groups-rings-fields

[rotund aurora]

Without axiom of choice you have fields without an algebraic closure

[chilly radish]

Consider where you are right now. Now consider if it would have seemed trivial to you a year ago

You were struggling with much simpler things (which is no shame, just for context), it's just that you need to be able to assess the level of someone and what would be immediate for them/not immediate if you want to be an effective helper

In their mind taking the alg closure might be a mental leap to take if they're being asked about the sizes of fields of a given characteristic

Anyways this is a talk for #math-pedagogy or discussion more than here. I'd be happy to continue there if anyone wants to

[untold cloud]

Hi, guys, is it true that G/H \cong N, then G\cong H\oplus N, where the isomorphism is group isomorphism

[coral shale]

use latex bot

from what I can tell of your Q, unlikely

very odd of you to use H for your normal subgroup

and N for something else

[chilly radish]

Absolutely not

You can find very elementary examples

Try doing so

Hint: ||Look at quotients of Z||

[untold cloud]

Thanks! I see, but suppose i have a group homomorphism f: G\ to H, under what condition for f, we have G = ker f \oplus im f

[lethal dune]

if you are talking about ⊕ are you assuming groups are abelian?

I that case, I'm rephrasing the question to
When should a map 0 → kerf → G → imf→ 0 split

This splits exactly when you have a section imf →G or G → kerf

I'll leave it up to you to figure out what that means

[untold cloud]

Thanks!

[solar shore]

hey all

so this question is asking me to calculate the number of elements in $\mathbb Z_3 \oplus \mathbb Z_3$, and i got 8

[cloud walrus]

blanket

[solar shore]

could anyone give a hint as to how you would go about it if it was $\mathbb Z_3 \oplus \mathbb Z_3 \oplus \cdots \oplus \mathbb Z_3$?

[cloud walrus]

blanket

[solar shore]

(n times)

[wraith cargo]

This is incorrect

[solar shore]

it is? how so

[wraith cargo]

So the element of the sum will be of the form (a,b) with a and b in Z3

[solar shore]

right

and lcm(|a|, |b|) = 3

[wraith cargo]

Now how many such combinations of element are there

[solar shore]

there would be a total of 9 possible elements in Z_3 + Z_3

[wraith cargo]

Yes

That's ur answer

[solar shore]

9 elements?

OH

[wraith cargo]

Yep

[solar shore]

i forgot to mention

they were supposed to be of order 3

[wraith cargo]

Ah lol

[solar shore]

sorry i misinformed you

i was like... how are there 9 LOL

but yeah, order 3 specifically

what im seeing is that every element that's not the identity is an element of order 3

or at least thats what i want to hypothesize

[wraith cargo]

I feel like that's true

[solar shore]

yeah because

like for the first one, i did directly calculate it

[wraith cargo]

I think you can show this inductively

[solar shore]

hmm okay

[wraith cargo]

On an arbitrary number of summands

[solar shore]

i will go about that then

so assume $\mathbb Z_3 \oplus \mathbb Z_3 \cdots \oplus \mathbb Z_3$ has $3^n - 1$ elements of order 3

[cloud walrus]

blanket

[solar shore]

show that its true for n + 1?

[wraith cargo]

Yeah

[solar shore]

gotcha okay

thank you! i appreciate it a ton

[solar shore]

before i go, could i ask you one more thing?

[wraith cargo]

Sure

[solar shore]

so im currently in abstract algebra rn (obviously) and we take the next part next semester, but i was planning on doing some studying over the summer

our first part covers some group theory, while our second part goes over ring theory and some other stuff

do you have any ideas as to what else they would cover in the next part of an abstract algebra class?

to at least look at and maybe get an idea of

[wraith cargo]

Probably just ring theory

[solar shore]

gotcha

someone suggested to look at finite fields as well

[wraith cargo]

Idk if they'll cover modules
Probably look at Artins book
That's a popular UG AA book

[solar shore]

gonna keep this in mind then, thank you

[wraith cargo]

There might be some field theory in there too
Realistically algebra classes vary on depth

[solar shore]

got it

ill see how i handle ring theory then during the summer

[wraith cargo]

So i cant say what you'll cover like 100%

[solar shore]

ye

i just got other things to study during the quarter so i was thinkin that i should get some level of exposure to not be completely blindsided lol

[tough raven]

Let K = F(α) be a simple algebraic extension of a field. There is a poset Field of intermediate fields between F and K ordered by inclusion, and a poset Poly of polynomials p in K[X] with α as a root and dividing m_{α,F}, the minimal polynomial of α over F, ordered by divisibility.

There are maps L -> m_{α,L} and
p -> F(p) := the field extension of F generated by the coefficients of F
in both directions between these two posets, which are anti-monotone. Do they form a Galois connection?

To show this, we need to show that $F(p) \subseteq L \iff m_{\alpha, L} \mid p$ for all intermediate fields $L$ and appropriate polynomials $p$.
The LHS is equivalent to $p$ having coefficients in $L$, i.e., $p \in L[X]$, so the left-to-right direction is trivial. But I haven't figured out how to prove the left-to-right direction.
Is it true? If not, is there a subset of either poset on which it becomes true?

[cloud walrus]

Raghuram

[white oxide]

is it true that a in Zn is a unit iff a is coprime with n?

[rotund aurora]

think about it

[white oxide]

yea i proved it

[rapid relic]

https://cdn.discordapp.com/attachments/767097743398535199/1095099222094786601/Screen_Shot_2023-04-10_at_5.33.11_PM.png

i cant figure out the distinct elements. so for this: my thought process is that this is Z[i]/<2i>

[white oxide]

bro super reacted

[rapid relic]

any b multiple of 2 is gonna be absorbed by <2i>

i think b can either be 0 or 1

but im confused as to what a can be--my instinct is all integers because you cant get a multiple of 2i by adding real valued integers

is this correct? if not why not/what am i missing

[south patrol]

wait so like Z[i]/(2i)?

[rapid relic]

im not familiar with that notation

[south patrol]

Hm

well what does <2i> mean here

In itself this seems like slightly dodgy notation in the question

[rapid relic]

thats what im confused about. typically we've discussed that notation in the context of a cyclic group generated by 2i

so like <2>={0,2,4,....}

so im inclined to think the same for <2i> but that just makes me more confused

[craggy lichen]

is this true for any n?

if so, under which phi

[wraith cargo]

I believe this is isomorphic to (2) as an ideal of Z[i]

So the quotient would be Z2[i]

[rapid relic]

ahhhh i see

tyty

[south patrol]

Wait isomorphic to (2) as an ideal wdym 👀

[wraith cargo]

sorry I meant that they're literally the same ideal lol

doesn't phi have to be a map K^x -> Aut(SL(n,K))?

[prisma ibex]

sorry I'm being goofy

but this map is basically going to be a splitting of the determinant

like this semidirect product is just witnessing the extension 0->SL_n(K)->GL_n(K)->K*->0 where the last projection is the determinant

[wraith cargo]

Tbh I'd wager that phi is something like

[prisma ibex]

you're picking a splitting and then conjugating to get the action

[wraith cargo]

conjugation by diagonal matrices who's entries are the same element of K^x

[prisma ibex]

this isn't a splitting 🙂

but this is almost the right idea yes

you could take the splitting K*->GL_n(K) sending x to the diagonal matrix with entries x,1,...,1

and then define the action K*->Aut(SL_n(K)) by conjugating by this matrix

[wraith cargo]

ah I wasn't talking about a splitting
I was thinking of what phi could be concretely in the semidirect product

actually is there some connection between how you choose a splitting and what phi is?

