groups-rings-fields

[simple mulch]

nvm, I see why

I mean not why, but I guess

hum ok, thanks

[delicate orchid]

what was the thing you spotted?

[simple mulch]

I was counting wrong the characters lol

Sorry

[delicate orchid]

np I was just curious

[untold basin]

Thx

For G a group, what is explicitly G/G ?

is it just {g.G | g in G} ?

[next obsidian]

Yeah

[untold basin]

oh

[next obsidian]

It’s gonna be trivial

[untold basin]

That's why I'm asking

[next obsidian]

Like, there’s only one coset

[untold basin]

My prof says G/G is trivial

[next obsidian]

Yeah

[untold basin]

I thought it means = {1_G}

[next obsidian]

Sort of?

The element is the coset G

Because gG = G for any g in G

[untold basin]

ok makes sense

I have a big potatoe that I haven't cut

ty

[next obsidian]

🥔

[untold basin]

x)

[stark sigil]

G/G is ambiguous

If you do springer theory

[next obsidian]

[stark sigil]

If you ask a springer theorist what G/G is, they'll say the groupoid of conjugacy classes of G with each conjugacy class having the stabilizer of any element of that conjugacy class as its automorphism group

[next obsidian]

Good thing intro group theory doesn’t involve any springer theorists

[stark sigil]

Lusztig must have taught intro group theory at least once

Springer theory

In slightly more seriousness, the second G in G/G standing for conjugation action is actually a thing in more than just springer theory

Sometimes people might notate it G//G for distinction

[unreal portal]

wanna make sure I have it right, the $[\rho_i, \rho_j]=1$ means that $\rho_i$ and $\rho_j$ commute if $|i-j|>1$, right?

[cloud walrus]

cgodfrey

[chilly ocean]

yes

as well as the other elements

[unreal portal]

oh, right, yeah

[sinful mirage]

how they get 5.7?

I have two extra terms

ah nvm

[prisma shuttle]

this is true

the way u want to thing about it

is that G/G (assuming this is like a modding out oerpation) is where u make new ring/group where everything in G is set to the identity

but here u are setting everything in G equal to the identity

so G/G isomoprhic to {1_G}

if your prof said G/G = G I think he might have meant G\G = G

which is extending G to G

or essentially doing nothing

[untold basin]

ok I think I understand intuitively

[untold basin]

But is there any formal argument for this ?

I don't have this in my course, my prof said that Z/Z is isomorphic to {0}

[prisma shuttle]

i mean wut i just did is essneitally a formal arguemnt

like by definition

when u do R/I for a ring R and an ideal I

[untold basin]

With my def I find {G}

[prisma shuttle]

all the elements in I are set to the identity in R/I

[prisma shuttle]

wut def does yoru prof use for R/I

[untold basin]

I'm working with groups right now

[prisma shuttle]

oh ok

[untold basin]

I guess it's the same

[prisma shuttle]

yea its the same idea

rings are basically just abelian groups with a multiplication oepration adde dot them

so it deosn't really matter

like they have the same gist

[untold basin]

G/H := {g.H | g in G}

[prisma shuttle]

oh wait wait

ignore wut i said above

this is very diffeent when its for groups

[untold basin]

ok damn

[prisma shuttle]

ok i see wut your prof is doing

basically like from yoru def

wait i think it shoudl be G/G = G

[untold basin]

I don't think so

[prisma shuttle]

why not

gG = G for all g\in G

[untold basin]

Can you explain how you got this ?

[prisma shuttle]

so shouldn't it just be G?

[untold basin]

I have {G} not G

[prisma shuttle]

i gotta go do smth rn

[untold basin]

ok np

[prisma shuttle]

but i'll be back later to post my explanation

and i'll ping u

[untold basin]

I'll be sleeping xD

it's midnight here

ty btw

[mighty spade]

It’s the set G, but in the factor group, so it’s both that and the trivial element of the factor group.

[prisma shuttle]

could u elaborate

am i misunderstanding smth

[mighty spade]

Probably. The definition of the identity in the group G/H is H.

[narrow marsh]

How was this calculated

There is a group action of G on a set M

g_1, g_2 \in G

and f is a function on M

[chilly ocean]

G acts on functions on M by (gf)(x) = f(g^{-1}x), the rest follows from basic group stuff

[narrow marsh]

im not sure i follow

shouldnt the first line be g1*(g2f)(x)=(g1*f)(g2^-1(x))

[chilly ocean]

no

g_1 is acting on g_2f

[narrow marsh]

right so you have g1*(g2f)(x)=(g2(g1^-1(f(x)) then

right?

[prisma shuttle]

@untold basin wut i was saying about the rings was actually corrrect

because G serves as teh idetntiy of G/G

so thus the whole group is identity in G/G

[narrow marsh]

@chilly ocean

[chilly ocean]

no

[narrow marsh]

ok

[chilly ocean]

an element g acts on a function f on M by sending it to the function g.f defined by (g.f)(x) = f(g^{-1}.x). in your picture, you have g_1.(g_2.f). unpack what this means

g_1 is acting on the function g_2.f, so it sends x to (g_2.f)(g_1^{-1}.x). now this equals...

[narrow marsh]

holy fuck im an idiot

thanks

||tfw when you havent done algebra in over a year||

by the way, (g_1g_2) means the group action on these 2 elements right

[next obsidian]

Like

The element g1g2

Is acting on that function

[narrow marsh]

i mean, im asking if g1g2 is just shorthand for g1 x g2 where x is the binary operation defined on G

im 99% sure it is but just want to double check

[next obsidian]

Yes

Lol

[narrow marsh]

thanks lmao

as i said, i havent seen any algebra in 1 year

this is from some QM notes im reading rn

based on representations of lie algebras

[west violet]

oh yah np

[west violet]

quantum mechanics?

[narrow marsh]

yeah

So a representation goes to GL(V) which is isomorphic to GL(n,V). It is true that n=dim(V) right

also, this one book says that GL(V) is isomorphic to GL(n,C). Is it V or C?

[next obsidian]

Should be GL(n,C)

If V is an n-dimensional vector space then GL(V) is non canonically isomorphic to GL(n,C)

The idea is that after you pick a matrix a map V -> V is given by an n by n matrix

And the map being invertible is exactly the matrix being invertible

[stark sigil]

GL(n,V) doesn't make much sense too

[unreal portal]

suppose I have a free group $F$ on $n$ generators and a free group $G$ on $m$ generators together with a homomorphism $\varphi: F\to G$. Now suppose I have a finitely generated group $H=F/\langle r_1, \dots, r_s\rangle$. Is it true that $\varphi(H)=G/\langle\varphi(r_1), \dots, \varphi(r_s)\rangle$?

[cloud walrus]

cgodfrey

[unreal portal]

It makes sense to me that it should, but I just want to make sure

[tribal moss]

No -- suppose phi maps everything to the identity in G.

[unreal portal]

oh that's true

oh, in this particular case I also know that $\varphi$ is injective, should've mentioned that

[cloud walrus]

cgodfrey

[obsidian sleet]

R is a PID

i know how to get the GCD entry but man getting the LCM is bonkers i have no clue what to do

ive tried all sorts of things but i can't get the 0s to work out the way i want

pls ping me if you have a suggestion for how i should approach dis ty !

[lethal dune]

I was thinking like maybe we can take $U = \m{y/lcm(x, y) & \ & x/gcd(x, y)}$?

[lethal dune]

@obsidian sleet

then maybe take V = id_2?

although the right might not be unitial

don't have anything beyond this

[obsidian sleet]

let me try this

[lethal dune]

also U is invertible because det(U) = 1

[obsidian sleet]

yes

[lethal dune]

oh wait, R is unitial

[obsidian sleet]

yes

[lethal dune]

then ig it's done?

