#groups-rings-fields

406252 messages · Page 510 of 407

next obsidian
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That means uhhh

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all finite ones are f.g. implies Noeth

latent anvil
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yes

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So I'm taking

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Showing its finitely generated

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And what I want is for A^m -> I to be surjective

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Snake says it suffices for A^n -> A to be surjective

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Working on that now

next obsidian
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So you're trying to show that I is f.g. right?

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Oh lol

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Brendan

latent anvil
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?

next obsidian
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I want to tell u ur being dumb

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because ur being dumb

latent anvil
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lol

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I'll think about it

next obsidian
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no

latent anvil
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oh

next obsidian
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look through your book again

latent anvil
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fine just go ahead and tell me

next obsidian
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at the theorems

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2.6

latent anvil
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oh

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yes

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I did this

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Lmao

next obsidian
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yeah

latent anvil
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What was your proof?

next obsidian
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I did it not contrapositive

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I said let A be non-noetherian

latent anvil
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Oh hang on a sec

next obsidian
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I an ideal not f.g.

latent anvil
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About that lemma

next obsidian
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yeah

latent anvil
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I didn't understand the proof

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But I just did it myself

next obsidian
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Oh, I did it manually

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yeah

latent anvil
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and it made sense

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Year lol

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Okay go ahead

next obsidian
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anyway it's literally almost the same

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I not f.g.

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exists cuz not Noeth

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look at 0 -> I -> A -> A/I -> 0

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A/I is f.g. if f.p. by 2.6 I is f.g.

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lol

latent anvil
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this is not a proof by contrapositive lol

next obsidian
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I mean

latent anvil
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this is you saying "assume not" at the start of a direct proof

next obsidian
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Nonono

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I mean

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your proof is the contrapositive of

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"Show for A non-noetherian there is a f.g. module not of f.p."

latent anvil
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oh right I forgot what was actually in the book

next obsidian
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Which is what the book said

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yeah

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I feel like what you proved is the

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more natural statement

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and I did the contrapositive of it

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but according to the book haha

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Also I remembered something

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remember how once you thought Noeth was equivalent to acc for prime ideals?

latent anvil
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Yes

next obsidian
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How funny that no, it's the other condition about Noeth involving ideals for which it suffices to check on primes

latent anvil
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Well that's not a coincidence

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I was misremembering the correct statement

next obsidian
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Is it in AM?

latent anvil
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Which I'd heard shortly before

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No I don't remember where

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Anyways section 3 done

next obsidian
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Nice!

latent anvil
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Just 2 to go before Hartshorne

next obsidian
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yup

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section 4 problems are realllly easy

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since you've done AG

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it's a ton of stuff about Spec and stuff and most of my proofs are like "have done b4" or "follows by correspondence thm"

latent anvil
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Nice

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oh lol

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Localization is great

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should I do sheaf stuff

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I should probably do sheaf stuff

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ugh

next obsidian
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Sheaf stuff got easier

latent anvil
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yeah I'll do the problems I haven't

next obsidian
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I think u did most of the hard sheaf stuff in Hartshorne

latent anvil
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I just want to enscheme myself

next obsidian
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my solutions for section 4 lol

latent anvil
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Yo don't post that

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I do not want spoilers

next obsidian
solemn rain
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what does upside down product mean

fair obsidian
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Disjoint union

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its when you take the product but go the wrong way

solemn rain
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disjoint union?

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isnt it just the big cup

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$$/bigcup$$

cloud walrusBOT
old hollow
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$$\bigcup f$$

cloud walrusBOT
solemn rain
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yea is that the same as product upsidie down?

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okay

upbeat burrow
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So do we use upside down sum for repeated subtraction?

latent anvil
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no

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you use it for repeated coaddition

fair obsidian
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Is coaddition a "commutative" comultiplication? So like the image of the map $\Delta: A \to A \otimes A$ lies in the symmetric tensors?

cloud walrusBOT
fair obsidian
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or maybe it should be called a cocommutative comultiplication

latent anvil
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lol

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okay so

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that's not exactly what cocommutative means

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it's okay to have Δ(x) = a (×) b + b (×) a

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Commutative is the same as m ° τ = m where τ : A (×) A -> A (×) A is the "twist" isomorphism τ(a (×) b) = b (×) a

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cocommutative is τ ° Δ = Δ

fair obsidian
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Right

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Isn't that what having each Δ(x) be symmetric means? I don't get your example

latent anvil
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oh wait you're totally right lol I forgot the defintion of symmetric tensor

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I was thinking it was the thing spanned by all a (×) a but I'm totally wrong

fair obsidian
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Ah ok

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That's like the ideal you use for the alternating algebra iirc

latent anvil
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then yeah that's the right definition of cocommutative

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for a coalgebra

fair obsidian
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Is the condition cocommutative ever used?

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I guess it would be weird if you had a commutative multiplication but non-cocommutative comultiplication. Maybe impossible for a bialgebra, I really don't know much about this stuff other than the definitions

latent anvil
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I couldn't really say, I learned a very small slice of hopf algebra stuff for a project

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you can have a hopf algebra with any combination of the properties iirc

fair obsidian
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Oh ok

latent anvil
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so a group ring is a hopf algebra with comultiplication just g |-> g (×) g, and this is cocommutative, but taking G nonabelian we get k[G] not a commutative hopf algebra

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you can get both with G commutative

fair obsidian
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That kinda makes sense

latent anvil
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I want to say if you have an affine algebraic group G, then the coordinate ring of G will be a hopf algebra which is always commutative and is cocommutative iff G is a commutative group

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maybe a (ring of functions on a) lie group would be a simpler example? It should work too

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the idea is that multiplication dualizes to become comultiplication, but multiplication is just like f |-> f (×) f

upbeat burrow
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I’ve never heard of cocommutative 😐

latent anvil
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and finally sweedler's 4d hopf algebra is neither

fair obsidian
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Thanks I'll look into those examples

latent anvil
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oh lol Wikipedia has a table with this info

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https://en.m.wikipedia.org/wiki/Hopf_algebra

Hopf algebra

In mathematics, a Hopf algebra, named after Heinz Hopf, is a structure that is simultaneously an (unital associative) algebra and a (counital coassociative) coalgebra, with these structures' compatibility making it a bialgebra, and that moreover is equipped with an antiautomor...

