groups-rings-fields

[junior harbor]

Because if a + Z = Z + a for the fact integers commute, then a + S = S + a

[chilly ocean]

Okay wait here's how you make cosets

a + H or left coset of H, where H≤Z, where say, H is a group of even Integers, then the left cosets are:
a = 0:
0+H = {...-2,0,2,...}
a = 1:
1+H = {...-3,-1,1,3,...}

Do that for right cosets too

I gtg for dinner

[junior harbor]

Ight, thanks for the help. Later can ya help me with quotient groups?

[chilly ocean]

Sure

[pastel cliff]

when does xy even have finite order

[delicate orchid]

G is finite

[pastel cliff]

ok yes

but do you just have to define the order xy yourself

and maybe some induction shebang

[delicate orchid]

it's defined to be n

right there :iamlooking:

[pastel cliff]

im gonna try induction

definitely not the move

[delicate orchid]

consider constructing a homomorphism from G to D_2n

[pastel cliff]

im confused bc the generators of D_2n dont have order 2

one does

oh wait

xy would map to the second generator in D_2n

[prisma shuttle]

yes thats correct

[untold basin]

I passed my algebra exam

Thanks to all of you

[delicate orchid]

I will be taking full credit

[prisma shuttle]

guys does anyone understand wut this mean

if the top part is tldr, R is a subring of the ring S, and \alpha is an element of S

i don't get the red part

wut does it mean to view M as an R[\alpha] module

isn't M and a R module

and also wut does R[\alpha] mean

[next obsidian]

Faithful means that the annihilator of M is 0

R[alpha] means take the map R[x] -> S which sends f(x) to f(alpha), and take the image

It’s all polynomial in alpha with coefficients in R

You know how to interpret this as an element of S because S is an R-algebra

[tribal moss]

But according to the text, M was assumed to be an R-module rather than an S-module, so it's not a priori clear how multiplication with alpha should work.

Perhaps it's a typo and M should have been an S-module from the beginning.

[chilly ocean]

"carrier"?

this isn't a standard term, you'll have to elaborate

the addition is the one you expect, and the multiplication is (n + r)(m + s) = nm + ns + mr + rs. the multiplicative identity is 1 + 0.

in tuple notation (less suggestive), (n, r)(m, s) = (nm, ns + mr + rs) and the multiplicative identity is (1, 0)

[next obsidian]

I think they mean the underlying set

[chilly ocean]

cartesian product, but that's for them to tell us

[simple mulch]

what is a symmetric generating set for a group/subgroup?

btw. If I have G a group, N a normal subgroup of G and H a subgroup of G. Then the direct product G = NH implies that every g in G can be written uniquely as g = nh for n in N and h in H. Does it also need to satisfy g = hn?

[chilly ocean]

Direct products and inner semi products are different.
What you're asking about is inner semi product

[simple mulch]

How so?

[chilly ocean]

Inner semi product is defined as G = NH = HN, and we can see NH or HN is a subgroup too.
This type of product is called frobenius product, and this is a subgroup generated by both N and H.

Secondly

[simple mulch]

The normal subgroup N = {(0,y,z)} and the subgroup H = {(x,0,0)} form a semi direct product?

[chilly ocean]

Direct products can be thought of as "Cartesian Products of groups"

[simple mulch]

Because NH would be different from HN

since NH implies (x,y,z) and HN results in (x,y,z+xy)

[chilly ocean]

Actually this one is a direct product

In a direct product, say you have two groups G and H

With same or different binary operations

[simple mulch]

(and indeed N cap H = (0,0,0)

[chilly ocean]

Then G×H is called direct product where we do the Cartesian Product G×H, st (g,h) belong to G×H, and the binary operations (say G has ° and H has •) are defined componentwise
(g,h).(i,j) = (g°i,h•j)

Here (i,j) also belongs to G×H

And G×H must satisfy axioms for being a group

[simple mulch]

we are not doing that

?

[chilly ocean]

Yeah we're not doing that here, I was just saying direct and semi direct products are different.

[simple mulch]

Oh yeah thanks for the reminder!

The questions are still up

[cloud walrus]

leaf

[wicked zephyr]

Why people care about characterizing projective, injective and flat $R$-modules¡'

?

[cloud walrus]

Potitov06

[prisma shuttle]

ex: projective modules allow you to generalize the characteristics of free modules

so essentially it allows u to use properties of free modules to more general modules

if thats wut u meant

oh also

in this proof here

of the lemma

does anyone know what the $\delta_{ij}$ stands for

[cloud walrus]

JustKeepRunning

[prisma shuttle]

because it was not defined previously in teh book

so ima assuming that means its like a common symbol that stands for smth

?

[sand cradle]

Kronecker delta

1 if i=j

0 if i is not equal to j

[sand cradle]

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

[junior harbor]

What is a quotient group? The book I'm reading isn't quite getting it through

[gusty cave]

It's like a group consisting of cosets
Ie if you have G/N it's just all the closets with the operation being aN*bN=abN

Like for example the quotient group Z/4Z has the elements 4Z, 1+4Z, 2+4Z, 3+4Z

[chilly ocean]

It's a group you obtain after some identifications have been made

Formally, you let G/N to be the set of equivalence classes of the equivalence relation x ~ y iff xy^-1 is in N, N being a normal subgroup of G.
Then on G/N we can introduce a group structure with product defined as [x]*[y] = [xy], where [z] denotes the equivalence class of the element z in G.

You can prove that this definition doesn't depend on the choice of x and y, so the operation on G/N is well-defined

Why we care about such equivalence relation, is because those kinds of relations are the only ones "preserving" the group structure

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

and by that I mean congruences

one can prove that for groups, those are precisely equivalence relations of the form x ~ y iff xy^-1 is in N for some fixed normal subgroup N.

Congruences are kind of equivalence relations, that defining operations on equivalence classes in the same way, it can turn out to be the object of the same type if and only if the equivalence class is a congruence

So if we want to obtain a group by quotienting, this can only be done using a normal subgroup

that's why they are important

This is a quite remarkable property of groups, and similar thing happens for rings. But for general kinds of structures, it's not enough to consider some special subset

You need to work with congruences

That's the case for example, for monoids and semigroups. If you want to consider quotients of those, you need to consider congruences instead.

[chilly ocean]

Also why they usually don't cover what congruences are in the theories of groups and rings, it's simply not needed. It does give you a better picture though, imo

[junior harbor]

Hm, and where did the G go in this?

[tribal moss]

That's where a and b come from.

[olive edge]

I'm in $\mathbb{Q}(\sqrt{2})$ and I want to know if $x^2 - 3$ has a solution, I would do:

$$(a + \sqrt{2}b)^2 - 3 = a^2 + 2\sqrt{2}ab + 2b^2 - 3 = 0$$

Apparently I can kill the $2ab$, but I'm not sure why?

https://media.discordapp.net/attachments/496784958430380033/986627160087347240/943816131746082836.png?width=1932&height=494

[delicate bloom]

if it weren't, it would imply sqrt(2) is rational

cause you could then just solve for it in terms of all the integers in the equation

[olive edge]

Why would it imply sqrt(2) is rational?

[delicate bloom]

solve for sqrt(2) in that equation

a,b are rational numbers

[olive edge]

Ah right

Ok that's very clear

[delicate bloom]

cool

[olive edge]

Awesome, thanks mero

[delicate bloom]

you're welcome 👍

[chilly ocean]

Yesterday we saw what were cosets. I hope you now know what are left and right cosets from our previous example.

Now we use these invariant subgroups to divide the whole group into different cosets.
Only the normal Subgroups of a group have the property to do this. Since for a normal/invariant subgroup N, its left and right cosets are equal, we define G/N as a set of all N + a, a is in G.
(N+a = a+N, so it doesn't really matter.)
So basically a Quotient group is a set of all cosets of a normal subgroup, such that the normal Subgroup divides the underlying set of a group into equal disjoint cosets.

In terms of equivalence classes as Blitz said, consider a relation R. A relation is always a subset of the Cartesian Product of two sets say A and B. A relation is then defined as a set of ordered-pairs (a,b) where they can be defined as per your wish.

So if xy^-1 is in N, which is a Normal Subgroup of G, and x and y^-1 (y inverse) are in G, you can define the Equivalence (Congruence) Relation (Blitz used ~ to denote their equivalence Relation).
R = {(x,y) | xy^-1 belongs to N}
And R is a subset of the Cartesian Product of the underlying set of G and G.
So if (x,y) is in R, we say "R(x,y)" or "x R y" or "x is Related to y".

This is how you formally define a binary relation.

An Equivalence class [x] is basically a set of all y values for a specific chosen x that are related to x in relation R.

(Say you have {(0,1),(0,2)}, then [0] = {1,2})

So G/N in other words is also a Group of all such equivalence classes.
And as Blitz correctly said, we can again induce a group structure on G/N, with a product operation specifically defined as [x]*[y] = [xy]. [xy] will be in G (follows closure property of being a Group). This is one of the notable properties of a Quotient Group.

Example of Quotient group is the famous Z/nZ or "Integers mod n"

So $G/N = {N+a| a \in G, N \trianglelefteq G} = {[x] : | :[x] = {y | (x,y) \in R, xy^{-1} \in N}}$

[cloud walrus]

Pencil/Idris

[chilly ocean]

$N \trianglelefteq G$ means N is a Normal Subgroup of G, where N may be equal to G

[cloud walrus]

Pencil/Idris

[chilly ocean]

So you can say [x] = N+a here.

[junior harbor]

Ok, that does clear it up. Thank you pencil

[chilly ocean]

Welcome

[dusty sapphire]

suppose F[x] is polynomial ring, where F is finite field, then number of ideals of ring is finite or not? How to check that?

