groups-rings-fields

[wooden ember]

lemme just try to write out what was going wrong with this last time i tried

might end up seeing my mistake

[rustic crown]

(here all tensor products are over R, where A, B are R-alg)

[wooden ember]

if $a\otimes b = a'\otimes b'$ and $\alpha\otimes \beta = \alpha'\otimes \beta'$, saying that $(a\otimes b)(\alpha\otimes \beta) = (a'\otimes b')(\alpha'\otimes \beta')$ is equivalent to saying that for all bilinear maps $\phi: A\oplus B \to N$ we have $\phi(a\alpha,b\beta)=\phi(a'\alpha',b'\beta')$ when $\phi(a,b)=\phi(a',b'), ~~\phi(\alpha,\beta)=\phi(\alpha',\beta')$

but this just seems wrong to me

[rustic crown]

right, if you just think of bilinearity this may seem very mysterious

think in terms of tetra-linearity?

the multiplication (A⊗B) x (A⊗B) --> (A⊗B)

should be a 4-linear map A x B x A x B --> (A⊗B)

[wooden ember]

right but why should i need to think in terms of 4 linear maps?

shouldnt i be able to check well definedness like this by taking sums of simple tensors instead

oh also typo

[cloud walrus]

𝓛ittle ℕarwhal ✓

[rustic crown]

that is true, but i never thought verifying well-definedness directly

[wooden ember]

i agree what you're suggesting sounds more elegant, it's just i cant figure out where this is going wrong

[rustic crown]

yea doesn't look like a wrong start, lets think about it

i feel one has to do something like
consider maps phi(- * x, - * y) and phi(- * x', - * y')
we want to say these are the same maps assuming x⊗y = x' ⊗y'

since multiplication in A and B are bilinear, these maps stay bilinear

[wooden ember]

hmm yeah

[rustic crown]

basically somehow connect "distributivity" as it's just bilinearity to the outer bilinear structure

[wooden ember]

yeah they are the same map cause you get a map from A oplus B to the module of bilinear maps from A oplus B to N by sending x,y to phi(-*x,-*y) no?

so we have like psi(x,y)=phi(-*x,-*y)=phi(-*x',-*y')=psi(x',y')

and then we conclude

which i guess justifies talking about tetra linearity

[rustic crown]

wait i didn't understand how you got the middle equality

[wooden ember]

i get the middle one from the outside ones

like psi(x,y)=psi(x',y')

from x otimes y = x' otimes y'

[rustic crown]

ah okie

[wooden ember]

and then that gives middle

[rustic crown]

wait, but to get that map psi you need the middle equality, no?

oh nvm

[wooden ember]

i dont think so?

[rustic crown]

you don't

yee all good

[wooden ember]

okay thanks it makes more sense now why you should think tetra linearly

[rustic crown]

i think what you just did there is combine the tetra linearity into tensor hom adjunction with two bilinearity

[wooden ember]

not sure i see where i used tensor hom

[rustic crown]

(A⊗B) ⊗ (A⊗B) --> N
((x⊗y) ⊗ (a, b)) --> phi(xa,yb)

the data of this map is same as

(A⊗B) --> Hom((A⊗B), N) = Bilin(A, B; N)
(x⊗y) --> phi(- * x, - * y)

[wooden ember]

right i see

first map lives in $Hom((A\otimes B)\otimes (A\otimes B), N)$ and second lives in $Hom(A\otimes B, Hom(A\otimes B, N))$?

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

is that what you're saying?

[rustic crown]

yee

[wooden ember]

coolio

[rustic crown]

[wooden ember]

thanks for the help 👍

so with your technique id use associativity of tensor product right?

[rustic crown]

yep

[wooden ember]

cause i want a map A otimes B otimes A otimes B to A otimes B and i can get that fom A otimes A otimes B otimes B to A otimes B

[rustic crown]

or i think it's also nice to verify the universal property when you take product of n things at once

[wooden ember]

which is given by the bilinear maps associated to the algebras A and B

[rustic crown]

you're also using commutativity

[wooden ember]

fair enough

[rustic crown]

oh btw a warning

so even though A⊗B has maps from both A and B

A --> A⊗B by sending a --> a⊗1

A⊗B isn't really the coproduct

unless you're working in the category of commutative R-algebras

because the elements a⊗1 and 1⊗b always commute in A⊗B

so tensor product is like taking direct product in groups, even though the algebras themselves may be non-commutative, the way we're putting them together doesn't let A and B interact much

[wooden ember]

wdym here?

[rustic crown]

(a⊗1)(1⊗b) = a⊗b = (1⊗b)(a⊗1)

[wooden ember]

ah right

[rustic crown]

compare the situation to groups with direct product and semi-direct product

the semi-direct product inherently introduces non-commutativity

[wooden ember]

right yeah

[rustic crown]

but direct product keeps the multiplication in both groups separate

[wooden ember]

could i view this as the identity from A oplus B to A oplus B not being bilinear?

because if it were then id have an isomophism between A otimes B and A oplus B right?

[rustic crown]

not sure i understand

[wooden ember]

it's not a very useful comment haha nvm

[rustic crown]

.<

[rustic crown]

btw i realized while talking about non-commutative algebras that i sadly changed the order here :p, should have been x*- and y*- or something

[wooden ember]

right fair enough

[chilly ocean]

Hello, can anyone tell me what Conrad means with Aut(L) for a field L in these notes? https://kconrad.math.uconn.edu/blurbs/galoistheory/artinschreier.pdf

The way I was taught Galois theory was always to start with a field extension L\K and to then set Gal(L\K) as the K-algebra automorphisms of L.
My best guess is that Aut(L) is the group of ring automorphisms of L, and in that case Gal(L\K) should be a subgroup of Aut(L).

[sharp sonnet]

probably just field automorphisms
you can show that any field automorphism fixes the prime subfield, say K of L, and then you can think of this as Gal(L/K)

[chilly ocean]

sure, that makes sense

thank you

[lethal dune]

Can anyone help me see why this formula makes sense?
M is not assumed to be right H modules

[rustic crown]

i'm guessing that tensor without subscript means tensor over the field k

so isn't this just the associativity of tensor product?

$k[G]\otimes_{k[H]} (M \otimes_k Res(N)) \cong (k[G]\otimes_{k[H]} M) \otimes_k N $

texit is sad?

k[G]⊗_k[H] (M ⊗_k N) = (k[G]⊗_k[H] M) ⊗_k N

hopefully that is legible

[lethal dune]

If the tensor is over k then it’s fine ig

I was thinking it’s over kH

[rustic crown]

yea that would be weird

as you say M isn't a right k[H]-module

[lethal dune]

Also why is M x Res(N) = MxN

My keyboard sucks so I’m not writing otimes, hopefully it’s clear from the context

[rustic crown]

oh because Res(N) is just N right

you're just restricting the action

as k-vector spaces nothing changes

[lethal dune]

Ok I see

The action does not matter in this tensor product

[rustic crown]

yea

[lethal dune]

Nice tks

[rustic crown]

[lethal dune]

[formal ermine]

what does $\operatorname{dim}_{\bK}(\bL)$ mean

[cloud walrus]

i believe in mathemagic

[toxic zephyr]

i feel stupid asking this but can someone explain quotient groups in terms of quotient rings? i understand the latter pretty well but I'm getting confused by the former, even though it seems simpler. like, what would it look like to mod out the dihedral group by r?