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so basically hopf algebras are just groups but in fancier categories

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that's how I think of these examples

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you can categorify the definition of "hopf algebra" to a "hopf algebra object" in any (symmetric?) monoidal category

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in a cartesian monoidal category (monoidal product = categorical product) this reduced down the definition of a group object, and the comultiplication Δ : G -> G×G is just the diagonal map

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now if you have a functor F : C -> D which preserves monoidal products, it'll send hopf algebra objects to hopf algebra objects. Taking C to be cartesian monoidal and D = k-Vect we get that F sends group objects to hopf algebras

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this generates the examples of group ring, functions on an affine algebraic group, cohomology group of a lie group, etc

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neat stuff imo

fair obsidian
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Right I saw that a "hopf algebra object" in the category of sets is just a group

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I'll think about the thing you said about functors and how you can create group rings, ...

latent anvil
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well, there's a free vector space functor Set -> Vect

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and F(V × W) is canonically isomorphic to F(V) (×) F(W)

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so it sends group objects (groups) to hopf algebras

fair obsidian
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Yeah makes sense

latent anvil
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that's the group ring k[G]

fair obsidian
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Oh yeah that's really nice

latent anvil
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Suppose F/k is some field extension and F/k({αi}) is algebraic. Does {αi} contain a transcendence basis?

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Okay I can Zorn up to a subset which is maximal among algebraically independent subsets of the αi. If F/k(this subset) isn't algebraic, there's some element x in F which isn't algebraic. But x satisfies a polynomial in all the αi, so there's some αi which is not algebraic over k(my subset). Adding this to my subset gives a larger one, contradiction

latent anvil
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oh good matsumura is claiming things again and not explaining

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Let p < q be primes of A = k[x1,...,xn]. Then tr.deg_k κ(q) < tr.deg_k κ(p)

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why is this lol

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The book says we have a surjective map k[X]/p -> k[X]/q, sure, and so the weak inequality tr.deg_k κ(q) <= tr.deg_k κ(p) is obvious

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I don't see why this is obvious lol

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hmm what if like

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take a transcendence basis for κ(p)

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wlog all the elements are actually in k[X]/p by scaling

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then if you push them through the map and adjoin to k maybe κ(q) is algebraic over that?

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yeah it is

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every element of k[X]/q is algebraic over it by pulling back to k[X]/p

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quotients of algebraic elements are algebraic so you get the whole field of fractions

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groovy

solemn rain
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are exact sequences hard?

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im in that section of atiyah and im just getting fucked

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and idk why is thiis shit useful

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they were easy in dummit and foote wtf

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and wtf there are so many theorems

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am i supposed to memorize this shit?

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like the theorems arent easy ilke

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like an easy theorem for me would be something thats general and strong like doesnt want necessary conditions

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but this shit has ' suppose this suppose this ssuppose this suppose this then this + this is that '

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is this normal

carmine fossil
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Group is 0

solemn rain
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how can you define some a homomoprhism from 0 to a module

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i dont see it

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0 ---> 0?

chilly ocean
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{0}

carmine fossil
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How do you define a morphism from a trivial group?

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To a non trivial group

solemn rain
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idk thats what im asking

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sends 0 to 0?

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yea

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should be

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yea so ker of that would be 0

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no not ker

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img of that would be 0 which means ker is 0

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which means f is injective yea

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then what im sayign is right ?

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yea

carmine fossil
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Yes

solemn rain
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cool\

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now wtf is this thoerem

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2.9

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how am i supposed to figure that out

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it doesnt even make any sense

carmine fossil
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It's pretty complicated,and aluffi made it an exercise

chilly ocean
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why does it not make sense

solemn rain
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;like there is no

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'naturalism'

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like it just seems random as fuck

carmine fossil
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They are all R module homomorphisms,with a special condition

chilly ocean
solemn rain
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yea idk

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how to saay it but

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like you know

carmine fossil
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Unnatural?

solemn rain
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when you see a theorem you go yea thats obv but i wonder wahts the proof

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this is not

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this is just weird

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yea

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unnatural

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totatlly unintuitive in any sense

carmine fossil
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it's pretty intuitive if you write it down and understand what it means(The statement to prove is)

solemn rain
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can u prove it

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for me

carmine fossil
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Define a morphism from Hom(M*,N) to Hom(M,N) by
$a(\phi)=\phi v$ and morphism from Hom(M,N) to Hom(M',N) by $b(\phi)=\phi u$

cloud walrusBOT
carmine fossil
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Now show a and b satisfy those conditions

solemn rain
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okay

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supposing the sequence M' --> M --> M'' --> 0 is exact

carmine fossil
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Yes

solemn rain
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with u:M'-->M and v:M-->M''

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now all i literally know

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literally

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is that ker(v) = img(u)

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and img(v) = 0

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?

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no

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nvm

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M''-->0 is exact

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so ker(0 map) = img(v)

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wtf is ker(0map) umm

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0

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so img(v) is 0?

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is that right

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lmao okay

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nvm

carmine fossil
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No,The entire set

solemn rain
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whats the entire setr

carmine fossil
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Entirety of M"

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img(v)=M"

solemn rain
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why

carmine fossil
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Because ker of that last one is M"

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You are mapping a non trivial group to a trivial group

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So, Everything gets mapped to identity

solemn rain
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yea obv

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lmfao yea

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okay

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okay so now define phi: Hom(M'',N) --> Hom(M,N) by phi(f) = f(v)

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what does f(v) mean lmfao

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v is a map

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composite?

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okay yea composite nvm

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and now defiine phi': Hom(M,N) --> Hom(M',N) by phi(f) = f(u)

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basically all i want to show is ker(f(u)) = img(f(v))

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right?

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and we have 0 --> Hom(M,N) aswell

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so ker(f(u)) must equal 0 ?

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as in img(0) ?

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im sorry if all of this wrong but its hard for me

carmine fossil
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right?
@solemn rain yes

solemn rain
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lmao okay

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i want to show ker(f(u)) = 0

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and img(that) = ker of that

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okay

carmine fossil
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The first one is easy

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And the second is not

solemn rain
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which first one

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and whnich sedc

carmine fossil
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i want to show ker(f(u)) = 0
@solemn rain first

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(Ker phi is 0 ,which means fu=0 implies f is 0 map),is what you want to prove

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Not what you stated

solemn rain
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so i want to prove that the set of all gs such that g(f(v)) = 0 is 0

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right?

