#groups-rings-fields

406252 messages · Page 688 of 407

terse crystal
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I forgot about details. Like we should prove that E={Σab:a from K_1,b from K_2} is an extension of K_1 containing all elements of K_2, and any another field E’ having this property E must be contained in E’ right? We can easily see that E contains K_1 and K_2, and it is contained in E’ for any E’ having such property, I forgot how to prove E is a field

tender mist
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Eh, yeah, I was thinking that closure for the inverse is problematic if we don't have algebraic extensions

terse crystal
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Wait a second,,, it is defined first as K_1(K_2) (=K_2(K_1) ) right?

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Then any element in K_1(K_2) is of the form Σab where a from K_1 and b from K_2

tender mist
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Mm, we need fractions of those, otherwise we just have K_1[K_2] right?

terse crystal
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Yeah so I wrote () not []

tender mist
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So we want to prove that K_1[K_2] = K_1(K_2)

terse crystal
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Oh …

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Yeah…

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Oh that’s why you said algebraic…

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At least one of them should be

tender mist
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Yeah, otherwise I don't see how

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Mm right

terse crystal
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Me neither 😂

tender mist
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Oh it's not true in general

terse crystal
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Thanks you!got it.

chilly ocean
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tbh never knew proof

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why r(p^n)=p

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for p a prime ideal and r being the radical of an ideal

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my idea is that

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x in r(p^n) if x^m in p^n

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not 💯 where to go sadly

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im guessing its some argument where m<n and then m>=n

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wait a second

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is it just that p^n a subset of p

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so x^m in p

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and then we use argument that x or x^m-1 in p

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i mean the definition that it is intersection of all prime ideals containing a might work at making it immediately obvious

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Also is R-p a saturated set. R a ring, p a prime ideal

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I think so because the property of prime means xy not in p implies x not in p, y not in p

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So this means xy in A-p implies x,y in A-p

lavish nexus
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why is it by Yoneda?
homD(B,F-) representable with (U, u) means there's a natural isomorphism between hom_D(B,F-) and hom_C(U,-)

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isomorphism meaning for every A in ob(C) there is a bijection between hom_D(B,FA) and hom_C(U, A)

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I don't see anything to do with Yoneda

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context

unknown.png
broken stirrup
tribal moss
broken stirrup
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what am i supposed to show exactly?

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question is a bit unclear for me

tribal moss
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Do you know what the standard Z-module structure on an abelian group is?

broken stirrup
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yep

tribal moss
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So a Z-module structure is a map Z × A -> A satisfying certain rules.

broken stirrup
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any abelian group is trivially a Z-module by the action n*a=na

tribal moss
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You're being asked to prove that his is the only map Z × A -> A that satisfies those rules.

broken stirrup
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oh in other words, abelian groups can't be seen as different modules right?

tribal moss
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That's another way of putting it, yes.

broken stirrup
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thanks you Troposphere

frank fiber
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Hi, in gap C2:C4 means a semidirect product, but here C2 is acting on C4 or C4 is acting on C2?

lavish nexus
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on the second to last line $\eta_N \zeta_N = 1_{R\otimes N}$

cloud walrusBOT
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Iteribus

230754993756438528.png
lavish nexus
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oh never mind I got it

chilly ocean
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assumption is R is commutative

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mb

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idk how rules for operations on ideals generalize for non commutative rings

lavish nexus
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what does this double arrow mean

next obsidian
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Does it mean that’s it’s a surjection?

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If g:N -> N’’ was a surjection to begin with, then this map is also surjective

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So it would make sense at least

delicate orchid
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it usually means surjection

next obsidian
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Yeah but it normally looks a lil different

delicate orchid
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although the arrow heads are usually touching for a surjection so it's a bit unusual

lavish nexus
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yeah that makes sense

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thank you

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like they're just lazy and didn't write the ->0 part

chilly ocean
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oh lol silly question incoming

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what is relationship between field of fractions of a local ring (R,m) and R/m

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I think we learned that K(R)/m is called a residue field or something

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but i dont remember

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no wait wait

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residue field is R/m

next obsidian
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both are useful in different situations

chilly ocean
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im angry

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im stuck on atiyah macdonald 5.2

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been stuck for 3 hrs

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and homework is due in 18 lol

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but really due in 4

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since i need to sleep

next obsidian
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What’s 5.2 lol

chilly ocean
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quick question

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when is it okay to look up answers vs make half assed attempts

next obsidian
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¯_(ツ)_/¯

chilly ocean
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profs are aware when you look up answers

next obsidian
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For learning purposes, after a few hours IMO. According to your school’s grading policy, I dunno

chilly ocean
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ok

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5.2 says

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A subring of B, and B integral over A, and f:A-> O a ring homomorphism where O is algebraically closed field

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Show that f can be extended to B->O

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afaik f(A) is a subfield

next obsidian
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It isn’t

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It won’t be a field in general

chilly ocean
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wait holdup

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f(A) iso A/ker f

next obsidian
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Yes

chilly ocean
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and ker f should be maximal?

next obsidian
#

No

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Why is the image a field?

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Consider Z -> Q

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The image is just Z

chilly ocean
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mmm

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yea

next obsidian
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The point here is that everything in B is a root of a polynomial in A

chilly ocean
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yeah

next obsidian
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You can push that polynomial through your map f and get a polynomial in O

chilly ocean
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yes okay but there is problem

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what happens to x^i in B under that map f

next obsidian
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x^i is a formal variable

chilly ocean
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wait

next obsidian
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It just maps to x^i

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I just want to turn this polynomial into one with coefficients in O

chilly ocean
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x in B integral over A if x^ia_i=0 a_i in A

next obsidian
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Not quite

chilly ocean
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a_n

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forgot constant term

next obsidian
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It should be x^n = Sum_0^n-1 a_ix^i

chilly ocean
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oh monic

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rip

next obsidian
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But this is sort of besides the point

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If b in B satisfies the polynomial g(x)

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You can just apply f to each coefficient

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And get a polynomial in O

chilly ocean
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yes

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oh okay

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oh shit

next obsidian
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And then what’s special about O?

