#groups-rings-fields

1 messages · Page 128 of 1

delicate orchid
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you just do the lil x_i -> x_ix_i+1 and then merk x_i+1 from the tuple

rocky cloak
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About the bar resolution?

real sparrow
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fundamental group of the klein bottle is the semidirect product of Z and Z

delicate orchid
rocky cloak
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Sounds like topology

delicate orchid
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it is a little bit but I hope that's ok

real sparrow
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its still an abstract aklgebra

agile burrow
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I mean, if you have a discrete group then you can construct a simplicial complex whose simplices are given by tuples of elements and show that it's contractible

delicate orchid
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yeah that's the one

agile burrow
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Then G acts freely and whatever other adjectives you need so that the quotient is a K(G, 1)

delicate orchid
#

right which is trivial as it's contractible?

agile burrow
#

Yeah, so the simplicial chain groups and boundary maps are precisely the bar resolution, providing a free ZG-resolution for Z

delicate orchid
#

there we go that's the connection KEK

agile burrow
#

Yeah, in practice this is how you might construct various resolutions for your group. Find a contractible space X which G acts freely and (properly discontinuously?) on. The quotient is a K(G, 1) by covering space theory and the simplicial/cellular chain complex of the universal cover becomes a free ZG-module resolution of Z

south patrol
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Uwu

real sparrow
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umu

delicate orchid
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ok new project: learn topology properly

south patrol
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no

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Topology is cringe

wraith cargo
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man do I love myself some resolutions 🚬

delicate orchid
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step 1: wtf is an open set...

real sparrow
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topology is too preccise, look into homotopy theory

delicate orchid
real sparrow
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where the disk is equal to the point

delicate orchid
#

can't answer that one now can you

agile burrow
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And there are variations of this theme. Cyclic groups act on the circle, which isn't quite contractible. But one can stitch these chain complexes together to get a ZG resolution of Z of infinite length. This explicitly demonstrates the 2-periodicity of the cohomology of cyclic groups

delicate orchid
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A_n -> B_n-1 type beat

real sparrow
delicate orchid
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I'm memeing about a homotopy of a chain complex

south patrol
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Uwu

wraith cargo
delicate orchid
real sparrow
#

homotopy theory kinda is a simpler version oif topology

wraith cargo
#

I think we should teach 1st year UGs sheaves so all their algebra intuition revolves around them and see how they fare in the real world

delicate orchid
#

also I'm still confused when someone says cohomology without refering to an explicit cohomological functor - you mean simplex cohomology I presume

real sparrow
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it focuses on more of the actually intrest8ing properties of topology

agile burrow
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I'm just talking about homology of the chain complex, but I'll be more explicit about it

delicate orchid
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right yeah ok

real sparrow
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no descrete topologies as well

wraith cargo
south patrol
delicate orchid
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sorry my homological algebra is dog water

south patrol
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How is it simpler

wraith cargo
delicate orchid
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I can see it now I think - in the case where it's contractable the chain complex would just be 0 right

south patrol
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Homotopy theory isn't even just topological spaces

agile burrow
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Homotopic to 0, yes

delicate orchid
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same thing KEK

real sparrow
south patrol
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No it isn't

real sparrow
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🤔

delicate orchid
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it's a subfield of MODEL THEORY

wraith cargo
delicate orchid
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ohh

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no no

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up to homotopy

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yes yes yes

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maybe

wraith cargo
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this is like a Weibel problem
Like in an abelian category TFAE

A chain complex is exact
H_n(C) = 0 for all n
The chain complex is homotopic to the 0 complex

south patrol
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Lol

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Whitehead thm

delicate orchid
wraith cargo
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Whitehead
Oh no
Oh nono no no

delicate orchid
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and then these are nice for the circle so we can do the ol Z_n/C_n garbage

wraith cargo
#

that name brings bad luck

agile burrow
delicate orchid
south patrol
formal ermine
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that's smart, thanks

wraith cargo
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ok lol

delicate orchid
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now that I believe without question

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le funny H^2 has arrived

wraith cargo
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How one adjective can ruin ur career

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Yeah I was thinking of these

image.png image.png
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"google how to avoid feelings of incompetence"

south patrol
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Yeah this has reminded me that I planned to git better at hom alg this month lol

delicate orchid
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so much to read so little time

agile burrow
delicate orchid
wraith cargo
delicate orchid
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too many distractions

rocky cloak
agile burrow
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It's just because you have to run through some details of checking the group acts trivially on homology and stuff, but if I actually bothered writing out the diagrams, you'd be able to see pretty clearly how to glue the resolutions together

wraith cargo
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I love global/tor dimensions!!!!

south patrol
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Well I was gonna review / relearn weibel 1-4 mostly I think

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Main goal rn is cohomology of dglas and stuff

wraith cargo
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Potato wanna learn K theory together

south patrol
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I did k theory a lot last year oop

wraith cargo
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Fucc
OK looks like I'm alone
me and my K-book

agile burrow
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Like I'm starting with this exact sequence arising from the action of C_n on the circle, but then I can just replace the first Z with another 0 -> Z -> ZG -> ZG and repeat indefinitely to get a free resolution

image.png
south patrol
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Well, wrote my bachelor thesis (equivalent) last year on it and stuff actually

wraith cargo
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algebraic or topological?

south patrol
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Though had to rush through some stuff rippies

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Oh I meant topological

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Did you mean alg

wraith cargo
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ye

south patrol
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Noice

wraith cargo
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I am going to read the K-book as someone mentioned that after I finish Weibel

south patrol
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OK maybe I will learn a bit but also busy w other tings devastation

rocky cloak
wraith cargo
wraith cargo
south patrol
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Maybe more topology and homotopy theory will crush me this summer

wraith cargo
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like in a sexy way?

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I have like 10 books on my reading list lmao

south patrol
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Lol what

wraith cargo
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I will literally finish when I'm like 70

delicate orchid
south patrol
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I have a few books but they are too long

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Lol

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And papers to read

delicate orchid
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better than having nothing to read

south patrol
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Eh

wraith cargo
# south patrol I have a few books but they are too long

I wanna finish Eisenbud, Hartshorne, Weibel hom alg, Weibel K theory, Neukirch alg NT, Washington cyclotmoic fields, McCleary spectral sequences book, Some more stuff on infty cats and uhhh I think there's something else but I forgor

south patrol
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Nothing to read means can touch grass

wraith cargo
south patrol
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Nice

agile burrow
wraith cargo
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reminder to self add Hall's lie algebra book to reading pile

delicate orchid
agile burrow
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Rip yeah I should've said X = L/K and then done X/G

delicate orchid
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I will investigate these spooky K thingimabobs further

agile burrow
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K being maximal compact subgroup or being shorthand for K(G, n) lol

delicate orchid
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like I know what K(G, 1) is

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it's some space with \pi_n = 0 for all n > 1

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but like... who CARES?? this is what we must find out