[prisma ibex]

yes

[wraith cargo]

in our algebra class we only showed an SES splits for some map into the automorphism group

[prisma ibex]

not every sequence splits

but the ones that do are described by semidirect products in this way

[wraith cargo]

yeah I wasn't saying that sorry if there was a misinterpretation

what I meant was
We showed that if we take a splitting SES then the middle group is a semidirect product of the two outer groups for some map into the automorphism group but we never showed if we can determine which map

[prisma ibex]

the map is always given by conjugating by the splitting

[wraith cargo]

ahhhh

ok interesting

Wikipedia gives a clear construction interesting

[solar shore]

i need some help with this question

i got the first part right; (m + 2)(n + 1) - 1

but they got 2(n + 1) for the second one

im not sure where they come from

[sinful mirage]

how does group homomorphism propertyof gamma follow from exactness?

[lethal dune]

hmm why does it split?

[agile burrow]

I think it's just a set theoretic section

[lethal dune]

I mean has a section*

[sinful mirage]

it has a section since exactness, and in particular g is surjective iff g admits a section

[lethal dune]

okay that doesn't necessarily behave well with homomorphisms

[sinful mirage]

the definition of gamma makes sense because of this

but idk why this is a group homo

[lethal dune]

so set theoretic section

[agile burrow]

Let x, y in Q. Then g'(x) g'(y) = g'(xy) f(n) for some n in N since g'(x) g'(y) and g'(xy) both map onto xy, hence their difference lies in (the image of) N as a subgroup of G.
Apply beta to both sides, then apply g tilde

By commutativity of the diagram, the f(n) dies

[lethal dune]

or you can use the universal property of cokernels

[agile burrow]

That works too

[sinful mirage]

why do g'(x) and g'(y) and g'(xy) map into xy?

[agile burrow]

g'(x) * g'(y) maps onto xy since g is a group homomorphism

You get g(g'(x)) * g(g'(y)) = xy by the definition of a section

[sinful mirage]

g(g'(x)) * g(g'(y))=g(g'(x)g'(y))

and for the left hand side, we get xy

so g(g'(x)g'(y))=xy

but idk why this hels

we have that g(g'(x) g'(y))=g(g'(xy)) in particular

[agile burrow]

yes, so g'(x) * g'(y) * g'(xy)^-1 lies in the kernel of g

[sinful mirage]

but we can't apply g^{-1} to this

[agile burrow]

which is the image of f

[sinful mirage]

note that g^{-1} need not exist

[agile burrow]

g'(xy)^-1 is the inverse of g'(xy) as an element of G

[south patrol]

ye i think this is bad terminology on the part of the writer right lol

[sinful mirage]

hmm i am very lsot

[south patrol]

if you say a group hom has a section, i'm definitely assuming that's a group hom

[sinful mirage]

this holds

what we do with this thing?

i don't really understand what @agile burrow is trying to say

[agile burrow]

Do you see why this shows g'(x) * g'(y) * g'(xy)^-1 is in the kernel of g?

[sinful mirage]

if i apply g to this equation, i get xy*(xy)^{-1}=1

but idk how to get there

[sinful mirage]

all i have is this so far

[agile burrow]

Lol, just apply g to the element g'(x) * g'(y) * g'(xy)^-1 and use the fact that g is a group homomorphism

[sinful mirage]

yes,that's what i did

[sinful mirage]

but what does that have to do with this?

[agile burrow]

It's just another way of saying that g'(x) * g'(y) * g'(xy)^-1 lies in the kernel

[sinful mirage]

sure, g'(x) g'(y) g'(xy)^{-1} is in kernel of g. right

kernel of g=image of f.

exactness

[agile burrow]

Right, so g'(x) * g'(y) * g'(xy)^-1 = f(n) for some n

Rearranging yields g'(x) * g'(y) = f(n) * g'(xy)

[sinful mirage]

ohhh

[agile burrow]

The rest should follow suit

[sinful mirage]

$\beta(g'(x) g'(y))=\beta(g'(xy) f(n))$

[cloud walrus]

ProphetX

[sinful mirage]

now I can use that beta is a homomorphism?

$\beta(g'(x)) \beta(g'(y))=\beta(g'(xy)) \beta(f(n))$

[cloud walrus]

ProphetX

[sinful mirage]

commutativity where?

[agile burrow]

$\beta(f(n)) = \tilde{f}(\alpha(n))$

[cloud walrus]

walter

[sinful mirage]

ah lol i think i see the trick

is it that $\tilde{g}(\tilde{f}(\alpha(n)))=1$ by exactness??

[cloud walrus]

ProphetX

[agile burrow]

that's right

[sinful mirage]

ok that makes sense. what does it mean for this to be unique?

what do they mean unique?is there just this one concrete map? or up to some iso?

[agile burrow]

There is only one way for this map to be defined such that the diagram commutes

[sinful mirage]

unique in what sense

[agile burrow]

One concrete map

[sinful mirage]

how could one see this?

(btw @agile burrow it's really funny because this is literally part of proof central extensions <=>proj reps haha)

[agile burrow]

oh lol

[sinful mirage]

this thing tells us that a linear rep of the central extension induces a unique proj. rep.

[agile burrow]

If the diagram is to commute, then $\gamma'(g(x)) = \tilde{g}(\beta(x))$. Since $g$ is surjective, every element of $Q$ is of the form $g(x)$ for some $x \in G$. This forces us to define $\gamma'$ by taking a preimage of an element and looking where it maps to in $\tilde{Q}$.

[cloud walrus]

walter

[sinful mirage]

I agree up to the point every element of Q is of the form g(x)

why does this force us to define gamma as such?

[agile burrow]

Because we need the first equation to hold for every x in G

And every y in Q is of the form y = g(x) for some x in G

Maybe the key point is that the definition of gamma is independent of the set-theoretic section you choose

[sinful mirage]

how so?

commutativity imposes that

[agile burrow]

We're demanding that gamma makes the diagram commute

We need to define gamma such that the equation holds

[sinful mirage]

holds for all x,no?

i'm confused by what you mean

diagram commutes iff $\gamma' \circ g = \tilde{g} \circ \beta$ independent of $x$

[cloud walrus]

ProphetX

[agile burrow]

Right, so now we need to define gamma as a function from Q -> \tilde{Q}

Since every element of Q has the form g(x), the fact that we want the diagram to commute forces us to define gamma in this way

[sinful mirage]

oh I think I see. is the idea that commutativity implies that the map $\gamma:Q \to \tilde{Q}$ must map $g(x)$ to $\tilde{g}(\beta(x))$

[cloud walrus]

ProphetX

[sinful mirage]

now, in principle,this fixes gamma on elements of the form $g(x)$. but we know that g is surjective, that is, every element of $Q$ is of the form $g(x)$

[cloud walrus]

ProphetX

[sinful mirage]

hence i have no other choice?

[agile burrow]

Right

[sinful mirage]

ok perfect

[white oxide]

how does this paragraph show what's underlined in blue?

is it because it has to contain the additive identity which is M since it's a subgroup?

