groups-rings-fields

[solar wyvern]

so I'll think about that first I guess

@thorny slate do we know what Gal(F_p^alg/F_p) is

@oblique river

[somber bramble]

what’s the opposite of a follow-up? cause to me it is not at all obvious that that is galois, but we’ve basically not talked about infinite extensions

so excuse my ignorance if it’s something super basic

[thorny slate]

@solar wyvern by ^alg you mean alg closure?

it's generated by the limits of the frobenius

i'll let you work out what that is

[oblique river]

(usually people will talk about K^sep instead of K^alg because K^alg/K is not galois whereas K^sep/K is)

[bleak abyss]

If only the two notions coincided... that'd be perfect

[solar wyvern]

@oblique river oooh

sorry I'm still not 100% clear on separable extensions but I read up a bit and think I know what they are

[oblique river]

haven't you been asking about separable extensions for like the past week though?

[solar wyvern]

yeah I'm retarded

if k^alg is the algebraic closure of k and E is some intermediate extension then [E:F]_s is the number of ways you can extend an embedding σ:k→k^alg to σ*: E →k^alg

does that work?

@oblique river it has been a pretty bad week, and apologies for the incoherent questions

[oblique river]

if k is the base field then there should only be one embedding k --> k^alg right?

but in any case i think the definition is just "number of ebneddings E --> k^alg

[solar wyvern]

idk why but conrad's notes and lang both define it as the number of ways you can extend an embedding (of the base field to a field containing its closure) to an embedding from an intermediate field E to a field containing its closure

I think the point was that for two embeddings from k to k^alg, the set of extensions of either is in bijection, so that indeed you only have to consider the embeddings E to k^alg

[oblique river]

oh i see

makes sense

[solar wyvern]

but once that's been shown it makes sense to only consider the embeddings

[wind steeple]

It's only the numbers of K-morphisms from L to Omega where Omega is an algebraic closure of K. ([L:K]_s) such morphisms exists by extending Id_K. And we can see that the extension is separable iff [L:K]_s=[L:K]. I don't think that its fine defining it as the number of way you can extend an embedding. Because it doesn't seem natural

[solar wyvern]

If you have a separable extension E/k and k⊂F⊂E, is F/k and E/F separable?

[wind steeple]

yes

the reciprocal is true too

E needs to be algebraic

[solar wyvern]

@oblique river all embeddings from a finite field F to an (algebraic) extension are induced by applying frobenius to your coefficients, right?

[oblique river]

applying frobenius shouldnt cause you to leave your field

so im not sure what you mean

[solar wyvern]

Can't you consider elements of a finite field as a vector field over another finite field and apply frobenius to the coefficents?

[oblique river]

yes you can but how is that goign to give you an embedding

i dont understand

are you conjugating yoru embedding by frobenius?

[solar wyvern]

I don't know

can you do that?

[oblique river]

do what?

can you jsut say what you want more formally?

if E/F is an extension of finite fields

then Gal(E/F) is generated by frobenius

[solar wyvern]

if you have an embedding K/F and K⊂E, can you extend it to E/F somehow

[oblique river]

"embedding K/F"?

[solar wyvern]

a ring map F→K

[oblique river]

so you have F in K in E

so it sounds like F is already embedded into K

and K is embedded into E

so F is embedded into E

[solar wyvern]

by your embedding composed with the inclusion from K to E?

[oblique river]

yes

F --> K --> E

also if you are considering F as your base field then F is automatically embedded into any extensino of F

[solar wyvern]

maybe I'm just doing this all wrong but if F is the base field then and A,B two fields over F then isn't an embedding t: A→B a F-linear map?

@oblique river

[oblique river]

yes

[solar wyvern]

then if A,B are algebraic over F you can identify elements α∈A,β∈B with the unique monic polynomials generating ker eval_α and ker eval_β respectively?

[oblique river]

sure

but if you apply frob to the coefficients

thats basically the same as applying frob to ther oots

[solar wyvern]

and frob cyclicly permutates roots?

[oblique river]

yes

every galois group of finite fields is cyclic generated by frobenius

that's like the most important fact about galois extensions of finite fields

and everything else you need to know basically follows from that

[solar wyvern]

What about Gal(F^{alg}/F)?

[oblique river]

F^{alg} isn't a finite field

but in any case it's isomorphic to \hat{Z}

the profinite completion of Z

generated by frobenius

it's the inverse limit over n of the groups Z/nZ

I guess the other most important fact about finite fields is that there is a unique extension of degree n for all n

[solar wyvern]

for each finite field A there's a finite field B such that [B:A] = [B:A]_s = n?

[oblique river]

first of all, finite fields are perfect so every extension is separable

so there's no real need to talk about separability degree

but yes you are right

like I said, for any finite field and any positive integer there is a unique field extension with that degree

and the galois group is Z/nZ, generated by frobenius

(which is a bit stronger than just saying it's cyclic of order n. It's cyclic of order n but it also comes with a "preferred" generator)

sry can you fix that formatting

haha

[solar wyvern]

can you see it now?

[oblique river]

it still has weird italics

[solar wyvern]

huh

[oblique river]

and i'm not sure what you're asking

but in any case, yes, F_{q^n} is the degree-n extension of F_{q}

and the galois group is generated by the q-th power frobenius

[solar wyvern]

I mean you don't really have a ton of options 🤷

[oblique river]

yeah

the theory is very rigid

there's really not much going on

[solar wyvern]

even if you consider trancendental extensions of finite fields not much is different, right?

[oblique river]

i mean there is almost no reason to consider something like F(x)/F

but if you're talking about using a transcendental extension of F as your base field

i.e. E/F(x)

then everything is different

[solar wyvern]

then things happen

nonseparable extensions

[oblique river]

F(x) is no longer perfect, there can be inseparable extensions

non-cyclic extensions

yeah the theory is more like the theory over Q

in its complexity

(in fact the theories are so similar that people group them under the term "global field")

[solar wyvern]

global fields are completions of local fields or something?

[oblique river]

uhhh

the other way lol

[solar wyvern]

fugg

😦

[oblique river]

Q_p is a local field

which is a completion of Q

Q is not a completion of Q_p

[solar wyvern]

what are global fields defined as then?

[oblique river]

you're not going to be happy with this, but:

a global field is, by definition, a finite extension of Q or a finite extension of F_q(t)

[solar wyvern]

I'm okay with this

[oblique river]

that was my point earlier

they are so similar that people group them under the same umbrella

[solar wyvern]

so this is dumb but just a yes/no question but are all extensions of Q separable? even trancendental ones

[oblique river]

separability only happens in characteristic p

period

also

the word separable implies algebraic

nonalgebraic things aren't separable or inseparable

separable is only a word that applies to algebraic things

[solar wyvern]

so no one really cares about things like separability for trancendental extensions?