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you've given me some good motivation already

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in some way this space captures the properties of the group homotopically

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which, is kinda obvious from the definition? but still cool

agile burrow
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Yeah I mean I care because cohomology of K(G, 1) is group cohomology

delicate orchid
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right yeah cause you make your funny chain complex

agile burrow
# delicate orchid in some way this space captures the properties of the group homotopically

Actually this is a really good point and kind of historical. Hurewicz was the first to work with these "aspherical spaces" (which is what he called spaces with \pi_n = 0 for n > 1). In particular, he showed that any two aspherical spaces are homotopy equivalent iff they have isomorphic fundamental groups. So in particular, a K(G, 1) is quite literally a homotopy-theoretic version of G and (co)homology of this space can be thought of as (co)homology of the group itself since these are homotopy-invariant properties of a space.

delicate orchid
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:iwon: chat I did a thing

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yeah I completely believe that result

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how can the loops be looped onto the loops if they loop differently

agile burrow
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Ofc the modern viewpoint via derived functors and such is probably a cleaner presentation, but the thing we were talking about regarding construction the free ZG-resolution from the K(G, 1) essentially shows the equivalence between the two definitions of group homology

delicate orchid
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yeah but I don't understand Ext so that presentation suckssss

agile burrow
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Valid

formal ermine
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can someone recommend me a good resource to learn about lie theory stuff? I just want an overview and not go too indepth

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like maybe a short 20-30 page exposition would be nice

barren sierra
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It's longer than that

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But Naive Lie Theory by Stillwell is pretty light

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Very cute little book

barren sierra
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nvm did it myself

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and generation of extraspecial groups

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although that last one is in GAP so maybe there's a better way to do this using that

real sparrow
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the domain of a monomorphism is a subgroup of the codomain?

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the codomain of an epimorphism is a quotient of the domain?

south patrol
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Not necessarily

real sparrow
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example

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/explain

south patrol
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Wait which category are we working over

real sparrow
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groups

delicate orchid
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is this just up to iso as well

real sparrow
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yes

south patrol
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Okay then yes this is true

delicate orchid
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first one is definitely true then

south patrol
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Second is non-trivial it seems

delicate orchid
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non-trivial in that I can't immediately see it at 1:30am

real sparrow
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im seeing problems where a monomorphism seems to imply an epimorphism, as inverses of each other.

delicate orchid
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obviously true for abelian groups

real sparrow
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wich i guess is fine if you restrict it to normal subgroups

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im thinking more in terms of constructive homotopy theory.

delicate orchid
real sparrow
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whats a section

delicate orchid
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a right inverse to an epimorphism

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sneezed TWICE while typing that

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forbidden knowledge

real sparrow
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is there always an inverse monomorphism to an epimorphism, and vice versa

delicate orchid
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no, I don't think so

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if that was the case then every exact sequence would split

real sparrow
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because in homotopy type theory, a monomorphism is defined as a homomorphism with a left inverse homomorphism/

white oxide
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can somebody help me verify that U(V) is a subgroup of GL(V)? namely how to show closure

subgroup.JPG
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i kinda forgot my lin alg lol

real sparrow
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and right inverse for epimorphism

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all maps are homomorphisms

delicate orchid
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so it's a subset! woo hooo

real sparrow
#

so i guess in homotopy type theory, a monomorphism only maps a normal subgroup

delicate orchid
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inverses, obvs
associativity, obvs

real sparrow
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guess that can be usefull in group theory

delicate orchid
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closure, uhhh <XYw, XYv> = <Yw, Yv> = <w, v>? does that work?

delicate bloom
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Works for me

white oxide
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oh right i could just treat Tv as a vector duh

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thanks wew

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good man

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wait i'm slow how are the inverses of unitary operators unitary

next obsidian
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Apply ^-1 to the equation defining unitary

white oxide
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oh I forgot U could do that

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how is it that a complex number is unitary iff its conjugate is equal to its inverse? if it's something using the properties of operators then i'll just go review my lin alg lmfao since i seemingly forgot all this shit

how.JPG
upper pivot
#

do you know the identification of Gl1(C) with C*?

white oxide
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yup i don't remember anything about operators on inner product spaces lol

white oxide
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namely just scalar multiplication

upper pivot
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yes, now do you know what unitary means

white oxide
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well i wrote <zv, zw> = <v, w>

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trying to see how that implies that the conjugate of z equals the inverse of z

upper pivot
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ok, what is this inner product in C^1

white oxide
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im confused

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wdym

upper pivot
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well you wrote <v,w>, the inner product of two things in say C^n right

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what is the definition of this

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and what is it specifically in C^1

white oxide
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well it's just the euclidean inner product

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oh wait

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<x, y> where x = (x1, x2, .., xn) y = (y1, y2, ..., yn) is equal to x1conjugate times y1 + x2 conjugate + y2 and so on right

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or the yjs are conjugate my bad

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ah i got it

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thanks

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well i'm a bit confused because this only pertains to the case where V = F^n right

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since it uses the definition of the euclidean inner product which is only defined on F^n

paper aurora
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could someone please help me with understanding this proof? L is a lattice, or rank 2 Z-module, additive subgroup of C with R-basis. the context is that p is a prime, L'' is a sublattice of L with index p^{n+1}. the problem is to count the number of sublattices L' of L that contain L'', such that [L':L''] = p^n and [L:L'] = p. this quantity is what they are calling 'a' in this excerpt. the argument says there can only be one such lattice, but i dont understand the last part when it says L' is completely determined by L''. why is this claim true, just because of the conclusion that the images of L' and L'' in L/pL are the same subgroup ? how can a condition about the image of a lattice in L/pL translate to a condition about the actual lattice ?

image.png
white oxide
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I'm still not sure I understand why the codomain is the codomain in $\psi: G \to GL_n(\mathbb{C})$; so I understand that since $\varphi: G \to GL(V)$, $V \simeq \mathbb{C}^n$ then $\varphi$ is essentially a representation $G \to GL_n(\mathbb{C})$ - but if $\psi_g(v) = T\varphi_gT^{-1}(v)$, how is it that $T\varphi_gT^{-1} \in GL_n(\mathbb{C})$? Do we just view the vector space isomorphism $T$ and $T^{-1}$ as a matrix?

understand.JPG
cloud walrusBOT
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okeyokay (analysis is MID)

418474799208071180.png
white oxide
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But at the same time $T$ was defined to be from $V \to \mathbb{C}^n$...

cloud walrusBOT
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okeyokay (analysis is MID)

418474799208071180.png
white oxide
#

I'm so fucking confused lmfao

rotund dragon
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hence you can talk about T in matrix form

white oxide
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yea true but I’m confused about why that’s the case if T is explicitly defined to be from V to C^n

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Do we just view it under isomorphism or smt

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If that’s even a phrase

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ohh i think i see

white oxide
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and T was defined from V to C^n just to give phi: G --> Gln(C)

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so the composition is in GLn(C)

rocky cloak
rocky cloak
paper aurora
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thank you again, i am done with that proof now

void cosmos
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is there a short proof of why

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an element in Q cannot be written as a linear combination of Z

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elements?