[formal ermine]

do you want intuition or the formal definition

[toxic zephyr]

probably intuition

[chilly ocean]

the latter is a particular case of the former

where you have x+J where J is an ideal for the latter, you have xH for the former, here in multiplicative notation

H being a normal subgroup

The process is the same - we consider an operation on the sets of the form xH

just like you did on the sets of the form x+J

This operation being (xH)(yH) = xyH

[toxic zephyr]

i wonder if part of my confusion is just that i don't really understand what a normal subgroup is... are they basically like the ideals of a group? when is a subgroup generated by an element normal? or is that not something you can always do in a group, in the way you can just create a principal ideal from any element of a ring?

[chilly ocean]

alr, I'll say it how I understand it, but be aware that you might not get it

[formal ermine]

a normal subgroup is a subgroup N where $gNg\inv = N$ (for all g in G)

[cloud walrus]

i believe in mathemagic

[formal ermine]

this definition naturally arises from when you attempt to define a group structure on G/N

[chilly ocean]

What you really are doing is dividing by an equivalence relation, and asking when the operation inherited on the equivalence classes forms a group

normal subgroups are what represents those admissible equivalence relations

and you can prove this!

ideals for rings are precisely the same in this sense

they represent those equivalence relations but for rings this time

in general we won't have any particular subsets that do so, but you can still divide by an equivalence relation

this is the comfyness of rings and groups

[toxic zephyr]

hopefully i'm not completely misunderstanding you, but are you sort of saying that the way we define ideals is such that the operations on a quotient ring actually make sense?

[chilly ocean]

yes, this is the only way you can do this

it has to be a quotient by an ideal

but coming back to groups again
If H is a normal subgroup of G, then you can see that H is normal when H commutes with every element of G, xH = Hx.
So when you're doing (xH)(yH) interpreting it as the set {xh_1yh_2 : h_i are in H}, you have xHyH = xyH^2 = xyH

so you want this commutativity to be able to get y on the front here

so maybe that helps a little for why it should be normal subgroups

[toxic zephyr]

okay that actually makes a LOT of sense. like for $(a+I)(b+I)=ab+I$ we need closure under multiplication and the ideal property $aI,Ib=I$.

[cloud walrus]

nilpotent nix

[chilly ocean]

aI and Ib will only be subsets of I in general, but yeah

[toxic zephyr]

right right

[toxic zephyr]

okay the definition seems a lot more motivated now

[chilly ocean]

You can try this:
For which equivalence relations ~ we can define product on G/~ such that for two equivalence classes [a], [b] of ~, [a][b] = [ab]?

and G/~ forms a group

you'll find that the answer is precisely, when a ~ b iff ab^-1 is in H for some normal subgroup H of G

same for rings but with [a][b] = [ab] and [a]+[b] = [a+b]

note that then the equivalence classes are precisely the classical cosets, that is elements of G/H

[toxic zephyr]

i'll definitely try this.
the thing i'm still wondering though is when we can say that the subgroup generated by an element is normal?

[chilly ocean]

Like, imagine you have some elements $G_0$ and you consider $H = \langle G_0\rangle$ the group generated by $G_0$. Then $H$ is normal iff $xgx^{-1}\in H$ for all $g\in G_0$

[cloud walrus]

Blitz

[chilly ocean]

what I deleted was wrong btw

so here you get that subgroup generated by g is normal iff xgx^-1 is a power of g for every x

[toxic zephyr]

so the subgroup generated by r in the dihedral group would be normal right?

[chilly ocean]

yeah because s will reduce if you conjugate by something with s

so there will be some power of r left

[toxic zephyr]

okay that makes sense. and then Dn/<r> would just be {e<r>,s<r>}?

[chilly ocean]

yep

[toxic zephyr]

okay okay i think i'm starting to understand it haha

thank you so much @chilly ocean this was incredibly helpful

[glossy crag]

Want to double-check something: if R is a UFD, K is the quotient field, then if f,g are coprime in R[x], they are coprime in K[x].

Let p be a common divisor in K[x] and write f=f'p and g=g'p. Factoring out content we get f=c(f')c(p)f''p* and g=c(g')c(p)g''p* with f'',g'',p* in R[x] and since content is unique up to units of R, we get that c(f')c(p) and c(g')c(p) are in R (since content of f and g is), so f=f'''p* and g=g'''p* with f''',g''',p* in R[x], i.e. p* is a common divisor in R[x]. Since (f,g)_R=1, this means p* is a unit of R, therefore p is in K, i.e. (f,g)_K=1.

[rustic crown]

yee that looks correct

[glossy crag]

Thanks, just wanted to be sure.

[rustic crown]

btw, i find that version of Gauss' lemma a little cumbersome to use. you can look at this cuter version

Say D is a UFD and F its fraction field. Let f and g be two polynomials in D[x], then

f divides g in D[x]

if and only if

f divides g in F[x] and
c(f) divides c(g) in D

so in this light, Gauss lemma is literally comparing the mysterious D[x] to the nicer UFDs F[x] and D. And because of this, it's not then surprising that D[x] also becomes a UFD.

(sorry i was just bored and wanted to message something )

the forward direction follows by definition and that content is multiplicative. and for backward, if you write g = f * h for some h in F[x], then c(h) = c(g)/c(f) in D, so h is also in D[x]

in the above version you save some usual "clearing the denominator" type argument

[rustic crown]

so with this, the proof of your statement is a little easier to write down. if f and g had a common divisor p in K[x], then you could without loss of generality assume it lies in R[x] and is primitive by replacing p with p/c(p)

now it's clear that c(p) divides both c(f) and c(g), as c(p) = 1, therefore p divides both f and g in R[x] :3

[rustic crown]

while i said so many things, maybe one more remark. it's best not to use the notation (f, g)_R for gcd(f, g)_R
the left could be confused with the ideal generated by them and in general UFD coprime doesn't imply that the ideal is (1)

[pastel cliff]

how do you prove that a ring is a factorization domain

im being askede to show that F[x^2, x^3] is a factorization domain but not a ufd

the not a ufd part is easy, there are non prime irreducibles

but can i just take for granted that it's a f.d.

[tender bough]

is there a more constructive proof? what does g_-alpha look like?

More specifically, if v in g_alpha, can we construct some vector in g_-alpha in terms of v?

[rustic crown]

show it noetherian

existence of factorizations is equivalent to the fact that any chain of divisors will eventually end at an irreducible

or said in terms of ideals, any increasing chain of principal ideals will stabilize

[formal ermine]

what does $\operatorname{dim}_\bL(\bK)$ mean

[cloud walrus]

i believe in mathemagic

[rustic crown]

what are L and K?

[formal ermine]

fields

[rustic crown]

if you have a map L --> K of fields

then K gets the structure of an L-vector space

and you can take the dimension of this

sorry i'm a little sleepy :p

[formal ermine]

what's the dimension of a vector space?

the amount of basis vectors?

[pastel cliff]

:O

,ti det

[cloud walrus]

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[pastel cliff]

,ti det#7067

[cloud walrus]

This user hasn't set their timezone! Ask them to set it using ,ti --set.

[formal ermine]

,ti sebb

[cloud walrus]

The current time for stμ₂dying is 06:10 PM (EST) on Sat, 03/12/2022.
illuminator3 is 6 hours ahead, at 12:10 AM (CET) on Sun, 04/12/2022.

[pastel cliff]

not here #bots

i just got lazy

[formal ermine]

you live on the east coast?

or wait

[cloud walrus]

Member selection timed out.