<@&286206848099549185>

[delicate orchid]

It’s the result about preimages being ideals from a quick glance

Also that was not 15 mins

[white oxide]

ye but how do we know that it's an ideal containing M

@delicate orchid

nvm got it

[tough raven]

So this is actually trivial, except that I forgot to actually check that the maps between Field and Poly are order-reversing. So the question comes down to:
given polynomials p, q in K[X] with α as a root, is it true that if p divides q then F(q) is a subfield of F(p)? If not, on what subset of these polynomials is it true?

[white oxide]

how do we know that fbar(x) = gbar(x)hbar(x)?

[broken stirrup]

Isn't it true by definition of fbar(x), gbar(x) and hbar(x). In other words, reducing coefficients of fbar(x) modulo p is the same as, reducing coefficients of gbar,hbar and then multiplying them

[kind jacinth]

Question: for an injective homomorphism to exist between 2 groups, must they have the same order?

say Z_7 to Z_12. They dont have the same order so the only possible mapping is all elements of Z_7 to Z_12

to the identity of z_12*

[hot lake]

No

[kind jacinth]

then what do they mean by this example?
Example 4.1.9 Suppose that we wish to determine all possible homomor-phisms φ from Z7 to Z12. Since the kernel of φ must be a subgroup of Z7,there are only two possible kernels, {0} and all of Z7. The image of a subgroupof Z7 must be a subgroup of Z12. Hence, there is no injective homomorphism;otherwise, Z12 would have a subgroup of order 7, which is impossible. Consequently, the only possible homomorphism from Z7 to Z12 is the one mapping all elements to zero

[hot lake]

What is it an example of

The whole thing relies on the non existence of a subgroup of order 7 inside Z12

I assume that is a consequence of whatever is being discussed before the example

[broken stirrup]

kernel is a subgroup of a group. By Langrage's theorem then only possible orders are 1 and 7. If it's 1 then image of Z_7 must be of order 7 such that it's subgroup of Z_12 but it's not possible by Lagrange's theorem again. Therefore Ker is all of Z_7

[hot lake]

Also it looks like at this point you are supposed to know the subgroups of Zn because they also just state that Z7 only has 2 subgroups

[craggy lichen]

if we have U(1) = complex numbers of norm 1, and we have a differentiable group homomorphism f: U(1) --> C^x, how do we show that f is a solution to y' = ky for some complex number k?

[proven halo]

f being a group homomorphism means f(e^{i(t+h)}) = f(e^{it}*e^{ih}) = f(e^{it}) f(e^{ih}) for all x, h in R.
Then [f(e^{i(t + h)}) - f(e^{it})]/h = [f(e^{it})f(e^{ih}) - f(e^{it})]/h = f(e^{it})[f(e^{ih}) - 1]/h.
If F: R -> C^x is F(t) = f(e^{it}) then [F(t+h) - F(t)]/h = F(t)[F(h) - 1]/h.

So if f is differentiable, then F is also differentiable and so F'(t) = lim h->0 [F(t+h) - F(t)]/h = F(t) lim h->0 [F(h) - 1]/h
So you can define k = lim h->0 [F(h) - 1]/h, then F'(t) = kF(t).

[lavish spoke]

something stupid: is it true that if R is a commutative ring with unity and ideals I, J containing another ideal K then (I \cap J)/K = I/K \cap J/K? Is it true taking the intersection of arbitrarily many ideals containing I? One inclusion is obvious but I'm blanking on showing that if an equivalence class has a representative in I and a representative in J that it has a representative in I \cap J, if that's even true

this was the specific thing - idk if there's something more specific that would make this true

[south patrol]

Ye, this follows from the correspondence theorem

[lavish spoke]

the one that there's a one-to-one correspondence between the ideals of R/K and the ideals of R containing K?

how does that come into play here?

[south patrol]

Yes, and it's order preserving

So the largest ideal of R contained in I and J corresponds to the largest ideal of R/K contained in I/K and J/K

[lavish spoke]

ohhhh

i see

cheers

[primal tusk]

anyone know of an orthogonal linear transformation in R^2 with infinite order?

[rustic crown]

rotation by some irrational multiple of pi?

[primal tusk]

im trying to do this

i guess that works thats a lot harder than i was thinking

i thought it was something obvious that i wasnt seeing

wait but it has to be composed of two finite order transformations

[rustic crown]

yea you can compose two weird reflections to get a weird irrational rotation

reflections are going to be the problem since SO(2) is just rotations and is abelian

like pick your favorite reflection f and a rotation r of infinite order, and consider the element rf and f

[primal tusk]

wow thats tricky

i never would have thought of that

thank you

[south patrol]

Lol I was about to say this is wrong until i saw the 2

So used to SO(3) and SU(2) lol

[rustic crown]

hehe :p

[lethal dune]

Imagine a matrix group being abelian

[rustic crown]

hewwo ryuu

[formal ermine]

hi det

[lethal dune]

konichiwa

[rustic crown]

hii illu

[primal tusk]

im trying to use this theorem to prove this

and i wrote down all the possible groups

including Z5 X Z5 X Z2 X Z2

and im pretty sure this is the Klein 4-group

or well it is contained in that form i mean

[formal ermine]

klein four is Z2 x Z2

[primal tusk]

oh

how do you know?

[formal ermine]

because that's what it's defined to be

[primal tusk]

sorry i mean a subgroup of an abelian group of order 100 that is the Klein 4-group

I guess im struggling to figure out which structure theorem composition is isomorphic to my original abelian group

[rustic crown]

it's either Z/25 x Z/2 x Z/2 or Z/5 x Z/5 x Z/2 x Z/2

but the problem is only asking you to show it contains a copy of Z/2 x Z/2

[primal tusk]

wait why nt Z/25 x Z/4 ?

[rustic crown]

because then there is an element of order 4

(0,1)

[primal tusk]

how would i know that they arnt isomorphic to that group though?

like without what the question says

[rustic crown]

there are 4 abelian groups of order 100 up to isomorphism.

so yea without using "no elements of order 4" you can't say anything :p

[primal tusk]

wait i just reread the question and completely missed that part lol i thought it just said abelian group of order 100 not the no elements of order 4 part

now it makes sense

[south patrol]

lol

Show that the centre of U(n) is { λ I_n }

I need to try this tbh maybe it is trivial tho

[rustic crown]

yea commuting with E_{ij}+E_{ji} should tell a lot about the matrix

[south patrol]

That isn't in U(n) though right 👀

or am i misunderstanding lol

Oh wait aha it's easy i think

Unitary matrices are diagonalisable and any element in the centre is similar only to itself

:)

[rustic crown]

whut

what was U(n)

[south patrol]

unitary matrices

[rustic crown]

isn't A^-1 = A = A*

A = E_{ij}+E_{ji}

E_{ij} is the matrix with 1 in (i,j)th spot and 0s everywhere else

oh lol i see now

this entire time i had E_{ij}+E_{ji}+sum_{k not i, j}(E_{k,k}) this in mind

ig a better way to write it is
id + E_{ij} + E_{ji} - E_{ii} - E_{jj}

[south patrol]

Yeah sure ookie :)

[south patrol]

But yeah ye i think this is coolest

Perhaps overkill

[formal ermine]

what determinant do you guys think the matrix has

like the one in the movie

[south patrol]

Oh

lol

It is a determinant which is so small that it causes errors SINGULAR MATRIX

[formal ermine]

it must be non zero

otherwise they would not be able to escape the matrix

[rustic crown]

[formal ermine]

how long did it take you guys to get good at the basics of algebra? I feel like I still have a skill issue

[elder wave]

you do

[void cosmos]

i am not good by any means but ik there is only one answer

and its to do more problems

develop your toolbox of arguments

and have your intuition guide you when to use which argument (and how to tinker them for your own problem)

untill u create ur own arguments and solutions then game over

lmfao

thats for the harder problems , the rest are all just " do u know what the definition is"

[lethal dune]

ye indeed the coolest

but how would you show that the only matrix similar to itself is the scalar ones?