[oblique river]

that's like asking "so no one really cares about things like separability for covering spaces of manifolds"

it's just literally not a word that applies in the context of transcendental extensions

[solar wyvern]

tfw your professor lies to u

[oblique river]

an element alpha is separable over K if the minimal polynomial of alpha is separable

if alpha is transcendental then there is no notion of separability

because there is no minimal polynomial

i guess you could think about like

if E / F is transcendental, then there is necessarily a transcendence basis {t_i} such that E / F(t_i) is algebraic

then you could ask if that extension is separable

but that just reduces to the finite extension case

which is basically the content of this: https://en.wikipedia.org/wiki/Separable_extension#Separability_of_transcendental_extensions

[solar wyvern]

there's a thing on there which says E/F is separable iff F^{1/p} ⊗_F E is reduced

does that not work if E/F is trancendental?

[oblique river]

that's still true

but it's just equivalent to the definition i gave above

[solar wyvern]

[oblique river]

let me step back for a sec

what is your goal?

like you've been asking about separability for a while now

and imo separability is a super technical and uninteresting thing

and you should only really get involved with it if you need it for somethign else

are you trying to do transcendental algebraic geometry?

[solar wyvern]

i guess I should ask...what should I know about separability

for like a galois theory class

[oblique river]

almost nothing

a galois theory class is only going to be concerned with finite extensions (at least for 95% of the time)

and you should know that it's a technical assumption that we need for galois-ness because it ensures that we have enough automorphisms

and it makes other stuff work, but you really should just let it stay under the hood

you should know that it's never a problem in characteristic 0 or for finite fields

[solar wyvern]

is there any point where I might care about it?

[oblique river]

"are you trying to do transcendental algebraic geometry?"

if you're doing any kind of geometry over finite fields it will come up

like elliptic curves or something

but not for a course in galois theory

i'm not trying to say that separability isn't important. it's just not important for a first course in galois theory

it's super technical

and not enlightening at all

[solar wyvern]

it sorta came up in vakil so I was just curious

[oblique river]

imo it's one of those things that you should only read deeply about if you absolutely have to

[solar wyvern]

so basically once I already know basic galois theory pretty well already?

[topaz solar]

Separability isn't a key topic for intro to galois

[solar wyvern]

for some reason separability qs are a huge part of our psets tho

[oblique river]

i will admit i'm not an algebraic geometer, but i am a number theorist, and literally all i ever need to know about separability is that "it means you're not taking a pth root in characteristic p"

[topaz solar]

i mean it shows up a bit but tbh i doubt you'll see it for a good while

[oblique river]

yes darkrifts is correct

[solar wyvern]

but you're probably right that I don't need anything fancy I guess?

[oblique river]

it's really not something that you should be caring about for a first course in galois theory

[topaz solar]

I'm no algebraic geometer, but separability is largely separable from your academics for now

[oblique river]

lol true

[solar wyvern]

yeah i guess if I flunk out because of this I'll be pretty separated from academics

[topaz solar]

lmao

[oblique river]

are you taking a galois theory class right now?

[solar wyvern]

yeah

am i doomed

[oblique river]

no

[topaz solar]

I'd be doomed, but you're fine

[solar wyvern]

@oblique river I guess one reason I was interested in separability is that separable extensions have a nice characterization in that they're in bijection with the ways you can extend an embedding (and this is invariant of your initial embedding), but for inseparable you have ways of extending an embedding....up to some multiplicity weirdness (not sure this is even invariant of the embedding you choose)?

[oblique river]

yeah, for me i kinda think about that in terms of automorphisms

like if you have one embedding K --> E (all relative to some base field F)

then you can pre-compose by some element of Aut(K/F)

and so (under a normality assumption) that will give you all of the embeddings K --> E

so if you're separable, i.e. if Aut(K/F) is large enough, then you have lots of embeddings

[solar wyvern]

ummm can you elaborate on the "if Aut(K/F) is large enough" part?

[oblique river]

if K/F is normal, i.e. is a splitting field

then Aut(K/F) is in bijection with {embeddings K --> F^{alg}}

you're saying you think of separability in terms of how big the right side is

I think of it in terms of how big the left side is

[solar wyvern]

which side is which

[oblique river]

the side on the left side of the line is the left side

:)

left side = Aut(K/F) and right side = {embeddings}

which is how you were saying you think of separability

[solar wyvern]

yeah

Sorta wondered about this and hopefully there's a easy answer but for K/F normal (and separable i guess, so Galois), if you restrict the the automorphisms of K to F, will they actually all fix F?

Does $\theta \in Aut K: \theta|{F} \in Aut F \implies \theta|{F} = \id_{Aut F}$

[cloud walrus]

flimflam:

[solar wyvern]

@oblique river

[oblique river]

when you say "automorphisms of K" what do you mean?

if you're talking about K/F normal and separable, it seems like you mean something like Aut(K/F)

in which case the answer is yes, by definition

[solar wyvern]

k-linear embeddings of K into itself (which may not fix F)

where k is the base field

[oblique river]

assuming k is the prime subfield of K (i.e. Q or F_p depending on char K)

then no

if K / k is Galois, and F is a subfield of K

then K/F is also Galois. If we assume that F/k is also Galois, then the "restrict to F" map gives a homomorphism Gal(K/k) --> Gal(F/k) that is surjective

(this is part of the "extending field embeddings" discussion: any automorphism F/k can be extended to one of K/k that restricts to the automorphism of F that you started with)

that's kinda the whole point of talking about extending automorphisms, so you can prove things like "that map on galois groups is surjective"

[solar wyvern]

F/k might fail to be normal and therefore not be Galois?

[oblique river]

yes

which is why I assumed that it was galois, equivalently, that Gal(K/F) is a normal subgroup of Gal(K/k)

[solar wyvern]

(but it would be separable)

[oblique river]

of course; subextensions of separable extensions are always separable

[solar wyvern]

yep

but this isn't so for normal extensions

[oblique river]

correct

[solar wyvern]

is the compositum of normal extensions E/F,E'/F normal over the intersection E⋂E'?

[oblique river]

compositum of normal extensions is normal, period

(normal over F, that is)

so it's certainly normal over E cap E'

[solar wyvern]

What about if you have normal extensions E/F,E/F', will E/(F⋂F') be normal? sorry if these seem like gratuitous questions

[oblique river]

nvm im dumb hahahahahaha

that is normal over Q

let me try again.......

actually it might be true

I can prove it if there is some subfield k over which E is galois

[solar wyvern]

lang mentions that normal (and thus galois?) extensions don't form a distinguished class (sep, finite, fg, and alg all do), but I'm wondering if it just fails on the subextensions thing (which is the first criterion, whereas pushouts are the second)

and not pushouts (EE'/E⋂E')/pullbacks (E/(F⋂F'))

[oblique river]

i cant think of a proof or counterexample off the top of my head

i'll think about it more as i go to bed (which is going to be soon)

[solar wyvern]

thanks a lot for your help!

