my taste:
Let R be the regular representation of G. R contains all irreps of G as a direct summand. Hence, R(g)=Id. But that's impossible since R is faithful. Contradiction.

What do you think? Is R faithful as I suspect?

Do you know another simpler proof?

(G is finite, for that matter)

[glossy yoke]

R is definitely faithful.

i'm a little scared of what might happen for fields of characteristic > 0. i think you can have weird situations where there are no non-trivial representations.

err

no non-trivial irreducible representations.

[weary terrace]

Thanks 🙂

[clever mountain]

A book I have states the number of homomorphisms from Z/8Z to Z/24Z
is 8.
I think this is wrong; I got that there can be 4 homomorphisms.
|Z/8Z/ker(phi)| = |im(phi)|,
so we can only have |im(phi)| = 1, 2, 4, 8.

Am I missing something?

[mild laurel]

Just because the sizes of images are the same for two homomorphisms does not mean they're the same @clever mountain

Consider the homomorphisms from Z to Z that map 1 to 1 and that map 1 to -1, both of their images are all of Z, but these aren't the same maps

[clever mountain]

Oh darn true, thanks

[mild laurel]

Also, you should be careful because there can also be multiple subgroups of the same size in general (although not here since Z/24Z is cyclic)

[clever mountain]

So one could perhaps argue as follows:

Given phi from Z/8 to Z/24 such that e.g. |ker(phi)| = 2,
we can always construct a new homomorphism f from Z/8 to Z/24
such that
f(x) = -phi(x), for all x in Z/8.

This has the same kernel as phi.
So initially I had 4 homomorphism, but each gives rise to another one with same kernel.
That gives us 8. Am I sure there exist any more homomorphisms than 8? Hm no, I would have to argue that somehow...

[mild laurel]

@clever mountain Be careful. Why is it true that this new homomorphism f is different from phi?

[clever mountain]

@mild laurel Hm. If they were the same, f(x) = phi(x) for all x in Z8.
That means phi(x) = -phi(x), or, phi(x) + phi(x) = 0.
The only element of order 2 in Z24 is the element 12.
So phi(x) = 12.
Hence f(x) = 12 for all x in Z24, which is false.

[mild laurel]

@clever mountain Does phi(x) + phi(x) = 0 necessarily imply that phi(x) has order 2?

[clever mountain]

Hm I suppose no, phi(x) could be the identity

But otherwise, if phi(x) is not the identity,
if we find that phi(x)+phi(x) = 0, then 2
is a multiple of its order. And since phi(x) is not the identity,
the order must be two (2 is prime).

[mild laurel]

That's right

But go back to the original question, then why is it true that this new homomorphism f is different from phi?

[clever mountain]

I need to show their images are distinct.
If f(x) = phi(x) for all x in Z8
then either (1) phi(x) is the identity, or (2) phi(x) = 12, who has order 2.

(1) This implies ker(f) = ker(phi) is trivial, so f and phi are both trivial homomorphisms i.e. the same. No new homomorphism then.

(2) This implies f(x) = 12 for all x in Z8, including 0. This is false

so there must exist x in Z8 such that f(x) =!= phi(x)

[mild laurel]

I don't see why these cases have to hold, its possible that phi(x) is 0 for some x, and 12 for other x

[clever mountain]

Hm, fair. Need to think more carefully over this one

[simple mulch]

what do you think?

[hidden haven]

G won't be {0}

s_0 is a fixed element of S

The definition of G is that it's the set of vectors which are 0 on their s_0 coordinate

Other coordinates can be whatever

[simple mulch]

but if G is the set of vectors which are 0?

[hidden haven]

Then yes, but here G isn't that

G depends on s_0 here

So better notation would be something like G_(s_0)

And they are asking if all G_(s_0)s are subspaces

[simple mulch]

so what is in G_(s_0) is f(s_0)?

