groups-rings-fields

[covert vector]

just pick an arbitrary element x \in A

and work with x

[neon tapir]

Ah

you are right

Thank you

[covert vector]

np

[neon tapir]

ok

so

I have to prove that A(1-i) + A = A

A = Z[i]

If I prove that

We can inverse (1-i) in Z[i]

We have A(1-i) = A

[thorny slate]

that doesn't imply A(1-i) = A

[neon tapir]

(1-i) is an ideal

[thorny slate]

for example A(2) + A = A

but A(2) is not A

[neon tapir]

Ah

Ok

So how can i prove

that A(1-i) + A = A

Do you have a clue

[thorny slate]

it's clear from the definition

0 + a is in A(1-i) + A

you can write any a as 0 + a

[neon tapir]

Ah

ok

thank you

[thorny slate]

(1-i) isn't the whole ring

you do not have A(1-i) = A

[neon tapir]

Yes

I just understand

thank you

[thorny slate]

👍

[neon tapir]

Thank you very much @thorny slate

But

We have 0 in (1-i)

Because 0 is in A

That's it ?

So for any x in A

xA + A = A

[thorny slate]

yes

[heavy jasper]

I'm trying to follow a proof that the number of conjugates/order of conjugacy class of an element in a group is equal to the number of cosets of the centralizer. The first step they take is to let x,y be elements of the conjugacy class of a. They then say "By the definition of the conjugacy class of a, xax^{-1}=yay^{-1}", which I don't see why that follows from the definition of conjugacy class. i assume im missing somethings fairly simple but i cant see it

https://proofwiki.org/wiki/Number_of_Conjugates_is_Number_of_Cosets_of_Centralizer proof im trying to follow

[thorny slate]

lol wtf

looks like a big oversight

[heavy jasper]

ok cool cuz im sitting here thinking theres nothing about this in the definition that is immediately related to this

[thorny slate]

yea

[timid cipher]

C_a isn't the conjugacy class of a, but the centralizer of a

The centralizer of a is all elements x such that ax = xa, i.e. xax⁻¹ = a

[thorny slate]

C_a is supposed to be the conjugacy class of a

the proof is similar

to what's written

they just fucked up the wording

what they wrote implies that elements in the same coset conjugate to the same thing

and elements in different cosets conjugate to different things

the result follows immediately

[lucid drum]

;

[neon tapir]

Hi

Is C[x] a commutative field ?

[worthy kindle]

is it even a field to begin with?

[neon tapir]

C is a field

So i would say C[x] is a field

[worthy kindle]

actually, are you talking about the set of polynomials with complex coefficients?

[neon tapir]

Yep

sorry

[worthy kindle]

okay, so it's not a field with the usual operations at least

[neon tapir]

Ok tu parles français

[worthy kindle]

Oui mdr

[neon tapir]

Donc C[x] c'est pas un corps ?

[jagged gate]

trop de puissance

[worthy kindle]

Nop

[neon tapir]

lol

Pourtant

Ah

ok

tous les élts ne sont pas inversibles

Mais c'est un anneau intègre

[worthy kindle]

Oui, et l'ensemble des fractions rationnelles à coeff complexes, c'est un corps

[neon tapir]

Q[i] ?

[worthy kindle]

$\mathbb C(X)$

[cloud walrus]

Tuong:

[neon tapir]

ah

C(x) je connais pas

[worthy kindle]

un polynôme sur un autre polynôme

[neon tapir]

oh lol

merci

Donc pour trouver

Des idéaux maximaux de C[x]

Je dois trouver

les idéaux I tq

C[x]/I est isomorphe à C

[neon tapir]

Hi, I have an algebra question in #help-3

[worn ginkgo]

I've got a theorem that needs proving. It's an analogue of the dimension theorem, specifically on the asymptotic dimension

Let asdim denote the asymptotic dimension of a group, and let f be a group homomorphism from G to H. Show that
asdim(G) = asdim(ker f) + asdim(f(G))

has anyone encountered this before?

[oblique river]

i'm not familiar with the notion of asymptotic dimension

what's the idea?

[neon tapir]

Hi, I have a little algebra question in #help-1 😄

[bleak abyss]

Sounds like a geometric group theory thing

[neon tapir]

The class is called Algebra lol

[bleak abyss]

I was talking about the asymptotic dimension

Also, and I should've said this before, please don't post in other channels telling people about your question

[worn ginkgo]

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

Specifically look at the bit at the bottom on geometric group theory

[unkempt briar]

hey here, i'm having some trouble with tensor product

|+> and |-> are orthonormal vectors

and i'm struggling with :

at the bottom

i can't prove <w3|w3w2> = 0

i mean, i don't land at all on this

i got

<-|-> <+|+> |->|+>

ie : <-|-> <+|+> |w3>

btw, this book is pretty good, but full of typo error

[viscid canyon]

$\langle w_2|w_3\rangle=\langle w_3|w_2\rangle=0$ would make sense

[cloud walrus]

tubular:

[viscid canyon]

not sure that $|w_3w_2\rangle$ is even supposed to mean anything

[cloud walrus]

tubular:

[autumn slate]

Oh hey

Is this standard notation?

[smoky bluff]

I don't think so

I think it's just written J^k

It's written like that on the next line

[thorny slate]

the tensors are also the elements of the tensor product of copies and V and its dual

so that's the usual notation

[chilly ocean]

Hey do we prove lagrange's theorem using right cosets by picking a subgroup H on the right side of the rectangle representing the actual set G?

Also why arent quotient groups a subgroup of the main group?because they dont have the identity element?

[quiet cave]

Rectangle...?

The quotient group of G is a group whose elements are the (right) cosets of G

It doesn't make much sense to consider it a subgroup of G

[viscid canyon]

quotient groups are not subsets of G with group structure, but partitions of G with group structure

subgroups have to be subsets

lagrange's theorem is proven by bijecting any two cosets

[chilly ocean]

Uh i have a more visual cue tbh for this..u take a rectangle representing the set G and then u take the normal subgroup H having the identity element and then u represent this group as a square and then u keep duplicating this square with non overlapping left cosets until it completely fills the set G

If there are k such cosets then

The order of H times k=the order of G

Hence the order of H divides the order of G

And u can easily prove the size of each coset is the same

[oblique river]

how do you know that the sets are nonoverlapping and disjoint?

how do you know that it will completely fill G?

by the time you've actually proven those facts, you've basically proven that there is a bijection between all the cosets

[chilly ocean]

Its not hard to prove sets wont overlap

[thorny slate]

yeah, it's not hard to prove cosets work

[timber bay]

Looking at this one, im trying to find the 5 subgroups of this galois group. Im thinking itll be comprised of the identity, and isomorphisms, phi, that send each basis element(2^.5, 3^.5, 6^.5) to their conjugates and such

ive got 4 elements so far, one that sends each element to their conjugate and the identity.

not sure how to make these elements into 5 subgroups

we didnt even talk about what H is

[thorny slate]

the goup is K = Z2 x Z2

the subgroups are all, left, right, diagonal, null

where the generators are rt(2) -> -rt(2) and rt(3) -> -rt(3) respectively

[timber bay]

what do you mean theyre all left, right, diagonal null?

is this just another way of thinking of it?