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ig its obvious :d

south patrol
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As in, Z linear combination?

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This is a vague question as is

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Or do you mean showing that Q can't be a f.g. Z-module or smth

void cosmos
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yes

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not a projective Z-module

south patrol
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wait now i'm confused

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you say yes when i asked for two options and then meniton projective Z-modules

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😭

void cosmos
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Z is a PID

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so Q would be free

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given its projective

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so im asking

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if it can be

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if it can have a basis

south patrol
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But Q isn't a projective Z-module

void cosmos
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ik

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but

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if it were it would have to have a Z-basis

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so thats what i asked

void cosmos
south patrol
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Having a Z-basis is different from that but sure

void cosmos
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yea maybe i worded incorrectly

south patrol
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Well having no Z-basis is easier actually like

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as soon as you take two elements a/b and c/d they are Z-linearly dependent

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by clearing denominators

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and so any basis would have <= 1 elements

void cosmos
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yea

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thank you

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goti t

south patrol
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np

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More interestingly you can show Q isn't even a f.g. Z-algebra but yeah

void cosmos
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what does f.g mean

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finitely generated

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okay nvm

coral spindle
# void cosmos if it can have a basis

If you don't feel comfortable assuming that projective <=> free for Z-modules, there is such a thing as a projective basis whose presence is equivalent to projectivity. It's not too hard, iirc, to show that Q has no such thing.

rocky cloak
# void cosmos not a projective Z-module

I guess one way of showing that Q is not projective is to consider a projective precover Z^(Q) -> Q. If Q was projective this would split, so there would be an injective map Q -> Z^(Q). But all such maps are 0, hence not possible.

real sparrow
#

is are cosections equal to epimorphisms?

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In topology.

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Well honotopy theory

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Up to homotopy.

coral spindle
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Every cosection is an epimorphism, but not every epimorphism is a cosection.

rocky cloak
# real sparrow Up to homotopy.

In a triangulated category every epimorphism splits. So if "up to homotopy" means something like "in the stable homotopy category of spectra" or some other triangulated category, then yes.

real sparrow
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So epimorphisms are cosections.

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I'm looking into linear homotopy theory wich would imply they are diferent.

chilly ocean
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I'm currently reading a discrete math book. Can I start abstract algebra right after. And then possibly group theory?

south patrol
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I mean, group theory is part of abstract algebra and often where you'd start

summer path
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You don't really need discrete math for abstract algebra

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Though I guess previous knowledge of modular arithmetic can help

tender wharf
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True but some discrete math courses also introduce proofs and set theory simultaneously. If you already know it you can just do group theory tbh

void cosmos
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thank you for the other altenartives 😄

rocky cloak
real sparrow
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in linear type theory a morphism object is equal to its dual.

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(A->B)=(~B->~A)

real sparrow
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the dual morphism makes sense as you are exchanging weakening for coweakening of wich has the same order.
And contraction and cocontraction.

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categorical product describes the additive product.

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2 projections.

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The dual object of a morphism I don't know how to understand.

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it's something like a product in type theory tho.

coral spindle
real sparrow
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Where should this be

coral spindle
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In your own head???

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But anyway, it sounds like you're talking about category theory, so perhaps #category-theory

real sparrow
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No

white oxide
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Who’s deez

real sparrow
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well category theory and homotopy theory

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Woxh also connects to group theory.

agile burrow
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I am perplexed by your group theory takes lol

coral spindle
#

Can anyone point me to the nearest lake

real sparrow
agile burrow
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Idk I just can't tell what you're actually doing lol

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Anyway, today I will review root space decomposition and then try to work through sp_4(C)

coral spindle
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Oh the lie algebra not the algebraic group

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what a letdown ||jk||

agile burrow
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I'll reach algebraic group eventually

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Does the Lie algebra rep theory interact with the algebraic group rep theory in a nice way?

coral spindle
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Honestly not in any way that I know

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I mean, I care mainly about characterstic p so I couldn't really say

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but I guess the adjoint action makes the structure of the lie algebra vaguely relevant

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Yeah sorry I'm trying to think and I just don't know

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I'm trying to remember chevalley's classification of reps but brain smol

agile burrow
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Oh, that's interesting. Then I guess I have no idea what techniques are used for char p stuff. My very vague understanding is that rep theory of algebraic groups over C is relatively similar to the Lie group stuff

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But I figure I ought to learn more alg geo and scheme theory stuff before doing algebraic groups anyway

untold turret
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"an integral domain is integrally closed if and only if it is the intersection of all valuation rings containing it"

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why would one come to think this? what's the intuition?

formal ermine
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@coral spindle nuts

coral spindle
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Good to know

void cosmos
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the isomorphism from Hom(Z_n,A) to A[m] = {a | am = 0 } is given by the usual one right? f --> f(identity)

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having inverse a --> f_a =ax for x in Z_n

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correct?

coral spindle
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That's right, but I imagine you meant to write either Z/m or A[n] because otherwise n and m have nothing to do with each other ofc

void cosmos
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yes

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ur right

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ty

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now i wan to show that Hom(Z_n,Z_m) is Z_(n,m)

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now i see that i can use what i just said above both

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like with each side

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is this correct

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like i get some ring that is annihilated by n and also m

coral spindle
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Well you'd get Z/mZ [n]

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so I reckon you can use that

void cosmos
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yup

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so it just follows immedaitely thats what i wanted to make sure of

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immediately*

coral spindle
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well I mean you do need to argue why that ring is Z/(n,m)Z

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Sorry, *Z-module rather

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and I don't think that's really been done yet

void cosmos
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it seems too obvious that idk how to do it

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cant i just send a --> a mod (n,m)

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and this just should work

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?

untold turret
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"an integral domain is integrally closed if and only if it is the intersection of all valuation rings containing it"
why would one come to think this? what's the intuition?

delicate bloom
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I don't know if that statement makes sense, how do you take the intersection of different valuation rings?