[oblique river]

it's kind of an odd question for me because I'm always dealing with fixed base fields

[solar wyvern]

yeah, I don't really have a good sense of things yet so sorry if this is all over the place

[oblique river]

so like I said, if E/k is Galois and F and F' contain k and are subfields of E such that E/F and E/F' are Galois

then E/(F cap F') is also Galois

but that's not really exciting at all

because of course it's true: If E / F / k and E is Galois over k then it is Galois over F

(take my F = your F cap F')

so I feel like your statement should be false in general, and I want to construct a counterexample like

take some irreducible quintic over Q such that when you add a single root, the quintic factors into a linear factor (corresponding to that root you added) and two irreducible quadratic factors

then maybe if you add in one of the roots to oen of the quadratic factors, the other quadratic factor doesn't split

[solar wyvern]

okay, I'll take a shot at that too then

[oblique river]

for irreducible quintics, there are only 5 possible galois groups

S_5, A_5, D_5, C_5, and F_20

where F_20 is some weird-ass group

[solar wyvern]

F_20?

[oblique river]

I think it might be isomorphic to a semidirect product of Z/5Z and (Z/5Z)^*

idk

[solar wyvern]

frobenius group?

[oblique river]

yeah

so out of those 5 options, I think the F_20 one is most likely to give a counterexample

[solar wyvern]

[oblique river]

same

gl

[solar wyvern]

thanks

[somber bramble]

"Let E/F be some finite field extension. Then it'd either Galois, or it's not. Let's look at a third possibility"

(it was "not galois, but contained in a galois extension")

[oblique river]

@solar wyvern i just figured it out, it's true

let k = F cap F' and let M be the normal closure of E over k

then let A = Gal(M/F) and B = Gal(M/F') as subgroups of Gal(M/k)

and consider C = Gal(M/E). By assumption, C is normal in A and normal in B, because E is galois over F and F'

thus C is normal in AB = Gal(M/k)

and so E is galois over k

(and so M = E the whole time)

also I am shamelessly assuming separability so that I can use galois theory

[solar wyvern]

i am okay with this. surprised, but okay (i was also assuming separability throughout).

[oblique river]

i still feel weird about it

i feel like maybe i made a mistake that i'm too tired to see

[solar wyvern]

I should be able to double check this right?

I will try anyways tomorrow when I'm more awake

[oblique river]

same

good night for real now

(this seems to confirm it https://math.stackexchange.com/questions/1780998/galois-extension-of-intersection-of-fields but i'm still interested in the case where you only assume normality)

[hollow peak]

X^4+X+1 is irreducible in F2 right?

and so is X^3+X+1?

is X^n+X+1 always irreducible in F2 for n>=2?

[wind steeple]

tu trouveras pas de pol irréductible explicitement comme ça :smart:

[open mist]

@key if it was reducible it would have a root yeah? What's 0^n and 1^n 😃?

[mild laurel]

@open mist that's not true. Look at the picture just above

[open mist]

lovely, back to analysis for me

[uncut girder]

What's the difference between irreducible and prime?

[mild laurel]

Irreducible means that it cant be written as the product of two non-units. Prime means that if p | ab, then p | a or p | b.

In a domain prime implies irreducible

[stone forum]

https://math.stackexchange.com/questions/1871637/irreducible-but-not-prime-element

[solar wyvern]

prime implies some sort of unique prime factorisation doesn't it?

[topaz solar]

primes are fun

[flat dagger]

Does anyone know some good examples of applications of the first isomorphism theorem?

[oblique river]

for groups or rings or vector spaces or what?

[flat dagger]

Groups

[oblique river]

basically if you ever want to prove something like "G/H is isomorphic to K", the way to do that is find a surjective homomorphism G --> K whose kernel is H

and then by the first iso theorem, the image (which is K) is iso. to G / kernel (which is G / H)

[flat dagger]

Ah, okay.

[chilly ocean]

ok

hmmm

i have a colouring problem

this is my shape

Find the number of essentially different ways this can be done when there
is an unlimited supply of pieces in each of n colour.

it can be rotated and flipped

orbit counting theorem in D4?

@covert vector would it just be 1/|D4| * sum of |fix(g)|

[covert vector]

how does D4 act on this

oh nvm

i guess so havent done orbits in a while

[chilly ocean]

fair dinkum

eugh

this is an evil subject

[uncut girder]

I'm just trying to be sane here

Let w be the 3rd root of unity

So w-bar = w^2

Now for any a, b integers,
We have (a + bw)-bar = a -b -bw, right?

-bar means complex conjugate

[woven delta]

$\overline{a+bw}=\bar{a}+\bar{b}\bar{w}$

oof

how do you get that result @uncut girder

[mild laurel]

\overline

[woven delta]

yeah

[cloud walrus]

Liquid:
Compile Error! Click the reaction for details. (You may edit your message)

[uncut girder]

Because 1 + w + w^2 = 0

[woven delta]

um

[lone shard]

that's clever

[woven delta]

are a, b real?

[lone shard]

yes; theyre integers

[woven delta]

oh okay

sure

[uncut girder]

Just wanted to check with ppl

That this is actually a typo

[woven delta]

$1+\omega+\omega^2=0\implies \omega^2=-\omega-1$

[cloud walrus]

Liquid:
Compile Error! Click the reaction for details. (You may edit your message)

[woven delta]

so yeah, you're right

[chilly ocean]

quick urgent question

how does one work out the order of a permutation

@covert vector 😮

[covert vector]

write it in disjoint cycles

then take the lcm of their lengths

[chilly ocean]

disjoint cycle

for example

( 1 3 4 6 2 5)

in S_6

it is order 6?

[covert vector]

well that has order 6

[chilly ocean]

😄

i know shidd

[covert vector]

cuz it has length 6

[chilly ocean]

and it's an odd permutation

because 1 disjoint cycle

[covert vector]

no

it is odd

but not because it's 1 cycle

[chilly ocean]

ohhh

im awful with terminology 😦

[covert vector]

you need to write it as a produce of 2-cycles

then count how many are in the product

[chilly ocean]

its odd

because 5 2-cycles

[covert vector]

yes

[chilly ocean]

this module is a very random module

covers group theory and also modulos but also proofs by induction

[solar wyvern]

ab groups are Z modules

@chilly ocean

[chilly ocean]

what

what does that mean

[solar wyvern]

fixed

[oblique river]

@solar wyvern are you trying to confuse them???