We havent used Z2 notation at all in this class

[thorny slate]

those are the 5 groups

[timber bay]

oh so like diagonal sends rt 2 to rt 3 or something?

im not familiar with the notation

[thorny slate]

that's not an element

diagonal is generated by (1,1)

sends sqrt(2) to -sqrt(2)

and sqrt(3) to -sqrt(3)

[timber bay]

sends rt 6 to -rt 6

[thorny slate]

no

[timber bay]

is rt6 not a basis element?

[thorny slate]

f (rt(6)) = f(rt(2) rt3)) = f(rt(2)) f(rt(3)) = rt(6)

[timber bay]

oh yeah of course

what is a left or right subgroup

[thorny slate]

Z2 means Z/2Z

[timber bay]

yeah

[thorny slate]

the galois group here is Z2 x Z2

the left one is Z2 x 0

right is 0 x Z2

[timber bay]

im not exactly sure what that means

[thorny slate]

generated by (1,0) and (0,1) respectively

[timber bay]

none of this notation is familiar to me

[thorny slate]

you havent seen products of groups?

[timber bay]

i think a little

but not in this context at all

if i were to see other examples i m sure i would be familiar with them

[thorny slate]

the galois group is the product of Z2 generated by rt(2) -> -rt(2) and Z2 generated by rt(3) -> -rt(3)

this isn't obvious and needs proof

[timber bay]

so my elements of the groups are going to be different isom's that send elements to their conjugates and such?

[thorny slate]

yeah

the automorphisms form a group under conjugation

[timber bay]

so the only differences between the groups will be what they do to the basis elements rt3 and rt2?

[timber bay]

i only saw 4 groups still if that is the case

[thorny slate]

which one are you missing?

probably the full group

which contains both rt(2) -> -rt(2) and rt(3) -> -rt(3) separately

[timber bay]

ohh. i was just confusing myself i think.

well

itll be something like this

sorry if thats hard to read

does my class seem to be fairly difficult?

i struggle with this homework every week. I'm not sure how to catch up./

i feel like everyone else is getting it better than I do.

but that could be imposter syndrome

[somber bramble]

i struggle with this homework every week
that’s normal

homework isn’t supposed to be easy

it’s okay not to be able to do all of it, but try hard and spend a lot of time on it

a few hours per class and week at least

let’s say at least as much time as you spend in lectures

[thorny slate]

who knows

it seems you are struggling to connect concepts

but you get things after working through them for a while

so keep working at it

[timber bay]

thanks

what does F linear mean in the terms of extension fields? I'm not finding a good definition

[thorny slate]

a map g is F-linear if whenever a is in F then g(ab) = a g(b)

extension fields of a field F are F-algebras

[timber bay]

hmm

sounds like an isomorphism property but fixing a

fixing F

[thorny slate]

yes, this is equivalent

once you pick a copy of F

[timber bay]

so im having trouble finding the splitting field of x^5-1 in Q. The poly factors into (x-1)(x^4+x^3+x^2+x+1) which factors into -1/4(x-1)(-2x^2+rt5 x - x-2)(2x^2+rt5 x +x+2). From this, there are no other clean factors, really. the roots of those two polynomials involve like i*rt(10-2rt(5)) and im not sure how to go about finding the elemets so that E(alpha) is a splitting field.

i know i*rt5 is likely to be on there, which has degree 2

[thorny slate]

write out the roots

and the splitting field is just Q(a1, a2, a3, a4)

where the ai are the roots

there's gonna be some redundancy and some simplification

you don't know this now - but this extension is cyclic

so it will be of the form Q(g) for some g

you could try to find this element

[timber bay]

all the roots are of the form 1, or (-1)^(4/5) or (-1)^(2/5) or something

[thorny slate]

dont write them like that

but yes

write them in terms of square roots and so on

[timber bay]

even when the roots is something nasty like this?

the splitting field would just be rt5, i*rt(10-2rt5)

it feels wrong saying that when the stuff inside the square root is rational

[timber bay]

wait so the degree of the splitting field divides the splitting polynomial's power factorial? or just the power?

like the degree of the splitting field of x^5 - 1 divides 5 or 5!?

[thorny slate]

the factorial

[timber bay]

okay

[thorny slate]

when its irreducible

this one is not irreducible

it splits as 1 and 4

and the degree 4 one is irreducible

which tells you the degree d satisfies 4 | d | 4!

[timber bay]

so it divides 24

[thorny slate]

yeah

[timber bay]

i got 8 so that is good

[thorny slate]

for this extension you can find a single element that generates it

[timber bay]

oh you can?

i was getting 2

[thorny slate]

i just know this because reasons

in general it's hard to know

but I know this extension is cyclic

note that (i*rt(10-2rt5))^2 = - (10 + 2 rt(5))

so you can get rt(5) from it

[timber bay]

oh okay so that's just our generating element then

[thorny slate]

yeah

in general it might not be in your set

like

[timber bay]

yeah

[thorny slate]

it can be that you have F(a,b)

[timber bay]

ive seen examples

[thorny slate]

and there's some g

which isn't either of those

with F(a,b) = F(g)

[timber bay]

2 hw's ago

[thorny slate]

nice

I know it's cyclic because finite separable extensions are

in particular splitting fields of separable polys

wait i dont mean cyclic

I mean principal

[timber bay]

oh

is this all stuff ill get into later?

likely

[thorny slate]

sure

yeah

but this in particular is also cyclic

cuz extensions by p roots of unity are cyclic

[timber bay]

oh I think ive seen stuff like that

but didnt understand it

[thorny slate]

youll likely see some more

but maybe not a lot

[timber bay]

im just getting into separable extensions

trying to prove that if a, b are separable, then a+b, a-b, ab, and a/b are

man the index on my book sucks

i cant find anything useful

So with trying to prove that if a, b are separable, then ab is separable, isn't this incorrect? couldn't one of the roots of a be a root of b, and then ab would have multiplicity of one of the roots being 2?

[thorny slate]

wat

are a,b elements or polys?

[timber bay]

theyre elements of the closure of F

F a field

[thorny slate]

right

so what's "a root of b"?

[timber bay]

when they say that a and b are separable doesnt that imply that the polynomials a and b have all their roots multiplicity 1?