[proud bear]

i don't think so. (x), (x^2), (x^3),... should all be different ideals. all the nonzero elements in (x^n) should all have degree at least n because F is a field

[dusty sapphire]

okay thanks, what about R=Z[X]/(p(x)) where p(x) polynomial over Z[X] of degree 2022

[proud bear]

ideals of R are in bijection with ideals of Z[X] containing (p(X)). so what are the ideals containing (p(X))

[dusty sapphire]

(f(x). p(x)) for any f(x) in Z[X] right

[proud bear]

almost

[dusty sapphire]

so in this case this is also infinite

[proud bear]

(f(x).p(x)) is contained in (p(x))

[dusty sapphire]

ahh yes

is there any polynomial over Z[x] of degree n whose number of zeros are more n?

[proud bear]

no

[dusty sapphire]

so then p(x) have finite roots and some factors are there, so then ideal generated by factors of p(x) contain (p(x)) then number of ideals of R will be finite isn't?

[proud bear]

ok that's what i initially thought too

but there are also ideals of the form (f(x),n) where f(x) divides p(x) and n is some integer

so i think it's infinite as well

[dusty sapphire]

oh yes Z[X] is not PID so

[proud bear]

ye

[dusty sapphire]

is there exist irreducible polynomial over Q for any positive degree?

[north sand]

x^n-2

[delicate bloom]

proof: ||eisenstein||

[chilly ocean]

What does the notation $(\mathbb{Z} / n\mathbb{Z})^×$ signify?

[cloud walrus]

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

[agile burrow]

multiplicative group of integers mod n?

its elements are cosets of integers mod n which are coprime to n

[chilly ocean]

Ah makes sense, thank you

[chilly radish]

I just noticed an inconsistency in my notes and I wanna ask

by convention, does an order in a number field have to contain 1?

or do we also consider ideals of the ring of integers orders

for example

[tribal moss]

What does an "order" mean in this context?

[grand sigil]

the one from wikipedia says an order is a subring

so yeah probably?

unless there is a different definition

[olive edge]

I'm a bit confused about splitting fields & how to construct them.

I'm trying to construct one for:

$$x^6 - 8 \text{ in } \mathbb{Q}$$

And I've started by finding one root $\sqrt[6]{8} = \sqrt{2}$, then use roots of unity and get all my roots $\beta = e^{\frac{2i\pi}{6}}$, so $\beta^k \cdot \sqrt{2}$ for $k$ = 0, 1, 2, 3, 4, 5 are my roots.

Then chucked them all into the field:
$$\mathbb{Q}(\sqrt{2})(\beta \cdot \sqrt{2})\dots(\beta^5 \cdot \sqrt{2})$$

However I seem to be missing the last step to shrink this down to: $\mathbb{Q}(\sqrt{2}, \beta)$.

https://media.discordapp.net/attachments/496784958430380033/986857532423303208/943816131746082836.png?width=1806&height=1034

I think I'm generally confused about the process of even constructing them.

[grand sigil]

thats all

im pretty sure you did it

[olive edge]

Yeah but how do I shrink it?

I mean i guess it's just noticing the produce of some of them become each other.

But idk, not sure if I'd notice it.

[grand sigil]

i think by adjoining both root2 and beta that automatically includes the 6 elements adjoined individually

i dont think you need to adjoin the six roots individually

its more that you get a splitting field from those 6 roots

[olive edge]

Ah right, but those 6 roots are simply products of beta and root2

[grand sigil]

yeah

[olive edge]

Hmm I guess that works

[grand sigil]

how do you know it reduces to that?

is that the question?

i think you just need to check individually

[olive edge]

Nah, I think I understand it

Oh wait maybe not

I just thought a field which has beta and root2, will have to have all powers of beta and root2

And so no matter what, the field that contains all the roots is the same as the field that just contains beta and root2

[grand sigil]

i thought this was true also

[grand sigil]

yeah

thats how it works out

also ur latex hasnt loaded for me lol

[olive edge]

oh rip

anyway i think that makes sense, thanks

[chilly ocean]

Same here

[grand sigil]

@olive edge

I guess the last step is just recognizing that you can use beta and root 2 as generating elements for the splitting field

[olive edge]

yea

[strong hill]

Does anyone have a good place where I could find resources and cheat sheets relating to group theory?

[lethal dune]

Konrad notes are pretty helpful

[strong hill]

And where would I find those?

[lethal dune]

just google

[chilly radish]

A subring must either have or not have 1 depending on your definitions

[chilly radish]

It's a finitely generated Z-submodule in a number field containing a Q-basis of the field (so a lattice) which is also a subring (I think it doesn't have to have 1 at this point)

[chilly ocean]

https://mathworld.wolfram.com/NumberFieldOrder.html

I think "including 1" means it does have to have 1

or I guess they mean they include 1 as a generator

okay, it's still confusing

[sharp sonnet]

to me its a generalisation of a ring of integers, only dropping the requirement to be integrally closed, so i still want O \cap Z = Z, in particular i want to be able to write O = Z[a_1, ..., a_n]

i also probably want a distinction between orders and ideals in orders etc

would definitions like conductor and the ideal class group even work without having 1 in your order

who allows subrings without unit, absolutely deranged

[slate mortar]

Are quotient of euclidean rings always euclidean?

[sharp sonnet]

Z is euclidean domain, but Z/4Z isnt even integral domain

funnily enough its true if you quotient by prime ideals

since those are maximal and fields are euclidean domains

[delicate orchid]

it’s also true if you quotient by maximal ideals

[sharp sonnet]

thanks wew

[slate mortar]

thank you @sharp sonnet

weird username btw

[sharp sonnet]

thank you

[thorny knoll]

Wait, maximal ideals are prime, not the other way around in general.

[sharp sonnet]

we are in an euclidean domain

[thorny knoll]

Oops, gotcha, then those are equivalent, nevermind then

[chilly radish]

@sharp sonnet yea that makes sense, just in one point in my notes we referred to ideals of the ring of integers as orders. That was probably just a mistake on the professor's part and really it's just that they are lattices that are also closed under multiplication

[civic tapir]

In the following definition of the rank of an abelian group, is torsion of G/F included so as to ensure we have the largest possible free abelian subgroup?

[cedar path]

I assume the concept of torsion is as given in the decomposation of Abelian groups: every Abelian group can be uniquely (up to isomorphism) decomposed the direct sum of two parts, where the first part is the 'free part' and the second one is the 'torsion' part which is, by defination, not free.

so, a Corollary: the torsion of G/F is G/F.

as you can again use the decomposation to G/F, since any subgroup of an abelian group is again abelian.

since the torsion of G/F is itself, we are done, as we've got the largetst possible free abelian subgroup

@civic tapirhope this helps, and hope I'm correct on this...

[cedar path]

and now we can say that the rank r is well defined, since the uniqueness of the decomposation.

[civic tapir]

@cedar path Ah you're right, that helps a lot, thanks!

[cedar path]

remark: the free part is by defination, a group which is free generated. But the 'torsion' part is obtained by no other defination, but only because: they're the complement of the 'free part'.

[cedar path]

you're welcome

[cedar path]

or we can say, an element of G is a torsion element if it has finite rank.

Then we define the torsion subgroup of G to be the set of all torsion elements.

One can verify: the torsion elements form a group.

In this way of speaking, if G/F is torsion, then every elements of it has finite order. So the torsion of G/F is itself.

[cedar path]

so if rank(G) = 0, then G is finite.

[cedar path]

oh sorry

I made a mistake on this. G is not necessarily finite

https://math.stackexchange.com/questions/1221327/example-of-infinite-torsion-abelian-group

[worthy haven]

I'm reading a paper and I've run into this definition, what is the definition of the curly C on the right? like just C(I) with no sigma

[chilly ocean]

$\mathscr{C}$?

[cloud walrus]

TTerra

[chilly ocean]

fuck you texit

[delicate orchid]

$\mathscr{L}$

[cloud walrus]

Wew "Boötes Void" Tbh ⊗

[delicate orchid]

hmmm

[worthy haven]

sorrynot wondering about the latex, wondering what it means, but yeah it's a mathscr

[chilly ocean]

what it means is the definition you see :^)

[worthy haven]

the definition defines C^(sigma)(I)

the definition of C^(sigma)(I) references C(I)

I do not know what C(I) is

[chilly ocean]

what is the paper?

[worthy haven]

https://www.sciencedirect.com/science/article/pii/S0021869305800365

[celest cairn]

Have I proven this correctly?

[rustic crown]

the squaring of (sqrt(8) + sqrt(11)) doesn't look right

maybe you meant 22?

[chilly ocean]

i tried looking at some of the references given around where they used that notation and couldn't find it at a quick glance :(

[worthy haven]

Ty for trying! I did the same and also didn’t see anything so I assumed it was standard

[chilly ocean]

mathematicians make their content accessible challenge literally impossible

[celest cairn]

Whoops… yea.

[leaden scaffold]

What is the following type of group $^2B_2(8)$

[cloud walrus]

matthew

[charred flume]

$$
\text { Describe the group of the polynomial } x^{4}-5 x^{2}+6 \in \mathbb{Q}[x] \text { over } \mathbb{Q} \text {. }
$$

[cloud walrus]

Ji

[charred flume]

What does this question mean?

Does it mean the group of the automorphisms?

[latent vector]

i think it is asking for the galois group

[vast quiver]

the Galois group of a polynomial $f\in F[x]$ is the Galois group of a splitting field $K$ over $F$, which is the group of $F$-automorphisms of $K$.