[solar wyvern]

no

[oblique river]

why talk about Z-modules then :P

[solar wyvern]

they're nice examples of modules besides vector space stuff

[oblique river]

why are you talking about modules at all though, they're obviously in a more intro class and haven't gotten to modules yet

[solar wyvern]

I'm sorry

[oblique river]

(did you read "modulos" as "modules"? hahaha)

[solar wyvern]

oooooooh

yes

[oblique river]

hahahahaha

also you're fine, there are people in this server who are really bad though

it's like "oh you're learning about group theory? let me flex on you with some words and jargon that you have never heard about before"

[chilly ocean]

Let G be any group order 17

then if g in G is not e, then why must it have order 17?

[somber bramble]

if an element has order n, then the elements {e, a, a^2, ... , a^(n-1)} form a subgroup

take it from here, @chilly ocean

[chilly ocean]

the reason give is something to do with |G|

ord(g) must divide |G|=17 ?

[somber bramble]

yes, because of the statement I just said

[chilly ocean]

oh

yeah

sorry

brainfart

🤦

order(g)=|H| ?

its a bad way of describing it

but that's sorta it

?

[fickle brook]

no

unless you say what H is

[chilly ocean]

|G|=m|H|for some integer m where m is the number of left cosets in H

because 17 is prime, H is either the trivial subgroup or G

[fickle brook]

introduce your objects properly. before you can claim ord(g) = |H|, you must say what H is.

[chilly ocean]

for g in G and x in X
orb(x)={y in X : y= g✶x}
stab(x)={g in G : g✶x=x}
fix(g)={x in X : g✶x=x}
send_x(y)={g in G : g✶x=y}

[fickle brook]

... what are you on about now

[chilly ocean]

just definitions

😛

[fickle brook]

...what do they have to do with the subject at hand?

[chilly ocean]

haha

these are all part of the same module for me

stab(x) must contain the identity element because stab(x)={g in G : g✶x=x}
and e*x=x

[fickle brook]

are you just hopping randomly from problem to problem or what

[random crag]

?

[chilly ocean]

yes

i am

[random crag]

whats the question?

-_-

[chilly ocean]

am i mentally stable in numbers and groups

😄

if y is in orb(x) , then x is in orb(y)
This is true because orb(x)={y in X : y=g✶x}

y=g✶x
x=g^-1✶y

so x is in orb(y)

because g^-1 is in G

But I do have a question now @random crag

why is the sum of |fix(g)|=sum of |stab(x)|

I can kinda see it but cant explain it formally

[fickle brook]

are G and X both meant to be finite

[chilly ocean]

yes

[fickle brook]

consider the set $F = { (g,x) \in G \times X \mid g * x = x}$

[chilly ocean]

ok

[cloud walrus]

Ann:

[chilly ocean]

oh wait, yeah I think I get it

fix(g) and stab(x) are just the set of pairs of (g,x) that produce g*x=x

right?

well one is the set of g

and the other the set of x

[fickle brook]

there are two ways to count the elements of F

[chilly ocean]

|fix(g)| and |stab(x)|

😄

you either count the x in X or the g in G

[fickle brook]

sorta.

[chilly ocean]

boy i love colouring problems

a^(p-1) == 1 (mod p)

only if a and p are coprime and p is prime

chinese remainder theorem

if x==a1 (mod m1) and x==a2 (mod m2)
where a1 and a2 are any integer and m1&m2 are coprime
then their solution is in the form (mod m1m2)

[hollow peak]

Does |G|=|G/H||H| even if H isn't normal?

[chilly ocean]

isn't that legrange's theorem?

|G|=m|H|

[hollow peak]

it is, but is |G/H|=m?

[chilly ocean]

i would assume so

well actually

idk

does |G|/|H| = m

=|G/H|

[hollow peak]

the question is does |G|/|H|=|G/H| even if H isn't normal?

[somber bramble]

yes, provided the numbers make sense (ie aren't infinite)

[hollow peak]

yes G is a finite group

ok ty

[somber bramble]

G/H is just the set of cosets

you have [G:H] many cosets, each with H elements

[hollow peak]

what's [G:H]?

[somber bramble]

the number of cosets of H in G :P

it's called the index

[hollow peak]

what's a coset? xD

[somber bramble]

...how can you know of normal subgroups and not know what a coset is?

[hollow peak]

oh okay

I do not know the name in english

[somber bramble]

oh

[hollow peak]

i just looked on wikipedia

[somber bramble]

so yea, basically you divide G into |G|/|H| many sets of size |H|

[hollow peak]

yes i was not sure if it was a partition

but it makes sense

[somber bramble]

and call the collection of those sets G/H

and if H is normal, you can give G/H a group structure

[hollow peak]

yes

thanks

[chilly ocean]

Is it possible to have two six-cycles whose composition is a cycle of
even length?

[hollow peak]

no

[chilly ocean]

why

[hollow peak]

a six cycle has signature -1

and the signature is a morphism

[chilly ocean]

sgn(ab)=sgn(a)sgn(b)

[hollow peak]

yes

[chilly ocean]

wait

but wouldnt 2 six cycles have signature -1

[fickle brook]

no

(-1)*(-1) = +1

[chilly ocean]

^

[fickle brook]

even cycles have signature -1, odd cycles have signature +1

[chilly ocean]

wait what

shi

shit

thank god for you Ann

😛

so the square of any cycle is always odd?

[hollow peak]

if by square you mean for composition then yes

[chilly ocean]

yes

😄

[fickle brook]

uh

[hollow peak]

i meant even*

not odd

(odd=impair?)

[chilly ocean]

wait

it's +1 if even

-1 if odd

@fickle brook

I'm certain of it

[fickle brook]

@hollow peak oui

[hollow peak]

a n-cycle has signature -1 if n is even

[fickle brook]

there's a bit of a discrepancy here

in that an even-length cycle is odd as a permutation

& vice versa

[chilly ocean]

permutation a
sgn(a)=1 if it is even ( made up of an even number of transpositions)

[hollow peak]

a transposition has signature -1

and a n cycle is the composition of n-1 transposition

so a n cycle has signature (-1)^(n-1)

[chilly ocean]

ok

random tangent

fermat's little theorem

a^(p-1)=1 (mod p) if p is prime and gcd(a,p)=1 (a and p are coprime)

5 always divides (n^4 -1) as long as n is a positive integer that is not a multiple of 5 because

n^4 -1 = 5m

n^4== 1 (mod 5) by fermat's little theorem

[hollow peak]

if p is prime then, all non 0 element verify a^(p-1)=1 in Z/pZ

[chilly ocean]

okie dokie

[chilly ocean]

what exactly is the subgroup S_A

rotations and reflections from O_2 such that they don't change the elements?