[thorny slate]

you just said they are elements

not polynomials

you mean their minimal polys

[timber bay]

well theyre separable. so doesnt that mean theyre polynomials?

[thorny slate]

separable elements are roots of a separable poly

[timber bay]

this is all very new to me. i missed class and we went over separable elements

oh okay

[timber bay]

anybody know any field extensions where {E:F} = 1 and [E:F] = n for n>0

[thorny slate]

whats {E:F}

[timber bay]

index

number of automorphisms

ooh

i was expecting it to be something ^n

[thorny slate]

galois extensions are exactly the ones where {} = []

[timber bay]

yeah

[thorny slate]

and these are normal and separable

so for a counterexample you want purely inseparable extensions

[timber bay]

wait what is F_p represent

[thorny slate]

field of p elements

[timber bay]

oh okay

so like Z/pZ?

[thorny slate]

yes

wait I don't mean F_p^n

I mean adding pth roots

hang on what do I mean

yeah I mean pth roots of new elements

so like

[F_p(t) : F_p(t^p)]

my bad

the other one is separable

[timber bay]

oh okay

it wouldnt be [F_p(t^p) : F_p(t)] right?

[thorny slate]

no

[timber bay]

okay

[thorny slate]

F_p(t) is bigger

[timber bay]

i think i see why

yeah

[solemn yew]

Hi can anyone tell me what G:H means in terms of groups? Here H is a subgroup of G. I was reading about Lagrange's Theorem and one of the corollaries was: If G is a finite group and H is a subgroup of G, then |G:H| =|G|/|H|. I'm not sure what G:H is here.

Please tag me if you answer, thanks!

[thorny slate]

cosets

@solemn yew

[solemn yew]

G:H is supposed to represent all cosets generated by H?

[thorny slate]

yeah

[solemn yew]

oh wow, that makes a lot of sense

thanks a lot!

[thorny slate]

👍

[viscid canyon]

usually G:H is not an object

"|G:H|" is a piece of notation that's meant to evoke the feeling of the size of a ratio but specifically avoid the quotient "G/H" which comes with the explicit restriction that H needs to be a normal subgroup for the quotient to be a group

when dealing with field extensions instead of groups, say L over K, the notation [L:K] refers to the dimension of L as a K-vector space, which is not even a ratio of the sizes of the fields anymore but a comparison of their relative size using the appropriate size-measuring technology

[thorny slate]

it's understood to mean cosets

but the notation is not very standard

[viscid canyon]

i've seen G/H to refer to coset spaces but never G:H

[thorny slate]

it's hardly used as a set

[viscid canyon]

[ - : - ] has always been atomic to me

[thorny slate]

sometimes people want to avoid G/H since it's not a group quotient

yeah I dunno it's weird

most people don't need notation for cosets

while [G:H] is very common

[sweet pagoda]

(4 – 5)² = (6 – 5)²

until here is correct

but 4-5 does not equal 6-5

[chilly ocean]

yes it does

if u thik about it

into deep thinking

[worthy geode]

Nope.

As sqrt(x²)=|x|

[blissful yarrow]

I'm trying to show that the left and right cosets of a kernel are equal

Let $f:G \to H$ be a homomorphism and let $K = \ker(f)$.
We know that $K$ is a subgroup of $G$.
For some $g$ in $G$ and some $k$ in $K$, consider $gk$ in the left coset $gK$.
We have
\begin{align*}
f(gk) = f(g)f(k) = f(g)e_H = f(g).
\end{align*}
Observe that this is the same if we consider the right coset:
\begin{align*}
f(kg) = f(k)f(g) = e_H f(g) = f(g).
\end{align*}
Since $g$ and $k$ are arbitrary, both $gK$ and $Kg$ are the set of all elements that map to $f(g)$.
Then $gK = Kg$ for all $g$.

[cloud walrus]

LiberaVeritas:

[blissful yarrow]

But couldn't there be some a not in gK or Kg such that f(a) = f(g)?

[sour plume]

Assume that a has the property that f(a) = f(g). Then f(a)f(g)^(-1) = e, and by homomorphism properties this means f(a * g^(-1)) = e, so a*g^(-1) is in the kernel of f

but that means that a is in gK

[blissful yarrow]

Ah thanks

Is there an easier way to do this?

[sour plume]

Your proof you mean? I think if you use the idea that I wrote down, that might help

[blissful yarrow]

I mean proving that the cosets of a kernel are equal

[sour plume]

Yeah, I know

Mhmmm let's see

I mean, it's basically the same way you usually prove that two sets are equal if you don't have a super direct way

Or actually let me think

it's 3AM almost

[blissful yarrow]

Oh wow

Europe?

[sour plume]

Yeah, Germany

Oki! So, you'd like to prove that gK = Kg are equal. You usually show equalities of sets by showing that the one set is a subset of the other one; so here, that an element ga in gK is also in Kg and vice versa

So you'll want that there exists some x in the kernel of f with ga = xg

That should show you what x has to be, and then you just have to show that it's simultaneously in the kernel

That's a fairly simple way to do this I think

(I edited some statements I did initially so it might look different now)

[blissful yarrow]

Oh so it ends up using conjugation after all, but now I see how to do it more directly than what I first had in mind

Danke

[sour plume]

Kein Problemi

[worldly robin]

1 + 1

[silent plover]

As a physics that is equivalent to e

[chilly ocean]

hi, there's a 3 book volume of books known as "problems in mathematical analysis", it has quite nice and challenging problems for undergrad analysis. I was wondering if there was an equivalent problem book for algebra? or at least a book with nice exercises

[somber bramble]

What are some tricks to show that (algebraic number) is not an element of (finite extension of ℚ), e.g. √3 is not in ℚ(√2) or ³√2 is not in ℚ(exp(2πi/3))

(these are two specific problems that came up on my algebra homework, I could do the rest of the exercises under these assumptions, but couldn’t show them)

(but having some general bag of tricks would be useful I reckon)

also, I forgot my notes at home and need a fact check on my memory:
Let E be a finite extension of F. Then $|\mathrm{Aut}(E/F)| \geq \dim_F(E)$, correct? (in particular, not the other way around)

[cloud walrus]

Sascha Baer:

[somber bramble]

I know it was one of th etwo and this is the one I’d need to finish this exercise :P

it looks wrong tho

[sharp sonnet]

your inequality should be the other way around

at least for simple, algebraic field extensions, not even sure if it holds in general

For your first question, you can write down elements of finite field extensions explicitly

i.e. an element of ℚ(√2) is of the form a+b√2

with a,b rational numbers

you can show that √3 can't be written this way

[bleak abyss]

One thing which helps is to show that when you adjoin each of them to Q, you don't get an iso/one isn't a subfield of the other

Well I guess the process of actually doing that often boils down to what you guys are already doing