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That is useful here

chilly ocean
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algebraically closed

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damn

next obsidian
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Right

chilly ocean
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i love you man

next obsidian
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This makes it a little hard but

chilly ocean
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not assuming gender

next obsidian
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Let’s assume B = A[b] for a second

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Try to show that you can extend f to B

chilly ocean
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b attached?

next obsidian
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In this case

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Like

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assume that B is generated by one element over A

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The point is you only have to define the extension on one element

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This is easy to do using the previous observation

chilly ocean
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yeah

next obsidian
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In general B won’t be generated by one element, but you can use Zorn’s

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Namely consider sub rings of B

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Which extend f

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Show by Zorn’s that there’s a maximal subring extending f

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And then because you can extend f by one new element

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This maximal subring extending f has to be B

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Call this maximal subring C

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If C ≠ B you could take a b not in C

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And extend f to C[b]

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And then C isn’t maximal

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Does that make sense?

chilly ocean
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yeah

next obsidian
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This is usually how you have to show maps extend because it’s usually easy to extend a map by one element, but hard to do it with multiple

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So you use Zorn’s to handle the rest after you show you can extend by one

chilly ocean
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So like A[b_1] in A[b_1,b_2] in … A[b_1,…,b_n] in …

next obsidian
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Well we can’t do that because B might not be finitely generated

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That’s why you need to appeal to Zorn’s

chilly ocean
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wait

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is there a problem adjoining countable generators

next obsidian
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Yea

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You can’t induct up to countable infinity

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Only up to arbitrarily high finite

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Also, there’s no reason B is even Countably generated

chilly ocean
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yeah

next obsidian
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(You might be able to use transfinite induction, but idfk, too hard. Just Zorn’s)

chilly ocean
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i would like to know what you mean by fant induct to countable infinity

next obsidian
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Induction only lets you prove statements for n in N

chilly ocean
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yeah

next obsidian
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When something is generated by a countably infinite number of things this is too small to ever reach it

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You only ever get to prove a statement for a finite n in N

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The idea behind induction is you establish the base case to get it started then if you know it for n, you know it for n + 1

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So the point is if you need the statement for m in N, well m = (m -1) + 1

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So if you knew it for m - 1 you have it for m

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But m - 1 = (m - 2) + 1

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So if you knew it for m - 2 you’re done

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…

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Eventually you only need to know it for 0

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But you already proved it for 0

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This won’t ever get to countable infinity, because countable infinity is not + 1 for any finite number

chilly ocean
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yeah

next obsidian
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That’s all I mean that you can’t induct up to countable infinity

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It can never reach it

chilly ocean
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So is adjoining countable elements never defined as a ring? Because using an inductive argument we know for sure that adjoining finite elements is a ring. But we cant use the same argument for a countably inf amount of elements?

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Because I know you are allowed to have polynomial rings with infinite indeterminants

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So id like to think that adjoining infinite amount of generators is also allowed,m

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Or I guess maybe its not that you cant

next obsidian
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You can definitely adjoin infinitely many variables

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You don’t have to define these things inductively so there’s no issue with that

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All you need to do is make sense of a polynomial ring in arbitrary many variables

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And then if you want to take specific elements, not formal variables, you just plug those elements into the polynomials

chilly ocean
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oh okay that is clear af now

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i wonder if it is worth it to learn about these constructions from category theory pov

next obsidian
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These aren’t really categorical

chilly ocean
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ik

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but you can probably find a definition

next obsidian
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I mean you could throw in the word “free algebra” I guess

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But these are all concrete algebraic constructions

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If you don’t already understand them as is, trying to abstract them won’t do you any good

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As a professor once told me, you can’t just abstract all your problems away

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At some point, you have to do math

chilly ocean
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i c

chilly ocean
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@next obsidian did u do this exercise b4

next obsidian
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No

chilly ocean
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that was quick thinking of u damn

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v impressed

next obsidian
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I’ve done similar stuff

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And I’ve done a lot of commutative algebra

chilly ocean
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i actually enjoy zorn lemma type arguments

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the whole constructing the right chain

next obsidian
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Also I think the transfinite induction thing works lmfao

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And it’s pretty easy, but you’d have to know what that is

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Zorn’s is way easier

chilly ocean
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Ive heard of transfinite induction in analysis class

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but the mechanics of it are kinda weird imo

next obsidian
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Yeah it lets you actually induct up to infinite things

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The point here is just that if you have an ordinal which is the union of all smaller ones

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We have to define it for that

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But let me translate that

chilly ocean
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iirc you induct on an inductive argument

next obsidian
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Suppose S is a set of generators, and we know that S = U S_i for all proper subsets S_i

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Suppose we’ve defined the extension on A[S_i] for all S_i

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Then we have to define it on A[S]

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But we can do this like… piecewise!

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Namely like, if b is in A[S], t exists in some A[S_i]

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And then you can send that to where it’s sent in A[S_i]

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And blah blah this is well-defined

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You’ve probably seen this before in analysis, if you defined a map on X and Y and they’re the same on X\cap Y

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You can define a map on X\cup Y

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By just saying “pick a set it lives in, map it to whatever that map does”

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Then it’s well-defined because they agree on the intersection

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In fact to do your Zorn’s argument

chilly ocean
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havent used it on sets ever

next obsidian
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You had to do this

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You had to show if you can extend f to a chain {B_i}

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You needed an upper bound

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So I assume you defined it on U_i B_i

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By doing exactly this

chilly ocean
next obsidian
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Yes

chilly ocean
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it came up in atiyah macdonald

next obsidian
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Why the fuck do

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Oh

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Okay

chilly ocean
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also im taking diff top rn

next obsidian
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Okay

chilly ocean
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so the fundemental is there

next obsidian
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But yes

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This is the idea behind that

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Lmao

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The point is you can define a function on a cover of your thing

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As long as they agree on intersections

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Actually idk if the transfinite induction works, I’m not sure that you can force them to agree on intersections

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Like they’ll agree over A

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But maybe you extended them differently

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Whatever, Zorn’s is just way easier lol

chilly ocean
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yeah

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also v pretty

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im impressed that your first instinct was zorns

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its been used every chapter in this book so far

next obsidian
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I mean when you’ve done a few similar arguments before

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It becomes the first thing to go to

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I can extend by one element so I know if I can produce a maximal thing it’ll be forced to be all of B

chilly ocean
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i wish there was a book like atiyah macdonald but for alg top

next obsidian
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So just use Zorn’s to get a maximal thing

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I don’t like Atiyah Macdonald lmfao

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I prefer Matsumura

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But that book is harder

chilly ocean
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yeah def

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when im more mature ill paruse

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i also have to prepare a presentation tomorrow

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its on the going down theorem

next obsidian
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I mean do you plan to do algebraic geometry or algebraic number theory?