[delicate orchid]

that's not the statement potato said (I hope - similarity is an equivalence relation and so every matrix is similar to itself)

[lethal dune]

ye lol*

[delicate orchid]

they're essentially saying the scalar matrices are central as their conjugacy classes only contain one element each

[south patrol]

ye i said "It is similar only to itself" rather than "Only it is similar to itself"

[south patrol]

idk was kinda just gradually getting intuition + found it kinda snowballs a bit lol

like i kinda sucked at intro group theory a bit i feel whereas it seems to get better 😳 but you definitely seem to be beyond the basics lol

like impressively so

[delicate orchid]

yeah then they throw Z_p-graded algebras at you

[south patrol]

I'm not saying these either lol

[delicate orchid]

tbh I didn't read any of your messages

[south patrol]

LOL

I'm basically saying like

let A in U(n) be central

There is P such that PAP^-1 is diagonal

but PAP^-1 = A by centrality

so A is diagonal

and then ig you just gotta show that all the diagonal entries coincide

[rustic crown]

centrality only tells you A commutes with stuff from U(n) right

unless A is like self adjiont and you can choose P to be a unitary

[south patrol]

Oh yeah I'm dumb mb

[delicate orchid]

gulp

[south patrol]

I saw smth like this once and must've got confused

[formal ermine]

powotato

hi

[south patrol]

Hi!!!!

[white oxide]

Can somebody give me a hint for the following question? Let $f(x)$ and $g(x)$ be cubic polynomials with integer coefficients such that $f(a) = g(a)$ for four integer values of $a$. Prove that $f(x) = g(x)$.

[cloud walrus]

okeyokay

[delicate orchid]

lagrange interpolation :pack:

[formal ermine]

look at f(x) - g(x)

[delicate orchid]

ok that's way quicker than lagrange interpolation

[south patrol]

Now show that this doesn't work in general rings or fields

[delicate orchid]

but I didn't have to think about my response :pack:

[south patrol]

mwah ha ha

[formal ermine]

I hate lagrange interpolation with a passion

[delicate orchid]

it's so easy stop crying

[south patrol]

I don't even knwo what Lagrange interpolation is really lol

[formal ermine]

splines

[delicate orchid]

u do the product of the dudes on the top and then discriminate one of the dudes on the bottom

[south patrol]

I mean I saw smth in analysis for approximating but they only introduced it to say it was shit and not good enough for Weierstrass

Oh that one

Yeah sure

[delicate orchid]

fuckin pure mathematicians smh

[south patrol]

Kinda cringe cause it involves division

[delicate orchid]

next u will tell me u can't even construct representations of C_m \wr S_n using nothing but gramm-schmit smh

[formal ermine]

I took a numerical analysis course once and I am so glad I didn't take the exam

[delicate orchid]

I quite like numerical analysis ngl

[formal ermine]

the prof required them to remember the cg method by heart lmaooo

[formal ermine]

I only like splines

I hate the recursive formula though

[delicate orchid]

if u ain't modelling PDEs what r u doing

[formal ermine]

I only remember something something energy norm

[elder wave]

i LOVE numerical analysis

[silent oxide]

that sounds weirdly like stockholm syndrome

[elder wave]

no idea what you're talking about

it's my all time favorite subject, i am incredibly good at it

wait this is abstract alg

[south patrol]

Lol gramm schmidt is cool

Shows that there is a homotopy equivalence between the two types of Stiefel manifolds of each flavour

[delicate orchid]

lemon and lime for me

[south patrol]

Fr

[white oxide]

how do we know that f(x) - g(x) is of degree 3? because couldn't the coefficients of x^3 for both f(x) and g(x) cancel each other out if they're the same (my intuition tells me that if they are the same then they are the same polynomial but ig i have to prove that)

i.e., if $f(x) = a_3x^3 + a_2x^2 + a_1x + a_0$ and $g(x) = b_3x^3 + b_2x^2 + b_1x + b_0$ with $a_3 = b_3$ is it true that $f(x) = g(x)$?

[cloud walrus]

okeyokay

[delicate orchid]

we know it's of degree 3 or less because how are you going to get an x^4, x^5... term by subtracting two cubics lol

[south patrol]

oh lol i think I just like assumed they were monic lol

[white oxide]

what does that mean again

[delicate orchid]

wait I'm confused why does them not being monic invalidate this idea

thanks for that one "i"lluminator"3"

[sonic coral]

coefficient on the highest degree term is 1

[white oxide]

oh ok

thx

[south patrol]

No I mean

Being monic would make it be of degree 2 or less

It doesn't affect the problem lol

Just my brain assumes polynomials are monic often lol

[formal ermine]

do non monic polynomials even exist

[south patrol]

No

[formal ermine]

shouldn't it work in any field

[south patrol]

||0 and x(x-1)...(x-p+1)||

well okay maybe there is a snag if you say they have to have the same degree

but you can just chuck on like x^N for big N or smth lmao and it'll work still

[delicate orchid]

"smth lmao" cope

[formal ermine]

if what has the same degree?

[south patrol]

Oh i mean

the example I gave is two distinct polynomials agreeing everywhere

But the original question had two polys of the same degree

so i'm just saying you can fit my example to that if you want by just chucking in some big power of x

[formal ermine]

I'm really confused right now

[south patrol]

idk

lol

[formal ermine]

if a degree n polynomial has n + 1 roots in a euclidean polynomial ring then it's the zero polynomial no?

[south patrol]

No

Only in characteristic 0

wait wdym by "euclidean" lol

[formal ermine]

oh I see

yeah ok I see where I went wrong

[south patrol]

Yeah dw

It's cool

[formal ermine]

R[x] euclidean iff R field

[south patrol]

sure

agreed yeah just not heard someone say "euclidean polynomial ring" rather than poly ring over a field lol

Or a better example is just like lol

well x^{p} - x in Z/p[x]

which is the same thing i guess (multiplied by x-1))

[formal ermine]

doesn't that have p roots?

[south patrol]

ye

but it isn't the 0 poly

which is cool

[formal ermine]

it's of degree p

[south patrol]

wdym

[formal ermine]

can you give me a polynomial of degree n that has n + 1 roots

[south patrol]

oh wait lol maybe that is a good point you have made there

yes you're correct

I

AAA

misunderstood

thank you

[formal ermine]

ok I thought I was going insane

[south patrol]

yes then

lol

I

[formal ermine]

I believe it fails if the polynomial ring isn't euclidean?