It's tricky to avoid circular reasoning in Galois theory lmao since you wanna use the later theorems

Sometimes you can use linear algebra in the easy cases

e.g. cube root of 2 not in Q(exp(2pi i/3)) because the latter is degree 2 (solves x^2 + x + 1) and the former lives in a degree 3 extension

But yeah you often have to work directly with algebraic numbers to infer the shinier facts

[somber bramble]

yea, I figured it should be the other way round, which is annoying because it would be real nice if it was that way round

you can show that √3 can't be written this way
yea but that’s the point I’m stuck on

how does one show such a thing

we’ve never proven an explicit theorem like “for n, m coprime, √n is not a + b√m for any rational a,b”

even though I would strongly suspect it is true

[solar wyvern]

try proving it

[somber bramble]

I don’t wanna :/

I’ve ended up deep inside “I don’t enjoy this part of mathematics” territory

algebra is weird like that

group theory is awesome, but this sorta stuff is definitely not my thing

(for reference: I don’t eat corn)

[solar wyvern]

eat ur veggies.

[somber bramble]

I do. but not corn

corn’s too sweet

I don’t like it

[solar wyvern]

i bet u don't eat fruits.

[somber bramble]

I do, but pretty much only in winter cause citrus fruits are great and that’s when they’re in season

the occasional apple is nice I guess

but I pretty much have to be reminded that that would be a nice idea :P

[chilly ocean]

Focus is on point👌 💦

[untold sapphire]

@somber bramble you can show that they have distinct minimum polynomials I think, and so cannot be conjugate

[somber bramble]

conjugate here meaning? generating the same field extension?

[untold sapphire]

as in two elements are conjugate <==> they are roots of some min polynomial, which generates the field extension

since both of them will be roots of polynomials of order 2, there are almost no options

i.e. the conjugate of sqrt(3) is -sqrt(3)

this won't work for higher roots, but for elements which are square roots it will I think

I also hated trying to prove that there isn't some element that generates both sqrt(3) and sqrt(2), that's the brute force approach

Have you done Galois Groups yet?

[somber bramble]

so uh, I don't know how to answer this question because my prof calls what some authors call the galois group the automorphism group of field extensions, and reserves "galois group" for a yet to be defined concept

i'm in week two of a course on galois theory

[untold sapphire]

The Galois group of L/K is the group of automorphisms which fix your base field K, or are K-linear if you prefer

[somber bramble]

yea we’re not using that terminology

my prof reserves “galois group” for only specific ones of those

[untold sapphire]

Specific ones of what?

[somber bramble]

we haven’t defined it yet though

automorphism groups

[bleak abyss]

Where L is a Galois extension of K, presumably

[untold sapphire]

Yes, that's what I said

They are only automorphisms that fix the base field

[somber bramble]

yea that’s what the assistant said, but we haven’t defined what a galois extension is yet

we’ve defined the automorphism group for general field extensions

[bleak abyss]

Well, what I mean is more like, if L = Q(\sqrt[3]{2}), then Aut(L/K) is trivial

[untold sapphire]

Anyway, your question becomes much easier in that language

[bleak abyss]

But my NT professor would call that the automorphism group, not the Galois group

[somber bramble]

yea but we’re developing that language

we barely have any tools yet

only four hours in

in terms of lecture time

[untold sapphire]

Sure, I'm just saying don't worry too much

[somber bramble]

ah, in that case glad to hear that :P

[untold sapphire]

@bleak abyss K being Q you mean?

[bleak abyss]

Yeah

[untold sapphire]

Yeah I wouldn't call that a Galois group

[wheat loom]

Can anyone take a second to explain 3rd dimension eigenvectors?

[fickle brook]

explain what?

[oblique river]

(That question probably belongs in #linear-algebra )

[thorny slate]

his question was surprisingly about the eigenfunctions of the schrodinger equation

[chilly ocean]

Sturm Louiville Linear operators

)))

[scenic harness]

The orbit stabilizer theorem is just an application of the first isomorphism theorem right?

[somber bramble]

as far as I can tell, we did not use it at all in our proof
additionally, while it’s a similar statement, you’re dealing with group actions here, so you don’t have a group hom

but just a function

(in particular, orbit stabilizer gives you a bijection but not a hom or anything between G/Stab(t) and Orb(t))

[scenic harness]

Actions are homs though.

For a certain x
We have
f:G->Bij(Gx).
Ker f = sth complicated
I guess so, thank you!

[viscid canyon]

viewing an action of G on X as a hom a : G -> Sym(X), first iso tells us G/ker(a) is isomorphic to a(G) = { p in Sym(X) : p = a(g) for some g in G }

this tells you that faithful actions are morally deserving of the name "faithful", because they have no redundancy

for you to learn that G/Stab(x) was isomorphic to anything through first iso you wold need to find a hom f : G -> something such that ker f = Stab(x)

and that something would have to be a group

Orb(x) does not have a group structure in general, because it is merely a subset of the G-set X

so there is no reason to expect that anything containing Orb(x) could be a natural codomain to this hypothetical hom that kills Stab(x)

[somber bramble]

the issue is ofr course that Stab(x) isn’t necessarily normal, so G/Stab(x) isn’t always a group

and therefore not isomorphic to anything

[scenic harness]

That's a good way to think about it.

Also besides the usual understanding of faithful i like to understand that anything that is not injective is not faithful. The reason is if so y=f(x1)=f(x2). And functions are just a set of pairs (x,y). The problem is that if f isn't faithful then there's a y that has two partners x1 and x2. That is pretty much the definition of unfaithful. And injective function is one in which every one in that is in a couple only had a single partner.

[viscid canyon]

yeah the assumed normality was hidden in the fact that i stipulated ker f = Stab(x) which i could've made more clear

imo the fact that there isn't even a good candidate for the codomain is more fundamental than that you can't make it a kernel

[timber bay]

so im given a field extension generated by 2 elements. if im trying to find a primitive element, would it generally work to add them? or multiply? is there a technique to this?

[viscid canyon]

well first of all your extension has to be simple

e.g. finite degree and separable

if your simple extension is written F(a,b)/F and the base field F is infinite, then you can use the fact that at most finitely many F-linear combinations of a and b will fail to be primitive elements

aka just do trial and error and eventually you will succeed

@timber bay

[timber bay]

it is simple because it's extended by 2 elements.

[thorny slate]

simple means 1 element

[timber bay]

well what does it mean if it's extended by a finite number of elements

I know this is a term

[thorny slate]

uhhh finite?

if they're algebraic

[timber bay]

hmm

[viscid canyon]

a and b were both transcendental then the extension would not be simple

extreme example but hey

[thorny slate]

anyway tubular gave the precise answer: a + cb for some c will work

if your field is infinite

[timber bay]

yeah I'm thinking of the wrong thing.