Dec 3, 2022, 11:10 PM

[formal ermine]

somewhere in land

[pastel cliff]

east coast

[formal ermine]

ah ye

[rustic crown]

yee

(also uniqueness of factorizations is equivalent to non-zero primes = irreducibles)

[pastel cliff]

there's like

a map of ideas and definitions here

and i need to make that map

[rustic crown]

it's like 12 midnight

[pastel cliff]

bc i forgor those things

[formal ermine]

you were the one who lives in germany, no?

[rustic crown]

for 2 years at least ig

[formal ermine]

epic

anyhow

is there any intuition to dim_L(K)?

[rustic crown]

like Q --> Q(sqrt2) is an extension of fields

and any element above can be written as linear combi of {1, sqrt2}

so dimension as a Q-vector space is 2

what more intuition do you want :p

[rotund aurora]

How do you not know that and are studying field theory? Pick a linear algebra book please (you will enjoy it)

[rustic crown]

yee

[formal ermine]

I'm in an algebra 1 course but didn't know linear algebra was a perquisite

[pastel cliff]

i mean

it isn't necessarily

depends on how course is taught

[rotund aurora]

field theory builds on basic lin alg concepts from the begining

its fields over fields

lol

[rustic crown]

wait how long is your algebra1 course?

[formal ermine]

I noticed

[formal ermine]

2 more months

[pastel cliff]

oh i havent studied field theory explicitly like that

[rotund aurora]

Like he can pick most of the concepts in the fly I believe, but still, he can just pick a lin alg book

[rustic crown]

and they're doing groups, rings, fields, everything?

[pastel cliff]

my basic algebra classs barely got to rings

[rotund aurora]

And then, you look at field elements as linear operators, and so on

[formal ermine]

yes and galois

[pastel cliff]

and my big boy algebra class isnt gonna get to fields

that's too much for alg1

[elder wave]

That’s standard treatment in germany I guess

We do the same

[pastel cliff]

yall built diff

[elder wave]

Except intro AG instead of Galois

[pastel cliff]

ive liked my baby steps teaching in algebra

[rotund aurora]

Anything with the name Galois > anything else. I think this criterion is pretty accurate, not sure

[rustic crown]

my alg1 in undergrad was just linear algebra then alg2 was just group theory, alg3 was rings and fields, alg4 was modules and galois theory

[formal ermine]

we did modules within 2 lectures

[rustic crown]

tf

[pastel cliff]

same here

[elder wave]

I had LA 1 and LA 2 before

[pastel cliff]

that's wack

[elder wave]

We did modules in LA 2

[pastel cliff]

idek what a module is

[rotund aurora]

what year are you, if you dont mind me asking? @formal ermine

[elder wave]

Then more advanced module theory in Algebra

[formal ermine]

10th

[pastel cliff]

high school?

[formal ermine]

yes

[pastel cliff]

[rotund aurora]

wait what?

[rustic crown]

[formal ermine]

they let me take 1 uni course per semester

I did numerical analysis last semester

[elder wave]

We do this in HS in germany

[rotund aurora]

ahhh ok ok

[elder wave]

Dunno about y’all

[rustic crown]

germany is so cool

i didn't even know this >.<

[south patrol]

ye

[rotund aurora]

I thought you were in uni

[elder wave]

Im surprised you stuck with math after doing numA

[rotund aurora]

Well then, its nice that you get exposed to confusion

[elder wave]

[south patrol]

German curriculum seems hot

[formal ermine]

lmaoooooo

[rotund aurora]

dude for a second I thought you were being taught this in high school

[formal ermine]

my prof was soooooooo nice

it was her first semester at the uni

and she did at least 10 arithmetic errors per lecture

[elder wave]

She just like me fr

[south patrol]

Meirl

[rustic crown]

me uwu

[pastel cliff]

[uncut girder]

Me too

[formal ermine]

what does the notation Q --> Q(sqrt2) mean

[rustic crown]

bro but is germany allowed to do this?

[rustic crown]

inclusion map

[formal ermine]

what's that

[elder wave]

Yes

[rustic crown]

like when A is a subset of B, so you define the function i : A --> B given by i(a) = a

[formal ermine]

does --> mean something different than ->

[rustic crown]

nu

[formal ermine]

okk

[south patrol]

nu metal

[rustic crown]

other places should copy germany >.<

[rotund aurora]

yes

[south patrol]

gute idee

[rustic crown]

it would be so much nicer if i learned more abstract alg by now >.<

[rotund aurora]

In my country the school system is so rigid. It rewards the avarage

[rustic crown]

it was the same in my undergrad institute

[formal ermine]

and then?

[rustic crown]

so it explains the notation

.<

like my keyboard doesn't have the subseteq symbol

so i jsut use the inclusion map

.<

[rotund aurora]

in field theory tho, its very usual that we use arrows with a hook $\hookrightarrow$

[cloud walrus]

Croqueta

[formal ermine]

injective?

[rotund aurora]

because field homomorphisms are either injections or trivial

[rustic crown]

only injective!!!

[rotund aurora]

(try to figure out why if you didnt know)

[south patrol]

Uh

[rotund aurora]

my bad

[rustic crown]

0 is NOTTT a field

[south patrol]

Reality can be what I want

[rustic crown]

(oh lol i thought you were gonna start a war :p)

[rotund aurora]

haha

[formal ermine]

what does the notation Q(sqrt(2)) mean?

I swear I remember it from somewhere

it's on the tip of my tongue

[rustic crown]

take rationals, throw in sqrt(2)

now make it closed under +, -, *, /

make a field out of it

[formal ermine]

throw in as?

[rotund aurora]

algebraic copy

[rustic crown]

the smallest subfield of C containing both Q and sqrt(2)

[formal ermine]

what's a subfield

[rotund aurora]

bruh

C is a beast

[rustic crown]

[elder wave]

I think you know a lot of AA lol

[rotund aurora]

I prefer to close it under the operation, cuz the complex (real) numbers are weird

[formal ermine]

is it just Q adjoint sqrt(2) or?

[rotund aurora]

definitionally speaking I mean

[rotund aurora]

yes

[formal ermine]

ahhhh

[rotund aurora]

mate (dont know your name sorry), remember how we constructed the complex numbers?

Try constructing Q(sqrt 2) in the same manner

[formal ermine]

yes

[rotund aurora]

From quotiening in Q[x] by some ideal

[formal ermine]

Q[x]/(x^2-2)

[rotund aurora]

obtain Q[sqrt 2]

yes

why is it a field though and not just a ring?

[formal ermine]

does Q(x) just mean Q[x] but for fields

[rotund aurora]

Q(x) is the field of fractions of Q[x]

[formal ermine]

ohhhhhhhhh

we always used Q(some ring) for that

[pastel cliff]

is a ufd necessarily a pid?

[rotund aurora]

but in this case, Q[sqrt 2] is already a field, so we just write Q[sqrt 2]=Q(sqrt 2)

[rustic crown]

nope

[rotund aurora]

wait im delusional

[pastel cliff]

[rustic crown]

k[x,y],Z[x] is ufd but not pid

[pastel cliff]

no keep it

that was hilarious

[rotund aurora]

oh yeh yeh, I had the set inclusion in my mind reversed

[rustic crown]

,av croqueta

[cloud walrus]

Croqueta#3385's Avatar

Click here to view the image.