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thats the kernel no?

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so to recap define phi: Hom(M'',N) --> Hom(M,N)

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by phi(f) = f(v)

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and define Phi bar : Hom(M,N) --> Hom(M',N) by phi(f) = f(u)

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i know ker(phi) = img(0 map) = 0

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thats a given right?

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no

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lmfao

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i want to show that

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ker(phi) = 0

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= img(phi bar)

carmine fossil
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Img(phi)=ker(phi bar)

solemn rain
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yea

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= 0

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wait why

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what

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wait

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i mixed them upo didnt i

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no i didnt?

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yea nvm

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i did

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ker(phi bar) = img(phi)

carmine fossil
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Yes

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Continue

solemn rain
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okay so ker(phi bar) = { x | x(phi bar) = 0 } = {x | x(f(v))=0}

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right?

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lmfao im stuck

carmine fossil
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This gets confusing with words

solemn rain
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x is a homomoprhism

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does it have to send identity to identity?

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no right

carmine fossil
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Yes

solemn rain
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its a moduje

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module 1

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okay

carmine fossil
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All group homomorphisms do that

solemn rain
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yea

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okay

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so f(v) must be 0

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right?

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this is so fucking confusing

carmine fossil
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Use symbols

solemn rain
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i am

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that wont help

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its just confusing

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x is ah omo

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it sends ech ident to iden

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so f(v) must be identity

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f(v) must be identity function

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help im dying

carmine fossil
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okay so ker(phi bar) = { x | x(phi bar) = 0 } = {x | x(f(v))=0}
@solemn rain phi bar(x)=0 which means xu=0

solemn rain
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@solemn rain phi bar(x)=0 which means xu=0
@carmine fossil yeda

carmine fossil
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What is f here?

solemn rain
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yea

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idk

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yea i fucked that up

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i fucked up kernel def

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phi bar (x) = 0

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which is x(u) = 0

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now whats next

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x must be 0

carmine fossil
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That is for phi

solemn rain
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as u is not 0 lmao?

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is this right argu

carmine fossil
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We are talking about phi bar here

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u is surjective

solemn rain
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okay

carmine fossil
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So xu=0 implies x=0

solemn rain
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why

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cuz u is not 0

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right?

carmine fossil
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u is surjective

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And x(u) is 0 for all inputs

solemn rain
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so u is nonzero

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?

carmine fossil
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Range of u is M"

solemn rain
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yea

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not 0

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right?

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so x must be 0

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thats what i mean

carmine fossil
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x(element a)=0 if a is in range of u

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And that is all elements in M"

solemn rain
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okayyyyyyyyyy

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lmfao

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okay

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brainlet time

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now Im(phi) = { x in Hom(M,N) | x =phi(h) for some h in Hom(M'',N) }

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i want to show thats 0

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let me try

carmine fossil
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Actually, you have to show im(phi)=ker(phi bar)

solemn rain
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i just showed that thats 0

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tho

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so its the asme as showing this as 0?

carmine fossil
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You showed ker(phi)=0

solemn rain
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no i fucking did not

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what the fuck

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i showed ker(phi bar)= 0 bro

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i wanted to show ker(phi bar ) = im(phi)

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and im(phi) was the kernel of the zero map

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so 0

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whats wrong

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this is so frustrating like hell

quiet cave
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The lack of TeX is infuriating

carmine fossil
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Same

solemn rain
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i dont kno

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how to use tex

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can u use tex for me?

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iam really confused here

cloud walrusBOT
solemn rain
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okay

cloud walrusBOT
carmine fossil
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Required to prove

solemn rain
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let me draw the sequence

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0 --> Hom(M'',N) --> Hom(M,N) --> Hom(M',N)

carmine fossil
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Yes

solemn rain
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for all A-modules N

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yes

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i want to show ker(phi) is img of 0 map

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which is 0

carmine fossil
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Yes

solemn rain
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and i wantt oshow img(phi bar) = ker(phi)

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lmfao finally

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ookay

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so showing ker(phi) is 0 map

carmine fossil
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uh

solemn rain
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what

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no

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yea yea i got them mixed up

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cuz i have like brain damage

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i want toshow ker(phi bar) = img(phi)

carmine fossil
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Yes

solemn rain
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now

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lets start with ker(phi)

carmine fossil
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Use $\phi$ for representing that

solemn rain
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ker(phi) = { x in Hom(M'',N) | phi(x) = 0 }

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phi(x) = 0 --> x(v) = 0

cloud walrusBOT
solemn rain
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now we know M'-->M-->M''-->0 is exact

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and u is from M' to M and v is from M to M''

carmine fossil
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Yes

solemn rain
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so im(v) = M''

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as im(v) = ker(0 map)

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right?

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so v is surjective

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v is not a zero map

carmine fossil
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Yes

solemn rain
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so x(v) = 0 would imply x is 0

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hence ker(phi) is 0

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done

carmine fossil
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Now,next part

solemn rain
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yea

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we want to show ker(phi bar) = img(phi)

carmine fossil
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Yea

solemn rain
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now ker(phi bar) = {x in Hom(M,N) | phi bar(x) = 0}

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img(phi) = { y in Hom(M,N) | y =phi(h) for some h in Hom(M'',N) }

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now let x be in ker(phi bar)

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now what lmfao

carmine fossil
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Well, That's the tough part

solemn rain
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should i say x(u) = 0 --> x(u) = phi(0) --> in img(phi)

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or is that trolling

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lmfao

carmine fossil
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Img(phi) is in ker(phi bar)

solemn rain
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lmfao yea lmfao

carmine fossil
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You have to show the converse

solemn rain
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cuz of waht i just said?

carmine fossil
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or is that trolling
@solemn rain Just trolling

solemn rain
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okay so

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i want to show if x=phi(h) for some h in Hom(M'',N) then phi bar(x) = 0

carmine fossil
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The other way around

solemn rain
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okay

carmine fossil
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If phi bar(x)=0 x=phi(h)

solemn rain
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if phi bar (x) = 0 --> x = phi(h) for some h in Hom(M'',N)