[hidden haven]

It is all the f such that f(s_0) = 0

So for example f(s_1) could be 1 for some s_1 ≠ s_0

Then that f could still be there in G

G doesn't contain f(s_0), it contains f itself

But the condition for containing f depends on f(s_0)

[simple mulch]

Oh I see

G is composed by functions

[hidden haven]

Yep

[simple mulch]

but those functions need to have 0

as an image for some s_0

[hidden haven]

Not some

[simple mulch]

in this case, for that specific s_0

[hidden haven]

Yes

s_0 is fixed

[simple mulch]

interesting

But then, taking x,y in G

[wooden ember]

I m trying to show that for any nontrivial normal subgroup $H$ of a nilpotent group $G$, then $H\cap Z(G)$ is non trivial. To do this I tried assuming there was a group $H$ such that $H\cap Z(G) = 1$ and then using the upper central series and induction, show that $H\cap Z_i(G)=1$ for all $i$, which would be a contradiction since the nilpotence of $G$ would give $H\cap G=1$ but im struggling on the inductive step

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

oop thought the convo was over

ill reiterate later

[simple mulch]

itsok

go on

[wooden ember]

nah finish what you were doing im about to watch a movie anyways

[simple mulch]

its ok, ask it and go for the movie xd

thanks tho

[delicate orchid]

👂 go on?

[simple mulch]

oh ok

so we take f,y in G

[hidden haven]

Narwhal lives on this server anyway

[simple mulch]

but when we take f,g in G. We take f and g such that f(s_0) = 0 and g(s_0) = 0 ?

No right?

We take the entire functions

[hidden haven]

Yeah f and g would then be elements of G

Assuming they satisfy the condition

[simple mulch]

I see, what would the zero vector be in here? a function with only one element zero?

[hidden haven]

Every element 0

Because then adding it to any other function does nothing

[simple mulch]

So for any g in G, the zero vector is g(s_0)

[hidden haven]

No, the zero vector is the function g for which g(s) = 0 for all s in S

Like the vector space is the set of functions

The vectors are the functions themselves

[simple mulch]

and that function is in g because it satisfies g(s_0) = 0 since for all s in S we have g(s) = 0

[hidden haven]

Yep

I would recommend going through the definition of F(S, F) again btw

It seems that that might be the source of confusion

[simple mulch]

yeah I have it in here xd

[hidden haven]

Cool yeah, the vectors in this space are the functions, not their values at specific points

[simple mulch]

ok so now taking arbitrary f,g in G

We know that f + g in F(S,F) (because it is a vector space) and thus f + g in G

[delicate orchid]

Hold up hold up

[hidden haven]

You also have to check if f+g satisfies the condition for being in G

[simple mulch]

yeah for that s_0 its 0 + 0

[delicate orchid]

I.e. (f+g)(s_0) = 0

Yup

[simple mulch]

So a better answer would be
Take f,g in G. Since G subset F(S,F) we see that f + g in F(S,F). Now we check the predicate of G. Note that for s_0 we have (f+g)(s_0) = f(s_0) + g(s_0) = 0. Therefore f + g in G

[hidden haven]

Yep

[simple mulch]

The part of the s_0 is a bit messy

[hidden haven]

Why

[simple mulch]

idk

[delicate orchid]

Take f,g in G, then (f+g)(s_0) = f(s_0)+g(s_0) = 0+0 = 0 so f+g in G

Yeah

It’s fine

[hidden haven]

Like it doesn't seem formal?

[delicate orchid]

Absolutely no clue what a predicate is tbh

[hidden haven]

If thats the case then you should go step by step and figure out which part you think is problematic

[simple mulch]

I guess

[hidden haven]

Because us saying that it is fine and you just accepting that won't really be helpful

[simple mulch]

its is

[hidden haven]

Any true/false thing

[delicate orchid]

you also need to check G is closed under scalar multiplication to prove it’s a subspace

[delicate orchid]

Ah ok

[simple mulch]

Regarding scalar

we take c in F

then cf(s_0) = c0 = 0

this cf in G

[hidden haven]

Nice

[delicate orchid]

Epic win 😎😎😎😎

[simple mulch]

you're beasts

thanks

[hidden haven]

np

[delicate orchid]

The whole set of elements that annihilate on a certain set forming a sub-structure thing is really swag and general

Variety moment

[hidden haven]

Non abelian logic when

[delicate orchid]

Never

Okkkk fineee maybe in a bit... idkkkk 🙄😏

[weary terrace]

Does anybody here knows about McKay graphs?
It's a way to present a represntation via a graph for which the vertices are all the irreps and edges are multiplicities.