[pastel cliff]

INTERIOR CROCODILE ALLIGATOR

[rotund aurora]

I have this image in my mind

but Idk why I read $\subseteq$ as $\implies$

[formal ermine]

"above"?

[cloud walrus]

Croqueta

[rustic crown]

usually people draw field maps like
Q(sqrt(2))
|
Q

[pastel cliff]

oh but a ufd must be a pid right

[rustic crown]

above here just means Q(sqrt2)

[rustic crown]

nu

pids are ufds

not the other way

[rotund aurora]

All Euclidean domains are PIDs, all PIDs are UFDs

wait what am I saying

[rustic crown]

[rotund aurora]

thats what I wanted to say

[pastel cliff]

that's what i meant too lmao

[rotund aurora]

I think I read it like that, thats why I said "yeah"

[rustic crown]

you can't just write the opposite of what you mean >.<

[elder wave]

[formal ermine]

ok so Q(sqrt 2) has the basis {1, sqrt 2}. where does Q come into play

[pastel cliff]

im just mixing myself up

[pastel cliff]

bc of this

[rustic crown]

the coefficients for 1 and sqrt(2) would come from Q

[formal ermine]

ahhh

dim_Q(Q(sqrt 2)) = 2?

[pastel cliff]

@rotund aurora that picture is wrong isn't it

am i losing my mind

[rotund aurora]

I think implies arrows work better for mental pictures, instead of set inclusions

[rustic crown]

like Q(sqrt(2), sqrt(3)) over Q(sqrt(2)) will have basis {1, sqrt(3)} as the scalars now come from Q(sqrt(2)), so this also has dimension 2

[pastel cliff]

oh wait

[rotund aurora]

I empathise

[pastel cliff]

wow im having a moment

[rustic crown]

[formal ermine]

dim_K(L) = min. amount of elements from L needed to represent L via linear combinations with coefficients in K?

ok epic

that's what I was looking for

thanks!

[rotund aurora]

hold on!!!

Why do you know thats well defined?

[rustic crown]

the subring of Q[x] where the constant term is an integer is a weird ring

[elder wave]

☝️

You rn

[pastel cliff]

👊 🖐️

he hit the whip!!!!!!

[rotund aurora]

wait I think maybe saying "min" is well-defined, but it is not obvious that all bases have the same cardinality

thats what I wanted to say

[formal ermine]

if $\bL/\bK$ is a field extension then $[\bK : \bL] = 1$, yes?

[cloud walrus]

i believe in mathemagic

[rotund aurora]

I think that doesnt make sense

[pastel cliff]

maybe a weird question

[rotund aurora]

You usually write L/K to mean K --> L, i.e., that L is an extension of K. So in that case, I dont think you can generally talk about K being a vector space over L

[pastel cliff]

but is the fact that a gcd exists in a euclidean domain a consequence of the fact that they exist in UFD's

[rustic crown]

[K:L] is just another notation for dim_L(K)

[rotund aurora]

^

[formal ermine]

yes I know

[rustic crown]

but we just saw an example where the dimension was 2 >.<

[pastel cliff]

my class has gone ED -> PID's -> UFD's

and we proved that gcd's exist in ED's and that seemed to be like the defining characteristic of an ED

[formal ermine]

L/K is a field extension means K subseteq L, no?

[rotund aurora]

Yes

[pastel cliff]

but now im realizing that they must also always exist in a UFD

[rotund aurora]

You can think of it as L "lying over" K

And you read [L:K] from left to right as the dimension of L over K

[rustic crown]

wait me no understand

[pastel cliff]

my thing?

[rustic crown]

if L is the extension why you looking at [K:L]?

[rotund aurora]

thats what I pointed out

[formal ermine]

because I was wondering what it's equal to

[rustic crown]

i'm too sleepy >.< sowwy

[rotund aurora]

👆

[formal ermine]

"K being a vector space over L"?

[rotund aurora]

thats what we have been talking about all the time

[rustic crown]

defining?

[rotund aurora]

K is a field, but if it contains another field L, then you can trivially think of K as a vector space with scalar field L

[pastel cliff]

liek

[formal ermine]

idk what your words mean :c

[pastel cliff]

maybe rather it characterizes an ED

the fact that the euclidean algo works

like it's in the name lol

[rustic crown]

oh euclidean algo is very different than just existence of gcd >.<

you also have
ED -> PID -> UFD -> Bezout domains -> GCD domains etc

[rotund aurora]

I was gonna type an example of the real numbers as a vector space over Q, because I think thats a baffling example. But I think that would confuse you more

I could give you the basic concepts of linear algebra, but its better if you pick a book (seriously)

[formal ermine]

[R : Q] = inf?

[rotund aurora]

You can see the pinned messages in #book-recommendations or ask there

[rotund aurora]

uncountable!

[formal ermine]

I'll do linalg after this course

but I got no time to do it now

[barren sierra]

lin alg after an algebra course?

[rotund aurora]

(Im trying to help btw, idk if Im sounding rude. But linear algebra is important, thats what Im trying to say)

[formal ermine]

yes I know you're just trying to help

[rotund aurora]

He is in high school

[formal ermine]

I greatly appreciate that

[barren sierra]

ah

[rotund aurora]

The basic definitions of linear algebra are not difficult. I would suggest that you pick a book (or some notes) and look at the basic definitions: vector spaces, linear independence, bases, etc. It wont take you a lot of time

also, if you go to classes, I think you can just ask the profs there, they will probably point you to good resources, to learn some of the material that you might be lacking

[formal ermine]

isn't a vector space just a module but instead of a ring it's a field

[rotund aurora]

hahaa

this is so weird

[elder wave]

Lmao

[rotund aurora]

usually people say "modules are like vector spaces but over a ring"

[rotund aurora]

yes

[formal ermine]

what does "K is a vector space over L" mean? K is an L-module?

[rotund aurora]

yes!

[formal ermine]

what operation do we use

[rotund aurora]

what do you think?

K is already a field (i.e., it already has defined operations), and L is a subfield of K

[formal ermine]

I'd guess addition? because a field under multiplication isn't an abelian group

[rotund aurora]

for vectors, yes

[rustic crown]

imagine learning modules before vector spaces and doing field theory in between
this is scary

[rotund aurora]

what about scalar multiplication (i..e, multiplying the elements of the ring (in this case your field L) by the elements of the module (in your case K))?

[formal ermine]

oh nvm I was thinking something wrong here

yeah it's obviously just multiplication

[rotund aurora]

yeah thats it

[formal ermine]

r.m = r * m

[rotund aurora]

you could sometimes define some other operations if you wanted, but if you are interested in studying K, then it makes sense to just look it in that way, which is a natural one

the point now is that vector spaces are nice

thats why you look at K/L in that way

if you knew linear algebra you would know

[formal ermine]

wouldn't an L-module over K just have the basis { 1 }

(is my terminology with L-module over K correct here? I want to say K is an L-module)

[rotund aurora]

no

You would say K is a module/vector space over L or K is an L-module

both are fine

[rotund aurora]

no

you have already given examples yourself

It is one tho when K=L

but not in any other case

[formal ermine]

ok I think I'm mixing up L and K

so we have L/K meaning K subseteq L. then if K is an L-Module it would have the basis { 1 } because we can write any m in M as l * 1 for m = l in L?