[sour plume]

Almost

[chilly ocean]

😮

[sour plume]

The ones that don't change the set

[chilly ocean]

the entire set

oh yeah

[sour plume]

But maybe that's what you meant

[chilly ocean]

A not a

ok

so it is a subgroup because
sg1) rot 0 is in S_A so S_A=/= empty set

hmm

idk how to show sg2 for it

but i feel it is obvious

[sour plume]

Yeah, you just use the definitions basically

You can work with equations of sets in similar ways as you would with elements

[chilly ocean]

rot(a)ref(b)=ref(a+b)

[sour plume]

But if that troubles you you can just look at single elements of the set-equation

And see how that equation translates to single elements

[chilly ocean]

wait

there are no reflections in S_A are there

[sour plume]

Sure there might be

[chilly ocean]

what sort of reflection would keep f(A)=A

[sour plume]

If A was a circle or a square for example

[chilly ocean]

i feel like im missing something here

[sour plume]

There are sets with -A = A

That are not empty

Think about what that means for a moment

[chilly ocean]

i don't get it

sorry

[sour plume]

If a set contains the negative of all elements in it

[chilly ocean]

can we use an example

I'm still a lil confuzzled

[sour plume]

Like the set consisting only of plus and minus 1

[chilly ocean]

okie

[sour plume]

That's invariant if you mirror it

[chilly ocean]

invariant?

meaning that it's the same set?

[sour plume]

It doesn't change

Ye

The elements all change

[chilly ocean]

ok

i never really thought of that lol

[sour plume]

But the set stays the same

That's how the two notions are different

You could ask that the elements don't change or that the whole set doesn't

But the former is extremely restrictive

[chilly ocean]

the former is just the identity

no?

[sour plume]

It's a bit weird to think about it for the first time tho

Yeah

If you're not just looking at the origin

[chilly ocean]

I think I understand this Q

just wanna confirm stuff

H_1 is not a subgroup because it does not have the identity element id where id(1)=1

H_2
SG1) id(1)=1 therefore id is in H_2 so H_2=/= empty set

SG2) let a(1)=1 & b(1)=1 where a,b are in H_2
ab(1)=a(b(1))=a(1)=1 so ab is in H_2

SG3) what do I do?

oh shit

erm

let a(1)=1 where a is in H_2
a^1(a(1))=a^-1(1)
a a^1(1)=id(1)=1=a^-1(1)

a^1 is in H_2

what about H_3

id^2=id so it is not the empty set

sg2 hmmmm

[somber bramble]

Assume k is a field of characteristic p such that the frobenius homomorphism φ is not surjective. If y is an element which is not in the image of φ (i.e. there is no x with x^p = y), then I claim the polynomial X^p - y is irreducible. It’s clear it does not have a root, but how can I argue that it doesn’t split otherwise?
note: I don’t know if this statement is actually true. but it’s the only idea I have for this proof

(I need to show that frobenius hom is surjective iff every poly of the form g(X^p) is reducible)

[chilly ocean]

@somber bramble you wouldn't by happen to know anything about proving sg2 doesn't work for H_3 in the image above would you?

[somber bramble]

I’m busy with my own stuff rn

[chilly ocean]

sorry

[sour plume]

I don't know what Sn is but sg2 is not true in general for this group I'm pretty sure

Or actually

Lol

No it is true but it's a little tricky

Try to show that the subgroup must be commutative

Or wait

Ah I'm so confused

[chilly ocean]

so it's clear H_3 is not a subgroup because sg2 fails , right?

but finding an example of this is a bit hard

[sour plume]

Yeah

Basically you need two 180° rotations which don't commute

Can that happen? I'm not sure right now

[chilly ocean]

rotations?

oh yeah its S_N

S_n

whoops

as long as I get like 60% on the exam I'm fine lol

here's a random one

i)true because groups must be closed

ii)true because a group must have an identity element

iii) and iv) idk how to explain

v)true because all h in H must also be in G

vi)true because legrange's theorem says |G|=m|H| . ie |H| must be a factor of |G|

vii)true because legranges theorem says |G|=m|H| where m is the number of left cosets in H, and if |G|=15 and |H|=5 then m must be 3 as 3*5=15

viii) I have no idea

ix) no idea

x) false G can be abelian if a(.)b=b(.)a

[somber bramble]

(reminder that I posted a question above)

(which immediately got drowned)

[oblique river]

@somber bramble if it was reducible, then it would have to factor. What would the constant terms have to look like?

[somber bramble]

hm, so we’d have $(X^k + \dots + a)(X^l + \dots + b)$, where $ab = -y$ and all the other products of the coefficients would have to cancel, except for the initial one

[cloud walrus]

Sascha Baer:

[oblique river]

i only care about the constant terms, and you can say more about a and b

how does the polynomial factor over the algebraic closure (or even over its splitting field)

[somber bramble]

into (X-a)^p, right?

where a is a (the) p-th root of y

[oblique river]

you tell me :P

[somber bramble]

ooh, and then the constant terms would have to be a^k and a^l (for some k, l summing to p), but we’ve established that a doesn’t exist in k, so that’s impossible

wait, is it impossible

just cause a isn’t in k doesn’t mean a^k isn’t in k

(…I’m bad at variable names)

[oblique river]

(maybe don't use k for an integer and a field :P)

you're on the right track

in this case it is true that "a not in k implies a^n not in k when p doesn't divide n"

[somber bramble]

oh, but if a^m was in k, for m coprime p, then we can get a in k as well, by uh, raising that to the right power

[oblique river]

yes

[somber bramble]

aight, I think I can take it from here

I’ll have to fill in some details but it seems doable

thanks :)

[oblique river]

np

[somber bramble]

actually, got any hints for the converse (if frobenius is surjective then for g(X) is irreducible, g(X^p) is reducible)?

I couldn’t really get anywhere there either

[oblique river]

what is g(X^p)^(1/p)?

(raising to the 1/p power is taking inverse frobenius. here you will use that it is surjective.)

[uncut girder]

For part a please tell me if this proof is correct:

Let a1, ..., an be an F-linear basis of K1.
Then K1K2 ⊆ K2(a1, ..., an) and is a finite extension of K2 of degree at most n.
So [K1K2 : F] = [K1K2 : K2][K2 :F]  ≤ n[K2 :F] = [K1 : F][K2 : F].