[hot lake]

that doesn't sound right

[weary terrace]

If you're referring to me, it's a bit complicated than that. I just added extra info for those who might know it by other names.

[hidden haven]

For the induction step, assume that there is some h ∈ H ∩ Z_(i+1)(G) but not in Z_i(G). Consider the function g ↦ [h,g]. Where does its image lie?

From reading the wikipedia page for nilpotent groups it seems that you should always keep this function in mind

Because nilpotence says that this function is always nilpotent

for all h

In the whole group not just some normal subgroup H

[thorny flame]

is group (U_18) cyclic?

U_18 being the group of modulo 18 to multiplication

[delicate orchid]

do you mean the group of units mod 18?

in which case yes iff you can prove it by finding a single element that generates the whole group ;)

[wooden ember]

i thought i could do it without thinking about the commutator, guess ill have to anyways. Thanks ill give it a thought

[thorny flame]

How do I prove this?

U_8 being all numbers between 1-7 under multiplication function

[hidden haven]

Figure out what the 2 groups are and show an explicit isomorphism

Write down all there elements

And the multiplication tables

You should be able to do this with partially drawn multiplication tables

Just figure out as much of the group structure as you need to prove this

[split escarp]

I think this should work:
||\phi: {1,3,5,7}\rightarrow {1,5,7,11} with \phi(1)=1, \phi(3)=5,\phi(5)=7, \phi(7)=11||

[oblique river]

step 1 is to make sure you have the definition of U_8 correct :^)

[wooden ember]

as buncho pointed out, multiplication for all elements between 1 and 7 would not form a group: U_8 is restricted in the sense that every element must have an inverse mod 8, and similarly for U_12

[cursive temple]

is 0 the only element with euclidean norm 0 in a euclidean domain

this feels like defn wrangling

[dusty river]

Degree 0 polynomials are a counterexample

over a field

[cursive temple]

owie

im doing this

\delta(a_m) = 0 for all sufficiently large m

and then i can pick n to be the first one such that \delta(a_(n+1)) = 0

[hidden haven]

Yeah and the claim is that a_n is then the gcd

It doesn't need to be 0 for that

[cursive temple]

But i need a_(n+1) = 0 unless thats a typo

[hidden haven]

Ye you will get divisibility

Like try this with a couple small integers

Ah I see what you mean

oof

So yeah if delta(a_n) is 0 for some n

Then inductively the delta of a_n+1 is less than this or a_n+1 is just 0

At each step you strictly reduce delta

[cursive temple]

Yee infinite descent

I just dont see how the condition on delta(a_n) gives us info on a_n

[hidden haven]

It tells you gcd(a_n-1, a_n)

Because from a_n+1 = 0, you have divisibility of that pair

Then backtrack and prove that gcd remains the same at each step

[cursive temple]

wait can we go back a bit

where do we get a_{n+1} = 0

i get delta(a_{n+1}) = 0

[hidden haven]

Delta reduces strictly at each step

So it can't stay constant

[cursive temple]

oh

lol

[hidden haven]

Unless you just have 0

[cursive temple]

okay yeah

thanks a lot

ill be fine from here i think

[hidden haven]

Np

[novel parrot]

anyone got a good book for commute algebra?

[potent briar]

how can i find the multiplicative inverse of a field's elements
that defines addition like (a, b) + (c, d) = (a + c, b + d)
and multiplication (a, b) * (c, d) = (a*c - b*d, a*d + b*c)
the additive identity is (0, 0)
the multiplicative identity is (1, 0)
tried doing (a * c - b * d, a * d + b * c) = (1, 0) and finding (c, d) in terms of (a, b)
is that the correct idea

[novel parrot]

yes

but you used + for multiplication too?

[potent briar]

:p

[novel parrot]

yeah just make it equal to 1 and 0

[delicate orchid]

chat isn't this just ||complex multiplication|| chat

if it isn't then yeah you've got the right idea

[potent briar]

it's isomorphic to C1

yeah

[next obsidian]

If you know a good bit of algebra (like, already are comfortable with the tensor product, it helps to know some homological algebra, Ext, Tor) then Matsumura’s commutative ring theory

If not… 🤢 Atiyah-MacDonald

But don’t…

Just use Reid

[novel parrot]

i was gonna choose mcdonald

is homological algebra really necessary
for matsumura?