[cloud walrus]

Joseph

[celest cairn]

@rustic crown Apologies for the bad lighting, but I belive I fixed it. I also used slightly clearer notation.

[rustic crown]

okie, i'll take a quick look

[celest cairn]

Ok thanks 🙂

It’s my first proof and I’m only in 11th grade. Wanted to learn abstract algebra for fun.

[rustic crown]

that should be -38 * alpha^2

[celest cairn]

Shoot…

[charred flume]

tnx

[celest cairn]

Ok well the whole proof is wrong now since -38 isn’t prime. @rustic crown

[rustic crown]

yea, but it doesn't work with 19 either

you want to find a prime which divides 9, but whose square doesn't divide 9

and sadly there is no such prime, only prime divisor of 9 is 3 and square of 3 does divide 9

there's a nice trick which you can use to change the polynomial slightly, such that you could apply Eisenstein's criterion to it

tell me if you would like a hint @celest cairn

[celest cairn]

Ok, and sure!

[rustic crown]

so our polynomial is K(x) = x^4 - 38x^2 + 9, right? the thing to notice is that K(x) is irreducible if and only if K(x+1) is. Can you show K(x+1) is irreducible?

[celest cairn]

I’ll give it a try, might take me a while though since I just got into this subject.

[rustic crown]

okie, good luck uwu

[celest cairn]

Thanks for the help 🙂

[rustic crown]

np

[charred flume]

is the flow of my proof correct?

i'll prove the last line later

but am i on the right track?

[thorny knoll]

you forgot to mention the negative roots respectively, but apart from that it's all correct so far :)

[charred flume]

ohh rightt thanks

[red imp]

I can't possible percieve how this could be the case

like surely N_k is always isomorphic to F_2 given that it's just a group with two generators

[delicate orchid]

there are free subgroups of F_2 that are isomorphic to every other finitely generated free group (including the free group on countably many generators)

[hidden haven]

The quantifiers are so wrong on that

[delicate orchid]

yeah F_2/N_k should be ZxZ/k by that presentation for a start

nevermind, misread it

the fact that it's saying prove rather than construct makes me think of funny universal property memes

could be a start?

moldi don't stare bleak me

[hidden haven]

I imagine you would just find k+1 generators

[delicate orchid]

of N_k?

[hidden haven]

probably

[delicate orchid]

they'd have to be free

[hidden haven]

Ye

[delicate orchid]

bobbicals do you know about Schreier transversals/generators

[red imp]

nope

[delicate orchid]

_ _

[red imp]

the fact that F_2 contains all the other free groups broke me at first but I think I see it now

cos you can take longer and longer words as generators

[delicate orchid]

yurr

the exact generators are of the form b^nab^{-n} iirc

[hidden haven]

This should be easy using covering theory lol

[red imp]

yes covering theory sounds appealing

[hidden haven]

Take cover of figure eight corresponding to this subgroup

And should that it has F_k+1 as its fundamental group

I am just not sure how to visualize normal closure with covering spaces

You have the cover corresponding to the subgroup H and you want to find the cover for the normal closure of H. It will be the smallest Galois covering that also covers the cover of H

[red imp]

is this cover just the cover that goes around the first ring once and the second ring k times

[hidden haven]

Ye

Figure 8 covering itself

[red imp]

my covering space skillz are very bad, how do you relate the fundamental group of the cover S^1 to the fundamental group of the figure 8

[hidden haven]

My guess is that the smallest Galois covering over this would be a circle with k circles attached to it at its boundary

[hidden haven]

S^1 doesn't cover the figure 8

[red imp]

oh yeah it'll be a bouquet of circles, not S^1

ahhhhh yes a bouquet of k+1 circles

[hidden haven]

This is figure 8 covering itself, with one circle covering itself via identity and the other one by a k fold map

[hidden haven]

Yes, but bouqueted at different points

In the sense that there can't be a point at which more than 2 lines intersect

Since covering maps are local homeomorphisms and there are no such points on the figure 8

[red imp]

oh yeah

so k+1 circles sitting in a line and kissing

okay I will see where this leads me, thanks

[hidden haven]

I think it will be one central circle and k circles kissing it's boundary

I mean that's a Galois cover of the figure 8

That also covers the cover corresponding to <a, b^k>

Just need to check that it's the smallest such cover

So that it is the cover corresponding to the normal closure

[lament dawn]

Does anyone have a large list of outer automorphism groups

wiki only has like 13

[charred flume]

yooo how to prove X^2 + X -1 is irreducible in Z_3[X]

[hidden haven]

Check if it has roots

This works for any quadratic or cubic polynomial because of the factor theorem

[charred flume]

uhmm yeah my problem is how to factor this polynomial

ahh i get it now, just found a way

[chilly ocean]

Why the linear group SL(n,F), a group of n by n matrices with determinant 1 under the matrix operation over a field F a "special" linear group?
I mean, it is a subgroup of the general linear group, but what is so "special" about this Subgroup

they got determinant 1, seems pretty special to me

[cedar path]

Special is by definition what opposed to general

[chilly ocean]

wikipedia says "These elements are "special" in that they form an algebraic subvariety of the general linear group – they satisfy a polynomial equation (since the determinant is polynomial in the entries)."

though you could just as well call any of the other matrix groups special

{I} is pretty special.

i wonder where the term "special" for this group first showed up

no reference is given for that wikipedia excerpt, by the way, so i'm assuming it's just another "these things satisfy this nice property, so let's use this nondescriptive english word to refer to them"

[cedar path]

and the proof of this fact is:

1. consider the function det over the manifold GL(n). It's clearly a smooth funcion.
2. compute the differential of det. It's called Jaboci formula: d(det) is non-zero at the non-vanishing points.
3. Using the implicit funciton theorem, conclude that the level set at a for a non-zero is a submanifold of codimension 1. In particular, when a=1, the level set is the special linear group.

[chilly ocean]

it's just the zero set of a polynomial lol

if you're working over R or C or something, and if you want to prove things about manifolds, then yeah

[cedar path]

zero set of a polynomial of high degree and multiple variables can be very complicated. You know this is the topic of Algebraic geometry. For example there can be singularities involved. But what we want to show here, is the zero set of det polynomial-1 has no singularities. Otherwise, it's not a manifold.

[chilly ocean]

"algebraic subvariety" just means zero set of polynomial

which is what wikipedia wrote

[cedar path]

for example, this is not allowed

[chilly ocean]

i know

[cedar path]

so that's why one has to do the step 2

https://en.wikipedia.org/wiki/Jacobi's_formula

[chilly ocean]

let me rephrase

[cedar path]

yeah, I mean as smooth submanifolds.

[chilly ocean]

and i am talking about algebraic varieties

as was the article i mentioned

[cedar path]

if one only consider algebraic things it's done.

[chilly ocean]

that is what is being done

we're doing algebra, not differential geometry

though you could still talk about smooth algebraic varieties i guess

[cedar path]

okay, so let my argument be the complement of yours but not related

[chilly ocean]

(via jacobi criterion)

[chilly ocean]

How does it do that though

what do you mean?

Like how do they satisfy a polynomial equation

Ah wait

(since the determinant is polynomial in the entries)

So the determinant forms a polynomial = 1

Makes sense

strictly speaking we're looking at the polynomial det - 1 and its zero set, but yeah

if you're not sure of it being a polynomial, look at that long ass formula for det involving permutations

👍

[cedar path]

and to check it's a submanifold we need something more....

[chilly ocean]

I'm not into algebraic geometry yet

Surprisingly I love how this topic connects linear algebra, abstract algebra and algebraic geometry :o

"i love how this topic connects..." is something you might find yourself saying a lot from now on

[untold sapphire]

IIRC if a curve has singularities then we can just embed it in a higher dimensional space and still view it as an algebraic variety

[cedar path]

you mean blowup?

[untold sapphire]

I don't think it's using anything that complicated, but it might be a blow up - Miranda covers in his early chapters

On my phone but found the part

This probably is a blow up tbh

My geometry is a bit weak

But Miranda is a really accessible point into AG

[fickle brook]

ok so this has been SEVERELY bugging me because i just cannot understand the difference between an affine algebra and a group algebra

going to translate this shortly

this is what the screenshot says:

We want to show that for diagonalizable groups the affine algebra K[G] of the group G is isomorphic to the group algebra of its character group X*(G):

K[G] = K[X*(G)].

Remark. Of course, the notation K[G] in the left and right hand sides of this formula have completely different meanings:

• on the left is the affine algebra - i.e. a subalgebra of the algebra of functions,
• on the right is the group algebra.
  The affine algebra depends contravariantly on G. The group algebra depends covariantly on X*(G) - but, of course, the character group X*(G) itself depends on G contravariantly.

---

and its come to my attention that

put simply i have no fucking idea what "affine algebra" or "group algebra" even mean in this context

or what the difference between them is. i'd like to know.

like ok the affine algebra is a subalgebra in the algebra of functions. but WHAT subalgebra is it?

each of these seems to have a basis indexed by elements of the group but i plain and simple dont understand the difference

and google is not helping either

and im growing increasingly desperate and increasingly distressed about this

<@&286206848099549185>

[chilly radish]

Are you sure it's not just a diff name for it

Affine algebra usually means finitely generated as an algebra

[fickle brook]

literally says "affine algebra" and "group algebra" have COMPLETELY DIFFERENT MEANINGS

[hidden haven]

https://math.stackexchange.com/questions/317228/affine-algebra-of-an-algebraic-group
From this it seems that the affine algebra is the coordinate ring of a group scheme? I am not completely sure

It is the representing object of the scheme as a functor apparently

This might be the functor of points thing

Viewing affine schemes as representables out of k-Alg

[fickle brook]

is that the only definition

i wish there was something non functorial

cause this one just

MAKES NO SENSE TO ME

i just CANNOT grasp it

[hidden haven]

Idk lol you'll need to find someone who can translate between functor of points and sheaves

I think Arianna was the who told me about functor of points so you might have some luck asking her

[fickle brook]

whomst?