[viscid canyon]

you can upper bound the number c's you have to try if you know the degrees of a and b

but your first or second try is likely to work

[somber bramble]

Can someone help me here? I thought I now knew how to procede but I’m still stuck.

So, knowlegde I have:
•I’ve found the roots, they’re ³√2 times the third roots of unity.
•I know that Aut(E/ℚ) is isomorphic to a subgroup of S₃ (because three roots)
•I know that deg(f) divides |Aut(E/ℚ)|, so it’s isomorphic to either S₃ or a subgroup of size 3
•I also know that |Aut(E/ℚ)| ≤ dim_ℚ(E), but this doesn’t seem useful

so I have to somehow show now that |Aut(E/ℚ)| can’t hvae order 3

I’m not really sure how the heck to procede, from the hint I can show that dim(E) = 6, but I don’t see how that helps

I may be missing another important theorem, but I feel like the assistant would’ve told us if there was another one we should be aware of

[thorny slate]

you need to find an element of order 2

there's a very quick way to check this, actually, I don't know if you're familiar with it

but the discriminant of the polynomial is important

hey

the hint says so

I didn't even look at it

[somber bramble]

I don’t really understand how the hint helps, it just tells me how to find the dimension of E

[thorny slate]

all you need to show is that Q(e^(2pi/3)) is not E

and then you win

[somber bramble]

how does that help?

[thorny slate]

because that's contained in E

and contains Q

[somber bramble]

yea, but how does that tell me anything about the size of the automorphism group?

I don’t need the dimension of E, I need the size of aut

[thorny slate]

how much do you know about galois extesions right now?

[somber bramble]

four hours of classes

total

basically nothing

[thorny slate]

oh

[somber bramble]

this is exercise sheet 2

[thorny slate]

do you know that Gal(F/K) <= [F:K]?

with equality if the extension is galois?

[somber bramble]

first bit yes, we haven’t defined galois extensions

[thorny slate]

okay so you have to be more tricky

find an element of order 2

in Gal(E/Q)

[somber bramble]

mhm, I guess that would work

[thorny slate]

do it by using the intermediate extension

[somber bramble]

so basically I have to show that swapping two roots is an automorphism, then I’m done

[thorny slate]

which is of degree 3

because if you use this intermediate extension, then the other one will be degree 2

[somber bramble]

because I already know that the size of aut is ≥ 3

[thorny slate]

and it will show you how to build your automorphism

[somber bramble]

well I know the automorphism has to permute the roots

[thorny slate]

yeah you can just do it directly like that

[somber bramble]

so I can just work with them

[thorny slate]

there's a special automorphism of order 2 that you can use

a very well known one

[somber bramble]

conjugation I presume

[thorny slate]

yeah

show that it permutes 2 of the roots

and fixes Q

so it's part of Gal(E/Q)

[somber bramble]

already done that, essentially

[thorny slate]

oh

then you're done

arent you

[somber bramble]

yea, just have to write it out

yep, I think so

[thorny slate]

ok cool

[somber bramble]

just have to write it all out

I think the hint makes sense given more theory that we haven’t had

but with the state we’re at it’s misleading

[thorny slate]

its basically

1. the order is 3 or 6
2. conjugation is an order 2 element so it's 6

[somber bramble]

yea

[thorny slate]

yeah the hint makes sense because the order of the group is the same as the order of the extension

in the case of galois extensions, which this is

you can identify subgroups of Gal with subextensions

as you'll soon see

[somber bramble]

well I guess I do have to show that conjugation is indeed a field automorphism of E, for which it would be hella useful to actually have constructed E directly

(I know what it ought to be, namely ℚ(³√2, ζ), where ζ is the third root of unity listed in the hint)

(in which case it becomes obvious)

which pretty much amounts to checking that if z is in E, so is its conjugate

since i know it’s a field isom on ℂ

[thorny slate]

if you wanna get ζ explicitly

note that your roots are

³√2, ζ ³√2 , ...

divide the second by the first

and you get it

but you can just show that conj( ζ ³√2 ) = ζ^2 ³√2

and you know that this fixes Q

so it's an element of Gal

you can also just give an explicit presentation for your group

³√2 -> ζ ³√2 is the other generator

and see whether it commutes with conj

and that they form S3 and not Z3 x Z2

[timber bay]

correct me if im wrong. If I'm trying to determine if an element is a primitive element would I find the minimal polynomial of that element and see if the first two elements are roots of that poly?

[thorny slate]

no

you can do one of two things

a) show that both elements a,b are in k(c)

b) show that the minpoly of c has the degree of k(a,b)/k

a,b have no reason to be roots of minpoly(c)

in fact they never will

since their degree is smaller

[timber bay]

so if i show that c can be produced by operating on a, b, that's how I'd show a,b in K(c)?

[thorny slate]

opposite

you want K(a,b) in K(c)

so you have to produce a,b with c

you will always start with some element c in K(a,b)

[timber bay]

oh yeah that makes sense

thats kind of what i have done before

are you an undergrad @thorny slate

[thorny slate]

na

im a first year

[timber bay]

nice

pure math?

[thorny slate]

yeah

interested in AG

[timber bay]

what are you thinking?

ag?

[thorny slate]

algebraic geometry

[timber bay]

ohh

well you seem pretty good at algebra

[thorny slate]

yeah I better be lol

you need a lot of algebra in ag

[timber bay]

im thinking about doing a number theory independent study next year

i dont think algebra is my thing

[thorny slate]

you need a lot of algebra for number theory xd

unless you analytic number theory I guess?

[timber bay]

and my school doesnt offer a number theory class

yeah thats what i was thinking lol

[thorny slate]

algebra is pretty important tho

just keep at it

the first time I learned galois theory I was dying

[timber bay]

yeah

i really enjoy it. i think the stuff's really cool.

[thorny slate]

for sure

it's really fun

[timber bay]

are you wanting to teach?

[thorny slate]

i mean it's part of the job

I enjoy it

[timber bay]

i mean as a career

[thorny slate]

I'm not a TA this year

cuz I got a fellowship

yeah it's part of the job

as a mathematician

you usually have to teach

[timber bay]

there is research mathematics

and lots of stuff really

[thorny slate]

research mathematicians usually have to teach

[timber bay]

physics

[thorny slate]

their main focus is research but you still have to put some hours in

[timber bay]

"usually"

[thorny slate]

teaching is part of the profession

usually means almost always

unless you're already tenured

or something

good teaching helps the tenure case

[timber bay]

i think i'd like to teach primarily

not positive what level

i'd like to bring abstract concepts in math to a fairly simple level.