[next obsidian]

Yeah lol

I mean there’s a like appendix

But I have never had a great experience using a little appendix for stuff that was pretty necessary

Like there’s a lot of diagrams, the snake lemma is used, the five lemma gets used quite a bit, and once you hit depth like

[novel parrot]

only Ext and Tor ?

[next obsidian]

It’s really really important

[novel parrot]

is needed?

[next obsidian]

I mean if you get up to Ext and Tor you probably know enough I think

[novel parrot]

not any of the cohomoligcal stuff?

ok

[next obsidian]

Like depends what you mean

You don’t need say

Chain homotopy

But like knowing the definition of cohomology is kinda basic

[novel parrot]

i see

Ext and Tor is covered in a single section of D and F

so i guess ill learn that first

and then matsumura

[next obsidian]

I mean what’s your goal?

[novel parrot]

geometry

[next obsidian]

For Commutative Algebra

Hmmmmmmmmm

Maybe I need to go back on my A-M is bad thing

[novel parrot]

why

[next obsidian]

Well

I think ideally youd just use Matsumura

But A-M has a really streamlined thing

So like…

If you want to get to geometry fast

Just doing A-M and then going for it is probably faster

I think it just sucks 10x more

[novel parrot]

but does A-M cover local stuff

its only like 2 or 3 pages

[next obsidian]

But I’m of the opinion as I learn more and more Commutative algebra

That you need it for AG

And like it’s better to know a lot going in

And wdym local?

Like localization?

Or local rings

[novel parrot]

both

[next obsidian]

I mean localization isn’t all that deep in and of itself

It just is used everywhere

So you end up kinda developing a lot for it

For local rings I think A-M does stuff about like Noetherian local rings

The end has dimension theory

I guess my thing is

realistically you’d want to do 1-9 and then 13-15 of Matsumura

But it might be faster to do all of A-M

I just cannot stand A-M and like

It doesn’t work for me because I literally cannot make progress because I hate it so much

[novel parrot]

whats the name of the matsumura book?

[next obsidian]

Commutative ring theory

Commutative algebra is ancient and like, just get the newer one lol

Honestly tho

Maybe Reid is actually the best for someone who wants to do geometry

I think it covers like the same as A-M and I like the exposition 10x more

And it’s also already kinda geometric so it’s nice in that sense

Then eventually you’ll need more commutative algebra then I think Matsumura is how you should continue

IMO

[novel parrot]

what is your opinion on chapter 15 of dummit and foote?

[potent briar]

ok so i got that a = c / c^2 + d^2 and b = d / c^2 + d^2

but then i tried an example

(2, 3) * (2 / 13, 3 / 13)

and it didnt give me (1, 0)

🤔

[novel parrot]

your a and b are wrong

[delicate orchid]

that's just... the entire thing

[terse crystal]

-3/13

[terse crystal]

-d not d

[potent briar]

i think Chapter was right

i flipped them

[viscid pewter]

stupid question

wait how does it go

ok wait for a start if G/H is cyclic does there exist K in G such that K is iso to G/H

i want to say yes

but blanking on exactly why

i mean it has to exist right, just get gh such that h is 1 so it's basically exactly its own representative??

well that's not quite right otherwise why not do it for non-cyclic groups

but there has to be an element with the same order, or... what

jesus i'm fairly sure i've proven this before and forgotten it, it's so goddamn basic

[mild laurel]

If you look at the map from G to G/H, then the preimage of a generator under G/H must have order some multiple of |G/H|, so by raising this element to an appropriate power, you get something with order |G/H| which generates the K you want

[viscid pewter]

that sounds right

lemme work it through

oh yeah it's completely obvious ty

why do i only do maths when i'm dog tired

well 1. so it can't get in the way of me enjoying myself too much and 2. so i can blame being tired when i'm bad at it when in fact i'm just bad

ok i leave now

[frank fiber]

let $P$ ands $B$ spaces with a G-action and $f:P \rightarrow B$ a map, what means that $f$ factors uniquely through the orbit space $P/G$?