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phi bar(x) = 0 --> x(u) = 0

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now u is injective thats all i know

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here we go again

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im stuck

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fucking bullshit

carmine fossil
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Now,You use the first isomorphism theorem

solemn rain
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yeaa thats the trick

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obv

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how could i have not known

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how do i use first iso thoerem

carmine fossil
solemn rain
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okay

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i got it

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i think]

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i should quit this shit

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tysm for spending the time

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hhelping me @carmine fossil

carmine fossil
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What is the book proof?

solemn rain
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it doesntr prove it

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it just says its easy

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and uses first iso theorem

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bnut it doesnt do it on its own

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it uses the composition which idk why

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wtf

carmine fossil
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Screenshot

solemn rain
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holy fuck theres a square

next obsidian
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Let $p,q$ be primes of an integral domain, with $p\subseteq q$. Then we have a natural inclusion $A_q\hookrightarrow A_p$ given by the localization map. Is it then true that $pA_p$, the maximal ideal of $A_p$ is equal to $pA_q$?

cloud walrusBOT
next obsidian
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Clearly $pA_q = pA_p\cap A_q$, but do we need to do the intersection?

cloud walrusBOT
next obsidian
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Actually, I think this matters not for my purposes, I bet you have to do the intersection tho TBH

next obsidian
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I think it matters again lmfao

upbeat burrow
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Atiyah Macdonald still gives me nightmares sometimes.

next obsidian
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Why

upbeat burrow
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I struggled a lot in that class. I found it insanely difficult.

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I recall the first assignment being like 30 pages of writing.

next obsidian
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Oof

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AM can suck my weiner Matsumura gang

upbeat burrow
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Much higher workload than I was used to, and I found the content difficult. The exercises in the book are great to do, but I found DF taught me the material better.

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DF carried me through AM haha

next obsidian
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I actually still need to know if what I said is true RIP

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I suspect it’s false but

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It’s true when p = 0 lol

latent anvil
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Do you still want me to look at this?

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This seems very very very false

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hmmm

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maybe this is true

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no no no take A = k[x, y] and p = (x), q = (x, y). Then pA_p contains x/y, but x/y isn't in pA_q since if it was, we'd have x/y = ax/t for not in (x, y). But we can cancel x and cross multiply to get t = ay, which is in q

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@next obsidian

next obsidian
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Rip

solemn rain
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@upbeat burrow yea im struggling with it now

prime jackal
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Where can I start learning abstract algebra, any websites you recommend? (I don't want to buy any books until later on)

solemn rain
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you can check out james cook on yuotube

next obsidian
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Don’t buy a book

#

Find this thing called libgen

prime jackal
#

Ok

next obsidian
#

And get a book there

solemn rain
#

oh

#

okay haha

prime jackal
#

No, Libgen is not legal.

solemn rain
#

check out utube then

prime jackal
#

Ok.

#

Yeah better.

latent anvil
#

are you at a university?

prime jackal
#

College yeah.

latent anvil
#

check your university library, they might have a (digital) copy of an algebra book

prime jackal
#

Ok.

latent anvil
#

I anti-recommend youtube videos and websites

#

I think it's very hard to learn this stuff without a textbook, and especially without doing problems

prime jackal
#

Alright.

#

Any book you could suggest then?

latent anvil
#

Abstract Algebra by Dummit and Foote

prime jackal
#

Alright I'll check it out!

latent anvil
#

Have fun! It's a great subject

prime jackal
#

Thanks!

woven obsidian
#

I'm having some trouble with the "only if part"

#

I don't really have any experience working with analytic functions

hot lake
#

they are functions that are locally the limit of their taylor series

woven obsidian
#

Yes, that's the extent of my knowledge

#

I've written up the formula for such an f, vanishing at P

#

I guess I could factor out (x-ai) to get a sum, but not sure how to do it in a natural way

latent anvil
#

Sorry, what's the only if part?

woven obsidian
#

Showing that any function vanishing has the form sum (xi-a1)gi

latent anvil
#

Ah gotcha

#

so

hot lake
#

if you can show it for polynomials, it's basically the same thing except that for analytic functions you add "stuff converges" every now and then

latent anvil
#

hmm I have confused myself

#

You can write f as a series near p and get it in that form

#

but why would that extend to the whole domain? Some kind of identity theorem maybe? Or I'm just being and idiot

hot lake
#

"gi analytic functions near a1...an"

latent anvil
#

oh by "near" do they mean germs?

hot lake
#

they mean on some open containing a1...an

latent anvil
#

ah gotcha

hot lake
#

"analytic functions near P form a ring" I think this means the ring of germs

latent anvil
#

yeah that makes more sense

#

It's just not very clear imo

woven obsidian
#

Hmm

#

I guess I can factor out (xi-ai) without sabotaging the "analyticity" because of absolute convergence or something

latent anvil
#

well

#

something like that

woven obsidian
#

It just feels like I can write it in many different ways

latent anvil
#

You can

woven obsidian
#

But I would expect some kind of unique decomposition

latent anvil
#

But it's the same in k[x1,...,xn]

#

For the ideal (xi-ai)

woven obsidian
#

Ah

hot lake
#

nah the decomposition isn't unique

latent anvil
#

Unique decomposition should be thought of as freeness

#

It's very very weird when an ideal is few

#

*free

#

I want to say you can't have a free ideal of rank > 1 but you might need noetherianness for that

hot lake
#

if you have f(x,y) = xy you have a choice of factoring x or factoring y ; or split it up and do a mix of both

woven obsidian
#

Unique decomposition should be thought of as freeness
Haven't encountered those terms yet, but I'll keep that in mind

latent anvil
#

oh

#

Well

woven obsidian
#

It's the first chapter of Miles Reid's undergraduate commutative algebra

latent anvil
#

Free means you have a basis

#

Like for vector spaces

woven obsidian
#

Looking forward to getting some more "machinery"

latent anvil
#

basis = linearly independent spanning set

#

Which is exactly the same as having a unique decomposition

woven obsidian
#

Ah

latent anvil
#

but this will pretty much never happen. If I <= A is an ideal and I is free then I is principal

woven obsidian
#

I really need to get a better "arbitrary ring" than Z when thinking

latent anvil
#

Yeah haha definitely

#

So my gotos are like

#

Z

#

Fields

#

polynomial rings over fields

#

Z/n

#

Quotients of polynomial rings over field (usually specific ones like k[x, y] /(y^2 - x^3)