[thorny slate]

nice

I mean you can focus on teaching if that's your thing

a lot of teaching universities that don't do a lot of research focus on that

and it's less hard to get a job there

[timber bay]

nice

id like to do research for sure

geometry seems cool. ive been doing geometry research for over a year now

[thorny slate]

all math is important

I think

it's all very connected

[timber bay]

i have this theory that no math is important

im about to make a breakthrough, so look for me in the news

so i am trying to show that (rt(5)+rt(2)) is a primitive element of Q(rt(2),rt(5)). i found that (rt(5)+rt(2))^3 = 17rt(2) + 11rt(5) so im guessing by that, it is a primitive element

[bleak abyss]

all math is important
set theory

Jan

[thorny slate]

@timber bay yeah

cuz then you have rt(5)+rt(2) and 17rt(2) + 11rt(5)

with easy linear aligebra you can get rt(5) and rt(2)

[oblique river]

@timber bay sorry, I'm forgetting, but how much galois theory do you know? do you know the fundamental theorem of galois theory yet?

because there is a "pure thought" (i.e. no computation required) method to show that rt(2) + rt(5) is a primitive element

if you know some galois theory

I guess in any case I can record the argument here and when you get back you can read it:

basically, you know Q(rt(2), rt(5)) is Galois over Q and the Galois group is C2 x C2 where the elements are all rt(2) --> ±rt(2) and rt(5) --> ±rt(5)

clearly the only one of those automorphisms that fixes rt(2) + rt(5) is the trivial one (that fixes both rt(2) and rt(5))

and thus by fundamental theorem of galois theory, rt(2) + rt(5) can't lie in any proper subfield of Q(rt(2), rt(5)) and thus is a generator for that field

[timber bay]

I just learned the fundamental theorem today actually

are you saying there's no primitive element? I'm finding one that seems to work.

[oblique river]

what? no I'm saying that rt(2) + rt(5) is a primitive element

@timber bay the general idea is the following: Suppose L/K is galois and let a be an element of L

then K(a) is some subfield of L (and a is a primitive element of L/K exactly when L = K(a))

by the fundamental theorem of galois theory, we can determine what K(a) is by looking at which automorphisms in Gal(L/K) fix that field

which we can check by looking at which automorphisms fix a

if the only automorphism in Gal(L/K) that fixes a is the identity

then you know that only the identity fixes K(a)

so by fundamental theorem

K(a) = L

and thus a is a primitive element

[oblique river]

(@ me if you wanna ask something, i'm gonna mute stuff for a while)

[timber bay]

sorry @oblique river fell asleep. i see what youre saying. i misintepreted something. thanks!

is there a method to showing that a polynomial is irred in a certain field? maybe looking at what it factors into and making sure none of those factors have roots in the field?

[timber bay]

wait is this necessarily true?

i know that if E is a splitting field that's finite and separable then the size of the galois group is equal to the degree

[timber bay]

will the 8 elements in this just be the basis {1, alpha, ...alpha^7}?

oh wait

{0,1,a,a^2,a+a^2, 1+a,1+a^2, 1+a+a^2} that's our basis

[thorny slate]

uhhh

that's not true is it

[Q(cuberoot(2)):Q] = 3

but Gal is order 1

{1, alpha, ...alpha^7} is not a basis

alpha^3 = -\alpha - 1

and that's not a basis, that's the whole list

of 8 elements

the basis is of size 3

as you are (Z/2Z)^3

[timber bay]

damn i just did a 64 element multiplication table and its wrong

i said a^3 = a+1 on accident

[thorny slate]

The basis is 1, a, a^2

You dont need to do a table

[timber bay]

yeah i get what youre sayinng

[oak kiln]

ima take abstract in the summver

[timber bay]

im supposed to do a multiplication table for the question

[oak kiln]

and good recommended books

[timber bay]

it didnt use basis' i was just thinking of the wrong terms

that sounds difficult @oak kiln

[oak kiln]

no good books

[timber bay]

dummit and foote

[thorny slate]

Dammit and fooke

[timber bay]

Drumstick and Fugue

[timber bay]

So ive made this multiplication table, and I'm supposed to find roots of f(x) in terms of alpha.

and im not exactly sure what that means

id think the roots are the elements that add up to 0 but that doesnt get me far

[thorny slate]

why would you think that

[oblique river]

could you explain why "because finite fields are perfect, F(a) is a splitting field for f" is true?

i think i'm just being dumb

but I don't see how those things are related

I happen to know (and can prove independently) that all extensions of finite fields are in fact splitting fields

but I don't see the claim of why it follows from being perfect

[thorny slate]

yeah I see that it's separable but I don't have an easy reason to say it's normal

anyway perfection in char p is related to the automorphism x -> x^p

so they want you to consider applying that to a

and seeing what happens

[oblique river]

yeah

ok maybe I see the relation now

[thorny slate]

because of how they're all roots of x^q - x?

it's almost equivalent to the classification of finite fields I guess

[oblique river]

like it's always true that if a is a root then a^p will be as well, that doesnt require the perfect hypothesis, but maybe yoiu want to guarantee that a^p is not equal to a, or a^(p^2) is not equal to a^p or something

but that's just saying that a is not in F_p

yeah im not entirely sure

[timber bay]

@thorny slate i thought that because it says to express my answer in terms of alpha, really. i could maybe see it being like alpha^2 or something

[thorny slate]

so try

[timber bay]

im not sure how to try that

thats my problem

[thorny slate]

plug it in?

[timber bay]

plug it into f(x)?

[thorny slate]

yes

[timber bay]

ohh

so its going to be something that when i plug it in, im left with 0, and its likely one of those 8 elements?

im guessing

ill try it

[thorny slate]

yes it's one of the 8 elements, but the automorphism suggest you should try powers of a

of the form a^(2^k)

[timber bay]

is there an upper bound on k?

3?

[thorny slate]

yeah 3

cuz a^8 = a

because the multiplicative group has 7 elements

[timber bay]

of course k = 2,1 didnt work

[thorny slate]

what are you talking about

they do

[timber bay]

yeah there has to be 2 roots

my calculation said no

ill try again

[thorny slate]

do it here

it's like 3 lines

[timber bay]

a^6 +a^2 + 1

(-a-1)^2 +a^2+1

a^2+2a+1+a^2+1

=/=0

[thorny slate]

it is 0

you're in characteristic 2

[timber bay]

what exactly does that imply?

[thorny slate]

that 2 = 0

so a + a = 0

[timber bay]

yeah

[thorny slate]

a^2+2a+1+a^2+1 = 2(a^2 + a + 1) = 0

[timber bay]

ah

man

ok lemme try 2

k = 2

ok yeah they both work

and the automorphisms are just going to send these elements to each other?