[somber bramble]

@oblique river oh my god, I had the decomposition written down and didn't see it

[hollow comet]

Let k be an algebraically closed field, f a polynomial in n variables over k. I want to prove that the map $\tau: D(f) \to X_f: (x_1,\ldots,x_n) \mapsto (x_1,\ldots,x_n,f(x_1,\ldots,x_n)^{-1})$ is continuous with respect to the Zariski topology

[cloud walrus]

edelopo:

[hollow comet]

I have no idea how to do it

I have tried just following the definition but I get stuck

Actually what I want to prove is that its inverse (which is just the first n projections) is a homeomorphism, I don't know if it can be done without proving directly that tau is continuous

I'm trying the simpler case where I just take $x \mapsto (x, x^{-1})$ and I can't even do that one

[cloud walrus]

edelopo:

[hollow comet]

<@&286206848099549185>

[fickle brook]

what's X_f

[hollow comet]

Oh, sorry

[fickle brook]

genuine question tho

i just wanna know lol

[hollow comet]

$X_f = { (x_1,\ldots,x_{n+1})\ |\ f(x_1,\ldots,x_n) x_{n+1} = 1 }$

[cloud walrus]

edelopo:

[hollow comet]

So it is like saying R\{0} is homeomorphic to a hyperbola

[fickle brook]

hmm

what's the defn of zariski topology

zero sets of polynomials are declared closed?

[hollow comet]

Yes

[fickle brook]

hmmmmm

well topologically, continuous iff preimage of closed set is closed

[hollow comet]

I tried going down that path but got stuck

Now I'm trying to prove preimage of open is open

[fickle brook]

sounds harderto me

[hollow comet]

And we know that the D(h) form a basis for the topology

I proved that in the first part of the exercise, so maybe it is there as a hint

[mild laurel]

Is it worth it to learn homological algebra before reading Atiyah MacDonald? I realize that it's definitely not necessary, but he makes references to the Tor functor in the exercises and I was wondering if it would be nice to know

[oblique river]

@mild laurel probably not on a first read-through

[chilly ocean]

Is it accurate to say that the conjugation class of an element is it's orbit under conjugation, and the centralizer being the elements stabilizer under conjugation?

[somber bramble]

yes, I think that’s right. at least it sounds right to me

[chilly ocean]

Nice, I sadly have to speedrun this algebra course so I just want to make sure I didnt miss some important distinction somehow

[somber bramble]

what’s WR?

[chilly ocean]

Idk, my goal is a PB of a bit under a month

[bleak finch]

This is almost a competition math question but...

Suppose a)b = s remainder p (this means grade-school long division) and c)d = q remainder r. What is ac)bd?

[bleak finch]

New question

What is (a mod b) x (c mod d)?

[mild laurel]

@oblique river Do you mean on a first read of Atiyah MacDonald?

[raw moth]

you probably want to use / instead of ) @bleak finch

and yes, this should be in competition math or elementary number theory

[covert vector]

@mild laurel lol i tried (only a little) learning homological algebra for the same reason, when encountering those exercises in AM

not worth it tbh, imo

granted I also don't know much about ext/tor, so I can't rate its usefulness

but atiyah macdonald has tons of interesting stuff, would be lame to put it on hold

[mild laurel]

Yeah that's how I feel. Seems like there's as much to learn from the exercises as there is from the actual text itself and it'd be nice to know that stuff

homological algebra seems like something I'd like to learn eventually anyways, but it also seems like something that might be better to learn after learning some algebraic geometry

[topaz solar]

homological seems interesting but beyond me

[oblique river]

@mild laurel yes that's what I mean

it kind of has a different feel than commutative algebra

of course there are links but I think you should just focus on the commutative algebra part

[somber bramble]

prove it by induction

also that’s linalg

[uncut girder]

True

I'll move there

[uncut girder]

Give me a hint

For finding the inverse of a formal Laurent series

Sum_{i>=n_0} a_i (x-a)^i

Coefficients in C

I know this much:

In the last 3 lines I'm trying to find conditions for the inverse of f

[woven delta]

@uncut girder didn't I show this to you once?

[uncut girder]

No

[woven delta]

Maybe I showed this to Hindy then

Finite tailed Laurent series?

[uncut girder]

Yes

[woven delta]

You can get rid of the tail

In an obvious way

I think that's the first step usually

[uncut girder]

Multiply by (x-a)^(-n_0)

[woven delta]

Yeah

And then you get a system of equations you have to solve

[uncut girder]

Does it have a nice solution

[woven delta]

The units of the formal power series are the power series with a 1 in it

ie a_0 +a_1(x-b)+....

Where a_0 is nonzero

It does have a nice solution

Find the inverse of that

And then you can convert any (nonzero) finite tailed laurent series into a formal power series of that form

[uncut girder]

Interesting

[woven delta]

Like I'll give you an example

1+x+x^2+...

We need some b_0+b_1x+... So that when they multiply together you get 1

So b_0=1

Then that gives you b_0+b_1=0

So you can get b_1

The point is you have a sequence of linear equations with one free variable in each, so they all have solutions

[somber bramble]

and you can solve them in sequence

[woven delta]

Yes

[somber bramble]

rather easily, too

it’s quite useful at times

[woven delta]

Well

[somber bramble]

well it can get annoying, but it’s never hard

[woven delta]

Not generally

[somber bramble]

hm? you can write a general formula for the coefficients

[woven delta]

Oh yeah, I guess you can

[somber bramble]

I should have it somewhere even

[woven delta]

I just prove that solutions exist and that's enough for me

@uncut girder does that make sense

[somber bramble]

there you go

[woven delta]

Yeah, that looks right

[somber bramble]

that’s assuming they’re both power series, i.e. no coefficients with 1/x^k

but you can just multiply stuff out so that’s w.l.o.g

[woven delta]

Yeah, we already discussed that

As long as we're only dealing with finite tailed Laurent series

[uncut girder]

I didnt think of reducing to power series

[somber bramble]

there’s clearly a mistake in my formula, i should only range to k-1

[woven delta]

Oof

[somber bramble]

but otherwise it’s fine

[woven delta]

I like the ring of formal power series a lot

It has the property that every proper ideal is ring isomorphic

(if you consider rings to not have 1)

(which I do)

[somber bramble]

smh rings without 1 don’t deserve to have an i in them

[woven delta]

I don't like the name rng

[somber bramble]

I‘ve seen some author use ryng for when we’re being agnostic as to whether it has a 1

[woven delta]

Why would we call rings with 1 unital rings

If we called rings without 1 rngs

[somber bramble]

and then ring and rng if we do care

well I call rings with a 1 “ring”

[woven delta]

I usually say ring with unity

[uncut girder]

That's really cool

[woven delta]

That actually also follows from our procedure @uncut girder

[uncut girder]

I mean wats cool is how the inverses work

[woven delta]

Ah ok

Yeah it's nice

We had to figure out the inverses for a hw from grad alg 1 in the fall

More specifically we had to figure out all the ideals of the ring of formal power series over a field k

But that just came down to figuring out the inverses

[oblique river]

in my experience, whenever people say "ring" they mean "with unity"

very very very rarely do people (in number theory or adjacent algebraic fields) consider rings without unit

in general terminology comes out of necessity

because people so rarely consider rings without unity , it doesnt make sense really to call those "rings"

[mild laurel]

Yeah all rings without unity are isomorphic to an ideal of a ring with unity so it seems weird to study rings without unity

[bleak abyss]

Wait really?