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like for instance let's take the integers Z, then I can believe there's an injective map into every valuation ring of p-adic integers, sure

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then for every prime p=1 mod 4 you'll have two elements in each Z_p that are roots to x^2+1, so I would maybe dubiously try to say that some of these are the "same" or something...? I don't really see a way to directly compare elements here

coral spindle
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At best we could compare the algebraic elements by identifying them in the algebraic closure

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I assume that's what's going on

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As is so often the case, context is being omitted.

delicate bloom
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yeah that sounds reasonable

rocky cloak
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Presumably it just means the valuation rings thar are subrings of the field of fractions. And I guess the intuition is that valuation rings are integrally closed and intersections of integrally closed rings is integrally closed, so it should at least contain the closure

untold turret
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It's a theorem from Atiyah-Macdonald, and I'm trying to get a feel for it because I understand the mechanics of the proof but not the big picture

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jagr2808 right, this is AM's definition. Thanks for the suggestion, i'll think about it

barren sierra
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so Burnside's Lemma gives a quick check as to if some groups are solvable

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are there any similarly nice little checks that a group action is primitive?

south patrol
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hm any nice resources for looking at totally derived (or hyperderived?) functors? I mean stuff like \mathbb RHom where we induce functors between derived categories

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I can't seem to find too much but probably looking at the wrong place lol

hidden haven
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Categorical homotopy theory by Riehl also has some conceptual explanation of where the definition comes from if you are goodly comfortable with cat theory in which case I recommend it

south patrol
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Okay cool thank you!

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Yeah I realised G&M is probs a good idea lol i think when I first looked at it, it was a bit too homotopical for what I was doing but now it seems perfect xd

formal ermine
void cosmos
#

what does type mean here

image.png
tribal moss
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I don't thing "type" is supposed to be a technical term there.

void cosmos
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yea i guess

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but i just dont understand what he means

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all elements of this form? or what

tribal moss
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It would be more conventional to say "of the following forms".

void cosmos
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can more than one element have the same form like this?

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can i say " elements that CAN be expressed of like this for elements x1,x2..?

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"

tribal moss
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Yes.

void cosmos
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got it

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tysm

tribal moss
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More formally you could say you're looking for the union of

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{ (x+x', y) - (x,y) - (x',y) | x, x' in M, y in N }
and
{ (x, y+y') - (x,y) - (x,y') | x in M, y, y' in N)
etc

rocky cloak
# formal ermine bump

Maybe not exactly what you're asking, but introduction to Lie algebras by Erdmann and Wildon gives a simple introduction to lie algebras including representation theory and classification through root systems.

void cosmos
#

and then now D would be the module of all sums r_i * d_i where r_i is in the ring and d_i is in one of those sets

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is that correct>

tribal moss
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Yes.

void cosmos
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ty

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im sorry but by image does he mean like the projection image?

image.png
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like the actual coset?

tribal moss
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Yes.

formal ermine
void cosmos
#

cool

formal ermine
#

for a bit of context: I have an exam in two weeks on an overview of newton okounkov bodies. one of the applications is its correspondence with bases in the classification of finitely dimensional irreducible complex sl_n+1 reps. so I'm looking for some short expository text that gives me a broad overview of lie algebras, so that I know more than 0 about them in the exam

static temple
formal ermine
#

my written french is ok, understanding... well not so

static temple
# formal ermine my written french is ok, understanding... well not so

https://www.youtube.com/watch?v=AZcQRDCJgxk&feature=youtu.be

YouTube
Scientia Egregia
Théorie de Lie, Épisode I : des racines et des poids

Premier volet d'une trilogie centrée principalement sur la théorie de Lie, mais au cours de laquelle on pourra voir tout un tas de choses en algèbre, en géométrie, en combinatoire et en physique. Le plan général pour cette trilogie est :

▶ Play video
agile burrow
formal ermine
static temple
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but you can activate the subtitles

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it gives a good overview, but you will need to understand the proofs after that

formal ermine
agile burrow
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I haven't actually read it myself so idk if it is lol, but just googling "rep theory of sl(2, C)" should give similar things

hollow mica
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If $G$ is a set and $\cdot$ is a binary operation such that:
\begin{itemize}
\item $\cdot$ is associative
\item $\exists 1 \in G$ (not necessarily unique) such that $\forall a \in G ~ 1 * a = a$
\item $\forall a \in G$ there exists a $b \in G$ (not necessarily unique) such that $a * b = b * a = 1$
\end{itemize}
Basically $G$ is a group less the existence of a two-sided identity (only on one side). Are inverses still unique here? The proof for groups I could produce requires a two-sided identity, but I am wondering if maybe there is another way to prove it without it.

cloud walrusBOT
agile burrow
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Such an identity is necessarily two-sided since a * 1 = a * (b * a) = (a * b) * a = 1 * a = a

hollow mica
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oh nice

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so it's just a group

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now if we said that the identity was two-sided but that only one-sided (suppose left) inverses are guaranteed to exist, could the same left inverse serve as a right inverse? (hence a group again)

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I don't think so

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I'm just trying to see how much you can fuck with the group axioms

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because I've heard that in most texts they're redundant (like the one-sided inverse thing above) just so the presentation isn't annoying

void cosmos
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yo i dont understand the construction of M_0? why didnt he just take the submodule generated by all x_is in the first coordinate and y_is in the second? then (x_i,y_i) would be in M_0 tensor N_0

image.png
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is that what he did?

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ops this is taken

south patrol
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what would N_0 be

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do you mean like you could suppose M_0 is generated by the {x_i (x) y_j }?

void cosmos
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N_0 would be generated by all the y_is

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that are also in D

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so it goes to 0 when tensoring

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is that what he did?

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i dont like the notation of tensor products at all i think the cosets one is much better :d

south patrol
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That doesn't quite make sense

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Wait no it does

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But you wouldn't be guaranteed that it vanishes in that submodule

void cosmos
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yea cuz thats what i want

south patrol
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Because an element vanishing in the tensor product is witnessed by some other elements

agile burrow
void cosmos
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is the theorem statement not visible?

tribal moss
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He's saying that if xi otimes yi = 0, then (xi,yi) must be in D, and therefore there is a finite combination of the, hmm, original D-elements, that yields (xi,yi).

void cosmos
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its 2.13

south patrol
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I know, 2.13 is what I'm talking about lol

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Sorry, I typo'd

void cosmos
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that yields (xi,xi)? whats that

south patrol
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I'm saying that you need extra elements to guarantee they vanish in the tensor product

tribal moss
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Sorry, I meant (xi,yi).

south patrol
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I can give you an example of what I mean i guess

void cosmos
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yes please

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examples is all i need rn

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i am so lost in this

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hardest math topic byfar

hollow mica
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but in my proposed structure, where only left-inverses are guaranteed to exist, there very well could be no right-inverse for some element, right?

void cosmos
agile burrow
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Doesn't my argument show that a left inverse also acts as a right inverse

void cosmos
south patrol
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Wait lmao

tribal moss
void cosmos
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elements of D meaning elements of D after being like written in the form of the genertors

south patrol
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My example I was gonna give is explained just above 2.13

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Pretty much

void cosmos
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generators*

south patrol
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So is the question basically just like what is M_0 i guess sure

void cosmos
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can u walk through me it plesae?