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If not there’s no point in going through Matsumura

chilly ocean
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im honestly not sure if i enjoy algebraic number theory, but it seems like it is one of the only fields that uses algebraic topology sorta often

next obsidian
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🤨

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When the hell are you using AT in ANT

chilly ocean
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atleast things come up like group cohomology

next obsidian
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I mean that’s just homological algebra

chilly ocean
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which is only thing ik that comes up

next obsidian
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There’s a formalism which kind of puts like the fundamental group in a number theoretic context

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But I don’t think that’s really “AT”

chilly ocean
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wait what?

next obsidian
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It might help to know AT for intuition but you aren’t like doing singular homology or whatever

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Yeah

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Look up like algebraic fundamental group

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Or Grothendieck Galois theory

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Or something like that

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This is kinda what anabelian geometry is about too

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Like number theory is “sorta” about studying Galois groups

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Like Gal(Q-bar/Q)

chilly ocean
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oh shit

next obsidian
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And this has an interesting similarity to like covering spaces

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And fundamental groups

chilly ocean
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it just generalizes the fact useful for the correspondence to deck transformations of universal covers

next obsidian
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Yeah that’s a big part of it

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There’s a formalism involving “Galois connections”

chilly ocean
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i really guess i never thought of applying anything to schemes

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or just spec X

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since idk wtf a scheme actually is

next obsidian
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I mean this stuff is hard

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Lol

chilly ocean
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ye

next obsidian
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Making it work is complicated, idk how any of it works

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I’m attending a conference at the end of April about this stuff

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Idk how much I’ll understand tho opencry

chilly ocean
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🙌

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how do you attend conferences as undergrad?

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just sign up?

next obsidian
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I applied for funding

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And they gave me some

chilly ocean
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wait wtf

lethal dune
next obsidian
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Yeah I mean most conferences have funding intended for early career ppl

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Mainly grad students and early post docs

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But I applied and said I’m a graduating senior

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Idk if you’d have much luck if you’re like a sophomore, but I think maybe the organizers are vaguely familiar with me

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¯_(ツ)_/¯

chilly ocean
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o thats cool

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keep winning

next obsidian
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I think in general ppl get funding when they apply tho so

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It might just be that if you write something that seems vaguely competent they’ll fund you

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I’ve only applied to one thing tho so idfk

chilly ocean
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also i have unrelated question

next obsidian
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Yeh

chilly ocean
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idk if ill like algebraic nt because im taking a seminar in it right now, and im not super interested in the results outside of curiosity

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i do like topics covered in AT however

next obsidian
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😬

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Uh huh

chilly ocean
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but my problem is that idk any other fields that apply topics covered other than diff top/geo

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is geometry what i should be looking at?

next obsidian
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I don’t know that geometry really uses AT a ton

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At least in algebraic geometry, the classical AT doesn’t work well I think

chilly ocean
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sucks like hell

next obsidian
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The topologies are incredibly fucked so I think it usually just returns garbage

chilly ocean
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and i enjoy commutative algebra right now

next obsidian
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I think in differential geometry it’s better though

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I don’t really know a lot outside of AG and commutative algebra

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But like, there’s obstructions to certain things in diff geo which live in cohomology classes I think

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Maybe lookup what a chern class is????

chilly ocean
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oh yeah

next obsidian
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But like I think you know you can extend stuff when that thing is 0

chilly ocean
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thats last topic going to be covered in my diff top class

next obsidian
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So if you can show there’s trivial cohomology

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Then the obstruction is 0

chilly ocean
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and apparently they relate to polynomials?

next obsidian
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Which is to say you’re unobstructed

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¯_(ツ)_/¯

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I think there’s more rich interactions with differential geometry and AT is my point

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Also if you want

chilly ocean
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im sorta headed in that direction anyways lol

next obsidian
#

AT naturally leads to like “higher” math

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Blah blah homotopy higher categories

chilly ocean
next obsidian
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Which is finding uses in other fields

chilly ocean
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that shit has me depressed ngl

next obsidian
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So it could be an introduction to that stuff if you want to go that direction

chilly ocean
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i dont even know where to start with that

next obsidian
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Yeah idk lol

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I think you’d like

chilly ocean
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like how you called it “higher” math

next obsidian
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Want to learn about suspensions or some shit, start talking about spectra and model categories maybe

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Idfk, i don’t know any AT

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Honestly I wouldn’t worry about it

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I’ve avoided it until I’ve needed it

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And even then I’ve only needed the most literal basics of 2-categories

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For stacks

chilly ocean
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those are used in ag?

next obsidian
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Yeah

chilly ocean
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i hate s words

next obsidian
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To define stacks you need some 2-categorical notions

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(Or at least as I learned it)

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But this really isn’t that bad

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It’s just like being explicit with natural transformations and stuff

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I don’t know anything infinity-categorical

chilly ocean
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explicit!

next obsidian
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I don’t even know a bullshit definition of the stuff

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If it doesn’t interest you I think you can just focus on doing what you find interesting

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And if it ever comes up, you should be able to recognize it

chilly ocean
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yeah i like that strategy

next obsidian
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And you’ll likely be in a situation where it’s easy to learn what you need then and fhere

chilly ocean
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i feel like moving based on curosity just hurts me

next obsidian
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Like for me, when I needed the 2-categorical stuff

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The text covered it

chilly ocean
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what book?

next obsidian
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And I just learnt that and immediately used it

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Well I used notes by Jarod Alper on stacks and moduli

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But that requires knowing quite a bit of AG

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But another book on the derived category I’m looking at introduces a bit of 2-categorical stuff

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Because it uses it when talking about derived functors

chilly ocean
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ive heard of those

next obsidian
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And so it’s a situation where like “this is the convenient way to talk about this, so we do”

chilly ocean
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i dont know what they are used for

next obsidian
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But you just learn a little bit right there

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Do you know what the tensor product is?

chilly ocean
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ye

next obsidian
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So okay you’ve done AT it sounds

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So you know why exact sequences are nice

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They’re useful!