[rotund aurora]

If K is an L-module and L is not equal to K, then that module won't align with our interests in field theory

if it is even possible to give such a module

because scalar multiplication is exotic

[formal ermine]

ah right

l * m won't always be in M

[rotund aurora]

In K you have 1, which is your identity. If K is an L-module and if x is in L and not in K, then you would have x*1 in K

if you can define such a scalar multiplication I mean

so x*1 wont be equal to x

(that assuming you can even define such a scalar product *, which in general wont be possible I think)

[formal ermine]

ye

okk

[chilly ocean]

wut

if K is an L module

x*1 would be in K

[rotund aurora]

• is a hypothetical scalar multiplication

not the field operation

[chilly ocean]

yeah and

if K is an L module

Scalar mult is

LxK -> K

[rotund aurora]

yes

[chilly ocean]

Yes

So x*1 is in K

[rotund aurora]

x*1 is in K, thats what I said

thats what I said

[chilly ocean]

oh

[rotund aurora]

but x*1 not equal to x, because x is not in K

[chilly ocean]

yes

🤣

i might have read

[rotund aurora]

thats what I was trying to say, hopefully its not too confusing

[chilly ocean]

not in K

[rotund aurora]

dont worry

[chilly ocean]

;-;

more important question is

why you guys using L as a subfield of K

that feels so wrong ;-;

L/K >>> K/L

[rotund aurora]

I dont even write field extensions as L/K, I prefer to use K<L

and actually, my preference for letters is F<K<E

[formal ermine]

K is a subfield of L here

[chilly ocean]

yeah

usually

L is an extension of K

[chilly ocean]

yeah fr

[formal ermine]

yeah that's why I got confused a couple of messages ago

cuz my lecture uses L/K too

[rotund aurora]

like, when you just have one extension like L/K, I think the / might be more aesthetic, but when you have towers and fields with many adjoint variables

[chilly ocean]

yeahhh

like its fine for one field

but eve then

not really the best option

[chilly ocean]

yeaaahh

other way feels super wrong

[formal ermine]

do we have $\bQ(\sqrt{2}) \cong \bQ \oplus \bQ$?

[cloud walrus]

i believe in mathemagic

[rotund aurora]

as vector spaces over Q yes

[formal ermine]

yes

[rotund aurora]

As fields, direct sums dont make a lot of sense I think

[chilly ocean]

yeah

[formal ermine]

wait

no

I don't get this then

let me translate it

[chilly ocean]

I mean there's the direct product of fields, where u can define a field structure on but that is a different one to the original one

But thats not a fields

[rotund aurora]

The problem is

[chilly ocean]

How would you define multiplication there

Which has inverses ^

[rotund aurora]

you would also say $\bQ(\sqrt 3)\approx \bQ\oplus \bQ)$ as vector spaces

[cloud walrus]

Croqueta

[rotund aurora]

and $\bQ(\sqrt 2)$ and $\bQ(\sqrt 3)$ are not isomorphic as fields (they are as vector spaces)

[formal ermine]

Let $\bM/\bL$ and $\bL/\bK$ be two field extensions. Then the following holds $$[\bM : \bK] = [\bM : \bL] \cdot [\bL : \bK]$$. Proof: If either of them are infinite, so is the product. We can safely assume that they are therefore finite. We have $$\bM \cong \underbrace{\bL \oplus \ldots \oplus \bL}_{[\bM : \bL] \text{-times}}$$

[cloud walrus]

i believe in mathemagic

Croqueta

[formal ermine]

what does the L oplus ... oplus L mean

[rotund aurora]

In this case, they are talking about M as a vector space if Im not mistaken

they are simply saying that M has dimension [M:L] over L (which is the definition of [M:L])

[formal ermine]

ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

I think I got it

so L oplus ... oplus L is just { (x, y, z, ...) | x,y,z,... in L } which is isomorphic to M under phi : (x, y, z, ...) -> b1x + b2y + b3z + ... for some basis elements b1, b2, b3 in M?

[rotund aurora]

yes

its isomorphic to M as vector spaces

[formal ermine]

AAAAA

this makes so much sense now

[rotund aurora]

For every vector space V over a field F, there exists an n such that V is isomorphic to F^n

this is just saying that every vector space has a basis. So you can think of vector spaces of dimension n over F as the vector space F^n, thats why vector spaces are nice

But F^n wont be a field in general. For example, think about R^2 (R is the real numbers). How would you define an inverse of (0,1)? This is not zero, so it should be invertible. Multiplication will be something weird, not the natural componentwise operations.

[formal ermine]

yesss

thank you so much! I finally get it

[rotund aurora]

Linear algebra is beautiful

[glossy crag]

Thanks, that's pretty neat

[safe holly]

I recently learned that polynomial rings are free objects in the category of R-algebras:

$$Hom_{R-ALG}(R[X], S) \cong Hom_{SET}(X, \hat{S})$$

Ie, $R$-algebra morphisms from $R[X]$ to $S$ correspond to length $|X|$ sequences in the underlying set of $S$.

Can we reflect this result into the category of Rings?

[cloud walrus]

Mars Industrial

[safe holly]

I am thinking I can show

$$Hom_{RING}(R[X], S) \cong Hom_{RING}(R, S) \times Hom_{SET}(X, \hat{S}) $$

[cloud walrus]

Mars Industrial

[pastel cliff]

bro is typing the magna carta

[safe holly]

am not, I am just having to revise because I don't know wtf I am doing lol

Okay, I think I can show

$$ Hom_{RING}(R[X], S) \cong \bigcup_{\alpha \in Hom_{RING}(R, S)} Hom_{R-ALG}(R[X], (S, \alpha))$$

[cloud walrus]

Mars Industrial

[next obsidian]

I think it’s equivalent to a choice of algebra structure on S and then an element of S

Namely, if you consider R -> R[x] -> S this makes S an R-algebra

So looking only at what the coefficients do

And then for that fixed choice of coefficients you then take a single element S for what x does

Maybe that’s what you wrote?

Ah it is

I didn’t see the thing you unioned over

[safe holly]

To what I wrote, let $f \in Hom_{RING}(R[X], S)$. Restrict $f|_R = f . j_R$, where $j_R : R \to R[X]$ is the canonical embedding.

Now $(S, f|_R)$ is an R-algebra, and so is $(R[X], j_R)$. But now $f$ is just a morphism from $R[X]$ to $(S, f|_R)$

And to go the other way, we can just use the forgetful functor to turn algebra morphisms into ring morphisms

[cloud walrus]

Mars Industrial

[next obsidian]

This is interesting it’s almost like a map into S and a choice of base point in some sense

[pastel cliff]

^this is interesting

[safe holly]

right, just an R map, and a base point (really a X-point, for arbitrary set X). And in fact the whole point of R[X] is that it lets you add this |X| level of freedom to an ordinary R map

[next obsidian]

Yes

[safe holly]

To finish the proof, I think we can do:

$$\bigcup_{\alpha \in Ring(R, S)} RAlg(R[X], (S, \alpha))$$
$$\cong \coprod_{\alpha \in Ring(R, S)} Set(X, \hat{S})$$
$$ \cong Ring(R, S) \times Set(X, \hat{S})$$

[cloud walrus]

Mars Industrial

[safe holly]

The second step feels kind of illegal. I'm appling the free object property in RAlg, which makes all dependence on $\alpha$ go away. So I turned it into a disjoint union to keep the actual structure

[pastel cliff]

fuck you math

[chilly ocean]

normal

[pastel cliff]

idk if you're saying that this kinda thing is normal or joking about the fact that "normal" gets the same kinda treatment with meanings getting lumped on it

[safe holly]

there are always n-versions of famous lemmas, and you can always ignore all of them

[pastel cliff]

but either way

[chilly ocean]

both

[safe holly]

you don't have to worry about what random crappy authors choose to call gauss' lemma, it's better to focus on collecting tools that work on the kinds of problems you care about, and agressively yet dispassionately ignore everything else

[pastel cliff]

aggressively yet dispassionately ignore everything else

i like that lol

ok i was talking about this before but i'll ask again - what exactly does Frac(R) look like

i understand that it's localizing a ring R at everything except 0

and localizing is "adding inverses" to elements of a subset of R

but if R isn't a field and you localize at pretty much all of R but not every element has an inverse how are giving elements inverses

[safe holly]

is it like giving elements formal inverses? Like when we construct the free group with generators x, y, z, first we have to magic up a neutral element, then add x^-1, y^-1, and z^-1 to the mix, then do a whole bunch more tedious stuff besides. Are we treating the inverse-free elements like generators for some larger set we're building?