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nah i got it now from tropo

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but still examples of like ocmputing stuff

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i would appreciate

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like take R be Z and take 4 submodules

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modules*

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such that a tneos product is zero in one tensor product but nonzero in the other

south patrol
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No it's generated by the x_i and all the stuff appearing as first coords as stuff in D

void cosmos
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yea

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why and hto

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why and tho*

hollow mica
south patrol
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Is D just the elements that are mapped to zero by M x N -> M (x) N

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I assume so

void cosmos
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no

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D is a submodule of the free module generated by the direct product

agile burrow
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You had the hypothesis that every element has a left-inverse. Under that assumption, I showed that a left inverse b for c is also a right inverse for c

void cosmos
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generated by some relations basicaly

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the relations that make bi-linear stujff

hollow mica
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Oh I see idk why I thought that was something extra

void cosmos
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i just want to prove that this actually works

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ie the diagram works

south patrol
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Okay sorry yes kernel of A^{(M x N)} -> M (x) N yes lol

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Anyway

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Oh also that isn't the direct product, it's a direct sum but yeah

void cosmos
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in my mind they are the same honestly

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other than the inclusion and projection

south patrol
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Well they aren't but yeah

void cosmos
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like when dealing with diagrams

south patrol
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They have (potentially) different underlynig sets

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which is pretty important

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Anyway uh

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So the point is like say I've got some like (m,n) which gives m (x) n = 0 in M (x) N

tribal moss
south patrol
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Yeah sure thanks

void cosmos
south patrol
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Oh okay now I see what you mean overall lol

void cosmos
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what do i mean

south patrol
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hm

void cosmos
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this section is going to hurt so bad

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honestlly

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why is 2 tensor x = 0 in Z tensor Z/2

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given x is nonzero

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over Z as the ring

south patrol
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So there are a couple of things like

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One way to see this is that if M, N are R-modules then rm (x) n = m (x) rn for any r in R, m in M, n in N

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That just comes from the definition

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Using that, hopefully it's clear why 2 (x) x = 0 for any x

tribal moss
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2 tensor x = 1 tensor 2x because (2,x)-(1,2x) is in D.

void cosmos
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right

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fuck i forgot that

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yea

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and 2 is 0

south patrol
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But let's look at this from the perspective of the universal property like

void cosmos
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okay cool

void cosmos
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cuz most of the isomorphisms come from it anyways

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like quotient topologies

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the point is that bi-linear maps on MxN become linear maps

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on the tensor product

south patrol
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if we have any billinear map Z x Z/2 -> M (for any Z-module M)

void cosmos
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or not become but factor through them

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ok

south patrol
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Well this'll end up being the same but note that f(2,1) = 2f(1,1) = f(1,2) = f(1,0) = 0 f(1,0) = 0

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Like we can just play around with scalars lol

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Using linearity

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Just cause we aren't necessarily working with fields we can end up with weird things like this

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In a similar vein, try to think about what Z/nZ (x)_Z Q looks like

void cosmos
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let me try some elements

formal ermine
void cosmos
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for example like (1,n)

south patrol
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That's a good place to start

void cosmos
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would this be n(1,1) which is (nm1) which is (0,1) which is 0?

south patrol
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yup

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This is the key observation

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Now can you show that (1,x) = 0 for any x in Q?

void cosmos
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yes i think

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cuz i can divide now

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like i can stick the n multiple

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so say

south patrol
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Yup

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In fact, this shows that uh

void cosmos
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(1,x) = (1,nx/n) which is n(1,x/n)

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which is 0

south patrol
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Identifying Z-modules with abelian groups lol

void cosmos
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after (n,x/n)

south patrol
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Tensoring with Q kills all elements of finite order besides identity

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(Torsion elements)

void cosmos
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anything with torsion

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ig yea

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but u just opened my mind to something

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that

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i would be able to do every same thing

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if i would look at M x N

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and then f being a bnilinear map

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like iw ould get the same properties same thing

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and so if i can find a bilinear map from MxN to something

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then the tensor product would be isomoprhic to that somehting aswell

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rihgt?

south patrol
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Not necessarily no

void cosmos
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isnt that the diagram ? 😦

south patrol
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Well im not entirely sure what you mran

void cosmos
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like the diagram

coral spindle
void cosmos
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ur so smart lol

south patrol
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M (x) N is determined by a universal property yes

void cosmos
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fuck

south patrol
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But it is like

void cosmos
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satisifes*

south patrol
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The key point is that bijection between bilinear maps out of M x N and linear maps out of the tensor product

delicate orchid
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obvious by the relators you're quotienting by :pack: rip bozo

void cosmos
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yea but im just lost and fuzzy so its not that obvious to me 😦

void cosmos
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but can u show like

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with an explciit example

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i wanna understand this

image.png
south patrol
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Yeah sure so like

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Firstly can you see that if I'm given f: M x N -> P there's at most one way to define the map M (x) N -> P

void cosmos
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no 😐

south patrol
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Like we are forced to send m (x) n to f(m,n)

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By the diagram

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and since M is generated by those elements that determines the map uniquely

void cosmos
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we already have the diagram?

south patrol
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I'm saying that if the map is well defined it must be given by this

void cosmos
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i meant like T = C/D

south patrol
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To fit into the diagram

void cosmos
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and i want to show that the diagram commutes with this T

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are we using the diagram as the definition

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or the construction

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sorry im lost

south patrol
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Well the uniqueness part is half of the statement

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Okay I will back track then

tribal moss
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Another way to look at the whole construction is that D consists of exactly those elements a1(x1,y1)+...+ak(xk,yk) of C where the rules of "bilinear map" force a1·f(x1,y1) + ... + ak·f(xk,yk) to be 0 for every bilinear f : M×N -> P.

south patrol
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A tensor product T of M and N is characterised by the fact that for any bilinear map f: M x N -> P there exists a unique map M (x) N -> P fitting into that diagram

void cosmos
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okayy

south patrol
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I am showing that our construction satisfies that proprrty

void cosmos
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thats was not what i started with

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i am doing the reverse

south patrol
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by first saying that if it exists it is unique

void cosmos
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like i have the construction

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okay okay

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i got you

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forget what i said

south patrol
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And yeah now we gotta show it is well defined

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So like

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We define A^{(M x N)} -> P by (m,n) -> f(m,n)

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And that is a linear map and perfectly well defined

void cosmos
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whats f

south patrol
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We started with a bilinear map f: M x N -> P

void cosmos
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okay

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now u want to show it factors through T right

south patrol
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Define the map as I did above

void cosmos
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ie the diagram

south patrol
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And yes

void cosmos
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what map

south patrol
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But like

void cosmos
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okay

south patrol
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The one I did above lol

void cosmos
south patrol
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Yup

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Now like

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D is contained in the kernel

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As you can hopefully see

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Check on generators of D

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Like (ax, y) - (x, ay) is sent to f(ax,y) - f(x,ay)

void cosmos
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okay so the kernel of this is all (m,n) such that f(m,n) = 0 , now D is generated by say (x1+x2,y1)-(x1,y1)-(x2,y2)

south patrol
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Which is just zero by bilinearity

void cosmos
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so now yea f(this stuff) would be 0

south patrol
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And similarly for what you just said yeah

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Yes

void cosmos
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since the behind the minus sign will be the same lol

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okay

south patrol
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Yeah

void cosmos
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okay so D is contained in the kernel

south patrol
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Yup

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So it factors through the quotient by D

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By a general fact

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Like if f: X -> Y is a map of R modules and Z contained in kernel of f then we can factor through as X -> X/Z -> Y

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Just like in the first iso theorem I mean

void cosmos
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yea and this would have no problems cuz it would vanish on all of D anyways

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ie it will be well defined

south patrol
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So this means we get a well defined map out of C/D

void cosmos
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yea

south patrol
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If im remembering notation right

void cosmos
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yea

south patrol
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And it is hopefully clear it fits into the diagram

void cosmos
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yea cuz basically f annihilates all relations anyways

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its like a map defined on C

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ig?