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If you have an exact sequence
0 -> M -> N -> L -> 0

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Then if you tensor with say K another module

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You only have
M (x) K -> N (x) K -> L (x) K -> 0 exact

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What the fuck man!

chilly ocean
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ye

next obsidian
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But you can “derive the tensor product”

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And what I mean is you get these things called Tor

chilly ocean
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yeah

next obsidian
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Namely Tor^A_i(K,M)

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And then you can extend that exact sequence to the left

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With like

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Tor^1(K,M) -> Tor^1(K,N) -> Tor^1(K,L) -> M (x) K ->…

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And it even goes further and further

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You get an entire LES like the LES in cohomology you’ve probably seen

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And this lets you analyze stuff further

chilly ocean
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are derived functors, functor on functors that extend to right or left LES

next obsidian
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Yeah

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Well

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Idk if it’s a functor on a functor

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So if you have a left exact functor

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Or a right exact functor

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It’s an associated sequence of functors which extend exactness

chilly ocean
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o

next obsidian
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Like what it really is, it’s a collection L^iF

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Such that if 0 -> M -> N -> L -> 0 is exact

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You have

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-> L^iF(M) -> L^iF(N) -> L^iF(L) -> L^i-1(M) -> … exact

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So think the LES in cohomology

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The thing is

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L^0F(M) = F(M)

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So like in the lowest degrees it’s just F

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This is the left derived functor of F

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Because like it goes to the left like what happens in Tor

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If F is left exact you get the right derived functors

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R^iF

chilly ocean
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o

next obsidian
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And these like go to the right

chilly ocean
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explains tor and ext

next obsidian
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So Tor = L^iF

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And Ext = R^iHom

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Yeah exactly

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Those are two special cases

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The most useful ones IMO

chilly ocean
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ive always been bad at knowing when to apply tor and ext

next obsidian
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Just whenever

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Idk

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Hahaha

chilly ocean
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i dont have good intuition for homological algebra stuff either rly

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when first starting AT i came in with assumption that visual arguments will be used a lot

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only to get cucked

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but i ended up liking the new way of thinking

next obsidian
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Yeah idk

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I do algebra pretty formally

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So I just think of it algebraically and don’t concern myself with that sort of stuff

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My intuition just comes from having dealt with the objects for a long time and having done a lot of algebra lmao

chilly ocean
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yeah thats the one thing i like about algebraic number theory

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but sadly i dont care about numbers outside of curiosity

next obsidian
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Lol

chilly ocean
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i need some bonafide career advice sooner or later. I like commutative algebra a lot because I like algebra. I also like topology because the ways they can be generalized and how they model lots of situations. I assumed I would like AT but it sorta removed the explicit aspect that I like about algebra and topologies dont really matter as much at a certain point it seems

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Ive tried to talk to profs about it also

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and i cant seem to get any actual advice besides trying out courses

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atleast im positive that i dont care about algebraic number theory because of prof advice

next obsidian
#

¯_(ツ)_/¯

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There’s commutative algebra used in combinatorics lmao

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Or you could try your hand at complex geometry

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I don’t know if complex geometry uses commutative algebra explicitly, but it uses sheaf theory if that’s how you want to approach it

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And because of GAGA you can use algebraic geometry techniques

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And that’s gonna use a lot of commutative algebra

chilly ocean
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loo

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ill check out combinatorics and complex geo ig

#

SE should give good book advice

wooden ember
#

Am I having a stroke beret

#

Here*

#

The functor taking K vector spaces to the free vector space of same dimension on K is naturally isomorphic to the identity by just looking at the functor diagram no?

#

But I thought natural isomorphisms were meant to signify independence of basis

next obsidian
#

That’s not even a functor

#

I mean, how can you even make it a functor?

#

Also every vector space is the free vector space of dimension itself

#

So you (I think) are fixing a specific vector space of every dimension

#

To map your vector spaces to

wooden ember
#

Which is my point

next obsidian
#

But…

#

No

#

Literally every vector space is free

#

They all have bases

#

Free means that maps from it are isomorphic to maps from the basis as a set into the target

wooden ember
#

Yeah but I mean mapping to the set Kx…xK

#

(With vector space structure)

next obsidian
#

Yeah you’re fixing a vector space of every dimension then

wooden ember
#

How so

next obsidian
#

But that doesn’t mean some abstract V isn’t free

#

You’re picking a specific vector space of every dimension

wooden ember
next obsidian
#

But if V has dimension m, it’s free of dimension m

wooden ember
#

Ok i see what you mean

next obsidian
#

But I don’t see how you make this functorial

wooden ember
#

Why would this disagree with a functor though

next obsidian
#

Given an abstract map V -> L

#

How do you define the map from K^n -> K^m

wooden ember
#

Map it to a matrix map in some basis?

next obsidian
#

That isn’t unique

#

You have to pick a basis

wooden ember
#

Oh right we can’t universally take a basis for each space

#

We have to pick for every space

next obsidian
#

Yes

chilly ocean
#

stupid quick question A subring of B and B integral over A. Is B the integral closure of A in B?

#

i think yes

next obsidian
wooden ember
#

Makes sense thanks

next obsidian
chilly ocean
#

is it proper to say integral closure is a local property?

next obsidian
#

What do you mean?

#

Integral closure isn’t really a property unless you mean like A being integrally closed in its field of fractions or something

chilly ocean
#

integrally closed is a property though

#

like a set can be integrally closed if each of its elements is integral over blah

next obsidian
#

If you mean integrally closed in your field of fractions then yes

#

You can show a few things

chilly ocean
#

yeah thats what id mean by local propert part

next obsidian
#

Let A’ be the integral closure of A in B

chilly ocean
#

but ig not always field of fractions

#

just localization

next obsidian
#

Then S^-1A’ = (S^-1A)’ where the latter is taken inside of S^-1B

#

For the field of fractions specifically, A = \Cap A_m over all m maximal ideals

#

And from this one can show A is integrally closed iff A_m is for all maximal ideals iff A_p for all prime ideals

#

And I think you can probably extract iff S^-1A for all multiplicatively closed sets S

#

In general tho, I’m not really sure how you want to frame it being a local property