I knew that aside from maybe harmonic analysis rings always have unit but I didn't know that those without are isomorphic to those with unity

[mild laurel]

Isomorphic to an ideal in a ring with unity

It's actually a cool exercise to show this

[void dragon]

I learnt something new.

[oblique river]

can you just do something like

if R doesn't have a unit

consider R + Z with some multiplication operation to make (r,0) * (0,1) = (r,0) or something

then (0,1) is the unit of the ring

(and then (R,0) is an ideal in this new ring that is isomorphic to R)

[somber bramble]

okay maschke's theorem was neat

[wary sphinx]

Can anyone clarify the meaning of a ring ideal written as <2,x>, or some other comma-separated, multiple-element nonsense?

It keeps showing up in my text and I can't find where it's actually introduced

Nevermind, found it on stack exchange.

[somber bramble]

for future reference, post the answer if you've found it in case someone reads this and wonders the same

it's the ideal generated by those elements, which I think one can represent as all R-linear combinations of those elements, or as the intersection of all ideals containing those elements

[wary sphinx]

Actually on second look, I'm not sure if there's a difference between (r,s) and <r,s>. I think what you're describing is
(r,s) = {xr+sy | r, s in R}.

But a different thing I'm looking at is
<2,x> has powers of x tossed in, not just linear. https://math.stackexchange.com/questions/36169/show-that-langle-2-x-rangle-is-not-a-principal-ideal-in-mathbb-z-x

[somber bramble]

well, yes

here, R = ℤ[x]

so it would be ℤ[x]-linear combinations

e.g. (3x³ + 2x² - 4)*x + (5x⁵ - 3x)*2

where the polynomials in the parens are the coefficients

[stray nexus]

hello

i was newly introduced to modern algebra, could someone explain galois theory

[fading sparrow]

it's kinda hard to explain if we don't know where to start @stray nexus

what is your question?

otherwise I suggest you read it up in books, there are plenty...

[stray nexus]

so i know what group theory is, but i dont know what field theory is. and galois theory is supposed to somehow connect those two theories

im searching for videos about field theory atm

but im still confused

havent found a good video yet

[alpine warren]

find a good book 😛

I had topics in algebra by herstein recommended to me

[fading sparrow]

Abstract Algebra by Thomas W. Judson and Stephen F. Austin looks good i suppose

it covers the basics

or abstract algebra by Irwin Kra

[stray nexus]

thank you for the book reccomendations^^

[woven delta]

Topics in algebra is nice

Dummit and Foote is good as well, also artin

Lang if you think you're a big boy

@stray nexus

These are just the standard recommendations though

[uncut girder]

@stray nexus galios theory concerns field extentions. It turns out field extensions are closely related to polynomials in an indeterminate variable x. The polynomial is like a relation, and the roots of the polynomial are new elements that satisfy this relation. This is just the first step in a long line of steps before you can talk about the correspondence between field theory and group theory, but hope this helps you get a sense of the start.

[void dragon]

Lang seems to get recommended a lot here. Now I'm tempted to open it

[full blaze]

is Fraleigh a good abstract book

[onyx mirage]

Is it possible to prove a set can not be a group without knowing the binary operator?

[fickle brook]

no

[onyx mirage]

Not in the general case

So I wouldn't be able to prove that the set of positive and negative odd integers in Z together with 0 do not form a subgroup?

Question in the book is phrased pretty poorly if that's the case

[fickle brook]

can you show the exact problem statement

[onyx mirage]

Alright if I cite it exactly?

[fickle brook]

that's exactly what i want you to do

[onyx mirage]

(a) ...
(c)...
(d)  the set of positive and negative odd integers in Z together with 0

a-c do not give relevant info

I don't see any super group to compare too, unless I can assume Z to be the additive group

[fickle brook]

just about any time Z is brought up as a group you endow it with addition as the operation

[onyx mirage]

The 0 element kind of hinted at it

I was able to solve c without knowledge of what the binary operator might've been, so figured it was an interesting question to ask whether there was some general theorem behind it

!t,rep @fickle brook

I guess we could always biject Z to Z in such a way our operation seems like it's additive

[fickle brook]

not quite

there are groups of countable cardinality that aren't isomorphic to Z

[onyx mirage]

🤔

[fringe nexus]

Every cyclic one is isomorphic to z though

[onyx mirage]

and every subgroup of a cyclic group is cyclic as well 👀

[uncut girder]

Yes

[mild laurel]

And every subgroup of a free group is free

[onyx mirage]

What's the symbol used to denote a mathematical object has a particular property, if there is any?

[somber bramble]

I don’t think there is any general one (how would you even notate that in general with a symbol?) but there might be stuff for more specific things

if you’re thinking of sth particular

[fickle brook]

don't chase symbols

[onyx mirage]

@fickle brook I feel like it's a common enough expression it would be nice to have a symbol for

[fickle brook]

as sascha said there exist symbols for particular relationships and/or properties

[onyx mirage]

But there is no "has a" symbol like comp sci has for xml diagrams?

[fickle brook]

no

[onyx mirage]

pity

[fickle brook]

what are you trying to say rn anyway

[uncut girder]

So

The nilradical of a commutative ring is the intersection of all prime ideals.

The Jacobson radical of a commutative ring is the intersection of maximal ideals.

Now since all maximal ideals are prime, we can conclude the Jacobson radical is contains the nilradical, right?

So taking intersections of sub families reverses the order of inclusion.

[mild laurel]

Do you mean to say contains or is contained?

[uncut girder]

Contains

[mild laurel]

Yeah that's true

[uncut girder]

Ok 👌

[uncut girder]

Why are radicals called radicals btw

[mild laurel]

Cause exponents are radicals?

[uncut girder]

Fractional exponents

I guess the nilradical is the set of nth roots of 0

[onyx mirage]

radix means root in Latin @uncut girder

[uncut girder]

Interesting

Radical means "forming the root"

Nilradical would mean forming the root of 0.
Jacobson radical would mean forming the root of Jacobson.

[onyx mirage]

I should do more abs alg

Probably one of my favorite fields in maths

[uncut girder]

"In ring theory, a branch of mathematics, a radical of a ring is an ideal of "not-good" elements of the ring."

So radicals can also be interpreted as badguys

[onyx mirage]

Explains why all the good girls like abs alg

[uncut girder]

I'm having trouble proving in A[x] where A is commutative ring, the Jacobson radical equals the nilradical.