#

Infinite product of copies of Z/2Z

#

The localization k[x]_(x) and more generally local rings

#

this is the ring of all rational function f/g with g(0) ≠ 0

woven obsidian
#

That's some good example to play around with

latent anvil
#

yeah haha you build up a library

#

if I want to get really fucked up

#

continuous functions on the unit interval

#

That ring is awful

#

Way too big

woven obsidian
#

Oh, I had never thought of that as a ring

#

But I can see it is

#

I'm still feeling very lost in this course

latent anvil
#

That sucks

woven obsidian
#

Yeah, hopefully it will get better

#

Right now it feels like you need to find a "neat trick" for most exercises

solemn rain
#

EXACTLY

chilly ocean
#

EXACTLY

carmine fossil
#

EXACT

solemn rain
#

fuck exact sequences

#

im never gonna see them again right?

carmine fossil
#

Apparently,They are pretty common

solemn rain
#

i mean they have their own fields ig

#

like homological alg

#

hopefully i am not going to see that much haha

carmine fossil
#

What about Topology?

solemn rain
#

idk

#

i am supposd to learn opoint-set

#

but this shit boring

carmine fossil
#

They are also apparently very common in topology

solemn rain
#

and no pictures so meh

chilly ocean
#

@solemn rain yo homeboy where did you learn group stuff from

solemn rain
#

i told you

#

dummit and foote

#

homeboy

#

and im p bad at math so u shouldnt ask me

#

or maybe uir asking so u avoid those sources

#

yea clever

chilly ocean
#

fuck groups man I don't know if I should gamble and not relearn them or just try to learn them cuz they might be important for my course

#

id rather just learn fields and rings better and assume groups dont exist

carmine fossil
#

You don't understand rings without groups

chilly ocean
#

ofc you do

carmine fossil
#

Rings are literally an extension of groups

chilly ocean
#

I mean I just dont remember all the sylov stuff, the izers and other hard words

#

which dont appear at rings as far as Ive seen

carmine fossil
#

k

chilly ocean
#

rings have better vibe you know im sayin

carmine fossil
#

Field modules are better

#

Much better

solemn rain
#

vector spaces defined on groups 😍

teal perch
#

all of the above are based on groups

#

group gang

chilly ocean
#

is it weird my course covered groups rings and a bit of fields and never mentioned modules?

carmine fossil
#

You never did la,ig

teal perch
#

modules will come later probs

chilly ocean
#

I did a lot of la

carmine fossil
#

Then,you dealt with lots of modules

chilly ocean
#

sure, but not from like abs algebra pov

teal perch
#

modules are generally ring theory specific

#

so unless ur taking a course on ring theory

solemn rain
#

godel

#

do u know rings

chilly ocean
#

yes at least I used to

#

a bit

solemn rain
#

modules are vector spaces on rings

#

you can also think of them rings acting on abelian groups

chilly ocean
#

yes I know that but like ive never heard about them in algebra course

#

maybe in the other one Im gonna take

solemn rain
#

yea you can call them higher ig lmfao

chilly ocean
#

fuck tbh I never understood how group action worked

#

I kinda got it very very late when I didnt need to know it anymore

carmine fossil
#

Vectors are modules over a field

solemn rain
#

a group G acts on a set S

chilly ocean
#

in diff eqs course

solemn rain
#

a group action is a map GxS-->S g*s such that

#

1*s = s for all s

chilly ocean
#

wow thank you mo2men very cool!

solemn rain
#

g*(hs)=(gHs)

#

thats it

#

really

chilly ocean
#

yeah I think it kinda clicked to me when I learned about sth called 'flow' in differnetial equation and you could kinda define it as a group action and it made a lot of sense if you wanted to imagine it

fair obsidian
#

I'm just learning some module stuff too. Question: is there an abelian group which when looked at as an R-module over any ring R (and any choice of scalar multiplication by R) isn't generated by a single element?

carmine fossil
#

Non cyclic Vector spaces?

fair obsidian
#

Do you have a specific example though?

carmine fossil
#

Take space of polynomials degree <=2

fair obsidian
#

So like R^3 right?

carmine fossil
#

Yes

fair obsidian
#

You can take the polynomial ring R[t] and have it's action on R^3 be have constant polynomials just scale as usual and t be the linear map that shifts the coordinates down

#

So like t*(a, b, c) = (0, a, b)

#

And then R^3 under that action is generated by (1,0,0)

#

I think it can't be a vector space because you can use linear maps like that

carmine fossil
#

Ok,Try Taking a ring and the product as a ring action on the abelian group

fair obsidian
#

Like on itself?

carmine fossil
#

Yes

#

Yea,That can be generated by a single element too

fair obsidian
#

Yeah

carmine fossil
#

What about Z/2Z on R?

#

The action is just product

fair obsidian
#

I don't think that's well defined as it would make R characteristic 2

#

Z acting on R works since R isn't a cyclic group

#

But I want a group on which no suitable ring and action can be found

#

And for R you can just take it as an R vector space

carmine fossil
#

Maybe there is an equivalent statement of axiom of choice which says that is always possible?

fair obsidian
#

Right so I think by assuming choice you get that every vector space has a basis, and so you can pick a big enough polynomial ring to be able to permute the basis vectors to make it "cyclic"

#

But I'm hoping there's a weird enough abelian group which can't be made into a vector space on which this fails

#

Right so I think by assuming choice you get that every vector space has a basis, and so you can pick a big enough polynomial ring to be able to permute the basis vectors to make it "cyclic"
Oh you should just be able to pick the ring of linear maps T:V to V and use that as the action, the endomorphism ring

carmine fossil
#

Will your method work with a commutative R module,instead of a vector space?

fair obsidian
#

I don't know enough about modules to say for sure

#

But I think I can reword the question.

#

Since an action of a ring R on an abelian group G is just an action by a subring of the group endomorphisms on G, we might as well just consider the action of the full endomorphism ring on G

#

So the question becomes: Is there an abelian group G so that given any element x of G there's some element y of G which is not of the form f(x) for any group homomorphism f:G to G.

oblique river
#

do you really want y = f(x) or just y can be generated by x?