#

If you want to phrase it as like A_m being integrally closed in B_m for all m in MSpec (or Spec)

next obsidian
#

And the other direction maybe follows in a way similar to when B = Frac(A) but I’m not sure

chilly ocean
#

ye is true

#

i can see intuitively why A=cap A_m

#

but i didnt come across this in AM

next obsidian
#

¯_(ツ)_/¯

#

It’s in Matsumura somewhere

chilly ocean
#

thats what i meant pmuch

#

but it can just be any submonoid

next obsidian
#

Yeah idk

chilly ocean
#

woah im confused again

#

integral domain is integrably closed if it is integrably closed in its field of fractions

#

give example of Z in Q

#

How can Z be integrably closed in Q. by the definition dont you need Q to be subring of Z and then every element of Z can be written as root of polynomial in Q?

next obsidian
#

Yeah any UFD is integrally closed

chilly ocean
#

i should sleep

next obsidian
#

It’s the other way around

#

The integral closure of Z inside of Q is Z

#

That’s the definition of integrally closed

chilly ocean
next obsidian
#

You have it reversed

#

Z integrally closed in Q means the integral closure of Z inside of Q is Z

#

It doesn’t say anything about Q inside of Z

#

That isn’t their confusion

next obsidian
chilly ocean
#

I thought the definition was that A subring of B and B is integrably closed in A if elements of B written as monic polynomial in A coefficients? I thought you need requirement of A being subring of B. So when it saying you need X integrally closed in field of fractions im confused because the field of fractions of X wont be a subring of X usually. Unless it means that the canonical morphism f:X->Frac(X) is integrally closed where this means Frac(X) integrally closed in f(X)?

next obsidian
#

No

#

Reread the definition of integrally closed

#

There’s two notions

#

For both of these A < B.

B is integral over A if every element of B is the root of a monic polynomial with coefficients in A.

The integral closure of A in B is the set of elements of B integral over A. Then, A is integrally closed in B if A = its integral closure in B

#

It means that anything in B integral over A is already in A. Aka it’s closed under taking integral elements

chilly ocean
#

ah okay

#

I conflated “integrally closed” with “integral over”

#

thanks for putting me in my place c dawg

#

or ig in otherwords if A subring B then the maximal subset integral over A is the integral closure and a set is integrally closed over a superset if its the integral closure of the superset

#

wait the second clause is weird

#

a ring X is integrally closed in Y if it equals its integral closure in Y

#

just paraphrasing

latent vector
#

can someone give me a hint to show that if E/F is the splitting field for f(x) in F[x], then if f is irreducible, then Gal(E/F) acts transitively on the roots of f

#

i know i'm supposed to use

#

@next obsidian

next obsidian
#

Fix a root alpha

#

And consider the product of (x - sigma(alpha)) over all sigma in the Galois group

latent vector
#

that is exactly f

#

@next obsidian

next obsidian
#

Then you know that the Galois group is transitive on the roots

#

Any root of f must appear as sigma(alpha)

latent vector
#

oh

#

then i guess i don't realy understand why that product is f

next obsidian
#

That’s where the work is to be done

latent vector
#

it's the same problem. showing that there exists a sigma that takes alpha to any other root...

next obsidian
#

No

#

You just need to show that this polynomial is alpha’s minimal polynomial

latent vector
#

since the action of sigma is on the roots, we have that this polynomial is irreducible over F[x] (IF it even is in F[x]).

#

and one of the sigmas has to be the identity

#

so i think i just have to show that it is in F[x]?

chilly ocean
#

y does this hold

Screenshot_20220317-183935_Office.jpg
hot lake
#

It should be 1=f(b)=ab

latent vector
next obsidian
#

Yes

#

I don’t see why that argument for it being irreducible means anything tho

#

You know that it has the same degree as f and alpha is a root of it

latent vector
#

ah

#

that's true.

next obsidian
#

So you just need that it’s in F[x]

#

Figure that out and you’re done

#

This is where you need that the extension is Galois

#

I can’t say anymore without just solving it for you

latent vector
#

uhh my textbook/class has never defined a galois extension

#

we are using rotman's galois theory

next obsidian
#

Wtf is your definition of Gal(E/F) then lmao

latent vector
next obsidian
#

Hurb

#

Do you know stuff about intermediate fields and subgroups of the Galois group?

latent vector
#

I know some stuff about intermediate fields

next obsidian
#

Like fixed fields and stuff?

latent vector
#

wdym fixed field

#

probably not then lmfao

next obsidian
#

Do you know about a correspondence between subgroups of Gal(E/F)

#

And intermediate field extensions of E and F

latent vector
#

I know like

#

the fundamental theorem of galois theory

next obsidian
#

Yes

#

That

latent vector
#

yeah

next obsidian
#

Use that

#

What does it mean for the polynomial to be in F[x]?

latent vector
#

that it's coefficients are in F

next obsidian
#

And according to the fundamental theorem of Galois theory

#

What’s that equivalent to?

latent vector
#

i have no clue tbh

next obsidian
#

Well figure that out first

#

And then apply it here

#

I really can’t say anything else

#

What’s your statement of the fundamental theorem?

latent vector
#

wait

#

that

next obsidian
#

Not that one

#

This isn’t what you need

#

You need this one

latent vector
#

So for every intermediate extension

#

F subset of K subset of E

#

There is a unique subgroup of Gal(E/F) that fixes K?

next obsidian
#

Yeah

#

But god damn it, this argument is bad I think

#

I was trying to prove it for a different kind of extension ugh

latent vector
#

I was able to prove the converse way easier

next obsidian
#

I think you do need to use the theorem you originally stated

latent vector
#

and that was part ii

#

yeah

#

my first idea was

#

to use that lemma with F = F'

next obsidian
#

Yeah

latent vector
#

so that F(a1) = F(a2) where a1,a2 are roots of f

#

then try to extend THAT to E

#

but like

#

from Q's perspective you can't tell the difference between \pm sqrt(2) in Q(sqrt2)

#

but if we overextend to R

#

then we can tell the difference

#

that's sort of my intuition

#

but i think the key is that

next obsidian
#

So like

latent vector
#

these extensions are simple

next obsidian
#

You have an isomorphism

#

F(a1) -> F(a2) sending a1 to a2 already yeah?