[obsidian sleet]

yeah the R wil be unital since thats the situation i care about

yueah

nice lol

im in pain

[lethal dune]

thing is 1/gcd(x, y) may not exist

oh it does

F

lmao

[obsidian sleet]

it does its all fine

(d) = (x,y) so whatever

d/x is fine

[lethal dune]

ye x/gcd = d

[obsidian sleet]

man i should have seen this

very clever

[lethal dune]

[obsidian sleet]

i was trying to do the thing like you know

where u solve a rubik's cube face by face

but this does it all at once

[lethal dune]

oof

[obsidian sleet]

i tried to get gcd entry and then the lcm entry

but thats very oog

but tyyy

i would have liked a hint instead this is a cute solution ahaha

maybe i can find another

surely theres symmetry

[lethal dune]

I was thinking out loud

[obsidian sleet]

u r too smart

it seems U and V can be interchanged

neat

idk how this is supposed to help me for smith normal form for pid but im trusting this worksheet

[tribal niche]

slight sidenote but what is this tex font lol @lethal dune

[wooden ember]

@magic owl will the blue book seminars be recorded? I obviously don’t have the prereqs to attend them now but I’d like to look them over once I do

[lethal dune]

eulervm

[lethal dune]

maybe this was not the intended solution

I believe SNF is what we are supposed to use

[wooden ember]

Is this a sound argument for computing $Gal(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})$? I take the minimal poly $x^3-2$ of $\sqrt[3]{2}$ and look at its splitting field, $\mathbb{Q}(\sqrt[3]{2}, \omega_3)$. The galois group of this $Q$-extension is $\mathbb{Z}/3\mathbb{Z}$ since it's an extension by separable elements of degree 3 into a splitting field. Then i note that i can uniquely extend every $\sigma\in Gal(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})$ to a $\sigma' \in Gal(\mathbb{Q}(\sqrt[3]{2}, \omega_3)/\mathbb{Q})$ so that $Gal(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}) \leq Gal(\mathbb{Q}(\sqrt[3]{2}, \omega_3)/\mathbb{Q})$. Finally i note that the automorphism $\sqrt[3]{2}\to \sqrt[3]{2}\omega$, $\omega \to \omega$ is an automorphism that doesnt fix $\mathbb{Q}(\sqrt[3]{2})$ and so $Gal(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})$ must be a proper subgroup of $\mathbb{Z}/3\mathbb{Z}$, ie $Gal(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}) = 0$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[iron vessel]

Correct me if I'm wrong, but since $[\mathbb{Q}(\sqrt[3]{2}, \zeta_3):\mathbb{Q}] = 6 $, and thus $|Gal(\mathbb{Q}(\sqrt[3]{2}, \zeta_3)/\mathbb{Q})| = 6$, and we know that $Gal(\mathbb{Q}(\sqrt[3]{2}, \zeta_3)/\mathbb{Q}) \leq S_3 \implies Gal(\mathbb{Q}(\sqrt[3]{2}, \zeta_3)/\mathbb{Q}) \cong S_3$ as they have the same cardinality. Then, as $Gal(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}) \leq Gal(\mathbb{Q}(\sqrt[3]{2}, \zeta_3)/\mathbb{Q})$ of order 3, then we obtain that $Gal(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}) \cong A_3$.

[cloud walrus]

Évariste Galois

[wooden ember]

oh fuck of course that galois group is of order 6 yeah

though Gal(Q(2^1/3)/ Q) is not of order 3 i dont think (i believe the conclusion that it's trivial remains correct), so that part is wrong im pretty sure

[iron vessel]

yeah i meant that $Gal(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})$ is of order 3 and is a subgroup of $Gal(\mathbb{Q}(\sqrt[3]{2}, \zeta_3)/\mathbb{Q})$

[cloud walrus]

Évariste Galois

[wooden ember]

but it's not of order 3

at least i dont think so

[iron vessel]

well $|Gal(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})| = [\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3$

[cloud walrus]

Évariste Galois

[wooden ember]

it's not a splitting field

so you cant give it the order of degree, or at least not with a theorem i know of

because that wouldve been enough to conclude otherwise since there's only one group of order 3

which is why i went to look at it as embedded into the splitting field of x^3-2

[iron vessel]

Would you agree that $\mathbb{Q}(\sqrt[3]{2})$ is a subfield of $\mathbb{Q}(\sqrt[3]{2},\zeta_3)$?

[wooden ember]

yeah ofc

[cloud walrus]

Évariste Galois

[wooden ember]

that was my approach

[iron vessel]

and it is of degree 3 over Q

[wooden ember]

sure

[iron vessel]

then using Galois correspondence

[wooden ember]

im not sure that's a theorem weve seen yet but go on

[iron vessel]

so there is a subgroup of the galois group of the splitting field that is of order 3 and corresponds to $\mathbb{Q}(\sqrt[3]{2})$.

[cloud walrus]

Évariste Galois

[iron vessel]

The fundamental theorem of Galois theory

[wooden ember]

so looked up the theorem online

definitely havent seen it but after reading it, what makes you say this subgroup corresponds to Q(2^1/3) and not Q(omega_3)

[iron vessel]

because Q(omega_3) is of degree 2 over Q

[wooden ember]

i see

[iron vessel]

but take everything with a grain of salt man im also in my first galois theory class.

[wooden ember]

could you provide a non trivial element of Gal(Q(2^1/3)/Q)?

[wooden ember]

we've only just started so we dont have a lot of theorems available sorry

[iron vessel]

Check the second example: https://en.wikipedia.org/wiki/Fundamental_theorem_of_Galois_theory

its exactly your question

[wooden ember]

yeah exactly

they say the subgroup of order 3 corresponds to Q(zeta_3)

and apparently Q(2^1/3) corresponds to a subgroup of order 2

[iron vessel]

Fair enough, i see it too. But thats super weird.

[wooden ember]

you sure you didnt get this mixed up

[iron vessel]

for any pth primitive root of unity, $[\mathbb{Q}(\zeta_p):\mathbb{Q}] = p-1$

[cloud walrus]

Évariste Galois

[wooden ember]

like dont you need to divide by it

no yeah the degree is right but doesnt it give you the index

not the order

[iron vessel]

wait I think my mistake is because when you have a galois correspondence, the diagram switches

like it is inverted

thats why

[wooden ember]

well anyways we'll see this theorem later so ima have to figure out another way to solve this one

[iron vessel]

How are you doing Galois theory without the fundamental theorem?

Like what other theorems have you seen?

[wooden ember]

i mean honestly i cant think of a single automorphism of Q(2^1/3) that isnt trivial

[wooden ember]

some fundamental properties like the galois group acting on roots or finding the order when the extension is a splitting field by separable elements

that's it really we just got into it

[iron vessel]

Ah ok so you just started

So another way is once you know the degree of the splitting field is of order 6 over Q, then you can explicitly look at the automorphisms

because any $\sigma \in Gal(L/K)$ is defined by $\sigma(\sqrt[3]{2})$ and $\sigma(\zeta_3)$

[cloud walrus]

Évariste Galois

[wooden ember]

yeah sure

but it seems to me like as long as you fix Q(2^1/3) you need to have sigma(2^1/3)=2^1/3

[iron vessel]

seems to me too to be honest

[wooden ember]

i think what the theorem is saying is that if I fix Q(2^1/3) i get a subgroup of order 2

and if i fix Q(zeta) i get one of order 3

[iron vessel]

like what sigma fixes Q but not \cbrt(2)

[wooden ember]

but this doesnt mean there is a non trivial automorphism of Q(2^1/3)

its just that we switch zeta_3 and zeta_3^2

[iron vessel]

This is how the theorem is stated in my book

[wooden ember]

right the extension isnt normal

[iron vessel]

2 is precisely what I meant. In fact it says "inclusion reversing".

[iron vessel]

Q(\cbrt(2)) isnt, right

[wooden ember]

2 only gives us a correspondence with subgroups of Gal(splitting field) but that doesnt necessarily tell us about Gal(intermediate field)

[wooden ember]

yeah i dont think so

we havent really covered normal extensions but the definition came up in an earlier exercise

[iron vessel]

well but Q(cbrt(2)) is an intermediate field of the extension Q(cbrt(2), zeta_3)

[wooden ember]

sure

but it just tells us how it corresponds to subgroups of the galois group of that extension

not the extension we're interested in

at least thats how i understand it

[iron vessel]

What extension are you interested in?