[hidden haven]

@golden pasture

[fickle brook]

well i guess now that you've pinged her i don't need to do it a second time

[golden pasture]

wat

[golden pasture]

huh

[fickle brook]

is that a "huh this makes no sense to me" or a "huh idk what youre asking"

[golden pasture]

ngl im confused too

[hidden haven]

Does this definition make sense in context? Coordinate ring of the group as an affine scheme?

[fickle brook]

not really

i mean like

ok

coordinate ring of the group

is that the same as the ring of functions on the group

[hidden haven]

Oh that's actually also what the functor of points thing should translate to lol

[fickle brook]

if not then i'll flip

[hidden haven]

Ye ring of polynomial functions

[fickle brook]

ring of polynomial functions right

so that's what

[golden pasture]

i dont see any functor of points here

[fickle brook]

that's the affine algebra?

[golden pasture]

i dont even see anything close to a scheme

[fickle brook]

ring of poly funcs on group as affine variety = affine algebra

yes?

[hidden haven]

The SE I linked treats affine schemes as representable functors

[hidden haven]

Ye that's what I'm guessing

[golden pasture]

i mean sure but

that feels super overkill for smt that looks like

just some group algebra

[hidden haven]

Lol idk what exactly functor of points is then, because I assumed that whole approach of spaces = functors was called that

[golden pasture]

it prob is trying to say smtsmt

[chilly ocean]

K[G] is really a confusing notation and anyone would think it's a group ring of G over K

[golden pasture]

G is finite here right

[golden pasture]

oh

monkas i thought it was saying that

[golden pasture]

er

sure but

i thought K[G] was referring to the group algebra

[hidden haven]

Lmao

[fickle brook]

nobody said that

okay but what is the group algebra then

[hidden haven]

You take character group and then the usual group algebra

Do you know the usual group algebra?

[fickle brook]

no

i don't think so

unless it's the one where you consider functions of finite support instead of poly funcs

but at this point i'm seriously doubting all of my knowledge here

and i must assume i know ABSOLUTELY NOTHING AT ALL and EVERYTHING I'VE KNOWN SO FAR IS A LIE

[hidden haven]

This should be easy to find online, you define the algebra to be the vector space with basis the group, and define an obvious multiplication

Obvious in the sense that it just bilinearly extends the group multiplication

Bilinearly/distributively

[golden pasture]

ngl idk whats happening there is K[G] some object representing some functor

[fickle brook]

[golden pasture]

notation wtf moments

[fickle brook]

is it possible to not involve functors in this

literally the word functor is going to make me fucking cry

im too pea-brained for such abstraction

[hidden haven]

Coordinate ring of G, which is the same as the algebra representing G as a functor of points scheme

According to the SE post

[golden pasture]

oh right

that thing

which is effectively the group algebra

[fickle brook]

affine algebra = poly funcs
group algebra = finite-support funcs

yes??

[golden pasture]

um

[chilly ocean]

You can see this example of a group algebra.
Afaik, and according to Wikipedia, a group algebra is a commutative group ring, or a group over a commutative unital ring

[golden pasture]

it feels so wrong to say finitely supported functions

[chilly ocean]

In other words, a group Algebra is a set of all linear combinations of k in field K and s in group S as:
k1s1+k2s2+...+knsn if S is finite

[golden pasture]

ok first clarification first

what exactly is G here

it is a group scheme right

ah wait

my brain just unshutdown itself

G is a affine group scheme

so K[G] here basically means like
"spec K[G] = G"

you could think of it that way

meanwhile X*(G) is some group so K[X*(G)] is the group algebra - i.e. the vector space over K with basis X*(G) and multiplication defined in the obvious way

my brain is too fried from doing pde in chinese

rip

[chilly ocean]

Ah wait yes

If the ring in group ring R[G] is replaced with a field K, forming K[G], this is literally just a vector space over K

[hidden haven]

Yes

[chilly ocean]

It's a K-algebra

I don't like the world literally, because it's not exactly one and the same thing

[spice whale]

how is this true?
if we have a glide reflection $m = t_v \rho_\theta r$ then doesn't $m^2 = t_v \rho_\theta r t_v \rho_\theta r = t_v \rho_\theta t_{rv} \rho_{-\theta} r r = t_v t_{\rho_\theta r v} \rho_\theta \rho_{-\theta} = t_{v + \rho_\theta r v}$

[cloud walrus]

ally 🌈

[spice whale]

not $t_{2v}$

[cloud walrus]

ally 🌈

[spice whale]

where have i gone wrong

[spice whale]

<@&286206848099549185>

[chilly radish]

Given an infinite separable extension, it's obvious that there are infinitely many embeddings of it into its algebraic closure right? Like otherwise you could take finite subextensions of arbitrary degree, find however many embeddings you want and extend those to distinct embeddings of the infinite extension (Say by Zorn's Lemma), so you can find arbitrarily many embeddings i.e. there are infinitely many. Is my reasoning correct?

[agile burrow]

this seems correct to me

[spice whale]

i figured it out
v is on the glide line so rho_theta r v = v

[chilly ocean]

Is lie bracket [-,-] defined as [-,-]: g×g -> g, where [X,Y] = XY-YX, or could it represent any alternating bilinear map that satisfies the Jacobi Identity?

not every lie algebra has as its lie bracket a commutator

I see

[hidden haven]

That's how you would define the Lie bracket on an algebra, but not every Lie algebra needs to come from an algebra

[chilly ocean]

XY and YX might not even make sense in gneeral

it makes sense if you're working in an algebra, as moldilocks says

Well here g is a vector space

no multiplication of its elements is defined in general, so no writing XY and YX for you without more information

Alright, thank you both 👍

Ah it depends upon context, for vector spaces it is defined as the cross product X×Y

in R^3, sure

but "cross product" is not defined in general

I see

R^3 with the cross product is indeed a lie algebra

Thanks again

isomorphic to the lie algebra of the lie group SO(3), which should say something geometrically

something something rotations of R^3

3D rotations in R^3 right

around the origin yeah

👍

[wise igloo]

I see stuff like $G/H$ written a lot but do people ever write $\frac GH$?

[cloud walrus]

gmod

[north sand]

I've seen it a few times. can't remember where. and I absolutely hate it

[runic hemlock]

Wikipedia has a convention of writing $\frac{R}{I}$ instead of $R/I$ when considering a quotient of a ring by an ideal. Ig it kinda looks better

[cloud walrus]

Porphyrion

[wise igloo]

why do you hate it? it looks like a fraction like $\frac15$ unless you write that as $1/5$ for some odd reason

[cloud walrus]

gmod

[wise igloo]

I see, I agree - it makes more sense imo

[runic hemlock]

well actually it's only a partial convention

some articles don't use it

[broken stirrup]

I'm trying to prove that every degree 2 extension is normal. I know the fact that a subgroup of an index 2 is normal. How can i use this to prove that extension of degree 2 is normal

Fundamental theorem of Galois theory should do the trick but I'm not exactly sure how

[north sand]

well can't really explain it. but I just don't think it looks good. I just like G/H more for some reason. maybe because you often include it in sentences and it looks better in that case. in a text I would write 1/5 instead of a fraction. or i'd probably write 0.2

[wise igloo]

I see

makes sense

[north sand]

here the fraction stands out completely while R/I fits in nicely

[agile burrow]

fundamental theorem feels a bit overkill for this, but part of the theorem is that the fixed field of a normal subgroup is normal

but I think it's more insightful or natural to prove it just by consider how an irreducible quadratic factors

if you have an irreducible polynomial x^2 + ax + b with a root p in your extension, what can you say about the other root q in relation to how your polynomial factors?

[cursive temple]

the fraction notation for quotients is pretty good to get rid of brackets

the butterfly lemma in the normal / notation would look like $((A \cap C) B)/((A \cap D) B) \cong ((C \cap A) D)/((C \cap B) D)$

[cloud walrus]

Pappa

[cursive temple]

i guess you dont necessarily need the brackets around the (A \cap C)B- terms

[chilly radish]

I do this whenever the quotient is just really long or I have stacked quotients, otherwise I use \sfrac

[chilly ocean]

In some contexts it's useful to quotient both from the right and the left

And G/H keeps a sense from which side we do

That's why it should be prefered

But normal fraction is ok too. Just not always

For rings it doesn't matter, addition is commutative

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

[low void]

Does someone know why a discrete subgroup $\Gamma \subset \mathbb{R}^n$ that is cocompact (i. e. $\mathbb{R}^n/\Gamma$ is compact) is a lattice?

[cloud walrus]

Anton.

[low void]

https://math.stackexchange.com/questions/3619086/cocompact-discrete-subgroups

Someone linked a script, but it just says that this is a corollary of another prop. and my case is less general

[broken stirrup]

Yes, it's pretty straightforward like this, its just -a-p and hence splits

but was looking for alternative solution in which i use Galois theory.