#

Because both are just F[x]/(f(x))

latent vector
#

yea

next obsidian
#

So now you can embed both of these into E the splitting field

latent vector
#

oh fuck

#

i'm a dumbass

next obsidian
#

Right

latent vector
#

thank you for your hlep

#

it's 6am and i am severely sleep deprived

next obsidian
#

I hate books that use Gal

#

For Aut(E/F)

#

Gal should be reserved for Galois extensions

#

Even then, my original argument was bad

latent vector
#

if i knew what a galois extension was i would agree

south patrol
#

Oh yes rotman moment

oblique leaf
#

can anyone give me an example of when K1 and K2 are algebraic extensions of F, but the tensor product of K1 and K2 over F is not a field?

next obsidian
#

Take C and C over R

#

The tensor product becomes C^2

#

Try to compute it yourself and see if you can do it

sour plume
#

man fields are so weird

#

is there any definition of a field which does not rely on the thousands of axioms in which they are usually defined

delicate orchid
#

abelian group acting on another abelian group where the action is distributive

oblique leaf
#

oooh i see

#

that tensor product isn't an integral domain

#

Thanks!

sour plume
#

then i can dig it, otherwise i don't know what field is represented by (Z \ 0, *) acting on (Z,+)

delicate orchid
#

(Z\0, *) is not a group

sour plume
#

oh whoops

#

wow i've not done math in a month and it shows

delicate orchid
#

and you're right on the extra condition

#

I also am a fellow fields hater

sharp sonnet
#

"is there a way to define fields without the usual axioms?"
"yes, you can do the usual axioms but in more complicated language"

sour plume
#

yeah i'm also not suuuper satisfied with it

#

i can accept abelian groups, i think, and things acting on abelian groups

#

idk, fields seem so incredibly central to many math things that it'd be really cool to have a really succinct definition

#

i mean i guess this is nice

unknown.png
#

a field is a "simple" commutative ring with identity then

#

and a commutative ring with identity is a set with two operations and a whole bunch of commuting diagrams

#

this is far from a solution to my issue but i guess it moves the goalpost to "why commutative rings with identity"

next obsidian
#

Wait

sour plume
#

maybe my question is stupid, too, because the only fields i really care about are R and C, and these can be classified in a lot cooler ways

next obsidian
#

What’s wrong with commutative ring with every nonzero element invertible?

sour plume
#

yeah i guess this is okay if you already appreciate commutative rings

#

i'm still soul-searching if i do

next obsidian
#

Well you could become Bourbaki

#

And let field mean division ring

south patrol
#

lol

next obsidian
#

And drop the commutative assumption

sour plume
#

yeah i guess commutativity feels a bit specific, too

#

when you define commutativity in terms of diagrams you always need to invent this annoying swap-function for example

sour plume
devout crow
sour plume
#

that is not too bad actually

next obsidian
#

This map doesn’t exist

#

as x^-1 doesn’t always exist

#

You could work this for defining a field by first supposing inverses exist then restricting to the group of units

sour plume
#

i guess you need to once again require some nonzeroness when applying this to a field

devout crow
#

yeah i should have said this is for groups

sour plume
#

which is also really annoying when trying to define a field, you always have to put in "uwu but only for nonzero things" somewhere because things break if you don't

oblique leaf
#

Does anyone know how to show that if K1 and K2 are algebraic extensions of F and [K1K2:F]=[K1:F][K2:F], then $$K1\bigotimes_F K2$$ is a field?

cloud walrusBOT
#

alyosha

561715686062161930.png
oblique leaf
#

I have struggled for days now, and can only show it assuming K1 and K2 are finite extensions

lethal dune
cloud walrusBOT
lethal dune
#

is there any fault in this argument?

#

it didn't use [K1K2:F] = [K1:F][K2:F] so I'm guessing it's wrong

oblique leaf
#

its wrong because I now know the counterexample with C over R

#

C tensor C over R is not an integral domain

lethal dune
#

hmm right

oblique leaf
#

one issue i can see is that i dont know what the inverse would be of a sum of elementary tensors

#

also i never knew that the inverse of a tensor b is a^-1 tensor b^-1

#

anyway it doens't matter that much

lethal dune
#

oh yea lol right

#

such a dumb mistake

chilly ocean
#

anyone know of any quick uses of going down theorem

#

or an explicit example of it being used

tribal moss
#

The [K1K2:F] = [K1:F][K2:F] assumption is not strong enough if we allow it to degenerate to infinity = infinity·infinity.

oblique leaf
#

would it be strong enough if we respect multiplication of cardinal numbers?

tribal moss
#

No, because cardinal arithmetic has aleph0 = aleph0·aleph0 too.

oblique leaf
#

okok and your example contradicts that as well correct

tribal moss
#

Yup.

oblique leaf
#

sorry i dont know anything about cardinal numbers

#

but thanks so much i think you are right!

tribal moss
#

Transfinite cardinal arithmetic is easy as long as it's just addition and multiplication. You have a+b = a·b = max(a,b) as soon as one of a and b is infinite (except for the trivial a·0 and 0·b).

#

Exponentiation is weird though. Almost everything becomes undecidable.

coral shale
#

The 2nd is a corollary of the 1st?

#

ie. the 1st is the stronger result (ignoring the stuff specific to fields)?

#

Or well idk how to phrase my thoughts lol

#

I'm just tryna compare the ring and field versions and see which properties were necessary to conclude those statements

#

Eg. for 1.18, I see in the proof it uses the fact K[t] is a Euclidean Domain, but now I'm thinking this isn't necessary surely

#

1.18 proof

unknown.png
#

2.3.5 proof

frail perch
#

Question about rep theory: So I know that given a representation of a group G, we have that the tensor square representation decomposes as the sum of 2 subrepresentations (the symmetric and alternating square representations)

What I dont get is, it seems that it is always explictly assumed that the characteristic of the field is 0, and I dont exactly see where this condition is actually used (I can see why one would want to get rid of char 2 though, but I dont see the issue with char p for p>2)

chilly ocean
#

Near 7th to last sentence starting with By5.14 it says y is integral over p2

#

5,14 states

971BBC37-EF43-4422-B687-085A9F1FA4F7.jpg
#

I dont see how this is being applied

chilly ocean
#

nvm

#

figured out that radical of Bp2 is extension of p2

#

i meant that extension of radical of p2 is Bp2

terse crystal
frail perch
#

That I understand, what I dont understand is why the insistence on the assumption that char =0

terse crystal
#

Oh i see… just to get rid of 2 probably?😂

#

I also see no reason why it can’t be odd p. I think it can, since 1/2 is well defined…

coral shale
#

Well in particular, I'm trying to understand this part of the proof of 2.3.5 rn

#

I haven't rlly seen much of modules apart from the definition

#

So this usage of matrices just feels sus

#

idk

#

In particular A is a matrix with coefficients in R

#

but aI_n is a matrix with coefficientsin R[a]

#

I'm uh ??? kinda

#

So we are working in GL(n, R[a]) to figure out the last bits?