[south patrol]

Ignored the n and n+1 thing

Yes

alugybjahlguyjahbghbag

[delicate orchid]

x^p-x has infinity roots mod p because I don't care about equivalence classes anymore :devourerofsouls:

[lethal dune]

[oblique river]

What in the

[formal ermine]

hey buncho

[oblique river]

Howdy

[south patrol]

Hey

[rustic crown]

[broken stirrup]

has anyone studied free groups from D.J. Robinson's A course in the Theory of Groups?

[chilly radish]

Z[x]

[lethal dune]

2x=0 has 2 roots in Z/6, laughs in non integral domains

[kind jacinth]

what if i have Q( (6)^1/4) ?

what is this set composed of?

is it {a + b* (6)^(1/4) + c*(6)^(1/4) + d*(6)^(1/4)}?

[delicate bloom]

did you mean to square and cube the terms with coefficients c and d? otherwise you can just factor out 6^(1/4) out

[kind jacinth]

no the coefficients arent raised to the power of anyhting

[summer path]

I think what they're saying is that currently you have a + 6^(1/4) * (b+c+d)

[kind jacinth]

ohh, then what does the set look like?

like how would i write it in the form like in the image

[summer path]

Pretty much what Mero said here

Currently it doesn't have 6^(2/4), etc

[kind jacinth]

ohh so i do it like {a + b*(6)^(1/4) + c*(6)^(2/4) + d*6^(3/4)}?

[summer path]

Yes

[charred bison]

is $[\bQ(2^{\frac13})\colon\bQ]=2$?

[cloud walrus]

CoolShot

[formal ermine]

no

[summer path]

Bruh

[formal ermine]

you saw nothing tubular

[summer path]

[formal ermine]

I had a brain fart

[summer path]

Try writing out a minimal polynomial

[charred bison]

right its degree 3

[south patrol]

ye fam

[charred bison]

so if an extension over Q has deg 2, is it necessary that it has the form Q(√a) for some a in Q, √a in Q?

[formal ermine]

yes

[south patrol]

Yes. This is basically the quadratic formula

[formal ermine]

quadratic extensions are normal

[charred bison]

normal?

[south patrol]

lol

[formal ermine]

there are equivalent conditions for being normal

[south patrol]

Why is being normal relevant here?

[formal ermine]

one of them being that you're the splitting field of a family of polynomials

[south patrol]

Anyway yes so this is basically the statement that uh

[summer path]

Assuming you're learning about field theory, you'll learn about it soon enough

[south patrol]

If x^2 + bx + c is a polynomial with discirminant Δ then it splits in Q(sqrt(Δ))

[wooden ember]

there is a more general version of this which is a consequence of the primitive element theorem: any finite extension of Q is generated by a single element

[charred bison]

damn ill have to go read some more algebra

[south patrol]

my condolences

[warm shoal]

hey

J is the kernel of h right?

and B is the image?

[wraith cargo]

yeah

[warm shoal]

bet

[warm shoal]

So for the first part

I have f: A --> Abar defined by f(a) = pi(a)

and then to prove that's a hom. is it actually that trivial?

[wraith cargo]

Well I mean you have to prove it satisfies the definition of a hom

[warm shoal]

shouldn't (a+N)^n = 0?

isn't that what a nilpotent element is

[fleet pelican]

it is nilpotent in R/N, you are looking in R

[warm shoal]

i j don't understand how (a+N)^n = 0 + n

[south patrol]

0 + N is the zero element of R/N

[round sandal]

What is Z/60Z localized at the prime ideal (2)?

[fleet pelican]

What do you want to know?

[round sandal]

I'm seeing that it would be Z/4Z, but this doesn't feel right.

[fleet pelican]

rings with zero divisors have a special sort of localization.

[wooden ember]

keep in mind the correspondence between ideals of the localization and ideals contained in the localizing prime

in this case you have a decent few ideals contained in (2), more than there are ideals in Z/4Z at least

[round sandal]

This is more related to ag probably, but I guess the following is then not true? The stalk O_{X, (2)} is isomorphic to the localization of the underlying ring Z/60Z at (2).

X is here just the spectrum of Z/60Z

It's likely a typo and there should be Z/2Z?

[formal ermine]

but 2 * 3 * 5 = 30

[warm shoal]

can i get a straight definition for characteristic of an integral domain

is it n * 1 = 0

or n * a = 0 for all a in A

(n being smallest int)

@south patrol

[chilly ocean]

The characteristic of $A$ is the smallest positive integer $n$ such that $na = 0_A$ for all $a\in A$. If such an $n$ does not exist, we say that $A$ has characteristic zero.

[cloud walrus]

0090

[chilly ocean]

Does that answer your question?

[wooden ember]

Yeah no sorry I’m an idiot lol

That correspondence is for prime ideals haha

[formal ermine]

characteristics in domains take on special values

[wooden ember]

Yeah Z/4Z is right (in general taking Z/nZ and localizing at p gives you Z/p^kZ where k=v_p(n), can you see why?)

[tender wharf]

n is a bit more limited than any integer

[tender wharf]

this is characteristic of a ring

[delicate orchid]

those two definitions are equivalent. If n*1 = 0 then n*a = n*1*a = 0*a = 0

the reverse implication is obvious

[tender wharf]

If your ring has unity sure

[delicate orchid]

of course it does

[pastel cliff]

wew lads what the fuck

[delicate orchid]

it's a ring

[tender wharf]

Some authors

[delicate orchid]

some authors are NERDS

[delicate orchid]

hello!

[tender wharf]

Surely if you write a math textbook you're a nerd

[delicate orchid]

this is true

[south patrol]

lol

yeah sebb just learnt through a non-unital textbook

cringe

[round sandal]

Not sure why. Could you elaborate a bit on this?

[formal ermine]

why only these weird kind of elements

[elder wave]

what

[formal ermine]

?

[elder wave]

ah i didn't get what your initial question was

now i do

[delicate orchid]

haven't you asked this before, illuminator

oh right lol yeah

[formal ermine]

no

[delicate orchid]

if you put an "= 0" at the end (which is what the quotient is actually doing) and do a little re-arranging you may recognise them as the properties of a particular kind of function

[formal ermine]

yes I know

but like

why like that

[delicate orchid]

this question is not well posed really

why like what

[formal ermine]

oh

I see it

thanks

[delicate orchid]

you want a multilinear map out of - ok cool

[wooden ember]

Either do it by the universal property of localization or show that the characteristic map Z-> (Z/nZ)_(p) is surjective and has kernel p^v_p(n)

[formal ermine]

is finding a concrete module for the output of a tensor product hard

[wooden ember]

I feel like there’s probably a way to show that’s also more generalizable to give some result on discrete valuations but I’m not sure

[wooden ember]

Can be pretty ugly yeah

Sometimes the tensor product description is best

[formal ermine]

like Z otimes Z/2Z over Z

[wooden ember]

No that one can be handled

[elder wave]

tensoring with the ring itself just gives you the original module

[wooden ember]

In general if you’re doing things like R/I tensor M you get nice results

[formal ermine]

oh yeah I saw that one question in #category-theory

[elder wave]

i was about to refer to that one

lol

[wooden ember]

And there’s a decent few results that allow you to handle some products

[delicate orchid]

if u can slam the universal properties together it is nice

make a square

[wooden ember]

Like if you know one of your modules is flat for example you can usually figure things out