[agile burrow]

fair enough

[broken stirrup]

part of the theorem is that the fixed field of a normal subgroup is normal

I have missed this part of theorem appearantly xD

or Hungerford didn't mention it

[agile burrow]

yeah, at least that's how i learned it

[broken stirrup]

oh nevermind

thank you

[agile burrow]

i can try to outline the proof here

Suppose F/k is a Galois extension with Galois group G, let H be a normal subgroup of G and let E denote the subfield of F fixed by H. Then G acts on the set of embeddings of E into the algebraic closure of k by embedding E and then restricting the automorphism of F to E, which necessarily sends elements to their image in another embedding

moreover, this action is transitive, which essentially follows from uniqueness of splitting fields

so the idea is that given an embedding i, the stabilizer of i under the action of G is equal to the subgroup of G containing the automorphisms of F which fix i(E)

a basic fact from group theory is that stabilizers of elements in the same orbit are conjugate, hence if H is normal, then H is equal to all of its conjugates

thus, i(E) = E for all inclusions, hence E is normal (and Galois)

[wise igloo]

oh damn this is interesting, never seen this before

thanks guys

[cloud walrus]

lirmirit

[fathom pasture]

From what I understand, if we talk about objects in 2D, those objects would be chiral if they have no reflectional symmetry (i.e. they are different from their mirror symmetry)

How would this concept apply to general objects?

[broken stirrup]

thank you so much 🙂

[chilly radish]

What is your exact definition of a lattice

I can give you the general outline

[low void]

Choose the one you like best: discrete subgroup of IR^n?

[low void]

Thanks 🙏

[broken stirrup]

why algebraic closure is the largest algebraic extension?

[chilly radish]

So iirc discrete + cocompact implies you're a full lattice.
The general idea that I know is by contrapositive, proving a non lattice is not cocompact.
First convince yourself cocompact is equivalent to having a (pre)compact subset K such that K+L=R^n for your group L.

Suppose you have a discrete subgroup with non-full span, say L with span_R(L)=W<R^n.

By isomorphism suppose WLOG that W is R^m. For any compact set K, you can find v whose m+1st coordinate is greater than all those of K, then v is not an element of K+L (since L is inside R^m it cannot change the m+1st coordinate)

There are surely other proofs, this is the most elementary one I know of

[chilly radish]

You should try proving this, to see if you understand what the algebraic closure is

[low void]

Thank you, I'll look at it tomorrow, it's really late here.

[chilly radish]

Np

@ me if you have questions

[broken stirrup]

oh that's really simple i see

[chilly radish]

You got it?

[broken stirrup]

yes

[chilly radish]

👍

[broken stirrup]

i mean its an algebraic extension which is algebraically closed but then we can show that since it's algebraically closed, there is no larger algebraic extension

[delicate orchid]

wtf... these people need to be stopped

[chilly radish]

exactly

[gusty cave]

That notation is whack I don't think I've ever seen division right to left be written like that

[broken stirrup]

So in other words, K doesn't have any proper algebraic extension but itself. But that's just the definition so K must be closed

or am i missing something? why is this an exercise? xd

[olive edge]

what book is this?

[broken stirrup]

Hungerford's Algebra

[olive edge]

ah nice thanks

[chilly ocean]

Seems pretty much correct

Let Kⁿ be the maximal algebraic field extension of K contained in F.
Then Kⁿ is a subfield of F

If Kⁿ = K, it implies that K has those elements of F that are algebraically closed over K
K is also a subfield of F.

Hence K is algebraically closed 👍

[low void]

Really appreciate your help! Understood.

[chilly radish]

No problem!

Glad I could help

fwiw there are definitely nicer ways to do this

but this is the way I thin kabout this

[low void]

👑

[chilly ocean]

A weak order is a broad class of such orders where any one of the relations x<y or y<x holds (strong connectedness), or none of them hold (transitivity of incompatibility) making x and y incomparable; with some additional properties if any.

Total pre-orders are weak orders that are transitive, reflexive and have strong connectedness.

Is my understanding correct

[junior harbor]

How could you do the 3rd exercise? I did it but idk if it's correct

[sharp sonnet]

post your solution

[junior harbor]

I am 99% sure it is incorrect/doesn't prove anything, if you still want it tho

Yeah thinking again it just doesn't do much lol

Especially since U doesn't commute under addition

[hidden haven]

Are your rings not defined to have a multiplicative identity?

Because that would always be left cancellable

So that seems like a redundant condition

[junior harbor]

That wouldn't do much about it not commuting under addition tho

Or do you not need that for a ring?

[hidden haven]

Just asking what the definition of a ring is for you

[junior harbor]

Well, they are abelian under addition, are closed under multiplication (a*b is in the group if a and b are) and distributive laws hold (namely n(a + b) = na + nb and same for right multiplication)

[hidden haven]

Ah so multiplicative identity doesn't necessarily exist? Interesting

Wait multiplication needs to be associative right

[junior harbor]

According to the book it needs to be a semi-group under multiplication, so not necessarily has an identity

[junior harbor]

Not sure, probably?

Yeah probably

[hidden haven]

I see

[chilly ocean]

is this Jacobson?

[junior harbor]

Yeah

[hidden haven]

https://math.stackexchange.com/questions/833270/rings-with-noncommutative-addition
This argument should work with 1 replaced with c

[chilly ocean]

hah. I read this book

[tight flare]

did you like it? I was thinking of picking it up as well

[chilly ocean]

notation is bad but yes

[junior harbor]

Ok so I think I found a way to show that the set U commutes under addition

$$ca = cb$$
$$ca + cb - cb = cb + ca - ca$$
$$(ca + cb) - cb = (cb + ca) - ca$$
Since $cb = ca$:
$$(ca + cb) - ca = (cb + ca) - ca$$
$$ca + cb - ca = cb + (ca - ca)$$
$$ca + cb - ca = cb$$

[cloud walrus]

Duhon

[junior harbor]

At least when c is involved, so if c is a zero then U doesn't necessarily commute

Is any of this even correct?

I mean ig you can repeat the same steps with a = b

[hidden haven]

You're starting with the assumption that ca = cb?

You need to show that a+b = b+a without this assumption

[junior harbor]

Gotcha

[junior harbor]

I just cancelled the c in the last step to say $b + a - b = a$, which cannot be true unless they commute in order to do $b - b + a = a$

[cloud walrus]

Duhon

[junior harbor]

That any good now?

[lament dawn]

is SL2Z index two in GL2Z

[delicate orchid]

applying first iso to the determinant we get that GL_n(Z)/SL_n(Z) iso to C_2 so yeah I think so

[celest cairn]

Hi, does anyone have any good abstract algebra book suggestions that have wording and notation that is easily understood?

[junior harbor]

Try that one book by Artin

I think it's just called "Algebra"

Easy to find pdfs online, tho official/hardcover versions might be hard to find outside of the US or other English speaking countries

[lament dawn]

Lang's Algebra

jk, but see pinned messages in #book-recommendations

or, actually, I think #books is off the ground now?

[broken stirrup]

You could try Thomas Judson or classical book from Dummit and Algebra ( the former one is simpler)

[celest cairn]

Ok! Thanks guys! 🙂

[vague granite]

Hi guys

I think I've spent the whole evening trying to tackle question b)ii) 😅

anybody got any ideas?

[tribal moss]

Hmm, if it is true at all, then my gut feeling would be that you can probably make one by defining f(s) by transfinite recursion. Something like, in each step,
1) if there is an s and g such that f(s) is already defined but f(g·s) isn't, then pick such a combination and define f(g·s) to be g*f(s).
2) otherwise, if there's any f(s) such that f(s) hasn't yet been defined, pick the smallest such s (according to a fixed bijection between S and an initial ordinal), and define it to be the smallest s' that is not yet in the domain of f.
3) otherwise, you're done.
The trick would be in proving that when both actions are free, then whenever you reach case 2, then f already satisfies f(g·s)=g*f(s) for all s in its domain so far and all g.

[broken stirrup]

would K(u_1,...,u_n) be splitting field over K for corresponding separable polynomials of u_i?

[tribal moss]

Wait, I missed that everything is finite, that makes it much easier.

For a free action, each orbit is in bijection with G once you pick an element of the orbit to represent 1.

Since S and G are finite, there are the same finite number of orbits according to · as according to *.

Just pick some way to match them up.

^^^ @vague granite

[vague granite]

I ended up reading something like you pick an s’ to write every s as gs’ and then f(gs’)=g*s’?

[tribal moss]

Yeah, that works if |S|=|G| so there is just one orbit. But if there are multiple orbits, then you need to pick a representative of each orbit on each side of f.

[fallow plume]

So the Galois Correspondence Thm tells us that if a field extension corresponds to an S_n group, then it has an A_n normal subgroup corresponding to a normal field extension of degree 2 (reiterating to make sure I'm not slipping up with basic logic). is there any kind of relationship between this new intermediate field extension and the base field/field extension themselves? obviously this implies there's an irreducible quadratic in this field that will induce the full field extension, but does this quadratic have any strong relationship with the original polynomial or so on?

I'm guessing that a lot of this is, more or less, a case by case basis, depending on what polynomial you choose the answer may be obvious what subfields are fixed by a particular subgroup and what not, but in general it's seemingly random as the Galois group only tells us about the structure of the automorphisms together, not the automorphisms themselves. is this reasonable?

[fallow plume]

figured a bit of it out.
elements of A_n preserve the sqrt of the discriminant
so the normal subfield that it predicts is exactly K(delta) where delta^2 = Delta = discrim of the polynomial

[upbeat fulcrum]

can someone help me with this one?