#

because if that's all you want you can look at G = Z/2Z x Z/3Z. No homomorphism will take (0,1) to (1,0)

#

but that group is generated by a single element as a module over the ring Z/2Z x Z/3Z

fair obsidian
#

But you can choose the element (1,1) to generate instead

oblique river
#

I see, any element

#

so in that case I'd suggest considering Q/Z

fair obsidian
#

Oh ok

#

Because any element has finite order, and you can always find an element which has a higher order

#

Cool

#

So something like the infinite direct sum of groups Z/nZ over the naturals n should also work

oblique river
#

mhm

#

seems right

next obsidian
#

Idk about the thing in the image but

#

you can just show that (gh)^|g||h| = 1

vestal snow
#

Yeah that was the easy part

next obsidian
#

and then if anything lower works you get some contradiction because of coprimality stuf

vestal snow
#

The converse was also easy

#

I was just being dumb

next obsidian
#

ah

vestal snow
#

Actually wait

#

I still don't know if I'm allowed to use stuff about divisibility

#

it's really frustrating because i've done this stuff in the undergrad version of this class and I don't know if using that is allowed or not

woven obsidian
#

How do we know that the extension field will be algebraic?

bleak abyss
#

Depends a bit on m lol

#

m is just some random maximal ideal?

#

Wait I guess this might be Zariski lemma

#

Yeah it should follow from Zariski lemma

#

@woven obsidian

woven obsidian
#

Never heard of it, will look it up

#

m is just supposed to be a maximal ideal

bleak abyss
#

Zariski lemma says that if K is a field that's finitely generated as a k-algebra, then K is finite over k

woven obsidian
#

And finite field extension implies algebraic?

latent anvil
#

this is exactly zariaki's lemma

#

yeah aoi

#

If finite, the set {1,α,α^2,α^3,...} is linearly independent over k for any α in F

#

@woven obsidian

bleak abyss
#

Yeah that's a really cute argument

#

Idk why it's so simple but it's so nice

leaden finch
#

can someone help me out with this fill in the blank proof

woven obsidian
#

Thanks, I'll have to look into that more. We haven't really talked about field extension other than k[x]/(irreducible polynomial) at all

tribal pasture
#

So for this question, my approach is that we will be able to show that Z[X] is a subring of A (just take products and sums of a). Now I have to show that it is indeed isomorphic, but I don't know how to do that.

One strategy would be like to prove that a = 10 as in 1+1+1... But that invokes the knowledge that A = Z which is true, but it's like the info isn't arriving naturally.

📎 unknown.png
knotty mason
#

maybe try R=Z[X]

#

that would give a unique homomorphism phi : A -> Z[X] with phi(a) = X

tribal pasture
#

Yes, that is what I did to prove that Z[X] is a subring of A

knotty mason
#

I don't think that makes sense

#

you mentioned that you constructed an inclusion map f : Z[X] -> A

tribal pasture
#

I meant that there is a subring in A isomorphic to Z[X]

knotty mason
#

f is defined by mapping f(0) = 0, f(1) = 1, f(X) = a

tribal pasture
#

I took the phi function, and then noted that phi(sums of a and products of a ) = sums of X and products of X which generates Z[X]. As such, there is a subring in A isomorphic to Z[X].

#

Now to conclude that A itself is that subring, I have to kind of show that a = sum of 1 or something.

#

The idea would be to use that if it is not then the homomoprhism isnt unique but i dunno how to formalise that

knotty mason
#

phi is a map from A to Z[X], if that is injective, then you can consider A a subring of Z[X]

#

what you're saying seems to be the reverse of that

#

and I don't think it's correct

tribal pasture
#

I mean there is a subring in A which is isomoprhic to Z[X].

knotty mason
#

You can show that Z[X] is a subring of A by constructing an inclusion map f : Z[X] -> A

#

this wont involve phi

tribal pasture
#

Okay thats fine too, but I still have to prove the isomorphism

tribal pasture
knotty mason
#

I realized an alternative way

#

we can show that (Z[X],X) is also initial in the sense that there exists a unique homomorphism from it to any (R,r in R)

#

then you have unique homomorphisms between (A,a) and (Z[X],X) and composing them must give the identity, since the homomorphisms from (A,a) to (A,a) and (Z[X],X) to (Z[X],X) are also unique

steady axle
#

Does anyone know cup products

oblique river
#

@steady axle what kind of cup products? cup products in group/galois cohomology? in the cohomology of manifolds?

steady axle
#

I am sorry. I meanr cup product in group cohomology

#

Thats the only one I knew

oblique river
#

ah okay

#

if you have a specific question about cup products I could maybe give some tips

steady axle
#

I have difficulty following use of cup product in proof of a result

#

I don't understand why the equality in the last line holds

#

The middle equality, others are fine

oblique river
#

let me take a look

steady axle
#

Let me know if any notation is not clear.

oblique river
#

maybe i'm confused, but the first equality in the last line, should that be "= delta(\bar{sigma} - 1) cup ..."?

#

oh no wait

steady axle
#

No

oblique river
#

sorry I think I see now

#

so yeah I think the first equality is just saying that cup product respects taking delta on one side and delta^\vee^(-1) on the other

steady axle
#

We r shuffling delta

oblique river
#

yeah

#

and then the second one I think is using the previous paragraph along with the fact that when the degrees are right, the cup product pairing is jsut evaluating a homomorphism

steady axle
#

Yeah that works without any -1 factor issue bcoz of -2 th tate cohomology grp

#

and then the second one I think is using the previous paragraph along with the fact that when the degrees are right, the cup product pairing is jsut evaluating a homomorphism
I had figured out this except why is cup product evaluation of homomorphism

oblique river
#

so I think it just comes down to the tensor product map underlying this

#

in general, the cup product is a map
H^i(G, A) \otimes H^j(G, B) --> H^(i+j)(G, A \otimes B)

steady axle
#

Umm for both cohomology groups of 0th order the cup product is just tensor product then we also have description of cup product in terms of cocycles but how does it help