[thorny slate]

yeah

a -> a^2 -> a^4 -> a^8 = a

[timber bay]

so three elements also because the minpoly has degree 3? or would that only work if its a different kind of field

wait 4 considering the id?

it says there are two roots. would this be a mistake, since a^8 is also a root

as well as a^2, a^4

[thorny slate]

a^8 is a

so that's your 3 roots

a, a^2, a^4

[timber bay]

ohh i see

thanks for all the help

this one's been alright

[thorny slate]

👍

[somber bramble]

"Does anyone here know how this proof goes in Rotman's book? I'm sure he screws it up one way or another but it might be good to re-read it" - Algebra prof

he really doesn't like Rotman

[thorny slate]

Lol

The proofs are usually fine

I dont remember errors

[somber bramble]

there was one in the proof of the infinitude of primes iirc

which is just embarassing

[thorny slate]

lmfao

[somber bramble]

(idr if it was that one or the uniqueness of prime factorization)

(one of those two)

[thorny slate]

ok yeah I remember a mistake in one book

[somber bramble]

(in Modern Advanced Algebra)

[thorny slate]

in some elementary shit before the actual content

he does that stuff fast

so there are errors

but in the actual content there aren't

[somber bramble]

I’m not a huge fan of the book anyway, but then I hate most math books

math book authors seem to assume people have perfect memory or enjoy flipping back to look what “theorem 2.4.1” was every ten seconds

like, if you’re gonna be referring to a theorem a lot, at least give it a name or sth

rotman in particular also refers back to exercises way too much

“proof by exercise you couldn’t/didn’t have time to solve” is not a convincing proof method, imo

[thorny slate]

solve all the exercises

[somber bramble]

lolno, I have ever so slightly too many other things on my hand

among them official problem sets, the book is just a reference

but a reference which points you to exercises for proofs is a shit reference

cause while that’s good for self-study in your own time, i foyu just need to review a proof that’s just annoying

[thorny slate]

:^)

[solar wyvern]

broke: refer to the exercise
woke: most of the material is contained in the exercises

seriously tho I'm not sure references are supposed to teach students how to prove basic stuff

[wise grove]

If you're referencing something, you should know that you'll only need the statement of a thm or lemma

[fringe nexus]

when computing the square of the determinant, are there different definitions? In this, he doesn't multiply it by a_n^(2n-2) or is he assuming a_n = 1?

[uncut girder]

Can someone help me understand this last sentence?

Im j* ⊂ Ker  δ

I don't get why this is true

Context: 0 -> A -> B -> C -> 0 is a short exact sequence of chain complexes.

[thorny slate]

diagram chasing

look at the definition of δ

[uncut girder]

I still don't get it

I can follow the definition of δ

But how is δb=0?

[thorny slate]

cuz it is

let me draw it for you

okay so

an element in Hn(B)

is an element in Bn in ker(d)

so when you pull it down using δ

recall that δ is like g-1 o d o f-1

d kills it

so it's 0

[solemn yew]

hi everyone. I'm having trouble understanding the definitions of stabalizer of a point, and orbit of a point. Related to groups, specifically cosets and lagrange's theorem.

[fickle brook]

okay so

the stabilizer of a point is a subgroup in the acting group, consisting of all the elements whose action leaves your point fixed

and the orbit of a point is a subset of the set being acted on, consisting of all the elements reachable from your point by the action of a group element

@solemn yew is there a particular problem you are stuck on rn?

[solemn yew]

No just the definitions. It's like my brain refuses to understand them.

I think I have the prerequisite knowledge to understand them

Never mind. I'll try again later with a fresh mind

Thank you

[earnest valley]

Groups can't stop acting weird 🙄 .

[somber bramble]

basically, you have a group acting on a set, i.e. every group element gives you a bijection (“symmetry”) of the set. now you pick out a special point, call it t.
then you can look at two things:
-when you act with group elements on t, which points can it land on? this is the orbit
-when you act with group elements on t, which of them do nothing to it? this is the stabilizer

as for the coset bit, lets say you have a group G and a subgroup H
then you can make a group action where H acts on G by left multiplication in G
the (left) cosets of H in G are then precisely the orbits of the elements in G

[uncut girder]

@thorny slate oh thanks

[uncut girder]

Im stuck on part a

[thorny slate]

how mcuh tech do you have?

the conjugacy classes of an abelian group are singletons

so the algebra C[G] decomposes into components of dimension 1

[uncut girder]

No tech at all

This is the first hw

Of representation theory of finite groups

[thorny slate]

well then uhh

G maps into GL(V)

since G is abelian it maps into GL(V)ab

which is k*

represented by scalar matrices

or something like that

or like

assume some g doesn't act as a scalar

and reach a contradiction

as in, you may assume G is cyclic in proving this

[bleak abyss]

If we're assuming complex then the point is gonna be to use that it's an irrep

Let's say g is in G and v is an eigenvector with eigenvalue lambda

I'll say t because I'm on my phone and don't wanna TeX

[thorny slate]

yeah it has have only one eigenvalue

since its irrep

[bleak abyss]

Actually not necessarily finite makes this difficult

Because matrices of finite order are diagonalizable

[chilly ocean]

I TeX on my phone all the time

[bleak abyss]

And the point is that since you're abelian you can do this simultaneously for everybody

[thorny slate]

yeah you reduce to cyclic, if it's Z/nZ it's easy, but for Z I don't see it clearly

[chilly ocean]

Too bad idk what you guys are talking about or I'd transcribe it for u

[thorny slate]

what are the irreps of Z

I mean this says it's just 1 -> c I

but how do you see that

is this even true

[bleak abyss]

Oh wait a second here's an idea, stated in a fancy way it's Schur's lemma

Which I don't think needs finiteness

The point is that any G-hom V->V is a multiple of the identity, and because G is abelian, rho(g):V->V is a G-hom

And this is easy to expand out

@uncut girder here's the idea

[thorny slate]

hol up G

[uncut girder]

V is a finite dimensional vector space over C

[thorny slate]

why is it a multiple of the identity

is it cuz it's a division ring

[bleak abyss]

No, it's just Schur's lemma, an eigenspace is a subrep

I mean maybe that translates into what you're saying but this is how I see it

[thorny slate]

so why can't you have an incomplete eigenspace

[a 1 0]
[0 a 1]
[0 0 a]

[bleak abyss]

It's a subrep and we're an irrep

[thorny slate]

what's the subrep here

[bleak abyss]

The eigenspace

Choose an eigenvector. There's an eigenspace

[uncut girder]

Oh

[bleak abyss]

That guy has to be a subrep of the irrep so it has to be the whole thing

[thorny slate]

oh are they all dim 1

that's dumb

[bleak abyss]

Wait what?

[uncut girder]

Where does commutativity enter?