[smoky cypress]

Since 0 is never a root you only need to consider p on the multiplicative group, which is cyclic. Then notice any m relatively prime to p^k - 1, x^m is bijective on the cyclic group. Therefore conclude that you can replace x^m-1 with x^{gcd(m,p^k -1)} - 1, so assume m divides p^k - 1. Lastly, remember that for cyclic group, how many elements satisfy x^m-1 when m divides the order of the group

@upbeat fulcrum

[chilly radish]

Let $k$ be a field. I know that for a finite dimensional division algebra $D$ with center $k$, its dimension is a square, and the dimension of any maximal subfield is the root of the dimension of $D$. Moreover, if $K$ is maximal, then $C_D(K)=K$ (This is an iff). Does the same result hold for a general CSA? If $A\cong M_n(D)$ is a CSA, then $\operatorname{dim}_k(A) = m^2n^2$ if $\operatorname{dim}_k(D)=m^2$, then firstly, is it true that for a maximal subfield $C_A(K)=K$? And if not, does it still hold that $[K:k]=mn$?

[cloud walrus]

ShiN

[upbeat fulcrum]

I already prove the theorem, I was hoping for the help in (1)

this one

[broken stirrup]

does anyone know where is this from?

[low void]

Why is this "clearly" the case? Am I missing some standard group theory?

[noble dust]

Both lattices are free abelian groups of rank n

You can find bases of both lattices that are stacked (see smith's normal form) so you get necessarily that the index is finite

Another way to see it is that if you choose a basis for $\Gamma'$, you can find a representative of any $x \in \Gamma/\Gamma'$ that belongs to the fundamental parallelogram of $\Gamma'$. There's only a finite number of points of $\Gamma$ in this parallelogram

[cloud walrus]

Digiteraat

[long obsidian]

Is there any situation in which the Galois group of the characteristic polynomial of a matrix is interesting?

[delicate bloom]

you can think of the matrix as representing multiplication by an element on a field extension represented by a vector space over your base field, so it'd end up being the galois group of this field extension I think

[low void]

Thank u

[long obsidian]

Ahhhhhh okay thank you

[wraith obsidian]

Is there an abstract way to see that Z cannot be a profinite group? Like along the lines of „finite groups have property P and P is preserved by limits“

[sour plume]

i was trying to go for an argument along the lines of "profinite groups are exactly the Hausdorff, compact & totally disconnected topological groups and Z cannot have such a topology", but it is indeed possible to equip Z with a topology that makes it Hausdorff, compact & totally disconnected (identify it with {0} \cup {1/n : n in N} as a subset of the reals), but I supposed this means that the set can't have a group structure compatible with the topology

so uh that's not an answer, but a very interesting question

[rustic crown]

i think one way would be to say that it's compact hausdorff which means we can put a Haar measure on it, and so the group can't be countable.

as translation invariance will force each singleton to have the same measure, but then they all need to add up to like 1

[broken stirrup]

anyone studied character degree graphs?

I would like to ask something

[grand sigil]

how do i show x-a is prime in C[x]

im thinking taking p*q=r(x-a)

where p,q,r are polynomials and a is complex

and then using a degree argument

degree of p=0 and q=1 or vice versa

and degree 0 cant be in (x-a) ideal

so (x-a) is prime?

[rustic crown]

you could use the fact that f(a) = 0 if and only if (x-a) divides f(x)

[sour plume]

I like this! Good one

[delicate bloom]

interesting example, cause it meets all the requirements except the group structure. Another way that meets all but one is the image of the integers in any p-adic integers is hausdorff, totally disconnected, and a topological group but not compact (although the integers are dense in the p-adic integers, and the p-adic integers are compact)

I wonder if it's possible to meet everything except hausdorff or everything except totally disconnected too

[junior harbor]

Ok so I'm not sure if this makes sense, but if $I$ is the set of all integers, then $I$ is an ideal of the factor group $I/I$ right? Just so I can condense the 2 concepts

[cloud walrus]

Duhon

[grand sigil]

So what if R is K[x]/(x^2)

and we choose the prime ideal (x)

if we take f = a1x+a0

then quotients with x give a0 and localizing gives a0/1

so f(x)=a0/1

However my book says it should be 0.

[chilly radish]

Gonna bump this. I didn't have time to ask my prof

[wraith obsidian]

Oh I like that
I need to remember that CompHaus is preserved.
So you used the property that a haar measure has finite measure on compact sets, right?

[uncut girder]

I'm not sure if this answers your question or is at all relevant, but here's a statement from Guillot chapter 6 that may be useful

Commutant means centralizer

[chilly radish]

This is actually exactly what I need

Or

Hmm

[uncut girder]

The proofs are in the book also

[chilly radish]

Well if it's a maximal subalgebra it's in particular a maximal field

I'll check it out thanks!

[uncut girder]

I was just reading this section of the book recently 😅

[chilly radish]

How convenient

I still wonder if there's a simpler way to do this for my case but probably not

@uncut girder is this a corollary of double centraliser in the book?

[chilly radish]

Nvm I thought about this for a sec and it's actually simple. I just needed the shift in perspective from only thinking about fields to thinking about simple commutative subalgebras in general (Which is the proper generalisation going from division algebras to CSAs)

Thanks!

[uncut girder]

Its a corollary of whats called "the commutant theorem" in the book

[chilly radish]

Most likely the double centraliser theirem

B''=B

for simple subalgebra of CSA

[chilly radish]

Alright yea

That's the one

Cool

[uncut girder]

Do you have any intuition about this? I'm still trying to process the statement of this theorem

[chilly radish]

Hmm, not really. The only intuition I really have is the fact that for maximal subfields (and commutative subalgebras) it gives the square root of the dimension, and the fact take taking the centraliser is like asymmetric.

Centralisers are weird, if you're starting with smth noncommutative you aren't even guaranteed your original algebra will be inside, but if you take it twice it has to be inside since you're forcing it.

All this to say I don't have much intuition for why this is true, or why, besides formally for the proof, we need a CSA. Maybe look into balanced modules tho.

[grand sigil]

bumping #groups-rings-fields message

[chilly ocean]

show what book says

[grand sigil]

,rotate

[chilly ocean]

f = x.

[chilly ocean]

a_1 = 1 and a_0 = 0.

[rustic crown]

Yep.
Both are pretty simple to check. We can write the limit as an equalizer, and since finite groups with discrete topology are cHaus, so are their products and so the Equalizer is a closed subspace of a product and so stays cHaus.

And yep we needed it's translation invariant, finite on compacts and countable additive.

[broken stirrup]

How can i show that character degree graphs of A_n where n>=15 are complete graphs ?

I know that vertices ( set of all prime divisors of degrees of irreducible characters are is the set of all the primes <= n )

but how do i show that product of any two prime divisors ( vertices ) divides some degree of a character so that there is a corresponding edge?

[hallow kettle]

Is there any theorems that say something about the subgroup of non abelian group (alternating group, dihedral group, symmetric group) are normal subgroup?

[chilly ocean]

Idrk such theorems about them, but for symmetric groups, the kernel of the group homomorphism f: S_n -> {+1,-1} where {+1,-1} is a group under multiplication, where +1 is the Identity; and f returns +1 for even permutations, the kernel of this homomorphism, that is, the set of all even permutations that map to +1 (the Identity) is the normal subgroup of S_n under the group operation of S_n (composition of functions)

And all subgroups of a symmetric group are automorphism groups or permutation groups

[hidden haven]

The kernel of any homomorphism is normal, and conversely any normal subgroup is the kernel of some homomorphism. This is why normal subgroups are defined

[chilly radish]

I'll add that the converse is true in a trivial sense (Although maybe less so if you define normal subgroups as kernels of homomorphisms). if you have a normal subgroup N, then G/N is a group, so the quotient map G->G/N is a homomorphism with kernel N

[hidden haven]

That is the general case by the first isomorphism theorem

[chilly radish]

sure

just saying it's not an especially deep connection

[chilly ocean]

Also $Z_{p^k}[i]$ is basically a polynomial ring with integer coefficients from the field $Z_{p^k}$ right where elements in the ring are of the form a+bi?

[cloud walrus]

Pencil/Idris

[grand sigil]

lol

i read it as letting x be arbitrary element of R

that was weird on my part

[unreal night]

A primitive element is a generator of the multiplicative group of a finite field. Is there a term for the generators of its additive group?

[hidden haven]

The additive group of a field F can only be cyclic if the field has prime order

Because otherwise it properly contains a cyclic group (the one generated by 1 assuming char F > 0) which isn't contained in any larger cyclic group

If char = 0 then it contains ℚ which isn't cyclic

So yes, the name would be ±1 then lmao

[unreal night]

ah right, thanks!

[devout crow]

If $R = M_2(\mathbb{F})$ and $R^n \cong R^m$ as $R$-modules, by restriction of scalars can we say this implies $R^n \cong R^m$ as $\mathbb{F}$-modules, and therefore that $n=m$?

[cloud walrus]

Average J∘du=du∘j enjoyer

[chilly ocean]

.

[devout crow]

yes, but the ring of integers mod p^k is not a field (for k > 1)

[chilly ocean]

Oh so a commutative ring

[devout crow]

yep

[chilly ocean]

[chilly ocean]

Yeah

Z_p^k is a field because it is a commutative ring with unity 😅 for k = 1

[chilly radish]

what

it's not a field

there are zero divisors

[chilly radish]

yes

because then the dimension over F will be 4n = 4m

and F is embedded in R

[west sinew]

Can someone help me unpack this example of localization of a ring? In particular, I don't understand how ``$A_P$ can be identified with the ring of all rational functions on $k^n$ which are defined at almost all points of $V$".