[cloud walrus]

Or x1

[mild laurel]

Is there any more context? I don't see how the usual idea of factoring through makes sense here

[dusty river]

There is a natural quotient map q: P → P/G. That statement is saying that there exists a unique map g: P/G → B such that f = gq

The action of G on P/G would be trivial, so the action of G on image of f should also be trivial for this to hold, and this should also be sufficient for the existence of a unique factorisation

[mild laurel]

You mean P/G not P/B right? And I think you mean f = gq and not gp

[dusty river]

Yes, ty

[fossil shuttle]

homological algebra is.... good

[untold cloud]

Hi guys, might be a silly question. R is a ring, and r\in R. what does it mean by 2r? Is it r+r, + is the group operation? Or if I say 2\in R, then 2r is 2r where * is the second operation? And whether 2r = r+r ?

[gritty sparrow]

It means r+r

And in some way, 2 is in every ring if you define 2 as 1+1

And in that way of viewing things, 2r really is (1+1)*r as well

[untold cloud]

Oh ,thanks!

[untold cloud]

inverse, sry

[urban acorn]

you can also define the symbol "2" when referring to an element of R to mean 1+1 where 1 is the identity of R

then 2r really is a multiplication rather than a notational shorthand

[hidden haven]

[golden pasture]

tbf you could prove without diagram chase

[thorny flame]

How do I prove that for group G and H
Which are [G:H] = 2
than for every g^2 in g than it also exists in H?

sorry for every g in G than it also exists as g^2 in H?

[hidden haven]

Quotient by H

[stone fulcrum]

[G:H] = 2 means that H is normal and G/H exists. What group is G/H iso to?

[thorny flame]

I don't know

[stone fulcrum]

It's gotta be a group of order 2

As there's only the two cosets

[south patrol]

Sorry if I'm being silly but what does this also mean lol

Do you mean each g in G can be written as g = h^2 for some h in H?

[stone fulcrum]

For all g ∈ G, g² ∈ H

[south patrol]

Ah

[thorny flame]

more like h = g^2 in H

[south patrol]

Ye sure

[fossil shuttle]

yeah the snake lemma took me a long time to understand the proof of the first time i read it

but it gets much easier in context because when you're talking about concrete chains and boundaries in a space rather than elements of an arbitrary module, it's easier to follow. it's just easier to parse "the boundary of sigma is in A" when you read that as a geometric statement rather than an algebraic one

[cursive temple]

is F(t)[x] = F[t, x]?

i havent seen this notation before, so im a bit confused

yeah its not

thanks

[potent briar]

if the integers with the usual operations are not a field

(no inverse)

what are they

[chilly ocean]

Ring

[potent briar]

oh i see

[delicate orchid]

a very nice ring at that

[long beacon]

a very initial ring at that

[bleak perch]

Hey I have a question, the goal is to prove or disprove that 1 + 103x^2 is a unit in Z_2021[x], but I am leaning towards it being false because the gcd(103, 2021) = 1, is this logic correct?

[glossy yoke]

Yeah. there is a general result about units in R[x]. You need the constant term to be a unit and all the other coefficients to be nilpotent (meaning that a^k = 0 for some k).

R is a commutative ring with 1.

[urban acorn]

it's an early exercise in atiyah-macdonald, I haven't seen much yet but I think that book has great exercises

[delicate orchid]

I remember it being deceptively hard

like you'd think it's just "oh since x^n doesn't have an inverse it has to be nilpotent" but actually showing the if and only if is quite tricky

[uneven jackal]

sorry for the ping, but I wanted to add there's a property that says there's a unique ring homomorphism from Z to any ring, so you can define unambiguously 1,2,3... etc in any ring. you can then naturally define 3a, for example. and it is equal by distributivity to (1+1+1)a=a+a+a

[uncut girder]

C[x,y]/(y^2 - x^3) is an integral domain that's finitely generated as a C-algebra but that's not integrally closed in its field of fractions
An example of an integral element that's not in C[x,y]/(y^2 - x^3) but is in its field of fractions is y/x. It satisfies the monic polynomial t^3 - y.
Question: what is the integral closure of C[x,y]/(y^2 - x^3) in its field of fractions?