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😄

south patrol
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Well relations isn't relevant to moving to C

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Just we have a function into P defined on M x N

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So we can get a linear map out of A^(M x N) into P

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In fact, as I think Troposphere mentioned earlier, that's motivation for how we construct T

void cosmos
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yea it shows from how everything just follows immedaitely from the defintion

south patrol
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So like we know that a map M x N -> P of sets corresponds to a linear map A^(M xN) -> P

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And so to change our focus from the maps of sets to bilinear naps

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We have to quotient out accordingly and make A^(M xN) smaller

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Since we need to reduce the number of maps we get lol since not all maps of sets are bilinear

void cosmos
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okay so

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from MxN to M tensor N the map is defined as (m,n) --> m tensor n

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and then let f:MxN --> H be a bi linear map

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then f induces a map f bar from C/D to H such that f bar of ( m tensor n ) is f

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is that correct?

delicate orchid
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yeah the maps agree, cause of the triangle dude

void cosmos
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yea

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cool

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tysm potato and tropo

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first theorem for me

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lol

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let R be a ring with identity and A be an R-module

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then R tensor A is isomorphic to A

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to prove this can i define a map from R x A to A by sending (r,a) to ra

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this is bilinear

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so now the map from R tensor A to A sends r tensor a to ra aswell

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but now this is only a homomorphism

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how can i find an inverse

south patrol
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Well you can actually just show that A satisfies the same universal property as R (x)_A A

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But yeah the inverse is actually nice here lol

formal ermine
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it's pretty much the first thing you can think of

south patrol
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So you want to define an A-linear map A -> R (x)_A A

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And you have to send 1 -> 1 (x) 1 to be an inverse

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So uh

formal ermine
void cosmos
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yea potato just said it

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it works

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im so stupid haha

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thank you guys

south patrol
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No not dumb but np

formal ermine
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you wouldn't wanna know how long it took me to get a grasp of tensor products and I still don't fully understand them

delicate orchid
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this ^

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luckily I can just kinda black box them as "ohhh le multiplication" or "ohhh le functor"

formal ermine
delicate orchid
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like if you asked me to compute what a tensor product was actually isomorphic to I'd shit the bed

formal ermine
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I'm just saying "oh fuck this universal property I will understand it when I have to use it"

south patrol
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Lol

void cosmos
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yea i think this is the hardest math for me

delicate orchid
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other than C_n \otimes C_m that's a cool one

void cosmos
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i have ever seen

formal ermine
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I've never related to wew this much

delicate orchid
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you've definitely done harder moamen

south patrol
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It will eventually be ok im sure

agile burrow
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Lol extension of scalars

void cosmos
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yea

delicate orchid
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I learnt this in my third year of undergrad

agile burrow
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Those tricky tensors

delicate orchid
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and I've seen you chat stuff about things I've either learnt this year or never heard of opencry

south patrol
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We didn't even do it in our comm alg course

void cosmos
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and it turns out this thing is a functor so i now i have to prove its properties when dealign with exact sequences

south patrol
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For some reason

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Lol

delicate orchid
south patrol
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Bleak

delicate orchid
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weed?

south patrol
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Well it was still lots of content

void cosmos
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they are like page 20 in AM

south patrol
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Like the most of our courses

void cosmos
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this shit is stupid hairy

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who came up with this

agile burrow
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Mr tensor

void cosmos
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no way

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let me google this guy

south patrol
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Mr tensor

void cosmos
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??

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it was in the 180s?

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1890s*

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Tensor calculus was developed around 1890 by Gregorio Ricci-Curbastro under the title absolute differential calculus, and originally presented by Ricci-Curbastro in 1892.

south patrol
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This is just a universal property and those take a while getting used to

void cosmos
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wtf this is new af its like 2002

south patrol
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Tensor calculus is p different lol

void cosmos
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yea okay wait

south patrol
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This textbook is much older than that lol

delicate orchid
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oh yeah tensor calc is like

south patrol
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I'm p sure

void cosmos
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oh its 1938

delicate orchid
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fuck all to do with tensor products hahaha

south patrol
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Not that old a textbook

void cosmos
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by some guy called whitney

south patrol
void cosmos
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yo look at this guy

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-1

The tensor product is actually a very simple concept.

It goes back to Babylonian times when people realised that two edges describes an area. Intuitively they realised that geometric area was bilinear but all this wasn't formalised until the twentieth century.

Mathematicians, being mathematicians, generalised by allowing the edges to take values in any vector space, as well as the attendant geometric area.

That formalisation took so long shouldn't be such a surprise. After arithmetic was invented well before Babylonian times and yet was only formalised with Peano's axioms in the early 20th C. I'd also add that it has taken the subject so far away from its geometric roots that its difficult to see the geometry of a tensor.

Its worth adding that the tensors of general relativity, which made the subject famous, are actually fields of tensors, that is tensor fields. Moreover, the tensors they use are of a special kind, comprising of a tensor power of a vector space tensored with a tensor power of its dual space.

delicate orchid
void cosmos
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from the depths of stackexchange

agile burrow
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I mean grassman had like exterior product and stuff

void cosmos
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yea its whitney and bourbaki

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1940s

agile burrow
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There's also Kronecker product of matrices but idk if Kronecker invented that

void cosmos
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it seems to me that all algebra is either noether or bourbaki

south patrol
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Lol

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Hilbert

void cosmos
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yea thats advanced

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i did cite a theorem of his tho

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in an exam

south patrol
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But ye lots of bourbaki

agile burrow
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He had a lot of those

delicate orchid
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fun exercise, prove that the Kronecker product corresponds exactly to the image of a pair of matrices under the canonical map into the tensor product of two vector spaces

agile burrow
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At least 90

south patrol
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Isn't that basically the whole point

delicate orchid
south patrol
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And I don't think the map is canonical lol

void cosmos
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yo bourbaki is actually 10 people

south patrol
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After picking a basis this is

void cosmos
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lol

south patrol
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Well

void cosmos
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what a group of nerds