#

This is precisely the direct sum and tensor product, isn't it?

hot lake
#

no

#

I+I = I

#

and that's different from a direct sum of I and I

#

tensor product should also be different

#

though if the intersection of I and J is {0} then I+J is a direct sum of I and J

#

there is also a natural map from the tensor product of I and J to IJ

#

but I'm not sure when that would be an isomorphism

coral shale
#

👌 ty

hot lake
#

if R = Z/4Z and I=J=2 Z/4Z then IJ = {0} but I tensor J should be nontrivial

#

but R is not an integral domain so uh

coral shale
#

What would I tensor J be for your example . . .

hot lake
#

do you have a good reason to be thinking that

coral shale
#

no I don't.

#

I am just trying to internalise concepts by relating them to other ones

coral shale
hot lake
#

well uuh

#

2 tensor 2 should be nonzero

coral shale
#

I tensor J = span{(0, 0), (0, 2), (2, 0), (2, 2)} = {(0, 0), (0, 2), (2, 0), (2, 2)}
Where
(a, b)(c, d) := (ac, bd)
(a, b) + (c, d) := (a+c, b+d)

#

Right?

hot lake
#

the tensor product isn't a ring you can't multiply its elements together

#

and no not that addition

coral shale
hot lake
#

the rules are (a * c) + (b * c) = (a+b) * c and (a * b) + (a * c) = a * (b+c)

#

and (r * a) * b = a * (r * b) ( = r * (a * b) to make it into an R-module)

#

so for example, (0,2) + (0,2) = (0+0,2) = (0,2) so we must have (0,2) = 0

#

so it's the span of (2,2)

#

and (2,2) + (2,2) = (4,2) = (0,2) = 0

#

so as an R-module it's basically the same thing as I and J already

#

it has a nonzero element that when you add it to itself it's 0

coral shale
#

ok thanks

hot lake
#

maybe it's true that when R is an integral domain then I tensor J = IJ though I would have to think on it a little

coral shale
#

urgh

#

why is this obvious?

#

(im so done today)

#

x -> 1+x

#

ofc

#

nvm

south patrol
#

In fact, both are isomorphic to F_2

coral shale
#

yh thats the next line

south patrol
#

Sure

coral shale
#

sigh

#

uh

#

the natural f : R/I -> R/J will always be a homomorphism

#

f(r + I) = r + J

#

but what's important is to show this is a bijection

#

right?

#

wait no

#

r + I -> r + J might not be well defined. . .

south patrol
coral shale
south patrol
#

this is just 1st isomorphism theorem

coral shale
#

I'm following this proof

south patrol
#

Or consider the evaluation map F2[x] -> F2[x] sending x to x+1

#

I think that's what you said earlier anyway lol

coral shale
#

well yeah like how do i put it

#

this is 'obvious' for this very specific example

#

but I don't see how I could do it if we had other polynomials

#

and I had to deduce if they were isomorphic or not

south patrol
#

Yeah so the way I've seen it done in general is that you can factorise the polynomials, use CRT to get an isomorphism and then apply this

#

That way you only have to worry about quotients of F[x] by maximal ideals i suppose

dull root
#

I have this basic fact that I am not seeing.

If I'm in the lattice Z^2, and I pick a parallelogram which has 4 lattice points as its vertices such that no other lattice pound exists inside or on the boundary of the parallelogram, then the area of the parallelogram is 1.

Equivalently, if I pick the 2 sides of this parallelogram as column vectors then the resultant 2x2 matrix has determinant 1. How do I see the determinant is 1, I'm having trouble translating the condition that there are no other lattice points on the parallelogram to "bound" the coordinates

coral shale
#

hmmmmm

#

Pick's theorem?

dull root
#

Ah yes, pick's theorem works here

coral shale
#

0 + 4/2 - 1 = 1 KEK

dull root
#

But what if I had another lattice on the plane

coral shale
#

as for intuition, who knows.

#

wym

dull root
#

If I had another lattice, then it is the linear image of Z^2, however, if I took a similar parallelogram on that lattice it is not true that its preimage in Z^2 is also a parallelogram with 4 lattice pts and no other right? For example, I can shrink or blow up Z^2 by a factor of 1/2 or 2?

coral shale
#

uhhh can you define lattice for me

#

you just me a square grid of points right?

#

Picks theorem holds for those

#

i dont get what u mean

dull root
#

We defined a lattice in the plane as a nonsingular linear transform of the integer lattice Z^2, the one we are familiar with. If I know pick's theorem works for Z^2, is there an easy way to see it works for such linear transmations of Z^2?

coral shale
#

so a linear transform..

#

you could shear the lattice

#

stretch it in one coordinate

#

and you claim the same statement as above works after the transform?

#

well what to say - the linear transform of the space can be described with a matrix

#

in particular the determinant of that matrix determines how the area of any shape is scaled

#

Any linear transformation preserves ratio of areas between any pair of shapes

dull root
#

ah I see. I also have a similar problem.

If I have a lattice on the plane where the area of the fundamental parallelogram(area of parallelogram determined by the two basis vectors of lattice) is 1, I want to show that there exists two lattice points which are a distance at most $\sqrt{\frac{2}{\sqrt 3}}$ apart.