[next obsidian]

I think you want to do some bullshit like

I think this ring can be rewritten as the subring of C[t] generated by t^2 and t^3 (send y to t^3 and x to t^2 to see this)

And then the integral closure is C[t] I think

This is also geometric, if you take integral closure you fix the singularity that y^2 - x^3 has and I believe it just becomes a line

If you actually care about the like “physical” integral closure of this ring inside its canonical field of fractions you can try to port everything back over idfk

[oblique river]

I think you can just write it as C[sqrt(x)]

If you want to write it "in terms of" x and y

The field of fractions of C[x,y]/(y^2 - x^3) is C(sqrt(x)) where y = (sqrt(x))^3

[errant zephyr]

Hey guys , im a little stuck on this question, I've managed to mostly prove that H is a group I've proved the operation is a closed binary operation, that its associative and the identity exists and its phi(the identity in G)

Im kind of stuck on figuring out how to ascertain what the inverse of each element in H under the operation 'dagger'. Can anyone help?

As well as this the part underneath with using the above construction: since I've proven that phi is an isomorphism, in this case would phi of the star operation (which is now addition) become multiply such that the two negative x's multiply to become a positive real? (sorry if i worded that badly but thank you in advance!)

[void cosmos]

whats the identity in H

@errant zephyr

[errant zephyr]

phi(identity in G) so phi(e)

[void cosmos]

okaay

[errant zephyr]

I think I figured out the inverse

[void cosmos]

so you want h * k = phi(phi^-1(h) * phi^-1(k)) = e and then find k

[errant zephyr]

but my math might be way off let me send lmao

[void cosmos]

---> phi^-1(h)*phi^-1(k) = phi^-1(e)

this is inside the group G

okay send it

[void cosmos]

im sorry

or nvm

[errant zephyr]

??

[void cosmos]

how did you go

from 1st line to 2nd line

in last paragraph

phi(g^-1)=phi(g)^-1

like

in the proof you should like

remark that you are in the group G

(phi(g))^-1 is not an element in G

ops yea i read wrong

yea ur right

[errant zephyr]

ahhh phew i thought i had to start again LOL

my brain cells were like pls no

thank you!!!

for all of the help

[void cosmos]

i didnt do much yea haha

so yea the inverse of each element

is the inverse image of the inverse of the element thats supposed to be coming from in the original image in the group

does make sense

so for the "construction" part

just find out what f^-1 is

and write out explicitly the operation

f(f^-1(x)+f^-1(y))

[errant zephyr]

okay will try that! thanks so much :)

[eager willow]

do Newton's identities extend to symmetric power series? i.e. if a power series in n variables is the same under permutation of variables is it a power series of the elementary symmetric polynomials?

[oblique river]

Yes this is true

At least Im pretty sure… The fact about polynomials is true degree-by-degree so you should be able to just assemble that into a power series

The only issue would be if there was some kind of convergence problem?

But there shouldnt be since youre in finitely many variables

Yeah I think this is true

[sharp dirge]

x is nilpotent

i want to show 1+x is unit

can i just do 1=1+x^n = (1+x)(x^n-1 -... +1)

[eager willow]

@oblique river that's what I was thinking, the powers can still be separated into homogeneous terms and the usual proof would apply

Thanks :)

[oblique river]

It’s probably simpler to do it as 1 - x^n so that you dont have to worry about the signs on the right side. But yes that’s fine.

[oblique river]

Np

[sharp dirge]

oh bruh

my teacher just told me its wrong

[oblique river]

Oh, yeah of course, you cant actually factor like that lmao

That only works when n is odd

Sorry

Your teacher is correct

My mistake

[eager willow]

Well it's just a matter of x vs -x

And one is nilpotent iff the other is nilpotent

[sharp dirge]

yea ...

i thought so too?