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smart tho ngl

delicate orchid
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you know EXACTLY which map I mean

south patrol
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Which one

delicate orchid
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the canonical one

south patrol
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Say I give you vector spaces V and W over F of dimensions m and n which canonical isomorphism between V (x) W and F^(mn) are we using

delicate orchid
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you know EXACTLY which one

void cosmos
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yo tensor product is associative tf

delicate orchid
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and distrubutes like multiplication over direct sums

south patrol
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Most operations we deal with are here

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But yes this is particularly nice

delicate orchid
void cosmos
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does it have a similar property with direct sums like the Hom one?

delicate orchid
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they're both additive iirc

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so yeah

void cosmos
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like Hom(sum A_i , B ) = product Hom(A_i,B)

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this is was the only good thing i saw

south patrol
void cosmos
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so its a functor too

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right?

south patrol
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Wdym

void cosmos
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like tensor product sends things from direct products to

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normal modules haha

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nvm

south patrol
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But yeah it's additive as wew said

celest cairn
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Random question:
Does the discriminant of a biquadratic polynomial help determine its Galois group, or is this only the case for cubics?

south patrol
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Yes it does

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Though biquadratic can have a couple of meanings

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But yes in fact discriminant always gives info on Galois group regardless of degree

delicate orchid
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one of the coolest things in maths for me was realising you could make rings using the tensor product and direct sum

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the first big leap in generality, the first peep down the rabbit hole

south patrol
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For example if we embed the Galois group of a separable irreducible polynomial of degree n with the set of permutations of the roots then it is contained in A_n iff the discriminant is a square in the base field

void cosmos
delicate orchid
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ok they're semirings but just take the grothendieck completion you nerd

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actual nerds

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shut up nerd

void cosmos
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they are just the same

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rings

south patrol
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Lol

void cosmos
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that you know about

south patrol
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Wdym

void cosmos
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like there is an exercise proving Z_n tensor Z_m is just Z_(n,m)

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"just like Hom tf"

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its all al ie man

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a lie*

delicate orchid
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not what we're saying

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you can make rings where the elements are modules

south patrol
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Lol

void cosmos
delicate orchid
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for example, the representation ring

south patrol
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Ye

void cosmos
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so and the tensor product is additive?

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so ur action is the tensor product or what

south patrol
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So for example fix a field k

delicate orchid
south patrol
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We consider the set of isomorphism classes of (finite dimensional) vector spaces V

void cosmos
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yea

south patrol
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You can put addition and miltiplication by ike

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[V] + [W] = [V (+) W]

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And similarly with tensor product

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That gives us a semiring

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Like no additive or multiplicative inverses unfortunately

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But everything else works nicely

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Like associativity etc

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And in fact this is basically just N

void cosmos
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yea i get you now

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yea this is like a step up

delicate orchid
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it's N if V is a vector space lol

void cosmos
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this is inhuman mann

delicate orchid
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oh nvm u said fix a field

south patrol
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Ye

delicate orchid
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and that's how you get characters

south patrol
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But here one interesting direction you can go is like

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Vector spaces parameterised in a suitable sense by topological spaces lol

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And that gives you lots of info

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And stuff

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So like this idea wew and I mentioned is v useful

delicate orchid
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yeah and there's a nice way to get inverses too

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for modules over semisimple rings you literally just add in negatives as formal classes

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slightly more complicated for other rings but it amounts to the same thing

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you want a thing X such that V+W+X = V

void cosmos
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yoo , an element in (A tensor B) tensor C would look like sum(h tensor c_i where h is in A tensor B which which that gives us sum(sum(a_i tensor b_i) tensor c_i , now if we send the element t(sum(a_i tensor b_i),c_i) to sum(a_i tensor (b_i tensor c_i)) this would be a bi-linear map from (A tensor B) x C to A tensor ( B tensor C)

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which gives a homomorphism

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and doing it the other way around gives us the inverse?

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proving the tensor is associative.

south patrol
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You can think about universal properties purely

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Rather than worrying about a given construction

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Show that the tensor product A tensor (B tensor C) is euivalently just the "triple" tensor product

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Which is mentioned in AMD im p sure

delicate orchid
#

very similar universal property thing shows it's commutative also

south patrol
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Ye

formal ermine
void cosmos
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Is what i said incorrect

delicate orchid
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dunno I didn't read it, too many letters

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I think I believe it

hollow mica
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it's common to define the degree of the 0 polynomial as -infty so as to preserve the identity deg(fg) = deg(f) + deg(g). we do a similar thing with the characteristic of rings. if in a ring R it is never the case that 1 + 1 ... + 1 = 0, we say that char(r) = 0. Though there probably isn't a connection between these two instances, why is it that defining edges cases "in the other direction" often plays nicely with the properties of a theory?

tribal moss
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The characteristic of a ring is the generator of the kernel of the (unique) ring homomophism from Z. So characteristic 0 is not really a special case, as far as definitions go.

hollow mica
#

Ah these things aren't related then

rocky cloak
hollow mica
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I meant like what the intuitive value of what a quantity should be vs the value we actually set it to be. so for those rings it's infinity vs 0, and for the degree of the 0 polynomial it's 0 vs negative infinity. Of course I'm just saying hogwash

rocky cloak
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Well, for polynomials the degree acts a bit like a logarithm, and log(0) = -inf, so I guess that's how you would expect.

Characteristics of rings are more "ordered" in terms of their divisors, and every number is a divisor of 0. So in that sense 0 is the "biggest" number.

delicate bloom
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by polynomial I guess I should say, laurent polynomial

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$ord(f(x^{-1})) = - deg(f(x))$

cloud walrusBOT
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merosity

622029249053982720.png
delicate bloom
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so you'd be doing some flipping here, although I'm not sure if this is entirely related to the kind of flipping you're thinking about or not

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in addition to deg(fg)=deg(f)+deg(g) you also get deg(f+g) <= max(deg(f), deg(g)), with equality when deg(f) != deg(g). Point of saying some of that is to say you get exactly everything of that transferring over to analogous stuff about orders, so they're like two sides of the same coin there

maiden shuttle
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Why, according to the Correspondence theorem of rings, does the size of the ideal seem to be linked to the completeness of the algebraic structure of the quotient ring, and algebraic structure link to order structure?

coral spindle
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the completeness...?

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Dunno what you mean exactly

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The correspondence theorem is not as deep as you seem to be implying lmao, it's essentially saying something about divisibility. You could see it as relating the structure of divisibility in the quotient to the structure of divisibility in the original ring.

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And idk that just seems like, obvious that these things would be linked

maiden shuttle
#

If the definition of the prime ideal can also remind me of 0 (ab=0 iff a=0 or b=0), I can vaguely feel the relationship with the definition of the integral domain, then for the maximum ideal, I really can't see anything related to the definition of the field. So I started thinking about the relationship between the maximal ideal definition and the field definition

coral spindle
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When you quotient by an ideal, this can be thought of as forcing the ideal to be zero. With the correspondence theorem, this is saying that if we quotient by a maximal ideal, in the quotient the zero ideal will be maximal, i.e. we have a field.