[wooden ember]

Q(cbrt(2))

[wooden ember]

ie what automorphisms fix our intermediary field

[iron vessel]

yeah but similarly, Gal(cbrt(2)) is a subgroup of Gal(cbrt(2),zeta)

[wooden ember]

yeah but it doesnt necessariyl have the order of the above corresponding subgroup

that's just the subgroup fixing the intermediary field

but we dont know about its restriction to the intermediary field

i coudl argue like this

[iron vessel]

from the theorem it even says it is an inclusion reversing bijection, so Gal(cbrt(2),zeta) is a subgroup of Gal(cbrt(2))

[wooden ember]

Gal(cbrt(2),zeta) acts on {2^1/3, zeta2^1/3, zeta^2*2^1/3}

let's look at how the subgroup Gal(cbrt(2)) acts on this

since it fixes Q(cbrt(2)) it must send 2^1/3 to 2^1/3

[iron vessel]

thats only for sigma in Gal(Q(cbrt(2),zeta)/Gal(Q(cbrt(2))

[wooden ember]

but then it restricts to the identity on Q(cbrt(2))

wait i need to think for a sec

[iron vessel]

Maybe it is simpler if you use the fact that $S_3 \cong D_6$

[cloud walrus]

Évariste Galois

[wooden ember]

what's wrong with this argument

[iron vessel]

Oh nothing is but its not really an argument yet

[wooden ember]

my point is that if i want to extend sigma from Gal(cbrt(2)) to sigma' in Gal(cbrt(2), zeta) it must fix 2^1/3

but then restricting back to Q(cbrt(2)) we see that sigma is trivial

[iron vessel]

yeah, and there is precisely 3 such element in D_6

[wooden ember]

2*

[iron vessel]

2 yeah lol

[wooden ember]

identity and involution of zeta_3 and zeta_3^2

but in both cases they fix Q(2^1/3)

so if they're extending sigma then sigma must be trivial

so Gal(Q(2^1/3)/Q) is trivial

[iron vessel]

yeah

but it is of order 2 still

[wooden ember]

no?

we just said that there are two ways to embed it into Gal(splitting field) is all

not that it itself is of order 2

[iron vessel]

oh you meant over Q

[wooden ember]

yes, not Gal(Q(2^1/3,zeta)/ Q(2^1/3))

that would be 2 i agree

[iron vessel]

fair enough

[wooden ember]

ok coolio thanks a bunch for your help

this has given me a much better picture of what's going on

[iron vessel]

i think its really just because it is not a galois extension

as it is not normal

and no worries, hope i didnt make you even more confused lol

[wooden ember]

i suppose a much simpler argument couldve just been to say: sigma in Gal(Q(2^1/3)/Q) is determined by sigma(2^1/3). We must have sigma(2^1/3)^3-2 = 0 since sigma fixes Q and hence sigma(2^1/3) = 2^1/3 since we're working in Q(2^1/3)

so Gal(Q(2^1/3)/Q) is trivial

[iron vessel]

I don't see why that shouldn't work

[wooden ember]

what kind of easy to use calculator for expanding polynomials involving roots of unity would you recommend

im getting sick of expanding all these expressions and making mistakes

[iron vessel]

hmm i personally dont really use any calculator, but any 'expand calculator' may work

i think, especially with roots of unity, you build techniques to work with them

over time

[wooden ember]

where am i going wrong here? Im trying to compute the minimal polynomial of $\omega_6+i$. To do that i note that $\mathbb{Q}(\omega_6, i) = \mathbb{Q}(\omega_{12})$ which is the splitting field of $x^4-x^2+1$. The galois group of the extension $\mathbb{Q}\subset\mathbb{Q}(\omega_{12})$ is the klein 4 group and generated by the automorphisms $\omega_{12}\to\omega_{12}^{-1}$ and $\omega_{12} \to -\omega_{12}$. So to compute the minimal polynomial, i look at the orbit of $\omega_6+i$ under this group and multiply all the factors to get $$(x-(\omega_6+i))(x-(\omega_6^5+i))(x-(\omega_6^5-i))(x-(\omega_6-i))$$ but this doesnt give me a polynomial over $\mathbb{Q}$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

here omega_i is i-th primitive root of unity

[iron vessel]

My first attempt would be a little "bruteforcy", as in, you know that its minimum polynomial is of degree 4. So I would start by computing x^4

[wooden ember]

note the exercise wasnt phrased like this

it was first getting the galis group of the extension Q(omega_6,i) and then computing a bunch of minimal polynomials

(doing this for a few extensions)

(for reference)

im on 3

[iron vessel]

Let f be the minimum polynomial of $\zeta + i$. I recall that for a galois extension, the minimum polynomial may be written as $\prod_{\sigma \in Gal(\mathbb{Q}(\zeta + i)/\mathbb{Q})} \sigma(\zeta + i)$.

[cloud walrus]

Évariste Galois

[wooden ember]

that's what im doing yeah (idk about galois extensions but in my cases it worked)

oh wait youre ranging over Gal(Q(zeta+i)/Q)

hmm

[iron vessel]

Yeah I would try and find what it is

primitive element theorem

[wooden ember]

idk i feel like the method they want us to go for is ranging over the galois group of the splitting field

[iron vessel]

yeah but ranging over a simple extension in general it is easier I would think

[wooden ember]

the splitting field is simple in this case

since it's Q(omega_12)

[iron vessel]

Then maybe try something like $\prod_{\sigma \in Gal(\mathbb{Q}(\zeta_a)/\mathbb{Q})} \sigma(\zeta + i)$ with $\zeta_a$ as omega12

[cloud walrus]

Évariste Galois

[wooden ember]

that's what i did

but it didnt give a polynomial over Q

[iron vessel]

what did it give?

[wooden ember]

.

the big expression at the bottom

[iron vessel]

did you expand it?

[wooden ember]

when expanding you get (x^2-x+2i+sqrt(3))(x^2-x+2i-sqrt(3)) which isnt over Q

(the complex part wont disappear in the expansion)

[iron vessel]

Even if you expand those degree 2 polys?

[wooden ember]

yeh

youd need the complex part to be along sqrt(3)

[iron vessel]

$\zeta_6 + i = (-1 + 2i + \sqrt{3}i)/2$

[cloud walrus]

Évariste Galois

[wooden ember]

yeah

uh no you mean+1 right

not -1

[iron vessel]

oh yeah

Ill try it right now

[wooden ember]

thanks

waitso 2i+sqrt(3) is wrong, i realised i made the same mistake twice in a row again

should be 2-sqrt(3)i and 2+sqrt(3)i

which is in Q but when i look at the splitting field in wolfram it gives something i believe different

gives the polynomial x^4-2x^3+5x^2-4x+7

[iron vessel]

Does it work when you plug in the root?

I got something else, i got x^4 -2x^3 + 5x^2 -4x + 1

like a difference of 6 lmao

[wooden ember]

did you remember to take i^2 = -1

[iron vessel]

yeah

[wooden ember]

because if you multiply sqrt(3)*sqrt(3) without taking into accoutn i^2 you get +1

and not +7

[iron vessel]

all it takes is to plug it in and see

[wooden ember]

true that

ill check

just need to find a calculator for it lol

yeah +1 is correct

gotta look for my mistake

[iron vessel]

What calculator did you use to check it?

[wooden ember]

wolfram

[iron vessel]

ok

[wooden ember]

i dont get it ive got intermediary factoring of (x^2-x+2-sqrt(3)i)(x^2-x+2+sqrt(3)i)

and 2+sqrt(3)i has square norm of 7

dont see where i went wrong in getting the factoring either...

well anyways thanks a bunch, ill figure this out once i get a brain

[iron vessel]

No worries, sorry I couldnt be more helpful, I am also trying to do my topology hw in the meantime and im braindead

[wooden ember]

ahhh i figured it out

forgot to multiply by i at one point in ym mental calculatios

[iron vessel]

lmao

cheeky i

[wooden ember]

so you get (x^2-x+2+sqrt(3))(x^2-x+2-sqrt(3)) which gets you what you want

wooo

[iron vessel]

Nicely done man

[wooden ember]

now my only issue is why wolfram alpha tells me the splitting field of this polynomial is weird as heck

[iron vessel]

What does it say it is?

[iron vessel]

Also splitting fields are unique up to isomorphism

and, though im not sure, this may just be omega_12

[wooden ember]

oh and anyways it might not be the same splitting field

since we're working with zeta_6+i

not zeta_6 and i separately

there's no reason we should get back omega_12

[iron vessel]

true

[wooden ember]

well whatever im content with my answer

now to compute more shit

[untold basin]

There is no ring morphism between Q and Z because if I suppose that there exist one, I have :
1_Z = f(1_Q) = f(2 * 1/2) = f(2) * f(1/2) = 2 * f(1/2)
That tells me that 2 is invertible in (Z, *) which is a contradiction

Is it okay for a proof ?

[delicate orchid]

seems ok to me

[untold basin]

ok ty

[narrow marsh]

oops i meant GL(n,F) instead of GL(n,V)

[languid meteor]

how do you identify when you have a direct product as a subgroup of your group?

is it when you have a product of two normal subgroups with trivial intersection?

[delicate orchid]

I believe that would suffice as long as their product is itself a group

[hidden haven]

Shouldn't be necessary though

[hidden haven]

Product of a subgroup with a normal subgroup is always a group right?