[eager willow]

it is still true WLOG but ugly, and better to use 1+x directly like what you originally thought

[oblique river]

Your argument would be fine if you specified that n were odd

So that you could actually factor 1 + x^n

[sharp dirge]

if i make it -x

[oblique river]

Or you could do it that way, sure

And factor 1 - x^n

[eager willow]

I don't think $n$ needs to be odd to factor $(1+x)(1-x+x^2+\dots+ (-x)^{n-1}$ just be careful with the $\pm$ and it's fine

[cloud walrus]

datorangeguy

[oblique river]

Take n = 2

1 + x^2 doesnt factor

[eager willow]

(1+x)(1-x) = 1 -x^2

Oh he had a typo I see nkw

[oblique river]

Yes, but they were trying to factor 1 + x^n

[eager willow]

No just show 1+x is a unit

So it's fine to factor 1 -(-x)^n

[oblique river]

Yes that is one option

The other option is to just say n is odd and then factor 1 + x^n

Either is fine.

[eager willow]

Ok we're in agreement then

[untold cloud]

Hi, guys, according to the definition of ring of p-adic integers on my lecture note, why i can have the highlighted part?

[next obsidian]

This is just saying you pull out as much p as possible from a

Like… p not dividing ab just means it doesn’t divide a or b

So if it doesn’t divide b then you can put b in the bottom

And if it divides a then you can pull another factor of p out

[untold cloud]

thanks, the p doesn't divide ab really confuses me, is it better or at least correct to write (p^e a)/b with p cannot divide a and b

[next obsidian]

No it’s equicalwnt

Because p is prime

[terse crystal]

Those are equivalent

[next obsidian]

I mean it is correct

And maybe better because it’s… clearer?

But one is true iff the other is

[untold cloud]

Oh, thanks

[novel parrot]

why is empty and whole space open

becz they algebraic set

[chilly ocean]

ye

[novel parrot]

what do they mean here?

@chilly ocean

[next obsidian]

??

The empty set is closed as it corresponds to Z(1)

And the entire space is closed as Z(0)

So their complements are open

Which are… each other

[novel parrot]

oh right

sorry

[next obsidian]

No need to apologize

I just was ??ing what you meant in response to ledog

[novel parrot]

i wasnt sure what they mean

meant

[next obsidian]

They meant “ye” in response to “becz they algebraic set”

I think

[novel parrot]

alright

[chilly ocean]

ye

what book is this btw?

[novel parrot]

robin harthshorne geometry

[fossil vapor]

I'm wondering about the situation of being given a polynomial equation p(X)=0 with integer coefficients (all a_i are in Z in the polynomials expanded form):
Is there a simple to compute upper bound on the (absolute value of) the largest solution p(L)=0 with L in Z, which can be computed in terms of its coefficients a_i? In the best case this avoids going through a norrmal form.
I'm thinking of something along the lines of "the largest solution is smaller than the product of the coefficients."
I haven't been able to cook up a counter-example since I don't rule out that the polynomial has non-integer solutions, say p(10/3)=0, and so form p(X)=\prod(a_i - X) isn't as useful it seems.
The motivation for the question is how the brute force solution search can be restricted, given an integer coefficient polynomial equation to solve.

[mild laurel]

The rational root theorem tells you that any integer solution must divide the constant term. So the absolute value of the constant term gives you an upper bound if thats all you want, but it really gives a lot more than that

[next obsidian]

I think they want a bound for all solutions?

Not just integer ones

[chilly ocean]

oh, i read it as upper bound on the integer solutions

[golden pasture]

sounds more like an analysis problem 🤔

[fossil vapor]

No I think that should suffice. I wans't sure if rational solution b and c in
a(X-b)(X-c)=0
could make the constant term being smaller than some other solution, but I suppose the theorem negates this

[golden pasture]

probably doable by spamming the triangle inequality very hard

like the simplest case would b smt like

say you have $f(x)=\sum_{i=0}^nf_ix^i$, and let $F$ be maxima of $f_i$

$$|f(x)|\leq\sum_{i=0}^n|f_i||x|^i\leq F\frac{|x|^{d+1}-1}{|x|-1}$$

[cloud walrus]

ari 亲

[golden pasture]

vaguely like this

but what you want really is

the magnitude of f_nx^n term is bigger than all other terms

smt liddat

that gives you your bound

[fossil vapor]