[pastel cliff]

not sure i follow

the best way i understand it rn is actually like we're reducing the ring with the equivalence relation we get from localizing

but idk if that's accurte

[safe holly]

@pastel cliff, wait, the elements of S^-1R are just pairs (r, s) in RxS, where for the purposes of arithmetic, r acts like a numerator and s acts like a denominator. Ie, so we have nice stuff like (r1, s1) * (r2, s2) = (r1*r2, s1, s2), just like these things are trying to pretend they are rational numbers

[pastel cliff]

*numerator lol

[safe holly]

can we pretend these are like rational numbers? Step 1 - make this bigger ring of rational pairs. Step 2. Quotient together equivalent pairs, where (a, b) ~ (c, d) if (a, b) = (cf, df) for some factor f.

wait, I guess that gets us where you are, without much way of seeing how these should be different from rational numbers

@pastel cliff it seems you feel we are reducing R, so you are not sure where we get extra elements from. But are you seeing that to localize R relative to S, first we make R bigger by passing to RxS? And then we do something to reduce that?

Does your sense of ominous foreboding start before or after that point?

[pastel cliff]

ohhhh i guess that's a better way to think of it

i keep forgetting that S^-1R is just R x S

and the equivalence relation exists on that

[safe holly]

RxS/~

[pastel cliff]

but it does reduce RxS right

[safe holly]

where ~ is.. {verbally vomits}

[pastel cliff]

ok my main concern is where noninvertible elements of R end up after making it over to R x S

do they get lumped into the equivalence class of some invertible element and thus become invertible that way

i tried thinking about this with Q and/or R but i dont think that works

or wait

eh no nvm

[safe holly]

I guess an illustrative example is what happens to the noninvertible elements of the integers (aka all of them) when we do this to obtain the rationals

[pastel cliff]

oh so Frac(Z)?

[pastel cliff]

ig that shows that this is wrong

[safe holly]

Yeah, I think there they get lumped according to numerator/denominators which are relatively prime

almost it's like the real interesting story is happens once we being to add invertible elements - does each invertible element get mapped to a unique equivalence class, or can they get they get squished together?

[pastel cliff]

favorite thing ive read today

proof by big gun

[thorn delta]

i am a fan of nuking the mosquito myself

[safe holly]

this seems so fanciful and whimsical

[chilly ocean]

[pastel cliff]

unrelated but noetherian-ness tells us that any collection of ideals of a ring has a maximal element - is this enough to assert that the number of ideals containing some other ideal must be finite...?

i would think not because of Z

but i feel like something of that nature could be true

[chilly ocean]

number of ideals containing (0) :^)

[safe holly]

wait doesn't Z show us that the # of ideals containing v is finite? If w > v, then w divides v. But v has finite # of divisors

except (0), that example is cursed

[safe holly]

there's gotta be a way to finagle this from the ascending chain condition, I've seen all sorts of BS pulled from that hat

[pastel cliff]

that's what i thought yeah

[pastel cliff]

[small bramble]

I don't get why the last statement here follows? Sorry, noob at abstract algebra

[thorn delta]

[F : Fp] = n means that the dimension of F is n as a Fp vector space. In other words, its the direct sum of n copies of Fp

[safe holly]

@pastel cliff suppose v is a sub-ideal of infinite ideals$ w_1, w_2, \dots$. Then construct ideals $u_k = w_1 \cap w_2 \dots \cap w_k$. Then $u_k$ looks like it might be an infinite chain descending down to v. But that is illegal!

[cloud walrus]

Mars Industrial

[safe holly]

is that convincing? I think that could fail if there is M such that u_k = v for all v > M.

Like say v = (24). If w_1 = (4) and w_2 = (3), then already u_2 = v. Is there some kind of way we can backtrack and pick a better w, such that u_k > u_k+1?

[wooden ember]

Take Z[x]: the ideals (p,x) satisfy no inclusion relations for primes p and each contain (x) if I’m not mistaken

[wooden ember]

I don’t think this works in non artinian rings

[safe holly]

thank you! I was wondering if polynomial rings give us a counterexample, but could not visualize or manipulate its ideals well enough to see!

[lethal dune]

ok forgive me but what's the definition of Normal subgroup? do we require gHg' ⊆ H or it's an equally?
Like is ⟨b⟩ normal in F(a,b) / ⟨aba'=b²⟩

[obtuse bear]

the usual definition is gHg ^-1 ⊆ H for all g in G, but this implies the opposite containment so in fact equality does hold: replace g with g^-1 to get g^-1 H g ⊆ H, which is equivalent to H ⊆ gHg^-1

[lethal dune]

not true for the example I gave

[rustic crown]

equality

basically you want the quotient map G --> G/H to be a homomorphism, whatever this forces your H to be should be the definition of normal

that means two things, G/H is a group and the map preserves the operation

because of the map, the group structure on G/H is forced on you

but like Bungo says, since you're asking that condition for every g, in particular g', you get the reverse inclusion as well

[rustic crown]

ahh i see your question now

i think you made a boo boo with the definition of the normalizer

N_G(H) = {g in G : gHg' = H}

here you can't make that a subseteq

because a priori you don't know if for every g in N_G(H), the inverse g' also lies in N_G(H)

i don't think we can say anything about a'ba, which is the whole problem

[lethal dune]

yes

this example actually comes up in covering spaces

it apparently gives an example of a covering transformation which is not a deck transformation

[rustic crown]

i should really learn some algtop >.<

[lethal dune]

for that I require that H to be not normal

[wooden ember]

What are you specializing in btw

Cause i used to think it was comm alg until you said you didn’t know comm alg

[rustic crown]

i'm trying for alg geo and arith geo right now

and i'm learning comm alg as i need it in ag

[wooden ember]

Arith geo is what ngroupoid does right?

[rustic crown]

dunno

ngroupoid teach me arigeo

[rustic crown]

what about you, are you also gonna do some AG?

[wooden ember]

Idk yet it’s too early to tell

Im doing a decent amount of algebra focused courses this year and next year I’m doing more analysis focused ones

[rustic crown]

i just know AG would be useful for anything lol, so might as well learn it :p it's also fun

[wooden ember]

Yeah it does seem pretty fun

[rustic crown]

complex analysis was the only analysis course i liked :p

[wooden ember]

Complex analysis is awesome

Tbf I say analysis focused but idk if differential geometry counts as analysis

It’s just geometry ig

[rustic crown]

yea

[wooden ember]

But yeah I’ve still got plenty of time before I decide what I wanna focus on 👍

[rustic crown]

i was doing a masters to get some extra time to decide >.<

[wooden ember]

Yeah I’ll probably do a masters too

That’s the more standard route in Europe

[rustic crown]

idk how my batchmates from UG made a decision so quick

they literally had a 3 year UG and then directly a PhD

[wooden ember]

Isn’t that what usually happens in the us?