[delicate orchid]

Moldi do you think I have the foggiest clue of what this mumbo jumbo is

I'll think about it though

[next obsidian]

Product of two subgroups is always a subgroup

[delicate orchid]

oh yeah it's union which isn't always

[next obsidian]

You just can’t build it up from the two constituents in any realistic way really

[hidden haven]

Oh ye product is defined by taking all words

[next obsidian]

But even when one is normal you still really can’t, when their intersection is trivial you have a semi direct product

But in general it’s kinda fooked

[delicate orchid]

https://cdn.discordapp.com/emojis/957820487130570792.webp?size=80&quality=lossless

[next obsidian]

I think you could apply second iso to try and get at it

But it’s still pretty cr.inge

Wait maybe GH is defined just to be {gh, g in G, h in H}

That definitely isn’t a subgroup always, it is iff GH = HG

But you can just close it under products

I WAS WRONG MOLDI I WAS WRONG

I was too used to rings where you always close the product of ideals under sums

I’m sorry

[delicate orchid]

which is true if one of the groups is normal

probably

[next obsidian]

I think so

Maybe

Maybe not

Kekw

[delicate orchid]

anyway point is we're looking for some kind of product so it's guaranteed to be the case that one of them is normal

[next obsidian]

It’s guaranteed when the product is the entire group

[delicate orchid]

nah I think it's always guaranteed

[languid meteor]

ok cool so I guess if I have a simple group

the only direct product subgroups are like 1 x N where N is a subgroup

[delicate orchid]

let H and N be two subgroups of G with N normal, then GN = NG, restricting to H we still have NH = HN

[delicate orchid]

for direct products both groups need to be normal

nevermind I'm silly I was gonna say those are the only semidirect products subgroups but semidirects with the trivial group is just the direct product with the trivial group

[hidden haven]

💩

[untold basin]

Can someone help me with this ?
(X) intersection (X+1) = (X²+X)

(P) principal ideal

[wooden ember]

Write an element of (X) intersection (X+1) as P*X = Q*(X+1)

What can you say about X and Q

(Or symmetrically about P and (X+1))

[untold basin]

Can I say that P * X = Q * (X+1) <=> P * X = Q*X + Q <=> X * (P-Q) = Q.
I deduct that Q is in (X)
So I come back to the fact that S = Q * (X+1) and S = P * X (the arbitrary element which is in (X) and (X+1))
And I can tell that S = Q * (X+1) = T * X * (X+1) = T * (X²+X)

[chilly ocean]

@untold basin yeah. It doesn't matter that we are in Q[X] though, also bad notation

[untold basin]

I know. I like to put details so that people can fully correct me

And wdym by bad notation

[chilly ocean]

Using Q in two different ways

[untold basin]

oh okay I see

[wise igloo]

if I consider myself knowledgeable in a first course of group theory, what concepts should I be proficient in?

[chilly ocean]

groups, morphisms, subgroups, quotients, group actions, fundamental theorem of fg abelian groups, maybe some sylow stuff but that's not terribly important

just to name off some stuff

[wise igloo]

morphisms meaning iso and homo right

[chilly ocean]

yes

included in that would be the isomorphism theorems

you should know the first and fourth by heart, other two aren't that important lol

[untold basin]

And put the first in bio xd

[wise igloo]

oh ok

[sharp sonnet]

is the fourth one correspondence

[chilly ocean]

yes

[wise igloo]

what other theorems/definitions should I definitely commit to memory?

[chilly ocean]

"lattice isomorphism theorem" or something

[sharp sonnet]

i see it sometimes called third or fourth, was just wondering if i have knowledge of basic group theory

i would add that you should be familiar with some common examples of groups

[chilly ocean]

if you can't write down all of the sylow theorems on the spot then you don't know basic group theory sweetie

[wise igloo]

fuck

[next obsidian]

Know the groups of order <= 8 up to isomorphism

[sharp sonnet]

i see it so often that people cant produce their favorite group having a certain property

[chilly ocean]

every group is a lie group

[wise igloo]

lie stuff is later stuff right

[sharp sonnet]

they're just matrix groups

[chilly ocean]

lie groups are differential geometry lmao ignore that message

[wise igloo]

ok lmao

[next obsidian]

What if you had a group but it was also a manifold but it was also a group = Lie group

[wise igloo]

does anyone have a good resource for practice problems for group theory?

[untold basin]

Serge Lang ?

[next obsidian]

Open up dummit and foote

To any of the first 150 or so pages

Maybe even more

[wise igloo]

oh ok ty

[prisma shuttle]

yes d&f has some really good group theory stuff

theres also some partial sols u cna find online

like theres this one website i remember had some of them

at least for the groups part

but then the rings it was a missing a bucnh

but quizlet plus also has like textbook solutions for d&f that are really good

so i would def recommend that

[mighty spade]

yes, serge lang has a ton of group theory problems from all sorts of different perspectives

i'm the biggest serge lang fan alive probably

[untold basin]

My prof harassed my class last year with Serge Lang xd

[chilly ocean]

my condolences

[mighty spade]

i've worked through all his undergrad books, and I plan to go through his complex analysis and real and functional analysis once I finally complete algebra

he's probably my favorite author

[mighty spade]

how so?

only problem I have with serge lang is that he calls integral domains entire rings

[untold basin]

He was suggesting us to look at Serge Lang whenever we finished to prove smth

[mighty spade]

which book?

Algebra?

[untold basin]

yes

[mighty spade]

I have a sentimental attachment to Algebra, because I spent the three weeks I was in the hospital working on Part 1 and the Field Extensions and Galois theory chapters

that and part one of munkres

[untold basin]

damn

[mighty spade]

it's crazy how productive you become when you don't have access to electronics, and just have an empty room with a chalkboard and two math books

and chalk

[untold basin]

I feel what you mean

A lot

It's like a natural thing to progress in these conditions

With a little bit of curiosity and you become really good

[mighty spade]

I really hope that when I go to CMU this summer and next school year, I'll be able to replicate that by taking classes that actually match up with what i'm learning/interested in

[untold basin]

May your dreams come true

[strong yacht]

Does anyone have some resource for calculating discriminants? I'm completely stuck on this

Oh I should clarify and say, I don't see how they came up with the hint that it's a linear combo of r^4 and s^3

[tribal niche]

Everyone and their cat at my uni will make fun of you for this opinion

[mighty spade]

you're telling me they won't be like "wow, this guy learned algebra and analysis from Lang, he must be cool and old school"

you're telling me they'll instead be like "serge lang books suck, what a loser"

[chilly ocean]

Don't make fun of people please, let's respect each other

[tribal niche]

Oh no I wasn't make fun of them, I'm saying that people at my uni hate lang lol

[tribal niche]

I think most people dislike lang at my school

[chilly ocean]

I think it's just because they see his Algebra book as hard

So they'd see it more as an accomplishment I presume, if they care at all

[tribal niche]

Twas joke bc ppl dislike that text for various reasons even tho it's required for classes

[tribal niche]

Nah it's more because it's too dense and is not extremely good at any particular topic. Sort of like a "jack of all master of none" kind of deal for a textbook

[languid meteor]

i keep trying to take the inner product of a class function (which is actually an irreducible character) restricted to a subgroup and an irreducible character of that subgroup

but I don't get an integer - does anyone know whats happening here?

can I not do this? I'm almost certain im not making a calculation error

[verbal harness]

I'm learning me some group theory (amateur, not college course). And I notice that the Four Axioms are a big deal. But I feel like I've just simplified them down to three somehow, and I'm wondering if somebody can help me determine if I am in fact simply insane instead?

In particular "all products of binary functions upon group members will always be a group member" and "all actions have an inverse" I figured out a way to merge into one, I think equally expressive axiom instead. "All chains of composition can always be reversed by a single action", aka the "no matter how circuitous the path, you're always no more than one step away from where you started".

I am convinced that the first two axioms I named can be completely derived from the third.

[chilly ocean]

could you state that in more precise terms

it's kind of vague

[verbal harness]

Sorry, also I mixed up one of the two axioms I'm trying to replace. 😛 One mo

OK, sorry nevermind. My problem is that I was confusing an axiom from Category theory with Group theory. Category theory has "any chain of actions can be replaced by one action", and I momentarily thought that was a group theory axiom. My bad! 😮

Would be nice though, then you'd always be able to solve a rubick's cube in one move lol 😉

[tribal moss]

Beware that "action" has a technical meaning in group theory, which doesn't seem to be what you're wanting to talk about.