[uncut girder]

Omg it's so frinking simple

Just had to use x

I can't believe I spent hours on this

[mild laurel]

Atiyah MacDonald? @uncut girder

[uncut girder]

Yes

Slow grind through the exercises

There are 28 exercises in chapter 1

I'm meeting with my prof on Monday and trying to do as many exercises as possible

[mild laurel]

Yeah I've also been reading through the book with some friends

It's been a lot of stuff like that, taking hours for a problem, then realizing it's super straightforward

[uncut girder]

Right????!?

[hollow comet]

@uncut girder AM is very very nice but I find it somewhat dry. It is super condensed. My suggestion would be to keep on reading it but maybe have a side book that you can refer to when something is not clear

For instance AM leaves all of the geometry to the exercises, but it is very illuminating for some concepts

[uncut girder]

Wdym geometry

Were doing commutative rings

[hollow comet]

Yes, but all of the stuff you are studying has a counterpart in algebraic geometry

For example take the ideal quotient (I : J)

Reading that from AM left me very confused

I see the point but...

But the cool thing is that it corresponds to set difference

So if you have two affine k-varieties X, Y, then (I(X) : I(Y)) = I(X\Y)

Maximal ideals correspond to points in affine space

And then you generalize to Spec(R)

I don't know

I guess there can be arguments for both

But I feel that keeping the geometry close is quite helpful

[mild laurel]

Yeah it's described in the exercises pretty well

And it's pretty cool discovering it for myself rather than just having it told

[bleak abyss]

I've heard there's a book by Altman and Kleiman called "A Term of Commutative Algebra" that's basically Atiyah-Macdonald but more modern

[uncut girder]

I can't find this on Amazon

[fringe nexus]

https://libgen.is/book/index.php?md5=616CAEFCB793AAFE6B3335A583872A8B

[bleak abyss]

Actually don't get it from libgen, the most recent copy is available freely

There are so many circling around on the internet

[chilly ocean]

Yeah, I found a copy on my first Google Search

[bleak abyss]

But from my investigations I think the most recent one is this: http://www.centerofmathematics.com/wwcomstore/index.php/commalg.html

It's slightly too big for me to send the pdf directly

But it's free to download if you make an account on this website

[uncut girder]

So all of AM's exercises are in this book with solutions in the back

It would be nice if I could find it in my library

[onyx mirage]

I'm using the book (abstract algebra) by Dave S Dummit

any opinions on it?

[uncut girder]

is big

[full blaze]

dummit of anykind is good

[onyx mirage]

Big is beautiful

Have any recommendations for other books on topics Dummit has written on?

[uncut girder]

dummit is dummy thicc

thats actually what dummit stands for

[bleak abyss]

Well the point is that you read it, get bored, fall asleep, drop it on your toe, and say "Dammit my foot"

[uncut girder]

loooooool

[uncut girder]

Does AM have typos

;_; sad if it has typos has you're trying to self study it

[mild laurel]

There's a MO post on errata for AM

[mellow lake]

Anybody need any help with abstract algebra?

[somber bramble]

just hang around this channel and questions will pop up

[unreal shore]

what are called polynomials modulo n

polynomials with coefficient in $\mathbb{Z}_{n}$

[cloud walrus]

boilhats:

[unreal shore]

cause I want to search something about them but I can't find it's wikipedia page or relevant search results

[fringe nexus]

its just polynomials modulo n

[mild laurel]

$\mathbb{Z}_n[x]$

[cloud walrus]

Zopherus:

[woven delta]

$\mathbb{Z}[x]/(n)$

[cloud walrus]

Liquid:
Compile Error! Click the reaction for details. (You may edit your message)

[mild laurel]

$n\mathbb{Z}$ isn't an ideal in $\mathbb{Z}[x]$

[cloud walrus]

Zopherus:

[woven delta]

That's true

[cloud walrus]

Liquid:
Compile Error! Click the reaction for details. (You may edit your message)

[inner acorn]

What is e^x mapping between?

R -> R+ ??

[daring sage]

Question, i'm trying to construct an isomorphism between $\qty(\mathbb{R},+) and \qty(\mathbb{R^+,\times})$

[cloud walrus]

barrackollama:

[solar wyvern]

u know there's an analysis channel for analysis questions 😢

[daring sage]

yeh

this is an algebra question

[inner acorn]

actually probably more analysis now I think about it xD

[bleak abyss]

So, prove in general that a bijective homomorphism is an isomorphism

[marble wagon]

each of these is each other's answer kinda

[inner acorn]

although you're proving it's an isomorphism

[bleak abyss]

Or just use the inverse that you know exists

[inner acorn]

I mean

[daring sage]

isn't that circular?

yeah i proved that a bijective homomorphism is an isomorphism

i'm trying to prove that e^x is bijective

[bleak abyss]

If you have a function with an inverse it's bijective

[inner acorn]

another way is by showing e^x is strictly increasing

[daring sage]

ah i haven't thought of that

[bleak abyss]

And log(x) is a function that's just known to be a thing

[solar wyvern]

wonder if there might be an inverse exp function...maybe some sort of logarithm

[daring sage]

and i haven't proved the inverse exists

[inner acorn]

xD

[daring sage]

so i don't feel comfortable using it

[solar wyvern]

(do it then)

[daring sage]

i'm doing that by proving it's a bijection lmfao

[bleak abyss]

If this is an algebra class I'm sure you've proven before that log is a thing that exists lmao

[daring sage]

it's not a class

[inner acorn]

I like this route

[daring sage]

i'm self teaching algebra

[bleak abyss]

Like I don't think the point of the exercise is constructing log so much as understanding that e^x is a hom

[inner acorn]

showing exp(x) is a bijection

[bleak abyss]

Oh

[inner acorn]

then showing ln(x) is the inverse

although, the increasing argument only works for real variables

but that's what you asked soo

[bleak abyss]

This is the real case lmao

[inner acorn]

yeah

[daring sage]

I'm pretty unfamiliar with what the "rules" are for proofs

Like... if i know the inverse can i just use it? or do i need to prove that i can? idfk

[inner acorn]

If you know it sure

[bleak abyss]

I mean that all sorta depends on you

Self-teaching is sorta awkward in math

[inner acorn]

but sounded like you didn't prove it o: so you could prove it, or find another method (or both)

[solar wyvern]

I mean, if you really want to do this thoroughly, you'd construct the exp function

[daring sage]

agreed lmao

[solar wyvern]

on Q

[inner acorn]

if you're coming from an algebraic perspective

[solar wyvern]

extend to R

[daring sage]

i construct the exp using a taylor series

cause im a physicist

[inner acorn]

it's a good habit to show injective/surjective