If we take two lattice points with minimal distance, we can shift them so one of the points is 0, so $u = (x_1,y_1)$ is a lattice point with minimal distance. Then I can find another lattice point $v = (x_2,y_2)$ such that u and v generate the entire lattice, so the area of the parallelogram determined by u,v is 1. Then 1 is the determinant of of the 2x2 matrix where u and v are column vectors ie $1 = \lvert x_1y_2 - x_2y_1 \rvert$. How do I see the distance of u and v is the bound we need?

cloud walrusBOT
#

MasakaBakana

303052934621560832.png
dull root
#

I just don't see where this distance is coming from... it is slightly larger than 1, so intuitively it seems there because the areas are 1

coral shale
#

Would Hurwitz help you by any chance

dull root
#

whats that?

coral shale
#

theorem.

dull root
#

Hmm, it somewhat resembles it, but it seems to have a root 5 in thm. But I think the idea here is supposed to be more elementary, I just can't see it lol

coral shale
#

no clue sry, try drawing a picture and do some geometry ig

lavish nexus
#

all of the s_i 's are in S, and a_i's are in R
so a'_i is some product of a_i with the s_i's and is in R
then on the fifth line from the bottom we have t_1u is a root of an equation all of whose coefficients are in R
hence t_1u is in R' the integral closure
but t_1 = product of s_i so is in S. Doesn't this provide us already with the s' we're looking for in line 6?
and what is the point of t_2?

coral shale
coral shale
#

Can I???

#

(I don't know why, if yes)

#

At least, I don't find it obvious why pandaOhNo

#

Is there a more general result I should be seeing here

#

(Oh just spotted an error from line 1 to line 2, but ignore for now)

bleak abyss
coral shale
#

yh yh

south patrol
#

either the evaluation hom Z12[x] -> Z_12[x] sending x -> x+6, or both iso to Z/12Z by the factor theorem too :)

coral shale
#

Surely I can't just do this for any 2 polynomials of the same degree

#

within Z_n[x]

#

or you're saying I can??

south patrol
#

not for the same degree no, but these are linear so it's just the factor theorem or equivalent

coral shale
#

Could you elaborate what you mean by factor thm

south patrol
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Oh as in the evaluation map at, say, -6 has kernel (6+x) and is surjective

coral shale
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oh

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so we are sending -6 -> 0

south patrol
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mapping x -> -6

coral shale
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x -> -6 -> 0 -> x ?

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or no. . .

south patrol
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i'm confused what you mean

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we're taking teh evaluation hom Z_12[x] -> Z_12 which sends x -> -6

coral shale
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thinking

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So rather than justifying this step

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this instead

south patrol
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yup you can do that

coral shale
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No, I'm asking which you are doing

south patrol
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Nice thing to remember

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Oh

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I'm doing the latter

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No need to shift

coral shale
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Ok, I'll think about evaluation homomorphism

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Yes, I agree it is surjective, hence

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By 1st iso, we have Z_12[x]/ker f = Z_12

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And ker f = (x + 6)

south patrol
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indeed

coral shale
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ok ok clocked it

south patrol
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:)

coral shale
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Now why does this not work for non-linears, thinking. . .

south patrol
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But yeah evaluation homomorphisms are really nice with their universal property

coral shale
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uhhh for non-linears, the root might not even be in Z_n ?

south patrol
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i.e. given a homomorphism A -> B and some b in B, it extends uniquely to a map A[x] -> B sending x -> b

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Yeah, the same argument doesn't work

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You could have an irreducible polynomial, or smth like x^2, etc

coral shale
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Suppose it has a root, we can use the same argument?

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We again will have a surjective evaluation map

south patrol
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Not with e.g. R[x]/(x^2) and stuff

coral shale
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x -> 0, what could go wrong. . .

south patrol
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kernel is (x), not (x^2)

coral shale
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ohhh

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ok thanks for going through this stuff with me

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makes sense now lol

south patrol
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But you can also (for e.g. commutative rings) use the chinese remainder theorem to split stuff like if you want though.

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Like C[x]/(x-2)(x-3)(x-4) is isomorphic to C[x]/(x-2) oplus C[x]/(x-3) etc

coral shale
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this?

south patrol
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which is in term isomorphic to C x C x C

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yeah

coral shale
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i haven't got there yet but ok

south patrol
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Oh sure

coral shale
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Or I haven't, my class has, I've almost caught up

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To prove, it looks like isomorphism thm ? 🤔

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Well I will try when I have time 😅

south patrol
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Yeah it is the first isomorphism theorem

lavish nexus
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Let A be an ideal of R, S a multiplicative submonoid of R

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why is A_S = R_S iff A \cap S is nonempty?

next obsidian
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A_S is an ideal of R_S, to be equal means that the former contains an invertible element

upbeat swift
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If someone could help, that would be greatly appreciated:
Describe the divisors of zero in F(R).
and describe the invertible elements in F(R)
(F(R) is reals under functions)
We are barely learning rings in my class and I am confused about these two parts specifically

next obsidian
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Containing an invertible element becomes exactly the statement that A and S share an element

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F(R) = {f:R->R}?

upbeat swift
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yes

next obsidian
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Well if f•g = 1

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First of all, do you know what 1 means in F(R)?

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Namely, what function is it?

lavish nexus
upbeat swift
next obsidian
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Right perfect

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So suppose f•g = 1

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What can we say about g(x) in terms of f(x)?

upbeat swift
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g(x) = 1/f(x)

next obsidian
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Awesome, so g = 1/f

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When does this g exist?

upbeat swift
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when f does not equal to 0?

next obsidian
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Well can you say what you mean?

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Do you mean f is not the zero function?

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Because that’s the function f(x) = 0 for all x

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You need to know more about f in order for 1/f to exist

upbeat swift
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I'm not so sure, that's what I am stuck on

next obsidian
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Well we need the formula

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g(x) = 1/f(x) to be a valid formula for all x in R

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This works as long as…?

upbeat swift
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f(x) has no zeroes?

next obsidian
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Yeah exactly

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So the set of units are exactly the functions with no zeroes

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Because then 1/f exists

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And conversely if 1/f exists, then f can’t have any zeroes

upbeat swift
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and this is for the divisors of 0?

next obsidian
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No this is for units

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Or sorry

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Invertible elements

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Unit means the same thing as invertible element

upbeat swift
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ah okay

next obsidian
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So now we want to analyze the zero divisors

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So let’s just start with a function that’s non-zero everywhere

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Is it possible for f to be a zero divisor?

upbeat swift
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I'm not so sure

next obsidian
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Well let’s write out the definition of a zero divisor

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Let’s suppose it were a zero divisor, then what?

upbeat swift
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is the zero divisor a product of two functions that equal 0?

next obsidian
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Well that’s close

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It’s an element (non-zero) such that there is a non-zero element with which the product is zero