[verbal harness]

nods, I thought I had that usage right as well. But "action" is not the binary operation between group elements, but instead (lemme know if this description sounds correct) it's the manifestation of each group element as a transformation on some underlying set?

Such as expressing a group element as a way of transforming a geometric figure symmetrically.

[tribal moss]

"Action" usually means an operation that takes a group element and a member of some other set and returns a (usually different) member of the other set.

[verbal harness]

That sounds like what I said. Take a group element from Z/5Z and use it to describe "a rotation of a pentagon" and you are taking one orientation of the pentagon and returning a different one.

[tribal moss]

Ah, okay, I misunderstood what you meant by "underlying set".

[verbal harness]

If "action" is the term of art for describing a group element that way, I'd love a term of art for "a set compatible with this group's actions" too. 🙂

[tribal moss]

Anyway, there are several different ways to state axioms for groups.
The fact that the group operation always produces an element of the group is sometimes stated as an axiom, but sometimes considered implicit in the very concept of a binary operation.
The associative law is almost always an axiom.
The existence and properties of an identity and inverses can be axiomatized in different ways.

[verbal harness]

Yeah, I think we always wind up with four though. I definitely wouldn't agree with seeing closure as implicit because so many other magma don't require it.

The only reason I thought I'd seen a way to boil it down to three is because I momentarily confused a Category Theory axiom in there. bonks head

[tribal moss]

Hmm, as far as I know, closure is the only property a magma does require.

[verbal harness]

Is it? I didn't realize. I haven't looked that closely but always figured a magma was just "here's a set, maybe find some ways to mess with the elements, I'll be back after lunch" lol 😉

And then applying different restrictions yields different kinds of magma with different properties and exciting new names like rings and uh.. quasiloops.

Maybe I get caught up there just because "the binary operation(s) that are necessary to strictly define your type of beasty" don't have to be congruent to "the binary operations you can try to apply to it anyway". Like, I can choose to try to subtract Natural Numbers, which is a group along with whatever other holes it fits through, I just can't expect that every pair of inputs will yield an output that's still in the set.

Because subtraction isn't involved in it's qualification to be called most of those things.

[tribal moss]

Natural numbers (with addition) are not a group, just a monoid.

[verbal harness]

Yep I was just seeing that too lol, you beat me to the correction. but that's just a poorly named example of what I'm getting at. 🙂 Perhaps a better example is Z is a group under addition, and a field under addition and multiplication. But I can still try to divide or take roots in it and sometimes get an answer that is also in the set.

It's not complete to division or roots, but those weren'

[tribal moss]

Sorry for doing it again, but you mean ring, not field, there.

[verbal harness]

t the operations that qualified it as a group or a field either.

Dang, I am just fulla malaprops today. Your well meaning corrections are very much appreciated though, thank you Troph. 😊

I like to explore the ideas in Abstract Algebra, but I guess constantly confusing the terminology and forgetting which axioms belong to which theories has got to make communicating to anyone else about it a real problem.

For example, I've worked out an informal conjecture that every verbal orientation problem such as figuring out the best words to use to unambiguously describe directions (stage left = house right, port is left when looking at the prow but right when looking aft, right hand rule, etc) can be mapped to an associated group, and that recognizing which problems map to the same group can help one lift orientation terms from one problem to solve another.

[tribal moss]

It'll come with practice.

[verbal harness]

I drive up to the pumps, there is one lane on each side of the pump. I ask my wife "which side is the gas for this car again?" she says "Left", so I go to the left lane and get into trouble. Problem there is "which side of the car is the gas cap on" and "which side of these pumps should I drive up to" have directly conflicting answers. I worked out that port/starboard solves both problems though.

They are orientation terms with assumptions about the object being oriented baked right in. port is left .. of the vessel. When facing forward. It's also the right side of whatever you're trying to dock to, and vice versa. Also, most boats moor to port and most cars have gas caps on the port side.

But the idea that this maps into group theory is an instinctive one, and I'm pretty far away from being able to clearly name an example let alone write out any rigorous proofs for it. In the above example, I can't decide if the Klein group or if something of order 8 really covers the idea properly. I just know that my instincts say "there is a group for this", so I'm following that hunch to see where it leads me. 🙂

[tribal moss]

If you want a general theory, I suspect you might be looking for appropriate subgroups of O(3), the group of rotations and reflections in space.

[verbal harness]

Well I feel like what I'm aiming for is more general than that. Like, it's not always about spatial rotations, those are just some easy examples to come up with. But it does align more with Symmetries which I know to be isomorphic to groups. It certainly wouldn't be a subroup of O(3) in particular to ask "how does the right hand rule manifest in >3 dimensions" for example, and it's hard to get any spatial traction at all to questions like "how do I describe the bit endianness that a particular CPU architecture works with".

They come up with terms like "little endian and big endian" or "machine order and network order" but then you get situations where byte order and bit order don't match. Or byte, bit, word, and nibble order don't match and it gets pretty hairy pretty fast. 😉

But those generally all map to Z/2Z^N, for N bit structures, I think.

hmm, for 2^N bit structures.

[unreal portal]

are there any resources on finitely generated groups where the generators are themselves words? For example, the group $G=\langle b_1b_2b_1^-1, b_2^-1b_1b_2,|,r_1, \dots, r_s\rangle$ for some relations in terms of the $b_i$?

[cloud walrus]

cgodfrey

[cloud walrus]

rye

[unreal portal]

your concern is definitely right, if a=9 then it's not a field because it's reducible as (x-1)(x-9). You may also have to be more specific about what you mean by no homomorphism to Z_11, there's certainly at least the trivial homomorphism sending everything to 0

[fallow plume]

Note: If you define rings necessarily have identity, the last part isn't true

[verbal harness]

@tribal moss After some thought, I've been able to both rally and amplify my original statement, and make it more rigorous.
If my conjecture is right, then not only can Completeness and Identity be combined into one axiom, but maybe all four group axioms can be lumped into just one axiom. Lemme know if it sounds like the following holds water? 🙂
Conjecture: If G is a set, and * is a binary operation on that set, then (G, *) is a group iff:
∃I∈G: ∀A∈G: ∀B∈G: ∃C∈G: A*B*C = I = C*A*B

[chilly ocean]

latex pls

[verbal harness]

Uh oh.. it eats my asterisks. I know how to unicode but I'm not sure how to latex. 😮

[fallow plume]

put \ in front of your asterisks

it'll escape it and they'll send

granted, you should still learn LaTeX sooner than later :P

[verbal harness]

Edited for escaping the asterisks.

[fallow plume]

For reference, the LaTeX is:
If $G$ is a set and $\cdot$ is a binary operation on that set, then $(G, \cdot)$ is a group iff:
$$\exists I \in G, \forall A \in G, \forall B \in G, \exists C \in G, A\cdot B\cdot C = I = C \cdot A \cdot B$$

[cloud walrus]

Thomas

[fallow plume]

you use single $'s for inline blocks, and you use double dollar signs for "math blocks"

Just like how you use ` and ``` almost.

[verbal harness]

Roger that, ty Thomas. 🙂 I wonder if my conjecture holds or not though. I expect probably not, but if not then learning why not will help me understand a lot of things better.

[fallow plume]

other than that it's a matter of just Googling the other symbols like \cdot

np

[verbal harness]

IIUC then the single statement above should completely encapsulate completeness, invertibility, existence of identity, and associativity without requiring commutativity.

[lethal dune]

@obsidian sleet you still here?

I found another sol ( this one is the intended one lol)

[obsidian sleet]

oh lmao yes ?

wait but dont tell me

do u have hints

[lethal dune]

the problem with the last approach was that x/lcm(x,y) may not be defined

[obsidian sleet]

oh why

hold on hmm

[lethal dune]

so you use SNF decomposotion algorithm to reduce it to SNF

I have

[obsidian sleet]

but like i can write x/lcm as d/x

[lethal dune]

so you just follow the algorithm

[obsidian sleet]

wait ryu the sad thing is that

[lethal dune]

d/ x is not defined in PID

[obsidian sleet]

i need this result to prove existence of SNF

[lethal dune]

ye like you can peek into #bots where I have been trying out

[fallow plume]

ya hate to see it

[lethal dune]

thr matrices are given there explicitly

[obsidian sleet]

d/x not defined in pid but i thought that (x,y) = (d) means that x = pd oihhhhhhhhhhhh

im in pain

ok

[fallow plume]

just to clarify, I is identity?