#

in general, the cup product is a map
H^i(G, A) \otimes H^j(G, B) --> H^(i+j)(G, A \otimes B)
Yes with some properties relating delta morphism

oblique river
#

if i = 0, then an element of H^0(G, A) is just an element a \in A, and a \cup b = a \otimes b

#

but often times when we're dealing with cup products, we might actually secretly have a map A \otimes B --> C

#

and we can compose with that map and then the cup product turns into
H^i(G, A) \otimes H^j(G, B) --> H^(i+j)(G, C)

#

in this case, A = I_G and B = Hom(I_G, Q/Z)

#

and there is a very natural pairing A \otimes B --> Q/Z given by a \otimes f --> f(a)

#

and that's what's happening here

#

the cup product we care about technically lives in H^(-1)(G, I_G \otimes Hom(I_G, Q/Z))

#

but we are just composing with the map which sends it to H^(-1)(G, Q/Z)

#

and the "evaluation of the homomorphism" part is coming from this tensor product

#

not necessarily from the cup product

#

you can think of the element in H^(-1)(G, I_G) as an element of (some subquotient of) I_G

#

and the element in H^0(G, Hom(I_G, Q/Z)) as an element of (some subquotient of) Hom(I_Q, Q_Z)

#

and so we can pair them I_G \otimes Hom(I_G, Q/Z) ---> Q/Z

#

and just get an element of Q/Z out

#

which is the evaluation of the homomorphism on the element

steady axle
#

if i = 0, then an element of H^0(G, A) is just an element a \in A, and a \cup b = a \otimes b
Doesn't this also require j to be 0 too

oblique river
#

no

#

H^(-1)(G,A) is a subquotient of A

#

meaning "a subgroup of a quotient of A"

#

and H^0(G,A) is "a quotient of a subgroup of A"

#

when you take the tate cohomology

#

with the little vees on top

#

which I haven't written

#

normal H^0(G,A) is a subgroup of A, and the tate H^0 is a quotient of that

#

normal H^(-1)(G,A) is a quotient of A, and tate H^(-1) is a subgroup of that

steady axle
#

Yes H^-1 is ker Nm / I_G A

oblique river
#

yes

#

H^(-1) is basically just H_0

#

which is some kind of mirrored version of H^0

#

which is just to say that H^(-1) behaves more like H^0 than H^1

steady axle
#

Nice I think get it now

#

Doesn't this also require j to be 0 too
@steady axle I think this misconception of mine was the culprit

#

Ur explaination was great thanks

#

May I ask at what stage r u in ur math education / career

oblique river
#

I have a phd in number theory

steady axle
#

Awesome

oblique river
#

and part of the title of my dissertation was "...cup products in galois cohomology"

steady axle
oblique river
#

hi shamrock hehe

latent anvil
#

hello

#

I've been meaning to learn group cohomology

#

But never get around to it

oblique river
#

yeah, for me it was "learn it once I need to know it" as opposed to like... learning it preemptively

#

bert, what is the context you're learning this in?

steady axle
#

Class field theory

oblique river
#

ah gotcha

#

yeah that was kinda where I learned it first too

latent anvil
#

My first course on algebra actually did some group cohomology

#

The dude running it was a number theorist who really liked finite group theory and he wanted to prove schur zassenhaus

#

So he put enough homological algebra and group cohomology on the problem sets to do that

oblique river
#

oh that seems like an adventure

latent anvil
#

and then nobody actually got to the result because it was too hard

steady axle
#

This result is essential for proving that the local artin amps can be extended to abelian closure

latent anvil
#

I was 1.5 weeks from the end and gave up 😢

oblique river
#

:(

#

yeah i'm not surprised

#

also bert glad I could help

#

:)

next obsidian
#

I did it up until like

#

the group cohomology

#

I theoretically learned why Ext is independent of resolution

latent anvil
#

I think the thing that killed me was tensor products

#

Like, Thomas wanted to define the bar resolution

#

so he used tensor products

knotty mason
#

tensor products are cruel

latent anvil
#

And part 3 of 6 on the problem was "learn about tensor products"

bleak abyss
#

Lmao o boi

latent anvil
#

Tensor products are great, but it's a bit much to learn in that context

#

Lemme find the homework

bleak abyss
#

Lol your algebra class sounded fun

latent anvil
#

It was great

#

Best part of freshman year

#

The TA running it was insane

bleak abyss
#

Insane in what way?

knotty mason
#

what's ZG?

bleak abyss
#

Z-linear combinations of elements of G

#

It's a ring where addition and multiplication is exactly what you'd expect

latent anvil
#

Insane in that he covered basics homological algebra/group cohomology, intro ring theory, and galois theory in 10 weeks

#

In a first course

next obsidian
#

Also insane as in

#

Challenge Problem 153 about S^|G|

#

was on our like week 3 hw

#

and was only put on there "because we hadn't defined subgroups yet"

#

and on hw 1 where we just learned about groups he said "prove |g| divides |G|" which is just Lagrange

knotty mason
#

homological algebra/group cohomology, intro ring theory, and galois theory

#

In that order? catAngery

#

S^|G| was tough, i never got that

latent anvil
#

no, homological/ring theory was concurrent for weeks 0-5 and group cohomology/Galois theory was the second half

kindred mist
#

Shouldn't this say algebraically closed field? I think I am missing something

📎 unknown.png
latent anvil
#

I think it should be fine? Take e.g. F(x) = x^2 + 1 in Q[x] and a = 1. Then x^2 + 1 = ((x-1)+1)^2 + 1 = (x-1)^2 + 2(x-1) + 2

#

@kindred mist

#

Like I think you can just replace all the instances xi with xi - ai and use the binomial theorem a lot

kindred mist
#

ah, ok I was discounting the fact that it can be any finite sum

#

my bad, thanks

latent anvil
#

No worries I had the same reaction haha

#

But this is a cool result because it shows that m = (x1 - a1,...,xn - an) is the ideal of functions vanishing at the point (a1,...,an)

kindred mist
#

ok I can see how this would be useful (at least roughly speaking)

#

(this was an exercise from Fulton's AG pdf - the start of it of course)

latent anvil
#

Well it is useful but it's even more about like, painting the general picture of AG

#

Like everything starts with this correspondence between varieties and prime ideals/affine algebraic sets and radical ideals