[rustic crown]

until a year ago i didn't even know how to properly cross a road in india

[wooden ember]

Ah you’re in India

[rustic crown]

i was :3

crossing roads here is so easy

literally you can walk without looking at the street

[wooden ember]

Lmao yeah I’m used to that too

[lethal dune]

what did I just hear

[rustic crown]

[formal ermine]

cross the autobahn with your eyes closed

[lethal dune]

make a run for it and pray

[rustic crown]

bro i was so surprised that the cars would wait for you to cross the road even when there is no traffic light

[formal ermine]

lmao

[rustic crown]

i always feel like they can go on ahead >.<

[lethal dune]

sorry what

[formal ermine]

yeah it's like that in germany

[rustic crown]

we awkwardly stare at each other for 2 seconds, and then i have to make a move

[lethal dune]

🫥

[formal ermine]

det have you ever used the öpnvs?

[rustic crown]

nu

whut that

also what's autobahn

[formal ermine]

öffentlicher personennahverkehr

busses n stuff

[lethal dune]

[formal ermine]

the big roads with no speed limit

[lethal dune]

just call it bus then wtf?

[formal ermine]

it's also trains but like where the busses drive

[rustic crown]

i was searching for that sticker "oh so i see now, (understood nothing)"

but didn't find it >.<

where it go

[lethal dune]

purged

[formal ermine]

either your bus comes 10 mins early and you gotta stand in the rain waiting for the next one

[rustic crown]

oh it was there

[formal ermine]

or your bus comes 5 mins late, resulting in you missing your connection

there's no inbetween

[rustic crown]

i used to be scared travelling more than a kilometer in india

here travelling is so easy >.<

[formal ermine]

why

[rustic crown]

the morning busses are usually on time

[wooden ember]

In Netherlands they’re on time 😎

[rustic crown]

because roads are scary for me >.<

[lethal dune]

so how far is your uni from where you stay?

[rustic crown]

i always arrive 5 min before at the bus stop, cause i don't trust my walking speed

[rustic crown]

currently like 3.5ish km i think

but if you only count the walking distance it's like 100m + bus + 100m

[lethal dune]

I see

[formal ermine]

oh wow

[lethal dune]

and how many roads do you have to cross on your way

[rustic crown]

the only way i crossed roads in india was to wait for someone else and follow their shadow >.<

[formal ermine]

my school is 3.2km away, my uni 7 km

[rustic crown]

only one :3

[lethal dune]

for me it's literally on the other side of the road

[rustic crown]

i reach my uni in like 20min usually

25 is an upperbound

#abstract-chill uwu

[lethal dune]

I go there for lunch/dinner as well

food is good

[wooden ember]

I basically live in my uni 😎

[rustic crown]

so cool

[wooden ember]

Very handy

[rustic crown]

i wish they did that here

[lethal dune]

hostels

[rustic crown]

i hate to go out of my room

[wooden ember]

It’s just the student dorms are very close to campus

[chilly ocean]

I also hate to go out of my room. Way too cold out there

But if I could I'd be hopping around

can't stand sitting here

The cold never bothered me anyway

frozen reference

Does anyone have intuition for projective/injective modules?

[lethal dune]

if you do AG then apparently projective modules are vector bundles over something. Idk the details and I'm yet to read AG

[chilly ocean]

yeah. They are kind of like the canonical modules in which you either embedd or are a quotient of

[south patrol]

it's time to see what calculus i can do, to test the limits and break through

[chilly ocean]

Oh that's a really nice intuition! It matches with the intuition for the particular case of free groups too. So how do projective modules and free modules differ in general?

[rustic crown]

projectives are direct summands of free

and injectives are direct summands of cofree

[chilly ocean]

Does this also hold for the category of all groups?

[rustic crown]

oh projective objects in Grp?

[chilly ocean]

or something analogous

[chilly ocean]

yeah

[rustic crown]

never thought about those

because the main use of projective/injective objects is to get projective/injective resolutions

and because of those nice extensions/lifting properties the quasi-isomorphisms which one cares about, become an easier to understand condition which is that of homotopy-equivalence

[chilly ocean]

surely projective/injective resolutions can be defined in Grp too

[rustic crown]

right

[slender hamlet]

Wait, cofree module is a thing?

Oh, direct products of Hom_Z(R, Q/Z)

(According to wolframalpha)

[rustic crown]

have to think about this... so the same proof shows that free groups are projectives, with the fact that epimorphisms of groups are surjective, but i only see that projectives are retracts of free, and not exactly direct "summands".

[rustic crown]

yee

so like Z is a projective Z-module, and projecitves are closed under taking arbitrary direct sums

and for any projective you have a surjectiveion Z^{⊕A} --> A

and moreover if you base change your situation by tensoring with R⊗_Z then you get other free modules

similarly, you can make a parallel for the situation of injectives

[slender hamlet]

Yes that is true

[chilly ocean]

Well in groups not every semidirect product is a product, so maybe it would be semidirect summands instead of direct summands?

[rustic crown]

Q/Z is an injective Z-mdoule
and you can use hom to base change

[slender hamlet]

Yeah we get injective cogenerator that way

Hom_Z(R, Q/Z)

[rustic crown]

and you also you have that other thing

if A is injective then you can embed A into (Q/Z)^{Hom(A, Q/Z)}

similar to how you can projective from Z^{⊕Hom(Z, A)} --> A

[rustic crown]

and using the injectivity of A you also get that it's a split injection

[rustic crown]

do you get semi-direct product(ands?) directly?

say G is prjective and F a free group that surjects to G

so we can form an exact seq
1 --> K --> F --> G --> 1

and the F --> G admits a section

[chilly ocean]

In Grp, split extensions are equivalent to semidirect products

[rustic crown]

ahh rightt

[chilly ocean]

same in Ab too, just all of those are products in Ab

[rustic crown]

nice

[south raptor]

hi, can someone help me with question 1.2? I managed to do 1.1 without a problem but I literally have no idea about 1.2

from 1.1(3) I know that the iff statement is true for one way but I don't know how to show the other way around

[chilly ocean]

what s your definition of R-module?

[chilly ocean]

For each r in R the law of composition gives you a function from M to M

maybe you can try using that

[chilly ocean]

what book is that, looks old

[south raptor]

Module Theory: an approach to linear algebra, T. S. Blyth

[formal ermine]

what's a trivial intermediate field? either the field itself or the extension field?

[pastel cliff]

a bit silly to ask but "quotienting" by an equivalence relation just means applying it to a ring right

still thinking about the localization stuff from yesterday, in particular R x S/~

[elder wave]

not sure if this is what you mean but you quotient by one by declaring two elements in the ring to be equal if they are equal under said equivalence relation

[pastel cliff]

yeah i think so

like R x S/~ just means we're defining ~ on RxS

[elder wave]

when localising on some multiplicative subset S you basically add multiplicative inverses for elements in S

think about fractions and how you would usually reduce them

[pastel cliff]

am i wrong in saying that every element in S must have in inverse then? since it's mult. closed

[elder wave]

no that's correct, the point of the construction is that every element of S becomes a unit

this means that we get a local ring when localizing at a prime ideal (why?) and local rings have nice properties

[formal ermine]

why is it K[M] subseteq K(M)?

aren't they defined the same?