[obsidian sleet]

u Win

[lethal dune]

x/d is defined

like you use SNF algo to determine the matrices

[obsidian sleet]

yeah x/d is fine

but yes ur right

d/x not so

[olive edge]

sorry guys, just wanna check if my reasoning is right, if I have a field, theres no way I could map to a ring with less elements. Because homomorphism needs to be injective for fields to rings right?

[lethal dune]

by more do you mean field is finite?

[olive edge]

yes

both finite

[lethal dune]

then yes

[olive edge]

awesome thanks

[obsidian sleet]

right the kernel is either trivial or the whole field

whole field kernel means u have zero map so it wont even be unital homo

very sad

[olive edge]

very sad indeed

crap frgot metal is a mod

[lethal dune]

good nickname tho

[chilly radish]

It's not encoding associativity. In fact as you wrote it right now it's not even well-defined. \cdot is a binary operation, if you're not a priori assuming associativity writing a*b*c doesn't mean anythiny

[verbal harness]

It means that if f(a,b) = a*b, then f(f(a,b),c). And in the full statement, f(f(a,b),c) = I = f(c,f(a,b)).
I found a hole in my assumptions regarding closure and inverses described in the discord-thread-thing above, but I haven't yet found a hole in the associativity implication.

[chilly radish]

I don't see how this implies associativity

[untold basin]

Hello

I don't understand why X² + 1 doesn't have roots in F_3

People on internet just plug 0,1 and 2 and they don't obtain 0

But F3 doesn't have a unique way to be written right ?

[sharp sonnet]

ok? renaming the elements wont change anything though (other than the names of the elements)

[untold basin]

I'm misinterpreting smth

1 min

[untold basin]

I meant that in F3, [2] = [-1]

wait it doesn't change anything

lol

[sharp sonnet]

yes

this renaming respects addition and multiplication

so you can work with whatever representative you want

[sullen island]

hello, i understand that there is exactly one copy of $k$ in the decomposition of the group ring kG, but why does that also imply there is exactly one copy of k as a composition factor when we consider the composition series of kG?

[untold basin]

ty

[cloud walrus]

xy

[sullen island]

like for instance, if M2 is of codimension one in M1, we could have a composition factor M1/M2 that might be isomorphic to k (which we are identifying as the trivial module) as well, right?

[sullen island]

<@&286206848099549185>

[chilly ocean]

F_3 has only 3 elements, 0, 1 and 2

You could write them as 0+3Z, 1+3Z, 2+3Z

[sullen island]

that question was already answered lol

[chilly ocean]

this tag is pretty much exclusively for people from the math help channels

[hidden haven]

If you have a decomposition into a direct sum of simple modules, then the composition factors can only be the simple modules that appear in that direct sum (with multiplicity)

That is because any composition series you write will be a sub-sum of the one you have in the decomposition

By simplicity

[sullen island]

sorry, what is "sub-sum" ?

[hidden haven]

By sub-sum I mean you only delete some of the summands

[hidden haven]

Ye I just made that up lmao

[hidden haven]

Like every summand either appears fully in the sub-sum or not at all

[sullen island]

so lets say we have k + S1 + S2, where S_i are simple

and suppose k+S1 has codimension one in k + S1 + S2

youre saying (k+S1+S2)/(k+S1) can never be the trivial module?

[hidden haven]

Ye because it is S2

[sullen island]

ohhh ohhhhhh

okay, i think i get it, thank you!

[untold basin]

Yes that's what I noticed

What is the first reflex I should have when I want to show that two rings are not isomorphic ?

[chilly ocean]

In topology we usually check that two spaces aren't homeomorphic by looking at their properties

there's compactness, connectedness...

maybe there are some properties of one ring which are not included in the other

like the tensor product with Q being different

etc etc

[untold basin]

In my case it is R[X] not isomorphic to C[X]

I guess I can just say
Suppose R[X] is isomorphic to C[X].
Since (R is a field => R is a domain ) and (C is a field => C is a domain)
Thus R[X]* = R* and C[X] = C*
We have R[X] is isomorphic to C[X] => R[X]* = R* isomorphic to C* = C[X]* which is a contradiction

Is it good ?

[south patrol]

That does seem to work, though you can sort of cut out the middle man (since you are assuming R* not iso to C*) and just do it more directly

e.g. note R[x] has no elements a such that a^2 + 1 = 0

[untold basin]

I had this intuition but I don't really know how to write it properly

[south patrol]

In my mind isomorphic is basically the same but different labels lol

[untold basin]

euhm wait

yes that's what I wanted to say

xd

[south patrol]

But sure, to write it more properly, sps we have a homomorphism φ: C[X] -> R[X]; then φ(i)^2 + 1 = φ(i^2) + φ(1) = φ(i^2 + 1) = φ(0)= 0

[chilly ocean]

Yeah good job

[untold basin]

Okay I see, thanks

[untold basin]

ty

[south patrol]

np

[odd flame]

Hi, I have a question and i found a solution online to it but im having a hard time understanding the solution itself

[chilly ocean]

post

[odd flame]

Q2

https://math.stackexchange.com/questions/3690924/prove-that-x−1-and-y−1-are-irreducible-in-r

im not familiar with norm maps

[untold basin]

This is exactly the objects I have to understand for my exam

Which question blocks you

[odd flame]

2a

I dont understand it

[untold basin]

I'm searching aswell

[untold basin]

We can prove that (x-1) is maximal

[untold basin]

Can someone help with 2a pls ?

I got stuck too :c

[odd flame]

Or atleast a little help with understanding the solution 🙂

[chilly ocean]

suppose that x-1 = pq with p, q elements of R
That means X-1 = p'q'+r(X^2+Y^2-1) for some p', q', r

treating both sides as polynomials in Y, we must have r = 0

[untold basin]

I wanted to do that but don't you find that elements of R are hard to manipulate ?

[chilly ocean]

that's an equation in R[X, Y]

[untold basin]

It's polynomials in R[X,Y] mod (X²+Y²-1)

Wait

Do I have an equality of equivalence classes

So the representors are in relation

And I obtain your equation ?

[chilly ocean]

this conclusion is wrong

[untold basin]

Don't spoil us too much pls c:

I think I have an idea now

[untold basin]

come back pls xD

[untold basin]

If anyone can help with 2.a pls I'm going to ☠️

[untold basin]

Let G be a group and n >= 1 an integer.
If G has an element of order n then there exists an injective group morphism Z/nZ -> G

can someone help pls ?

[chilly ocean]

what have you tried

what are your thoughts

[untold basin]

I know a morphism which is f_k : Z -> G : x |-> x^k for k an integer

[chilly ocean]

hmmm

i'm not sure about that phrasing

[untold basin]

Wait pls I'm kinda tired

My bad

It is

e_x : Z -> G : k |-> x^k

For x an element in G

Ker(e_g) = nZ where g is the element of order n from hypothesis

So I have Z/nZ isomorphic to Im(e_x) = <g>

[chilly ocean]

you have solved the exercise

[untold basin]

????

Wait what

[chilly ocean]

think about what you just wrote

[untold basin]

But <g> isnt G

And what I tried is

[chilly ocean]

okay, but it's a subgroup of G

[untold basin]

I mapped to G but it isnt an application

Yes

[chilly ocean]

elaborate on this

application?

[untold basin]

Function that is defined everywhere

[chilly ocean]

how, precisely, did you map to G?

i want you to think about why "isomorphism Z/nZ -> <g>" gives you "injection Z/nZ -> G"

[untold basin]

Ok 1 min

[chilly ocean]

Hi, a quick question, what do they mean here by "all polynomials
having constant term 0"

[chilly ocean]

the first term is zero

the term with no x

the a_0 term

the x^0 term

that thing

Oh alright. thanks

[untold basin]

Are u mapping like this ? <g> -> G : g^k |-> g^k ? xd

[chilly ocean]

yes

[untold basin]

Lol

[chilly ocean]

this is just the inclusion mapping

[untold basin]

That os bullshit

[chilly ocean]

it's injective, so...

[untold basin]

Why do I

Not Think simply

[untold basin]

Ty

[chilly ocean]

Suppose I have an ideal I of A

and I want to calculate the quoitent of A\otimes Z[x,y] by I

is this the same as

A/I \otimes Z[x,y]

[vestal snow]

Let B be a finite A-algebra and I be an ideal of B with preimage J. Then is it true that the completion of B is finite over the completion of A?