Okay I was really just looking for a crude bound

This seems to work

def eval(pol, x):
    return sum(coeff * x ** order for order, coeff in enumerate(pol[::-1]))

def is_root(pol, x):
    return eval(pol, x) == 0

def int_pol_find(pol):
    c = abs(pol[-1])
    candidates = (x for x in range(-c, c)
        if x in [-1, 0, 1] or c % abs(x) == 0
    )
    sol = [x for x in candidates if is_root(pol, x)]
    return sol


print("6 X^2 - 27 X + 12 = 6 (X - 4) (X - 1/2)")
print(int_pol_find([6, -27, 12]))

print("2 (X-3) (X-7) (X-9)")
print(int_pol_find([2, -38, 222, -378]))

[simple mulch]

Hey guys this ok? I could use the reasoning that 0x = 0 in W but we don't know that 0 in F (?)

[dusty river]

That proof of 0 in W doesn't work

if 0+x is in W, and x is in W, you can't say that 0 is too

but 0 is in F yes

or well it is the additive identity of F

And you can show that (0 of F) scalar multiplied by any vector = zero vecotr

[simple mulch]

But how do you know that 0 is in F?

[dusty river]

it is one of the field axioms

that there is an additive identity

Also you probably have to prove that if x is in W then -x is also in W

[simple mulch]

I see

So it would be much easier by just noticing that 0 in F. Since whenever a in F then ax in W and because 0 in F then 0x = 0 in W for all x in W

[dusty river]

Yep

much easier and also actually possible

[simple mulch]

valid

[novel parrot]

hi

[chilly ocean]

hi

[novel parrot]

i need help again

🙂

in part c) if coefficient of y^2 and xy are 0

will it be iso to ring in part a)

and iso to ring in b) otherwise

[golden pasture]

not rllly

you want to ||write as a matrix|| and ||try to diagonalize or smt||

[fossil shuttle]

use like, the jordan normal form or something yeah

i don't remember exactly what it is

you can't always diagonalize right

[golden pasture]

yup

thats vaguely why you get the other case

[novel parrot]

ok that question is beyond me

im struggling to find subsets of algebraic sets

how is it generally done?

[fossil shuttle]

what are you talking about

[novel parrot]

Another question

If i have an algebraic set

idrk how to find subsets

for example when $Y = V(x^2-yz,xz-x)$

[cloud walrus]

ActiveChapter

[novel parrot]

how can i find subsets?

[wraith obsidian]

Given a finite permutation group G of known order, are there obvious lower bounds to the size of generating sets?

For instance, if I know my group is nonabelian it cannot be cyclic, so a single generator is impossible. But how can one check whether e.g. n=2 does not work as well?

[oblique river]

That’s hard

You shouldnt be able to say anything just knowing the order except in some very small cases when you can just enumerate all groups

[south patrol]

By permutation groups, are we talking symmetric groups S_n? If so they all have generating sets size two I thought

[oblique river]

Or like, “a group of order pq is 2-generated for any primes p and q”

[south patrol]

Beyond the first few lol

[wraith obsidian]

Finite groups
Specifically, I was wondering whether for rubik's-like groups (i.e. nxnxn) there's an obvious way to see whether the generating set of „turn a side by 90deg“ can be improved

(for me, permutation group usually means subgroup of Sₙ)

[south patrol]

Ah, sure

Although any finite group is a subgroup of a symmetric group

[oblique river]

All finite groups are permutation groups

[wraith obsidian]

„obvious“ meaning „not reliant on finding a decomposition of that gorup as semidirect products“ or sth like that

[oblique river]

Often people use “permutation group” to mean a specific subgroup of S_n

[south patrol]

Sure thanks

[oblique river]

For example the klein four group can be embedded in S_4 in multiple ways, and which embedding you have can be important

[wraith obsidian]

Well, I was mimicking the nomenclature that treats finite group algorithms as „permutation group algorithms“ because you only work with such a concrete embedding → Sₙ there

but fair

[oblique river]

Yes i think that’s standard terminology

In any case I dont think you can really say much in general

[wraith obsidian]

I see.

thanks anyway

[next obsidian]

This is massively vague like, you can grab a random ass point in there and that’s a singleton subset

If you want algebraic subsets you can always add in another equation