[cloud walrus]

adi

[west sinew]

for ref. the textbook is Atiyah & MacDonald's Comm. Algebra

[inland otter]

well recall that an ideal p of the polynomial ring defines a subset of k^n, ie. the subset of all points on which all elements of p vanish

for instance suppose p is maximal, then this subset is simply a point.
Localising at p then is (in essence) just adding inverses to all the polynomials which do not vanish at this point

so if for example p= (x,y), then the polynomial xy+1 does not vanish "at" p.
inverting it gives you 1/xy+1, this rational function is defined on all of k^n except on V(xy+1), ie. is defined at allmost all points

[west sinew]

Just to clarify on your example: Do you mean to take the prime ideal (x, y) in the parent ring R[x, y]? In which case the variety defined by it is just the point (0, 0), at which the polynomial xy+1 does not vanish. And is it so that (1/xy+1) is the 'inverse' subsequently supplied to it in the localized ring?

Finally, the text talks about being defined at almost all points of V, whereas you spoke of being defined on all points of k^n except V. Is this an error in the text? Or is it because you're referring to different varieties? I'm not clear on this point.

[inland otter]

As for the first point, that is what i was referring to yes and yes xy+1 gets mapped to a unit in the localised ring whose inverse is 1/(xy+1).
What the text means is the same as what I mean.
I dont know how much you know about varieties but you can define a topology on k^n whose closed sets are exactly given by sets of the form V(I).
Turns out that in this topology open sets (complements of the V(I)) are always dense.
As 1/xy+1 is defined on k^n\ V(xy+1) it is defined on an open set and thus on a dense subset of k^n, which justifies the usage of almost all points in the text.

[west sinew]

Thank you

[thorny knoll]

Any ideas how to go about this exercise? I can't see how I could prove the hint.

[chilly ocean]

what are your thoughts on the question?

[opaque pilot]

Suppose phi is a non zero map

So there is an x in Q such that phi(x) is non zero

Im guessing the contradiction is to get to the conclusion that phi is not a homomorphism

[chilly ocean]

it's not the way i'd have done it, but why not give it a shot and see if it works?

[opaque pilot]

Ok, so phi(x) = a for some integer a

im not sure..

[chilly ocean]

it's only been a few minutes

[chilly ocean]

phi(x) = n*phi(x/n), so phi(x) is divisible by all natural numbers

it's probably easier to show that such a homomorphism must be zero than to show that if a function Q -> Z is non-zero it's not a homomorphism

but i wouldn't give up so quickly

[opaque pilot]

hm, Ill try both ways

[chilly ocean]

or you can just use what blitz wrote, if you didn't feel like playing around with the problem

[opaque pilot]

its been a few days since I looked at this, I think ill play around with it abit just to re familiarize with the definitions

[chilly ocean]

that would be a good idea

just mess around with the assumption that phi: Q -> Z is a homomorphism and see what you can get from that

[chilly ocean]

Hi

Why do we require that an action is transitive when defining blocks?

could you provide more context?

Well I see that on wikipedia the definition is what is our characterization and vice versa

We’ve defined block on a transitive action B = B^g or their intersection is empty

And I’m wondering why do we need action to be transitive there

[chilly ocean]

where on wikipedia?

[winter yoke]

I've been stuck on the following problem, any ideas? $f \in S = R[x,y]$ is a 420 polynomial if we can write $f(x,y)=g(x^4,y^{20})$ for some $g$ in $S,$ and we define $T(f)$ as the corresponding $g.$ If $f, g, h$ are 420 polynomials and $f=rg+sh$ for some $r, s \in S,$ prove $T(f)=uT(g)+vT(h)$ for some $u, v \in S.$

[cloud walrus]

aadfg

[delicate orchid]

weed polynomial

[agile burrow]

weed polynomial

[chilly ocean]

weed polynomial

[chilly radish]

Weed polynomial

[quartz solstice]

I've been pondering on a question about groups and I can't find any information on it since it's really specific, but are there any procedures to find how to transform permutation A into permutation B?

an example that explains my question better would be:

assume group G are the integers under addition, let's assume permutation A is the number 4 and permutation B is the number 9. Is there any algorithm that tells us the steps we need to take to transform the number 4 into 9?

and in another sense, if we have a set of actions A that act over a system S, is there an algorithm that tells us the actions we need to take in sequence to take a specific permutation of system S into another permutation of system S?

in this specific example, we could make the system a number, and we have a set of actions that increases the number by 1 and another action that decreases it by 1 {+, -}, does anybody have any idea on how to take a permutation in this system into another permutation in terms of the actions we have available automatically?

[winter yoke]

Ok, but any ideas? It's a real problem, no joke

[chilly ocean]

To prove inverses map to inverses in a group homomorphism, is it fine to do something like this (as an example):

(G1 has ° operation and G2 has + operation)

f: G1 -> G2; g1 = g2

f(-g1) = f(-g1 ° 0)
=> f(-g1 ° 0) = f(-g1) + f(0) = -g2 + 0
=> f(-g1) + f(g1) = -g2 + 0 + f(g1)
=> f(-g1) + f(g1) = -g2 + g2 + 0
=> f(-g1) + f(g1) = -g2 + g2 = 0

what is g_2 supposed to be?

element in G2

which

f(g_1)? f(-g_1)?

f(g1) = g2

g1 maps to g2

and -g_1 is supposed to be the inverse of g_1? likewise for g_2?

just so i understand your notation

Yes

so in the second line, you wrote f(-g_1) + f(0) = -g_2 + 0. presumably you did this using f(-g_1) = -g_2?

Yes

you can't do that, since that's the thing you're trying to prove

Wait

Oh

Yeah right

Thanks I'll try again

if you want a hint: ||apply f to both sides of g_1 ° (-g_1) = 0||

Got it
I really need to get good at writing proofs lol

it's one of those skills that takes a while to develop

[celest cairn]

Hi, I have a question.
In this equation, I don't understand how I got my x and y values. I used a Diophantine equation calculator that showed me some of the steps. How did I get x=38 and y=-133?
These are obviously the correct answers because when I plug the values in, I get 19, which is what the equation was originally equal to.

[wooden ember]

I mean you can always invert A and then apply B? But I’m not sure that’s what you may have meant by steps?

[north sand]

You just multiplied 2 and -7 by 19

You have 1=2*25-7*7. So then you multiplied both sides by 19

[celest cairn]

Ah ok thanks 🙂

[hallow kettle]

this is all subgroups of dihedral group d12 (the hexagon), how to find the subgroups D6 and (Z_2)^2?

[rustic crown]

identify a triangle inside of the hexagon. now any symmetry of the triangle would give you a symmetry of the hexagon

for Z/2 + Z/2, pick your favorite reflection and a 180 degree rotation

[wooden ember]

does this mean linear transformations of complex projective space preserve circles? That's mad

wait no that's sounds totally wrong i must be misinterpreting

i think im getting confused by the fact that they talk about C^2 and C when clearly the projective complex plane is hiding somewhere

is it more like, a linear transformation of C^2 induces a non linear transformation of CP?

yeah cause the induced map is more a map of the riemann sphere which is CP right? I guess this belongs more in #real-complex-analysis though

[hot abyss]

i can follow this proof until the very last paragraph where it says “in cases 2,3, and 4, we repeat the argument one or more times”, what would we even be repeating there?

also epsilon is the identity permutation

i actually do understand the proof, and how it would lead to a contradiction, i just don’t get that one part

[sweet echo]

I assume the part that says "the last occurrence of x is one position further left than it was at the start"? You repeat the argument to keep pushing the x to the left

[dull ginkgo]

Let E_1(n) be the matrices that permute rows in a nxn matrix
Let E_2(n) be the matrices that add varying rows in an nxn matrix
E_1(n) and E_2(n) have trivial intersection afaik, so what is their semidirect product in GL(n,F)

They just use F's additive and multiplicative identities so

[west sinew]

What does it mean to say that an ideal "meets" a subset of a ring?

[chilly ocean]

non-empty intersection

[somber thorn]

any ideas on how to do this?

i've already tried $b^{2n}$ but it seems to be wrong

[cloud walrus]

sean

[chilly ocean]

what makes you say b^{2n}?

can you prove this by induction?

[somber thorn]

never mind sorry i got it, just needed to think a bit 👍

[shadow coyote]

i don't understand here why it "clearly suffices to prove that if G is finite, abelian, then G admits a cyclic tower"

[proud bear]

did you figure it out

[shadow coyote]

if we have a cyclic tower on G_i/G_(i+1) then we can append {e} to the end, and then pull it back into a cyclic refinement between G_i and G_(i+1) [using the homomorphism from G_i -> (G_i/G_(i+1))]

and then i think we can do this between each pair of consecutive subgroups in the abelian tower

i'm not 100% sure if it works

[proud bear]

this looks correct i think

[chilly ocean]

Why is that true

[sharp sonnet]

if a_2 is in (a_1), then there is some r such that a_2 = ra_1
but a_1 = a_2c, so a_2 = ra_1 = a_2cr and cr=1, so c is a unit

[chilly ocean]

Tq

[dull ginkgo]

I don't know much about abstract algebra and this might be more linear algebra but

Let V be a vector space of dimension n with scalar field F
Let V_B be the set of full-rank basises of V, a subset of P(V)

Let GL(n,F) act on P(V) such that the output is the image of the subset under the partial function of the action.

GL(n,F) acting on elements of V_B is transitive afaik since there is only one change of basis matrix for 2 basis, but my question is: is this action free? in other words, does GL(n,F) permute the set of basises?

idk the permutation group of V_B is isomorphic to GL(n,F) tho

[north sand]

when we consider the bases as sets then we can change the "order" of them without changing the basis. stuff like $\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \cdot \left{\begin{pmatrix} 1 \ 0 \end{pmatrix}, \begin{pmatrix} 0 \ 1 \end{pmatrix}\right} = \left{\begin{pmatrix} 1 \ 0 \end{pmatrix}, \begin{pmatrix} 0 \ 1 \end{pmatrix}\right}$

[cloud walrus]

Denascite

[chilly ocean]

Hi, does anyone get this? I need a hand.