[untold basin]

hello

1. Determine the order of G
2. Determine the center of G : Z(G)

I found that order of G is 8 with ord(x) = ord(y) = 4

So 1) is ok

But for 2) I struggle a bit because I found that
Since Z(G) is a subgroup of G then |Z(G)| divides 8 so m := |Z(G)| is in {1,2,4,8}
If m is 8 then G is abelian. Contradiction because xy =/= yx
if m is 4 then |G/Z(G)| = 2 so G/Z(G) is cyclic so G is abelian. Contradiction
But with m = 2 I don't really know what to say

[delicate orchid]

try and find a non-trivial element in the centre would be the easiest way imo

[untold basin]

I guess that's why they help me with yxy = x

I feel like I'm going to spend 2 hours to find it xd

ty

[untold basin]

this is killing me

[untold basin]

y²

[delicate orchid]

lemme check

[untold basin]

y² = 4 * Id

sooooooooooo

idk

[delicate orchid]

yeah that's -identity

so that should commute with everything

[untold basin]

nice !!!!!!!!!!!!!!!!!!!!!!!!!

[untold basin]

thanks

[inland otter]

Consider the polynomial $p=x^2+x+1$ over $F_2$ and let $F_4\simeq F_2(\alpha)$ be the field extension where $\alpha$ is a root of $p$.

I had an exercise in which I had to show that the matrix representation of the Frobenius endomorphism over $F_2(\alpha)$ (wrt the basis ${ 1, \alpha }$ was not diagonalisable.

Is there any conceptual reason for this or is it just a random fact? I tried googling it but couldn't find anything relevant.

[cloud walrus]

dadaurs

[terse crystal]

Just consider the λ matrix $\begin{pmatrix}λ+1&1\0&λ+1\end{pmatrix}$ I think

[cloud walrus]

Cogwheels of the mind

[terse crystal]

It’s equivalent to $\begin{pmatrix}1&0\0&(λ+1)^{2}\end{pmatrix}$ but any diagonalizable matrix is equivalent to one of the following three: $\begin{pmatrix}1&0\0&λ\end{pmatrix}$, $\begin{pmatrix}λ&0\0&0\end{pmatrix}$ , $\begin{pmatrix}λ+1&0\0&0\end{pmatrix}$

[cloud walrus]

Cogwheels of the mind

[delicate orchid]

actually using the lambda character and not just latexing it

[terse crystal]

The first one isn’t equivalent to any three of them

That’s because I still can’t remember how to spell them😂

[hidden haven]

Idk how conceptual you want it to be, but F_4 has 4 elements, 0,1,a,a+1. The Frobenius endomorphism fixes 0 and 1, and exchanges a and a+1. So 1 is an eigenvector, but since a and a+1 are not scalar multiples of each other (over F_2) so there is no other eigenvector. This explicit calculation also gives you the Jordan form of the matrix, as
1 1
0 1

Sort of a brute force solution from the fact that there are only 4 elements lol

[terse crystal]

I forgot there is a fourth one which is 0 matrix… doesn’t matter, can’t be equivalent to 0 too

[gritty sparrow]

I think i have a more conceptual proof (based on moldi's answer). First consider the frobenius endomorphism on $F_{p^2}$ as an $F_{p}$ vector space and call it T. Then we see that $T^2=I$ since $x^{p^2}=x$ for all x. therefore T satisfies the polynomial $t^2-1$ which is the same as $(t-1)(t+1)$. As T satisfies neither factors, that polynomial must be the minimal polynomial, and hence also the characteristic polynomial. If $char(p) \neq 2$, then the polynomial is separable, so T is diagonalisable. If char(p)=2 then we see that if it is diagonalisable it must be the identity since all its eigenvalues are 1, which is false. Hence T is not diagonalisable.

[cloud walrus]

chmonkeynumber1fan

[inland otter]

oooooh this is really nice thanks

[gritty sparrow]

Np

[sinful mirage]

Is there any inf dim semisimple lie algebra?

[gritty sparrow]

Take an infinite direct sum of a simple lie algebra

[untold basin]

Hello

I struggle a bit for 2nd one

It says "Show that Z[X] contains an infinite number of maximal ideals"

I did the first one so we know that f is a ring morphism and Ker(f) = (X,p) with p a prime number

[untold basin]

I did

So I have (X,p) maximal

[hidden haven]

This works for the 2nd part as well

[untold basin]

But idk what to do next

[hidden haven]

Ah

[untold basin]

I came back to the def of maximal

And if I suppose that there is a finite number of maximal ideals

[hidden haven]

Show that if q ≠ p is a different prime, then (X, q) ≠ (X, p)

There are infinitely many primes

[untold basin]

Why is it helpful ?

[hidden haven]

Because then for each prime number you have a maximal ideal

[hidden haven]

If there are infinitely many primes, there are infinitely many maximal ideals

Since the maximal ideal corresponding to each prime is distinct

[untold basin]

FUck my life man

bro

kill me pls

[hidden haven]

[runic hemlock]

I'm trying to show that for any $n>1$, there is a finite Galois extenstion $K/\bQ$ contained in $\bR$ so that $\text{Gal}(K/\bQ)\cong \bZ/n$.

[cloud walrus]

Porphyrion

[runic hemlock]

The previous part of the problem was to show that there is a subfield of the cyclotomic field $\bQ(\zeta_n)$ that is contained in $\bR$ and $[\bQ(\zeta_n):K]=2$.

[cloud walrus]

Porphyrion

[runic hemlock]

So maybe you're supposed to use it

[next obsidian]

I think using cyclotomic extensions and composites thereof you can get any cyclic Galois group (this lets you solve the abelian inverse Galois problem)

Maybe you can combine these to like, get Z/2nZ and then use this K to get Z/nZ?

[runic hemlock]

if you get Z/2nZ using composites of cyclotomic extenstions, just taking the intersection with R gives Z/nZ, so that works

but how do you get Z/(2nZ)?

[next obsidian]

It’s like ummm

[strong yacht]

Is this theorem true?

[next obsidian]

I just remember you can get any finite abelian groups using cyclotomic extensions and then sub fields there of

[wooden ember]

had this exact same exercise where we had to compute eigenvalues and check diagonalizability of the forbenius automorphism over F_4 and F_8. Didnt really get the point either

[next obsidian]

So maybe you need to go to sub fields in order to get Z/2nZ

[runic hemlock]

yes, just take K=Q(zeta_n) n R

[strong yacht]

Maybe I misunderstood

There is always a quadratic sub extension yes, but not always contained in R I thought

[runic hemlock]

that's an extension of Q tho

i meant that Q(zeta_n) is a quad extenstion of K

[strong yacht]

Ahhhhh ok

[inland otter]

I know you had the same exercise, we’re in the same course lol (im the dude who recommended you take it)

Yeah I discussed the exercise a bit further and the result seems to be kind of useless I guess you can generalize it to extensions F_p^2n:F_p^n and show that in char 2 the matrix will never be diagonalizable while in higher characteristics it will be.

You can also easily show that the eigenvalues will always be roots of unity, which, again, is an utterly useless fact.

I asked the TA’s, they didn’t have a clue about why this exercise is even remotely interesting, ill try and ask the prof on wednesday.

[wooden ember]

ah i see, shame there's not more to it. I didnt know you were on this discord lmao

[long obsidian]

What the heck would I see in a commutative algebra class?

[chilly ocean]

lots of commutative algebra

take a look at the contents of some commutative algebra books, e.g. atiyah-macdonald

[long obsidian]

But like what are the core concepts. Like for instance in my smooth manifolds class a big take away was the generalized Stokes' theorem. Is there a Stokes'theorem in commutative algebra

[chilly ocean]

i feel like commutative algebra is more of a "foundational" course in that, rather than building up to a few big and very important theorems, it gives you a bunch of the tools you need to go on to study other things like algebraic geometry, algebraic number theory, etc. (which may have their own "big theorems")

i guess the nullstellensatz counts

that's a massive one

(and i can't think of any other )

[mighty spade]

But isn’t that assigned as an exercise in many books?

[strong yacht]

I came back to your problem. The first part of your question does help actually, it is suggesting the chain $\mathbb{Q} \subset K \subset Q(\zeta_p)$, where $K=\mathbb{Q}(\zeta_p+ \zeta_p^{-1})$ (presumably this is the $K$ you found to answer your last question). Then $Gal(\mathbb{Q}(\zeta_p)/\mathbb{Q}) = C_{p-1}$, and $Gal(\mathbb{Q}(\zeta_p)/K) = C_2$, so $Gal(K/\mathbb{Q}) = C_{p-1} / C_2 \cong C_{(p-1)/2}$ (this is by the Galois correspondence and of course that subgroups of cyclic groups are cyclic and cyclic groups are abelian so normal). Now all that's left is to pick $p \equiv 1 \mod 2n$.