[elder wave]

generally you shouldn't expect to find a "nice" form

[wooden ember]

But it’s sometimes best to leave it in tensored form anyways

[delicate orchid]

generally you shouldn't care about what it is up to isomorphism really

[wooden ember]

It’s indicative of the type of structure your module has

[delicate orchid]

that's the tricky part, you have to work with it without actually knowing all the details of what "it" is

[wooden ember]

(Like you can read from the notation that it’s a « free » extension to a module over whatever ring you’re working with, or a free way to get a certain algebra etc)

[delicate orchid]

or be like me and only work with vector spaces :blazedtothe9s:

[formal ermine]

lol

[elder wave]

be careful though they messed up their tex a bit

R/I \otimes M is iso M/IM and not M

[formal ermine]

oh I didn't read the proof

[elder wave]

yeah they prove the right thing but the initial statement is wrong

[formal ermine]

oh lmao

[delicate orchid]

uhhh but then you cancel the Ms? 🤣

[elder wave]

1/I

[cloud walrus]

AoiKunie

[woven obsidian]

Specifically I want to check how the prime ideal P = (\sqrt(2)) splits

On the one hand the minimal polynomial of $\sqrt{6}$ is $X^2-6 = X^2 \mod P$, so it seems like one should have $P = P_1^2$

[cloud walrus]

AoiKunie

[woven obsidian]

For some prime ideal P_1

On the other hand, we might as well write our extension as $\mathbb{Q}(\sqrt{2}) \subseteq \mathbb{Q}(\sqrt{2}) (\sqrt{3})$. The minimal polynomial is $X^2-3 \mod P$ which does not give a ramified prime

Why do I get different results?

[cloud walrus]

AoiKunie

[woven obsidian]

Am I missing something?

[wooden ember]

How did you get X^2-6 = X^2 mod P

Wait nvm

[wooden ember]

Okay I think X^2-3 is (X-3)^2=(X+3)^2 in your second example because again 6=0 mod P

So -3 = 3 mod P

So you still get a ramified prime

[woven obsidian]

Ahh

[wooden ember]

Wait no what am I saying

Jesus

I need to rest

[woven obsidian]

Q[sqrt(2)]/P is just F2 right?

[wooden ember]

Okay what we do have is 3 = 9 mod P

And then my decomposition holds

So not entirely wrong

But still you couldn’t straight away factor

[woven obsidian]

Wait so

[wooden ember]

Sounds about right

[wooden ember]

X^2-3 = X^2-9 = (X-3)(X+3) = (X-3)^2

[woven obsidian]

Wasn't your factorization correct?

Oh lol I see

[wooden ember]

Yeah but in my head I was doing sqrt(3)=3 without justification haha

So stroke of luck

[woven obsidian]

So if I were to write it out

(Sqrt(2))=(sqrt(2),sqrt(6))^2

It "looks" weird but I guess it has something to do with how the ring of integers look

[wooden ember]

Yeah it does look a bit weird 🤔

I would think that (sqrt(2),sqrt(6)) would just be (sqrt(2)) again

Since sqrt(3) is integral

[woven obsidian]

Right, that's reallt weird

Is the criterion stated incorrectly?

[wooden ember]

I don’t think so

[woven obsidian]

Hmm

But something must be wrong, right?

[wooden ember]

Yeah I’m thinking

I’ll let you know if I think of something but right now I’m blank

Idk where we’re going wrong

Im absolutely sure P should ramify do I don’t think that’s it

Maybe it’s the criterion that has a mistake but I doubt it

[woven obsidian]

Hmm

There can't be such a thing as an "idempotent ideal" in a ring of integers right, because of unique factorization?

[wooden ember]

Im not so absolutely sure of this anymore haha

Okay so I don’t think P ramifies cause unless I’m mistaken the discriminant ideal should divide (3)Z[sqrt(2)], and P does not divide (3)Z[sqrt(2)]

And I think the issue may be that the criterion is missing a condition

Namely that v_P(disc(theta))= v_P(D_(O_L | O_K)) where

In our case the first valuation is 2

But the discriminant ideal I’m pretty sure gets valuation 1

[woven obsidian]

Hmm I see

[wooden ember]

Does that sound right to you?

[woven obsidian]

Not sure I'm understanding the notation, what is v_P?

[wooden ember]

Highest power of P dividing an ideal

The valuation at P

They never write that condition anywhere?

I mean I’m not sure that it’s a necessary condition

But I have a variant of this criterion in my course and it makes use of this condition

[woven obsidian]

No this is all there is to it, but there has been several errors in the rest of the text so a missing condition would not surprise me

[wooden ember]

We haven’t seen the proof of the criterion yet though so I’m completely sure how it plays a role

[wooden ember]

I’d suggest asking your professor just to be sure

Cause I’m not to be trusted on this subject quite yet I’m also in the process of learning it

[woven obsidian]

Yeah I will, we haven't really talked about relative discriminants

[formal ermine]

is there heterological algebra

[wooden ember]

Maybe ask on #advanced-number-theory too

[woven obsidian]

I guess the criterion is true over Q though?

[wooden ember]

Yeah if your ring of integers is a PID the condition should be true I think (so over Q you get E and it works out fine)

[wooden ember]

Okay not sure about this but it does hold if O_L is a monogenic extension of O_K

Not too sure what the ring of integers of Q(sqrt(2),sqrt(6)) over Q(sqrt(2)) is though so can’t check that here

[woven obsidian]

I think I have seen the proof of the criterion over Q in another course. Then I think you look at O_K/P and write it as (Z[X]/(H(x)))/P = F_p[X]/(H(x)) and then use CRT

[wooden ember]

Yeah that sounds about right

[woven obsidian]

Hmm so I guess something goes wrong if you try to do the same thing over some other O_K

[white oxide]

how does this exercise imply that f(x)g(x)bar = f(x)barg(x)bar = 0?

[agile burrow]

The exercise implies $\bar{f}(x) \bar{g}(x) = \overline{f(x) g(x)}$. The fact that p divides each of the coefficients of $f(x) g(x)$ implies that $\overline{f(x) g(x)} = 0 \mod p$

[cloud walrus]

walter

[agile burrow]

hi det

[rustic crown]

hewwo walter

[karmic moat]

dumb question, mind blanking: if $\sigma$ is frobenius automorphism on $\mathbb K$ and $\alpha \in \mathbb K$, is $\sigma^j(\alpha)$ the composition $j$ times or multiplication $j$ times

[cloud walrus]

anamono for anamono

[white oxide]

ah thank you!!

[rustic crown]

it can mean both things, but more likely the composition.

[karmic moat]

ok dope thanks

[rustic crown]

[formal ermine]

does there exist a ring such that a polynomial over it has more roots than its degree

[lethal dune]

Z/6

[formal ermine]

what polynomial?

[lethal dune]

2x

[formal ermine]

I think my code is broken then

import numpy as np

for n in range(2, 1000):
    roots = 0

    for k in range(0, n):
        if (2 * k) % n == 0:
            roots = roots + 1
    
    if roots > n:
        print(n, "has", roots, "roots")
        break

I ran this on a couple of polynomials but it never found anything

oh lmao

I'm stupid

roots > n

should be like roots > polynomial degree 🤦

[lethal dune]

Not degree

Lol

[formal ermine]

yeah lmao

[lethal dune]

[formal ermine]

wait is 43 prime

[lethal dune]

does 43 | 42!+1?