[chilly ocean]

K[M] is the ring, and K(M) is the field

[formal ermine]

R must be a ring here?

[chilly ocean]

sounds about right

[formal ermine]

ah ok thanks

it didn't specify that in the text lol

[chilly ocean]

in general when we talk about K[A] we mean the of polynomials in A with coefficients in K

but A don't have to be independent variables or anything

like how K[x] is the ring of polynomials in x

[formal ermine]

yes

[formal ermine]

why do we have K(M) subseteq Q(K[M])?

[chilly ocean]

K(M) is the smallest field containing both K and M right

actually idk if this isn't an equality

[formal ermine]

this is an equality

but they are saying "we obviously have ..."

but I don't get why it's obvious

[chilly ocean]

why did they write subset then

oh right

Q(K[M]) is a field

so what they're saying is that since Q(K[M]) is a field, contains K and M, we have K(M) subset of Q(K[M])

I think the point here is that they want to prove the equality, you just didn't write enough context

(to confuse us perhaps?)

[formal ermine]

"We obviously have ...." this is the entire context

[formal ermine]

ah ok this sounds simpler than I thought

[chilly ocean]

no way lol

there's probably stuff after it

[formal ermine]

"Because K(M) is the smallest intermediate field that contains M, we even have K(M) = Q(K[M])"

[chilly ocean]

alright

[formal ermine]

actually I don't understand why K(M) = Q(K[M])

I now get why it's subseteq but not why it's equal

[chilly ocean]

an element of K[M] will be of the form k_1m_1 + ... + k_nm_n

where k_i are in K and m_i are in M

right

[formal ermine]

uh why

[chilly ocean]

well now we get into another loop

[formal ermine]

wait

is the generated ring by K and M over L the same as the polynomial ring of M over K?

[chilly ocean]

I actually said something wrong

an element of K[M] will be of the form k_0+k_1m_1 + ... + k_nm_n
where k_i are in K and m_i are in M

actually I said it wrong again

lol

they'll be sums of elements of the form, some element of k times non-negative powers of elements of M

I'm thinking of modules for some reason and we have rings here

my bad

[formal ermine]

so this is correct?

[chilly ocean]

I'm not sure what that terminology entails to be 100% sure to say if it's correct

all I can say is that since M and K both exist in L, they obey the same laws as laws in L

so there's no independence going on

now let me get back on topic

[formal ermine]

this is just the same as { sum_i k_i m_i^n | n in N_0, m_i in M } ?

[chilly ocean]

since Q(K[M]) consists of quotients of two elements in K[M], we need to have it in K(M)

[chilly ocean]

well no because

If you have something like Q[sqrt(2), sqrt(3)] for example
there will be sqrt(6) there for example

as a product of sqrt(2) and sqrt(3) which are in your "M"

so those are more like how you have polynomials with multiple variables

xy+zx, x^2y

and so on

[formal ermine]

right

[formal ermine]

but this definition is essentially just the same as a polynomial ring?

[chilly ocean]

it isn't because you have powers of the exact same element

[pastel cliff]

bumping this for myself cuz i need to get to this

[formal ermine]

I don't get it

[pastel cliff]

but im looking at this cuz i finally realized how UFD's and the idea of localization come into place

[formal ermine]

what do the elements of K[M] look like

[elder wave]

this might be a bit tricky if you don't know how the ideals look after localising, if you need a hint for the prime ideal part let me know

[chilly ocean]

well it'd be notationally heavy

but fortunately we aren't using what they look like idk why I said that

[formal ermine]

?

[chilly ocean]

I always like to know what I am working with you know

[chilly ocean]

no

that was wrong

[pastel cliff]

i have a proof in prof's lecture notes, but im working backwars

like he introduced localizations then went on to talk about UFD's and i thought it was kinda random

[formal ermine]

ok

so why is K(M) = Q(K[M])

[chilly ocean]

@formal ermine just read this

[pastel cliff]

but, if i understand correctly, if we have a UFD (call it R), then usnig Frac(R)[x] and the nice properties we get from localizaton we can show that R[x] is also a UFD

[formal ermine]

ah

[pastel cliff]

so im going back to the localization stuff i glossed over before

[formal ermine]

wait

no

I don't get it

so an element of Q(K[M]) looks like a/b for a,b in K[M]

[chilly ocean]

yeah

[formal ermine]

why does it need to be in K(M)

[chilly ocean]

because a, b are in K(M)

K(M) is a field

[formal ermine]

ah and because division is closed

so we have K(M) subseteq Q(K[M]) and Q(K[M]) subseteq K(M) therefore K(M) = Q(K[M])

ok thanks!

[chilly ocean]

should have said that earlier, that would save us some time

[formal ermine]

L/K is simple if [L : K] is prime?

my reasoning is that if [L : K] is prime it has no intermediate fields, therefore the smallest field that contains a in L but not in K must be L

[rustic crown]

yee, that works

(in fact as long as you have finitely many intermediate fields, the extension is simple, assume the extension is finite for safety lol, idr the exact hypothesis)

[formal ermine]

why does $\bK[x]$ contain a polynomial with $g(a) = a\inv$ even though $a \not \in \bK$?

[cloud walrus]

i believe in mathemagic

[rustic crown]

you need a to be algebraic for that

[formal ermine]

this is before the definition of an algebraic number

the context is:

Let L/K be a field extension and a in L. Then the following statements are equivalent:
(1) K[a] is a field
(2) There exists a polynomial f(x) in K[x] with f(x) \neq 0 and f(a) = 0
(3) The degree [K[a] : K] < infty is finite

[formal ermine]

the proof for (1) => (2) uses this

[rustic crown]

and these are the equivalent definitions for a to be algebraic over K

[formal ermine]

yes

this is Proposition 4.1.5

and Definition 4.1.6 says that any element a in L is algebraic if it fulfills one of those conditions

[pastel cliff]

this is talking about localizations - what is that homomorphism it's referring to?

[rustic crown]

yee so do you agree that a^{-1} lives in K[a]?

[formal ermine]

yes

yeh for a = 0 we can just choose f(x) = x

[rustic crown]

yea

and what is K[a]?

it's the set of polynomials evaluated at a

[formal ermine]

the smallest intermediate ring that contains a

[rustic crown]

with coefficnets in K

[elder wave]

the embedding, r - -> r/1

[rustic crown]

nu, that's K(a)

this is K[a]

[formal ermine]

I am so confused

[elder wave]

for every element in s this maps to an invertible element

[formal ermine]

ring?

[rustic crown]

what's your definition of K[a]?

yee!

[formal ermine]

yes

I just mixed up those two lol

[elder wave]

maybe i shouldn't call this embedding since it's only injective for Domains

but you get the point

[formal ermine]

wut

blitz told me it isn't

[rustic crown]

whut

[formal ermine]

.

[pastel cliff]

then noninvertible elements obv dont have an inverse

[rustic crown]

okie so i'm not saying K[a] is the polynomial ring

[pastel cliff]

and get lumped in with invertible elements...?

wait that was terribly phrased

[elder wave]

[rustic crown]

i'm saying K[a] = {g(a) : g in K[x]}

[elder wave]

sorry but that sounded funny

[pastel cliff]

the water is wet