[cloud walrus]

Greenman

[chilly ocean]

it's literally one of the most important theorems in commutative algebra. it literally makes algebraic geometry work

[mighty spade]

true, i agree

[chilly ocean]

just because it's an exercise in some books doesn't detract from its sheer significance

[runic hemlock]

Wow, really cool! Thanks

[mighty spade]

I wouldn't say that it's something you build up to though. By the time you're taking a commutative algebra course, you should probably already know the basic facts about rings, fields, modules, groups, and algebraic extensions. You can introduce the nullstellensatz as an exercise within the first month, or even week, of a commutative algebra or algebraic geometry course.

i'm not disputing its relevance, I'm just saying that it is more like the basic limit theorems in analysis than something like stokes' theorem

[chilly ocean]

okay

[next obsidian]

CA big theorems:
Hauptidealsatz, Cohen structure theorem, Serre’s homological criterion for regular rings.

Also: homological conjectures

The latter are kind of grey because not all of them are theorems yet lol

[chilly ocean]

Minimal:smallest=maximal:x
What's x

"largest"

is this an algebra question or a vocabulary question?

[chilly ocean]

Algebra@chilly ocean

you need to include a LOT more context

(in all of your questions)

[chilly ocean]

hoi

[wild solar]

I saw this statement today but the instructor didn't give a proof(I don't remember if it was because it's trivial). BUt I ran into trouble when I was trying to prove it. Could anyone tell me if my idea is correct?

I was just using the definition of an element being algebraic

[next obsidian]

There’s no way to write something down of the form
a_nx^n + … + a_0 = 0 where the a_i are rational numbers unless all a_i = 0

This just shows x doesn’t satisfy any non-trivial polynomial over Q

[wild solar]

what do you mean x doesn't satisfy any polynomial over Q?

[next obsidian]

What’s your definition of an algebraic element?

Isn’t it just that there exists f(t) in Q[t] such that it is a root of f?

Satisfy here means “is a root”

[wild solar]

oh, yes

ohhhh

okay

let me think

[next obsidian]

What is written in that photo is equivalent to saying a is a root of c_0 + … + c_nt^n

But you can just directly apply that definition with a = x

[wild solar]

so it doesn't matter that here x is the quotient of two polynomials? Is this what you mean

[next obsidian]

x is just x

x lives inside of Q(x) as the formal variable x

It’s like, x/1 if you really want to write it as a quotient of something

But regardless, it’s just x, the only way you can have
a_0 + … + a_nx = 0 is if all the a_I are 0

This is just by definition of what it means to be equal inside a polynomial ring

[wild solar]

oh okay that makes sense, I just assumed that the instructor means x is any element in Q(x) and so its a quotient of two elements of Q[x], but I guess the reasoning is similar...?

[next obsidian]

Right

I mean there are elements of Q(x) which are algebraic, but those are just the constants which are already in Q

In general if you had a rational function f(x)/g(x) you’d want to like, write these out so that the bottom and top are coprime

Meaning you factor them as much as you can, then remove any duplicates that appear on top and bottom

Then plug this into a polynomial
f(t) = a_0 + … + a_nt^n

Clear the denominators by multiplying by g(x)^n

Then conclude again that all the a_i are 0

[wild solar]

Wait now I'm confused. I thought the instructor means ANY element of Q(x) is not algebraic, but are you saying what he actually means is the x/1 is not algebraic and other elements can be...?

[next obsidian]

They mean that x/1 is not algebraic

[wild solar]

ohhhhh

[next obsidian]

x is a specific element

[wild solar]

that clears things up

[next obsidian]

But you can show that the only elements in Q(x) that are algebraic are just the constants

Any other element isn’t algebraic by a similar sort of argument

Which I detailed above

[wild solar]

okay I understand now... Thank you very much! 🙂

[lapis temple]

Hey. can anyone help me with understanding tensor product of modules? I am not able to grasp the concept

[chilly ocean]

Is there a injective module equivalent of free module?

Like free modules are a special kind of projective modules

What kind of modules would correspond to injective modules, if any

[chilly ocean]

Historical reasons? Or like explaining what the definition says. Or motivation?

[lapis temple]

lets start with the historical reasons and the motivation. Then we can discuss the definition with more ease

[untold basin]

You want a full course XD

[hidden haven]

@fossil shuttle your time has come

[delicate orchid]

Just seen someone write “closed under inverses” as “retains a passage to the negative” which I thought was nifty

[fair quartz]

im trying to wrap my head around the introductory chapters of functional analysis
could someone explain this section from an example for bounded operators to me?

V is said to be a Banach Space

im also given this, but i think i understood this part

[chilly ocean]

Limited operators?

[fair quartz]

ah bounded

had to look up the proper translation 😂

[fair quartz]

my main problem rn is just deciphering what V is supposed to be
or rather what this C does and how the interval from 0 to 1 makes any sense in complex space

[hidden haven]

It is the vector space of all continuous functions from [0,1] to the complex numbers

That is standard notation

[fair quartz]

ohhh okay

that makes a lot more sense

[hidden haven]

It becomes a normed linear space with the definition of norm that you posted

[fair quartz]

i have never seen this notation tbh

[hidden haven]

Btw functional analysis should go in #advanced-analysis

[fair quartz]

oh my bad!

thanks 😄

[south patrol]

how can you discuss isomorphisms (a type of homomorphism) without using the concept of homomorphisms?

[chilly ocean]

Given the commutative ring i.e. the cyclic group of order 2, $\mathbb{F}2$ would the polynomial ring, $\mathbb{F}{2}[1]$ only contain the single element, 1, or would it contain every sum of 1, i.e. ${\sum_{i=1}^{n} 1 : n \in \bN - {0}}$

[hidden haven]

Some books do that

[chilly ocean]

bro

[hidden haven]

\bF is not a thing lol

[chilly ocean]

noticed

[cloud walrus]

simp

[chilly ocean]

there we are

or for the last bit

[hidden haven]

The latter, but that is just F_2 itself

[chilly ocean]

just the natural numbers

wait wouldn't it just infact be not - 0

since the coefficients are 0 and 1, the first polynomial would be 0 right

how old is this book

"one-to-one" is standing for injectivity and surjectivity here

[chilly ocean]

It's not "one-to-one" but "one-to-one correspondence", it's a very subtle difference

https://en.wikipedia.org/wiki/Bijection

[chilly ocean]

fine

[dull root]

Do I have the correct understanding of Galois extensions:

Let L be a Gal ext of K with galois group G. Then for any polynomial f in K[x]:

1. It is true that for any g \in G, g maps a root of f to a root of f.

2. if f was irreducible, then for any two roots of f, say a, b, there is a group element that maps a to b. Equivalently, G acts transtiively on the roots of f.

Is it true that G acting transitively on the roots of f iff f is irreducible over K?

That is, a way to see a polynomial f is reducible is to find two roots a,b of f such that no group elements maps a to b

[rigid cave]

I think that the implication "G acts transitively -> f irreducible" is false, take for example x^2 in Q. But I'm not sure about the other direction

[dull root]

Yea that was a a simple example. The other direction is def true tho

[rigid cave]

oh okay I see. How do you prove it?

oh I found a MSE post

[strong yacht]

As stated, the claim is false. Q(i) is a Galois extension with complex conjugation as the non-trivial element of its Galois group, but does this act transitively on the irreducible X^2-2? No

What you want to say is the Galois group OF f acts transitively on the roots of f iff f irreducible

[dull root]

I see, so we can't have any galois ext. We want an the splitting field of an irreducible polynomial f for the statement ot be true

[rigid cave]

doesn't my example show that this isn't true?

[strong yacht]

It's not a separable polynomial, so I don't think we want to talk about its Galois group

[wooden ember]

Not separable ?

[strong yacht]

My brain might be lagging lol

[wooden ember]

it splits over Q(sqrt(2))

and Gal(Q(sqrt(2))/Q) acts transitively on the roots

[strong yacht]

We're talking about X^2, Tokidoki's example

[wooden ember]

ah okay my bad

i was a bit confused lol

[strong yacht]

No problem, I was doubtful of my sanity for a second there

[wooden ember]

lmao relatable

[rigid cave]

oh right lol