[formal ermine]

,w is 43 prime

[cloud walrus]

Results provided by WolframAlpha

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

Requested by illuminator3#0001

[formal ermine]

ah

[elder wave]

#bots

[solar shore]

i feel like im trippin up right now, i have no idea how to complete this sentence

i know its not identity but

is it in H?

[agile burrow]

That's right

Remember that the elements in G/H are cosets, so they have the form gH. In particular, the identity element in the factor group G/H is the coset H

[solar shore]

right yep

okay thank you!

[formal ermine]

how would I show that $\bZ/m\bZ \otimes_\bZ \bZ/n\bZ = 0$ if $m$ and $n$ are coprime

[formal ermine]

like I'm not sure where to begin

or what the proof would even look like

would I show that x otimes y is always 0?

[south patrol]

Lol

Atiyah macdonald moment

So uh it suffices to show that 1 tensor 1 = 0

Try to write 1 tensor 1 in another way and use properties of tensors to force it to be zero

[formal ermine]

yeah so my idea of showing x otimes y is always 0 wasn't so far off innit

but yeah lemme try

[south patrol]

Tbf doesn't matter ye if you do general or not

[formal ermine]

right

ok so we somehow need to use the fact that n in invertible in Zm and vice versa

but I can't seem to figure it out

hmmm wait I think I have an idea

ok that was too easy lmao

does this work

,, 1 \otimes 1 = (nn\inv) \otimes 1 = n(n\inv \otimes 1) = n\inv \otimes n = n\inv \otimes 0 = 0

[south patrol]

Beautiful

Gg

[formal ermine]

I haven't solved a math problem in so long

probably been like 3 months

[south patrol]

Sadge

[formal ermine]

how does tensoring the ses help us here?

M isn't necessarily flat

[elder wave]

This is the exact problem we talked about earlier

TP is right exact

[south patrol]

Ye that's all that matters

[formal ermine]

lol

I forgor

[elder wave]

That suffices here

[formal ermine]

wdym

[elder wave]

The tensor product is right exact

Always

As a functor

[formal ermine]

ohhhh

I see

[south patrol]

funky

Cause it has a right adjoint lol

Actually that is the argument given in atiyah macdonald ig lol

[formal ermine]

I couldn't be arsed reading the proof of prop 2.18 lol

[south patrol]

Lel

[formal ermine]

okay so we should obviously use iso innit

we have M ->> A/a otimes M

surjective

which is cash money

now we need to show that the kernel is aM

[south patrol]

Like what is a tensor M

Wat nah

Ignore me

Wait nah

[formal ermine]

it just screams iso theorem to me

[south patrol]

Ye

I can't remember exactly what I did but ye

[formal ermine]

are tensors and localization used in algebraic number theory

[south patrol]

They r like very important techniques in algebra in general, yes

[formal ermine]

for this

do I need to know what f looks like

or can I just guess

I can see why a is in the kernel

[pastel cliff]

simple question but can i get a nudge pls

im rusty havent done algebra in like 2 weeks and im now behind

[south patrol]

This should be immediate from the definition of T1 tensor T2

So check how that is defined

[pastel cliff]

the defn's feel a little funky

i have universal property onne

i get the idea that the tensor product is trying to extend a bilinear map to a homomorphism

though that's not a definition

this picture

[formal ermine]

yeah nah still can't seem to figure it out. like a vanishes because of the A/a, but is that all to it? I think my issue comes from not understanding aM. is it just identifying a with a submodule of M or something? like ik it's just a linear combination

[pastel cliff]

actually im guessing i need to think of this through the elementary tensors first

so we have a map R^2 x R^2 -> R^2 \otimes R^2...?

or thisss

https://www.math3ma.com/blog/the-tensor-product-demystified

super helpful distillation

[formal ermine]

OH I THINK I SEE IT

[formal ermine]

saving this for later

[pastel cliff]

the links at the bottom are also nice

[formal ermine]

can't quite put it into words but like

we wanna know when f otimes 1 is 0. this happens when either f is 0 or when 1 is 0. 1 is zero if the input is zero. f is zero when a. so it's zero when it's an element of aM?

wait no the end part doesn't quite sound right

ker f = a

ker 1 = 0

[pastel cliff]

ok so if im understanding

tensor products are just... products... but spicy

which i guess makes sense from the name

ive been trying to begin my understanding with the universal property but that hasnt been working

where's the det button when you need it

if you spam enough

maybe det will spawn in

[formal ermine]

I could use some det in my life right now too

[pastel cliff]

i know he doesn't mind pings but i always feel bad

ok i have thisi

but this feels silly

[formal ermine]

this is frustrating me more than it should

it looks so trivial

[pastel cliff]

mine def is simple lol

[formal ermine]

to summarize my question again.

I have a map f : A ->> A/a, what is ker f otimes 1 (i.e. kernel of M ->> A/a otimes M)?

I know ker f is a

why is ker f otimes 1 = aM

[pastel cliff]

my homework

not a textbook

just problems prof gives us

illum do i do it

do i hit the det button

[south patrol]

But ye use dis ting

Like

[pastel cliff]

potat button

🥔

[south patrol]

(f tensor g)(u tensor v) = f(u) tensor g(v)

That is enough to do ur problem right

[delicate orchid]

yeah it's this

[south patrol]

Or do you want to know why u can do dat

[delicate orchid]

do le kronecker product

[pastel cliff]

would be nice

[pastel cliff]

didnt define this

[delicate orchid]

you literally wrote the definition 10 messages ago

[pastel cliff]

wew thannk god you're back

the homestuck fan let you go

[delicate orchid]

^ will forever remain ignorant

[pastel cliff]

[delicate orchid]

I don't know of anyway of explaining the connection between the tensor product of spaces to the kronecker product without invoking le V \otimes V' meme

but if you tensor the two matrices as linear maps you get the result of the kronecker product

[pastel cliff]

ah okie

[delicate orchid]

you might be able to show this for 2x2s in general considering that's kinda what it's asking you to do here

T_1 maps e_1 -> e_1 and e_2 -> 2e_1-e_2 etc. etc.

then you literally tensor everything and use the "algebraic" properties (u know the ones) to simplify it down

[pastel cliff]

okay i'll be right back mr lads

[delicate orchid]

specifically, T_1 \otimes T_2(e_1\otimes e_1) = T_1(e_1)\otimesT_2(e_1) iirc

there we go

[formal ermine]

uwu?

[rustic crown]

[pastel cliff]

the det button

[rustic crown]

[pastel cliff]

can i ping in like

30 min

getting dinner

[rustic crown]

i'm me was just about to sleep >.<

,ti

[cloud walrus]

The current time for det is 02:41 AM (CEST) on Fri, 14/04/2023.

[pastel cliff]

oop

i'll bother you tm then

[rustic crown]

[south patrol]

owo

gm

is det french

[white oxide]

im confused, what exactly do they mean by the line in blue? that if the content is not zero, we can just divide it by the content of f(x) to get it to 1?

my bad sorry to interrupt

[south patrol]

hm well i find this comment a bit odd actually

[white oxide]

tfw the clearly is not clear

[south patrol]

yeah lol

But you can just say like, if f isn't primitive, then f = kg for some k in Z (a non-unit!) and g in Z[x]

so f is immediately just reducible over Z

How much have you done on content?