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Same thing with prime ideals, primary ideals, etc.

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Exercise: prove all this carefully.

delicate orchid
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If you want to go way back you could consider a local ring - the unique maximal ideal in a local ring contains all non-units, so it makes sense that quotienting out by this gets you a field

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A similar-ish thing holds in all other rings. But if you want an actual rigorous reason it’s what Mr. Terrible flat said

coral spindle
#

Thank u mr ladz

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Actually I don't know your pronouns

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Is mr ok?

delicate orchid
#

I can be whatever u want bby

coral spindle
summer path
coral shale
#

Sir wew

coral spindle
#

Queen wew?

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Imperator wew?

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wew-denka?

delicate orchid
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I prefer male pronouns but as long as it’s clear who you’re referring to I don’t mind

coral spindle
#

<@&268886789983436800>

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So fast ivyoarFlushed

fair quartz
#

id like to understand/prove this identity
could you guys suggest some approach to take or some identity that helps to get there?

image.png
#

( U is unitary btw)

coral spindle
#

What is U?

fair quartz
#

ah a unitary operator

coral spindle
#

I mean I think this depends on the particular operator, no?

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Oh wait no

delicate orchid
#

also not really AA lol

fair quartz
#

our script just says that it should work for any unitary operator

delicate orchid
#

This just boils down to the fact that the only place on the complex plane where the complex conjugate is the inverse is S^1 rigjt

coral spindle
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Oh I thought this was a hilbert space thing

delicate orchid
#

Am I getting it confused with a unitary element

coral spindle
#

I honestly don't know

fair quartz
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oh what id give to still have any idea of what particular space we are currently working with in our lecture 😂

coral spindle
#

OK but you have an inner product space at least?

fair quartz
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that seems to be all the example is giving

image.png
delicate orchid
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You have to for it to be unitary

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Oh does this mean it’s finite dimensional?

fair quartz
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it has to be a hilbert space we are always working with them

fair quartz
coral spindle
delicate orchid
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But I can’t remember it as I am a group theorist opencry

fair quartz
# coral spindle wat is B

id love to know that too, the notation just started to appear in our script

i think it just denotes some banach space? (assuming that makes sense)

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ah no

#

its the uhhh
looks up the translation

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dual space!

delicate orchid
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We know for unitary matrices that eigenvalues must have an absolute value of one

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How do I make this infinite dimensional

coral spindle
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I'm being stoopy give me a sec

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0 < <Uv, Uv> = <kv, kv> = k k^* . <v,v> right

formal ermine
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algebraists try doing analysis (colorized)

delicate orchid
#

Let x be an eigenvector of U
<Ux, Ux> = <ax,ax> = aa*<x, x>
But we know this is equal to <x,x>, so aa* = 1

coral spindle
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Yeah

delicate orchid
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That’s how it goes for matrices anyway

coral spindle
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Wait why is it equal to <x,x> sorry

delicate orchid
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Definition of unitary

coral spindle
#

Oh right unitary dudoy

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@fair quartz there ya go

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I was so close lmao

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Always use all the information you're given lmao

delicate orchid
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Now all you need to do is prove a* = a^-1 <=> a \in S^1

coral spindle
#

Weeeeelllllll

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You know what, Ludicrous can do this, what am I typing for

delicate orchid
coral spindle
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Yeah I reckon not

summer path
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This isn't even really analysis

delicate orchid
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It’s a more sinister third option

summer path
#

It's like lin alg with slightly mysterious notation

coral spindle
#

It's........ elementary 😱

delicate orchid
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Elementary operators

fair quartz
#

i see, thanks!

fair quartz
delicate orchid
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Circle

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⭕️

fair quartz
#

ahh unit circle?

delicate orchid
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Yur

fair quartz
#

thx!

formal ermine
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S^2 would be the b a l l

delicate orchid
#

It’s not explicitly the embedding into C but I was being lazy

inland otter
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Let G be a finite group and let A be a G-module which is also injective as an abelian group. Is it still true the A is acyclic for the G-fixed points functor?

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Oh no wait fields of characteristic 0 aren't acyclic under the Galois action

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nvm

coral spindle
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Is every finite extension of a local field also local? I know zilch about local fields.

tribal moss
summer path
#

Sqrt(2) is already in R, or am I missing something

coral spindle
#

So sqrt(2) isn't in Q_5, right

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If we looked at Q_5(sqrt(2)) would that then give us another p-adic

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Yeah ok I just needed to read more carefully

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A finite extension of a p-adic field is explicitly listed there

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goooooooooood to know.

coral spindle
rocky cloak
#

The characteristic 0 cases seem to be closed under finite extensions, but the characteristic p case only lists F((t)), so the question is what are the finite extensions of F((t))

This MSE suggests they should be closed under finite extensions

https://math.stackexchange.com/questions/3424134/finite-extension-of-the-field-of-formal-laurent-series-over-a-finite-field

Mathematics Stack Exchange
Finite extension of the field of formal Laurent series over a finit...

Let $K=\mathbb{F}q((T))$ and $L/K$ be a finite extension of degree n, then is it true that $L \cong \mathbb{F}{q^n}((T))$ as fields?
Thanks!

coral spindle
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Plus I didn't mention that I only cared about the p-adics, so that was my bad

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Yeah

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Thanks all happy

fair quartz
coral spindle
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So we've shown that any eigenvalue is in S^1

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And the spectrum consists exactly of the eigenvalues

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so.............................

fair quartz
elder wave
#

I mean

coral spindle
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What's the definition of the spectrum

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What definition do you know

elder wave
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Yes i'm wondering as well now

coral spindle
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Because as far as I'm concerned if you were just working with something without knowing the definition, that's on you lmao

fair quartz
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i know the one with lambda*identity matrix - operator must not be invertible

coral spindle
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Right OK

south patrol
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so what is an eigenvalue to you

fair quartz
#

and i find that definition very unintuitive to work with

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i did notice the resemblance with the eigenvalue equation

coral spindle
#

I am inclined to agree

fair quartz
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but i never managed to draw the connection between finding eigenvalues and something being not invertible

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it almost seemed to me as if it was the opposite, that youd find the eigenvalues in the set of resolvents (is that the name? xD)

coral spindle
#

OK. As it happens, a linear map is invertible iff it is surjective and injective

delicate orchid
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the definition of an eigenvalue involves something not being invertible
I am concerned about your course now lmfao

coral spindle
#

So we're looking only at the finite-dimensional case, I guess this doesn't really matter, but

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A - lambda I being non-invertible means there is some nonzero vector v such that (A - lambda I)v = 0

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I.e., Av - lambda*v = 0

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I.e., Av = lambda*v

fair quartz
fair quartz
delicate orchid
#

"the eigenvalue equation" is this not exactly the determinant of something being equal to 0

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aka something not being invertible