[thorny slate]

elements in the centralizer of the the representation are exactly the G maps from V to itself

if you are abelian your centralizer is everything

[bleak abyss]

@uncut girder so the point is that showing its a subrep uses that we're abelian. Fix g, and let V be the eigenspace associated to a given eigenvalue t of rho(g).

[uncut girder]

Oh I have a statement I could use:
If V is an irreducible representation over C
And f:V->V is a homomorphism of representations
Then f=lambda I for some lambda in C

[thorny slate]

lmfao

[bleak abyss]

Then for any h in G and v in V, rho(g)(rho(h)v) = rho(h)(rho(g)v) = rho(h)tv = t(rho(h)v), who rho(h)v is in V, meaning V is a non-zero subrep

Oh so that's Schur's lemma, the point is that rho(g) is a G-hom for any g because G is abelian

[uncut girder]

Ah, that final equation is the commutativity

[thorny slate]

yeah

that condition says, if you have one single representation, that
r(h) r(g) = r(g) r(h)

that is

r(g-1 h g) = r(h)

so g should be in the centralizer

if you want it to work for all h

if everything commutes of course it works

[oak kiln]

jacobian carrying pty in abstract always

😎

[uncut girder]

Is the trivial representation an irreducible representation

[somber bramble]

I mean, yea. it’s in a 1-Dimensional vector space so it can’t be reducible

cause there’s no fixed subspaces that aren’t the whole thing

[solar wyvern]

so um

WTS that for $T \in \End_F V$ over an alg closed field, eigenvalues of $T$ map to eigenbalues of $g(T)$ \
$f_T$ splits since $F$ alg closed. Then for an eigenvalue-eigenvector pair $(\lambda,w_\lambda)$ such that $w_\lambda \in W_j$ is a 1D $T$-invariant summand in $V = \bigoplus W_j$,
we want to show that for $g \in F[t]$, there exists an morphism between $v_\lambda$ to $g(v)$ in the corresponding 1D $g(T)$-invariant subspace $U_j:V = \bigoplus U_j$ such that
[
g(T)(g(v_\lambda)) = g(\lambda)(g(v_\lambda))
]

Wondering if I just have to do the obvious thing and show g carries over everything nicely. Also wondering if I've lost it.

@simple valley

on a scale of 0 to 1 how much sense does this make

@chilly ocean @thorny slate

[simple valley]

sorry, the 5D sphere said no

[solar wyvern]

rup

[cloud walrus]

flimflam:

[solar wyvern]

tbh it's already all sorts of jank because g acts on EndoV, V and F

[somber bramble]

simply juding by how much stuff there is in that paragraph and the fact you didn’t just call it “eigenstuff“ like a sensible human, yes, you’ve lost it

[solar wyvern]

eigenstuff

[somber bramble]

it’s the umbrella-term for, well, eigenstuff :P

wait, waht does it even mean for eigenvalues of T to map to eigenvalues of g(T)?

also you never defined g in that paragraph, is it in reference to a previous thing?

[solar wyvern]

it's just any polynomial

my bad

[somber bramble]

also I’m not quite following the stuff you wrote but are you assuming that there’s a basis of eigenvectors cause that’s not true in general

oh, a polynomial, that’s not what I expected but I guess then stuff makes sense

and you want to show, what, that g(T) and T have the same eigenstuff?

well, can’t be the same but like

λ eigenvalue of T ⇒ g(λ) eigenvalue of g(T) perhaps?

that would sound reasonable to me, it’s certainly true for constants

yea I think that sounds like a true statement. what I’d show is this:
v eigenvector of T ⇒ v eigenvector of g(T)
then show the statement for monomials
then show the statement for linear combinations of monomials

no silliness with direct sums needed

…assuming that’s what you even wanna show

[solar wyvern]

for T diag eigenvectors just elements of those 1D subspaces tho?

[somber bramble]

not all T are diagonalizable

I would just ignore that possibility

also that’s not even a coherent sentence

[solar wyvern]

I think I probably want to use the fact that k[T] is a (End_k V)-module and a k-module

and I guess a k[t] module

[somber bramble]

if the statment is indeed what I thouhgt it is, then no, you really really don0t

because it’s like

simple

Let $g = a_n X^n + \dots + a_1 X + a_0$ and let $v$ be an eigenvector of $T$ with eigenvalue $\lambda$. Then \begin{align*}g(T)(v) &= (a_n T^n + \dots + a_1 T + a_0 I)v\ &= a_n T^n (v) + \dots + a_1 T(v) + a_0 v\ &= a_n \lambda^n v + \dots + a_1 \lambda v + a_0 v\ &= g(\lambda) v\end{align*}

missing the intermediate step that $T^k v = \lambda^k v$ which is an easy proof by induction

[cloud walrus]

Sascha Baer:

Sascha Baer:

[uncut girder]

Um

I have a basic question

[somber bramble]

if the statement you wanted to prove is anything different, then you really need to work on your way of expressing yourself

[uncut girder]

Are there nonzero complex numbers x, y
Such that x+y=-xy

[somber bramble]

uh well, you can put that into a system of equations I guess
$$a + bi + c + di = - (a + bi)(c + di)$$

[cloud walrus]

Sascha Baer:

[somber bramble]

separate by real/imaginary and see if you can find any solutions

so \begin{align*}a + c &= -ac + bd \ b + d &= -bc - ad\end{align*}

[cloud walrus]

Sascha Baer:

[somber bramble]

since there are four variables and only two equations you should probably be able to find solutions just by trial and error tbh

I’d be very surprised if there was none

there seems to be a whole curve of them

[uncut girder]

Example?

[somber bramble]

y = 1
x = - 1/2

y = i
x = -i/(i+1)

literally just plug in any value you like for y in there and you get an x that does it

and as far as I can tell, this’ll work in any field

(but ones of characteristic 2 will have no nontrivial solution)

[solar wyvern]

@chilly ocean lapse of sanity

[somber bramble]

I need to show that $\mathbb{F}_{p^m} / \mathbb{F}_p$ is a galois extension and find the galois group. I’ve managed to show that $F: x \mapsto x^p$ is an automorphism of this extension. My guess is now that the automorphism group is exactly the powers of $F$, that is ${id, F, F^2, \dots, F^{m-1}}$. To that end I need to show that all those functions are actually distinct. But I’m stuck, any hints?

[cloud walrus]

Sascha Baer:

[somber bramble]

I’ve considered trying to show that $F$ fixes only $\mathbb{F}_p$, which would guarantee the extension to be galois, but doesn’t a priori tell me that the galois group consists only of powers of $F$

[cloud walrus]

Sascha Baer:

[somber bramble]

and I also don’t really know how to do that

[thorny slate]

X^p - X has exactly p roots

[somber bramble]

thanks, that actually helped a lot

I think I can take it from there

yep, that did it, big thanks