Because it seems this proof is longer than necessary if you already know a bit about content

[white oxide]

uh it was introduced on the page before that lol, only used content to prove that the product of two primitive polynomials is primitive

[south patrol]

have you defined content only for polys in Z[x]?

[white oxide]

is it just the gcd of the coefficients

[south patrol]

ye for Z[x]

[white oxide]

what would it be for like Q

[south patrol]

Okay so actually the proof kinda uses the idea of content for a poly in Q[x]

ok lemme use cont(-) for content lol

given f in Q[x] and N > 0 with Nf in Z[x], we define cont(f) = cont(Nf)/N

it's easy to see that is well-defined basically

As a corollary of product of primitives being primitive, we actually have that cont(fg) = cont(f)cont(g) for any f,g in Z[x]

that then carries over to Q[x]

further, we have that cont(f) is in Z if and only if f is in Z[x], which is nice

but yeah these are like extra details you don't really need lol

just with them, you can write it as like: sps f = gh with f primitive. by multiplying g by a factor and dividing h by it, we may assume g has content 1. Then taking contents, h also has content 1. So g,h in Z[x]

[white oxide]

ah i see i'll have to digest that

thank u

[languid minnow]

I've been playing around with this family of fields over the past few months or so, and I'm pretty I've identified them as fields of formal Laurent series. I only really dabble in doing this sort of stuff and I was mostly leveraging intuitions and techniques from algebra and calculus (and a bit from the little familiarity with topology, coding, and some other stuff) as they came in handy, so I didn't really have the language or the conceptual toolkit to understand my field as such initially. I also actually was first exploring what I think (?) were polynomial rings modulo prime ideals, initially just checking out extensions of the binary field where addition and subtraction are the same (x+x=0), and eventually that lead me into stumbling into the infinite fields I been checking out more recently.

Anyways, I was curious if anyone had any recommended reading/watching/listening if I want to understand fields of formal Laurent series?

[frigid lark]

A wild det appeared
Det used baby doll eyes

[charred crescent]

are ideals of commutative rings closed under subtraction?

[solar shore]

trying to figure out subgroups of order 12 for the last row

[frigid lark]

Yes, cause all ideals are closed under subtraction

Any clues for part b?

For part a I got $Z(t) = \frac{1}{(1-t)(1-qt)}$, I have no idea if that is correct or not, cause I assumed $e^{\text{log}Z(t)} = Z(t)$

[cloud walrus]

Parrot Tea

[formal ermine]

bump

[grand cliff]

This was in our pset

solution is ridiculous, not really sure how we were supposed to get it :\

Solution (spoilered)

someone tell me if I'm overreacting

I mean, it's not THAT hard, but it requires making quite a few connections (which I certainly didn't make rip)

[elder wave]

because you can pull the scalar to the other side

[agile burrow]

You have a natural SES $0 \to (a) \to A \to A/(a) \to 0$. Tensoring is right exact so tensoring with $M$ yields $(a) \otimes M \to A \otimes M \to A/(a) \otimes M \to 0$. Presumably you know that $A \otimes M$ is isomorphic to $M$. The image of $(a) \otimes M$ in $M$ under this isomorphism is precisely $aM$, but this is then equal to the kernel of the map $M \to A/(a) \otimes M$, the latter of which is surjective by right-exactness.

[elder wave]

not sure if walter is typing up a more elaborate answer rn

[cloud walrus]

walter

[agile burrow]

lol good call timo

[elder wave]

indeed

[agile burrow]

I like yours better

[elder wave]

right the fact that the map that identifies A \otimes M with M is am solves this as well

forgor about that for a sec

[agile burrow]

Sorry, what is degree in the context of this problem?

[grand cliff]

degree refers to the number of elements in the set that permutation group is acting on

And then subdegree refers to the number of orbits in the induced action of a stabiliser

[formal ermine]

why is the image of it equal to aM

[elder wave]

.

[formal ermine]

wdym with this

[elder wave]

the isomorphism $A \otimes M \to M$ is $f(a \otimes m)=am$

[cloud walrus]

Timo

[formal ermine]

ah I see

thanks

[elder wave]

yw

[formal ermine]

thanks walter too

I forgot that the original ses is left exact too

[agile burrow]

it is a very nice SES

[formal ermine]

yeah

[elder wave]

indeed

[agile burrow]

if you look at the first Tor group you get the a-torsion in M

I think

[agile burrow]

I'll spend a bit more time thinking about this, I find the given solution a bit odd. Presumably I'm having an easier time because now I know one way to solve it lol. But my thinking is as follows:
If $|G|$ is even, there's an element $g$ of order 2 which permutes elements $a$ and $b$. If $[G : G_{(a, b)}]$ is even (where $G_{(a, b)}$ is the set of elements such that $ga = a$ and $gb = b$), then orbit-stabilizer implies that $|Ga|$ or $|G_a b|$ is even (where $Ga$ refers to the orbit of $a$ under the action of $G$).
But $[G : G_{(a, b)}] = [G : G_{{a, b}}] \cdot [G_{{a, b}} : G_{(a, b)}]$, where $G_{{a, b}}$ is the set of elements such that $ga = a$ and $gb = b$ or $ga = b$ and $gb = a$. The latter index in the product is equal to 2, hence the index of $G_{(a ,b)}$ is even and we get a contradiction.

I think for a pset problem it isn't totally far-fetched

[cloud walrus]

walter

[grand cliff]

Yeah, your solution looks a lot more natural tbh (and it's closer to what I was doing)

The flavor of the intended solution is so my professor

[agile burrow]

LOL I was gonna say

I feel like solutions to group theory problems say so much about what kind of math people do

[grand cliff]

yeah lmao

[south patrol]

ok so like any simple module V over a f.d. commutative C-algebra A is one dimensional, but how badly does this break over other fields?

I guess that if you view C as an R-algebra then C as a simple C-module has real dimension 2

[void cosmos]

let F be a field and p(x) be an irreducible polynomial over F, with E being its splitting field

suppose there exists a in E such that both a and a+1 are roots of p(x)

prove that char(F) != 0

any hints?

[formal ermine]

I'm guessing that you have to do a proof by contradiction using the fact that splitting fields in char 0 are separable

[south patrol]

Weird/interesting question tbh

[void cosmos]

its from a quali exam

[void cosmos]

fuck i didnt know that

[formal ermine]

it's interesting and boring at the same time

[formal ermine]

proof: exercise

[south patrol]

Yeah I'd just do what illum said lol basically

If you know like galois theory this falls out quickly

By contraposition again

[formal ermine]

oh ic

[void cosmos]

yea ig i can do that

from the derivative

[formal ermine]

yip

[void cosmos]

okay but what contradiction do i get

[chilly ocean]

can someone help me with this please?

[delicate orchid]

main trick is to try not to cry. Luckily for you that presentation is a semi-direct product of cyclic groups so you already know 75% of the conjugation straight away

[south patrol]

actually no i just had a walk lol

found a way more elementary solution to moamen's thing lol

try to contradict minimality of the minimal polynomial

||p(x) - p(x-1) also annihilates a but is of lower degree so it vanishes||

[delicate orchid]

no worries you don't need to know anything special about the semidirect product