#

and this is the first step towards that

#

The nullstellensatz is basically a generalization of this result: if k is an algebraically closed field and I some ideal of k[x1,...,xn], and we define V(I) = { (a1,...,an) in k^n : f(a1,...,an) = 0 for all f in I }, then the ideal of functions vanishing on V(I) is the radical of I

#

so in particular if m = (x1 - a1,...,xn - an) then V(m) = {(a1,...,an)} and so the ideal of functions vanishing at that point is r(m) = m

kindred mist
#

Nice, ok yeah I'll try prove that then (the original thing I posted that is), I was watching this video series on com. alg. and Nullstellensatz is the last video iirc, so I definitely like having stepping stones

latent anvil
#

the nullstellensatz is awesome

kindred mist
#

I found this other pdf that had like 6 different versions and one with I(V(I)) = r(I) was the strongest version, so I can try to prove my way through those weaker ones first

latent anvil
#

Haha yeah there's so so many

#

You should not try to prove the nullstellensatz right now

#

I mean

#

You can try

#

But the weak versions are still hard

#

And usually proven with other hard comm alg results

kindred mist
#

I am about 1/2 way through the video series on com. alg. so yeah I'm not really there yet

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I remember doing NAK and tensor products and primary ideals, not much farther than that as of now though

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(I am only a first year masters student though so no rush I guess)

latent anvil
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Yeah for sure

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I'm working through matsumura's book right now

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Wanted to get through the first 5 sections before trying Hartshorne again

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but the last problem of section 5 has been really really hard

kindred mist
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was Hartshorne an AG or com. alg. or...?

latent anvil
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I'm on day 3 or so?

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Hartshorne is the canonical book to learn scheme theory (modern AG). I'm not saying it's good but it's what nearly everybody recommends

kindred mist
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ah I see, I think Ponnen points to that to prep for his "Rational Points on Varieties" text

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there is a seminar for this Ponnen text on my campus atm but I can't really contribute (it is like year 2 AG), but I just like to pop in it and listen

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The commutative algebra seminar on the other hand I do a lot of problems for which is fun

latent anvil
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That makes sense

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Ooh sounds fun

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I'm just sitting around waiting for my classes to start right now

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Doing this comm alg stuff as a distraction

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I won't be taking a class on anything like it until spring

kindred mist
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Ah mine started a few weeks ago, this is why I have gotten less alg./AG done lately, also because of teaching classes

latent anvil
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Ah yeah I decided not to TA this year because of how shit it was in spring (i.e. while online)

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I'm an undergrad so it's not like, required by my program or anything

kindred mist
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ah see I am a TA but what I didnt know until recently (like mid august) is that I am the actual teacher this semester, so thats fun

latent anvil
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Oh wow

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That's not fun to find out on short notice lol

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I mean teaching is cool but not having time to prep is a little ehhhh

kindred mist
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I was kinda suprised bc normally teaching at a university youd have a masters first

latent anvil
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Oh really? I think at my school any PhD student can teach a course

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Maybe they need to be a candidate though, I'm not sure

kindred mist
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I just thought that was the case in general in the US

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well I am a "PhD student" in the sense that I am in a fully funded program and I am expected to make it into the PhD level courses (as I plan to of course)

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funny that I am like the only person without a masters in my program though

latent anvil
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Are you in the US?

kindred mist
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mhmm

latent anvil
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Oh weird

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I didn't think it was very common to come in with a masters

kindred mist
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I didnt either, maybe just something about my uni.

latent anvil
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I don't think any of the first years at my school last year had one

vestal snow
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We can't use Lagrange

latent anvil
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My hint is to think in terms of combinatorics

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Hm that's not super helpful

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Okay hint 2, let m' be the number of elements satisfying g^2 = 1. Then m' = m + 1 (why?). You should try and show n - m' is even instead of showing n - m is odd

vestal snow
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Hmm okay

latent anvil
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Sorry I'm trying not to spoil the answer

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I can give more hints if you're stuck

vestal snow
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no its fine

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I want to try and do it myself first

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I was doing it by cases

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But I thought that there must be a simpler way

latent anvil
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Yeah the proof I have in mind is pretty elegant

next obsidian
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Thx Thomas

vestal snow
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I think I got it

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g is not equal to g^-1 whenever g^2 is not e

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That's thr trick right?

next obsidian
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yee

latent anvil
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Yeah exactly

vestal snow
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Thanks

next obsidian
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$$
\begin{tikzcd}
(A\times_C B)\times_B D\arrow{d} \arrow{r} & A \times_C B\arrow{d}\arrow{r} & A\arrow{d}\
D\arrow{r} & B\arrow{r} & C
\end{tikzcd}
$$

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@latent anvil

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The fact that the two squares are pullbacks means the rectangle is so you get that $(A\times_C B)\times_B C = A\times_C D$

cloud walrusBOT
latent anvil
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Should D be C?

next obsidian
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No that should be D

latent anvil
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I don't get what D is

next obsidian
#

Oh

cloud walrusBOT
next obsidian
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Sorry, this

latent anvil
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ah okay

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So yeah it's a pullback

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Oh I see

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You do this the other way

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Get the same diagram of stuff over C

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Wait no

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The other diagram will be stuff over B

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Oh you're not doing the general case

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Just the one where we want to eliminate B

next obsidian
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I think there’s some condition that lets you do this in general but I’m not sure what it is

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Luckily for projective space C = Spec Z so there’s only one way to be over it so we don’t need to worry about the maps

slate forum
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How do I show for an algebraically closed field k, that if f and g are coprime in k[x,y], then they are coprime in k(x)[y]?

sullen tinsel
#

Say you have 10 shares of TAZ with an average share price of $10. The current share price is $5. How many shares of TAZ at $5 would you need to purchase to bring the average share price down to $6. What formula would you use for this.

cloud walrusBOT
sullen tinsel
#

I need help with that

latent anvil
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@sullen tinsel not here

sullen tinsel
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Please

latent anvil
sullen tinsel
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They sent me here actually

latent anvil
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they did not lol

slate forum
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It may be meant for me

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It was in beta right?

latent anvil
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Sorry are you asking Taz?

slate forum
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Yeah

sullen tinsel
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Yes

slate forum
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That's why taz is here lmao

sullen tinsel
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😂😂

latent anvil
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oh lmao I see