[south patrol]

Not algebra, #probability-statistics might be better to ask ig

[lapis trail]

a ring is boolean if and only if both operators distribute over each other. Fact or fiction?

[celest cairn]

Is there a way of solving this Diophantine equation using the Euclidean algorithm?

[sharp sonnet]

#elementary-number-theory

[lethal dune]

there are other axioms as well

[lapis trail]

But boolean implies both operations distribute over the other?

[chilly ocean]

Fact.
In fact, all rings have operators that distribute over each other. It is one of the axioms for a set equipped with two operations to be a ring
If you define a Boolean operation V as:
x V y = x + y + xy
And . for usual ring multiplication (x.y is abbreviated as xy here),
x.y V x.z = xy + xz + xxyz = xy + xz + xyz = x(y+z + yz) = x.(y V z)

And since Boolean rings are commutative
x.(y V z) = (y V z).x
For any x,y,z in Boolean Ring R (this makes R also a Boolean Lattice)

Distrubutivity holds.

And as ryu sama correctly said, distributivity isn't the only axiom for a set equipped with two operations to be a ring. With distrubutivity, a ring must be an abelian group under ring addition, and a semi-group under ring multiplication, though some people and authors define the ring to have the multiplicative Identity making it be a monoid under multiplication.

In general if we only see a Boolean Ring, it always has two elements only, since it is an associative algebra over GF(2).
Great example could be Integers modulo 2.
In this also distributivity holds.

x.x V x.y = x V xy = x + xy + xxy = xx + xxy + xxxy = x(x + xy + xxy) = x(x V xy) = x(xx V xy) = (xx V xy)x, for x and y in R.

And for a Boolean ring, xx or x² = x due to idempotency
That is why above xxyz = xyz (for Boolean Lattice)

[chilly ocean]

Indeed

[chilly ocean]

Wikipedia has also nicely explained this, and even nLab. You can also check them out

x+yz = (x+y)(x+z) ?

Not necessarily

Easy calculation shows that for a Boolean ring this is equivalent to xy = xz

If x = 1 and y =/= z then this isn't satisfied

Any non-trivial Boolean ring doesn't have this property

[muted sun]

Let $R=K^{n\times n}$ be the Ring of $n\times n$ Matrices over a field $K\newline$
interpret $R$ as a (Left)module over itself$\newline$

[cloud walrus]

~Martin

[muted sun]

I am looking for a decomposition of this module into a direct sum of indecomposable sub modules
I found this in my script:
R is a euklidian ring and M a free R-module
then there are indecomposable elements $f_1,\dots ,f_s \in R$ and $t,n_1,\dots ,n_s \geq 0$
such that
$M \cong R^t \oplus R/(f_1^{n_1})\oplus\dots\oplus R/(f_s^{n_s})$

[cloud walrus]

~Martin

[muted sun]

i have found theorems that say that this is possible, but not really a way to do it

[agile burrow]

are you asking for a general procedure to compute this, or are you asking about the case you wrote above, where you view R as a module over itself?

[muted sun]

the case i wrote above

[agile burrow]

R is not a Euclidean domain, it isn't even an integral domain

[agile burrow]

also, I'm not sure what theorem you're referring to here. It looks kinda like the classification of finitely generated modules over a PID (since euclidean domains are PIDs) but you also state the hypothesis that M is a free module, in which case there are no modules over the form R/(f) since those would be torsion submodules of M

[chilly radish]

I'm trying to prove that the group of units of the ring of integers of $\mathbb Q(\xi)$ where $\xi$ is a primitve $p$-th root of unity for an odd prime $p$ is a direct product of the set ${\pm \xi^j:0\leq j \leq p-1}$ and a (multiplicative) free abelian group $U\subseteq \mathbb R$ of rank $\frac{p-3}{2}$.
\\
I used Dirichlet's Unit Theorem to show that it's a product of the roots of unity and a lattice, and then I showed the roots of unity are exactly $\pm \xi^j$, now all I have to do is show that $U\subseteq \mathbb R$. The hint I have says to consider the totally real subfield $L = \mathbb Q(\xi + \bar \xi)$ and to compare the rings of integers. I found this field is totally real, with degree $\frac{p-1}{2}$ and hence its group of units is just ${\pm 1}U'$ where $U'\subseteq \mathbb R$ is a free abelian group of rank $\frac{p-3}{2}$, but I can't seem to show that $U=U'$. Any hints?

[cloud walrus]

ShiN

[twin ore]

∂∂𝑥 and ∂∂𝑦, are they elements of a Ring or a Vector Space??

[chilly ocean]

I'm not claiming x + yz = (x+z)(x+y). Of course this is not true for non trivial Boolean rings and y≠z.

I'm specifically saying for x(y V z) = xy V xz, for any Boolean ring (more specifically a Boolean Lattice)

If we only see a Boolean ring with two elements, basically an associative algebra over GF(2), then x(x V y) = xx V xy = x V xy also holds true.

[slate mortar]

Why do we have that matrices over a field are nilpotent if and only if its characteristic polynomial is x^n?

[hidden haven]

Nilpotent implies that the matrix satisfies x^m = 0 for some m, so its minimal polynomial divides this so must itself be of the form x^k. All factors of the characteristic polynomial appear in the minimal polynomial, so that must also have the form x^n

[slate mortar]

I don't get the end?

The minimal polynomial divides the characteristic polynomial right?

[hidden haven]

If you go to the algebraic closure, every root of the characteristic polynomial is a root of the minimal polynomial

ie every eigenvalue is a root of the minimal polynomial

You can use Jordan form to prove it

There's also a more elementary argument

Which you should be able to find easily by googling something like eigenvalues are roots of the minimal polynomial

[rustic crown]

you can simplify the argument a little if we don't talk about minimal polynomial at all. if a is an eigenvalue of A with eigenvector v, then 0 = A^n v = a^n v which means a = 0. so only root of the char poly is 0.

but yea this requires you to go the algebraic closure or at least the splitting field of the char poly

[slate mortar]

I think I do get why all factors of the characteristic polynomial appear in the minimal polynomial by working in the splitting field of the characteristic polynomial and using https://math.stackexchange.com/a/101284
To conclude from there, we use that x divides the characteristic polynomial. Is this correct?

Oops I don't think so, I'm lost

[hidden haven]

We use the fact that x is the only irreducible divisor of the characteristic polynomial, because it is the only irreducible divisor of the minimal polynomial

But use det's argument instead

Much simpler

[slate mortar]

Oh yes of course, thank you so much!

[slate mortar]

True, but I wanted to understand yours too haha

Thank you so much @hidden haven and @rustic crown

[hidden haven]

Np

[wooden ember]

Can’t you do something much simpler? If a polynomial vanishes on A then you can evaluate that polynomial at A, multiply by an eigenvector on the right and factor out the eigenvalue vector so you’re left with the polynomial evaluated at the eigenvalue

So in particular minimal polynomial vanishes on all eigenvalues

[hidden haven]

Works, but if you accept Jordan form then it's a one liner

The minimal polynomial is the gcd of the minimal polynomials of all the Jordan blocks, while the characteristic polynomial is their product

[wooden ember]

« If you accept Jordan form » is a big if lol considering this could be a result you would prove when you learn about eigenvalues

[hidden haven]

Hence

[wise igloo]

what are the prerequisites to studying differential algebra? it looks interesting but idk if I can learn it yet

should I be familiar with algebras and stuff?

[delicate orchid]

did you get inspired by the talk lol

[wise igloo]

no I was taking a shit last night and thought like "differential geometry exists, what about differential algebra?" so I looked it up and yeah

I just forgot to ask until now

[delicate orchid]

based

[hidden haven]

You should probably know what an algebra is if you want to study algebras with extra structure

But there shouldn't be much needed beyond that

[wise igloo]

alright ty

[chilly ocean]

You use those in symbolic integration actually

[wise igloo]

use what?

[chilly ocean]

differential algebras

I was reading a book called Symbolic Integration I by Bronstein out of boredom once

that's how I know

[wise igloo]

oh I aee

see*

[atomic python]

i've been asked to find a subfield, $L = K(l_1, ... l_m)$ of a finite field extension $E = K(e_1, ..., e_n)$ such that the minimial polynomials of $\forall i, e_i$ are equal over $L$ and over $K$.

[cloud walrus]

Quarky

[atomic python]

this seems impossible to me
my reasoning so far is suppose $E$ is a degree $t$ extension and $L$ is a degree $t$ extension. $[E:K] = [E:L][L:K]$. Some theorem tells us that $E$ is isomorphic to $K$ over a minimal polynomial, and same for $L$. But clearly the degrees can't be equal without messing up the multiplicativity of the field extension degrees.

[cloud walrus]

Quarky

[atomic python]

i'm guessing that what I'm going wrong here with is "what minimal polynomial am i quotienting by"

but it isn't clear

[chilly radish]

What you said is only true for a simple extension

If you.mean that E is K[x] quotient by some maximal ideal (m)

[atomic python]

okay, time to stare at it some more

[broken stirrup]

I've already shown that (a) > (b) > (c)

Now I want to show that (c) > (a) but got stuck i think

[rapid slate]

Is there a simple proof of the fact that the group of units of the subring O of the quadratic extension field is infinite?

I could see it for special cases like √2,√3 etc but not generally

[delicate bloom]

look up pell's equation

[signal rain]

In this problem, what does the notation R - {0,1} mean?

[south patrol]

• is an alternative to \setminus (in this context)

[signal rain]

Ah, okay, thank you!

[south patrol]

Np

[rapid slate]

I mean an algebraic proof of it